549
GRAVITY DAMS
Example 16.1 For the profile of a gravity dam shown in Fig. 16.8, compute principal stresses for usual loading and vertical stresses for extreme loading at the heel and toe of the base of the dam. Also determine factors of safety against overturning and sliding as well as shear-friction factors of safety for usual loading and extreme loading (with drains inoperative) conditions. Consider only downward earthquake acceleration for extreme loading condition. Sediment is deposited to a height of 15 m in the reservoir. Other data are as follows: Coefficient of shear friction, µ = 0.7 (usual loading) = 0.85 (extreme loading) Shear strength at concrete-rock contact, C = 150 × 104 N/m2 = 2.4 × 104 N/m3
Weight density of concrete 8 4 15
0.75 1
h = 96
¢ Ww
0.15 30 1
0.3 Gallery
4.5
V¢
h¢ = g
8 63.75 76.25 (a) Profile
4.8 g 38 Drains inoperative
96 Drains operative
(b) Uplift pressure head diagram
Fig. 16.8 Profile and uplift pressure diagram for the gravity dam of Example 16.1
550
IRRIGATION AND WATER RESOURCES ENGINEERING
= 1 × 104 N/m3
Weight density of water
αh = 0.1; αv = 0.05 Solution: Computation of Stresses: (i) Usual loading combination (normal design reservoir elevation with appropriate dead loads, uplift (with drains operative), silt, ice, tail-water, and thermal loads corresponding to usual temperature): Resultant vertical force = ΣW = sum of vertical forces at sl. nos. 1, 2 (i), 3 (i), and 4(ii) of Table 16.1. = (8584.50 + 394.88 – 2000.68 + 32.48) × 104 = 7011.18 × 104 N Resultant horizontal force = ΣH = sum of horizontal forces at sl. nos. 2 (ii) and 4 (i) of Table 16.1. = (– 4567.50 – 153.00) × 104 = – 4720.50 × 104 N Moment about toe of the dam at the base = ΣM = sum of moments at sl. nos. 1,2, 3 (i), and 4 of Table 16.1. = [418302.75 + 27091.99 – 147334.50 – 96183.57 + 1662.88] × 104 = 203539.55 × 104 Nm Distance of the resultant from the toe, y =
ΣM 203539.55 × 10 4 = = 29.03 m ΣW . × 104 701118
∴ Eccentricity, e = 38.125 – 29.03 = 9.10 m (The resultant passes through the downstream of the centre of the base). Using Eqs. (16.12) and (16.13) σyD =
. × 10 4 . ΣW 6 × 910 6e 701118 = 1+ 1+ b b 76.25 76.25
LM N
OP Q
LM N
OP Q
= 157.79 × 104 N/m2
6 × 9.10 6e 7011.18 × 10 4 ΣW = 1− 1− 76.25 76.25 b b = 26.11 × 104 N/m2
σyU =
LM N
OP Q
LM N
Using Eq. (16.16), the major principal stress at the toe, σ1D = σyD sec2 φD – p′ tan2 φD ∴
σ1D = 157.79 × 104 × 1.5625 – 9 × 104 × 0.5625 = 239.66 × 104 N/m2
Using Eq. (16.21), shear stress at the toe, (τyx)D = (σyD – p′) tan φD ∴
(τyx)D = (157.79 – 9) × 104 × 0.75 = 111.59 × 104 N/m2
OP Q
Dead load Wc
Water load Vertical Head-water Ww
Tail-water Ww′
Horizontal Head-water Wl Tail-water Wl′
Uplift force, U Drains operative
Drains inoperative
2. (i) (a)
(b)
(ii) (a) (b)
3. (i)
(ii) 1 × 0.5 × 87 × 76.25 × 1 1 × 9 × 76.25 × 1
1 × 38 × 4.8 × 1 1 × 58 × 4.8 × 0.5 × 1 1 × 29 × 71.45 × 0.5 × 1 1 × 9 × 71.45 × 1
1 × 0.5 × 96 × 96 × 1.0 1 × 0.5 × 9 × 9 × 1.0
1 × 66 × 4.5 × 1.0 1 × 0.5 × 30 × 4.5 × 1.0 1 × 0.5 × 9.0 × 6.75 × 1.0
1 × 30 × 4.5 × 0.5 × 2.4 1 × 100 × 8 × 2.4 1 × 85 × 63.75 × 0.5 × 2.4
(3)
(2)
(1)
1.
