Headworks of Storage Plants

Purpose of a spillway

A spillway is a structure that safely passes surplus water (floods) from the reservoir to the downstream river without damaging the dam. It controls the reservoir level, protects the dam (especially an earthfill dam) from overtopping, and dissipates the energy of the released water so that the toe and river bed are not scoured. It is provided with a gate or an ungated crest at or above the full supply level.

Types of spillways (by key features)

Ogee (overflow) spillway

The crest has the shape of the underside of the nappe of a sharp-crested weir (USBR profile), so that the pressure on the crest is nearly atmospheric at the design head. The downstream face is sloping and ends in a curved bucket or stilling basin.

        FSL   ____  crest gates / piers
  ~~~~~~~~~~~/ ~~\
 Reservoir  | ogee  \
            |crest   \  d/s face (slope)
            |          \
            |           \___ bucket / stilling basin
          --+--------------==========--> river
             Dam body (concrete gravity)

• Discharge: $Q = C\,L\,H^{3/2}$, with $C\approx2.2$ (SI) for the design head, $L$ the effective crest length and $H$ the head over crest.
• At heads below the design head the nappe sticks to the crest (pressure above atmospheric); at heads above it, negative pressure and cavitation can occur.
• Advantages: efficient, cheap in a concrete dam, and gives a large capacity. A stilling basin or bucket dissipates the energy.

For a homogeneous earth dam with a horizontal filter (drain) at the toe, Casagrande's method treats the top flow line (phreatic line) as a basic parabola with its focus at the starting point $F$ of the filter. The seepage follows Darcy's law with the Dupuit assumption.

Sketch and notation

        B' (corrected entry)
 FRL ~~~~~|.\
          |  `.  phreatic line
   h      |    `-.
          |        `--._
 ---------+---------------`-.----
 A       E           d    F=======  filter
          |<--- d ---->|<-- horizontal drain -->

• $h$ = height of water above the dam base (head), $d$ = horizontal distance from the corrected entry point $B$ to the filter start $F$.
• $B$ is the corrected entry point: it lies $0.3\,\Delta$ horizontally inside the dam (towards the downstream) from the point where the water surface meets the u/s face, where $\Delta$ is the horizontal projection of the wetted part of the u/s face.

Equation of the phreatic line

Put the origin at the focus $F$. A parabola with focus at the origin and the focal length $y_0/2$ (vertex at $x=-y_0/2$) is

$$y^2 = 2y_0x + y_0^2$$

where $y_0$ is the ordinate of the parabola at the focus ($x=0$).

Seepage discharge (Dupuit)

At any section of the dam the flow per unit width is $q = k\,i\,A$, with the hydraulic gradient $i=dy/dx$ and the flow area $A=y\times1$:

$$q = k\,y\,\frac{dy}{dx}$$

Differentiating the parabola: $2y\,dy = 2y_0\,dx$, so $y\,\dfrac{dy}{dx} = y_0$. Therefore

$$q = k\,y_0$$

Finding $y_0$

The parabola passes through $B$, where $x = d$ and $y = h$:

$$h^2 = 2y_0d + y_0^2 \;\Rightarrow\; y_0^2 + 2dy_0 - h^2 = 0$$ $$y_0 = \sqrt{d^2+h^2} - d$$ $$\boxed{q = k\left(\sqrt{d^2+h^2}-d\right)}$$

Drawing the phreatic line

1. Mark the corrected entry point $B$ at $0.3\Delta$ from the water-line intersection and measure $d$ and $h$.
2. Calculate $y_0$ and $q$.
3. Find points of the parabola from $y=\sqrt{2y_0x+y_0^2}$ for several values of $x$ and plot them.
4. Join the parabola smoothly to the u/s face at $B$, making it perpendicular to the u/s face (an equipotential line).
5. At the filter start the line meets the horizontal drain tangentially; with a horizontal filter no correction at the exit is needed (the line ends at $F$, with the exit face being the filter).

Statement

For a gravity dam the resultant of all forces on any horizontal section should cut the base within its middle third; i.e. the eccentricity $e\le B/6$ from the centre. Then there is no tension (at the heel when the reservoir is full, at the toe when it is empty).

Derivation

Consider a horizontal section of width $B$ and unit length. Let $\Sigma V$ be the total vertical load and $R$ the resultant, cutting the base at a distance $e$ from the centre of the base.

The resultant is equivalent to a vertical force $\Sigma V$ at the centroid of the base and a moment $M=\Sigma V\cdot e$. The stress at the base is (axial plus bending):

$$\sigma = \frac{\Sigma V}{A} \pm \frac{M\,y}{I}$$

with $A = B\times1$, $I = \dfrac{B^3}{12}$, and $y=\dfrac{B}{2}$ at the extreme edge:

$$\sigma = \frac{\Sigma V}{B} \pm \frac{\Sigma V\,e\,(B/2)}{B^3/12} = \frac{\Sigma V}{B}\left(1 \pm \frac{6e}{B}\right)$$

So

$$\sigma_{max} = \frac{\Sigma V}{B}\left(1+\frac{6e}{B}\right), \qquad \sigma_{min} = \frac{\Sigma V}{B}\left(1-\frac{6e}{B}\right)$$

The minimum stress is at the heel when the resultant lies toward the toe (reservoir full), and it becomes tensile if

$$1-\frac{6e}{B} < 0 \;\Rightarrow\; e > \frac{B}{6}$$

To avoid tension: $\sigma_{min}\ge0$, hence

$$e \le \frac{B}{6}$$

The eccentricity $e=B/6$ means the resultant passes through the limit of the middle third, a distance $B/3$ from the toe. Thus the resultant must lie within the middle third of the base ($B/3$ to $2B/3$ from the toe). At the limit $e=B/6$, the stress at the heel is zero and at the toe $\sigma_{max}=2\Sigma V/B$.

Where $e$ comes from

$$\bar x = \frac{\Sigma M_R - \Sigma M_O}{\Sigma V}\ (\text{from the toe}),\qquad e = \frac{B}{2}-\bar x$$

Why it is used

• It keeps the entire base in compression, so there is no crack at the heel to admit water (which would raise uplift).
• The dam is economical because the compressive strength of concrete is used well.
• It also gives a factor of safety against overturning of about 3 or more at the limit position.

Common types of crest gates

1. Vertical lift (fixed wheel / roller) gate
2. Radial (tainter) gate
3. Drum gate
4. Stoney gate (roller-train gate)
5. Flap (hinged) gate
6. Rolling (cylinder) gate

Radial (tainter) gate

A radial gate has a curved skin plate that is a section of a cylinder, supported by horizontal girders and radial arms. The arms are hinged at a trunnion pin on the piers downstream of the gate, and the centre of the arc is at the trunnion, so the water thrust passes through the trunnion pin. The gate is hoisted by wire ropes or hydraulic cylinders and rotates about the trunnion.

           trunnion pin (O)
              o
             /|\   radial arms
            / | \
           /  |  \
   ~~~~~~~/~~~|~~~\  skin plate (arc)
 Water -->    |    ) <-- gate
 ___________ sill ____________
 Section through spillway pier

