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{
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"name": "",
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"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter10:Columns"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.10.1, Page No:369"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable Decleration\n",
"Le=7 #Effective length in m\n",
"P=450 #Applied axial Load in kN\n",
"FOS=3 #Factor of safety \n",
"sigma_pl=200*10**6 #Stress allowable in Pa\n",
"E=200*10**9 #Youngs Modulus in Pa\n",
"end_cond=0.7 #End Condition factor to be multiplied\n",
"\n",
"#Calculations\n",
"Pcr=P*FOS #Critical Load in kN\n",
"A=Pcr*sigma_pl**-1*10**9 #Area in mm^2\n",
"\n",
"#Part 1\n",
"I1=10**15*(Pcr*Le**2)*(pi**2*E)**-1 #Moment of Inertia Required in mm^4\n",
"#From table selecting appropriate Section W250x73\n",
"\n",
"#Part 2\n",
"I2=10**15*(Pcr*end_cond**2*Le**2)*(pi**2*E)**-1 #Moment of Inertia Required in mm^4\n",
"#From table selecting appropriate Section W200x52\n",
"\n",
"#Lightest Section that meets these criterion is W250x58 section\n",
"\n",
"\n",
"#Result\n",
"print \"From the above computation we select W250x58 section\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"From the above computation we select W250x58 section\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.10.2, Page No:375"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variabel Decleration\n",
"E=200*10**9 #Youngs Modulus in Pa\n",
"sigma_yp=380*10**6 #Stress allowable in Pa\n",
"Le=10 #Length in m\n",
"end_cond=0.5 #Support condition factor to be ,ultiplied to length\n",
"A=15.5*10**-3 #Area in m^2\n",
"\n",
"#Calculations\n",
"Cc=sqrt((2*pi**2*E)*sigma_yp**-1) #Slenderness Ratio\n",
"\n",
"#Part 1\n",
"S_R1=142.9 #Slenderness ratio \n",
"sigma_w=(12*pi**2*E)/(23*S_R1**2) #Allowable Working Stress in Pa\n",
"P=sigma_w*A #Maximum Allowable Load in kN\n",
"\n",
"#Part 2\n",
"S_R2=79.37 #Slenderness ratio \n",
"N=5*3**-1+((3*S_R2)/(8*Cc))-(S_R2**3*(8*Cc**3)**-1) #Factor Of Safety\n",
"\n",
"sigma_w2=(1-(S_R2**2*0.5*Cc**-2))*(sigma_yp*N**-1) #Allowable working Stress in Pa\n",
"P2=sigma_w2*A #Allowable Load in kN\n",
"\n",
"#Part 3\n",
"S_R3=55.56 #Slenderness Ratio\n",
"N3=5*3**-1+((3*S_R3)/(8*Cc))-(S_R3**3*(8*Cc**3)**-1) #Factor Of Safety\n",
"\n",
"sigma_w3=(1-(S_R3**2*0.5*Cc**-2))*(sigma_yp*N3**-1) #Allowable working Stress in Pa\n",
"P3=sigma_w3*A #Allowable Load in kN\n",
"\n",
"#Result\n",
"print \"The results for Part 1 are\"\n",
"print \"Maximum Allowable Load P=\",round(P*10**-3),\"kN\"\n",
"print \"Part 2\"\n",
"print \"Maximum Allowable Load P=\",round(P2*10**-3),\"kN\"\n",
"print \"Part 3\"\n",
"print \"Maximum Allowable Load P=\",round(P3*10**-3),\"kN\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The results for Part 1 are\n",
"Maximum Allowable Load P= 782.0 kN\n",
"Part 2\n",
"Maximum Allowable Load P= 2161.0 kN\n",
"Part 3\n",
"Maximum Allowable Load P= 2710.0 kN\n"
]
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.10.3, Page No:383"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"import numpy as np\n",
"\n",
"#Variabel Decleration\n",
"E=29*10**6 #Youngs Modulus in psi\n",
"sigma_yp=36*10**3 #Stress in psi\n",
"L=25 #Length in ft\n",
"A=17.9 #Area in in^2\n",
"Iz=640 #Moment of inertia in in^4\n",
"Sz=92.2 #Sectional Modulus in in^3\n",
"P=150*10**3 #Load in lb\n",
"e=4 #Eccentricity in inches\n",
"\n",
"#Calculations\n",
"\n",
"#Part 1\n",
"a=1.09836\n",
"sigma_max=P*A**-1+(P*e*Sz**-1)*a #Maximum Stress in psi\n",
"\n",
"#Part 2\n",
"#After simplification we get the equation to compute N\n",
"N=2.19 #Trial and Error yields\n",
"\n",
"#Part 3\n",
"v_max=e*((np.cos(sqrt((P*L**2*12**2)*(4*E*Iz)**-1)))**-1-1)\n",
"\n",
"#Result\n",
"print \"The maximum compressive stress in the Column is\",round(sigma_max,2),\"psi\"\n",
"print \"The factor of safety is\",N\n",
"print \"The maximum lateral dfelection is\",round(v_max,3),\"in\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The maximum compressive stress in the Column is 15527.57 psi\n",
"The factor of safety is 2.19\n",
"The maximum lateral dfelection is 0.393 in\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.10.4, Page No:384"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable Decleration\n",
"Le=7 #Effective Length in m\n",
"N=2 #Factor of Safety\n",
"h_max=400 #Maximum depth in mm\n",
"E=200 #Youngs Modulus in GPa\n",
"sigma_yp=250 #Maximum stress in yielding in MPa\n",
"P1=400 #Load 1 in kN\n",
"P2=900 #Load 2 in kN\n",
"x1=75 #Distance in mm\n",
"x2=125 #Distance in mm\n",
"\n",
"#Calculations\n",
"e=(P2*x2-P1*x1)*(P1+P2)**-1 #Eccentricity in mm\n",
"P=N*(P1+P2) #Applied load after factor of safety is considered in kN\n",
"\n",
"#Part 1 is not computable\n",
"I=415*10**-6 #Moment of inertia from the table in mm^4\n",
"\n",
"#Part 2\n",
"P_cr=pi**2*E*10**9*I*Le**-2 #Critical load for buckling in kN\n",
"FOS=P_cr*10**-3/(P1+P2) #Factor of safety against buckling in y-axis\n",
"\n",
"\n",
"#Result\n",
"print \"The critical load for buckling is\",round(P_cr*10**-3),\"kN\"\n",
"print \"The factor of safety is\",round(FOS,1)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The critical load for buckling is 16718.0 kN\n",
"The factor of safety is 12.9\n"
]
}
],
"prompt_number": 27
}
],
"metadata": {}
}
]
}
|
