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Revised list of TBCs
Diffstat (limited to 'Strength_Of_Materials_by_B_K_Sarkar')
-rwxr-xr-xStrength_Of_Materials_by_B_K_Sarkar/chapter_10_som.ipynb689
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-{
- "metadata": {
- "name": "chapter 10 som.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 10:Analysis Of Framed Structures"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 10.1,Page No.249"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "#Consider Equilibrium of joint A\n",
- "#As there are no Load applied at A members AC and AB have nothing to Balance \n",
- "#So they are null members\n",
- "F_AB=0\n",
- "F_AC=0\n",
- "\n",
- "#Consider Equilibrium of joint B\n",
- "\n",
- "#Applying the summation of horizontal forces we get\n",
- "F_DB=4*(cos(45*pi*180**-1))**-1\n",
- "\n",
- "#Applying the summation of vertical forces we get\n",
- "F_BC=F_DB*sin(45*pi*180**-1)\n",
- "\n",
- "#Consider Equilibrium of joint B\n",
- "\n",
- "#Applying the summation of vertical forces we get\n",
- "F_CE=4*(sin(45*pi*180**-1))**-1\n",
- "\n",
- "#Applying the summation of horizontal forces we get\n",
- "F_DC=F_CE*cos(45*pi*180**-1)\n",
- "\n",
- "#Result\n",
- "print\"The Forces in Each members are as follows:F_AB\",F_AB,\"KN\"\n",
- "print\" :F_AC\",F_AC,\"KN\"\n",
- "print\" :F_DB\",round(F_DB,2),\"KN(compression)\"\n",
- "print\" :F_BC\",round(F_BC,2),\"KN(Tension)\"\n",
- "print\" :F_CE\",round(F_CE,2),\"KN(Tension)\"\n",
- "print\" :F_DC\",round(F_DC,2),\"KN (compression)\" "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Forces in Each members are as follows:F_AB 0 KN\n",
- " :F_AC 0 KN\n",
- " :F_DB 5.66 KN(compression)\n",
- " :F_BC 4.0 KN(Tension)\n",
- " :F_CE 5.66 KN(Tension)\n",
- " :F_DC 4.0 KN (compression)\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 10.2,Page No.250"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "#Taking moment at Pt A we get\n",
- "R_B=100*8*4**-1\n",
- "\n",
- "#Applying the summation of vertical forces we get\n",
- "R_AV=-R_B\n",
- "\n",
- "#Applying the summation of horizontal forces we get\n",
- "R_H=100\n",
- "\n",
- "#joint B\n",
- "\n",
- "#Applying the summation of vertical forces we get\n",
- "F_CB=R_B\n",
- "\n",
- "#Applying the summation of horizontal forces we get\n",
- "F_AB=0 #As there is no force to balance in horizontal direction\n",
- "\n",
- "#joint A\n",
- "\n",
- "#Applying the summation of horizontal forces we get\n",
- "F_AC=R_H*(cos(45*pi*180**-1))**-1\n",
- "\n",
- "#Applying the summation of vertical forces we get\n",
- "F_AD=200-F_AC*sin(45*pi*180**-1)\n",
- "\n",
- "#joint C\n",
- "\n",
- "#Applying the summation of vertical forces we get\n",
- "F_EC=200-F_AC*cos(45*pi*180**-1)\n",
- "\n",
- "#Applying the summation of horizontal forces we get\n",
- "F_DC=F_AC*cos(45*pi*180**-1)\n",
- "\n",
- "#joint D\n",
- "\n",
- "#Applying the summation of horizontal forces we get\n",
- "F_DE=F_DC*(cos(45*pi*180**-1))**-1\n",
- "\n",
- "#DF and EF are null members at this joint as each member individually has nothing to balance\n",
- "F_DF=0\n",
- "F_EF=0\n",
- "\n",
- "#Result\n",
- "print\"The Forces in Each members are as follows:F_AB\",F_AB,\"KN\"\n",
- "print\" :F_CB\",F_CB,\"KN(compressive)\"\n",
- "print\" :F_AC\",round(F_AC,2),\"KN(Tensile)\"\n",
- "print\" :F_AD\",F_AD,\"KN(Tensile)\"\n",
- "print\" :F_EC\",F_EC,\"KN(Compressive)\"\n",
- "print\" :F_DC\",F_DC,\"KN(compressive)\"\n",
- "print\" :F_DE\",round(F_DE,2),\"KN(Tensile)\"\n",
- "print\" :F_DF\",F_DF,\"KN\"\n",
- "print\" :F_EF\",F_EF,\"KN\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Forces in Each members are as follows:F_AB 0 KN\n",
- " :F_CB 200.0 KN(compressive)\n",
- " :F_AC 141.42 KN(Tensile)\n",
- " :F_AD 100.0 KN(Tensile)\n",
- " :F_EC 100.0 KN(Compressive)\n",
- " :F_DC 100.0 KN(compressive)\n",
- " :F_DE 141.42 KN(Tensile)\n",
- " :F_DF 0 KN\n",
- " :F_EF 0 KN\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 10.3,Page No.252"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "#taking moment at pt A we get\n",
- "R_D=(90*6+120*3)*9**-1 #Reaction at Pt D\n",
- "\n",
- "#Joint D\n",
- "\n",
- "#Applying the summation of vertical forces we get\n",
- "F_GD=100*(sin(60*pi*180**-1))**-1\n",
- "\n",
- "#Applying the summation of horizontal forces we get\n",
- "F_DC=F_GD*cos(60*pi*180**-1)\n",
- "\n",
- "#Joint G\n",
- "\n",
- "#Applying the summation of vertical forces we get\n",
- "F_GC=F_GD\n",
- "\n",
- "#Applying the summation of horizontal forces we get\n",
- "F_FG=F_GD*cos(60*pi*180**-1)+F_GC*cos(60*pi*180**-1)\n",
- "\n",
- "#joint C\n",
- "\n",
- "#Applying the summation of vertical forces we get\n",
- "F_FC=(115.5*sin(60*pi*180**-1)-90)*(sin(60*pi*180**-1))**-1\n",
- "\n",
- "#Applying the summation of horizontal forces we get\n",
- "F_CB=F_DC+F_GC*cos(60*pi*180**-1)+F_FC*cos(60*pi*180**-1)\n",
- "\n",
- "#joint F\n",
- "\n",
- "#Applying the summation of vertical forces we get\n",
- "F_FB=F_FC\n",
- "\n",
- "#Applying the summation of horizontal forces we get\n",
- "F_EF=F_FG+F_FC*cos(60*pi*180**-1)+F_FB*cos(60*pi*180**-1)\n",
- "\n",
- "#Joint B\n",
- "\n",
- "#Applying the summation of vertical forces we get\n",
- "F_EB=(120-F_FB*sin(60*pi*180**-1))*(sin(60*pi*180**-1))**-1\n",
- "\n",
- "#Applying the summation of horizontal forces we get\n",
- "F_BA=F_CB+F_FB*cos(60*pi*180**-1)-F_EB*cos(60*pi*180**-1)\n",
- "\n",
- "#Joint E\n",
- "\n",
- "#Applying the summation of vertical forces we get\n",
- "F_AE=F_EB\n",
- "\n",
- "#Result\n",
- "print\"Forces in Each members are as follows:F_GD\",round(F_GD,1),\"KN(compression)\"\n",
- "print\" :F_DC\",round(F_DC,2),\"KN(Tension)\" \n",
- "print\" :F_GC\",round(F_GC,1),\"KN(Tension)\" \n",
- "print\" :F_FG\",round(F_FG,1),\"KN(compression)\"\n",
- "print\" :F_FC\",round(F_FC,1),\"KN(compression)\"\n",
- "print\" :F_CB\",round(F_CB,2),\"KN(Tension)\"\n",
- "print\" :F_FB\",round(F_FB,1),\"KN(Tension)\"\n",
- "print\" :F_EF\",round(F_EF,2),\"KN(compression)\"\n",
- "print\" :F_EB\",round(F_EB,2),\"KN(Tension)\"\n",
- "print\" :F_BA\",round(F_BA,2),\"KN(Tension)\"\n",
- "print\" :F_AE\",round(F_AE,2),\"KN(compression)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Forces in Each members are as follows:F_GD 115.5 KN(compression)\n",
- " :F_DC 57.74 KN(Tension)\n",
- " :F_GC 115.5 KN(Tension)\n",
- " :F_FG 115.5 KN(compression)\n",
- " :F_FC 11.6 KN(compression)\n",
- " :F_CB 121.26 KN(Tension)\n",
- " :F_FB 11.6 KN(Tension)\n",
- " :F_EF 127.05 KN(compression)\n",
- " :F_EB 126.99 KN(Tension)\n",
- " :F_BA 63.55 KN(Tension)\n",
- " :F_AE 126.99 KN(compression)\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 10.4,Page No.253"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "#JOint D\n",
- "\n",
- "#Applying the summation of vertical forces we get\n",
- "F_1=6*sin(30*pi*180**-1)**-1\n",
- "\n",
- "#Applying the summation of horizontal forces we get\n",
- "F_5=F_1*cos(30*pi*180**-1)\n",
- "\n",
- "#Joint C\n",
- "\n",
- "#Resolving forces perpendicular to plane\n",
- "F_6=10*cos(30*pi*180**-1)\n",
- "\n",
- "#Resolving forces parallel to plane\n",
- "F_2=F_1+10*cos(60*pi*180**-1)\n",
- "\n",
- "#Joint E\n",
- "\n",
- "#Applying the summation of vertical forces we get\n",
- "F_7=(8+F_6*sin(60*pi*180**-1))*(sin(60*pi*180**-1))**-1\n",
- "F_4=F_5+F_6*cos(60*pi*180**-1)+F_7*cos(60*180**-1*pi)\n",
- "\n",
- "#Resolving forces perpendicular to plane\n",
- "F_3=F_7*sin(60*pi*180**-1)\n",
- "\n",
- "#Resolving forces parallel to plane\n",
- "F_8=F_2+F_7*cos(30*pi*180**-1)\n",
- "\n",
- "#Result\n",
- "print\"Forces in Each members are as follows:F_1\",round(F_1,2),\"KN(Tension)\"\n",
- "print\" :F_5\",round(F_5,2),\"KN(compression)\"\n",
- "print\" :F_6\",round(F_6,2),\"KN(compression)\"\n",
- "print\" :F_2\",round(F_2,2),\"KN(Tension)\"\n",
- "print\" :F_7\",round(F_7,2),\"KN(Tension)\"\n",
- "print\" :F_4\",round(F_4,2),\"KN(compression)\"\n",
- "print\" :F_3\",round(F_3,2),\"KN(compression)\"\n",
- "print\" :F_8\",round(F_8,2),\"KN(Tension)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Forces in Each members are as follows:F_1 12.0 KN(Tension)\n",
- " :F_5 10.39 KN(compression)\n",
- " :F_6 8.66 KN(compression)\n",
- " :F_2 17.0 KN(Tension)\n",
- " :F_7 17.9 KN(Tension)\n",
- " :F_4 23.67 KN(compression)\n",
- " :F_3 15.5 KN(compression)\n",
- " :F_8 32.5 KN(Tension)\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 10.5,Page No.256"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "BC=6 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "AB=2*BC*(3**0.5)**-1\n",
- "\n",
- "#Taking moment about B we get\n",
- "R_A=-(-2000*3-1000*6)*(12*(3**0.5)**-1)**-1 #reaction at the roller support A\n",
- "\n",
- "#The resultant of all the three Loads is 4000 N acting at right angle to BC at D\n",
- "\n",
- "#Resolving it vertically we have\n",
- "V=4000*sin(60*pi*180**-1)\n",
- "\n",
- "#Resolving it horizontal we have\n",
- "H=4000*cos(60*pi*180**-1)\n",
- "\n",
- "#Applying the summation of vertical forces we get\n",
- "R_B_v=V-R_A\n",
- "\n",
- "#Applying the summation of horizontal forces we get\n",
- "R_B_h=H\n",
- "R_B=((R_B_v)**2+(R_B_h)**2)**0.5\n",
- "\n",
- "tan_theta=R_B_v*R_B_h**-1\n",
- "\n",
- "#Joint B\n",
- "\n",
- "#Applying the summation of vertical forces we get\n",
- "F_BD=1000*(3**0.5)*2\n",
- "\n",
- "#Applying the summation of horizontal forces we get\n",
- "F_BE=R_B_h+F_BD*cos(30*pi*180**-1)\n",
- "\n",
- "#Joint D\n",
- "F_DE=2000 #N\n",
- "F_CD=F_BD\n",
- "\n",
- "#Consider equilibrium of truss to the Left of section 2-2\n",
- "F_CE=R_A*AB*(sin(30*pi*180**-1)*6)**-1\n",
- "\n",
- "#Joint A\n",
- "\n",
- "#Applying the summation of vertical forces we get\n",
- "F_AC=R_A*(sin(60*pi*180**-1))**-1\n",
- "\n",
- "#Applying the summation of horizontal forces we get\n",
- "F_AE=F_AC*cos(60*pi*180**-1)\n",
- "\n",
- "#Result\n",
- "print\"Forces in Each members are as follows:F_BD\",round(F_BD,2),\"KN(compression)\"\n",
- "print\" :F_BE\",round(F_BE,2),\"KN(Tension)\"\n",
- "print\" :F_DE\",round(F_DE,2),\"KN(compression)\"\n",
- "print\" :F_CD\",round(F_CD,2),\"KN(compression)\"\n",
- "print\" :F_CE\",round(F_CE,2),\"KN(Tension)\"\n",
- "print\" :F_AC\",round(F_AC,2),\"KN(compression)\"\n",
- "print\" :F_AE\",round(F_AE,2),\"KN(Tension)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Forces in Each members are as follows:F_BD 3464.1 KN(compression)\n",
- " :F_BE 5000.0 KN(Tension)\n",
- " :F_DE 2000.0 KN(compression)\n",
- " :F_CD 3464.1 KN(compression)\n",
- " :F_CE 4000.0 KN(Tension)\n",
- " :F_AC 2000.0 KN(compression)\n",
- " :F_AE 1000.0 KN(Tension)\n"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 10.6,Page No.258"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Taking moment of the Forces about the hinge A\n",
- "P=1000*2**0.5*1.2*(0.9)**-1\n",
- "\n",
- "#Let R_AH be the Horizontal component of the reaction at A\n",
- "R_AH=P-1000*2**0.5\n",
- "R_A=((R_AH)**2+(1000*2**0.5)**2)**0.5\n",
- "\n",
- "#Resolving the forces vertically we get\n",
- "R_AV=1000*2**0.5 #vertical component of the reaction at A\n",
- "\n",
- "#joint A\n",
- "\n",
- "#Resolving vertically we get\n",
- "F_BA=1000*2**0.5*(sin(30*pi*180**-1))**-1\n",
- "\n",
- "#Resolving horizontally we get\n",
- "F_AD=2000*2**0.5*3**0.5*2**-1-1000*2**0.5*3**-1 #N\n",
- "\n",
- "#Joint C\n",
- "\n",
- "BD=1.2*sin(30*pi*180**-1)\n",
- "BE=0.6*sin(30*pi*180**-1)\n",
- "ED=0.6*cos(30*pi*180**-1)\n",
- "CE=0.9-0.52\n",
- "\n",
- "theta=arctan(BE*CE**-1)*(180*pi**-1)\n",
- "\n",
- "F_CB=P*(sin(38.29*pi*180**-1))**-1\n",
- "\n",
- "#Resolving vertically\n",
- "F_CD=F_CB*cos(theta*pi*180**-1)\n",
- "\n",
- "#Joint D\n",
- "\n",
- "#Resolving horizontally\n",
- "F_DB=(F_AD-1000*2**0.5)*(cos(60*pi*180**-1))**-1\n",
- "\n",
- "#Result\n",
- "print\"The Pull in chain is\",round(P,2),\"N\"\n",
- "print\"Force in the each members are as follows:F_BA\",round(F_BA,2),\"KN(compressive)\"\n",
- "print\" :F_AD\",round(F_AD,2),\"KN(Tensile)\"\n",
- "print\" :F_CB\",round(F_CB,2),\"KN(compression)\"\n",
- "print\" :F_CD\",round(F_CD,2),\"KN(Tensile)\"\n",
- "print\" :F_DB\",round(F_DB,2),\"KN(compressive)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Pull in chain is 1885.62 N\n",
- "Force in the each members are as follows:F_BA 2828.43 KN(compressive)\n",
- " :F_AD 1978.09 KN(Tensile)\n",
- " :F_CB 3043.08 KN(compression)\n",
- " :F_CD 2388.46 KN(Tensile)\n",
- " :F_DB 1127.74 KN(compressive)\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 10.7,Page No.261"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Calculations\n",
- "\n",
- "theta=arctan(1*2**-1)*(180*pi**-1) #Radian\n",
- "\n",
- "#Taking moment about A\n",
- "R_EH=10*8*4**-1\n",
- "\n",
- "#Horizontal component of reaction at A\n",
- "R_AH=20 #KN\n",
- "\n",
- "#Applying the summation of horizontal forces we get\n",
- "F_AB=20*cos(theta*pi*180**-1)**-1\n",
- "\n",
- "#Applying the summation of vertical forces we get\n",
- "R_AV=10*5**0.5*sin(theta*pi*180**-1)\n",
- "\n",
- "#Vertical Reaction at E\n",
- "R_EV=0\n",
- "\n",
- "#Joint C\n",
- "\n",
- "#Applying the summation of vertical forces we get\n",
- "F_DC=10*sin(theta*pi*180**-1)**-1\n",
- "\n",
- "#Applying the summation of horizontal forces we get\n",
- "F_CB=F_DC*cos(theta*pi*180**-1)\n",
- "\n",
- "#Joint D\n",
- "\n",
- "#Applying the summation of vertical forces we get\n",
- "F_DB=F_DC*sin(theta*pi*180**-1)\n",
- "\n",
- "#Applying the summation of horizontal forces we get\n",
- "F_DE=F_DC*cos(theta*pi*180**-1)\n",
- "\n",
- "#Joint E\n",
- "\n",
- "#Applying the summation of vertical forces we get\n",
- "F_EB=R_EV*sin(theta*pi*180**-1)\n",
- "\n",
- "#Result\n",
- "print\"Forces in Each members are as follows:F_AB\",round(F_AB,2),\"KN(Tensile)\"\n",
- "print\" :F_DC\",round(F_DC,2),\"KN(compression)\"\n",
- "print\" :F_CB\",round(F_CB,2),\"KN(Tensile)\"\n",
- "print\" :F_DB\",round(F_DB,2),\"KN(Tensile)\" \n",
- "print\" :F_DE\",round(F_DE,2),\"KN(compression)\"\n",
- "print\" :F_EB\",round(F_EB,2),\"KN\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Forces in Each members are as follows:F_AB 22.36 KN(Tensile)\n",
- " :F_DC 22.36 KN(compression)\n",
- " :F_CB 20.0 KN(Tensile)\n",
- " :F_DB 10.0 KN(Tensile)\n",
- " :F_DE 20.0 KN(compression)\n",
- " :F_EB 0.0 KN\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 10.8,Page No.262"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F_c=20 #KN #Force at C\n",
- "F_d=5 #KN #Force at D\n",
- "F_e=15 #KN #Force at E\n",
- "F_f=10 #KN #Force at F\n",
- "L_CD=3.6 #m #Length of CD\n",
- "L_DE=3.6 #m #Length of DE\n",
- "L_EF=4.8 #m #Length of EF\n",
- "L_AD=L_BE=3.6 #m #Length of AD & BE\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Let R_A and R_B be the reactions at pts at A and B\n",
- "\n",
- "#Taking moment at A\n",
- "R_B=-(-F_f*(L_DE+L_EF)+F_c*L_CD-F_e*L_DE)*(L_DE)**-1\n",
- "R_A=50-R_B\n",
- "\n",
- "#Considering section 1-1 through members AB,DB,DE and taking F.B.D of left side of section 1-1\n",
- "\n",
- "#Taking moment at B\n",
- "sigma_1=(F_d*L_DE+F_c*(L_CD+L_DE)-R_A*L_DE)*L_AD**-1 #Force i member DE\n",
- "\n",
- "#Taking moment @ D\n",
- "sigma_3=(F_c*L_CD)*L_AD**-1 #KN #force in member AB\n",
- "\n",
- "\n",
- "#Consider triangle DBE\n",
- "theta=arctan(L_BE*L_DE**-1)*(180*pi**-1)\n",
- "\n",
- "#Taking moment @ A\n",
- "sigma_2=(-sigma_1*L_AD+F_c*L_CD)*(L_AD*cos(theta*pi*180**-1))**-1 #Force in member F_DE\n",
- "\n",
- "#Now considering section 2-2 passing through members AB,AD,CD and taking left hand F.B.D\n",
- " \n",
- "#Taking moment @C\n",
- "sigma_5=(R_A*L_CD-sigma_3*L_AD)*L_CD**-1 #Force in member AD\n",
- "\n",
- "#Taking moment @A=0\n",
- "sigma_4=F_c*L_CD*L_AD**-1 #Force in member CD\n",
- "\n",
- "\n",
- "#Result\n",
- "print\"Force in member CD is\",round(sigma_4,2),\"KN(Compressive)\"\n",
- "print\"Force in member AD is\",round(sigma_5,2),\"KN(Tensile)\"\n",
- "print\"Force in member BD is\",round(sigma_2,2),\"KN(Compression)\"\n",
- "print\"Force in member AB is\",round(sigma_1,2),\"KN(Tension)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Force in member CD is 20.0 KN(Compressive)\n",
- "Force in member AD is 11.67 KN(Tensile)\n",
- "Force in member BD is 9.43 KN(Compression)\n",
- "Force in member AB is 13.33 KN(Tension)\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [],
- "language": "python",
- "metadata": {},
- "outputs": []
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file
diff --git a/Strength_Of_Materials_by_B_K_Sarkar/chapter_11_som.ipynb b/Strength_Of_Materials_by_B_K_Sarkar/chapter_11_som.ipynb
deleted file mode 100755
index 25a93e73..00000000
--- a/Strength_Of_Materials_by_B_K_Sarkar/chapter_11_som.ipynb
+++ /dev/null
@@ -1,402 +0,0 @@
-{
- "metadata": {
- "name": "chapter 11 som.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter no 11:Combined Direct And Bending Stresses"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 11.1,Page no.273"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P=10 #KN #Load\n",
- "e=0.06 #m #eccentricity\n",
- "b=0.240 #m #width of column\n",
- "d=0.150 #m #depth of column\n",
- "\n",
- "#Calculations\n",
- "\n",
- "sigma_d=P*(b*d)**-1 #KN/m**2\n",
- "M=P*e #KN*m #Moment due to eccentricity\n",
- "Z=(d*(b)**2)*6**-1 #mm**3\n",
- "\n",
- "sigma_b=M*Z**-1 #KN/m**2\n",
- "\n",
- "sigma_CD=sigma_d+sigma_b\n",
- "sigma_AB=sigma_d-sigma_b\n",
- "\n",
- "#Result\n",
- "print\"Stress at face CD is\",round(sigma_CD,2),\"KN/m**2\"\n",
- "print\"Stress at face AB is\",round(sigma_AB,2),\"KN/m**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Stress at face CD is 694.44 KN/m**2\n",
- "Stress at face AB is -138.89 KN/m**2\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 11.2,Page no.274"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "d=2 #cm #Diameter of specimen\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Let P be the Load on the section\n",
- "\n",
- "A=pi*4**-1*d**2 #cm**2 #Area of section\n",
- "I=pi*64**-1*d**4 #cm**4 #M.I of the section\n",
- "y=d*2**-1 #cm\n",
- "Z=I*y**-1 #cm**3 #Section modulus\n",
- "#M=P.e #Moment\n",
- "\n",
- "#Stress due to direct load\n",
- "#sigma_d=(4*P)*(pi*d**2)**-1 #N/cm**2\n",
- "\n",
- "#stress due to moment\n",
- "#sigma_b=(32*P*e)*(pi*d**3)**-1 N/cm**2\n",
- "\n",
- "#Maximum stress\n",
- "#sigma_r_max=(((4*P)*(pi*d**2)**-1)+((32*P*e)*(pi*d**3)**-1))\n",
- "\n",
- "#Mean stress \n",
- "#sigma_r_mean=((4*P)*(pi*d**2)**-1) \n",
- "\n",
- "#Since the maximum stress is 20% greater than the mean stress \n",
- "#(((4*P)*(pi*d**2))+((32*P*e)*(pi*d**3)))=1.2*4*P*(pi*d**2)**-1\n",
- "\n",
- "#After substituing values and simplifyinf we get\n",
- "\n",
- "e=0.2*d*8**-1 #cm #distance of line of thrust from the axis\n",
- "\n",
- "#Result\n",
- "print\"The distance of line of thrust from the axis is\",round(e,2),\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The distance of line of thrust from the axis is 0.05 cm\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 11.3,Page no.274"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "A=300 #cm**2 #Area of column\n",
- "e=5 #cm #eccentricity\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#sigma_d=P*A**-1 #Direct compressive stress\n",
- "#M=P*e #Bending Moment\n",
- "Z=((20**4-10**4)*(6*20)**-1) #cm**3 #Section modulus\n",
- "\n",
- "#sigma_b=M*Z**-1=P*250**-1 \n",
- "\n",
- "#Now sigma_d+sigma_b=60*10**2\n",
- "\n",
- "#P*300**-1+P*250**-1=6000\n",
- "\n",
- "#After simplifying we get\n",
- "P_1=6000*300*250*550**-1 #N #Load\n",
- "\n",
- "#sigma_b-sigma_d=300 \n",
- "\n",
- "P_2=300*300*250*50**-1 #N #Load \n",
- "\n",
- "#Result\n",
- "print\"The maximum load column can carry\",round(P_2,2),\"N\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum load column can carry 450000.0 N\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 11.4,Page no.275"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "D=40 #cm #External diameter of column\n",
- "d=30 #cm #Internal diameter of column\n",
- "e=20 #cm #Eccentricity\n",
- "P=150 #KN #Load\n",
- "\n",
- "#calculations\n",
- "\n",
- "A=pi*4**-1*(D**2-d**2) #cm**2 #Area of the column\n",
- "Z=pi*32**-1*((D**4-d**4)*D**-1) #cm**3 #Section modulus\n",
- "M=P*10**3*e #N*cm #Moment\n",
- "\n",
- "sigma_r_max=((P*10**3*A**-1)+(M*Z**-1)) #N/cm**2 #Max stress \n",
- "sigma_r_min=((P*10**3*A**-1)-(M*Z**-1)) #N/cm**2 #Min stress \n",
- "\n",
- "#Result\n",
- "print\"Max intensities of stress in the section is\",round(sigma_r_max,2),\"N/cm**2\"\n",
- "print\"Min intensities of stress in the section is\",round(sigma_r_min,2),\"N/cm**2(tension)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Max intensities of stress in the section is 971.3 N/cm**2\n",
- "Min intensities of stress in the section is -425.63 N/cm**2(tension)\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 11.5,Page no.277"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "b=4 #m #width of pier\n",
- "d=3 #m #depth of pier\n",
- "e_x=1 #m #distance from y axis\n",
- "e_y=0.5 #m #distance from x axis\n",
- "P=80 #KN #Load\n",
- "\n",
- "#Calculations\n",
- "\n",
- "A=b*d #m**2 #Area of pier\n",
- "I_x_x=b*d**3*12**-1 #m**4 #M.I about x-x axis\n",
- "I_y_y=d*b**3*12**-1 #m**4 #M.I about y-y axis\n",
- "M_x=P*e_y #KN*m #Moment about x-x axis\n",
- "M_y=P*e_x #KN*m #Moment about y-y axis\n",
- "\n",
- "x=2 #m #Distance between y-y axis and corners A and B\n",
- "y=1.5 #m ##Distance between x-x axis and corners A and D\n",
- "\n",
- "#Part-1\n",
- "#Stress developed at each corner\n",
- "\n",
- "\n",
- "sigma_A=P*A**-1+M_x*y*I_x_x**-1-M_y*x*I_y_y**-1 #KN/m**2 #stress at A\n",
- "sigma_B=P*A**-1+M_x*y*I_x_x**-1+M_y*x*I_y_y**-1 #KN/m**2 #stress at B\n",
- "sigma_C=P*A**-1-M_x*y*I_x_x**-1+M_y*x*I_y_y**-1 #KN/m**2 #stress at C\n",
- "sigma_D=P*A**-1-M_x*y*I_x_x**-1-M_y*x*I_y_y**-1 #KN/m**2 #stress at D\n",
- "\n",
- "#Part-2\n",
- "#Let f be the additional load that should be placed at centre\n",
- "\n",
- "#sigma_c=F*A**-1 #KN/m**2 #compressive stress\n",
- "\n",
- "#For no tension in pier section, compressive stress is equal to tensile stress\n",
- "sigma_c=10 #KN/m**2\n",
- "F=sigma_c*A #KN\n",
- "\n",
- "#Part-3\n",
- "\n",
- "sigma=F*A**-1 #KN/m**2 #stress due to additional load of 120 KN\n",
- "\n",
- "sigma_A_1=sigma_A+10 #stress at A\n",
- "sigma_B_1=sigma_B+10 #stress at B\n",
- "sigma_C_1=sigma_C+10 #stress at C\n",
- "sigma_D_1=sigma_D+10 #stress at D\n",
- "\n",
- "#Result\n",
- "print\"Stress at each corner are as follows:stress_A\",round(sigma_A,2),\"KN/m**2\"\n",
- "print\" :stress_B\",round(sigma_B,2),\"KN/m**2\"\n",
- "print\" :stress_C\",round(sigma_C,2),\"KN/m**2\"\n",
- "print\" :stress_D\",round(sigma_D,2),\"KN/m**2(tensile)\"\n",
- "\n",
- "print\"Additional load that should be placed at centre is\",round(F,2),\"KN\"\n",
- "\n",
- "print\"Stresses at the corners with the additional load in centre are as follows:Stress_A_1\",round(sigma_A_1,2),\"KN/m**2\"\n",
- "print\" :Stress_B_1\",round(sigma_B_1,2),\"KN/m**2\"\n",
- "print\" :Stress_C_1\",round(sigma_C_1,2),\"KN/m**2\"\n",
- "print\" :Stress_D_1\",round(sigma_D_1,2),\"KN/m**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Stress at each corner are as follows:stress_A 3.33 KN/m**2\n",
- " :stress_B 23.33 KN/m**2\n",
- " :stress_C 10.0 KN/m**2\n",
- " :stress_D -10.0 KN/m**2(tensile)\n",
- "Additional load that should be placed at centre is 120.0 KN\n",
- "Stresses at the corners with the additional load in centre are as follows:Stress_A_1 13.33 KN/m**2\n",
- " :Stress_B_1 33.33 KN/m**2\n",
- " :Stress_C_1 20.0 KN/m**2\n",
- " :Stress_D_1 0.0 KN/m**2\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 11.11.6,Page no.278"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "#d=Diameter of rod\n",
- "P=500 #KN\n",
- "e=0.75 #cm #eccentricity\n",
- "\n",
- "#calculation\n",
- "\n",
- "#A=pi*d**2*4**-1 #cm**2 #Area of rod\n",
- "#sigma_d=P*A**-1 #KN/cm**2 #stress due to direct load\n",
- "\n",
- "#After substituting value and simplifying we get,\n",
- "#sigma_d=2000*(pi*d**2)**-1 #KN/cm**2 \n",
- "\n",
- "M=P*e #Kn*cm #Moment\n",
- "\n",
- "#Z=pi*d**3*32**-1 #cm**3 #section modulus\n",
- "#sigma_b=M*Z**-1 #KN/cm**2 #Stress due to moment\n",
- "\n",
- "#After substituting value and simplifying we get,\n",
- "#sigma_b=12000*(pi*d**3)**-1 #KN/cm**2\n",
- "\n",
- "#Max stress \n",
- "#sigma=sigma_d+sigma_b \n",
- "\n",
- "#After substituting value and simplifying we get,\n",
- "#2000*(pi*d**2)**-1+12000*(pi*d**3)**-1=12.5\n",
- "\n",
- "#After simplifying we get,\n",
- "#d**3-53.05*d-318.3=0\n",
- "\n",
- "#From Synthetic Division we get d**2+4.73*d-42.918\n",
- "a=1\n",
- "b=-4.73\n",
- "c=-42.918\n",
- "\n",
- "X=b**2-(4*a*c)\n",
- "\n",
- "d_1=(-b+X**0.5)*(2*a)**-1\n",
- "d_2=(-b-X**0.5)*(2*a)**-1\n",
- "\n",
- "#Result\n",
- "print\"The minimum diameter of tie rod is\",round(d_1,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The minimum diameter of tie rod is 9.33 cm\n"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file
diff --git a/Strength_Of_Materials_by_B_K_Sarkar/chapter_12_som.ipynb b/Strength_Of_Materials_by_B_K_Sarkar/chapter_12_som.ipynb
deleted file mode 100755
index fd200d5e..00000000
--- a/Strength_Of_Materials_by_B_K_Sarkar/chapter_12_som.ipynb
+++ /dev/null
@@ -1,425 +0,0 @@
-{
- "metadata": {
- "name": "chapter 12 som.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 12:Propped Cantilever"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 12.1,Page No.286"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=6 #m #Length of Beam\n",
- "L_1=4 #m #Length of Beam with udl Load\n",
- "w=10 #KN/m #u.d.l\n",
- "\n",
- "#Calculation\n",
- "\n",
- "#Deflection of cantileverat C due to udl on AB\n",
- "y_c=w*L_1**4*8**-1+w*L_1**3*6**-1*(L-L_1) \n",
- "\n",
- "#Deflection of cantileverat C due to prop reaction alone\n",
- "#y_c_2=R_c*L**3*3**-1\n",
- "\n",
- "#Since both Deflection are Equal\n",
- "#y_c=y_c_2\n",
- "\n",
- "R_c=y_c*(6**3)**-1*3 #Reaction at C\n",
- "\n",
- "#Result\n",
- "print\"The Reaction at End C is\",round(R_c,3),\"KN\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Reaction at End C is 7.407 KN\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 12.2,Page No.286"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=10 #m #Length\n",
- "b=15 #cm #Width\n",
- "d=40 #cm #Depth\n",
- "y_c=1.5*10**-2 #m #Deflection\n",
- "E=120*10**9 \n",
- "y=0.2\n",
- "sigma=10*10**6 #Bending stress\n",
- "\n",
- "#Calculations\n",
- "\n",
- "I=b*d**3*12**-1*10**-8 #cm #M.I\n",
- "\n",
- "#From Deflection at the centre of cantilever we get\n",
- "w=y_c*192*E*I*(L**4)**-1*10**-3 #udl distributed over the cantilever\n",
- "\n",
- "#From Bending Moment Equation we get\n",
- "w_2=sigma*I*y**-1*16*(L**2)**-1*10**-3 #udl distributed over the cantilever\n",
- "\n",
- "#Result\n",
- "print\"The safe uniformly Distributed Load is\",round(w_2,2),\"KN/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The safe uniformly Distributed Load is 6.4 KN/m\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 12.3,Page No.287"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=6 #m #span of beam\n",
- "w=30*10**3 #KN/m #u.d.l\n",
- "P=160*10**3 #N #concentrated Load\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Consider a section at a distance x from the fixed end A and B.M at x\n",
- "#M_x=R_b*(6-x)-30*2**-1*(6-x)**2-160*(3-x)\n",
- "\n",
- "#E*I*d**2y*(dx**2)**-1=-M_x=-R_b*(6-x)+15*(6-x)+160*(3-x)\n",
- "\n",
- "#Now Integrating above term we get\n",
- "#E*I*dy*(dx)**-1=R_b*2**-1*(6-x)**2-5*(6-x)**3-80*(3-x)**2+C_1 (Equation 1)\n",
- "\n",
- "#Now on Integrating we get\n",
- "#E*I*y=-R_b*6**-1*(6-x)**3+5*4**-1*(6-x)**2+80*3**-1*(3-x)**3+C_1*x+C_2 (Equation 2)\n",
- "\n",
- "#At x=0,dy*dx**-1=0\n",
- "#substituting in equation 1 we get\n",
- "#C_1=1800-R_b\n",
- "\n",
- "#At x=0,y=0\n",
- "#substituting in equation 2 we get\n",
- "#C_2=36*R_b-2340\n",
- "\n",
- "#At x=6,y=0\n",
- "R_b=72**-1*(10800-2340)\n",
- "\n",
- "#At x=0\n",
- "x=0\n",
- "M_x=R_b*(6-x)-30*2**-1*(6-x)**2-160*(3-x)\n",
- "\n",
- "#Result\n",
- "print\"Bending Moment at A is\",round(M_x,2),\"KNm\"\n",
- "print\"The Reaction at B\",round(R_b,2),\"KN\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Bending Moment at A is -315.0 KNm\n",
- "The Reaction at B 117.5 KN\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 12.4,Page No.288"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=4 #span of beam\n",
- "w_1=20*10**3 #Nm #u.d.l\n",
- "w_2=30*10**3 #Nm #u.d.l\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#consider a section at a distance x from A and B.M at this section is \n",
- "#M_x=R_b*(3-x)-10*x**2+90*x-195\n",
- "\n",
- "#Now integrating above equation we get\n",
- "#E*I*dy*(dx)**-1=-R_b(3*x-x**2*2**-1)+10*x**3*3**-1-45*x**2+195*x+C_1\n",
- "\n",
- "#again on Integrating we get\n",
- "#E*I*y=-R_b*(3*x**2*2**-1-x**3*6**-1)+10*x**4*12**-1-15*x**3+195*x**2*2**-1+C_1*x+C_2\n",
- "\n",
- "#At x=0,dy*(dx)**-1=0 Therefore C_1=0\n",
- "\n",
- "#At x=0,y=0 Therefore C_2=0\n",
- "\n",
- "#At x=3m, y=0\n",
- "x=3\n",
- "C_1=0\n",
- "C_2=0\n",
- "R_b=-(-10*x**4*12**-1+15*x**3-195*x**2*2**-1-C_1*x-C_2)*(3*x**2*2**-1-x**3*6**-1)**-1\n",
- "\n",
- "#result\n",
- "print\"Load taken by prop is\",round(R_b,2),\"KN\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Load taken by prop is 60.0 KN\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 12.5,Page No.289"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=2 #m #Span of beam\n",
- "w=10 #KN/m #u.d.l\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Downward deflection at B(of Beam AB) due to u.d.l of 10 KN/m is\n",
- "Y_B_1=w*L**4*8**-1 \n",
- "\n",
- "#Upward deflection at B due to reaction at C alone is\n",
- "#Y_B_2=R_c*8*3**-1\n",
- "\n",
- "#Net downward deflection of cantilever at AB at B\n",
- "#Y_B=Y_B_1-Y_B_2\n",
- "\n",
- "#Downward Deflection of Beam CD at C due to the reaction\n",
- "#R_c=R_c*(3*E*I)**-1\n",
- "\n",
- "#since both deflection at C and B are equal\n",
- "R_c=20*(1*3**-1+8*3**-1)**-1\n",
- "\n",
- "#Result\n",
- "print\"Reaction at C is\",round(R_c,2),\"KN\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Reaction at C is 6.67 KN\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 12.8,Page No.292"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=8 #m #span\n",
- "W=24*10**3 #N/m #U.D.L\n",
- "y=2*10**-2 #m #deflection\n",
- "E=20*10**9\n",
- "I=10**-5 #m**4\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#The Downward deflection at C Due to u.d.l\n",
- "#Y_c=5*W*L**3*(384*E*I)**-1\n",
- "\n",
- "#The Upward Deflection at C due to prop Reaction P \n",
- "#Y_c_1=P*L**3*(48*E*I)**-1\n",
- "\n",
- "#Since the prop is at the same level as end supports\n",
- "#Y_c_1=Y_c\n",
- "P_1=5*W*8**-1*10**-3 #KN\n",
- "\n",
- "#The reaction at A and B is equal\n",
- "R_a=R_b=(24-15)*2**-1\n",
- "\n",
- "#Shear Force at B\n",
- "V_B=4.5 #KN\n",
- "\n",
- "#Shear Force at C\n",
- "V_C1=4.5-24*2**-1\n",
- "V_C2=4.5-24*2**-1+15\n",
- "\n",
- "#Shea rForce at A\n",
- "V_A=-4.5 #KN\n",
- "\n",
- "#B.M at C due to u.d.l\n",
- "M_C1=W*L*8**-1*10**-3 #KN*m\n",
- "\n",
- "#B.M due to only prop reaction P=15 KN\n",
- "P=15\n",
- "M_C2=-P*L*4**-1 #KN*m\n",
- "\n",
- "#B.M at D\n",
- "M_D=4.5*1.5-24*8**-1*1.5**2*2**-1\n",
- "\n",
- "#In second case prop sinks by 2 cm\n",
- "#Y_c-Y_c_1=2 \n",
- "\n",
- "#So Further simplifying and sunstituting values in above equation we get\n",
- "P=-(2*100**-1-(5*W*L**3*(384*E*I)**-1))*(L**3*(48*E*I)**-1)**-1\n",
- "\n",
- "#Let Each end reaction be X\n",
- "X=(24-14.625)*2**-1\n",
- "\n",
- "#Result\n",
- "print\"prop reaction is\",round(P_1,2),\"KN\"\n",
- "print\"The End Reaction is\",round(X,2),\"KN\"\n",
- "\n",
- "#Plotting the SHear Force Diagram\n",
- "\n",
- "X1=[0,4,4,8,8]\n",
- "Y1=[V_B,V_C1,V_C2,V_A,0]\n",
- "Z1=[0,0,0,0,0]\n",
- "plt.plot(X1,Y1,X1,Z1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Force in kN\")\n",
- "plt.show()\n",
- "\n",
- "#Plotting the Bendimg Moment Diagram\n",
- "\n",
- "X2=[0,4,4]\n",
- "Y2=[0,M_C1,0]\n",
- "Z2=[0,0,0]\n",
- "plt.plot(X2,Y2)\n",
- "plt.xlabel(\"Lenght in m\")\n",
- "plt.ylabel(\"Bending Moment in kN.m\")\n",
- "plt.show()"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "prop reaction is 15.0 KN\n",
- "The End Reaction is 4.69 KN\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
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- "text": [
- "<matplotlib.figure.Figure at 0x56bdc90>"
- ]
- },
- {
- "metadata": {},
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- "text": [
- "<matplotlib.figure.Figure at 0x5651370>"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [],
- "language": "python",
- "metadata": {},
- "outputs": []
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file
diff --git a/Strength_Of_Materials_by_B_K_Sarkar/chapter_13_som.ipynb b/Strength_Of_Materials_by_B_K_Sarkar/chapter_13_som.ipynb
deleted file mode 100755
index 6f724b86..00000000
--- a/Strength_Of_Materials_by_B_K_Sarkar/chapter_13_som.ipynb
+++ /dev/null
@@ -1,694 +0,0 @@
-{
- "metadata": {
- "name": "chapter 13 som.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Chapter No.13:Shear Stress in Beams"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 13.2,Page No.301"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "#W=10*w #KN/m #u.d.l\n",
- "sigma=805*10**6 #Pa #Bending stress\n",
- "Tou=0.85*10**6 #Pa #Shear stress\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#M=W*L**2*10**-4*8**-1 #Max B.M\n",
- "#F=W*L*10**-2*2**-1 #Max S.F\n",
- "#y=h*2**-1 #depth\n",
- "#A-b*h #Area of c/s\n",
- "\n",
- "#Now using relation we get\n",
- "#sigma=M*h*(2*I)**-1 #Bending stress\n",
- "\n",
- "#AFter substituitng values we get\n",
- "#805*10**6=w*l**2*h*(16*10**5*I)**-1 #Equation 1\n",
- "\n",
- "#Again using the relation we get \n",
- "#tou=F*A*y_bar*(I*b)**-1 #shear atress\n",
- "\n",
- "#AFter substituitng values we get\n",
- "#0.85*10**6=w*L*h**2*(16*10**5*I)**-1 #Equation 2\n",
- "\n",
- "#Dividing equation 1 & 2 we get\n",
- "#L*h**-1=10\n",
- "#Let L*h**-1=Z\n",
- "z=10\n",
- "\n",
- "#Result\n",
- "print\"The Ratio of span to depth ratio is\",round(z,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Ratio of span to depth ratio is 10.0\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 13.3,Page No.302"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=2 #m #span\n",
- "w=20*10**3 #N/m #u.d.L\n",
- "b=12.5 #cm #width of Flange\n",
- "t=2.5 #cm #flange thickness\n",
- "w_t=2.5 #cm #web thickness\n",
- "D=20 #cm #Overall depth\n",
- "w_d=17.5 #m #Depth of web\n",
- "\n",
- "#Calculations\n",
- "\n",
