287plf 335plf page 4 university of california

Page 4 UNIVERSITY OF CALIFORNIA, BERKELEY
Department of Civil Engineering
CE 120 - Instructor: J. P. Moehle - Spring 201 5
To solve for Lc: 1)Set |+Mmax|=|-Mmax| or 2)Iterate.
1) Setting |+Mmax|=|-Mmax:
+Mmax and –Mmax varies inversely with change of Lc. Therefore, minimum value for design moment
will be obtained when |+Mmax|=|-Mmax|.
w/8*(50-Lc) 2 = 25wLc
(50-Lc) 2 = 200Lc (Note: does not depend on w)
Lc=8.58ft.
2) Alternatively Lc can be found by Iteration:
Lc (ft) Mb (k-ft) Mc (k-ft) Me(k-ft)
Maximum Absolute Design
Moment (k-ft)
0.00 233.44 0.00 233.44
233.44
1.00 224.19 -18.68 224.19
224.19
2.00 215.14 -37.35 215.14
215.14
3.00 206.27 -56.03 206.27
206.27
4.00 197.58 -74.70 197.58
197.58
5.00 189.08 -93.38 189.08
189.08
6.00 180.77 -112.05 180.77
180.77
7.00 172.65 -130.73 172.65
172.65
8.00 164.71 -149.40 164.71
164.71
9.00 156.96 -168.08 156.96
168.08
8.10 163.93 -151.27 163.93
163.93
8.20 163.15 -153.14 163.15
163.15
8.30 162.37 -155.00 162.37
162.37
8.40 161.59 -156.87 161.59
161.59
8.50 160.82 -158.74 160.82
160.82
8.60 160.04 -160.61 160.04
160.61
8.70 159.27 -162.47 159.27
162.47
Lc= 8.6ft gives minimum design moment of 160.6k-ft.
[158.1k-ft]

UNIVERSITY OF CALIFORNIA, BERKELEY Page 5
Department of Civil Engineering
CE 120 - Instructor: J. P. Moehle - Spring 2013