Dam Engineering

A gravity dam resists the water thrust and other forces mainly by its own weight. It can fail in four ways; each one gives a stability criterion that the design must satisfy for all load combinations (empty reservoir, full reservoir, earthquake, flood).

     |\           W = self weight
 P-> | \          P = water thrust
     |  \         U = uplift
 ~~~~|   \        R = resultant
     | W  \
     |  \  \
heel |___R__\ toe
     ^ U ^ ^ ^

1. Overturning about the toe

• Mode: the horizontal forces (water pressure, silt, wave, earthquake) and uplift produce a moment that tends to rotate the dam about its toe.
• Criterion: factor of safety against overturning

$$FS_o = \frac{\sum M_R}{\sum M_O} \geq 1.5\ (\text{usually } 2 \text{ to } 3)$$

In practice, a dam that satisfies the no-tension and crushing criteria will not overturn.

2. Sliding (shear failure)

• Mode: the dam slides along the base or along a weak joint/seam in the foundation when the horizontal force exceeds the frictional and shear resistance.
• Criteria:
  • Factor of safety against sliding (friction only): $FS_s = \dfrac{\mu \sum V}{\sum H} \geq 1$ (usually 1 to 1.5).
  • Shear friction factor (friction plus shear strength $q$ of the joint over base width $B$): $SFF = \dfrac{\mu \sum V + B q}{\sum H} \geq 3$ to 5.

3. Compression or crushing

• Mode: the vertical stress at the toe (full reservoir) or heel (empty reservoir) exceeds the allowable compressive strength of concrete or rock.
• Criterion:

$$p_{max} = \frac{\sum V}{B}\left(1 + \frac{6e}{B}\right) \leq \text{allowable stress}$$

The principal stress at the toe, $\sigma_1 = p_v\sec^2\alpha - p'\tan^2\alpha$, must also be below the allowable value.

4. Tension cracking

• Mode: if the resultant falls outside the middle third of the base ($e > B/6$), tension develops at the heel. Concrete is weak in tension, so a crack opens, uplift enters the crack, the effective base width reduces and stresses at the toe rise, which can lead to failure.
• Criterion (middle-third rule):

$$e \leq \frac{B}{6} \quad\Rightarrow\quad p_{min} = \frac{\sum V}{B}\left(1 - \frac{6e}{B}\right) \geq 0$$

A small tension (e.g. up to about 500 kN/m²) is sometimes allowed for extreme load cases such as earthquakes.

Summary of criteria

Elementary profile

The elementary profile of a gravity dam is the right-angled triangle with a vertical upstream face, zero top width and base width $B$, with water standing up to its apex. It is the theoretical shape that resists water pressure with the least material, because water pressure is zero at the top and increases linearly with depth, exactly like the triangle's width. Under empty reservoir the resultant passes through the inner middle-third point, and under full reservoir (with proper base width) through the outer middle-third point, so there is no tension. A practical profile is made by adding a top width, freeboard and other details to it.

       apex (water level)
        |\
        | \
  P --> |  \     H
        | W \
        |    \
   heel |_____\ toe
        <--B-->
        ^ uplift ^

Forces (per metre length)

Let $S_c$ = specific gravity of concrete, $\gamma_w$ = unit weight of water, $C$ = uplift intensity factor.

• Self weight: $W = \frac{1}{2}BH S_c \gamma_w$, acting at $B/3$ from the heel.
• Water pressure: $P = \frac{1}{2}\gamma_w H^2$, acting at $H/3$ above the base.
• Uplift: $U = \frac{1}{2}C\gamma_w H B$, acting at $B/3$ from the heel.

Base width for no tension

For no tension, the resultant must pass through the outer middle-third point (distance $B/3$ from the toe, i.e. $2B/3$ from the heel). Taking moments about this point: $W$ and $U$ both have lever arm $B/3$, and $P$ has lever arm $H/3$.

$$\begin{aligned} (W - U)\frac{B}{3} &= P\frac{H}{3} \\ \frac{1}{2}\gamma_w BH(S_c - C)\,B &= \frac{1}{2}\gamma_w H^2 \cdot H \\ B^2(S_c - C) &= H^2 \\ B &= \frac{H}{\sqrt{S_c - C}} \end{aligned}$$

