How is the base width B = 71 m obtained?

30 m flare run + 5 m crest + 36 m d/s run (0.6 x 60)

Which results and formulae give the factors of safety?

FSo = ΣM_R/ΣM_O = 2.35 and FSs = μΣV/ΣH = 1.61, both safe

FSo = ΣH/ΣV = 0.44; FSs = ΣM_O/ΣM_R

FSo = μΣV/ΣH and FSs = ΣM_R/ΣM_O, both below 1

How is uplift treated when finding ΣV?

Subtracted from the downward weight (ΣV = 58,977.7 - 11,318.3)

Why is there no tension, and how do the base stresses compare with allowables?

e = 7.61 m < B/6 = 11.83 m, so heel stress is 240 kN/m² (compressive) and toe stress 1103 kN/m² < 2943 kN/m²

e = 7.61 m > B/6, so tension appears at the heel

e = 35.5 m, so the toe stress is zero

Stresses are not compared because uplift is neglected

2082 Bhadra · 12 marks

Gravity dam stability (SG 2.4, RL 0-70)

Short answer

For the 71 m base, 65 m head dam, overturning FS = 2.35 (>1.5), sliding FS = 1.61 (>1.0). Eccentricity e = 7.61 m < B/6 = 11.83 m, so no tension. Max compressive stress at toe = 1103 kN/m² (< 2943 kN/m² allowable); heel stress = 240 kN/m² (compression). The dam is stable.

A concrete gravity dam with a specific gravity of 2.4 has the dimensions as shown in the figure below. Neglect all other forces except self-weight, hydrostatic force, and uplift pressure. The friction coefficient between the base and foundation of the dam is 0.70, the coefficient of uplift pressure is 0.5, and the allowable crushing strength and tensile strength of dam materials are 30 kgf/cm² and 5 kgf/cm², respectively. Check the stability of the dam. [Figure: dam section with base at RL 0 m and crest at RL 70 m, crest width 5 m; water level at 65 m; upstream face vertical from the crest down to RL 30 m, then flaring outward with a 1.0 : 1.0 slope to the base at RL 0 m; downstream face vertical for the top 10 m, then sloping at 0.6 : 1.0 to the base.]

Assumptions

Base at RL 0 m. U/s face: vertical from the crest (RL 70) down to RL 30, then a 1:1 flare out to the heel (horizontal projection 30 m). D/s face: vertical for 10 m (to RL 60), then 0.6 (H) : 1 (V) to the toe (horizontal projection $0.6\times60=36$ m).
Base width $B=30+5+36=71$ m. Water depth $H_w=65$ m, no tail water. Per metre length of dam.
$\gamma_c = 2.4\times9.81=23.54$ kN/m³, $\gamma_w=9.81$ kN/m³.
Uplift: triangular, intensity at the heel $=0.5\,\gamma_wH_w = 0.5\times9.81\times65 = 318.8$ kN/m², zero at the toe (uplift coefficient 0.5).
Allowable crushing $30$ kgf/cm² $=2943$ kN/m²; allowable tension $5$ kgf/cm² $=490.5$ kN/m².

 RL70  +-----+
       |     |
 RL65 ~|~~~~~|  water level
       |     |  d/s vertical to RL60
       |     \
 RL30  |      \  0.6 : 1
      /|        \
 RL0 /_|__________\
    heel          toe   B = 71 m

Forces and moments about the toe

Force	Magnitude (kN/m)	Arm from toe (m)	Moment about toe (kN·m/m)
Vertical (resisting)
Flare triangle (30 × 30)	10,594.8	51.00	540,335
Rectangle 5 × 70	8,240.4	38.50	317,255
D/s triangle (36 × 60)	25,427.5	24.00	610,260
Water on u/s face	14,715.0	57.50	846,112
ΣW (downward)	58,977.7		ΣM_R = 2,313,963
Horizontal / uplift (overturning)
Water thrust $P_w=\frac12\gamma_w H_w^2$	20,723.6	21.67	449,012
Uplift part 1 (upward)	11,318.3	47.33	535,732
Uplift U	11,318.3
Total overturning			ΣM_O = 984,744

(The water thrust acts at $H_w/3 = 21.67$ m above the base. Uplift is a triangle of area $\tfrac12\times318.8\times71$ acting at $B/3$ from the heel.)

Stability checks

$\Sigma V = 58,977.7-11,318.3 = 47,659.4$ kN/m; $\Sigma H = 20,723.6$ kN/m
$\Sigma M_R = 2,313,963$ kN·m/m; $\Sigma M_O = 984,744$ kN·m/m

Overturning: $FS_o = \dfrac{\Sigma M_R}{\Sigma M_O} = \dfrac{2,313,963}{984,744} = 2.35 > 1.5$ — safe.

Sliding: $FS_s = \dfrac{\mu\,\Sigma V}{\Sigma H} = \dfrac{0.70\times47,659.4}{20,723.6} = 1.61 > 1.0$ — safe (friction only).

Location of resultant: $\bar x = \dfrac{\Sigma M_R-\Sigma M_O}{\Sigma V} = \dfrac{1,329,219}{47,659.4} = 27.89$ m from the toe.

$e = B/2-\bar x = 35.5-27.89 = 7.61$ m $< B/6 = 11.83$ m — resultant is inside the middle third, so no tension.

Base stresses:

\sigma_{toe} = \frac{\Sigma V}{B}\left(1+\frac{6e}{B}\right) = 671.3\,(1+0.643) = 1102.9\ \text{kN/m}^2

\sigma_{heel} = 671.3\,(1-0.643) = 239.6\ \text{kN/m}^2

Maximum compressive stress $=1103$ kN/m² $< 2943$ kN/m² (allowable) — safe against crushing.
Minimum stress $=240$ kN/m² is positive (compression) — no tension, and it is greater than $-490.5$ kN/m² in any case.
Principal stress at the toe (no tail water): $\sigma_1=\sigma_{toe}(1+\tan^2\phi_d) = 1102.9\,(1+0.6^2)=1500$ kN/m² $<2943$ kN/m² — safe.

Conclusion: The dam is safe against overturning (2.35), sliding (1.61), tension and crushing.

✓ Answer checked by Claude Opus 5.5 (high effort)

Gravity dam forces and stability checks (2081 Baishakh) · Hydropower Engineering, chapter 4
Gravity dam concrete volume from hydrostatic thrust · Hydropower Engineering, chapter 4
Gravity dam stresses with drainage gallery uplift · Hydropower Engineering, chapter 4
Gravity dam equilibrium and overtopping discharge coefficient · Hydropower Engineering, chapter 4
Gravity dam stability with seismic load · Hydropower Engineering, chapter 4
Gravity dam stability and shear stress at toe (2081) · Hydropower Engineering, chapter 4
Elementary profile and 40 m gravity dam stability · Hydro Power, chapter 4
Sliding and overturning check of gravity dam · Hydro Power, chapter 4
Forces on gravity dam and stability requirements · Hydro Power, chapter 4
Stability of trapezoidal gravity dam (RL 250 m) · Hydro Power, chapter 4

All Hydropower Engineering chapter 4 questions →

Gravity dam stability (SG 2.4, RL 0-70)

Assumptions

Forces and moments about the toe

Stability checks

Related questions