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{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 13: Inelastic Action"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 13.13.1, Page No:461"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable Decleration\n",
"d=150 #Depth of the web in mm\n",
"wf=100 #Width of the flange in mm\n",
"df=20 #Depth of the flange in mm\n",
"t=20 #Thickness of the web in mm\n",
"\n",
"#Calculations\n",
"y_bar=10**-3*(((wf*df*(d+df*0.5))+(d*t*d*0.5))/(wf*df+d*t)) #Distance of Neutral Axis in m\n",
"#Simplfying the computation\n",
"a=wf*df**3*12**-1\n",
"b=wf*df*((d+df*0.5)-y_bar*10**3)**2\n",
"c=t*d**3*12**-1\n",
"f=t*d*((d*0.5)-y_bar*10**3)**2\n",
"I=(a+b+c+f)*10**-12 #Moment of inertia in mm^3\n",
"\n",
"#Limit Moment\n",
"yp=(wf*df+d*t)/(2*t) #Plastic Neutral Axis in mm\n",
"Myp=I/y_bar #Yielding will start at moment without the stress term to ease computation\n",
"mom=10**-9*((t*yp**2*0.5)+(wf*df*(d-yp+10))+(t*25**2*0.5)) #Sum of 1st moments\n",
"Ml_Myp=mom*Myp**-1 #Ratio\n",
"\n",
"#Result\n",
"print \"The ratio ML/Myp=\",round(Ml_Myp,3)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The ratio ML/Myp= 1.765\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 13.13.2, Page No:467"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"import numpy as np\n",
"\n",
"#Variable Decleration\n",
"E_st=200 #Youngs Modulus of Steel in GPa\n",
"sigma_st_yp=290 #Yielding Stress in MPa\n",
"E_al=70 #Youngs Modulus of Aluminium in GPa\n",
"sigma_al_yp=330 #Yielding Stresss of Aluminium in MPa\n",
"A_st=900 #Area of steel rod in mm^2\n",
"A_al=600 #Area of Aluminium rod in mm^2\n",
"L_st=350 #Length of the steel rod in mm\n",
"L_al=250 #Length of the aluminium rod in mm\n",
"\n",
"#Calculations\n",
"#Limit Load\n",
"P_st=sigma_st_yp*A_st*10**-3 #Load in limiting condition in kN\n",
"P_al=sigma_al_yp*A_al*10**-3 #Load in limiting condition in kN\n",
"P_L=P_st+2*P_al #Total Loading in kN\n",
"\n",
"#Elastic Unloading\n",
"#Solving for Pst and Pal using matri approach\n",
"A=np.array([[1,2],[L_st*(E_st*A_st)**-1,-L_al*(E_al*A_al)**-1]])\n",
"B=np.array([P_L,0])\n",
"C=np.linalg.solve(A,B) #Loading in kN\n",
"\n",
"#Residual Stresses\n",
"P_res_st=C[0]-P_st #Residual Load in kN\n",
"P_res_al=C[1]-P_al #Residual Load in kN\n",
"sigma_st=P_res_st/A_st #residual Stress in Steel in MPa\n",
"sigma_al=P_res_al/A_al #residual Stress in Aluminium in MPa\n",
"\n",
"\n",
"#Result\n",
"print \"The Residual stresses are as follows\"\n",
"print \"Sigma_st=\",round(sigma_st*10**3,1),\"MPa and sigma_al=\",round(sigma_al*10**3,1),\"MPa\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Residual stresses are as follows\n",
"Sigma_st= 151.5 MPa and sigma_al= -113.6 MPa\n"
]
}
],
"prompt_number": 22
}
],
"metadata": {}
}
]
}
|
