1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
|
{
"metadata": {
"name": "",
"signature": "sha256:087506e84c2c9ece60ca0c23e6e5e2f5e309357e67157657330423d5e748517d"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 24 :Theories of Failure"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem 24.10,page no.1019"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#Given\n",
"#Variable declaration\n",
"P=9*1000 #Axial pull in N\n",
"F=4.5*1000 #Shear force in N \n",
"sigmat_star=225 #Elastic limit in tension in N/sq.mm\n",
"Sf=3 #Factor of safety \n",
"mu=0.3 #Poisson's ratio \n",
"sigma3=0 #third principle stress\n",
"\n",
"#Calculation\n",
"sigmat=sigmat_star/Sf \n",
"sigma=(P/(math.pi/4))\n",
"tau=float(str(F/(math.pi/4))[:6])\n",
"sigma1=float(str((tau)+int(round(math.sqrt((sigma/2)**2+tau**2),0)))[:7])\n",
"sigma2=float(str((tau)-int(round(math.sqrt((sigma/2)**2+tau**2),0)))[:8])\n",
"d=round(((((sigma1-sigma2)**2+(sigma2-sigma3)**2+(sigma3-sigma1)**2)/(2*(sigmat**2)))**(1/4)),3) \n",
"\n",
"#Result\n",
"print\"Diameter of the bolt =\",d,\"mm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Diameter of the bolt = 14.217 mm\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem 24.12,page no.1027"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"#Given\n",
"#Variable declaration\n",
"d=1.2 #Diameter in m\n",
"p=1.5 #Internal pressure in MN/sq.m\n",
"sigmat_star=200 #Yield stress in MN/sq.m\n",
"Sf=3 #Factor of safety\n",
"\n",
"#Calculation\n",
"sigmat=sigmat_star/Sf #Permissible stress in simple tension in MN/sq.m\n",
"\n",
"#case(i):Thickness on the basis of Maximum principal stress theory\n",
"t1=((p*d)/2)/sigmat*1e3\n",
"\n",
"#case(ii):Thickness on the basis of Maximum shear stress theory\n",
"t2=((p*d)/2)/sigmat*1e3\n",
"\n",
"#case(iii):Thickness on the basis of Maximum shear strain energy theory\n",
"t3=round(math.sqrt((((p*d/2)**2)+((p*d/4)**2)-((p*d/2)*(p*d/4)))/(sigmat**2)),4)\n",
"\n",
"#Result\n",
"print \"Thickness of plate on the basis of maximum principal stress theory =\",\"%.1fmm\"%t1\n",
"print \"Thickness of plate on the basis of maximum shear stress theory =\",\"%.1fmm\"%t2\n",
"print \"Thickness of plate on the basis of maximum shear strain energy theory =\",\"%.4fmm\"%t3"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Thickness of plate on the basis of maximum principal stress theory = 13.5mm\n",
"Thickness of plate on the basis of maximum shear stress theory = 13.5mm\n",
"Thickness of plate on the basis of maximum shear strain energy theory = 0.0117mm\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}
|
