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authorTrupti Kini2016-09-27 23:30:25 +0600
committerTrupti Kini2016-09-27 23:30:25 +0600
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Added(A)/Deleted(D) following books
A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter25_7.ipynb A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter26_7.ipynb A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter27_7.ipynb A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter28_7.ipynb A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter29_7.ipynb A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter30_7.ipynb A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter31_7.ipynb A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter32_7.ipynb A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter33_7.ipynb A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter34_7.ipynb A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter35_7.ipynb A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter36_7.ipynb A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter37_7.ipynb A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter38_7.ipynb A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter39_7.ipynb A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/screenshots/chapter29example32_7.png A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/screenshots/chapter29example33_7.png A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/screenshots/chapter32example30_7.png A RCC_Theory_and_Design_by_M._G._Shah_and_C._M._Kale/Chapter1.ipynb A RCC_Theory_and_Design_by_M._G._Shah_and_C._M._Kale/Chapter10.ipynb A RCC_Theory_and_Design_by_M._G._Shah_and_C._M._Kale/Chapter11.ipynb A RCC_Theory_and_Design_by_M._G._Shah_and_C._M._Kale/Chapter12.ipynb A RCC_Theory_and_Design_by_M._G._Shah_and_C._M._Kale/Chapter13.ipynb A RCC_Theory_and_Design_by_M._G._Shah_and_C._M._Kale/Chapter14.ipynb A RCC_Theory_and_Design_by_M._G._Shah_and_C._M._Kale/Chapter15.ipynb A RCC_Theory_and_Design_by_M._G._Shah_and_C._M._Kale/Chapter16.ipynb A RCC_Theory_and_Design_by_M._G._Shah_and_C._M._Kale/Chapter17.ipynb A RCC_Theory_and_Design_by_M._G._Shah_and_C._M._Kale/Chapter18.ipynb A RCC_Theory_and_Design_by_M._G._Shah_and_C._M._Kale/Chapter19.ipynb A RCC_Theory_and_Design_by_M._G._Shah_and_C._M._Kale/Chapter2.ipynb A RCC_Theory_and_Design_by_M._G._Shah_and_C._M._Kale/Chapter3.ipynb A RCC_Theory_and_Design_by_M._G._Shah_and_C._M._Kale/Chapter4.ipynb A RCC_Theory_and_Design_by_M._G._Shah_and_C._M._Kale/Chapter5.ipynb A RCC_Theory_and_Design_by_M._G._Shah_and_C._M._Kale/Chapter7.ipynb A RCC_Theory_and_Design_by_M._G._Shah_and_C._M._Kale/Chapter8.ipynb A RCC_Theory_and_Design_by_M._G._Shah_and_C._M._Kale/Chapter9.ipynb A RCC_Theory_and_Design_by_M._G._Shah_and_C._M._Kale/screenshots/11.4.png A RCC_Theory_and_Design_by_M._G._Shah_and_C._M._Kale/screenshots/13.1.png A RCC_Theory_and_Design_by_M._G._Shah_and_C._M._Kale/screenshots/15.4.png A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.10_9.ipynb A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.2_9.ipynb A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.3_9.ipynb A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.4_9.ipynb A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.5_9.ipynb A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.6_9.ipynb A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.7_9.ipynb A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.8_9.ipynb A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.9_9.ipynb A Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/BMD_1.JPG A Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/SFD2_1.JPG A Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S_F_D_1_1.JPG A f_by_Gayakwad/abhishek.ipynb A f_by_Gayakwad/screenshots/blank1.png A f_by_Gayakwad/screenshots/blank1_(another_copy).png A f_by_Gayakwad/screenshots/blank1_(copy).png A sample_notebooks/IshitaTewari/Chapter08.ipynb
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diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.10_9.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.10_9.ipynb
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+++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.10_9.ipynb
@@ -0,0 +1,280 @@
+{
+ "metadata": {
+ "name": "chapter no.10.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 10:Theory of Failures"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.10.10.1,Page No.401"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P_e=300 #N/mm**2 #Elastic Limit in tension\n",
+ "FOS=3 #Factor of safety\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "P=12*10**3 #N Pull \n",
+ "Q=6*10**3 #N #Shear force\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let d be the diameter of the shaft\n",
+ "\n",
+ "#Direct stress\n",
+ "#P_x=P*(pi*4**-1*d**3)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#P_x=48*10**3\n",
+ "\n",
+ "#Now shear stress at the centre of bolt\n",
+ "#q=4*3**-1*q_av\n",
+ "#After substituting values and further simplifying we get\n",
+ "#q=32*10**3*(pi*d**2)**-1\n",
+ "\n",
+ "#Principal stresses are\n",
+ "#P1=P_x*2**-1+((P_x*2**-1)**2+q**2)**0.5\n",
+ "#After substituting values and further simplifying we get\n",
+ "#p1=20371.833*(d**2)**-1\n",
+ "\n",
+ "#P2=P_x*2**-1-((P_x*2**-1)**2+q**2)**0.5\n",
+ "#After substituting values and further simplifying we get\n",
+ "#P2=-5092.984*(d**2)**-1\n",
+ "\n",
+ "#q_max=((P_x*2**-1)**2+q**2)**0.5\n",
+ "\n",
+ "#From Max Principal stress theory\n",
+ "#Permissible stress in Tension\n",
+ "P1=100 #N/mm**2 \n",
+ "d=(20371.833*P1**-1)**0.5\n",
+ "\n",
+ "#Max strain theory\n",
+ "#e_max=P1*E**-1-mu*P2*E**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#e_max=21899.728*(d**2*E)**-1\n",
+ "\n",
+ "#According to this theory,the design condition is\n",
+ "#e_max=P_e*(E*FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d2=(21899.728*3*300**-1)**0.5 #mm\n",
+ "\n",
+ "#Max shear stress theory\n",
+ "#e_max=shear stress at elastic*(FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d3=(12732.421*6*300**-1)**0.5 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Diameter of Bolt by:Max Principal stress theory\",round(d,2),\"mm\"\n",
+ "print\" :Max strain theory\",round(d2,2),\"mm\"\n",
+ "print\" :Max shear stress theory\",round(d3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of Bolt by:Max Principal stress theory 14.27 mm\n",
+ " :Max strain theory 14.8 mm\n",
+ " :Max shear stress theory 15.96 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.10.10.2.Page No.402"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "M=40*10**6 #N-mm #Bending moment\n",
+ "T=10*10**6 #N-mm #TOrque\n",
+ "mu=0.25 #Poissoin's ratio\n",
+ "P_e=200 #N/mm**2 #Stress at Elastic Limit\n",
+ "FOS=2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let d be the diameter of the shaft\n",
+ "\n",
+ "#Principal stresses are given by\n",
+ "\n",
+ "#P1=16*(pi*d**3)**-1*(M+(M**2+T**2)**0.5)\n",
+ "#After substituting values and further simplifying we get\n",
+ "#P1=4.13706*10**8*(d**3)**-1 ............................(1)\n",
+ "\n",
+ "#P2=16*(pi*d**3)**-1*(M-(M**2+T**2)**0.5)\n",
+ "#After substituting values and further simplifying we get\n",
+ "#P2=-6269718*(pi*d**3)**-1 ..............................(2)\n",
+ "\n",
+ "#q_max=(P1-P2)*2**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#q_max=2.09988*10**8*(d**3)**-1\n",
+ "\n",
+ "#Max Principal stress theory\n",
+ "#P1=P_e*(FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d=(4.13706*10**8*2*200**-1)**0.33333 #mm \n",
+ "\n",
+ "#Max shear stress theory\n",
+ "#q_max=shear stress at elastic limit*(FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d2=(2.09988*10**8*4*200**-1)**0.33333\n",
+ "\n",
+ "#Max strain energy theory\n",
+ "#P_3=0\n",
+ "#P1**2+P2**2-2*mu*P1*P2=P_e**2*(FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d3=(8.62444*10**12)**0.166666\n",
+ "\n",
+ "#Result\n",
+ "print\"Diameter of shaft according to:MAx Principal stress theory\",round(d,2),\"mm\"\n",
+ "print\" :Max shear stress theory\",round(d2,2),\"mm\"\n",
+ "print\" :Max strain energy theory\",round(d3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of shaft according to:MAx Principal stress theory 160.52 mm\n",
+ " :Max shear stress theory 161.33 mm\n",
+ " :Max strain energy theory 143.2 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.10.10.3,Page No.403"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "f_x=40 #N/mm**2 #Internal Fliud Pressure\n",
+ "d1=200 #mm #Internal Diameter\n",
+ "r1=d1*2**-1 #mm #Radius\n",
+ "q=300 #N/mm**2 #Tensile stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Lame's Equation we have,\n",
+ "\n",
+ "#Hoop Stress\n",
+ "#f_x=b*(x**2)**-1+a ..........................(1)\n",
+ "\n",
+ "#Radial Pressure\n",
+ "#p_x=b*(x**2)**-1-a .........................(2)\n",
+ "\n",
+ "#the boundary conditions are\n",
+ "x=d1*2**-1 #mm \n",
+ "#After sub values in equation 1 and further simplifying we get\n",
+ "#40=b*100**-1-a ..........................(3)\n",
+ "\n",
+ "#Max Principal stress theory\n",
+ "#q*(FOS)**-1=b*100**2+a ..................(4)\n",
+ "#After sub values in above equation and further simplifying we get\n",
+ "\n",
+ "#From Equation 3 and 4 we get\n",
+ "a=80*2**-1\n",
+ "#Sub value of a in equation 3 we get\n",
+ "b=(f_x+a)*100**2\n",
+ "\n",
+ "#At outer edge where x=r_0 pressure is zero\n",
+ "r_0=(b*a**-1)**0.5 #mm\n",
+ "\n",
+ "#thickness\n",
+ "t=r_0-r1 #mm\n",
+ "\n",
+ "#Max shear stress theory\n",
+ "P1=b*(100**2)**-1+a #Max hoop stress\n",
+ "P2=-40 #pressure at int radius (since P2 is compressive)\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(P1-P2)*2**-1\n",
+ "\n",
+ "#According max shear theory the design condition is\n",
+ "#q_max=P_e*2**-1*(FOS)**-1\n",
+ "#After sub values in equation we get and further simplifying we get\n",
+ "#80=b*(100**2)**-1+a\n",
+ "#After sub values in equation 1 and 3 and further simplifying we get\n",
+ "b2=120*100**2*2**-1\n",
+ "\n",
+ "#from equation(3)\n",
+ "a2=120*2**-1-a\n",
+ "\n",
+ "#At outer radius r_0,radial pressure=0\n",
+ "r_02=(b2*a2**-1)**0.5\n",
+ "\n",
+ "#thickness\n",
+ "t2=r_02-r1\n",
+ "\n",
+ "#Result\n",
+ "print\"Thickness of metal by:Max Principal stress theory\",round(t,2),\"mm\"\n",
+ "print\" :Max shear stress thoery\",round(t2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness of metal by:Max Principal stress theory 41.42 mm\n",
+ " :Max shear stress thoery 73.21 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.2_9.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.2_9.ipynb
new file mode 100644
index 00000000..c7846d57
--- /dev/null
+++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.2_9.ipynb
@@ -0,0 +1,2794 @@
+{
+ "metadata": {
+ "name": "chapter no.2.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 2:Simple Stresses And Strains"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.1,Page No.14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "P=45*10**3 #N #Load\n",
+ "E=200*10**3 #N/mm**2 #Modulus of elasticity of rod\n",
+ "L=500 #mm #Length of rod\n",
+ "d=20 #mm #Diameter of rod\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A=pi*d**2*4**-1 #mm**2 #Area of circular rod\n",
+ "p=P*A**-1 #N/mm**2 #stress\n",
+ "e=p*E**-1 #strain \n",
+ "dell_l=(P*L)*(A*E)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"The stress in bar due to Load is\",round(p,5),\"N/mm\"\n",
+ "print\"The strain in bar due to Load is\",round(e,5),\"N/mm\"\n",
+ "print\"The Elongation in bar due to Load is\",round(dell_l,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The stress in bar due to Load is 143.23945 N/mm\n",
+ "The strain in bar due to Load is 0.00072 N/mm\n",
+ "The Elongation in bar due to Load is 0.36 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.2,Page No.15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ " \n",
+ "A=15*0.75 #mm**2 #area of steel tape\n",
+ "P=100 #N #Force apllied\n",
+ "L=30*10**3 #mm #Length of tape\n",
+ "E=200*10**3 #N/m**2 #Modulus of Elasticity of steel tape\n",
+ "AB=150 #m #Measurement of Line AB \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "dell_l=P*L*(A*E)**-1 #mm #Elongation\n",
+ "l=L+dell_l*10**-3 #mm #Actual Length \n",
+ "AB1=AB*l*L**-1 #m Actual Length of AB\n",
+ "\n",
+ "#Result\n",
+ "print\"The Actual Length of Line AB is\",round(AB1,2),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Actual Length of Line AB is 150.0 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.3,Page No.15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Let y be the yield stress\n",
+ "\n",
+ "y=250 #N/mm**2 #yield stress\n",
+ "FOS=1.75 #Factor of safety\n",
+ "P=140*10**3 #N #compressive Load\n",
+ "D=101.6 #mm #External diameter\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "p=y*(FOS)**-1 #N/mm**2 #Permissible stress\n",
+ "A=P*p**-1 #mm**2 #Area of hollow tube\n",
+ "\n",
+ "#Let d be the internal diameter of tube\n",
+ "d=-((A*4*(pi)**-1)-D**2)\n",
+ "X=d**0.5\n",
+ "t=(D-X)*2**-1 #mm #Thickness of steel tube\n",
+ "\n",
+ "#result\n",
+ "print\"The thickness of steel tube is\",round(t,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The thickness of steel tube is 3.17 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.4,Page No.16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=25 #mm #diameter of steel\n",
+ "d2=18 #mm #Diameter at neck\n",
+ "L=200 #mm #length of stee\n",
+ "P=80*10**3 #KN #Load \n",
+ "P1=160*10**3 #N #Load at Elastic Limit\n",
+ "P2=180*10**3 #N #Max Load\n",
+ "L1=56 #mm #Total Extension\n",
+ "dell_l=0.16 #mm #Extension\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A=pi*d**2*4**-1 #Area of steel #mm**2\n",
+ "\n",
+ "p=P1*A**-1 #Stress at Elastic Limit #N/mm**2\n",
+ "Y=P*L*(A*dell_l)**-1 #Modulus of elasticity\n",
+ "\n",
+ "#Let % elongation be x\n",
+ "x=L1*L**-1*100 \n",
+ "\n",
+ "#Percentage reduction in area\n",
+ "#Let % A be a\n",
+ "a=((pi*4**-1*d**2)-(pi*4**-1*d2**2))*(pi*4**-1*d**2)**-1*100\n",
+ "\n",
+ "#Ultimate tensile stress\n",
+ "sigma=P2*A**-1 #N/mm**2\n",
+ "\n",
+ "#result\n",
+ "print\"Stress at Elastic limit is\",round(p,2),\"N/mm**2\"\n",
+ "print\"Young's Modulus is\",round(Y,2),\"N/mm**2\"\n",
+ "print\"Percentage Elongation is\",round(a,2)\n",
+ "print\"Percentage reduction in area is\",round(P2,2)\n",
+ "print\"Ultimate tensile stress\",round(sigma,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress at Elastic limit is 325.95 N/mm**2\n",
+ "Young's Modulus is 203718.33 N/mm**2\n",
+ "Percentage Elongation is 48.16\n",
+ "Percentage reduction in area is 180000.0\n",
+ "Ultimate tensile stress 366.69 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.5,Page No.16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "matplotlib.pyplot\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=20 #mm #Diameter of bar\n",
+ "d2=14.7 #mm #Diameter at neck \n",
+ "L=200 #mm #guage Length \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,10,20,30,40,50,60]\n",
+ "Y1=[0,32,64,95,127,160,190]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Extension in divisions\")\n",
+ "plt.ylabel(\"Load in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "A=pi*4**-1*d**2 #mm**2 #Area of Bar\n",
+ "A2=pi*4**-1*d2**2\n",
+ "\n",
+ "P=45 #KN #Load obtained from graph\n",
+ "dell=0.143 #mm #Divisions\n",
+ "\n",
+ "#Modulus of Elasticity\n",
+ "E=P*L*(dell*A)**-1 \n",
+ "\n",
+ "BL=93*10**3 #N #Breaking Load\n",
+ "\n",
+ "#Nominal stress at Breaking point\n",
+ "sigma=BL*A**-1 #KN/mm**2 \n",
+ "\n",
+ "#True stress at breaking Point\n",
+ "sigma1=BL*A2**-1\n",
+ "\n",
+ "#Percentage Elongation \n",
+ "dell_l=(A-A2)*A**-1*100\n",
+ "\n",
+ "#Result\n",
+ "print\"The Value of ELongation is\",round(E,2),\"N/mm**2\"\n",
+ "print\"The Nominal stress at the Breaking Point\",round(sigma,2),\"KN/mm**2\"\n",
+ "print\"The True stress at the Breaking Point\",round(sigma1,2),\"KN/mm**2\"\n",
+ "print\"The Percentage Reduction in Area is\",round(dell_l,2)"
+ ],
+ "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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gwAFOnDjBRx99xNatWys876vr+8tf/kKzZs2Ii4vDVHG/iq+u7bL09HT279/P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\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x555ced0>"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Value of ELongation is 200.33 N/mm**2\n",
+ "The Nominal stress at the Breaking Point 296.03 KN/mm**2\n",
+ "The True stress at the Breaking Point 547.97 KN/mm**2\n",
+ "The Percentage Reduction in Area is 45.98\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.6,Page No.19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=40*10**3 #N #Load \n",
+ "L1=160 #mm #Length of Bar1\n",
+ "L2=240 #mm #Length of bar2\n",
+ "L3=160 #mm #Length of bar3\n",
+ "d1=25 #mm #Diameter of Bar1\n",
+ "d2=20 #mm #diameter of bar2\n",
+ "d3=25 #mm #diameter of bar3\n",
+ "dell_l=0.285 #mm #Total Extension of bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "E=P*4*(dell_l*pi)**-1*(L1*(d1**2)**-1+L2*(d2**2)**-1+L3*(d3**2)**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"The Young's Modulus of the material\",round(E,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Young's Modulus of the material 198714.72 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.7,Page No.19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E1=2*10**5 #N/mm**2 #modulus of Elasticity of material1\n",
+ "E2=1*10**5 #N/mm**2 #modulus of Elasticity of material2\n",
+ "P=25*10**3 #N #Load \n",
+ "t=20 #mm #thickness of material\n",
+ "b1=40 #mm #width of material1\n",
+ "b2=30 #mm #width of material2\n",
+ "L1=500 #mm #Length of material1\n",
+ "L2=750 #mm #Length of material2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A1=b1*t #mm**2 #Area of materila1\n",
+ "A2=b2*t #mm**2 #Area of material2\n",
+ "\n",
+ "dell_l1=P*L1*(A1*E1)**-1 #Extension of Portion1\n",
+ "dell_l2=P*L2*(A2*E2)**-1 #Extension of portion2\n",
+ "\n",
+ "#Total Extension of Bar is\n",
+ "dell_l=dell_l1+dell_l2\n",
+ "\n",
+ "#Result\n",
+ "print\"The Total Extension of the Bar is\",round(dell_l,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Total Extension of the Bar is 0.39 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.8,Page No.20"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1000 #mm #Length of Bar\n",
+ "l=400 #mm #Length upto which bire is drilled \n",
+ "D=30 #mm #diameter of bar\n",
+ "d1=10 #mm #diameter of bore\n",
+ "P=25*10**3 #N #Load\n",
+ "dell_l=0.185 #mm #Extension of bar\n",
+ "\n",
+ "#Calculations \n",
+ "\n",
+ "L1=L-l #Length of bar above the bore\n",
+ "L2=400 #mm #Length of bore\n",
+ "\n",
+ "A1=pi*4**-1*D**2 #Area of bar\n",
+ "A2=pi*4**-1*(D**2-d1**2) #Area of bore\n",
+ "\n",
+ "E=P*dell_l**-1*(L1*A1**-1+L2*A2**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"The Modulus of ELasticity is\",round(E,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Modulus of ELasticity is 200735.96 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.11,Page No.23"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "t=10 #mm #Thickness of steel\n",
+ "b1=60 #mm #width of plate1\n",
+ "b2=40 #mm #width of plate2\n",
+ "P=60*10**3 #Load\n",
+ "L=600 #mm #Length of plate\n",
+ "E=2*10**5 #N/mm**2\n",
+ " \n",
+ "#Calculations\n",
+ "\n",
+ "#Extension of taperong bar of rectangular section\n",
+ "dell_l=P*L*(t*E*(b1-b2))**-1*log(b1*b2**-1)\n",
+ "\n",
+ "A_av=(b1*t+b2*t)*2**-1 #Average Area #mm**2\n",
+ "dell_l2=P*L*(A_av*E)**-1 \n",
+ "\n",
+ "#PErcentage Error\n",
+ "e=(dell_l-dell_l2)*(dell_l)**-1*100\n",
+ "\n",
+ "#Result\n",
+ "print\"The Percentage Error is\",round(e,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Percentage Error is 1.35\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.12,Page No.23"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1.5 #m #Length of steel bar\n",
+ "L1=1000 #m0 #Length of steel bar 1\n",
+ "L2=500 #m #Length of steel bar 2 \n",
+ "d1=40 #Diameter of steel bar 1\n",
+ "d2=20 #diameter of steel bar 2\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "P=160*10**3 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A1=pi*4**-1*d1**2 #Area of Portion 1\n",
+ "\n",
+ "#Extension of uniform Portion 1\n",
+ "dell_l1=P*L1*(A1*E)**-1 #mm\n",
+ "\n",
+ "#Extension of uniform Portion 2\n",
+ "dell_l2=4*P*L2*(pi*d1*d2*E)**-1 #mm\n",
+ "\n",
+ "#Total Extension of Bar\n",
+ "dell_l=dell_l1+dell_l2\n",
+ "\n",
+ "#Result\n",
+ "print\"The Elongation of the Bar is\",round(dell_l,2),\"mm\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Elongation of the Bar is 1.27 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.14,Page No.25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Portion AB\n",
+ "L_AB=600 #mm #Length of AB\n",
+ "A_AB=40*40 #mm**2 #Cross-section Area of AB\n",
+ "\n",
+ "#Portion BC\n",
+ "L_BC=800 #mm #Length of BC\n",
+ "A_BC=30*30 #mm #Length of BC\n",
+ "\n",
+ "#Portion CD\n",
+ "L_CD=1000 #mm #Length of CD\n",
+ "A_CD=20*20 #mm #Area of CD\n",
+ "\n",
+ "P1=80*10**3 #N #Load1\n",
+ "P2=60*10**3 #N #Load2\n",
+ "P3=40*10**3 #N #Load3\n",
+ "\n",
+ "E=2*10**5 #Modulus of Elasticity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "P4=P1-P2+P3 #Load4\n",
+ "\n",
+ "#Now Force in AB\n",
+ "F_AB=P1\n",
+ "\n",
+ "#Force in BC\n",
+ "F_BC=P1-P2\n",
+ "\n",
+ "#Force in CD\n",
+ "F_CD=P4\n",
+ "\n",
+ "#Extension of AB\n",
+ "dell_l_AB=F_AB*L_AB*(A_AB*E)**-1\n",
+ "\n",
+ "#Extension of BC\n",
+ "dell_l_BC=F_BC*L_BC*(A_BC*E)**-1\n",
+ "\n",
+ "#Extension of CD\n",
+ "dell_l_CD=F_CD*L_CD*(A_CD*E)**-1\n",
+ "\n",
+ "#Total Extension\n",
+ "dell_l=dell_l_AB+dell_l_BC+dell_l_CD\n",
+ "\n",
+ "#Result\n",
+ "print\"The Total Extension in Bar is\",round(dell_l,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Total Extension in Bar is 0.99 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.15,Page No.26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=800 #mm #Length of bar\n",
+ "F1=30*10**3 #N #Force acting on the bar\n",
+ "F2=60*10**3 #N #force acting on the bar\n",
+ "L=800 #mm #Length of bar\n",
+ "d=25 #mm #diameter of bar \n",
+ "L_AC=275 #mm #Length of AC\n",
+ "L_CD=150 #mm #Length of CD\n",
+ "L_DB=375 #mm #Length of DB\n",
+ "E=2*10**5 #Pa #Modulus of elasticity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P be the Reaction on tne Bar from support at A\n",
+ "\n",
+ "#Shortening of Portion AC\n",
+ "#dell_l_AC1=P*L_AC*(A*E)**-1\n",
+ "\n",
+ "#Shortening of Portion CD\n",
+ "#dell_l_CD1=(30+P)*L_CD*(A*E)**-1\n",
+ "\n",
+ "#Extension of Portion DB\n",
+ "#dell_l_DB1=(30-P)*L_DB*(A*E)**-1\n",
+ "\n",
+ "#Total Extensions=1*(A*E)**-1*(P*L_AC-(30+P)*L_CD+(30-P)*L_DB)\n",
+ "#As Supports are unyielding,Total Extensions=0\n",
+ "\n",
+ "#After substituting values in above equation and Further simplifying we get\n",
+ "P=(30*375-150*30)*800**-1\n",
+ "\n",
+ "#Reaction of support A\n",
+ "R_A=P\n",
+ "\n",
+ "#Reaction of support B\n",
+ "R_B=30-P\n",
+ "\n",
+ "#Cross-sectional Area\n",
+ "A=pi*4**-1*d**2\n",
+ "\n",
+ "#Stress in Portion AC\n",
+ "sigma1=P*10**3*A**-1 #N/mm**2\n",
+ "\n",
+ "#Stress in Portion CD\n",
+ "sigma2=(30+P)*10**3*A**-1 #N/mm**2\n",
+ "\n",
+ "#Stress in Portion DB\n",
+ "sigma3=(30-P)*10**3*A**-1 #N/mm**2\n",
+ "\n",
+ "#Shortening of Portion AC\n",
+ "dell_l_AC2=P*10**3*L_AC*(A*E)**-1 #mm \n",
+ "\n",
+ "#Shortening of Portion CD\n",
+ "dell_l_CD2=(30+P)*10**3*L_CD*(A*E)**-1 #mm \n",
+ "\n",
+ "#Extension of Portion DB\n",
+ "dell_l_DB2=(30-P)*10**3*L_DB*(A*E)**-1 #mm \n",
+ "\n",
+ "#result\n",
+ "print\"The Reactios at two Ends are:R_A\",round(R_A,2),\"KN\"\n",
+ "print\" :R_B\",round(R_B,2),\"KN\"\n",
+ "print\"Stress in Portion AC\",round(sigma1,2),\"N/mm**2\"\n",
+ "print\"Stress in Portion CD\",round(sigma2,2),\"N/mm**2\"\n",
+ "print\"Stress in Portion DB\",round(sigma3,2),\"N/mm**2\"\n",
+ "print\"Shortening of Portion AC\",round(dell_l_AC2,3),\"mm\"\n",
+ "print\"Shortening of Portion CD\",round(dell_l_CD2,3),\"mm\"\n",
+ "print\"Shortening of Portion DB\",round(dell_l_DB2,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Reactios at two Ends are:R_A 8.44 KN\n",
+ " :R_B 21.56 KN\n",
+ "Stress in Portion AC 17.19 N/mm**2\n",
+ "Stress in Portion CD 78.3 N/mm**2\n",
+ "Stress in Portion DB 43.93 N/mm**2\n",
+ "Shortening of Portion AC 0.024 mm\n",
+ "Shortening of Portion CD 0.059 mm\n",
+ "Shortening of Portion DB 0.082 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.19,Page No.29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ " \n",
+ "h=4 #m #height of Pillars\n",
+ "P=20 #KN #Load at M\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_A,P_B,P_C,P_D be the forces introduced in the Pillars\n",
+ "#Sun of All Vertical Forces\n",
+ "#P_A+P_B+P_C+P_D=20 ....................(1)\n",
+ "\n",
+ "#Sum of moment about AB, we get\n",
+ "#P_D+P_C=12 ....................(2)\n",
+ "\n",
+ "#Sum of Moment about AD\n",
+ "#P_C+P_B=8 ....................(3)\n",
+ "\n",
+ "#Let dell_l_A,dell_l_B,dell_l-C,dell_l_D be the deformations of Pillars A,B,C,D respectively\n",
+ "#Diagonals AC and BD will remain straight Lines even after the Load is applied.\n",
+ "#Deflection of central Point is given by (dell_l_A+dell_l_C)*2**-1 & (dell_l_B+dell_l_D)*2**-1\n",
+ "\n",
+ "#dell_l_A+dell_l_C=dell_l_B+ell_l_D\n",
+ "#P_A*L*(A*E)**-1+P_C*L*(A*E)**-1=P_B*L*(A*E)**-1+P_D*L*(A*E)**-1\n",
+ "\n",
+ "#Since Pillars are identical in Length,cross-sectional area,material Property\n",
+ "#P_A+P_C=P_B+P_D ..............(4)\n",
+ "\n",
+ "#From Equations 1 and 4 we get\n",
+ "#P_B+P_D=10 ....................(5)\n",
+ " \n",
+ "#Substracting Equation 3 from Equation 2 we get\n",
+ "#P_D-P_B=4 ....................(6)\n",
+ "\n",
+ "#Adding Equation 5 and 6 we get\n",
+ "\n",
+ "P_D=14*2**-1\n",
+ "P_C=12-P_D\n",
+ "P_B=8-P_C\n",
+ "\n",
+ "#Now substituting values of P_B,P_C,P_D in equation1 we get\n",
+ "P_A=20-(P_B+P_C+P_D)\n",
+ "\n",
+ "#Result\n",
+ "print\"The Forces Developed in the Pillars are:P_A\",round(P_A,2),\"KN\"\n",
+ "print\" :P_B\",round(P_B,2),\"KN\"\n",
+ "print\" :P_C\",round(P_C,2),\"KN\"\n",
+ "print\" :P_D\",round(P_D,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Forces Developed in the Pillars are:P_A 5.0 KN\n",
+ " :P_B 3.0 KN\n",
+ " :P_C 5.0 KN\n",
+ " :P_D 7.0 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.20,Page No.31"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "sigma=150 #N/mm**2 #Stress\n",
+ "P=40*10**3 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt P_A.P_B,P_C,P_D be the forces developed in wires A,B,C,D respectively\n",
+ "\n",
+ "#Let sum of all Vertical Forces=0\n",
+ "#P_A+P_B+P_C+P_D=40 ..........................(1)\n",
+ "\n",
+ "#Let x be the distance between each wires\n",
+ "#sum of all moments=0\n",
+ "#P_B*x+P_C*2*x+P_D*3*x=40*2*x\n",
+ "\n",
+ "#After further simplifying we get\n",
+ "#P_B+2*P_C+3*P_D=80 ..........................(2)\n",
+ "\n",
+ "#As the equations of statics ae not enough to find unknowns,Consider compatibilit Equations\n",
+ "\n",
+ "#Let dell_l be the increse in elongation of wire\n",
+ "\n",
+ "#dell_l_B=dell_l_A+dell_l\n",
+ "#dell_l_C=dell_l_A+2*dell_l\n",
+ "#dell_l_D=dell_l_A+3*dell_l\n",
+ "\n",
+ "#Let P1 be the force required for the Elongation of wires,then\n",
+ "#P_B=P_A+P1 ]\n",
+ "#P_C=P_A+2*P1 ]\n",
+ "#P_D=P_A+3*P1 ] ................................(3) \n",
+ "\n",
+ "#from Equation (3) and (1) we get\n",
+ "#2*P_A+3*P1=20 ................................(4)\n",
+ "\n",
+ "#from Equation (3) and (2) we get\n",
+ "#6*P_A+14*P1=80 \n",
+ "\n",
+ "#subtracting 3 times equation (4) from (3) we get\n",
+ "P1=20*5**-1\n",
+ "\n",
+ "#from Equation 4 we get\n",
+ "P_A=(80-14*P1)*6**-1\n",
+ "P_B=P_A+P1\n",
+ "P_C=P_A+2*P1 \n",
+ "P_D=P_A+3*P1\n",
+ "\n",
+ "#Let d be the diameter required,then\n",
+ "d=(P_D*10**3*4*(pi*150)**-1)**0.5\n",
+ "\n",
+ "#result\n",
+ "print\"The Required Diameter is\",round(d,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Required Diameter is 11.65 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.21,Page No.32"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=20*10**3 #N #Load\n",
+ "d=6 #mm #diameter of wire\n",
+ "E=2*10**5 #N/mm**2 \n",
+ "L_BO=4000 #mm #Length of BO\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let theta be the angle between OA and OB and also between OC and OB\n",
+ "theta=30\n",
+ "\n",
+ "#Let P_OA,P_OB,P_OC be the Forces introduced in wires OA,OB,OC respectively\n",
+ "#Due to symmetry P_OA=P_OC (same angles)\n",
+ "\n",
+ "#Sum of all Vertical Forces=0\n",
+ "#P_OA*cos(theta)+P_OB+P_OC*cos(theta)=P\n",
+ "\n",
+ "#After further simplifyinf we get\n",
+ "#2*P_OA*cos(theta)+P_OB=20 ...............(1)\n",
+ "\n",
+ "#Let oo1 be the extension of BO\n",
+ "#oo1=L_A1o1*(cos(theta))**-1\n",
+ "\n",
+ "#From relation we get\n",
+ "#P_OB*L_BO=P_OA*L_AO*(cos(theta))**-1\n",
+ "\n",
+ "#But L_AO=L_BO*(cos(theta))**-1\n",
+ "\n",
+ "#After substituting value of L_AO in above equation we get\n",
+ "#P_OB=0.75*P_OA .......................(2)\n",
+ "\n",
+ "#substituting in Equation 1 we get\n",
+ "#2*P_OA*cos(theta)+0.75*P_OA=20\n",
+ "\n",
+ "P_OA=20*(2*cos(theta*pi*180**-1)+0.75)**-1\n",
+ "\n",
+ "P_OB=0.75*P_OA\n",
+ "\n",
+ "A=pi*4**-1*d**2 \n",
+ "\n",
+ "#Vertical displacement of Load\n",
+ "dell_l_BO=P_OB*10**3*L_BO*(A*E)**-1\n",
+ " \n",
+ "#Result\n",
+ "print\"Forces in each wire is:P_OA\",round(P_OA,2),\"KN\"\n",
+ "print\" :P_OB\",round(P_OB,2),\"KN\"\n",
+ "print\"Vertical displacement of Loadis\",round(dell_l_BO,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Forces in each wire is:P_OA 8.06 KN\n",
+ " :P_OB 6.04 KN\n",
+ "Vertical displacement of Loadis 4.27 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.22,Page No.34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_s=L_a=L=500 #mm #Length of bar\n",
+ "A_a=50*20 #mm #Area of aluminium strip\n",
+ "A_s=50*15 #mm #Area of steel strip\n",
+ "P=50*10**3 #N #Load\n",
+ "E_a=1*10**5 #N/mm**2 #Modulus of aluminium \n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of steel\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_a and P_s br the Load shared by aluminium and steel strip\n",
+ "#P_a+P_s=P ..................(1)\n",
+ "\n",
+ "#For compatibility condition,dell_l_a=dell_l_s\n",
+ "#P_a*L_a*(A_a*E_a)**-1=P_s*L_s*(A_s*E_s)**-1 .....(2)\n",
+ "\n",
+ "#As L_a=L_s we get\n",
+ "#P_s=1.5*P_a .................(3)\n",
+ " \n",
+ "#From Equation 1 and 2 we get\n",
+ "P_a=P*2.5**-1\n",
+ "\n",
+ "#Substituting in equation 1 we get\n",
+ "P_s=P-P_a\n",
+ "\n",
+ "#stress in aluminium strip \n",
+ "sigma_a=P_a*A_a**-1\n",
+ "\n",
+ "#stress in steel strip\n",
+ "sigma_s=P_s*A_s**-1\n",
+ "\n",
+ "#Now from the relation we get\n",
+ "dell_l_a=dell_l_s=P_s*L_s*(A_s*E_s)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Stress in Aluminium strip is\",round(sigma_a,2),\"N/mm**2\"\n",
+ "print\"Stress in steel strip is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"The Extension of the bar is\",round(dell_l_s,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in Aluminium strip is 20.0 N/mm**2\n",
+ "Stress in steel strip is 40.0 N/mm**2\n",
+ "The Extension of the bar is 0.1 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.23,Page No.35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D_s=20 #mm #Diameter of steel\n",
+ "D_Ci=20 #mm #Internal Diameter of Copper\n",
+ "t=5 #mm #THickness of copper bar\n",
+ "P=100*10**3 #N #Load\n",
+ "E_s=2*10**5 #N/mm**2 #modulus of elasticity of steel\n",
+ "E_c=1.2*10**5 #N/mm**2 #Modulus of Elasticity of Copper\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A_s=pi*4**-1*D_s**2 #mm**2 #Area of steel\n",
+ "D_Ce=D_s+2*t #mm #External Diameterof Copper Tube\n",
+ "\n",
+ "A_c=pi*4**-1*(D_Ce**2-D_Ci**2) #mm**2 #Area of Copper\n",
+ "\n",
+ "#From static Equilibrium condition\n",
+ "#Let P_s and P_c be the Load shared by steel and copper in KN\n",
+ "#P_s+P_c=100 ....................................(1)\n",
+ "\n",
+ "#From compatibility Equation,dell_l_s=dell_l_c\n",
+ "#P_s*L*(A_s*E_s)**-1=P_c*L*(A_c*E_c)**-1\n",
+ "\n",
+ "#Substituting values in above Equation we get\n",
