K12 For Timber Designers' Manuel By me.xls
Transcript of K12 For Timber Designers' Manuel By me.xls
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It follows : 17.2 TIMBER STRESS GRADES FOR GLULAM COLUMNS
E = EMEAN X K20
= 10800 X 1.07 = 11556
c,par = grade x K3 X K17
= 7.9 X 1 X 1.04 = 8.216
= 7.9 X 1.25 X 1.04 = 10.27
= 7.9 X 1.5 X 1.04 = 12.324
= 7.9 X 1.75 X 1.04 = 14.378
E/c,par = 11556 / 8.22 = 1407
= 11556 / 10.3 = 1125
= 11556 / 12.3 = 937.7
= 11556 / 14.4 = 803.7
FOR SERVICE CLASS 3
E = EMEAN X K2 X K20
= 10800 X 0.8 X 1.07 = 9244.8
c,par = grade x K2 X K3 X K17
= 7.9 X 0.6 X 1 X 1.04 = 4.9296
= 7.9 X 0.6 X 1.25 X 1.04 = 6.162
= 7.9 X 0.6 X 1.5 X 1.04 = 7.3944
= 7.9 X 0.6 X 1.75 X 1.04 = 8.6268
E/c,par = 9244.8 / 4.93 = 1875
= 9244.8 / 6.16 = 1500
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= 9244.8 / 7.39 = 1250
= 9244.8 / 8.63 = 1072
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TABLE 15.1 DETAMINATION OF E/ FOR SERVICES CLASSES 1 AND 2 AND 3
E c//g E/c//g c//g E/c//g c//g E/c//g c//g
C14 4600 5.2 885 6.5 708 7.8 590 9.1
C16 5800 6.8 853 8.5 682 10.2 569 11.9
C18 6000 7.1 845 8.875 676 10.65 563 12.425
C22 6500 7.5 867 9.375 693 11.25 578 13.125
C24 7200 7.9 911 9.875 729 11.85 608 13.825
TR26 7400 8.2 902 10.25 722 12.3 602 14.35
C27 8200 8.2 1000 10.25 800 12.3 667 14.35
Kempas 16000 19.4 825 24.25 660 29.1 550 33.95
Teak 7400 13.4 552 16.75 442 20.1 368 23.45
E c//g E/c//g c//g E/c//g c//g E/c//g c//g
C14 3680 3.12 1179 3.9 944 4.68 786 5.46
C16 4640 4.08 1137 5.1 910 6.12 758 7.14
C18 4800 4.26 1127 5.325 901 6.39 751 7.455
C22 5200 4.5 1156 5.625 924 6.75 770 7.875
C24 5760 4.74 1215 5.925 972 7.11 810 8.295
TR26 5920 4.92 1203 6.15 963 7.38 802 8.61
C27 6560 4.92 1333 6.15 1067 7.38 889 8.61
Kempas 12800 11.64 1100 14.55 880 17.46 733 20.37
Teak 5920 8.04 736 10.05 589 12.06 491 14.07
SERVICE CLASSES 1 AND 2
SERVICE CLASSES 1 AND 3
Long Medium Short Very
Long Medium Short Very
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E/c//g
505
487
483
495
521
516
571
471
316
E/c//g
674
650
644
660
694
688
762
628
421
Short
Short
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Long Medium Short Very Short Long Medium
(N/mm2)
5 0.976 0.976 0.976 0.975 4566 0.976 0.976
10 0.952 0.952 0.952 0.952 1141 0.952 0.952
15 0.929 0.928 0.928 0.928 507 0.929 0.929
20 0.906 0.905 0.904 0.903 285 0.907 0.906
25 0.883 0.881 0.879 0.878 183 0.884 0.883
30 0.859 0.857 0.854 0.851 127 0.862 0.860
35 0.836 0.831 0.827 0.822 93 0.840 0.837
40 0.811 0.805 0.799 0.792 71 0.817 0.813
45 0.786 0.778 0.769 0.759 56 0.794 0.788
50 0.761 0.749 0.737 0.725 46 0.771 0.763
55 0.734 0.720 0.704 0.688 38 0.748 0.738
60 0.707 0.689 0.670 0.650 32 0.724 0.71265 0.680 0.658 0.635 0.612 27 0.701 0.685
70 0.652 0.626 0.600 0.573 23 0.676 0.658
75 0.624 0.594 0.564 0.535 20 0.652 0.631
80 0.595 0.562 0.530 0.498 18 0.628 0.604
85 0.567 0.531 0.496 0.464 16 0.604 0.576
90 0.540 0.501 0.465 0.431 14 0.579 0.550
95 0.513 0.472 0.434 0.400 13 0.556 0.523
100 0.487 0.445 0.406 0.372 11 0.532 0.498
105 0.462 0.418 0.380 0.346 10 0.509 0.473
110 0.438 0.394 0.355 0.323 9 0.487 0.450
115 0.415 0.371 0.333 0.301 9 0.466 0.427
120 0.393 0.349 0.312 0.281 8 0.445 0.405
125 0.373 0.329 0.293 0.263 7 0.425 0.385
130 0.353 0.310 0.275 0.247 7 0.406 0.366
135 0.335 0.293 0.259 0.231 6 0.388 0.347
140 0.318 0.277 0.244 0.217 6 0.370 0.330
145 0.302 0.262 0.230 0.205 5 0.354 0.314
150 0.287 0.248 0.217 0.193 5 0.338 0.299
155 0.273 0.235 0.205 0.182 5 0.324 0.284
160 0.260 0.223 0.195 0.172 4 0.309 0.271
165 0.248 0.212 0.184 0.163 4 0.296 0.259170 0.236 0.201 0.175 0.155 4 0.284 0.247
175 0.225 0.192 0.166 0.147 4 0.272 0.236
180 0.215 0.183 0.158 0.140 4 0.260 0.225
185 0.206 0.174 0.151 0.133 3 0.250 0.215
190 0.197 0.166 0.144 0.126 3 0.240 0.206
195 0.188 0.159 0.137 0.121 3 0.230 0.197
200 0.181 0.152 0.131 0.115 3 0.221 0.189
205 0.173 0.146 0.125 0.110 3 0.213 0.182
Service Classes 1 and 2 ServiceEuler
Stress
K12 and Euler Stress Value FOR C24 glulam
E/c,par1,407 1,125 938 804 1,875 1,500