Without uplift ($C = 0$): $B = H/\sqrt{S_c}$. For $S_c = 2.4$, $B = 0.645H$; with $C = 1$, $B = 0.84H$.

Base width for no sliding

For no sliding, the frictional resistance must at least equal the horizontal force (μ = coefficient of friction):

$$\begin{aligned} \mu (W - U) &\geq P \\ \mu \cdot \frac{1}{2}\gamma_w BH(S_c - C) &= \frac{1}{2}\gamma_w H^2 \\ B &= \frac{H}{\mu (S_c - C)} \end{aligned}$$

Without uplift: $B = H/(\mu S_c)$.

Design base width

The base width adopted is the larger of the two values. Example: $S_c = 2.4$, $C = 1$, $\mu = 0.7$ gives $B = 0.845H$ (no tension) and $B = 1.02H$ (no sliding), so $B = 1.02H$ would be adopted. For usual values of $\mu$ (0.65–0.75), the sliding criterion often governs.

The limiting height of the elementary profile (from allowable stress $f$) is $H = f/[\gamma_w(S_c - C + 1)]$; a dam taller than this is a "high" gravity dam.

Elementary profile

The elementary profile of a gravity dam is a right-angled triangle with a vertical upstream face, zero top width, and water up to the apex. It is the ideal shape to resist water pressure (which grows linearly with depth) with minimum material and no tension in the base.

Data (per metre length)

H = 40 m, B = 28 m, $\gamma_c$ = 24 kN/m³, $\gamma_w$ = 10 kN/m³, C = 0.6, μ = 0.7. Upstream face vertical; water up to the crest. Moments are taken about the toe.

        |\
  P --> | \   40 m
  13.33 |W \
   m    |   \
   heel |____\ toe
        <-28 m->

Forces and moments

$$\begin{aligned} \sum V &= 13{,}440 - 3{,}360 = 10{,}080\ \text{kN} \\ \sum H &= 8{,}000\ \text{kN} \\ \sum M_R &= 250{,}880\ \text{kN·m} \\ \sum M_O &= 106{,}667 + 62{,}720 = 169{,}387\ \text{kN·m} \end{aligned}$$

(a) Position of resultant and eccentricity

$$\begin{aligned} \bar{x} &= \frac{\sum M_R - \sum M_O}{\sum V} = \frac{250{,}880 - 169{,}387}{10{,}080} = 8.085\ \text{m from toe} \\ e &= \frac{B}{2} - \bar{x} = 14 - 8.085 = 5.915\ \text{m} \end{aligned}$$

Since $e = 5.915 > B/6 = 4.667$ m, the resultant lies outside the middle third (towards the toe), so tension develops at the heel.

(b) Factor of safety against sliding

$$FS_s = \frac{\mu \sum V}{\sum H} = \frac{0.7 \times 10{,}080}{8{,}000} = 0.882$$

$FS_s < 1$: the section is unsafe in sliding (by friction alone).

(c) Maximum vertical stress at the toe

$$\begin{aligned} p_{toe} &= \frac{\sum V}{B}\left(1 + \frac{6e}{B}\right) = \frac{10{,}080}{28}\left(1 + \frac{6\times5.915}{28}\right) \\ &= 360 \times 2.2676 = 816.3\ \text{kN/m}^2 \end{aligned}$$

At the heel: $p_{heel} = 360(1 - 1.2676) = -96.3$ kN/m² (tension).

(d) Factor of safety against overturning

$$FS_o = \frac{\sum M_R}{\sum M_O} = \frac{250{,}880}{169{,}387} = 1.48$$

Answer: Resultant at 8.085 m from the toe (eccentricity e = 5.915 m from the base centre); FS against sliding = 0.88 (unsafe); maximum toe stress = 816.3 kN/m² (heel tension 96.3 kN/m²); FS against overturning = 1.48 (just below the usual 1.5). The 28 m base is less than the no-tension width $H/\sqrt{S_c - C} = 40/\sqrt{2.4-0.6} = 29.8$ m, which explains the heel tension.

Geometry and assumptions (per metre length)

• Water depth $h_1$ = 65 m, water surface 5 m below crest, so dam height = 70 m.
• Top width 12 m; downstream face vertical for the top 15 m, then sloping from 55 m height down to the toe.
• Base width $B$ = 10 + 40 = 50 m (drain holes 10 m from heel). Tailwater depth $h_2$ = 5 m.
• Downstream slope: horizontal 50 − 12 = 38 m over vertical 55 m, so $\tan\alpha = 38/55 = 0.6909$ (α measured from vertical).
• Uplift with drains (usual IS/textbook rule): heel $= \gamma_w h_1$; at drains $= \gamma_w[h_2 + \frac{1}{3}(h_1 - h_2)]$; toe $= \gamma_w h_2$.
• Silt, wave and earthquake forces are ignored. Moments are taken about the toe.

      12 m
     +----+       crest (70 m)
  ~~~|~~~~|  5 m
     |    |  15 m vertical d/s
 65m |    +     (55 m)
 wat |  W1  \
 er  |    W2  \
     |          \  ~~~ tailwater 5 m
heel +-----------\ toe
     <10>|<---40--->
       drains   B = 50 m