+ "#P_s=1.3333*P_C \n",
+ "\n",
+ "#Now Substituting value of P_s in Equation (1),we get\n",
+ "P_c=100*2.3333**-1 #KN\n",
+ "P_s=100-P_c #KN\n",
+ "\n",
+ "#Stress in steel\n",
+ "sigma_s=P_s*10**3*A_s**-1 #N/mm**2 \n",
+ "\n",
+ "#Stress in copper\n",
+ "sigma_c=P_c*10**3*A_c**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Two material are:sigma_s\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\" :sigma_c\",round(sigma_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses Developed in Two material are:sigma_s 181.89 N/mm**2\n",
+ " :sigma_c 109.14 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.24,Page No.36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "A_C=230*400 #mm #Area of column\n",
+ "D_s=12 #mm #Diameter of steel Bar\n",
+ "P=600*10**3 #N #Axial compression\n",
+ "#E_s*E_c=18.67\n",
+ "n=8 #number of steel Bars\n",
+ "\n",
+ "#Calculations \n",
+ "\n",
+ "A_s=pi*4**-1*D_s**2*n #Area of steel #mm**2 \n",
+ "A_c=A_C-A_s #mm**2 #Area of concrete\n",
+ "\n",
+ "#From static Equilibrium condition\n",
+ "#P_s+P_c=600 .........(1)\n",
+ "\n",
+ "#Now from compatibility Equation dell_l_s=dell_l_c we get,\n",
+ "#P_s*L*(A_s*E_s)**-1=P_c*L*(A_c*E_c)**-1\n",
+ "\n",
+ "#Substituting values in above Equation we get\n",
+ "#P_s=0.1854*P_c\n",
+ "\n",
+ "#Now Substituting value of P_s in Equation (1),we get\n",
+ "P_c=600*1.1854**-1\n",
+ "P_s=600-P_c\n",
+ "\n",
+ "#Stress in steel\n",
+ "sigma_s=P_s*10**3*A_s**-1 #N/mm**2\n",
+ "\n",
+ "#Stress in copper\n",
+ "sigma_c=P_c*10**3*A_c**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Two material are:sigma_s\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\" :sigma_c\",round(sigma_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses Developed in Two material are:sigma_s 103.72 N/mm**2\n",
+ " :sigma_c 5.56 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.25,Page No.36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=200*10**3 #N #Load\n",
+ "A_a=1000 #mm**2 #Area of Aluminium\n",
+ "A_s=800 #mm**2 #Area of steel\n",
+ "E_a=1*10**5 #N/mm**2 #Modulus of Elasticity of Aluminium\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of ELasticity of steel\n",
+ "sigma_a1=65 #N/mm**2 #stress in aluminium\n",
+ "sigma_s1=150 #N/mm**2 #Stress in steel\n",
+ "\n",
+ "#Calculations \n",
+ "\n",
+ "#Let P_a and P_s be the force in aluminium and steel pillar respectively\n",
+ "\n",
+ "#Now,sum of forces in Vertical direction we get\n",
+ "#2*P_a+P_s=200 .........................................(1)\n",
+ "\n",
+ "#By compatibility Equation dell_l_s=dell_l_a we get\n",
+ "#P_s=1.28*P_a ..........................................(2)\n",
+ "\n",
+ "#Now substituting value of P_s in Equation 1 we get\n",
+ "P_a=200*3.28**-1 #KN\n",
+ "P_s=200-2*P_a #KN\n",
+ "\n",
+ "#Stress developed in aluminium\n",
+ "sigma_a=P_a*10**3*A_a**-1 #N/mm**2 \n",
+ "\n",
+ "#Stress developed in steel\n",
+ "sigma_s=P_s*10**3*A_s**-1 #N/mm**2 \n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#Let sigma_a1 and sigma_s1 be the stresses in Aluminium and steel due to Additional LOad\n",
+ "\n",
+ "P_a1=sigma_a1*A_a #Load carrying capacity of aluminium\n",
+ "P_s1=1.28*P_a1\n",
+ "\n",
+ "#Total Load carrying capacity \n",
+ "P1=2*P_a1+P_s1 #N \n",
+ "\n",
+ "P_s2=sigma_s1*A_s #Load carrying capacity of steel\n",
+ "P_a2=P_s2*1.28**-1\n",
+ "\n",
+ "#Total Load carrying capacity\n",
+ "P2=2*P_a2+P_s2\n",
+ "\n",
+ "#Additional Load\n",
+ "P3=P1-P\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Each Pillar is:sigma_a\",round(sigma_a,2),\"N/mm**2\"\n",
+ "print\" :sigma_s\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"Additional Load taken by pillars is\",round(P3,2),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses Developed in Each Pillar is:sigma_a 60.98 N/mm**2\n",
+ " :sigma_s 97.56 N/mm**2\n",
+ "Additional Load taken by pillars is 13200.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.26,Page No.37"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=500 #mm #Length of assembly\n",
+ "D=16 #mm #Diameter of steel bolt\n",
+ "Di=20 #mm #internal Diameter of copper tube\n",
+ "Do=30 #mm #External Diameter of copper tube\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity of steel\n",
+ "E_c=1.2*10**5 #N/mm**2 #Modulus of Elasticity of copper\n",
+ "p=2 #mm #Pitch of nut\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_s be the Force in bolt and P_c be the FOrce in copper tube\n",
+ "#P_s=-P_s\n",
+ "\n",
+ "dell=1*4**-1*2 #Quarter turn of nut total movement\n",
+ "\n",
+ "#dell=dell_s+dell_c\n",
+ " \n",
+ "#Area of steel\n",
+ "A_s=pi*4**-1*D**2\n",
+ "\n",
+ "#Area of copper\n",
+ "A_c=pi*4**-1*(Do**2-Di**2)\n",
+ "\n",
+ "#dell=P*L*(A_s*E_s)**-1+P*L*(A_c*E_c)**-1\n",
+ "P=dell*(1*(A_s*E_s)**-1+1*(A_c*E_c)**-1)**-1*L**-1 #LOad\n",
+ "\n",
+ "P_s=P*A_s**-1\n",
+ "P_c=P*A_c**-1\n",
+ "\n",
+ "#result\n",
+ "print\"stress introduced in bolt is\",round(P_s,2),\"N/mm**2\"\n",
+ "print\"stress introduced in tube is\",round(P_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "stress introduced in bolt is 107.91 N/mm**2\n",
+ "stress introduced in tube is 55.25 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.27,Page No.39"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=20 #mm #Diameter of Bolts\n",
+ "Di=25 #m #internal Diameter\n",
+ "t=10 #mm #Thickness of bolt\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "E_c=1.2*10**5 #N/mm**2 #Modulus of copper\n",
+ "p=3 #mm #Pitch\n",
+ "theta=30 #degree\n",
+ "L_c=500 #Lengh of copper \n",
+ "L_s=600 #Length of steel\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_s be the Force in each bolt and P_c be the FOrce in copper tube\n",
+ "#From Static Equilibrium condition\n",
+ "#P_c=2*P_s\n",
+ "\n",
+ "#As nut moves by 60 degree.If nut moves by 360 degree its Longitudinal movement is by 3 mm\n",
+ "dell=theta*360**-1*p\n",
+ "\n",
+ "#From Compatibility Equaton we get\n",
+ "#dell=dell_c+dell_s\n",
+ "\n",
+ "\n",
+ "A_s=pi*4**-1*Di**2 #mm**2 #Area of steel\n",
+ "A_c=pi*4**-1*(45**2-Di**2) #mm**2 #Area of copper\n",
+ "\n",
+ "#Force introduced in steel\n",
+ "P_s=0.5*(2*L_c*(A_c*E_c)**-1+L_s*(A_s*E_s)**-1)**-1 #N\n",
+ "P_s2=P_s*A_s**-1\n",
+ "\n",
+ "#Force introduced in copper \n",
+ "P_c=2*P_s*A_c**-1 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress introduced in bolt is\",round(P_s2,2),\"N/mm**2\"\n",
+ "print\"stress introduced in tube is\",round(P_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress introduced in bolt is 74.4 N/mm**2\n",
+ "stress introduced in tube is 66.43 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.28,Page No.40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=9 #m #Length of rigid bar\n",
+ "L_b=3000 #Length of bar\n",
+ "A_b=1000 #mm**2 #Area of bar\n",
+ "E_b=1*10**5 #N/mm**2 #Modulus of Elasticity of brasss bar\n",
+ "L_s=5000 #mm #Length of steel bar\n",
+ "A_s=445 #mm**2 #Area of steel bar\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of elasticity of steel bar\n",
+ "P=3000 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From static equilibrium Equation of the rod after appliying Load is\n",
+ "#P_b+P_s=P ......................(1)\n",
+ "\n",
+ "#P_b=1.8727*P_s ..................(2)\n",
+ "\n",
+ "#NOw substituting equation 2 in equation 1 we get\n",
+ "P_s=P*2.8727**-1\n",
+ "P_b=P-P_s\n",
+ "\n",
+ "d=P_s*L*P**-1 \n",
+ "\n",
+ "#Result\n",
+ "print\"Distance at which Load applied even after which bar remains horizontal is\",round(d,2),\"m\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance at which Load applied even after which bar remains horizontal is 3.13 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.29,Page No.41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "A_b=1000 #MM**2 #Area of brass bar\n",
+ "E_b=1*10**5 #N/mm**2 #Modulus of Elasticity of brass\n",
+ "A_s=600 #N/mm**2 #Area of steel rod\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of eLasticity of steel bar\n",
+ "P=10*10**2 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_b be the tensile force in brass bar and P_s be the compressive force in steel bar\n",
+ "#Now taking moment about A we get static Equilibrium condition as\n",
+ "#P_b+2*P_s=27500 ......................................(1)\n",
+ "\n",
+ "#Now from deformed shape we get\n",
+ "#dell_s=2*dell_b\n",
+ "\n",
+ "#P_s*L_s*(A_s*E_s)**-1=P_b*L_b*(A_b*E_b)**-1\n",
+ "#Further simplifying we get\n",
+ "#P_s=1.2*P_b .........................................(2)\n",
+ "\n",
+ "#Now substituting equation 1 in equation 2 we get\n",
+ "P_b=27500*3.4**-1\n",
+ "P_s=1.2*P_b \n",
+ "\n",
+ "#Tensile stress in brass bar \n",
+ "sigma_b=P_b*A_b**-1\n",
+ "\n",
+ "#compressive stress in steel bar\n",
+ "sigma_s=P_s*A_s**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Compressive Stress in Bar is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"tensile Stress in Bar is\",round(sigma_b,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Compressive Stress in Bar is 16.18 N/mm**2\n",
+ "tensile Stress in Bar is 8.09 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.30,Page No.44"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=12.6 #m #Length of rail\n",
+ "t1=24 #Degree celsius\n",
+ "t2=44 #degree celsius\n",
+ "alpha=12*10**-6 #Per degree celsius\n",
+ "E=2*10**5 #N/mm**2 #Modulus of ELasticity\n",
+ "gamma=2 #mm #Gap provided for Expansion\n",
+ "sigma=20 #N/mm**2 #Stress\n",
+ "\n",
+ "#Calculations \n",
+ "\n",
+ "t=t2-t1 #Temperature Difference\n",
+ "\n",
+ "#Free Expansion of the rails\n",
+ "dell=alpha*t*L*1000 #mm \n",
+ "\n",
+ "#When no expansion joint is provided then\n",
+ "p=dell*E*(L*10**3)**-1\n",
+ "\n",
+ "#When a gap of 2 mm is provided,then free expansion prevented is\n",
+ "dell_1=dell-gamma\n",
+ "p2=dell_1*E*(L*10**3)**-1\n",
+ "\n",
+ "#When stress is developed,then gap left is\n",
+ "gamma2=-(sigma*L*10**3*E**-1-dell)\n",
+ "\n",
+ "#Result\n",
+ "print\"The minimum gap between the two rails is\",round(dell,2),\"mm\"\n",
+ "print\"Thermal Developed in the rials if:No expansionn joint is provided:p\",round(p,2),\"N/mm**2\"\n",
+ "print\" :If a gap of is provided then :p2\",round(p2,2),\"N/mm**2\"\n",
+ "print\"When stress is developed gap left between the rails is\",round(gamma2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum gap between the two rails is 3.02 mm\n",
+ "Thermal Developed in the rials if:No expansionn joint is provided:p 48.0 N/mm**2\n",
+ " :If a gap of is provided then :p2 16.25 N/mm**2\n",
+ "When stress is developed gap left between the rails is 1.76 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.31,Page No.45"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "t=20 #degree celsius\n",
+ "E_a=70*10**9 #N/mm**2 #Modulus of Elasticicty of aluminium\n",
+ "alpha_a=11*10**-6 #per degree celsius #Temperature coeff of aluminium\n",
+ "alpha_s=12*10**-6 #Per degree celsius #Temperature coeff of steel\n",
+ "L_a=1000 #mm #Length of aluminium \n",
+ "L_s=3000 #mm #Length of steel\n",
+ "E_a=7*10**4 #N/mm**2 #Modulus of Elasticity of aluminium\n",
+ "E_s=2*10**5 #N/mm*2 #Modulus of Elasticity of steel\n",
+ "A_a=600 #mm**2 #Area of aluminium\n",
+ "A_s=300 #mm**2 #Area of steel\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Free Expansion \n",
+ "dell=alpha_a*t*L_a+alpha_s*t*L_s\n",
+ " \n",
+ "#support Reaction\n",
+ "P=dell*(L_a*(A_a*E_a)**-1+L_s*(A_s*E_s)**-1)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Reaction at support is\",round(P,2),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Reaction at support is 12735.48 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.33,Page No.48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=25 #mm #Diameter of Brass\n",
+ "De=50 #mm #External Diameter of steel tube\n",
+ "Di=25 #mm #Internal Diameter of steel tube\n",
+ "L=1.5 #m #Length of both bars\n",
+ "t1=30 #degree celsius #Initial Temperature\n",
+ "t2=100 #degree celsius #final Temperature\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of ELasticity of steel bar\n",
+ "E_b=1*10**5 #N/mm**2 #Modulus of Elasticity of brass bar\n",
+ "alpha_s=11.6*10**-6 #Temperature Coeff of steel\n",
+ "alpha_b=18.7*10**-6 #Temperature coeff of brass bar\n",
+ "d=20 #mm #diameter of pins\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "t=t2-t1 #Temperature Difference\n",
+ "A_s=pi*4**-1*(De**2-Di**2) #mm**2 #Area of steel\n",
+ "A_b=pi*4**-1*D**2 #mm**2 #Area of brass\n",
+ "\n",
+ "#Let P_b be the tensile force in brass bar and P_s be the compressive force in steel bar\n",
+ "#But from Equilibrium of Forces \n",
+ "#P_b=P_s=P\n",
+ "\n",
+ "#Let dell=dell_s+dell_b\n",
+ "dell=(alpha_b-alpha_s)*t*L*1000\n",
+ "\n",
+ "P=dell*(1*(A_s*E_s)**-1+1*(A_b*E_b)**-1)**-1*(L*1000)**-1\n",
+ "P_b=P_s=P\n",
+ "\n",
+ "#Stress in steel\n",
+ "sigma_s=P_s*A_s**-1\n",
+ "\n",
+ "#Stress in Brass\n",
+ "sigma_b=P_b*A_b**-1\n",
+ "\n",
+ "#Area of Pins\n",
+ "A_p=pi*4**-1*d**2\n",
+ "\n",
+ "#Since,the force is resisted by two cross section of pins\n",
+ "tou=P*(2*A_p)**-1\n",
+ " \n",
+ "#Result\n",
+ "print\"Stress in steel bar is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"Stress in Brass bar is\",round(sigma_b,2),\"N/mm**2\"\n",
+ "print\"Shear Stresss induced in pins is\",round(tou,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in steel bar is 14.2 N/mm**2\n",
+ "Stress in Brass bar is 42.6 N/mm**2\n",
+ "Shear Stresss induced in pins is 33.28 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.34,Page No.49"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b_s=60 #mm #width of steel Bar\n",
+ "t_s=10 #mm #thickness of steel Bar\n",
+ "b_c=40 #mm #width of copper bar\n",
+ "t_c=5 #mm #thickness of copper bar\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity of steel bar\n",
+ "E_c=1*10**5 #N/mm**2 #Modulus of Elasticity of copper bar\n",
+ "alpha_s=12*10**-6 #Per degree celsius #Temperature coeff of steel bar\n",
+ "alpha_c=17*10**-6 #Per degree celsius #Temperature coeff of copper bar\n",
+ "L_s=L_c=L=1000 #mm #Length of bar\n",
+ "t=80 #degree celsius\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A_s=b_s*t_s #Area of steel bar\n",
+ "A_c=b_c*t_c #Area of copper bar\n",
+ "\n",
+ "#Let P_s be the tensile force in steel bar and P_c be the compressive force in copper bar\n",
+ "#The equilibrium of forces gives \n",
+ "#P_s=2*P_c\n",
+ "\n",
+ "#Let dell=dell_s+dell_b\n",
+ "dell=(alpha_c-alpha_s)*t\n",
+ "\n",
+ "P_c=dell*(2*(A_s*E_s)**-1+1*(A_c*E_c)**-1)**-1\n",
+ "P_s=2*P_c\n",
+ "\n",
+ "#Stress in copper \n",
+ "sigma_c=P_c*A_c**-1\n",
+ "\n",
+ "#Stress in steel \n",
+ "sigma_s=P_s*A_s**-1\n",
+ "\n",
+ "#Change in Length of bar\n",
+ "dell_2=alpha_s*t*L+P_s*L_s*(A_s*E_s)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Stress in copper is\",round(sigma_c,2),\"N/mm**2\"\n",
+ "print\"Stress in steel is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"the change in Length is\",round(dell_2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in copper is 30.0 N/mm**2\n",
+ "Stress in steel is 20.0 N/mm**2\n",
+ "the change in Length is 1.06 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.35,Page No.50"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=2*10**5 #N #Weight\n",
+ "L=1 #m #Length of each rod\n",
+ "A_c=A_s=A=500 #mm**2 #Area of each rod\n",
+ "t=40 #degree celsius #temperature\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity of steel rod\n",
+ "E_c=1*10**5 #N/mm**2 #modulus of Elastictiy of copper rod\n",
+ "alpha_s=1.2*10**-5 #Per degree Celsius #temp coeff of steel rod\n",
+ "alpha_c=1.8*10**-5 #Per degree Celsius #Temp coeff of copper rod\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_s be the force in each one of the copper rods and P_s be the force in steel rod\n",
+ "#2*P_c+P_s=P .....................(1)\n",
+ "\n",
+ "#Extension of copper bar=Extension of steel bar\n",
+ "#P_s*L*(A_s*E_s)**-1=P_c*L*(A_c*E_c)**-1\n",
+ "#after simplifying above equation we get\n",
+ "#P_s=2*P_c ........................(2)\n",
+ "\n",
+ "#Now substituting value of P_s in Equation 1 we get\n",
+ "P_c=P*4**-1\n",
+ "P_s=2*P_c\n",
+ "\n",
+ "#Now EXtension due to copper Load\n",
+ "dell_1=P_c*L*1000*(A_c*E_c)**-1\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#Due to rise of temperature of40 degree celsius\n",
+ "\n",
+ "#As bars are rigidly joined,let P_c1 be the compressive forccesdeveloped in copper bar and P_s1 be the tensile force in steel causing changes\n",
+ "#P_s1=2*P_c1\n",
+ "\n",
+ "#dell_s+dell_c=(alpha_c-alpha_s)*t*L .......................................(3)\n",
+ "#P_s1*L*(A_s*E_s)**-1+P_c1*L*(A_c*E_c)**-1=(alpha_c-alpha_s)*t*L ................(4)\n",
+ "#After substituting values in above equation and further simplifying we get,\n",
+ "P_c1=(alpha_c-alpha_s)*t*L*(2*(A_s*E_s)**-1+1*(A_c*E_c)**-1)**-1 #.................(5)\n",
+ "P_s1=2*P_c1\n",
+ "\n",
+ "#Extension of bar due to temperature rise\n",
+ "dell_2=alpha_s*t*L+P_s1*L*(A_s*E_s)**-1\n",
+ "\n",
+ "#Amount by which bar will descend\n",
+ "dell_3=dell_1+dell_2\n",
+ "\n",
+ "#Load carried by steel bar\n",
+ "P_S=P_s+P_s1\n",
+ "\n",
+ "#Load carried by copper bar\n",
+ "P_C=P_c-P_c1\n",
+ "\n",
+ "#Part-3\n",
+ "\n",
+ "#Let P_c1_1=P_c #For convenience\n",
+ "#Rise in temperature if Load is to be carried out by steel rod alone\n",
+ "P_c1_1=P_c\n",
+ "\n",
+ "#From equation 5 \n",
+ "t=P_c1_1*(2*(A_s*E_s)**-1+1*(A_c*E_c)**-1)*(alpha_c-alpha_s)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Extension Due top copper Load\",round(dell_1,2),\"mm\"\n",
+ "print\"Load carried by each rod:P_s\",round(P_s,2),\"N\"\n",
+ "print\" :P_c\",round(P_c,2),\"N\"\n",
+ "print\"Rise in Temperature of steel rod should be\",round(t,2),\"degree Celsius\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Extension Due top copper Load 1.0 mm\n",
+ "Load carried by each rod:P_s 100000.0 N\n",
+ " :P_c 50000.0 N\n",
+ "Rise in Temperature of steel rod should be 333.33 degree Celsius\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.36,Page No.53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "t=40 #degree celsius #temperature\n",
+ "A_s=400 #mm**2 #Area of steel bar\n",
+ "A_c=600 #mm**2 #Area of copper bar\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity of steel bar\n",
+ "E_c=1*10**5 #N/mm**2 #Modulus of Elasticity of copper bar\n",
+ "alpha_s=12*10**-6 #degree celsius #Temperature coeff of steel bar\n",
+ "alpha_c=18*10**-6 #degree celsius #Temperature coeff of copper bar\n",
+ "L_c=800 #mm #Length of copper bar\n",
+ "L_s=600 #mm #Length of steel bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_s be the tensile force in steel bar and P_c be the compressive force in copper bar\n",
+ "#Static Equilibrium obtained by taking moment about A\n",
+ "#P_c=2*P_s\n",
+ "\n",
+ "#From property of similar triangles we get\n",
+ "#(alpha_c*Lc-dell_c)*1**-1=(alpha_s*L_s-dell_s)*2**-1\n",
+ "#After substituting values in above equations and further simplifying we get\n",
+ "P_s=(2*alpha_c*L_c-alpha_s*L_s)*t*(L_s*(A_s*E_s)**-1+4*L_c*(A_c*E_c)**-1)**-1\n",
+ "P_c=2*P_s\n",
+ "\n",
+ "#Stress in steel rod\n",
+ "sigma_s=P_s*A_s**-1 #N/mm**2 \n",
+ "\n",
+ "#Stress in copper rod\n",
+ "sigma_c=P_c*A_c**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress in steel rod is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"STress in copper rod is\",round(sigma_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in steel rod is 35.51 N/mm**2\n",
+ "STress in copper rod is 47.34 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.37,Page No.61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=20 #mm #Diameter of bar\n",
+ "P=37.7*10**3 #N #Load\n",
+ "L=200 #mm #Guage Length \n",
+ "dell=0.12 #mm #Extension\n",
+ "dell_d=0.0036 #mm #contraction in diameter\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Area of bar\n",
+ "A=pi*4**-1*d**2\n",
+ "\n",
+ "#Let s and dell_s be the Linear strain and Lateral strain\n",
+ "s=dell*L**-1\n",
+ "dell_s=dell_d*d**-1\n",
+ "mu=dell_s*s**-1 #Poissoin's ratio \n",
+ "\n",
+ "#dell=P*L*(A*E)**-1\n",
+ "E=P*L*(dell*A)**-1 #N/mm**2 #Modulus of Elasticity of bar\n",
+ "\n",
+ "#Modulus of Rigidity\n",
+ "G=E*(2*(1+mu))**-1 #N/mm**2\n",
+ "\n",
+ "#Bulk Modulus \n",
+ "K=E*(3*(1-2*mu))**-1 #N/mm**2\n",
+ "\n",
+ "#result\n",
+ "print\"Poisson's ratio is\",round(mu,2)\n",
+ "print\"The Elastic constant are:E\",round(E,2)\n",
+ "print\" :G\",round(G,2)\n",
+ "print\" :K\",round(K,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Poisson's ratio is 0.3\n",
+ "The Elastic constant are:E 200004.71\n",
+ " :G 76924.89\n",
+ " :K 166670.59\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.38,Page No.62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=100 #mm #Diameter of circular rod\n",
+ "P=1*10**6 #N #Tensile Force\n",
+ "mu=0.3 #Poisson's ratio\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus \n",
+ "L=500 #mm #Length of rod\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Modulus of Rigidity\n",
+ "G=E*(2*(1+mu))**-1 #N/mm**2\n",
+ "\n",
+ "#Bulk Modulus \n",
+ "K=E*(3*(1-2*mu))**-1 #N/mm**2\n",
+ "\n",
+ "A=pi*4**-1*d**2 #mm**2 #Area of Circular rod\n",
+ "#Let sigma be the Longitudinal stress\n",
+ "sigma=P*A**-1 #N/mm**2 \n",
+ "\n",
+ "s=sigma*E**-1 #Linear strain\n",
+ "e_x=s\n",
+ "\n",
+ "#Volumetric strain\n",
+ "e_v=e_x*(1-2*mu)\n",
+ "\n",
+ "v=pi*4**-1*d**2*L\n",
+ "#Change in VOlume\n",
+ "dell_v=e_v*v\n",
+ "\n",
+ "#Result\n",
+ "print\"Bulk Modulus is\",round(E,2),\"N/mm**2\"\n",
+ "print\"Modulus of Rigidity is\",round(G,2),\"N/mm**2\"\n",
+ "print\"The change in Volume is\",round(dell_v,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Bulk Modulus is 200000.0 N/mm**2\n",
+ "Modulus of Rigidity is 76923.08 N/mm**2\n",
+ "The change in Volume is 1000.0 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.39,Page No.62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=500 #mm #Length of rectangular cross section bar\n",
+ "A=20*40 #mm**2 #Area of rectangular cross section bar\n",
+ "P1=4*10**4 #N #Tensile Force on 20mm*40mm Faces\n",
+ "P2=2*10**5 #N #compressive force on 20mm*500mm Faces\n",
+ "P3=3*10**5 #N #Tensile Force on 40mm*500mm Faces\n",
+ "E=2*10**5 #N/mm**2 #young's Modulus \n",
+ "mu=0.3 #Poisson's Ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_x,P_y,P_z be the forces n x,y,z directions\n",
+ "\n",
+ "P_x=P1*A**-1\n",
+ "P_y=P2*A**-1\n",
+ "P_z=P3*A**-1\n",
+ "\n",
+ "#Let e_x,e_y,e_z be the strains in x,y,z directions\n",
+ "e_x=1*E**-1*(50+mu*20-15*mu)\n",
+ "e_y=1*E**-1*(-mu*50-20-mu*15)\n",
+ "e_z=1*E**-1*(-mu*50+mu*20+15)\n",
+ "\n",
+ "#Volumetric strain\n",
+ "e_v=e_x+e_y+e_z\n",
+ "\n",
+ "#Volume\n",
+ "V=20*40*500 #mm**3\n",
+ "#Change in Volume \n",
+ "dell_v=e_v*V #mm**3\n",
+ "\n",
+ "#Result\n",
+ "print\"The change in Volume is\",round(dell_v,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The change in Volume is 36.0 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.41,Page No.65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=2.1*10**5 #N/mm**2 #Young's Modulus \n",
+ "G=0.78*10**5 #N/mm**2 #Modulus of Rigidity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Now using the relation\n",
+ "#E=2*G*(1+mu)\n",
+ "mu=E*(2*G)**-1-1 #Poisson's ratio\n",
+ "\n",
+ "#Bulk Modulus \n",
+ "K=E*(3*(1-2*mu))**-1 #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"The Poisson's Ratio is\",round(mu,2)\n",
+ "print\"The modulus of Rigidity\",round(K,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Poisson's Ratio is 0.35\n",
+ "The modulus of Rigidity 227500.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.42,Page No.65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "G=0.4*10**5 #N/mm**2 #Modulus of rigidity\n",
+ "K=0.75*10**5 #N/mm**2 #Bulk Modulus \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Young's Modulus\n",
+ "E=9*G*K*(3*K+G)**-1\n",
+ "\n",
+ "#Now from the relation\n",
+ "#E=2*G(1+2*mu)\n",
+ "mu=E*(2*G)**-1-1 #POissoin's ratio \n",
+ "\n",
+ "#result\n",
+ "print\"Young's modulus is\",round(E,2),\"N/mm**2\"\n",
+ "print\"Poissoin's ratio is\",round(mu,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Young's modulus is 101886.79 N/mm**2\n",
+ "Poissoin's ratio is 0.27\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.43,Page No.65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b=60 #mm #width of bar\n",
+ "d=30 #mm #depth of bar\n",
+ "L=200 #mm #Length of bar\n",
+ "A=30*60 #mm**2 #Area of bar\n",
+ "A2=30*200 #mm**2 #Area of bar along which expansion is restrained\n",
+ "P=180*10**3 #N #Compressive force\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#The bar is restrained from expanding in Y direction\n",
+ "P_z=0\n",
+ "P_x=P*A**-1 #stress developed in x direction\n",
+ "\n",
+ "#Now taking compressive strain as positive\n",
+ "#e_x=P_x*E**-1-mu*P_y*E**-1 .......................(1)\n",
+ "#e_y=-mu*P_x*E**-1+P_y*E**-1 ....................(2)\n",
+ "#e_z=-mu*P_x*E**-1-mu*P_y*E**-1 ......................(3)\n",
+ "\n",
+ "#Part-1\n",
+ "#When it is fully restrained\n",
+ "e_y=0\n",
+ "P_y=30 #N/mm**2 \n",
+ "e_x=P_x*E**-1-mu*P_y*E**-1\n",
+ "e_z=-mu*P_x*E**-1-mu*P_y*E**-1\n",
+ "\n",
+ "#Change in Length \n",
+ "dell_l=e_x*L #mm\n",
+ "\n",
+ "#Change in width\n",
+ "dell_b=b*e_y\n",
+ "\n",
+ "#change in Depth\n",
+ "dell_d=d*e_z\n",
+ "\n",
+ "#Volume of bar\n",
+ "V=b*d*L #mm**3\n",
+ "#Change in Volume\n",
+ "e_v=(e_x+e_y+e_z)*V #mm**3\n",
+ "\n",
+ "#Part-2\n",
+ "#When 50% is restrained\n",
+ "\n",
+ "#Free strain in Y direction\n",
+ "e_y1=mu*P_x*E**-1\n",
+ "\n",
+ "#As 50% is restrained,so\n",
+ "e_y2=-50*100**-1*e_y1\n",
+ "\n",
+ "#But form Equation 2 we have e_y=-mu*P_x*E**-1+P_y*E**-1 \n",
+ "#After substituting values in above equation and furthe simplifying we get\n",
+ "P_y=e_y2*E+d\n",
+ "\n",
+ "e_x2=P_x*E**-1-mu*P_y*E**-1 \n",
+ "e_z2=-mu*P_x*E**-1-mu*P_y*E**-1\n",
+ "\n",
+ "#Change in Length \n",
+ "dell_l2=e_x2*L #mm\n",
+ "\n",
+ "#Change in width\n",
+ "dell_b2=b*e_y2\n",
+ "\n",
+ "#change in Depth\n",
+ "dell_d2=d*e_z2\n",
+ "\n",
+ "#Change in Volume\n",
+ "e_v2=(e_x2+e_y2+e_z2)*V #mm**3\n",
+ "\n",
+ "#REsult\n",
+ "print\"Change in Dimension of bar is:dell_l\",round(dell_l,2),\"mm\"\n",
+ "print\" :dell_b\",round(dell_b,4),\"mm\"\n",
+ "print\" :dell_d\",round(dell_d,2),\"mm\"\n",
+ "print\"Change in Volume is\",round(e_v,2),\"mm**3\"\n",
+ "print\"Changes in material when only 50% of expansion can be reatrained:dell_l2\",round(dell_l2,2),\"mm\"\n",
+ "print\" :dell_b2\",round(dell_b2,4),\"mm\"\n",
+ "print\" :dell_d2\",round(dell_d2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Change in Dimension of bar is:dell_l 0.09 mm\n",
+ " :dell_b 0.0 mm\n",
+ " :dell_d -0.01 mm\n",
+ "Change in Volume is 93.6 mm**3\n",
+ "Changes in material when only 50% of expansion can be reatrained:dell_l2 0.1 mm\n",
+ " :dell_b2 -0.0045 mm\n",
+ " :dell_d2 -0.01 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.44,Page No.72"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=10*10**3 #N #Load\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus\n",
+ "d2=12 #mm #Diameter of bar1\n",
+ "d1=16 #mm #diameter of bar2\n",
+ "L1=200 #mm #Length of bar1\n",
+ "L2=500 #mm #Length of bar2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let A1 and A2 be the cross Area of Bar1 & bar2 respectively\n",
+ "A1=pi*4**-1*d1**2 #mm**2\n",
+ "A2=pi*4**-1*d2**2 #mm**2\n",
+ "\n",
+ "#Let p1 and p2 be the stress in Bar1 nad bar2 respectively\n",
+ "p1=P*A1**-1 #N/mm**2\n",
+ "p2=P*A2**-1 #N/mm**2\n",
+ "\n",
+ "#Let V1 nad V2 be the Volume of of Bar1 and Bar2\n",
+ "V1=A1*(L1+L1)\n",
+ "V2=A2*L2\n",
+ "\n",
+ "#Let E be the strain Energy stored in the bar\n",
+ "E=p1**2*(2*E)**-1*V1+p2**2*V2*(2*E)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"The Strain Energy stored in Bar is\",round(E,2),\"N-mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Strain Energy stored in Bar is 1602.6 N-mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.45,Page No.73"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Bar-A\n",
+ "d1=30 #mm #Diameter of bar1\n",
+ "L=600 #mm #length of bar1\n",
+ "\n",
+ "#Bar-B\n",
+ "d2=30 #mm #Diameter of bar2\n",
+ "d3=20 #mm #Diameter of bar2\n",
+ "L2=600 #mm #length of bar2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Area of bar-A\n",
+ "A1=pi*4**-1*d1**2\n",
+ "\n",
+ "#Area of bar-B\n",
+ "A2=pi*4**-1*d2**2\n",
+ "A3=pi*4**-1*d3**2\n",
+ "\n",
+ "#let SE be the Strain Energy\n",
+ "#Strain Energy stored in Bar-A\n",
+ "#SE=p**2*(2*E)**-1*V\n",
+ "#After substituting values and simolifying further we get\n",
+ "#SE=P**2*E**-1*0.4244\n",
+ "\n",
+ "#Strain Energy stored in Bar-B\n",
+ "#SE2=p1**2*V1*(2*E)**-1+p2**2*V2*(2*E)**-1\n",
+ "#After substituting values and simolifying further we get\n",
+ "#SE2=0.6897*P**2*E**-1\n",
+ "\n",
+ "#Let X be the ratio of SE in Bar-B and SE in Bar-A\n",
+ "X=0.6897*0.4244**-1\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#When Max stress is produced is same:Let p be the max stress produced\n",
+ "\n",
+ "#Stress in bar A is p throughout \n",
+ "#In bar B:stress in 20mm dia.portion=p2=p\n",
+ "\n",
+ "#Stress in 30 mm dia.portion\n",
+ "#p1=P*A2*A3**-1\n",
+ "#After substituting values and simolifying further we get\n",
+ "#p1=4*9**-1*p\n",
+ "\n",
+ "#Strain Energy in bar A\n",
+ "#SE_1=p**2*(2*E)**-1*A1*L1\n",
+ "#After substituting values and simolifying further we get\n",
+ "#SE_1=67500*p**2*pi*E**-1\n",
+ "\n",
+ "#Strain Energy in bar B\n",
+ "#SE_2=p1**2*V1*(2*E)**-1+p2**2*V2*(2*E)**-1\n",
+ "#After substituting values and simolifying further we get\n",
+ "#SE_2=21666.67*pi*p**2*E**-1\n",
+ "\n",
+ "#Let Y be the Ratio of SE in bar B and SE in bar A\n",
+ "Y=21666.67*67500**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Gradually applied Load is\",round(X,2)\n",
+ "print\"Gradually applied Load is\",round(Y,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Gradually applied Load is 1.63\n",
+ "Gradually applied Load is 0.32\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.46,Page No.74"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables \n",
+ "\n",
+ "W=100 #N #Load\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus \n",
+ "h=60 #mm #Height through Load falls down\n",
+ "L=400 #mm #Length of collar\n",
+ "d=30 #mm #diameter of bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A=pi*4**-1*d**2 #mm**2 #Area of bar\n",
+ "\n",
+ "#Instantaneous stress produced is\n",
+ "p=W*A**-1*(1+(1+(2*A*E*h*(W*L)**-1))**0.5)\n",
+ "\n",
+ "#Now the EXtension of the bar is neglected in calculating work doneby the Load,then\n",
+ "P=(2*E*h*W*(A*L)**-1)**0.5\n",
+ "\n",
+ "#Let percentage error be denoted by E1\n",
+ "#Percentage error in approximating is\n",
+ "E1=(p-P)*p**-1*100\n",
+ "\n",
+ "#Instantaneous Extension produced is\n",
+ "dell_l=round(P,3)*E**-1*L\n",
+ "\n",
+ "#Result\n",
+ "print\"The Instantaneous stress is\",round(p,2),\"N/mm\"\n",
+ "print\"Percentage Error is\",round(E1,2)\n",
+ "print\"The Instantaneous extension is\",round(dell_l,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Instantaneous stress is 92.27 N/mm\n",
+ "Percentage Error is 0.15\n",
+ "The Instantaneous extension is 0.18 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.47,Page No.75"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=20 #mm #Diameter of steel bar\n",
+ "L=1000 #mm #Length of bar\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus \n",
+ "p=300 #N/mm**2 #max Permissible stress\n",
+ "h=50 #mm #Height through which weight will fall\n",
+ "w=600 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#ARea of steel bar\n",
+ "A=pi*4**-1*d**2\n",
+ "\n",
+ "#Instantaneous extension is\n",
+ "dell_l=p*L*E**-1 #mm \n",
+ "\n",
+ "#Work done by Load \n",
+ "#W=W1*(h+dell_l)\n",
+ "\n",
+ "#Volume of bar\n",
+ "V=round(A,2)*L\n",
+ "#Let E1 be the strain Energy\n",
+ "E1=p**2*(2*E)**-1*V\n",
+ "\n",
+ "#Answer in Book for Strain Energy is Incorrect \n",
+ "\n",
+ "#Now Equating Workdone by Load to strain Energy \n",
+ "W1=E1*51.5**-1\n",
+ "\n",
+ "#Now when w=600 N\n",
+ "#Let W2 be the Work done by the Load\n",
+ "#W2=w(h2*dell_l)\n",
+ "\n",
+ "h=E1*w**-1-dell_l\n",
+ "\n",
+ "#Result\n",