Working mechanism

• Closed: the skin plate rests on the sill and the seals at the sides and bottom keep the opening watertight.
• Opened: ropes attached to the lower part of the gate lift it by rotation about the trunnion, leaving a gap at the bottom for the discharge. Gate position is adjustable to regulate flow.
• Water thrust is carried through the arms to the trunnion and from there into the pier; only a small hoisting force is needed because the thrust passes through the hinge (little friction).

Advantages: low hoisting force, no grooves in the piers, compact, good flow control. Disadvantages: large piers needed to take the trunnion loads, and a larger spillway bay length is required.

Vertical lift gate (for comparison)

A flat steel gate sliding or rolling in grooves in the piers, lifted vertically by a gantry hoist. It is simple, but the lifting force is large because of friction from the water thrust and grooves in the piers.

An intake of a storage plant draws water from the reservoir into the tunnel or penstock. It must operate at all reservoir levels from the minimum operating level (MOL) to the full supply level (FSL), prevent entry of debris and air, and be able to shut off the flow.

Types of intakes

1. Tower (free-standing) intake: A vertical tower in the reservoir connected to the dam crest by a bridge, with ports at several levels. Used for earth and rockfill dams.
2. Intake in the dam body (dam-mounted): The intake structure is built on the upstream face of a concrete dam, with the penstock passing through the dam; economical and compact.
3. Submerged (inclined or horizontal) intake: A rock-cut intake at the abutment connected with a tunnel; used in rocky valleys with a long tunnel.
4. Shaft (gate shaft) intake: A vertical shaft in the abutment with the gate chamber and connected to the tunnel.
5. Floating / multilevel intake: Draws the best-quality (warmer, cleaner) water from different levels.

General arrangement of a dam-mounted intake

   FSL ~~~~~~~~~~~
   MOL ~~~~~~~~~~~~~      Gate hoist house
        |  Trash rack      |  Air vent
        |  ___________     |
 Res.   | |intake mouth|-- Service gate
        | |bell mouth  |        |
        |_|____________|--------+--> Penstock
          Dam body (concrete)       to turbine

Components:

• Trash rack: keeps out floating and large debris.
• Bellmouth transition: smooth entrance to reduce head loss.
• Gates: a service (control) gate and an emergency (stoplog) gate.
• Air vent: supplies air behind the gate when it closes and prevents vacuum.
• Sediment control: the intake sill is set above the dead storage level.