Uplift pressures

• Heel: $10 \times 65 = 650$ kN/m²
• Drains: $10 \times [5 + (65 - 5)/3] = 10 \times 25 = 250$ kN/m²
• Toe: $10 \times 5 = 50$ kN/m²

Forces and moments about the toe

(Tailwater wedge width on the slope $= 5 \times 0.6909 = 3.455$ m.)

$$\begin{aligned} \sum U &= 10{,}500\ \text{kN} \\ \sum V &= 20{,}160 + 25{,}080 + 86.4 - 10{,}500 = 34{,}826.4\ \text{kN} \\ \sum H &= 21{,}125 - 125 = 21{,}000\ \text{kN} \\ \sum M_R &= 1{,}522{,}708\ \text{kN·m} \\ \sum M_O &= 457{,}708 + 352{,}500 = 810{,}208\ \text{kN·m} \end{aligned}$$

Position of resultant

$$\begin{aligned} \bar{x} &= \frac{1{,}522{,}708 - 810{,}208}{34{,}826.4} = 20.459\ \text{m from toe} \\ e &= 25 - 20.459 = 4.541\ \text{m} < B/6 = 8.333\ \text{m (no tension)} \end{aligned}$$

(a) Maximum vertical stresses

$$\begin{aligned} \frac{\sum V}{B} &= \frac{34{,}826.4}{50} = 696.53\ \text{kN/m}^2, \quad \frac{6e}{B} = 0.545 \\ p_{v,toe} &= 696.53(1 + 0.545) = 1076.1\ \text{kN/m}^2 \\ p_{v,heel} &= 696.53(1 - 0.545) = 316.9\ \text{kN/m}^2 \end{aligned}$$

(b) Major principal stress at the toe

With tailwater pressure at the toe $p' = 10 \times 5 = 50$ kN/m², $\tan^2\alpha = 0.4774$, $\sec^2\alpha = 1.4774$:

$$\begin{aligned} \sigma_1 &= p_v \sec^2\alpha - p'\tan^2\alpha \\ &= 1076.1 \times 1.4774 - 50 \times 0.4774 \\ &= 1565.9\ \text{kN/m}^2 \end{aligned}$$

(c) Shear stress on a horizontal plane near the toe

$$\begin{aligned} \tau &= (p_v - p')\tan\alpha \\ &= (1076.1 - 50) \times 0.6909 = 708.9\ \text{kN/m}^2 \end{aligned}$$

Answer: Vertical stress at toe = 1076.1 kN/m², at heel = 316.9 kN/m² (both compressive); major principal stress at toe = 1565.9 kN/m²; shear stress near toe = 708.9 kN/m².

The dam is checked for full reservoir against overturning, sliding, tension and compression (crushing), per metre length.

Data and assumptions

• Upstream water depth $h_1 = 250 - 200 = 50$ m; tailwater depth $h_2 = 210 - 200 = 10$ m.
• Dam height is not given, so the crest is taken at the maximum reservoir level: $H = 50$ m.
• Upstream face vertical; top width 6 m; downstream face slopes straight from the crest to the toe, so $\tan\alpha = (40-6)/50 = 0.68$.
• $\gamma_w = 9.81$ kN/m³, $\gamma_c = 24$ kN/m³. Coefficient of friction (not given) assumed $\mu = 0.7$.
• Uplift with drainage gallery (IS 6512 rule): heel $\gamma_w h_1$, at gallery $\gamma_w[h_2 + \frac{1}{3}(h_1 - h_2)]$, toe $\gamma_w h_2$.

Uplift intensities: heel $= 9.81 \times 50 = 490.5$ kN/m²; gallery $= 9.81 \times (10 + 40/3) = 9.81 \times 23.33 = 228.9$ kN/m²; toe $= 98.1$ kN/m².

     6 m
    +--+      El. 250 (crest = MRL)
    |  |\
 P1 |W1| \  W2         50 m
 -> |  |  \
    |  |   \  ~~ TW El. 210 (10 m)
heel+--+----\ toe   El. 200
    <10>|<- 30 ->
    gallery   B = 40 m

Forces and moments about the toe

$$\begin{aligned} \sum U &= 8{,}502\ \text{kN} \\ \sum V &= 7{,}200 + 20{,}400 + 333.5 - 8{,}502 = 19{,}431.5\ \text{kN} \\ \sum H &= 12{,}262.5 - 490.5 = 11{,}772\ \text{kN} \\ \sum M_R &= 731{,}191\ \text{kN·m}, \quad \sum M_O = 415{,}835\ \text{kN·m} \end{aligned}$$

(i) Overturning

$$FS_o = \frac{731{,}191}{415{,}835} = 1.76 > 1.5 \quad \text{(safe)}$$

(ii) Sliding

$$FS_s = \frac{\mu\sum V}{\sum H} = \frac{0.7 \times 19{,}431.5}{11{,}772} = 1.16 > 1 \quad \text{(safe)}$$

(iii) Tension

$$\begin{aligned} \bar{x} &= \frac{731{,}191 - 415{,}835}{19{,}431.5} = 16.23\ \text{m from toe} \\ e &= 20 - 16.23 = 3.77\ \text{m} < B/6 = 6.67\ \text{m} \end{aligned}$$

The resultant is within the middle third, so no tension.

(iv) Compression

$$\begin{aligned} \frac{\sum V}{B} &= 485.8\ \text{kN/m}^2, \quad \frac{6e}{B} = 0.566 \\ p_{toe} &= 485.8(1 + 0.566) = 760.6\ \text{kN/m}^2 \\ p_{heel} &= 485.8(1 - 0.566) = 211.0\ \text{kN/m}^2 \end{aligned}$$

Major principal stress at the toe ($p' = 98.1$ kN/m²):

$$\sigma_1 = p_v\sec^2\alpha - p'\tan^2\alpha = 760.6(1.4624) - 98.1(0.4624) = 1066.9\ \text{kN/m}^2$$

These exceed the permissible 300 kN/m², so the dam fails in crushing as the data stand.