+ "print\"The Max Lodad which can Fall from a height of 50 mm on the collar is\",round(W1,2),\"N\"\n",
+ "print\"the Max Height from which a 600 N Load can fall on the collar is\",round(h,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Max Lodad which can Fall from a height of 50 mm on the collar is 1372.54 N\n",
+ "the Max Height from which a 600 N Load can fall on the collar is 116.31 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.48,Page No.76"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D_s=30 #mm #Diameter of steel rod\n",
+ "d=30 #mm #Internal Diameter of copper tube\n",
+ "D=40#mm #External Diameter of copper tube\n",
+ "E_s=2*10**5 #N/mm**2 #Young's Modulus of Steel rod\n",
+ "E_c=1*10**5#N/mm**2 #Young's Modulus of copper tube\n",
+ "P=100 #N #Load\n",
+ "h=40 #mm #height from which Load falls\n",
+ "L=800 #mm #Length \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Area of steel rod\n",
+ "A_s=pi*4**-1*D_s**2\n",
+ "\n",
+ "#Area of copper tube\n",
+ "A_c=pi*4**-1*(D**2-d**2)\n",
+ "\n",
+ "#But Dell_s=dell_c=dell\n",
+ "#p_s*E_s**-1*L=p_c*L*E_c\n",
+ "#After simplifying furthe we get\n",
+ "#p_s=2*p_c\n",
+ "\n",
+ "#Now Equating internal Energy to Workdone we get\n",
+ "p_c=(2*P*h*L**-1*(4*A_s*E_s**-1+A_c*E_c**-1))**0.5\n",
+ "p_s=2*p_c\n",
+ "\n",
+ "#Result\n",
+ "print A_s\n",
+ "print\"STress produced in steel is\",round(p_s,2),\"N/mm**2\"\n",
+ "print\"STress produced in copper is\",round(p_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "706.858347058\n",
+ "STress produced in steel is 0.89 N/mm**2\n",
+ "STress produced in copper is 0.44 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.49,Page No.77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "dell=0.25 #mm #Instantaneous Extension\n",
+ "\n",
+ "#Bar-A\n",
+ "b1=25 #mm #width of bar\n",
+ "D1=500 #mm #Depth of bar\n",
+ "\n",
+ "#Bar-B\n",
+ "b2_1=25 #mm #width of upper bar\n",
+ "b2_2=15 #mm #Width of Lower Bar\n",
+ "L2=200 #mm #Length of upper bar\n",
+ "L1=300 #mm #Length of Lower bar\n",
+ "\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus of bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Strain\n",
+ "e=dell*D1**-1 \n",
+ "\n",
+ "#Load\n",
+ "p=e*E\n",
+ "\n",
+ "#Area of bar-A\n",
+ "A=pi*4**-1*25**2\n",
+ "\n",
+ "#Volume of bar-A\n",
+ "V=A*D1\n",
+ "\n",
+ "#Let E1 be the Energy of Blow\n",
+ "#Energy of Blow\n",
+ "E1=p**2*(E)**-1*V\n",
+ "\n",
+ "#Let p2 be the Max stress in bar B When this blow is applied.\n",
+ "#the max stress occurs in the 15mm dia. portion,Hence, the stress in 25 mm dia.portion is\n",
+ "#p2*pi*4**-1*b2_2**2*(pi*4**-1*b2_2**2=0.36*p\n",
+ "\n",
+ "#Strain Energy of bar B\n",
+ "#E2=p**2*(2*E)**-1*v1+1*(2*E)**-1*(0.36*p2)**2*v2\n",
+ "#After substituting values and Further substituting values we get\n",
+ "#E2=0.1643445*p2**2\n",
+ "\n",
+ "#Equating it to Energy of applied blow,we get\n",
+ "p2=(12271.846*0.1643445**-1)**0.5\n",
+ "\n",
+ "#Stress in top portion\n",
+ "sigma=0.36*p2\n",
+ "\n",
+ "#Extension in Bar-1\n",
+ "dell_1=p2*E**-1*L1\n",
+ "\n",
+ "#Extension in Bar-2\n",
+ "dell_2=0.36*p2*E**-1*L2\n",
+ "\n",
+ "#Extension of bar\n",
+ "dell_3=dell_1+dell_2\n",
+ "\n",
+ "#Result\n",
+ "print\"Instantaneous Max stress is\",round(sigma,2),\"N/mm**2\"\n",
+ "print\"extension in Bar is\",round(dell_3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Instantaneous Max stress is 98.37 N/mm**2\n",
+ "extension in Bar is 0.51 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.3_9.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.3_9.ipynb
new file mode 100644
index 00000000..47f0aab5
--- /dev/null
+++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.3_9.ipynb
@@ -0,0 +1,1601 @@
+{
+ "metadata": {
+ "name": "chapter no.3.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3:Shear Force And Bending Moment Diagrams in Statically Determinate Beams"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.1,Page No.100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "matplotlib.pyplot\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_AC=L_CD=1 #m #Length of AC & CD\n",
+ "L_DB=1.5 #m #Lengh of DB\n",
+ "L=3.5 #m #Length of Beam\n",
+ "F_B=10 #KN #Force at pt B\n",
+ "F_C=F_D=20 #KN #Force at pt C & D\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "R_A=F_C+F_D+F_B #KN #Force at support A \n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At pt B\n",
+ "V_B1=0 #KN \n",
+ "V_B2=F_B #KN\n",
+ "\n",
+ "#S.F At pt D\n",
+ "V_D1=V_B2 #KN\n",
+ "V_D2=V_D1+F_D #KN\n",
+ "\n",
+ "#S.F At pt C \n",
+ "V_C1=V_D2 #KN\n",
+ "V_C2=V_D2+F_C #KN\n",
+ "\n",
+ "#S.F At Pt A\n",
+ "V_A1=V_C2 #KN\n",
+ "V_A2=V_C2-R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=0 #KN.m\n",
+ "\n",
+ "#B.M AT Pt D\n",
+ "M_D=F_B*L_DB #KN.m\n",
+ "\n",
+ "#B.M At pt C\n",
+ "M_C=F_B*(L_DB+L_CD)+F_D*L_CD #KN.m\n",
+ "\n",
+ "#B.M At pt A\n",
+ "M_A=F_B*L+F_D*(L_CD+L_AC)+F_C*L_AC\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_DB,L_DB,L_CD+L_DB,L_CD+L_DB,L_CD+L_DB+L_AC,L_CD+L_DB+L_AC]\n",
+ "Y1=[V_B1,V_B2,V_D1,V_D2,V_C1,V_C2,V_A1,V_A2]\n",
+ "Z1=[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",
+ "Y2=[M_B,M_D,M_C,M_A]\n",
+ "X2=[0,L_DB,L_DB+L_CD,L_AC+L_CD+L_DB]\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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+ "text": [
+ "<matplotlib.figure.Figure at 0x581d130>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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pqQavCQgIEADwi1/84he/rPgKCAiw6e+yYlJMDQ0NuPfee/H999+jZ8+eGDhw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\no9ggiIjIKDYIIiIy6v8BsxKV3Pt7cnUAAAAASUVORK5CYII=\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5655f50>"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.2,Page No.101"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "matplotlib.pyplot\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "w1=10 #KN/m #u.d.L\n",
+ "F_D=20 #KN #Force at pt D\n",
+ "F_C=30 #KN #Force at pt C\n",
+ "L_DB=4 #m #Length of DB\n",
+ "L_CD=L_AC=2 #m #Length of AC & CD\n",
+ "L=8 #m #Length of Beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A And R_B be the Reactions at pt A and B \n",
+ "#R_A+R_B=90 \n",
+ "#Now Taking moment at A,M_A we get\n",
+ "R_A=(w1*L_DB*(L_DB*2**-1)+F_D*L_DB+F_C*(L_CD+L_DB))*L**-1\n",
+ "R_B=90-R_A\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At Pt B\n",
+ "V_B1=0 #KN\n",
+ "V_B2=R_B #KN\n",
+ "\n",
+ "#S.F At pt D\n",
+ "V_D1=R_B-w1*L_DB #KN\n",
+ "V_D2=V_D1-F_D #KN\n",
+ "\n",
+ "#S.F at Pt C\n",
+ "V_C1=V_D2 #KN\n",
+ "V_C2=V_C1-F_C \n",
+ "\n",
+ "#S.F at PT A\n",
+ "V_A1=V_C2 #KN\n",
+ "V_A2=V_C2+R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=0 #KN.m\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D=-R_B*L_DB+w1*L_DB*L_DB*2**-1 #KN.m\n",
+ "\n",
+ "#B.M At PT C\n",
+ "M_C=-R_B*(L_DB+L_CD)+w1*L_DB*(L_DB*2**-1+L_CD)+F_D*L_CD #KN.m\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=-R_B*L+w1*L_DB*(L_DB*2**-1+L_CD+L_AC)+F_D*(L_CD+L_AC)+F_C*L_AC\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_DB,L_DB,L_CD+L_DB,L_CD+L_DB,L_CD+L_DB+L_AC,L_CD+L_DB+L_AC]\n",
+ "Y1=[V_B1,V_B2,V_D1,V_D2,V_C1,V_C2,V_A1,V_A2]\n",
+ "Z1=[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",
+ "Y2=[M_B,M_D,M_C,M_A]\n",
+ "X2=[0,L_DB,L_DB+L_CD,L_AC+L_CD+L_DB]\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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Pni0aGhr0vo4HrxERkcSmlo+IiMi8WApERCRhKRARkYSlQEREEpYCERFJWApE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\nICIiCUuBiIgk/w9P4ODmR+1IBgAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5650890>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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Gt27dTE9mRiwY9ovtQ+zHlSuAv780HCkgQO00tkGRgnHr1i2kp6ejpKQEdXV1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\nx6hRo1BfX48XX3wRgYGBasd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+ "text": [
+ "<matplotlib.figure.Figure at 0x56444f0>"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.3,Page No.102"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "matplotlib.pyplot\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_DB=L_CD=1.5 #m #Length of DB & CD\n",
+ "L_AC=3 #m #Length of AC\n",
+ "F_D=80 #KN #Force at Pt D\n",
+ "w=40 #KN/m #u.v.l\n",
+ "L=6 #Length of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A and R_B be the Reactions at Pt A & B respectively\n",
+ "#R_A+R_B=140 \n",
+ "#Taking moment at B we get,M_B\n",
+ "R_A=(1*2**-1*L_AC*w*(1*3**-1*L_AC+(L_CD+L_DB))+F_D*L_DB)*L**-1\n",
+ "R_B=140-R_A\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F at B\n",
+ "V_B1=0 #KN\n",
+ "V_B2=R_B #KN\n",
+ "\n",
+ "#S.F At D\n",
+ "V_D1=V_B2 #KN\n",
+ "V_D2=V_D1-F_D #KN\n",
+ "\n",
+ "#S.F at C\n",
+ "V_C=V_D2 #KN\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A1=V_C-1*2**-1*w*L_AC #KN\n",
+ "V_A2=V_A1+R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=0 #KN.m\n",
+ "\n",
+ "#B.M At D\n",
+ "M_D=-R_B*L_DB\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=F_D*L_CD-R_B*(L_DB+L_CD)\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A=F_D*(L_CD+L_AC)-R_B*L+1*2**-1*w*L_AC*(1*3**-1*L_AC)+R_A\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_DB,L_DB,L_DB+L_CD,L_DB+L_CD+L_AC,L_DB+L_CD+L_AC]\n",
+ "Y1=[V_B1,V_B2,V_D1,V_D2,V_C,V_A1,V_A2]\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_DB,L_CD+L_DB,L_AC+L_CD+L_DB]\n",
+ "Y2=[M_B,M_D,M_C,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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+ "text": [
+ "<matplotlib.figure.Figure at 0x56f78d0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x583c830>"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.4,Page No.104"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "matplotlib.pyplot\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "M_D=120 #KN.m #B.M at Pt D\n",
+ "F_C=40 #KN #Force at Pt C\n",
+ "w1=20 #KN.m\n",
+ "L_DB=1.5 #m #Length of DB\n",
+ "L_CD=1.5 #m #Length of CD\n",
+ "L_AC=3 #m #Length of AC\n",
+ "L=6 #m #Length of Beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A And R_B be the Reactions at pt A and B \n",
+ "#R_A+R_B=100\n",
+ "#Now Taking Moment At Pt B We get,M_B\n",
+ "R_A=-(M_D-F_C*(L_CD+L_DB)-w1*L_AC*(L_AC*2**-1+L_CD+L_DB))*L**-1\n",
+ "R_B=100-R_A\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At Pt B\n",
+ "V_B1=0\n",
+ "V_B2=R_B\n",
+ "\n",
+ "#S.F at Pt D\n",
+ "V_D=V_B2 #KN\n",
+ "\n",
+ "#S.F At Pt C\n",
+ "V_C1=V_D #KN\n",
+ "V_C2=V_C1-F_C\n",
+ "\n",
+ "#S.F At Pt A\n",
+ "V_A1=V_C2-w1*L_AC #KN\n",
+ "V_A2=V_A1+R_A\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=0 #KN.m\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D1=M_B-R_B*L_DB #KN.m\n",
+ "M_D2=M_B+M_D-R_B*L_DB\n",
+ "\n",
+ "#B.M At Pt C\n",
+ "M_C=M_D-R_B*(L_CD+L_DB)\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=M_D-R_B*L+F_C*L_AC+w1*L_AC*L_AC*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=[0,0,L_DB,L_DB+L_CD,L_DB+L_CD,L_DB+L_CD+L_AC,L_DB+L_CD+L_AC]\n",
+ "Y1=[V_B1,V_B2,V_D,V_C1,V_C2,V_A1,V_A2]\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_D1,M_D2,M_C,M_A]\n",
+ "X2=[0,L_DB,L_DB,L_CD+L_DB,L_AC+L_CD+L_DB]\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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+ "text": [
+ "<matplotlib.figure.Figure at 0x5944a90>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5716fd0>"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.5,Page No.105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "matplotlib.pyplot\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F_C=20 #KN #Force at Pt C\n",
+ "F_D=40 #KN #Force at pt D\n",
+ "w=20 #KN.m #u.d.l \n",
+ "L_AD=L_DB=2 #m #Length of AD & DB\n",
+ "L_BC=1 #m #Length of BC\n",
+ "L=5 #m #Length of Beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt R_A and R_B be the reactions at A & B respectively\n",
+ "#R_A+R_B=100 \n",
+ "#Now Taking Moment at B,M_B we get\n",
+ "R_A=-(F_C*L_BC-F_D*L_DB-w*L_AD*(L_AD*2**-1+L_DB))*(L_AD+L_DB)**-1\n",
+ "R_B=100-R_A\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At pt C\n",
+ "V_C1=0 #KN\n",
+ "V_C2=-F_C #KN\n",
+ "\n",
+ "#S.F At PT B\n",
+ "V_B1=V_C2 #KN\n",
+ "V_B2=V_C2+R_B #KN\n",
+ "\n",
+ "#S.F At Pt D\n",
+ "V_D1=V_B2 #KN\n",
+ "V_D2=V_D1-F_D #KN\n",
+ "\n",
+ "#S.F At Pt A\n",
+ "V_A1=V_D2-w*L_AD #KN\n",
+ "V_A2=V_A1+R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt C\n",
+ "M_C=0 \n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=F_C*L_BC\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D=F_C*(L_BC+L_DB)-R_B*L_DB\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=F_C*L-R_B*(L_DB+L_AD)+F_D*L_AD+w*L_AD*L_AD*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=[0,0,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_C1,V_C2,V_B1,V_B2,V_D1,V_D2,V_A1,V_A2]\n",
+ "Z1=[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",
+ "Y2=[M_C,M_B,M_D,M_A]\n",
+ "X2=[0,L_BC,L_BC+L_DB,L_BC+L_DB+L_AD]\n",
+ "Z2=[0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\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 0x592b8f0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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Zb7/9lh07drB8+XLy8/N58cUXmTVrlrt+NSKaPpLAUlFRwZdffsnI\nkSOd91VWVgLGm/WZE1a7du3K4cOHAfj8889JSUkBcPYoONvw4cMB6NGjBx9++KHzfldOkKntmufr\n0KED3bp1A6Bbt24kJiYCEB0dTUlJSZ3XEXGVkoIElNOnT3PFFVdQWFhY48+bNm3q/PrMm/r5dYPz\n3+wvueQSAJo0aUJVVVW9Y6rpmuc7cw0w+iWceU5QUFCDrilSG00fSUC5/PLL6dChAx988AFgvAn/\n61//uuhz+vTpw4oVK3A4HBw+fJi8vLw6r9O8eXPnUcXn0xmUYmVKCuLXTpw4Qbt27Zy3//3f/+Xd\nd99l0aJFxMbGEh0dfU4z97Pn+898PWLECMLCwoiKiuK+++6jR48etGjR4oJr2Ww253OGDh3KypUr\niYuLIz8/v9bH1XbNml67tu8D+UhocT8dnS3igp9//pnLLruM//znPyQkJPDFF1/QqlUrs8MScTvV\nFERcMGTIEMrLy6msrGTatGlKCOK3NFIQEREn1RRERMRJSUFERJyUFERExElJQUREnJQURETESUlB\nRESc/j+iaB8fTwtkoQAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x593a350>"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.6,Page No.107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "matplotlib.pyplot\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_BC=L_EB=L_AD=1 #m #Length of spans BC,ED,AD\n",
+ "L_ED=2 #m #Length of ED\n",
+ "w=60 #KNm #u.d.l\n",
+ "F_C=20 #KN Pt Load at C\n",
+ "L=5 #m #Span of beam \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A & R_B be the reactions at A & B respectively\n",
+ "#R_A+R_B=80 \n",
+ "#Taking Moment At A,we get M_A\n",
+ "R_B=(F_C*L+1*2**-1*L_ED*w*(2*3**-1*L_ED+L_AD))*(L_AD+L_ED+L_EB)**-1\n",
+ "R_A=80-R_B\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C1=0 #KN\n",
+ "V_C2=-F_C #KN\n",
+ "\n",
+ "#S.F At B\n",
+ "V_B1=V_C2 #KN\n",
+ "V_B2=V_C2+R_B #KN \n",
+ "\n",
+ "#S.F aT E\n",
+ "V_E=V_B2 #KN\n",
+ "\n",
+ "#S.F AT D\n",
+ "V_D=V_B2-1*2**-1*L_ED*w #KN\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A1=V_D #KN \n",
+ "V_A2=V_D+R_A\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M at C\n",
+ "M_C=0 #KN.m\n",
+ "\n",
+ "#B.M at B\n",
+ "M_B=F_C*L_BC #KN.m\n",
+ "\n",
+ "#B.M at E\n",
+ "M_E=F_C*(L_EB+L_BC)-R_B*L_EB #KN.m\n",
+ "\n",
+ "#B.M at D\n",
+ "M_D=F_C*(L_ED+L_EB+L_BC)-R_B*(L_ED+L_EB)+1*2**-1*L_ED*w*1*3**-1*L_ED #KN.m\n",
+ "\n",
+ "#B.M at A\n",
+ "M_A=1*2**-1*L_ED*w*(2*3**-1*L_ED+L_AD)-R_B*(L_AD+L_ED+L_EB)+F_C*L\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_BC,L_BC,L_EB+L_BC,L_ED+L_EB+L_BC,L_AD+L_ED+L_EB+L_BC,L_ED+L_EB+L_BC+L_AD]\n",
+ "Y1=[V_C1,V_C2,V_B1,V_B2,V_E,V_D,V_A1,V_A2]\n",
+ "Z1=[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_BC,L_BC+L_EB,L_EB+L_BC+L_ED,L_EB+L_BC+L_ED+L_AD]\n",
+ "Y2=[M_C,M_B,M_E,M_D,M_A]\n",
+ "Z2=[0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\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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Cr9fzLAHAgQMHUFpaig8++AAbNmzAvn37Ot1OUUUhJCSkwx1Vq6qqEBoaKjAR\nuYsLFy5g5syZuPfee5GWliY6jlu45pprMH36dHz++eeiowjxySefoKioCGFhYcjIyMCePXswd+5c\n0bGEue666wAAQ4YMwR133GHzvnKKKgpjxoxBeXk5Kisr0dLSgq1bt8JgMIiORYJJkoQHHngAWq22\ny1umeIPTp0+jsbERAHD+/Hns3r1bbg71NqtWrUJVVRUqKirwzjvvYMqUKXj99ddFxxLi3Llz+OWX\nXwAAv/76K0pKSmxeuaioouDr64v169dj6tSp0Gq1+OMf/+i1V5hkZGRgwoQJOH78OIYPH45NmzaJ\njiTMgQMH8MYbb+Cjjz6CXq+HXq9HcXGx6FhC1NXVYcqUKdDpdIiPj0dqaiqSkpJEx3IL3jz8bDKZ\nMGnSJPnfxYwZM5CSktLptoq6JJWIiJxLUWcKRETkXCwKREQkY1EgIiIZiwIREclYFIiISMaiQERE\nMhYF8ijOvr3F888/j/Pnzzv8eO+//75X3wqe3Af7FMij9OvXT+7cdIawsDB8/vnnuPbaa11yPCJX\n45kCebzvvvsO06ZNw5gxY3DzzTfj2LFjAID77rsPf/3rX3HTTTdhxIgRKCgoAGC9y+iCBQsQHR2N\nlJQUTJ8+HQUFBVi3bh1qa2uRmJjYoUv4scceg06nw/jx4/HDDz9cdvxFixZhxYoVAIB///vfmDx5\n8mXbbN68GQsXLuwyV3uVlZWIiopCVlYWRo4ciXvuuQclJSW46aabEBkZic8++6z3Hxx5J4nIgwQF\nBV322pQpU6Ty8nJJkiTp4MGD0pQpUyRJkqTMzEwpPT1dkiRJKisrk8LDwyVJkqTt27dLt912myRJ\nklRfXy8NHDhQKigokCRJkjQajXTmzBn5d6tUKmnnzp2SJEnSkiVLpKeffvqy4587d06KiYmR9uzZ\nI40cOVL6/vvvL9tm8+bN0kMPPdRlrvYqKiokX19f6euvv5YsFosUFxcn3X///ZIkSVJhYaGUlpbW\n7WdF1Blf0UWJyJmamprw6aefYvbs2fJrLS0tAKz3wmm7o2p0dDRMJhMAYP/+/UhPTwcAeU0CW/z9\n/TF9+nQAQFxcHHbv3n3ZNn379sXLL7+MSZMmYe3atQgLC+sys61clwoLC0NMTAwAICYmBrfccgsA\nIDY2FpWVlV0eg8gWFgXyaBaLBQMGDEBpaWmn7/v7+8vPpd+n11QqVYf770tdTLv5+fnJz318fNDa\n2trpdl9377KBAAABP0lEQVR++SWGDBli96JQneW6VEBAQIdjt+3TVQ6i7nBOgTxa//79ERYWhnff\nfReA9Qv2yy+/7HKfm266CQUFBZAkCSaTCXv37pXf69evH86ePdujDCdPnsRzzz0nL3DS2X3suyo8\nRK7EokAe5dy5cxg+fLj8eP755/Hmm2/i1VdfhU6nQ2xsbIfF29vfTrnt+cyZMxEaGgqtVos5c+bg\nhhtuwDXXXAMAePDBB3HrrbfKE82X7n/p7ZklSUJ2djbWrFmD4OBgvPrqq8jOzpaHsGzta+v5pfvY\n+tmbbxNNvcNLUok68euvv+Lqq6/GmTNnEB8fj08++QRDhw4VHYvI6TinQNSJGTNmoLGxES0tLXj8\n8cdZEMhr8EyBiIhknFMgIiIZiwIREclYFIiISMaiQEREMhYFIiKSsSgQEZHs/wFJvODf5hYcpQAA\nAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x58a6f70>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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UlD0DZoHZoDk7OxvPPfec/SqwAoNm98HQmcjAloBZINmcwq5du5rcTwEAHnnkEeuqkgCb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\nkeDZfAAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x567bcf0>"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.7,Page No.109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "matplotlib.pyplot\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_BC=1 #m #Length of BC\n",
+ "L_DB=2 #m #Length of DB\n",
+ "L_AD=4 #m #Length 0f AD\n",
+ "M_D=30 #KN.m #Moment at D\n",
+ "w=45 #KN/m #u.d.l\n",
+ "L=7 #m #Span of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_B & R_A be the Reactions at B & A respectively\n",
+ "#R_B+R_A=180+P ............(1)\n",
+ "\n",
+ "#Now Taking Moment about A,we get\n",
+ "#R_B=7*P+390 ...............(2)\n",
+ "\n",
+ "#Since R_A & R_B Are Equal\n",
+ "#2*R_B=180+P ...................(3)\n",
+ "\n",
+ "#From equation 1 and 3 we get\n",
+ "#3*(180+P)=7P+390\n",
+ "#After simplifying Further above equation we get\n",
+ "P=150*4**-1 #KN\n",
+ "R_A=R_B=(180+P)*2**-1\n",
+ "F_C=P\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C1=0 #KN\n",
+ "V_C2=-P #KN\n",
+ "\n",
+ "#S.F At B\n",
+ "V_B1=V_C2 #KN\n",
+ "V_B2=V_C2+R_B #KN \n",
+ "\n",
+ "#S.F At D\n",
+ "V_D=V_B2 #KN\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A1=V_D-w*L_AD #KN\n",
+ "V_A2=V_A1+R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M at C\n",
+ "M_C=0 #KN.m \n",
+ "\n",
+ "#B.M at B\n",
+ "M_B=F_C*L_BC #KN.m\n",
+ "\n",
+ "#B.M at D\n",
+ "M_D1=F_C*(L_BC+L_DB)-R_B*L_DB #KN.m\n",
+ "M_D2=M_D1+M_D\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A=w*L_AD*L_AD*2**-1+M_D-R_B*(L_AD+L_DB)+P*L\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_BC,L_BC,L_DB+L_BC,L_DB+L_BC+L_AD,L_DB+L_BC+L_AD]\n",
+ "Y1=[V_C1,V_C2,V_B1,V_B2,V_D,V_A1,V_A2]\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_DB+L_BC,L_DB+L_BC,L_AD+L_DB+L_BC]\n",
+ "Y2=[M_C,M_B,M_D1,M_D2,M_A]\n",
+ "Z2=[0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\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 0x582e990>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x592a7f0>"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.8,Page No.110"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "matplotlib.pyplot\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=6 #m #Span Of beam\n",
+ "w=30 #KN/m #u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Due to Symmetry\n",
+ "#Let R_B and R_C be the reactions at B & C Respectively\n",
+ "R_B=R_C=w*L*2**-1 #KN\n",
+ "\n",
+ "#Let a be the overhang.The Max -ve moment occurs at the support and max +ve moment at middle of the beam\n",
+ "#Now Equating these two equations we get\n",
+ "#30*a*a*2**-1=90*(3-a)-w*L*2**-1*L*4**-1\n",
+ "#After simplifying we get an equation as\n",
+ "#a**2+6*a-9=0\n",
+ "x=1\n",
+ "y=6\n",
+ "z=-9\n",
+ "\n",
+ "p=y**2-4*x*z\n",
+ "\n",
+ "a1=(-y+p**0.5)*2**-1\n",
+ "a2=(-y-p**0.5)*2**-1\n",
+ "\n",
+ "#Now Length cannot be negative,so taking a1 into Consideration\n",
+ "\n",
+ "L_CD=L_AB=a1\n",
+ "L_BC=L-2*a1\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At D\n",
+ "V_D=0\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C1=V_D-w*L_CD #KN\n",
+ "V_C2=V_C1+R_C #KN\n",
+ "\n",
+ "#S.F At B\n",
+ "V_B1=-w*(L_BC+L_CD)+R_C\n",
+ "V_B2=V_B1+R_B\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A=round(V_B2,2)-round(w*L_AB,2)\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At D\n",
+ "M_D=0\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=w*L_CD*L_CD*2**-1 #KN.m\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=w*(L_BC+L_CD)*(L_BC+L_CD)*2**-1-R_C*L_BC*L_BC*2**-1\n",
+ "\n",
+ "#B.M At A\n",
+ "X=w*L*L*2**-1\n",
+ "Y=-R_C*(L_AB+L_BC)-R_B*L_AB\n",
+ "M_A=X+Y\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_CD,L_CD,L_CD+L_BC,L_CD+L_BC,L_CD+L_BC+L_AB]\n",
+ "Y1=[V_D,V_C1,V_C2,V_B1,V_B2,V_A]\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_CD,L_BC+L_CD,L_AB+L_BC+L_CD]\n",
+ "Y2=[M_D,M_C,M_B,M_A]\n",
+ "Z2=[0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\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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lXLhwAQBw6dIlHDhwwKleBejt7Q0/Pz+UlpYCAA4ePIiwsLAub6vKm9f6yt3d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\nAElFTkSuQmCC\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5710ff0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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LUVxcrNtAqS48PDxqdH5WVhby8vIQEBAAAJg4cSJ27NjBhEGkEDc3ID5e7BWe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\nMQcOHJASExMlb29vtUPRi6y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+ "text": [
+ "<matplotlib.figure.Figure at 0x5822270>"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.9,Page No.112"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "matplotlib.pyplot\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F_F=6 #KN #Force at F\n",
+ "w1=w2=w=3 #KN.m #u.d.l\n",
+ "M_D=24 #KN.m \n",
+ "L_AB=L_CD=L_DE=L_EF=4 #m #Length of AB,CD,DE,EF\n",
+ "L_BC=2 #m #Length of BC\n",
+ "L=18 #m #Span of Beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt R_B and R_E be the Reactions at B & E respectively\n",
+ "#R_B+R_E=42\n",
+ "\n",
+ "#Taking Moment At Pt B,M_B\n",
+ "R_E=(F_F*(L_BC+L_CD+L_DE+L_EF)+w*(L_CD+L_DE)*((L_CD+L_DE)*2**-1+L_BC)-w*L_AB*L_AB*2**-1-M_D)*(L_BC+L_CD+L_DE)**-1\n",
+ "R_B=42-R_E #KN\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F aT F\n",
+ "V_F1=0 #KN \n",
+ "V_F2=-F_F #KN\n",
+ "\n",
+ "#S.F at E\n",
+ "V_E1=V_F2 #KN\n",
+ "V_E2=V_E1+R_E #KN\n",
+ "\n",
+ "#S.F aT C\n",
+ "V_C=V_E2-w*(L_CD+L_DE) #KN\n",
+ "\n",
+ "#S.F at B\n",
+ "V_B1=V_C #KN \n",
+ "V_B2=V_C+R_B #KN\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A=V_B2-w*L_AB #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At F\n",
+ "M_F=0\n",
+ "\n",
+ "#B.M At E\n",
+ "M_E=F_F*L_EF #KN.m\n",
+ "\n",
+ "#B.M At D\n",
+ "M_D1=F_F*(L_DE+L_EF)-R_E*L_DE+w*L_DE*L_DE*2**-1 #KN.m\n",
+ "M_D2=M_D1-M_D\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=F_F*(L_CD+L_DE+L_EF)-R_E*(L_CD+L_DE)+w*(L_CD+L_DE)*(L_CD+L_DE)*2**-1-M_D\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=F_F*(L_BC+L_CD+L_DE+L_EF)-R_E*(L_BC+L_CD+L_DE)-M_D+w*(L_CD+L_DE)*((L_CD+L_DE)*2**-1+L_BC)\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A=w*L_AB*L_AB*2**-1-R_B*L_AB+w*(L_CD+L_DE)*((L_CD+L_DE)*2**-1+L_BC+L_AB)-R_E*(L_AB+L_BC+L_CD+L_DE)+F_F*L-M_D\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_EF,L_EF,L_EF+L_DE+L_CD,L_EF+L_DE+L_CD+L_BC,L_EF+L_DE+L_CD+L_BC,L_EF+L_DE+L_CD+L_BC+L_AB]\n",
+ "Y1=[V_F1,V_F2,V_E1,V_E2,V_C,V_B1,V_B2,V_A]\n",
+ "Z1=[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_EF,L_DE+L_EF,L_DE+L_EF,L_CD+L_DE+L_EF,L_CD+L_DE+L_EF+L_BC,L_CD+L_DE+L_EF+L_BC+L_AB]\n",
+ "Y2=[M_F,M_E,M_D1,M_D2,M_C,M_B,M_A]\n",
+ "Z2=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\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 0x56eccf0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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VxMfHIzs7G/369cOvv/6KcePG4bHHHnNknEROixVL5KzMziBqa2uRmJiIMWPG\n4IEHHkC/fv0AAMHBwdBoNA4LkMiZsccSOTOzCaJhEmCbb6KmY8USOTuzS0zHjx+Ht7c3AODGjRvG\n7+t/JiLzamqAlBT2WCLnZjZB1NbWOjIOIpfyyiuAhwcrlsi5sZ6CSGbLlwO7drHHEjk/Dl8iGbFi\niVwJEwSRTNhjiVyNxV5MRGQZK5bIFTFBENmIFUvkqhRJEDNnzkRISAgiIiIwcuRIXL161XhfWloa\nevTogeDgYGODQCI1Y8USuSpFEkRiYiJ+/vln/PjjjwgKCkJaWhoAoKCgAOvXr0dBQQGys7Mxbdo0\n1NXVKREikVXqK5bYY4lckSIJIiEhwXgIUWxsLIqLiwEAWVlZGD9+PDw9PeHn54fAwEAcOnRIiRCJ\nLKqvWNq6lRVL5JoU34NYtWoVhgwZAgAoLS2Fr6+v8T5fX1+UlJQoFRqRWeyxRO7AbpPihIQElJeX\n33H7/PnzkZSUBACYN28evLy8kJqaavZ12BiQ1IYVS+Qu7JYgdu7cedf716xZg23btuG///2v8TYf\nHx8UFRUZfy4uLoaPj4/J57/11lvG7+Pi4hAXF2dTvETWYMUSOZPc3Fzk5uY2+/lWHTkqt+zsbLz6\n6qvIy8trdEJdQUEBUlNTcejQIZSUlGDQoEE4ffr0HbMIHjlKcmrKkaMvvigtL339NTelyfnY5chR\nub344oswGAzGs60feughZGZmQqfTISUlBTqdDi1btkRmZiaXmEg12GOJ3I0iMwhbcQZBcrJmBpGT\nA6SmSo/hpjQ5K6eYQRA5E/ZYIneleJkrkZqxYoncGRMEkRmsWCJ3xwRBZAZ7LJG74x4EkQmsWCJi\ngiC6A0+FI5IwQRA1wIolor9xD4LoL6xYImqMCYIIrFgiMoUJggjA99+zYonodkwQ5Pbuvx/o14+n\nwhHdjr2YiIjcRFM/OzmDICIik5ggiIjIJCYIIiIyiQmCiIhMYoIgIiKTmCCIiMgkJggiIjKJCYKI\niExigiAiIpOYIIiIyCQmCCIiMokJgoiITGKCICIik5ggiIjIJCYIIiIyiQmCiIhMUiRBvPHGG4iI\niEBkZCQGDhyIoqIi431paWno0aMHgoODsWPHDiXCIyIiKJQgZs2ahR9//BHHjh1DcnIy3n77bQBA\nQUEB1q9fj4KCAmRnZ2PatGmoq6tTIsRmyc3NVTqEOzAm6zAm66kxLsZkH4okCG9vb+P3VVVV6Ny5\nMwAgKyvgK/nsAAAKZklEQVQL48ePh6enJ/z8/BAYGIhDhw4pEWKzqHFAMCbrMCbrqTEuxmQfih3R\n/vrrr+PTTz/FPffcY0wCpaWl6Nevn/Exvr6+KCkpUSpEIiK3ZrcZREJCAsLDw+/42rp1KwBg3rx5\nOHfuHKZMmYLp06ebfR2NRmOvEImI6G6Ewn7//XcRGhoqhBAiLS1NpKWlGe8bPHiwOHjw4B3PCQgI\nEAD4xS9+8YtfTfgKCAho0uezIktMhYWF6NGjBwBp3yEqKgoAMGzYMKSmpmLGjBkoKSlBYWEh+vbt\ne8fzT58+7dB4iYjckSIJYs6cOTh58iRatGiBgIAAfPjhhwAAnU6HlJQU6HQ6tGzZEpmZmVxiIiJS\niEYIIZQOgoiI1MfprqTOzs5GcHAwevTogYyMDKXDQVFREeLj4xEaGoqwsDAsWbJE6ZCMamtrERUV\nhaSkJKVDMaqsrMTo0aMREhICnU6HgwcPKh0S0tLSEBoaivDwcKSmpuLPP/90eAxPPfUUtFotwsPD\njbddvnwZCQkJCAoKQmJiIiorKxWPaebMmQgJCUFERARGjhyJq1evKh5TvUWLFsHDwwOXL192aEx3\ni2vp0qUICQlBWFgYZs+erXhMhw4dQt++fREVFYU+ffrg8OHDd38RWzaYHa2mpkYEBASIs2fPCoPB\nICIiIkRBQYGiMZWVlYmjR48KIYS4fv26CAoKUjymeosWLRKpqakiKSlJ6VCMJk2aJFauXCmEEOLW\nrVuisrJS0XjOnj0r/P39xc2bN4UQQqSkpIg1a9Y4PI7vvvtO5Ofni7CwMONtM2fOFBkZGUIIIdLT\n08Xs2bMVj2nHjh2itrZWCCHE7NmzVRGTEEKcO3dODB48WPj5+YlLly45NCZzceXk5IhBgwYJg8Eg\nhBDi/Pnzisc0YMAAkZ2dLYQQYtu2bSIuLu6ur+FUM4hDhw4hMDAQfn5+8PT0xLhx45CVlaVoTF27\ndkVkZCQAoG3btggJCUFpaamiMQFAcXExtm3bhqlTp0KoZBXx6tWr2LNnD5566ikAQMuWLdG+fXtF\nY2rXrh08PT1RXV2NmpoaVFdXw8fHx+Fx/OMf/0DHjh0b3bZlyxZMnjwZADB58mRs3rxZ8ZgSEhLg\n4SF9bMTGxqK4uFjxmABgxowZePfddx0aS0Om4vrwww8xZ84ceHp6AgDuv/9+xWN64IEHjLO+yspK\ni2PdqRJESUkJunXrZvxZbRfS6fV6HD16FLGxsUqHgldeeQULFiww/mNWg7Nnz+L+++/HlClTEB0d\njWeeeQbV1dWKxnTffffh1VdfRffu3fHggw+iQ4cOGDRokKIx1auoqIBWqwUAaLVaVFRUKBxRY6tW\nrcKQIUOUDgNZWVnw9fVFr169lA6lkcLCQnz33Xfo168f4uLi8MMPPygdEtLT043jfebMmUhLS7vr\n49Xz6WEFNVc0VVVVYfTo0Vi8eDHatm2raCxff/01unTpgqioKNXMHgCgpqYG+fn5mDZtGvLz83Hv\nvfciPT1d0ZjOnDmD999/H3q9HqWlpaiqqsK6desUjckUjUajqvE/b948eHl5ITU1VdE4qqurMX/+\nfGM/NwCqGfM1NTW4cuUKDh48iAULFiAlJUXpkPD0009jyZIlOHfuHN577z3jbN4cp0oQPj4+jTq/\nFhUVwdfXV8GIJLdu3cKoUaMwYcIEJCcnKx0O9u/fjy1btsDf3x/jx49HTk4OJk2apHRY8PX1ha+v\nL/r06QMAGD16NPLz8xWN6YcffsDDDz+MTp06oWXLlhg5ciT279+vaEz1tFotysvLAQBlZWXo0qWL\nwhFJ1qxZg23btqkikZ45cwZ6vR4RERHw9/dHcXExYmJicP78eaVDg6+vL0aOHAkA6NOnDzw8PHDp\n0iVFYzp06BBGjBgBQPr3Z6nXnVMliN69e6OwsBB6vR4GgwHr16/HsGHDFI1JCIGnn34aOp3uri1D\nHGn+/PkoKirC2bNn8fnnn+Of//wnPvnkE6XDQteuXdGtWzecOnUKALBr1y6EhoYqGlNwcDAOHjyI\nGzduQAiBXbt2QafTKRpTvWHDhmHt2rUAgLVr16ril4/s7GwsWLAAWVlZaN26tdLhIDw8HBUVFTh7\n9izOnj0LX19f5OfnqyKZJicnIycnBwBw6tQpGAwGdOrUSdGYAgMDkZeXBwDIyclBUFDQ3Z9grx10\ne9m2bZsICgoSAQEBYv78+UqHI/bs2SM0Go2IiIgQkZGRIjIyUmzfvl3psIxyc3NVVcV07Ngx0bt3\nb9GrVy8xYsQIxauYhBAiIyND6HQ6ERYWJiZNmmSsOnGkcePGiQceeEB4enoKX19fsWrVKnHp0iUx\ncOBA0aNHD5GQkCCuXLmiaEwrV64UgYGBonv37sax/vzzzysSk5eXl/HvqSF/f39FqphMxWUwGMSE\nCRNEWFiYiI6OFrt371YkpoZj6vDhw6Jv374iIiJC9OvXT+Tn59/1NXihHBERmeRUS0xEROQ4TBBE\nRGQSEwQREZnEBEFERCYxQRARkUlMEEREZBITBLk0e7c98fPzM9leOi8vDwcOHDD5nK1bt6qiVT2R\nJYqcKEfkKPbuX6TRaEz2/tm9eze8vb3x0EMP3XFfUlKSqs7oIDKHMwhyO2fOnMFjjz2G3r1749FH\nH8XJkycBAE8++SRefvll9O/fHwEBAdi4cSMAoK6uDtOmTUNISAgSExMxdOhQ432AdChMTEwMevXq\nhZMnT0Kv1+Ojjz7Ce++9h6ioKOzdu7fR+69ZswYvvvjiXd+zIb1ej+DgYEyZMgU9e/bEE088gR07\ndqB///4ICgqyfOgLUTMxQZDbefbZZ7F06VL88MMPWLBgAaZNm2a8r7y8HPv27cPXX3+N1157DQDw\n1Vdf4ffff8cvv/yCTz/9FAcOHGg0M7n//vtx5MgRPP/881i4cCH8/Pzwr3/9CzNmzMDRo0fxyCOP\nNHr/22c1pt7zdmfOnMG///1v/Prrrzh58iTWr1+Pffv2YeHChZg/f75cfzVEjXCJidxKVVUVDhw4\ngDFjxhhvMxgMAKQP7vqGeCEhIcbzF/bu3Wts1azVahEfH9/oNes7dkZHR+Orr74y3m5NFxtz73k7\nf39/Y2PD0NBQ45kVYWFh0Ov1Ft+HqDmYIMit1NXVoUOHDjh69KjJ+728vIzf13/A377PcPsHf6tW\nrQAALVq0QE1NTZNjMvWet6t/DwDw8PAwPsfDw6NZ70lkDS4xkVtp164d/P398eWXXwKQPpCPHz9+\n1+f0798fGzduhBACFRUVxnbJd+Pt7Y3r16+bvI/9MclZMEGQS6uurka3bt2MX++//z7WrVuHlStX\nIjIyEmFhYdiyZYvx8Q33B+q/HzVqFHx9faHT6TBx4kRER0ebPEu74alvSUlJ2LRpE6KiorBv3z6z\njzP3nqZe29zPajppjlwL230TWeGPP/7Avffei0uXLiE2Nhb79+9XxaE0RPbEPQgiKzz++OOorKyE\nwWDA3LlzmRzILXAGQUREJnEPgoiITGKCICIik5ggiIjIJCYIIiIyiQmCiIhMYoIgIiKT/h/FhIMx\nfyRzHAAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x56e1390>"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.10,Page No.114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "matplotlib.pyplot\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_DC=L_BA=2 #m #Length of BA & DC\n",
+ "L_CB=1 #m #Length of CB\n",
+ "F_A=10 #KN #Force at pt A\n",
+ "F_B=20 #KN #Force at pt B\n",
+ "w=4 #KN.m #u.d.l\n",
+ "L=5 #m #Length of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_D be the reactions at Pt D\n",
+ "R_D=F_B+F_A+w*L_DC #KN\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F at Pt A\n",
+ "V_A1=0 #KN\n",
+ "V_A2=F_A #KN\n",
+ "\n",
+ "#S.F At Pt B\n",
+ "V_B1=V_A2\n",
+ "V_B2=F_B+F_A\n",
+ "\n",
+ "#S.F at Pt C\n",
+ "V_C=F_B+F_A #KN \n",
+ "\n",
+ "#S.F At Pt D\n",
+ "V_D1=V_B2+w*L_DC\n",
+ "V_D2=F_B+F_A+w*L_DC-R_D\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=0\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=F_A*L_BA\n",
+ "\n",
+ "#B.M at Pt C\n",
+ "M_C=F_B*L_CB+F_A*(L_BA+L_CB) #KN\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D1=F_A*L+F_B*(L_CB+L_DC)+w*L_DC*L_DC*2**-1\n",
+ "M_D2=(F_A*L+F_B*(L_CB+L_DC)+w*L_DC*L_DC*2**-1)-M_D1\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_BA,L_BA,L_BA+L_CB,L_BA+L_CB+L_DC,L_BA+L_CB+L_DC]\n",
+ "Y1=[V_A1,V_A2,V_B1,V_B2,V_C,V_D1,V_D2]\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_A,M_B,M_C,M_D1,M_D2]\n",
+ "X2=[0,L_BA,L_CB+L_BA,L_CB+L_BA+L_DC,L_CB+L_BA+L_DC]\n",
+ "Z2=[0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\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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e3542bZr69++vhIQEzZ49WyNGjNAFF1wgSbrxxht15ZVXen+5e/LxJ0/9axiG\n8vLy9Oc//1kxMTFatWqV8vLyvMNNvo71tX3yMb6+tvMUxDh7XM4JWzp8+LAiIiL0ww8/aPTo0dq8\nebP69OljdSwgKBjjhy1NnjxZNTU1amho0L333kvpw1Y44wcAm2GMHwBshuIHAJuh+AHAZih+ALAZ\nih8AbIbiBwCb+T+gqVKkQGpm/QAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x593bb30>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5655c70>"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.11,Page No.115"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "matplotlib.pyplot\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "w=20 #KN/m #u.v.l\n",
+ "F_C=40 #KN #Force at Pt C\n",
+ "M_D=40 #KN.m #Moment at pt D\n",
+ "L_AB=3 #m #Length of AB\n",
+ "L_BC=1 #m #Length of BC\n",
+ "L_CD=L_DE=2 #m #Length of CD & DE\n",
+ "L=8 #8 #Length of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A & R_E be the Reactions at A & E respectively\n",
+ "#R_A+R_E=70\n",
+ "\n",
+ "#Taking Moments At Pt A we get,M_A\n",
+ "R_E=(F_C*(L_AB+L_BC)+1*2**-1*L_AB*w*2+40)*L**-1\n",