Intake tower (in the reservoir)

  bridge    ____________
 =========|  hoist     |
 Dam crest|            |
          |  upper port|--
          |  mid port  |--  gates
          |  lower port|--
          |____________|
             |
        Tunnel/penstock

The ports at different levels let the operator choose the water level of the draw-off at varying reservoir levels.

Given data and assumptions: homogeneous dam without any drain; foundation impervious; tail water ignored. Casagrande's method (basic parabola with focus at the d/s toe) is used.

• Dam height $= 290-235 = 55$ m; water head $h = 285-235 = 50$ m; free board $= 5$ m.
• $k = 5.1\times10^{-4}$ cm/s $= 5.1\times10^{-6}$ m/s.

 290 ______5 m______
     /|              \
  3:1/ |   homogeneous \ 2:1
    /  |    dam        \
 285 ~~|                 \
    /  |                  \
 235 ///////////////////////////
     |<--------- 280 m ------->|

Step 1: Geometry

$$\begin{aligned} \text{Base width } B &= 3(55)+5+2(55) = 280\ \text{m} \\ \text{Wetted horizontal projection } b &= 3\times 50 = 150\ \text{m} \\ \text{Entry correction } 0.3b &= 45.0\ \text{m} \\ d &= B-0.7b = 280-105.0 = 175.0\ \text{m} \end{aligned}$$

Here $d$ is the horizontal distance from the corrected entry point to the d/s toe (the focus $F$).

Step 2: Basic parabola

$$y_0=\sqrt{h^2+d^2}-d=\sqrt{50^2+175.0^2}-175.0=7.00\ \text{m}$$ $$y=\sqrt{y_0^2+2y_0x}$$

with $x$ measured from the focus (d/s toe) towards the u/s side:

At $x=d=175.0$ m the parabola reaches $y=h=50$ m, which is the corrected entry point. The u/s end of the actual line is made to meet the u/s face at right angles.

Step 3: Exit correction (d/s face)

The d/s slope angle is $\alpha=\tan^{-1}(1/2)=26.57^\circ$. The parabola meets the d/s face at distance

$$a+\Delta a=\frac{y_0}{1-\cos\alpha}=\frac{7.00}{1-\cos 26.57^\circ}=66.3\ \text{m}$$

(measured along the slope from the toe). Casagrande's factor for $\alpha\approx 30^\circ$ is $\Delta a/(a+\Delta a)\approx0.36$, so $\Delta a=23.9$ m and the line leaves the face at $a=42.5$ m along the slope from the toe. The last part of the line is drawn tangent to the parabola and blended to this exit point.

Step 4: Discharge

$$q=k\,y_0=5.1\times10^{-6}\times 7.00=3.571 \times 10^{-5}\ \text{m}^3/\text{s per m}$$

Answer: the phreatic line is the parabola $y=\sqrt{49.04+14.01\,x}$ (x from d/s toe, with entry and exit corrections), $y_0=7.00$ m; seepage $q=3.571 \times 10^{-5}\ \text{m}^3/\text{s}$ per metre length of dam ($\approx 3.09$ m³/day per m).

Given data and assumptions: homogeneous dam on an impervious foundation, no drain, tail water neglected. Casagrande's basic parabola with focus at the d/s toe.

• Dam height $= 300-278 = 22$ m; water head $h = 297.5-278 = 19.5$ m; $k = 5\times10^{-4}$ cm/s $=5\times10^{-6}$ m/s.

Step 1: Geometry

$$\begin{aligned} B &= 3(22)+4.5+2(22)=114.5\ \text{m}\\ b &= 3\times 19.5 = 58.5\ \text{m},\quad 0.3b=17.55\ \text{m}\\ d &= B-0.7b = 114.5-40.95 = 73.55\ \text{m} \end{aligned}$$

Step 2: Basic parabola

$$y_0=\sqrt{h^2+d^2}-d=\sqrt{19.5^2+73.55^2}-73.55=2.54\ \text{m}$$ $$y=\sqrt{y_0^2+2y_0x}=\sqrt{6.46+5.08\,x}$$

Coordinates ($x$ from the d/s toe, $y$ above the base):

At $x=d$, $y=h=19.5$ m (corrected entry point A, located $0.3b=17.55$ m from the water-line on the u/s side).

Step 3: Corrections

• Entry: the line starts at right angles to the u/s face, joining the parabola smoothly at A.
• Exit: $\alpha=\tan^{-1}(1/2)=26.57^\circ$, so $a+\Delta a=\dfrac{y_0}{1-\cos\alpha}=24.1$ m. With $\Delta a/(a+\Delta a)\approx0.36$, $\Delta a=8.7$ m and the line leaves the d/s face at $a=15.4$ m along the slope from the toe.

