As shown in the Figure, a rigid beam with negligible weight is pinned at one end and attached to two vertical rods. The beam was initially horizontal before the load W = 35.11 kips was applied. For the steel rod: A = 0.5 in^2, L = 90 in. For the bronze rod: A = 2 in^2, L = 30 in. Determine the following:

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
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PROBLEM 2:
As shown in the Figure, a rigid beam with negligible weight is pinned at one end and attached to two
vertical rods. The beam was initially horizontal before the load W = 35.11 kips was applied. For the steel
rod: A = 0.5 in^2, L = 90 in. For the bronze rod: A = 2 in^2, L = 30 in. Determine the following:
1. Force in the steel rod
2. Force in the bronze rod
3. Stress in the steel rod
4. Stress in the bronze rod
5. Vertical movement of W, OW
A
Bronze
E = 12 x 106 psi
←3 ft →
7. Vertical movement at the steel rod, ost
6. Vertical movement at the bronze rod, obr
8 ft
8. Vertical reaction at the pinned end (include sign)
9. Shearing stress in the pin at A assuming the
diameter of the pin is 1.5 inches (assume double shear)
E = 29 x 105 psi
>
Steel
W
4 ft
lbs
lbs
psi
psi
in
in
in
lbs
psi
Transcribed Image Text:5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 PROBLEM 2: As shown in the Figure, a rigid beam with negligible weight is pinned at one end and attached to two vertical rods. The beam was initially horizontal before the load W = 35.11 kips was applied. For the steel rod: A = 0.5 in^2, L = 90 in. For the bronze rod: A = 2 in^2, L = 30 in. Determine the following: 1. Force in the steel rod 2. Force in the bronze rod 3. Stress in the steel rod 4. Stress in the bronze rod 5. Vertical movement of W, OW A Bronze E = 12 x 106 psi ←3 ft → 7. Vertical movement at the steel rod, ost 6. Vertical movement at the bronze rod, obr 8 ft 8. Vertical reaction at the pinned end (include sign) 9. Shearing stress in the pin at A assuming the diameter of the pin is 1.5 inches (assume double shear) E = 29 x 105 psi > Steel W 4 ft lbs lbs psi psi in in in lbs psi
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