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diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter09_1.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter09_1.ipynb new file mode 100755 index 00000000..ac59e6fc --- /dev/null +++ b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter09_1.ipynb @@ -0,0 +1,285 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:19ebf66d9da8e61964dee3081a6c3e2bb440c25ec82a0b4b2dc44b82d335cf92" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter09:Composite Beams" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.9.1, Page No:346" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable Decleration\n", + "n=20 #Modular Ratio\n", + "sigma_wd=8*10**6 #Maximum bending stress in wood in Pa\n", + "sigma_st=120*10**6 #Maximum bending stress in steel in Pa\n", + "\n", + "#Cross Sectional Details\n", + "Awd=45 #Area of wood in mm^2\n", + "y_wd=160 #Neutral Axis of from bottom of the wooden section in mm\n", + "Ast=15 #Area of steel in mm^2\n", + "y_st=5 #Neutral Axis of the Steel section in mm\n", + "#Dimensions\n", + "ww=150 #width of wooden section in mm\n", + "dw=300 #depth of wooden section in mm\n", + "ws=75 #width of steel section in mm\n", + "ds=10 #depth of steel section in mm\n", + "\n", + "#Calculations\n", + "y_bar=(Awd*y_wd+Ast*y_st)*(Ast+Awd)**-1 #Location of Neutral axis from the bottom in mm\n", + "#Moment of inertia \n", + "I=(ww*dw**3*12**-1)+(ww*dw*(y_wd-y_bar)**2)+(n*ws*ds**3*12**-1)+(n*ws*ds*(y_bar-y_st)**2) #mm^4\n", + "c_top=dw+ds-y_bar #Distance from NA to top fibre in mm\n", + "c_bot=y_bar #Distance from NA to bottom fibre in mm\n", + "\n", + "#The solution will be in different order \n", + "M1=sigma_wd*I*10**-12*c_top**-1 #Maximum Bending Moment in N.m\n", + "M2=sigma_st*I*10**-12*c_bot**-1 #Maximum Bending Moment in N.m\n", + "M=min(M1,M2) #Maximum allowable moment in N.m\n", + "\n", + "#Result\n", + "print \"The Maximum Allowable moment that the beam can support is\",round(M,1),\"kN.m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Maximum Allowable moment that the beam can support is 25.8 kN.m\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.9.2, Page No:351" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable Decleration\n", + "dw=8 #Depth of wooden section in inches\n", + "da=0.4 #Depth og aluminium section in inches \n", + "w=2 #Width of the section in inches\n", + "n=40*3**-1 #Modular Ratio\n", + "Ewd=1.5*10**6 #Youngs modulus of wood in psi\n", + "Eal=10**7 #Youngs Modulus of aluminium in psi\n", + "V_max=4000 #Maximum shear in lb\n", + "b=24 #Inches\n", + "L=72 #Length in inches\n", + "P=6000 #Load on the beam in lb\n", + "\n", + "#Calculations\n", + "I=w*dw**3*12**-1+2*(n*w*da**3*12**-1+n*da*4.2**2) #Moment of Inertia in in^4\n", + "\n", + "#Part 1\n", + "Q=(w*dw*0.5)*2+(n*da)*(dw*0.5+da*0.5) #First Moment in in^3\n", + "tau_max=V_max*Q*I**-1*w**-1 #Maximum Shear Stress in psi\n", + "\n", + "#Part 2\n", + "delta_mid=(P*b)*(48*Ewd*I)**-1*(3*L**2-4*b**2)\n", + "\n", + "#Result\n", + "print \"The maximum shear stress allowable is\",round(tau_max),\"psi\"\n", + "print \"The deflection at the mid-span is\",round(delta_mid,4),\"in\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum shear stress allowable is 281.0 psi\n", + "The deflection at the mid-span is 0.0968 in\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.9.3, Page