Answer: FS overturning = 1.76 (safe); FS sliding = 1.16 (safe, μ = 0.7 assumed); e = 3.77 m < B/6 (no tension); toe stress 760.6 kN/m² and principal stress 1066.9 kN/m² > 300 kN/m² (unsafe in compression). A permissible strength of 300 kN/m² is very low for concrete; if 3000 kN/m² was intended, the dam is safe in all four checks.

Data (per metre length)

H = 35 m, B = 26 m, uplift coefficient C = 0.5, μ = 0.75. Unit weight of concrete $= 2400$ kgf/m³ $= 2400 \times 9.81/1000 = 23.544$ kN/m³; water $= 9.81$ kN/m³. Upstream face vertical, water up to the top. Uplift varies from $C\gamma_w H$ at the heel to zero at the toe. Moments are about the toe.

        |\
  P --> | \    35 m
        | W\
   heel |___\ toe
        <26 m>

Forces

$$\begin{aligned} \sum V &= 10{,}712.5 - 2{,}231.8 = 8{,}480.7\ \text{kN} \\ \sum M_R &= 185{,}683.7, \quad \sum M_O = 108{,}784.7\ \text{kN·m} \end{aligned}$$

a) Eccentricity

$$\begin{aligned} \bar{x} &= \frac{185{,}683.7 - 108{,}784.7}{8{,}480.7} = 9.067\ \text{m from toe} \\ e &= \frac{26}{2} - 9.067 = 3.933\ \text{m} \end{aligned}$$

$e < B/6 = 4.333$ m, so the resultant lies in the middle third (no tension).

b) Factor of safety against sliding

$$FS_s = \frac{\mu\sum V}{\sum H} = \frac{0.75 \times 8{,}480.7}{6{,}008.6} = 1.06$$

c) Maximum compressive (crushing) stress

$$\begin{aligned} p_{toe} &= \frac{\sum V}{B}\left(1 + \frac{6e}{B}\right) = \frac{8{,}480.7}{26}\left(1 + \frac{6 \times 3.933}{26}\right) \\ &= 326.18 \times 1.9075 = 622.2\ \text{kN/m}^2 \ (\approx 63.4\ \text{t/m}^2) \end{aligned}$$

Heel stress $= 326.18(1 - 0.9075) = 30.2$ kN/m² (compression). Major principal stress at toe, with $\tan\alpha = 26/35$: $\sigma_1 = 622.2 \times (1 + 0.5518) = 965.5$ kN/m².

d) Factor of safety against overturning

$$FS_o = \frac{185{,}683.7}{108{,}784.7} = 1.71$$

Answer: Resultant at 9.067 m from the toe (e = 3.933 m from centre); FS sliding = 1.06; maximum vertical stress at toe = 622.2 kN/m² (principal stress 965.5 kN/m²); FS overturning = 1.71. The dam is safe in overturning and tension; FS against sliding is just above 1, so the margin is small.

Data and assumptions (per metre length)

• Dam height $= 1530 - 1470 = 60$ m; upstream water depth $h_1 = 1525 - 1470 = 55$ m; tailwater depth $h_2 = 6$ m.
• Top width 6 m, base 60 m, upstream face vertical, downstream face straight from crest to toe: $\tan\alpha = 54/60 = 0.9$.
• $\gamma_w = 9.81$, $\gamma_c = 24$ kN/m³; μ (not given) assumed 0.7.
• "Full uplift": full hydrostatic uplift at the heel; with the gallery, IS 6512 rule: head at gallery $= h_2 + \frac{1}{3}(h_1 - h_2) = 6 + 49/3 = 22.33$ m.

Uplift intensities: heel 539.6, gallery 219.1, toe 58.9 kN/m².

      6
    +--+  1530
    |  |\
 55 |  | \     60 m
 m  |W1|W2\
    |  |   \  ~~ TW 6 m
    +--+----\ 1470
    <10>|<-50->
       gallery  B = 60 m

Forces and moments about the toe

$$\begin{aligned} \sum U &= 10{,}742\ \text{kN} \\ \sum V &= 8{,}640 + 38{,}880 + 158.9 - 10{,}742 = 36{,}937\ \text{kN} \\ \sum H &= 14{,}837.6 - 176.6 = 14{,}661\ \text{kN} \\ \sum M_R &= 1{,}892{,}799, \quad \sum M_O = 690{,}420\ \text{kN·m} \end{aligned}$$

Overturning

$$FS_o = \frac{1{,}892{,}799}{690{,}420} = 2.74 > 1.5 \quad \text{(safe)}$$

Sliding

$$FS_s = \frac{0.7 \times 36{,}937}{14{,}661} = 1.76 > 1 \quad \text{(safe)}$$

Tension

$$\begin{aligned} \bar{x} &= \frac{1{,}892{,}799 - 690{,}420}{36{,}937} = 32.55\ \text{m from toe} \\ e &= 30 - 32.55 = -2.55\ \text{m} \end{aligned}$$

The resultant is 2.55 m from the centre on the heel side; $|e| < B/6 = 10$ m, so no tension.

