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

.) the observation by semmelweis that more mothers died of childbed fever when their doctors also worked in the morgue

energy can be converted from one form to another, but it cannot be created or destroyed in ordinary chemical or physical means. do not understand what you are asking but hope this