A gas pressure difference is applied to the legs of a U-tube manometer filled with a liquid with specific gravity of 1.7. The ma
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The amount of settlement that would occur at the end of 1.5 year and 5 year are 7.3 cm and 13.14 cm respectively.
<h3>How to determine the amount of settlement?</h3>For a layer of 3.8 m thickness, we were given the following parameters:
U = 50% = 0.5.
Sc = 7.3 cm.
For Sf, we have:
Sf = Sc/U
Sf = 7.3/0.5
Sf = 14.6
Therefore, Sf for a layer of 38 m thickness is given by:
Sf = 14.6 × 38/3.8
Sf = 146 cm.
At 50%, the time for a layer of 3.8 m thickness is: = 1.5 year.
At 50%, the time for a layer of 38 m thickness is:
= 1.5 × (38/3.8)²
= 150 years.
For the thickness of 38 m, U₂ is given by:
The new settlement after 1.5 year is:
Sc = U₂Sf
Sc = 0.05 × 146
Sc = 7.3 cm.
For time, t₂ = 5 year:
The new settlement after 5 year is:
Sc = U₂Sf
Sc = 0.09 × 146
Sc = 13.14 cm.
Read more on clay layer here: brainly.com/question/22238205
Answer:
W = 112 lb
Explanation:
Given:
- δb = 0.025 in
- E = 29000 ksi (A-36)
- Area A_de = 0.002 in^2
Find:
Compute Weight W attached at C
Solution:
- Use proportion to determine δd:
δd/5 = δb/3
δd = (5/3) * 0.025
δd = 0.0417 in
- Compute εde i.e strain in DE:
εde = δd / Lde
εde = 0.0417 / 3*12
εde = 0.00116
- Compute stress in DE, σde:
σde = E*εde
σde = 29000*0.00116
σde = 33.56 ksi
- Compute the Force F_de:
F_de = σde *A_de
F_de = 33.56*0.002
F_de = 0.0672 kips
- Equilibrium conditions apply:
(M)_a = 0
3*W - 5*F_de = 0
W = (5/3)*F_de
W = (5/3)* 0.0672 = 112 lb
