Respuesta :
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, [tex]{V_u}=276\ N=0.276 KN[/tex]
Shear stress at failure, [tex]\tau_f=\frac{V_u}{Area\ of\ the\ specimen}[/tex]
Area = [tex]\frac{\pi D^2}{4}[/tex]
or
Area = [tex]\frac{\pi (0.063)^2}{4}=0.00311 m^2[/tex]
thus,
Shear stress at failure, [tex]\tau_f=\frac{0.276}{0.00311}=88.745 KN/m^2[/tex]
a) [tex]\tau_f=C'+ \sigma'\tan\phi'[/tex] ............(1)
where,
C' = cohesion
for sand C' = 0
∅' = angle of friction
σ' = Normal stress
on substituting the values we get,
[tex]88.745=0+ 150\times\tan\phi'[/tex]
or
[tex]\tan\phi'=0.591[/tex]
or
[tex]\phi'=30.58^o[/tex]
b) Shear force required at the time of failure with normal stress 200 KN/m² can be calculated by using the equation 1
[tex]\tau_f=C'+ \sigma'\tan\phi'[/tex]
on substituting the values, we get
[tex]\tau_f=0+ 200\tan30.58^o[/tex]
or
[tex]\tau_f=118.185 KN/m^2[/tex]
shear force will be = 118.185 × area
or
shear force will be = 118.185 × 0.00311 = 0.36755 KN = 367.55 KN
