Gravity Dam Design Example PDF

 

 

GRAVITY DAMS

549

 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

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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 16.8 Profile and uplift pressure diagram for the gravity dam of 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 (normal design reservoir elevation elevation with appropriate dead 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 = Σ H = 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 = Σ M  =  = 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

Σ M  203539.55 55 × 10 4 = 2903 = Distance of the resultant from the toe,  y = . m ΣW  7011.18 × 104 ∴

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)

ΣW  L 6 e O 7011.18 × 10 4 = 1+ σ yD = b MN b PQ . 7625 4

6 × 9.10 O L M1 + 7625 P . N Q

2

= 157.79 × 10  N/m 6 e O 7011.18 × 10 4 ΣW L − = 1 σ yU = 76.25 b MN b PQ = 26.11 × 104 N/m2

6 × 9.10 O L M1 − 76.25 P N Q

Using Eq. (16.16), the major principal stress at the toe, σ1 D = σ yD sec2 φ D – p′ tan2  φ D ∴

σ1 D = 157.79 × 104 × 1.5625 – 9 × 104 × 0.5625 = 239.66 × 104 N/m2

Using Eq. (16.21), shear stress at the toe, (τ (τ yx) D = (σ (σ yD – p′) tan φ D ∴

(τ yx) D = (157.79 – 9) × 104 × 0.75 = 111.59 × 104 N/m2

 

 

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GRAVITY DAMS

553

Using Eq. (16.19), with  p e = 0, the minor principal stress at the heel  

σ1U  = σ yU  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, (τ (τ yx)U  = (σ ( σ yU  – p′) tan φU  ∴

(τ yx)U  = – (26.11 – 96.00) × 10 4 × 0.15   = 10.48 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

(4 (418 1830 302 2.7 .75 5 + 27 2709 091 1.9 .99 9 + 16 1662 62.8 .88 8) × 104 (147334.50 + 96183.57) × 10 4

 = 1.84

Σ H  = ΣW  =

4720.55 55 × 10 4

 = 0.67

7011.18 × 104

Shear-friction factor of safety (with drains operative),  F s =

=

Cb × 1 + µΣW 

Σ H 

150 × 104 × 76.25 × 1 + 0.7(70 7011.18 × 104 ) 4720.50 50 × 104

 = 3.46

(ii) Extreme loading combination combination (usual loading combination 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  y =

Now, ∴

Σ M  ΣW 

=

4484  3.5 53 3 × 10 4 6560.59 59 × 10 4

. m = 6835

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, σ yD =

  6 e I   = 656 0.59 × 104 F 1 + 6  ×  31.29 I  GH 1 + J  G J  H  K  b K  7625 . . 7625

ΣW  F 

b

= 297.89 × 104 N/m2 and

 

σ yU   =

  6 e I  656 0.59 × 10 4 F  6  ×  31.29 I  J  G 1 − 7625 G 1 − b J  K  H  K   = H  . 76.25

ΣW F 

b = – 125.81 × 104 N/m2

Factor of safety against overturning = Sliding factor

= =

=

(418302.75 + 27091.99 + 1662.88) × 10 4 (147334.50 + 194763.72 + 28183.58 + 19321.38) × 10 4 Σ H  ΣW 

( 4567.50 + 153 + 858.4 45 5 + 491.19) × 10 4 6560.59 59 × 10 4 607014 .  = 0.925 656059 .

Shear-friction factor of safety,  F s = = =

 = 1.15

Cb × 1 +  µΣW  Σ H 

150 × 10 4 × 76.25 × 1 + 0.7 × 6560.59 × 104 6070.1 14 4 × 10 4 16029 160 29.91 607014 .

 = 2.64

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