Compression

$$\begin{aligned} \frac{\sum V}{B} &= 615.6\ \text{kN/m}^2, \quad \frac{6|e|}{B} = 0.255 \\ p_{heel} &= 615.6(1 + 0.255) = 772.7\ \text{kN/m}^2 \\ p_{toe} &= 615.6(1 - 0.255) = 458.5\ \text{kN/m}^2 \end{aligned}$$

Principal stress at toe $= 458.5(1.81) - 58.9(0.81) = 782.2$ kN/m². All stresses are well below 3000 kN/m² (safe).

Answer: FS overturning = 2.74, FS sliding = 1.76, e = 2.55 m (within middle third, no tension), maximum stress = 772.7 kN/m² at heel < 3000 kN/m². The dam is safe. (If drains are ignored and full uplift acts over the whole base, FS overturning = 1.98 and FS sliding = 1.42; still safe.)

Geometry and assumptions (per metre length)

• Dam height $= 13 + 20 = 33$ m; water depth $h = 30$ m; no tailwater.
• Top width 6 m; downstream face vertical for 13 m, then slopes 15 m horizontally over 20 m height; base $B = 21$ m.
• $\gamma_w = 9.81$ kN/m³, $\gamma_c = 24$ kN/m³, μ = 0.7, shear strength $q = 1400$ kN/m² (taken per m² of base).
• Uplift is taken as full (C = 1), varying from $\gamma_w h$ at heel to zero at toe.

      6 m
    +----+
    |    |  13 m
 30 | W1 +
 m  |     \  W2   20 m
 -> |      \
heel+-------\ toe
    <6>|<-15->
       B = 21 m

Forces and moments about the toe

$$\begin{aligned} \sum V &= 4{,}752 + 3{,}600 - 3{,}090.2 = 5{,}261.9\ \text{kN} \\ \sum M_R &= 121{,}536, \quad \sum M_O = 44{,}145 + 43{,}262 = 87{,}407\ \text{kN·m} \end{aligned}$$

Overturning

$$FS_o = \frac{121{,}536}{87{,}407} = 1.39$$

Sliding

$$\begin{aligned} FS_s &= \frac{\mu\sum V}{\sum H} = \frac{0.7 \times 5{,}261.9}{4{,}414.5} = 0.83 \\ SFF &= \frac{\mu\sum V + Bq}{\sum H} = \frac{3{,}683.3 + 21 \times 1400}{4{,}414.5} = 7.49 \end{aligned}$$

Results

(Without uplift: $\sum V = 8{,}352$ kN, $\sum M_O = 44{,}145$ kN·m.)

Answer: With full uplift, FS overturning = 1.39 (slightly below 1.5) and friction FS = 0.83, but the shear friction factor = 7.49 > 5, so the dam is safe against sliding when the shear strength of the joint is counted. Without uplift, FS overturning = 2.75 and SFF = 7.98 (safe). To improve overturning safety, drainage holes should be provided to reduce uplift.

Data

• Upstream water depth $h_1 = 250 - 200 = 50$ m; tailwater depth $h_2 = 210 - 200 = 10$ m.
• Base width $B = 40$ m; drainage gallery 10 m from the vertical upstream face. $\gamma_w = 9.81$ kN/m³. Values are per metre length of dam.

Hydrostatic thrust

$$\begin{aligned} P_1 &= \frac{1}{2}\gamma_w h_1^2 = \frac{1}{2}\times9.81\times50^2 = 12{,}262.5\ \text{kN} \ (\text{at } 16.67\ \text{m above base}) \\ P_2 &= \frac{1}{2}\gamma_w h_2^2 = \frac{1}{2}\times9.81\times10^2 = 490.5\ \text{kN} \ (\text{at } 3.33\ \text{m above base}) \\ P_{net} &= 12{,}262.5 - 490.5 = 11{,}772\ \text{kN}\ (\text{downstream}) \end{aligned}$$

Uplift pressure

Net seepage head $= h_1 - h_2 = 40$ m. It is reduced by 50% at the gallery:

$$h_g = h_2 + 0.5(h_1 - h_2) = 10 + 20 = 30\ \text{m}$$

 heel      gallery              toe
  |<- 10 ->|<------- 30 -------->|
  |\       |                     |
  | \      |                     |
  |  \_____|_____________        |
  |        |             -----___|
  | 490.5  | 294.3          98.1 |

$$\begin{aligned} U_1 &= \frac{490.5 + 294.3}{2} \times 10 = 3{,}924\ \text{kN} \\ U_2 &= \frac{294.3 + 98.1}{2} \times 30 = 5{,}886\ \text{kN} \\ U &= 3{,}924 + 5{,}886 = 9{,}810\ \text{kN} \end{aligned}$$

Its line of action (moments about heel) is at 15.33 m from the heel.

For comparison, without the gallery: $U = \frac{490.5 + 98.1}{2}\times40 = 11{,}772$ kN, so the gallery reduces uplift by about 17%.

Answer: Net hydrostatic thrust = 11,772 kN/m (upstream 12,262.5 kN/m, tailwater 490.5 kN/m); uplift force = 9,810 kN/m acting 15.33 m from the heel.

