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{
"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": {}
}
]
}
|
