Civil Engineering - GATE Exam Questions - Discussion

17. 

A 40° slope is excavated to a depth of 10 m in a deep layer of saturated clay of unit weight 20 kN.m3; the relevant shear strength parameters are cu = 70 kN/m2 and φu = 0. The rock ledge is at a great depth. The Taylor's stability coefficient for φu = 0 and 40° slope angle is 0.18. The factor of safety of the slope is :

[A]. 2.0
[B]. 2.1
[C]. 2.2
[D]. 2.3

Answer: Option A

Explanation:

No answer description available for this question.

Desinerd said: (Aug 4, 2014)  
Cu->Shear strength parameter 70 KN/m2.

The formula to be used is : F.S = Cu/Ns*Gamma symbol*H.

Where,

F.S = factor of safety.
Cu=read as C sub u = Undrained shear strength parameter = 70KN/m2.
Gamma Symbol = γ = unit weight of saturated clay = 20 KN/m3.
H = Depth/Height = 10m.
Ns = read as N sub s = Taylor's stability Co-efficient = 0.18.

F.S thus works out to be = 1.94444,which approx ~ 2.

Vls said: (Apr 26, 2016)  
This can also be calculated by the graph.

Rihaan said: (Aug 25, 2016)  
Which formula is applied here?

Karthik said: (Sep 12, 2016)  
F.S = Cu/Ns * γ * H.
= 70 * 20 * 10/(0.18 * 40 * 100).
= 1.94.

Prosanta Bittel said: (Feb 8, 2017)  
According to me,

F.S = Cu/(Ns * γ * H).
= 2.08.

Asaithambi said: (Feb 9, 2017)  
Shear strength parameters are cu = 70 kN/m2,
Saturated clay of unit weight = 20 kN/.m3,
Coefficient for φu = 0.18,
Excavated to a depth =10m,
Factor of safety=(Cu/(coefficient*h*unit weight).
= (70/(0.18*10*20).
= 2.

Jeganathan said: (Feb 1, 2020)  
Stability number sn = c/(FrH).

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