Analytical Chem Soil HW Solutions 28

Analytical Chem Soil HW Solutions 28 - 2θ2 =(90°...

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2θ2 = (90° + φ'apparent) + (ω1 + 18° ) ⇒ θ2 = 0.5 × (108° + ω1 + φ'apparent) 2θ3 = (90° + φ'apparent) + (ω4 + 18° ) ⇒ θ3 = 0.5 × (108° + ω4 + φ'apparent) 2θ4 = (90° + φ'apparent) + (ω4 + 18° ) + 2(180° - 18° - ω4) ⇒ θ4 = 0.5 × (432° - ω4 + φ'apparent) The values of ω1, ω4 and θ1 to θ4 forφ'apparent = 21°, 24°, 27° and 30° are detailed in the table below. φ'apparent ω1 ω4 θ1 θ2 θ3 θ4 21 24 27 30 120.43 130.56 137.10 141.83 16.72 23.28 28.05 31.94 94.29 90.72 88.98 88.06 124.72 131.28 136.05 139.92 166.28 162.72 160.95 160.01 59.57 49.44 42.90 38.12 These values are used to construct the graph of apparent angle of shearing resistance φ'apparent against orientation of the clay laminations θ shown in Figure Q2.8b: note that for orientations of the laminations θ between 32° and 88°, and between 140° and 160°, the value of φ'apparent is equal to φ' for the silt, 30°. Figure Q2.8b: apparent effective angle of friction against angle of lamination inclination Note that unless you are very confident with geometry and trigonometry, this problem is probably much more easily addressed by drawing out the four individual Mohr circles to scale and measuring off the angles θ1 to θ4. The principles, and hopefully the answers, are however the same. 28 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online