CE%203400%20HW19-Solution%20Fall%202009

CE 3400 HW19-Sol - CE 235 Homework No 20 SOLUTION Shear Strength of Clay Problem 1 Part(a 1 = 3 P 20 = 20 2 A 2.8 4 = 23.25 psi c= 1 3 23.25 20 = =

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CE 235 - Homework No. 20 - SOLUTION Shear Strength of Clay Problem 1 : Part(a) () 13 2 P2 0 20 23.25 psi A 2.8 4 σ=σ+ = + = π 23.25 20 c 1.63 psi 234 psf 22 σ−σ == Part (b) Same as Part (a) because P A σ−σ= does not change. Problem 2 : For eth clay: s e wG 0.52 2.70 1.40 = s w G e 2.70 1.40 62.4 107pcf 1e 11 . 4 0 ++ γ= γ = = ( )( ) ( ) ( )( ) v p ' 10 115 5 128 62.4 25 107 62.4 =σ = + + v ' 1150 328 1115 2593 psf σ= + + = ( ) v c c ' 0.32 2593 830 psf p =σ= = ( ) u q 2c 2 830 1660 psf =
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Problem 3 : See the Mohr-Coulomb diagram below. I chose to ignore the test with zero confining pressure because it gave a lower strength and typically does.
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This note was uploaded on 06/21/2010 for the course CE 3400 taught by Professor Brentrosenblad during the Fall '09 term at Missouri (Mizzou).

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CE 3400 HW19-Sol - CE 235 Homework No 20 SOLUTION Shear Strength of Clay Problem 1 Part(a 1 = 3 P 20 = 20 2 A 2.8 4 = 23.25 psi c= 1 3 23.25 20 = =

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