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210 0.166 0.139 0.120 0.105 3 0.204 0.174
215 0.159 0.134 0.115 0.101 2 0.197 0.167
220 0.153 0.128 0.110 0.097 2 0.190 0.161
225 0.147 0.123 0.106 0.093 2 0.183 0.155
230 0.142 0.118 0.102 0.089 2 0.176 0.149
235 0.137 0.114 0.098 0.085 2 0.170 0.144
240 0.132 0.110 0.094 0.082 2 0.164 0.138245 0.127 0.106 0.090 0.079 2 0.158 0.134
250 0.122 0.102 0.087 0.076 2 0.153 0.129
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Short Very Short
(N/mm2)
0.976 0.976 3653
0.952 0.952 913
0.929 0.928 406
0.905 0.905 228
0.882 0.881 146
0.858 0.856 101
0.833 0.830 75
0.808 0.804 57
0.782 0.776 45
0.755 0.746 37
0.727 0.716 30
0.698 0.684 250.669 0.652 22
0.639 0.619 19
0.609 0.587 16
0.579 0.554 14
0.549 0.522 13
0.520 0.492 11
0.492 0.462 10
0.465 0.435 9
0.439 0.408 8
0.415 0.384 8
0.392 0.361 7
0.370 0.339 6
0.350 0.319 6
0.331 0.301 5
0.313 0.284 5
0.296 0.268 5
0.281 0.253 4
0.266 0.240 4
0.253 0.227 4
0.240 0.215 4
0.229 0.204 30.218 0.194 3
0.207 0.185 3
0.198 0.176 3
0.189 0.168 3
0.180 0.160 3
0.173 0.153 2
0.165 0.146 2
0.158 0.140 2
1,072
Classes 3Euler
Stress
1,250
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0.152 0.134 2
0.146 0.128 2
0.140 0.123 2
0.134 0.118 2
0.129 0.114 2
0.124 0.109 2
0.120 0.105 20.115 0.101 2
0.111 0.098 1
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Long Medium Short Very Short Long Medium
(N/mm2)
5 0.976 0.975 0.975 0.975 6322 0.976 0.976
10 0.952 0.951 0.951 0.951 1580 0.952 0.952
15 0.928 0.927 0.926 0.926 702 0.928 0.928
20 0.903 0.902 0.900 0.898 395 0.905 0.904
25 0.878 0.875 0.872 0.868 253 0.881 0.879
30 0.851 0.846 0.841 0.835 176 0.856 0.853
35 0.823 0.815 0.806 0.797 129 0.831 0.825
40 0.793 0.781 0.768 0.754 99 0.804 0.796
45 0.761 0.744 0.725 0.706 78 0.777 0.765
50 0.727 0.704 0.680 0.654 63 0.748 0.732
55 0.691 0.662 0.632 0.601 52 0.718 0.698
60 0.654 0.619 0.583 0.548 44 0.687 0.66265 0.616 0.575 0.536 0.498 37 0.655 0.626
70 0.578 0.533 0.490 0.451 32 0.623 0.589
75 0.540 0.492 0.448 0.409 28 0.591 0.553
80 0.504 0.454 0.409 0.371 25 0.559 0.517
85 0.469 0.418 0.374 0.337 22 0.527 0.483
90 0.437 0.385 0.342 0.307 20 0.497 0.451
95 0.406 0.355 0.314 0.280 18 0.468 0.421
100 0.378 0.328 0.289 0.257 16 0.440 0.392
105 0.352 0.304 0.266 0.236 14 0.414 0.366
110 0.328 0.282 0.246 0.217 13 0.389 0.342
115 0.306 0.262 0.227 0.201 12 0.366 0.320
120 0.286 0.243 0.211 0.186 11 0.344 0.299
125 0.268 0.227 0.196 0.173 10 0.324 0.280
130 0.251 0.212 0.183 0.161 9 0.306 0.263
135 0.236 0.199 0.171 0.150 9 0.289 0.247
140 0.222 0.186 0.160 0.140 8 0.273 0.233
145 0.209 0.175 0.150 0.131 8 0.258 0.219
150 0.197 0.165 0.141 0.123 7 0.244 0.207
155 0.186 0.155 0.133 0.116 7 0.231 0.196
160 0.176 0.146 0.125 0.109 6 0.219 0.185
165 0.167 0.138 0.118 0.103 6 0.208 0.175170 0.158 0.131 0.112 0.098 5 0.198 0.166
175 0.150 0.124 0.106 0.092 5 0.188 0.158
180 0.143 0.118 0.101 0.088 5 0.179 0.150
185 0.136 0.112 0.095 0.083 5 0.171 0.143
190 0.129 0.107 0.091 0.079 4 0.163 0.136
195 0.123 0.102 0.086 0.075 4 0.156 0.130
200 0.118 0.097 0.082 0.072 4 0.149 0.124
205 0.112 0.093 0.079 0.068 4 0.143 0.119
K12 FOR KEMPUS
E/c,par825 660 550 471 1,100 880
Service Classes 1 and 2Euler
Stress
Service
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210 0.108 0.088 0.075 0.065 4 0.137 0.114
215 0.103 0.085 0.072 0.062 3 0.131 0.109
220 0.099 0.081 0.069 0.060 3 0.126 0.104
225 0.095 0.078 0.066 0.057 3 0.121 0.100
230 0.091 0.075 0.063 0.055 3 0.116 0.096
235 0.087 0.072 0.061 0.053 3 0.112 0.092
240 0.084 0.069 0.058 0.050 3 0.108 0.089245 0.081 0.066 0.056 0.048 3 0.104 0.085
250 0.078 0.064 0.054 0.047 3 0.100 0.082
320.1 0.049 0.040 0.033 0.029 2 0.063 0.052
121.9 0.279 0.237 0.205 0.181 11 0.336 0.292
130.8 0.249 0.210 0.181 0.159 9 0.303 0.261
178.1 0.145 0.120 0.103 0.089 5 0.183 0.153
346.5 0.042 0.034 0.029 0.025 1 0.054 0.044
166.3 0.164 0.137 0.117 0.102 6 0.206 0.173
436 0.027 0.022 0.018 0.016 1 0.035 0.029
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Short Very Short
(N/mm2)
0.975 0.975 5057
0.951 0.951 1264
0.927 0.927 562
