Short-term factor of safety of a slope on a circular failure surface: 16-Civ-B3 Geotechnical Design, May 2014, Q8

Question 8: Short-term factor of safety of a slope on a circular failure surface (24 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

[Figure not reproduced: Figure 2 redrawn. Moments are taken about the centre O: the weight W acts at a horizontal offset d = 2.0 m, and the undrained strength acts over the whole 15.5 m arc at the constant lever arm R = 5.0 m. See the official exam paper.]

Given. The slope and trial circle of Figure 2, with every quantity needed for a total-stress analysis printed on the figure.

Question 8 — data read from Figure 2
Quantity	Symbol	Value
Radius of the trial circle	R	5.0 m
Length of the failure arc	L_a	15.5 m
Weight of the sliding mass, per metre run	W	530 kN/m
Horizontal offset of the centroid of W from the vertical through O	d	2.0 m
Vertical offset of the centroid of W below O (not needed)	—	3.0 m
Undrained shear strength (uniform)	c_u	45 kPa
Bulk unit weight	γ_t	19 kN/m³
Slope height and gradient	—	4.5 m at 4 vertical to 5 horizontal
Effective-strength parameters (long term, not used here)	c', φ'	0 and 36 degrees

Find. The short-term (undrained) factor of safety of the slope on the failure surface drawn.

Approach. Short term means undrained, so the analysis is a total-stress phi_u = 0 mass-procedure analysis: take moments about the centre of the circle, with the whole sliding mass as one free body. The disturbing moment is the weight of the mass times its horizontal offset from the centre; the resisting moment is the constant undrained shear strength acting over the whole arc at the constant lever arm R.

Confirm that the phi_u = 0 mass procedure applies. The question asks for the short-term factor of safety and states that the undrained shear strength is uniform. With phi_u = 0 the shear strength on the failure surface is $$\tau_f = c_u = 45\ \text{kPa}\quad\text{everywhere, independent of normal stress,}$$ so the normal forces on the arc, which all pass through the centre O, need never be evaluated. No slice division is required and the whole mass can be treated as a single free body — this is Fellenius' mass procedure, and for phi_u = 0 it is an exact statement of moment equilibrium, not an approximation.
Resisting moment about O. The undrained shear strength acts tangentially over the entire arc length, and every element of it is at the same lever arm R from the centre: $$M_R = c_u L_a R = 45(15.5)(5.0) = 3487.5\ \text{kN}\cdot\text{m per metre run}.$$
Disturbing moment about O. The only disturbing force is the weight of the sliding mass, which acts vertically through its centroid. Its lever arm about O is the horizontal distance from the vertical through O to that centroid, marked 2 m on the figure: $$M_D = W d = 530(2.0) = 1060\ \text{kN}\cdot\text{m per metre run}.$$ The 3.0 m vertical offset shown on the figure locates the centroid but plays no part in the moment of a vertical force.
Factor of safety. Taking moments about O, $$FS = \frac{M_R}{M_D} = \frac{c_u L_a R}{W d} = \frac{3487.5}{1060} = \boxed{FS = 3.29}$$
Interpret the result and check it is credible. The mobilised shear stress needed for equilibrium is $\tau_{mob} = c_u/FS = 45/3.29 = 13.7$ kPa, i.e. only about 30 per cent of the strength available, and the slope is comfortably stable in the short term on this surface. The circle analysed is a toe circle constrained above the rock at 3.0 m below the toe; because the rock is shallow, deeper and more critical circles are physically excluded, which is part of why the factor of safety is high.
Note the two checks a complete answer should mention. First, a tension crack: the theoretical depth of the zone that can carry no shear is $$z_0 = \frac{2c_u}{\gamma} = \frac{2(45)}{19} = 4.74\ \text{m},$$ which exceeds the 4.5 m slope height, so on this stiff a clay the crest could crack right through. The question directs the analysis at the surface drawn, whose arc length is given, so 3.29 stands as the answer; but a cracked-crest circle with the crack full of water would need to be checked before the slope was signed off. Second, the long-term case: once the excess pore pressures dissipate the strength is governed by $c^{\prime} = 0$, $\phi^{\prime} = 36^{\circ}$, and stability then depends entirely on the effective normal stress and the seepage regime. For a slope at 4 vertical to 5 horizontal the face angle is $\tan^{-1}(4/5) = 38.7^{\circ}$, which is steeper than phi' = 36 degrees, so the long-term factor of safety on a shallow, fully drained surface falls below unity: the long-term case, not the short-term case, governs this slope.