Step 4: Discharge

$$q=k\,y_0=5\times10^{-6}\times2.54=1.271 \times 10^{-5}\ \text{m}^3/\text{s per m}$$

Answer: $y_0=2.54$ m, phreatic line $y=\sqrt{6.46+5.08\,x}$; discharge $q=1.271 \times 10^{-5}$ m³/s per metre ($\approx1.098$ m³/day per m).

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

(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.

Assumptions (from the figure description)

• Dam height $=5+40+20=65$ m. Crest width 10 m. U/s face vertical down to 20 m above the base, then flared with a horizontal run of 15 m to the heel. Water level 5 m below the crest, so $H_w=60$ m. D/s face vertical for the top 10 m, then sloping (35 m horizontal run over 55 m height) to the toe. Base width $B=15+10+35=60$ m.
• Tail water depth $H_t=8$ m. $\gamma_c=24$ kN/m³, $\gamma_w=10$ kN/m³, $\mu=0.75$.
• Uplift: full hydrostatic head at the heel ($\gamma_wH_w=600$ kN/m²) and at the toe ($\gamma_wH_t=80$ kN/m²). The drainage gallery (assumed with its centre 5 m from the heel) cuts the uplift at the gallery to $80+\frac13(600-80)=253.3$ kN/m². Earthquake is neglected.

 crest 10 m
  +--------+
 ~|~~5 m~~~|  water level
  |        |  d/s vertical 10 m
  |  40 m  \
  |         \  slope: 35 m run in 55 m
  |  20 m    \
   \__________\_ tail water 8 m
  15     10     35   B = 60 m

Forces and moments about the toe

(The tail water also acts on the sloping d/s face; its vertical component is included as a resisting load and its horizontal thrust acts at $H_t/3$ above the base.)

Stability

• $\Sigma V=50,003.6-11,300.0=38,703.6$ kN/m; $\Sigma H=18,000.0-320.0=17,680.0$ kN/m
• $\Sigma M_R=1,741,699$ kN·m/m (including tail water); $\Sigma M_O=779,167$ kN·m/m

Overturning: $FS_o=2.24$ — safe (> 1.5).

Sliding (friction only): $FS_s=\mu\Sigma V/\Sigma H = 0.75\times38,703.6/17,680.0=1.64$ — safe (> 1).

Shear-friction factor: $SFF=\dfrac{\mu\Sigma V+\tau B}{\Sigma H}=\dfrac{0.75\times38,703.6+2200\times60}{17,680.0}=9.1$ — very safe (> 3 to 5).

Resultant and stresses: $\bar x=(962,532)/38,703.6=24.87$ m from the toe; $e=30-24.87=5.13$ m; $B/6=10$ m (inside the middle third).

$$\sigma_{toe}=645.1\,(1+0.513)=976.0\ \text{kN/m}^2,\quad \sigma_{heel}=645.1\,(1-0.513)=314.1\ \text{kN/m}^2$$

Both are compressive and below the allowable 2500 kN/m².

Shear stress on a horizontal plane near the toe and heel

Face slopes: d/s $\tan\phi_d=35/55=0.636$; u/s (flare) $\tan\phi_u=15/20=0.75$. Water pressures at the faces: $p'=\gamma_wH_t=80$ kN/m² at the toe and $p=\gamma_wH_w=600$ kN/m² at the heel.

$$\tau_{toe}=(\sigma_{toe}-p')\tan\phi_d=(976.0-80)\times0.636=570.2\ \text{kN/m}^2$$ $$\tau_{heel}=(p-\sigma_{heel})\tan\phi_u=(600-314.1)\times0.75=214.4\ \text{kN/m}^2$$

Both are much smaller than the average shear strength of 2200 kN/m². Principal stresses: $\sigma_{1,toe}=\sigma_{toe}\sec^2\phi_d-p'\tan^2\phi_d=1338.9$ kN/m², $\sigma_{1,heel}=\sigma_{heel}\sec^2\phi_u-p\tan^2\phi_u=153.3$ kN/m².

Conclusion: the dam is safe against overturning (2.24), sliding (1.64, shear-friction factor 9.1), tension (no tension: minimum stress 314 kN/m² is compressive) and crushing (maximum 976 kN/m² < 2500 kN/m²).

