2 k/ft E 1= 1,600 in.4 D 1= 1,600 in.- 10 ft + 40 k- 10 ft A B 30 ft -30 ft- E = 29,000 ksi (a) Frame 1%3D800 in.4 / = 800 in.4
2 k/ft E 1= 1,600 in.4 D 1= 1,600 in.- 10 ft + 40 k- 10 ft A B 30 ft -30 ft- E = 29,000 ksi (a) Frame 1%3D800 in.4 / = 800 in.4
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
Determine the member end moments for the frame shown in Fig. 16.13(a) by using the moment-distribution method.
Transcribed Image Text:2 k/ft
E
1= 1,600 in.4 D
1= 1,600 in.-
10 ft
+ 40 k-
10 ft
A
B
30 ft
-30 ft-
E = 29,000 ksi
(a) Frame
1%3D800 in.4
/ = 800 in.4
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