+ "R_A=70-R_E\n",
+ "\n",
+ "#shear Force Calculations\n",
+ "\n",
+ "#S.F At Pt E\n",
+ "V_E1=0\n",
+ "V_E2=R_E #KN\n",
+ "\n",
+ "#S.F aT pt D\n",
+ "V_D=V_E2\n",
+ "\n",
+ "#S.F At PT C\n",
+ "V_C1=V_D\n",
+ "V_C2=V_D-F_C #KN\n",
+ "\n",
+ "#S.F At Pt A\n",
+ "V_A1=V_C2-(1*2**-1*w*L_AB)\n",
+ "V_A2=V_A1+R_A\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt E\n",
+ "M_E=0\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D1=M_E-R_E*L_DE\n",
+ "M_D2=M_D1+M_D\n",
+ "\n",
+ "#B.M At Pt C\n",
+ "M_C=-R_E*(L_DE+L_CD)+M_D\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=-R_E*(L_DE+L_CD+L_BC)+M_D+F_C*L_BC\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=-R_E*L+M_D+(1*2**-1*L_AB*w*2)+F_C*(L_BC+L_AB)\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_CD+L_DE,L_CD+L_DE,L_CD+L_DE+L_AB,L_CD+L_DE+L_AB]\n",
+ "Y1=[V_E1,V_E2,V_D,V_C1,V_C2,V_A1,V_A2]\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_E,M_D1,M_D2,M_C,M_B,M_A]\n",
+ "X2=[0,L_DE,L_DE,L_CD+L_DE,L_DE+L_CD+L_BC,L_AB+L_BC+L_CD+L_DE]\n",
+ "Z2=[0,0,0,0,0,0]\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 0x5662db0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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PZcuW8dRTT5GZmcnOnTtp1KgRI0aMsPs+cvTnxbp1U4fwLFyoO4nzTCaYMkWt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\nd/r0aduBO+VVUFCA2Wxm+/b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6ePR5C6Jyk56CqHT27dtHt27daNu2LXfddRdpaWkAPPbY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\nGhPHAAAAAElFTkSuQmCC\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x593b530>"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.12,Page No.116"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "matplotlib.pyplot\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F_G=10 #KN #Force at Pt G\n",
+ "F_B=F_E=15 #KN #Force at Pt B & E\n",
+ "w=20 #KN/m #U.d.L\n",
+ "L_FG=L_EF=L_DE=L_CD=L_BC=L_AB=1 #m #Lengths of FG,EF,DE,CD,BC,AB respectively\n",
+ "L=6 #m #Length of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt R_F & R_A be the Reactions at E & A respectively\n",
+ "#R_F+R_A=60\n",
+ "\n",
+ "#Taking Moment At Pt A,M_A\n",
+ "R_F=(F_G*L+F_E*(L_AB+L_BC+L_CD+L_DE)+w*L_CD*(L_AB+L_BC+L_CD*2**-1)+F_B*L_AB)*(L_AB+L_BC+L_CD+L_DE+L_EF)**-1\n",
+ "R_A=60-R_F\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At G\n",
+ "V_G1=0 #KN \n",
+ "V_G2=F_G #KN\n",
+ "\n",
+ "#S.F At F\n",
+ "V_F1=V_G2 #KN\n",
+ "V_F2=V_F1-R_F\n",
+ "\n",
+ "#S.F At E\n",
+ "V_E1=V_F2 #KN\n",
+ "V_E2=V_F2+F_E\n",
+ "\n",
+ "#S.F At D\n",
+ "V_D=V_E2\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C=V_E2+w*L_CD\n",
+ "\n",
+ "#S.F At B\n",
+ "V_B1=V_C\n",
+ "V_B2=V_B1+F_B\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A1=V_B2\n",
+ "V_A2=V_B2-R_A\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt G\n",
+ "M_G=0\n",
+ "\n",
+ "#B.M At F\n",
+ "M_F=F_G*L_FG \n",
+ "\n",
+ "#B.M At E\n",
+ "M_E=F_G*(L_FG+L_EF)-R_F*L_EF\n",
+ "\n",
+ "#B.M At D\n",
+ "M_D=F_G*(L_FG+L_EF+L_DE)-R_F*(L_EF+L_DE)+F_E*L_DE\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=F_G*(L_FG+L_EF+L_DE+L_CD)-R_F*(L_EF+L_DE+L_CD)+F_E*(L_DE+L_CD)+w*L_CD*L_CD*2**-1\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=F_G*(L_FG+L_EF+L_DE+L_CD+L_BC)-R_F*(L_EF+L_DE+L_CD+L_BC)+F_E*(L_DE+L_CD+L_BC)+w*L_CD*(L_CD*2**-1+L_BC)\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A=F_G*L-R_F*(L_EF+L_DE+L_CD+L_BC+L_AB)+F_E*(L_DE+L_CD+L_BC+L_AB)+F_B*L_AB+w*L_CD*(L_CD*2**-1+L_BC+L_AB)\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_FG,L_FG,L_FG+L_EF,L_FG+L_EF,L_FG+L_EF+L_DE,L_FG+L_EF+L_DE+L_CD,L_FG+L_EF+L_DE+L_CD+L_BC,L_FG+L_EF+L_DE+L_CD+L_BC,L_FG+L_EF+L_DE+L_CD+L_BC+L_AB,L_FG+L_EF+L_DE+L_CD+L_BC+L_AB]\n",
+ "Y1=[V_G1,V_G2,V_F1,V_F2,V_E1,V_E2,V_D,V_C,V_B1,V_B2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0,0,0,0,0,0]\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_FG,L_EF+L_FG,L_EF+L_FG+L_DE,L_EF+L_FG+L_DE+L_CD,L_EF+L_FG+L_DE+L_CD+L_BC,L_EF+L_FG+L_DE+L_CD+L_BC+L_AB]\n",
+ "Y2=[M_G,M_F,M_E,M_D,M_C,M_B,M_A]\n",
+ "Z2=[0,0,0,0,0,0,0]\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 0x593bf90>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5671870>"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.13,Page No.117"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "matplotlib.pyplot\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "L_AB=L_BC=L_CD=L_DE=L_EF=1 #m #LEngth of AB,BC,CD,DE,EF respectively\n",
+ "M_A=50 #KN/m #Moment at A\n",
+ "w=5 #KN/m #u.v.l\n",
+ "F_D=10 #KN\n",
+ "w2=5 #KN/m #u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_B & R_E be the Reactions at B and E respectively\n",
+ "#R_B+R_E=20\n",
+ "\n",
+ "#Taking Moment At Pt B,M_B\n",
+ "R_E=(w2*L_EF*(L_EF*2**-1+L_DE+L_CD+L_BC)+w*L_BC*2**-1*2*3**-1+50+F_D*(L_BC+L_CD))*3**-1\n",
+ "R_B=17.5-R_E #KN\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At F\n",
+ "V_F=0\n",
+ "\n",
+ "#S.F aT E\n",
+ "V_E1=-w2*L_EF #KN\n",
+ "V_E2=V_E1+R_E\n",
+ "\n",
+ "#S.F at D\n",
+ "V_D1=R_E-w2*L_EF #KN\n",
+ "V_D2=V_D1-F_D #KN\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C=V_D2\n",
+ "\n",
+ "#S.F aT B\n",
+ "V_B1=-L_BC*w*2**-1-F_D+R_E-w2*L_EF\n",
+ "V_B2=V_B1+R_B\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M at F\n",
+ "M_F=0 #KN.m\n",
+ "\n",
+ "#B.M At E\n",
+ "M_E=w2*L_EF*L_EF*2**-1 #KN.m\n",
+ "\n",
+ "#B.M at D\n",
+ "M_D=-R_E*L_DE+w2*L_EF*(L_EF*2**-1+L_DE) #KN.m\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=F_D*L_CD*R_E*(L_CD+L_DE)+w2*L_EF*(L_EF*2**-1+L_DE+L_CD) #KN.m\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=F_D*(L_CD+L_BC)-R_E*(L_BC+L_CD+L_DE)+w2*L_EF*(L_EF*2**-1+L_BC+L_CD+L_DE)+1*2**-1*L_BC*w*2*3**-1\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A1=w*L_EF*(L_EF*2**-1+L_AB+L_BC+L_CD+L_DE)-R_E*(L_AB+L_BC+L_CD+L_DE)+F_D*(L_AB+L_BC+L_CD)+1*2**-1*L_BC*w*(2*3**-1*L_BC+L_AB)-R_B*L_AB\n",
+ "M_A2=M_A1+M_A\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_EF,L_EF,L_DE+L_EF,L_DE+L_EF,L_CD+L_DE+L_EF,L_CD+L_DE+L_EF+L_BC,L_CD+L_DE+L_EF+L_BC]\n",
+ "Y1=[V_F,V_E1,V_E2,V_D1,V_D2,V_C,V_B1,V_B2]\n",
+ "Z1=[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_EF,L_DE+L_EF,L_CD+L_DE+L_EF,L_CD+L_DE+L_EF+L_BC,L_CD+L_DE+L_EF+L_BC+L_AB,L_CD+L_DE+L_EF+L_BC+L_AB]\n",
+ "Y2=[M_F,M_E,M_D,M_C,M_B,M_A1,M_A2]\n",
+ "Z2=[0,0,0,0,0,0,0]\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 0x5705710>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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LlixREhMTLY8fMWKEsm/fvjuex4ZfSQi7euUVRXnxRa1TuIdr1xRFr1eUb77R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\nQmCC\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5942910>"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.4_9.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.4_9.ipynb
new file mode 100644
index 00000000..e2cbe575
--- /dev/null
+++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.4_9.ipynb
@@ -0,0 +1,1661 @@
+{
+ "metadata": {
+ "name": "chapter no.4.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 4:Stresses in Beams"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.1,Page no.130"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=5000 #mm #Length of Beam\n",
+ "a=2000 #mm #Length of start of beam to Pt Load\n",
+ "b=3000 #mm #Length of Pt load to end of beam\n",
+ "A=150*250 #m**2 #Area of beam \n",
+ "b=150 #mm #Width of beam\n",
+ "d=250 #mm #Depth of beam\n",
+ "sigma=10#N/mm**2 #stress\n",
+ "l=2000 #m #Load applied from one end\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*b*d**3 #m**4\n",
+ "\n",
+ "#Distance from N.A to end\n",
+ "y_max=d*2**-1 #m\n",
+ "\n",
+ "#Section Modulus\n",
+ "Z=1*6**-1*b*d**2 #mm**3\n",
+ "\n",
+ "#Moment Carrying Capacity\n",
+ "M=sigma*Z #N-mm\n",
+ "\n",
+ "#Let w be the Intensity of the Load in N/m,then Max moment\n",
+ "#M_max=w*L**2*8**-1 #N-mm\n",
+ "#After substituting values and further simplifying we get\n",
+ "#M_max=w*25*100*8**-1\n",
+ "\n",
+ "#EQuating it to moment carrying capacity,we get max intensity load\n",
+ "w=M*(25*1000)**-1*8*10**-3\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#Let P be the concentrated load,then max moment occurs under the load and its value\n",
+ "#M1=P*a*b*L**-1 #N-mm\n",
+ "\n",
+ "#Equting it to moment carrying capacity we get\n",
+ "P=M*1200**-1*10**-3 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Intensity of u.d.l it can carry\",round(w,3),\"KN-m\"\n",
+ "print\"MAx concentrated Load P apllied at 2 m from one end is\",round(P,3),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Intensity of u.d.l it can carry 5.0 N-mm\n",
+ "MAx concentrated Load P apllied at 2 m from one end is 13.021 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.2,Page no.131"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=70 #mm #External Diameter\n",
+ "t=8 #mm #Thickness of pipe\n",
+ "L=2500 #mm #span \n",
+ "sigma=150 #N/mm**2 #stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Internal Diameter \n",
+ "d=D-2*t #mm\n",
+ "\n",
+ "#M.I Of Pipe\n",
+ "I=pi*64**-1*(D**4-d**4) #mm**4\n",
+ "\n",
+ "y_max=D*2**-1 #mm\n",
+ "Z=I*(y_max)**-1 #mm**3\n",
+ "\n",
+ "#Moment Carrying capacity\n",
+ "M=sigma*Z #N*mm\n",
+ "\n",
+ "#Max moment int the beam occurs at the mid-span and is equal to\n",
+ "#m=P*L*4**-1\n",
+ "\n",
+ "#Equating Max moment to moment carrying capacity we get,\n",
+ "#M=P*2.5*L*4**-1\n",
+ "#After substituting and simplifying we get\n",
+ "P=4*M*(L)**-1*10**-3 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Max concentrated load that can be applied at the centre of span is\",round(P,3),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max concentrated load that can be applied at the centre of span is 5.22 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.3,Page no.132"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flanges Dimension\n",
+ "b1=180 #mm #Width\n",
+ "d1=10 #mm #Thickness\n",
+ "\n",
+ "D=500 #mm #Overall depth\n",
+ "t=8 #mm #Thickness of web\n",
+ "\n",
+ "#Plate Dimensions\n",
+ "b2=240 #mm #Width\n",
+ "t2=12 #mm #Thickness\n",
+ "\n",
+ "sigma=150 #N/mm**2 #Stress\n",
+ "L=3000 #mm #span\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of centroid from bottom fibre\n",
+ "y_bar=(b2*t2*(D+t2*2**-1)+b1*d1*(D-t1*2**-1)+(D-2*t1)*t*D*2**-1+(b1*t1*t1*2**-1))*(b2*t2+b1*d1+b1*d1+(D-2*d1)*t)**-1\n",
+ "\n",
+ "#M.I of section\n",
+ "I=(1*12**-1*b2*t2**3+b2*t2*(D+t2*2**-1-y_bar)**2+1*12**-1*b1*d1**3+b1*d1*(D-t1*2**-1-y_bar)**2+1*12**-1*b1*t1**3+b1*t1*(t1*2**-1-y_bar)**2+1*12**-1*t*(D-2*t1)**3+t*(D-2*t1)*(D*2**-1-y_bar)**2)\n",
+ "\n",
+ "#Section Modulus\n",
+ "Z=I*(y_bar)**-1 #mm**3\n",
+ "\n",
+ "#Moment or Resistance\n",
+ "M=sigma*Z\n",
+ "\n",
+ "#Let Load on Cantilever be w/m Length \n",
+ "#Max M.I produced\n",
+ "#M_max=w*L**2**-1 \n",
+ "\n",
+ "#Now Equating Moment of resistance to Max moment,we get Max load\n",
+ "#4.5*w=M\n",
+ "#After rearranging and further simplifying we get\n",
+ "w=M*4.5**-1*10**3*10**-9\n",
+ "\n",
+ "#Result\n",
+ "print\"Moment of Resistance is\",round(M,2),\"KN-mm\"\n",
+ "print\"Load the section can carry is\",round(w,3),\"KN/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Moment of Resistance is 198770121.83 KN-mm\n",
+ "Load the section can carry is 44.171 KN/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.4,Page no.134"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flange (Top)\n",
+ "b1=80 #mm #Width \n",
+ "t1=40 #mm #Thickness\n",
+ "\n",
+ "#Flange (Bottom)\n",
+ "b2=160 #mm #width\n",
+ "t2=40 #mm #Thickness\n",
+ "\n",
+ "#web\n",
+ "d=120 #mm #Depth\n",
+ "t3=20 #mm #Thickness\n",
+ "\n",
+ "D=200 #mm #Overall Depth\n",
+ "sigma1=30 #N/mm**2 #Tensile stress\n",
+ "sigma2=90 #N/mm**2 #Compressive stress\n",
+ "L=6000 #mm #Span\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of centroid from bottom fibre\n",
+ "y_bar=(b1*t1*(D-t1*2**-1)+d*t3*(d*2**-1+t2)+b2*t2*t2*2**-1)*(b1*t1+d*t3+b2*t2)**-1 #mm\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*b1*t1**3+b1*t1*(D-t1*2**-1-round(y_bar,2))**2+1*12**-1*t3*d**3+t3*d*(d*2**-1+t2-round(y_bar,2))**2+1*12**-1*b2*t2**3+b2*t2*(t2*2**-1-round(y_bar,2))**2\n",
+ "\n",
+ "#Extreme fibre distance of top and bottom fibres are y_t and y_c respectively\n",
+ "\n",
+ "y_t=y_bar #mm\n",
+ "y_c=D-y_bar #mm\n",
+ "\n",
+ "#Moment carrying capacity considering Tensile strength \n",
+ "M1=sigma1*I*y_t**-1*10**-6 #KN-m\n",
+ "\n",
+ "#Moment carrying capacity considering compressive strength \n",
+ "M2=sigma2*I*y_c**-1*10**-6 #KN-m\n",
+ "\n",
+ "#Max Bending moment in simply supported beam 6 m due to u.d.l\n",
+ "#M_max=w*L*10**-3*8**-1\n",
+ "#After simplifying further we get\n",
+ "#M_max=4.5*w\n",
+ "\n",
+ "#Now Equating it to Moment carrying capacity, we get load carrying capacity\n",
+ "w=M1*4.5**-1 #KN/m\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Uniformly Distributed Load is\",round(w,3),\"KN/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Uniformly Distributed Load is 5.096 KN/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.5,Page no.136"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flanges\n",
+ "b=200 #mm #Width\n",
+ "t=25 #mm #Thickness \n",
+ "\n",
+ "D1=500 #mm #Overall Depth\n",
+ "t2=20 #mm #Thickness of web\n",
+ "\n",
+ "d=450 #mm #Depth of web\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Consider,Element of Thickness \"y\" at Distance \"dy\" from N.A \n",
+ "#Let Bending stress \"sigma_max\"\n",
+ "\n",
+ "#Stress on the element \n",
+ "#sigma=y*(D*2**-1)*sigma_max ..............(1)\n",
+ "\n",
+ "#Area of Element\n",
+ "#A=b*dy .................................(2)\n",
+ "\n",
+ "#Force on Element \n",
+ "#F=y*250**-1*sigma_max*b*dy\n",
+ "\n",
+ "#Let M be the Moment of resistance\n",
+ "#M=y*250**-1*sigma_max*b*dy*y\n",
+ "\n",
+ "#Moment of Resistance of top flange be M1\n",
+ "def integrand(y, b, D):\n",
+ " return b*y**2*D**-1\n",
+ "b=200 \n",
+ "D=250\n",
+ "\n",
+ "X = quad(integrand, 225, 250, args=(b,D))\n",
+ "\n",
+ "Y=2*X[0]\n",
+ "\n",
+ "#M1=Y*sigma\n",
+ "\n",
+ "#Now Moment of Inertia I section is\n",
+ "X=b*D1**3\n",
+ "Y=(b-t2)*d**3\n",
+ "I=(X-Y)*12**-1*10**-8\n",
+ "\n",
+ "#Moment acting on the entire section\n",
+ "#since sigmais the value at y=250\n",
+ "y_max=250\n",
+ "Z=I*10**8*y_max**-1\n",
+ "#M=sigma*Z \n",
+ "#After Simplifying Further we get\n",
+ "#M2=Z*sigma\n",
+ "\n",
+ "#Percentage Moment resisted by Flanges\n",
+ "P1=2258333.3*(2865833.3)**-1*100\n",
+ "\n",
+ "#Percentage Moment resisted by web\n",
+ "P2=100-P1\n",
+ "\n",
+ "#Result\n",
+ "print\"Percentage Moment resisted by Flanges\",round(P1,2),\"%\"\n",
+ "print\"Percentage Moment resisted by web\",round(P2,2),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percentage Moment resisted by Flanges 78.8 %\n",
+ "Percentage Moment resisted by web 21.2 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.6,Page no.137"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flanges\n",
+ "b1=200 #mm #Width\n",
+ "t1=10 #mm #Thickness\n",
+ "\n",
+ "#Web\n",
+ "d=380 #mm #Depth \n",
+ "t2=8 #mm #Thickness\n",
+ "\n",
+ "D=400 #mm #Overall Depth\n",
+ "sigma=150 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Area\n",
+ "A=b1*t1+d*t2+b1*t1 #mm**2\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*(b1*D**3-(b1-t2)*d**3)\n",
+ "\n",
+ "#Bending Moment\n",
+ "M=sigma*I*(D*2**-1)**-1\n",
+ "\n",
+ "#Square Section\n",
+ "\n",
+ "#Let 'a' be the side\n",
+ "a=A**0.5\n",
+ "\n",
+ "#Moment of Resistance of this section\n",
+ "M1=1*6**-1*a*a**2*sigma\n",
+ "\n",
+ "X=M*M1**-1\n",
+ "\n",
+ "#Rectangular section\n",
+ "#Let 'a' be the side and depth be 2*a\n",
+ "\n",
+ "a=(A*2**-1)**0.5\n",
+ "\n",
+ "#Moment of Rectangular secction\n",
+ "M2=1*6**-1*a*(2*a)**2*sigma\n",
+ "\n",
+ "X2=M*M2**-1\n",
+ "\n",
+ "#Circular section\n",
+ "#A=pi*d1**2*4**-1\n",
+ "\n",
+ "d1=(A*4*pi**-1)**0.5\n",
+ "\n",
+ "#Moment of circular section\n",
+ "M3=pi*32**-1*d1**3*sigma\n",
+ "\n",
+ "X3=M*M3**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Moment of resistance of beam section\",round(M,2),\"mm\"\n",
+ "print\"Moment of resistance of square section\",round(X,2),\"mm\"\n",
+ "print\"Moment of resistance of rectangular section\",round(X2,2),\"mm\"\n",
+ "print\"Moment of resistance of circular section\",round(X3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Moment of resistance of beam section 141536000.0 mm\n",
+ "Moment of resistance of square section 9.58 mm\n",
+ "Moment of resistance of rectangular section 6.78 mm\n",
+ "Moment of resistance of circular section 11.33 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.7,Page no.139"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=12 #KN #Force at End of beam\n",
+ "L=2 #m #span\n",
+ "\n",
+ "#Square section \n",
+ "b=d=200 #mm #Width and depth of beam\n",
+ "\n",
+ "#Rectangular section\n",
+ "b1=150 #mm #Width\n",
+ "d1=300 #mm #Depth\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max bending Moment\n",
+ "M=F*L*10**6 #N-mm\n",
+ "\n",
+ "#M=sigma*b*d**2\n",
+ "sigma=M*6*(b*d**2)**-1 #N/mm**2\n",
+ "\n",
+ "#Let W be the central concentrated Load in simply supported beam of span L1=3 m\n",
+ "#MAx Moment\n",
+ "#M1=W*L1*4**-1\n",
+ "#After Further simplifying we get\n",
+ "#M1=0.75*10**6 #N-mm\n",
+ "\n",
+ "#The section has a moment of resistance\n",
+ "M1=sigma*1*6**-1*b1*d1**2\n",
+ "\n",
+ "#Equating it to moment of resistance we get max load W\n",
+ "#0.75*10**6*W=M1\n",
+ "#After Further simplifying we get\n",
+ "W=M1*(0.75*10**6)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum Concentrated Load required to brek the beam\",round(W,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum Concentrated Load required to brek the beam 54.0 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.8,Page no.140"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=3 #m #span\n",
+ "sigma_t=35 #N/mm**2 #Permissible stress in tension\n",
+ "sigma_c=90 #N/mm**2 #Permissible stress in compression\n",
+ "\n",
+ "#Flanges\n",
+ "t=30 #mm #Thickness\n",
+ "d=250 #mm #Depth\n",
+ "\n",
+ "#Web\n",
+ "t2=25 #mm #Thickness\n",
+ "b=600 #mm #Width\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let y_bar be the Distance of N.A from Extreme Fibres\n",
+ "y_bar=(t*d*d*2**-1*2+(b-2*t)*t2*t2*2**-1)*(t*d*2+(b-2*t)*t2)**-1\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=(1*12**-1*t*d**3+t*d*(d*2**-1-y_bar)**2)*2+1*12**-1*(b-2*t)*t2**3+(b-2*t)*t2*(t2*2**-1-y_bar)**2\n",
+ "\n",
+ "#Part-1\n",
+ "\n",
+ "#If web is in Tension\n",
+ "y_t=y_bar #mm\n",
+ "y_c=d-y_bar #mm\n",
+ "\n",
+ "#Moment carrying caryying capacity From consideration of tensile stress\n",
+ "M=sigma_t*I*(y_bar)**-1 #N-mm\n",
+ "\n",
+ "#Moment carrying caryying capacity From consideration of compressive stress\n",
+ "M1=sigma_c*I*(y_c)**-1 #N-mm\n",
+ "\n",
+ "#If w KN/m is u.d.l in beam,Max bending moment\n",
+ "#M=wl**2*8**-1\n",
+ "#After further simplifyng we get\n",
+ "#M=1.125*w*10**6 N-mm\n",
+ "w=M*(1.125*10**6)**-1 #KN\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#If web is in compression\n",
+ "y_t2=178.299 #mm\n",
+ "y_c2=71.71 #mm \n",
+ "\n",
+ "#Moment carrying caryying capacity From consideration of tensile stress\n",
+ "M2=sigma_t*I*(y_t2)**-1 #N-mm\n",
+ "\n",
+ "#Moment carrying caryying capacity From consideration of compressive stress\n",
+ "M3=sigma_c*I*(y_c2)**-1 #N-mm\n",
+ "\n",
+ "#Moment of resistance is M2\n",
+ "\n",
+ "#Equating it to bending moment we get\n",
+ "#M2=1.125*10**6*w2\n",
+ "#After further simplifyng we get\n",
+ "w2=M2*(1.125*10**6)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Uniformly Distributed Load carrying capacity if:web is in Tension\",round(w,2),\"KN\"\n",
+ "print\" :web is in compression\",round(w2,3),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Uniformly Distributed Load carrying capacity if:web is in Tension 73.21 KN\n",
+ " :web is in compression 29.446 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.9,Page no.141"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b1=200 #mm #Width at base\n",
+ "b2=100 #mm #Width at top\n",
+ "\n",
+ "L=8 #m Length\n",
+ "P=500 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Consider a section at y metres from top\n",
+ "\n",
+ "#At this section diameter d is\n",
+ "#d=b2+y*L**-1*(b1-b2)\n",
+ "#After Further simplifying we get\n",
+ "#d=b2+12.5*y #mm\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "#I=pi*64**-1*d**4\n",
+ "\n",
+ "#Section Modulus \n",
+ "#Z=pi*32**-1*(b1+12.5*y)**3\n",
+ "\n",
+ "#Moment \n",
+ "#M=5*10**5*y #N-mm\n",
+ "\n",
+ "#Let sigma be the fibre stress at this section then\n",
+ "#M=sigma*Z\n",
+ "#After sub values in above equation and further simplifying we get\n",
+ "#sigma=5*10**5*32*pi**-1*y*((b2+12.5*y)**3)**-1\n",
+ "\n",
+ "#For sigma to be Max,d(sigma)*(dy)**-1=0\n",
+ "#16*10**6*pi**-1*((b2+12.5*y)**-3+y*(-3)*(b2+12.5*y)**-4*12.5)\n",
+ "#After Further simplifying we get\n",
+ "#b2+12.5*y=37.5*y\n",
+ "#After Further simplifying we get\n",
+ "y=b2*25**-1 #m\n",
+ "\n",
+ "#Stress at this section\n",
+ "sigma=5*10**5*32*pi**-1*y*((b2+12.5*y)**3)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress at Extreme Fibre is max\",round(y,2),\"m\"\n",
+ "print\"Max stress is\",round(sigma,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress at Extreme Fibre is max 4.0 m\n",
+ "Max stress is 6.04 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.10,Page no.143"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "H=10 #mm #Height\n",
+ "A1=160*160 #mm**2 #area of square section at bottom\n",
+ "L1=160 #mm #Length of square section at bottom\n",
+ "b1=160 #mm #width of square section at bottom\n",
+ "A2=80*80 #mm**2 #area of square section at top\n",
+ "L2=80 #mm #Length of square section at top\n",
+ "b2=80 #mm #Width of square section at top\n",
+ "P=100 #N #Pull\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Consider a section at distance y from top.\n",
+ "#Let the side of square bar be 'a'\n",
+ "#a=L2+y*(H)**-1*(b1-b2)\n",
+ "#After further simplifying we get\n",
+ "#a=L2+8*y\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "#I=2*1*12**-1*a*(2)**0.5*(a*((2)**0.5)**-1)**3\n",
+ "#After further simplifying we get\n",
+ "#I=a**4*12**-1\n",
+ "\n",
+ "#Section Modulus \n",
+ "#Z=a**4*(12*a*(2)**0.5)**-1\n",
+ "#After further simplifying we get\n",
+ "#Z=2**0.5*a**3*(12)**-1 #mm**3\n",
+ "\n",
+ "#Bending moment at this section=100*y N-mm\n",
+ "#M=100*10**3*y #N-mm\n",
+ "\n",
+ "#But\n",
+ "#M=sigma*Z\n",
+ "#After sub values in above equation we get\n",
+ "#sigma=M*Z**-1\n",
+ "#After further simplifying we get\n",
+ "#sigma=1200*10**3*(2**0.5)**-1*y*((80+80*y)**3)**-1 .......(1)\n",
+ "\n",
+ "#For Max stress df*(dy)**-1=0\n",
+ "#After taking Derivative of above equation we get\n",
+ "#df*(dy)**-1=1200*10**3*(2**0.5)**-1*((80+8*y)**-3+y(-3)*(80+8*y)**-4*8)\n",
+ "#After further simplifying we get\n",
+ "y=80*16**-1 #m\n",
+ "\n",
+ "#Max stress at this level is\n",
+ "sigma=1200*10**3*(2**0.5)**-1*y*((80+8*y)**3)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Bending stress is Developed at\",round(y,3),\"m\"\n",
+ "print\"Value of Max Bending stress is\",round(sigma,3),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Bending stress is Developed at 5.0 m\n",
+ "Value of Max Bending stress is 2.455 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.12,Page no.147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b=200 #mm #Width of timber \n",
+ "d=400 #mm #Depth of timber\n",
+ "t=6 #mm #Thickness\n",
+ "b2=200 #mm #width of steel plate\n",
+ "t2=20 #mm #Thickness of steel plate\n",
+ "M=40*10**6 #KN-mm #Moment\n",
+ "#Let E_s*E_t**-1=X\n",
+ "X=20 #Ratio of Modulus of steel to timber\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#let y_bar be the Distance of centroidfrom bottom most fibre\n",
+ "y_bar=(b*d*(b+t)+t2*b2*t*t*2**-1)*(b*d+t2*b2*t)**-1 #mm\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*b*d**3+b*d*(b+t-round(y_bar,3))**2+1*12**-1*t2*b2*t**3+b2*t2*t*(round(y_bar,3)-t*2**-1)**2\n",
+ "\n",
+ "#distance of the top fibre from N-A\n",
+ "y_1=d+t-y_bar #mm\n",
+ "\n",
+ "#Distance of the junction of timber and steel From N-A\n",
+ "y_2=y_bar-t #mm\n",
+ "\n",
+ "#Stress in Timber at the top\n",
+ "Y=M*I**-1*y_1 #N/mm**2\n",
+ "\n",
+ "#Stress in the Timber at the junction point\n",
+ "Z=M*I**-1*y_2\n",
+ "\n",
+ "#Coressponding stress in steel at the junction point\n",
+ "Z2=X*Z #N/mm**2 \n",
+ "\n",
+ "#The stress in Extreme steel fibre \n",
+ "Z3=X*M*I**-1*y_bar\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress in Extreme steel Fibre\",round(Z3,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in Extreme steel Fibre 69.67 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.13,Page no.149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Timber size\n",
+ "b=150 #mm #Width\n",
+ "d=300 #mm #Depth\n",
+ "\n",
+ "t=6 #mm #Thickness of steel plate\n",
+ "l=6 #m #Span\n",
+ "\n",
+ "#E_s*E_t**-1=20 \n",
+ "#m=E_s*E_t**-1\n",
+ "m=20 \n",
+ "sigma_timber=8 #N/mm**2 #Stress in timber\n",
+ "sigma_steel=150 #N/mm**2 #Stress in steel plate\n",
+ "\n",
+ "#Let m*t=Y\n",
+ "Y=m*t #mm\n",
+ "L=(2*t+b)*m #mm #Width of flitched beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Due to synnetry cenroid,the neutral axis is half the depth\n",
+ "I=(1*12**-1*L*t**3+L*t*(b+t*2**-1)**2)*2+1*12**-1*(Y+b+Y)*d**3 #mm**4\n",
+ "\n",
+ "y_max1=150 #mm #For timber\n",
+ "y_max2=156 #mm #For steel\n",
+ "\n",
+ "#stress in steel\n",
+ "f_t1=1*m**-1*sigma_steel #N/mm**2\n",
+ "\n",
+ "#Moment of resistance\n",
+ "M=f_t1*(I*y_max2**-1)\n",
+ "\n",
+ "#load\n",
+ "w=8*M*(l**2)**-1*10**-6 #KN/m\n",
+ "\n",
+ "#Result\n",
+ "print\"Load beam can carry is\",round(w,2),\"KN/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Load beam can carry is 19.1 KN/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.14,Page no.151"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=6000 #mm #Span of beam\n",
+ "W=20*10**3 #N #Load\n",
+ "sigma=8 #N/mm**2 #Stress\n",
+ "b=200 #mm #Width of section\n",
+ "d=300 #mm #Depth of section\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#let x be the distance from left side of beam\n",
+ "\n",
+ "#Bending moment\n",
+ "#M=W*2**-1*x #Nmm .......(1)\n",
+ "\n",
+ "#But M=sigma*Z ..........(2)\n",
+ "\n",
+ "#Equating equation 1 and 2 we get\n",
+ "#W*2**-1*x=sigma*Z ............(3)\n",
+ "\n",
+ "#Section Modulus \n",
+ "#Z=1*6*b*d**2 ...............(4)\n",
+ "\n",
+ "#Equating equation 3 and 4 we get\n",
+ "#b*d**2=3*W*x*sigma**-1 .............(5)\n",
+ "\n",
+ "#Beam of uniform strength of constant depth\n",
+ "#b=3*W*x*(sigma*d**2) \n",
+ "\n",
+ "#When x=0\n",
+ "b=0\n",
+ "\n",
+ "#When x=L*2**-1\n",
+ "b2=3*W*L*(2*sigma*d**2)**-1 #mm\n",
+ "\n",
+ "#Beam with constant width of 200 mm\n",
+ "\n",
+ "#We have\n",
+ "#d=(3*W*x*(sigma*d)**-1)**0.5\n",
+ "#thus depth varies as (x)**0.5\n",
+ "\n",
+ "#when x=0\n",
+ "d1=0\n",
+ "\n",
+ "#when x=L*2**-1\n",
+ "d2=(3*W*L*(2*sigma*200)**-1)**0.5 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Cross section of rectangular beam is:\",round(b2,2),\"mm\"\n",
+ "print\" :\",round(d2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Cross section of rectangular beam is: 250.0 mm\n",
+ " : 335.41 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.15,Page no.154"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=800 #mm #Span\n",
+ "n=5 #number of leaves\n",
+ "b=60 #mm #Width\n",
+ "t=10 #mm #thickness\n",
+ "sigma=250 #N/mm**2 #Stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#section Modulus\n",
+ "Z=n*6**-1*b*t**2 #mm**3\n",
+ "\n",
+ "#from the relation\n",
+ "#sigma*Z=M ...................(1)\n",
+ "#M=P*L*4**-1\n",
+ "#sub values of M in equation 1 we get\n",
+ "P=sigma*Z*4*L**-1*10**-3 #KN #Load\n",
+ "\n",
+ "#Length of Leaves\n",
+ "L1=0.2*L #mm\n",
+ "L2=0.4*L #mm\n",
+ "L3=0.6*L #mm\n",
+ "L4=0.8*L #mm\n",
+ "L5=L #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Load it can take is\",round(P,2),\"KN\"\n",
+ "print\"Length of leaves:L1\",round(L1,2),\"mm\"\n",
+ "print\" :L2\",round(L2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Load it can take is 6.25 KN\n",
+ "Length of leaves:L1 160.0 mm\n",
+ " :L2 320.0 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.16,Page no.161"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "matplotlib.pyplot\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=20*10**3 #N #Shear Force\n",
+ "\n",
+ "#Tee section\n",
+ "\n",
+ "#Flange\n",
+ "b=100 #mm #Width\n",
+ "t=12 #mm #Thickness\n",
+ "\n",
+ "#Web\n",
+ "d=88 #mm #Depth\n",
+ "t2=12 #mm #Thicknes\n",
+ "\n",
+ "D=100 #mm #Overall Depth\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of C.G from Top Fibre\n",
+ "y=(b*t*t*2**-1+t2*d*(d*2**-1+t))*(b*t+d*t2)**-1 #mm \n",
+ "\n",
+ "#Moment Of Inertia\n",
+ "I=1*12**-1*b*t**3+b*t*(y-t*2**-1)**2+1*12**-1*t2*d**3+t2*d*(t+d*2**-1-y)**2 #mm**4\n",
+ "\n",
+ "#shear stress at bottom Flange\n",
+ "\n",
+ "#Area above this level\n",
+ "A=b*t #mm**2\n",
+ "\n",
+ "#C.G of this area from N-A\n",
+ "y2=y-t*2**-1\n",
+ "\n",
+ "#Stress at bottom of flange\n",
+ "sigma=F*A*y2*(b*I)**-1 #N/mm**2 \n",
+ "\n",
+ "#sigma2 at same level but in web where width is 12 mm\n",
+ "sigma2=F*A*y2*(t2*I)**-1 #N/mm**2 \n",
+ "\n",
+ "#To find shear stress at N-A\n",
+ "X=t*b*(y-t*2**-1)+t2*(y-t2)*(y-t2)*2**-1 #mm**3\n",
+ "\n",
+ "sigma3=F*X*(t2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Shear stress at top and bottom fibre is zero\n",
+ "#sigma4 and sigma5 are top and bottom fibre shear stress\n",
+ "sigma4=sigma5=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,t,t,y,D]\n",
+ "Y1=[sigma4,sigma,sigma2,sigma3,sigma5]\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"
+ ],
+ "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 0x5667930>"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.17,Page no.163"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "matplotlib.pyplot\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=40*10**3 #N #shear Force\n",
+ "\n",
+ "#I-section\n",
+ "\n",
+ "#Flanges\n",
+ "b=80 #mm #Width of flange\n",
+ "t=20 #mm #Thickness\n",
+ "\n",
+ "#Web\n",
+ "d=200 #mm #Depth\n",
+ "t2=20 #mm #Thickness\n",
+ "\n",
+ "#Flange-2\n",
+ "b2=160 #mm #Width\n",
+ "t3=20 #mm #Thickness\n",
+ "\n",
+ "D=240 #mm #Overall Depth\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of N-A from Top Fibre \n",
+ "y=(b*t*t*2**-1+d*t2*(t+d*2**-1)+b2*t3*(t+d+t3*2**-1))*(b*t+d*t2+b2*t3)**-1 #mm\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*b*t**3+b*t*(y-(t*2**-1))**2+1*12**-1*t2*d**3+t2*d*(y-(t+d*2**-1))**2+1*12**-1*b2*t3**3+t3*b2*((d+t+t3*2**-1)-y)**2 #mm**4\n",
+ "\n",
+ "#Shear stress bottom of flange\n",
+ "sigma=F*b*t*(y-t*2**-1)*(b*I)**-1 #N/mm**2\n",
+ "\n",
+ "#At same Level but in web\n",
+ "sigma2=F*b*t*(y-t*2**-1)*(t2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#for shear stress at N.A\n",
+ "X=b*t*(y-t*2**-1)+t2*(y-t)*(y-t)*2**-1 #mm**3\n",
+ "sigma3=F*X*(t2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Shear stress at bottom of web\n",
+ "\n",
+ "X=b2*t3*((D-y)-t3*2**-1) #mm**3\n",
+ "\n",
+ "#Stress at bottom of web\n",
+ "sigma4=F*X*(t2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Stress at Lower flange\n",
+ "sigma5=F*X*(b2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force Diagram is the result\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,20,20,140,220,220,240]\n",
+ "Y1=[0,sigma,sigma2,sigma3,sigma4,sigma5,0]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length in mm\")\n",
+ "plt.ylabel(\"Shear Force in N\")\n",
+ "plt.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force Diagram is the result\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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OnYL/+i/TF3DLLWZPgXbt7K5KRHyRwsAGJ0+afoBXXjHNQOvXQ1iY3VWJiC8r1gj183MN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\n/w/qZz1xEBCKMwAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5683070>"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.18,Page no.164"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=30*10**3 #N #Shear Force\n",
+ "\n",
+ "#Channel Section\n",
+ "d=400 #mm #Depth of web \n",
