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
247
248
249
250
251
252
253
254
255
256
257
258
259
260
|
{
"metadata": {
"name": "",
"signature": "sha256:88c8946645e4f3f31d0d74d88b4d147ad878a660c1c0e5aff6962c72b58f187c"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"chapter 4:Shear stress in Beams and Related Problems "
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.1 page number 365"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"shear_v = 3000 #N - Transmitted vetical shear \n",
"shear_al = 700 #N - The maximum allowable \n",
"#We will divide this into two parts\n",
"l_1 = 50.0 #mm \n",
"l_2 = 200.0 #mm \n",
"b_1 = 200.0 #mm \n",
"b_2 = 50.0 #mm\n",
"A_1 = l_1* b_1 #mm2 - area of part_1\n",
"y_1 = 25.0 #mm com distance \n",
"A_2 =l_2*b_2 #mm2 - area of part_1\n",
"y_2 = 150.0 #in com distance \n",
"y_net = (A_1*y_1 +A_2*y_2)/(A_1+A_2) #mm - The com of the whole system\n",
"c_max = (4-y_net) #mm - The maximum distace from com to end\n",
"c_min = y_net #mm - the minimum distance from com to end\n",
"I_1 = b_1*(l_1**3)/12 + A_1*((y_1-y_net)**2) #Parallel axis theorm\n",
"I_2 = b_2*(l_2**3)/12 + A_2*((y_2-y_net)**2)\n",
"I_net = I_1 + I_2 #mm4 - the total moment of inertia\n",
"Q = A_1*(-y_1+y_net) #mm3\n",
"q = shear_v*Q/I_net #N/mm - Shear flow\n",
"d = shear_al/q # The space between the nails \n",
"print \"The minimal space between the nails \",round(d,0) ,\"mm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The minimal space between the nails 42.0 mm\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.2 pagenumber 365"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given \n",
"l = 6 #m -length of the beam \n",
"p = 3 #KN-m _ the load applied\n",
"R_a = l*p/2 #KN -The reaction at a, Since the system is symmetry \n",
"R_b = l*p/2 #KN -The reaction at b \n",
"l_s = 10 #mm - The length of the screw \n",
"shear_al = 2 #KN - The maximum load the screw can take \n",
"I = 2.36*(10**9) #mm2 The moment of inertia of the whole system\n",
"#We will divide this into two parts\n",
"l_1 = 50.0 #mm \n",
"l_2 = 50.0 #mm \n",
"b_1 = 100.0 #mm \n",
"b_2 = 200.0 #mm\n",
"A_1 = l_1* b_1 #in2 - area of part_1\n",
"y_1 = 200.0 #mm com distance \n",
"A_2 =l_2*b_2 #mm2 - area of part_1\n",
"y_2 = 225.0 #in com distance\n",
"Q = 2*A_1*y_1 + A_2*y_2 # mm3 For the whole system\n",
"q = R_a*Q*(10**3)/I #N/mm The shear flow \n",
"d = shear_al*(10**3)/q #mm The space between the nails\n",
"print \"The minimal space between the nails \",round(d,0),\"mm\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The minimal space between the nails 123.0 mm\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.6 page number 376"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"#we will divide this into two equal parts and other part\n",
"l = 10.0 # in - The height \n",
"t = 0.1 # in - The width\n",
"b = 5.0 #mm- The width of the above part \n",
"A = t* b #in2 - area of part\n",
"y_net = l/2 # The com of the system \n",
"y_1 = l # The position of teh com of part_2\n",
"I_1 = t*(l**3)/12 #in4 The moment of inertia of part 1\n",
"I_2 = 2*A*((y_1-y_net)**2) #in4 The moment of inertia of part 2 \n",
"I = I_1 + I_2 #in4 The moment of inertia \n",
"e = (b**2)*(l**2)*t/(4*I) #in the formula of channels\n",
"l_sc = e - t/2 #in- The shear centre \n",
"print \"The shear centre from outside vertical face is \",l_sc ,\"in\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The shear centre from outside vertical face is 1.825 in\n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.8 page number 387"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given \n",
"dia = 10.0 #mm - The diameter of the cylinder \n",
"c = dia/2 #mm - the radius of the cylinder \n",
"A = 3.14*(c**2) #mm2 The area of the crossection \n",
"y = 4*c/(3*3.14) #mm The com of cylinder \n",
"I = 3.14*(c**4)/4 #mm4 - The moment of inertia of the cylinder\n",
"j = 3.14*(dia**4)/32 #mm4\n",
"T = 20.0 #N.m - The torque \n",
"V = 250.0 #N - The shear \n",
"M = 25.0 #N-m The bending moment \n",
"Q = A*y/2 #mm\n",
"stress_dmax = 4*V/(3*A) #V*Q/(I*d) #Mpa The direct maximum stress\n",
"stress_tmax = T*c*(10**3)/j #-Mpa The torsion maximum stress\n",
"stress_total = stress_dmax + stress_tmax #Mpa The total stress\n",
"print \"The direct maximum stress\",round(stress_dmax,2),\"Mpa\"\n",
"print \"The torsion maximum stress\",round(stress_tmax,2),\"Mpa\"\n",
"print \"The total stress\",round(stress_total,2),\"Mpa\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The direct maximum stress 4.25 Mpa\n",
"The torsion maximum stress 101.91 Mpa\n",
"The total stress 106.16 Mpa\n"
]
}
],
"prompt_number": 38
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.9 page number 393"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"dia = 15 #mm - The diameter of the rod\n",
"h = 0.5 #mt - The freely falling height \n",
"A = 3.14*(dia**2)/4 #mm2 The area of the crossection\n",
"E = 200 #Gpa -Youngs modulus\n",
"L = 750 #mm - The total length of the rod\n",
"G = 80 #gpa - Shear modulus \n",
"N = 10 #number of live coils\n",
"d = 5 #mm the diameter of live coil \n",
"m = 3 # the mass of freely falling body\n",
"H = 500 #mm -from mass to spring \n",
"F= m*9.81 #Kg the force due to that mass\n",
"#e = e_rod + e_spr\n",
"#e_rod\n",
"e_rod = p*L*(10**-3)/(A*E) #mm The elongation due to freely falling body\n",
"#e_spr\n",
"e_spr = 64*F*(dia**3)*N*(10**-3)/(G*(d**4)) #mm The elongation due to spring\n",
"e = e_rod + e_spr #mm The total elongation \n",
"p_dyn =F*(1+pow((1+(2*H/e)),0.5))\n",
"Stress_max = p_dyn/A #MPa - The maximum stress in the system \n",
"print \"The maximum stress in the system \",round(Stress_max,2),\"Mpa\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The maximum stress in the system 4.84 Mpa\n"
]
}
],
"prompt_number": 43
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}
|
