Solved problem of the "at-rest" lateral earth pressure upon an unyielding wall
Articles > Solved problem of the "at-rest" lateral earth pressure upon an unyielding wallQuestion:
Fin the lateral "at-rest" force F0 on the wall and its location with respect to the top of the wall.
Given: Sand #1 has a unit weight of 105 pcf, c =0 and friction angle of 30; sand #2 has a unit weight of 122 pcf, c=0 and friction angle of 30.
Solution:
From Jaky's empirical relation: k0 = 1 - sinφ' = 1-sin30 = 0.5
at z = 0 feet σ'h = 0 ksf because there is no surcharge loading upon the surface of sand #1
at z = 10 feet σ'h = K0σ'v = 0.5(0.105 kcf)(10ft)=0.525 ksf
at z = 20 feet σ'h = K0σ'v = 0.5[ (0.105 kcf)(10ft) + (0.122 - 0.0624)10]=0.823 ksf
σw = γwh = (0.0624 pcf) (10ft) = 0.624 ksf
F0 = F1 + F2 + F3 + F4 = 1/2 (0.525)(10) + (0.525)(10) + 1/2 (0.302)(10) + 1/2 (0.624)(10) = 2.63 + 5.25 + 1.49 + 3.12 = 12.5 kip/ft
z = 13.8ft from the top of the wall
Read also:
- At-Rest State
- Rankine's Lateral Earth Pressure
- Coulomb's Lateral Earth Pressure
- All Types of Foundation Materials
- Anchored Sheet Pile Walls Penetrating Sandy Soils
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