Question

Following data are given for a direct shear test conducted on dry sility sand: Specimen dimensions: diameter 71 mm; height 25 mm Normal stress: 150 kN/m2 Shear force at failure: 276 N a. Determine the angle of friction, ф. b. For a normal stress of 200 kN/m2, what shear force is required to cause failure?

Answer 1

Answer a

Normal stress at failure =¶= 150 KN/m2

Shear stress = s= 276*4*1000/(3.142*71*71) KN/m2 = 69.71 KN/m2

Thus S = ¶ tan¢ thus angle of friction = taninverse(69.71/150) = 24.925°

Answer b

Normal stress = 200 KN/m2 thus shear stress at failure = 200*tan(24.925) = 92.943 KN/m2 thus shear force at failure = 92.943*1000*3.142*71*71/4000000 N = 367.979 N

2u.25 0 O A =チ6.84 2.942 0 0