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{
"metadata": {
"name": "chapter_10.ipynb"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 10:Theory of Failures"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example No.10.10.1,Page No.401"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"from math import sin, cos, tan, pi, radians\n",
"\n",
"#Initilization of Variables\n",
"\n",
"P_e=300 #N/mm**2 #Elastic Limit in tension\n",
"FOS=3 #Factor of safety\n",
"mu=0.3 #Poissoin's ratio\n",
"P=12*10**3 #N Pull \n",
"Q=6*10**3 #N #Shear force\n",
"\n",
"#Calculations\n",
"\n",
"#Let d be the diameter of the shaft\n",
"\n",
"#Direct stress\n",
"#P_x=P*(pi*4**-1*d**3)**-1\n",
"#After substituting values and further simplifying we get\n",
"#P_x=48*10**3\n",
"\n",
"#Now shear stress at the centre of bolt\n",
"#q=4*3**-1*q_av\n",
"#After substituting values and further simplifying we get\n",
"#q=32*10**3*(pi*d**2)**-1\n",
"\n",
"#Principal stresses are\n",
"#P1=P_x*2**-1+((P_x*2**-1)**2+q**2)**0.5\n",
"#After substituting values and further simplifying we get\n",
"#p1=20371.833*(d**2)**-1\n",
"\n",
"#P2=P_x*2**-1-((P_x*2**-1)**2+q**2)**0.5\n",
"#After substituting values and further simplifying we get\n",
"#P2=-5092.984*(d**2)**-1\n",
"\n",
"#q_max=((P_x*2**-1)**2+q**2)**0.5\n",
"\n",
"#From Max Principal stress theory\n",
"#Permissible stress in Tension\n",
"P1=100 #N/mm**2 \n",
"d=(20371.833*P1**-1)**0.5\n",
"\n",
"#Max strain theory\n",
"#e_max=P1*E**-1-mu*P2*E**-1\n",
"#After substituting values and further simplifying we get\n",
"#e_max=21899.728*(d**2*E)**-1\n",
"\n",
"#According to this theory,the design condition is\n",
"#e_max=P_e*(E*FOS)**-1\n",
"#After substituting values and further simplifying we get\n",
"d2=(21899.728*3*300**-1)**0.5 #mm\n",
"\n",
"#Max shear stress theory\n",
"#e_max=shear stress at elastic*(FOS)**-1\n",
"#After substituting values and further simplifying we get\n",
"d3=(12732.421*6*300**-1)**0.5 #mm\n",
"\n",
"#Result\n",
"print\"Diameter of Bolt by:Max Principal stress theory\",round(d,2),\"mm\"\n",
"print\" :Max strain theory\",round(d2,2),\"mm\"\n",
"print\" :Max shear stress theory\",round(d3,2),\"mm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Diameter of Bolt by:Max Principal stress theory 14.27 mm\n",
" :Max strain theory 14.8 mm\n",
" :Max shear stress theory 15.96 mm\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example No.10.10.2.Page No.402"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"from math import sin, cos, tan, pi, radians\n",
"\n",
"#Initilization of Variables\n",
"\n",
"M=40*10**6 #N-mm #Bending moment\n",
"T=10*10**6 #N-mm #TOrque\n",
"mu=0.25 #Poissoin's ratio\n",
"P_e=200 #N/mm**2 #Stress at Elastic Limit\n",
"FOS=2\n",
"\n",
"#Calculations\n",
"\n",
"#Let d be the diameter of the shaft\n",
"\n",
"#Principal stresses are given by\n",
"\n",
"#P1=16*(pi*d**3)**-1*(M+(M**2+T**2)**0.5)\n",
"#After substituting values and further simplifying we get\n",
"#P1=4.13706*10**8*(d**3)**-1 ............................(1)\n",
"\n",
"#P2=16*(pi*d**3)**-1*(M-(M**2+T**2)**0.5)\n",
"#After substituting values and further simplifying we get\n",
"#P2=-6269718*(pi*d**3)**-1 ..............................(2)\n",
"\n",
"#q_max=(P1-P2)*2**-1\n",
