CIE444 - HW1 solution

# 44kn w ws ww 1544 329 1873kn w 329 w w 981

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Unformatted text preview: 19 = 0.581m 3 Ws W → 2.71 = Gs = s 3 Vs γ w (0.581m )(9.81kN / m 3 ) Ww W → 0.213 = w.c. = w × 100% Ws 15.44kN W = Ws + Ww = 15.44 + 3.29 = 18.73kN W 3.29 γw = w → 9.81 = Vw Vw n= Vv × 100% V → 0.419 = → Vv = 0.419 m 3 → Ws = 15.44kN → Ww = 3.29kN → V w = 0.335m 3 Va = Vv − V w = 0.419 − 0.335 = 0.084 m 3 Volume (m3) Weight (kN) 0 18.72 3.29 15.43 Air Water Solids 0.083 0.336 1.000 0.581 Step-2: Compute the required parameters V 0.335 × 100% = 80.0% Degree of saturation: S = w × 100% = 0.335 + 0.084 Vv Moist unit weight: γ= W 18.72kN = = 18.72kN / m 3 V 1m 3 Page 5 of 7 PROBLEM #4 (25%) The standard method of measuring t...
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## This note was uploaded on 10/11/2013 for the course CIE 444 taught by Professor Dr.aboujaoude during the Fall '13 term at Lebanese American University.

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