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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 3: Elements of the Theory of Plasticity"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Example 3.1, True Stress and True Strain, Page No. 76"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
"Engineering Stress at maximum load = 99852.1 psi\n",
"True Fracture Stress = 112785 psi\n",
"True Strain at fracture = 0.344939\n",
"Engineering strain at fracture = 0.411903\n"
]
}
],
"source": [
"from math import pi\n",
"from math import log\n",
"from math import exp\n",
"\n",
"#variable declaration\n",
"D_i=0.505;\n",
"L=2;\n",
"P_max=20000;\n",
"P_f=16000;\n",
"D_f=0.425;\n",
"\n",
"#calculation\n",
"E_St= P_max*4/(pi*D_i**2);\n",
"T_fr_St= P_f*4/(pi*D_f**2);\n",
"e_f=log(D_i**2/D_f**2);\n",
"e=exp(e_f)-1;\n",
"\n",
"#result\n",
"print('\\nEngineering Stress at maximum load = %g psi\\nTrue Fracture Stress = %g psi\\nTrue Strain at fracture = %g\\nEngineering strain at fracture = %g')%(E_St,T_fr_St,e_f,e);\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Example 3.2, Yielding Criteria for Ductile Metals, Page No. 78"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
"Since the calculated value of sigma0 = 224.054 MPa, which is less than the yield strength of the aluminium alloy\n",
"Thus safety factor is = 2.23161\n"
]
}
],
"source": [
"\n",
"from math import sqrt\n",
"\n",
"#variable declaration\n",
"sigma00=500;\n",
"sigma_z=-50;\n",
"sigma_y=100;\n",
"sigma_x=200;\n",
"T_xy=30;\n",
"T_yz=0;\n",
"T_xz=0;\n",
"\n",
"#calculation\n",
"sigma0=sqrt((sigma_x-sigma_y)**2+(sigma_y-sigma_z)**2+(sigma_z-sigma_x)**2+6*(T_xy**2+T_yz**2+T_xz**2))/sqrt(2);\n",
"s=sigma00/sigma0;\n",
"\n",
"#result\n",
"print('\\nSince the calculated value of sigma0 = %g MPa, which is less than the yield strength of the aluminium alloy\\nThus safety factor is = %g')%(sigma0,s);\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Example 3.3, Tresca Criterion, Page No. 81"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
"Since the calculated value of sigma0 = 250 MPa, which is less than the yield strength of the aluminium alloy\n",
"Thus safety factor is = 2\n"
]
}
],
"source": [
"\n",
"\n",
"#variable declaration\n",
"sigma00=500;\n",
"sigma_z=-50;\n",
"sigma_y=100;\n",
"sigma_x=200;\n",
"T_xy=30;\n",
"T_yz=0;\n",
"T_xz=0;\n",
"\n",
"#calculation\n",
"sigma0=sigma_x-sigma_z;\n",
"s=sigma00/sigma0;\n",
"\n",
"#result\n",
"print('\\nSince the calculated value of sigma0 = %g MPa, which is less than the yield strength of the aluminium alloy\\nThus safety factor is = %g')%(sigma0,s);\n"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"### Example 3.4, Levy-Mises Equation, Page No. 91"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
"Plastic Strain = 0.199532\n"
]
}
],
"source": [
"from math import sqrt\n",
"\n",
"#variable declaration\n",
"r_t=20;\n",
"p=1000;\n",
"\n",
"#calculation\n",
"sigma1=p*r_t;\n",
"sigma1=sigma1/1000; #conversion to ksi\n",
"sigma=sqrt(3)*sigma1/2;\n",
"e=(sigma/25)**(1/0.25);\n",
"e1=sqrt(3)*e/2;\n",
"\n",
"#result\n",
"print('\\nPlastic Strain = %g')%(e1);\n"
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 2",
"language": "python",
"name": "python2"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 2
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython2",
"version": "2.7.9"
}
},
"nbformat": 4,
"nbformat_minor": 0
}
|