Data (per metre length)

• Dam height 50 m, freeboard 5 m, so water depth $h = 45$ m (no tailwater).
• Top width 4.5 m, base 42 m, upstream face vertical; downstream slope $\tan\alpha = 37.5/50 = 0.75$.
• $\gamma_w = 10$, $\gamma_c = 25$ kN/m³; $\mu = \tan 37° = 0.754$; uplift coefficient C = 0.8.
• Crushing strength $= 10$ kg/cm² $= 10 \times 9.81 \times 10^4$ N/m² $= 981$ kN/m².

     4.5
    +--+   ---
    |  |\   5 m FB
 45 |W1| \
 m  |  |W2\    50 m
 -> |  |   \
heel+--+----\ toe
    <-- 42 m -->

Forces and moments about the toe

$$\begin{aligned} \sum V &= 5{,}625 + 23{,}437.5 - 7{,}560 = 21{,}502.5\ \text{kN} \\ \sum M_R &= 809{,}531, \quad \sum M_O = 363{,}555\ \text{kN·m} \end{aligned}$$

(i) Overturning

$$FS_o = \frac{809{,}531}{363{,}555} = 2.23 > 1.5 \quad \text{(safe)}$$

(ii) Sliding

$$FS_s = \frac{0.754 \times 21{,}502.5}{10{,}125} = 1.60 > 1 \quad \text{(safe)}$$

(iii) Tension

$$\begin{aligned} \bar{x} &= \frac{809{,}531 - 363{,}555}{21{,}502.5} = 20.74\ \text{m from toe} \\ e &= 21 - 20.74 = 0.26\ \text{m} < B/6 = 7\ \text{m} \end{aligned}$$

No tension (safe).

(iv) Crushing

$$\begin{aligned} p_{toe} &= \frac{21{,}502.5}{42}\left(1 + \frac{6\times0.26}{42}\right) = 511.97 \times 1.037 = 530.9\ \text{kN/m}^2 \\ p_{heel} &= 511.97 \times 0.963 = 493.0\ \text{kN/m}^2 \\ \sigma_1 &= p_{toe}\sec^2\alpha = 530.9 \times 1.5625 = 829.6\ \text{kN/m}^2 \end{aligned}$$

Both are below 981 kN/m² (safe).

Recommendation: The dam is safe in all four checks. The resultant is almost at the centre of the base, so stresses are nearly uniform; the section is on the heavy side and the downstream slope could be steepened to save concrete, after re-checking with earthquake and silt loads.

Geometry and data (per metre length)

• Water depth 30 m, water surface 1 m below crest, so dam height $= 31$ m.
• Top width 2 m; downstream face vertical for 1.5 m, then slopes 0.75H : 1V for the remaining $31 - 1.5 = 29.5$ m, giving horizontal projection $0.75 \times 29.5 = 22.125$ m.
• Base width $B = 2 + 22.125 = 24.125$ m. No tailwater.
• $\gamma_w = 9.81$, $\gamma_c = 24$ kN/m³; $\mu = \tan37° = 0.754$; uplift C = 0.8 (triangular, $0.8\gamma_w h$ at heel to zero at toe).

     2 m
    +--+    1 m above WL
    |  +    1.5 m vertical
 30 |   \
 m  |W1  \  0.75H:1V
 -> |  W2 \       29.5 m
heel+------\ toe
    <-24.125->

Forces and moments about the toe

$$\begin{aligned} \sum V &= 1{,}488 + 7{,}832.3 - 2{,}840.0 = 6{,}480.3\ \text{kN} \\ \sum H &= 4{,}414.5\ \text{kN} \\ \sum M_R &= 149{,}936, \quad \sum M_O = 89{,}822\ \text{kN·m} \end{aligned}$$

Safety against sliding

$$FS_s = \frac{\mu\sum V}{\sum H} = \frac{0.754 \times 6{,}480.3}{4{,}414.5} = 1.11 > 1$$

Safety against overturning

$$FS_o = \frac{149{,}936}{89{,}822} = 1.67 > 1.5$$

Check of resultant: $\bar{x} = (149{,}936 - 89{,}822)/6{,}480.3 = 9.28$ m from toe, $e = 12.06 - 9.28 = 2.79$ m $< B/6 = 4.02$ m, so there is no tension at the heel (toe stress 454.7 kN/m², heel stress 82.5 kN/m²).

Answer: FS against sliding = 1.11 and FS against overturning = 1.67; the dam is safe.

Recommendations: the sliding margin is small (just above 1), so:

• provide foundation drainage holes/gallery to reduce uplift,
• key the base into sound rock or give it a slight upstream slope to increase shear resistance,
• check the shear friction factor using the shear strength of the rock–concrete joint.

Data (per metre length)

• Dam height 40 m, free board 3 m, so upstream water depth $h_1 = 37$ m.
• Tailwater depth $h_2 = 5$ m, base width $B = 30$ m, $\gamma_w = 9.81$ kN/m³.
• Full uplift is assumed (C = 1): pressure varies linearly from $\gamma_w h_1$ at the heel to $\gamma_w h_2$ at the toe.

