6 k 6k 6k 4 k - C E 10 ft B AR - 5 ft--5 ft-– 5 ft- 5 ft- EA = constant E= 29,000 ksi A = 6 in.?
6 k 6k 6k 4 k - C E 10 ft B AR - 5 ft--5 ft-– 5 ft- 5 ft- EA = constant E= 29,000 ksi A = 6 in.?
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
Use Castigliano’s second theorem to determine the horizontal deflection at joint E of the truss shown in Fig. P7.8.

Transcribed Image Text:6 k
6k
6k
4 k
-
C
E
10 ft
B
AR
- 5 ft--5 ft-– 5 ft- 5 ft-
EA = constant
E= 29,000 ksi
A = 6 in.?
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