- "F=w*L*2**-1 #N #Max S.F\n",
- "a_1=b*t #Area of flange\n",
- "a_2=w_d*w_t #Area of web\n",
- "y_1=t*2**-1 #C.G of flange\n",
- "y_2=w_d*2**-1+t #C.G of web\n",
- "\n",
- "#C.G of c/s\n",
- "Y=(a_1*y_1+a_2*y_2)*(a_1+a_2)**-1 \n",
- "\n",
- "#M.I about N.A\n",
- "I=b*t**3*12**-1+b*t*(Y-y_1)**2+w_t*w_d**3*12**-1+w_t*w_d*(y_2-Y)**2 \n",
- "\n",
- "#Shear Stress in flange at the junction with web\n",
- "#Let tou(Shear stress)=S \n",
- "#Change in the notifications of Shear Stress For convenience\n",
- "S_1=(F*a_1*(Y-y_1)*10**-6)*(I*10**-8*b*10**-2)**-1*10**-3 \n",
- "\n",
- "#Shear Stress in web at the junction with flange \n",
- "S_2=(F*a_1*(Y-y_1)*10**-6)*(I*10**-8*w_t*10**-2)**-1*10**-3 \n",
- "\n",
- "#Max Shear Stres at N.A\n",
- "S_max=(F*(a_1*(Y-y_1)+(w_t*(Y-t))*((Y-t)*2**-1))*10**-6)*(I*10**-8*w_t*10**-2)**-1*10**-3\n",
- "\n",
- "#Result\n",
- "print\"The Max shear stress in the beam is\",round(X_max,2),\"KN/m**2\"\n",
- "\n",
- "print\"Shear stress distribution Diagram\"\n",
- "\n",
- "#Plotting the Shear stress distribution Diagram\n",
- "\n",
- "X_1=[0,2.5,2.5,4.58,15.42]\n",
- "Y_1=[0,S_1,S_2,S_max,0]\n",
- "Z_1=[0,0,0,0,0]\n",
- "plt.plot(X_1,Y_1,X_1,Z_1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Stress in kN/m**2\")\n",
- "plt.show()"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Max shear stress in the beam is 5644.64 KN/m**2\n",
- "Shear stress distribution Diagram\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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MmTKFn3/+2W4BhRA1T+PG8NlnUFgIw4bB+fNGJxLVqczCoZQiPz+f4uJiPv/8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\nHEIIISpFCocQQohK+f/7gGcS8RurtQAAAABJRU5ErkJggg==\n",
- "text": [
- "<matplotlib.figure.Figure at 0x57624b0>"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 13.4,Page No.303"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=100*10**3 #N #Shear Force\n",
- "I=11340*10**-8 #m**4 #M.I\n",
- "b=20 #cm #width of Flange\n",
- "t=5 #cm #thickness of flange\n",
- "w_d=20 #cm #Depth of web\n",
- "w_t=5 #cm #thickness of web\n",
- "\n",
- "#Calculations\n",
- "\n",
- "a_1=b*t #cm**2 #Area of flange\n",
- "a_2=w_d*w_t #cm**2 #Area of web\n",
- "y_1=t*2**-1 #cm #C.G of flange\n",
- "y_2=t+w_d*2**-1\n",
- "\n",
- "#C.G of C/s\n",
- "Y=(a_1*y_1+a_2*y_2)*(a_1+a_2)**-1\n",
- "\n",
- "#Shear Stress in flange at the junction with web\n",
- "#Let tou(Shear stress)=S \n",
- "#Change in the notifications of Shear Stress For convenience\n",
- "S_1=(F*a_1*(Y-y_1)*10**-6)*(I*b*10**-2)**-1*10**-3 \n",
- "\n",
- "#Shear Stress in web at the junction with flange \n",
- "S_2=(F*a_1*(Y-y_1)*10**-6)*(I*w_t*10**-2)**-1*10**-3 \n",
- "\n",
- "#Max Shear Stres at N.A\n",
- "S_max=(F*(a_1*(Y-y_1)+(w_t*(Y-t))*((Y-t)*2**-1))*10**-6)*(I*w_t*10**-2)**-1*10**-3\n",
- "\n",
- "#Result\n",
- "print\"Shear Stress in flange at the junction with web\",round(S_1,2),\"KN/m**2\"\n",
- "print\"Shear Stress in web at the junction with flange\",round(S_2,2),\"KN/m**2\"\n",
- "print\"Max Shear Stres at N.A\",round(S_max,2),\"KN/m**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Shear Stress in flange at the junction with web 2755.73 KN/m**2\n",
- "Shear Stress in web at the junction with flange 11022.93 KN/m**2\n",
- "Max Shear Stres at N.A 11642.97 KN/m**2\n"
- ]
- }
- ],
- "prompt_number": 38
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 13.5,Page No.304"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "from scipy.integrate import quad\n",
- "\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "D=50 #cm #Overall depth\n",
- "b=19 #cm #width of flange\n",
- "t=2.5 #cm #Thickness of Flange\n",
- "w_t=1.5 #cm #Web thickness\n",
- "w_d=45 #cm #web thickness\n",
- "F=400*10**3 #N #Shear Force\n",
- "I=64500*10**-8 #m**4 #M.I\n",
- "\n",
- "#Calculations (Part-1)\n",
- " \n",
- "a_1=b*t #cm**2 #Area of flange\n",
- "a_2=w_d*w_t #cm**2 #Area of web\n",
- "y_1=t*2**-1 #cm #C.G of flange\n",
- "y_2=t+w_d*2**-1\n",
- "\n",
- "#As section is symmetrical \n",
- "Y=D*2**-1 #cm\n",
- "\n",
- "#Shear Stress in flange at the junction with web\n",
- "#Let tou(Shear stress)=S \n",
- "#Change in the notifications of Shear Stress For convenience\n",
- "S_1=(F*a_1*(Y-y_1)*10**-6)*(I*b*10**-2)**-1*10**-3 \n",
- "\n",
- "#Shear Stress in web at the junction with flange \n",
- "S_2=(F*a_1*(Y-y_1)*10**-6)*(I*w_t*10**-2)**-1*10**-3 \n",
- "\n",
- "#Max Shear Stres at N.A\n",
- "S_max=(F*(a_1*(Y-y_1)+(w_t*(Y-t))*((Y-t)*2**-1))*10**-6)*(I*w_t*10**-2)**-1*10**-3\n",
- "\n",
- "#Calculations (Part-2)\n",
- "\n",
- "#consider a strip in the flange of thickness dy at a distance y from N.A \n",
- "\n",
- "#S=F*(b*(Y-y)*(Y+y)*2**-1*10**-6)*(I*b*10**-2)**-1\n",
- "#after substituting values we get\n",
- "#S=625-y**2*(3225*10**-8)**-1\n",
- "\n",
- "#shear force carried by small strip\n",
- "#F_1=625-y**2*(3225*10**-8)**-1*b*dy*10**-4\n",
- "\n",
- "#Now Integrating above Equation we get\n",
- "def integrand(y, a, b):\n",
- " return 625-y**2\n",
- "a =625\n",
- "b =-1\n",
- "I = quad(integrand, 22.5, 25, args=(a,b))\n",
- "\n",
- "#Shear force carried by one flange\n",
- "F_1=19*3225**-1*10**4*I[0]\n",
- "\n",
- "#Shear force carried by two flange\n",
- "F_2=2*F_1\n",
- "\n",
- "#Shear force carried by web\n",
- "F_3=F-F_2\n",
- "\n",
- "#Result\n",
- "print\"The shear Force int the section is\",round(S_max,2),\"Pa\"\n",
- "print\"Total Shear Force in the web is\",round(F_3,2),\"N\"\n",
- "\n",
- "\n",
- "print\"Shear stress distribution Diagram\"\n",
- "\n",
- "#Plotting the Shear stress distribution Diagram\n",
- "\n",
- "X_1=[0,2.5,2.5,25,47.5,47.5,50]\n",
- "Y_1=[0,S_1,S_2,S_max,S_2,S_1,0]\n",
- "Z_1=[0,0,0,0,0,0,0]\n",
- "plt.plot(X_1,Y_1,X_1,Z_1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Stress in kN/m**2\")\n",
- "plt.show()"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The shear Force int the section is 62338.5 MPa\n",
- "Total Shear Force in the web is 382202.84 N\n",
- "Shear stress distribution Diagram\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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LC/n2t7/NxIkTeeyxx5o08+7duwNQWVlJdXU1119/PRkZGSQlJQGQlJTE5s2b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\nUERExC0UKCIi4hYKFBERcYv/D6L3iyTClBJuAAAAAElFTkSuQmCC\n",
- "text": [
- "<matplotlib.figure.Figure at 0x5633a10>"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 13.6,Page No.305"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F=5*10**3 #N #shea Force\n",
- "b=20 #cm #width of Flange\n",
- "t=6 #cm #Thickness of flange\n",
- "w_d=20 #cm #depth of web\n",
- "w_t=6 #cm #thickness of web\n",
- "X=700 #N #Shear Looad\n",
- "\n",
- "#Calculations\n",
- "\n",
- "a_1=b*t #cm**2 #Area ofFlange\n",
- "a_2=w_d*w_t #cm**2 #Area of web\n",
- "y_1=t*2**-1 #cm #C.G of Flange\n",
- "y_2=t+w_d*2**-1 #cm #C.G of Web\n",
- "\n",
- "Y=(a_1*y_1+a_2*y_2)*(a_1+a_2)**-1\n",
- "\n",
- "#M.I about N.A\n",
- "I=b*t**3*12**-1+b*t*(Y-y_1)**2+w_t*w_d**3*12**-1+w_t*w_d*(y_2-Y)**2 \n",
- "\n",
- "#Shear Force per metre Length in Plane of contact of two Planks \n",
- "#Let Shear Force per metre Length=F_1 \n",
- "F_1=(F*a_1*(Y-y_1)*10**-6)*(I*10**-8)**-1\n",
- "\n",
- "#Spacing of nails\n",
- "s=X*F_1**-1*100\n",
- "\n",
- "#Result\n",
- "print\"The spacing of nails along the Length of beam is\",round(s,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The spacing of nails along the Length of beam is 2.6 cm\n"
- ]
- }
- ],
- "prompt_number": 74
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 13.7,Page No.306"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=3 #m #span\n",
- "d=5 #cm #depth of each plank\n",
- "b=15 #cm #width of plank\n",
- "d_1=1.9 #cm #Diameter of bolt\n",
- "s=12.5 #cm #spacing of bolt\n",
- "w=3.3*10**3 #N.m #u.d.l\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Shear Force at 1.5m from support\n",
- "F=w*1.5\n",
- "\n",
- "I=b*(5*d)**3*12**-1 #M.I\n",
- "A=pi*4**-1*d_1**2 #area of Bolt\n",
- "Y=5*d*2**-1 #C.G of beam\n",
- "y_1=d*2**-1 #c.G of top plank\n",
- "\n",
- "#Shear Force per metre Length \n",
- "F_1=F*b*d*(Y-y_1)*10**-6*(I*10**-8)**-1\n",
- "\n",
- "#Load carried by bolt\n",
- "W_1=F_1*s*10**-2\n",
- "\n",
- "#shear stress\n",
- "X_1=W_1*(round(A,3))**-1*10**+4\n",
- "\n",
- "#Shear Force per metre Length \n",
- "F_2=F*b*2*d*((d+y_1)-Y*10**-6)*(I*10**-8)**-1*10**-6\n",
- "\n",
- "#Load carried by bolt\n",
- "W_2=F_2*s*10**-2\n",
- "\n",
- "#shear stress\n",
- "X_2=W_2*(A*10**-4)**-1*10**-3\n",
- "\n",
- "#Reult\n",
- "print\"Shear stress in a bolt Located at 1.5 m from support is\",round(X_2,2),\"KN/m**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Shear stress in a bolt Located at 1.5 m from support is 12570.13 KN/m**2\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 13.8,Page No.307"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\\\n",
- "\n",
- "b=15 #cm #width of plank\n",
- "t=2.5 #cm #thickness of planf\n",
- "F_1=1250 #N #Shear Force\n",
- "F_2=5*10**3 #shaear force transmitted by screw\n",
- "d=15 #cm #Depth of plank\n",
- "D=20 #cm #Overall depth\n",
- "\n",
- "#Calculations\n",
- "\n",
- "Y=D*2**-1 #C.G of beam\n",
- "y_1=t*2**-1 #C.G of flange\n",
- "\n",
- "I=((b*D**3)-(D*2**-1*b**3))*12**-1 #cm**4 #M.I\n",
- "\n",
- "#Shear Stress in the Flange at 7.5 cm from N.A\n",
- "X_1=F_2*b*t*(Y-y_1)*10**-6*(I*10**-8*d*10**-2)**-1*10**-3\n",
- "\n",
- "#Shear Stress in the web at 7.5 cm from N.A\n",
- "X_2=X_1*d*(2*t)**-1\n",
- "\n",
- "#shear stress at N.A\n",
- "X_max=F_2*(b*t*(Y-y_1)+2*t*d*2**-1*d*4**-1)*10**-6*(I*10**-8*2*t*10**-2)**-1*10**-3\n",
- "\n",
- "#horizontal shear force per pitch length to the shearing strength of two bolts we have\n",
- "#X_h=X_2*10**3*2*t*10**-2*p\n",
- "\n",
- "#Equating horizontal shear force per pitch length to the shearing strength of two bolts we have\n",
- "p=F_1*2*(X_2*10**3*2*t*10**-2)**-1*10**2\n",
- "\n",
- "#Result\n",
- "print\"The Min spacing of screw along the beam is\",round(p,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Min spacing of screw along the beam is 10.95 cm\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 13.9,Page No.308"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=4 #m #span\n",
- "w=80*10**3 #N/m #u.d.l\n",
- "D=35 #cm #Overall depth\n",
- "b=15 #cm #width of Flange\n",
- "t=2.5 #cm #Thickness of flange\n",
- "w_d=30 #cm #Depth of web\n",
- "w_t=1.2 #cm #thickness of web\n",
- "\n",
- "#Calculations\n",
- "\n",
- "R_a=R_b=160 #KN #Reactions at supports\n",
- "\n",
- "#Shear FOrce at 1m from left support\n",
- "F=R_a*10**3-w\n",
- "\n",
- "M=R_a*10**3-w*2**-1 #B.M at 1m From support\n",
- "\n",
- "I=(b*D**3-((b-w_t)*w_d**3))*12**-1 #cm**4\n",
- "\n",
- "y=w_d*2**-1\n",
- "sigma=M*I**-1*y #N/m**2\n",
- "\n",
- "#Shear stress in Flange at the junction with web\n",
- "X_1=w*b*t*(w_d*2**-1+t*2**-1)*10**-6*(I*10**-8*b*10**-2)**-1*10**-3\n",
- "\n",
- "#Shear stress in web at the junction with Flange \n",
- "X_2=X_1*15*1.2**-1\n",
- "\n",
- "#Result\n",
- "print\"The Magnitude of Bending is\",round(sigma,2),\"N/m**2\"\n",
- "print\"Shear stress in web at the junction with Flange\",round(X_1,2),\"KN/m**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Magnitude of Bending is 79.84 N/m**2\n",
- "Shear stress in web at the junction with Flange 1441.64 KN/m**2\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 13.10,Page No.309"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "b=25 #cm #width of top FLange\n",
- "t=5 #cm #thickness of top Flange\n",
- "D=35 #cm #Depth of overall section\n",
- "w_d=25 #cm #depth of web\n",
- "w_t=5 #cm #thickness of web\n",
- "t_1=5 #cm #thickness of bottom Flange\n",
- "b_1=15 #cm #width of bottom Flange\n",
- "sigma=17.5*10**6\n",
- "F=100*10**3 #N #S.F\n",
- "\n",
- "#Calculations\n",
- "\n",
- "a_1=b*t #area of top flange\n",
- "a_2=w_d*w_t #area of web\n",
- "a_3=b_1*t_1 #area of bottom Flange\n",
- "y_1=t*2**-1 #C.G of top flange\n",
- "y_3=D-(t_1*2**-1) #C.G of bottom Flange \n",
- "y_2=D*2**-1 #c.G of Web\n",
- "\n",
- "Y=(a_1*y_1+a_2*y_2+a_3*y_3)*(a_1+a_2+a_3)**-1\n",
- "\n",
- "I=b*t**3*12**-1+b*t*(Y-y_1)**2+w_t*w_d**3*12**-1+w_t*w_d*(D*2**-1-Y)**2+b_1*t_1**3*12**-1+b_1*t_1*(y_3-Y)**2\n",
- "\n",
- "M=sigma*I*10**-8*(Y*10**-2)**-1 #B.M\n",
- "\n",
- "#Shear Stress in upper Flange at the junction with web\n",
- "S_1=F*b*t*(Y-y_1)*10**-6*(I*10**-8*b*10**-2)**-1*10**-3\n",
- "\n",
- "#Shear Stress in web at the junction with upper Flange \n",
- "S_2=S_1*b*t**-1\n",
- "\n",
- "#Max shear stress at the N.A\n",
- "S=F*(b*t*(Y-y_1)+w_t*(Y-t)*(Y-t)*2**-1)*10**-6*(I*10**-8*w_t*10**-2)**-1*10**-3\n",
- "\n",
- "#Shear Stress in Lower Flange at the junction with web\n",
- "S_3=F*(a_3*(D-Y-t_1*2**-1))*10**-6*(I*10**-8*b_1*10**-2)**-1*10**-3\n",
- "\n",
- "#Shear Stress in web at the junction with Lower Flange \n",
- "S_4=S_3*b_1*t_1**-1\n",
- "\n",
- "#Result\n",
- "print\"The Bending Moment section can take is\",round(M,2),\"N-m\"\n",
- "print\"The shear stress Distribution Diagram\"\n",
- "\n",
- "#Plotting the Shear stress distribution Diagram\n",
- "\n",
- "X_1=[0,5,5,15.19,30,30,35]\n",
- "Y_1=[0,S_1,S_2,S,S_3,S_4,0]\n",
- "Z_1=[0,0,0,0,0,0,0]\n",
- "plt.plot(X_1,Y_1,X_1,Z_1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Stress in kN/m**2\")\n",
- "plt.show()\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Bending Moment section can take is 57821.07 N-m\n",
- "The shear stress Distribution Diagram\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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Dx1mEcDft2sH778OuXfDss3pHI8T5LrrkSE2Lo3bLw17ffPMNM2fOpKKiAl9fX9544w2q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\noJRixowZBAUF8eCDD9qOx8b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- "text": [
- "<matplotlib.figure.Figure at 0x5641790>"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [],
- "language": "python",
- "metadata": {},
- "outputs": []
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file
diff --git a/Strength_Of_Materials_by_B_K_Sarkar/chapter_14_som.ipynb b/Strength_Of_Materials_by_B_K_Sarkar/chapter_14_som.ipynb
deleted file mode 100755
index 636bc8f6..00000000
--- a/Strength_Of_Materials_by_B_K_Sarkar/chapter_14_som.ipynb
+++ /dev/null
@@ -1,971 +0,0 @@
-{
- "metadata": {
- "name": "chapter 14 som.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 14:Dams and Retaining Walls"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 14.1,Page No.325"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "b=2 #m #width \n",
- "FOS=1.5 #Factor of safety\n",
- "#rho_mason=2.5*rho_w\n",
- "mu=0.5 #coeffeicient of friction\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Let L=1 m (length of dam)\n",
- "L=1\n",
- "#W=b*H*L*rho\n",
- "#After substituting values and Further simplifying we get\n",
- "#W=2*H*rho\n",
- "\n",
- "#Total Pressure\n",
- "#P=W*H**2*2**-1\n",
- "\n",
- "x_bar=b*2**-1 #Distance of Line of action of W from waterface\n",
- "\n",
- "#Distance of pt where resultant cuts the base measured from Line of action\n",
- "#x=P*W**-1*H*3**-1\n",
- "#After substituting values and Further simplifying we get\n",
- "#x=H**2*30**-1\n",
- "\n",
- "#x_bar+x=2*b*3**-1\n",
- "#After substituting values and Further simplifying we get\n",
- "#1+H**2*30**-1=2*b*3**-1\n",
- "H=(30*(2*b*3**-1-1))**0.5 #height of dam\n",
- "\n",
- "#Frictional Resistance offered at the base\n",
- "#F=mu*W\n",
- "#After substituting values and Further simplifying we get\n",
- "#F=3.16*rho\n",
- "\n",
- "#Total Lateral Pressure\n",
- "#P=W*H**2*2**-1\n",
- "#P=4.99*W\n",
- "\n",
- "#Factor of safety against sliding\n",
- "#FOS1=F*P**-1=3.16*4.99**-1*rho_mason*rho_w**-1\n",
- "FOS1=3.16*4.99**-1*2.5\n",
- "\n",
- "#FOS1>FOS\n",
- "\n",
- "#Result\n",
- "print\"Dam is safe against sliding\",round(FOS1,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Dam is safe against sliding 1.58\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 14.2,Page No.327"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "D=2 #m #External Diameter\n",
- "d=1.5 #m #Internal Diameter\n",
- "P=1600 #N/m**2 #N/m**2 #Wind Pressure\n",
- "W=19200 #N/m**2 #Weight of masonry\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Let H be max height of dam\n",
- "\n",
- "#W2=pi*4**-1*(D**2-d**2)*H*W #weight of chimney\n",
- "#W2=26400*H\n",
- "\n",
- "#Eccentricrty\n",
- "x=(D**2+d**2)*(8*D)**-1\n",
- "\n",
- "#P2=H*D*P #Lateral thrust of wind on chimney\n",
- "#P2=3200*H\n",
- "\n",
- "#Now by using the relation we get P*W**-1=x*(H*2**-1)**-1\n",
- "#After substituting values and Further simplifying we get\n",
- "H=0.39*2*26400*3200**-1\n",
- "\n",
- "#result\n",
- "print\"The Height of Dam is\",round(H,2),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Height of Dam is 6.44 m\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 14.3,Page No.327"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "rho_w=10 #KN/m**3 #Density of water\n",
- "rho_mason=22.4 #KN/m**3 #Density of mason\n",
- "H=6 #m #height of dam\n",
- "a=1 #m #width of top\n",
- "b=4 #m #bottom width\n",
- "h=5.5 #m #Weight of water depth \n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Let L=1 m (length of dam)\n",
- "L=1\n",
- "\n",
- "#weight of dam\n",
- "W=(a+b)*2**-1*H*a*rho_mason\n",
- "\n",
- "#Lateral thrust\n",
- "P=rho_w*h**2*a*2**-1\n",
- "\n",
- "#distance of Line of action of W measured from vertical face\n",
- "x_bar=(b**2+b*a+a**2)*(3*(b+a))**-1\n",
- "\n",
- "#distance of pt where resultant cuts the base\n",
- "x=P*W**-1*h*3**-1\n",
- "\n",
- "#Eccentricity\n",
- "e=x_bar+x-b*2**-1\n",
- "\n",
- "#Stress at Pt B\n",
- "sigma1=W*b**-1*(1-6*e*b**-1)\n",
- "\n",
- "#stress at Pt C\n",
- "sigma2=W*b**-1*(1+6*e*b**-1)\n",
- "\n",
- "#Result\n",
- "print\"Max stress intensities at the base is\",round(sigma2,2),\"KN/m**2\"\n",
- "print\"Min stress intensities at the base is\",round(sigma1,2),\"KN/m**2\"\n",
- "\n",
- "#Plotting the Shear Force Diagram\n",
- "\n",
- "X1=[0,L,L]\n",
- "Y1=[sigma2,sigma1,0]\n",
- "Z1=[0,0,0]\n",
- "plt.plot(X1,Y1,X1,Z1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Force in kN\")\n",
- "plt.show()\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Max stress intensities at the base is 112.38 KN/m**2\n",
- "Min stress intensities at the base is 55.62 KN/m**2\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x55942b0>"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 14.4,Page No.329"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "H=10 #m #height od dam\n",
- "a=2 #m #top width\n",
- "b=5 #m #bottom width\n",
- "W=25 #KN/m**3 #weight of mason\n",
- "rho_w=10 #KN/m**3 #density of water\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Let L=1 m (length of dam)\n",
- "L=1\n",
- "\n",
- "#weight of dam\n",
- "W2=(b+a)*H*L*W*2**-1\n",
- "\n",
- "##Lateral thrust\n",
- "P=rho_w*H**2*L*2**-1 \n",
- "\n",
- "#Resultant thrust\n",
- "R=(P**2+W**2)**0.5\n",
- "\n",
- "#Distance of Line of action from vertical base\n",
- "x_bar=(b**2+b*a+a**2)*(3*(b+a))**-1\n",
- "\n",
- "##distance of pt where resultant cuts the base\n",
- "x=P*W2**-1*H*3**-1\n",
- "\n",
- "#Eccentricity\n",
- "e=x_bar+x-b*2**-1\n",
- "\n",
- "#Stress at Pt B\n",
- "sigma1=W2*b**-1*(1-6*e*b**-1)\n",
- "\n",
- "#stress at Pt C\n",
- "sigma2=W2*b**-1*(1+6*e*b**-1)\n",
- "\n",
- "#Result\n",
- "print\"The Resultant Thrust on the base is\",round(R,2),\"KN\"\n",
- "\n",
- "#Plotting the Shear Force Diagram\n",
- "\n",
- "X1=[0,L,L]\n",
- "Y1=[-sigma2,-sigma1,0]\n",
- "Z1=[0,0,0]\n",
- "plt.plot(X1,Y1,X1,Z1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Force in kN\")\n",
- "plt.show()\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Resultant Thrust on the base is 500.62 KN\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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KXKQmq9IxEH+lBHJ5p05Zc1W98AJMmACPPgpXXml3VCJiN69MZSKBwRhrXY4H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\nSGUdO3aM9evXM2zYMPe2wsJ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- "text": [
- "<matplotlib.figure.Figure at 0x5657d50>"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 14.5,Page No.330"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "H=30 #m #height of Dam\n",
- "h=29 #m #height of water\n",
- "rho_w=9810 #N/m**3 \n",
- "rho_mason=22560 #N/m**3\n",
- "sigma1=0 #KN/m**3\n",
- "sigma2=880 #KN/m**3\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Let L=1 m (length of dam)\n",
- "L=1\n",
- "\n",
- "#weight of dam\n",
- "#W=(a+b)*2**-1*L*H*rho_mason*10**-3\n",
- "#After substituting values and Further simplifying we get\n",
- "#W=338.4*(a+b) #equation1\n",
- "\n",
- "#Pressure at B=0, Sinc etension at base has just been avoided\n",
- "\n",
- "#Eccentricity\n",
- "e=b*6**-1 #as sigma1=0\n",
- "\n",
- "#Pressure at C\n",
- "#sigma2=W2*b**-1*(1+6*e*b**-1)\n",
- "#After substituting values and Further simplifying we get\n",
- "#W=440*b\n",
- "\n",
- "#From equation1,440*b=338*(a+b)\n",
- "#b=3.33*a\n",
- "\n",
- "#the distance of C.Gof dam\n",
- "#x_bar=(b**2+b*a+a**2)*(3*(b+a))**-1\n",
- "#After substituting values and Further simplifying we get\n",
- "#x_bar=1.187*a\n",
- "\n",
- "#distance of pt where resultant cuts the base\n",
- "#x=P*W2**-1*H*3**-1\n",
- "#After substituting values and Further simplifying we get\n",
- "#x=27.214*a**-1\n",
- "\n",
- "#Now x_bar+x=2*3**-1*b\n",
- "#After substituting values and Further simplifying we get\n",
- "a=(27.17*(2.22-1.187)**-1)**0.5\n",
- "b=3.33*a\n",
- "\n",
- "#Result\n",
- "print\"The top width dam is\",round(a,2),\"m\"\n",
- "print\"The bottom width dam is\",round(b,2),\"m\"\n",
- "\n",
- "\n",
- "#Plotting the Shear Force Diagram\n",
- "\n",
- "X1=[0,L,L]\n",
- "Y1=[sigma2,sigma1,0]\n",
- "Z1=[0,0,0]\n",
- "plt.plot(X1,Y1,X1,Z1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Force in kN\")\n",
- "plt.show()\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The top width dam is 5.13 m\n",
- "The bottom width dam is 17.08 m\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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Izs4mLi7uwiMUEbe54gpYvRruuw9GjYLZs+EiNpiUJqzWJJGcnMzHH39Mhw4d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\nhIiI1EhJQkREavT/G7kEgAv8kjUAAAAASUVORK5CYII=\n",
- "text": [
- "<matplotlib.figure.Figure at 0x5650490>"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 14.6,Page No.332"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "H= 4 #m #height of dam\n",
- "a=1 #m #Top width\n",
- "b=3 #m #bottom width\n",
- "rho1=9810 #N/m**3 #weight of water\n",
- "rho2=19620 #n/m**3 #Weight of mason\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Let L=1 m (length of dam)\n",
- "L=1\n",
- "\n",
- "#weight of dam\n",
- "W=(a+b)*2**-1*L*H*rho2*10**-3\n",
- "\n",
- "##Lateral thrust\n",
- "P=rho1*H**2*L*2**-1*10**-3 \n",
- "\n",
- "#Distance of Line of action from vertical base\n",
- "x_bar=(b**2+b*a+a**2)*(3*(b+a))**-1\n",
- "\n",
- "#distance of pt where resultant cuts the base\n",
- "x=P*W**-1*H*3**-1\n",
- "\n",
- "#Eccentricity\n",
- "e=x_bar+x-b*2**-1\n",
- "\n",
- "#Stress at Pt B\n",
- "sigma1=W*10**3*b**-1*(1-6*e*b**-1)\n",
- "\n",
- "#stress at Pt C\n",
- "sigma2=W*10**3*b**-1*(1+6*e*b**-1)\n",
- "\n",
- "#Stresses at the base when resorvoir is empty\n",
- "\n",
- "e2=x_bar-b*2**-1\n",
- "\n",
- "#Minus sign indicates sigma_b>sigma_c\n",
- "\n",
- "#Stress at C\n",
- "sigma2_2=W*10**3*b**-1*(1+6*round(e2,2)*b**-1)\n",
- "\n",
- "#Stress at Pt B\n",
- "sigma1_2=W*10**3*b**-1*(1-6*round(e2,2)*b**-1)\n",
- "\n",
- "#result\n",
- "print\"When the Reservoir is full :sigma1\",round(sigma1,2),\"KN/m**2\"\n",
- "print\" :sigma2\",round(sigma2,2),\"KN/m**2\"\n",
- "print\"When the Reservoir is empty:sigma1_2\",round(sigma1_2,2),\"KN/m**2\"\n",
- "print\" :sigma2_2\",round(sigma2_2,2),\"KN/m**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "When the Reservoir is full :sigma1 26160.0 KN/m**2\n",
- " :sigma2 78480.0 KN/m**2\n",
- "When the Reservoir is empty:sigma1_2 96268.8 KN/m**2\n",
- " :sigma2_2 8371.2 KN/m**2\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 14.7,Page No.333"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "H=8 #m #height of dam\n",
- "h=7.5 #m #Height of water\n",
- "a=1 #m #top width\n",
- "mu=0.6 #Coeffeicient of friction\n",
- "rho_mason=22.4 #KN/m**3 #weight of mason\n",
- "rho_w=9.81 #KN/m**3 #density of water\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#weight of dam\n",
- "#W=(a+b)*2**-1*L*H*rho2*10**-3\n",
- "#After substituting values and further simplifying we get\n",
- "#W=89600*(b+1)\n",
- "\n",
- "#Distance of Line of action from vertical base\n",
- "#x_bar=(b**2+b*a+a**2)*(3*(b+a))**-1\n",
- "#After substituting values and further simplifying we get\n",
- "#x_bar=(1+b+b**2)*(3*(1+b))**-1\n",
- "\n",
- "#Lateral thrust\n",
- "P=rho_w*h**2*2**-1\n",
- "\n",
- "#Min width to avoid tension at base\n",
- "#Z=x_bar+P*W**-1*h*3**-1\n",
- "#Z=2*3**-1*b\n",
- "#After substituting values and further simplifying we get\n",
- "#b**2+b-24.09=0\n",
- "a=1\n",
- "b=1\n",
- "c=-24.09\n",
- "\n",
- "X=b**2-4*a*c\n",
- "\n",
- "b1=(-b+X**0.5)*(2*a)**-1\n",
- "b2=(-b-X**0.5)*(2*a)**-1\n",
- "\n",
- "#Thus width cannot be negative,b1 is considered\n",
- "\n",
- "#Min width to avoid sliding\n",
- "#mu*W>P\n",
- "#After substituting values and further simplifying we get\n",
- "b=P*10**3*(mu*89600)**-1-1\n",
- "\n",
- "#Result\n",
- "print\"The Min bottom width is\",round(b,2),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Min bottom width is 4.13 m\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 14.8,Page No.334"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "H=10 #m #height of dam\n",
- "a=1 #m #top width\n",
- "b=7 #m #Bottom width\n",
- "rho_mason=19620 #N/m**3 #weight of mason\n",
- "rho_w=9810 #N/m**3 #density of water\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Lateral thrust\n",
- "P=rho_w*H**2*2**-1\n",
- "\n",
- "#weight of dam\n",
- "W=(rho_w*H*2**-1*a)+(rho_mason*(a+b)*2**-1*H)\n",
- "\n",
- "#Taking Moment at B,M_B=0\n",
- "x_bar=((rho_w*H*2**-1*1*3**-1)+(rho_mason*H*2**-1*2*3**-1)+(rho_mason*H*1.5)+(rho_mason*H*5*11*2**-1*3**-1))*W**-1\n",
- "\n",
- "#Now using relation we get\n",
- "x=P*W**-1*H*3**-1\n",
- "\n",
- "#Eccentricity\n",
- "e=x_bar+x-b*2**-1\n",
- "\n",
- "#Max stress\n",
- "sigma_max=W*b**-1*(1+6*round(e,2)*b**-1)\n",
- "\n",
- "#Min stress\n",
- "sigma_min=W*b**-1*(1-6*round(e,2)*b**-1)\n",
- "\n",
- "#Result\n",
- "print\"The Max stresses on the base is\",round(sigma_max,2),\"N/m**2\"\n",
- "print\"The Min stresses on the base is\",round(sigma_min,2),\"N/m**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Max stresses on the base is 228372.8 N/m**2\n",
- "The Min stresses on the base is 9870.06 N/m**2\n"
- ]
- }
- ],
- "prompt_number": 47
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 14.10,Page No.337"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "H=6 #m #height of dam\n",
- "a=1 #m #top width\n",
- "b=3 #m #Bottom width\n",
- "rho_mason=22 #KN/m**3 #weight of mason\n",
- "rho_earth=16 #KN/m**3 #density of water\n",
- "phi=30 #Degree #angle of repose\n",
- "mu=0.5 #Coeffecient of friction\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Let Length of dam ,L=1 m\n",
- "L=1 #m\n",
- "\n",
- "#weight of dam\n",
- "W=(a+b)*2**-1*L*H*rho_mason\n",
- "\n",
- "#Lateral thrust\n",
- "P=rho_earth*H**2*L*2**-1*((1-sin(30*pi*180**-1))*(1+sin(phi*pi*180**-1))**-1)\n",
- "\n",
- "#Distance of Line of action from vertical base\n",
- "x_bar=(b**2+b*a+a**2)*(3*(b+a))**-1\n",
- "\n",
- "#distance of pt where resultant cuts the base\n",
- "x=P*W**-1*H*3**-1\n",
- "\n",
- "#Eccentricity\n",
- "e=x_bar+x-b*2**-1\n",
- "e_max=b*6**-1\n",
- "\n",
- "#stress at toe\n",
- "sigma2=W*10**3*b**-1*(1+6*round(e,2)*b**-1)*10**-3\n",
- "\n",
- "#Safe agaainst bearing\n",
- "\n",
- "#Frictional Resistance \n",
- "F=mu*W\n",
- "\n",
- "#Hence F>P,so it is safe against sliding\n",
- "\n",
- "#Result\n",
- "print\"Safe against bearing as well as sliding\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Safe against bearing as well as sliding\n"
- ]
- }
- ],
- "prompt_number": 69
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 14.11,Page No.338"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "H=8 #m #height of dam\n",
- "a=1 #m #top width\n",
- "b=4.5 #m #Bottom width\n",
- "rho_mason=24 #KN/m**3 #weight of mason\n",
- "rho_earth=20 #KN/m**3 #density of water\n",
- "phi=30 #Degree #angle of repose\n",
- "mu=0.5 #Coeffecient of friction\n",
- "BC=120 #KN/m**2\n",
- "\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Let Length of dam ,L=1 m\n",
- "L=1 #m\n",
- "\n",
- "#weight of dam\n",
- "W=(a+b)*2**-1*L*H*rho_mason\n",
- "\n",
- "#Rankine's coeff earth pressure\n",
- "K=((1-sin(30*pi*180**-1))*(1+sin(phi*pi*180**-1))**-1)\n",
- "\n",
- "#Lateral thrust\n",
- "P=rho_earth*H**2*L*2**-1*round(K,2)\n",
- "\n",
- "#Distance of Line of action from vertical base\n",
- "x_bar=(b**2+b*a+a**2)*(3*(b+a))**-1\n",
- "\n",
- "#distance of pt where resultant cuts the base\n",
- "x=P*W**-1*H*3**-1\n",
- "\n",
- "#Eccentricity\n",
- "e=x_bar+x-b*2**-1\n",
- "\n",
- "#Pressure at heel B\n",
- "sigma1=W*b**-1*(1-6*round(e,2)*b**-1)\n",
- "\n",
- "#Pressure at heel C\n",
- "sigma2=W*b**-1*(1+6*round(e,2)*b**-1)\n",
- "\n",
- "#sigma2>120 #KN/m**2,so it is unsafe against bearing capacity of the soil\n",
- "\n",
- "#result\n",
- "print\"Unsafe against the bearing capacity of soil\" \n",
- "\n",
- "#Plotting the Shear Force Diagram\n",
- "\n",
- "X1=[0,L,L]\n",
- "Y1=[sigma2,sigma1,0]\n",
- "Z1=[0,0,0]\n",
- "plt.plot(X1,Y1,X1,Z1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Force in kN\")\n",
- "plt.show()\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Unsafe against the bearing capacity of soil\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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pKYm4uDj8/PwICQkhLCyM1NTUS892ImUQHAyJibBnj1lL8cADphbUm2+anfBEqpoiK+23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\nzhxISzMrtrt3h75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- "text": [
- "<matplotlib.figure.Figure at 0x54faf90>"
- ]
- }
- ],
- "prompt_number": 86
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 14.12,Page No.340"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "H=6 #m #height of dam\n",
- "a=1.5 #m #top width\n",
- "b=3.5 #m #Bottom width\n",
- "rho_s=16 #KN/m**3 #density of soil\n",
- "rho_mason=22.5 #KN/m**3 #density of mason\n",
- "phi=30 #Degree #angle of repose\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Let Length of dam ,L=1 m\n",
- "L=1 #m\n",
- "\n",
- "#weight of dam\n",
- "W=(a+b)*2**-1*L*H*rho_mason\n",
- "\n",
- "#Rankine's coeff earth pressure\n",
- "K=((1-sin(30*pi*180**-1))*(1+sin(phi*pi*180**-1))**-1)\n",
- "\n",
- "#Lateral thrust\n",
- "P=rho_s*H**2*L*2**-1*K\n",
- "\n",
- "#Distance of Line of action from vertical base\n",
- "x_bar=(b**2+b*a+a**2)*(3*(b+a))**-1\n",
- "\n",
- "#distance of pt where resultant cuts the base\n",
- "x=P*W**-1*H*3**-1\n",
- "\n",
- "#Eccentricity\n",
- "e=x_bar+x-b*2**-1\n",
- "\n",
- "#Pressure at heel B\n",
- "sigma1=W*b**-1*(1-6*round(e,3)*b**-1)\n",
- "\n",
- "#Pressure at heel C\n",
- "sigma2=W*b**-1*(1+6*round(e,3)*b**-1)\n",
- "\n",
- "#Result\n",
- "print\"The Max Intensities of soil at the wall is\",round(sigma2,2),\"KN/m**2\"\n",
- "print\"The Min Intensities of soil at the wall is\",round(sigma1,2),\"KN/m**2\"\n",
- "\n",
- "#Plotting the Shear Force Diagram\n",
- "\n",
- "X1=[0,L,L]\n",
- "Y1=[sigma2,sigma1,0]\n",
- "Z1=[0,0,0]\n",
- "plt.plot(X1,Y1,X1,Z1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Force in kN\")\n",
- "plt.show()\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Max Intensities of soil at the wall is 118.91 KN/m**2\n",
- "The Min Intensities of soil at the wall is 73.95 KN/m**2\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x555a0f0>"
- ]
- }
- ],
- "prompt_number": 113
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 14.13,Page No.341"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "H=6 #m #height of dam\n",
- "a=1 #m #top width\n",
- "b=3 #m #Bottom width\n",
- "rho_s=18 #KN/m**3 #density of soil\n",
- "rho_mason=24 #KN/m**3 #density of mason\n",
- "alpha=20\n",
- "phi=30\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Let Length of dam ,L=1 m\n",
- "L=1 #m\n",
- "\n",
- "a2=cos(alpha*pi*180**-1)\n",
- "b2=(cos(alpha*pi*180**-1)-((cos(alpha*pi*180**-1)**2-cos(phi*pi*180**-1)**2))**0.5)\n",
- "c2=(cos(alpha*pi*180**-1)+((cos(alpha*pi*180**-1)**2-cos(phi*pi*180**-1)**2)**0.5))\n",
- "\n",
- "X=a2*b2*c2**-1\n",
- "\n",
- "#Total Pressue on the wall\n",
- "P=rho_s*H**2*2**-1*X\n",
- "\n",
- "#The Horizontal component of pressure\n",
- "P_H=P*cos(20*pi*180**-1)\n",
- "\n",
- "#The Vertical component of pressure\n",
- "P_V=P*sin(20*pi*180**-1)\n",
- "\n",
- "#weight of wall\n",
- "W=(a+b)*H*rho_mason*2**-1\n",
- "\n",
- "#TotaL Weight \n",
- "W1=W+P_V\n",
- "\n",
- "#Taking moment of vertical Loads about B,M_B=0\n",
- "x_bar=(rho_mason*a*H*0.5+rho_mason*H*2)*W1**-1\n",
- "\n",
- "x=P_H*W1**-1*H*3**-1\n",
- "\n",
- "#eccentricity\n",
- "e=x_bar+x-b*2**-1\n",
- "\n",
- "#Stress at the toe at C\n",
- "sigma_max=W1*b**-1*(1+6*round(e,2)*b**-1)\n",
- "\n",
- "#Stress at the heel at B\n",
- "sigma_min=W1*b**-1*(1-6*round(e,2)*b**-1)\n",
- "\n",
- "#Result\n",
- "print\"Pressure at the base of the wall:Pressure at the heel\",round(sigma_min,2),\"KN/m**2\"\n",
- "print\" :Pressure at the toe\",round(sigma_max,2),\"KN/m**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pressure at the base of the wall:Pressure at the heel 37.84 KN/m**2\n",
- " :Pressure at the toe 184.76 KN/m**2\n"
- ]
- }
- ],
- "prompt_number": 155
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file
diff --git a/Strength_Of_Materials_by_B_K_Sarkar/chapter_15_som.ipynb b/Strength_Of_Materials_by_B_K_Sarkar/chapter_15_som.ipynb
deleted file mode 100755
index 8ae049c0..00000000
--- a/Strength_Of_Materials_by_B_K_Sarkar/chapter_15_som.ipynb
+++ /dev/null
@@ -1,657 +0,0 @@
-{
- "metadata": {
- "name": "chapter 15.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter no.15:Thin Cyclindrical Shell"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 15.1,Page no.351"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "D=0.8 #m #Diameter of Shell\n",
- "L=3 #m #Length of shell\n",
- "t=0.01 #m #thickness of metal\n",
- "E=200*10**9 #Pa \n",
- "p=2.5*10**6 #Pa #Internal Pressure\n",
- "m=4 #Poisson's ratio\n",
- "\n",
- "#Calculation\n",
- "\n",
- "sigma_1=p*D*(2*t)**-1 #N/m**2 #Hoop stress\n",
- "sigma_2=p*D*(4*t)**-1 #N/m**2 #Longitudinal stress\n",
- "\n",
