Advanced Methods of Structural Analysis

566 Appendix

Example. For beam shown below to find the bending moments at supportsB
andC.

B

lll

D

A C

q l/^3 l/^3

l/ 2 P^1 P^2 P^2

MB = −0.117ql^2 − 0.075P 1 l + 0.044P 2 l

MC = −0.033ql^2 − 0.075P 1 l + 0.178P 2 l

Two span beams with different spans:

The bending moment at supportBand
maximum bending moment at the first
and second spans may be calculated by
formulaM D kql^2. Parameterskis
presented in TableA.12. A l^1 B l^2 C

q 1 q 2

Ta b l e A. 1 2 Bending moments due to uniformly distributed load

l 1 :l 2

Loadq 1 is applied in

the first span only

Loadq 2 is applied in

the second span only

Loadq 1 Dq 2 Dqis applied
both in the first and second
span
MB M 1 max MB M 2 max MB M 1 max M 2 max
1.0 0:063 0.095 0:063 0.095 0:125 0.070 0.070
1.1 0:079 0.114 0:060 0.096 0:139 0.090 0.065
1.2 0:098 0.134 0:057 0.097 0:153 0.111 0.059
1.3 0:119 0.155 0:054 0.098 0:174 0.133 0.053
1.4 0:143 0.178 0:052 0.099 0:195 0.157 0.047
1.5 0:169 0.203 0:050 0.100 0:219 0.183 0.040
1.6 0:197 0.228 0:048 0.101 0:245 0.209 0.033
1.7 0:227 0.256 0:046 0.102 0:274 0.237 0.026
1.8 0:260 0.285 0:045 0.103 0:305 0.267 0.019
1.9 0:296 0.315 0:043 0.103 0:339 0.298 0.013
2.0 0:333 0.347 0:042 0.104 0:375 0.330 0.008
2.2 0:416 0.415 0:039 0.106 0:455 0.398 0.001
2.4 0:508 0.488 0:037 0.107 0:545 0.473 a
2.6 0:610 0.570 0:035 0.108 0:645 0.553 a
2.8 0:722 0.655 0:033 0.109 0:755 0.639 a
3.0 0:844 0.743 0:031 0.110 0:875 0.730 a

Factor q 1 l 22 q 1 l 22 q 2 l^22 q 2 l 22 ql^22 ql^22 ql 22
aWithin the second span the bending moments are negative