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diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter12_2.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter12_2.ipynb new file mode 100755 index 00000000..9163c9cf --- /dev/null +++ b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter12_2.ipynb @@ -0,0 +1,319 @@ +{ + "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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