Force computations (104 newtons)
Type of load
Sl. No.
– 4003.13
– 3316.88 – 686.25
– 2000.68
– 182.40 – 139.20 – 1036.03 – 643.05
+ 394.88
+ 297.00 + 67.50 + 30.38
+ 8584.50
+ 162.00 + 1920.00 + 6502.50
(4)
Vertical forces (downward + ve) (104 newtons)
– 4567.50
– 4908.00 + 40.50
(5)
Horizontal forces (upstream + ve) (104 newtons)
Magnitude of forces
50.83 38.13
73.85 74.65 47.63 35.73
32.00 3.00
76.25 – 2.25 = 74.00 76.25 – 1.50 = 74.75 (9 × 0.75)/3 = 2.25
73.25 67.75 42.50
(6)
Moment arm (metres)
Table 16.1 Computation of forces and moments for unit length of dam section
13470.24 10391.28 49345.87 22976.18
(Contd.)...
– 194763.72
– 168597.01 – 26166.71
– 96183.57
– – – –
– 147334.50
–147456.00 + 121.50
+ 27091.99
+21978.00 +5045.63 + 68.36
+ 418302.75
+ 11866.50 + 130080.00 + 276356.25
(7)
Moment about the toe (anticlockwise + ve) (104 Nm)
GRAVITY DAMS
551
1Hydrodynamic
Tail-water1
(b)
(a)
Hydrodynamic force Head-water (c1=cm for y=h)
(ii)
(i)
5.
(ii)
load Earthquake forces Inertial horizontal force due to weight of the dam
Load due to sediment deposit Ws Excess horizontal pressure Excess vertical
4.
(i)
(2)
(1)
– 491.19
– 2.76
– 488.43
50.00 28.33
10.00
76.25 – 4.5/3 = 74.75
15/3 = 5.00
(6)
force due to tail -water has been considered negative for the most critical condition.
At the base c = cm = 0.73 (for φU = 0) Vpe = 0.726 (0.73 × 0.1 × 1 × 96) × 96 Mpe = 0.299 (0.73 × 0.1 × 1 × 96) × (96)2 At the base c = cm = 0.47 (for φD = tan–1 (0.75)) Vpe = 0.726 (0.47 × 0.1 × 1 × 9) × 9 Mpe = 0.299 (0.47 × 0.1 × 1 × 9) × (9)2
– 858.45
– 192.00 – 650.25
1920.00 × 0.1 6502.50 × 0.1
– 153
(5)
– 16.20
32.48
(4)
162.00 × 0.1
1 × 0.5 × 15 × 2.25 × 1.925
1 × 0.5 × 15 × 15 × 1.36
(3)
19321.38
– 10.25
– 19311.13
– 28183.58
– 9600.00 – 18421.58
– 162.00
+ 1662.88
+ 2427.88
– 765.00
(7)
552 IRRIGATION AND WATER RESOURCES ENGINEERING
553
GRAVITY DAMS
Using Eq. (16.19), with pe = 0, the minor principal stress at the heel σ1U = σyU sec2 φU – p′ tan2 φU = 26.11 × 104 × 1.0225 – 96 × 104 × 0.0225 = 24.54 × 104 N/m2 Using Eq (16.22), shear stress at the heel, (τyx)U = (σyU – p′) tan φU ∴
(τyx)U = – (26.11 – 96.00) × 104 × 0.15 = 10.48 × 104 N/m2
Further, major principal stress at the heel = p = 96 × 104 N/m2 and minor principal stress at the toe = p′ = 9.0 × 104 N/m2 Factor of safety for overturning = = Sliding factor
= =
stabilising moment overturning moment
(418302.75 + 27091.99 + 1662.88) × 10 4 (147334.50 + 96183.57) × 10 4
= 1.84
ΣH ΣW 4720.55 × 10 4 701118 . × 104
= 0.67
Shear-friction factor of safety (with drains operative), Fs = =
Cb × 1 + µΣW ΣH
150 × 104 × 76.25 × 1 + 0.7(701118 . × 104 ) 4720.50 × 104
= 3.46