- "e_1=1*E**-1*(sigma_1-sigma_2*m**-1) #Hoop strain\n",
- "e_2=1*E**-1*(sigma_2-sigma_1*m**-1) #Hoop strain\n",
- "\n",
- "d=e_1*D*100 #cm #Increase in Diameter\n",
- "l=e_2*L*100 #cm #Increase in Length\n",
- "\n",
- "dell_v=2*e_1+e_2 #Volumetric strain\n",
- "V=dell_v*pi*4**-1*D**2*L*10**6 #cm**3 #Increase in Volume\n",
- "\n",
- "#Result\n",
- "print\"Change in Diameter is\",round(d,3),\"cm\"\n",
- "print\"Change in Length is\",round(l,3),\"cm\"\n",
- "print\"Change in Volume is\",round(V,2),\"cm**3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Change in Diameter is 0.035 cm\n",
- "Change in Length is 0.037 cm\n",
- "Change in Volume is 1507.96 cm**3\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 15.2,Page no.352"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "D=0.8 #m #iameter of water main\n",
- "h=100 #m #Pressure head\n",
- "w=10*10**3 #N/m**3 #Weight of Water\n",
- "sigma_t=20*10**6 #MPa #Permissible stress\n",
- "\n",
- "#Calculation\n",
- "\n",
- "p=w*h #N/m**2 #Pressure of inside the main\n",
- "t=p*D*(2*sigma_t)**-1*100 #m #Thcikness of metal\n",
- "\n",
- "#Result\n",
- "print\"The Thickness of metal is\",round(t,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Thickness of metal is 2.0 cm\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 15.3,Page no.352"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "p=2*10**6 #MPa #Steam Pressure\n",
- "t=0.02 #m #thickness of boiler plate\n",
- "sigma_t=120*10**6 #MPa #Tensile stress\n",
- "sigma_l=120*10**6 #MPa #Longitudinal stress\n",
- "rho=0.90 #% #Efficiency of Longitudinal joint\n",
- "rho_e=0.40 #% #Efficiency of circumferential joint\n",
- "\n",
- "#Calculations\n",
- "\n",
- "D_1=sigma_t*2*t*rho*p**-1 #Diameter of boiler \n",
- "D_2=sigma_l*4*t*rho_e*p**-1 #Diameter of boiler \n",
- "\n",
- "#Max diameter of boiler is equal to minimum value of diameter\n",
- "\n",
- "#Result\n",
- "print\"Maximum diameter of boiler is\",round(D_2,2),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum diameter of boiler is 1.92 m\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 15.4,Page no.352"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=0.9 #m #Length of cyclindrical shell\n",
- "D=0.2 #m #Internal Diameter\n",
- "t=0.008 #m #thickness of metal\n",
- "dV=20*10**-6 #m**3 #Additional volume\n",
- "E=200*10**9 #Pa \n",
- "m=1*0.3**-1 #Poissoin's ratio\n",
- "\n",
- "#Calculations\n",
- "\n",
- "V=pi*4**-1*D**2*L #Volume of cyclinder\n",
- "\n",
- "#Let X=2*e_1+e_2\n",
- "X=dV*V**-1 #Volumetric strain (Equation 1)\n",
- "\n",
- "#e_1=p*D*(2*E*t)**-1*(1-1*(2*m)**-1) #Circumferential strain\n",
- "#e_2=p*D*(2*E*t)**-1*(1*2**-1-1*(2*m)**-1) #Circumferential strain\n",
- "\n",
- "#substituting above values in equation 1 we get\n",
- "p=X*E*t*(D*((1-1*(2*m)**-1)+(1*4**-1-1*(2*m)**-1)))**-1*10**-3 #KN/m**2 #Pressure exerted by fluid\n",
- "sigma_t=p*D*(2*t)**-1 #KN/m**2 #hoop stress\n",
- "\n",
- "#Result\n",
- "print\"Pressure Exerted by Fluid on the cyclinder is\",round(p,2),\"KN/m**2\"\n",
- "print\"Hoop stress is\",round(sigma_t,2),\"KN/m**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pressure Exerted by Fluid on the cyclinder is 5956.68 KN/m**2\n",
- "Hoop stress is 74458.45 KN/m**2\n"
- ]
- }
- ],
- "prompt_number": 41
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 15.5,Page no.353"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "t=0.015 #m #Thickness of plate\n",
- "sigma_t=120*10**6 #Pa #tensile stress\n",
- "sigma_l=120*10**6 #Pa #Longitudinal stress\n",
- "rho=0.7 #% #Efficiency of longitudinal joints\n",
- "rho_l=0.3 #% #Efficiency of circumferential joints\n",
- "p=2*10**6 #Pa #Internal pressure\n",
- "D=1.5 #m #shell diameter\n",
- "\n",
- "#Calculations (Part-1)\n",
- "\n",
- "D_1=sigma_t*2*t*rho*p**-1 #m \n",
- "D_2=sigma_l*4*t*rho_l*p**-1 #m \n",
- "\n",
- "#Thus max diameter of shell is min of above two cases\n",
- "\n",
- "#Calculations (Part-2)\n",
- "\n",
- "p_1=sigma_t*2*t*rho*D**-1*10**-6 #MPa\n",
- "p_2=sigma_l*4*t*rho_l*D**-1*10**-6 #MPa\n",
- "\n",
- "#Thus Internal pressure is min of above two cases\n",
- "\n",
- "#Result\n",
- "print\"Max Permissible Diameter of shell is\",round(D_2,2),\"m\"\n",
- "print\"Max Permissible Internal Pressure is\",round(p_2,2),\"MPa\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Max Permissible Diameter of shell is 1.08 m\n",
- "Max Permissible Internal Pressure is 1.44 MPa\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 15.6,Page no.354"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=3 #m #Length\n",
- "D=1 #m #Internal Diameter\n",
- "t=0.015 #m #thickness\n",
- "p=1.5*10**6 #Pa #Internal pressure\n",
- "E=200*10**9 #Pa \n",
- "m=1*0.3**-1 #Poissoin's ratio\n",
- "\n",
- "#Calculations\n",
- "\n",
- "sigma_t=p*D*(2*t)**-1*10**-6 #MPa #Hoop stress\n",
- "sigma_l=p*D*(4*t)**-1*10**-6 #MPa #Longitudinal stress\n",
- "\n",
- "dD=(p*D**2*(2*t*E)**-1*(1-1*(2*m)**-1))*10**2 #cm #Change in Diameter\n",
- "dL=p*D*L*(2*t*E)**-1*(1*2**-1-1*m**-1)*10**2 #cm #Change in Length\n",
- "\n",
- "V=pi*4**-1*D**2*L #Volume \n",
- "dV=p*D*(2*t*E)**-1*(5*2**-1-2*(m)**-1)*V*10**6 #cm #Change in Volume\n",
- "\n",
- "#Result\n",
- "print\"The circumferential stresses induced is\",round(sigma_t,2),\"MPa\"\n",
- "print\"The Longitudinal stresses induced is\",round(sigma_l,2),\"MPa\"\n",
- "print\"The change in dimension are:D is\",round(dD,3),\"cm\"\n",
- "print\" :L is\",round(dL,4),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "0.02125\n",
- "The circumferential stresses induced is 50.0 MPa\n",
- "The Longitudinal stresses induced is 25.0 MPa\n",
- "The change in dimension are:D is 0.021 cm\n",
- " :L is 0.015 cm\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 15.7,Page no.355"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=0.9 #m #Length of cyclinder\n",
- "D=0.4 #m #Diameter \n",
- "t=0.006 #m #thickness\n",
- "p=5*10**6 #Pa #Pressure\n",
- "E=100*10**9\n",
- "m=3 #Poissoin's ratio\n",
- "k=2.6*10**9 #Pa #Bulk modulus\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Let X=dV_1*V_1**-1\n",
- "X=p*(0.4-2*0.006)*(2*t*E)**-1*(5*2**-1-2*m**-1) #Volumetric strain\n",
- "dV_1=round(X,5)*pi*4**-1*0.388**2*L #cm**3 #Increase in volume of cyclinder\n",
- "V_1=pi*4**-1*0.388**2*L #VOlume\n",
- "dV_2=p*k**-1*V_1 #DEcrease in volume of oil due to increase in pressure\n",
- "\n",
- "dV=(dV_1+dV_2)*10**6 #Resultant additional space \n",
- "\n",
- "#Result\n",
- "print\"additional quantity of oil to be pumped is\",round(dV,2),\"cm**3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "additional quantity of oil to be pumped is 519.62 cm**3\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 15.8,Page no.356"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "A=1600*(3600)**-1 #Kg/sec #Amount of steam generated\n",
- "v=0.24 #m**3/kg #specific volume of steam\n",
- "sigma_t=4*10**6 #MPa #Tensile stress\n",
- "V_1=30 #m/s #Velocity of steam\n",
- "p=1*10**6 #Pa #Steam pressure\n",
- "\n",
- "#Calculation\n",
- "\n",
- "V=A*v #m**3/s #volume of steam \n",
- "D=(V*(pi*4**-1*V_1)**-1)**0.5*100 #Diameter of pipe\n",
- "t=p*D*(2*sigma_t)**-1 #Thicknes of pipe\n",
- "\n",
- "#Result\n",
- "print\"Diameter of boiler is\",round(D,2),\"cm\"\n",
- "print\"Thickness of steel plpe is\",round(t,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Diameter of boiler is 6.73 cm\n",
- "Thickness of steel plpe is 0.84 cm\n"
- ]
- }
- ],
- "prompt_number": 43
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 15.9,Page no.359"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P=14*10**3 #N #Axial pull\n",
- "dL=0.0084 #cm #Elongation\n",
- "L=0.25 #m #Length\n",
- "p=7*10**6 #Internal pressure\n",
- "dL_2=0.0034 #cm #Longation\n",
- "d=0.0475 #m #Internal diameter \n",
- "D=0.05 #m #External Diameter\n",
- "m=0.25\n",
- "\n",
- "#Calculation\n",
- "\n",
- "t=(D-d)*2**-1 #thickness od tube\n",
- "A=pi*4**-1*(D**2-d**2) #Area of tube\n",
- "sigma=P*A**-1 #stress\n",
- "e=dL*(L)**-1 #strain\n",
- "E=sigma*e**-1 #Modulus of Elasticity\n",
- "sigma_1=p*d*(2*t)**-1 #Hoop stress\n",
- "sigma_2=p*d*(4*t)**-1 #Longitudinal stress\n",
- "\n",
- "m=-(sigma_1*(dL_2*L**-1*E-sigma_2)**-1) #POissoin's ratio\\\n",
- "\n",
- "#Let X=1*m**-1\n",
- "X=1*m**-1 #Poissoin's ratio\n",
- "\n",
- "#Result\n",
- "print\"The value of Poissoin's ratio is\",round(X,3)\n",
- " "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of Poissoin's ratio is 0.277\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 15.10,Page no.357"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "D=0.8 #m #Diameter\n",
- "t=0.01 #m #Thickness \n",
- "p=5*10**6 #Pa #Pressure\n",
- "m=1*0.25**-1\n",
- "E=200*10**9 #Pa\n",
- "\n",
- "#Calculations\n",
- "\n",
- "sigma_1=5*10**6*0.8*(4*0.01)**-1 #stress\n",
- "sigma_2=sigma_1\n",
- "e_1=sigma_1*E**-1-sigma_2*(m*E)**-1 #strain\n",
- "e_v=3*e_1\n",
- "V=4*3**-1*pi*(D*2**-1)**3 #m**3 tress\n",
- "dell_v=e_v*V*10**6 #cm**3\n",
- "\n",
- "#Result\n",
- "print\"Volume of additional Fluid\",round(dell_v,3),\"cm**3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Volume of additional Fluid 301.593 cm**3\n"
- ]
- }
- ],
- "prompt_number": 37
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 15.11,Page no.358"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "d=0.3 #m #Diameter\n",
- "D=0.003 #m #Diameter of steel wire\n",
- "t=0.006 #m #thickness\n",
- "sigma_w=8*10**6 #Pa #Stress\n",
- "p=1*10**6 #Pa #Internal pressure\n",
- "E_s=200*10**9 #Pa #Modulus of Elasticity for steel\n",
- "E_c=100*10**9 #Pa #Modulus of Elasticity for cast iron\n",
- "m=1*0.3**-1\n",
- "\n",
- "#Calculations\n",
- "\n",
- "sigma_p=(sigma_w*pi*2**-1*d)*(2*t)**-1 #compressive hoop stress\n",
- "sigma_l=p*d*(4*t)**-1 #Longitudinal stress\n",
- "\n",
- "#when internal presure is apllied Let sigma_w_1=Tensile in wire and sigma_p_1=tensile hoop in wire\n",
- "#sigma_p_1*2*t+sigma_w_1*2*d**-1*pi*4**-1*d**2=p*D\n",
- "\n",
- "#After substituting values and further simplifying we get\n",
- "#1.2*sigma_p_1+0.471*sigma_w_1=3000 Equation 1\n",
- "\n",
- "#1*E_c**-1(sigma_p_1-sigma_1*m**-1+sigma_p)=1*E_s**-1(sigma_w_1-sigma_w)\n",
- "\n",
- "#After substituting values and further simplifying we get\n",
- "#sigma_p_1-0.5*sigma_w_1=1.36*10**6 \n",
- "#sigma_p_1=0.5*sigma_w_1-3.39*10**6 Equation 2\n",
- "\n",
- "#From Equation 2 substituting value of sigma_p_1 in Equation 1\n",
- "\n",
- "\n",
- "sigma_w_1=(40.68*10**3+0.3*10**6)*(10.71238*10**-3)**-1\n",
- "sigma_p_1=0.5*sigma_w_1-3.39*10**6\n",
- "\n",
- "#Let X=sigma_p_1 and Y=sigma_w_1\n",
- "X=sigma_p_1*10**-6 #MPa #Stresses in pipe\n",
- "Y=sigma_w_1*10**-6 #MPa #Stresses in wire\n",
- "\n",
- "#Result\n",
- "print\"Stress in the pipe is\",round(X,2),\"MN/m**2\"\n",
- "print\"Stress in the wire is\",round(Y,2),\"MN/m**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Stress in the pipe is 12.51 MN/m**2\n",
- "Stress in the wire is 31.8 MN/m**2\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 15.12,Page no.359"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "D=0.038 #m #External Diameter\n",
- "d=0.035 #m #Internal Diameter\n",
- "d_1=0.0008 #m #Steel wire diameter\n",
- "p=2*10**6 #pa #Pa #Internal Pressure\n",
- "sigma_t_1=7*10**6 #Pa #Circumferential stress\n",
- "#E_s=1.6*E_s\n",
- "m=0.3\n",
- "\n",
- "#Calculation\n",
- "\n",
- "t=(D-d)*2**-1 #m Thickness \n",
- "\n",
- "#sigma_t*2*t=pi*d*2**-1*sigma_w\n",
- "#From Above equation we get\n",
- "\n",
- "#sigma_t=0.419*sigma_w (Equation 1)\n",
- "\n",
- "sigma_w_1=(p*d-sigma_t_1*2*t)*(2*d_1**-1*pi*4**-1*d_1**2)**-1 #stress in wire\n",
- "sigma_l=p*d*(4*t)**-1 #Longitudinal stress in tube\n",
- "\n",
- "#Now Equating equations of strain in tube and wire we get\n",
- "sigma_w=-(1.6*(sigma_t_1-sigma_l*m)-sigma_w_1)*1.67**-1*10**-6\n",
- "\n",
- "#Result\n",
- "print\"The Tension at which wire must have been wound is\",round(sigma_w,2),\"MPa\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Tension at which wire must have been wound is 20.0 MPa\n"
- ]
- }
- ],
- "prompt_number": 30
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file
diff --git a/Strength_Of_Materials_by_B_K_Sarkar/chapter_16_som.ipynb b/Strength_Of_Materials_by_B_K_Sarkar/chapter_16_som.ipynb
deleted file mode 100755
index 177d5ab6..00000000
--- a/Strength_Of_Materials_by_B_K_Sarkar/chapter_16_som.ipynb
+++ /dev/null
@@ -1,524 +0,0 @@
-{
- "metadata": {
- "name": "chapter 16 som.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 16:Riveted Joints"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 16.1,Page no.366"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "t=1 #cm #thickness of plates\n",
- "sigma_t=150 #MPa #Working stress\n",
- "sigma_c=212.5 #MPa #crushing stress\n",
- "sigma_s=94.5 #MPa #shearing stress\n",
- "\n",
- "#Calculation (Part-1)\n",
- "\n",
- "#P_s=pi*4**-1*d**2*sigma_s #N #Shearing strength\n",
- "#After substituting values and further simplifying we get\n",
- "#P_s=pi*4**-1*d**2*94.5*10**6 #N \n",
- "\n",
- "#P_c=d*t*sigma_c #N #crushing strength\n",
- "#After substituting values and further simplifying we get\n",
- "#P_c=d*1*10**-2*212.5*10**6 #N \n",
- "\n",
- "#P_t=(p-d)*t*sigma_t #N #Strength of plate in tearing\n",
- "#After substituting values and further simplifying we get\n",
- "#P_t=(p-d)*1*10**-2*150*10**6\n",
- "\n",
- "#Now comparing strengths \n",
- "#P_s=P_c \n",
- "#pi*4**-1*d**2*94.5*10**6=d*1*10**-2*212.5*10**6\n",
- "d=1*10**-2*212.5*10**6*(pi*4**-1*94.5*10**6)**-1 #m #Diameter of rivet\n",
- "\n",
- "#Now comparing strengths \n",
- "#P_t=P_c\n",
- "#(p-d)*1*10**-2*150*10**6=d*1*10**-2*212.5*10**6\n",
- "#Afte further simplifying equation we get\n",
- "#(p-d)=1.4166*d\n",
- "p=(1.4166*d+d) #m #Pitch length of rivet\n",
- "\n",
- "P=p*sigma_t*10**6*t*10**-2 #N #Strength of solid plate #Answer for strength of solid plate is incorrect in textbook \n",
- "\n",
- "rho=(p-d)*p**-1*100 #Efficiency of the joint #Notification has been changed\n",
- "\n",
- "#Calculation (Part-2)\n",
- "\n",
- "#P_s=2*pi*4**-1*d**2*sigma_s #N #Shearing strength\n",
- "#After substituting values and further simplifying we get\n",
- "#P_s=2*pi*4**-1*d**2*94.5*10**6 #N \n",
- "\n",
- "#P_c=2*d*t*sigma_c #N #crushing strength\n",
- "#After substituting values and further simplifying we get\n",
- "#P_c=2*d*1*10**-2*212.5*10**6 #N \n",
- "\n",
- "#P_t=(p-d)*t*sigma_t #N #Strength of plate in tearing\n",
- "#After substituting values and further simplifying we get\n",
- "#P_t=(p-d)*1*10**-2*150*10**6\n",
- "\n",
- "#Now comparing strengths \n",
- "#P_s=P_c \n",
- "#2*pi*4**-1*d**2*94.5*10**6=2*d*1*10**-2*212.5*10**6\n",
- "d=1*10**-2*212.5*10**6*(pi*4**-1*94.5*10**6)**-1 #m #Diameter of rivet\n",
- "\n",
- "#Now comparing strengths \n",
- "#P_t=P_c\n",
- "#(p-d)*1*10**-2*150*10**6=2*d*1*10**-2*212.5*10**6\n",
- "#Afte further simplifying equation we get\n",
- "#(p-d)=2.833*d\n",
- "p_1=(2.833*d+d) #m #Pitch length of rivets in shearing strength of plate #Notification for pitch length has been changed\n",
- "\n",
- "rho_2=(p_1-d)*p_1**-1*100 #Efficiency of the joint #Notification has been changed\n",
- "\n",
- "#Result \n",
- "print\"The Efficiency of joint in single rivet is\",round(rho,2),\"%\"\n",
- "print\"The Efficiency of joint in double rivet is\",round(rho_2,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Efficiency of joint in single rivet is 58.62 %\n",
- "The Efficiency of joint in double rivet is 73.91 %\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 16.2,Page no.367"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "p=7.5 #cm #Pitch of rivets\n",
- "t=1.5 #cm #Thickness of plate\n",
- "d=2.5 #cm #diameter of rivets\n",
- "sigma_t=400 #MPa #Working stress\n",
- "sigma_c=640 #MPa #crushing stress\n",
- "sigma_s=320 #MPa #shearing stress\n",
- "n=2 #No. of rivets\n",
- "\n",
- "#Calculation\n",
- "\n",
- "P_t=(p-d)*t*10**-4*sigma_t*10**6*10**-3 #N #Strength of plate in tearing\n",
- "P_s=n*pi*4**-1*d**2*10**-4*sigma_s*10**6*10**-3 #N #Shearing strength\n",
- "P_c=n*d*t*10**-4*sigma_c*10**6*10**-3 #N #crushing strength\n",
- "\n",
- "#Thus Minimum force that will rapture the joint is least of P_t,P_s,P_c i.e P_t\n",
- "\n",
- "#Result\n",
- "print\"Minimum force that will rapture the joint is\",round(P_t,2),\"N\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Minimum force that will rapture the joint is 300.0 N\n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 16.3,Page no.367"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "d_1=2 #cm #Diameter of rivets\n",
- "p_1=6 #cm #Pitch of rivet\n",
- "d_2=3 #cm #Diameter of rivet\n",
- "p_2=8 #cm #Pitch of rivet\n",
- "sigma_t=120 #MPa #Working stress\n",
- "sigma_c=160 #MPa #crushing stress\n",
- "sigma_s=90 #MPa #shearing stress\n",
- "t=1.2 #cm #thickness of plate\n",
- "n=2 #No. of rivets\n",
- "\n",
- "#Calculation (part-1)\n",
- "\n",
- "P_t=(p_1-d_1)*t*10**-4*sigma_t*10**6 #N #Strength of plate in tearing\n",
- "P_s=n*pi*4**-1*d_1**2*10**-4*sigma_s*10**6 #N #Shearing strength\n",
- "P_c=n*d_1*t*10**-4*sigma_c*10**6 #N #crushing strength\n",
- "P=p_1*t*10**-4*sigma_t*10**6 #N #Strength of solid per pitch length\n",
- "\n",
- "rho_1=P_s*(P)**-1*100 #% #Efficiency of the joint\n",
- "\n",
- "#Calculation (part-2)\n",
- "\n",
- "P_t_2=(p_2-d_2)*t*10**-4*sigma_t*10**6 #N #Strength of plate in tearing\n",
- "P_s_2=n*pi*4**-1*d_2**2*10**-4*sigma_s*10**6 #N #Shearing strength\n",
- "P_c_2=n*d_2*t*10**-4*sigma_c*10**6 #N #crushing strength\n",
- "P_2=p_2*t*10**-4*sigma_t*10**6 #N #Strength of solid per pitch length\n",
- "\n",
- "rho_2=P_t_2*(P_2)**-1*100 #% #Efficiency of the joint\n",
- "\n",
- "#Result\n",
- "print\"First joint has higher Efficiency i.e\",round(rho_1,2),\"% than second joint\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "First joint has higher Efficiency i.e 65.45 % than second joint\n"
- ]
- }
- ],
- "prompt_number": 60
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 16.4,Page no.368"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "t=18 #mm #thickness of plates\n",
- "sigma_t=100 #MPa #Tensile stress #Notification has been changed\n",
- "sigma_s=70 #MPa #Shearing stress #Notification has been changed\n",
- "\n",
- "#Calculations\n",
- "\n",
- "d=6*t**0.5 #mm #Diameter of rivet #Answer is in correct in textbook\n",
- "s=pi*4**-1*d**2*10**-6*sigma_s*10**6 #N #Strength of one rivet in single shear #Answer is in correct in textbook\n",
- "\n",
- "#Consider strip of joint equal to pitch p\n",
- "\n",
- "#S_1=(p-d)*t*10**-3*sigma_t*10**6 #Strength of plate against tearing along 1-1\n",
- "#After substituting values and further simplifying we get\n",
- "#S_1=1800*p-45900 (Equation 1)\n",
- " \n",
- "#S_2=(p-d)*t*10**-3*sigma_t*10**6+s #Strength of plate against tearing along 1-1\n",
- "#After substituting values and further simplifying we get\n",
- "#S_1=1800*p-56050.64 (Equation 2)\n",
- "\n",
- "#But the value of Equation 2 is smaller than Equation 1\n",
- "\n",
- "#Strength of rivets in single shear is\n",
- "S=4*s\n",
- "\n",
- "#Equating Equation 2 to shearing value\n",
- "#1800*p-56050.64=S\n",
- "p=(S+56050.64)*18000**-1 #cm #Pitch of rivet\n",
- "\n",
- "#Result\n",
- "print\"Diameter of rivets is\",round(d,2),\"mm\"\n",
- "print\"Pitch of rivet is\",round(p,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "35625.6606917\n",
- "Diameter of rivets is 25.46 mm\n",
- "Pitch of rivet is 11.03 cm\n"
- ]
- }
- ],
- "prompt_number": 77
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 16.5,Page no.369"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "t=12 #mm #Thickness of plate\n",
- "d=24 #mm #Diameter of rivets\n",
- "sigma_t=120 #MPa #stress in tension\n",
- "sigma_s=200 #MPa #stress in double shear\n",
- "sigma_b=200 #MPa #stress in Bearing\n",
- "n=1 #No. of rivet\n",
- "\n",
- "#Calculation\n",
- "\n",
- "#P_t=(p-d)*t*10**-4*sigma_t*10**6 #N #Strength of plate in tearing\n",
- "#After further simplifying we get\n",
- "#P_t=(p-24)*14400 #N \n",
- "\n",
- "P_s=n*pi*4**-1*d**2*10**-6*sigma_s*10**6 #N #Shearing strength of rivet in double shear\n",
- "\n",
- "P_b=n*d*10**-3*t*10**-3*sigma_b*10**6 #N #Bearing strength per pitch length\n",
- "\n",
- "#Now Equating P_t to P_s or P_b whichever is small\n",
- "#(p-24)*14400=P_b\n",
- "p=P_b*14400**-1+24*10**-1 #cm #Pitch of rivet\n",
- "p_min=2.5*d*10**-1 #cm #Minimum pitch\n",
- "\n",
- "#Now adopting 6.4 cm pitch\n",
- "\n",
- "rho=(p-d*10**-1)*p**-1*100\n",
- "\n",
- "#Result\n",
- "print\"Pitch of rivet is\",round(p,2),\"cm\"\n",
- "print\"Efficiency of joint is\",round(rho,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pitch of rivet is 6.4 cm\n",
- "Efficiency of joint is 62.5 %\n"
- ]
- }
- ],
- "prompt_number": 90
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 16.6,Page no.370"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "t=12 #mm #thickness of plate\n",
- "d=18 #mm #Diameter of rivet\n",
- "p=8 #cm #pitch of rivet\n",
- "sigma_t=460 #MPa #Tensile stress\n",
- "sigma_s=320 #MPa #shearing stress\n",
- "sigma_b=640 #MPa #bearing stress\n",
- "n=2 #No. of rivet\n",
- "\n",
- "#Calculation\n",
- "\n",
- "P_t=(p-d*10**-1)*t*10**-1*10**-4*sigma_t*10**6 #N #Strength of plate in tearing\n",
- "P_s=n*2*pi*4**-1*d**2*10**-6*sigma_s*10**6 #N #Shearing strength of rivet pr pitch length\n",
- "P_b=n*d*10**-3*t*10**-3*sigma_b*10**6 #N #Bearing strength per pitch length\n",
- "\n",
- "#The joint will fail at a pull of P_b\n",
- "\n",
- "S=p*t*sigma_t*10**6*10**-5 #N #strength of solid plate\n",
- "rho=P_b*S**-1*100 #Efficiency of joint\n",
- "\n",
- "#Result\n",
- "print\"Pull per pitch length at which joint will fail is\",round(P_b,2),\"N\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pull per pitch lenght at which joint will fail is 276480.0 N\n"
- ]
- }
- ],
- "prompt_number": 102
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 16.7,Page no.370"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "W=270 #KN #Load \n",
- "t=14 #mm #thickness of plate\n",
- "b=20 #cm #width of plate\n",
- "d=20 #mm #diameter of rivet\n",
- "sigma_s=70 #MPa #shear stress\n",
- "sigma_b=190 #MPa #stress in bearing\n",
- "sigma_t=110 #MPa #stress in tensile\n",
- "\n",
- "#Calculation\n",
- "\n",
- "S_1=1.75*pi*4**-1*b**2*10**-4*sigma_s*10**6 #strength of one rivet in double shear\n",
- "S_2=20*10**-3*t*10**-3*sigma_b*10**6\n",
- "\n",
- "n=W*10**3*S_1**-1\n",
- "\n",
- "#Adopt 7 rivets\n",
- "\n",
- "#The plates may tear along section 1-1\n",
- "W_1=(20-4)*10**-2*t*10**-3*sigma_t*10**6 #N #Permissible Load\n",
- "\n",
- "#The plates may tear along section 2-2,at the same time shearing the 4 rivets along 1-1 \n",
- "W_2=(20-2*2)*10**-2*t*10**-3*sigma_t*10**6+2*S_1 #N #Permissible Load\n",
- "\n",
- "#The plates may tear along section 3-3,at the same time shearing the rivets along 1-1 and 2-2\n",
- "W_3=(20-3*2)*10**-2*t*10**-3*sigma_t*10**6+4*S_1 #N #Permissible Load\n",
- "\n",
- "W_s=7*S_1 #N #Load to shear all the rivets \n",
- "W_c=7*S_2 #N #Load to crush all the rivets\n",
- "\n",
- "W_4=b*10**-2*t*10**-3*sigma_t*10**6 #N #Load carried by solid plate\n",
- "\n",
- "rho=W_1*W_4**-1*100 #% #Efficiency of joint\n",
- " \n",
- "#Result\n",
- "print\"Efficiency of joint is\",round(rho,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Efficiency of joint is 80.0 %\n"
- ]
- }
- ],
- "prompt_number": 131
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 16.8,Page no.371"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "D=1.5 #cm #Diameter of boiler\n",
- "rho=75 #% #Efficiency of joint\n",
- "sigma_t=85 #MPa #stress in tension\n",
- "sigma_s=70 #MPa #stress in shear\n",
- "P=1 #MPa #Steam Pressure #Notification has been changed\n",
- "\n",
- "#Calculation\n",
- "\n",
- "t=P*10**6*D*(2*sigma_t*10**6*rho*10**-2)**-1*100\n",
- "\n",
- "#Adopt 12 mm thickness of plate\n",
- "t_1=12 #mm \n",
- "d=6*t_1**0.5\n",
- "\n",
- "#Adopt 21 mm diameter of rivet\n",
- "d_1=21 #mm\n",
- "\n",
- "#P_t=(p-d_1*10**-1)*t*10**-1*10**-4*sigma_t*10**6 #N #Strength of plate in tearing\n",
- "#After substituting values and further simplifying we get\n",
- "#P_t=(p-2.1)*10200 #N \n",
- "\n",
- "P_s=1.875*pi*4**-1*d_1**2*10**-6*2*sigma_s*10**6\n",
- "\n",
- "#(p-d_1*10**-1)*10200=P_s\n",
- "p=P_s*10200**-1+d_1*10**-1\n",
- "\n",
- "#Result\n",
- "print\"pitch of plate is\",round(p,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "pitch of plate is 11.01 cm\n"
- ]
- }
- ],
- "prompt_number": 111
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file
diff --git a/Strength_Of_Materials_by_B_K_Sarkar/chapter_17_som.ipynb b/Strength_Of_Materials_by_B_K_Sarkar/chapter_17_som.ipynb
deleted file mode 100755
index f99e1c75..00000000
--- a/Strength_Of_Materials_by_B_K_Sarkar/chapter_17_som.ipynb
+++ /dev/null
@@ -1,443 +0,0 @@
-{
- "metadata": {
- "name": "chapter 17 som.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 17:Welded Joints"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no.17.1,Page no.379"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of Variables\n",
- "\n",
- "b=12 #cm #width of steel plates\n",
- "t=1 #cm #thickness of steel plates\n",
- "sigma=75 #MPa #stress\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#The maximum Load which the plate can carry\n",
- "P=b*t*10**-6*sigma*10**6 #N \n",
- "\n",
- "#Length of weld for static loading\n",
- "\n",
- "#size of weld is equal to thickness of plate\n",
- "S=t #cm\n",
- "\n",
- "#P=2**0.5*l*S*sigma\n",
- "\n",
- "#After substituting values and simplifying above equation, we get, \n",
- "l=((P)*(2**0.5*S*sigma)**-1) #cm\n",
- "\n",
- "#add 1.25 to allow start and stop of weld run\n",
- "L=l+1.25 #cm \n",
- "\n",
- "#Length of weld for Dynamic loading\n",
- "\n",
- "#The stress concentration factor for transverse fillet weld is 1.5\n",
- "\n",
- "sigma_2=sigma*1.5**-1 #MPa #Permissible tensile stress\n",
- "\n",
- "#P=2**0.5*l_2*S*sigma_2 \n",
- "\n",
- "#After substituting values and simplifying above equation, we get,\n",
- "l_2=((P)*(2**0.5*S*sigma_2)**-1) #cm\n",
- "\n",
- "#add 1.25 cm\n",
- "l_3=l_2+1.25 #cm \n",
- "\n",
- "#Result\n",
- "print\"Length of weld for static loading\",round(L,2),\"cm\"\n",
- "print\"Length of weld for Dynamic loading\",round(l_3,2),\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Length of weld for static loading 9.74 cm\n",
- "Length of weld for Dynamic loading 13.98 cm\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no.17.2,Page no.380"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of Variables\n",
- "\n",
- "d=6 #cm #diameter of rod\n",
- "L=40 #cm #Length of steel plate\n",
- "P=12 #KN #Load\n",
- "sigma=180 #MPa #Allowable stress\n",
- "\n",
- "#Calculations\n",
- "\n",
- "A=pi*4**-1*d**2 #cm**2 #Area of rod\n",
- "M=P*10**3*L #Ncm\n",
- "\n",
- "F=M*A**-1 #N/cm #Force per unit cm of weld at top and bottom\n",
- "\n",
- "V_s=P*10**3*(pi*d)**-1 #N/cm #vertical shear\n",
- "\n",
- "R=(F**2+V_s**2)**0.5 #N/cm #resultant Load\n",
- "\n",
- "S=R*(sigma)**-1*10**-2 #cm #size of weld\n",
- "\n",
- "#Result\n",
- "print\"size of weld is\",round(S,2),\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "size of weld is 0.94 cm\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no.17.3,Page no.380"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of Variables\n",
- "\n",
- "b=12 #cm #width of plate\n",
- "S=t=1 #cm #thickness of plate\n",
- "P=50 #KN #load\n",
- "sigma_s=60 #MPa #shear stress\n",
- "\n",
- "#Calculations (part-1)\n",
- "\n",
- "#Under static Loading\n",
- "\n",
- "#P=2**0.5*l*S*sigma_s\n",
- "\n",
- "l=((P*10**3)*(2**0.5*S*sigma_s*10**-4*10**6)**-1) #cm \n",
- "\n",
- "#add 1.25 cm to start and stop weld run\n",
- "\n",
- "L=l+1.25 #cm #length of weld\n",
- "\n",
- "#Calculations (part-2)\n",
- "\n",
- "#Under Fatigue load\n",
- "\n",
- "#stress concentration factor for parallel fillet weld is 2.7\n",
- "\n",
- "sigma_s_2=sigma_s*2.7**-1 #MPa #permissible shear stress\n",
- "\n",
- "#P=2**0.5*l_2*S*sigma_s_2\n",
- "\n",
- "l_2=((P*10**3)*(2**0.5*S*sigma_s_2*10**-4*10**6)**-1) #cm\n",
- "\n",
- "#add 1.25 cm \n",
- "\n",
- "l_3=l_2+1.25 #cm #length of weld\n",
- "\n",
- "#Result\n",
- "print\"Length of weld Under static Loading is\",round(L,3),\"cm\"\n",
- "print\"Length of weld Under Ftigue Loading is\",round(l_3,3),\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Length of weld Under static Loading is 7.143 cm\n",
- "Length of weld Under Ftigue Loading is 17.16 cm\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no.17.4,Page no.381"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of Variables\n",
- "\n",
- "sigma_t=100 #MPa #tensile stress\n",
- "P=170 #KN #Load\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#For equal stress in the welds A and B, the load shared by the fillet welds will be proportional to size of weld\n",
- "\n",
- "#t_a=0.7*s #Effective throat thickness of weld A in upper plate\n",
- "#s=size of weld \n",
- "\n",
- "#t_b=1.05*s #Effective throat thickness of weld B in lower plate\n",
- "\n",
- "#For weld A\n",
- "#P_1=l*t*sigma_t \n",
- "\n",
- "#After substituting values and simplifying above equation, we get,\n",
- "#P_1=84000*s #N (equation 1)\n",
- "\n",
- "#P_2=l*t_2*sigma_t\n",
- "\n",
- "#After substituting values and simplifying above equation, we get,\n",
- "#P_2=126000*s #N (equation 2)\n",
- "\n",
- "#After adding equation 1 and 2, we get,\n",
- "#P=210000*s (equation 3)\n",
- "\n",
- "#Now equating total forces of the fillets to load on the plates\n",
- "s=P*10**3*210000**-1 #cm\n",
- "\n",
- "#Result\n",
- "print\"size of end fillet is\",round(s,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "size of end fillet is 0.81 cm\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no.17.5,Page no.381"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of Variables\n",
- "\n",
- "L_1=30 #cm #length of longitudinal weld\n",
- "L_2=16 #cm #length of transverse weld\n",
- "#t=0.7*s #Effective thickness of weld \n",
- "sigma_t_1=100 #MPa #working stress for transverse welds\n",
- "sigma_t_2=85 #MPa #working stress for longitudinal welds\n",
- "P=150 #KN #load\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#For transverse weld\n",
- "#P_1=L_1*t*10**-4*sigma_t_1*10**6 \n",
- "\n",
- "#After substituting values and simplifying above equation, we get,\n",
- "#P_1=112000*s #N\n",
- "\n",
- "#For longitudinal weld\n",
- "#P_2=L_2*t*10**-4*sigma_t_2*10**6\n",
- "\n",
- "#Total force of resistance of weld\n",
- "#P=P_1+P_2 #N\n",
- "\n",
- "#after adding we get,\n",
- "#P=290500*s #N\n",
- "\n",
- "#Now equating total forces of resistance to pull of the joint\n",
- "s=P*10**3*290500**-1 #cm\n",
- "\n",
- "#Result \n",
- "print\"size of weld is\",round(s,3),\"cm\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "size of weld is 0.516 cm\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no.17.6,Page no.382"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of Variables\n",
- "\n",
- "P=200 #KN #Load carried by the angle \n",
- "S=0.6 #mm #size of weld\n",
- "b=4.46 #cm #Distance of centre of gravity of the angle from the top shorter leg\n",
- "a=10.54 #cm #Distance of centre of gravity of the angle from the top edge of the angle\n",
- "sigma_s=102.5 #MPa #shear stress\n",
- "#l_1=Length of the top weld\n",
- "#l_2=length of the bottom weld\n",
- "#L=l_1+l_2 #cm #total length weld\n",
- "\n",
- "#Using the relation\n",
- "#P=L*0.7*S*sigma_s\n",
- "\n",
- "#After substituting values and simplifying we get\n",
- "L=(P*10**3)*(0.7*S*sigma_s*10**-4*10**6)**-1 #cm (equation 1)\n",
- "\n",
- "#Using the relation\n",
- "l_1=(L*b)*(a+b)**-1 #cm\n",
- "\n",
- "#substituting this value in equation 1 we have,\n",
- "l_2=L-l_1 #cm \n",
- "\n",
- "#Result\n",
- "print\"Distance of centre of gravity of the angle from the top edge of the angle\",round(l_2,2),\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Distance of centre of gravity of the angle from the top edge of the angle 32.64 cm\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no.17.7,Page no.383"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of Variables\n",
- "\n",
- "P=12 #KN #Load\n",
- "sigma_s=75 #N/mm**2 #shear stress\n",
- "e=12 #cm\n",
- "r_1=2.5 #cm\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#A=(2*S*l)*(2)**0.5\n",
- "#sigma_s=P*A**-1 #MPa #shear stress\n",
- "\n",
- "#After substituting values and simplifying we get\n",
- "#sigma_s=16.97*S**-1 #MPa\n",
- "\n",
- "#I_g=S*l*(3*b**2+l**2)*(6)**-1 #cm**4 #Polar moment of Inertia of weld\n",
- "\n",
- "#After substituting values and simplifying we get\n",
- "#I_g=180.833*S #cm**4\n",
- "r_2=((8*2**-1)**2)+((5*2**-1)**2)**0.5 #cm #max radius of weld\n",
- "\n",
- "#sigma_s_2=P*e*r_2*I_g**-1 #MPa #shear stress due to bending moment\n",
- "\n",
- "cos_theta=r_1*r_2**-1\n",
- "\n",
- "#Now using the relation\n",
- "#sigma_s=(sigma_s_1**2+sigma_s_2**2+2sigma_s_1*sigma_s_2*cos_theta\n",
- "\n",
- "S=(2363.8958*5625**-1)**0.5 #cm #size of the weld\n",
- "\n",
- "#Result\n",
- "print\"size of the weld\",round(S,3),\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "size of the weld 0.648 cm\n"
- ]
- }
- ],
- "prompt_number": 11
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file
diff --git a/Strength_Of_Materials_by_B_K_Sarkar/chapter_1_som.ipynb b/Strength_Of_Materials_by_B_K_Sarkar/chapter_1_som.ipynb
deleted file mode 100755
index ff3fcb22..00000000
--- a/Strength_Of_Materials_by_B_K_Sarkar/chapter_1_som.ipynb
+++ /dev/null
@@ -1,438 +0,0 @@
-{
- "metadata": {
- "name": "chapter 1 som.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 1:Centre Of Gravity\n"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 1.1,Page No.8"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of Variables\n",
- "\n",
- "#Rectangle-1\n",
- "a_1=37.5 #cm**2 \n",
- "y_1=26.25 #cm \n",
- "\n",
- "#Rectangle-2\n",
- "a_2=50 #cm**2 \n",
- "y_2=15 #cm \n",
- "\n",
- "#Rectangle-3\n",
- "a_3=150 #cm**2 \n",
- "y_3=2.5 #cm \n",
- "\n",
- "\n",
- "#Calculation\n",
- "\n",
- "\n",
- "Y_bar=(a_1*y_1+a_2*y_2+a_3*y_3)*(a_1+a_2+a_3)**-1 #cm \n",
- "\n",
- "#Result\n",
- "print\"The centroid of the section is\",round(Y_bar,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroid of the section is 8.88 cm\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 1.2,Page No.9"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "#Area-1\n",
- "a_1=6 #cm**2 \n",
- "x_1=3 #cm\n",
- "y_1=0.5 #cm\n",
- "\n",
- "#Area-2\n",
- "a_2=6 #cm**2\n",
- "x_2=2.671 #cm\n",
- "y_2=3 #cm\n",
- "\n",
- "#Area-3\n",
- "a_3=16 #cm**2\n",
- "x_3=1 #cm\n",
- "y_3=5 #cm\n",
- "\n",
- "\n",
- "#Calculation\n",
- "\n",
- "\n",
