I can't seem to find the shear stress at point b. Here's my work for reference.

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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I can't seem to find the shear stress at point b. Here's my work for reference.

54
Homework 12
1. F= P. CL2.L3) = 13.5 (9.3f+- 14.4ft) = -1807,92 lb
Normal Sheor stress Cequal to all points)
▷do-di - €
A
D
K
K
D
K
K
e
Torsional shear stress Cequal to all Points).
▷T = F • ²2/2/2/2 = 1807.92-16.
13,017.024 1boft
J = 3/2 (9.5₁4-8.88 in") = 189.2 in 1
DC torsional = 13,017.024 16.12:0
31189.2 in CIO.
a. sin
D
Normal
D
Tnorman
Corsional
Shed
B
|CB| =
AJ
2318
1807.9216
(9.52-8,882)
ос - МС
→ do-di= 2€ di = do-26=9.5-2(0.31) = 8.88 in
D
Tc = √√√ 202² + 3921.882
T
1.4
= 202 psisi
14.4FF
1354
e
T20
PED
202² + 3921.882 = 3927.1 psi
202-3921.88 = 3719.88 psi
2:1807.92 (32 +93). 122. (9.5
द4 (9.54-8.88५)
2
20273921.88 psp 301
27
3
TERUS
-39926.9 psi
A
24
Transcribed Image Text:54 Homework 12 1. F= P. CL2.L3) = 13.5 (9.3f+- 14.4ft) = -1807,92 lb Normal Sheor stress Cequal to all points) ▷do-di - € A D K K D K K e Torsional shear stress Cequal to all Points). ▷T = F • ²2/2/2/2 = 1807.92-16. 13,017.024 1boft J = 3/2 (9.5₁4-8.88 in") = 189.2 in 1 DC torsional = 13,017.024 16.12:0 31189.2 in CIO. a. sin D Normal D Tnorman Corsional Shed B |CB| = AJ 2318 1807.9216 (9.52-8,882) ос - МС → do-di= 2€ di = do-26=9.5-2(0.31) = 8.88 in D Tc = √√√ 202² + 3921.882 T 1.4 = 202 psisi 14.4FF 1354 e T20 PED 202² + 3921.882 = 3927.1 psi 202-3921.88 = 3719.88 psi 2:1807.92 (32 +93). 122. (9.5 द4 (9.54-8.88५) 2 20273921.88 psp 301 27 3 TERUS -39926.9 psi A 24
An advertising sign located along a city street is cantilevered from a pole as shown below. The pressure due to
the wind impinging on the sign is uniformly distributed over its area and equal to p = 13.5 lb/ft² in the x-
direction. The pole is fabricated from a tube with an outside diameter of do= 9.5 in. and a wall thickness of
t = 0.31 in. Determine the stresses at points Cand Blocated at the base of the pole.
Given:
.
• L₁=32 ft
L₂=9.3 ft
11
L3
go=
pr
|TC=
14.4 ft
3.927
OB-0
L2
L₁
X
Z
L3 →
y
D
A
B
✓ 100%
ksi @ 100%
- 100%
XOX
Transcribed Image Text:An advertising sign located along a city street is cantilevered from a pole as shown below. The pressure due to the wind impinging on the sign is uniformly distributed over its area and equal to p = 13.5 lb/ft² in the x- direction. The pole is fabricated from a tube with an outside diameter of do= 9.5 in. and a wall thickness of t = 0.31 in. Determine the stresses at points Cand Blocated at the base of the pole. Given: . • L₁=32 ft L₂=9.3 ft 11 L3 go= pr |TC= 14.4 ft 3.927 OB-0 L2 L₁ X Z L3 → y D A B ✓ 100% ksi @ 100% - 100% XOX
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