0.902 0.901 316
0.876 0.874 202
0.849 0.845 140
0.819 0.813 103
0.787 0.778 79
0.753 0.739 62
0.716 0.698 51
0.677 0.655 42
0.637 0.610 350.596 0.565 30
0.555 0.522 26
0.516 0.480 22
0.478 0.442 20
0.443 0.406 17
0.410 0.374 16
0.380 0.344 14
0.352 0.318 13
0.326 0.293 11
0.303 0.272 10
0.282 0.252 10
0.263 0.235 9
0.246 0.219 8
0.230 0.204 7
0.216 0.191 7
0.203 0.179 6
0.190 0.168 6
0.179 0.158 6
0.169 0.149 5
0.160 0.140 5
0.151 0.133 50.143 0.126 4
0.136 0.119 4
0.129 0.113 4
0.123 0.107 4
0.117 0.102 4
0.111 0.097 3
0.106 0.093 3
0.102 0.089 3
Euler
Stress
733 628
Classes 3
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0.097 0.085 3
0.093 0.081 3
0.089 0.078 3
0.085 0.074 2
0.082 0.071 2
0.079 0.068 2
0.076 0.066 20.073 0.063 2
0.070 0.061 2
0.044 0.038 1
0.256 0.228 9
0.228 0.202 7
0.132 0.115 4
0.037 0.032 1
0.149 0.131 5
0.024 0.021 1
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Long Medium Short Very Short Long Medium
(N/mm2)
5 0.975 0.975 0.975 0.975 4566 0.975 0.975
10 0.951 0.951 0.951 0.950 1141 0.951 0.951
15 0.926 0.925 0.924 0.923 507 0.927 0.927
20 0.900 0.898 0.895 0.892 285 0.902 0.901
25 0.872 0.867 0.862 0.856 183 0.876 0.873
30 0.841 0.832 0.822 0.812 127 0.849 0.843
35 0.806 0.792 0.776 0.759 93 0.819 0.810
40 0.768 0.747 0.723 0.698 71 0.787 0.773
45 0.726 0.696 0.664 0.631 56 0.753 0.733
50 0.680 0.642 0.602 0.563 46 0.716 0.690
55 0.633 0.587 0.541 0.499 38 0.677 0.644
60 0.584 0.532 0.484 0.441 32 0.637 0.59865 0.537 0.481 0.432 0.390 27 0.596 0.551
70 0.491 0.434 0.386 0.346 23 0.556 0.507
75 0.449 0.392 0.346 0.308 20 0.517 0.465
80 0.410 0.355 0.311 0.275 18 0.479 0.426
85 0.375 0.322 0.280 0.247 16 0.444 0.391
90 0.343 0.292 0.254 0.223 14 0.411 0.358
95 0.315 0.267 0.230 0.202 13 0.381 0.329
100 0.290 0.244 0.210 0.184 11 0.353 0.303
105 0.267 0.224 0.192 0.168 10 0.327 0.280
110 0.246 0.206 0.177 0.154 9 0.304 0.259
115 0.228 0.190 0.163 0.142 9 0.283 0.240
120 0.212 0.176 0.150 0.131 8 0.264 0.223
125 0.197 0.163 0.139 0.121 7 0.247 0.208
130 0.184 0.152 0.130 0.113 7 0.231 0.194
135 0.172 0.142 0.121 0.105 6 0.216 0.181
140 0.161 0.133 0.113 0.098 6 0.203 0.170
145 0.151 0.124 0.105 0.092 5 0.191 0.159
150 0.142 0.117 0.099 0.086 5 0.180 0.150
155 0.133 0.110 0.093 0.081 5 0.170 0.141
160 0.126 0.103 0.087 0.076 4 0.160 0.133
165 0.119 0.097 0.082 0.071 4 0.152 0.126170 0.112 0.092 0.078 0.067 4 0.144 0.119
175 0.106 0.087 0.074 0.064 4 0.136 0.113
180 0.101 0.083 0.070 0.060 4 0.130 0.107
185 0.096 0.078 0.066 0.057 3 0.123 0.101
190 0.091 0.074 0.063 0.054 3 0.117 0.097
195 0.087 0.071 0.060 0.052 3 0.112 0.092
200 0.083 0.067 0.057 0.049 3 0.107 0.088
205 0.079 0.064 0.054 0.047 3 0.102 0.084
K12 FOR TEAK
E/c,par552 442 368 316 736 589
Service Classes 1 and 2Euler
Stress
Service
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210 0.075 0.061 0.052 0.045 3 0.097 0.080
215 0.072 0.059 0.050 0.043 2 0.093 0.076
220 0.069 0.056 0.047 0.041 2 0.089 0.073
225 0.066 0.054 0.045 0.039 2 0.086 0.070
230 0.063 0.052 0.043 0.038 2 0.082 0.067
235 0.061 0.049 0.042 0.036 2 0.079 0.065
240 0.058 0.047 0.040 0.035 2 0.076 0.062245 0.056 0.046 0.038 0.033 2 0.073 0.060
250 0.054 0.044 0.037 0.032 2 0.070 0.057
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Short Very Short
(N/mm2)
0.975 0.975 3653
0.951 0.951 913
0.926 0.925 406
0.899 0.897 228
0.869 0.866 146
0.836 0.830 101
0.799 0.788 75
0.758 0.741 57
0.711 0.688 45
0.661 0.632 37
0.610 0.575 30
0.558 0.520 250.508 0.468 22
0.462 0.421 19
0.419 0.380 16
0.381 0.343 14
0.347 0.310 13
0.316 0.282 11
0.289 0.257 10
0.265 0.235 9
0.244 0.215 8
0.225 0.198 8
0.208 0.183 7
0.192 0.169 6
0.179 0.157 6
0.166 0.146 5
0.155 0.136 5
0.145 0.127 5
0.136 0.119 4
0.128 0.112 4
0.120 0.105 4
0.113 0.099 4
0.107 0.093 30.101 0.088 3
0.096 0.083 3
0.091 0.079 3
0.086 0.075 3
0.082 0.071 3
0.078 0.068 2
0.074 0.064 2
0.071 0.062 2
Euler
Stress
491 421
Classes 3
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0.068 0.059 2
0.065 0.056 2
0.062 0.054 2
0.059 0.051 2
0.057 0.049 2
0.055 0.047 2
0.052 0.045 20.050 0.044 2