- "X_bar=(a_1*x_1+a_2*x_2+a_3*x_3)*(a_1+a_2+a_3)**-1 #cm\n",
- "Y_bar=(a_1*y_1+a_2*y_2+a_3*y_3)*(a_1+a_2+a_3)**-1 #cm\n",
- "\n",
- "\n",
- "#Result\n",
- "print\"The centre of gravity of section is\",round(X_bar,2),\"cm\"\n",
- "print\"The centre of gravity of section is\",round(Y_bar,2),\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centre of gravity of section is 1.79 cm\n",
- "The centre of gravity of section is 3.61 cm\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 1.3,Page no.10"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "#Area-1\n",
- "a_1=93.75 #cm**2 \n",
- "y_1=6.25 #cm\n",
- "\n",
- "#Area-2\n",
- "a_2=93.75 #cm**2 \n",
- "y_2=6.25 #cm\n",
- "\n",
- "#Area-3\n",
- "a_3=375 #cm**2 \n",
- "y_3=9.375 #cm\n",
- "\n",
- "#Area-4\n",
- "a_4=353.43 #cm**2\n",
- "y_4=6.366 #cm\n",
- "\n",
- "\n",
- "#Calculation\n",
- "\n",
- "Y_bar=(a_1*y_1+a_2*y_2+a_3*y_3-a_4*y_4)*(a_1+a_2+a_3-a_4)**-1 #cm\n",
- "\n",
- "\n",
- "#Result\n",
- "print\"The centre of gravity lies at a distance of \",round(Y_bar,2),\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centre of gravity lies at a distance of 11.66 cm\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 1.4,Page no.10\n",
- "\n"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "\n",
- "a_1=36*pi #cm**2 #Area of Quadrant of a circle\n",
- "x_1=16/pi #cm \n",
- "y_1=16*pi**-1 #cm\n",
- "\n",
- "\n",
- "a_2=18*pi #cm**2 #Area of the semicircle\n",
- "x_2=6 #cm\n",
- "y_2=8*pi**-1 #cm\n",
- "\n",
- "\n",
- "#Calculation-1\n",
- "\n",
- "X_bar=(a_1*x_1-a_2*x_2)*(a_1-a_2)**-1 #cm\n",
- "\n",
- "#Calculation-2\n",
- "#To calculate Y_bar,taking AB as the Reference line\n",
- "\n",
- "Y_bar=(a_1*y_1-a_2*y_2)*(a_1-a_2)**-1 #cm\n",
- "\n",
- "#Result\n",
- "\n",
- "print\"The centre of gravity is \",round(X_bar,2),\"cm\"\n",
- "print\"The centre of gravity is\",round(Y_bar,2),\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centre of gravity is 4.19 cm\n",
- "The centre of gravity is 7.64 cm\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 1.5,Page no.11"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "#Circle-1 \n",
- "a_1=100*pi #cm**2\n",
- "x_1=10 #cm\n",
- " \n",
- "#Square-2 \n",
- "a_2=50 #cm**2\n",
- "x_2=15 #cm\n",
- " \n",
- "#Calculation\n",
- "\n",
- "X_bar=(a_1*x_1-a_2*x_2)*(a_1-a_2)**-1 #cm\n",
- "\n",
- "\n",
- "#Result\n",
- "print\"The centre of gravity is\",round(X_bar,2),\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centre of gravity is 9.05 cm\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 1.6,Page no.12\n"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#intilization of variables \n",
- "\n",
- "#Rectangle-1\n",
- "a_1=51200 #mm**2 \n",
- "x_1=160 #mm\n",
- "y_1=80 #mm\n",
- "\n",
- "#Triangle-2\n",
- "a_2=6400 #mm**2\n",
- "x_2=80*3**-1 #mm\n",
- "y_2=320*3**-1 #mm\n",
- "\n",
- "#Semicircle-3\n",
- "a_3=1250*pi #mm**2\n",
- "x_3=210 #mm\n",
- "y_3=(160-(4*50-(3*pi)**-1)) #mm\n",
- "\n",
- "\n",
- "#Calculation\n",
- "\n",
- "X_bar=(a_1*x_1-a_2*x_2-a_3*x_3)*(a_1-a_2-a_3)**-1 #mm\n",
- "Y_bar=(a_1*y_1-a_2*y_2-a_3*y_3)*(a_1-a_2-a_3)**-1 #mm\n",
- "\n",
- "#Result\n",
- "print\"The centroid of the given area is\",round(X_bar,2),\"mm\"\n",
- "print\"The centroid of the given area is\",round(Y_bar,2),\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroid of the given area is 176.07 mm\n",
- "The centroid of the given area is 87.34 mm\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 1.8,Page no.12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "\n",
- "alpha=pi/2 #degree #In case of semicircle\n",
- "\n",
- "#Semicircle-1\n",
- "r_1=20 #cm #radius of semicircle \n",
- "y_1=4*r_1*(3*pi)**-1 #cm #distance from the base\n",
- "a_1=(pi*r_1**2)*2**-1 #cm**2 #area of semicircle\n",
- "\n",
- "#Semicircle-2\n",
- "r_2=16 #cm #radius of semicircle\n",
- "y_2=4*r_2*(3*pi)**-1 #cm #distance from the base\n",
- "a_2=(pi*r_2**2)*2**-1 #cm**2 #area of semicircle\n",
- "\n",
- "#Calculations\n",
- "\n",
- "\n",
- "Y_bar=(a_1*y_1-a_2*y_2)*(a_1-a_2)**-1 #cm #centroid\n",
- "\n",
- "\n",
- "#Result\n",
- "print\"The centroid of the area is \",round(Y_bar,2),\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The centroid of the area is 11.51 cm\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no1.12,Page no.16"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "#Right Circular Cyclinder\n",
- "#m_1=(16*pi*h*rho_1) #gm \n",
- "#y_1=4+h*2**-1 #cm\n",
- "\n",
- "#Hemisphere\n",
- "#m_2=256*pi*rho_1 #gm \n",
- "y_2=2.5 #cm \n",
- "\n",
- "Y_bar=4 #cm\n",
- "r=4 #cm\n",
- "\n",
- "#Calculation\n",
- "\n",
- "#Y_bar=(m_1*y_1+m_2*y_2)*(m_1+m_2)**-1 #cm #Centroid\n",
- "h=(402.114*25.132**-1)**0.5\n",
- "\n",
- "#Result\n",
- "print\"The value of h is\",round(h,2),\"cm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of h is 4.0 cm\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file
diff --git a/Strength_Of_Materials_by_B_K_Sarkar/chapter_2_som.ipynb b/Strength_Of_Materials_by_B_K_Sarkar/chapter_2_som.ipynb
deleted file mode 100755
index 16ba879f..00000000
--- a/Strength_Of_Materials_by_B_K_Sarkar/chapter_2_som.ipynb
+++ /dev/null
@@ -1,465 +0,0 @@
-{
- "metadata": {
- "name": "chapter 2 som.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 2:Moment Of Inertia"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 2.1,Page no.29"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of Variables\n",
- " \n",
- "#Rectangle-1\n",
- "b_1=10 #cm #width of Rectangle-1\n",
- "d_1=2 #cm #breadth of Rectangle-1\n",
- "a_1=40 #cm**2 #Area of Rectangle-1\n",
- "y_1=1 #cm #Distance of centroid-1\n",
- "\n",
- "#Rectangle-2\n",
- "b_2=2 #cm #width of Rectangle-2\n",
- "d_2=10 #cm #breadth of Rectangle-2\n",
- "a_2=20 #cm**2 #Area of rectangle-2\n",
- "y_2=7 #cm #Distance of centroid-2\n",
- "\n",
- "#Rectangle-3\n",
- "b_3=20 #cm #width of Rectangle-3\n",
- "d_3=2 #cm #breadth of Rectangle-3\n",
- "a_3=20 #cm**2 #Area of rectangle-3\n",
- "y_3=13 #cm #Distance of centroid-3\n",
- "\n",
- "\n",
- "\n",
- "#Calculation\n",
- "Y_bar=((a_1*y_1+a_2*y_2+a_3*y_3)*(a_1+a_2+a_3)**-1) #cm #centre of gravity of section\n",
- "\n",
- "Y_1=4.5 #cm #Distance of centroid of rectangle 1 to C.G\n",
- "Y_2=1.5 #cm #Distance of centroid of rectangle 2 to C.G\n",
- "Y_3=7.5 #cm #Distance of centroid of rectangle 3 to C.G\n",
- "\n",
- "I_x_x_1=b_1*d_1**3*12**-1+a_1*Y_1**2 #moment of inertia of rectangle 1 about centroidal x-x axis of the section\n",
- "I_x_x_2=b_2*d_2**3*12**-1+a_2*Y_2**2 #moment of inertia of rectangle 2 about centroidal x-x axis of the section\n",
- "I_x_x_3=b_3*d_3**3*12**-1+a_3*Y_3**2 #moment of inertia of rectangle 3 about centroidal x-x axis of the section\n",
- "I_x_x=I_x_x_1+I_x_x_2+I_x_x_3 #cm**4 \n",
- "\n",
- "#Result\n",
- "print\"Moment of Inertia of the section is\",round(I_x_x,2),\"cm**4\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Moment of Inertia of the section is 2166.67 cm**4\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 2.2,Page no.31"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of Variables\n",
- "\n",
- "#Rectangle-1\n",
- "b_1=2 #cm #width of Rectangle-1\n",
- "d_1=12 #cm #breadth of Rectangle-1\n",
- "a_1=24 #cm**2 #Area of Rectangle-1\n",
- "y_1=6 #cm #Distance of centroid-1\n",
- "\n",
- "#Rectangle-2\n",
- "b_2=6 #cm #width of Rectangle-2\n",
- "d_2=2 #cm #breadth of Rectangle-2\n",
- "a_2=12 #cm**2 #Area of rectangle-2\n",
- "y_2=1 #cm #Distance of centroid-2\n",
- "\n",
- "#Rectangle-3\n",
- "b_3=2 #cm #width of Rectangle-3\n",
- "d_3=12 #cm #breadth of Rectangle-3\n",
- "a_3=24 #cm**2 #Area of rectangle-3\n",
- "y_3=6 #cm #Distance of centroid-3\n",
- "\n",
- "#Calculation\n",
- "Y_bar=((a_1*y_1+a_2*y_2+a_3*y_3)*(a_1+a_2+a_3)**-1) #cm #centre of gravity of section\n",
- "\n",
- "Y_1=6 #cm #Distance of centroid of rectangle 1 to base \n",
- "Y_2=1 #cm #Distance of centroid of rectangle 2 to base\n",
- "Y_3=6 #cm #Distance of centroid of rectangle 3 to base\n",
- "\n",
- "I_x_x_1=b_1*d_1**3*12**-1+a_1*Y_1**2 #moment of inertia of rectangle 1 about centroidal x-x axis of the section\n",
- "I_x_x_2=b_2*d_2**3*12**-1+a_2*Y_2**2 #moment of inertia of rectangle 2 about centroidal x-x axis of the section\n",
- "I_x_x_3=b_3*d_3**3*12**-1+a_3*Y_3**2 #moment of inertia of rectangle 3 about centroidal x-x axis of the section\n",
- "I_x_x=I_x_x_1+I_x_x_2+I_x_x_3 #cm**4 \n",
- "\n",
- "\n",
- "#Result\n",
- "print\"Moment of Inertia of the section is\",round(I_x_x,2),\"cm**4\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Moment of Inertia of the section is 2320.0 cm**4\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no2.3,Page no.32"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of Variables\n",
- "\n",
- "#Rectangle-1\n",
- "b_1=12 #cm #width of Rectangle-1\n",
- "d_1=2 #cm #breadth of Rectangle-1\n",
- "a_1=24 #cm**2 #Area of Rectangle-1\n",
- "y_1=1 #cm #Distance of centroid-1\n",
- "\n",
- "#Rectangle-2\n",
- "b_2=2 #cm #width of Rectangle-2\n",
- "d_2=6 #cm #breadthof Rectangle-2\n",
- "a_2=12 #cm**2 #Area of rectangle-2\n",
- "y_2=5 #cm #Distance of centroid-2\n",
- "\n",
- "#Rectangle-3\n",
- "b_3=5 #cm #width of Rectangle-3\n",
- "d_3=2 #cm #breadth of Rectangle-3\n",
- "a_3=10 #cm**2 #Area of rectangle-3\n",
- "y_3=9 #cm #Distance of centroid-3\n",
- "\n",
- "#Calculation\n",
- "Y_bar=((a_1*y_1+a_2*y_2+a_3*y_3)*(a_1+a_2+a_3)**-1) #cm #centre of gravity of section\n",
- "\n",
- "Y_1=2.78 #cm #Distance of centroid of rectangle 1 to C.G \n",
- "Y_2=1.22 #cm #Distance of centroid of rectangle 2 to C.G\n",
- "Y_3=5.22 #cm #Distance of centroid of rectangle 3 to C.G \n",
- "\n",
- "I_x_x_1=b_1*d_1**3*12**-1+a_1*Y_1**2 #moment of inertia of rectangle 1 about centroidal x-x axis of the section\n",
- "I_x_x_2=b_2*d_2**3*12**-1+a_2*Y_2**2 #moment of inertia of rectangle 2 about centroidal x-x axis of the section\n",
- "I_x_x_3=b_3*d_3**3*12**-1+a_3*Y_3**2 #moment of inertia of rectangle 3 about centroidal x-x axis of the section\n",
- "I_x_x=I_x_x_1+I_x_x_2+I_x_x_3 #cm**4 \n",
- "\n",
- "\n",
- "#Result\n",
- "print\"Moment of Inertia of the section is\",round(I_x_x,2),\"cm**4\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Moment of Inertia of the section is 523.16 cm**4\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no2.4,Page no.33"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "D=10 #cm #diameter of circle\n",
- "b=4 #cm #width of rectangle \n",
- "d=4 #cm #breadth of rectangle\n",
- "Y=1 #cm #Distance of centroid of rectangle 1 to C.G\n",
- "a=16 #cm**2 #area of rectangle\n",
- "\n",
- "#Calculations\n",
- "\n",
- "I_x_x_1=pi*64**-1*(D**4) #cm**4 #moment of inertia of circle about x-x axis\n",
- "I_x_x_2=b*d**3*12**-1+a*Y**2 #cm**4 #moment of inertia of rectangle about x-x axis\n",
- "I_x_x=I_x_x_1-I_x_x_2 #cm**4 #Total moment of inertia of the section\n",
- "\n",
- "#Result\n",
- "print\"Total moment of inertia of the section is\",round(I_x_x,2),\"cm**4\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total moment of inertia of the section is 453.54 cm**4\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 2.5,Page no.33"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "b_1=10 #cm #Breadth of the triangle\n",
- "h=9 #cm #Height of triangle\n",
- "b_2=2 #cm #width of rectangle\n",
- "d=3 #cm #Depth of rectangle\n",
- "\n",
- "#Triangle ABC-1\n",
- "a_1=45 #cm**2 #Area of triangle\n",
- "y_1=3 #cm #C.G of triangle\n",
- "\n",
- "#Rectanglar hole-2\n",
- "a_2=6 #cm**2 #Area of rectangle\n",
- "y_2=4.5 #cm #C.G of rectangle\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Using relations\n",
- "Y_bar=((a_1*y_1-a_2*y_2)*(a_1-a_2)**-1) #cm\n",
- "\n",
- "I_1=b_1*h**3*36**-1+a_1*(y_1-Y_bar)**2 #cm**4 #M.I of triangle ABC about x-x passing through C.G of section\n",
- "I_2=b_2*d**3*12**-1+a_2*(y_2-Y_bar)**2 #cm**4 #M.I of rectangular hole about x-x passing through C.G of section\n",
- "I=I_1-I_2 #cm**4 #M.I of whole section about x-x passing through the C.G \n",
- "\n",
- "I_3=b_1*h**3*12**-1 #cm**4 #M.I of triangle ABC about the base BC\n",
- "I_4=b_2*d**3*12**-1+a_2*y_2**2 #cm**4 #M.I of Rectangular hole about the base BC\n",
- "\n",
- "I_5=I_3-I_4 #cm**4 #M.I of the whole section about the base BC\n",
- "\n",
- "#Result\n",
- "print\"M.I of whole section about x-x passing through the C.G\",round(I,2),\"cm**4\"\n",
- "print\"M.I of the whole section about the base BC is\",round(I_5,2),\"cm**4\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "M.I of whole section about x-x passing through the C.G 182.42 cm**4\n",
- "M.I of the whole section about the base BC is 481.5 cm**4\n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no2.6,Page no.34"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "\n",
- "#Notifications has been changed as per requirement\n",
- "\n",
- "\n",
- "h=8 #cm #height of triangle\n",
- "b=8 #cm #breadth of triangle or diameter semicircle\n",
- "d=4 #cm #diameter of circle enclosed\n",
- "\n",
- "#Calculations\n",
- "\n",
- "I_1=b*h**3*12**-1 #cm #moment of inertia of the triangle ABC about the axis AB\n",
- "I_2=pi*b**4*128**-1 #cm ##moment of inertia of the semicircle about the axis AB\n",
- "I_3=pi*d**4*64**-1 #cm #moment of inertia of circle about the circle about the axis\n",
- "\n",
- "I=I_1+I_2-I_3 #cm #Moment of Inertia of the shaded area about the axia AB\n",
- "\n",
- "#Result\n",
- "print\"Moment of Inertia of the shaded area is\",round(I,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Moment of Inertia of the shaded area is 429.3 cm\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no.2.12,Page no.38"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "#Rectangle\n",
- "a_1=600 #cm**2 #Area of the Rectangle\n",
- "y_1=15 #cm #C.G of Rectangle\n",
- "b=20 #cm #width of rectangle\n",
- "d=30 #cm #depth of rectangle\n",
- "D=15 #cm #Diameter of circle\n",
- "\n",
- "#Circle\n",
- "a_2=176.7 #cm**2 #Area of the circle\n",
- "y_2=20 #cm #C.G of the circle\n",
- " \n",
- "#Calculation\n",
- "\n",
- "Y_bar=((a_1*y_1-a_2*y_2)*(a_1-a_2)**-1) #cm #Distance of C.G From the AB\n",
- "Y_bar_1=2.1 #cm\n",
- "Y_bar_2=7.1 #cm\n",
- "\n",
- "I_1=b*d**3*12**-1 #cm**4 #M.I of the rectangle about its C.G and parallel to x-x axis\n",
- "I_2=I_1+a_1*Y_bar_1**2\n",
- "I_3=pi*D**4*64**-1+a_2*Y_bar_2**2 #cm**4 #M.I of circular section about x-x axis\n",
- "\n",
- "I=I_2-I_3 #cm**4 #M.I of the section about x-x axis\n",
- "\n",
- "#Result\n",
- "print\"M.I of the section about x-x axis\",round(I,2),\"cm**4\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "M.I of the section about x-x axis 36253.5 cm**4\n"
- ]
- }
- ],
- "prompt_number": 50
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no.2.13,Page no.38"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "d=90 #cm #Diameter of grindstone\n",
- "t=10 #cm #thickness of grindstone\n",
- "rho=0.0026 #Kg/cm**3 #Density\n",
- "\n",
- "#calculations\n",
- "\n",
- "#M=Mass of grindstone=Volume *Density=Area*Thickness*Density\n",
- "M=pi*4**-1*d**2*t*rho #Kg \n",
- "R=d*2**-1 #cm #radius\n",
- "I_g=M*R**2*2**-1 #Kg*m**2\n",
- "\n",
- "k=R*(2**0.5)**-1 #cm #Radius of gyration\n",
- "\n",
- "#Result\n",
- "print\"Radius of gyration is\",round(k,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Radius of gyration is 31.82 cm\n"
- ]
- }
- ],
- "prompt_number": 6
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file
diff --git a/Strength_Of_Materials_by_B_K_Sarkar/chapter_3_som.ipynb b/Strength_Of_Materials_by_B_K_Sarkar/chapter_3_som.ipynb
deleted file mode 100755
index 8ecddb1d..00000000
--- a/Strength_Of_Materials_by_B_K_Sarkar/chapter_3_som.ipynb
+++ /dev/null
@@ -1,948 +0,0 @@
-{
- "metadata": {
- "name": "chapter 3 som.ipynb",
- "signature": "sha256:f204f91e23b9b059e7969e16bcc2c68fd35a3a7bdfec7a765456a9a04f17dab5"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 3:Stresses And Strains"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 3.1,Page no.54"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of Variables\n",
- "\n",
- "P=40 #mm #Force applied to stretch a tape\n",
- "L=30 #m #Length of steel tape\n",
- "A=6*1 #mm #Cross section area\n",
- "E=200*10**9*10**-6 #KN/m**2 #Modulus of Elasticity\n",
- "\n",
- "#Calculations\n",
- "\n",
- "sigma_L=(P*L*10**3)*(A*E)**-1 #mm \n",
- "\n",
- "#Result\n",
- "print\"The Elongation of steel tape is\",round(sigma_L,4),\"mm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Elongation of steel tape is 1.0 mm\n"
- ]
- }
- ],
- "prompt_number": 65
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 3.2,Page no.54"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of VAriables\n",
- "\n",
- "\n",
- "#D=(D_0-2) #cm #Inside Diameter of cyclinder\n",
- "#A=(pi*(D_0-1)) #cm**2 #Area of cross-section\n",
- "#L=(pi*(D_0-1)*5400) #N #Crushing load for column\n",
- "F=6 #Factor of safety\n",
- "T=1 #cm #wall thickness of cyclinder\n",
- "\n",
- "#S=L*F**-1\n",
- "#After Simplifying,we get\n",
- "S=600*10**3\n",
- "\n",
- "#Calculations\n",
- "\n",
- "D_0=(S*F)*(pi*54000)**-1+1 #cm #Outside diameter of cyclinder\n",
- "\n",
- "#Result\n",
- "print\"The outside Diameter of cyclinder is\",round(D_0,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The outside Diameter of cyclinder is 22.22 cm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 3.3,Page no.56"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math \n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P=800 #N #force applied to steel wire\n",
- "L=150 #m #Length of steel wire\n",
- "E=200 #GN/m**2 #Modulus of Elasticity\n",
- "d=10 #mm #Diameter of steel wire\n",
- "W=7.8*10**4 #N/m**3 #Weight Density of steel\n",
- "#A=(pi*4**-1)*(d)**2 #m**2\n",
- "\n",
- "#After simplifying Area,we get\n",
- "A=7.85*10**-5 #m**2\n",
- "\n",
- "#calculation (Part-1)\n",
- "\n",
- "#Elongation Due to 800N Load \n",
- "dell_L_1=(P*L*10**-3)*(A*E*10**9*10**-6)**-1 #mm\n",
- "\n",
- "#calculation (Part-2)\n",
- "\n",
- "#Elongation due to Weight of wire \n",
- "dell_L_2=((pi*4**-1)*150*W*L*10**-3)*(2*A*E*10**7)**-1 #mm\n",
- "\n",
- "#calculation (Part-3)\n",
- "\n",
- "#Total Elongation of wire\n",
- "dell_L_3=dell_L_1+dell_L_2\n",
- "\n",
- "\n",
- "#Result\n",
- "print\"The Elongation due to 800N Load is\",round(dell_L_1,2),\"mm\"\n",
- "print\"The Elongation due to Weight of wire is\",round(dell_L_2,2),\"mm\"\n",
- "print\"Total Elongation of wire is\",round(dell_L_3,2),\"mm\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Elongation due to 800N Load is 7.64 mm\n",
- "The Elongation due to Weight of wire is 4.39 mm\n",
- "Total Elongation of wire is 12.03 mm\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 3.4,Page no.55"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Intilization of variables\n",
- "\n",
- "d=10 #mm #Diameter of Punching Hole\n",
- "t=4 #mm #Thickness of Mild Steel Plate\n",
- "tou=320 #N/mm**2 #Shear Strength of mild Steel\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Force Required for punching the hole\n",
- "P=tou*pi*d*t #N \n",
- "\n",
- "#Area of punch in contact with the plate surface\n",
- "A=(pi*4**-1*d**2) #mm*2\n",
- "\n",
- "#Compressive stress\n",
- "sigma_c=P*A**-1 #N/mm*2\n",
- "\n",
- "#Result\n",
- "print\"Force Required for punching the hole is\",round(P,2),\"N\"\n",
- "print\"Compressive stress is\",sigma_c,\"N/mm*2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Force Required for punching the hole is 40212.39 N\n",
- "Compressive stress is 512.0 N/mm*2\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 3.6,Page no.57"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P=200*10**3 #N\n",
- "L_1=0.10 #mm #Length of of portin AB\n",
- "L_2=0.16 #mm #Length of of portin BC\n",
- "L_3=0.12 #mm #Length of of portin CD\n",
- "E=200*10**9 #N\n",
- "d_1=0.1 #cm\n",
- "d_2=0.08 #cm\n",
- "d_3=0.06 #cm\n",
- "A_1=(pi*4**-1)*(0.1)**2 #mm**2\n",
- "A_2=(pi*4**-1)*(0.08)**2 #mm**2\n",
- "A_3=(pi*4**-1)*(0.06)**2 #mm**2\n",
- "\n",
- "#Calculations\n",
- "\n",
- "dell_L_1=(P*L_1*10**3)*(A_1*E)**-1 #mm\n",
- "dell_L_2=(P*L_2*10**3)*(A_2*E)**-1 #mm\n",
- "dell_L_3=(P*L_3*10**3)*(A_3*E)**-1 #mm\n",
- "dell_L=dell_L_1+dell_L_2+dell_L_3 #mm\n",
- "\n",
- "#Result\n",
- "print\"Total Elongation of steel bar is\",round(dell_L,3),\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total Elongation of steel bar is 0.087 mm\n"
- ]
- }
- ],
- "prompt_number": 67
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 3.7,Page no.57"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "#from F.B.D,we get\n",
- "P_1=50 #KN\n",
- "P_2=20 #KN\n",
- "P_3=40 #KN\n",
- "\n",
- "d=0.02 #mm #Diameter of steel bar\n",
- "L_1=0.4 #mm\n",
- "L_2=0.3 #mm\n",
- "L_3=0.2 #mm\n",
- "E=210*10**9 #N\n",
- "\n",
- "#After simplifying Area,we get\n",
- "A=pi*10**-4 #m**2 #Area of cross section\n",
- "\n",
- "#Calculations\n",
- "\n",
- "sigma_AB=P_1*1000*A #N/m**2\n",
- "sigma_BA=P_2*1000*A #N/m**2\n",
- "sigma_CD=P_3*1000*A #N/m**2\n",
- "dell_L=((P_1*L_1+P_2*L_2+P_3*L_3)*(A*E)**-1)*10**6 #mm\n",
- "\n",
- "#Result\n",
- "print\"Total Elongation of Steel bar is\",round(dell_L,3),\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total Elongation of Steel bar is 0.515 mm\n"
- ]
- }
- ],
- "prompt_number": 44
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 3.8,Page no.58"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "#R_a+R_c=25 #KN #R_a,R_b are reactions at supports A and C respectively\n",
- "L_ab=2 #m\n",
- "L_bc=3 #m\n",
- "\n",
- "#Calculation\n",
- "\n",
- "#From F.B.D,we get\n",
- "#dell_L_AB=(R_a*L_AB)*(A*E)**-1 #Elongation of portion AB\n",
- "#dell_L_BC=(R_c*L_BC)*(A*E)**-1 #Compression of portion BC\n",
- "\n",
- "#After simplifying above equations we get,\n",
- "#R_a=(1.5)*R_c #KN\n",
- "#R_a+R_c=25 #KN\n",
- "#Solving the above simultaneous equations using matrix method\n",
- "A=np.array([[1,-1.5],[1,1]]) #Here the coefficients of the first equations of unknowns are setup\n",
- "B=np.array([0,25]) #Here the RHS of both equations are setup\n",
- "C=np.linalg.solve(A,B)\n",
- "\n",
- "#print C[0] #Prints the first element in the vector C\n",
- "#print C[1] #Prints the second element in the vector C\n",
- "\n",
- "#Result\n",
- "print\"The reaction at support A is\",round(C[0],2),\"KN\"\n",
- "print\"The reaction at support C is\",round(C[1],2),\"KN\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reaction at support A is 15.0 KN\n",
- "The reaction at support C is 10.0 KN\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 3.9,Page no.59"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "#P is the force acting on the bar BC compressive in nature and force on AB is (100-P) Tensile in nature\n",
- "E=200*10*9 #N\n",
- "A_1=3*10**-4 #cm**2 #Area of AB\n",
- "A_2=4*10**-4 #cm**2 #Area of BC\n",
- "L=1.5 #cm #Length of bar\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#The total elongation of bar\n",
- "#(((100-P)*10**3*1.5)*(3*10**-4*E)**-1)-((P*10**3*1.5)*(4*10**-4*E)**-1)=0 \n",
- "\n",
- "#The total elongation of bar is limited to 1\n",
- "#(25-0.4375*P)*10**-4=1*10**-3\n",
- "\n",
- "#After simplifying above equation we get,\n",
- "P=-(10-25)*0.4375**-1 #KN #Total elongation of bar\n",
- "F_AB=100-P #KN #force in AB\n",
- "F_BC=P #KN #Force in BC\n",
- "sigma_AB=(((F_AB)*(3*10**-4)**-1)*10**-3) #KN #Stress in AB\n",
- "sigma_BC=((F_BC)*(4*10**-4)**-1*10**-3) #KN #Stress in Bc\n",
- "\n",
- "\n",
- "#Result\n",
- "print\"F_AB\",round(F_AB,2),\"KN\"\n",
- "print\"F_BC\",round(F_BC,2),\"KN\"\n",
- "print\"sigma_AB\",round(sigma_AB,2),\"KN\"\n",
- "print\"sigma_BC\",round(sigma_BC,2),\"KN\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "F_AB 65.71 KN\n",
- "F_BC 34.29 KN\n",
- "sigma_AB 219.05 KN\n",
- "sigma_BC 85.71 KN\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 3.12,Page no.61"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P=500 #KN #Safe Load\n",
- "d=20 #mm #steel rod diameter\n",
- "n=4 #number of steel rod\n",
- "sigma_c=4 #N/mm**2 #stress in concrete \n",
- "#E_S*E_c**-1=15\n",
- "\n",
- "\n",
- "#Calculations\n",
- "\n",
- "A_s=4*pi*4**-1*d**2 #mm**2 #Area os steel rod\n",
- "sigma_s=15*sigma_c #N/mm**2 #stress in steel\n",
- "\n",
- "#P=sigma_s*A_s+sigma_c*A_c \n",
- "\n",
- "#After substituting and simplifying above equation we get,\n",
- "\n",
- "A_c=(P*10**3-sigma_s*1256)*(sigma_c)**-1 #mm**2 #Area of the concrete\n",
- "X=(A_s+A_c)**0.5 #mm #Total cross sectional area\n",
- "P_s=A_s*sigma_s*10**-3 #KN #Load carried by steel\n",
- "\n",
- "#Result\n",
- "print\"Load carried by steel is\",round(P_s,2),\"KN\"\n",
- "print\"stress induced in steel is\",round(sigma_s,3),\"KN\"\n",
- "print\"cross sectional area of column is\",round(X,2),\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Load carried by steel is 75.4 KN\n",
- "stress induced in steel is 60.0 KN\n",
- "cross sectional area of column is 327.74 mm\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 3.13,Page no.62"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "A_s=500 #mm**2\n",
- "E_s=200000\n",
- "E_al=80000\n",
- "A_al=1000\n",
- "\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#(P_al*L_al)*(A_al*E_al)**-1+(P_s*L_s)*(A_s*E_s)**-1=1*2**-1 \n",
- "\n",
- "P=1*1000**-1*((A_s*E_s*A_al*E_al)*(A_s*E_s+A_al*E_al)**-1) #N\n",
- "P_s=P_al=P #N\n",
- "sigma_t=P_s*A_s**-1 #N/mm**2 #Tensile stress in bolt\n",
- "sigma_c=P_al*A_al**-1 #N/mm**2 #Compressive stress in Aluminium tube\n",
- "\n",
- "#result\n",
- "print\"Tensile stress in bolt is\",round(sigma_t,2),\"N/mm**2\"\n",
- "print\"Compressive stress in Aluminium tube is\",round(sigma_c,2),\"N/mm**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Tensile stress in bolt is 88.89 N/mm**2\n",
- "Compressive stress in Aluminium tube is 44.44 N/mm**2\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 3.14,Page no.63"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "A=1600 #mm**2 #Area of the Bar\n",
- "P=480*10**3 #N #Load\n",
- "dell_L=0.4 #mm #Contraction of metal bar\n",
- "L=200 #mm #Length of metal bar\n",
- "sigma_t=0.04 #mm #Guage Length\n",
- "t=40\n",
- "\n",
- "#Calculations\n",
- "\n",
- "sigma_L=dell_L*L**-1\n",
- "E=((P*L)*(A*dell_L)**-1*10**-3) #N/mm**2 #Young's Modulus \n",
- "m=t*sigma_t**-1*sigma_L\n",
- "\n",
- "\n",
- "#Result\n",
- "print\"The value of Young's Modulus is\",round(E,2),\"N/mm*2\"\n",
- "print\"The value of Poissoin's ratio is\",round(m,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of Young's Modulus is 150.0 N/mm*2\n",
- "The value of Poissoin's ratio is 2.0\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 3.15,Page no.63"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "A_s=0.003848 #m**2 #Area of steel bar\n",
- "A_al=0.003436 #m**2 #Area of Aluminium tube\n",
- "E=220*10*9 #N #Young's modulus of steel\n",
- "E=70*10*9 #N #Young's modulus of aluminium\n",
- "P=600*10**3 #N #Load applied to the bar\n",
- "#dell_L_al-dell_L_s=0.00015 #mm #difference between strain in aluminium bar and steel bar\n",
- "\n",
- "#Calculations\n",
- "\n",
- "\n",
- "#Let the aluminium tube be compressed by dell_L_al and steel bar by by dellL_s\n",
- "#dell_L_al=sigma_al*E_al**-1*L_al\n",
- "#dell_L_s=sigma_s*E_s**-1*L_s\n",
- "\n",
- "#After substituting and simplifying above equation we get,\n",
- "#((sigma_al*70**-1)-(sigma_s*220**-1))=300000 #(equation 1)\n",
- "\n",
- "#After simplifying above equation we get,\n",
- "#sigma_al=17462.165*10**4-1.1199*sigma_s #(equation 2)\n",
- "\n",
- "#Now substituting sigma_al in equation(1)\n",
- "#((17462.165*10**4-1.1199*sigma_s)*(70)**-1)-(sigma_s*220**-1)=300000\n",
- "\n",
- "#After simplifying above equation we get,\n",
- "\n",
- "sigma_s=-((300000-249.4594*10**4)*0.0205444**-1)*10**-6 #MN/m**2 #stress developed in steel bar\n",
- "#sigma_al=17462.165*10**4-1.1199*sigma_s\n",
- "sigma_al=(17462.165*10**4-1.1199*106822005.02)*10**-6\n",
- "\n",
- "\n",
- "#Result\n",
- "print\"stress developed in steel bar is\",round(sigma_s,2),\"MN/m**2\"\n",
- "print\"stress developed in aluminium bar is\",round(sigma_al,2),\"MN/m**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "stress developed in steel bar is 106.82 MN/m**2\n",
- "stress developed in aluminium bar is 54.99 MN/m**2\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 3.16,Page no.64"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "E=200 #GN/m**2 #Modulus of elasticity\n",
- "alpha=11*10**-6 #per degree celsius #coeffecient o flinear expansion of steel bar\n",
- "L=6 #m #Length of rod\n",
- "\n",
- "\n",
- "#Calculations \n",
- "\n",
- "#(Part-1) #IF the walls do not yield\n",
- "\n",
- "t=58 #degree celsius #Fall in temperature #(t=80-22)\n",
- "dell=alpha*t #strain\n",
- "sigma=E*10**9*dell*10**-6 #MN/m**2 #Stress\n",
- "A=pi*4**-1*6.25*10**-4 #mm**2 #Area of wall and rod\n",
- "P=sigma*10**6*A*10**-3 #KN #Pull Exerted\n",
- "\n",
- "#(Part-2) #IF the walls yield together at the two ends is 1.15 mm\n",
- "\n",
- "L_2=L*(1-alpha*t) #m #Length of rod at 22 degree celsius\n",
- "L_3=L-L_2 #m #Decrease in Length \n",
- "\n",
- "#As the walls yield by 1.5 mm, actual decrease in length is\n",
- "L_4=L_3-0.0015 #m \n",
- "dell_2=L_4*L**-1 #strain \n",
- "P_2=E*10**9*dell_2*A*10**-3 #KN\n",
- "\n",
- "#Result\n",
- "print\"Pull Exerted by the rod:when walls do not yield\",round(P,2),\"KN\"\n",
- "print\" :when total yield together at two ends is 1.5 mm\",round(P_2,2),\"KN\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pull Exerted by the rod:when walls do not yield 62.64 KN\n",
- " :when total yield together at two ends is 1.5 mm 38.09 KN\n"
- ]
- }
- ],
- "prompt_number": 43
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 3.17,Page no.65"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "D=4.5 #cm #External Diameter of tube\n",
- "d=3 #cm #Internal diameter of tube\n",
- "t=3 #mm #thickness of tube\n",
- "t_1=30 #degree celsius\n",
- "t_2=180 #degree celsius #when metal heated\n",
- "L=30 #cm #Original LEngth\n",
- "alpha_s=1.08*10**-5 #Per degree celsius #coefficient of Linear expansion of steel tube\n",
- "alpha_c=1.7*10**-5 #Per degree celsius #coefficient of Linear expansion of copper tube\n",
- "E_s=210 #GPa #Modulus of Elasticity of steel \n",
- "E_c=110 #GPA #Modulus of Elasticity of copper\n",
- "\n",
- "#Calculation\n",
- "\n",
- "#For Equilibrium of the system, Total tension in steel=Total tension in copper\n",
- "\n",
- "#sigma_s*A_s=sigma_c*A_c (equation 1)\n",
- "\n",
- "A_c=pi*4**-1*d**2 #cm**2 #Area of copper\n",
- "A_s=pi*4**-1*(D**2-d**2) #cm**2 #Area of steel\n",
- "\n",
- "#simplifying equation 1\n",
- "#sigma_s=1.785*sigma_c\n",
- "\n",
- "T=t_2-t_1 #change in temperature\n",
- "\n",
- "#Actual expansion of steel=Actual expansion of copper\n",
- "#alpha_s*T*L+sigma_s*E_s**-1*L=alpha_c*T*L-sigma_c*E_c**-1*L\n",
- "\n",
- "#After substituting values in above equation and simplifying we get\n",
- "\n",
- "sigma_c=(930*10**5*1.7591**-1)*10**-6 #MN/m**2 #Stress in copper\n",
- "sigma_s=1.785*sigma_c #MN/m**2 #Stress in steel\n",
- "\n",
- "#Increase in Length of either component\n",
- "L_2=(alpha_s*T+sigma_s*10**6*(E_s*10**9)**-1)*L\n",
- "\n",
- "#Result\n",
- "print\"stress in copper bar is\",round(sigma_c,2),\"MN/m**2\"\n",
- "print\"stress in steel bar is\",round(sigma_s,2),\"MN/m**2\"\n",
- "print\"Increase in Length is\",round(L_2,3),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "stress in copper bar is 52.87 MN/m**2\n",
- "stress in steel bar is 94.37 MN/m**2\n",
- "Increase in Length is 0.062 cm\n"
- ]
- }
- ],
- "prompt_number": 54
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 3.18,Page no.66"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "t_1=15 #degree celsius #temperature of steel bar\n",
- "t_2=315 #degree celsius #raised temperature \n",
- "E_s=210 #GPa #Modulus of Elasticity of steel bar\n",
- "E_c=100 #GPa #Modulus of Elasticity of copper bar\n",
- "dell_L=0.15 #cm #Increase in Length of bar\n",
- "\n",
- "#Calculation\n",
- "\n",
- "#For Equilibrium of the system, Tension in steel bar = Tension in copper bar\n",
- "#sigma_s*A_s = sigma_c*2*A_c\n",
- "#sigma_S=2*sigma_c\n",
- "\n",
- "#Actual expansion of steel = Actual expansion of copper\n",
- "#L*alpha_s*T+sigma_s*E_s**-1*L = L*alpha_c*T-sigma_c*E_c**-1*L (Equation 1)\n",
- "\n",
- "T=t_2-t_1 #per degree celsius #change in temperature\n",
- "\n",
- "#After substituting values in above equation and simplifying we get\n",
- "sigma_c=(1650*10**5*1.9524**-1)*10**-6 #MN/m**2 #Stress in copper\n",
- "sigma_s=2*sigma_c #MN/m**2 #Stress in steel\n",
- "\n",
- "#Actual Expansion of steel bar \n",
- "#L*alpha_s*T+sigma_s*E_s**-1*L = L*alpha_c*T-sigma_c*E_c**-1*L \n",
- "#After substituting values in above equation and simplifying we get\n",
- "L=0.15*10**-2*0.0044048**-1 #m\n",
- "\n",
- "#Result\n",
- "print\"Stress in copper bar is\",round(sigma_c,2),\"MN/m**2\"\n",
- "print\"Stress in steel bar is\",round(sigma_s,2),\"MN/m**2\"\n",
- "print\"Original Length of bar is\",round(L,2),\"m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Stress in copper bar is 84.51 MN/m**2\n",
- "Stress in steel bar is 169.02 MN/m**2\n",
- "Original Length of bar is 0.34 m\n"
- ]
- }
- ],
- "prompt_number": 60
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 3.19,Page no.67"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=100 #m #Length of rod\n",
- "A=2 #cm**2 #cross sectional area\n",
- "rho=80 #KN/m**3\n",
- "\n",
- "#Calculatiom\n",
- "W=A*10**-4*L*rho #KN\n",
- "\n",
- "sigma_s=10+1.6 #KN #Rod experiencing max tensile stress when it is at top performing upstroke\n",
- "sigma_s_2=sigma_s*10**3*200**-1 #N/mm**2 #corresponding stress at this moment\n",
- "\n",
- "sigma_c=1 #KN ##Rod experiencing max compressive stress at its lower end,free from its own weight\n",
- "sigma_c_2=sigma_c*10**3*200**-1 #corresponding stress at this moment\n",
- "\n",
- "#Result\n",
- "print\"Tensile stress in bar\",round(sigma_s_2,2),\"N/mm**2\"\n",
- "print\"Compressive stress in bar\",round(sigma_c_2,2),\"N/mm**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Tensile stress in bar 58.0 N/mm**2\n",
- "Compressive stress in bar 5.0 N/mm**2\n"
- ]
- }
- ],
- "prompt_number": 64
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 3.20,Page no.68"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "sigma=0.012 #strain\n",
- "P=150 #KN #Total Load on the Post\n",
- "E=1.4*10**4 #N/mm**2 #modulus of elasticity\n",
- "#b be the width of the post in mm\n",
- "#2b is the longer dimension of the post in mm\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#We know,\n",
- "#sigma=(P*(A*E)**-1) \n",
- "\n",
- "#After substituting values and simplifying, we get\n",
- "b=((150*10**3)*(0.012*1.4*2*10**4)**-1)**0.5\n",