Assumptions and data

• RL of base $=305$ m; top $=390$ m (the printed 1390 m is read as 390 m); dam height $=85$ m. Maximum water level 385 m, so $H_w=385-305=80$ m. Tail water 311 m, so $H_t=6$ m.
• Crest width 5 m; u/s face vertical; d/s face vertical from 390 to 380 m, then sloping to the toe. Base width $B=56$ m, so the d/s slope runs 51 m horizontally over 75 m (about 2 : 3 as drawn), $\tan\phi_d=51/75=0.68$.
• $\gamma_w=10$, $\gamma_c=24$ kN/m³, $\mu=0.70$, allowable stress 2500 kN/m², shear strength 2000 kN/m².
• Uplift: $\gamma_wH_w=800$ kN/m² at the heel, $\gamma_wH_t=60$ kN/m² at the toe; the gallery (8 m from the u/s face) reduces the pressure there to $60+\frac13(800-60)=306.7$ kN/m². Earthquake is neglected.

 390 +----+
     |    | crest 5 m
 385~|~~~~|  MWL (H_w = 80 m)
     |    |
     |    |  vertical to 380
     |    \
     |     \  slope 2 : 3
     |      \
     | G     \
 311 |        \~~ tail water 6 m
 305 +---------\
     |<-- B = 56 m -->|

Forces and moments about the toe

Stability checks

• $\Sigma V = \Sigma W - U = 56,222.4 - 13,226.7 = 42,995.7$ kN/m
• $\Sigma H = 32,000.0 - 180.0 = 31,820.0$ kN/m
• $\Sigma M_R = 2,106,826$ kN·m/m (tail water thrust moment included); $\Sigma M_O = 1,344,711$ kN·m/m

Overturning: $FS_o = \Sigma M_R/\Sigma M_O = 2,106,826/1,344,711 = 1.57$ — safe (> 1.5).

Sliding (friction): $FS_s = \mu\Sigma V/\Sigma H = 0.7\times42,995.7/31,820.0 = 0.95$ — less than 1: not safe by friction alone.

Shear-friction factor: $SFF = (\mu\Sigma V + \tau B)/\Sigma H = (0.7\times42,995.7 + 2000\times56)/31,820.0 = 4.47$ — safe (> 3 to 5).

Resultant: $\bar x = (\Sigma M_R-\Sigma M_O)/\Sigma V = 762,115/42,995.7 = 17.73$ m from the toe; $e = B/2-\bar x = 28-17.73 = 10.27$ m; $B/6 = 9.33$ m — outside the middle third (tension at the heel).

$$\sigma_{toe} = \frac{\Sigma V}{B}\left(1+\frac{6e}{B}\right) = 767.8\,(1+1.101) = 1613.0\ \text{kN/m}^2$$ $$\sigma_{heel} = 767.8\,(1-1.101) = -77.4\ \text{kN/m}^2$$

Toe stress 1613 kN/m² compared with the allowable 2500 kN/m²: safe against crushing. Heel stress is -77 kN/m² (tensile): tension occurs at the heel.

Shear and principal stress at the toe ($\tan\phi_d=0.680$, $p'=60.0$ kN/m²):

$$\tau_{toe} = (\sigma_{toe}-p')\tan\phi_d = (1613.0-60.0)\times0.680 = 1056.0\ \text{kN/m}^2$$ $$\sigma_{1,toe} = \sigma_{toe}\sec^2\phi_d - p'\tan^2\phi_d = 1613.0\,(1+0.462) - 60.0\times0.462 = 2331.1\ \text{kN/m}^2$$

Shear stress near the toe: $\tau_{toe}$ is given above and is far smaller than the average shear strength of 2000 kN/m².

Conclusion: overturning and the shear-friction factor are satisfactory; the friction-only sliding factor (0.95) is below 1, which is covered by the shear strength of the concrete-rock contact (SFF above). The resultant lies slightly outside the middle third, so a small tension of about 77 kN/m² appears at the heel; it is small, but a better drainage or a slightly wider base would remove it.

Assumptions (figure not available in full)

• Dam height 50 m; crest width 5 m; water level 2 m below the crest, $H_w=48$ m; tail water $H_t=5$ m.
• U/s face vertical from the crest to 15 m above the base, then flared at 1 : 1 to the heel (15 m horizontal run). D/s face runs straight from the d/s edge of the crest to the toe; total base width $B=57$ m gives a d/s run of $57-15-5=37$ m over 50 m ($\tan\phi_d=0.74$).
• $\gamma_c=24$, $\gamma_w=9.81$ kN/m³, $\mu=0.70$, uplift coefficient $K=0.45$: heel ordinate $=\gamma_wH_t+K\gamma_w(H_w-H_t)=238.9$ kN/m², toe ordinate $=\gamma_wH_t=49.1$ kN/m² (the gallery near the heel is the reason for this reduced uplift).
• Allowable stresses: foundation 50 kgf/cm² $=4905$ kN/m²; dam body 10.5 kgf/cm² $=1030$ kN/m².

