{ "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": {} } ] }