(a) Without drainage gallery

Heel: $9.81 \times 37 = 362.97$ kN/m²; toe: $9.81 \times 5 = 49.05$ kN/m².

$$U = \frac{362.97 + 49.05}{2} \times 30 = 6{,}180.3\ \text{kN/m}$$

(b) With drainage gallery 6 m from upstream face

By the usual rule (IS 6512), the uplift head at the line of drains is the tailwater head plus one-third of the difference:

$$h_d = h_2 + \frac{1}{3}(h_1 - h_2) = 5 + \frac{32}{3} = 15.67\ \text{m} \Rightarrow p_d = 153.69\ \text{kN/m}^2$$

heel  drain                    toe
 |<-6->|<-------- 24 --------->|
 |\    |                       |
 | \   |                       |
 |  \__|______                 |
 |     |      -------______    |
362.97 153.69             49.05

$$\begin{aligned} U_1 &= \frac{362.97 + 153.69}{2} \times 6 = 1{,}550.0\ \text{kN/m} \\ U_2 &= \frac{153.69 + 49.05}{2} \times 24 = 2{,}432.9\ \text{kN/m} \\ U &= 1{,}550.0 + 2{,}432.9 = 3{,}982.9\ \text{kN/m} \end{aligned}$$

Answer: Uplift force = 6,180.3 kN per metre without drains; 3,982.9 kN per metre with the drainage gallery at 6 m. The gallery cuts uplift by about 36%.

A gravity dam is a solid concrete or masonry dam that resists all external forces by its own weight. The forces acting on it are:

          crest
   ~~~~~ +---+  <- wave pressure Pw
   |     |   |\
 Pi ->   |   | \  <- wind
   |     | W |  \    W = self weight
 P1 ->   |   |   \   (+ EQ: aW horizontal)
   |     |   | W2 \
 Ps ->   |   |     \  ~~~~ tailwater
 (silt)  |   |      \  <- P2
         +---+-------+ toe
  heel   ^  ^  ^  ^  ^
         U (uplift)

Forces acting on a gravity dam

1. Self weight (W): the main stabilising force; $W = \gamma_c \times$ area per metre length, acting at the centroid. Divided into rectangles and triangles for calculation.
2. Water pressure:
   • Upstream horizontal: $P_1 = \frac{1}{2}\gamma_w h_1^2$ at $h_1/3$ above base.
   • Vertical weight of water on a sloping upstream face (if any).
   • Tailwater: $P_2 = \frac{1}{2}\gamma_w h_2^2$ acting upstream, plus weight of water on the downstream slope.
3. Uplift pressure (U): water seeping through the foundation and joints pushes upward. It varies from $C\gamma_w h_1$ at heel to $\gamma_w h_2$ at toe; drainage galleries reduce it (head at drains $= h_2 + \frac{1}{3}(h_1 - h_2)$).
4. Silt pressure (Ps): silt deposited against the upstream face: $P_s = \frac{1}{2}\gamma_{sub} h_s^2 K_a$, $K_a = \frac{1-\sin\phi}{1+\sin\phi}$, at $h_s/3$.
5. Wave pressure (Pw): due to wind waves; wave height $h_w = 0.032\sqrt{VF} + 0.763 - 0.271F^{1/4}$ for fetch $F < 32$ km; $P_w = 2\gamma_w h_w^2$, acting $3h_w/8$ above still water level.
6. Earthquake forces:
   • Inertia force on the dam $= \alpha_h W$ horizontally (and $\alpha_v W$ vertically) at the centroid.
   • Hydrodynamic pressure of water: Westergaard/Zanger, e.g. $P_e = 0.726\,p_e\,h$ with $p_e = C_m\alpha_h\gamma_w h$.
7. Ice pressure: in cold regions, about 250 kN/m² over the ice thickness.
8. Wind pressure: small; usually neglected.

Stability requirements

The dam must be safe for all load combinations (empty reservoir, full reservoir, flood, earthquake):

Also, the foundation must be safe in bearing and the dam must have adequate freeboard to avoid overtopping.

A gravity dam stays in place by its own weight. The forces acting on it are grouped into stabilising forces (self weight, tailwater) and destabilising forces (water pressure, uplift, silt, waves, earthquake).

          crest
   ~~~~~ +---+  <- wave pressure Pw
   |     |   |\
 Pi ->   |   | \  <- wind
   |     | W |  \    W = self weight
 P1 ->   |   |   \   (+ EQ: aW horizontal)
   |     |   | W2 \
 Ps ->   |   |     \  ~~~~ tailwater
 (silt)  |   |      \  <- P2
         +---+-------+ toe
  heel   ^  ^  ^  ^  ^
         U (uplift)

1. Self weight (W)

Main resisting force. $W = \gamma_c \times$ cross-sectional area (per metre length), acting through the centroid. The section is split into rectangles and triangles.

2. Water pressure

• Upstream: $P_1 = \frac{1}{2}\gamma_w h_1^2$, horizontal, at $h_1/3$ above the base. If the upstream face is sloping, the weight of water above the slope acts vertically downward.
• Tailwater: $P_2 = \frac{1}{2}\gamma_w h_2^2$ acting upstream, and the weight of water on the downstream slope.