Question 8 — results
Quantity	Value
Resisting moment about O, c_uL_aR	3487.5 kN·m/m
Disturbing moment about O, Wd	1060 kN·m/m
Short-term factor of safety	3.29
Mobilised shear strength	13.7 kPa of the 45 kPa available
Theoretical tension-crack depth, 2c_u/γ	4.74 m
Slope face angle	38.7 degrees, against φ' = 36 degrees

Check: the numbers are taken from the figure as printed. The arc length of 15.5 m, the weight of 530 kN/m and its 2.0 m horizontal offset are all annotated on Figure 2, and the answer above uses them exactly as printed, which is what the candidate is expected to do. The figure does not close geometrically, however. A 15.5 m arc on a 5.0 m radius subtends 15.5/5.0 = 3.10 radians, or 178 degrees, a near-semicircle, whereas the circle drawn runs from the toe-level ground to the crest through only about 140 degrees. Solving for the one circle that reproduces the printed arc length, the area W/γ = 530/19 = 27.9 m² and the 2.0 m lever arm simultaneously (4V:5H face, 4.5 m high) returns R = 6.35 m, with its centre 4.27 m above toe level and its deepest point 2.08 m below the toe, clear of the rock at 3.0 m and matching the drawing. The printed R = 5.0 m is therefore the value in error. On that self-consistent circle FS = 45(15.5)(6.35)/(530 × 2.0) = 4.18 against 3.29 as printed. Either way the slope is comfortably stable in the short term, so the conclusion does not depend on which value is taken; 3.29 is the lower and the one reported.

Topic. Undrained (total-stress) stability of a slope by the phi_u = 0 mass procedure on a circular failure surface.

Key relations. $FS = \dfrac{M_R}{M_D} = \dfrac{c_u L_a R}{W d}$ with $L_a = R\theta$ (theta in radians); $\tau_{mob} = c_u/FS$; tension-crack depth $z_0 = 2c_u/\gamma$; Taylor's stability number $m = \dfrac{c_u}{FS\,\gamma H}$ for a first estimate without a trial circle.

Why this works. Every normal force on a circular arc points at the centre of the circle, so it has no moment about that centre. In a phi_u = 0 material the shear strength does not depend on those normal forces either, so moment equilibrium about O involves only two things: the weight of the mass and the cohesion on the arc. That is why the undrained circular analysis is statically determinate while the drained one is not, and why the drained case needs the method of slices with an assumption about interslice forces. The whole of Bishop's simplified method exists to handle the complication that disappears when phi is zero.

Common pitfall. Measuring the lever arm of W along the slope or to the centroid itself rather than horizontally from the vertical through O — the moment of a vertical force depends only on its horizontal offset. A second is using the chord instead of the arc length for L_a, which under-estimates the resisting moment by up to a third on a deep circle. A third is quoting a comfortable short-term factor of safety without checking the long-term drained case, which on a steep slope in a frictional soil is usually the one that fails.

Source. Das, Principles of Geotechnical Engineering, 9th ed., Sections 15.4–15.7 (mass procedure, phi = 0 condition, Taylor's charts); Craig, Soil Mechanics, 8th ed., Ch. 9; CFEM 4th ed., Ch. 6.

Question 8 of 9: Short-term factor of safety of a slope on a circular failure surface

Question 8: Short-term factor of safety of a slope on a circular failure surface (24 marks)