Question

A direct shear test, when conducted on a remolded sample of clay, gave the following data at the time of failure: of = 189.1 kN/m2 os = 63.8 kN/m2 C = 11.6 kN/m2 Diameter of soil sample = 5 cm (cylindrical sample) Height of soil sample = 3 cm A. Draw the Mohr Circle and show the essential values on that. SA B. Determine the angle of internal friction. C. Determine the normal and shear forces at the tirhe of failure. D. What would be the shear stress if the normal stress on the failure plane was 170 kN/m??

Answer 1

Solution: Page - Given that r = 189, 1KN/m" = 63.8 kN/m² c' = 11.6 kN/m² 3an. Dia. of soil sample, d = 5cm Height of soil sample, n = 3am 1160 6 = 63.8 kN/m2 0 =189. I KN/m² Radius of Mohr's Circle, R = 189. 1-63.8 = 62.650 kn/m? centre of Mona's chile, C = (1891 +63.8, 0) = (126.05, 0) (3) we know that - Oido tent (uB + $) +2c fan(us+ 8) → 189.1 = 63.8tan? (us+ p) +2x116 tan/u5+) = 63.8 tum (45 + b) + 23.2tan (45+) – 189,1207

Pone > clearly thine" is a quadratic in thn (4B + D). =) Tuus, tan lus + 4) = -23.2 + 123.92-4X 33.3K-18 2x63.8 = 1.549,-1.913 Negreeting the negative nalue of tanfus + 2) tan(us+ 8) = 1.549 ») 45°= tan"/549) = 57.159° ») ^ <24.369° Thus angle of internal fichion, 8=24,309²

@ Page - Again drawing the mour's Cluele - V R = 62.65 0=24,3099 Á (63.80) o (126.45,0) 1189.1. ol Thus, Point 's' represents the condition of choil at failure. Coordinate of D= (126.45-asino, Rios) =(126.45-62.65 Säso(21:30), 62,65 LOS (24:30) => 66., t.) = (100.660,57.095) -) 6,=100.660, T = 57.095 Thus, At failure - Normal forte, FN = GI XA d² = 100.660* . x6.05) = 0.198 KN FN = 197.645N.

Page & Sheere force at failure, Fs = T, X z d² = 57.095x xlos mo = 112.106N (d) If Normal stress aut failure, 8,= 170 kr/m² Then, to per conlumbs Theory - T = ctotano => t = 11.6 +170 tan(24.309) Tie 88.390KN)m? Tunes, Shear stress of failure, , = 88.390 kW/m