- "q=2*b #mm #Longer dimension of post\n",
- "\n",
- "#Result\n",
- "print\"Width of post is\",round(b,2),\"mm\"\n",
- "print\"Longer dimension of post is\",round(q,2),\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Width of post is 21.13 mm\n",
- "Longer dimension of post is 42.26 mm\n"
- ]
- }
- ],
- "prompt_number": 6
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file
diff --git a/Strength_Of_Materials_by_B_K_Sarkar/chapter_4_som.ipynb b/Strength_Of_Materials_by_B_K_Sarkar/chapter_4_som.ipynb
deleted file mode 100755
index 87824640..00000000
--- a/Strength_Of_Materials_by_B_K_Sarkar/chapter_4_som.ipynb
+++ /dev/null
@@ -1,2019 +0,0 @@
-{
- "metadata": {
- "name": "chapter 4 som.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter No.4:Shearing Force And Bending Moment"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 4.4.1,Page No.89"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "F_B=-10 #KN #Force at pt B\n",
- "F_D=-20 #KN #Force at pt D\n",
- "w_CB=5 #KN/m #u.d.l at CB\n",
- "w_AE=40 #KN/m #u.d.l at AE\n",
- "L_ED=L_CB=2 #m #Length of ED & CB\n",
- "L_CD=1 #m #Length of CD\n",
- "L_AE=3 #m #Length of AE\n",
- "L=8 #m #span of beam\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Shear Force Calculations\n",
- "\n",
- "#Shear Force at B\n",
- "V_B=F_B #KN\n",
- "\n",
- "#Shear Force at C\n",
- "V_C=F_B-(w_CB*L_CB)\n",
- "\n",
- "#Shear Force at D\n",
- "V_D1=V_C\n",
- "V_D2=V_C+F_D\n",
- "\n",
- "#Shear Force at E\n",
- "V_E=V_D2\n",
- "\n",
- "#Shear Force at A\n",
- "V_A=V_D-(w_AE*L_AE)\n",
- "\n",
- "#Bending Moment Calculations\n",
- "\n",
- "#Bending Moment at B\n",
- "M_B=0\n",
- "\n",
- "#Bending Moment at C\n",
- "M_C=F_B*L_CB-w_CB*L_CB**2*2**-1\n",
- "\n",
- "#Bending Moment at D\n",
- "M_D=F_B*(L_CB+L_CD)-w_CB*L_CB*(L_CB*2**-1+L_CD)\n",
- "\n",
- "#Bending Moment at E\n",
- "M_E=F_B*(L_CB+L_CD+L_ED)-w_CB*L_CB*(L_CB*2**-1+L_CD+L_ED)+F_D*L_ED\n",
- "\n",
- "#Bending Moment at A\n",
- "M_A=F_B*L-w_CB*L_CB*(L_CB*2**-1+L_CD+L_ED+L_AE)+F_D*(L_AE+L_ED)-w_AE*(L_AE**2*2**-1)\n",
- "\n",
- "#Result\n",
- "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
- "\n",
- "#Plotting the Shear Force Diagram\n",
- "\n",
- "X1=[0,L_CB,L_CB+L_DC,L_CB+L_DC,L_CB+L_DC+L_ED,L_CB+L_DC+L_ED+L_AE,L_CB+L_DC+L_ED+L_AE]\n",
- "Y1=[V_B,V_C,V_D1,V_D2,V_E,V_A,0]\n",
- "Z1=[0,0,0,0,0,0,0]\n",
- "plt.plot(X1,Y1,X1,Z1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Force in kN\")\n",
- "plt.show()\n",
- "\n",
- "#Plotting the Bendimg Moment Diagram\n",
- "\n",
- "Y2=[M_B,M_C,M_D,M_E,M_A]\n",
- "X2=[0,L_CB,L_CB+L_DC,L_CB+L_DC+L_ED,L_CB+L_DC+L_ED+L_AE]\n",
- "Z2=[0,0,0,0,0]\n",
- "plt.plot(X2,Y2)\n",
- "plt.xlabel(\"Lenght in m\")\n",
- "plt.ylabel(\"Bending Moment in kN.m\")\n",
- "plt.show()"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Shear Force and Bending Moment Diagrams are the results\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x59f9130>"
- ]
- },
- {
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Xwk8/wYIFWqexbpWOkpo6dSpBQUGEh4ejKAp79+4lMjKSmzdv4i1z7IUQ9YCT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\ngQMH6NKlS6nj7u7upKamkpq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- "text": [
- "<matplotlib.figure.Figure at 0x5549230>"
- ]
- }
- ],
- "prompt_number": 86
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 4.4.2,Page No.90"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "w_CB=1 #KN/m #u.d.l on Length CB\n",
- "F_D=2 #KN #Pt Load at D\n",
- "L_AD=L_DC=1 #m #Length of AD & DC\n",
- "L_CB=2 #m #Length of CB\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Shear Force at B\n",
- "V_B=0 #KN\n",
- "\n",
- "#Shear Force at C\n",
- "V_C=-(w_CB*L_CB)\n",
- "\n",
- "#Shear Force at D\n",
- "V_D1=V_C\n",
- "V_D2=V_C-F_D\n",
- "\n",
- "#Shear Force at A\n",
- "V_A=V_D2 \n",
- "\n",
- "#Bending Moment Calculations\n",
- "\n",
- "#Bending Moment at B\n",
- "M_B=0\n",
- "\n",
- "#Bending Moment at C\n",
- "M_C=-w_CB*L_CB**2*2**-1\n",
- "\n",
- "#Bending Moment at D\n",
- "M_D=-w_CB*L_CB*(L_CB*2**-1+L_DC)\n",
- "\n",
- "#Bending Moment at A\n",
- "M_A=-w_CB*L_CB*(L_CB*2**-1+L_DC+L_AD)-F_D*L_AD\n",
- "\n",
- "#Result\n",
- "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
- "\n",
- "#Plotting the Shear Force Diagram\n",
- "\n",
- "X1=[0,L_CB,L_CB+L_DC,L_CB+L_DC,L_CB+L_DC+L_AD,L_CB+L_DC+L_AD]\n",
- "Y1=[0,V_C,V_D1,V_D2,V_A,0]\n",
- "Z1=[0,0,0,0,0,0]\n",
- "plt.plot(X1,Y1,X1,Z1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Force in kN\")\n",
- "plt.show()\n",
- "\n",
- "#Plotting the Bendimg Moment Diagram\n",
- "\n",
- "Y2=[M_B,M_C,M_D,M_A]\n",
- "X2=[0,L_CB,L_CB+L_DC,L_CB+L_DC+L_AD]\n",
- "Z2=[0,0,0,0]\n",
- "plt.plot(X2,Y2)\n",
- "plt.xlabel(\"Length in m\")\n",
- "plt.ylabel(\"Bending Moment in kN.m\")\n",
- "plt.show()"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Shear Force and Bending Moment Diagrams are the results\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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mzJiBqKioXu8rYTytxaiUsXzkkUfw3HPPwc3N/Nd9f8bSKRKHrYsBr82mjl5E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\nEZEkTBxERCQJEwcREUnCxEFERJL8P6KhBCKbfJ61AAAAAElFTkSuQmCC\n",
- "text": [
- "<matplotlib.figure.Figure at 0x59c7d50>"
- ]
- },
- {
- "metadata": {},
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- "text": [
- "<matplotlib.figure.Figure at 0x5549cb0>"
- ]
- }
- ],
- "prompt_number": 85
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 4.4.3,Page No.91"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "AC=5000 #N/m #u.v.l\n",
- "L_AB=4 #m #Length of AB\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Consider a section at Distance x from B\n",
- "#DB=x\n",
- "#By similar triangles (triangle ABC and BDE) we get\n",
- "\n",
- "#Shear Force at x\n",
- "#F_x=-DB*DE*2**-1\n",
- "#After substituting values in above equation we get\n",
- "#F_x=625*x**2\n",
- "\n",
- "#shear Force at B where x=0\n",
- "V_B=0\n",
- "\n",
- "#shear Force at A where x=L_AB=4\n",
- "V_A=625*L_AB**2\n",
- "\n",
- "#Bending Moment Calculation\n",
- "\n",
- "#M_x=DB*DE*DB*3**-1*2**-1\n",
- "#Substituting values in above equation we get\n",
- "#M_x=-625*x**3*3**-1\n",
- "\n",
- "#Bending Moment at B where x=0\n",
- "M_B=0\n",
- "\n",
- "#Bending Moment at A where x=L_AB=4\n",
- "M_A=-625*L_AB**3*3**-1\n",
- "\n",
- "#Result\n",
- "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
- "\n",
- "#Plotting the Shear Force Diagram\n",
- "\n",
- "X1=[0,L_AB]\n",
- "Y1=[V_B,V_A]\n",
- "Z1=[0,0]\n",
- "plt.plot(X1,Y1,X1,Z1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Force in kN\")\n",
- "plt.show()\n",
- "\n",
- "#Plotting the Bendimg Moment Diagram\n",
- "\n",
- "Y2=[M_B,M_A]\n",
- "X2=[0,L_AB]\n",
- "Z2=[0,0]\n",
- "plt.plot(X2,Y2)\n",
- "plt.xlabel(\"Length in m\")\n",
- "plt.ylabel(\"Bending Moment in kN.m\")\n",
- "plt.show()"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Shear Force and Bending Moment Diagrams are the results\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x59a61f0>"
- ]
- },
- {
- "metadata": {},
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AERERxMfHU1BQQHp6Omlpadb2bTc3N1JSUlBKsXz5coYOHWp9TVxcHACrVq2i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\njE+fPr1K76P79Joq1nJ38eJ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- "text": [
- "<matplotlib.figure.Figure at 0x557b350>"
- ]
- }
- ],
- "prompt_number": 84
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 4.4.4,Page No.92"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "F_C=F_D=F_E=30 #KN #Pt Load at C,D,E respectively\n",
- "L_AE=L_ED=L_DC=1.5 #m #Length of AE,ED,DC respectively\n",
- "L_CB=0.5 #m #Length of CB\n",
- "L_AC=4.5 #m #Length of AC\n",
- "L_AD=3 #m #Length of AD\n",
- "w=10 #KN/m #u.d.l\n",
- "L=5 #m #Length of beam\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Shear Force Calculations\n",
- "\n",
- "#Shear Force at B\n",
- "V_B=0 #KN\n",
- "\n",
- "#Shear Force at C\n",
- "V_C1=-w_CB*L_CB\n",
- "V_C2=-w*L_CB-F_C #KN\n",
- "\n",
- "#Shear Force at D\n",
- "V_D1=-w*(L_DC+L_CB)-F_C*L_DC\n",
- "V_D2=-w*(L_DC+L_CB)-F_C-F_D #KN\n",
- "\n",
- "#Shear Force at E\n",
- "V_E1=-w*(L_DC+L_CB+L_ED)-F_C*(L_DC+L_ED)\n",
- "V_E2=-F_C-F_D-F_E-w*(2*L_ED+L_CB)\n",
- "\n",
- "#Shear Force at A\n",
- "V_A=-w*L-F_C-F_D-F_E\n",
- "\n",
- "#Bending Moment Calculations\n",
- "\n",
- "#Bending Moment at B\n",
- "M_B=0\n",
- "\n",
- "#Bending Moment at C\n",
- "M_C=-w*L_CB**2*2**-1\n",
- "\n",
- "#Bending Moment at D\n",
- "M_D=-w*(L_DC+L_CB)**2*2**-1-F_C*L_DC\n",
- "\n",
- "#Bending Moment at E\n",
- "M_E=-w*(L_DC+L_CB+L_ED)**2*2**-1-F_C*(L_ED+L_DC)-F_D*L_ED\n",
- "\n",
- "#Bending Moment at A\n",
- "M_A=-w*L**2*2**-1-F_C*L_AC-F_D*L_AD-F_E*L_AE\n",
- "\n",
- "#Result\n",
- "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
- "\n",
- "#Plotting the Shear Force Diagram\n",
- "\n",
- "X1=[0,L_CB,L_CB,L_CB+L_DC,L_CB+L_DC,L_CB+L_DC+L_ED,L_CB+L_DC+L_ED,L_CB+L_DC+L_ED+L_AE,L_CB+L_DC+L_ED+L_AE]\n",
- "Y1=[V_B,V_C1,V_C2,V_D1,V_D2,V_E1,V_E2,V_A,0]\n",
- "Z1=[0,0,0,0,0,0,0,0,0]\n",
- "plt.plot(X1,Y1,X1,Z1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Force in kN\")\n",
- "plt.show()\n",
- "\n",
- "#Plotting the Bendimg Moment Diagram\n",
- "\n",
- "X2=[0,L_CB,L_CB+L_DC,L_CB+L_DC+L_ED,L_CB+L_DC+L_ED+L_AE,L_CB+L_DC+L_ED+L_AE]\n",
- "Y2=[M_B,M_C,M_D,M_E,M_A,0]\n",
- "Z2=[0,0,0,0,0,0]\n",
- "plt.plot(X2,Y2)\n",
- "plt.xlabel(\"Length in m\")\n",
- "plt.ylabel(\"Bending Moment in kN.m\")\n",
- "plt.show()"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Shear Force and Bending Moment Diagrams are the results\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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EqVOniIuLA2DixImsXr26WWr1Zt72P4M76bu4SN/FRd7+XbRpYz/L6pln4MMP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\nAAAAAElFTkSuQmCC\n",
- "text": [
- "<matplotlib.figure.Figure at 0x59f97d0>"
- ]
- },
- {
- "metadata": {},
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FRUVcccUVwC/DZ9esWcO0adMICAjAz8+Pp556igEDBgC/DKs9deoUiYmJLFiw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\nlJQUIiIiCA8PZ86cOVbHsdQ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- "text": [
- "<matplotlib.figure.Figure at 0x59a25b0>"
- ]
- }
- ],
- "prompt_number": 88
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 4.4.5,Page No.93"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "w1=30 #KN/m #u.d.l on L_CB\n",
- "F_C=120 #KN #Pt Load at C\n",
- "w2=50 #KN/m #u.d.l on L_AD\n",
- "L_DC=L_CB=2 #m #Length of DC and CB respectively\n",
- "L_AD=4 #m #Length of AD\n",
- "L_AB=L=8 #m #Length of beam\n",
- "\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Let R_A & R_B be the reactions at A & B\n",
- "#R_A+R_B=380\n",
- "\n",
- "#Taking Moment at A\n",
- "#M_A=-R_B*L+F_C(L_DC+L_AD)+w1*L_CB*(L_CB*2**-1+L_DC+L_AD)+w2*L_AD**2*2**-1=0\n",
- "\n",
- "#After Rearranging the terms we get\n",
- "R_B=(F_C*(L_DC+L_AD)+w1*L_CB*(L_CB*2**-1+L_DC+L_AD)+w2*L_AD**2*2**-1)*L**-1\n",
- "R_A=380-R_B\n",
- "\n",
- "#Shear Force Calculations\n",
- "\n",
- "#Shear Force at B\n",
- "V_B=R_B\n",
- "\n",
- "#Shear Force at C\n",
- "V_C1=-w1*L_CB+R_B\n",
- "V_C2=R_B-w1*L_CB-F_C\n",
- "\n",
- "#Shear Force at D\n",
- "V_D=V_C2\n",
- "\n",
- "#Shear Force at A\n",
- "V_A=V_D-w2*L_AD\n",
- "\n",
- "#Point of contraflexure \n",
- "#Let E be the point EB=x\n",
- "#Shear Force at E\n",
- "#V_E=0=R_B-F_C-w1*L_CB-w2*(L_EB-L_DC-L_CB)\n",
- "L_EB=-((-R_B+F_C+w1*L_CB)*w2**-1-L_DC-L_CB)\n",
- "V_E=0\n",
- "\n",
- "#Bending Moment Calculations\n",
- "\n",
- "#Bending Moment at B\n",
- "M_B=0\n",
- "\n",
- "#Bending Moment at C\n",
- "M_C=R_B*L_CB-w1*L_CB**2*2**-1\n",
- "\n",
- "#Bending Moment at D\n",
- "M_D=R_B*(L_CB+L_DC)-w1*L_CB*(L_CB*2**-1+L_DC)-F_C*L_DC\n",
- "\n",
- "#Bending Moment at A\n",
- "M_A=0\n",
- "\n",
- "#Bending Moment at E\n",
- "L_ED=L_EB-(L_DC+L_CB) #m #Length of ED\n",
- "M_E=-w1*L_CB*(L_ED+L_DC+L_CB*2**-1)-F_C*(L_DC+L_ED)+R_B*L_EB\n",
- "\n",
- "#Result\n",
- "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
- "\n",
- "#Plotting the Shear Force Diagram\n",
- "\n",
- "X1=[0,L_CB,L_CB,L_CB+L_DC,L_CB+L_DC+L_AD,L_CB+L_DC+L_AD]\n",
- "Y1=[V_B,V_C1,V_C2,V_D,V_A,0]\n",
- "Z1=[0,0,0,0,0,0]\n",
- "plt.plot(X1,Y1,X1,Z1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Force in kN\")\n",
- "plt.show()\n",
- "\n",
- "#Plotting the Bendimg Moment Diagram\n",
- "\n",
- "X2=[0,L_CB,L_CB+L_DC,L_CB+L_DC+L_ED,L_CB+L_DC+L_AD]\n",
- "Y2=[M_B,M_C,M_D,M_E,M_A]\n",
- "Z2=[0,0,0,0,0]\n",
- "plt.plot(X2,Y2)\n",
- "plt.xlabel(\"Length in m\")\n",
- "plt.ylabel(\"Bending Moment in kN.m\")\n",
- "plt.show()"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Shear Force and Bending Moment Diagrams are the results\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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lZQFoP2X14YcfwtvbG1u3bsX06dMBtF9Wu2TJEjQ3N2PWrFl47733HB2biMij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\nQmCC\n",
- "text": [
- "<matplotlib.figure.Figure at 0x59f5d30>"
- ]
- },
- {
- "metadata": {},
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KuXh248wZdULrL3/ROokQru3ppyEwUF3RbKefHe2W2YLw66+/0rBhQ9PPHh4e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\nTiWEENp7+WXYu1e9hHT9ZXS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- "text": [
- "<matplotlib.figure.Figure at 0x5a39bd0>"
- ]
- }
- ],
- "prompt_number": 79
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 4.4.6,Page No.95"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "F_C=100 #KN #Pt Load at C\n",
- "F_E=50 #KN #Pt Load at E\n",
- "w=20 #KN/m\n",
- "L_AE=L_ED=L_DC=L_CB=2 #m #Length of AE,ED,DC,CB respectively\n",
- "L=8 #m #Length of Beam\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Let R_A & R_B be the reactions at A & B\n",
- "#R_A+R_B=190\n",
- "\n",
- "#Taking Moment at A\n",
- "#M_A=-R_B*L+F_C*(3*L_AE)+w*L_DC*(L_DC*2**-1+2*L_ED)+F_E*L_AE=0\n",
- "R_B=(F_C*(3*L_AE)+w*L_DC*(L_DC*2**-1+2*L_ED)+F_E*L_AE)*L**-1\n",
- "R_A=190-R_B\n",
- "\n",
- "#Shear Force Calculations\n",
- "\n",
- "#Shear Force at B\n",
- "V_B=R_B\n",
- "\n",
- "#Shear Force at C\n",
- "V_C1=R_B\n",
- "V_C2=R_B-F_C\n",
- "\n",
- "#Shear Force at D\n",
- "V_D=V_C2-w*L_DC\n",
- "\n",
- "#Shear Force at E\n",
- "V_E1=V_D\n",
- "V_E2=V_D-F_E\n",
- "\n",
- "#Shear Force at A\n",
- "V_A=V_E2\n",
- "\n",
- "#Point of contraflexure \n",
- "#Let F be the point BF=x\n",
- "#Shear Force at F\n",
- "#V_F=R_B-F_C-w*(L_BF-L_CB)\n",
- "L_FB=-((-R_B+F_C)*w**-1-L_CB)\n",
- "V_F=0\n",
- "\n",
- "#Bending Moment Calculations\n",
- "\n",
- "#Bending Moment at B\n",
- "M_B=0\n",
- "\n",
- "#Bending Moment at C\n",
- "M_C=R_B*L_CB\n",
- "\n",
- "#Bending Moment at D\n",
- "M_D=R_B*(L_CB+L_DC)-F_C*L_DC-w*L_DC**2*2**-1\n",
- "\n",
- "#Bending Moment at E\n",
- "M_E=R_B*(L_CB+L_DC+L_ED)-F_C*(L_ED+L_DC)-w*L_DC*(L_DC*2**-1+L_ED)\n",
- "\n",
- "#Bending Moment at A\n",
- "M_A=R_B*(L_ED+L_DC+L_AE+L_CB)-F_C*(L_ED+L_DC+L_AE)-w*L_DC*(L_DC*2**-1+L_ED+L_AE)-F_E*L_AE\n",
- "\n",
- "#Bending Moment at F\n",
- "L_FC=L_BF-L_CB\n",
- "M_F=R_B*L_FB-F_C*L_FC-w*L_FC**2*2**-1\n",
- "L_DF=L_DC-L_FC\n",
- "\n",
- "#Result\n",
- "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
- "\n",
- "#Plotting the Shear Force Diagram\n",
- "\n",
- "X1=[0,L_CB,L_CB,L_CB+L_DC,L_CB+L_DC+L_ED,L_CB+L_DC+L_ED,L_CB+L_DC+L_ED+L_AE]\n",
- "Y1=[V_B,V_C1,V_C2,V_D,V_E1,V_E2,V_A]\n",
- "Z1=[0,0,0,0,0,0,0,]\n",
- "plt.plot(X1,Y1,X1,Z1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Force in kN\")\n",
- "plt.show()\n",
- "\n",
- "#Plotting the Bendimg Moment Diagram\n",
- "\n",
- "X2=[0,L_CB,L_CB+L_FC,L_CB+L_DC,L_CB+L_DC+L_ED,L_CB+L_DC+L_AD]\n",
- "Y2=[M_B,M_C,M_F,M_D,M_E,M_A]\n",
- "Z2=[0,0,0,0,0,0]\n",
- "plt.plot(X2,Y2)\n",
- "plt.xlabel(\"Length in m\")\n",
- "plt.ylabel(\"Bending Moment in kN.m\")\n",
- "plt.show()"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Shear Force and Bending Moment Diagrams are the results\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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Hx6OpqQlNTU2YN28exo8fr3Qss6x1CLW2thYzZswAYBj6mTt3LqKiohRO1VJa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\nSBYWBhERyfL/OMwEVaZmLAEAAAAASUVORK5CYII=\n",
- "text": [
- "<matplotlib.figure.Figure at 0x59f37d0>"
- ]
- },
- {
- "metadata": {},
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- "text": [
- "<matplotlib.figure.Figure at 0x552a590>"
- ]
- }
- ],
- "prompt_number": 74
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 4.4.7,Page No.96"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "w=20 #KN/m #u.d.l on Length CB\n",
- "F_D= 50 #KN #Pt Load at D\n",
- "L_CB=5 #m #Length of CB\n",
- "L_DC=3 #M #Length of DC\n",
- "L_AD=2 #m #Length of AD\n",
- "L=10 #m #Length of Beam\n",
- "\n",
- "#Calculations\n",
- "\n",
- "theta=arctan(4*3**-1)*(180*pi**-1)\n",
- "F_DV=F_D*sin(theta*pi*180**-1) #Force at Pt D vertically\n",
- "F_DH=F_D*cos(theta*pi*180**-1) #Force at pt D horizontally\n",
- "\n",
- "#Let R_A & R_B be the reactions at A & B\n",
- "#R_A+R_B=140\n",
- "\n",
- "#Taking Moment at A\n",
- "#M_A=0=-R_B*L+w*L_CB*(L_CB*2**-1+L_DC+L_AD)+F_DV*L_AD\n",
- "R_B=(w*L_CB*(L_CB*2**-1+L_DC+L_AD)+F_DV*L_AD)*L**-1\n",
- "R_A=140-R_B\n",
- "\n",
- "#Shear Force Calculations\n",
- "\n",
- "#Shear Force at B\n",
- "V_B=R_B\n",
- "\n",
- "#Shear Force at C\n",
- "V_C=V_B-w*L_CB\n",
- "\n",
- "#Shear Force at D\n",
- "V_D1=V_C\n",
- "V_D2=V_C-F_DV\n",
- "\n",
- "#Shear Force at A\n",
- "V_A=V_D\n",
- "\n",
- "#Pt of Contraflexure\n",
- "#Let E be the pt And BE=x\n",
- "#V_E=0=R_B-w*x\n",
- "x=L_BE=R_B*w**-1\n",
- "\n",
- "#Bending Moment Calculations\n",
- "\n",
- "#Bending Moment at B\n",
- "M_B=0\n",
- "\n",
- "#Bending Moment at C\n",
- "M_C=R_B*L_CB-w*L_CB**2*2**-1\n",
- "\n",
- "#Bending Moment at D\n",
- "M_D=R_B*(L_CB+L_DC)-w*L_CB*(L_CB*2**-1+L_DC)\n",
- "\n",
- "#Bending Moment at A\n",
- "M_A=R_B*L-w*L_CB*(L_CB*2**-1+L_DC+L_AD)-F_DV*L_AD\n",
- "\n",
- "#Bending Moment at E\n",
- "M_E=R_B*L_BE-w*L_BE**2*2**-1\n",
- "\n",
- "#Result\n",
- "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
- "\n",
- "#Plotting the Shear Force Diagram\n",
- "\n",
- "X1=[0,L_CB,L_CB+L_DC,L_CB+L_DC,L_CB+L_DC+L_AD] \n",
- "Y1=[V_B,V_C,V_D1,V_D2,V_A]\n",
- "Z1=[0,0,0,0,0]\n",
- "plt.plot(X1,Y1,X1,Z1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Force in kN\")\n",
- "plt.show()\n",
- "\n",
- "#Plotting the Bendimg Moment Diagram\n",
- "\n",
- "X2=[0,L_BE,L_CB,L_CB+L_DC,L_CB+L_DC+L_AD]\n",
- "Y2=[M_B,M_E,M_C,M_D,M_A] \n",
- "Z2=[0,0,0,0,0,0]\n",
- "plt.plot(X2,Y2)\n",
- "plt.xlabel(\"Length in m\")\n",
- "plt.ylabel(\"Bending Moment in kN.m\")\n",
- "plt.show()"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Shear Force and Bending Moment Diagrams are the results\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x5692870>"
- ]
- },
- {
- "metadata": {},
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HwcHBuHTpErerPmHqVODwYWDDBvZvJlKKKbf/1MvJ51OnTmHlypX45ZdfMGPG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\ng48Y9u7di9deew0vv/yydro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- "text": [
- "<matplotlib.figure.Figure at 0x5687310>"
- ]
- }
- ],
- "prompt_number": 37
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 4.4.8,Page No.97"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "F_C=150 #KN #Pt LOad at C\n",
- "w=300 #KN #u.v.l\n",
- "L=6 #m #Length of beam\n",
- "L_AE=L_DC=L_CB=1 #m #Lengthof AE,DC,CB\n",
- "L_ED=3 #m #Length of ED\n",
- "L_Ed=2 #m\n",
- "L_dD=1 #m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Let R_A & R_B be the reactions at A & B\n",
- "#R_A+R_B=450\n",
- "\n",
- "#Taking Moment at A\n",
- "#M_A=0=R_B*L-F_C*(L_CD+L_ED+L_AE)-w*(2*3**-1*L_ED+L_AE)\n",
- "R_B=(F_C*(L_DC+L_ED+L_AE)+w*(2*3**-1*L_ED+L_AE))*L**-1\n",
- "R_A=450-R_B\n",
- "\n",
- "#Shear Force Calculations\n",
- "\n",
- "#Shear Force at B\n",
- "V_B=R_B\n",
- "\n",
- "#Shear Force at C\n",
- "V_C1=R_B\n",
- "V_C2=R_B-F_C\n",
- "\n",
- "#Shear Force at D\n",
- "V_D=V_C2\n",
- "\n",
- "#Shear Force at E\n",
- "V_E=V_D-w\n",
- "\n",
- "#Shear Force at A\n",
- "V_A=V_E\n",
- "\n",
- "#Pt of contraflexure\n",
- "#Let F be the pt and EF=x\n",
- "#Let w1 be the rate of Loading at D we get\n",
- "w1=w*2*3**-1\n",
- "#The rate of Loading at distance x is200*x*3**-1\n",
- "\n",
- "#V_F=0=-R_B+200*x*3**-1*x*2**-1\n",
- "#After substituting values and simplifying further we get\n",
- "L_EF=x=(R_A*3*100**-1)**0.5\n",
- "\n",
- "#Bending Moment Calculations\n",
- "\n",
- "#Bending Moment at B\n",
- "M_B=0\n",
- "\n",
- "#Bending Moment at C\n",
- "M_C=R_B*L_CB\n",
- "\n",
- "#Bending Moment at D\n",
- "M_D=R_B*(L_CB+L_DC)-F_C*L_DC\n",
- "\n",
- "#Bending Moment at E\n",
- "M_E=R_B*(L_CB+L_DC+L_ED)-F_C*(L_DC+L_ED)-w*L_Ed\n",
- "\n",
- "#Bending Moment at A\n",
- "M_A=0\n",
- "\n",
- "#Bending Moment at F\n",
- "M_F=R_A*(L_AE+L_EF)-200*x*3**-1*x*2**-1*x*3**-1\n",
- "\n",
- "L_FD=L_ED-L_EF\n",
- "\n",
- "\n",
- "#Result\n",
- "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
- "\n",
- "#Plotting the Shear Force Diagram\n",
- "\n",
- "X1=[0,L_CB,L_CB,L_CB+L_CD,L_CB+L_CD+L_ED,L_CB+L_CD+L_ED+L_AE,L_CB+L_CD+L_ED+L_AE] \n",
- "Y1=[V_B,V_C1,V_C2,V_D,V_E,V_A,0]\n",
- "Z1=[0,0,0,0,0,0,0]\n",
- "plt.plot(X1,Y1,X1,Z1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Force in kN\")\n",
- "plt.show()\n",
- "\n",
- "#Plotting the Bendimg Moment Diagram\n",
- "\n",
- "X2=[0,L_CB,L_CB+L_DC,L_FD+L_DC+L_CB,L_CB+L_DC+L_ED,L_CB+L_DC+L_ED+L_AE]\n",
- "Y2=[M_B,M_C,M_D,M_F,M_E,M_A] \n",
- "Z2=[0,0,0,0,0,0]\n",
- "plt.plot(X2,Y2)\n",
- "plt.xlabel(\"Length in m\")\n",
- "plt.ylabel(\"Bending Moment in kN.m\")\n",
- "plt.show()"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Shear Force and Bending Moment Diagrams are the results\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x5a94790>"
- ]
- },
- {
- "metadata": {},
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- "text": [
- "<matplotlib.figure.Figure at 0x561c550>"
- ]
- }
- ],
- "prompt_number": 132
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 4.4.9,Page No.99"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "M_C=40 #KNM #Moment at Pt C\n",
- "w=20 #KNm #u.d.l on L_AD\n",
- "L=10 #m #Length of beam\n",
- "L_CB=5 #m #Length of CB\n",
- "L_DC=1 #m #Length of DC\n",
- "L_AD=4 #m #Length of AD\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Let R_A & R_B be the reactions at A & B\n",
- "#R_A+R_B=80\n",
- "\n",
- "#Taking Moment at A\n",
- "#M_A=0=R_B*L-M-w*L_AD**2*2**-1\n",
- "R_B=(w*L_AD**2*2**-1+M_C)*L**-1\n",
- "R_A=80-R_B\n",
- "\n",
- "#Shear Force Calculations\n",
- "\n",
- "#Shear Force at B\n",
- "V_B=R_B\n",
- "\n",
- "#Shear Force at C\n",
- "V_C=V_B\n",
- "\n",
- "#Shear Force at D\n",
- "V_D=V_C\n",
- "\n",
- "#Shear Force at A\n",
- "V_A=V_D-w*L_AD\n",
- "\n",
- "#Pt of contraflexure\n",
- "#Let E be the pt and BE=x\n",
- "#V_E=0=R_B-w*(L_BE-L_DC-L_CB)\n",
- "L_BE=x=R_B*w**-1+L_DC+L_CB\n",
- "\n",
- "#Bending Moment Calculations\n",
- "\n",
- "#Bending Moment at B\n",
- "M_B=0\n",
- "\n",
- "#Bending Moment at C\n",
- "M_C1=R_B*L_CB\n",
- "M_C2=M_C1-M_C\n",
- "\n",
- "#Bending Moment at D\n",
- "M_D=R_B*(L_CB+L_DC)-M_C\n",
- "\n",
- "#Bending Moment at A\n",
- "M_A=R_B*L-M_C-w*L_AD**2*2**-1\n",
- "\n",
- "#Bending Moment at E\n",
- "L_ED=L_BE-(L_DC+L_CB)\n",
- "M_E=R_B*L_BE-M_C-w*L_ED**2*2**-1\n",
- "\n",
- "#Result\n",
- "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
- "\n",
- "#Plotting the Shear Force Diagram\n",
- "\n",
- "X1=[0,L_CB,L_CB+L_DC,L_CB+L_DC+L_AD,L_CB+L_DC+L_AD] \n",
- "Y1=[V_B,V_C,V_D,V_A,0]\n",
- "Z1=[0,0,0,0,0]\n",
- "plt.plot(X1,Y1,X1,Z1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Force in kN\")\n",
- "plt.show()\n",
- "\n",
- "#Plotting the Bendimg Moment Diagram\n",
- "\n",
- "X2=[0,L_CB,L_CB,L_CB+L_DC,L_CB+L_DC+L_ED,L_CB+L_DC+L_AD]\n",
- "Y2=[M_B,M_C1,M_C2,M_D,M_E,M_A] \n",
- "Z2=[0,0,0,0,0,0]\n",
- "plt.plot(X2,Y2,X2,Z2)\n",
- "plt.xlabel(\"Length in m\")\n",
- "plt.ylabel(\"Bending Moment in kN.m\")\n",
- "plt.show()"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Shear Force and Bending Moment Diagrams are the results\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x5687390>"
- ]
- },
- {
- "metadata": {},
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- "text": [
- "<matplotlib.figure.Figure at 0x5565a90>"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 4.4.10,Page No.100"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "w=10 #KNm #u.d.l on L_AD\n",
- "F_D=20 #KN #Pt Load at D\n",
- "M_C=240 #KNm #moment at Pt C\n",
- "L_DC=L_CB=2 #m #Length of DC and CB\n",
- "L_AD=4 #m #Length of AD\n",
- "L=8 #m #Length of Beam\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Let R_A & R_B be the reactions at A & B\n",
- "#R_A+R_B=60\n",
- "\n",
- "#Taking Moment at A\n",
- "#M_A=0=-R_B*L-M_C+F_D*L_AD+w*L_AD**2*2**-1\n",
- "R_B=-(M_C-F_D*L_AD-w*L_AD**2*2**-1)*L**-1\n",
- "R_A=60-R_B\n",
- "\n",
- "#Shear Force Calculations\n",
- "\n",
- "#Shear Force at B\n",
- "V_B=R_B\n",
- "\n",
- "#Shear Force at C\n",
- "V_C=V_B\n",
- "\n",
- "#Shear Force at D\n",
- "V_D1=V_B\n",
- "V_D2=V_D1-F_D\n",
- "\n",
- "#Shear Force at A\n",
- "V_A=V_D2-w*L_AD\n",
- "\n",
- "#Bending Moment Calculations\n",
- "\n",
- "#Bending Moment at B\n",
- "M_B=0\n",
- "\n",
- "#Bending Moment at C\n",
- "M_C1=R_B*L_CB\n",
- "M_C2=M_C+R_B*L_CB\n",
- "\n",
- "#Bending Moment at D\n",
- "M_D=R_B*(L_DC+L_CB)+M_C\n",
- "\n",
- "#Bending Moment at A\n",
- "M_A=R_B*L+M_C-w*L_AD**2*2**-1-F_D*L_AD\n",
- "\n",
- "#Result\n",
- "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
- "\n",
- "#Plotting the Shear Force Diagram\n",
- "\n",
- "X1=[0,L_CB,L_CB+L_DC,L_CB+L_DC,L_CB+L_DC+L_AD] \n",
- "Y1=[V_B,V_C,V_D1,V_D2,V_A]\n",
- "Z1=[0,0,0,0,0]\n",
- "plt.plot(X1,Y1,X1,Z1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Force in kN\")\n",
- "plt.show()\n",
- "\n",
- "#Plotting the Bendimg Moment Diagram\n",
- "\n",
- "X2=[0,L_CB,L_CB,L_CB+L_DC,L_CB+L_DC+L_AD,L_CB+L_DC+L_AD]\n",
- "Y2=[M_B,M_C1,M_C2,M_D,M_A,0] \n",
- "Z2=[0,0,0,0,0,0]\n",
- "plt.plot(X2,Y2,X2,Z2)\n",
- "plt.xlabel(\"Length in m\")\n",
- "plt.ylabel(\"Bending Moment in kN.m\")\n",
- "plt.show()"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Shear Force and Bending Moment Diagrams are the results\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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SExOVjmKXVqsVp0+fVjpGm9LT08XGjRuFENL/93PnzimcqHUNDQ1i8ODB4sSJ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\nrAAAAABJRU5ErkJggg==\n",
- "text": [
- "<matplotlib.figure.Figure at 0x5684b90>"
- ]
- },
- {
- "metadata": {},
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Lr+lu9SC9XmraOH0aKC1tu792DQgNBcLCAK227X7QIFv/ZPaJK68ROQ9zVl6z\nuyMFV1dX/P3vf8ekSZPQ0tKCBQsWdAgEY9zdpQ/8sDDgySfbtl+9KgXE6dPA//0f8N57wL//DQwe\n3DUoxowB3Nys+MMREdk5uztSMEdv12hubQUqKjoeUZw+DXz3HRAY2DUsRo2y7NS09oRHCkTOwyGP\nFGzBxQX43e+kW2Ji2/YbN4CyMikkSkuBvDzpvqWlY0iEhQEhIYCHh3I/AxGRNThlKBjzm98AUVHS\nrb26urajiYIC4K23gDNnpCOIzmHh5ydNUEVE5IgYCmbw8gLi4qSbQXMzcOFCW1i8+650X1cHaDRd\nw2L4cOXqJyIyF0PhDrm6AkFB0i0pqW37tWvAf/7TFhYffijd9+/fdaxCowG4/DUR2ROGgoUNGAD8\n4Q/SzUAUpbVUDYPaeXnA3/4mHWmo1V3DQq3uuwPbRGTfGAo2IAjA6NHSbfLktu2deyv+8Q/pMXsr\niEgpDAUFmeqtKC2VeivefVc6JcXeCiKyNoaCHRoyRFpv9aGH2ra1tgLl5W1h8eGHQFoa8P33ztdb\nQUTWw1BwEC4u0uWufn4deyuamqTeivbjFeytIKI75ZQdzc6gfW+F4b673orycuDUKXY0EzkDcz47\nGQpOpHNvheH+0UelFZqIqG9jKBARkcycz067W2SHiIiUw1AgIiIZQ4GIiGQMBSIikjEUiIhIxlAg\nIiIZQ4GIiGQMBSIikjEUiIhIxlAgIiIZQ4GIiGQMBSIikjEUiIhIxlAgIiIZQ4GIiGQMBSIikikS\nCmlpaVCpVNDpdNDpdMjJyZGfS09PR0BAAIKCgpCXl6dEeURETkuRUBAEAcuWLUNxcTGKi4uRkJAA\nACgrK8OePXtQVlaG3NxcLF68GK2trUqUaBH5+flKl2AW1mlZrNOyHKFOR6jRXIqdPupuSbjs7Gwk\nJyfDzc0NarUa/v7+KCwsVKA6y3CUfyis07JYp2U5Qp2OUKO5FAuFjRs3Ijw8HAsWLEB9fT0AoKam\nBiqVSn6NSqVCdXW1UiUSETkdq4VCXFwctFptl9uhQ4fw7LPPory8HCUlJRg5ciSWL19u9PsIgmCt\nEomIqDNRYeXl5WJoaKgoiqKYnp4upqeny89NmjRJPHnyZJf3+Pn5iQB444033njrwc3Pz8/kZ7Ir\nFFBbW4uRI0cCAA4cOACtVgsAmDZtGlJSUrBs2TJUV1fj/PnziI6O7vL+Cxcu2LReIiJnoUgovPji\niygpKYF5/s8nAAAIRElEQVQgCPD19cXmzZsBABqNBklJSdBoNHB1dcWmTZt4+oiIyIYEUezmMiAi\nInJKDtfRnJubi6CgIAQEBCAzM1Ppcro1f/58eHl5yafF7FVlZSViY2MREhKC0NBQZGVlKV1St27e\nvIlx48YhIiICGo0Gq1atUroko1paWqDT6TB16lSlSzFKrVYjLCwMOp2u29Oz9qK+vh6zZs1CcHAw\nNBoNTp48qXRJXZw9e1ZuwtXpdBg4cKDd/j9KT09HSEgItFotUlJS8Msvv3T/QksNGNtCc3Oz6Ofn\nJ5aXl4t6vV4MDw8Xy8rKlC6ri2PHjolFRUXyALq9qq2tFYuLi0VRFMWGhgYxMDDQLv88RVEUr1+/\nLoqiKN66dUscN26cePz4cYUr6t7atWvFlJQUcerUqUqXYpRarRZ//PFHpcswae7cueK2bdtEUZT+\n3uvr6xWu6PZaWlpEb29v8fvvv1e6lC7Ky8tFX19f8ebNm6IoimJSUpK4Y8eObl/rUEcKhYWF8Pf3\nh1qthpubG5544glkZ2crXVYXDzzwAAYPHqx0GSZ5e3sjIiICAODh4YHg4GDU1NQoXFX37r77bgCA\nXq9HS0sLhgwZonBFXVVVVeHjjz/GwoULu23OtCf2Xt/PP/+M48ePY/78+QAAV1dXDBw4UOGqbu/T\nTz+Fn58fRo8erXQpXQwYMABubm5oampCc3Mzmpqa4OPj0+1rHSoUqqurO/yBs7nNcioqKlBcXIxx\n48YpXUq3WltbERERAS8vL8TGxkKj0ShdUhd/+tOf8Ne//hUuLvb930oQBEycOBFjx47F1q1blS6n\nW+Xl5Rg+fDjmzZuHyMhILFq0CE1NTUqXdVvvv/8+UlJSlC6jW0OGDMHy5ctx7733YtSoURg0aBAm\nTpzY7Wvt+19vJ7wSyToaGxsxa9YsbNiwAR4eHkqX0y0XFxeUlJSgqqoKx44ds7tpBT766COMGDEC\nOp3O7n8LLygoQHFxMXJycvDmm2/i+PHjSpfURXNzM4qKirB48WIUFRXhnnvuQUZGhtJlGaXX63H4\n8GHMnj1b6VK6dfHiRaxfvx4VFRWoqalBY2Mjdu3a1e1rHSoUfHx8UFlZKX9dWVnZYVoM6rlbt25h\n5syZmDNnDhITE5Uux6SBAwdiypQp+Oqrr5QupYMvv/wShw4dgq+vL5KTk/HZZ59h7ty5SpfVLUOP\n0PDhwzFjxgy7nF9MpVJBpVLh97//PQBg1qxZKCoqUrgq43JychAVFYXhw4crXUq3vvrqK9x///0Y\nOnQoXF1d8dhjj+HLL7/s9rUOFQpjx47F+fPnUVFRAb1ejz179mDatGlKl+WwRFHEggULoNFosHTp\nUqXLMeqHH36Q58e6ceMGjhw5Ap1Op3BVHa1ZswaVlZUoLy/H+++/j4cffhjvvPOO0mV10dTUhIaG\nBgDA9evXkZeXZ5dXyXl7e2P06NE4d+4cAOl8fUhIiMJVGbd7924kJycrXYZRQUFBOHnyJG7cuAFR\nFPHpp58aPwVro8Fvi/n444/FwMBA0c/PT1yzZo3S5XTriSeeEEeOHCm6u7uLKpVK3L59u9Ildev4\n8eOiIAhieHi4GBERIUZERIg5OTlKl9VFaWmpqNPpxPDwcFGr1Yqvv/660iXdVn5+vt1effTtt9+K\n4eHhYnh4uBgSEmK3/4dEURRLSkrEsWPHimFhYeKMGTPs9uqjxsZGcejQoeK1a9eULuW2MjMzRY1G\nI4aGhopz584V9Xp9t69j8xoREckc6vQRERFZF0OBiIhkDAUiIpIxFIiISMZQICIiGUOBiIhkDAXq\n06w9bcf69etx48aNHu3v8OHDdjvtOxH7FKhP8/T0lDt4rcHX1xdfffUVhg4dapP9EVkbjxTI6Vy8\neBEJCQkYO3YsHnzwQZw9exYA8Mc//hEvvPACxo8fDz8/P+zfvx+ANEPr4sWLERwcjPj4eEyZMgX7\n9+/Hxo0bUVNTg9jYWEyYMEH+/n/+858RERGBP/zhD7h8+XKX/e/YsQPPPffcbffZXkVFBYKCgjBv\n3jyMGTMGTz75JPLy8jB+/HgEBgbi1KlT1vhjImdlwy5rIpvz8PDosu3hhx8Wz58/L4qiKJ48eVJ8\n+OGHRVEUxdTUVDEpKUkURVEsKysT/f39RVEUxQ8++ECcPHmyKIqieOnSJXHw4MHi/v37RVHsumCN\nIAjiRx99JIqiKK5YsUJ89dVXu+x/x44d4pIlS267z/bKy8tFV1dX8d///rfY2toqRkVFifPnzxdF\nURSzs7PFxMTEnv6xEBnlqnQoEdlSY2MjTpw40WGKY71eD0Camt0wU2xwcDDq6uoAAF988QWSkpIA\nQF7PwRh3d3dMmTIFABAVFYUjR47cth5j++zM19dXnhAuJCREngs/NDQUFRUVt90HUU8wFMiptLa2\nYtCgQSguLu72eXd3d/mx+OtwmyAIHdZIEG8zDOfm5iY/dnFxQXNzs8mauttnZ/379+/wfQ3vMXcf\nRObimAI5lQEDBsDX1xf79u0DIH0Il5aW3vY948ePx/79+yGKIurq6nD06FH5OU9PT1y7dq1HNdwu\nVIiUxlCgPq2pqQmjR4+Wb+vXr8euXbuwbds2REREIDQ0FIcOHZJf3351P8PjmTNnQqVSQaPR4Kmn\nnkJkZKS8XvDTTz+NRx55RB5o7vz+7lYL7Lzd2OPO7zH2NVckJEviJalEZrh+/Truuece/Pjjjxg3\nbhy+/PJLjBgxQumyiCyOYwpEZnj00UdRX18PvV6Pl19+mYFAfRaPFIiISMYxBSIikjEUiIhIxlAg\nIiIZQ4GIiGQMBSIikjEUiIhI9v+ebmxA5oKUQQAAAABJRU5ErkJggg==\n",
- "text": [
- "<matplotlib.figure.Figure at 0x54a1290>"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 4.4.11,Page No.101"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "F_C=5 #KN #Force at C\n",
- "w=2 #KNm #u.d.l on beam\n",
- "L_BC=3 #m #Length of BC\n",
- "L_AB=6 #m #Length of AB\n",
- "L=9 #m #Length of Beam\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Let R_A & R_B be the reactions at A & B\n",
- "#R_A+R_B=23\n",
- "\n",
- "#Taking Moment at A\n",
- "#M_A=0=F_C*L-R_B*L_AB+w*L**2*2**-1\n",
- "R_B=-(-F_C*L-w*L**2*2**-1)*L_AB**-1\n",
- "R_A=23-R_B\n",
- "\n",
- "#Shear Force Calculations\n",
- "\n",
- "#Shear Force at C\n",
- "V_C1=0\n",
- "V_C2=-F_C\n",
- "\n",
- "#Shear Force at B\n",
- "V_B=V_C2-w*L_BC**2*2**-1\n",
- "\n",
- "#Shear Force at A\n",
- "V_A=F_C*L+R_B*L_AB-w*L**2*2**-1\n",
- "\n",
- "#Pt of contraflexure\n",
- "#Let D be the pt And L_AD=x\n",
- "#V_D=0=R_A+w*L_AD\n",
- "L_AD=x=R_A*w**-1\n",
- "\n",
- "#Bending Moment Calculations\n",
- "\n",
- "#Bending Moment at C\n",
- "M_C=0\n",
- "\n",
- "#Bending Moment at B\n",
- "M_B=-F_C*L_BC-w*L_BC**2*2**-1\n",
- "\n",
- "#Bending Moment at A\n",
- "M_A=-F_C*L-w*L**2*2**-1+R_B*L_AB\n",
- "\n",
- "#Bending Moment at D\n",
- "L_DC=L-L_AD\n",
- "L_DB=L_DC-L_BC\n",
- "M_D=-R_A*L_AD+w*L_AD**2*2**-1\n",
- "\n",
- "#Result\n",
- "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
- "\n",
- "#Plotting the Shear Force Diagram\n",
- "\n",
- "X1=[0,L_BC,L_BC+L_AB,L_BC+L_AB] \n",
- "Y1=[V_C,V_B,V_A,0]\n",
- "Z1=[0,0,0,0]\n",
- "plt.plot(X1,Y1,X1,Z1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Force in kN\")\n",
- "plt.show()\n",
- "\n",
- "#Plotting the Bendimg Moment Diagram\n",
- "\n",
- "X2=[0,L_BC,L_BC+L_DB,L_BC+L_AB]\n",
- "Y2=[M_C,M_B,M_D,M_A] \n",
- "Z2=[0,0,0,0]\n",
- "plt.plot(X2,Y2,X2,Z2)\n",
- "plt.xlabel(\"Length in m\")\n",
- "plt.ylabel(\"Bending Moment in kN.m\")\n",
- "plt.show()\n",
- "\n",
- "#The Bending moment in book is incorrect"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Shear Force and Bending Moment Diagrams are the results\n"
- ]
- },
- {
- "metadata": {},
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- "text": [
- "<matplotlib.figure.Figure at 0x5a802b0>"
- ]
- },
- {
- "metadata": {},
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- "text": [
- "<matplotlib.figure.Figure at 0x587df90>"
- ]
- }
- ],
- "prompt_number": 56
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 4.4.12,Page No.102"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "F_C=5 #KN #Pt Load at C\n",
- "F_D=4 #KN #Pt Load at D\n",
- "L_BC=1.25 #m #Length of BC\n",
- "L_DB=1 #m #Length of DB\n",
- "L_AD=3 #m #Length of AD\n",
- "w=2 #KN/m #u.d.l\n",
- "L=5.25 #m #Length of beam\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Let R_A & R_B be the reactions at A & B\n",
- "#R_A+R_B=15\n",
- "\n",
- "#Taking Moment at A\n",
- "#M_A=0=F_C*L-R_B*(L_DB+L_AD)+F_D*L_AD+w*L_AD**2*2**-1\n",
- "R_B=-(-F_C*L-F_D*L_AD-w*L_AD**2*2**-1)*(L_DB+L_AD)**-1\n",
- "R_A=15-R_B\n",
- "\n",
- "#Shear Force Calculations\n",
- "\n",
- "#Shear Force at C\n",
- "V_C=-F_C\n",
- "\n",
- "#Shear Force at B\n",
- "V_B1=V_C\n",
- "V_B2=V_C+R_B\n",
- "\n",
- "#Shear Force at D\n",
- "V_D1=V_B2\n",
- "V_D2=V_B2-F_D\n",
- "\n",
- "#Shear Force at A\n",
- "V_A=-(w*L_AD)-F_D-F_C+R_B\n",
- "\n",
- "#Pt of contraflexure\n",
- "#Let E be the pt and BE=x\n",
- "#V_E=0=-F_C+R_B-F_D-w*(L_BE-L_DB)\n",
- "L_BE=x=-((F_C-R_B+F_D)*w**-1-L_DB)\n",
- "\n",
- "#Bending Moment Calculations\n",
- "\n",
- "#Bending Moment at C\n",
- "M_C=0\n",
- "\n",
- "#Bending Moment at B\n",
- "M_B=-F_C*L_BC\n",
- "\n",
- "#Bending Moment at D\n",
- "M_D=-F_C*(L_DB+L_BC)-R_B*L_DB\n",
- "\n",
- "#Bending Moment at A\n",