3. Uplift pressure (U)

Seepage water under the base and in the joints acts upward and reduces the effective weight. It varies linearly from $C\gamma_w h_1$ at the heel to $\gamma_w h_2$ at the toe; drains reduce it.

4. Silt (sediment) pressure

$P_s = \frac{1}{2}\gamma_{sub}h_s^2 K_a$, where $K_a = (1-\sin\phi)/(1+\sin\phi)$, acting at $h_s/3$ from the base.

5. Wave pressure

Wind waves strike the top of the dam. $P_w = 2\gamma_w h_w^2$ acting at $3h_w/8$ above the reservoir level; wave height from Molitor's formula.

6. Earthquake forces

• Horizontal inertia force on the dam $\alpha_h W$ (and vertical $\alpha_v W$).
• Hydrodynamic pressure of the reservoir water (Westergaard/Zanger method), acting at about $4h/(3\pi)$ above the base.

7. Ice and wind pressure

Ice thrust (cold regions) about 250 kN/m² on the ice thickness; wind pressure is small and usually neglected.

Dam

A dam is a barrier built across a river or stream to store water, raise its level or divert it. The stored water is used for hydropower, irrigation, water supply and flood control.

Types of dams

Examples in Nepal: Kulekhani (rockfill storage dam), Kaligandaki A (concrete gravity diversion dam).

Forces acting on a gravity dam

          crest
   ~~~~~ +---+  <- wave pressure Pw
   |     |   |\
 Pi ->   |   | \  <- wind
   |     | W |  \    W = self weight
 P1 ->   |   |   \   (+ EQ: aW horizontal)
   |     |   | W2 \
 Ps ->   |   |     \  ~~~~ tailwater
 (silt)  |   |      \  <- P2
         +---+-------+ toe
  heel   ^  ^  ^  ^  ^
         U (uplift)

1. Self weight (W): main stabilising force, $W = \gamma_c \times$ area, at the centroid.
2. Upstream water pressure: $P_1 = \frac{1}{2}\gamma_w h_1^2$ at $h_1/3$; plus water weight on a sloping upstream face.
3. Tailwater pressure: $P_2 = \frac{1}{2}\gamma_w h_2^2$ acting upstream, plus water weight on the downstream slope.
4. Uplift pressure (U): upward seepage pressure on the base, from $C\gamma_w h_1$ at heel to $\gamma_w h_2$ at toe; reduced by drainage galleries.
5. Silt pressure: $P_s = \frac{1}{2}\gamma_{sub}h_s^2K_a$.
6. Wave pressure: $P_w = 2\gamma_w h_w^2$ at $3h_w/8$ above water level.
7. Earthquake forces: inertia force $\alpha_h W$ on the dam and hydrodynamic water pressure (Westergaard/Zanger).
8. Ice and wind pressure: ice thrust in cold regions; wind usually neglected.

Elementary profile

The elementary profile of a gravity dam is a right-angled triangle with a vertical upstream face, zero top width and base width B, with water up to its apex (height H). Water pressure is zero at the top and grows linearly with depth, so a triangle is the ideal shape: it uses the least material and, if the base is wide enough, has no tension under full or empty reservoir. Practical profiles are made by adding top width and freeboard to it.

        |\
        | \
  P --> |  \      H
  (H/3) | W \
        |    \
   heel |__.__\ toe
        ^  ^   ^
        U (KγwH at heel -> 0 at toe)
        <--B-->

Forces per metre length

With $G$ = specific gravity of dam material, $K$ = uplift intensity coefficient:

$$\begin{aligned} W &= \frac{1}{2}BH\,G\gamma_w \quad (\text{at } B/3 \text{ from heel}) \\ P &= \frac{1}{2}\gamma_w H^2 \quad (\text{at } H/3 \text{ above base}) \\ U &= \frac{1}{2}K\gamma_w H B \quad (\text{at } B/3 \text{ from heel}) \end{aligned}$$

Proof of $B = H/\sqrt{G-K}$ (no-tension condition)

For no tension at the heel with full reservoir, the resultant must pass through the outer middle-third point, i.e. at $2B/3$ from the heel ($B/3$ from the toe). Taking moments about this point:

• W and U act at $B/3$ from the heel, so their lever arm is $2B/3 - B/3 = B/3$.
• P has lever arm $H/3$.

For the resultant to pass through this point, the net moment must be zero:

$$\begin{aligned} (W - U)\frac{B}{3} &= P\frac{H}{3} \\ \left(\frac{1}{2}BHG\gamma_w - \frac{1}{2}K\gamma_w HB\right)B &= \frac{1}{2}\gamma_w H^2 \cdot H \\ \frac{1}{2}\gamma_w B^2 H (G - K) &= \frac{1}{2}\gamma_w H^3 \\ B^2 (G - K) &= H^2 \\ B &= \frac{H}{\sqrt{G - K}} \end{aligned}$$

Hence proved. Without uplift ($K = 0$), $B = H/\sqrt{G}$; for concrete ($G = 2.4$) this gives $B = 0.645H$, and with full uplift ($K = 1$) $B = 0.845H$.

No-sliding condition (for comparison)

$\mu(W - U) = P$ gives $B = \dfrac{H}{\mu(G - K)}$. The larger of the two widths is adopted.