No:356" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "import numpy as np\n", + "\n", + "#Variable Decleration\n", + "b=300 #Breadth in mm\n", + "d=500 #Depth in mm\n", + "Ast=1500 #Area of steel in mm^2\n", + "n=8 #Modular Ratio\n", + "M=70*10**3 #Bending Moment in N.m\n", + "\n", + "#Calculations\n", + "#Let the LHS be C\n", + "C=2*n*Ast*b**-1*d**-1 #The LHS computation\n", + "h=np.roots([d**-2,C*d**-1,-C])\n", + "#Taking only real root\n", + "h=h[1] #mm\n", + "\n", + "sigma_co_max=(2*M)/(b*h*(d-h*3**-1)) #Maximum Compressive Stress in GPa\n", + "sigma_st_max=M/((d-h*3**-1)*Ast) #Maximum Stress in Steel in GPa\n", + "#Result\n", + "print \"The maximum stress in compression is\",round(sigma_co_max*10**3,2),\"MPa\"\n", + "print \"The maximum stress in streel is\",round(sigma_st_max*10**3,1),\"MPa\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum stress in compression is 6.39 MPa\n", + "The maximum stress in streel is 104.8 MPa\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.9.4, Page No:356" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable Decleration\n", + "sigma_co_w=12 #Maximum stress in compression in MPa\n", + "sigma_st_w=140 #Maximum stress in steel in MPa\n", + "M=90 #Moment in kN.m\n", + "n=8 #Modular Ratio \n", + "\n", + "#Calculations\n", + "#h=0.4068d\n", + "#bd^2=0.04266\n", + "b=(0.04266/(1.5**2))**0.3333 #Breadth in m \n", + "d=1.5*b #Depth in m\n", + "h=0.4068*d #Height in m\n", + "\n", + "#Area of steel\n", + "Ast=((M*10**3)/((d-h*3**-1)*sigma_st_w*10**3))*10**3 #Area of steel in mm^2\n", + "\n", + "#Result\n", + "print \"The dimensions of the beam are\"\n", + "print \"b=\",round(b*1000),\"mm and d=\",round(d*1000),\"mm\"\n", + "print \"Area of steel=\",round(Ast),\"mm^2\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The dimensions of the beam are\n", + "b= 267.0 mm and d= 400.0 mm\n", + "Area of steel= 1859.0 mm^2\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.9.5, Page No:357" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "import numpy as np\n", + "\n", + "#Variable Decleration\n", + "A1=75*10**3 #Area 1 in mm^2\n", + "A3=19.20*10**3 #Area 3 in m^2\n", + "w=750 #Width in mm\n", + "w1=350 #Width in mm\n", + "d=444.45 #Depth in mm\n", + "sigma_co_w=12*10**6 #Maximum Permissible Bending stress in concrete in Pa\n", + "sigma_st_w=140*10**6 #Maximum Permissible Bending stress in steel in Pa\n", + "n=8 #Modular Ratio\n", + "\n", + "#Calculations\n", + "#After simplfying the equation we get the following \n", + "H=np.roots([200,-200**2+A1+A3,-A1*50+100**2*200-600*A3])\n", + "h=max(H) #Depth of NA in mm\n", + "#Moment Of Inertia\n", + "I=w*h**3*3**-1-(w1*(h-100)**3*3**-1)+A3*d**2 #Moment of inertia in mm^4\n", + "\n", + "M1=sigma_co_w*I*h**-1*(10**-3)**4*10**3 #Largest Bending Moment in concrete in N.m\n", + "M2=sigma_st_w*I*(n*d)**-1*(10**-3)**4*10**3 #Largest Bending Moment in Steel in N.m\n", + "M=min(M1,M2) #Largest Bending Moment that can be supported safely in N.m\n", + "#Result\n", + "print \"The largest Bending Moment that can be supported is\",round(M*10**-3,1),\"kN.m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The largest Bending Moment that can be supported is 185.6 kN.m\n" + ] + } + ], + "prompt_number": 27 + } + ], + "metadata": {} + } + ] +}
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