Question 11 of 13: Which Triangular Dam Is More Likely to Slide?

Question 11: Which Triangular Dam Is More Likely to Slide? (5 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.

Dam A presents a vertical face to the water; Dam B presents a face sloped back so the water rests partly on top of the incline. No dimensions are given in the source figure, so the reasoning below is developed symbolically for a general triangular section of height H and base B, with an illustrative numeric example (H = 5 m, friction coefficient μ = 0.5) used only to show the SIZE of the effect.

Horizontal driving force (identical for both dams). The horizontal hydrostatic thrust on any submerged surface depends only on the vertical projection of the wetted face, which is the SAME height H for both dams regardless of whether that face is vertical or sloped: $$F_h = \tfrac12\rho_w g H^2$$ For the illustrative H = 5 m this gives $F_h = \tfrac12(1000)(9.81)(5^2) = 122.6$ kN/m — identical for Dam A and Dam B.

Vertical (normal) load — where the dams differ. Both dams have the same self-weight W (same triangular cross-section, just mirrored). But on Dam B, the water sits directly ON TOP of the sloped face, so its weight adds an extra downward force W_v equal to the weight of the water "wedge" resting on the incline; on Dam A the water face is vertical, so the hydrostatic pressure there is purely horizontal — it contributes NO vertical load at all. Illustratively (B=H=5 m, ρ_c=2400 kg/m³): $$W = \tfrac12\rho_c g\,B\,H = \tfrac12(2400)(9.81)(5)(5) = 294.3\ \text{kN/m (both dams)}$$ $$W_{v}= \tfrac12\rho_w g\,B\,H = \tfrac12(1000)(9.81)(5)(5) = 122.6\ \text{kN/m (Dam B only)}$$ $$N_A = W = 294.3\ \text{kN/m}, \qquad N_B = W+W_v = 416.9\ \text{kN/m}$$

Sliding resistance. Resistance to sliding is friction, $\mu N$, while the driving force $F_h$ is identical for both. Since $N_B > N_A$ (by the full weight of the water sitting on the incline), Dam B has MORE frictional resistance for the SAME driving thrust: $$\frac{\mu N_B}{\mu N_A} = \frac{416.9}{294.3} = \boxed{1.42}$$ i.e. Dam B can resist about 42% more sliding force than Dam A before slipping, for this illustrative geometry.

Conclusion. Dam A (the vertical-faced dam) is more likely to slide. Its water face contributes no vertical load to increase the normal (and hence frictional) reaction, while Dam B's sloped, water-covered face gains extra stabilizing weight from the water resting on it — for an identical horizontal thrust. This is precisely why real gravity dams are often given a sloped, not vertical, upstream face.