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CEE 366 - Wartman - Winter 2012 - Homework 3

CEE 366 - Wartman - Winter 2012 - Homework 3 - Homework 3...

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Homework 3 Solutions Problem 1: The figure below shows the soil profile at the site of an existing warehouse (ie, covers a large area) that causes a surface loading of 2000 psf. Draw (but don't hand in) the σ v and σ v ' profiles with depth. What is the total and effective stress at 48ft (in pcf)? Figure is not to scale, but all important data is shown | | | | | | | | | | 2000psf v v v v v --------------- 0ft Fine Silty Sand γ t = 116 pcf γ d = 110 pcf --------------- 12ft (GWT starts here) Clay γ t = 119 pcf ---------------- 38ft Dense sand γ t = 121 pcf ----------------48ft (end of data) Solution: Stresses due to unit weights Above the water table: Incremental stress: σ = σ’ = γ d * Δ z Below the water table Incremental stress: σ = γ sat * Δ z σ’ = γ b * Δ z Where γ b = γ sat γ water Soil Type Depth (ft) Unit weight (pcf) Total Stress (psf) Effective Stress (psf) Fine Silty Sand 12 110 1320 1320 Clay 38 119 4414 2792 Dense Sand 48 121 5624 3378
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Stresses due to all loads at 48ft σ = 5624 + 2000 7600 psf σ’ = 3378 + 2000 ≈
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