(ii) Extreme loading combination (usual loading combination with drains inoperative and the loading due to earthquake): The inertial and hydrodynamic forces and corresponding moments due to horizontal earthquake have been computed as shown in Table 16.1. The effect of vertical earthquake can be included in stability computations by multiplying the forces by (1 + αv) and (1 – αv) for upward and downward accelerations, respectively. Since the computation of hydrodynamic force involves the use of unit weight of water, the hydrodynamic force will also be modified by vertical acceleration due to earthquake. Further, the effect of earthquake on uplift forces is considered negligible. For ‘reservoir full’ condition, the downward earthquake acceleration results in more critical condition. Therefore, the following computations have been worked out for the downward earthquake acceleration only. Resultant vertical force with downward earthquake acceleration = (8584.50 + 394.88 + 32.48) × 104 × 0.95 – 2000.68 × 104 = 6560.59 × 104 N Resultant horizontal force with downward earthquake acceleration = (4567.50 + 153 + 858.45 + 491.19) × 104 = 6070.14 × 104 N
554
IRRIGATION AND WATER RESOURCES ENGINEERING
Resultant moment about the toe with downward acceleration = (418302.75 +27091.99 – 147334.50 + 1662.88 – 28183.58 – 19321.38) × 104 × 0.95 – 194763.72 × 104 = 44843.53 × 104 Nm Now,
y=
ΣM 44843.53 × 10 4 = 6.835 m = ΣW 6560.59 × 10 4
∴ Eccentricity, e = 38.125 – 6.835 = 31.29 m The resultant passes through the downstream side of the centre of the base. The value of e is more than b/6 i.e., 12.71 m. Therefore, there would be tensile stresses around the heel of the dam. The vertical stresses at the toe and heel with downward earthquake acceleration are, σyD =
ΣW 6 × 31.29 6e 6560.59 × 10 4 = 1+ 1+ b b 76.25 76.25
FG H
IJ K
FG H
IJ K
= 297.89 × 104 N/m2
6 × 31.29 6e 6560.59 × 10 4 ΣW = 1− 1− 76.25 76.25 b b 4 2 = – 125.81 × 10 N/m Factor of safety against overturning and
Sliding factor
FG H
σyU =
IJ K
FG H
IJ K
=
(418302.75 + 27091.99 + 1662.88) × 10 4 = 1.15 (147334.50 + 194763.72 + 28183.58 + 19321.38) × 10 4
=
ΣH ΣW
= =
( 4567.50 + 153 + 858.45 + 49119 . ) × 10 4 6560.59 × 10 4 607014 . = 0.925 6560.59
Shear-friction factor of safety, Fs = = =
Cb × 1 + µΣW ΣH
150 × 10 4 × 76.25 × 1 + 0.7 × 6560.59 × 104 607014 . × 10 4 16029.91 = 2.64 607014 .
16.7. FOUNDATION TREATMENT The foundation of a gravity dam should be firm and free of major faults which, if present, may require costly foundation treatment. The entire loose overburden over the area of the foundation to be occupied by the base of the dam should be removed. The dam itself must be based on the firm material which can withstand the loads imposed by the dam, reservoir, and other appurtenant structures. To consolidate the rock foundation and to make it an effective barrier