- "M_A=-F_C*L+R_B*(L_DB+L_AD)-F_D*L_AD-w*L_AD**2*2**-1\n",
- "\n",
- "#Bending Moment at E\n",
- "L_ED=L_BE-L_DB\n",
- "M_E=-F_C*(L_BC+L_BE)+R_B*L_BE-F_D*(L_BE-L_DB)-w*(L_BE-L_DB)**2*2**-1\n",
- "\n",
- "#Result\n",
- "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
- "\n",
- "#Plotting the Shear Force Diagram\n",
- "\n",
- "X1=[0,L_BC,L_BC,L_BC+L_DB,L_BC+L_DB,L_BC+L_DB+L_AD,L_BC+L_DB+L_AD] \n",
- "Y1=[V_C,V_B1,V_B2,V_D1,V_D2,V_A,0]\n",
- "Z1=[0,0,0,0,0,0,0]\n",
- "plt.plot(X1,Y1,X1,Z1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Force in kN\")\n",
- "plt.show()\n",
- "\n",
- "#Plotting the Bendimg Moment Diagram\n",
- "\n",
- "X2=[0,L_BC,L_BC+L_DB,L_BC+L_DB+L_ED,L_BC+L_DB+L_AD]\n",
- "Y2=[M_C,M_B,M_D,M_E,M_A] \n",
- "Z2=[0,0,0,0,0]\n",
- "plt.plot(X2,Y2,X2,Z2)\n",
- "plt.xlabel(\"Length in m\")\n",
- "plt.ylabel(\"Bending Moment in kN.m\")\n",
- "plt.show()\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Shear Force and Bending Moment Diagrams are the results\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x57a1c30>"
- ]
- },
- {
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KoijKpk2blCFDhtT1r4WoVnZ19hGRpa5fv47vv/8eI0eONN5WUlIC\nQJznU3Fya3BwMC5cuAAA2L17N0aNGgUAxl4JtXF3d8fAgQMBAFFRUUhNTb1tntpe88/8/f2NB5iF\nhoaiT58+AICwsDAYDIbbvgZRXbAokFMpLy9HixYtkJWVVeP33d3djZ8rf0y36XS6Kj0ElNtMw7m5\nuRk/d3FxQWlpqclMNb3mnzVs2LDK81Y8xtzXIDIX5xTIqTRr1gz+/v745JNPAIg34R9//PG2j+ne\nvTuSk5OhKAouXLiAXbt2Gb/n6emJa9eu1SnD7YoKkWwsCuTQioqK4Ovra/xYsmQJPvzwQ6xatQoR\nEREICwvD5s2bjfevfCR0xefDhw+Hj48PQkJC8NhjjyEyMtLY3/app57Cww8/bJxo/vPjazpi+s+3\n1/b5nx9T29eOdow1ycUlqURmKCwsRNOmTXH58mXExMTgu+++Q+vWrWXHIlId5xSIzDBo0CBcuXIF\nJSUleOWVV1gQyGFxpEBEREacUyAiIiMWBSIiMmJRICIiIxYFIiIyYlEgIiIjFgUiIjL6fy+kTf8G\nZiLiAAAAAElFTkSuQmCC\n",
- "text": [
- "<matplotlib.figure.Figure at 0x567dff0>"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 4.4.13,Page No.103"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "F_E=20 #KN #Pt Load at E\n",
- "F_C=30 #KN #Pt Load at C\n",
- "F_B=60 #KN #Pt Load at B\n",
- "L_AB=L_BC=L_CD=1.5 #m #Length of AB,BC,CD respectively\n",
- "L_DE=2.5 #m #Length od DE\n",
- "L_AD=4.5 #m #Length of AD\n",
- "L=7 #m #Length of beam\n",
- "w=30 #KN/m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#LEt R_A and R_D be the reactions at A and D\n",
- "#R_A+R_D=245\n",
- "\n",
- "#Taking moment at A\n",
- "#M_A=0=-R_D*(L_BC+L_AB+L_CD)+F_E*L+w*L_Ad**2*2**-1+F_C*(L_AB+L_BC)+F_B*L_AB\n",
- "R_D=-(-(F_E*L)-(w*L_AD**2*2**-1)-F_C*(L_AB+L_BC)-F_B*L_AB)*(L_BC+L_AB+L_CD)**-1\n",
- "R_A=245-R_D\n",
- "\n",
- "#Shear Force Calculations\n",
- "\n",
- "#Shear Force at C\n",
- "V_E1=0\n",
- "V_E2=-F_E\n",
- "\n",
- "#Shear Force at D\n",
- "V_D1=V_E2\n",
- "V_D2=V_E2+R_D\n",
- "\n",
- "#Shear Force at C\n",
- "V_C1=V_D2\n",
- "V_C2=V_D2-F_C-w*L_CD\n",
- "\n",
- "#Shear Force at B\n",
- "V_B1=V_C2\n",
- "V_B2=-F_E+R_D-F_C-w*(L_BC+L_CD)-F_B\n",
- "\n",
- "#Shear Force at A\n",
- "V_A=-F_E-F_C-F_B-w*L_AD+R_D\n",
- "\n",
- "#Pt of contraflexure\n",
- "#Let F be the pt and EF=x\n",
- "#V_F=-F_E-F_C+R_D-w*L_FE+w*L_DE\n",
- "L_FE=-(F_E+F_C-R_D-w*L_DE)*w**-1\n",
- "L_FD=L_FE-L_DE\n",
- "L_FC=L_FE-L_CD-L_DE\n",
- "\n",
- "#Bending Moment Calculations\n",
- "\n",
- "#Bending Moment at E\n",
- "M_E=0\n",
- "\n",
- "#Bending Moment at D\n",
- "M_D=-F_E*L_DE\n",
- "\n",
- "#Bending Moment at C\n",
- "M_C=-F_E*(L_CD+L_DE)+R_D*L_CD-w*L_CD**2*2**-1\n",
- "\n",
- "#Bending Moment at F\n",
- "M_F=-w*L_FD**2*2**-1-F_C*L_FC+R_D*L_FD-F_E*L_FE\n",
- "\n",
- "#Bending Moment at B\n",
- "M_B=-F_E*(L_DE+L_CD+L_BC)-F_C*L_BC+R_D*(L_CD+L_BC)-w*(L_BC+L_CD)**2*2**-1\n",
- "\n",
- "#Bending Moment at A\n",
- "M_A=-F_E*L+R_D*(L_AD)-F_C*(L_BC+L_AB)-F_B*L_AB-w*(L_AD)**2*2**-1\n",
- "\n",
- "#Bending Moment at F\n",
- "M_F=-F_E*L_FE+R_D*L_FD-F_C*L_FC-w*L_FD**2*2**-1\n",
- "\n",
- "#Pt of contraflexure\n",
- "#Let G be the pt and GE=y\n",
- "#M_G=-F_E*L_GE+R_D*(L_GE-L_DE)-F_C*(L_GE-L_DE)**2*2**-1\n",
- "#After substituting values and further simplifying we get\n",
- "#y**2-12.9+29.35=0\n",
- "a=1\n",
- "b=-12.9\n",
- "c=29.35\n",
- "\n",
- "X=b**2-4*a*c\n",
- "\n",
- "y1=(-b+X**0.5)*(2*a)**-1\n",
- "y2=(-b-X**0.5)*(2*a)**-1\n",
- "\n",
- "#Result\n",
- "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
- "\n",
- "#Plotting the Shear Force Diagram\n",
- "\n",
- "X1=[0,0,L_DE,L_DE,L_DE+L_CD,L_DE+L_CD,L_DE+L_CD+L_BC,L_DE+L_CD+L_BC,L_DE+L_CD+L_BC+L_AB,L_DE+L_CD+L_BC+L_AB] \n",
- "Y1=[V_E1,V_E2,V_D1,V_D2,V_C1,V_C2,V_B1,V_B2,V_A,0]\n",
- "Z1=[0,0,0,0,0,0,0,0,0,0]\n",
- "plt.plot(X1,Y1,X1,Z1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Force in kN\")\n",
- "plt.show()\n",
- "\n",
- "#Plotting the Bendimg Moment Diagram\n",
- "\n",
- "X2=[0,L_DE,L_DE+L_CD,L_DE+L_CD+L_FC,L_DE+L_CD+L_BC,L_DE+L_CD+L_BC+L_AB]\n",
- "Y2=[M_E,M_D,M_C,M_F,M_B,M_A] \n",
- "Z2=[0,0,0,0,0,0]\n",
- "plt.plot(X2,Y2,X2,Z2)\n",
- "plt.xlabel(\"Length in m\")\n",
- "plt.ylabel(\"Bending Moment in kN.m\")\n",
- "plt.show()\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Shear Force and Bending Moment Diagrams are the results\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x588f3b0>"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
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- "text": [
- "<matplotlib.figure.Figure at 0x54f18f0>"
- ]
- }
- ],
- "prompt_number": 71
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 4.4.14,Page No.105"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "L_DE=L_BC=2.5 #m #Length of DE & BC\n",
- "L_CD=L_AB=5 #m #Length of CD & AB\n",
- "F_C=F_B=80 #KN #Pt Load at C & B\n",
- "w1=16 #KN/m #u.d.l on L_DE\n",
- "w2=10 #KN/m #u.d.l on L_AB\n",
- "L=10 #m #Length of beam\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#LEt R_A and R_D be the reactions at A and D\n",
- "#R_A+R_D=250\n",
- "\n",
- "#Taking moment at A\n",
- "#M_A=0=w1*L_DE*(L_DE*2**-1+L_CD+L_BC+L_AB)-R_D*(L_CD+L_BC+L_AB)+F_C*(L_BC+L_AB)+F_C*(L_BC+L_AB)+F_B*L_AB+w2*L_AB**2*2**-1\n",
- "R_D=-(-w1*L_DE*(L_DE*2**-1+L_CD+L_BC+L_AB)-F_C*(L_BC+L_AB)-F_B*(L_AB)-w2*L_AB**2*2**-1)*(L_CD+L_BC+L_AB)**-1\n",
- "R_A=250-R_D\n",
- "\n",
- "#Shear Force Calculations\n",
- "\n",
- "#Shear Force at E\n",
- "V_E=0\n",
- "\n",
- "#Shear Force at D\n",
- "V_D1=-w1*L_DE\n",
- "V_D2=-w1*L_DE+R_D\n",
- "\n",
- "#Shear Force at C\n",
- "V_C1=V_D2\n",
- "V_C2=V_D2-F_C\n",
- "\n",
- "#Shear Force at B\n",
- "V_B1=V_C2\n",
- "V_B2=V_C2-F_B\n",
- "\n",
- "#Shear Force at A\n",
- "V_A1=V_B2-w2*L_AB\n",
- "V_A2=0\n",
- "\n",
- "#Bending Moment Calculations\n",
- "\n",
- "#Bending Moment at E\n",
- "M_E=0\n",
- "\n",
- "#Bending Moment at D\n",
- "M_D=-w1*L_DE**2*2**-1\n",
- "\n",
- "#Bending Moment at C\n",
- "M_C=R_D*L_CD-w1*L_DE*(L_DE*2**-1+L_CD)\n",
- "\n",
- "#Bending Moment at B\n",
- "M_B=-w1*L_DE*(L_DE*2**-1+L_CD+L_BC)+R_D*(L_CD+L_BC)-F_C*L_BC\n",
- "\n",
- "#Bending Moment at A\n",
- "M_A=-w1*L_DE*(L_DE*2**-1+L_CD+L_BC+L_AB)+R_D*(L_CD+L_BC+L_AB)-F_C*(L_BC+L_AB)-F_B*L_AB-w2*L_AB**2*2**-1\n",
- "\n",
- "#Result\n",
- "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
- "\n",
- "#Plotting the Shear Force Diagram\n",
- "\n",
- "X1=[0,L_FE,L_FE,L_DE+L_CD,L_DE+L_CD,L_DE+L_CD+L_BC,L_DE+L_CD+L_BC,L_DE+L_CD+L_BC+L_AB,L_DE+L_CD+L_BC+L_AB] \n",
- "Y1=[V_E,V_D1,V_D2,V_C1,V_C2,V_B1,V_B2,V_A1,V_A2]\n",
- "Z1=[0,0,0,0,0,0,0,0,0]\n",
- "plt.plot(X1,Y1,X1,Z1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Force in kN\")\n",
- "plt.show()\n",
- "\n",
- "#Plotting the Bendimg Moment Diagram\n",
- "\n",
- "X2=[0,L_DE,L_DE+L_CD,L_DE+L_CD+L_BC,L]\n",
- "Y2=[M_E,M_D,M_C,M_B,M_A] \n",
- "Z2=[0,0,0,0,0]\n",
- "plt.plot(X2,Y2,X2,Z2)\n",
- "plt.xlabel(\"Length in m\")\n",
- "plt.ylabel(\"Bending Moment in kN.m\")\n",
- "plt.show()"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Shear Force and Bending Moment Diagrams are the results\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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WfS666CKKioro2LEjubm5JCUl8cknnzRWRBER+bkMG8XGxhq7du1q8PXS0lIj\nPDzctf25554zpk+fXus+3bt3NwA99NBDDz0sPLp3727p53ejHXmcLcMwXM+//PJLOnbsSJs2bdi3\nbx/5+flcfPHFdOjQAV9fX3bu3MnAgQN59tlnufXWW2v9vM8++6ypoouItFq2jHmsW7eOoKAgduzY\nwZgxY4iPjwfgrbfeok+fPsTExDBx4kSefPJJOnToAMDSpUtJT08nLCyM0NBQRo8ebUd0EREBHMbp\nv/qLiIicBbe72urn2rRpE+Hh4YSFhbFw4UK749SqqKiIYcOGERkZSe/evXn00UftjlSv6upqYmJi\n6rzAwR0cOXKECRMm0KtXLyIiItixY4fdkWo1f/58IiMjiYqKYtKkSXz33Xd2RwJg6tSp+Pn5ERUV\n5dp2+PBh4uLi6NGjB6NGjeLIkSM2JjTVlvOOO+6gV69e9OnTh3HjxvH111/bmLD2jKc89NBDeHh4\ncPjwYRuSnamunEuWLKFXr1707t2bO++8s+EPsjRC4qaqqqqM7t27GwUFBUZlZaXRp08fIy8vz+5Y\nNZSVlRm7d+82DMMwjh49avTo0cMtc57y0EMPGZMmTTISEhLsjlKnKVOmGMuWLTMMwzC+//5748iR\nIzYnqqmgoMAICQkxTpw4YRiGYSQnJxsrVqywOZXp7bffNnJzc43evXu7tt1xxx3GwoULDcMwjAUL\nFhh33nmnXfFcasu5ZcsWo7q62jAMw7jzzjttz1lbRsMwjM8//9y48sorjeDgYOPQoUM2pftRbTm3\nbdtmjBw50qisrDQMwzC++OKLBj+nRRx55OTkEBoaSnBwMF5eXlx33XVkZWXZHasGf39/oqOjAWjX\nrh29evWitLTU5lS1Ky4u5rXXXiM9Pf2Mixrcyddff80777zD1KlTAfD09OSCCy6wOVVNvr6+eHl5\nUVFRQVVVFRUVFQQGBtodC4ChQ4fSsWPHM7Zt2LCB1NRUAFJTU91iQm5tOePi4vDwMH+EDRo0iOLi\nYjuiudSWEWDWrFk8+OCDNiSqXW05H3/8ce6++268vLwA6NKlS4Of0yLKo6SkhKCgINfXTqfT7ScR\nFhYWsnv3bgYNGmR3lFr9/ve/Z9GiRa5/nO6ooKCALl26kJaWRt++fbnpppuoqKiwO1YNnTp14g9/\n+APdunXjoosuokOHDowcOdLuWHUqLy/Hz88PAD8/P8rLy21O1LDly5dz1VVX2R2jhqysLJxOJ5dc\ncondUep+TK2EAAAGkUlEQVSVn5/P22+/zeDBg4mNjeWDDz5ocB/3/clggcPhsDuCJceOHWPChAk8\n8sgjtGvXzu44Nbzyyit07dqVmJgYtz3qAKiqqiI3N5eMjAxyc3Px8fFhwYIFdseqYe/evSxevJjC\nwkJKS0s5duwYq1atsjvWWXE4HG7/7+uBBx6gbdu2TJo0ye4oZ6ioqCAzM5O5c+e6trnrv6eqqiq+\n+uorduzYwaJFi0hOTm5wnxZRHoGBgRQVFbm+LioqOmM5E3fy/fffM378eCZPnkxSUpLdcWr13nvv\nsWHDBkJCQkhJSWHbtm1MmTLF7lg1OJ1OnE4nAwYMAGDChAlnrNLsLj744AMuu+wyOnfujKenJ+PG\njeO9996zO1ad/Pz8XKtDlJWV0bVrV5sT1W3FihW89tprblnGe/fupbCwkD59+hASEkJxcTH9+vXj\niy++sDtaDU6nk3HjxgEwYMAAPDw8OHToUL37tIjy6N+/P/n5+RQWFlJZWcnq1atJTEy0O1YNhmEw\nbdo0IiIiuP322+2OU6fMzEyKioooKCjghRdeYPjw4TzzzDN2x6rB39+foKAg9uzZA8DWrVuJjIy0\nOVVN4eHh7Nixg+PHj2MYBlu3biUiIsLuWHVKTExk5cqVAKxcudJtf8nZtGkTixYtIisri1/96ld2\nx6khKiqK8vJyCgoKKCgowOl0kpub65ZlnJSUxLZt2wDYs2cPlZWVdO7cuf6dGmM03w6vvfaa0aNH\nD6N79+5GZmam3XFq9c477xgOh8Po06ePER0dbURHRxuvv/663bHqlZ2d7dZXW3344YdG//79jUsu\nucQYO3asW15tZRiGsXDhQiMiIsLo3bu3MWXKFNdVLXa77rrrjICAAMPLy8twOp3G8uXLjUOHDhkj\nRowwwsLCjLi4OOOrr76yO2aNnMuWLTNCQ0ONbt26uf4tzZgxwy0ytm3b1vV3ebqQkBC3uNqqtpyV\nlZXG5MmTjd69ext9+/Y13nzzzQY/R5MERUTEshZx2kpERJqWykNERCxTeYiIiGUqDxERsUzlISIi\nlqk8RETEMpWHtAqNvQzM4sWLOX78+Dn/fhs3bnTbWwxI66Z5HtIqtG/fnqNHjzba54eEhPDBBx+4\nZuU29vcTsZuOPKTV2rt3L/Hx8fTv35/LL7+cTz/9FIAbb7yR2267jSFDhtC9e3fWrl0LwMmTJ8nI\nyKBXr16MGjWKMWPGsHbtWpYsWUJpaSnDhg1jxIgRrs+/5557iI6O5tJLL611PaPbb7+defPmAbB5\n82auuOKKGu9ZsWIFt9xyS725TldYWEh4eDhpaWn07NmT66+/ni1btjBkyBB69OjB+++//8v/4kSg\n5SxPIlKfdu3a1dg2fPhwIz8/3zAMw9ixY4cxfPhwwzAMIzU11UhOTjYMwzDy8vKM0NBQwzAMY82a\nNcZVV11lGIZhHDhwwOjYsaOxdu1awzCMGjf6cTgcxiuvvGIYhmHMnj3buP/++2t8/4qKCiMyMtLY\ntm2b0bNnT2Pfvn013rNixQpj5syZ9eY6XUFBgeHp6Wl8/PHHxsmTJ41+/foZU6dONQzDMLKysoyk\npKQG/65Ezoan3eUlYodjx46xfft2Jk6c6NpWWVkJmMuQn1oMsFevXq77Wbz77ruupar9/PwYNmxY\nnZ/ftm1bxowZA0C/fv144403arznvPPO46mnnmLo0KE88sgjhISE1Ju5rlw/FRIS4logMjIy0nXv\nkN69e1NYWFjv9xA5WyoPaZVOnjxJhw4d2L17d62vt23b1vXc+GFY0OFwnHE/BqOe4cJTd2QD8PDw\noKqqqtb3ffTRR3Tp0uWsb15WW66f8vb2PuN7n9qnvhwiVmnMQ1olX19fQkJCeOmllwDzB/FHH31U\n7z5Dhgxh7dq1GIZBeXk5b731luu19u3b880331jKsH//fh5++GF2797N66+/Tk5OTo331FdQInZS\neUirUFFRQVBQkOuxePFiVq1axbJly4iOjqZ3795s2LDB9f7T75536vn48eNxOp1ERERwww030Ldv\nX9c902+++WZGjx7tGjD/6f4/vRufYRikp6fz0EMP4e/vz7Jly0hPT3edOqtr37qe/3Sfur5297sC\nSvOhS3VFLPj222/x8fHh0KFDDBo0iPfee88tb+4j0tg05iFiwdVXX82RI0eorKzk3nvvVXFIq6Uj\nDxERsUxjHiIiYpnKQ0RELFN5iIiIZSoPERGxTOUhIiKWqTxERMSy/w/YyVmczHJjsQAAAABJRU5E\nrkJggg==\n",
- "text": [
- "<matplotlib.figure.Figure at 0x5639ad0>"
- ]
- },
- {
- "metadata": {},
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- "text": [
- "<matplotlib.figure.Figure at 0x58b75b0>"
- ]
- }
- ],
- "prompt_number": 95
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 4.4.15,Page No.105"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "L=8 #m #Length of beam\n",
- "L_AD=4 #m #Length of AD\n",
- "w=300 #KN #u.d.l\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Let R_A and R_C be the reactions at A and C\n",
- "#R_A+R_C=300\n",
- "\n",
- "#Taking moment at A\n",
- "#LEt x be the distance from Pt B L_CB=x\n",
- "#R_C*(L-L_CB)=300*L*2**-1\n",
- "#R_C=1200*(8-x)**-1\n",
- "#After substituting values and further simplifying we get\n",
- "#R_A=300-R_C\n",
- "#R_A=1200-300*x*(8-x)**-1\n",
- "\n",
- "#B.M at D\n",
- "#M_D=R_A*L_AD-w*2**-1*2=0\n",
- "\n",
- "#Now substituting value of R_A we get\n",
- "#M_D=4*1200-300*x*(8-x)**-1-300=0\n",
- "\n",
- "#Further on simplification we get\n",
- "L_CB=x=600*225**-1\n",
- "\n",
- "R_C=1200*(8-x)**-1\n",
- "R_A=(1200-300*x)*(8-x)**-1\n",
- "\n",
- "#Pt of contraflexure\n",
- "#Let E be the pt and BE=y\n",
- "#V_E=0=-R_A*2**-1*L_BE+R_C\n",
- "L_BE=R_C*(R_A*2**-1)**-1\n",
- "L_AE=L-L_BE\n",
- "L_AC=L-L_CB\n",
- "L_EC=L_BE-L_CB\n",
- "\n",
- "#Shear Force at B\n",
- "V_B=0\n",
- "\n",
- "#Shear Force at C\n",
- "V_C1=-w\n",
- "V_C2=-V_C1+R_C\n",
- "\n",
- "#Shear Force at A\n",
- "V_A=-w+R_C\n",
- "\n",
- "#B.M at C\n",
- "M_C=-w*L_CB\n",
- "\n",
- "#B.M at E\n",
- "M_E=-R_A*L_AE+w*L_AE\n",
- "\n",
- "#B.M at A\n",
- "M_A=0\n",
- "\n",
- "#Result\n",
- "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
- "\n",
- "#Plotting the Shear Force Diagram\n",
- "\n",
- "X1=[0,L_CB,L_CB,L_CB+L_AC,L_CB+L_AC] \n",
- "Y1=[V_B,V_C1,V_C2,V_A,0]\n",
- "Z1=[0,0,0,0,0]\n",
- "plt.plot(X1,Y1,X1,Z1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Force in kN\")\n",
- "plt.show()\n",
- "\n",
- "#Plotting the Bendimg Moment Diagram\n",
- "\n",
- "X2=[0,L_CB,L_CB+L_EC,L_CB+L_AC]\n",
- "Y2=[M_B,M_C,M_E,M_A] \n",
- "Z2=[0,0,0,0]\n",
- "plt.plot(X2,Y2,X2,Z2)\n",
- "plt.xlabel(\"Length in m\")\n",
- "plt.ylabel(\"Bending Moment in kN.m\")\n",
- "plt.show()"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Shear Force and Bending Moment Diagrams are the results\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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PJCUlUVRUBEB1dTWbNm2iurqa8vJyCgsLPRHp5ptvZt26ddTU1FBTU0N5ebkZ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\n",
- "text": [
- "<matplotlib.figure.Figure at 0x541cd90>"
- ]
- },
- {
- "metadata": {},
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Qr1+/VlthlYYJg6TS0ACMGGEo0nv6abmjIbItq/phnDhxApGRkZgxYwYKCwsR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\nmDCIiMgi/w/S42ZzlwktGgAAAABJRU5ErkJggg==\n",
- "text": [
- "<matplotlib.figure.Figure at 0x5513cb0>"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 4.4.16,Page No.107"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "%matplotlib inline\n",
- "\n",
- "#Initilization of variables\n",
- "F_C=250 #KN #Pt LOad at C\n",
- "M_D=120 #KNM #moment at Pt D\n",
- "w=50 #KN/m #u.d.l 0n L_AD\n",
- "L_DB=L_BC=2 #m #Length of DB & BC\n",
- "L_AD=4 #m #Length of AD\n",
- "L=8 #m #Length of beam\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#LEt R_A and R_D be the reactions at A and D\n",
- "#R_A+R_D=450\n",
- "\n",
- "#Taking moment at A\n",
- "#M_A=0=-R_B*(L_DB+L_AD)+M_D+F_C*L+w*L_AD**2*2**-1\n",
- "R_B=-(-M_D-F_C*L-w*L_AD**2*2**-1)*(L_DB+L_AD)**-1\n",
- "R_D=450-R_B\n",
- "\n",
- "#Shear Force Calculations\n",
- "\n",
- "#Shear Force at C\n",
- "V_C=-F_C\n",
- "\n",
- "#Shear Force at B\n",
- "V_B1=V_C\n",
- "V_B2=R_B-F_C\n",
- "\n",
- "#Shear Force at D\n",
- "V_D=V_B2\n",
- "\n",
- "#Shear Force at A\n",
- "V_A=-F_C+R_B-w*L_AD\n",
- "\n",
- "#Pt of contralfexure\n",
- "#Let E be the pt and CE=x\n",
- "#V_E=0=-F_C+R_B-w*(L_EC-L_DB-L_BC)\n",
- "L_EC=-((+F_C-R_B)*w**-1-L_DB- L_BC)\n",
- "L_ED=L_EC-L_DB-L_BC\n",
- "\n",
- "#Bending Moment Calculations\n",
- "\n",
- "#Bending Moment at C\n",
- "M_C=0\n",
- "\n",
- "#Bending Moment at B\n",
- "M_B=-F_C*L_BC\n",
- "\n",
- "#Bending Moment at D\n",
- "M_D1=-F_C*(L_BC+L_DB)+R_B*L_DB\n",
- "M_D2=M_D1-M_D\n",
- "\n",
- "#Bending Moment at E\n",
- "M_E=-F_C*L_EC+R_B*(L_ED+L_DB)-w*L_ED**2*2**-1-M_D\n",
- "\n",
- "#Bending Moment at A\n",
- "M_A=0\n",
- "\n",
- "#Pt of contraflexure\n",
- "#Let F be the pt and CF=y\n",
- "#M_F=0=- F_C*L_FC+R_B*(L_FC-L_BC)-M_D-w*(L_FC-L_DB-L_BC)\n",
- "#After substituting values and further simplifying we get equation as\n",
- "#y**2-14.8*y+54.5=0\n",
- "\n",
- "a=1\n",
- "b=-14.8\n",
- "c=54.4\n",
- "\n",
- "X=b**2-4*a*c\n",
- "\n",
- "y1=(-b+X**0.5)*(2*a)**-1\n",
- "y2=(-b-X**0.5)*(2*a)**-1\n",
- "\n",
- "#From above two equations y2 is taken into consideration\n",
- "\n",
- "#Result\n",
- "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
- "\n",
- "#Plotting the Shear Force Diagram\n",
- "\n",
- "X1=[0,L_BC,L_BC,L_BC+L_DB,L_BC+L_DB+L_AD,L_BC+L_DB+L_AD] \n",
- "Y1=[V_C,V_B1,V_B2,V_D,V_A,0]\n",
- "Z1=[0,0,0,0,0,0]\n",
- "plt.plot(X1,Y1,X1,Z1)\n",
- "plt.xlabel(\"Length x in m\")\n",
- "plt.ylabel(\"Shear Force in kN\")\n",
- "plt.show()\n",
- "\n",
- "#Plotting the Bendimg Moment Diagram\n",
- "\n",
- "X2=[0,L_BC,L_BC+L_DB,L_BC+L_DB,L_BC+L_DB+L_ED,L_BC+L_DB+L_AD,]\n",
- "Y2=[M_C,M_B,M_D1,M_D2,M_E,M_A] \n",
- "Z2=[0,0,0,0,0,0]\n",
- "plt.plot(X2,Y2,X2,Z2)\n",
- "plt.xlabel(\"Length in m\")\n",
- "plt.ylabel(\"Bending Moment in kN.m\")\n",
- "plt.show()"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Shear Force and Bending Moment Diagrams are the results\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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ELiLit0yfw3j55ZeZOHEiAKWlpdhsNtc+m81GSUlJne1Wq5WSkpJWj1VExJ8F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\nQ0REPPL/AYCLPN0YseBLAAAAAElFTkSuQmCC\n",
- "text": [
- "<matplotlib.figure.Figure at 0x55131f0>"
- ]
- },
- {
- "metadata": {},
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mly2TJwhbS2mp3IiXlianIf39rffaREowa+Pe3r17sWvXLmg0GvTq1Qvdu3e3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\niIjILP8Py78BVKsS0nAAAAAASUVORK5CYII=\n",
- "text": [
- "<matplotlib.figure.Figure at 0x540da10>"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [],
- "language": "python",
- "metadata": {},
- "outputs": [],
- "prompt_number": 2
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [],
- "language": "python",
- "metadata": {},
- "outputs": []
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file
diff --git a/Strength_Of_Materials_by_B_K_Sarkar/chapter_5_som.ipynb b/Strength_Of_Materials_by_B_K_Sarkar/chapter_5_som.ipynb
deleted file mode 100755
index f706786e..00000000
--- a/Strength_Of_Materials_by_B_K_Sarkar/chapter_5_som.ipynb
+++ /dev/null
@@ -1,740 +0,0 @@
-{
- "metadata": {
- "name": "chapter 5 som.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 5:Bending Stresses in Beams"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 5.1,Page no.121"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "b=100 #mm #width of timber joist\n",
- "d=200 #mm #depth of joist\n",
- "L=3 #m #Length of beam\n",
- "sigma=7 #KN/mm**2 #bending stress\n",
- "w_1=5 #KN/mm**2 #unit weight of timber\n",
- "\n",
- "#Calculations\n",
- "w=0.1*0.2*1*5*100 #N/m #self weight of the joist\n",
- "I_xx=1*12**-1*100*200**3 #mm**4 #M.I of section about N.A\n",
- "\n",
- "#M=W*L+w*L**2*2**-1 #Max Bending moment\n",
- "#Therefore,M=(3*W+450)\n",
- "\n",
- "#using the relation M*I**-1=sigma*y**-1,we get\n",
- "W=(((7*2*10**8)*(100*10**3*3)**-1)-450)*3**-1 #N #Max Load applied\n",
- "\n",
- "#Result\n",
- "print\"The Max value of Load applied is\",round(W,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Max value of Load applied is 1405.56\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 5.2,Page no.122"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "D=160 #mm #Overall Depth\n",
- "B=150 #mm #Width of Flange\n",
- "f_t=40 #mm #Flange thickness\n",
- "W_t=50 #mm #Web thickness\n",
- "sigma_t=20 #N/mm**2 #tension stress\n",
- "sigma_c=75 #N/mm**2 #compression stress\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Rectangle-1\n",
- "a_1=150*40 #mm**2 #Area of Rectangle-1\n",
- "y_1=40*2**-1 #mm #C.G of Rectangle-1\n",
- "\n",
- "#Rectangle-2\n",
- "a_2=120*50 #mm**2 #Area of Rectangle-2\n",
- "y_2=40+120*2**-1 #mm #C.G of Rectangle-2\n",
- "\n",
- "Y_bar=(a_1*y_1+a_2*y_2)*(a_1+a_2)**-1 #mm #Distance of C.G from the bottom flange\n",
- "I=1*12**-1*150*40**3+150*40*(60-40)**2+1*12**-1*50*120**3+50*120*(100-60)**2 #mm**4 #M.I of section about N.A\n",
- "y_t=60 #mm #Permissible tensile stress at the bottom face of flange from N.A\n",
- "y_c=100 #mm #Permissible tensile stress at the top face of flange from N.A\n",
- "\n",
- "#M=W*L*4**-1 #Max bending mooment at the centre\n",
- "\n",
- "#Using the relation M*I**-1=sigma_t*y_t**-1 we get\n",
- "W=(0.333*4*272*10**5)*(2.5*1000)**-1 #N #MAx central load\n",
- "\n",
- "#Result\n",
- "print\"The Max Bending Moment at the centre is\",round(W,2),\"N\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Max Bending Moment at the centre is 14492.16 N\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 5.3,Page no.123"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "b=10 #cm #width of beam\n",
- "d=20 #cm depth of beam\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#R_a and R_b are the reactions at A and B respectively.\n",
- "#Moment of all forces about A\n",
- "\n",
- "R_b=(4*4*4*2**-1-2*1.5)*(2)**-1 #KN \n",
- "#R_a+R_b=18 \n",
- "R_a=18-R_b\n",
- "\n",
- "#Consider a section at a distance x from A\n",
- "#M_x=9.25*x-2(x-1.5)-4*x*x*2**-1=7.25*x+3-2*x**2 \n",
- "\n",
- "#Taking derivative of above equation to find max value of M_x we get\n",
- "x=1.81 #m\n",
- "\n",
- "M=7.25*x+3-2*x**2 #kN*m \n",
- "I=b*d**3*12**-1 #cm**4 #M.I of the section\n",
- "y=10 \n",
- "sigma=M*I**-1*y*10**8*(10**2)**-1 #Max bending stress\n",
- "\n",
- "#Result\n",
- "print\"The Max Bending stress is\",round(sigma,2),\"KN/m**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Max Bending stress is 14355.45 KN/m**2\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 5.4,Page no.124"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "sigma=8 #N/mm**2 #Max bending stress\n",
- "W=5000 #N/m**2 #Load of floor\n",
- "A=450 #cm**2 #Area of joist\n",
- "L=5 #m #span of floor\n",
- "\n",
- "#Calculations\n",
- "#Let x be the centre to centre spacingof the joists\n",
- "\n",
- "#A_1=5*x**2 #m**2 #Area of floor between any two joists\n",
- "#W_1=5*x*W #N #total load supported by one interior joist\n",
- "#M=W_1*L*8**-1 #Max bending moment \n",
- "I=1*12**-1*b*(d*10**-2)**3*10**-2 #m**4 #M.I of joist\n",
- "y=0.15 #cm #Distance of of farthest fibre\n",
- "M=I*y**-1*sigma #N*m\n",
- "\n",
- "#Now equating to max bending moment we get\n",
- "x=(18000*8)*(25000*5)**-1\n",
- "\n",
- "#Result\n",
- "print\"The Max Bending Moment is\",round(x,2),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Max Bending Moment is 1.15 m\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 5.7,Page no.126"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "d=12 #m #depth of mast\n",
- "D_1=20 #cm #diameter at the base \n",
- "D_2=10 #cm #diameter at the top\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Consider section at a distance x cm below top of mast and y be the diameter at this section\n",
- "\n",
- "#triangle OAB and ODC are similar,we get\n",
- "#2*AB=x*120**-1\n",
- "#EB=y=10+x*120**-1\n",
- "#after simplifying we get, x=120*(y-10)\n",
- "\n",
- "#Z=pi*64**-1*y**4)*(y*32**-1)**-1 #Section modulus\n",
- "#After simplifying we get\n",
- "#Z=(pi*y**3)*(32)**-1\n",
- "\n",
- "#M=120*P(y-10) #bending moment at that section\n",
- "\n",
- "#From flexural formula we get,\n",
- "#sigma=M*Z**-1\n",
- "#After substituting and simplifying above equation we get,\n",
- "#sigma=3840*P*pi**-1*(1*y**2-1-10*y**3-1)\n",
- "\n",
- "#To find max value of sigma taking derivative of above equation we get\n",
- "y=15 #cm\n",
- "\n",
- "#Now substituting value of y in all equations with variable y\n",
- "x=120*(y-10)\n",
- "#sigma=3840*P*(15-10)*(pi*15**3)**-1 \n",
- "\n",
- "#After implifying above equation we get\n",
- "\n",
- "P=(3500*pi*15**3)*(3840*5)**-1 #N #Magnitude of load causing failure\n",
- "\n",
- "#Result\n",
- "print\"The Magnitude of Load is\",round(P,2),\"N\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Magnitude of Load is 1932.82 N\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 5.8,Page no.127"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=3 #m #span of beam\n",
- "t=20 #mm #Thickness of steel\n",
- "D=200 #mm #overall depth \n",
- "B=140 #mm #overall width\n",
- "b=100 #mm #width of inner rectangle\n",
- "d=160 #mm #depth of inner rectangle\n",
- "w=77 #KN/mm**2\n",
- "sigma=100 #N/mm**2 #Bending stress\n",
- "#Calculations\n",
- "V=((D*10**-3*B*10**-3)-(d*10**-3*b*10**-3)) #m**3 #Volume of rectangular box\n",
- "W=V*3*w #KN #Weight of Beam\n",
- "I=(B*D**3-b*d**3)*12**-1 #mm**4 #M.I of beam section\n",
- "\n",
- "#Now using the relation,M*I**-1=sigma*y**-1\n",
- "\n",
- "y=200 #mm #distance from farthest fibre\n",
- "M=I*sigma*2*y**-1 #N*mm\n",
- "\n",
- "#M=3000*W+2772*3000*2**-1\n",
- "#After sub values in above equation we get\n",
- "\n",
- "W=((59.2*10**6-2772*3000*2**-1)*(3000)**-1)*10**-3 #KN #Max concentrated Load at free end\n",
- "\n",
- "F=W+2.772*2**-1 #KN #shear force at half length\n",
- "\n",
- "#Result\n",
- "print\"The shear force at half length is\",round(F,2),\"KN\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The shear force at half length is 19.73 KN\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 5.9,Page no.128"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "B=24 #mm #width of beam section\n",
- "D=21.7 #mm #depth of beam section\n",
- "E_1=11440 #MN/m**2 #Modulus of Elasticity parallel grain\n",
- "E_2=2860 #MN/m**2 ##Modulus of Elasticity perpendicular grain\n",
- "sigma_1=8.57 #MN/m**2\n",
- "sigma_2=2.14 #MN/m**2\n",
- "L=1.2 #m #span of beam\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Ratio of smaller modulus to larger modulus is E_2:E_1=1:4\n",
- "#Dimension of transformed Beam section\n",
- "b=18 #mm #width of Beam section\n",
- "d=3.1 #mm #depth of beam section\n",
- "\n",
- "I=(1*12**-1*B*10**-3*(D*10**-3)**3)-(3*(1*12**-1*b*10**-3*(d*10**-3)**3)) #m**4 #M.I of transformed section\n",
- "y=21.7*10**-3*2**-1 \n",
- "M=I*sigma_1*10**6*y**-1 #N*m #Safe B.M\n",
- "P=4*M*L**-1 #N\n",
- "\n",
- "#Result\n",
- "print\"Safe value of Load is\",round(P,2),\"N\" "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Safe value of Load is 53.45 N\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 5.11,Page no.131"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "D=4 #cm #Outside diameter\n",
- "d=3 #cm #inside diamter\n",
- "L=2 #m #span of beam\n",
- "W=1000 #N #Max safe Load\n",
- "\n",
- "#Calculations\n",
- "\n",
- "I=pi*64**-1*(D**4-d**4) #cm**4 #M.I\n",
- "A=pi*4**-1*(D**2-d**2) #cm**2 #Area \n",
- "y=2 \n",
- "Z=I*y**-1 #cm**3 #Section modulus\n",
- "\n",
- "M=W*L*4**-1 #N*cm #Max bending moment\n",
- "\n",
- "#From Flexural Formula\n",
- "sigma=M*Z**-1 #N/cm**2\n",
- "\n",
- "#For Tubes\n",
- "#M.I about x-x axis\n",
- "I_1=4*(8.59+5.492*2**2) #cm**4 \n",
- "\n",
- "Z_1=122.32*4**-1 #cm**3 \n",
- "\n",
- "#M=W_1*200*4**-1 #N*cm\n",
- "#After substituting values we get\n",
- "#M=50*W_1 (equation 1)\n",
- "\n",
- "#Again from Flexural Formula\n",
- "M=sigma*Z_1\n",
- "\n",
- "#substitute value of M in equation 1\n",
- "\n",
- "W=11640*30.58*50**-1 #N\n",
- "\n",
- "#Result\n",
- "print\"Max central load is\",round(W,2),\"N\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Max central load is 7119.02 N\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 5.12,Page no.133"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "b=200 #mm #width of beam\n",
- "d=300 #mm #depth of beam\n",
- "t=12 #mm #thickness of beam\n",
- "E_s=220 #KN/m**2 #modulus of elasticity of steel\n",
- "E_w=11 #KN/m**2 #modulus of elasticity of timber\n",
- "sigma_s=115 #MN/m**2 #stress of steel\n",
- "sigma_w=9.2 #MN/m**2 #stress of timber\n",
- "L=2 #m #Span of beam\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#E_w*E_s**-1=1*20**-1 #ratio of Modulus of elasticity of timber to steel\n",
- "\n",
- "\n",
- "#(Part-1)\n",
- "b_1=b*20**-1 #mm #web thickness of transformed section\n",
- "stress=20*sigma_w #MN/m**2 #Allowable stress in web of equivalen beam\n",
- "#But allowable stress in flanges is sigma_s is 115 KN/m**2 and therefore taken into consideration\n",
- "\n",
- "\n",
- "d_1=324 #mm #depth of beam with thickness in consideration\n",
- "I=1*12**-1*0.2*0.324**3-2*1*12**-1*0.095*0.3**3 #m**4 #M.I of transformed section\n",
- "\n",
- "#Using Relation, M*I**-1=sigma*y**-1 we get\n",
- "\n",
- "#Part-2\n",
- "M_max=I*(324*10**-3*2**-1)**-1*sigma_s*10**6 #N*m #Max allowable Bending moment for steel section\n",
- "\n",
- "#Part-3\n",
- "#As beam is simply supported at the ends and the load is applied at the centre of beam\n",
- "#M_max=W*L*4**-1 #Max Bending moment \n",
- "W=M_max*4*L**-1 #N #Allowable stress Load\n",
- "\n",
- "#Result\n",
- "print\"Web thickness of Equivalent steel section is\",round(b_1,2),\"mm\"\n",
- "print\"Max Allowable bending moment for section is\",round(M_max,2),\"N*m\"\n",
- "print\"Allowable safe Load is\",round(W,2),\"N\"\n",
- " "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Web thickness of Equivalent steel section is 10.0 mm\n",
- "Max Allowable bending moment for section is 98935.78 N*m\n",
- "Allowable safe Load is 197871.56 N\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 5.13,Page no.135"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "d=10 #cm #distance between joists\n",
- "t=2 #cm #thickness of steel plate \n",
- "d_2=20 #cm #depth of beam\n",
- "sigma_t=8.5 #N/mm**2 #stress in timber \n",
- "E_s=2*10**5 #N/mm**2 #Modulus of elasticity of steel\n",
- "E_t=10**4 #N/mm**2 ##Modulus of elasticity of timber\n",
- "L=5 #cm #span of beam\n",
- "\n",
- "#calculation\n",
- "sigma=10*15**-1*sigma_t #stress in timber at distance of 10 cm from XX\n",
- "\n",
- "dell=sigma*E_t**-1 #strain in timber at 10 cm from XX\n",
- "\n",
- "sigma_s=dell*E_s #N/mm**2 #Max stress\n",
- "\n",
- "#For Timber\n",
- "Z_w=1*6**-1*10*30**2*2 #cm**3 #section modulus of timber\n",
- "M_w=sigma_t*100*Z_w #moment of resistance of timber\n",
- "\n",
- "#For steel\n",
- "Z_s=1*6**-1*2*20**2 #cm**3 #section modulus of steel\n",
- "M_s=sigma_s*Z_s*100 #moment of resistance of steel\n",
- "\n",
- "M=(M_w+M_s)*10**-5 #total moment of resistance\n",
- "\n",
- "#M=w*L**2*8**-1 #N*m #Max bending moment\n",
- "w=8*M*(L**2)**-1 #N/m #Max uniform distributed Load\n",
- "\n",
- "#Result\n",
- "print\"Moment of resistance is\",round(M,3),\"N*cm\"\n",
- "print\"Max uniform distributed Load\",round(w,3),\"N/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Moment of resistance is 40.611 N*cm\n",
- "Max uniform distributed Load 12.996 N/m\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 5.14,Page no.136"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "B=10 #cm #width of timber section\n",
- "D=15 #cm #depth of timber section\n",
- "b=10 #cm #width of steel plate\n",
- "t=12 #mm #thickness \n",
- "w=3 #KN/m #Uniformly distributed Load\n",
- "L=4 #m #Span of beam\n",
- "m=20 #Ratio of modulus of elasticity of steel to timber \n",
- "W=3 #KN/m #Load\n",
- "\n",
- "#Calculations\n",
- "\n",
- "y_1=15*2**-1 #C.G of timber \n",
- "y_2=1.2*2**-1 #C.G of steel plate\n",
- "b_s=10*m**-1 #cm #Equivalent width of steel\n",
- "Y_bar=(10*1.2*0.6+15*0.5*8.7)*(10*1.2+15*0.5)**-1 #cm #distance of C.G from bottom edge\n",
- "\n",
- "I=1*12**-1*10*(1.2)**3+10*1.2*(3.72-0.6)**2+1*12**-1*0.5*(15)**3+0.5*15*(7.5-3.72)**2\n",
- "M=W*10**5*L**2*8**-1 #N*m\n",
- "\n",
- "Y_bar_1=3.72 #cm #C.G from bottom edge\n",
- "Y_bar_2=16.2-Y_bar #cm #C.G from top edge\n",
- "\n",
- "sigma_1=(M*I**-1*Y_bar_1)*10**-2 #N/mm**2 #stress at bottom\n",
- "\n",
- "sigma_2=(M*I**-1*Y_bar_2)*10**-2 #N/mm**2 #stress at top\n",
- "\n",
- "sigma_max=sigma_2*m**-1 \n",
- "\n",
- "#The Answers in book for Moment of Inertia about x-x axis onwards are incorrect\n",
- "\n",
- "#Result\n",
- "print M\n",
- "print\"The Max Stress in steel is\",round(sigma_1,2),\"N/mm**2\"\n",
- "print\"The Max Stress in timber is\",round(sigma_max,2),\"N/mm**2\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "600000.0\n",
- "The Max Stress in steel is 60.98 N/mm**2\n",
- "The Max Stress in timber is 10.23 N/mm**2\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 5.15,Page no.137"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "B=20 #cm #width of timber\n",
- "D=30 #cm #depth of timber\n",
- "d=25 #cm #depth of steel plate\n",
- "b=1.2 #cm #width of steel plate\n",
- "sigma_s=90 #N/mm**2 #Bending stress in steel\n",
- "sigma_t=6 #N/mm**2 #Bending stress in timber\n",
- "m=20 #Ratio of modulus of elasticity of of steel to timber\n",
- "\n",
- "#Calculation\n",
- "\n",
- "#Equivalent width of wood section,when 1.2 cm wide steel plate is replaced by steel plate is\n",
- "b_1=1.2*20 #cm\n",
- "d_1=25 #cm #depth of wood section\n",
- "y_1=d*2**-1 #cm #C.G of timber section\n",
- "y_2=D*2**-1 #cm #C.G of steel section\n",
- "\n",
- "Y_bar=(2*d*b_1*y_1+D*B*y_2)*(2*d*b_1+D*B)**-1 #cm #Distance of C.G from Bottom edge\n",
- "I=B*D**3*12**-1+B*D*(y_2-Y_bar)**2+2*(b_1*d_1**3*12**-1+b_1*d_1*(Y_bar-y_1)**2) #M.I of equivalent timber section about N.A\n",
- "Y=30-Y_bar #distance of C.G from top of equivalent wood section\n",
- "\n",
- "#Thus max stress will occur at top and that in steel will occur at bottom\n",
- "#sigma_s=m*Y_bar*Y**-1*sigma_t \n",
- "\n",
- "#After simplifying we get\n",
- "#sigma_s=15.99*sigma_t\n",
- "\n",
- "sigma_t=sigma_s*15.99**-1 #N/mm**2 #Max stress in Equivalent timber section\n",
- "\n",
- "Z_t=I*Y**-1 #Section modulus of equivalent section\n",
- "M=sigma_t*Z_t*10**-5*100 #Moment of resistance of beam\n",
- "\n",
- "#Result\n",
- "print\"Position of N.A is\",round(Y_bar,2),\"cm\"\n",
- "print\"Moment of Resistance of beam is\",round(M,2),\"KN*m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Position of N.A is 13.33 cm\n",
- "Moment of Resistance of beam is 37.15 KN*m\n"
- ]
- }
- ],
- "prompt_number": 13
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file
diff --git a/Strength_Of_Materials_by_B_K_Sarkar/chapter_6_som.ipynb b/Strength_Of_Materials_by_B_K_Sarkar/chapter_6_som.ipynb
deleted file mode 100755
index 7960d97b..00000000
--- a/Strength_Of_Materials_by_B_K_Sarkar/chapter_6_som.ipynb
+++ /dev/null
@@ -1,914 +0,0 @@
-{
- "metadata": {
- "name": "chapter 6.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 6:Slope and Deflection"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 6.1,Page No.154"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "b=0.12 #m #Width of beam\n",
