May I ask why the horizontal component of CD was neglected in the calculation of forces in the last part

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
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May I ask why the horizontal component of CD was neglected in the calculation of forces in the last part?

EXPLAIN
Additional examples are given in this section. For more explanations on trusses and frames
read Chapter 4 of your textbook.
1. Determine the force in each member of
the crane truss shown.
Solution:
6"
For the given crane truss, the member
forces can be determined without solving
for the support reactions.
B
9'
12'
30°
We can start our solution at joint A, since
there are only two unknowns at joint A.
5200 lb
Solve for the angle between member AC and AB:
Cosine law:
x = V(6)² + (9)² – (2)(6)(9) cos 120°
x = 13.08 ft
Sine law:
X.
sin a
sin 120°
1200
13.08
a = 23.41°
B
9'
Transcribed Image Text:EXPLAIN Additional examples are given in this section. For more explanations on trusses and frames read Chapter 4 of your textbook. 1. Determine the force in each member of the crane truss shown. Solution: 6" For the given crane truss, the member forces can be determined without solving for the support reactions. B 9' 12' 30° We can start our solution at joint A, since there are only two unknowns at joint A. 5200 lb Solve for the angle between member AC and AB: Cosine law: x = V(6)² + (9)² – (2)(6)(9) cos 120° x = 13.08 ft Sine law: X. sin a sin 120° 1200 13.08 a = 23.41° B 9'
ule4.pdf
AD IZUTU.74 IN
BD
+K Ey = 0
BC +12010.74 sin 30° = 0
%3D
%3D
BC = -6005.37 lb
BC = 6005.37 lb (C)
FBD of joint C:
+1 EV = 0
-CD sin 56.57° + 6005.37 cos 30° - 13088.08 sin 23.41° = 0
CE
%3D
56.57°
23.41°
CD = 0.96 lb 0
30°
AC=13088.08 lb
-6005.37 sin 30° + 13088.08 cos 23.41° – CE = 0
CE = 9008.05 lb (T)
%3D
BC=6005.37 ib
CD
117
Transcribed Image Text:ule4.pdf AD IZUTU.74 IN BD +K Ey = 0 BC +12010.74 sin 30° = 0 %3D %3D BC = -6005.37 lb BC = 6005.37 lb (C) FBD of joint C: +1 EV = 0 -CD sin 56.57° + 6005.37 cos 30° - 13088.08 sin 23.41° = 0 CE %3D 56.57° 23.41° CD = 0.96 lb 0 30° AC=13088.08 lb -6005.37 sin 30° + 13088.08 cos 23.41° – CE = 0 CE = 9008.05 lb (T) %3D BC=6005.37 ib CD 117
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