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"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter1-Mohrs circle"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex1-pg12"
]
},
{
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"collapsed": false,
"input": [
"##Mohr's circle\n",
"#calculate center of the circle and maxi principal stress and pricipal stress and maxi shearing stresss\n",
"import numpy\n",
"from numpy.linalg import inv\n",
"import math\n",
"sigma=((40+80)/2.)\n",
"print'%s %.2f %s'%(\"center of the circle in MPa = \",sigma,\"\")\n",
"\n",
"##solution a\n",
"x=((80-40)**2.);\n",
"y=30**2.;\n",
"sigma1=60.+math.sqrt((.25*x)+y)\n",
"print'%s %.2f %s'%(\"maxi pricipal stress in MPa = \",sigma1,\"\");## print'%s %.2f %s'%laying result\n",
"sigma2=60.-math.sqrt((.25*x)+y)\n",
"print'%s %.2f %s'%(\"mini pricipal stress in MPa = \",sigma2,\"\");## print'%s %.2f %s'%laying result\n",
"theta1=((math.atan((30./20.))/2)*57.3)\n",
"print'%s %.2f %s'%(\"pricipal stresses in degree\",theta1,\"\");## print'%s %.2f %s'%laying result\n",
"theta2=(((math.atan(30/20.))+180.)/2.)*57.3\n",
"print'%s %.2f %s'%(\"pricipal stresses in degree\",theta2,\"\");## print'%s %.2f %s'%laying result\n",
"\n",
"##solution b\n",
"tau=math.sqrt((.25*x)+y)\n",
"print'%s %.2f %s'%(\"maxi shearing stress in MPa = \",tau,\"\");## print'%s %.2f %s'%laying result\n",
"theta3=theta1+45.\n",
"print'%s %.2f %s'%(\"stress in MPa = \",theta3,\"\");## print'%s %.2f %s'%laying result\n",
"theta4=theta2+45\n",
"print'%s %.2f %s'%(\"stress in MPa = \",theta4,\"\");## print'%s %.2f %s'%laying result\n",
"\n",
"##final solution in matrix form\n",
"p=([[80 ,30], [30 ,40]])\n",
"print(p) \n",
"q=([[sigma1, 0 ],[0 ,sigma2]])\n",
"print(q)\n",
"r=([[sigma ,-tau], [-tau ,sigma]])\n",
"print(r)\n"
],
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{
"output_type": "stream",
"stream": "stdout",
"text": [
"center of the circle in MPa = 60.00 \n",
"maxi pricipal stress in MPa = 96.06 \n",
"mini pricipal stress in MPa = 23.94 \n",
"pricipal stresses in degree 28.16 \n",
"pricipal stresses in degree 5185.16 \n",
"maxi shearing stress in MPa = 36.06 \n",
"stress in MPa = 73.16 \n",
"stress in MPa = 5230.16 \n",
"[[80, 30], [30, 40]]\n",
"[[96.05551275463989, 0], [0, 23.944487245360108]]\n",
"[[60.0, -36.05551275463989], [-36.05551275463989, 60.0]]\n"
]
}
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},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex2-pg17"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##Mohr's circle\n",
"#calculate radius of the circle in and orientation of the stress\n",
"import math\n",
"radius=((14+28)/2.)\n",
"print'%s %.2f %s'%(\"radius of the circle in degree = \",radius,\"\")\n",
"sigma1=(7+radius *math.cos(60/57.3))\n",
"print'%s %.2f %s'%(\" the circle in MPa = \",sigma1,\"\")\n",
"sigma2=(7-radius *math.cos(60/57.3))\n",
"print'%s %.2f %s'%(\" the circle in MPa = \",sigma2,\"\")\n",
"tau1=radius*math.sin(60./57.3)\n",
"print'%s %.2f %s'%(\" orientation of the stresses in MPa = \",tau1,\"\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"radius of the circle in degree = 21.00 \n",
" the circle in MPa = 17.50 \n",
" the circle in MPa = -3.50 \n",
" orientation of the stresses in MPa = 18.19 \n"
]
}
],
"prompt_number": 2
}
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}
]
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|