+ "t=10 #mm #THickness of web\n",
+ "t2=15 #mm #Thickness of flange\n",
+ "b=100 #mm #Width of flange\n",
+ "\n",
+ "#Rectangular Welded section\n",
+ "b2=80 #mm #Width\n",
+ "d2=60 #mm #Depth\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of Centroid From Top Fibre\n",
+ "y=(d*t*t*2**-1+2*t2*(b-t)*((b-t)*2**-1+10)+d2*b2*(d2*2**-1+t))*(d*t+2*t2*(b-t)+d2*b2)**-1 #mm\n",
+ "\n",
+ "#Moment Of Inertia of the section about N-A\n",
+ "I=1*12**-1*d*t**3+d*t*(y-t*2**-1)**2+2*(1*12**-1*t2*(b-t)**3+t2*(b-t)*(((b-t)*2**-1+t)-y)**2)+1*12**-1*d2**3*b2+d2*b2*(d2*2**-1+t-y)**2\n",
+ "\n",
+ "#Shear stress at level of weld\n",
+ "sigma=F*d*t*(y-t*2**-1)*((b2+t2+t2)*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Max Shear Stress occurs at Neutral Axis\n",
+ "X=d*t*(y-t*2**-1)+2*t2*(y-t)*(y-t)*2**-1+b2*(y-t)*(y-t)*2**-1\n",
+ "\n",
+ "sigma_max=F*X*((b+t)*I)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Shear stress in the weld is\",round(sigma,2),\"N/mm**2\"\n",
+ "print\"Max shear stress is\",round(sigma_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shear stress in the weld is 3.62 N/mm**2\n",
+ "Max shear stress is 4.48 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.19,Page no.165"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Wooden Section\n",
+ "b=300 #mm #Width\n",
+ "d=300 #mm #Depth\n",
+ "\n",
+ "D=100 #mm #Diameter of Bore\n",
+ "F=10*10**3 #N #Shear Force\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Moment Of Inertia Of Section\n",
+ "I=1*12**-1*b*d**3-pi*64**-1*D**4\n",
+ "\n",
+ "#Shear stress at crown of circle\n",
+ "sigma=F*b*D*(d*2**-1-D*2**-1)*(b*I)**-1\n",
+ "\n",
+ "#Let a*y_bar=X\n",
+ "X=b*d*2**-1*d*4**-1-pi*8**-1*D**2*4*D*2**-1*(3*pi)**-1 #mm**3\n",
+ "\n",
+ "#Shear Stress at Neutral Axis\n",
+ "sigma2=F*X*((b-D)*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Shearing Stress at Crown of Bore\",round(sigma,3),\"N/mm**2\"\n",
+ "print\"Shear Stress at Neutral Axis\",round(sigma2,3),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shearing Stress at Crown of Bore 0.149 N/mm**2\n",
+ "Shear Stress at Neutral Axis 0.246 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.20,Page no.166"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#flanges\n",
+ "b=200 #mm #width\n",
+ "t1=25 #mm #Thickness\n",
+ "\n",
+ "#web\n",
+ "d=450 #mm #Depth \n",
+ "t2=20 #mm #thickness\n",
+ "\n",
+ "D=500 #mm #Total Depth of section\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Moment Of Inertia of the section about N-A\n",
+ "I=1*12**-1*b*D**3-1*12**-1*(b-t2)*d**3 #mm**4\n",
+ "\n",
+ "#Consider an element in the web at distance y from y from N-A\n",
+ "#Depth of web section=225-y\n",
+ "\n",
+ "#C.G From N-A\n",
+ "#y2=y+(((D*2**-1-t)-y)*2**-1)\n",
+ "\n",
+ "#ay_bar for section at y\n",
+ "#Let ay_bar be X\n",
+ "#X=X1 be of Flange + X2 be of web above y\n",
+ "#X=b*t1*(D*2**-1-t1*2**-1)+t2*(d-t1)*(d-t1+y)*2**-1\n",
+ "#After Sub values and Further simplifying we get\n",
+ "#X=1187500+10*(225**2-y**2)\n",
+ "\n",
+ "#Shear stress at y\n",
+ "#sigma_y=F*(X)*(t2*I)**-1\n",
+ "\n",
+ "#Shear Force resisted by the Element\n",
+ "#F1=F*X*t2*dy*(t2*I)**-1\n",
+ "\n",
+ "#Shear stress resisted by web \n",
+ "#sigma=2*F*I**-1*(X)*dy\n",
+ "\n",
+ "#After Integrating above equation and further simplifying we get\n",
+ "#sigma=0.9578*F\n",
+ "\n",
+ "sigma=0.9578*100\n",
+ "\n",
+ "#Result\n",
+ "print\"Shear Resisted by web\",round(sigma,2),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shear Resisted by web 95.78 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.21,Page no.167"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Wooden Beam\n",
+ "\n",
+ "b=150 #mm #width\n",
+ "d=250 #mm #Depth\n",
+ "\n",
+ "L=5000 #mm #span\n",
+ "m=11.2 #N/mm**2 #Max Bending stress\n",
+ "sigma=0.7 #N/mm**2 #Max shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let 'a' be the distance from left support\n",
+ "#Max shear force\n",
+ "#F=R_A=W*(L-a)*L**-1 \n",
+ "\n",
+ "#Max Moment\n",
+ "#M=W*(L-a)*a*L**-1\n",
+ "\n",
+ "#But M=sigma*Z\n",
+ "#W*(L-a)*a*L**-1=m*1*6**-1*b*d**2 .....................(1)\n",
+ "\n",
+ "#In Rectangular Section MAx stress is 1.5 times Avg shear stress\n",
+ "F=sigma*b*d*1.5**-1\n",
+ "\n",
+ "#W*(L-a)*L**-1=F .....................(2)\n",
+ "\n",
+ "#Dividing Equation 1 nad 2 we get\n",
+ "a=m*6**-1*b*d**2*1.5*(sigma*b*d)**-1\n",
+ "\n",
+ "#Sub above value in equation 2 we get\n",
+ "W=(L-a)**-1*L*F*10**-3 #KN \n",
+ "\n",
+ "#Result\n",
+ "print\"Load is\",round(W,2),\"KN\"\n",
+ "print\"Distance from Left support is\",round(a,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Load is 21.87 KN\n",
+ "Distance from Left support is 1000.0 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.22,Page no.168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1000 #mm #span\n",
+ "\n",
+ "#Rectangular Section\n",
+ "\n",
+ "b=200 #mm #width\n",
+ "d=400 #mm #depth\n",
+ "\n",
+ "sigma=1.5 #N/mm**2 #Shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let AB be the cantilever beam subjected to load W KN at free end\n",
+ "\n",
+ "#MAx shear Force\n",
+ "#F=W*10**3 #KN\n",
+ "\n",
+ "#Since Max shear stress in Rectangular section\n",
+ "#sigma_max=1.5*F*A**-1 \n",
+ "#After sub values and further simplifyng we get\n",
+ "W=1.5*b*d*(1.5*1000)**-1 #KN\n",
+ "\n",
+ "#Moment at fixwed end\n",
+ "M=W*1 #KN-m\n",
+ "y_max=d*2**-1 #mm\n",
+ "\n",
+ "#M.I\n",
+ "I=1*12**-1*b*d**3 #mm**3\n",
+ "\n",
+ "#MAx Stress\n",
+ "sigma_max=M*10**6*I**-1*y_max\n",
+ "\n",
+ "#Result\n",
+ "print\"Concentrated Load is\",round(sigma_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Concentrated Load is 15.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.24,Page no.170"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=4000 #mm #span\n",
+ "\n",
+ "#Rectangular Cross-section\n",
+ "b=100 #mm #Width\n",
+ "d=200 #mm #Thickness\n",
+ "\n",
+ "F_per=10 #N/mm**2 #Max Bending stress\n",
+ "q_max=0.6 #N/mm**2 #Shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#If the Load W is in KN/m\n",
+ "\n",
+ "#Max shear Force\n",
+ "#F=w*l*2**-1 #KN\n",
+ "#After substituting values and further simplifying we get\n",
+ "#M=2*w #KN-m\n",
+ "\n",
+ "#Max Load from Consideration of moment\n",
+ "#M=1*6**-1*b*d**2*F_per\n",
+ "#After substituting values and further simplifying we get\n",
+ "w=(1*6**-1*b*d**2*F_per)*(2*10**6)**-1 #KN/m\n",
+ "\n",
+ "#Max Load from Consideration of shear stress\n",
+ "#q_max=1.5*F*(b*d)**-1 #N\n",
+ "#After substituting values and further simplifying we get\n",
+ "F=q_max*(1.5)*b*d #N\n",
+ "\n",
+ "#If w is Max Load in KN/m,then\n",
+ "#2*w*1000=8000\n",
+ "#After Rearranging and Further simplifying we get\n",
+ "w2=8000*(2*1000)**-1 #KN/m\n",
+ "\n",
+ "#Result\n",
+ "print\"Uniformly Distributed Load Beam can carry is\",round(w,2),\"KN/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Uniformly Distributed Load Beam can carry is 3.33 KN/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.5_9.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.5_9.ipynb
new file mode 100644
index 00000000..35226aed
--- /dev/null
+++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.5_9.ipynb
@@ -0,0 +1,1019 @@
+{
+ "metadata": {
+ "name": "chapter no.5.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 5:Deflections Of Beams By Double Integration Methods"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.2,Page No.192"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=3000 #mm #span of beam\n",
+ "a=2000 #mm\n",
+ "W1=20*10**3 #N #Pt Load Acting on beam\n",
+ "W2=30*10**3 #N #Pt Load Acting on beam\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus\n",
+ "I=2*10**8 #mm**4 #M.I\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Deflection at free End Due to W2\n",
+ "dell1=W2*L**3*(3*E*I)**-1 #mm\n",
+ "\n",
+ "#Deflection at free end Due to W1\n",
+ "dell2=W1*a**3*(3*E*I)**-1+(L-a)*W1*a**2*(2*E*I)**-1 #mm\n",
+ "\n",
+ "#Total Deflection at free end\n",
+ "dell=dell1+dell2 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflection at Free End is\",round(dell,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflection at Free End is 9.08 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.4,Page No.193"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus\n",
+ "I=180*10**6 #mm**4 #M.I\n",
+ "W1=20 #N/m #u.d.l\n",
+ "W2=20*10**3 #N #Pt load\n",
+ "L=3000 #m #Span of beam\n",
+ "a=2000 #m #Span of u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Displacement of free End due to 20 KN Pt load at free end\n",
+ "dell1=W2*L**3*(3*E*I)**-1 #mm\n",
+ "\n",
+ "#Displacement of free end due to u.d.l\n",
+ "dell2=W1*a**4*(8*E*I)**-1+(L-a)*W1*a**3*(6*E*I)**-1\n",
+ "\n",
+ "#Deflection at free end\n",
+ "dell=dell1+dell2 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"The Displacement of Free End of cantilever beam is\",round(dell,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Displacement of Free End of cantilever beam is 6.85 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.10,Page No.201"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=200*10**6 #KN/m**2 #Young's Modulus\n",
+ "I=15*10**-6 #m**4 #M.I\n",
+ "a=4000 #m \n",
+ "L_AB=6 #m #Span of beam\n",
+ "L_CB=2 #m #Length of CB\n",
+ "F_C=18 #KN #force at C\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the Reactions at A & B Respectively\n",
+ "#V_A+V_B=18\n",
+ "#Now taking moment at B,we get M_B\n",
+ "V_A=(F_C*L_CB)*L_AB**-1\n",
+ "V_B=18-V_A\n",
+ "\n",
+ "#Now Taking Moment at distance x\n",
+ "#M_x=6*x-18*(x-4)\n",
+ "#EI*d**2*y*(d*x**2)**-1=6*x-18*(x-4)\n",
+ "\n",
+ "#Now Integrating above equation,we get\n",
+ "#EI*dy*(dx)**-1=C1+3*x**2-9(x-4)**4\n",
+ "\n",
+ "#Again Integrating above equation we get\n",
+ "#EI*y=C2+C1*x+x**3-3*(x-4)**3\n",
+ "\n",
+ "#The Boundary conditions\n",
+ "x=0\n",
+ "y=0 #.....(a)\n",
+ "\n",
+ "x=6\n",
+ "y=0 #....(b)\n",
+ "\n",
+ "#From Boundary Condition(B.C) a we get\n",
+ "C2=0\n",
+ "\n",
+ "#From Boundary Condition(B.C) b we get\n",
+ "#6*C1+216-3*8\n",
+ "#After Further simplifying we get\n",
+ "C1=-(216-24)*6**-1\n",
+ "\n",
+ "#EI*y=-32*x+x**3-3*(x-4)**3\n",
+ "#EI*dy*(dx)**-1=-32+3*x**2-9(x-4)**4\n",
+ "\n",
+ "#For Max Deflection\n",
+ "#Assume it inthe Porion AC i.e x=4=a\n",
+ "#0=-32+3*x**2\n",
+ "x=(32*3**-1)**0.5\n",
+ "\n",
+ "#Value of Max deflection is\n",
+ "ymax=(-32*x+x**3)*(E*I)**-1 #mm\n",
+ "\n",
+ "#slope at mid-span\n",
+ "\n",
+ "#EI*(dy*(dx)**-1)_centre=-32+3*x**2\n",
+ "#at centre ,\n",
+ "x1=3 #m\n",
+ "\n",
+ "#Let (dy*(dx)**-1)_centre=X\n",
+ "X=-(-32+3*x1**2)*(E*I)**-1 #Radian\n",
+ "\n",
+ "#Deflection at Load Point\n",
+ "x2=4 #m\n",
+ "#EI*y_c=-32*x2+x2**3\n",
+ "\n",
+ "y_c=-(-32*x2+x2**3)*(E*I)**-1\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of Max Deflection\",round(ymax,4),\"mm\"\n",
+ "print\"SLope at mid-span\",round(X,4),\"radian\"\n",
+ "print\"Deflection at the Load Point is\",round(y_c,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of Max Deflection -0.0232 mm\n",
+ "SLope at mid-span 0.0017 radian\n",
+ "Deflection at the Load Point is 0.0213 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.11,Page No.203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_CB=2 #m #Length of CB\n",
+ "L_AC=4 #m #Length of AB\n",
+ "M_C=15 #KN.m #Moment At Pt C\n",
+ "F_C=30 #KN\n",
+ "L=6 #m Span of Beam\n",
+ "\n",
+ "#Let X=E*I\n",
+ "X=10000 #KN-m**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A and V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=30\n",
+ "\n",
+ "#Taking Moment a A,we get\n",
+ "V_B=(F_C*L_AC+M_C)*L**-1\n",
+ "V_A=30-V_B\n",
+ "\n",
+ "#Now Taking Moment at distacnce x from A\n",
+ "#M_x=7.5*x-30*(x-4)+15\n",
+ "\n",
+ "#By using Macaulay's Method\n",
+ "#EI*(d**2*x/dx**2)=M_x=7.5*x-30*(x-4)+15\n",
+ "\n",
+ "#Now Integrating above Equation we get\n",
+ "#EI*(dy/dx)=C1+7.5*x**2*2**-1-15*(x-4)**2+15*(x-4) ............(1)\n",
+ "\n",
+ "#Again Integrating above Equation we get\n",
+ "#EIy=C2+C1*x+7.5*6**-1*x**3-5*(x-4)**3+15*(x-4)**2*2**-1..........(2)\n",
+ "\n",
+ "#Boundary Cinditions\n",
+ "x=0\n",
+ "y=0\n",
+ "\n",
+ "#Substituting above equations we get \n",
+ "C2=0\n",
+ "\n",
+ "x=6 #m\n",
+ "y=0\n",
+ "\n",
+ "C1=-(7.5*6**3*6**-1-5*2**3+15*2**2*2**-1)*6**-1\n",
+ "\n",
+ "#EIy_c=C2+C1*x+7.5*6**-1*x**3-5*(x-4)**3+15*(x-4)**2*2**-1\n",
+ "#Sub values in Above equation we get\n",
+ "y_c=(93.3333*(X)**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"The Deflection at C\",round(y_c,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Deflection at C 0.0093 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.12,Page No.204"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_AC=L_CD=L_DB=2 #m #Length of AC,CD,DB\n",
+ "F_C=40 #KN #Force at C\n",
+ "w=20 #KN/m #u.d.l\n",
+ "L=6 #m #span of beam\n",
+ "\n",
+ "#Let E*I=X\n",
+ "X=15000 #KN-m**2\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=80\n",
+ "\n",
+ "#Taking Moment B,M_B\n",
+ "V_A=(F_C*(L_CD+L_DB)+w*L_DB*L_DB*2**-1)*L**-1 #KN\n",
+ "V_B=80-V_A #KN\n",
+ "\n",
+ "#Taking Moment at distance x from A\n",
+ "#M_x=33.333*x-40*(x-2)-20*(x-4)**2*2**-1\n",
+ "#EI*(d**2/dx**2)=33.333*x-40*(x-2)-10*(x-4)**2\n",
+ "\n",
+ "#Integrating above equation we get\n",
+ "#EI*(dy/dx)=C1+33.333*x**2*2**-1-20*(x-2)**2-10*3**-1*(x-4)**3\n",
+ "\n",
+ "#Again Integrating above equation we get\n",
+ "#EI*y=C2+C1*x+33.333*x**3*6**-1-20*3**-1*(x-2)**3-10*12**-1*(x-4)**4\n",
+ "\n",
+ "#At\n",
+ "x=0\n",
+ "y=0\n",
+ "C2=0\n",
+ "\n",
+ "#At\n",
+ "x=6\n",
+ "y=0\n",
+ "C1=-760*6**-1\n",
+ "\n",
+ "#Assuming Deflection to be max in portion CD and sustituting value of C1 in equation of slope we get\n",
+ "#EI*y=C2+C1*x+33.333*x**3*6**-1-20*3**-1*(x-2)**3-10*12**-1*(x-4)**4\n",
+ "#0=-126.667+33.333*x**2**-1-20*(x-2)**2\n",
+ "\n",
+ "#After rearranging and simplifying further we get\n",
+ "\n",
+ "#x**2-24*x+62=0\n",
+ "#From above equations\n",
+ "a=1\n",
+ "b=-24\n",
+ "c=62\n",
+ "\n",
+ "y=(b**2-4*a*c)**0.5\n",
+ "\n",
+ "x1=(-b+y)*(2*a)**-1\n",
+ "x2=(-b-y)*(2*a)**-1\n",
+ "\n",
+ "#Taking x2 into account\n",
+ "x=2.945 #m\n",
+ "C1=-126.667\n",
+ "C2=0\n",
+ "\n",
+ "y_max=(C2+C1*x+33.333*x**3*6**-1-20*3**-1*(x-2)**3)*X**-1 #mm\n",
+ "\n",
+ "#Max slope occurs at the ends\n",
+ "#At A,\n",
+ "#EI*(dy/dx)_A=-126.667\n",
+ "#At B\n",
+ "#EI*(dy/dx)_B=126.667+33.333*6**2*2**-1-20*4**2-10*2**3\n",
+ "#After simplifying Further we get\n",
+ "#EI*(dy/dx)_B=73.3273\n",
+ "\n",
+ "#Now Max slope is EI(dy/dx)_A=-126.667\n",
+ "#15000*(dy/dx)_=-126.667\n",
+ "\n",
+ "#Let Y=dy/dx\n",
+ "Y=-126.667*X**-1 #Radians\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum Deflection for Beam is\",round(y_max,4),\"mm\"\n",
+ "print\"Maximum Slope for beam is\",round(Y,4),\"radians\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum Deflection for Beam is -0.0158 mm\n",
+ "Maximum Slope for beam is -0.0084 radians\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.13,Page No.206"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=2*10**8 #KN/m**2\n",
+ "I=450*10**-6 #m**4\n",
+ "L_AC=1 #m #Length of AC\n",
+ "L_CD=3 #m #Length of CD\n",
+ "L_DB=2 #m #Length of DB\n",
+ "w=10 #KN/m #u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=30\n",
+ "\n",
+ "#Taking Moment at distance x from A\n",
+ "#M_x=17.5*x-10*(x-1)**2*2**-1+10*(x-4)**2*2**-1\n",
+ "#EI*(d**2/dx**2)=17.5*x-10*(x-1)**2*2**-1+10*(x-4)**2*2**-1\n",
+ "\n",
+ "#Now Integrating Above equation we get\n",
+ "#EI(dy/dx)=C1+17.5*x**2*2**-1-5*3**-1*(x-1)**2+5*3**-1*(x-4)**3\n",
+ "\n",
+ "#Again Integrating Above equation we get\n",
+ "#EI*y=C2+C1*x+17.5*x**3*6**-1-5*12**-1*(x-1)**4+5*12**-1*(x-4)**4\n",
+ "\n",
+ "#At \n",
+ "x=0\n",
+ "y=0\n",
+ "C2=0\n",
+ "\n",
+ "#At \n",
+ "x=6 \n",
+ "y=0\n",
+ "C1=(-17.5*x**3*6**-1+5*12**-1*(x-1)**4-5*12**-1*(x-4)**4)*x**-1\n",
+ "\n",
+ "# 1)Slope at A .i.e at x=0\n",
+ "#EI*(dy/dx)_A=C1=-62.708 #KN-m**2\n",
+ "#let (dy/dx)=X\n",
+ "X=C1*(E*I)**-1 #radiams\n",
+ "\n",
+ "#Deflection at mid-span\n",
+ "x=3 #m\n",
+ "#EI*y_centre=C1*x+17.5*x**3*6**-1-5*12**-1*(x-1)**2\n",
+ "y_centre=-(C1*x+17.5*x**3*6**-1-5*12**-1*(x-1)**4)*(E*I)**-1\n",
+ "\n",
+ "#Maximum Deflection\n",
+ "\n",
+ "#At point of Max deflection (dy/dx)=0\n",
+ "#Assuming it in portion CD\n",
+ "\n",
+ "#0=C1*x+17.5*x**2*2**-1-5*3**-1*(x-1)**3\n",
+ "\n",
+ "#Now Let\n",
+ "#F(x)=C1+17.5*x**2*2**-1-5*3**-1*(x-1)**3\n",
+ "\n",
+ "#Let F(x)=Y\n",
+ "#At \n",
+ "x=2.5\n",
+ "Y1=-(C1+17.5*x**2*2**-1-5*3**-1*(x-1)**3)\n",
+ "\n",
+ "#AT\n",
+ "x=3\n",
+ "Y2=-(C1+17.5*x**2*2**-1-5*3**-1*(x-1)**3)\n",
+ "\n",
+ "#At\n",
+ "x=2.9 #m\n",
+ "Y3=-(C1+17.5*x**2*2**-1-5*3**-1*(x-1)**3)\n",
+ "\n",
+ "#A curve may be plotted for (F(x) and the value for which F(x)=0 may be found\n",
+ "#For F(x)=0 for x=2.92 m\n",
+ "#Therefore y_max occur at x=2.92\n",
+ "\n",
+ "x=2.92 #m\n",
+ "y_max=(C1*x+17.5*x**3*6**-1-5*12**-1*(x-1)**4)*(E*I)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Slope at A\",round(X,6),\"mm\"\n",
+ "print\"Deflection at mid-span\",round(y_centre,6),\"mm\"\n",
+ "print\"Maxmimum Deflection is\",round(y_max,5),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Slope at A -0.000697 mm\n",
+ "Deflection at mid-span 0.001289 mm\n",
+ "Maxmimum Deflection is -0.00129 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.14,Page No.208"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_AC=LDE=L_EB=1 #m #Length of AC\n",
+ "L_CD=2 #m #Length of CD\n",
+ "E=200 #KN/mm**2\n",
+ "I=60*10**6 #mm**4 #M.I\n",
+ "F_C=20 #KN #Force at C\n",
+ "F_E=30 #KN #Force at E\n",
+ "w=10 #KN/m #u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "X=E*I*10**-6 #KN-m**2\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=70\n",
+ "\n",
+ "#Taking Moment at distance x from A\n",
+ "#M_x=34*x-20*(x-1)-10*(x-1)**2*2**-1+10*(x-3)**2*2**-1-30*(x-4)\n",
+ "#EI*(d**2y/dx**2)=34*x-20*(x-1)-10*(x-1)**2*2**-1+10*(x-3)**2*2**-1-30*(x-4)\n",
+ "\n",
+ "#Now Integrating Above equation,we get\n",
+ "#EI*(dy/dx)=C1+17*x**2-10*(x-1)**2-5*3**-1*(x-1)**3+5*3**-1*(x-3)**3-15*(x-4)**2\n",
+ "\n",
+ "#Again Integrating Above equation,we get\n",
+ "#EI*y=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4-5*(x-4)**3\n",
+ "\n",
+ "#At\n",
+ "x=0\n",
+ "y=0\n",
+ "C2=0\n",
+ "\n",
+ "#At \n",
+ "x=5 #m\n",
+ "y=0\n",
+ "C1=(-17*3**-1*x**3+10*3**-1*(x-1)**3+5*12**-1*(x-1)**4-5*12**-1*(x-3)**4+5*(x-4)**3)*5**-1\n",
+ "\n",
+ "#EI*y=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4-5*(x-4)**3\n",
+ "C2=0\n",
+ "C1=-78\n",
+ "x=1\n",
+ "y_c=(-78*x+17*3**-1*x)*(X)**-1\n",
+ "\n",
+ "#EI*y_D=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4\n",
+ "x=3\n",
+ "C1-78\n",
+ "C2=0\n",
+ "y_D=(C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4)*(X**-1)\n",
+ "\n",
+ "#EI*y_E=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4\n",
+ "x=4\n",
+ "C1-78\n",
+ "C2=0\n",
+ "y_E=(C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4)*X**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflections at C\",round(y_c,5),\"mm\"\n",
+ "print\"Deflections at D\",round(y_D,5),\"mm\"\n",
+ "print\"Deflections at E\",round(y_E,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflections at C -0.00603 mm\n",
+ "Deflections at D -0.00953 mm\n",
+ "Deflections at E -0.0061 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.15,Page No.209"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "matplotlib.pyplot\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=200 #KN/mm**2 #Modulus of Elasticity\n",
+ "I=300*10**6 #mm\n",
+ "L_AB=L_BC=L_CD=L_DE=1 #m #Length of AB,BC,CD,DE respectively\n",
+ "F_A=20 #KN #Force at A\n",
+ "F_C=10 #KN #Force at C\n",
+ "w=30 #KN/m #u.d.l\n",
+ "\n",
+ "#Let E*I=X\n",
+ "X=E*I*10**-6 #KN-2**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_E be the reactions at E\n",
+ "V_E=F_A+F_C+w*(L_BC+L_CD) #KN \n",
+ "\n",
+ "#Taking Moment at distance x\n",
+ "#EI*(d**2x/dy**2)=M=-20*x-30*(x-1)**2*2**-1-10*(x-2)+30*(x-3)**2*2**-1\n",
+ "\n",
+ "#Integrating above equation we get\n",
+ "#EI*(dy/dx)=C1-10*x**2-5*(x-1)**3-5*(x-2)**2+5*(x-3)**3\n",
+ "\n",
+ "#Again Integrating above equation\n",
+ "#EI*y=C2+C1*x-10*x**3*3**-1-5*(x-1)**4*4**-1-5*(x-3)**4*4**-1-5*3*(x-2)**3\n",
+ "\n",
+ "#At\n",
+ "#dy/dx=0\n",
+ "x=4 #m\n",
+ "C1=10*x**2+5*(x-1)**3+5*(x-2)**2-5*(x-3)**3\n",
+ "\n",
+ "#AT\n",
+ "x=4\n",
+ "y=0\n",
+ "C2=-C1*4+10*x**3*3**-1+5*(x-1)**4*4**-1-5*(x-3)**4*4**-1+5*3**-1*(x-2)**3\n",
+ "\n",
+ "#Max Deflection and Max slopes occurs at Free end in case of cantilever\n",
+ "y_max=y_A=C2*X**-1\n",
+ "\n",
+ "#EI*(dy/dx)_max=C1\n",
+ "#Let (dy/dx)=Y\n",
+ "Y=C1*X**-1 #radian\n",
+ "\n",
+ "#Now deflection at x=1 #m\n",
+ "C2=-913.333\n",
+ "C1=310\n",
+ "x=1\n",
+ "y_B=(C2+C1*x-10*x**3*3**-1)*X**-1\n",
+ "\n",
+ "#Now Deflection at x=2 #m\n",
+ "C2=-913.333\n",
+ "C1=310\n",
+ "x=2 #m\n",
+ "y_C=(C2+C1*x-10*x**3*3**-1-5*(x-1)**4*4**-1)*X**-1\n",
+ "\n",
+ "#Now Deflection at x=3 #m\n",
+ "C2=-913.333\n",
+ "C1=310\n",
+ "x=3 #m\n",
+ "y_D=(C2+C1*x-10*x**3*3**-1-5*(x-1)**4*4**-1-5*3**-1*(x-2)**3)*X**-1\n",
+ "\n",
+ "y_E=0\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Deflection for Beam\",round(y_A,4),\"mm\"\n",
+ "print\"Max Slope for beam\",round(Y,5),\"radians\"\n",
+ "\n",
+ "#Plotting the ELastic Curve\n",
+ "\n",
+ "Y2=[y_E,y_D,y_C,y_B,y_A]\n",
+ "X2=[L_AB+L_BC+L_CD+L_DE,L_AB+L_BC+L_CD,L_AB+L_BC,L_AB,0]\n",
+ "Z2=[0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in mm\")\n",
+ "plt.ylabel(\"Deflection in mm\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Deflection for Beam -0.0152 mm\n",
+ "Max Slope for beam 0.00517 radians\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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8yjGZmZkkJSWRmZlJSkoKM2bMoLKy8pxilcZbsQJGj4Y//QkWL1aSEWnvaq3R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\nL+mlAAAAAElFTkSuQmCC\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5627790>"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.16,Page No.211"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_BD=L_CB=L_AC=2 #m #Length of BD,CB,AC\n",
+ "F_C=40 #KN #Force at C\n",
+ "F_D=10 #KN Force at D\n",
+ "L=6 #m spna of beam\n",
+ "\n",
+ "#EI is constant in this problem\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B Respectively\n",
+ "#V_A+V_B=50\n",
+ "\n",
+ "#Taking Moment at Pt A\n",
+ "V_B=(F_D*L+F_C*L_AC)*(L_AC+L_CB)**-1\n",
+ "V_A=50-V_B\n",
+ "\n",
+ "#Now Taking Moment at distance x from A,M_x\n",
+ "#M_x=15*x-40*(x-2)+35*(x-4)\n",
+ "#EI*(d**2*y/dx**2)=15*x-40*(x-2)+35*(x-4)\n",
+ "\n",
+ "#Now Integrating above equation we get\n",
+ "#EI*(dy/dx)=C1+7.5*x**2-20*(x-2)**2+17.5(x-4)**2\n",
+ "\n",
+ "#Again Integrating above equation we get\n",
+ "#EI*y=C2+C1*x+2.5*x**2-20*3**-1*(x-2)**3+17.5*(x-4)**3*3**-1\n",
+ "\n",
+ "#At\n",
+ "x=0\n",
+ "y=0\n",
+ "#we get\n",
+ "C2=0\n",
+ "\n",
+ "#At\n",
+ "x=4 \n",
+ "y=0\n",
+ "#we get\n",
+ "C1=(2.5*4**3-20*3**-1*2**3)*4**-1\n",
+ "\n",
+ "#Now Deflection at C\n",
+ "x=2\n",
+ "C1=-26.667\n",
+ "C2=0\n",
+ "y_C=C2+C1*x+2.5*x**3\n",
+ "\n",
+ "#Now Deflection at D\n",
+ "C1=-21.667\n",
+ "C2=0\n",
+ "y_D=-26.667*6+2.5*6**3-20*3**-1*4**3+17.5*2**3*3**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflections Under Loads are:y_D\",round(y_D,4)\n",
+ "print\" :y_C\",round(y_C,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflections Under Loads are:y_D -0.002\n",
+ " :y_C -33.33\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.17,Page No.212"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_BC=L_EB=2 #m #Length of BC & EB\n",
+ "E=200*10**6 #KN/m**2 #Modulus of eLasticity\n",
+ "I=45*10**-6 #mm**4 #M.I\n",
+ "L_DE=3 #m #Length of DE\n",
+ "L_AD=1 #m #Length of AD\n",
+ "w=20 #KN/m #u.d.l\n",
+ "L=8 #m #span of beam\n",
+ "F_C=30 #KN #Force at C\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=90\n",
+ "\n",
+ "#Taking Moment at A,M_A\n",
+ "V_B=(w*L_DE*(L_DE*2**-1+L_AD)+F_C*L)*(L_AD+L_DE+L_EB)**-1\n",
+ "V_A=90-V_B\n",
+ "\n",
+ "#Taking Moment at distance x\n",
+ "#M_x=25*x-20*(x-1)**2*2**-1+20*(x-4)**2*2**-1+65*(x-6)\n",
+ "\n",
+ "#Integrating above equation we get\n",
+ "#EI*(d**2*y/dx**2)=25*x-10*(x-1)**2+10*(x-4)**2+65*(x-6)\n",
+ "\n",
+ "#again Integrating above equation we get\n",
+ "#EI*(dy/dx)=C1+25*x**2*2**-1-10*3**-1*(x-1)**3+10*3**-1*(x-4)**2+65*2**-1*(x-6)**2\n",
+ "\n",
+ "#again Integrating above equation we get\n",
+ "#EI*y=C2+C1*x+25*6**-1*x**3-10*12**-1*(x-1)**4+10*12**-1*(x-4)**4+65*6**-1*(x-6)**3\n",
+ "\n",
+ "x=0\n",
+ "y=0\n",
+ "#Sub these values in above equation,we get\n",
+ "C2=0\n",
+ "\n",
+ "x=6 #m\n",
+ "y=0\n",
+ "C1=-(25*6**-1*6**3-10*12**-1*5**4+10*12**-1*2**4)*6**-1\n",
+ "\n",
+ "#deflection at C is given by\n",
+ "x=8\n",
+ "y_c=(C2+C1*x+25*6**-1*x**3-10*12**-1*(x-1)**4+10*12**-1*(x-4)**4+65*6**-1*(x-6)**3)*(E*I)**-1\n",
+ "\n",
+ "#Assuming y is max in the portion DE,then\n",
+ "#(dy/dx)=0 for that point\n",
+ "\n",
+ "#0=-65.417+25*2**-1*x**2-10*3**-1*x(-1)**3\n",
+ "\n",
+ "#Let F(x)=-65.417+25*2**-1*x**2-10*3**-1*x(-1)**3\n",
+ "#Let z=F(x)\n",
+ "\n",
+ "#AT \n",
+ "x=3\n",
+ "z=-65.417+25*2**-1*x**2-10*3**-1*(x-1)**3\n",
+ "\n",
+ "x=2.5\n",
+ "z1=-65.417+25*2**-1*x**2-10*3**-1*(x-1)**3\n",
+ "\n",
+ "x=2.4\n",
+ "z2=-65.417+25*2**-1*x**2-10*3**-1*(x-1)**3\n",
+ "\n",
+ "#The assumption is max in portion DE\n",
+ "x=2.46\n",
+ "y_max=(-65.417*x+25*6**-1*x**3-10*12**-1*1.46**4)*(E*I)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflection at free end C\",round(y_c,4),\"mm\"\n",
+ "print\"Max Deflection between A and B\",round(y_max,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflection at free end C -0.0101 mm\n",
+ "Max Deflection between A and B -0.0114 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.18,Page No.213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_DB=L_AC=L_ED=2 #m #Length of DB & AC\n",
+ "L_CD=4 #m #Length of CD\n",
+ "L_CE=2 #m #Length of CE\n",
+ "F_A=40 #KN #Force at C\n",
+ "F_B=20 #KN #Force at A\n",
+ "E=200*10**6 #KN/mm**2 #Modulus of Elasticity\n",
+ "I=50*10**-6 #m**4 #M.I\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt V_C & V_D be the reactions at C & D respectively\n",
+ "#V_C+V_D=60\n",
+ "\n",
+ "#Taking Moment At D,M_D\n",
+ "V_C=-(-F_A*(L_AC+L_CE+L_ED)+F_B*L_DB)*L_CD**-1\n",
+ "V_D=60-V_C\n",
+ "\n",
+ "#Now Taking Moment at Distance x from A,\n",
+ "#M_x=-40*x+50*(x-2)+10*(x-6)\n",
+ "\n",
+ "#EI*(d**2*y/dx**2)=-40*x+50*(x-2)+10*(x-6)\n",
+ "\n",
+ "#Now Integrating above Equation we get\n",
+ "#EI*(dy/dx)=C1+20*x**2-25*(x-2)+5*(x-6)**2\n",
+ "\n",
+ "#Again Integrating above Equation we get\n",
+ "#EI*y=C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3\n",
+ "\n",
+ "#At\n",
+ "x=0\n",
+ "y=0\n",
+ "#C2+2*C1=-53.33 ...............(1)\n",
+ "\n",
+ "#At \n",
+ "x=6\n",
+ "y=0\n",
+ "#C2+6*C1=906.667 ...............(2)\n",
+ "\n",
+ "#Subtracting Equation 1 from 2 we get\n",
+ "C1=853.333*4**-1\n",
+ "C2=53.333-2*C1\n",
+ "x=0\n",
+ "y_A=(C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3)*(E*I)**-1\n",
+ "\n",
+ "#Answer For y_A is incorrect in textbook\n",
+ "\n",
+ "#At Mid-span\n",
+ "C1=853.333*4**-1\n",
+ "C2=53.333-2*C1\n",
+ "x=4\n",
+ "y_E=(C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3)*(E*I)**-1\n",
+ "\n",
+ "#Answer For y_E is incorrect in textbook\n",
+ "\n",
+ "#At B\n",
+ "C1=853.333*4**-1\n",
+ "C2=53.333-2*C1\n",
+ "x=8\n",
+ "y_B=(C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3)*(E*I)**-1\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflection relative to the level of the supports:at End A\",round(y_A,4),\"mm\"\n",
+ "print\" :at End B\",round(y_B,4),\"mm\"\n",
+ "print\" :at Centre of CD\",round(y_E,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflection relative to the level of the supports:at End A -0.08 mm\n",
+ " :at End B -0.0267 mm\n",
+ " :at Centre of CD 0.0107 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.6_9.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.6_9.ipynb
new file mode 100644
index 00000000..e3bdfbe1
--- /dev/null
+++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.6_9.ipynb
@@ -0,0 +1,1518 @@
+{
+ "metadata": {
+ "name": "chapter no.6.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter No.6:Torsion"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.1,Page No.225"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=10000 #mm #Length of solid shaft\n",
+ "d=100 #mm #Diameter of shaft\n",
+ "n=150 #rpm\n",
+ "P=112.5*10**6 #N-mm/sec #Power Transmitted\n",
+ "G=82*10**3 #N/mm**2 #modulus of Rigidity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "J=pi*d**4*(32)**-1 #mm**3 #Polar Modulus\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",
+ "\n",
+ "r=50 #mm #Radius\n",
+ "\n",
+ "q_s=T*r*J**-1 #N/mm**2 #Max shear stress intensity\n",
+ "Theta=T*L*(G*J)**-1 #angle of twist\n",
+ "\n",
+ "#Result\n",
+ "print\"Max shear stress intensity\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Angle of Twist\",round(Theta,3),\"radian\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max shear stress intensity 36.48 N/mm**2\n",
+ "Angle of Twist 0.089 radian\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.2,Page No.226"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=440*10**6 #N-m/sec #Power transmitted\n",
+ "n=280 #rpm\n",
+ "theta=pi*180**-1 #radian #angle of twist\n",
+ "L=1000 #mm #Length of solid shaft\n",
+ "q_s=40 #N/mm**2 #Max torsional shear stress\n",
+ "G=84*10**3 #N/mm**2 #Modulus of rigidity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#P=2*pi*n*T*(60)**-1 #Equation of Power transmitted\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #torsional moment\n",
+ "\n",
+ "#From Consideration of shear stress\n",
+ "d1=(T*16*(pi*40)**-1)**0.333333 \n",
+ "\n",
+ "#From Consideration of angle of twist\n",
+ "d2=(T*L*32*180*(pi*84*10**3*pi)**-1)**0.25\n",
+ "\n",
+ "#result\n",
+ "print\"Diameter of solid shaft is\",round(d1,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of solid shaft is 124.09 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.3,Page No.227"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "G=80*10**3 #N/mm**2 #Modulus of rigidity\n",
+ "q_s=80 #N/mm**2 #Max sheare stress\n",
+ "P=736*10**6 #N-mm/sec #Power transmitted\n",
+ "n=200\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",
+ "\n",
+ "#Now From consideration of angle of twist\n",
+ "theta=pi*180**-1\n",
+ "#L=15*d\n",
+ "\n",
+ "d=(T*32*180*15*(pi**2*G)**-1)**0.33333\n",
+ "\n",
+ "#Now corresponding stress at the surface is\n",
+ "q_s2=T*32*d*(pi*2*d**4)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Max diameter required is\",round(d,2),\"mm\"\n",
+ "print\"Corresponding shear stress is\",round(q_s2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max diameter required is 156.66 mm\n",
+ "Corresponding shear stress is 46.55 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.4,Page No.228"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=25 #mm #Diameter of steel bar\n",
+ "p=50*10**3 #N #Pull\n",
+ "dell_1=0.095 #mm #Extension of bar\n",
+ "l=200 #mm #Guage Length\n",
+ "T=200*10**3 #N-mm #Torsional moment\n",
+ "theta=0.9*pi*180**-1 #angle of twist\n",
+ "L=250 #mm Length of steel bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A=pi*4**-1*d**2 #Area of steel bar #mm**2\n",
+ "E=p*l*(dell_1*A)**-1 #N/mm**2 #Modulus of elasticity \n",
+ "\n",
+ "J=pi*32**-1*d**4 #mm**4 #Polar modulus\n",
+ "\n",
+ "G=T*L*(theta*J)**-1 #Modulus of rigidity #N/mm**2\n",
+ "\n",
+ "#Now from the relation of Elastic constants\n",
+ "mu=E*(2*G)**-1-1\n",
+ "\n",
+ "#result\n",
+ "print\"The Poissoin's ratio is\",round(mu,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Poissoin's ratio is 0.292\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.5,Page No.229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=6000 #mm #Length of circular shaft\n",
+ "d1=100 #mm #Outer Diameter\n",
+ "d2=75 #mm #Inner Diameter\n",
+ "R=100*2**-1 #Radius of shaft\n",
+ "T=10*10**6 #N-mm #Torsional moment\n",
+ "G=80*10**3 #N/mm**2 #Modulus of Rigidity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar Modulus \n",
+ "\n",
+ "#Max Shear stress produced\n",
+ "q_s=T*R*J**-1 #N/mm**2\n",