0.048 0.042 1
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UDL on Member = 0.44 kN/m
Axial Loading on Member
= 3453.262384
Length of Member = 4.62 m
Moment on Member = 1.173942 kN m
Trail section = 50 x 125
Z = 1.30E+05
m,a = 9.02E+00
c,a = 0.552521981
x = 320.0829892
y = 128.0331957
= 320.0829892
K12 = 0.03960829 For Medium
K3 = 1.25
c,g = 19.4
c,adm = 0.960501021
Check for very short term condition
K3 = 1.75
K12 = 0.028850779 320.083
c,adm = 0.979483934
c,a = 0.552521981
K7 = 1.10109106
For =
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m,g = 19.3
m,adm = 37.18935056
m,a = 9.02E+00
Euler Critical Stress
e = 2 N/mm2
Euler Co-efficient Keu = 0.984499194
Interaction Formula = 0.246248674 + 0.575244
= 0.821492187
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UDL on Member = 0.44 kN/m
Axial Loading on Member
= 2065.0652
Length of Member = 1.76 m
Moment on Member = 0.170368 kN m
Trail section = 50 x 125
Z = 1.30E+05
m,a = 1.31E+00
c,a = 0.330410432
x = 121.9363769
y = 48.77455074
= 121.9363769
K12 = 0.236904042 For Medium
K3 = 1.25
c,g = 19.4
c,adm = 5.74492303
Check for very short term condition
K3 = 1.75
K12 = 0.180663873 121.9364
c,adm = 6.133538499
For =
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c,a = 0.330410432
K7 = 1.10109106
m,g = 19.3
m,adm = 37.18935056
m,a = 1.31E+00
Euler Critical Stress
e = 11 N/mm2
Euler Co-efficient Keu = 0.991576092
Interaction Formula = 0.035481715 + 0.057513
= 0.09299518
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UDL on Member = 0.44 kN/m
Axial Loading on Member
= 3453.262384
Length of Member = 4.62 m
Moment on Member = 1.173942 kN m
Trail section = 50 x 125
Z = 1.30E+05
m,a = 9.02E+00
c,a = 0.552521981
x = 320.0829892
y = 128.0331957
= 320.0829892
K12 = 0.03960829 For Medium
K3 = 1.25
c,g = 19.4
c,adm = 0.960501021
Check for very short term condition
K3 = 1.75
K12 = 0.028850779 320.083
c,adm = 0.979483934
For =
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c,a = 0.552521981
K7 = 1.10109106
m,g = 19.3
m,adm = 37.18935056
m,a = 9.02E+00
Euler Critical Stress
e = 2 N/mm2
Euler Co-efficient Keu = 0.984499194
Interaction Formula = 0.246248674 + 0.575244
= 0.821492187
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Lets check deflection
2470 mm length Span Continuation of Cantilever Portion of 1820 mm is considered
1 Geomatric Properties
= 1.82 m
= 0.4 m
Raft Dimension
= 50 mm= 125 mm
= b x h
= mm2
= 1/12 x b x h3
= mm4
2 Loading
= 0.6 kNm-2
= 0.5 kNm-2
= (DL+IL) x JS xLe
= 0.8008 kN
3 K-Factors
= 1
= 1 for longterm
= 1
= 1 for notch fro
= 1
= (300 mm /h)0.11
K7 = 1.10109106
= 1
4 Grade Stress
Strength Class
= 19.3 Nmm-2
= 4.3 Nmm-2
no wa
= 2.3 Nmm-2
Reference was made to Ex 5.15 on page No:249 Analysis of Structures Vol-1 By Prof V.N. Vazirani, Dr M.
Dr S.K.Duggal
It is illustrated that Bending Moment of Inclind member could be written as in the form of
member
Reference was made to Example 4.2 Design of Main Beam from from Structural Timber Design By Abdy Ke
BS 5268: Part 2, Tables 14
Kempas Grade HS
Bending parallel to grain m.g.//
Compression perpenidcular to grain c.g.pp
Shear parallel to grain g.//
Bearing : assume 50 mm, but located>75 mm
from the end of the member (K4, Table 18)
Notch end effect (K5,Clause 2.10.4)
Form factor K6 (K6, Clause 2.10.5)
Depth factor (K7 Clause 2.10.6)
No Load sharing(K8, Clause 2.9) K8
Effective span Le
Spacing of Rafters, JS
Breadth of Section, bDepth of section, h
Dead Load, DL
Imposed Load, IL
Cross-sectional area, A
Second moment of area Ixx
8.1E+06
6.25E+03
Total Load, W
Load Duration(K#, Table 17) K3
Service class 1 (K2, Table 16) K2
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= Nmm-2
5 Bending stress
= W.Le/2
= 0.728728 kN-m
= bh2/6
= mm3
= M/ Zprovided
= Nm-2
= 19.3 x K2.K3.K6.K
21.25105746 Nm-2
6 Lateral stability
= 2.5End held in position for h/b=3, No Leteral Support for h/b = 2
7 Shear stress
= W(N+L)2/2Le
N=1290 mm Cantilever portion N = 1820 mm
L= Span length following Cantilever 1750 mm Span L = 2470 mm
W = 0.44 kN/m =N/mm
= W(N+L)2/2Le
= 1639.231579 N
3 Fv
2 b .
