# Civil Engineering - GATE Exam Questions - Discussion

### Discussion :: GATE Exam Questions - Section 1 (Q.No.17)

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

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).