        5 m
    +--------+
 ~~~|~~~~~~~~|  water (48 m)
    |        |\
    |        | \  d/s slope
    |        |  \
   /|        |   \ ~~ tail water 5 m
  /_|________|____\
  15 m flare      B = 57 m

Forces and moments about the toe

Checks

• $\Sigma V = \Sigma W - U = 36,950.3 - 8,205.8 = 28,744.5$ kN/m
• $\Sigma H = 11,301.1 - 122.6 = 11,178.5$ kN/m
• $\Sigma M_R = 1,209,574$ kN·m/m (tail water thrust moment included); $\Sigma M_O = 466,078$ kN·m/m

Overturning: $FS_o = \Sigma M_R/\Sigma M_O = 1,209,574/466,078 = 2.60$ — safe (> 1.5).

Sliding (friction): $FS_s = \mu\Sigma V/\Sigma H = 0.7\times28,744.5/11,178.5 = 1.80$ — safe (> 1).

Resultant: $\bar x = (\Sigma M_R-\Sigma M_O)/\Sigma V = 743,496/28,744.5 = 25.87$ m from the toe; $e = B/2-\bar x = 28.5-25.87 = 2.63$ m; $B/6 = 9.50$ m — inside the middle third (no tension).

$$\sigma_{toe} = \frac{\Sigma V}{B}\left(1+\frac{6e}{B}\right) = 504.3\,(1+0.277) = 644.1\ \text{kN/m}^2$$ $$\sigma_{heel} = 504.3\,(1-0.277) = 364.5\ \text{kN/m}^2$$

Toe stress 644 kN/m² compared with the dam-body allowable 1030 kN/m²: safe against crushing. Heel stress 364 kN/m² is compressive: no tension.

Shear and principal stress at the toe ($\tan\phi_d=0.740$, $p'=49.1$ kN/m²):

$$\tau_{toe} = (\sigma_{toe}-p')\tan\phi_d = (644.1-49.1)\times0.740 = 440.4\ \text{kN/m}^2$$ $$\sigma_{1,toe} = \sigma_{toe}\sec^2\phi_d - p'\tan^2\phi_d = 644.1\,(1+0.548) - 49.1\times0.548 = 970.0\ \text{kN/m}^2$$

Shear and principal stress at the heel ($\tan\phi_u=1.000$, $p=470.9$ kN/m²):

$$\tau_{heel} = (p-\sigma_{heel})\tan\phi_u = (470.9-364.5)\times1.000 = 106.4\ \text{kN/m}^2$$ $$\sigma_{1,heel} = \sigma_{heel}\sec^2\phi_u - p\tan^2\phi_u = 258.0\ \text{kN/m}^2$$

• Toe stress 644 kN/m² is also below the foundation allowable of 4905 kN/m².

Conclusion: sliding ($FS_s=1.80$), overturning ($FS_o=2.60$), tension (none) and crushing (below 1030 and 4905 kN/m²) are all satisfied, so the dam section is safe.

Data and assumptions

• Bed RL 470 m, top RL 554 m, so height $=84$ m; MWL 550 m, so $H_w=80$ m; no tail water.
• U/s face vertical. D/s face vertical from 554 to 545 m, then 0.67 : 1 to the toe, giving a horizontal run $0.67\times75=50.25$ m.
• Crest width is not given; 8 m is assumed, so $B=8+50.25=58.25$ m.
• $\gamma_c=24$ kN/m³ (assumed), $\gamma_w=9.81$ kN/m³, $\mu=0.75$.
• Uplift: 100% of $\gamma_wH_w=784.8$ kN/m² at the heel, 50% ($392.4$ kN/m²) at the gallery (8 m from the u/s face), zero at the toe.

 554 +---+          crest
 550 |~~~|  MWL  (H_w = 80 m)
     |   |
 545 |   \          d/s slope begins
     | G  \         0.67 : 1
     |     \
 470 +------\
     heel     toe

Forces and moments about the toe

Results

• $\Sigma V = \Sigma W - U = 61,353.0 - 14,567.9 = 46,785.1$ kN/m
• $\Sigma H = 31,392.0 = 31,392.0$ kN/m
• $\Sigma M_R = 2,389,982$ kN·m/m; $\Sigma M_O = 1,424,943$ kN·m/m

Overturning: $FS_o = \Sigma M_R/\Sigma M_O = 2,389,982/1,424,943 = 1.68$ — safe (> 1.5).

Sliding (friction): $FS_s = \mu\Sigma V/\Sigma H = 0.75\times46,785.1/31,392.0 = 1.04$ — safe (> 1).

Resultant: $\bar x = (\Sigma M_R-\Sigma M_O)/\Sigma V = 965,038/46,785.1 = 20.63$ m from the toe; $e = B/2-\bar x = 29.125-20.63 = 8.50$ m; $B/6 = 9.71$ m — inside the middle third (no tension).

$$\sigma_{toe} = \frac{\Sigma V}{B}\left(1+\frac{6e}{B}\right) = 803.2\,(1+0.875) = 1506.2\ \text{kN/m}^2$$ $$\sigma_{heel} = 803.2\,(1-0.875) = 100.1\ \text{kN/m}^2$$

Toe stress 1506 kN/m² compared with the allowable stress 2500 kN/m²: safe against crushing. Heel stress 100 kN/m² is compressive: no tension.

Shear and principal stress at the toe ($\tan\phi_d=0.670$, $p'=0.0$ kN/m²):

$$\tau_{toe} = (\sigma_{toe}-p')\tan\phi_d = (1506.2-0.0)\times0.670 = 1009.2\ \text{kN/m}^2$$ $$\sigma_{1,toe} = \sigma_{toe}\sec^2\phi_d - p'\tan^2\phi_d = 1506.2\,(1+0.449) - 0.0\times0.449 = 2182.4\ \text{kN/m}^2$$

Answer: maximum vertical stress at the toe = 1506 kN/m²; at the heel = 100 kN/m²; factor of safety against overturning = 1.68; against sliding = 1.04. (If the actual crest width differs from 8 m, replace the crest rectangle and recompute with the same table.)

Data and assumptions

• Crest width 8 m; water level 5 m below the crest; $H_w=85$ m, dam height $=90$ m; tail water $H_t=10$ m; base width $B=60$ m.
• U/s face vertical; the d/s face is taken as a straight slope from the d/s edge of the crest to the toe (horizontal run 52 m over 90 m, $\tan\phi_d=0.578$).
• $\gamma_c=24$ kN/m³ (given), $\gamma_w=9.81$ kN/m³, $\sigma_a=2500$ kN/m². Friction coefficient is not given; $\mu=0.70$ is assumed.
• Uplift: $\gamma_wH_w=833.9$ kN/m² at the heel, $\gamma_wH_t=98.1$ kN/m² at the toe, and at the gallery (10 m from the u/s face) $98.1+\frac13(833.9-98.1)=343.4$ kN/m².

  crest 8 m
   +--+
 5 m  |   freeboard
 ~~~~~|   water 85 m
      |\
      | \  d/s slope
  G   |  \
      |   \ ~~ tail water 10 m
      +----\
      B = 60 m