+ "\n",
+ "#Angle of twist\n",
+ "theta=T*L*(G*J)**-1 #Radian\n",
+ "\n",
+ "#Result\n",
+ "print\"MAx shear stress produced is\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Angle of Twist is\",round(theta,2),\"Radian\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "MAx shear stress produced is 74.5 N/mm**2\n",
+ "Angle of Twist is 0.11 Radian\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.6,Page No.229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=200 #mm #External Diameter of shaft\n",
+ "t=25 #mm #Thickness of shaft\n",
+ "n=200 #rpm\n",
+ "theta=0.5*pi*180**-1 #Radian #angle of twist\n",
+ "L=2000 #mm #Length of shaft\n",
+ "G=84*10**3 #N/mm**2\n",
+ "d2=d1-2*t #mm #Internal Diameter of shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar Modulus \n",
+ "\n",
+ "#Torsional moment\n",
+ "T=G*J*theta*L**-1 #N/mm**2 \n",
+ "\n",
+ "#Power Transmitted\n",
+ "P=2*pi*n*T*60**-1*10**-6 #N-mm\n",
+ "\n",
+ "#Max shear stress transmitted\n",
+ "q_s=G*theta*(d1*2**-1)*L**-1 #N/mm**2 \n",
+ "\n",
+ "#Result\n",
+ "print\"Power Transmitted is\",round(P,2),\"N-mm\"\n",
+ "print\"Max Shear stress produced is\",round(q_s,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power Transmitted is 824.28 N-mm\n",
+ "Max Shear stress produced is 36.65 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.7,Page No.230"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=3750*10**6 #N-mm/sec\n",
+ "n=240 #Rpm\n",
+ "q_s=160 #N/mm**2 #Max shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#d2=0.8*d2 #mm #Internal Diameter of shaft\n",
+ "\n",
+ "#J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar modulus\n",
+ "#After substituting value in above Equation we get\n",
+ "#J=0.05796*d1**4\n",
+ "\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",
+ "\n",
+ "#Now from Torsion Formula\n",
+ "#T*J**-1=q_s*R**-1 ......................................(1)\n",
+ "\n",
+ "#But R=d1*2**-1 \n",
+ "\n",
+ "#Now substituting value of R and J in Equation (1) we get\n",
+ "d1=(T*(0.05796*q_s*2)**-1)**0.33333\n",
+ "\n",
+ "d2=d1*0.8\n",
+ "\n",
+ "#Result\n",
+ "print\"The size of the Shaft is:d1\",round(d1,3),\"mm\"\n",
+ "print\" :d2\",round(d2,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The size of the Shaft is:d1 200.362 mm\n",
+ " :d2 160.289 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.8,Page No.231"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=245*10**6 #N-mm/sec #Power transmitted\n",
+ "n=240 #rpm\n",
+ "q_s=40 #N/mm**2 #Shear stress\n",
+ "theta=pi*180**-1 #radian #Angle of twist\n",
+ "L=1000 #mm #Length of shaft\n",
+ "G=80*10**3 #N/mm**2\n",
+ "\n",
+ "#Tmax=1.5*T\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional Moment\n",
+ "Tmax=1.5*T\n",
+ "\n",
+ "#Now For Solid shaft\n",
+ "#J=pi*32*d**4\n",
+ "\n",
+ "#Now from the consideration of shear stress we get\n",
+ "#T*J**-1=q_s*(d*2**-1)**-1\n",
+ "#After substituting value in above Equation we get\n",
+ "#T=pi*16**-1*d**3*q_s\n",
+ "\n",
+ "#Designing For max Torque\n",
+ "d=(Tmax*16*(pi*40)**-1)**0.33333 #mm #Diameter of shaft\n",
+ "\n",
+ "#For max Angle of Twist\n",
+ "#Tmax*J**-1=G*theta*L**-1 \n",
+ "#After substituting value in above Equation we get\n",
+ "d2=(Tmax*32*180*L*(pi**2*G)**-1)**0.25\n",
+ "\n",
+ "#For Hollow Shaft\n",
+ "\n",
+ "#d1_2=Outer Diameter\n",
+ "#d2_2=Inner Diameter\n",
+ "\n",
+ "#d2_2=0.5*d1_2\n",
+ "\n",
+ "# Polar modulus\n",
+ "#J=pi*32**-1*(d1_2**4-d2_2**4)\n",
+ "#After substituting values we get\n",
+ "#J=0.092038*d1_2**4\n",
+ "\n",
+ "#Now from the consideration of stress\n",
+ "#Tmax*J**-1=q_s*(d1_2*2**-1)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d1_2=(Tmax*(0.092038*2*q_s)**-1)**0.33333\n",
+ "\n",
+ "#Now from the consideration of angle of twist\n",
+ "#Tmax*J**-1=G*theta*L**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d1_3=(Tmax*180*L*(0.092038*G*pi)**-1)**0.25\n",
+ "\n",
+ "d2_2=0.5*d1_2\n",
+ "\n",
+ "#result\n",
+ "print\"Diameter of shaft is:For solid shaft:d\",round(d,2),\"mm\"\n",
+ "print\" :For Hollow shaft:d1_2\",round(d1_2,3),\"mm\"\n",
+ "print\" : :d2_2\",round(d2_2,3),\"mm\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of shaft is:For solid shaft:d 123.01 mm\n",
+ " :For Hollow shaft:d1_2 125.69 mm\n",
+ " : :d2_2 62.845 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.11,Page No.235"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=250*10**6 #N-mm/sec #Power transmitted\n",
+ "n=100 #rpm\n",
+ "q_s=75 #N/mm**2 #Shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Equation of Power we have\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",
+ "\n",
+ "#Now from torsional moment equation we have\n",
+ "#T=j*q_s*(d/2**-1)**-1\n",
+ "#After substituting values in above equation and further simplifying we get\n",
+ "#T=pi*16**-1**d**3*q_s\n",
+ "d=(T*16*(pi*q_s)**-1)**0.3333 #mm #Diameter of solid shaft\n",
+ "\n",
+ "#PArt-2\n",
+ "\n",
+ "#Let d1 and d2 be the outer and inner diameter of hollow shaft\n",
+ "#d2=0.6*d1\n",
+ "\n",
+ "#Again from torsional moment equation we have\n",
+ "#T=pi*32**-1*(d1**4-d2**4)*q_s*(d1/2)**-1\n",
+ "d1=(T*16*(pi*(1-0.6**4)*q_s)**-1)**0.33333\n",
+ "d2=0.6*d1\n",
+ "\n",
+ "#Cross sectional area of solid shaft\n",
+ "A1=pi*4**-1*d**2 #mm**2\n",
+ "\n",
+ "#cross sectional area of hollow shaft\n",
+ "A2=pi*4**-1*(d1**2-d2**2)\n",
+ "\n",
+ "#Now percentage saving in weight\n",
+ "#Let W be the percentage saving in weight\n",
+ "W=(A1-A2)*100*A1**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Percentage saving in Weight is\",round(W,3),\"%\"\n",
+ "print\"Size of shaft is:solid shaft:d\",round(d,3),\"mm\"\n",
+ "print\" :Hollow shaft:d1\",round(d1,3),\"mm\"\n",
+ "print\" : :d2\",round(d2,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percentage saving in Weight is 29.735 %\n",
+ "Size of shaft is:solid shaft:d 117.418 mm\n",
+ " :Hollow shaft:d1 123.031 mm\n",
+ " : :d2 73.818 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.12,Page No.237"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "d=100 #mm #Diameter of solid shaft\n",
+ "d1=100 #mm #Outer Diameter of hollow shaft\n",
+ "d2=50 #mm #Inner Diameter of hollow shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Torsional moment of solid shaft\n",
+ "#T_s=J*q_s*(d*2**-1)**-1 \n",
+ "#After substituting values in above equation and further simplifying we get\n",
+ "#T_s=pi*16*d**3*q_s ...............(1)\n",
+ "\n",
+ "#torsional moment for hollow shaft is\n",
+ "#T_h=J*q_s*(d1**4-d2**4)**-1*(d1*2**-1)\n",
+ "#After substituting values in above equation and further simplifying we get\n",
+ "#T_h=pi*32**-1*2*d1**-1*(d1**4-d2**4)*q_s ...........(2)\n",
+ "\n",
+ "#Dividing Equation 2 by 1 we get\n",
+ "#Let the ratio of T_h*T_s**-1 Be X\n",
+ "X=1-0.5**4\n",
+ "\n",
+ "#Loss in strength \n",
+ "#Let s be the loss in strength\n",
+ "#s=T_s*T_h*100*T_s**-1\n",
+ "#After substituting values in above equation and further simplifying we get\n",
+ "s=(1-0.9375)*100\n",
+ "\n",
+ "#Weight Ratio \n",
+ "#Let w be the Weight ratio\n",
+ "#w=W_h*W_s**-1\n",
+ "\n",
+ "A_h=pi*32**-1*(d1**2-d2**2) #mm**2 #Area of Hollow shaft\n",
+ "A_s=pi*32**-1*d**2 #mm**2 #Area of solid shaft\n",
+ "\n",
+ "w=A_h*A_s**-1 \n",
+ "\n",
+ "#Result\n",
+ "print\"Loss in strength is\",round(s,2)\n",
+ "print\"Weight ratio is\",round(w,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Loss in strength is 6.25\n",
+ "Weight ratio is 0.75\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.13,Page No.239"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "T=8 #KN-m #Torque \n",
+ "d=100 #mm #Diameter of portion AB\n",
+ "d1=100 #mm #External Diameter of Portion BC\n",
+ "d2=75 #mm #Internal Diameter of Portion BC\n",
+ "G=80 #KN/mm**2 #Modulus of Rigidity\n",
+ "L1=1500 #mm #Radial Distance of Portion AB\n",
+ "L2=2500 #mm #Radial Distance ofPortion BC\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "R=d*2**-1 #mm #Radius of shaft\n",
+ "\n",
+ "#For Portion AB,Polar Modulus\n",
+ "J1=pi*32**-1*d**4 #mm**4 \n",
+ "\n",
+ "#For Portion BC,Polar modulus \n",
+ "J2=pi*32**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "#Now Max stress occurs in portion BC since max radial Distance is sme in both cases\n",
+ "q_max=T*J2**-1*R*10**6 #N/mm**2 \n",
+ "\n",
+ "#Let theta1 be the rotation in Portion AB and theta2 be the rotation in portion BC\n",
+ "theta1=T*L1*(G*J1)**-1 #Radians\n",
+ "theta2=T*L2*(G*J2)**-1 #Radians\n",
+ "\n",
+ "#Total Rotational at end C\n",
+ "theta=(theta1+theta2)*10**3 #Radians\n",
+ "\n",
+ "#Result\n",
+ "print\"Max stress induced is\",round(q_max,2),\"N/mm**2\"\n",
+ "print\"Angle of Twist is\",round(theta,3),\"radians\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max stress induced is 59.6 N/mm**2\n",
+ "Angle of Twist is 0.053 radians\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.14,Page No.240"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "q_b=80 #N/mm**2 #Shear stress in Brass\n",
+ "q_s=100 #N/mm**2 #Shear stress in Steel\n",
+ "G_b=40*10**3 #N/mm**2 \n",
+ "G_s=80*10**3 \n",
+ "L_b=1000 #mm #Length of brass shaft\n",
+ "L_s=1200 #mm #Length of steel shaft\n",
+ "d1=80 #mm #Diameter of brass shaft\n",
+ "d2=60 #mm #Diameter of steel shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Polar modulus of brass rod\n",
+ "J_b=pi*32**-1*d1**4 #mm**4 \n",
+ "\n",
+ "#Polar modulus of steel rod\n",
+ "J_s=pi*32**-1*d2**4 #mm**4\n",
+ "\n",
+ "#Considering bras Rod:AB\n",
+ "T1=J_b*q_b*(d1*2**-1)**-1 #N-mm \n",
+ "\n",
+ "#Considering Steel Rod:BC\n",
+ "T2=J_s*q_s*(d2*2**-1)**-1 #N-mm\n",
+ "\n",
+ "#Max Torque that can be applied\n",
+ "T2\n",
+ "\n",
+ "#Let theta_b and theta_s be the rotations in Brass and steel respectively\n",
+ "theta_b=T2*L_b*(G_b*J_b)**-1 #Radians\n",
+ "theta_s=T2*L_s*(G_s*J_s)**-1 #Radians\n",
+ "\n",
+ "theta=theta_b+theta_s #Radians #Rotation of free end\n",
+ "\n",
+ "#Result\n",
+ "print\"Total of free end is\",round(theta,3),\"Radians\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total of free end is 0.076 Radians\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.15,Page No.241"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "G=80*10**3 #N/mm**2 #Modulus of Rigidity\n",
+ "d1=100 #mm #Outer diameter of hollow shft\n",
+ "d2=80 #mm #Inner diameter of hollow shaft\n",
+ "d=80 #mm #diameter of Solid shaft\n",
+ "d3=60 #mm #diameter of Solid shaft having L=0.5m\n",
+ "L1=300 #mm #Length of Hollow shaft\n",
+ "L2=400 #mm #Length of solid shaft\n",
+ "L3=500 #mm #LEngth of solid shaft of diameter 60mm\n",
+ "T1=2*10**6 #N-mm #Torsion in Shaft AB\n",
+ "T2=1*10**6 #N-mm #Torsion in shaft BC\n",
+ "T3=1*10**6 #N-mm #Torsion in shaft CD\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Now Polar modulus of section AB\n",
+ "J1=pi*32**-1*(d1**4-d2**4) #mm**4 \n",
+ "\n",
+ "#Polar modulus of section BC\n",
+ "J2=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Polar modulus of section CD\n",
+ "J3=pi*32**-1*d3**4 #mm**4\n",
+ "\n",
+ "#Now angle of twist of AB\n",
+ "theta1=T1*L1*(G*J1)**-1 #radians\n",
+ "\n",
+ "#Angle of twist of BC\n",
+ "theta2=T2*L2*(G*J2)**-1 #radians\n",
+ "\n",
+ "#Angle of twist of CD\n",
+ "theta3=T3*L3*(G*J3)**-1 #radians\n",
+ "\n",
+ "#Angle of twist\n",
+ "theta=theta1-theta2+theta3 #Radians\n",
+ "\n",
+ "#Shear stress in AB From Torsion Equation\n",
+ "q_s1=T1*(d1*2**-1)*J1**-1 #N/mm**2 \n",
+ "\n",
+ "#Shear stress in BC\n",
+ "q_s2=T2*(d*2**-1)*J2**-1 #N/mm**2 \n",
+ "\n",
+ "#Shear stress in CD\n",
+ "q_s3=T3*(d3*2**-1)*J3**-1 #N-mm**2\n",
+ "\n",
+ "#As max shear stress occurs in portion CD,so consider CD\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of twist at free end is\",round(theta,5),\"Radian\"\n",
+ "print\"Max Shear stress\",round(q_s3,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of twist at free end is 0.00496 Radian\n",
+ "Max Shear stress 23.58 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.16,Page No.242"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1000 #mm #Length of bar\n",
+ "L1=600 #mm #Length of Bar AB\n",
+ "L2=400 #mm #Length of Bar BC\n",
+ "d1=60 #mm #Outer Diameter of bar BC\n",
+ "d2=30 #mm #Inner Diameter of bar BC\n",
+ "d=60 #mm #Diameter of bar AB\n",
+ "T=2*10**6 #N-mm #Total Torque\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Polar Modulus of Portion AB\n",
+ "J1=pi*32**-1*d**4 #mm*4\n",
+ "\n",
+ "#Polar Modulus of Portion BC\n",
+ "J2=pi*32**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "#Let T1 be the torque resisted by bar AB and T2 be torque resisted by Bar BC\n",
+ "#Let theta1 and theta2 be the rotation of shaft in portion AB & BC\n",
+ "\n",
+ "#theta1=T1*L1*(G*J1)**-1 #radians\n",
+ "#After substituting values and further simplifying we get \n",
+ "#theta1=32*600*T1*(pi*60**4*G)**-1\n",
+ "\n",
+ "#theta2=T2*L*(J2*G)**-1 #Radians\n",
+ "#After substituting values and further simplifying we get \n",
+ "#theta2=32*400*T2*(pi*60**4*(1-0.5**4)*G)**-1 \n",
+ "\n",
+ "#Now For consistency of Deformation,theta1=theta2\n",
+ "#After substituting values and further simplifying we get \n",
+ "#T1=0.7111*T2 ..................................................(1)\n",
+ "\n",
+ "#But T1+T2=T=2*10**6 ...........................................(2)\n",
+ "#Substituting value of T1 in above equation\n",
+ "\n",
+ "T2=T*(0.7111+1)**-1\n",
+ "T1=0.71111*T2\n",
+ "\n",
+ "#Max stress in Portion AB\n",
+ "q_s1=T1*(d*2**-1)*(J1)**-1 #N/mm**2\n",
+ "\n",
+ "#Max stress in Portion BC\n",
+ "q_s2=T2*(d1*2**-1)*J2**-1 #N/mm**2 \n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Portion:AB\",round(q_s1,2),\"N/mm**2\"\n",
+ "print\" :BC\",round(q_s2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses Developed in Portion:AB 19.6 N/mm**2\n",
+ " :BC 29.4 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.17,Page No.243"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=80 #mm #External Diameter of Brass tube\n",
+ "d2=50 #mm #Internal Diameter of Brass tube\n",
+ "d=50 #mm #Diameter of steel Tube\n",
+ "G_b=40*10**3 #N/mm**2 #Modulus of Rigidity of brass tube\n",
+ "G_s=80*10**3 #N/mm**2 #Modulus of rigidity of steel tube\n",
+ "T=6*10**6 #N-mm #Torque\n",
+ "L=2000 #mm #Length of Tube\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Polar Modulus of brass tube\n",
+ "J1=pi*32**-1*(d1**4-d2**4) #mm**4 \n",
+ "\n",
+ "#Polar modulus of steel Tube\n",
+ "J2=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Let T_s & T_b be the torque resisted by steel and brass respectively\n",
+ "#Then, T_b+T_s=T ............................................(1)\n",
+ "\n",
+ "#Since the angle of twist will be the same\n",
+ "#Theta1=Theta2\n",
+ "#After substituting values and further simplifying we get \n",
+ "#Ts=0.360*Tb ...........................................(2)\n",
+ "\n",
+ "#After substituting value of Ts in eqn 1 and further simplifying we get \n",
+ "T_b=T*(0.36+1)**-1 #N-mm\n",
+ "T_s=0.360*T_b\n",
+ "\n",
+ "#Let q_s and q_b be the max stress in steel and brass respectively\n",
+ "q_b=T_b*(d1*2**-1)*J1**-1 #N/mm**2\n",
+ "q_s=T_s*(d2*2**-1)*J2**-1 #N/mm**2\n",
+ "\n",
+ "#Since angle of twist in brass=angle of twist in steel\n",
+ "theta_s=T_s*L*(J2*G_s)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Materials are:Brass\",round(q_b,2),\"N/mm**2\"\n",
+ "print\" :Steel\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Angle of Twist in 2m Length\",round(theta_s,3),\"Radians\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses Developed in Materials are:Brass 51.79 N/mm**2\n",
+ " :Steel 64.71 N/mm**2\n",
+ "Angle of Twist in 2m Length 0.065 Radians\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.18,Page No.245"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=60 #mm #External Diameter of aluminium Tube\n",
+ "d2=40 #mm #Internal Diameter of aluminium Tube\n",
+ "d=40 #mm #Diameter of steel tube\n",
+ "q_a=60 #N/mm**2 #Permissible stress in aluminium\n",
+ "q_s=100 #N/mm**2 #Permissible stress in steel tube\n",
+ "G_a=27*10**3 #N/mm**2 \n",
+ "G_s=80*10**3 #N/mm**2 \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Polar modulus of aluminium Tube\n",
+ "J_a=pi*32**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "#Polar Modulus of steel Tube\n",
+ "J_s=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Now the angle of twist of steel tube = angle of twist of aluminium tube\n",
+ "#T_s*L_s*(J_s*theta_s)**-1=T_a*L_a*(J_a*theta_a)**-1\n",
+ "#After substituting values in above Equation and Further simplifyin we get\n",
+ "#T_s=0.7293*T_a .....................(1)\n",
+ "\n",
+ "#If steel Governs the resisting capacity\n",
+ "T_s1=q_s*J_s*(d*2**-1)**-1 #N-mm\n",
+ "T_a1=T_s1*0.7293**-1 #N-mm\n",
+ "T1=(T_s1+T_a1)*10**-6 #KN-m #Total Torque in steel Tube\n",
+ "\n",
+ "#If aluminium Governs the resisting capacity \n",
+ "T_a2=q_a*J_a*(d1*2**-1) #N-mm\n",
+ "T_s2=T_a2*0.7293 #N-mm\n",
+ "T2=(T_s2+T_a2)*10**-6 #KN-m #Total Torque in aluminium tube\n",
+ "\n",
+ "#Result\n",
+ "print\"Steel Governs the torque carrying capacity\",round(T1,2),\"KN-m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Steel Governs the torque carrying capacity 2.98 KN-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.19,Page No.247"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=225*10**6 #N-mm/sec #Power Trasmitted\n",
+ "q_b=80 #N/mm**2 #Shear stress\n",
+ "n=200 #Rpm\n",
+ "q_k=100 #N/mm**2 #PErmissible stress in Keys\n",
+ "D=300 #mm #Diameter of bolt circle\n",
+ "L=150 #mm #Length of shear key\n",
+ "d=16 #mm #Diameterr of bolt\n",
+ "\n",
+ "#Calculations\n",
+ "T=60*P*(2*pi*n)**-1 #N-mm #Torque\n",
+ "\n",
+ "#Now From Torsion Formula\n",
+ "#T*J**-1=q_s*R**-1\n",
+ "#After substituting values we get\n",
+ "#T=pi*16*d**3*n\n",
+ "#After further simplifying we get\n",
+ "d1=(T*16*(pi*q_s)**-1)**0.33333\n",
+ "\n",
+ "#Let b be the width of shear Key\n",
+ "#T=q_k*L*b*R\n",
+ "#After simplifying further we get\n",
+ "b=T*(q_k*L*(d1*2**-1))**-1 #mm\n",
+ "\n",
+ "#Let n2 be the no. of bolts required at bolt circle of radius\n",
+ "R_b=D*2**-1 #mm \n",
+ "\n",
+ "n2=T*4*(q_b*pi*d**2*R_b)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Minimum no. of Bolts Required are\",round(n2,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum no. of Bolts Required are 4.45\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.20,Page No.250"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "T=2*10**6 #N-mm #Torque transmitted\n",
+ "G=80*10**3 #N/mm**2 #Modulus of rigidity\n",
+ "d1=40 #mm \n",
+ "d2=80 #mm\n",
+ "r1=20 #mm\n",
+ "r2=40 #mm\n",
+ "L=2000 #mm #Length of shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Angle of twist \n",
+ "theta=2*T*L*(r1**2+r1*r2+r2**2)*(3*pi*G*r2**3*r1**3)**-1 #radians\n",
+ "\n",
+ "#If the shaft is treated as shaft of average Diameter\n",
+ "d_avg=(d1+d2)*2**-1 #mm\n",
+ "\n",
+ "theta1=T*L*(G*pi*32**-1*d_avg**4)**-1 #Radians\n",
+ "\n",
+ "#Percentage Error\n",
+ "#Let Percentage Error be E\n",
+ "X=theta-theta1\n",
+ "E=(X*theta**-1)*100 \n",
+ "\n",
+ "#Result\n",
+ "print\"Percentage Error is\",round(E,2),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percentage Error is 32.28 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.21,Page No.252"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "G=80*10**3 #N/mm**2 \n",
+ "P=1*10**9 #N-mm/sec #Power\n",
+ "n=300 \n",
+ "d1=150 #mm #Outer Diameter\n",
+ "d2=120 #mm #Inner Diameter\n",
+ "L=2000 #mm #Length of circular shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm\n",
+ "\n",
+ "#Polar Modulus \n",
+ "J=pi*32**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "q_s=T*J**-1*(d1*2**-1) #N/mm**2 \n",
+ "\n",
+ "\n",
+ "#Strain ENergy\n",
+ "U=q_s**2*(4*G)**-1*pi*4**-1*(d1**2-d2**2)*L\n",
+ "\n",
+ "#Result\n",
+ "print\"Max shear stress is\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Strain Energy stored in the shaft is\",round(U,2),\"N-mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max shear stress is 81.36 N/mm**2\n",
+ "Strain Energy stored in the shaft is 263181.37 N-mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.22,Page No.254"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=12 #mm #Diameter of helical spring\n",
+ "D=150 #mm #Mean Diameter\n",
+ "R=D*2**-1 #mm #Radius of helical spring\n",
+ "n=10 #no.of turns\n",
+ "G=80*10**3 #N/mm**2 \n",
+ "W=450 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max shear stress \n",
+ "q_s=16*W*R*(pi*d**3)**-1 #N/mm**2\n",
+ "\n",
+ "#Strain Energy stored\n",
+ "U=32*W**2*R**3*n*(G*d**4)**-1 #N-mm\n",
+ "\n",
+ "#Deflection Produced\n",
+ "dell=64*W*R**3*n*(G*d**4)**-1 #mm\n",
+ "\n",
+ "#Stiffness Spring\n",
+ "k=W*dell**-1 #N/mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Max shear stress is\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Strain Energy stored is\",round(U,2),\"N-mm\"\n",
+ "print\"Deflection Produced is\",round(dell,2),\"mm\"\n",
+ "print\"Stiffness spring is\",round(k,2),\"N/mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max shear stress is 99.47 N/mm**2\n",
+ "Strain Energy stored is 16479.49 N-mm\n",
+ "Deflection Produced is 73.24 mm\n",
+ "Stiffness spring is 6.14 N/mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.23,Page No.255"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "K=5 #N/mm #Stiffness\n",
+ "L=100 #mm #Solid Length\n",
+ "q_s=60 #N/mm**2 #Max shear stress\n",
+ "W=200 #N #Max Load\n",
+ "G=80*10**3 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#K=W*dell**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#d=0.004*R**3*n ........(1) #mm #Diameter of wire\n",
+ "#n=L*d**-1 ........(2)\n",
+ "\n",
+ "#From Shearing stress\n",
+ "#q_s=16*W*R*(pi*d**3)**-1 \n",
+ "#After substituting values and further simplifying we get\n",
+ "#d**4=0.004*R**3*n .................(4)\n",
+ "\n",
+ "#From Equation 1,2,3\n",
+ "#d**4=0.004*(0.0785*d**3)**3*100*d**-1\n",
+ "#after further simplifying we get\n",
+ "d=5168.101**0.25\n",
+ "n=100*d**-1\n",
+ "R=(d**4*(0.004*n)**-1)**0.3333\n",
+ "\n",
+ "#Result\n",
+ "print\"Diameter of Wire is\",round(d,2),\"mm\"\n",
+ "print\"No.of turns is\",round(n,2)\n",
+ "print\"Mean Radius of spring is\",round(R,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of Wire is 8.48 mm\n",
+ "No.fo turns is 11.79\n",
+ "Mean Radius of spring is 47.83 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 54
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.24,Page No.255"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "m=5*10**5 #Wagon Weighing\n",
+ "v=18*1000*36000**-1 \n",
+ "d=300 #mm #Diameter of Beffer springs\n",
+ "n=18 #no.of turns\n",
+ "G=80*10**3 #N/mm**2\n",
+ "dell=225\n",
+ "R=100 #mm #Mean Radius\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Energy of Wagon\n",
+ "E=m*v**2*(9.81*2)**-1 #N-mm\n",
+ "\n",
+ "#Load applied\n",
+ "W=dell*G*d**4*(64*R**3*n)**-1 #N \n",
+ "\n",
+ "#Energy each spring can absorb is\n",
+ "E2=W*dell*2**-1 #N-mm\n",
+ "\n",
+ "#No.of springs required to absorb energy of Wagon\n",
+ "n2=E*E2**-1 *10**7\n",
+ "\n",
+ "#Result\n",
+ "print\"No.of springs Required for Buffer is\",round(n2,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "No.of springs Required for Buffer is 4.47\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.25,Page No.259"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b=180 #mm #width of flange\n",
+ "d=10 #mm #Depth of flange\n",
+ "t=10 #mm #Thickness of flange\n",
+ "D=400 #mm #Overall Depth \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "I_xx=1*12**-1*(b*D**3-(b-t)*(D-2*d)**3)\n",
+ "I_yy=1*12**-1*((D-2*d)*t**3+2*t*b**3)\n",
+ "\n",
+ "#If warping is neglected\n",
+ "J=I_xx+I_yy #mm**4\n",
+ "\n",
+ "#Since b/d>1.6,we get\n",
+ "J2=1*3**-1*d**3*b*(1-0.63*d*b**-1)*2+1*3**-1*t**3*(D-2*d)*(1-0.63*t*b**-1)\n",
+ "\n",
+ "#Over Estimation of torsional Rigidity would have been \n",
+ "T=J*J2**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Error in assessing torsional Rigidity if the warping is neglected is\",round(T,2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Error in assessing torsional Rigidity if the warping is neglected is 808.28\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.26,Page No.261"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=100 #mm #Outer Diameter\n",
+ "d2=95 #mm #Inner Diameter\n",
+ "T=2*10**6 #N-mm #Torque\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar Modulus\n",
+ "\n",
+ "#Shear stress\n",
+ "q_max=T*J**-1*d1*2**-1 #N/mm**2 \n",
+ "\n",
+ "#Now theta*L**-1=T*(G*J)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#Let theta*L**-1=X\n",
+ "X=T*J**-1\n",
+ "\n",
+ "#Now Treating it as very thin walled tube\n",
+ "d=(d1+d2)*2**-1 #mm\n",
+ "\n",
+ "r=d*2**-1 \n",
+ "t=(d1-d2)*2**-1\n",
+ "q_max2=T*(2*pi*r**2*t)**-1 #N/mm**2\n",
+ "\n",
+ "X2=T*(2*pi*r**3*t)**-1 \n",
+ "\n",
+ "#Result\n",
+ "print\"When it is treated as hollow shaft:Max shear stress\",round(q_max,2),\"N/mm**2\"\n",
+ "print\" :Angle of Twist per unit Length\",round(X,3)\n",
+ "print\"When it is very thin Walled Tube :Max shear stress\",round(q_max2,2),\"N/mm**2\"\n",
+ "print\" :Angle of twist per Unit Length\",round(X2,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "When it is treated as hollow shaft:Max shear stress 54.91 N/mm**2\n",
+ " :Angle of Twist per unit Length 1.098\n",
+ "When it is very thin Walled Tube :Max shear stress 53.57 N/mm**2\n",
+ " :Angle of twist per Unit Length 1.099\n"
+ ]
+ }
+ ],
+ "prompt_number": 72
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.7_9.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.7_9.ipynb
new file mode 100644
index 00000000..b75d886a
--- /dev/null
+++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.7_9.ipynb
@@ -0,0 +1,1328 @@
+{
+ "metadata": {
+ "name": "chapter no.7.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter No.7:Compound Stresses And Strains"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.1,Page No.269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "sigma1=30 #N/mm**2 #Stress in tension\n",
+ "d=20 #mm #Diameter \n",
+ "sigma2=90 #N/mm**2 #Max compressive stress\n",
+ "sigma3=25 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#In TEnsion\n",
+ "\n",
+ "#Corresponding stress in shear\n",
+ "P=sigma1*2**-1 #N/mm**2\n",
+ "\n",
+ "#Tensile force\n",
+ "F=pi*4**-1*d**2*sigma1\n",
+ "\n",
+ "#In Compression\n",
+ "\n",
+ "#Correspong shear stress\n",
+ "P2=sigma2*2**-1 #N/mm**2\n",
+ "\n",
+ "#Correspong compressive(axial) stress\n",
+ "p=2*sigma3 #N/mm**2 \n",
+ "\n",
+ "#Corresponding Compressive force\n",
+ "P3=p*pi*4**-1*d**2 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Failure Loads are:\",round(F,2),\"N\"\n",
+ "print\" :\",round(P3,2),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Failure Loads are: 9424.78 N\n",
+ " : 15707.96 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.2,Page No.270"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=25 #mm #Diameter of circular bar\n",
+ "F=20*10**3 #N #Axial Force\n",
+ "theta=30 #Degree #angle \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Axial stresses\n",
+ "p=F*(pi*4**-1*d**2)**-1 #N/mm**2\n",
+ "\n",
+ "#Normal Stress\n",
+ "p_n=p*(cos(30*pi*180**-1))**2\n",
+ "\n",
+ "#Tangential Stress\n",
+ "p_t=p*2**-1*sin(2*theta*pi*180**-1)\n",
+ "\n",
+ "#Max shear stress occurs on plane where theta2=45 \n",
+ "theta2=45\n",
+ "sigma_max=p*2**-1*sin(2*theta2*pi*180**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses developed on a plane making 30 degree is:\",round(p_n,2),\"N/mm**2\"\n",
+ "print\" :\",round(p_t,2),\"N/mm**2\"\n",
+ "print\"stress on max shear stress is\",round(sigma_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses developed on a plane making 30 degree is: 30.56 N/mm**2\n",
+ " : 17.64 N/mm**2\n",
+ "stress on max shear stress is 20.37 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.3,Page No.272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "theta=30 #degree\n",
+ "\n",
+ "#Stresses acting on material\n",
+ "p1=120 #N/mm**2\n",
+ "p2=80 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Normal Stress\n",
+ "P_n=(p1+p2)*2**-1+(p1-p2)*2**-1*cos(2*theta*pi*180**-1) #N/mm**2\n",
+ "\n",
+ "#Tangential stress\n",
+ "P_t=(p1-p2)*2**-1*sin(2*theta*pi*180**-1)\n",
+ "\n",
+ "#Resultant stress\n",
+ "P=(P_n**2+P_t**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Inclination to the plane\n",
+ "phi=arctan(P_n*P_t**-1)*(180*pi**-1)\n",
+ "\n",
+ "#Angle made by resultant with 120 #N/mm**2 stress\n",
+ "phi2=phi+theta #Degree\n",
+ "\n",
+ "#Result\n",
+ "print\"Normal Stress is\",round(P_n,2),\"N/mm**2\"\n",
+ "print\"Tangential Stress is\",round(P_t,2),\"N/mm**2\"\n",
+ "print\"Angle made by resultant\",round(phi2,2),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Normal Stress is 110.0 N/mm**2\n",
+ "Tangential Stress is 17.32 N/mm**2\n",
+ "Angle made by resultant 111.05 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.4,Page No.272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Direct Stresses\n",
+ "P1=60 #N/mm**2 \n",
+ "P2=100 #N/mm**2\n",
+ "\n",
+ "Theta=25 #Degree #Angle\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Normal Stress\n",
+ "P_n=(P1-P2)*2**-1+(P1+P2)*2**-1*cos(2*Theta*pi*180**-1) #N/mm**2\n",
+ "\n",
+ "#Tangential Stress\n",
+ "P_t=(P1+P2)*2**-1*sin(Theta*2*pi*180**-1) #N/mm**2\n",
+ "\n",
+ "#Resultant stress\n",
+ "P=(P_n**2+P_t**2)**0.5 #N/mm**2\n",
+ "\n",
+ "theta2=arctan(P_n*P_t**-1)*(180*pi**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses on the plane AC is:\",round(P_n,2),\"N/mm**2\"\n",
+ "print\" \",round(P_t,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses on the plane AC is: 31.42 N/mm**2\n",
+ " 61.28 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.6,Page No.278"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Stresses acting on material\n",
+ "p_x=180 #N/mm**2 \n",
+ "p_y=120 #N/mm**2\n",
+ "\n",
+ "q=80 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "theta=arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1) #degrees\n",
+ "theta2=theta*2**-1 #Degrees\n",
+ "theta3=theta+180 #Degrees\n",
+ "theta4=theta3*2**-1 #Degrees\n",
+ "\n",
+ "#Stresses\n",
+ "p_1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "p_2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of Principal stress is:\",round(p_1,2),\"N/mm**2\"\n",
+ "print\" \",round(p_2,2),\"N/mm**2\"\n",
+ "print\"Magnitude of max shear stress is\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of Principal stress is: 235.44 N/mm**2\n",
+ " 64.56 N/mm**2\n",
+ "Magnitude of max shear stress is 85.44 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.7,Page No.279"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#stresses\n",
+ "p_x=60 #N/mm**2\n",
+ "p_y=-40 #N/mm**2\n",
+ "\n",
+ "q=10 #N/mm**2 #shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Principal Stresses\n",
+ "p1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "p2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Inclination of principal stress to plane\n",
+ "theta=arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1)#Degrees\n",
+ "theta2=(theta)*2**-1 #degrees\n",
+ "\n",
+ "theta3=(theta+180)*2**-1 #degrees\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses are:\",round(p1,2),\"N/mm**2\"\n",
+ "print\" :\",round(p2,2),\"N/mm**2\"\n",
+ "print\"Max shear stresses\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses are: 60.99 N/mm**2\n",
+ " : -40.99 N/mm**2\n",
+ "Max shear stresses 50.99 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.8,Page No.280"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#stresses\n",
+ "p_x=-120 #N/mm**2\n",
+ "p_y=-80 #N/mm**2\n",
+ "\n",
+ "q=-60 #N/mm**2 #shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Principal Stresses\n",
+ "p1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "p2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Inclination of principal stress to plane\n",
+ "theta=arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1)#Degrees\n",
+ "theta2=(theta)*2**-1 #degrees\n",
+ "\n",
+ "theta3=(theta+180)*2**-1 #degrees\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses are:\",round(p1,2),\"N/mm**2\"\n",
+ "print\" :\",round(p2,2),\"N/mm**2\"\n",
