12.1 following data are given for a direct shear test conducted on dry sand: specimen dimensions: diameter= 63 mm; height= 25 mm normal stress: 150 kn/m2 shear force at failure: 276 n a. determine the angle of friction, φ’ b. for a normal stress of 200 kn/m2 , what shear force is required to cause failure?

a) 30.58°

b) 367.55 N

Explanation:

Given:

Diameter of the specimen, D = 63mm = 0.063 m

Height of the specimen, H = 25 mm

Shear force at failure,

Shear stress at failure,

Area =

or

Area =

thus,

Shear stress at failure,

a)      ............(1)

where,

C' = cohesion

for sand C' = 0

∅' = angle of friction

σ' = Normal stress

on substituting the values we get,

or

or

b) Shear force required at the time of failure with normal stress 200 KN/m² can be calculated by using the equation 1

on substituting the values, we get

or

shear force will be = 118.185 ×  area

or

shear force will be = 118.185 ×  0.00311 = 0.36755 KN = 367.55 KN

i'm thinking more like energy

33: i think it’s b