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
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
|
{
"metadata": {
"name": "",
"signature": "sha256:cfde82eb2eac726301e92e1b27c27d3628db6f9b0e8b5d7b0cd15017ca0006f7"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 9:Elastic stress analysis and design"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.4 pagenumber 465"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given \n",
"b = 40.0 #mm - The width of the beam crossection\n",
"h = 300.0 #mm - The length of the beam crossection \n",
"V = 40.0 #KN - The shear stress in teh crossection\n",
"M = 10.0 #KN-m - The bending moment on K----K crossection \n",
"c = h/2 #mm -The position at which maximum stress occurs on the crossection\n",
"I = b*(h**3)/12 #mmm4 - the moment of inertia \n",
"#Caliculations \n",
"\n",
"stress_max_1 = M*c*(10**6)/I #The maximum stress occurs at the end\n",
"stress_max_2 = -M*c*(10**6)/I #The maximum stress occurs at the end\n",
"y = 140 #mm The point of interest, the distance of element from com\n",
"n = y/(c) # The ratio of the distances from nuetral axis to the elements\n",
"stress_L_1 = n*stress_max_1 #The normal stress on elements L--L\n",
"stress_L_2 = -n*stress_max_1 #The normal stress on elements L--L\n",
"x = 10 #mm The length of the element\n",
"A = b*x #mm3 The area of the element \n",
"y_1 = y+x/2 # the com of element from com of whole system\n",
"stress_xy = V*A*y_1*(10**3)/(I*b) #Mpa - The shear stress on the element \n",
"#stresses acting in plane 30 degrees \n",
"o = 60 #degrees - the plane angle\n",
"stress_theta = stress_L_1/2 + stress_L_1*(math.cos(math.radians(o)))/2 - stress_xy*(math.sin(math.radians(o))) #Mpa by direct application of equations\n",
"stress_shear = -stress_L_1*(math.sin(math.radians(o)))/2 - stress_xy*(math.cos(math.radians(o))) #Mpa Shear stress\n",
" \n",
"print \"a)The principle stresses are \",round(stress_max_1,2),\"MPa,\",round(stress_max_2,2),\"Mpa\"\n",
"print \"b)The stresses on inclines plane \",round(stress_theta,2),\"Mpa noraml, \",round(stress_shear,2),\"Mpa shear \""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"a)The principle stresses are 16.67 MPa, -16.67 Mpa\n",
"b)The stresses on inclines plane 11.11 Mpa noraml, -7.06 Mpa shear \n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.5 page number 476"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"M = 10 #KN-m moment\n",
"v = 8.0 #KN - shear Stress \n",
"stress_allow = 8 #MPa - The maximum allowable stress\n",
"shear_allow_per = 1.4 #Mpa - The allowable stress perpendicular to grain\n",
"stress_allow_shear = 0.7 #MPa - The maximum allowable shear stress\n",
"#Caliculations \n",
"\n",
"S = M*(10**6)/stress_allow #mm3 \n",
"#lets arbitarly assume h = 2b\n",
"#S = b*(h**2)/6\n",
"h = pow(12*S,0.333) #The depth of the beam\n",
"b = h/2 #mm The width of the beam\n",
"A = h*b #mm2 The area of the crossection , assumption\n",
"stress_shear = 3*v*(10**3)/(2*A) #Mpa The strear stress \n",
"if stress_shear<stress_allow_shear:\n",
" print \"The stress developed \",round(stress_shear,2),\" is in allowable ranges for \",round(A,2),\"mm2 area\"\n",
"else:\n",
" print \"The stress developed\",stress_shear,\" is in non allowable ranges\",A,\"area\"\n",
"Area_allow = v*(10**3)/shear_allow_per #mm - the allowable area\n",
"print \"The minimum area is \",Area_allow ,\"mm2\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The stress developed 0.4 is in allowable ranges for 30077.85 mm2 area\n",
"The minimum area is 5714.28571429 mm2\n"
]
},
{
"metadata": {},
"output_type": "pyout",
"prompt_number": 36,
"text": [
"8.0"
]
}
],
"prompt_number": 36
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.6 page number 478"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"stress_allow = 24 #ksi - The maximum allowable stress\n",
"stress_allow_shear = 14.5 #ksi- The maximum allowable shear stress\n",
"M_max = 36 #k-ft The maximum moment\n",
"l = 16 #in-The length of the rod\n",
"w = 2 #k/ft - The force distribution on the rod\n",
"R_A = 6.4 #k - The reaction at A\n",
"R_B = 25.6 #k - the reaction at B\n",
"v_max = R_B-l*w #kips the maximum stress, from diagram\n",
"#W8x24 is used from the appendix table 3 and 4 \n",
"l =0.245 #in - W8x24 crossesction length\n",
"#Caliculations \n",
"\n",
"stress_xy = v_max/A #ksi the approximate shear stress \n",
"if stress_xy < stress_allow_shear:\n",
" print \"W8x24 gives the allowable ranges of shear stress\"\n",
"else:\n",
" print \"W8x24 doesnot gives the allowable ranges of shear stress\"\n",
"k = 7.0/8 #in the distance from the outer face of the flange to the webfillet\n",
"#at+kt should not exceed 0.75 of yeild stress\n",
"#a1t+2kt should not exceed 0.75 of yeild stress\n",
"Stress_yp = 36 #Ksi - The yeild stress\n",
"t = 0.245 #in thickness of the web\n",
"#support a \n",
"a = R_A/(0.75*Stress_yp*t)-k #in lengths of the bearings\n",
"#support b\n",
"a_1 = R_B/(0.75*Stress_yp*t)-2*k #in lengths of the bearings\n",
"print \"lengths of the bearing at A \",round(a,3),\"in\"\n",
"print \"lengths of the bearing at B\",round(a_1,3),\"in\"\n",
" \n",
"\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"W8x24 gives the allowable ranges of shear stress\n",
"lengths of the bearing at A 0.092 in\n",
"lengths of the bearing at B 2.12 in\n"
]
},
{
"metadata": {},
"output_type": "pyout",
"prompt_number": 44,
"text": [
"0.875"
]
}
],
"prompt_number": 44
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.8 page number 483"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#given \n",
"hp = 63000 #horse power\n",
"T = hp*20*(10**-3)/63 #k-in the torsion implies due to horse power\n",
"stress_allow_shear = 6 #ksi- The maximum allowable shear stress\n",
"M_ver = 6.72/2 #k-in the vertical component of the moment \n",
"M_hor = 9.10 #k-in the horizantal component of the moment \n",
"#Caliculations \n",
"\n",
"M = pow(((M_ver**2)+(M_hor**2)),0.5) #K-in The resultant \n",
"d = pow((16*(((M**2)+(T**2))**0.5)/(stress_allow_shear*3.14)),0.333) #in, The suggested diameter from derivation\n",
"print \"The suggested diameter is\",round(d,2),\"in\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The suggested diameter is 2.66 in\n"
]
}
],
"prompt_number": 49
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}
|