+ "print\"Max shear stresses\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses are: -36.75 N/mm**2\n",
+ " : -163.25 N/mm**2\n",
+ "Max shear stresses 63.25 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.9,Page No.282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#stresses\n",
+ "p_x=-40 #N/mm**2\n",
+ "p_y=80 #N/mm**2\n",
+ "\n",
+ "q=48 #N/mm**2 #shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=((((p_x-p_y)*2**-1)**2)+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Inclination of principal stress to plane\n",
+ "theta=arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1)#Degrees\n",
+ "theta2=(theta)*2**-1 #degrees\n",
+ "\n",
+ "theta3=(theta+180)*2**-1 #degrees\n",
+ "\n",
+ "#Normal Corresponding stress\n",
+ "p_n=(p_x+p_y)*2**-1+(p_x-p_y)*2**-1*cos(2*(theta2+45)*pi*180**-1)+q*sin(2*(theta2+45)*pi*180**-1) #Degrees\n",
+ "\n",
+ "#Resultant stress\n",
+ "p=((p_n**2+q_max**2)**0.5) #N/mm**2\n",
+ "\n",
+ "phi=arctan(p_n*q_max**-1)*(180*pi**-1) #Degrees\n",
+ "\n",
+ "#Inclination to the plane\n",
+ "alpha=round((theta2+45),2)+round(phi ,2)#Degree\n",
+ "\n",
+ "#Answer in book is incorrect of alpha ie41.25\n",
+ "\n",
+ "#Result\n",
+ "print\"Planes of max shear stress:\",round(p_n,2),\"N/mm**2\"\n",
+ "print\" \",round(q_max,2),\"N/mm*2\"\n",
+ "print\"Resultant Stress is\",round(p,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Planes of max shear stress: 20.0 N/mm**2\n",
+ " 76.84 N/mm*2\n",
+ "Resultant Stress is 79.4 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.10,Page No.283"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Stresses\n",
+ "p_x=50*cos(35*pi*180**-1)\n",
+ "q=50*sin(35*pi*180**-1)\n",
+ "p_y=0\n",
+ "\n",
+ "theta=40 #Degrees #Plane AB inclined to vertical\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Normal Stress on AB\n",
+ "p_n=(p_x+p_y)*2**-1+(p_x-p_y)*2**-1*cos(2*theta*pi*180**-1)+q*sin(2*theta*pi*180**-1)\n",
+ "\n",
+ "#Tangential Stress on AB\n",
+ "p_t=(p_x-p_y)*2**-1*sin(2*theta*pi*180**-1)-q*cos(2*theta*pi*180**-1) #N/mm**2\n",
+ "\n",
+ "#Resultant stress\n",
+ "p=(p_n**2+p_t**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Angle of resultant\n",
+ "phi=arctan(p_n*p_t**-1)*(180*pi**-1) #degrees\n",
+ "phi2=phi+theta #Degrees\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of resultant stress is\",round(p,2),\"N/mm**2\"\n",
+ "print\"Direction of Resultant stress is\",round(phi2,2),\"Degrees\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of resultant stress is 54.44 N/mm**2\n",
+ "Direction of Resultant stress is 113.8 Degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.12,Page No.285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Direct stresses\n",
+ "p_x=120 #N/mm**2 #Tensile stress\n",
+ "p_y=-100 #N/mm**2 #Compressive stress\n",
+ "p1=160 #N/mm**2 #Major principal stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let q be the shearing stress\n",
+ "\n",
+ "#p1=(p_x+p_y)*2**-1+((((p_x+p_y)*2**-1)**2)+q**2)**0.5\n",
+ "#After further simplifying we get\n",
+ "q=(p1-((p_x+p_y)*2**-1))**2-((p_x-p_y)*2**-1)**2 #N/mm**2\n",
+ "q2=(q)**0.5 #N/mm**2\n",
+ "\n",
+ "#Minimum Principal stress\n",
+ "p2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shearing stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Shearing stress of material\",round(q,2),\"N/mm**2\"\n",
+ "print\"Min Principal stress\",round(p2,2),\"N/mm**2\"\n",
+ "print\"Max shearing stress\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shearing stress of material 10400.0 N/mm**2\n",
+ "Min Principal stress -140.0 N/mm**2\n",
+ "Max shearing stress 150.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.14,Page No.291"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=40*10**3 #N #Shear Force\n",
+ "M=20*10**6 #Bending Moment\n",
+ "\n",
+ "#Rectangular section\n",
+ "b=100 #mm #Width\n",
+ "d=200 #mm #Depth\n",
+ "\n",
+ "x=20 #mm #Distance from Top surface upto point\n",
+ "y=80 #mm #Distance from point to Bottom\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "I=1*12**-1*b*d**3 #mm**4 #M.I\n",
+ "\n",
+ "#At 20 mm Below top Fibre\n",
+ "f_x=M*I**-1*y #N/mm**2 #Stress\n",
+ "\n",
+ "#Assuming sagging moment ,f_x is compressive p_x=f_x=-24 #N/mm**2\n",
+ "p_x=f_x=-24 #N/mm**2\n",
+ "\n",
+ "#Shearing stress\n",
+ "q=F*(b*I)**-1*(b*x*(b-x*2**-1)) #N/mm**2\n",
+ "\n",
+ "#Direct stresses\n",
+ "\n",
+ "p_y=0 #N/mm**2\n",
+ "\n",
+ "p1=(p_x+p_y)*2**-1+(((p_x+p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "p2=(p_x+p_y)*2**-1-(((p_x+p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Directions of principal stresses at a point below 20mm is:\",round(p1,2),\"N/mm**2\"\n",
+ "print\" \",round(p2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Directions of principal stresses at a point below 20mm is: 0.05 N/mm**2\n",
+ " -24.05 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.15,Page No.292"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=4000 #mm #Span\n",
+ "W1=W2=W3=2*10**3 #N #Load\n",
+ "\n",
+ "#SEction of beam\n",
+ "b=100 #mm #Width\n",
+ "d=240 #mm #Dept\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A and R_B be the reactions\n",
+ "R_A=R_B=(W1+W2+W3)*2**-1 #KN\n",
+ "\n",
+ "#Now at the section 1.5m from left support A\n",
+ "#Shear Force\n",
+ "F=R_A-W1 #KN\n",
+ "\n",
+ "#B.M\n",
+ "M=R_A*1.5-W1*0.5 #KN-m\n",
+ "\n",
+ "#M.I\n",
+ "I=1*12**-1*b*d**3 #mm**4\n",
+ "\n",
+ "#Bending stress\n",
+ "#f=M*I**-1*y\n",
+ "#After Sub values and further simplifying we get\n",
+ "#f=3.04*10**-2*y\n",
+ "\n",
+ "#As it varies Linearly\n",
+ "\n",
+ "#at distance 0 From NA \n",
+ "f1=0\n",
+ "#at distance 60 mm from NA\n",
+ "f2=1.823 #N/mm**2\n",
+ "#at distance 120 mm from NA\n",
+ "f3=3.646 #N/mm**2\n",
+ "\n",
+ "#Shearing stress\n",
+ "q=F*b*d*2**-1*d*4**-1*(b*I)**-1\n",
+ "\n",
+ "#At 60 mm above NA\n",
+ "q2=F*b*d*4**-1*(d*2**-1-d*8**-1)*(b*I)**-1\n",
+ "\n",
+ "#At 120 mm above NA\n",
+ "q3=0 \n",
+ "\n",
+ "#At NA element is under pure shear\n",
+ "p1=q #N/mm**2\n",
+ "p2=-q #N/mm**2 \n",
+ "\n",
+ "#Inclination of principal plane to vertical\n",
+ "#theta=2*q*0**-1\n",
+ "#Further simplifying we get\n",
+ "#theta=infinity\n",
+ "\n",
+ "#therefore\n",
+ "theta=90*2**-1 #degrees\n",
+ "theta2=270*2**-1 #degrees\n",
+ "\n",
+ "#At 60 mm From NA\n",
+ "p_x=-1.823 #N/mm**2 \n",
+ "p_y=0\n",
+ "q=0.0469 #N/mm**2\n",
+ "\n",
+ "#principal planes\n",
+ "P1=(p_x+p_y)*2**-1+(((p_x+p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2=(p_x+p_y)*2**-1-(((p_x+p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Principal planes inclination to hte plane of p_x is given by\n",
+ "theta3=(arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1))\n",
+ "theta4=theta3*2**-1#degrees\n",
+ "\n",
+ "theta5=theta3+180 #Degrees\n",
+ "\n",
+ "#At 120 mm From N-A\n",
+ "p_x2=3.646 #N/mm**2\n",
+ "p_y2=0 #N/mm**2\n",
+ "q2=0 #N/mm**2\n",
+ "\n",
+ "P3=p_x2 #N/mm**2\n",
+ "P4=0 #N/mm**2\n",
+ "\n",
+ "#Answer for P2 at 60 mm from NA is incorrect\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Planes at 60 mm from NA:\",round(p_x,2),\"N/mm**2\"\n",
+ "print\" \",round(p_y,2),\"N/mm**2\"\n",
+ "print\"Principal Stresses at 60 mm From NA\",round(P1,4),\"N/mm**2\"\n",
+ "print\" \",round(P2,4),\"N/mm**2\"\n",
+ "print\"Principal Planes at 60 mm from NA:\",round(p_x2,4),\"N/mm**2\"\n",
+ "print\" \",round(p_y2,4),\"N/mm**2\"\n",
+ "print\"Principal Stresses at 60 mm From NA\",round(P3,4),\"N/mm**2\"\n",
+ "print\" \",round(P4,4),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Planes at 60 mm from NA: -1.82 N/mm**2\n",
+ " 0.0 N/mm**2\n",
+ "Principal Stresses at 60 mm From NA 0.0012 N/mm**2\n",
+ " -1.8242 N/mm**2\n",
+ "Principal Planes at 60 mm from NA: 3.646 N/mm**2\n",
+ " 0.0 N/mm**2\n",
+ "Principal Stresses at 60 mm From NA 3.646 N/mm**2\n",
+ " 0.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.16,Page No.295"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=8000 #mm #Span of beam\n",
+ "w=40*10**6 #N/mm #udl\n",
+ "\n",
+ "#I-section\n",
+ "\n",
+ "#Flanges\n",
+ "b=100 #mm #Width\n",
+ "t=10 #mm #Thickness\n",
+ "\n",
+ "D=400 #mm #Overall Depth\n",
+ "t2=10 #mm #thickness of web\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A and R_B be the Reactions at A & B respectively\n",
+ "R_A=w*2**-1*L*10**-9 #KN\n",
+ "\n",
+ "#Shear force at 2m for left support\n",
+ "F=R_A-2*w*10**-6 #KN\n",
+ "\n",
+ "#Bending Moment\n",
+ "M=R_A*2-2*w*10**-6 #KN-m\n",
+ "\n",
+ "#M.I\n",
+ "I=1*12**-1*b*D**3-1*12**-1*(b-t)*(D-2*t2)**3 #mm**4\n",
+ "\n",
+ "#Bending stress at 100 mm above N_A\n",
+ "f=M*10**6*I**-1*b\n",
+ "\n",
+ "#Shear stress \n",
+ "q=F*10**3*(t*I)**-1*(b*t*(D-t)*2**-1 +t2*(b-t2)*145) #N/mm**2\n",
+ "\n",
+ "p_x=-197.06 #N/mm**2 \n",
+ "p_y=0 #N/mm**2\n",
+ "q=21.38 #N/mm**2\n",
+ "\n",
+ "#Principal Stresses\n",
+ "\n",
+ "P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses are:\",round(P1,2),\"N/mm**2\"\n",
+ "print\" \",round(P2,2),\"N/mm**2\"\n",
+ "print\"Max shear stress\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses are: 2.29 N/mm**2\n",
+ " -199.35 N/mm**2\n",
+ "Max shear stress 100.82 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.18,Page No.298"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=100 #mm #Diameter of shaft\n",
+ "M=3*10**6 #N-mm #B.M\n",
+ "T=6*10**6 #N-mm #Twisting Moment\n",
+ "mu=0.3\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max principal Stress\n",
+ "\n",
+ "P1=16*(pi*d**3)**-1*(M+(M**2+T**2)**0.5) #N/mm**2 \n",
+ "P2=16*(pi*d**3)**-1*(M-(M**2+T**2)**0.5) #N/mm**2 \n",
+ "\n",
+ "#Direct stress\n",
+ "P=round(P1,2)-mu*round(P2,2) #N/mm**2 \n",
+ "\n",
+ "#Result\n",
+ "print\"Principal stresses are:\",round(P1,2),\"N/mm**2\"\n",
+ "print\" :\",round(P2,2),\"N/mm**2\"\n",
+ "print\"Stress Producing the same strain is\",round(P,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal stresses are: 49.44 N/mm**2\n",
+ " : -18.89 N/mm**2\n",
+ "Stress Producing the same strain is 55.11 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.19,Page No.299"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=75 #mm #diameter \n",
+ "P=30*10**6 #W #Power transmitted\n",
+ "W=6 #N-mm/sec #Load\n",
+ "L=1000 #mm \n",
+ "N=300 #r.p.m\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#B.M\n",
+ "M=W*L*4**-1 #N-mm\n",
+ "T=P*60*(2*pi*N)**-1 #Torque transmitted\n",
+ "\n",
+ "#M.I\n",
+ "I=pi*64**-1*d**4 #mm**4\n",
+ "\n",
+ "#Bending stress\n",
+ "f_A=M*I**-1*(d*2**-1) #N/mm**2\n",
+ "\n",
+ "#At A\n",
+ "p_x=f_A\n",
+ "p_y=0\n",
+ "\n",
+ "#Polar Modulus\n",
+ "J=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Shearing stress\n",
+ "q=T*J**-1*(d*2**-1) #N/mm**2\n",
+ "\n",
+ "#Principal Stresses\n",
+ "P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Bending stress\n",
+ "p_x2=0\n",
+ "p_y2=0\n",
+ "\n",
+ "#Shearing stress\n",
+ "q2=T*J**-1*d*2**-1 #N/mm**2\n",
+ "\n",
+ "#Principal stresses\n",
+ "P3=(p_x2+p_y2)*2**-1+(((p_x2-p_y2)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "P4=(p_x2+p_y2)*2**-1-(((p_x2-p_y2)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max2=(((p_x2-p_y2)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Answer for Principal Stresses P1,P2 and Max stress i.e q_max is incorrect in Book\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses at vertical Diameter:P1\",round(P1,2),\"N/mm**2\"\n",
+ "print\" :P2\",round(P2,2),\"N/mm**2\"\n",
+ "print\"Max stress at vertical Diameter : \",round(q_max,2),\"N/mm**2\"\n",
+ "print\"Principal Stresses at Horizontal Diameter:P3\",round(P3,2),\"N/mm**2\"\n",
+ "print\" :P4\",round(P4,2),\"N/mm**2\"\n",
+ "print\"Max stress at Horizontal Diameter : \",round(q_max2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses at vertical Diameter:P1 11.55 N/mm**2\n",
+ " :P2 -11.51 N/mm**2\n",
+ "Max stress at vertical Diameter : 11.53 N/mm**2\n",
+ "Principal Stresses at Horizontal Diameter:P3 11.53 N/mm**2\n",
+ " :P4 -11.53 N/mm**2\n",
+ "Max stress at Horizontal Diameter : 11.53 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.20,Page No.302"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=100 #mm #External Diameter\n",
+ "d2=50 #mm #Internal Diameter\n",
+ "N=500 #mm #r.p.m\n",
+ "P=60*10**6 #N-mm/sec #Power\n",
+ "p=100 #N/mm**2 #principal stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#M.I\n",
+ "I=pi*(d1**4-d2**4)*64**-1 #mm**4\n",
+ "\n",
+ "#Bending Stress\n",
+ "#f=M*I*d1*2**-1 #N/mm**2\n",
+ "\n",
+ "#Principal Planes\n",
+ "#p_x=32*M*(pi*(d1**4-d2**4))*d1\n",
+ "#p_y=0\n",
+ "\n",
+ "#Shear stress\n",
+ "#q=T*J**-1*(d1*2**-1)\n",
+ "#After sub values and further simplifying we get\n",
+ "#q=16*T*d1*(pi*(d1**4-d2**4))*d1\n",
+ "\n",
+ "#Principal stresses\n",
+ "#P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "#After sub values and further simplifying we get\n",
+ "#P1=16*(pi*(d1**4-d2**4))*d1*(M+(M**2+t**2)**0.5) ...............(1)\n",
+ "\n",
+ "#P=2*pi*N*T*60**-1\n",
+ "#After sub values and further simplifying we get\n",
+ "T=P*60*(2*pi*N)**-1*10**-6 #N-mm\n",
+ "\n",
+ "#Again Sub values and further simplifying Equation 1 we get\n",
+ "M=(337.533)*(36.84)**-1 #KN-m\n",
+ "\n",
+ "#Min Principal stress\n",
+ "#P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "#Sub values and further simplifying we get\n",
+ "P2=16*(pi*(d1**4-d2**4))*d1*(M-(M**2+T**2)**0.5)*10**-11\n",
+ "\n",
+ "#Result\n",
+ "print\"Bending Moment safely applied to shaft is\",round(M,2),\"KN-m\"\n",
+ "print\"Min Principal Stress is\",round(P2,3),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Bending Moment safely applied to shaft is 9.16 KN-m\n",
+ "Min Principal Stress is -0.336 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.21,Page No.303"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=150 #mm #Diameter\n",
+ "T=20*10**6 #N #Torque\n",
+ "M=12*10**6 #N-mm #B.M\n",
+ "F=200*10**3 #N #Axial Thrust\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#M.I\n",
+ "I=(pi*64**-1*d**4)\n",
+ "\n",
+ "#Bending stress \n",
+ "f_A=M*I**-1*(d*2**-1) #N/mm**2\n",
+ "f_B=-f_A #N/mm**2\n",
+ "\n",
+ "#Axial thrust due to thrust\n",
+ "sigma=F*(pi*4**-1*d**2)**-1\n",
+ "\n",
+ "#At A\n",
+ "p_x=f_A-sigma #N/mm**2\n",
+ "\n",
+ "#At B\n",
+ "p_x2=f_B-sigma #N/mm**2\n",
+ "\n",
+ "p_y=0 #At A and B\n",
+ "\n",
+ "#Polar Modulus\n",
+ "J=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Shearing stress at A and B\n",
+ "q=T*J**-1*(d*2**-1) #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Principal Stresses\n",
+ "#At A\n",
+ "P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max1=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#At B\n",
+ "P1_2=(p_x2+p_y)*2**-1+(((p_x2-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2_2=(p_x2+p_y)*2**-1-(((p_x2-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max2=(((p_x2-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"MAx Principal Stresses:P1\",round(P1,2),\"N/mm**2\"\n",
+ "print\" :P2\",round(P2,2),\"N/mm**2\"\n",
+ "print\"Min Principal Stresses:P1_2\",round(P1_2,2),\"N/mm**2\"\n",
+ "print\" :P2_2\",round(P2_2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "MAx Principal Stresses:P1 45.1 N/mm**2\n",
+ " :P2 -20.2 N/mm**2\n",
+ "Min Principal Stresses:P1_2 14.65 N/mm**2\n",
+ " :P2_2 -62.18 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.22,Page No.311"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#strains\n",
+ "e_A=500 #microns\n",
+ "e_B=250 #microns\n",
+ "e_C=-150 #microns\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "theta=45 #Degrees\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "e_x=e_A=500\n",
+ "e_45=e_B=250\n",
+ "e_y=e_C=-150 \n",
+ "\n",
+ "#e_45=(e_x+e_y)*2**-1+(e_x-e_y)*2**-1*cos(2*theta)+rho_x_y*2**-1*sin(2*theta)\n",
+ "#After sub values and further simplifying we get\n",
+ "rho_x_y=(e_45-(e_x+e_y)*2**-1-(e_x-e_y)*2**-1*cos(2*theta*pi*180**-1))*(sin(2*theta*pi*180**-1))**-1*2\n",
+ "\n",
+ "#Principal strains are given by\n",
+ "e1=(e_x+e_y)*2**-1+(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 #microns\n",
+ "e2=(e_x+e_y)*2**-1-(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 #microns\n",
+ "\n",
+ "#Principal Stresses\n",
+ "sigma1=E*(e1+mu*e2)*(1-mu**2)**-1*10**-6 #N/mm**2\n",
+ "sigma2=E*(e2+mu*e1)*(1-mu**2)**-1*10**-6 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Strains are:e1\",round(e1,2),\"N/mm**2\"\n",
+ "print\" :e2\",round(e2,2),\"N/mm**2\"\n",
+ "print\"Principal Stresses are:sigma1\",round(sigma1,2),\"N/mm**2\"\n",
+ "print\" :sigma2\",round(sigma2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Strains are:e1 508.54 N/mm**2\n",
+ " :e2 -158.54 N/mm**2\n",
+ "Principal Stresses are:sigma1 101.31 N/mm**2\n",
+ " :sigma2 -1.31 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.23,Page No.313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Strains\n",
+ "e_A=600 #microns\n",
+ "e_B=-450 #microns\n",
+ "e_C=100 #micron\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "theta=240\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "e_x=e_A=600\n",
+ "\n",
+ "#e_A=(e_x+e_y)*2**-1+(e_x-e_y)*2**-1*cos(theta)+rho_x_y*2**-1*sin(theta)\n",
+ "#After sub values and further simplifying we get\n",
+ "#-450=(e_x+e_y)*2**-1-(e_x-e_y)*2**-1*(0.5)-0.866*2**-1*rho_x_y .....................(1)\n",
+ "\n",
+ "#e_C=(e_x+e_y)*2**-1+(e_x-e_y)*2**-1*cos(2*theta)+rho_x_y*2**-1*sin(2*theta)\n",
+ "#After sub values and further simplifying we get\n",
+ "#100=(e_x+e_y)*2**-1-0.5*(e_x-e_y)*2**-1*(0.5)-0.866*2**-1*rho_x_y .....................(2)\n",
+ "\n",
+ "#Adding Equation 1 and 2 we get equations as\n",
+ "#-350=e_x+e_y-(e_x-e_y)*2**-1 ...............(3)\n",
+ "#Further simplifying we get\n",
+ "\n",
+ "e_y=(-700-e_x)*3**-1 #micron \n",
+ "\n",
+ "rho_x_y=(e_C-(e_x+e_y)*2**-1-(e_x-e_y)*2**-1*cos(2*theta*pi*180**-1))*(sin(2*theta*pi*180**-1))**-1*2 #micron\n",
+ "\n",
+ "#Principal strains\n",
+ "e1=(e_x+e_y)*2**-1-(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 #microns\n",
+ "e2=(e_x+e_y)*2**-1+(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 #microns\n",
+ "\n",
+ "#Principal Stresses\n",
+ "sigma1=E*(e1+mu*e2)*(1-mu**2)**-1*10**-6 #N/mm**2\n",
+ "sigma2=E*(e2+mu*e1)*(1-mu**2)**-1*10**-6 #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses are:sigma1\",round(sigma1,2),\"N/mm**2\"\n",
+ "print\" :sigma2\",round(sigma2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses are:sigma1 -69.49 N/mm**2\n",
+ " :sigma2 117.11 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.8_9.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.8_9.ipynb
new file mode 100644
index 00000000..7d96f4c3
--- /dev/null
+++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.8_9.ipynb
@@ -0,0 +1,1531 @@
+{
+ "metadata": {
+ "name": "chapter no.8.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter No.8:Thin And Thick Cyclinders And Spheres"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.1,Page No.322"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=3000 #mm #Length\n",
+ "d1=1000 #mm #Internal diameter\n",
+ "t=15 #mm #Thickness\n",
+ "P=1.5 #N/mm**2 #Fluid Pressure\n",
+ "E=2*10**5 #n/mm**2 #Modulus of elasticity\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Hoop stress\n",
+ "f1=P*d1*(2*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Longitudinal Stress\n",
+ "f2=P*d1*(4*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(f1-f2)*2**-1 #N/mm**2\n",
+ "\n",
+ "#Diametrical Strain\n",
+ "#Let e1=dell_d*d**-1 .....................(1)\n",
+ "e1=(f1-mu*f2)*E**-1 \n",
+ "\n",
+ "#Sub values in equation 1 and further simplifying we get\n",
+ "dell_d=e1*d1 #mm\n",
+ "\n",
+ "#Longitudinal strain\n",
+ "#e2=dell_L*L**-1 ......................(2)\n",
+ "e2=(f2-mu*f1)*E**-1 \n",
+ "\n",
+ "#Sub values in equation 2 and further simplifying we get\n",
+ "dell_L=e2*L #mm\n",
+ "\n",
+ "#Change in Volume \n",
+ "#Let Z=dell_V*V**-1 ................(3)\n",
+ "Z=2*e1+e2\n",
+ "\n",
+ "#Sub values in equation 3 and further simplifying we get\n",
+ "dell_V=Z*pi*4**-1*d1**2*L\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Intensity of shear stress\",round(q_max,2),\"N/mm**2\"\n",
+ "print\"Change in the Dimensions of the shell is:dell_d\",round(dell_d,2),\"mm\"\n",
+ "print\" :dell_L\",round(dell_L,2),\"mm\"\n",
+ "print\" :dell_V\",round(dell_V,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Intensity of shear stress 12.5 N/mm**2\n",
+ "Change in the Dimensions of the shell is:dell_d 0.21 mm\n",
+ " :dell_L 0.15 mm\n",
+ " :dell_V 1119192.38 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.2,Page No.323"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=2000 #mm #Length\n",
+ "d=200 #mm # diameter\n",
+ "t=10 #mm #Thickness\n",
+ "dell_V=25000 #mm**3 #Additional volume\n",
+ "E=2*10**5 #n/mm**2 #Modulus of elasticity\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let p be the pressure developed\n",
+ "\n",
+ "#Circumferential Stress\n",
+ "\n",
+ "#f1=p*d*(2*t)**-1 #N/mm**2\n",
+ "#After sub values and further simplifying\n",
+ "#f1=10*p\n",
+ "\n",
+ "#f1=p*d*(4*t)**-1 #N/mm**2\n",
+ "#After sub values and further simplifying\n",
+ "#f1=5*p\n",
+ "\n",
+ "#Diameterical strain = Circumferential stress\n",
+ "#Let X=dell_d*d**-1 ................................(1)\n",
+ "#X=e1=(f1-mu*f2)*E**-1 \n",
+ "#After sub values and further simplifying\n",
+ "#e1=8.5*p*E**-1\n",
+ "\n",
+ "#Longitudinal strain\n",
+ "#Let Y=dell_L*L**-1 ......................................(2)\n",
+ "#Y=e2=(f2-mu*f1)*E**-1 \n",
+ "#After sub values and further simplifying\n",
+ "#e2=2*p*E**-1\n",
+ "\n",
+ "#Volumetric strain\n",
+ "#Let X=dell_V*V**-1 \n",
+ "#X=2*e1+e2\n",
+ "#After sub values and further simplifying\n",
+ "#X=19*p*E**-1\n",
+ "#After further simplifying we get\n",
+ "p=dell_V*(pi*4**-1*d**2*L)**-1*E*19**-1 #N/mm**2\n",
+ "\n",
+ "#Hoop Stress\n",
+ "f1=p*d*(2*t)**-1\n",
+ "\n",
+ "X=e1=8.5*p*E**-1\n",
+ "#Sub value of X in equation 1 we get\n",
+ "dell_d=8.5*p*E**-1*d\n",
+ "\n",
+ "Y=e2=2*p*E**-1\n",
+ "#Sub value of Y in equation 2 we get\n",
+ "dell_L=2*p*E**-1*L\n",
+ "\n",
+ "#Result\n",
+ "print\"Pressure Developed is\",round(p,2),\"N/mm**2\"\n",
+ "print\"Hoop stress Developed is\",round(f1,2),\"N/mm**2\"\n",
+ "print\"Change in diameter is\",round(dell_d,2),\"mm\"\n",
+ "print\"Change in Length is\",round(dell_L,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Pressure Developed is 4.19 N/mm**2\n",
+ "Hoop stress Developed is 41.88 N/mm**2\n",
+ "Change in diameter is 0.04 mm\n",
+ "Change in Length is 0.08 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.3,Page No.324"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=750 #mm #Diameter of water supply pipes\n",
+ "h=50*10**3 #mm #Water head\n",
+ "sigma=20 #N/mm**2 #Permissible stress\n",
+ "rho=9810*10**-9 #N/mm**3\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Pressure of water\n",
+ "P=rho*h #N/mm**2\n",
+ "\n",
+ "#Stress\n",
+ "#sigma=p*d*(2*t)**-1\n",
+ "#After further simplifying\n",
+ "t=P*d*(2*sigma)**-1 #mm \n",
+ "\n",
+ "#Result\n",
+ "print\"Thickness of seamless pipe is\",round(t,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness of seamless pipe is 9.197 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.4,Page No.326"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=2500 #mm #Diameter of riveted boiler\n",
+ "P=1 #N/mm**2 #Pressure\n",
+ "rho1=0.7 #Percent efficiency\n",
+ "rho2=0.4 #Circumferential joints\n",
+ "sigma=150 #N/mm**2 #Permissible stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Equating Bursting force to longitudinal joint strength ,we get\n",
+ "#p*d*L=rho1*2*t*L*sigma\n",
+ "#After rearranging and further simplifying we get\n",
+ "t=P*d*(2*sigma*rho1)**-1 #mm\n",
+ "\n",
+ "#Considering Longitudinal force\n",
+ "#pi*d**2*4**-1*P=rho2*pi*d*t*sigma\n",
+ "#After rearranging and further simplifying we get\n",
+ "t2=P*d*(4*sigma*rho2)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Thickness of plate required is\",round(t,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness of plate required is 11.9 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.5,Page No.326"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Boiler Dimensions\n",
+ "t=16 #mm #Thickness\n",
+ "p=2 #N/mm**2 #internal pressure\n",
+ "f=150 #N/mm**2 #Permissible stress\n",
+ "rho1=0.75 #Longitudinal joints\n",
+ "rho2=0.45 #circumferential joints\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Equating Bursting force to longitudinal joint strength ,we get\n",
+ "d1=rho1*2*t*f*p**-1 #mm\n",
+ "\n",
+ "#Considering circumferential strength \n",
+ "d2=4*rho2*t*f*p**-1 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Largest diameter of Boiler is\",round(d1,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Largest diameter of Boiler is 1800.0 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 52
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.6,Page No.329"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=250 #mm #Diameter iron pipe\n",
+ "t=10 #mm #Thickness\n",
+ "d2=6 #mm #Diameter of steel\n",
+ "p=80 #N/mm**2 #stress\n",
+ "P=3 #N/mm**2 #Pressure\n",
+ "E_c=1*10**5 #N/mm**2\n",
+ "mu=0.3 #poissoin's ratio\n",
+ "E_s=2*10**5 #N/mm**2\n",
+ "n=1 #No.of wires\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "L=6 #mm #Length of cyclinder\n",
+ "\n",
+ "#Force Exerted by steel wire at diameterical section\n",
+ "F=p*2*pi*d2**2*1*4**-1 #N\n",
+ "\n",
+ "#Initial stress in cyclinder\n",
+ "f_c=F*(2*t*d2)**-1 #N/mm**2\n",
+ "\n",
+ "#LEt due to fluid pressure alone stresses developed in steel wire be F_w and in cyclinder f1 and f2\n",
+ "f2=P*d*(4*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Considering the equilibrium of half the cyclinder, 6mm long we get\n",
+ "#F_w*2*pi*4**-1*d2**2*n+f1*2*t*d2=P*d*d2\n",
+ "#After further simplifying we get\n",
+ "#F_w+2.122*f1=79.58 . ......................................(1)\n",
+ "\n",
+ "#Equating strain in wire to circumferential strain in cyclinder \n",
+ "#F_w=(f1-mu*f2)*E_s*E_c**-1 #N/mm**2\n",
+ "#After further simplifying we get\n",
+ "#F_w=2*f1-11.25 ....................................(2)\n",
+ "\n",
+ "#Sub in equation in1 we get\n",
+ "f1=(79.58+11.25)*(4.122)**-1 #N/mm**2\n",
+ "F_w=2*f1-11.25 #N/mm**2\n",
+ "\n",
+ "#Final stresses\n",
+ "#1) In steel Wire\n",
+ "sigma=F_w+p #N/mm**2\n",
+ "\n",
+ "#2) In Cyclinder\n",
+ "sigma2=f1-f_c\n",
+ "\n",
+ "#Result\n",
+ "print\"Final Stresses developed in:cyclinder is\",round(sigma,2),\"N/mm**2\"\n",
+ "print\" :Steel is\",round(sigma2,2),\"N/mm**2\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Final Stresses developed in:cyclinder is 112.82 N/mm**2\n",
+ " :Steel is -15.66 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.7,Page No.332"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=750 #mm #Diameter of shell\n",
+ "t=8 #mm #THickness\n",
+ "p=2.5 #N/mm**2\n",
+ "E=2*10**5 #N/mm**2\n",
+ "mu=0.25 #Poissoin's ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Hoop stress\n",
+ "f1=f2=p*d*(4*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Change in Diameter\n",
+ "dell_d=d*p*d*(1-mu)*(4*t*E)**-1 #mm\n",
+ "\n",
+ "#Change in Volume\n",
+ "dell_V=3*p*d*(1-mu)*(4*t*E)**-1*pi*6**-1*d**3\n",
+ "\n",
+ "#Answer for Change in diameter is incorrect in book\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress introduced is\",round(f1,2),\"N/mm**2\"\n",
+ "print\"Change in Diameter is\",round(dell_d,2),\"N/mm**2\"\n",
+ "print\"Change in Volume is\",round(dell_V,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress introduced is 58.59 N/mm**2\n",
+ "Change in Diameter is 0.16 N/mm**2\n",
+ "Change in Volume is 145608.33 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.8,Page No.333"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=600 #mm #Diameter of sherical shell\n",
+ "t=10 #mm #Thickness\n",
+ "f=80 #N/mm**2 #Permissible stress\n",
+ "rho=0.75 #Efficiency joint\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max Pressure\n",
+ "p=f*4*t*rho*d**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Pressure is\",round(p,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Pressure is 4.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 60
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.9,Page No.333"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1000 #mm #Length of shell\n",
+ "d=200 #mm #Diameter\n",
+ "t=6 #mm #Thickness\n",
+ "p=1.5 #N/mm**2 #Internal Pressure\n",
+ "E=2*10**5 #N/mm**2\n",
+ "mu=0.25 #Poissoin's Ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Change in Volume of sphere\n",
+ "dell_V_s=3*p*d*(1-mu)*(4*t*E)**-1*pi*6**-1*d**3\n",
+ "\n",
+ "#Hoop stress\n",
+ "f1=p*d*(2*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Longitudinal stress\n",
+ "f2=p*d*(4*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Principal strain\n",
+ "e1=(f1-mu*f2)*E**-1\n",
+ "e2=(f2-mu*f1)*E**-1\n",
+ "\n",
+ "V_c=1000 #mm**3\n",
+ "\n",
+ "#Change in Volume of cyclinder\n",
+ "dell_V_c=(2*e1+e2)*pi*4**-1*d**2*L\n",
+ "\n",
+ "#Total Change in Diameter\n",
+ "dell_V=dell_V_s+dell_V_c #mm**3\n",
+ "\n",
+ "#Result\n",
+ "print\"Change in Volume is\",round(dell_V,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Change in Volume is 8443.03 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.10,Page No.337"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "matplotlib.pyplot\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=400 #mm #Internal Diameter\n",
+ "t=100 #mm #Thickness\n",
+ "p=80 #N/mm**2 #Fluid pressure\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Internal Radius\n",
+ "r1=d1*2**-1 #mm\n",
+ "\n",
+ "#Outer Radius\n",
+ "r_o=r1+t #mm\n",
+ "\n",
+ "p1=80 #N/mm**2\n",
+ "p2=0\n",
+ "\n",
+ "#Now From Lame's Euation\n",
+ "#p_x=b*(x**2)**-1-a\n",
+ "#at x=200 #mm \n",
+ "p_x=80 #N/mm**2\n",
+ "#80=b*(200**2)**-1-a ..........................(1)\n",
+ "\n",
+ "#at x=300 #mm\n",
+ "#p_x2=0\n",
+ "#0=b*(300**2)**-1-a ...........................(2)\n",
+ "\n",
+ "#Sub equation 2 from 1\n",
+ "#80=b*(200**2)**-1-b*(300**2)**-1\n",
+ "#After Further simplifying we get\n",
+ "b=(50000)**-1*(200**2*300**2*80)\n",
+ "\n",
+ "#From equation 2 we get\n",
+ "a=b*(300**2)**-1\n",
+ "\n",
+ "#Variation of radial pressure p_x;\n",
+ "#p_x=b*(x**2)**-1-a\n",
+ "#After sub values and further simplifying we get\n",
+ "\n",
+ "#Radial pressure Variation\n",
+ "#At \n",
+ "x=200 #mm\n",
+ "p_x=b*(x**2)**-1-a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x2=250 #mm\n",
+ "p_x2=b*(x2**2)**-1-a #N/mm**2\n",
+ "\n",
+ "#At \n",
+ "x3=300 #mm\n",
+ "p_x3=b*(x3**2)**-1-a #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Hoop stress Distribution\n",
+ "#Variation of F_x\n",
+ "\n",
+ "#At \n",
+ "x=200 #mm\n",
+ "F_x=b*(x**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x2=250 #mm\n",
+ "F_x2=b*(x2**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x3=300 #mm\n",
+ "F_x3=b*(x3**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Hoop stress is\",round(F_x,2),\"N/mm**2\"\n",
+ "print\"Min Hoop stress is\",round(F_x3,2),\"N/mm**2\"\n",
+ "print\"Plot of Hoop stress\"\n",
+ "\n",
+ "#Plotting Variation of hoop stress\n",
+ "\n",
+ "X1=[x,x2,x3]\n",
+ "Y1=[p_x,p_x2,p_x3]\n",
+ "Y2=[-F_x,-F_x2,-F_x3]\n",
+ "Z1=[0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Y2,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Radial Stress Distribution & Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Hoop stress is 208.0 N/mm**2\n",
+ "Min Hoop stress is 128.0 N/mm**2\n",
+ "Plot of Hoop stress\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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skdWFW7ZsQffu3XH8+HGMGzcOY8aMAaA9wnfKlCkIDg7GmDFjEBcXJ7WK4uLi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\nbm5uYtOmTUII3b27hBBCoVCIHTt2CCG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+ "text": [
+ "<matplotlib.figure.Figure at 0x549c530>"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.11,Page No.338"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "matplotlib.pyplot\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=300 #mm #Outside diameter \n",
+ "d2=200 #mm #Internal Diameter\n",
+ "p=14 #N/mm**2 #internal Fluid pressure\n",
+ "t=50 #mm #Thickness\n",
+ "r_o=150 #mm #Outside Diameter\n",