= 0.393415579 Nmm-2
= 2.3 x K2.K3.K5.K
= 2.3
8 Bearing stress
= W(N+L)2/2Le
( Ref:Example 5.8 pages (233-237) and page 1317 of APPENDIX from Analysis of structure Vol-1 by ProfDr M.M. Ratwani,Dr. S.K.Duggal )
( Ref:Example 5.8 pages (233-237) and page 1317 of APPENDIX from Analysis of structure Vol-1 by Prof
Dr M.M. Ratwani,Dr. S.K.Duggal )
=
Section modulus Zprovided
Applied bending stress
BS 5268; Part 2, Clause 2.10.8, and Table 19
Maximum depth to breadth ratio, h/b
Applied shear force from shear force diagram
Applied shear force Fv
Applied bending moment
Mean modulus of elasticity, load sharing Emean
1.E+05
5.60
Applied Load from shear force diagram
Permissible shear stress, no notch
Applied shear force Fv
19100
Applied shear stress a
Permissible bending stress
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1639.231579 N
Assume bearing width, bw, of 100 mm either side
bw = 100
= Fv/(b .bw)
= 3.E-03
= 4.3 x K2.K3.K4.K
= 4.3
= Fv/(b . c.a.pp)
9 Deflection
w = (DL+IL) x JS kN/m
= 0.44
L = 2570.019 mm
= 2.570019 m
N = 1820
1.82 m
I = Ixx
= 8138020.833 x 1.E+05
n = N/L
= 0.708165971
= WLe3N
24E I
OR
( Ref: pages (232-233) along with figure 12.16 of Timber Designers' Manuel 3rd Ed by E.C.Ozalton )
= 7.55 mm
= (1.2/Gbd)wx.dx
Shear Modulus =Shear stress/Shear strain i.e. say G=(/)
Poissson's ratio ( ) = Lateral strain /Longitudinal strain
( Ref:Example 5.8 pages (233-237) and page 1317 of APPENDIX from Analysis of structure Vol-1 by Prof
Dr M.M. Ratwani,Dr. S.K.Duggal )
( Ref:page 451 of Analysis of structure Vol-1 by Prof V.N.Vaziani, Dr M.M. Ratwani,Dr. S.K.Duggal )
( Ref:page 272 of Strength of Materials for Civil Engineers 2nd Ed by T.H.G.Megson )
Permissible bearing stress c.a.pp
Minimum required bearing width
Applied bearing stress c.a.pp
Defelection due to Bending m
Defelection due to Shearing
Defelection due to Bending m
(3n
3
+4n
2
-1)
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E=2G(1+)
Also G=Emean/16
(19.2/Ebd)wx.dx
= 0.117206933
= 7.67 mm
OR
= Le/180
= 10.11 mm
Lets Check with Interaction Formula
UDL on Member = 0.44 kN/m
Axial Loading on Member
= 512.4862134 N
Horizontal Length of Member X2 = 2.47 m
Horizontal Length of Cantilever X1 = 1.82 m
Inclined Length of Member l = 2.570019
Inclined Length of Cantilever a = 1.887432
Moment on Member = 0.070098719 kN m
Trail section = 50 x 125
Z = 1.30E+05
m,a = 0.538358161
c,a = 0.081997794
( Ref:page 18: 2.5.7 Additional Properties of Structural Timber Design by ABDY KERMANI )
( Ref:Deflection Limit from page 68 of Structural Engineer's Pocket Book by FIONA COBB )
( Ref:4th line of 1st pargraph from page 232 of Timber Designers' Manuel 3rd Ed by E.C.Ozalton
( Ref:page 249-251 of Analysis of structure Vol-1 by Prof V.N.Vaziani, Dr M.M. Ratwani,Dr. S.K.Dug
Defelection due to Shear s
Total Defelection due to Bending plus Shear T
Permissible deflection for Cantilevers adm
=
= ( 19 .2) .( M )
( b . h ) . E
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k5=he/h
fig 2(b) of the code
ne
.Ratwani,
Horizontal
ramani
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.K8
.N.Vaziani,
.N.Vaziani,
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mm3
.N.Vaziani,
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al )
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O.k
130.7651
O.k
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UDL on Member = 0.44 kN/m
Axial Loading on Member
= 512.4862134 N
Horizontal Length of Member X2 = 2.47 m
Horizontal Length of Cantilever X1 = 1.82 m
Inclined Length of Member l = 2.570019
Inclined Length of Cantilever a = 1.887432
Moment on Member = 0.070098719 kN m
Trail section = 50 x 125
Z = 1.30E+05
m,a = 0.538358161
c,a = 0.081997794
x = 178.0561394
y = 71.22245575
= 178.0561394
K12 = 0.120349401 For Medium
K3 = 1.25
c,g = 19.4
c,adm = 2.918472963 > 0.081998
K3 = 1.75
K12 = 0.089337591 For =
Check for very short term condition
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c,adm = 3.033011216
c,a = 0.081997794
K7 = 1.10109106
m,g = 19.3
m,adm = 37.18935056
m,a = 5.38E-01
Euler Critical Stress
e = 5 N/mm2
Euler Co-efficient Keu = 0.997795692
Interaction Formula = 0.014508118 + 0.028096
= 0.042604248 < 1
Lets check deflection
It follows : Example 4.2 Design of Main Beam from Structural Timber Design By Abdy Keramani
2470 mm length Span Continuation of Cantilever Portion
1 Geomatric Properties
= 2.47 m
= 0.4 m
Raft Dimension
= 50 mm
= 125 mm
= b x h
= 6.25 x 1.E+03
= 1/12 x b x h3
= mm4
2 Loading
= 0.6 kNm-2
= 0.5 kNm-2
= (DL+IL) x JS xLe
= 1.0868 kN
3 K-Factors
= 1
8.1E+06
Effective span
Breadth of Section, b
Depth of section, h
Cross-sectional area, A
Spacing of Rafters, JS
Second moment of area Ixx
Dead Load, DL
Imposed Load, IL
Total Load, W
Service class 1 (K2, Table 16) K2
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= 1 for longterm
Bearing : assume 50 mm, but located>75 mm
from the end of the member (K4, Table 18) = 1
= 1 for no notch
= 1
= (300 mm /h)0.11
K7 = 1.10109106
= 1.1
4 Grade Stress
BS 5268: Part 2, Tables 14
