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diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter12_1.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter12_1.ipynb deleted file mode 100755 index 9163c9cf..00000000 --- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter12_1.ipynb +++ /dev/null @@ -1,319 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:d8a1a6f05cd4df2c46b4f5147d4f831726de5041386c1f65ed2892779a5fb0fb" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 12:Special Topics" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.12.1, Page No:422" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "W=24*10**3 #Load in kips\n", - "E=29*10**6 #Youngs Modulus in psi\n", - "L=72 #length in inches\n", - "theta=30 #Angle in degrees\n", - "\n", - "#Calculations\n", - "L_ab=L/np.sin(theta*pi*180**-1) #Length of AB in inches\n", - "L_ac=L/np.sin((90-theta)*pi*180**-1) #Length of AC in inches\n", - "\n", - "#Applying the forces in x and y sum to zero\n", - "#Applying the Starin energy formula\n", - "#Applying Castiglinos theorem \n", - "delta_A=91.16*W*E**-1 #Displacement in inches\n", - "\n", - "#Result\n", - "print \"The displacement of point A is\",round(delta_A,4),\"in\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The displacement of point A is 0.0754 in\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.12.3, Page No:423" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "from scipy.integrate import quad\n", - "\n", - "\n", - "\n", - "#We will directly compute the integral as function cannot be declared\n", - "#Calculations\n", - "#Part 1\n", - "def integrand(x):\n", - " return (800*x-400*x**2)*(4-0.5*x)\n", - "I = quad(integrand, 0, 2)\n", - "delta=I[0] #Deflection in horizontal direction in N.m^3\n", - "\n", - "#Part 2\n", - "def inte1(x):\n", - " return x**2\n", - "I1=quad(inte1,0,4)\n", - "I2=quad(inte1,0,3)\n", - "def inte2(x):\n", - " return (4-0.5*x)*(4-0.5*x)\n", - "I3=quad(inte2,0,2)\n", - "\n", - "Q=-delta/(I1[0]+I2[0]+I3[0]) #Horizontal reaction in N\n", - "\n", - "\n", - "#Result\n", - "print \"The Horizontal deflection is\",round(delta),\"N.m^3\"\n", - "print \"The Horizontal reaction is\",round(Q,1),\"N\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The Horizontal deflection is 1867.0 N.m^3\n", - "The Horizontal reaction is -33.9 N\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.12.4, Page No:433" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "import numpy as np\n", - "\n", - "#NOTE:The figure mentions the unit of length as ft which is incorrect\n", - "#Variable Decleration\n", - "L=30 #Length in m\n", - "m=2000 #Mass in kg\n", - "v=2 #Velocity in m/s\n", - "E=10**5 #Youngs Modulus in MPa\n", - "A=600 #Area in mm^2\n", - "g=9.81 #Acceleration due to gravity in m/s^2\n", - "\n", - "#Calculations\n", - "k=E*A*L**-1 #Stifness of the cable in N/m\n", - "\n", - "#Applying the Work-Energy principle \n", - "delta_max=np.sqrt((0.5*m*v**2)*(0.5*k)**-1) #Maximum Displacement in m\n", - "\n", - "P_max=k*delta_max+m*g #Maximum force in N\n", - "\n", - "#Result\n", - "print \"The maximum force is\",round(P_max*10**-3,1),\"kN\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The maximum force is 146.1 kN\n" - ] - } - ], - "prompt_number": 51 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.12.5, Page No:434" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "b=0.060 #Breadth of the section in mm\n", - "d=0.03 #Depth of the section in mm\n", - "L=1.2 #Length in m\n", - "m=80 #Mass in kg\n", - "g=9.81 #Acceleration due to gravity in m/s^2\n", - "E=200*10**9 #Youngs Modulus in Pa\n", - "e=0.015 \n", - "h=0.01 #height in m\n", - "\n", - "#Calculations\n", - "#Part 1\n", - "I=b*d**3*12**-1 #Moment of Inertia in m^4\n", - "delta_st=m*g*L**3/(48*E*I) #Mid-span Displacement in m\n", - "n=1+np.sqrt(1+(2*h/delta_st)) #Impact Factor\n", - "\n", - "#Part 2\n", - "P_max=n*m*g #Maximum dynamic load in N at midspan\n", - "M_max=P_max*0.5*L*0.5 #Maximum moment in N.m\n", - "sigma_max=M_max*e/I #Maximum dynamic Bending Stress in Pa\n", - "\n", - "#Result\n", - "print \"The impact factor is\",round(n,3)\n", - "print \"The maximum dynamic Bending Moment is\",round(sigma_max*10**-6,1),\"MPa\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The impact factor is 5.485\n", - "The maximum dynamic Bending Moment is 143.5 MPa\n" - ] - } - ], - "prompt_number": 65 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.12.7, Page No:440" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable decleration\n", - "M=2.21 #Applied moment in kip.ft\n", - "d=3 #Diameter of the bar in inches\n", - "sigma_y=40 #Yield strength of the of steel in ksi\n", - "\n", - "#Calculations\n", - "#Part 1\n", - "sigma=32*M*12*(pi*d**3)**-1 #Maximum Bending Stress in ksi\n", - "T1=np.sqrt((sigma_y*0.5)**2-5**2)/(12*0.18863) #Maximum Allowable torque in kip.ft\n", - "\n", - "#Part 2\n", - "R=np.sqrt((sigma_y**2-5**2)*3**-1) #Maximum shear stress in ksi\n", - "T2=np.sqrt(R**2-5**2)/(12*0.18863) #Maximum safe torque in kpi.ft\n", - "\n", - "#Result\n", - "print \"Using the maximum shear stress theory T=\",round(T1,2),\"kip.ft\"\n", - "print \"Using the maximum sitrotion energy theory T=\",round(T2,2),\"kip.ft\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Using the maximum shear stress theory T= 8.56 kip.ft\n", - "Using the maximum sitrotion energy theory T= 9.88 kip.ft\n" - ] - } - ], - "prompt_number": 79 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.12.8, Page No:448" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "D=250 #Wideness in mm\n", - "b=20 #Thickness of the plate in mm\n", - "r=50 #Radius of the hole in mm\n", - "e=50 #Eccentricity in mm\n", - "sigma_max=150 #Maximum normal stress at the hole in MPa\n", - "kb=2 #Stress Concentraion factor \n", - "\n", - "#Calculations\n", - "A=b*(D-2*r)*10**-6 #Area in m^2\n", - "I=10**-12*(b*D**3*12**-1-(b*2**3*r**3*12**-1)) #Moment of inertia in m^4\n", - "#Simplfying computation\n", - "a=2*r*D**-1\n", - "kt=3-3.13*a+3.66*a**2-1.53*a**3 #Stress Concentration factor\n", - "#Simplfying computation\n", - "b=kt*A**-1\n", - "c=kb*r*r*10**-6*I**-1\n", - "P=10**3*sigma_max*(b+c)**-1 #Maximum Load in N\n", - "\n", - "#Result\n", - "print \"The maximum value of P is\",round(P,1),\"kN\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The maximum value of P is 157.8 kN\n" - ] - } - ], - "prompt_number": 106 - } - ], - "metadata": {} - } - ] -}
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