Design Procedure D y = 100 lb/ft³ ka = 0.3 kp = 3.0 Pp Z₁ Zp 4 2 Za Z Ap Solve for D Pa 1 2 3 4 0 Tendon / Rod Sheet pile (100×10×0.3)x10x0.5 = 1500 Anchor plate (100×10×0.3)xD = 300xD D lb 2 ft D lb 2 ft (100xDx0.3)x (100XDX3.0)x ΣMA=0 →1500 [10x + 300D [10 + 2] + 30D² [10 +D] + 300² [10+D] = 0 lb ft lb
Design Procedure D y = 100 lb/ft³ ka = 0.3 kp = 3.0 Pp Z₁ Zp 4 2 Za Z Ap Solve for D Pa 1 2 3 4 0 Tendon / Rod Sheet pile (100×10×0.3)x10x0.5 = 1500 Anchor plate (100×10×0.3)xD = 300xD D lb 2 ft D lb 2 ft (100xDx0.3)x (100XDX3.0)x ΣMA=0 →1500 [10x + 300D [10 + 2] + 30D² [10 +D] + 300² [10+D] = 0 lb ft lb
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![Design Procedure
D
Y = 100 lb/ft³
ka = 0.3
kp
= 3.0
Pp Z₁
Zp
4
2
Za
Z
Ap
Solve for D
Pa
1
2
3
4
0
Tendon / Rod
Sheet pile
(100×10×0.3)x10x0.5 = 1500
lb
ft
Anchor plate
(100×10×0.3)xD = 300xD
D lb
2 ft
D lb
2 ft
(100xDx0.3)x
(100XDX3.0)×
ΣM₁ = 0 →1500 [10x + 300D [10 + 2] + 30D² [10 +D] + 300² [10+D] = 0
lb
t](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4ba0e982-a700-4bff-b357-a163e1463777%2F8e68931f-d94c-431f-9fb5-51a5ef98e865%2Frxzwsms_processed.png&w=3840&q=75)
Transcribed Image Text:Design Procedure
D
Y = 100 lb/ft³
ka = 0.3
kp
= 3.0
Pp Z₁
Zp
4
2
Za
Z
Ap
Solve for D
Pa
1
2
3
4
0
Tendon / Rod
Sheet pile
(100×10×0.3)x10x0.5 = 1500
lb
ft
Anchor plate
(100×10×0.3)xD = 300xD
D lb
2 ft
D lb
2 ft
(100xDx0.3)x
(100XDX3.0)×
ΣM₁ = 0 →1500 [10x + 300D [10 + 2] + 30D² [10 +D] + 300² [10+D] = 0
lb
t
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