Kempas Grade HS Strength Class
= 19.3 Nmm-2
= 4.3 Nmm-2
no wa
= 2.3 Nmm-2
= Nmm-2
5 Bending stress( Ref:Example 5.8 pages (233-237) and page 1317 of APPENDIX from Analysis of structure Vol-1 by
= W x (L2-a
2)2
8.L2
L = 2470 mm
= 2.47 m
a = 1820 mm
1.82 m
= W x (L2-a
2)2
8.L2
= 0.070098719 kN-m
= bh2/6
= mm3
= M/ Zprovided
= Nm-2
= 19.3 x K2.K3.K6.K
23.37616321 Nm-2
6 Lateral stability
= 2.5
End held in position and compression edge held in line as by direct connection of sheating,
19100
1.30E+05
5.38.E-01
Load Duration(K#, Table 17) K3
Notch end effect (K5,Clause 2.10.4)
Form factor K6 (K6, Clause 2.10.5)
Depth factor (K7 Clause 2.10.6)
Load sharing appies with Purlin(K8, Clause 2.9)K8
Bending parallel to grain m.g.//
Compression perpenidcular to grain c.g.pp
Shear parallel to grain g.//
Mean modulus of elasticity, load sharing Emean
Applied bending moment
Applied bending moment
Section modulus Zprovided
Applied bending stress
Permissible bending stress
BS 5268; Part 2, Clause 2.10.8, and Table 19
Maximum depth to breadth ratio, h/b
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= 0.01 mm
= -0.11 mm
( Ref:Deflection Limit from page 68 of Structural Engineer's Pocket Book by FIONA COBB )
= Le/330 Or 14 mm
= 7.484848485 mm
= ( 19 .2) .( M )
( b . h ) . EDefelection due toShear s
Total Defelection due toShear plus Bending T
Permissible deflection adm
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O.k
178.0561
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O.k
mm2
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ne
rof V.N.Vaziani, Dr M.M. Ratwani,Dr. S.K.Duggal )
.K8
deck or joist
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rof V.N.Vaziani, Dr M.M. Ratwani,Dr. S.K.Duggal )
2.47 m
1.82 m
0.564474 m
0.44 kN/m=N/mm
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Lets check deflection
It follows : Example 4.2 Design of Main Beam from Structural Timber Design By Abdy Keramani
2470 mm length Span Continuation of Cantilever Portion
1 Geomatric Properties
= 4.81 m= 0.4 m
Raft Dimension
= 50 mm
= 150 mm
= b x h
= 7.5 x 1.E+03 mm2
= 1/12 x b x h3
= mm4
2 Loading
= 0.6 kNm-2
= 0.5 kNm-2
= (DL+IL) x JS xLe
= 2.1164 kN
3 K-Factors
= 1
= 1 for longterm
Bearing : assume 50 mm, but located>75 mm
from the end of the member (K4, Table 18) = 1
= 1 for no notch
= 1
= (300 mm /h)0.11
K7 = 1.079228237
= 1.1
4 Grade Stress
BS 5268: Part 2, Tables 14
Kempas Grade HS Strength Class
= 19.3 Nmm-2
= 4.3 Nmm-2
no wane
= 2.3 Nmm-2
= Nmm-2
5 Bending stress
( Ref:Example 5.8 pages (233-237) and page 1317 of APPENDIX from Analysis of structure Vol-1 by Prof V.N.V
19100
Depth factor (K7 Clause 2.10.6)
Load sharing appies with Purlin(K8, Clause 2.9)K8
Bending parallel to grain m.g.//
Compression perpenidcular to grain c.g.pp
Shear parallel to grain g.//
Mean modulus of elasticity, load sharing Emean
Imposed Load, IL
Total Load, W
Service class 1 (K2, Table 16) K2
Load Duration(K#, Table 17) K3
Notch end effect (K5,Clause 2.10.4)
Form factor K6 (K6, Clause 2.10.5)
Effective spanSpacing of Rafters, JS
Breadth of Section, b
Depth of section, h
Cross-sectional area, A
Second moment of area Ixx
1.4E+07
Dead Load, DL
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= W x (L2-a
2)
2
8.L2
L = 4810 mm
= 4.81 m
a = 0 mm
0 m
= W x (L2-a
2)
2
8.L2
= 1.2724855 kN-m
= bh2/6
= mm3
= M/ Zprovided
= Nm-2
= 19.3 x K2.K3.K6.K7.K8
22.91201546 Nm-2
6 Lateral stability
= 3
End held in position and compression edge held in line as by direct connection of sheating, deck or joi
7 Shear stress
= W(N+L)2/2Le kN
W = 0.44 kN/m =N/mm
= W(N+L)2/2Le
= 5212.730769 N
= 3 Fv
2 b .
= 1.042546154
= 2.3 x K2.K3.K5.K8
L=9'-9" Span following Cantilever
Applied shear force from shear force diagram
Applied shear force
Applied shear stress a
Applied shear stress
Permissible shear stress, no notch
6.79.E+00
Permissible bending stress
BS 5268; Part 2, Clause 2.10.8, and Table 19
Maximum depth to breadth ratio, h/b
Applied shear force from shear force diagram
N=6'-0" Cantilever portion
Applied bending moment
Applied bending moment
Section modulus Zprovided
1.88E+05
Applied bending stress
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= 2.53
8 Bearing stress
= 5.213 kN
Assume bearing width, bw, of 50 mm either side
bw = 100
= Fv/(b .bw)
= 1.042546154
= 4.3 x K2.K3.K4.K8
= 4.73
=
9 Deflection
= Emean
( Ref:Example 5.8 pages (233-237) and page 1317 of APPENDIX from Analysis of structure Vol-1 by Prof V.N.V
l = 4.81
a= 0
= (l2-a
2)/2l x= 2.405
= 0.5 w= 0.44
= 0
= wl4(m
4-2m
3(1-n
2)+m(1-2n
2))
24EI
= 11.42 mm
= 0.17 mm
= 11.59 mm
( Ref:Deflection Limit from page 68 of Structural Engineer's Pocket Book by FIONA COBB )
= Le/330 Or 14 mm
= 14.57575758 mm
( b . h ) . E
Total Defelection due toShear plus Bending T
Permissible deflection adm
x
m=x/l
n=a/l
m
Defelection due toShear s =
Permissible shearing stress
Minimum required bearing width
Load Sharing system E
Max Defelection due to Bending at x=(l2-a2)/2l m =
( 19 .2) .( M )
Applied Load from shear force
Applied bearing stress c.a.pp