Forces and moments about the toe

Results

• $\Sigma V = \Sigma W - U = 73,723.4 - 16,922.2 = 56,801.1$ kN/m
• $\Sigma H = 35,438.6 - 490.5 = 34,948.1$ kN/m
• $\Sigma M_R = 2,916,741$ kN·m/m (tail water thrust moment included); $\Sigma M_O = 1,658,912$ kN·m/m

Overturning: $FS_o = \Sigma M_R/\Sigma M_O = 2,916,741/1,658,912 = 1.76$ — safe (> 1.5).

Sliding (friction): $FS_s = \mu\Sigma V/\Sigma H = 0.7\times56,801.1/34,948.1 = 1.14$ — safe (> 1).

Resultant: $\bar x = (\Sigma M_R-\Sigma M_O)/\Sigma V = 1,257,829/56,801.1 = 22.14$ m from the toe; $e = B/2-\bar x = 30-22.14 = 7.86$ m; $B/6 = 10.00$ m — inside the middle third (no tension).

$$\sigma_{toe} = \frac{\Sigma V}{B}\left(1+\frac{6e}{B}\right) = 946.7\,(1+0.786) = 1690.4\ \text{kN/m}^2$$ $$\sigma_{heel} = 946.7\,(1-0.786) = 203.0\ \text{kN/m}^2$$

Toe stress 1690 kN/m² compared with the allowable stress 2500 kN/m²: safe against crushing. Heel stress 203 kN/m² is compressive: no tension.

Shear and principal stress at the toe ($\tan\phi_d=0.578$, $p'=98.1$ kN/m²):

$$\tau_{toe} = (\sigma_{toe}-p')\tan\phi_d = (1690.4-98.1)\times0.578 = 920.0\ \text{kN/m}^2$$ $$\sigma_{1,toe} = \sigma_{toe}\sec^2\phi_d - p'\tan^2\phi_d = 1690.4\,(1+0.334) - 98.1\times0.334 = 2221.9\ \text{kN/m}^2$$

Answer: (i) maximum vertical stresses: toe = 1690 kN/m², heel = 203 kN/m²; (ii) major principal stress at the toe = 2222 kN/m² (< 2500 kN/m²); (iii) $FS$ overturning = 1.76, $FS$ sliding = 1.14.