- "d=0.2 #m #Depth of beam\n",
- "dell=0.005 #m #Deflection\n",
- "E=2*10**5*10**6 #N/m**2 \n",
- "L=2.5 #m #Length of beam\n",
- "\n",
- "#Calculations\n",
- "\n",
- "I=b*d**3*12**-1 #m**4 #M.I of rectangular section\n",
- "w=8*E*I*dell*(L**4)**-1 #N/m #U.d.l\n",
- "\n",
- "#Let slope at free end be theta\n",
- "theta=w*L**3*(6*E*I)**-1 #Radian \n",
- "\n",
- "W=dell*3*E*I*(L**3)**-1*10**-3 #kN #Concentrated Load \n",
- "\n",
- "theta_2=W*L**2*(2*E*I)**-1 #Slope at free end\n",
- "\n",
- "#Result\n",
- "print\"Uniformly distributed Load beam should carry is\",round(w,2),\"N/m\"\n",
- "print\"Concentrated Load at free end is\",round(W,2),\"KN\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Uniformly distributed Load beam should carry is 16384.0 N/m\n",
- "Concentrated Load at free end is 15.36 KN\n"
- ]
- }
- ],
- "prompt_number": 73
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 6.2,Page No.155"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=6 #m #Length of beam\n",
- "y_b=1.5*10**-2 #m #Deflection\n",
- "E=2*10**7*10**4\n",
- "sigma=10*10**3*10**4\n",
- "#d=2*b\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Let w*I**-1=X #From Deflection at the free end Equation \n",
- "X=y_b*8*E*(L**4)**-1*10**-3 #Equation 1\n",
- "\n",
- "#Let w*b*I**-1=Y #From Max bending stress at the extreme fibre From N.A\n",
- "Y=sigma*2*(L**2)**-1 #Equation 2\n",
- " \n",
- "b=Y*X**-1 #width of beam #mm\n",
- "d=2*b #depth of beam #mm\n",
- " \n",
- "#Result\n",
- "print\"The Dimension of Beam are:b\",round(b,2),\"mm (width)\"\n",
- "print\" :d\",round(d,2),\"mm (depth)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Dimension of Beam are:b 300.0 mm (width)\n",
- " :d 600.0 mm (depth)\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 6.3,Page No.156"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=3 #m #Length of beam\n",
- "L_1=1.2 #m #Distance from fixed end\n",
- "d=0.25 #m #Depth of beam\n",
- "w=15*10**3 #N #U.d.L\n",
- "W=40*10**3 #N #Point Load\n",
- "E=2*10*10**4 #N/m**2 \n",
- "I=13500*10**-4 #M.I\n",
- "\n",
- "#Calculations\n",
- "\n",
- "y_b=W*L_1**3*(3*E*I)**-1+W*L_1**2*(2*E*I)**-1*(L-L_1)+w*L**4*(8*E*I)**-1 #Deflection at free end\n",
- "\n",
- "M=W*L_1+w*L*L*2**-1 #Max Bending moment at the fixed end A #Nm\n",
- "y=d*2**-1\n",
- "sigma_max=M*y*I**-1 #N/cm**2 #Max Bending stress at extreme fibre \n",
- "\n",
- "#Result\n",
- "print\"Deflection at the free end is\",round(y_b,4),\"cm\"\n",
- "print\"Max stress due to bending is\",round(sigma_max,2),\"N/cm**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Deflection at the free end is 0.8398 cm\n",
- "Max stress due to bending is 10694.44 N/cm**2\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 6.4,Page No.156"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "M=100*10**3 #N #Moment\n",
- "L=3 #m #Length\n",
- "d=0.15 #m #Width\n",
- "b=0.1 #m #width\n",
- "E=2.1*10**7*10**4 #N/cm**2 \n",
- "\n",
- "#Calculations\n",
- "\n",
- "I=b*d**3*12**-1 #cm**4 #M.I of beam section\n",
- "B_1=M*L*(E*I)**-1 #radian #Slope at B\n",
- "B_2=M*L**2*(2*E*I)**-1*10**2 #cm #Deflection at point B\n",
- "\n",
- "#Result\n",
- "print\"The slope at Point B is\",round(B_1,2),\"radian\"\n",
- "print\"The Deflection at point B is\",round(B_2,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The slope at Point B is 0.05 radian\n",
- "The Deflection at point B is 7.62 cm\n"
- ]
- }
- ],
- "prompt_number": 81
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 6.5,Page No.157"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "b=0.1 #m #width \n",
- "d=0.2 #m #depth\n",
- "L=2 #m #Length of beam\n",
- "L_1=1 #m #Length from free end\n",
- "E=210*10**9 \n",
- "W=1*10**3 #N #Concentrated Load\n",
- "w=2*10**3 #N/m\n",
- "\n",
- "#Calculations\n",
- "\n",
- "I=b*d**3*12**-1 #m**4 #M.I of the beam section\n",
- "\n",
- "#Slope at free end\n",
- "theta=W*L**2*(2*E*I)**-1+w*L**3*(6*E*I)**-1-w*(L-L_1)**3*(6*E*I)**-1 \n",
- "\n",
- "#Deflection at free end\n",
- "y_b=(W*L**3*(3*E*I)**-1+w*L**4*(8*E*I)**-1-w*(L-L_1)**4*(8*E*I)**-1-w*(L-L_1)**3*L_1*(6*E*I)**-1)*10**3 \n",
- "\n",
- "#Result\n",
- "print\"Slope at free end is\",round(theta,5),\"radian\"\n",
- "print\"Deflection at free end is\",round(y_b,2),\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Slope at free end is 0.00031 radian\n",
- "Deflection at free end is 0.43 mm\n"
- ]
- }
- ],
- "prompt_number": 59
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 6.6,Page No.158"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=10 #m #span of beam\n",
- "W=10*10**3 #N #Point Load\n",
- "a=6 #m #Distance from left end of beam to point Load\n",
- "b=4 #m ##Distance from right end of beam to point Load\n",
- "E=210*10**9 \n",
- "I=10**-4 #m #M.I of beam\n",
- "\n",
- "#Calculation \n",
- "\n",
- "#slope at left end is given by\n",
- "theta_A=W*b*(L**2-b**2)*(6*E*I*L)**-1 #radian\n",
- "\n",
- "#Deflection under Load is\n",
- "y_c=W*a*b*(L**2-a**2-b**2)*(6*E*I*L)**-1*10**3 #m\n",
- "\n",
- "#Maximum Deflection of the beam is\n",
- "y_max=W*a*(L**2-a**2)**1.5*(15.588457*E*I*L)**-1*10**3 #m\n",
- "\n",
- "#Result\n",
- "print\"slope at left end is\",round(theta_A,5),\"radian\"\n",
- "print\"Deflection under Load is\",round(y_c,2),\"m\"\n",
- "print\"#Maximum Deflection of the beam is\",round(y_max,2),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "slope at left end is 0.00267 radian\n",
- "Deflection under Load is 9.14 m\n",
- "#Maximum Deflection of the beam is 9.38 m\n"
- ]
- }
- ],
- "prompt_number": 72
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 6.7,Page No.158"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=5 #m #Length of beam\n",
- "w=40*10**3 #N #U.d.L \n",
- "y_max=0.01 #Deflection\n",
- "sigma_s=7*10**6 #Bending stress\n",
- "E=10.5*10**9\n",
- "\n",
- "#Calculation\n",
- "\n",
- "M=w*L*8**-1 #N*m #Max Bending moment \n",
- "\n",
- "#From equation of max deflection\n",
- "I=5*w*L**3*(y_max*384*E)**-1 #m**4 \n",
- "\n",
- "d=sigma_s*2*I*M**-1*10**2 #cm\n",
- "b=12*I*((d*10**-2)**3)**-1*10**2 #cm #Breadth\n",
- "\n",
- "#Result\n",
- "print\"Minimum value of breadth is\",round(b,2),\"cm\"\n",
- "print\"Minimum value of Depth is\",round(d,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Minimum value of breadth is 17.77 cm\n",
- "Minimum value of Depth is 34.72 cm\n"
- ]
- }
- ],
- "prompt_number": 94
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 6.8,Page No.159"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=6 #m #Length of beam\n",
- "d=0.15 #m #diameter\n",
- "y_max=1.035*10**-2 #m #Deflection\n",
- "E=210*10**9 \n",
- "\n",
- "#Calculations\n",
- "\n",
- "I=pi*64**-1*d**4 #M.I of Beam\n",
- "W=y_max*48*E*(L**3)**-1 #Point Load\n",
- "theta_A=3*y_max*L**-1\n",
- "theta_B=-theta_A\n",
- "\n",
- "#Result\n",
- "print\"The Heaviest central Point Load placed is\",round(W,2),\"N\"\n",
- "print\"Slope at supports are:theta_A\",round(theta_A,5),\"radian\"\n",
- "print\" :theta_B\",round(theta_B,5),\"radian\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Heaviest central Point Load placed is 483000000.0 N\n",
- "Slope at supports are:theta_A 0.00517 radian\n",
- " :theta_B -0.00517 radian\n"
- ]
- }
- ],
- "prompt_number": 103
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 6.9,Page No.160"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=14 #m #Lenth of steel girder\n",
- "E=210*10**9 #modulus of Elasticity of steel\n",
- "I=16*10**4*10**-8 #M.I of girder section\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#R_a+R_b=200 #R_a & R_b are the Reactions at supports A & B respectively\n",
- "\n",
- "#After taking moment at B We get\n",
- "R_a=(120*11+80*4.5)*14**-1 #KN\n",
- "R_b=200-R_a\n",
- "\n",
- "#After considering section at X-X at a distance x from left end A and taking B.M at X-X\n",
- "#M=120*x-120(x-3)-80*(x-9.5)\n",
- "\n",
- "#After Integrating twice we get\n",
- "#EI*dy*dx**-1=-60*x**2*+60(x-3)**2+40(x-9.5)**2+C_1 #slope \n",
- "\n",
- "#Again on Integrating we get\n",
- "#EI*y=-20*x**3+20(x-3)**3+40*3**-1*(x-9.5)**3+C_1*x+C_2 #Deflection\n",
- "\n",
- "#At A deflection is zero,i.e at x=0,y=0 \n",
- "#At B deflection is zero,i.e at x=14,y=0 So C_2=0\n",
- "\n",
- "C_1=-(-20*(14)**3+20*(11)**3+40*3**-1*(14-9.5)**3)*14**-1 #constant \n",
- "\n",
- "#Now Deflection at D i.e at x=3 m\n",
- "x=3\n",
- "y_D=1*(E*I)**-1*(-20*x**3+20*(x-3)**3+C_1*x)*10**3\n",
- "\n",
- "#Now Deflection at D i.e at x=9.5 m\n",
- "x=9.5\n",
- "y_C=1*(E*I)**-1*(-20*x**3+20*(x-3)**3+40*3**-1*(x-9.5)**3+C_1*x)*10**3\n",
- "\n",
- "#Result\n",
- "print\"Deflection under points of two Loads are i.e: at pt D\",round(y_D,4),\"m\"\n",
- "print\" : at pt C\",round(y_C,4),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Deflection under points of two Loads are i.e: at pt D 0.0156 m\n",
- " : at pt C 0.0199 m\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 6.10,Page No.161"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "E=200*10**9 #Pa \n",
- "I=20000*10**-8 #m**4\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Now Taking moment at B\n",
- "R_a=(1000*3*4.5+1000*2)*6**-1 #Reaction Force at pt A\n",
- "\n",
- "#On part BC u.d.l of 1KN/m is introduced both above and below \n",
- "#consider section at distance x i.e X-X and considering moment at section X-X\n",
- "\n",
- "#M=15500*x*6**-1-1000*x**2*2**-1-1000(x-4)+1000*2**-1*(x-3)**2\n",
- "#EI*d**2y*d**x=-M=15500*x*6**-1-1000*x**2*2**-1-1000(x-4)+1000*2**-1*(x-3)**2\n",
- "\n",
- "#Now Integrating above Equation we get Equation of slope\n",
- "#EI*dy*dx**-1=-15500*x**2*12**-1+1000*x**3*6**-1+1000*(x-4)**2*2**-1+1000*6**-1*(x-3)**3+C_1\n",
- "\n",
- "#Now Integrating above Equation we get Equation of Deflection\n",
- "#EI*y=-15500*x**3*36**-1+1000*x**4*24**-1+1000*(x-4)**3*6**-1+1000*24**-1*(x-3)**3+C_1*x+C_2\n",
- "\n",
- "#At x=0,deflection is zero,i.e y=0 C_2=0\n",
- "#At x=6,deflection is zero,i.e y=0 \n",
- "x=6\n",
- "C_1=-(-15500*x**3*36**-1+1000*x**4*24**-1+1000*(x-4)**3*6**-1+1000*(x-3)**4*24**-1)*x**-1 #Constant\n",
- "\n",
- "#Answer for constant C_1 is incorrect in Book\n",
- "\n",
- "#Now Deflection at C,put x=3 m\n",
- "x=3\n",
- "y_C=1*(E*I)**-1*(-15500*x**3*36**-1+1000*x**4*24**-1+1000*(x-4)**3*6**-1+1000*24**-1*(x-3)**3+C_1*x)*10**3\n",
- "\n",
- "#Now Deflection at D,put x=4 m\n",
- "x=4\n",
- "y_D=1*(E*I)**-1*(-15500*x**3*36**-1+1000*x**4*24**-1+1000*(x-4)**3*6**-1+1000*24**-1*(x-3)**3+C_1*x)*10**3\n",
- "\n",
- "#Answers for y_C & y_D are incorrect in book\n",
- "\n",
- "#Result\n",
- "print\"Deflection at pt C is\",round(y_C,2),\"mm\"\n",
- "print\"Deflection at pt D is\",round(y_D,2),\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Deflection at pt C is 0.22 mm\n",
- "Deflection at pt D is 0.15 mm\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 6.11,Page No.162"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=2.5 #m #Length of beam\n",
- "L_1=1.5 #m #Length from Fixed end\n",
- "W=50*10**3 #N #Load\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Case-1\n",
- "y=W*L**3*3**-1 #Deflection of the cantilever at free end\n",
- "\n",
- "#Case-2\n",
- "#Deflection of cantilever at free end is\n",
- "#y_1=W_1*L**3*3**-1+W_1*L_1**3*3**-1+W_1*L_1**3*3**-1*(L-L_1)\n",
- "#After substituting values in above equation and simplifying further we get\n",
- "\n",
- "#y_1=22.375*W_1*3**-1\n",
- "\n",
- "W_1=y*3*22.375**-1*10**-3 #Magnitude of equal Loads\n",
- "M_1=W*L*10**-3\n",
- "M_2=W_1*L+W_1*L_1\n",
- "\n",
- "#Let M_1=sigma_1*z and M_2=sigma_2*z\n",
- "#Dividing above two equations we get\n",
- "\n",
- "#Let X=sigma_1*sigma_2**-1\n",
- "X=M_2*M_1**-1*100\n",
- "\n",
- "#Result\n",
- "print\"Magnitude of equal Loads is\",round(W_1,2),\"KN\"\n",
- "print\"Max Bending stress is\",round(X,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnitude of equal Loads is 34.92 N\n",
- "Max Bending stress is 111.73 %\n"
- ]
- }
- ],
- "prompt_number": 113
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 6.12,Page No.163"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=4 #m #Length of Beam\n",
- "\n",
- "#calculations\n",
- "\n",
- "#Consider a section at distance x from A and B.M at this section is\n",
- "#M=P*(3-x)-10*x**2+90*x-195 \n",
- "\n",
- "#Now #EI*d**2*y*d**2*x=-P*(3-x)+10*x**2-90*x+195\n",
- "\n",
- "#On Integrating above equation we get \n",
- "#E*I*dy*dx**-1=-P*(3*x-x**2*2**-1)+10*x**3*2**-1-45*x**2+195*x+C_1\n",
- "\n",
- "#Again On Integrating above equation we get \n",
- "#E*I*y=-P*(3*x**2*2**-1-x**3*6**-1)+10*x**4*12**-1-15*x**3+195*x**2*2**-1+C_1*x+C_2\n",
- "\n",
- "#But at x=0,dy*dx**-1=0 we get ,C_1=0\n",
- "# x=0,y=0 we get ,C_2=0\n",
- "#At x=3 m,y=0\n",
- "x=3\n",
- "C_1=0\n",
- "C_2=0\n",
- "P=(10*x**4*12**-1-15*x**3+195*x**2*2**-1+C_1*x+C_2)*(3*x**2*2**-1-x**3*6**-1)**-1\n",
- "\n",
- "#Result\n",
- "print\"Load taken by prop is\",round(P,2),\"KN\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Load taken by prop is 60.0 KN\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 6.13,Page No.163"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=6 #m #Span of Beam\n",
- "sigma=100*10**6 #Pa #Bending stress\n",
- "E=210*10**9\n",
- "y=0.45 #m #Depth\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Taking moment at B\n",
- "R_a=20*6*3+6*40*2*2**-1\n",
- "\n",
- "#At a section x from A the rate of Loading=20+2*3**-1*x #KN/m\n",
- "#S.F=100-20*x-x**2*3**-1\n",
- "#M=100*x-10*x**2-x**3*9**-1 \n",
- "\n",
- "#Thus B.M will be max where S.F is zero,we get equation as\n",
- "#x**2+60*x-300=0\n",
- "a=1\n",
- "b=60\n",
- "c=-300\n",
- "\n",
- "X=b**2-4*a*c\n",
- "x_1=(-b+X**0.5)*(2*a)**-1\n",
- "x_2=(-b-X**0.5)*(2*a)**-1\n",
- "\n",
- "x=4.641\n",
- "M=100*x-10*x**2-x**3*9**-1 #KN*m #Max bending moment\n",
- "I=M*sigma**-1*y*1000*2**-1 #m**4 #M.I\n",
- "\n",
- "#E*I*d**2*y*(d*x**2)**-1=-100*x+10*x**2+x**3*9**-1\n",
- "\n",
- "#AFter Integrating above EquATION WE get\n",
- "#E*I*dy*(dx)**-1=-50*x**2+10*3**-1*x**3+x**4*36**-1+C_1\n",
- "#Again Integrating above EquATION WE get\n",
- "#E*I*y=-50*x**3*3**-1+10*12**-1*x**4+x**5*180**-1+C_1*x+C_2\n",
- "\n",
- "#At x=0,y=0 ,C_2=0\n",
- "#At x=6,y=0 \n",
- "x=6\n",
- "C_2=0\n",
- "C_1=-(-50*x**3*3**-1+10*12**-1*x**4+x**5*180**-1)*x**-1\n",
- "\n",
- "x=3 #m\n",
- "y=1*(E*I)**-1*(-50*x**3*3**-1+10*12**-1*x**4+x**5*180**-1+C_1*x+C_2)*1000*100\n",
- "\n",
- "#Result\n",
- "print\"The central Deflection is\",round(y,2),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The central Deflection is 0.76 m\n"
- ]
- }
- ],
- "prompt_number": 58
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 6.14,Page No.168"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=10 #m #Lenght of cantilever beam\n",
- "P_1=20*10**3 #N #Load at free end\n",
- "P_2=20*10**3 #N #Load at middle of beam\n",
- "E=200*10**9 #Pa \n",
- "I=20000*10**-8 #m**4\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Taking moment at pt B we get\n",
- "R_a=20*5*10**-1 #Force at pt A\n",
- "\n",
- "#Now B.M at b=0,at C=-100,at A=-300 KN*m\n",
- "\n",
- "#Now Area of B.M \n",
- "A_1=2**-1*5*100 #KN*m**2\n",
- "A_2=5*100 #KN*m**2\n",
- "A_3=2**-1*5*200 #KN*m**2\n",
- "\n",
- "#Total Area of B.M diagram is given by A\n",
- "A=A_1+A_2+A_3\n",
- "\n",
- "theta=A*10**3*(E*I)**-1 #radian\n",
- "\n",
- "x_1=2*3**-1*5\n",
- "x_2=3*2**-1*5\n",
- "x_3=5*3**-1*5\n",
- "M_1=A_1*x_1\n",
- "M_2=A_2*x_2\n",
- "M_3=A_3*x_3\n",
- "\n",
- "M=M_1+M_2+M_3 #Total moments of B.M about B\n",
- "\n",
- "y_B=M*10**3*(E*I)**-1 #Deflection a tfree end\n",
- "\n",
- "#REsult\n",
- "print\"Slope of cantilever at free end is\",round(theta,2),\"radian\"\n",
- "print\"Deflection of cantilever at free end is\",round(y_B,2),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Slope of cantilever at free end is 0.03 radian\n",
- "Deflection of cantilever at free end is 0.22 m\n"
- ]
- }
- ],
- "prompt_number": 65
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no.6.15,Page no.169"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Slope at A is Zero and deflection at C is zero According to Mohr's second theorem\n",
- "#Let A_1*x_1=Y\n",
- "Y=1*30**-1*80*4*(3*4**-1*4+2)\n",
- "P=200*27**-1 #Reaction at ens D\n",
- "\n",
- "#Result\n",
- "print\"The reaction at end C is\",round(P,2),\"KN\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reaction at end C is 7.41 KN\n"
- ]
- }
- ],
- "prompt_number": 72
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 6.16,Page No.171"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "E=200*10**9 #Pa\n",
- "I=2500*10**-8 #m**4\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Taking moment about A we get\n",
- "R_a=(30*5+30*1)*6**-1 #Reaction at pt A\n",
- "R_b=60-R_a #Reaction at pt B\n",
- "\n",
- "M_c=30*1 #B.M at C\n",
- "M_d=30*1 #B.M at D\n",
- "M_a=0 #B.M at a\n",
- "M_b=0 #B.M at b\n",
- "\n",
- "#For conjugate beam taking moment about B_dash\n",
- "R_a_dash=(30*2**-1*(5+1*3**-1)+30*4*3+30*2*3**-1*2**-1)*6**-1\n",
- "R_b_dash=150-R_a_dash\n",
- "\n",
- "y_e=1*(E*I)**-1*(R_a_dash*3-30*2*1-2**-1*1*30*(2+1*3**-1))*1000\n",
- "\n",
- "#Result\n",
- "print\"Deflection at the centre is\",round(y_e,2),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Deflection at the centre is 0.03 m\n"
- ]
- }
- ],
- "prompt_number": 69
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file
diff --git a/Strength_Of_Materials_by_B_K_Sarkar/chapter_7_som.ipynb b/Strength_Of_Materials_by_B_K_Sarkar/chapter_7_som.ipynb
deleted file mode 100755
index 7c6e418f..00000000
--- a/Strength_Of_Materials_by_B_K_Sarkar/chapter_7_som.ipynb
+++ /dev/null
@@ -1,642 +0,0 @@
-{
- "metadata": {
- "name": "chapter 7 som.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 7:Torsion"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 7.1,Page no.183"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "G=84 #Gpa #Modulus of Rigidity\n",
- "N=110 #no. of revolution\n",
- "#d*D**-1=0.6 #Ratio of inner diameter to outer diameter\n",
- "sigma_s=63 #MPa #shear stress\n",
- "L=3 #m #Length of shaft\n",
- "P=590 #KW #Power\n",
- "\n",
- "#Calculation\n",
- "\n",
- "#P=2*pi*N*T_mean*60000**-1 #KW #Power\n",
- "T_mean=P*60000*(2*pi*N)**-1 #N*m #Mean Torque\n",
- "\n",
- "#I_p=p*32**-1*(D**4-d**4)\n",
- "\n",
- "#After substituting value of d in above equation we get\n",
- "#I_p=0.0272*pi*D**4 #m**4 #Polar moment of Inertia\n",
- "\n",
- "T_max=1.2*T_mean #N*m #Max torque\n",
- "\n",
- "#Using Relation\n",
- "#T_max*T_p**-1=sigma_s*R**-1=G*theta*L**-1 \n",
- "\n",
- "#After substituting values and simplifying we get\n",
- "\n",
- "D=(5.7085*10**-3)**0.3333 #m #Diameter of shaft\n",
- "\n",
- "theta=1.4*pi*180**-1 #radians\n",
- "\n",
- "#theta=((T_max*L)*(G*10**9*I_p)) #radians\n",
- "\n",
- "#After substituting values and simplifying we get\n",
- "D_1=(1.0513*10**-3)**0.25\n",
- "\n",
- "#Result\n",
- "print\"The Minimum external diameter is\",round(D_1,2),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Minimum external diameter is 0.18 m\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 7.2,Page no.184"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P=295 #KW #Power\n",
- "N=100 #R.p.m \n",
- "sigma_s=80 #MPa #shear stress\n",
- "\n",
- "\n",
- "#Calculations\n",
- "\n",
- "T_mean=((P*60000)*(2*pi*N)**-1) #N*m\n",
- "\n",
- "#T_max=T_mean=(pi*D**3*sigma_s)*16**-1\n",
- "D=((T_mean*16)*(pi*sigma_s*10**6)**-1)**0.333 #m #Diameter of solid shaft\n",
- "\n",
- "#For hollow shaft\n",
- "#I_p_h=pi*32**-1*(D_1**4-d_1**4) (equation 1)\n",
- "\n",
- "#Now d_1=0.6*D_1\n",
- "#substituting above value in equation 1,we get,\n",
- "\n",
- "#I_p_h=0.0272*pi*D_1**4\n",
- "\n",
- "#For solid shaft\n",
- "#I_p_s=pi*32**-1*D**4\n",
- "\n",
- "#T and sigma_s being the same then I_p*R**-1 will be the same for the two shafts\n",
- "#Using relation I_p_h*R_1**-1=I_p_s*R**-1\n",
- "\n",
- "#Substituting values and simplifying we get\n",
- "\n",
- "D_1=(D**3*0.8704**-1)**0.3333333 #m #External diameter of hollow shaft\n",
- "d_1=0.6*D_1 #cm #Internal diameter of hollow shaft\n",
- "\n",
- "A_s=pi*4**-1*(D*10**2)**2 #cm**2 #Area of solid shaft\n",
- "A_h=pi*4**-1*(((D_1*10**2)**2)-((d_1*10**2)**2))\n",
- "\n",
- "W=(A_s-A_h)*A_s**-1*100 #Percentage #Percentage saving in weight\n",
- "\n",
- "\n",
- "#Result\n",
- "print\"Diameter of solid shaft is \",D,\"m\"\n",
- "print\"Percentage saving in weight is\",round(W,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Diameter of solid shaft is 0.121751208998 m\n",
- "Percentage saving in weight is 29.8 %\n"
- ]
- }
- ],
- "prompt_number": 54
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 7.7,Page no.188"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P_C=45 #KW Power aplled at C\n",
- "P_B=15 #KW Power taken off at B\n",
- "P_BA=30 #KW #Power transmitted across BA\n",
- "G=85 #GPa\n",
- "\n",
- "#Calculations (Part-1)\n",
- "\n",
- "#For BC\n",
- "P_1=45 #KW #Power across BC\n",
- "N_1=200 #r.p.m\n",
- "d_1=0.075 #m #diameter of shaft BC\n",
- "L_BC=2 #m #Length of shaft BC\n",
- "\n",
- "\n",
- "T_BC=60000*P_1*(2*pi*N_1)**-1 #N*m #Torque transmitted across BC\n",
- "sigma_s_BC=16*T_BC*((pi*(d_1)**3)**-1)*10**-6 #N/m**2 #max shear stress in BC\n",
- "I_p_BC=pi*32**-1*d_1**4 #m**4 #Polar M.I of BC\n",
- "theta_1=T_BC*L_BC*(G*10**9*I_p_BC)**-1 #Radian #Max angle of twist theta_1 in BC of B relative to C\n",
- "\n",
- "#For AB\n",
- "P_2=30 #KW #Power across AB\n",
- "N_2=200 #r.p.m\n",
- "d_2=0.05 #m #diameter of shaft AB\n",
- "L_BC=4 #m #Length of shaft AB\n",
- "\n",
- "\n",
- "T_AB=60000*P_2*(2*pi*N_2)**-1 #N*m #Torque transmitted across AB\n",
- "sigma_s_AB=16*T_AB*(pi*(d_2)**3)**-1*10**-6 #N/m**2 #max shear stress in AB\n",
- "I_p_AB=pi*32**-1*d_2**4 #m**4 #Polar M.I of AB\n",
- "theta_2=T_AB*L_BC*(G*10**9*I_p_AB)**-1 #Radian #Max angle of twist theta_1 in AB of A relative to B\n",
- "C=(theta_1+theta_2)*180*pi**-1 #radian #Angle of Twist of gear\n",
- "\n",
- "\n",
- "#Result\n",
- "print\"Angle of Twist of gear is\",round(C,2),\"Degree\"\n",
- "print\"The maximum shear stress developed in the shaft AB is\",round(sigma_s_AB,2)\n",
- "\n",
- " "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Angle of Twist of gear is 7.23 Degree\n",
- "The maximum shear stress developed in the shaft AB is 58.36\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 7.8,Page no.189"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L_BC=1.8 #m #Length of BC\n",
- "L_AB=1.2 #m #Length of AB\n",
- "sigma_s=70 #MPa #shear stress\n",
- "d_1=0.05 #m #diameter of BC\n",
- "d_2=0.1 #m #diameter of AB\n",
- "r_BC=0.025 #cm #Radius of BC\n",
- "\n",
- "#Calculations\n",
- "\n",
- "I_p_BC=pi*32**-1*d_1**4 #m**4 #Polar M.I of BC\n",
- "I_p_AB=pi*32**-1*d_2**4 #m**4 #Polar M.I od AB\n",
- "\n",
- "#For BC\n",
- "#theta_1=T*L_BC*(G*10**9*I_p_BC)**-1 #Angle of Twist of C relative to B\n",
- "#After substituting and simplifying value, we get\n",
- "\n",
- "#theta_1=3.4923*10**-5*T\n",
- "\n",
- "#For AB\n",
- "#theta_2=T*L_AB*(G*10**9*I_p_AB)**-1 #Angle of Twist of B relative to A\n",
- "#After substituting and simplifying value, we get\n",
- "\n",
- "#theta_2=1.45513*T\n",
- "\n",
- "#sigma_s=T*R*(I_P)**-1 #The max shear stress in BC\n",
- "\n",
- "#After substituting and simplifying value in above equation, we get\n",
- "\n",
- "T=sigma_s*10**6*I_p_BC*r_BC**-1 \n",
- "theta_1=3.4923*10**-5*T\n",
- "theta_2=1.45513*10**-6*T\n",
- "theta_c=theta_1-theta_2 #radian #total angle of twist\n",
- "\n",
- "#Result\n",
- "print\"Total angle of Twist is\",round(theta_c,3),\"radian\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total angle of Twist is 0.057 radian\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 7.9,Page no.190"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "D=0.05 #m #Diameter of shaft\n",
- "sigma_s_a=55 #MPa #shear stress of alloy\n",
- "sigma_s_s=80 #MPa #shear stress of steel\n",
- "P=185 #KW #Power \n",
- "\n",
- "#Calculations\n",
- "\n",
- "#For alloy shaft,\n",
- "#theta*L**-1=T*(G_A*I_p_A)**-1\n",
- "\n",
- "#For steel shaft,\n",
- "#theta*L*-1=I*(G_S*I_p_S)**-1\n",
- "\n",
- "#After substituting and simplifying we get\n",
- "d=(246.2*10**-8)**0.25 #m #Internal diameter of steel shaft\n",
- "\n",
- "T_1=pi*16**-1*D**3*sigma_s_s*10**6 #N*m #For alloy shaft max torque\n",
- "T_2=pi*16**-1*((D**4-d**4)*D**-1)*sigma_s_s*10**6 #N*m #For steel shaft max torque\n",
- "\n",
- "#Permissible torque,T_2\n",
- "\n",
- "#P=2*pi*N*T_2*(60000)**-1 \n",
- "\n",
- "#After substituting we get\n",
- "N=P*60000*(2*pi*T_2)**-1 #r.p.m #Speed\n",
- "\n",
- "#Result\n",
- "print\"The speed at which the shafts to be driven is\",round(N,0),\"r.p.m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The speed at which the shafts to be driven is 1485.0 r.p.m\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 7.10,Page no.190"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "sigma_s=90 #MPa #shear stress of steel\n",
- "sigma_d=60 #MPa #shear stress of duralumin\n",
- "G_d=28 #GPa #modulus of rigidity of duralumin\n",
- "G_s=84 #GPa #modulus of rigidity of steel\n",
- "L=1 #m #Length of shaft\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#theta*L**-1=sigma_s*(G_s*R_s)**-1=sigma_d*(G_d*R_d)**-1\n",
- "\n",
- "#After substituting and simplifying,we get,\n",
- "#D=2*d\n",
- "\n",
- "#T_s=pi*16**-1*d**3*sigma_s #N*m #torque of steel\n",
- "#T_d=pi*16*(((D**4-d**4)*D**4)**-1)*sigma_d #N*m #torque of duralumin\n",
- "\n",
- "#After substituting and simplifying above two equations,we get,\n",
- "\n",
- "#T_s=17.6714*10**6*d**3 #N*m\n",
- "#T_d=88.3572*d**3 #N*m\n",
- "\n",
- "#T=T_s+T_d #Total torque\n",
- "\n",
- "#T=106.02875*10**6*d**3 \n",
- "\n",
- "d=(700*(106.02875*10**6)**-1)**0.333 #m \n",
- "D=2*d #m\n",
- "R_s=d*2**-1 #m\n",
- "\n",
- "theta=(sigma_s*10**6*L*(G_s*10**9*R_s)**-1)*180*pi**-1 #degree #Angle of twist\n",
- "\n",
- "#Result\n",
- "print\"The Angle of Twist is\",round(theta,2),\"Degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Angle of Twist is 6.52 Degree\n"
- ]
- }
- ],
- "prompt_number": 36
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 7.11,Page no.191"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P=4415 #KW #Power transmitted\n",
- "N=110 #r.p.m\n",
- "sigma_s=75 #MPs #shear stress\n",
- "G=85 #GPa\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#D=2*d \n",
- "\n",
- "T=P*60000*(2*pi*N)**-1 #N*m #Torque Transmitted\n",
- "\n",
- "#T=pi*16**-1*(D**4-d**4)*D**-1*sigma_s #N*m\n",
- "\n",
- "#After substituting and simplifying above equations,we get,\n",
- "\n",
- "D=(T*16*pi**-1*(sigma_s*10**6)**-1)**(1*3**-1)\n",
- "d=D*2**-1\n",
- "X=5*(sigma_s*10**6)**2*(16*G*10**9)**-1\n",
- "\n",
- "#U*V**-1 #Energy stored\n",
- "#X=U*V**-1 #Energy stored #Notations has been changed\n",
- "\n",
- "#Result\n",
- "print\"Diameter of shaft is:D\",round(D,2),\"cm\"\n",
- "print\" :d\",round(d,2),\"cm\"\n",
- "print\"Energy stored per cubic meter is\",round(X,2),\"N/m**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Diameter of shaft is:D 0.3 cm\n",
- " :d 0.15 cm\n",
- "Energy stored per cubic meter is 20680.15 N/m**2\n"
- ]
- }
- ],
- "prompt_number": 65
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 7.12,Page no.192"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P=3680 #KW #Power transmitted\n",
- "N=110 #r.p.m \n",
- "X=20000 #N*m #Energy stored\n",
- "G=85 #GPa\n",
- "\n",
- "#Calculations\n",
- "\n",
- "\n",
- "#U*V**-1=X #Strain Energy per unit volume #Notification has been changed\n",
- "#X=sigma_s**2*(4*G)**-1*((D**2+d**2)*(D**2)**-1)\n",
- "\n",
- "T=P*60000*(2*pi*N)**-1 #N*m #Torque transmitted by shaft\n",
- "sigma_s=(20000*3*G*10**9)**(1*2**-1) #MPa #shear stress of shaft\n",
- "\n",
- "#T=pi*16**-1*((D**4-d**4)*D**-1)*sigma_s \n",
- "\n",
- "#After substituting and simplifying above equations,we get,\n",
- "\n",
- "d=((T*16*3**0.5)*(pi*8*sigma_s)**-1)**(1*3**-1)\n",
- "D=3**0.5*d \n",
- "\n",
- "#Result\n",
- "print\"Diameter of shaft is D=\",round(D,2),\"m\"\n",
- "print\" d=\",round(d,2),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Diameter of shaft is D= 0.29 m\n",
- " d= 0.17 m\n"
- ]
- }
- ],
- "prompt_number": 83
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 7.13,Page no.193"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "D=8 #cm #Diameter of bronze\n",
- "d=5 #cm #diameter of steel shaft\n",
- "R_b=4 #cm #Radius of bronze\n",
- "R_s=2.5 #cm #Radius of steel shaft\n",
- "sigma_b=40 #MPa #shear stress of bronze\n",
- "sigma_s=65 #MPa #shear stress of steel shaft\n",
- "N=500 #r.p.m\n",
- "G_s=85 #GPa #Modulus of rigidity of steel\n",
- "G_b=45 #GPa #Modulus of rigidity of bronze\n",
- "\n",
- "#Calculations\n",
- "\n",
- "I_p_s=pi*32**-1*(5*10**-2)**4 #m**4 #Polar M.I of Steel shaft\n",
- "I_p_b=pi*32**-1*((8*10**-2)**4-(5*10**-2)**4) #m**4 #Polar M.I of Bronze shaft\n",
- "\n",
- "#T*(G_b*I_p_b)**-1=T_s*(G_s*I_s)**-1\n",
- "\n",
- "#After substituting and simplifying above equations,we get\n",
- "\n",
- "#T_b=2.94*T_s\n",
- "\n",
- "T_b=I_p_b*sigma_b*10**6*(R_b*10**-2)**-1 #N*m #Torque carried by bronze\n",
- "T_s=I_p_s*sigma_s*10**6*(R_s*10**-2)**-1 #N*m #Torque carried by steel shaft\n",
- "T_s_1=T_b*2.94**-1 #N*m \n",
- "\n",
- "T=T_b+T_s_1 #N*m #Total Torque\n",
- "P=(2*pi*N*T*(60000)**-1) #KW #Power transmitted\n",
- "\n",
- "#Result\n",
- "print\"Power transmitted by compound shaft is\",round(P,2),\"KW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Power transmitted by compound shaft is 239.11 KW\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 7.14,Page no.194"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "d=10 #cm #Diameter of shaft\n",
- "r=5 #cm #radius of shaft\n",
- "P=100 #KW #Power \n",
- "N=120 #r.p.m \n",
- "n=6 \n",
- "L_k=14 #cm #Length of key\n",
- "B_k=2.5 #cm #width of key\n",
- "n=6 \n",
- "d_b=2 #cm #Diameter of bolt\n",
- "D_b=30 #cm #Diameter of bolt circle\n",
- "R_b=15 #cm #radius \n",
- "\n",
- "#Calculations\n",
- "\n",
- "T=(P*60000*(2*pi*N)**-1)*10**2 #N*m #Torque\n",
- "I_p=pi*32**-1*d**4 #Polar M.I of shaft\n",
- "sigma_s=T*r*(I_p)**-1 #N/cm**2\n",
- "sigma_k=T*(L_k*B_k*r)**-1 #N/cm**2\n",
- "sigma_b=T*4*(n*pi*d_b**2*R_b)**-1 #N/cm**2\n",
- "\n",
- "#Result\n",
- "print\"shear stress in shaft\",round(sigma_s,2),\"N/cmm**2\"\n",
- "print\" Key\",round(sigma_k,2),\"N/cm**2\"\n",
- "print\" bolts\",round(sigma_b,2),\"N/cm**2\" "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "shear stress in shaft 4052.85 N/cmm**2\n",
- " Key 4547.28 N/cm**2\n",
- " bolts 2814.48 N/cm**2\n"
- ]
- }
- ],
- "prompt_number": 90
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file
diff --git a/Strength_Of_Materials_by_B_K_Sarkar/chapter_8_som.ipynb b/Strength_Of_Materials_by_B_K_Sarkar/chapter_8_som.ipynb
deleted file mode 100755
index 08e71ab5..00000000
--- a/Strength_Of_Materials_by_B_K_Sarkar/chapter_8_som.ipynb
+++ /dev/null
@@ -1,1251 +0,0 @@
-{
- "metadata": {
- "name": "chapter 8 som.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 8:Springs"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 8.1,Page no.206"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "k=1 #KN/m #stiffness of spring\n",
- "P=45 #N #Maximum Load\n",
- "sigma_s=126 #MPa #Max shear stress\n",
- "L=4.5 #cm #Lenght of spring\n",
- "G=42 #GPa #Modulus of rigidity\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#sigma_s_max=16*P*R*(pi*d**3)**-1 #Max shear stress\n",
- "\n",
- "#After substituting values in above equation and simolifying we get\n",
- "#1000=42*10**9*d**4*(64*R**3*n)**-1 (#Equation 1)\n",
- "\n",
- "#R=0.175*10**6*pi*d**3 #Radius of spring (Equation 2)\n",
- "\n",
- "#L=n*d #solid length of spring\n",
- "#Thus simplifying above equation, n=L*d**-1\n",
- "\n",
- "#substituting value of n and R in (equation 1) we get,\n",
- "\n",
- "d=(42*10**9*(1000*64*4.5*10**-2*(0.175*pi)**3*(10**6)**3)**-1)**0.25*10**2 #cm #diameter of helical spring\n",
- "\n",
- "#substituting value d in (equation 2) we get,\n",
- "R=0.175*10**6*pi*(d)**3*10**-6*100 #cm #Radius of coil\n",
- "D=2*R #cm #Mean diameter of coil\n",
- "n=0.045*0.00306**-1 #Number of turns\n",
- "\n",
- "\n",
- "#Result\n",
- "print\"The Diameter of wire is\",round(d,3),\"cm\"\n",
- "print\"The Mean Diameter of coil is\",round(D,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Diameter of coil is 0.306 cm\n",
- "The Mean Diameter of coil is 3.15 cm\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 8.2,Page no.207"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "import numpy as np\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=15 #cm #Length of close coiled helical spring\n",
- "U=50 #N*m #Strain energy\n",
- "sigma_s=140 #MPa #Shear stress\n",
- "D=10 #cm #Mean coil diameter\n",
- "G=80 #GPa #Modulus of rigidity\n",
- "\n",
- "R=D*2**-1 #cm #Mean coil Radius\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#Let dell be the deflection of the spring when fully compressed\n",
- "# 0.15-dell=n*d (Equation 1)\n",
- " \n",
- "#U=(sigma_s)**2*V*(4*G)**-1 #Strain energy\n",
- "\n",
- "#After substituting values in above equation and simolifying we get\n",
- "V=50*4*80*10**9*((140*10**6)**2)**-1 #m**3 #Volume of spring\n",
- "\n",
- "#But V=pi*4**-1*d**2*2*pi*R*n\n",
- "#After substituting values in above equation and simolifying we get\n",
- "#n=3.308*10**-3*(d**2)**-1 #Number of turns\n",
- "\n",
- "#We know, T=P*R \n",
- "#Now substituting values in T and simolifying we get\n",
- "#P=549.7787*10**6*d**3 #Load\n",
- "\n",
- "#U=P*dell*2**-1\n",
- "#After substituting values in above equation and simolifying we get\n",
- "#dell=0.18189*10**-6*d**3 #Deflection \n",
- "\n",
- "#After substituting values in above equation and simolifying we get\n",
- "\n",
- "#d**3-22.0533*10**-3*d**2-1.21261*10**-6=0\n",
- "\n",
- "coeff=[1,-22.0533*10**-3,0,-1.21261*10**-6]\n",
- "d=np.roots(coeff) #Diameter of steel wire\n",
- "n=3.308*10**-3*((d[0])**2)**-1 #no.of coils\n",
- "\n",
- "#Result\n",
- "print\"Diameter of steel wire is\",d[0],\"m\"\n",
- "print\"number of coils\",n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Diameter of steel wire is (0.0241350343273+0j) m\n",
- "number of coils (5.67897110767+0j)\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 8.3,Page no.208"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "k=10 #KN/m #stiffness\n",
- "L=40 #cm #Length of coil when adjascent coil touch each other\n",
- "G=80 #GPa #Modulus of rigidity\n",
- "#dell=0.002*n #Max compression\n",
- "\n",
- "#Calculation\n",
- "\n",
- "#k=G*d**4*(8*D**3*n)**-1 #Stiffness\n",
- "#After substituting values in above equation and simolifying we get\n",
- "#d**4=D**3*n*10**-6 (Equation 1)\n",
- "\n",
