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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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.

Ask a homework question - tutors are online