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Lets Check with Interaction FormulaUDL on Member = 0.44 kN/m
= 0.55756767 N
= 4.81 m
= 0 m
= 5.0013
= 0
= 1.2724855 kN m
= 50 x 150
Z = 1.88E+05
m,a = 6.786589333
c,a = 7.43424E-05
x = 346.5002282
y = 115.5000761
= 346.5002282
K12 = 0.0339821 For Medium
K3 = 1.25
c,g = 19.4
c,adm = 0.824065919 > 7.43E-05 O.k
Trail section
Axial Loading on Member
Horizontal Length of Member X2
Horizontal Length of Cantilever X1
Inclined Length of Member l
Inclined Length of Cantilever a
Moment on Member
Check for very short term condition
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K3 = 1.75
K12 = 0.089337591 346.5002
c,adm = 3.033011216
c,a = 7.43424E-05
K7 = 1.079228237
m,g = 19.3
m,adm = 36.45093369
m,a = 6.79E+00
Euler Critical Stress
e = 1 N/mm2
Euler Co-efficient Keu = 0.999992432
Interaction Formula = 0.186185648 + 9.02E-05
= 0.186275862 < 1 O.k
For =
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ziani, Dr M.M. Ratwani,Dr. S.K.Duggal )
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t
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ziani, Dr M.M. Ratwani,Dr. S.K.Duggal )
m
m
m
kN/m=N/mm
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Lets check deflection
It follows : Example 4.2 Design of Main Beam from Structural Timber Design By Abdy Keramani
2470 mm length Span Continuation of Cantilever Portion
1 Geomatric Properties
= 2.31 m= 0.4 m
Raft Dimension
= 50 mm
= 150 mm
= b x h
= 7.5 x 1.E+03 mm2
= 1/12 x b x h3
= mm4
2 Loading
= 0.6 kNm-2
= 0.5 kNm-2
= (DL+IL) x JS xLe
= 1.0164 kN
3 K-Factors
= 1
= 1 for longterm
Bearing : assume 50 mm, but located>75 mm
from the end of the member (K4, Table 18) = 1
= 1 for no notch
= 1
= (300 mm /h)0.11
K7 = 1.079228237
= 1.1
4 Grade Stress
BS 5268: Part 2, Tables 14
Kempas Grade HS Strength Class
= 19.3 Nmm-2
= 4.3 Nmm-2
no wane
= 2.3 Nmm-2
= Nmm-2
5 Bending stress
( Ref:Example 5.8 pages (233-237) and page 1317 of APPENDIX from Analysis of structure Vol-1 by Prof V.N.V
Mean modulus of elasticity, load sharing Emean 19100
Form factor K6 (K6, Clause 2.10.5)
Depth factor (K7 Clause 2.10.6)
Load sharing appies with Purlin(K8, Clause 2.9)K8
Bending parallel to grain m.g.//
Compression perpenidcular to grain c.g.pp
Shear parallel to grain g.//
Dead Load, DL
Imposed Load, IL
Total Load, W
Service class 1 (K2, Table 16) K2
Load Duration(K#, Table 17) K3
Notch end effect (K5,Clause 2.10.4)
Effective spanSpacing of Rafters, JS
Breadth of Section, b
Depth of section, h
Cross-sectional area, A
Second moment of area Ixx
1.4E+07
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= W x (L2-a
2)
2
8.L2
L = 2310 mm
= 2.31 m
a = 0 mm
0 m
= W x (L2-a
2)
2
8.L2
= 0.2934855 kN-m
= bh2/6
= mm3
= M/ Zprovided
= Nm-2
= 19.3 x K2.K3.K6.K7.K8
22.91201546 Nm-2
6 Lateral stability
= 3
End held in position and compression edge held in line as by direct connection of sheating, deck or joi
7 Shear stress
= W(N+L)2/2Le kN
W = 0.44 kN/m =N/mm
= W(N+L)2/2Le
= 5212.730769 N
= 3 Fv
2 b .
= 1.042546154
= 2.3 x K2.K3.K5.K8
Applied shear stress
Permissible shear stress, no notch
N=6'-0" Cantilever portion
L=9'-9" Span following Cantilever
Applied shear force from shear force diagram
Applied shear force
Applied shear stress a
Applied bending stress
1.57.E+00
Permissible bending stress
BS 5268; Part 2, Clause 2.10.8, and Table 19
Maximum depth to breadth ratio, h/b
Applied shear force from shear force diagram
Applied bending moment
Applied bending moment
Section modulus Zprovided
1.88E+05
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= 2.53
8 Bearing stress
= 5.213 kN
Assume bearing width, bw, of 50 mm either side
bw = 100
= Fv/(b .bw)
= 1.042546154
= 4.3 x K2.K3.K4.K8
= 4.73
=
9 Deflection
= Emean
( Ref:Example 5.8 pages (233-237) and page 1317 of APPENDIX from Analysis of structure Vol-1 by Prof V.N.V
l = 2.31
a= 0
= (l2-a
2)/2l x= 1.155
= 0.5 w= 0.44
= 0
= wl4(m
4-2m
3(1-n
2)+m(1-2n
2))
24EI
= 0.61 mm
= 0.04 mm
= 0.65 mm
( Ref:Deflection Limit from page 68 of Structural Engineer's Pocket Book by FIONA COBB )
= Le/330 Or 14 mm
= 7 mm
( 19 .2) .( M )
( b . h ) . E
Total Defelection due toShear plus Bending T
n=a/l
Permissible deflection adm
m
Defelection due toShear s =
Load Sharing system E
Max Defelection due to Bending at x=(l2-a2)/2l m =
x
m=x/l
Applied Load from shear force
Applied bearing stress c.a.pp
Permissible shearing stress
Minimum required bearing width
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Lets Check with Interaction FormulaUDL on Member = 0.44 kN/m
= 0.267771584 N
= 2.31 m
= 0 m
= 2.399708
= 0
= 0.2934855 kN m
= 50 x 150
Z = 1.88E+05
m,a = 1.565256000
c,a
= 3.57029E-05
x = 166.2566472
y = 55.41888239
= 166.2566472
K12 = 0.136524546 For Medium
K3 = 1.25
c,g = 19.4
c,adm = 3.310720248 > 3.57E-05 O.k
Trail section
Axial Loading on Member
Horizontal Length of Member X2
Horizontal Length of Cantilever X1
Inclined Length of Member l
Inclined Length of Cantilever a
Moment on Member
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K3 = 1.75