- "#L=n*d, #After substituting values we get\n",
- "#n=0.4*d**-1 (Equation 2)\n",
- "\n",
- "#Again, d*D**-1=1*10**-1\n",
- "#After solving above ratios we get,D=10*d\n",
- "\n",
- "#After substituting values in Equation 1 And Equation 2 we get\n",
- "d=(10**3*0.4*10**-6)**0.5*100 #cm\n",
- "D=10*d #cm #Mean Diameter \n",
- "R=D*2**-1 #cm #Mean Radius\n",
- "n=0.4*(d*10**-2)**-1 #Number of turns\n",
- "dell=0.002*n*100 #Deflection\n",
- "\n",
- "#k=P*dell**-1 \n",
- "#after solving above equation we get\n",
- "P=k*10**3*dell*10**-2 #N #Load\n",
- "\n",
- "sigma_s_max=16*P*R*10**-2*(pi*(d*10**-2)**3)**-1 #N/m**2 #Max shear stress\n",
- "\n",
- "#Result\n",
- "print\"The wire diameter is\",round(d,2),\"cm\"\n",
- "print\"The Mean diameter is\",round(D,2),\"cm\"\n",
- "print\"Max Load applied is\",round(P,2),\"N\"\n",
- "print\"Max shear stress is\",round(sigma_s_max,0),\"N/m**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The wire diameter is 2.0 cm\n",
- "The Mean diameter is 20.0 cm\n",
- "Max Load applied is 400.0 N\n",
- "Max shear stress is 25464791.0 N/m**2\n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 8.4,Page no.209"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "G=80 #GPa #Modulus of rigidity\n",
- "P=1 #KN #Load\n",
- "dell=10 #cm #Deflection\n",
- "sigma_s=350 #MPa #Max shear stress\n",
- "rho=78000 #N/m**3 #Density of materials\n",
- "\n",
- "#Calculations\n",
- "\n",
- "U=P*1000*dell*10**-2*2**-1 #N*m #Energy stored\n",
- "\n",
- "#Again U=sigma_s**2*V*(4*G)**-1 \n",
- "#After substituting values in above equation and further simplifying we get\n",
- "V=50*4*80*10**9*((350*10**6)**2)**-1 #m**3 #Volume\n",
- "\n",
- "W=V*rho #N #Weight\n",
- "\n",
- "#Now T=P*R=pi*d**3*sigma_s*16**-1\n",
- "#After substituting values in above equation and further simplifying\n",
- "D=(10**6*16*(2*pi*350*10**6)**-1)**0.5*10**2 #cm #Mean diameter of coil\n",
- "\n",
- "k=P*10**3*(dell*10**-2)**-1 #stiffness\n",
- "\n",
- "#Also k=D*n**-1*10**6\n",
- "#After substituting values in above equation and further simplifying\n",
- "n=10**6*D*10**-2*k**-1 #number of turns\n",
- "\n",
- "#Result\n",
- "print\"The Value of weight is\",round(W,3),\"N\"\n",
- "print\"Mean coil diameter is\",round(D,2),\"cm\"\n",
- "print\"The number of Turns is\",round(n,4)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Value of weight is 10.188 N\n",
- "Mean coil diameter is 8.53 cm\n",
- "The number of Turns is 8.5297\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 8.5,Page no.210"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "d=6 #mm #Diameter of steel wire\n",
- "n=50 #number of turns\n",
- "D=5 #cm #Mean Diameter\n",
- "R=D*2**-1 #cm #Radius of coil\n",
- "G=80 #GPa #Modulus of Rigidity\n",
- "P=150 #KN #Load\n",
- "\n",
- "#Calculation\n",
- "\n",
- "#Dell=64*P*R**3*n*(G*d**4)**-1 #Deflection\n",
- "#After substituting values in above equation and simplifying we get\n",
- "#P=2073.6*dell #Gradually applied equivalent Load\n",
- "\n",
- "#loss of potential Energy of the weight=Gain of strain Energy of the spring\n",
- "#150*(0.05+dell)=P*dell*2**-1\n",
- "#After substituting values in above equation we get\n",
- "\n",
- "#dell**2-0.1446*dell-0.00723=0\n",
- "#Above Equation is in the form of ax^2+bx+c=0\n",
- "\n",
- "a=1\n",
- "b=-0.1446\n",
- "c=-0.00723\n",
- "\n",
- "#First computing value of b^2-4ac and store it in a variable say X\n",
- "X=b**2-(4*a*c)\n",
- "#now roots are given as x=(-b+X**0.5)/(2*a) and second root is negative sign before X\n",
- "\n",
- "\n",
- "dell_1=(-b+X**0.5)*(2*a)**-1*10**2\n",
- "dell_2=(-b-X**0.5)*(2*a)**-1*10**2\n",
- "\n",
- "P=2073.6*dell_1*10**-2 #N \n",
- "\n",
- "sigma=16*P*R*10**-2*(pi*(d*10**-3)**3)**-1 #N/m**2 #Max stress\n",
- "\n",
- "#Result\n",
- "print\"The Max Extension of the Spring is\",round(dell_1,2),\"cm\"\n",
- "print\"The Max stress is\",sigma,\"N/m**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Max Extension of the Spring is 18.39 cm\n",
- "The Max stress is 224797751.663 N/m**2\n"
- ]
- }
- ],
- "prompt_number": 46
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 8.6,Page no.209"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "W=200 #N #weight \n",
- "v=4 #m/s #velocity of spring\n",
- "sigma=600 #MPa #max allowable stress in spring\n",
- "G=80 #GPa #Modulus of rigidity\n",
- "rho=78000 #N/m**3 #density\n",
- "d=8 #mm #diameter of spring\n",
- "D=5 #cm #Mean Diameter of coil\n",
- "\n",
- "\n",
- "#Calculation \n",
- "\n",
- "E=W*v**2*(2*9.81)**-1 #N*m #Kinetic Energy #Notification has been changed\n",
- "\n",
- "#U=sigma_s**2*V*(4*G)**-1 #Strain Energy stored inthe spring\n",
- "\n",
- "#After substituting values in above equation and simplifying we get\n",
- "V=163.1*4*80*10**9*((600*10**6)**2)**-1 #Volume \n",
- "\n",
- "W=rho*V #N #Weight of spring\n",
- "\n",
- "#Now V=pi*4**-1*d**2*pi*D*n\n",
- "#After substituting values in above equation and simplifying we get\n",
- "n=0.000145*4*(pi**2*0.008**2*0.05)**-1 #number of turns of spring\n",
- "\n",
- "#T=P*R=pi*16**-1*d**3*sigma_s #Torsion\n",
- "#After substituting values in above equation and simplifying we get\n",
- "P=pi*0.008**3*600*10**6*(0.025*16)**-1 #N\n",
- "\n",
- "#Now U=P*dell*2**-1 \n",
- "#Again,After substituting values in above equation and simplifying we get\n",
- "dell=163.1*2*(2412.743)**-1*10**2 #cm\n",
- "\n",
- "#Result\n",
- "print\"The Max Deflection Produced is\",round(dell,2),\"cm\"\n",
- "print\"Number of coil are\",round(n,2),"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Max Deflection Produced is 13.52 cm\n",
- "Number of coil are 18.36\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 8.7,Page no.211"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "n=12 #number of coils\n",
- "d=3 #cm #mean diameter\n",
- "k=720 #N/m #stiffness of spring\n",
- "sigma_s=190 #MPa #Max shear stress\n",
- "G=80 #GPa #Modulus of rigidity\n",
- "D=3 #mm #Diameter of outer spring\n",
- "\n",
- "#Calculations\n",
- "R=D*2**-1 #mm #Radius of outer spring\n",
- "\n",
- "#Dell_1=64*P*(R*10**-3**3*n*(G*10**9*(d*10**-3)**4)**-1 #m #Extension of first spring\n",
- "#After substituting values and further simplifying we get\n",
- "#Dell_1=0.0004*P #m\n",
- "\n",
- "#Dell_2=64*P*(R*10**-3**3*n*(G*10**9*(d_1)**4)**-1 #m #Extension of first spring\n",
- "#After substituting values and further simplifying we get\n",
- "#Dell_2=3.24*10**-14*P*(d_1**4)**-1 #m #where d_1 is diameter of inner spring\n",
- "\n",
- "#Dell=Dell_1+Dell_2\n",
- "#After substituting values and further simplifying we get\n",
- "#dell=0.0004*P+3.24*10**-14*P*((d)**4)**-1\n",
- "\n",
- "#But dell=P*k**-1=P*720**-1\n",
- "\n",
- "#Now substituting value of dell in above equation we get\n",
- "d_1=(3.24*10**-14*(1*720**-1-0.0004)**-1)**0.25 #cm #diameter of inner spring\n",
- "\n",
- "#Now T=P*R=pi*d_1**3*dell_s*sigma_s*16**-1\n",
- "#simplifying above equation further \n",
- "#P=pi*d**3*sigma_s*(16*R)**-1 \n",
- "#Now substituting values and further simplifying we get\n",
- "P=pi*d_1**3*sigma_s*10**6*(16*R*10**-2)**-1 #N #Limiting Load\n",
- "\n",
- "dell=P*k**-1*10**2 #cm #Total Elongation\n",
- "\n",
- "#Result\n",
- "print\"Greatest Load that can be carried by composite spring is\",round(P,0),\"N\"\n",
- "print\"Extension in spring is\",round(dell,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Greatest Load that can be carried by composite spring is 34.0 N\n",
- "Extension in spring is 4.73 cm\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 8.8,Page no.212"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "#Outer spring\n",
- "n_1=10 #number of coils\n",
- "D_1=3 #cm #Diameter of coil\n",
- "d_1=3 #mm #diameter of wire\n",
- "dell_1=2 #cm #deflection of spring\n",
- "\n",
- "#Inner spring\n",
- "n_2=8 #number of coils\n",
- "\n",
- "G=80 #GPa #Modulus of rigidity\n",
- "\n",
- "#Calculation\n",
- "\n",
- "R_1=D_1*2**-1\n",
- "P_1=G*10**9*dell_1*10**-2*(d_1*10**-3)**4*(64*(R_1*10**-2)**3*n_1)**-1 #Load carried outer spring for compression of 2 cm\n",
- "\n",
- "P_2=100-P_1 #N #Load carried by inner spring\n",
- "k_2=P_2*0.01**-1 #N/m #stiffness of inner spring\n",
- "\n",
- "#D_2=D_1*10**-2-d_1*10**-3-2*dell_1*10**-2-d_2 #Diameter of inner spring\n",
- "#Further simplifying above equation we get\n",
- "#D_2=0.023-d_2\n",
- "\n",
- "#Now from stiffness equation of inner spring\n",
- "#k=G*d_2**4*(8*D_2**3*n_2)**-1\n",
- "#Now substituting values and further simplifying we get\n",
- "#d**4=(0.023-d)**3*312500**-1\n",
- "\n",
- "#As d is small compared with 0.023,as a first appromixation\n",
- "d_2_1=(0.023**3*312500**-1)**0.25 #m\n",
- "\n",
- "#Second Approximation\n",
- "d_2_2=((0.023-d_2_1)**3*312500**-1)**0.25 #m\n",
- "\n",
- "#Final approximation\n",
- "d_2_3=((0.023-d_2_2)**3*312500**-1)**0.25*100 #cm\n",
- "\n",
- "#Result\n",
- "print P_1\n",
- "print\"Stiffness of inner spring is\",round(k_2,2),\"N/m\"\n",
- "print\"Wire Diameter of inner spring is\",round(d_2_3,3),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "60.0\n",
- "Stiffness of inner spring is 4000.0 N/m\n",
- "Wire Diameter of inner spring is 0.231 cm\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 8.9,Page no.212"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L= 3 #m #Length of rod\n",
- "d_1=25*10**-3 #m #Diameter of rod\n",
- "n= 5 #no. of coils\n",
- "sigma=70*10**6 #MPa #instantaneous stress\n",
- "E=70*10**9 #Pa \n",
- "G=80*10**9 #Pa\n",
- "D=24*10**-2 #Spring diameter\n",
- "R=d_2*2**-1 #spring radius\n",
- "d=4*10**-2 #diameter of steel\n",
- "\n",
- "#Calculations\n",
- "\n",
- "dell_1=sigma*L*(E)**-1\n",
- "\n",
- "#Let P be the equivalent applied Load which will produce same stress of 70 MPa\n",
- "P=pi*4**-1*(d_1)**2*E*10**-3 #KN\n",
- "\n",
- "#deflection of the spring is given by\n",
- "dell_2=P*64*R**3*n*(G*d**4)**-1 \n",
- "\n",
- "#Now Loss of Potential Energy of the weight=strain energy stored in the rod and the spring\n",
- "#Height measured from top of uncompressed spring\n",
- "h=((P*dell_1*2**-1+P*dell_2*2**-1)*(5.5*10**3)**-1-(dell_1+dell_2))*10**2\n",
- "\n",
- "#Shear stress in the spring is given by\n",
- "sigma_s=16*P*R*(pi*d**3)**-1*10**-6 #MPa \n",
- "\n",
- "#Result\n",
- "print\"Height measured from top of uncompressed spring\",round(h,2),\"cm\"\n",
- "print\"max shearing stress is\",round(sigma_s,2),\"MPa\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Height measured from top of uncompressed spring 20.34 cm\n",
- "max shearing stress is 328.13 MPa\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 8.10,Page no.213"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=75 #cm #Legth of Leaf spring\n",
- "P=8 #KN #Load\n",
- "y_c=20 #mm #Deflection\n",
- "sigma=200 #MPa #Bending stress \n",
- "E=200 #GPa #modulus of Elasticity\n",
- "#b=12*t\n",
- "\n",
- "#Calculation\n",
- "\n",
- "#y_c=sigma*L**2*(4*E*t)**-1 \n",
- "#After substituting values and further simplifying we get\n",
- "t=200*10**6*(75*10**-2)**2*(4*200*10**9*0.02)**-1*10**2 #Thickness of plate\n",
- "\n",
- "b=12*t #width of plate\n",
- "\n",
- "#Now using relation we get\n",
- "#sigma=3*P*L*(2*n*b*t**2)**-1 \n",
- "#After substituting values and further simplifying we get\n",
- "n=3*8*10**3*0.75*(2*200*10**6*0.084*0.007**2)**-1\n",
- "\n",
- "#Y_c=L**2*(8*R)**-1\n",
- "R=(L*10**-2)**2*(8*y_c*10**-3)**-1 #m #Radius of spring\n",
- "\n",
- "#Result\n",
- "print\"The thickness of plate is\",round(t,2),\"cm\"\n",
- "print\"The width of plate is\",round(b,2),\"cm\"\n",
- "print\"The number of plate is\",round(n,2)\n",
- "print\"The Radius of plate is\",round(R,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The thickness of plate is 0.7 cm\n",
- "The width of plate is 8.44 cm\n",
- "The number of plate is 10.93\n",
- "The Radius of plate is 3.52\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 8.11,Page no.214"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=75 #cm #span of laminated steel spring\n",
- "P=7.5 #KN #Load\n",
- "y_c=5 #cm #Central Deflection\n",
- "sigma=400 #MPa #Bending stress\n",
- "E=200 #GPa #Modulus of Elasticity\n",
- "#b=12*t\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#y_c=3*P*L**3*(8*E*n*b*t**3)**-1 #Deflection\n",
- "#After substituting values and further simplifying we get\n",
- "#nt**4=9.887*10**-9 (Equation 1)\n",
- "\n",
- "#We Know sigma=3*P*L*(2*n*b*t**3)**-1 #bending stress\n",
- "#Again after substituting values and further simplifying we get\n",
- "#nt**3=1.757*10**-6 (Equation 2)\n",
- "\n",
- "#After Divviding (Equation 1) by (Equation 2) we have\n",
- "t=9.887*10**-9*(1.757*10**-6)**-1*10**2 #cm \n",
- "\n",
- "#substituting value of t in Equation 2) we get\n",
- "n=1.757*10**-6*((t*10**-2)**3)**-1 #Number of plates\n",
- "R=(L*10**-2)**2*(8*y_c*10**-2)**-1 #Radius of curvature\n",
- "\n",
- "#Result\n",
- "print\"The thickness of Plates is\",round(t,2),\"cm\"\n",
- "print\"The Number of Plates is\",round(n,2)\n",
- "print\"The Radius of Curvature of Plates is\",round(R,2),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The thickness of Plates is 0.56 cm\n",
- "The Number of Plates is 9.86\n",
- "The Radius of Curvature of Plates is 1.41 m\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 8.12,Page no.214"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=1.3 #m #Length of carriage spring\n",
- "b=10 #cm #width of spring\n",
- "t=12 #mm #thickness of spring\n",
- "sigma=150 #MPa #Bending stresses\n",
- "E=200 #GPa #Modulus of Elasticity\n",
- "U=120 #N*m #Strain Energy\n",
- "\n",
- "#Calculation\n",
- "\n",
- "#V=n*b*t*L #Volume of carriage spring\n",
- "#U=sigma**2*(6*E)**-1*V\n",
- "#After substituting values in above equation and further simplifying we get\n",
- "n=120*6*200*10**9*2*((150*10**6)**2*10*10**-2*12*10**-3*1.3)**-1\n",
- "\n",
- "sigma_1=(120*6*200*10**9*2*(9*0.1*0.012*1.3)**-1)**0.5*10**-6 #MPa #Actual Bending stress\n",
- "\n",
- "R=E*t*(2*sigma_1)**-1 #m \n",
- "\n",
- "#Result\n",
- "print\"The number of plates is\",round(n,2)\n",
- "print\"Radius of curvature is\",round(R,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The number of plates is 8.21\n",
- "Radius of curvature is 8.379\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 8.13,Page no.215"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "P=200 #N #Load\n",
- "h=10 #cm #Height of Load dropped\n",
- "n=10 #Number of turns\n",
- "b_1=5 #cm #width of plates\n",
- "t=6 #mm #thickness of plates\n",
- "L=75 #cm #Length of plates\n",
- "E=200 #GPa #Modulus of Elasticity\n",
- "\n",
- "#Calculaion\n",
- "\n",
- "#Let P be the equivalent gradually applied load whuch would cause the same stress as is caused by the impact Load\n",
- "#200(0.1+dell)=P*dell*2**-1 (Equation 1)\n",
- "\n",
- "#dell=3*P*L**3*(8*E*n*b*t**3)**-1 #Deflection\n",
- "#After substituting values in above equation and further simplifying we get\n",
- "#P=136533.33*dell\n",
- "\n",
- "#After substituting values of P in (equation 1) and further simplifying we get\n",
- "#200(0.1+dell)=136533.33*dell**2*2**-1\n",
- "\n",
- "#simplifying above equation we get\n",
- "#dell**2-2.93*10**-3*dell-2.93*10**-4=0\n",
- "#The Above Equation is in the form of ax**2+bx+c=0\n",
- "a=1\n",
- "b=-2.93*10**-3\n",
- "c=-2.93*10**-4\n",
- "\n",
- "#First computing value of b^2-4ac and store it in a variable say X\n",
- "X=b**2-(4*a*c)\n",
- "#now roots are given as x=(-b+X**0.5)/(2*a) and second root is negative sign before X\n",
- "\n",
- "dell_1=(-b+X**0.5)*(2*a)**-1\n",
- "dell_2=(-b-X**0.5)*(2*a)**-1\n",
- "\n",
- "#Now deflection cannot be negative so consider value of dell_1\n",
- "\n",
- "P=136533.33*dell_1\n",
- "sigma=3*P*L*10**-2*(2*n*b_1*10**-2*(t*10**-3)**2)**-1*10**-6 #MPa #Max instantaneous stress\n",
- "R=(L*10**-2)**2*(8*dell_1)**-1 #Radius of curvature\n",
- " \n",
- "#Result\n",
- "print\"Max instantaneous stress in plates is\",round(sigma,2),\"MPa\"\n",
- "print\"Radius of curvature of spring is\",round(R,2),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Max instantaneous stress in plates is 159.1 MPa\n",
- "Radius of curvature of spring is 3.77 m\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 8.14,Page no.215"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=70 #cm #Length of Longest plate\n",
- "P=3.5 #KN #central Load\n",
- "y_c=1.8 #cm #central deflection\n",
- "sigma=190 #MPa #allowable bending stress\n",
- "#b=12*t\n",
- "E=200 #GPa #Modulus of Elasticity\n",
- "\n",
- "#Calculation\n",
- "\n",
- "y_c=3*P*L**3*(8*n*E*b*t**3)**-1 #Deflection (#Equation 1)\n",
- "sigma=3*P*L*(2*n*b*t**2)**-1 #stress\n",
- "#Dividing Equation 1 by Equation 2 we get\n",
- "#y_c*sigma**-1=L**2*(4*E*t)**-1\n",
- "#After substituting values in above equation and further simplifying we get\n",
- "t=190*10**6*0.7**2*(1.8*10**-2*4*200*10**9)**-1*10**3 #thickness of plate\n",
- "b=12*t #Width of plates\n",
- "\n",
- "#sigma=3*2**-1*P*L*(n*b*t**2)**-1 #stress\n",
- "#After substituting values in above equation and further simplifying we get\n",
- "n=3*3.5*10**3*0.7*(2*190*10**6*0.077583*(6.465*10**-3)**2)**-1\n",
- "\n",
- "#Now sigma*y**-1=E*R**-1 \n",
- "#simplifying above equationwe get\n",
- "R=200*10**9*6.465*10**-3*(2*190*10**6)**-1 #Radius of Curvature\n",
- "a=L*10**-2*(2*n)**-1*10**3 #Overlap\n",
- "\n",
- "#Result\n",
- "print\"size of the plate is:\",round(b,2),\"mm\"\n",
- "print\" :\",round(t,2),\"mm\"\n",
- "print\"Overlap of plates is\",round(a,2),\"mm\"\n",
- "print\"Number of plates is\",round(n,2)\n",
- "print\"The Radius of curvature is\",round(R,3),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "size of the plate is: 77.58 mm\n",
- " : 6.47 mm\n",
- "Overlap of plates is 58.68 mm\n",
- "Number of plates is 5.96\n",
- "The Radius of curvature is 3.403 m\n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 8.15,Page no.216"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "alpha=30 #degree #helix angle\n",
- "dell=2.3*10**-2 #m #Vertical displacement\n",
- "W=40 #N #Axial Load\n",
- "d=6*10**-3 #steel rod diameter\n",
- "E=200*10**9 #Pa \n",
- "G=80*10**9 #Pa \n",
- "\n",
- "#Calculations\n",
- "\n",
- "#from equation of deflection of the spring under the Load we get\n",
- "#R**3*n=8.49*10**-4 \n",
- "\n",
- "#Let R**3*n=X\n",
- "X=8.49*10**-4 #Equation 1\n",
- "\n",
- "#from equation of angular rotation\n",
- "#R**2*n=8.1*10**-3\n",
- "\n",
- "#Let R**2*n=Y\n",
- "Y=8.1*10**-3 #Equation 2\n",
- "\n",
- "#After dividing equation 1 by equation 2 we get R \n",
- "#Let Z=R\n",
- "\n",
- "Z=X*Y**-1\n",
- "R=Z*10**2 #cm #Mean Radius\n",
- "\n",
- "#Result\n",
- "print\"Mean Radius of Open coiled spring of helix angle is\",round(R,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Mean Radius of Open coiled spring of helix angle is 10.48 cm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 8.16,Page no.217"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "n=10 #Number of coils\n",
- "sigma=100 #MPa #Bending stress\n",
- "sigma_s=110 #MPa #Twisting stress\n",
- "#D=8*d\n",
- "dell=1.8 #cm #Max extension of of wire\n",
- "E=200 #GPa #Modulus of Elasticity\n",
- "G=80 #GPa #Modulus od Rigidity\n",
- "\n",
- "#Calculation\n",
- "\n",
- "#M=W*R*sin_alpha=pi*d**3*sigma_1*32**-1 #(Equation 1) #Bending moment\n",
- "#As D=8*d \n",
- "#then R=D*2**-1\n",
- "#Therefore, R=4*d\n",
- "\n",
- "#Now substituting values in equation 1 we get\n",
- "#W*sin_alpha=2454369.3*d**2 (Equation 2)\n",
- "\n",
- "#T=W*R*cos_alpha=pi*d**3*sigma_s #Torque (Equation 3) \n",
- "#Now substituting values in equation 3 we get\n",
- "#W*cos_alpha=5399612.4*d**2 (Equation 4)\n",
- "\n",
- "#Dividing Equation 2 by Equation 4 we get,\n",
- "#tan_alpha=0.4545\n",
- "alpha=arctan(0.4545)*180*pi**-1\n",
- "\n",
- "#From Equation 2 we get\n",
- "#W=2454369.3*d**2*(sin24.443)**-1\n",
- "#W=5931241.1*d**2 (Equation 5)\n",
- "\n",
- "#dell=64*W*R**3*n*sec_alpha*(d**4)**-1*((cos_alpha)**2*G**-1+2*sin_alpha**2*E**-1)\n",
- "#Now substituting values in above equation we get\n",
- "#W=33140.016*d (Equation 6)\n",
- "\n",
- "#From Equation 5 and Equation 6 we get\n",
- "#5931241.1*d**2=33140.016*d\n",
- "#After simplifying above equation we get\n",
- "d=33140.016*5931241.1**-1 #m #Diameter of wire\n",
- "W=33140.016*d #N #MAx Permissible Load\n",
- "\n",
- "#Result\n",
- "print\"The Max Permissible Load is\",round(W,2),\"N\"\n",
- "print\"Thw Wire Diameter is\",round(d,6),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Max Permissible Load is 185.17 N\n",
- "Thw Wire Diameter is 0.005587 m\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 8.18,Page no.218"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "#Calculation\n",
- "\n",
- "n=10 #number of coils\n",
- "d=2*10**-2 #m #Diameter of wire\n",
- "D=12*10**-2 #m #Diameter of coiled spring\n",
- "R=0.06 #m #Radius of coiled spring\n",
- "dell=0.5*10**-2 #Deflection\n",
- "E=200*10**9 #Pa \n",
- "G=80*10**9 #Pa \n",
- "alpha=30 #degree\n",
- "\n",
- "#Calculations\n",
- "\n",
- "#beta=64*W*R**2*n*sinalpha*(d**4)**-1*(1*G**-1-2*E**-1)+64*T*R*n*secalpha*(d**4)**-1*(sin**2alpha*G**-1+2*cos**2alpha*E**-1)=0\n",
- "#From above equation anf simplifying we get\n",
- "\n",
- "#T=-6.11*10**-3*W\n",
- "\n",
- "#dell=64*W*R**3*n*sec(alpha)*(d**4)**-1*[(cos(alpha))**2*G**-1+2*(sin(alpha))**2*E**-1]+64*T*R**2*n*sin(alpha)*(d**4)**-1*[1*G**-1+2*E**-1]\n",
- "\n",
- "#After substituting Values and further simplifying we get\n",
- "#1.1847*10**-5*W+1.62*10**-4*T=0.005\n",
- "\n",
- "#Now substituting value of T in above equation we get\n",
- "#1.1847*10**-5*W-9.8982*10**-7*W=0.005\n",
- "W=0.005*(1.1847*10**-5-9.8982*10**-7)**-1 #N\n",
- "T=-6.11*10**-3*W #N*m\n",
- "\n",
- "#Result\n",
- "print\"The axial Load is\",round(W,2),\"N\"\n",
- "print\"Necesscary torque is\",round(T,2),\"N*m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The axial Load is 460.52 N\n",
- "Necesscary torque is -2.81 N*m\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 8.19,Page no.219"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "d=6 #mm #Diameter of steel wire\n",
- "n=1 #number of turns\n",
- "D=6.5 #cm #Mean of diameter\n",
- "G=80 #GPa #modulus of rigidity\n",
- "P_1=150 #Load\n",
- "p=1.5 #cm #Pitch of coil\n",
- "\n",
- "#Calculation \n",
- "\n",
- "R=D*2**-1\n",
- "#For one turn deflection is\n",
- "dell=p-d*10**-1 #cm \n",
- "\n",
- "#dell=64*P*R**3*n*(G*d**4)**-1 \n",
- "#Now, after simplifying further we get,\n",
- "P=dell*10**-2*G*10**9*(d*10**-3)**4*(64*(R*10**-2)**3*n)**-1 #N #Axial Load\n",
- "\n",
- "dell_2=dell*8 #cm #Total Displacement #Notification has been changed\n",
- "U=P*dell_2*10**-2*2**-1 #N-m #Strain Energy\n",
- "\n",
- "#Potential Energy given by 150N Load is\n",
- "#U=150*(h+0.072)\n",
- "\n",
- "#After simplifying above equation we get\n",
- "h=(U*P_1**-1-0.072)*10**2 #cm #Height from which 150 N load falls\n",
- "\n",
- "#Result\n",
- "print\"Axial Load is\",round(P,2),\"N\"\n",
- "print\"Height from which 150 N load falls is\",round(h,2),\"cm\" "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Axial Load is 424.72 N\n",
- "Height from which 150 N load falls is 2.99 cm\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem 8.20,Page no.219"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "alpha=30 #degree \n",
- "E=200*10**9 #Pa\n",
- "G=80*10**9 #pa \n",
- "\n",
- "#Calculations\n",
- "\n",
- "#For alpha=30 #Degree\n",
- "#dell=64*W*R**3*n*sec(alpha)*(d**4)**-1*(cos(alpha)**2*G**-1+2*sin(alpha)**2*E**-1)\n",
- "#Now substituting values in above equation we get\n",
- "\n",
- "#dell_1=64*W*R**3*n*(d**4)**-1*1330*(10**9)**-1 (equation 1)\n",
- "\n",
- "#For alpha=0 #Degree\n",
- "#dell=64*W*R**3*n*sec(alpha)*(d**4)**-1*(cos(alpha)**2*G**-1+2*sin(alpha)**2*E**-1)\n",
- "#Now substituting values in above equation we get\n",
- "\n",
- "#dell_2=64*W*R**3*n*(d**4)**-1*1250*(10**9)**-1 (equation 2)\n",
- "\n",
- "#subtracting equation 1 and equation 2 we get\n",
- "#Let dell_1-dell_2=X\n",
- "#X=64*W*R**3*n*(d**4)**-1*80*(10**9)\n",
- "\n",
- "#Let Y=X*dell_1**-1*100\n",
- "Y=80*1330**-1*100 #% under estimation of axial extension\n",
- "\n",
- "#Result\n",
- "print\"% under estimation of axial extension is\",round(Y,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "% under estimation of axial extension is 6.02\n"
- ]
- }
- ],
- "prompt_number": 12
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file
diff --git a/Strength_Of_Materials_by_B_K_Sarkar/chapter_9_som.ipynb b/Strength_Of_Materials_by_B_K_Sarkar/chapter_9_som.ipynb
deleted file mode 100755
index 318df44c..00000000
--- a/Strength_Of_Materials_by_B_K_Sarkar/chapter_9_som.ipynb
+++ /dev/null
@@ -1,502 +0,0 @@
-{
- "metadata": {
- "name": "chapter 9 som.ipynb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Chapter No.9:Columns And Struts"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 9.1,Page no.232"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "D=15 #cm #External Diameter\n",
- "t=2 #cm #Thickness\n",
- "L=6 #m #Length of cyclinder\n",
- "E=80*10**9 #Pa\n",
- "alpha=1*1600**-1 \n",
- "sigma_c=550*10**6 #Pa #compressive stress\n",
- "\n",
- "#Calculations\n",
- "\n",
- "d=D-2*t #m #Internal Diameter\n",
- "A=pi*4**-1*(D**2-d**2)*10**-4 #m**2 #Areaof Tube\n",
- "I=pi*64**-1*(D**4-d**4)*10**-4 #m**4 #M.I of tube\n",
- "k=(I*A**-1)**0.5 #m #Radius of Gyration\n",
- "\n",
- "P_e=pi**2*E*I*(L**2)**-1 #Euler's Load\n",
- "P_R=sigma_c*A*(1+alpha*(L*k**-1)**2)**-1 #According to Rankine's Formula\n",
- "\n",
- "#The Answer in Textbook is incorrect for P_R \n",
- "\n",
- "#Now again from Rankine's Formula\n",
- "#As K=I*A**-1,so substituting in below equation\n",
- "#Thus Stress calculated from Euler's Formula cannot exceed the yield stress of 550 MPa\n",
- "L=(pi**2*E*k**2*(550*10**6)**-1)**0.5*10**-2 #m #Length of cyclinder\n",
- "\n",
- "#Result\n",
- "print\"The Length of strut is\",round(L,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Length of strut is 1.76 cm\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 9.2,Page no.233"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "\n",
- "L=1.5 #m #Length of steelbar\n",
- "b=2 #cm #bredth of steelbar\n",
- "d=0.5 #cm #depth of steelbar\n",
- "sigma=320 #MPa #Yield point\n",
- "E=210 #GPa #modulus of Elasticity of steelbar\n",
- "\n",
- "#Calculations\n",
- "\n",
- "I_min=b*d**3*12**-1*10**-8 #m**4 #Moment of Inertia \n",
- "P=pi**2*E*10**9*I_min*(L**2)**-1 #N #N #Crippling Load\n",
- "\n",
- "#Let dell=Central Deflection\n",
- "\n",
- "#M=P*dell #Max Bending moment\n",
- "#After substituting value in above equation we get\n",
- "#M=191.9*dell\n",
- "\n",
- "A=b*d*10**-4 #m**2 #Area of steel bar\n",
- "sigma_1=P*A**-1*10**-6 #Mpa #Direct stress\n",
- "\n",
- "Z=b*d**3*10**-6 #Section modulus \n",
- "#sigma_2=M*Z**-1 #N/m**2 #Bending stress\n",
- "#After substituting value in above equation we get\n",
- "#sigma_2=dell*2302.8*10**6 #N/m**2 \n",
- "\n",
- "#sigma=sigma_1+sigma_2\n",
- "#Now substituting value of Bending stress and direct stress in above equation we get\n",
- "\n",
- "#320*10**6=1.919*10**6+2302.8*10**6*dell\n",
- "dell=((320*10**6-1.919*10**6)*(2302.8*10**6)**-1)*10**2 #cm #Central Deflection\n",
- "\n",
- "#Result\n",
- "print\"Maximum Central Deflection is\",round(dell,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum Central Deflection is 13.81 cm\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 9.3,Page no.233"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "dell=1 #cm #Deflection\n",
- "FOS=4 #Factor of safety\n",
- "E=210 #GPa #Modulus of Elasticity of steel bar\n",
- "W=40 #KN #Load \n",
- "\n",
- "#Flange Dimensions\n",
- "b=30 #cm #width of flange\n",
- "d=5 #cm #depth of flange\n",
- "\n",
- "#Web Dimensions\n",
- "d_1=100 #cm #Depth of web\n",
- "t_1=2 #cm #Thcikness of web\n",
- "\n",
- "#Calculations\n",
- "\n",
- "I_xx=(0.3*1.1**3-0.28*1**3)*12**-1 #m**4 #M.I about x-x axis\n",
- "I_yy=2*0.05*0.3**3*12**-1+1*0.02**3*12**-1 #m**4 #M.I about y-y axis\n",
- "\n",
- "#From the equation of deflection we get\n",
- "L=(dell*10**-2*384*E*10**9*I_xx*(5*40*10**3)**-1)**0.25 #m #Length of beam\n",
- "P=pi**2*210*I_yy*10**9*4*(L**2)**-1 #N #crippling Load\n",
- "S=P*4**-1 #N #Safe Load\n",
- "\n",
- "#Result\n",
- "print\"The Safe Load is\",round(S,2),\"N\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Safe Load is 2336127.78 N\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 9.5,Page no.235"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=4 #m #Length of column\n",
- "W=250 #KN #Safe Load\n",
- "FOS=5 #Factor of safety\n",
- "#d=0.8*D #Internal diameter is 0.8 times Extarnal Diameter\n",
- "sigma_c=550 #MPa #Compressive stress\n",
- "alpha=1*1600**-1 #constant\n",
- "\n",
- "#Calculations\n",
- "\n",
- "P=W*FOS #N #Crippling Load\n",
- "\n",
- "#A=pi*4**-1(D**2-d**2) #m**2 #Area of hollow cyclinder\n",
- "#After substituting value of d we get\n",
- "\n",
- "#A=pi*0.09*D**2\n",
- "\n",
- "#I=pi*64**-1*(D**4-d**4) #m**4 #Mo,ent Of Inertia\n",
- "#After substituting value of d we get d we get\n",
- "\n",
- "#I=0.009225*pi*D**4\n",
- "\n",
- "#K=(I*A**-1)**0.5 #Radius of Gyration\n",
- "#After substituting value of I and A and further simplifying we get\n",
- "#K=0.32*D\n",
- "\n",
- "#Now using the Relation we get\n",
- "#P=sigma_c*A*(1+alpha*(l_e*k)**2)**-1 #Rankines Formula\n",
- "#Now Substituting values in above equation we get\n",
- "#125*10**4=550*10**6*pi*0.09*D**2*(1+1*1600**-1*((2*0.32)**2)**-1)**-1\n",
- "\n",
- "#Further simplifying and rearranging we get\n",
- "#D**4-0.008038*D**2-0.0001962397=0\n",
- "\n",
- "a=1\n",
- "b=-0.008038\n",
- "c=-0.0001962397\n",
- "\n",
- "X=b**2-4*a*c\n",
- "\n",
- "D_1=((-b+X**0.5)*(2*a)**-1)**0.5*10**2\n",
- "D_2=((-b-X**0.5)*(2*a)**-1)**0.5\n",
- "\n",
- "#Thus Diameter cannot be negative, discard value of D_2\n",
- "d=0.8*D_1\n",
- "\n",
- "#Result\n",
- "print\"The Minimum Diameter is\",round(d,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Minimum Diameter is 10.91 cm\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 9.6,Page no.236"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "d=0.04 #m #Internal Diameter of tube\n",
- "D=0.05 #m #External Diameter of tube\n",
- "P_1=240*10**3 #N #Compressive Load\n",
- "P_2=158*10**3#N #Failure Load\n",
- "L=2 #m #Length of tube\n",
- "l=3 #m #Length of strut\n",
- "\n",
- "#Calculations\n",
- "\n",
- "A=pi*4**-1*(D**2-d**2) #m**2 #Areaof Tube\n",
- "I=pi*64**-1*(D**4-d**4) #m**4 #M.I of tube\n",
- "k=(I*A**-1)**0.5 #m #Radius of Gyration\n",
- "sigma_c=P*A**-1 #Pa #Compressive stress\n",
- "\n",
- "l_e=L*2**-1 #m #According to given condition i.e Both ends fixed\n",
- "\n",
- "#Now from crippling Load Equation we get\n",
- "alpha=((sigma_c*A*P_2**-1-1)*((l_e*k**-1)**2)**-1)*10**4\n",
- "\n",
- "#Now Crippling Load when L=3 m Is used as strut\n",
- "l_e_2=l*(2**0.5)**-1\n",
- "P_3=sigma_c*A*(1+alpha*10**-4*(l_e_2*k**-1)**2)**-1 \n",
- "\n",
- "\n",
- "print\"The Value of constant value alpha is\",round(alpha,2)\n",
- "print\"The Crippling Load of Tube is\",round(P_3,2),\"N\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Value of constant value alpha is 1.33\n",
- "The Crippling Load of Tube is 71954.46 N\n"
- ]
- }
- ],
- "prompt_number": 54
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 9.8,Page no.239"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "D=0.038 #m #External diameter\n",
- "d=0.035 #m #Internal diameter\n",
- "P=20*10**3 #N #Load\n",
- "E=210*10**9 #Pa \n",
- "e=0.002 #m #Eccentricity\n",
- "L=1.5 #m #Lenght of tube\n",
- "\n",
- "#Calculations\n",
- "\n",
- "A=pi*4**-1*(D**2-d**2) #m**2 column\n",
- "I=pi*64**-1*(D**4-d**4) #m**4 #M.I of column\n",
- "m=(P*(E*I)**-1)**0.5 \n",
- "\n",
- "#Let X=secmL*2**-1\n",
- "X=(1*(cos(m*L*2**-1))**-1)\n",
- "M=P*e*X #N-m #MAx Bending Moment\n",
- "sigma_1=P*A**-1*10**-6 #Pa #Direct stress\n",
- "sigma_2=M*0.019*I**-1*10**-6 #Pa #Bending stress\n",
- "\n",
- "sigma_c_max=(sigma_1+sigma_2) #MPa #Max compressive stress\n",
- "\n",
- "#Result\n",
- "print\"The Max stress developed is\",round(sigma_c_max,2),\"MPa\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Max stress developed is 246.79 MPa\n"
- ]
- }
- ],
- "prompt_number": 130
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 9.9,Page no.239"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=5 #m #Length of column\n",
- "D=0.2 #m #External Diameter\n",
- "d=0.14 #m #Internal diameter\n",
- "P=200*10**3 #N #Load\n",
- "e=0.015 #m #Eccentricity\n",
- "E=95 *10**9 #Pa \n",
- "\n",
- "#Calculations\n",
- "\n",
- "L_2=L*2**-1 #m #half length of column\n",
- "A=pi*4**-1*(D**2-d**2) #m**2 column\n",
- "I=pi*64**-1*(D**4-d**4) #m**4 #M.I of column\n",
- "m=(P*(E*I)**-1)**0.5 \n",
- "\n",
- "#Let X=secmL*2**-1\n",
- "X=(1*(cos(m*L_2*2**-1))**-1)\n",
- "M=P*e*X #N-m #MAx Bending Moment\n",
- "sigma_1=P*A**-1*10**-6 #Pa #Direct stress\n",
- "sigma_2=M*0.1*I**-1*10**-6 #Pa #Bending stress\n",
- "\n",
- "sigma_c_max=(sigma_1+sigma_2) #MPa #Max compressive stress\n",
- "\n",
- "print\"The Max stress developed is\",round(sigma_c_max,2),\"MPa\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Max stress developed is 17.65 MPa\n"
- ]
- }
- ],
- "prompt_number": 125
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Problem no 9.10,Page no.240"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Initilization of variables\n",
- "\n",
- "L=3 #m #Length of strut\n",
- "b=0.04 #m #Width of rectangle\n",
- "d=0.10 #m #Depth if rectangle\n",
- "P=100*10**3 #N #Axial thrust\n",
- "w=10*10**3 #N #Uniformly Distributed Load\n",
- "E=210*10**9 #Pa \n",
- "\n",
- "#Calculations\n",
- "\n",
- "A=b*d #m**2 #Area of strut\n",
- "I=b*d**3*12**-1 #m**4 #M.I \n",
- "m=(P*(E*I)**-1)**0.5 \n",
- "\n",
- "#Let X=secmL*2**-1\n",
- "X=(1*(cos(m*L*2**-1))**-1)\n",
- "\n",
- "M=w*E*I*P**-1*(X-1)*3**-1 #N*m #Max Bending Moment\n",
- "sigma_1=P*A**-1 #Pa #Direct stress\n",
- "sigma_2=M*0.05*I**-1 #Pa #Bending stress\n",
- "\n",
- "sigma_c_max=sigma_1+sigma_2 #Max compressive stress\n",
- "\n",
- "#If the Eccentricity of thrust is neglected\n",
- "M_2=w*L**2*(3*8)**-1 #Max Bending moment\n",
- "sigma_2_2=M_2*0.05*I**-1 #Pa #Bending stress\n",
- "\n",
- "sigma_c_max_2=(sigma_1+sigma_2_2)*10**-6 #Pa\n",
- "\n",
- "#Let Y=Percentage error\n",
- "Y=((sigma_c_max-sigma_c_max_2*10**6)*sigma_c_max**-1)*100\n",
- "\n",
- "#Result\n",
- "print\"Max stress induced is\",round(sigma_c_max_2,2),\"MPa\"\n",
- "print\"The Percentage Error is\",round(Y,3),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Max stress induced is 81.25 MPa\n",
- "The Percentage Error is 9.638 %\n"
- ]
- }
- ],
- "prompt_number": 98
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file