Question

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?

Answer 1

Answer:

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, ${V_u}=276\ N=0.276 KN$

Shear stress at failure, $\tau_f=\frac{V_u}{Area\ of\ the\ specimen}$

Area = $\frac{\pi D^2}{4}$

or

Area = $\frac{\pi (0.063)^2}{4}=0.00311 m^2$

thus,

Shear stress at failure, $\tau_f=\frac{0.276}{0.00311}=88.745 KN/m^2$

a)  $\tau_f=C'+ \sigma'\tan\phi'$    ............(1)

where,

C' = cohesion

for sand C' = 0

∅' = angle of friction

σ' = Normal stress

on substituting the values we get,

$88.745=0+ 150\times\tan\phi'$

or

$\tan\phi'=0.591$

or

$\phi'=30.58^o$

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

$\tau_f=C'+ \sigma'\tan\phi'$  

on substituting the values, we get

$\tau_f=0+ 200\tan30.58^o$  

or

$\tau_f=118.185 KN/m^2$

shear force will be = 118.185 ×  area

or

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