Contents preface (VII) introduction 1—37

Fig. 16.7 Elementary profile of a gravity dam

or	b =	h
s − c′
For c′ = 1,		b =	h
s − 1
and if uplift is ignored,	c′ = 0
∴	b =	h
s

For no-sliding requirement, one obtains µ (W_c – U) = P = W₁ coefficient.

or	µ (0.5 sρgbh – 0.5 ρghbc′) = 0.5 ρgh^2
or	b =		h
		µ (s − c′ )
For c′ = 1		b =		h
	µ (s − 1)
and for no uplift,	c′ = 0, and b =		h
µs

(16.25)
(16.26)

(16.27)
in which, µ is shear-friction

(16.28)
(16.29)

(16.30)

It is obvious that for satisfying the requirement of stability, the elementary profile of a gravity dam should have minimum base width equal to the higher of the base widths obtained from no-sliding and no-tension criteria.

	Again, for an elementary profile, ΣW = (Wc – U)
		1
or	ΣW =		bρgh (s − c′ )
2

548 IRRIGATION AND WATER RESOURCES ENGINEERING

		ΣW F		12exI
∴	σ =		G 1	±	J	(16.31)
	yx	b H		b^2 K

For no tension in the dam, e = b/6

Therefore, at the toe of the dam (i.e., x = b/2)
	σ		= 2 ΣW
	yD	b
	σyD = ρgh (s – c′)	(16.32)
and at the heel of the dam (i.e.,		x = – b/2)
	σyU = 0
Accordingly, the principal stress σ	1D	= σ sec^2	φ	D
		yD
∴	σ1D = ρgh (s – c′) [1 + (b/h)^2]

L

1

O

= ρgh (s – c′) M1

+

P

N

(s − c′ ) Q

= ρ gh (s – c′ + 1)

Similarly,

(τ )

= σ

tan φ

= ρgh (s – c′)

b

D

D

h

yx

yD

1

= ρgh (s – c′)	s − c′

= ρgh s − c′

The principal and shear stresses at the heel are, obviously, zero. Similarly, when the reservoir is empty, ΣW = 0.5 ρgbhs

σ_yD = 0

σ_1U = σ_yU = ^{2 ΣW} = ρ ghs
b

(16.33)

(16.34)
(16.35)

Sometimes, depending upon whether or not the compressive stress at the toe σ_1D exceeds the maximum permissible stress σ_m for the material of the dam, a gravity dam is called a ‘high’ or ‘low’ dam. On this basis, the limiting height h_l is obtained by equating the expression for σ_1D with σ_m. Thus,

σ_m = ρgh_l (s – c′ + 1)

or	hl =	σ m	(16.36)
ρg ( s − c′ + 1)

If the height of a gravity dam is less than h_l, it is a low dam; otherwise, it is a high dam.
16.6. DESIGN OF A GRAVITY DAM
An elementary profile is only an ideal profile which needs to be modified for adoption in actual practice. Modifications would include provision of a finite crest width, suitable freeboard, batter in the lower part of the upstream face, and a flatter downstream face. The design of a gravity dam involves assuming its tentative profile and then dividing it into a number of zones by horizontal planes for stability analysis at the level of each dividing horizontal plane. The analysis can be either two-dimensional or three-dimensional. The following example illustrates the two-dimensional method of analysis of gravity dams.

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 Weight density of concrete = 2.4 × 10⁴ N/m³

8

4. 
   
   

15

							0.75
h = 96						1
		0.15								W ¢
										w
30	1					0.3
									Gallery				V¢	h¢ = g

5. 
   8
   

									63.75
		76.25
									(a) Profile
		4.8										g
		38
96			Drains
											inoperative
											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
Weight density of water	= 1 × 10^4 N/m^3
Solution:	αh = 0.1; αv = 0.05
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 corre-sponding 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	=	20353955. × 10^4	= 29.03 m
ΣW	701118. × 10^4

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

6e O

701118. × 10^4

L

6 × 910.

O

^σyD

=

M1 +

P =

M^1

+

P

b

b

76.25

76.25

N

Q

N

Q

= 157.79 × 10^4 N/m^2

ΣW L

6e O

7011.18 × 10^4

L

6 × 9.10 O

^σyU

=

M1 −

P ^=

M1 −

P

b

76.25

76.25

N

b Q

N

Q

= 26.11 × 10^4 N/m^2

Using Eq. (16.16), the major principal stress at the toe,

σ

= σ

sec^2

φ

D

– p′ tan^2 φ

1D

yD

D

∴

σ

= 157.79 × 10^4 × 1.5625 – 9 × 10^4 × 0.5625

1D

= 239.66 × 10^4 N/m^2

Using Eq. (16.21), shear stress at the toe, (τyx)D = (σyD – p′) tan φD

∴

(τ )

D

= (157.79 – 9) × 10^4 × 0.75

yx

= 111.59 × 10⁴ N/m²

GRAVITY DAMS

	Moment about	the toe (anticlock-	wise + ve)	4(10 Nm)
dam section	Moment arm	(metres)