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Gravity dam stability with seismic load

Short answer

Overturning FS=1.83 (safe). Sliding friction alone FS=1.04, shear-friction factor SFF=3.50 (safe). Maximum vertical stress 641 kN/m², principal stress 1051 kN/m², both within allowable limits; no tension. Dam is stable.

Check the stability of the dam against overturning, sliding and material failure (stresses) with respect to the worst location, assuming that in addition to self weight, 25% of the mass of the dam will act as a horizontal component (from the upstream side), whereas 15% acts as an upward vertical component as seismic load and will act at the CG of the section. Assume the unit weight of concrete as 24 kN/m³, allowable compressive stress in foundation and concrete as 2,500 kN/m² and 3,000 kN/m², angle of friction between concrete and foundation as 36° and unit shear resistance between foundation and dam as 700 kN/m². [Figure: dam with crest width 6.0 m, top level 150.0, water level 135.0, base level 100.0; upstream face with slope 0.1 : 1; downstream slope 0.8 : 1; points A (heel) and B (toe).]

Data and assumptions

  • Base RL 100, crest RL 150: height 50 m. Water RL 135: Hw=35H_w=35 m. Crest width 6 m. U/s face slope 0.1 : 1 (leaning inwards going up), so the heel A is at x=0x=0 and the u/s edge of the crest at x=5x=5; d/s slope 0.8 : 1, so the toe B is at x=11+40=51x=11+40=51 m. Base B=51B=51 m.
  • The base (section A-B) is the worst location: it carries the largest load and moment.
  • γc=24\gamma_c=24 kN/m³, γw=9.81\gamma_w=9.81 kN/m³. No tail water. Hydrostatic thrust and uplift (full, triangular, no drainage: 343.4343.4 kN/m² at the heel to zero at the toe) act together with the seismic loads.
  • Seismic: horizontal force 0.25W0.25W from the upstream side (towards the downstream) and vertical force 0.15W0.15W upward, both at the centre of gravity (CG).
   6 m
  +----+  RL150
 /     |
/      |  RL135 ~~ water
/      |
A(heel) \___________  B(toe)
 0.1:1       0.8:1       B = 51 m

Self-weight and CG

Area =12(6+51)×50=1425=\frac12(6+51)\times50=1425 m². W=24×1425=34,200W=24\times1425=34,200 kN/m. CG: xˉ=19.05\bar x=19.05 m from the heel (B−xˉ=31.95B-\bar x=31.95 m from the toe), zˉ=18.42\bar z=18.42 m.

Seismic: Fh=0.25W=8,550F_h=0.25W=8,550 kN/m; Fv=0.15W=5,130F_v=0.15W=5,130 kN/m (upward), so the effective weight =0.85W=29,070=0.85W=29,070 kN/m.

Forces

ForceMagnitude (kN/m)Arm from toe (m)Moment about toe (kN·m/m)
Effective weight 0.85W29,07031.95928,710
Water over the u/s slope60149.8329,943
ΣM_R958,653
Water thrust 12γwHw2\frac12\gamma_wH_w^26,00911.67 (height)70,101
Uplift (upward)8,75534.00297,684
Seismic FhF_h8,55018.42 (height)157,500
ΣM_O525,285

ΣV=29,070+601−8,755=20,915\Sigma V=29,070+601-8,755=20,915 kN/m; ΣH=6,009+8,550=14,559\Sigma H=6,009+8,550=14,559 kN/m.

Overturning

FSo=ΣMRΣMO=958,653525,285=1.83FS_o=\frac{\Sigma M_R}{\Sigma M_O}=\frac{958,653}{525,285}=1.83

This is above 1.5 (a value of 1.1-1.5 is adequate for the seismic case), so the dam is safe against overturning.

Sliding

Friction coefficient μ=tan⁡36∘=0.727\mu=\tan36^\circ=0.727; shear resistance q=700q=700 kN/m².

FSfriction=μΣVΣH=0.727×20,91514,559=1.04FS_{friction}=\frac{\mu\Sigma V}{\Sigma H}=\frac{0.727\times20,915}{14,559}=1.04 SFF=μΣV+qBΣH=0.727×20,915+700×5114,559=3.50SFF=\frac{\mu\Sigma V+qB}{\Sigma H}=\frac{0.727\times20,915+700\times51}{14,559}=3.50

Friction alone is just above 1. Including the shear resistance of the foundation contact, SFF=3.50SFF=3.50, which is well above the minimum usually required for loading with earthquake, so the section is safe in sliding.

Material failure (stresses on the base)

xˉ=(ΣMR−ΣMO)/ΣV=20.72\bar x=(\Sigma M_R-\Sigma M_O)/\Sigma V=20.72 m from the toe; e=25.5−20.72=4.78e=25.5-20.72=4.78 m <B/6=8.5<B/6=8.5 m (no tension).

σtoe=20,91551(1+6×4.7851)=641 kN/m2 (<2500),σheel=179 kN/m2\sigma_{toe}=\frac{20,915}{51}\left(1+\frac{6\times4.78}{51}\right)=641\ \text{kN/m}^2\ (<2500),\qquad \sigma_{heel}=179\ \text{kN/m}^2

Both are compressive and less than the allowable 2500 kN/m² (foundation).

Major principal stress at the toe (no tail water, tan⁡ϕd=0.8\tan\phi_d=0.8):

σ1=σtoe(1+tan⁡2ϕd)=641×1.64=1051 kN/m2<3000 kN/m2\sigma_1=\sigma_{toe}(1+\tan^2\phi_d)=641\times1.64=1051\ \text{kN/m}^2<3000\ \text{kN/m}^2

so the concrete is also safe.

Conclusion: the dam is safe against overturning (FSo=1.83FS_o=1.83), sliding (SFF=3.50SFF=3.50) and material failure (maximum vertical stress 641 kN/m² < 2500 kN/m², principal stress 1051 kN/m² < 3000 kN/m², no tension).

✓ Answer checked by Claude Opus 5.5 (high effort)

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