Assumptions (from the figure)

• Crest width 10 m, height 110 m, base width 100 m, u/s face vertical, d/s face straight from the crest edge to the toe (run 90 m, $\tan\phi_d=90/110=0.818\approx0.8:1$). Water depth $H_w=94$ m, silt depth $h_s=20$ m. No tail water.
• $\gamma_c=25$, $\gamma_w=10$ kN/m³ (given). Uplift: $\gamma_wH_w=940$ kN/m² at the heel, one-third of it at the gallery (8 m from the heel), zero at the toe.
• Silt: submerged unit weight $10$ kN/m³, $\phi=30^\circ$ so $K_a=\frac13$: $P_s=\frac12K_a\gamma_sh_s^2$ acting at $h_s/3$.
• Earthquake: $\alpha_h=0.1$ acting towards the downstream (worst case). Inertia force $=\alpha_hW$ at the centre of gravity of the dam. Hydrodynamic pressure by Zangar: $p_e=C_m\alpha_h\gamma_wH_w$ with $C_m=0.735$, resultant $P_e=0.726\,p_eH_w$ at $0.412H_w$ above the base.

Calculations

Dam weights: $W_1=25\times10\times110=27{,}500$ kN, $W_2=25\times\frac12\times90\times110=123{,}750$ kN, total $W=151,250$ kN/m. Centre of gravity height $\bar z=40.00$ m.

$P_s=\frac12\times\frac13\times10\times20^2=666.7$ kN/m; $p_e=0.735\times0.1\times10\times94=69.1$ kN/m²; $P_e=0.726\times69.1\times94=4,715$ kN/m; $F_i=0.1\times151,250=15,125$ kN/m.

$$\Sigma V = W-U = 151,250-19,427 = 131,823\ \text{kN/m}$$ $$\bar x=\frac{\Sigma M_R-\Sigma M_O}{\Sigma V}=\frac{10,037,500-3,544,993}{131,823}=49.25\ \text{m from the toe},\quad e=50-49.25=0.75\ \text{m}$$

($B/6=16.67$ m: the resultant is inside the middle third.)

$$\sigma_{toe}=\frac{131,823}{100}\left(1+\frac{6\times0.75}{100}\right)=1,377.4\ \text{kN/m}^2,\quad \sigma_{heel}=1,318.2\left(1-0.045\right)=1,259.0\ \text{kN/m}^2$$

Principal stresses

Toe (no tail water, $p'=0$):

$$\sigma_1=\sigma_{toe}(1+\tan^2\phi_d)=1,377.4\,(1+0.669)=2,299.5\ \text{kN/m}^2,\quad \sigma_3=0$$

Shear stress $\tau_{toe}=\sigma_{toe}\tan\phi_d=1,127.0$ kN/m².

Heel (u/s face vertical, so no shear on the face and the face is a principal plane). The normal stress on the face is the total water, hydrodynamic and silt pressure $p=940+69.1+66.7=1075.8$ kN/m².

• Vertical principal stress: $\sigma_z=1,259.0$ kN/m²
• Horizontal principal stress: $\sigma_h=p=1,075.8$ kN/m²
• Major principal stress at the heel $=1,259.0$ kN/m², minor $=1,075.8$ kN/m².

Answer: toe — σ_z = 1,377 kN/m², major principal stress = 2,300 kN/m², τ = 1,127 kN/m²; heel — σ_z = 1,259 kN/m² (face normal stress 1,076 kN/m²).

Position of the resultant

Net moment about the toe divided by the vertical load gives the distance of the resultant from the toe:

$$\bar x=\frac{\Sigma M_R-\Sigma M_O}{\Sigma V}=\frac{2.5\times10^6-1.2\times10^6}{6\times10^4}=21.667\ \text{m}$$

Eccentricity from the centre of the base:

$$e=\frac{B}{2}-\bar x=27.5-21.667=5.833\ \text{m}\quad (<B/6=9.167\ \text{m},\ \text{so no tension})$$

Maximum and minimum vertical stresses

$$\sigma_{max}=\frac{\Sigma V}{B}\left(1+\frac{6e}{B}\right)=\frac{6\times10^4}{55}\left(1+\frac{6\times5.833}{55}\right)=1090.91\times1.6364=1,785.1\ \text{kN/m}^2\ (\text{at the toe})$$ $$\sigma_{min}=1090.91\times\left(1-0.6364\right)= 396.7\ \text{kN/m}^2\ (\text{at the heel})$$

Maximum principal stress at the toe

With no tail water, $p'=0$ and $\tan\phi_d=0.8$:

$$\sigma_1=\sigma_{max}\sec^2\phi_d=\sigma_{max}(1+\tan^2\phi_d)=1,785.1\times(1+0.8^2)=2,927.6\ \text{kN/m}^2$$

Answer: σ_max = 1,785 kN/m² (toe), σ_min = 397 kN/m² (heel), principal stress at the toe σ₁ = 2,928 kN/m².