+ "r2=100 #mm #Internal Diameter\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Lame's Equation\n",
+ "#p_x=b*(x**2)**-1-a #N/mm**2 ...................(1)\n",
+ "#F_x=b*(x**2)**-1+a #N/mm**2 ...................(2)\n",
+ "\n",
+ "#At \n",
+ "x=r2=100 #mm\n",
+ "p_x=14 #N/mm**2\n",
+ "\n",
+ "#Sub value of p_x in equation 1 we get\n",
+ "#14=(100)**-1*b-a ............................(3)\n",
+ "\n",
+ "#At\n",
+ "x2=r_o=150 #mm\n",
+ "p_x2=0 #N/mm**2\n",
+ "\n",
+ "#Sub value in equation 1 we get\n",
+ "#0=b*(150**2)**-1-a ......................(4)\n",
+ "\n",
+ "#From Equations 3 and 4 we get\n",
+ "#14=b*(100**2)**-1-b*(100**2)**-1\n",
+ "#After sub values and further simplifying we get\n",
+ "b=14*100**2*150**2*(150**2-100**2)**-1\n",
+ "\n",
+ "#From equation 4 we get\n",
+ "a=b*(150**2)**-1\n",
+ "\n",
+ "#Hoop Stress\n",
+ "#F_x=b*(x**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At \n",
+ "x=100 #mm\n",
+ "F_x=b*(x**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x2=125 #mm\n",
+ "F_x2=b*(x2**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x3=150 #mm\n",
+ "F_x3=b*(x3**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#If thin Cyclindrical shell theory is used,hoop stress is uniform and is given by\n",
+ "F=p*d2*(2*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Percentage error in estimating max hoop tension\n",
+ "E=(F_x-F)*F_x**-1*100 #%\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Hoop Stress Developed in the cross-section is\",round(F,2),\"N/mm**2\"\n",
+ "print\"Plot of Variation of hoop stress\"\n",
+ "\n",
+ "#Plotting Variation of hoop stress\n",
+ "\n",
+ "X1=[x,x2,x3]\n",
+ "Y1=[F_x,F_x2,F_x3]\n",
+ "Z1=[0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Radial Stress Distribution & Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Hoop Stress Developed in the cross-section is 28.0 N/mm**2\n",
+ "Plot of Variation of hoop stress\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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29vaQJAknTpx4bAP379/HiBEjMGzYMMycOdNwT41GA1dXV2RkZGDgwIEcMiIi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\nSUVORK5CYII=\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x54a6210>"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.12,Page No.339"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=300 #mm #Outside diameter \n",
+ "d2=200 #mm #Internal Diameter\n",
+ "p=12 #N/mm**2 #internal Fluid pressure\n",
+ "F_max=16 #N/mm**2 #Tensile stress\n",
+ "r_o=150 #mm #Outside Diameter\n",
+ "r2=100 #mm #Internal Diameter\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let p_o be the External Pressure applied.\n",
+ "#From LLame's theorem\n",
+ "#p_x=b*(x**2)**-1-a ..............(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#Now At\n",
+ "x=100 #mm\n",
+ "p_x=12 #N/mm**2\n",
+ "#sub in equation 1 we get\n",
+ "#12=b*(100**2)**-1-a . ..................(3)\n",
+ "\n",
+ "#The Max Hoop stress occurs at least value of x where\n",
+ "x=r1=100 #mm\n",
+ "#16=b*(100**2)**-1+a .......................(4)\n",
+ "\n",
+ "#From Equations 1 and 2 we get\n",
+ "#28=b*(100**2)**-1+b*(100**2)**-1\n",
+ "#After furhter Simplifying we get\n",
+ "b=28*100**2*2**-1\n",
+ "\n",
+ "#sub in equation 1 we get\n",
+ "a=-(12-(b*(100**2)**-1))\n",
+ "\n",
+ "#Thus At\n",
+ "x2=150 #mm\n",
+ "p_o=b*(x2**2)**-1-a\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum External applied is\",round(p_o,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum External applied is 4.22 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.13,Page No.340"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=160 #mm #Internal Diameter \n",
+ "r1=80 #mm #External Diameter\n",
+ "p1=40 #N/mm**2 #Internal Diameter\n",
+ "P_max=120 #N/mm**2 #Allowable stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Lame's Equation we have\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#At \n",
+ "x=r1=80 #N/mm**2 \n",
+ "#Sub in equation 1 we get\n",
+ "#120=b*(80**2)**-1+a ........................(3)\n",
+ "\n",
+ "#The hoop tension at inner edge is max stress\n",
+ "#Hence\n",
+ "#120=b*(80**2)**-1+a .............................(4)\n",
+ "\n",
+ "#From Equation 3 and 4 we get\n",
+ "b=160*80**2*2**-1 \n",
+ "\n",
+ "#Sub in equation 3 we get\n",
+ "a=-(40-(b*(80**2)**-1))\n",
+ "\n",
+ "#Let External radius be r_o.Since at External Surface is Zero,we get\n",
+ "#0=b*(r_o)**-1-a\n",
+ "#After Further simplifying we get\n",
+ "r_o=(b*a**-1)**0.5\n",
+ "\n",
+ "#Thickness of Cyclinder \n",
+ "t=r_o-r1 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Thickness Required is\",round(t,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness Required is 33.14 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.14,Page No.341"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=300 #mm #Outside diameter \n",
+ "d1=180 #mm #Internal Diameter\n",
+ "p=12 #N/mm**2 #internal Fluid pressure\n",
+ "p_o=6 #N/mm**2 #External Pressure\n",
+ "r_o=150 #mm #Outside Diameter\n",
+ "r=90 #mm #Internal Diameter\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Lame's Equation we have\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#At \n",
+ "x=r1=90 #N/mm**2 \n",
+ "p=42 #N/mm**2\n",
+ "#Sub in equation 1 we get\n",
+ "#42=b*(90**2)**-1-a ..............................(3)\n",
+ "\n",
+ "#At \n",
+ "x=r_o=150 #mm\n",
+ "p2=6 #N/mm**2\n",
+ "#sub in equation 1 we get\n",
+ "#6=b*(150**2)**-1-a ..............................(4)\n",
+ "\n",
+ "#From equations 3 and 4 weget\n",
+ "#36=b*(90**2)**-1-b2(150**2)**-1\n",
+ "#After further simplifying we get\n",
+ "b=36*90**2*150**2*(150**2-90**2)**-1\n",
+ "\n",
+ "#Sub value of b in equation 4 we get\n",
+ "a=b*(150**2)**-1-p_o\n",
+ "\n",
+ "#At \n",
+ "x=r1=90 #mm\n",
+ "F_x=b*(x**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At \n",
+ "x2=r_o=150 #mm \n",
+ "F_x2=b*(x2**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#Now if External pressure is doubled i.e p_o2=12 #N/mm**2 We have\n",
+ "p_o2=12 #N/mm**2\n",
+ "#sub in equation 4 we get\n",
+ "#12=b2*(150**2)**-1-a2 ..........................(5)\n",
+ "\n",
+ "#Max Hoop stress is to be 70.5 #N/mm**2,which occurs at x=r1=90 #mm\n",
+ "#Sub in equation 4 we get\n",
+ "#70.5=b*(90**2)**-1+a2 ................................(6)\n",
+ "\n",
+ "#Adding equation 5 and 6\n",
+ "#82.5=b2*(150**2)**-1+b*(90**2)**-1\n",
+ "#After furhter simplifying we get\n",
+ "b2=82.5*150**2*90**2*(150**2+90**2)**-1\n",
+ "\n",
+ "#Sub in equation 5 we get\n",
+ "a2=b2*(150**2)**-1-12 \n",
+ "\n",
+ "#If p_i is the internal pressure required then from Lame's theorem\n",
+ "p_i=b2*(r1**2)**-1-a2\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses int the material are:F_x\",round(F_x,2),\"N/mm**2\"\n",
+ "print\" :F_x2\",round(F_x2,2),\"N/mm**2\"\n",
+ "print\"Internal Pressure that can be maintained is\",round(p_i,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses int the material are:F_x 70.5 N/mm**2\n",
+ " :F_x2 34.5 N/mm**2\n",
+ "Internal Pressure that can be maintained is 50.82 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.15,Page No.344"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "matplotlib.pyplot\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "r1=200 #mm #Inner Radius\n",
+ "r2=250 #mm #Radius at common surface\n",
+ "r3=300 #mm #Outer radius\n",
+ "p=6 #N/mm**2 #Inital pressure\n",
+ "p2=80 #N/mm**2 #Pressure\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Inner Cyclinder:\n",
+ "\n",
+ "#From Lame's Equation we have\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#At \n",
+ "x=r1=200 #mm\n",
+ "p_x=0\n",
+ "#0=b1*(250**2)**-1-a1 .................(3)\n",
+ "\n",
+ "#At x=r2=250 #mm\n",
+ "p_x2=6 #N/mm**2\n",
+ "#6=b1*(250**2)-a1 ...................(4)\n",
+ "\n",
+ "#From Equation 3 and 4 we get\n",
+ "b1=6*200**2*250**2*(200**2-250**2)**-1\n",
+ "\n",
+ "#From equation 3 we get\n",
+ "a1=b1*(200**2)**-1\n",
+ "\n",
+ "F_200=b1*(200**2)**-1+a1\n",
+ "F_250=b1*(250**2)**-1+a1\n",
+ "\n",
+ "#For outer cyclinder \n",
+ "#From Lame's Equation we have\n",
+ "#p_x2=b2*(x**2)**-1-a2 ..........................(5)\n",
+ "#F_x2=b2*(x**2)**-1+a2 ...........................(6)\n",
+ "\n",
+ "\n",
+ "#At \n",
+ "x2=r2=250 #mm\n",
+ "p_x2=6 #N/mm**2\n",
+ "#6=b2*(250**2)**-1-a2 ...........................(7) \n",
+ "\n",
+ "#At\n",
+ "x3=300 #mm\n",
+ "#p_x2=0\n",
+ "#0=b2**2*(300**2)**-1-a2 .................................(8)\n",
+ "\n",
+ "#from equation 7 and 8 we get\n",
+ "b2=6*250**2*300**2*(300**2-250**2)**-1\n",
+ "\n",
+ "#sub in equation 8 we get\n",
+ "a2=b2*(300**2)**-1\n",
+ "\n",
+ "F_250_2=b2*(250**2)**-1+a2\n",
+ "F_300_2=b2*(300**2)**-1+a2\n",
+ "\n",
+ "#When Fluid is admitted\n",
+ "#Let Lame's equation be\n",
+ "#p_x3=b3*(x**2)**-1-a3 ..........................(5)\n",
+ "#F_x3=b3*(x**2)**-1+a3 ...........................(6)\n",
+ "\n",
+ "\n",
+ "#At x=200\n",
+ "p_x3=80 #N/mm**2\n",
+ "#80=b3*(200**2)**-1-a3 ................................(7)\n",
+ "\n",
+ "#At x=300 #mm\n",
+ "#p_x=0\n",
+ "#0=b3*(300**2)**-1-a3 ..............................(8)\n",
+ "\n",
+ "#from Equation 7 and 8 we get\n",
+ "b3=80*200**2*300**2*(300**2-200**2)**-1\n",
+ "\n",
+ "#From Equation 8 we get\n",
+ "a3=b3*(300**2)**-1\n",
+ "\n",
+ "#Hoop stresses \n",
+ "F_200_3=b3*(200**2)**-1+a3 #N/mm**2\n",
+ "F_250_3=b3*(250**2)**-1+a3 #N/mm**2\n",
+ "F_300_3=b3*(300**2)**-1+a3 #N/mm**2\n",
+ "\n",
+ "#Pressure at common surface\n",
+ "p_250=b3*(250**2)**-1-a3 #N/mm**2\n",
+ "\n",
+ "#final stress\n",
+ "f_200=F_200+F_200_3 #N/mm**2\n",
+ "f_250=F_250+F_250_3 #N/mm**2\n",
+ "f_300=F_250_2+F_250_3 #N/mm**2\n",
+ "f_300_2=F_300_2+F_300_3 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"final Hoop stress are:f_200\",round(f_200,2),\"N/mm**2\"\n",
+ "print\" :f_250\",round(f_250,2),\"N/mm**2\"\n",
+ "print\" :f_300\",round(f_300,2),\"N/mm**2\"\n",
+ "print\" :f_300_2\",round(f_300_2,2),\"N/mm**2\"\n",
+ "print\"Variation of Hoop stress and Radial stress\"\n",
+ "\n",
+ "#Final stresses\n",
+ "#Variation of hoop stress \n",
+ " \n",
+ "X1=[x,x2,x3,x3]\n",
+ "Y1=[f_200,f_250,f_300,f_300_2]\n",
+ "Z1=[0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Due to Fluid\n",
+ "#Variation of hoop stress \n",
+ " \n",
+ "X1=[x,x2,x3]\n",
+ "Y1=[F_200_3,F_250_3,F_300_3]\n",
+ "Z1=[0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "final Hoop stress are:f_200 174.67 N/mm**2\n",
+ " :f_250 128.83 N/mm**2\n",
+ " :f_300 189.43 N/mm**2\n",
+ " :f_300_2 155.27 N/mm**2\n",
+ "Variation of Hoop stress and Radial stress\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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MHz4cr7zyCmbPno1///vf1fIBnZeXp/l+8+bNmhVR0dHR+P7771FYWIjMzExcvHgRnTp1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\nVw+AnLO4d++eQY0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+ "text": [
+ "<matplotlib.figure.Figure at 0x548ecf0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x54b60b0>"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.16,Page No.348"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "do=200 #mm #Inner Diameter\n",
+ "r_o=100 #mm #Inner radius\n",
+ "d1=300 #mm #outer diameter\n",
+ "r1=150 #mm #Outer radius\n",
+ "d2=250 #mm #Junction Diameter\n",
+ "r2=125 #mm #Junction radius\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "p=30 #N/mm**2 #radial pressure\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#from Lame's Equation we get\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#Then from Boundary condition \n",
+ "#p_x=0 at x=100 #mm\n",
+ "#0=b1*(100**2)**-1-a1 .....................(3)\n",
+ "\n",
+ "#p_x2=30 #N/mm**2 at x2=125 #mm\n",
+ "#30=b1*(125**2)**-1-a1 ................................(4)\n",
+ "\n",
+ "#From equation 3 and 4 we get\n",
+ "b1=30*125**2*100**2*(100**2-125**2)**-1\n",
+ "\n",
+ "#From Equation 3 we get\n",
+ "a1=b1*(100**2)**-1\n",
+ "\n",
+ "#therefore Hoop stress in inner cyclinder at junction\n",
+ "F_2_1=b1*(125**2)**-1+a1 #N/mm**2\n",
+ "\n",
+ "#Outer Cyclinder\n",
+ "#p_x=b*(x**2)**-1-a ..........................(5)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(6)\n",
+ "\n",
+ "#Now at x=125 #mm\n",
+ "#p_x3=30 #N/mm**2\n",
+ "#30=b2*(125**2)**-1-a2 ..................................(7)\n",
+ "\n",
+ "#At x=150 #mm\n",
+ "#p_x4=0\n",
+ "#0=b2*(150**2)**-1-a2 ...................................(8)\n",
+ "\n",
+ "#From equations 7 and 8\n",
+ "b2=30*150**2*125**2*(150**2-125**2)**-1\n",
+ "\n",
+ "#From eqauation 8 we get\n",
+ "a2=b2*(150**2)**-1\n",
+ "\n",
+ "#Hoop stress at junction \n",
+ "F_2_0=b2*(125**2)**-1+a2 #N/mm**2\n",
+ "\n",
+ "rho_r=(F_2_0-F_2_1)*E**-1*r2\n",
+ "\n",
+ "#Result\n",
+ "print\"Shrinkage Allowance is\",round(rho_r,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shrinkage Allowance is 0.189 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.17,Page No.350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=500 #mm #Outer Diameter\n",
+ "r_o=250 #mm #Outer Radius\n",
+ "d1=300 #mm #Inner Diameter\n",
+ "r1=150 #mm #Inner Radius\n",
+ "d2=400 #mm #Junction Diameter\n",
+ "E=2*10**5 #N/mm**2 #Modulus ofElasticity\n",
+ "alpha=12*10**-6 #Per degree celsius\n",
+ "dell_d=0.2 #mm\n",
+ "dell_r=0.1 #mm\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let p be the radial pressure developed at junction\n",
+ "#Let Lame's Equation for internal cyclinder be\n",
+ "#p_x=b*(x**2)**-1-a ................................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...............................(2)\n",
+ "\n",
+ "#At \n",
+ "x=150 #mm \n",
+ "p_x=0\n",
+ "#Sub in equation 1 we get\n",
+ "#0=b*(150**2)**-1-a .........................(3)\n",
+ "\n",
+ "#At \n",
+ "x2=200 #mm\n",
+ "#p_x2=p\n",
+ "#p=b*(200**2)**-1-a ......................(4)\n",
+ " \n",
+ "#From Equation 3 and 4\n",
+ "#p=b*(200**2)**-1-b(150**2)**-1\n",
+ "#after further simplifying we get\n",
+ "#b=-51428.571*p\n",
+ "\n",
+ "#sub in equation 3 we get\n",
+ "#a1=-2.2857*p\n",
+ "\n",
+ "#therefore hoop stress at junction is\n",
+ "#F_2_1=-21428.571*p*(200**2)**-1-2.2857*p\n",
+ "#after Further simplifying we geet\n",
+ "#F_2_1=3.5714*p\n",
+ "\n",
+ "#Let Lame's Equation for cyclinder be \n",
+ "#p_x=b*(x**2)**-1-a .........................5\n",
+ "#F_x=b*(x**2)**-1+a .............................6\n",
+ "\n",
+ "#At \n",
+ "x=200 #mm\n",
+ "#p_x=p2\n",
+ "#p2=b2*(20**2)**-1-a2 ...................7\n",
+ "\n",
+ "#At\n",
+ "x2=200 #mm\n",
+ "p_x2=0\n",
+ "#0=b2*(250**2)**-1-a2 ....................8\n",
+ "\n",
+ "#from equation 7 and 8 we get\n",
+ "#p2=b2*(200**2)**-1-b2*(250**2)**-1\n",
+ "#After further simplifying we get\n",
+ "#p2=b2*(250**2-200**2)*(200**2*250**2)**-1\n",
+ "#b2=111111.11*p\n",
+ "\n",
+ "#from equation 7\n",
+ "#a2=b2*(250**2)**-1\n",
+ "#further simplifying we get\n",
+ "#a2=1.778*p\n",
+ "\n",
+ "#At the junctionhoop stress in outer cyclinder \n",
+ "#F_2_0=b2*(200**2)**-1+a2\n",
+ "#After further simplifying we get\n",
+ "#F_2_0=4.5556*p\n",
+ "\n",
+ "#Considering circumferential strain,the compatibility condition\n",
+ "#rho_r*r2**-1=1*E**-1*(F_2_1+F_2_0)\n",
+ "#where F_2_1 is compressive and F_2_0 is tensile\n",
+ "#furter simplifying we get\n",
+ "p=0.1*200**-1*2*10**5*(3.5714+4.5556)**-1\n",
+ "\n",
+ "#Let T be the rise in temperature required\n",
+ "#dell_d=d*alpha*T\n",
+ "#After sub values and further simplifying we get\n",
+ "d=250 #mm\n",
+ "T=dell_d*(d*alpha)**-1 #Per degree celsius\n",
+ "\n",
+ "#Result\n",
+ "print\"Radial Pressure Developed at junction\",round(p,2),\"N/mm**2\"\n",
+ "print\"Min Temperatureto outer cyclinder\",round(T,2),\"Per degree Celsius\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radial Pressure Developed at junction 12.3 N/mm**2\n",
+ "Min Temperatureto outer cyclinder 66.67 Per degree Celsius\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.18,Page No.355"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "matplotlib.pyplot\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=400 #mm #Outer Diameter\n",
+ "r_o=200 #mm #Outer radius\n",
+ "t=50 #mm #Thickness\n",
+ "r1=150 #mm #Internal Radius\n",
+ "p=50 #N/mm**2 #Internal Pressure\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#The Radial Pressure and hoop stress at any radial distance x are given by\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#Now at\n",
+ "x=150 #N/mm**2\n",
+ "p_x1=50 #N/mm**2\n",
+ "#Sub in equation 1 we get\n",
+ "#50=2*b*(150**3)**-1-a ...........................(3)\n",
+ "\n",
+ "#At x=200 #mm\n",
+ "p_x2=0\n",
+ "#0=2*b*(200**2)**-1-a ....................(4)\n",
+ "\n",
+ "#From equation 3 and 4 we get\n",
+ "#50=2*b*(150**3)**-1-2*b*(200**3)**-1\n",
+ "#After further simplifying we get\n",
+ "b=50*150**3*200**3*(200**3-150**3)**-1*2**-1\n",
+ "\n",
+ "#Sub in equation 3 we get\n",
+ "a=b*(200**3)**-1\n",
+ "\n",
+ "#Now At\n",
+ "x=150 #mm\n",
+ "F_x=b*(x**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x2=160 #mm\n",
+ "F_x2=b*(x2**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x3=170 #mm\n",
+ "F_x3=b*(x3**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x4=180 #mm\n",
+ "F_x4=b*(x4**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x5=190 #mm\n",
+ "F_x5=b*(x5**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x6=200 #mm\n",
+ "F_x6=b*(x6**3)**-1+a\n",
+ "\n",
+ "#Result\n",
+ "print\"Plot of Variation of hoop stress\"\n",
+ "\n",
+ "#Plotting Variation of hoop stress\n",
+ "\n",
+ "X1=[x,x2,x3,x4,x5,x6]\n",
+ "Y1=[F_x,F_x2,F_x3,F_x4,F_x5,F_x6]\n",
+ "Z1=[0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Plot of Variation of hoop stress\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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w3g2UlJRgzJgxCAsLw3vvvQcA8PT0hFqthpOTE7KyshAYGIhLly7VLIxzCERE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\nBgIREQEA/h+bezx5xlsz+QAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5669670>"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.9_9.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.9_9.ipynb
new file mode 100644
index 00000000..cecacb12
--- /dev/null
+++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.9_9.ipynb
@@ -0,0 +1,779 @@
+{
+ "metadata": {
+ "name": "chapter no.9.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 9:Columns And Struts"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.1,Page No.377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "L=5000 #mm #Length of strut\n",
+ "dell=10 #mm #Deflection\n",
+ "W=10 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Central Deflection of a simply supported beam with central concentrated load is\n",
+ "#dell=W*L**3*(48*E*I)**-1 \n",
+ "\n",
+ "#Let E*I=X\n",
+ "X=W*L**3*(48*dell)**-1 #mm\n",
+ "\n",
+ "#Euler's Load\n",
+ "#Let Euler's Load be P\n",
+ "P=pi**2*X*(L**2)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Critical Load of Bar is\",round(P,2),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Critical Load of Bar is 1028.08 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.2,Page No.377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=2000 #mm #Length of square column\n",
+ "E=12*10**3 #N/mm**2 #Modulus of Elasticity\n",
+ "sigma=12 #N/mm*2 #stress\n",
+ "W1=95*10**3 #N #Load1\n",
+ "W2=200*10**3 #N #Load2\n",
+ "FOS=3\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Euler's Formula\n",
+ "#P=pi**2*E*I*(L**2)**-1 .........(1)\n",
+ "\n",
+ "#Working Load\n",
+ "#W=P*(FOS)**-1\n",
+ "\n",
+ "#Part-1\n",
+ "\n",
+ "#At W1=95*10**3 #N\n",
+ "#W1=P*(3*L**2)**-1\n",
+ "\n",
+ "#Let 'a' be the side of the square\n",
+ "#I=1*12**-1*a**4\n",
+ "\n",
+ "#sub value of I in Equation 1 and further rearranging we get\n",
+ "a=(W1*3*12*L**2*(pi**2*E)**-1)**0.25 #mm\n",
+ "\n",
+ "#From Consideration of direct crushing\n",
+ "#sigma*a**2=W1\n",
+ "#After Reaaranging the above equation we get\n",
+ "a2=(W1*(sigma)**-1)**0.5 #mm\n",
+ "\n",
+ "#required size is 103.67*103.67 i.e a*a\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#At W2=200*10**3 #N\n",
+ "#W2=P*(3*L**2)**-1\n",
+ "#After substituting values and further Rearranging the above equation we get\n",
+ "a3=(W2*3*12*L**2*(pi**2*E)**-1)**0.25 #mm\n",
+ "\n",
+ "#From consideration of direct compression,size required is\n",
+ "a4=(W2*sigma**-1)**0.5\n",
+ "\n",
+ "#required size is 129.10*129.10 i.e a4*a4\n",
+ "\n",
+ "#Result\n",
+ "print\"For W1 Load Required size is\",round(a*a,2),\"mm**2\"\n",
+ "print\"For W2 Load Required size is\",round(a4*a4,2),\"mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "For W1 Load Required size is 10747.38 mm**2\n",
+ "For W1 Load Required size is 16666.67 mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.3,Page No.378"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flange \n",
+ "b=100 #mm #Width\n",
+ "\n",
+ "D=80 #mm #Overall Depth\n",
+ "t=10 #mm #Thickness of web and flanges\n",
+ "L=3000 #mm #Length of strut\n",
+ "E=200*10**3 #N/mm**2 #Modulus of Elasticity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let centroid be at depth y_bar from top fibre\n",
+ "y_bar=(b*t*t*2**-1+(D-t)*t*((D-t)*2**-1+t))*(b*t+(D-t)*t)**-1 #mm \n",
+ "\n",
+ "#M.I at x-x axis\n",
+ "I_x=1*12**-1*b*t**3+b*t*(y_bar-t*2**-1)**2+1*12**-1*t*((D-t))**3+t*((D-t))*((((D-t)*2**-1)+t)-y_bar)**2\n",
+ "\n",
+ "#M.I at y-y axis\n",
+ "I_y=1*12**-1*t*b**3+1*12**-1*(D-t)*t**3 #mm**3\n",
+ "\n",
+ "#Least M.I\n",
+ "I=I_y\n",
+ "\n",
+ "#Since both ends are hinged\n",
+ "#Feective Length=Actual Length\n",
+ "L=l=3000 #mm\n",
+ "\n",
+ "#Buckling Load \n",
+ "P=pi**2*E*I*(l**2)**-1*10**-3 #KN\n",
+ "\n",
+ "#Result\n",
+ "print\"The Buckling Load for strut of tee section\",round(P,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Buckling Load for strut of tee section 184.05 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.4,Page No.379"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=400 #mm #Overall Depth\n",
+ "\n",
+ "#Flanges\n",
+ "b=300 #mm #Width\n",
+ "t=50 #mm #Thickness\n",
+ "\n",
+ "t2=30 #mm #Web Thickness\n",
+ "\n",
+ "dell=10 #mm #Deflection\n",
+ "w=40 #N/mm #Load\n",
+ "FOS=1.75 #Factor of safety\n",
+ "E=2*10**5 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#M.I at x-x axis\n",
+ "I_x=1*12**-1*(b*D**3-(b-t2)*b**3) #mm**4\n",
+ "\n",
+ "#Central Deflection\n",
+ "#dell=5*w*L**4*(384*E*I)**-1\n",
+ "#After sub values in above equation and further simplifying we get\n",
+ "L=(dell*384*E*I_x*(5*w)**-1)**0.25\n",
+ "\n",
+ "#M.I aty-y axis\n",
+ "I=I_y=1*12**-1*t*b**3+1*12**-1*b*t2**3+1*12**-1*t*b**3 #mm**4\n",
+ "\n",
+ "#Both the Ends of column are hinged\n",
+ "\n",
+ "#Crippling Load\n",
+ "P=pi**2*E*I*(L**2)**-1 #N\n",
+ "\n",
+ "#Safe Load\n",
+ "S=P*(FOS)**-1*10**-3 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe Load if I-section is used as column with both Ends hhinged\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe Load if I-section is used as column with both Ends hhinged 4123.29 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.5,Page No.381"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=200 #mm #External Diameter\n",
+ "t=20 #mm #hickness\n",
+ "d=200-2*t #mm #Internal Diameter\n",
+ "E=1*10**5 #N/mm**2\n",
+ "a=1*(1600)**-1 #Rankine's Constant\n",
+ "L=4.5 #m #Length\n",
+ "sigma=550 #N/mm**2 #Stress\n",
+ "FOS=2.5\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=pi*D**4*64**-1-pi*d**4*64**-1\n",
+ "\n",
+ "#Both Ends are fixed\n",
+ "\n",
+ "#Effective Length\n",
+ "l=1*2**-1*L*10**3 #mm\n",
+ "\n",
+ "#Euler's Critical Load\n",
+ "P_E=pi**2*E*I*(l**2)**-1\n",
+ "\n",
+ "A=pi*4**-1*(D**2-d**2) #mm*2\n",
+ "\n",
+ "k=(I*A**-1)**0.5\n",
+ "\n",
+ "#Rankine's Critical Load\n",
+ "P_R=sigma*A*(1+a*(l*k**-1)**2)**-1\n",
+ "\n",
+ "X=P_E*P_R**-1 \n",
+ "\n",
+ "#Safe Load using Rankine's Formula\n",
+ "S=P_R*(FOS)**-1*10**-3 #KN\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe Load by Rankine's Formula is\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe Load by Rankine's Formula is 1404.36 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.6,Page No.382"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=3000 #mm #Length of column\n",
+ "W=800*10**3 #N #Load\n",
+ "a=1*1600**-1 #Rankine's constant\n",
+ "FOS=4 #Factor of safety\n",
+ "sigma=550 #N/mm**2 #stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Effective Length\n",
+ "l=L*2**-1 #mm \n",
+ "\n",
+ "#Let d1=outer diameter & d2=inner diameter\n",
+ "#d1=5*8**-1*d2\n",
+ "\n",
+ "#M.I\n",
+ "#I=pi*64**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "#Area of section\n",
+ "#A=pi4**-1*(d1**2-d2**2) #mm**2\n",
+ "\n",
+ "#k=(I*A**-1) \n",
+ "#substituting values in above equation \n",
+ "#k=1*16**-1*(d1**2-d2**2)\n",
+ "#after simplifying further we get\n",
+ "#k=0.2948119.d1\n",
+ "\n",
+ "#X=l*k**-1\n",
+ "#substituting values in above equation and after simplifying further we get\n",
+ "#X=5087.9898*d1**-1\n",
+ "\n",
+ "#Crtitcal Load\n",
+ "P=W*FOS #N\n",
+ "\n",
+ "#From Rankine's Load\n",
+ "#P2=sigma*A*(1+a*(X)**2)**-1\n",
+ "#substituting values in above equation and after simplifying further we get\n",
+ "#d1**4-12156618*d1**4-1.96691*10**8=0\n",
+ "#Solving Quadratic Equation we get\n",
+ "#d1**2-12156618*d1-196691000=0\n",
+ "a=1\n",
+ "b=-12156.618\n",
+ "c=-196691000\n",
+ "\n",
+ "Y=b**2-4*a*c\n",
+ "\n",
+ "d1_1=((-b+Y**0.5)*(2*a)**-1)**0.5 #mm\n",
+ "d1_2=((-b-Y**0.5)*(2*a)**-1) #mm\n",
+ "\n",
+ "d2=5*8**-1*d1_1\n",
+ "\n",
+ "#Result\n",
+ "print\"Section of cast iron hollow cylindrical column is:d1_1\",round(d1_1,2),\"mm\"\n",
+ "print\" :d2 \",round(d2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Section of cast iron hollow cylindrical column is:d1_1 146.16 mm\n",
+ " :d2 91.35 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.7,Page No.383"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Let X=(P*A**-1) #Average Stress at Failure \n",
+ "Lamda_1=70 #Slenderness Ratio\n",
+ "Lamda_2=170 #Slenderness Ratio\n",
+ "X1=200 #N/mm**2 \n",
+ "X2=69 #N/mm**2 \n",
+ "\n",
+ "#Rectangular section\n",
+ "b=60 #mm #width\n",
+ "t=20 #mm #Thickness\n",
+ "\n",
+ "L=1250 #mm #Length of strut\n",
+ "FOS=4 #Factor of safety\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Slenderness ratio\n",
+ "#Lamda=L*k**-1\n",
+ "\n",
+ "#The Rankine's Formula for strut\n",
+ "#P=sigma*A*(1+a*(L*k**-1)**-1\n",
+ "\n",
+ "#From test result 1,\n",
+ "#After sub values in above equation we get and further simplifying we get\n",
+ "#sigma_1=200+980000*a ...................(1)\n",
+ "\n",
+ "#From test result 2,\n",
+ "#After sub values in above equation we get and further simplifying we get\n",
+ "#sigma_2=69+1994100*a ...................(2)\n",
+ "\n",
+ "#Substituting it in equation (1) we get\n",
+ "a=131*1014100**-1 \n",
+ "\n",
+ "#Substituting a in equation 1\n",
+ "sigma_1=200+980000*a #N/mm**2\n",
+ "\n",
+ "#Effective Length \n",
+ "l=1*2**-1*L #mm\n",
+ "\n",
+ "#Least of M.I\n",
+ "I=1*12**-1*b*t**3 #mm**4\n",
+ "\n",
+ "#Area \n",
+ "A=b*t #mm**2 \n",
+ "\n",
+ "k=(I*A**-1)**0.5\n",
+ "\n",
+ "#Slenderness ratio\n",
+ "Lamda=l*k**-1\n",
+ "\n",
+ "#From Rankine's Ratio\n",
+ "P=sigma_1*A*(1+a*(Lamda)**2)**-1\n",
+ "\n",
+ "#Safe Load\n",
+ "S=P*(FOS)**-1*10**-3 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Constant in the Formula is:a \",round(a,6)\n",
+ "print\" :sigma_1\",round(sigma_1,2)\n",
+ "print\"Safe Load is\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Constant in the Formula is:a 0.000129\n",
+ " :sigma_1 326.6\n",
+ "Safe Load is 38.98 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.8,Page No.385"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=200 #mm #Depth\n",
+ "b=140 #mm #width\n",
+ "\n",
+ "#Plate\n",
+ "b2=160 #mm #Width\n",
+ "t2=10 #mm #Thickness\n",
+ "\n",
+ "L=l=4000 #mm #Length\n",
+ "FOS=4 #Factor of safety\n",
+ "sigma=315 #N/mm**2 #stress\n",
+ "a2=1*7500**-1 \n",
+ "I_xx=26.245*10**6 #mm**4 #M.I at x-x\n",
+ "I_yy=3.288*10**6 #mm**4 #M.I at y-y\n",
+ "a=3671 #mm**2 #Area\n",
+ "k_x=84.6#mm\n",
+ "k_y=29.9 #mm\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Total Area\n",
+ "A=a+2*t2*b2 #mm**2\n",
+ "\n",
+ "#M.I\n",
+ "I=I_yy+2*12**-1*t2*b2**3 #mm**4\n",
+ "\n",
+ "k=(I*A**-1)**0.5 #mm\n",
+ "\n",
+ "#Let X=L*k**-1\n",
+ "X=L*k**-1\n",
+ "\n",
+ "#Appliying Rankine's Formula\n",
+ "P=sigma*A*(1+a2*(X)**2)**-1 #N\n",
+ "\n",
+ "#Safe Load\n",
+ "S=P*(FOS)**-1*10**-3 #KN\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe axial Load is\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe axial Load is 220.93 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.9,Page No.389"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=200*10**3 #N/mm**2 #Modulus of elasticity\n",
+ "sigma=330 #N/mm**2 #Stress\n",
+ "a=1*7500**-1 #Rankine's constant\n",
+ "A=5205 #mm**2 #area of column\n",
+ "I_xx=59.431*10**6 #mm**4 #M.I at x-x axis\n",
+ "I_yy=8.575*10**6 #mm**24#M.I at y-y axis\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Total M.I\n",
+ "I=I_xx+I_yy #mm**4\n",
+ "\n",
+ "#Area of compound Section \n",
+ "A2=2*A #mm**2\n",
+ "\n",
+ "k=(I*A2**-1)**0.5 #mm\n",
+ "\n",
+ "#Equating Euler's Load to Rankine's Load we get\n",
+ "#pi**2*E*I*(L**2)**-1=sigma*A*(1+a*(L*k)**2)**-1\n",
+ "#After Substitt=uting values and further simplifying we get\n",
+ "L=(39076198*(1-0.7975432)**-1)**0.5*10**-3 #m\n",
+ "\n",
+ "#Result\n",
+ "print\"Length of column for which Rankine's formula and Euler's Formula give the same result is\",round(L,2),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Length of column for which Rankine's formula and Euler's Formula give the same result is 13.89 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.10,Page No.387"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "sigma=326 #N/mm**2 #stress\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "FOS=2 #Factor of safety\n",
+ "a=1*7500**-1 #Rankine's constant\n",
+ "D=350 #mm #Overall Depth \n",
+ "\n",
+ "#Cover plates\n",
+ "b1=500 #mm #width\n",
+ "t1=10 #mm #Thickness\n",
+ "\n",
+ "d=220 #mm #Distance between two channels\n",
+ "\n",
+ "L=6000 #mm #Length of column\n",
+ "\n",
+ "A=5366 #mm**2 #Area of Column section \n",
+ "I_xx=100.08*10**6 #mm**4 #M.I of x-x axis\n",
+ "I_yy=4.306*10**6 #mm**4 #M.I of y-y axis\n",
+ "C_yy=23.6 #mm #Centroid at y-y axis\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Symmetric axes are the centroidal axes is\n",
+ "\n",
+ "#M.I of Channel at x-x axis\n",
+ "I_xx_1=2*I_xx+2*(1*12**-1*b1*t1**3+b1*t1*(D*2**-1+t1*2**-1)**2)\n",
+ "\n",
+ "#M.I of Channel at y-y axis\n",
+ "I_yy_1=2*(I_yy+A*(d*2**-1+C_yy)**2)+2*12**-1*t1*b1**3\n",
+ "\n",
+ "#As I_yy<I_xx\n",
+ "#So\n",
+ "I=I_yy_1 #mm**4 \n",
+ "\n",
+ "A2=2*A+2*t1*b1 #Area of channel\n",
+ "\n",
+ "k=(I*A2**-1)**0.5 #mm\n",
+ "\n",
+ "#Critical Load\n",
+ "P=sigma*A2*(1+a*(L*k**-1)**2)**-1 \n",
+ "\n",
+ "#Safe Load\n",
+ "S=P*2**-1*10**-3 #KN\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe Load carrying Capacity is\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe Load carrying Capacity is 2717.35 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 67
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.11,Page No.390"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "I=4.085*10**8 #mm**4 #M.I\n",
+ "A=20732.0 #mm**2 #area of column\n",
+ "f_y=250 #N/mm**2 \n",
+ "L=6000 #mm #Length of column\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "k=(I*A**-1)**0.5 #mm\n",
+ "lamda=L*k**-1 #Slenderness ratro\n",
+ "\n",
+ "#From Indian standard table\n",
+ "lamda_1=40 \n",
+ "sigma_a_c_1=139 #N/mm**2\n",
+ "lamda_2=50 \n",
+ "sigma_a_c_2=132 #N/mm**2 \n",
+ "\n",
+ "#Linearly interpolating between these values for lambda=42.744\n",
+ "\n",
+ "sigma_a_c_3=sigma_a_c_1-2.744*10**-1*(sigma_a_c_1-sigma_a_c_2)\n",
+ "\n",
+ "#Safe Load carrying capacity of column\n",
+ "P=sigma_a_c_3*A*10**-3\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe Load carrying capacity is\",round(P,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe Load carrying capacity is 2841.93 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
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generated by cgit (git 2.34.1) at 2024-12-19 15:16:47 +0000