K12 = 0.101707005 166.2566
c,adm = 3.452952825
c,a = 3.57029E-05
K7 = 1.079228237
m,g = 19.3
m,adm = 36.45093369
m,a = 1.57E+00
Euler Critical Stress
e = 6 N/mm2
Euler Co-efficient Keu = 0.999999047
Interaction Formula = 0.042941492 + 1.08E-05
= 0.042952276 < 1 O.k
Check for very short term condition
For =
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t
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ziani, Dr M.M. Ratwani,Dr. S.K.Duggal )
m
m
m
kN/m=N/mm
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Lets check deflection
It follows : Example 4.2 Design of Main Beam from Structural Timber Design By Abdy Keramani
6050 mm length Span Continuation of Cantilever Portion = 0.00 mm
1 Geomatric Properties
= 6.05 m
= 0.4 m
Raft Dimension
= 50 mm
= 200 mm
= b x h
= 10 x 1.E+03
= 1/12 x b x h3
= mm4
2 Loading
= 0.6 kNm-2
= 0.5 kNm-2
= (DL+IL) x JS xLe
= 2.662 kN
3 K-Factors
= 1
= 1 for longterm
Bearing : assume 50 mm, but located>75 mm
from the end of the member (K4, Table 18) = 1
= 1 for no notch
= 1
= (300 mm /h)0.11
K7 = 1.045610747
= 1.1
4 Grade Stress
BS 5268: Part 2, Tables 14
Kempas Grade HS Strength Class
= 19.3 Nmm-2
= 4.3 Nmm-2
no wa
= 2.3 Nmm-2
= Nmm-2
5 Bending stress
( Ref:Example 5.8 pages (233-237) and page 1317 of APPENDIX from Analysis of structure Vol-1 by
= W x (L2-a
2)2
8.L2
L = 6050 mm
Applied bending moment
Load sharing appies with Purlin(K8, Clause 2.9)K8
Bending parallel to grain m.g.//
Compression perpenidcular to grain c.g.pp
Shear parallel to grain g.//
Mean modulus of elasticity, load sharing Emean 19100
Total Load, W
Service class 1 (K2, Table 16) K2
Load Duration(K#, Table 17) K3
Notch end effect (K5,Clause 2.10.4)
Form factor K6 (K6, Clause 2.10.5)
Depth factor (K7 Clause 2.10.6)
Depth of section, h
Cross-sectional area, A
Second moment of area Ixx
3.3E+07
Dead Load, DL
Imposed Load, IL
Effective span
Spacing of Rafters, JS
Breadth of Section, b
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= 6.05 m
a = 0 mm
0 m
= W x (L2-a
2)2
8.L2
= 2.0131375 kN-m
= bh2/6
= mm3
= M/ Zprovided
= Nm-2
= 19.3 x K2.K3.K6.K
22.19831617 Nm-2
6 Lateral stability
= 4
End held in position and compression edge held in line as by direct connection of sheating,
7 Shear stress
= W(N+L)2/2Le kN
W = 0.44 kN/m =N/mm
= W(N+L)2/2Le
= 5212.730769 N
= 3 Fv
2 b .
= 0.781909615
= 2.3 x K2.K3.K5.K
= 2.53
8 Bearing stress
= 5.213 kNApplied Load from shear force
Applied shear force from shear force diagram
Applied shear force
Applied shear stress a
Applied shear stress
Permissible shear stress, no notch
Permissible bending stress
BS 5268; Part 2, Clause 2.10.8, and Table 19
Maximum depth to breadth ratio, h/b
Applied shear force from shear force diagram
N=6'-0" Cantilever portion
L=9'-9" Span following Cantilever
Applied bending moment
Section modulus Zprovided
3.33E+05
Applied bending stress
6.04.E+00
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Assume bearing width, bw, of 50 mm either side
bw = 100
= Fv/(b .bw)
= 1.042546154
= 4.3 x K2.K3.K4.K
= 4.73
=
9 Deflection
= Emean
( Ref:Example 5.8 pages (233-237) and page 1317 of APPENDIX from Analysis of structure Vol-1 by
l =
a=
= (l2-a
2)/2l x=
= 0.5 w=
= 0
= wl4(m
4-2m
3(1-n
2)+m(1-2n
2))
24EI
= 12.06 mm
= 0.20 mm
= 12.26 mm
( Ref:Deflection Limit from page 68 of Structural Engineer's Pocket Book by FIONA COBB )
= Le/330 Or 14 mm
= 18.33333333 mm
Lets Check with Interaction FormulaUDL on Member = 0.44 kN/m
= 0.701306529 N
Permissible deflection adm
m
Defelection due toShear s = ( 19 .2) .( M )
( b . h ) . E
Total Defelection due toShear plus Bending T
Max Defelection due to Bending at x=(l2-a2)/2l m =
x
m=x/l
n=a/l
Applied bearing stress c.a.pp
Permissible shearing stress
Minimum required bearing width
Load Sharing system E
Axial Loading on Member
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= 6.05 m
= 0 m
= 6.292487584
= 0
= 2.0131375 kN m
= 50 x 200
Z = 3.33E+05
m,a = 6.039412500
c,a = 7.01307E-05
x = 435.9563281
y = 108.989082
= 435.9563281
K12 = 0.021752194 For Medium
K3 = 1.25
c,g = 19.4
c,adm = 0.527490711 > 7.01E-05
K3 = 1.75
K12 = 0.015758838
c,adm = 0.535012561
c,a = 7.01307E-05
K7 = 1.045610747
Trail section
Check for very short term condition
For =
Horizontal Length of Member X2
Horizontal Length of Cantilever X1
Inclined Length of Member l
Inclined Length of Cantilever a
Moment on Member
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m,g = 19.3
m,adm = 35.31550299
m,a = 6.04E+00
Euler Critical Stress e = 1 N/mm2
= 0.999998006
= 0.171013409 + 0.000133
= 0.17114636 < 1
Euler Co-efficient Keu
Interaction Formula
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mm2
ne
rof V.N.Vaziani, Dr M.M. Ratwani,Dr. S.K.Duggal )
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.K8
deck or joist
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rof V.N.Vaziani, Dr M.M. Ratwani,Dr. S.K.Duggal )
6.05 m
0 m
3.025 m
0.44 kN/m=N/mm
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O.k
435.9563
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O.k