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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": [
"Chapter1-Introduction"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex1-pg16"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##calculate the\n",
"## initialization of variables\n",
"import math\n",
"## part (a)\n",
"a=700. ## M Pa from figure 1.8\n",
"b=100. ## M Pafrom figure 1.8\n",
"m=1/6. ## from figure 1.8\n",
"Y=450. ## M Pa from figure 1.9\n",
"##calculations\n",
"sigma_u=a+m*b\n",
"## results\n",
"print('\\n part (a) \\n')\n",
"print\"%s %.2f %s\"%(' The ultimate strength is sigma = ',sigma_u,' M Pa')\n",
"print\"%s %.2f %s\"%('\\n and the yield strength is Y = ',Y,'M Pa')\n",
"\n",
"## part (b)\n",
"c1=62. ## from figure 1.8\n",
"d1=0.025 ## from figure 1.8\n",
"c2=27. ## from figure 1.10a\n",
"d2=0.04 ## from figure 1.10a\n",
"## calculations\n",
"U_f1=c1*b*d1*10**6\n",
"U_f2=c2*b*d2*10**6\n",
"## results\n",
"print('\\n part (b)')\n",
"print\"%s %.2e %s\"%('\\n The modulus of toughness for alloy steel is Uf = ',U_f1,' N/m^2')\n",
"print\"%s %.2e %s\"%('\\n and structural steel is Uf = ',U_f2,' N/m^2')\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" part (a) \n",
"\n",
" The ultimate strength is sigma = 716.67 M Pa\n",
"\n",
" and the yield strength is Y = 450.00 M Pa\n",
"\n",
" part (b)\n",
"\n",
" The modulus of toughness for alloy steel is Uf = 1.55e+08 N/m^2\n",
"\n",
" and structural steel is Uf = 1.08e+08 N/m^2\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex2-pg16"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##calculate the permanet strain\n",
"## initialization of variables\n",
"import math\n",
"sigma=500. ## Stress M Pa\n",
"eps=0.0073 ## Strain\n",
"sigma_A=343. ## M Pa from figure 1.9\n",
"eps_A=0.00172 ## from figure 1.9\n",
"## part (a)\n",
"E=sigma_A/eps_A\n",
"\n",
"## part (B)\n",
"eps_e=sigma/E\n",
"eps_p=eps-eps_e\n",
"## results\n",
"print(' part (a) \\n')\n",
"print\"%s %.2f %s\"%(' The modulus of elasticity of the rod is E = ',E/1000,' G Pa')\n",
"print('\\n part (b)')\n",
"print\"%s %.4f %s\"%('\\n the permanent strain is = ',eps_p,'')\n",
"print\"%s %.4f %s\"%('\\n and the strain recovered is =',eps_e,'')\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" part (a) \n",
"\n",
" The modulus of elasticity of the rod is E = 199.42 G Pa\n",
"\n",
" part (b)\n",
"\n",
" the permanent strain is = 0.0048 \n",
"\n",
" and the strain recovered is = 0.0025 \n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex3-pg19"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##calculate the diameter\n",
"## initialization of variables\n",
"import math\n",
"D=25. ## kN\n",
"L=60. ## kN\n",
"W=30. ##kN\n",
"Y=250. ## M Pa\n",
"safety=5./3. ## AISC, 1989\n",
"## calculations\n",
"Q=(D+L+W)*10**3. ## converted to N\n",
"A=safety*Q/Y\n",
"r=math.sqrt(A/math.pi)+0.5 ## additional 0.5 mm is for extra safety\n",
"d=1.8*r ## diameter\n",
"## results\n",
"print('Part (a) \\n ')\n",
"print\"%s %.2f %s %.2f %s \"%('A rod of ',d,' mm'and ' in diameter, with a cross sectional area of ',math.pi*(d**2./4.),' mm^2, is adequate')\n",
"## The diameter is correct as given in the textbook. Area doesn't match due to rounding off error and partly because it's a design problem.\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Part (a) \n",
" \n",
"A rod of 29.02 in diameter, with a cross sectional area of 661.39 mm^2, is adequate \n"
]
}
],
"prompt_number": 14
}
],
"metadata": {}
}
]
}
|