"#After substituting values and further simplifying we get\n",
"#q_max=2.09988*10**8*(d**3)**-1\n",
"\n",
"#Max Principal stress theory\n",
"#P1=P_e*(FOS)**-1\n",
"#After substituting values and further simplifying we get\n",
"d=(4.13706*10**8*2*200**-1)**0.33333 #mm \n",
"\n",
"#Max shear stress theory\n",
"#q_max=shear stress at elastic limit*(FOS)**-1\n",
"#After substituting values and further simplifying we get\n",
"d2=(2.09988*10**8*4*200**-1)**0.33333\n",
"\n",
"#Max strain energy theory\n",
"#P_3=0\n",
"#P1**2+P2**2-2*mu*P1*P2=P_e**2*(FOS)**-1\n",
"#After substituting values and further simplifying we get\n",
"d3=(8.62444*10**12)**0.166666\n",
"\n",
"#Result\n",
"print\"Diameter of shaft according to:MAx Principal stress theory\",round(d,2),\"mm\"\n",
"print\" :Max shear stress theory\",round(d2,2),\"mm\"\n",
"print\" :Max strain energy theory\",round(d3,2),\"mm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Diameter of shaft according to:MAx Principal stress theory 160.52 mm\n",
" :Max shear stress theory 161.33 mm\n",
" :Max strain energy theory 143.2 mm\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example No.10.10.3,Page No.403"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"from math import sin, cos, tan, pi, radians\n",
"\n",
"#Initilization of Variables\n",
"\n",
"f_x=40 #N/mm**2 #Internal Fliud Pressure\n",
"d1=200 #mm #Internal Diameter\n",
"r1=d1*2**-1 #mm #Radius\n",
"q=300 #N/mm**2 #Tensile stress\n",
"\n",
"#Calculations\n",
"\n",
"#From Lame's Equation we have,\n",
"\n",
"#Hoop Stress\n",
"#f_x=b*(x**2)**-1+a ..........................(1)\n",
"\n",
"#Radial Pressure\n",
"#p_x=b*(x**2)**-1-a .........................(2)\n",
"\n",
"#the boundary conditions are\n",
"x=d1*2**-1 #mm \n",
"#After sub values in equation 1 and further simplifying we get\n",
"#40=b*100**-1-a ..........................(3)\n",
"\n",
"#Max Principal stress theory\n",
"#q*(FOS)**-1=b*100**2+a ..................(4)\n",
"#After sub values in above equation and further simplifying we get\n",
"\n",
"#From Equation 3 and 4 we get\n",
"a=80*2**-1\n",
"#Sub value of a in equation 3 we get\n",
"b=(f_x+a)*100**2\n",
"\n",
"#At outer edge where x=r_0 pressure is zero\n",
"r_0=(b*a**-1)**0.5 #mm\n",
"\n",
"#thickness\n",
"t=r_0-r1 #mm\n",
"\n",
"#Max shear stress theory\n",
"P1=b*(100**2)**-1+a #Max hoop stress\n",
"P2=-40 #pressure at int radius (since P2 is compressive)\n",
"\n",
"#Max shear stress\n",
"q_max=(P1-P2)*2**-1\n",
"\n",
"#According max shear theory the design condition is\n",
"#q_max=P_e*2**-1*(FOS)**-1\n",
"#After sub values in equation we get and further simplifying we get\n",
"#80=b*(100**2)**-1+a\n",
"#After sub values in equation 1 and 3 and further simplifying we get\n",
"b2=120*100**2*2**-1\n",
"\n",
"#from equation(3)\n",
"a2=120*2**-1-a\n",
"\n",
"#At outer radius r_0,radial pressure=0\n",
"r_02=(b2*a2**-1)**0.5\n",
"\n",
"#thickness\n",
"t2=r_02-r1\n",
"\n",
"#Result\n",
"print\"Thickness of metal by:Max Principal stress theory\",round(t,2),\"mm\"\n",
"print\" :Max shear stress thoery\",round(t2,2),\"mm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Thickness of metal by:Max Principal stress theory 41.42 mm\n",
" :Max shear stress thoery 73.21 mm\n"
]
}
],
"prompt_number": 3
}
],
"metadata": {}
}
]
}
|
