The document discusses the analysis of a gravity dam, including calculating stresses and checking stability, for both an empty reservoir and full reservoir condition. It provides numerical examples of determining vertical stresses, principal stresses, and shear stresses at the toe and heel of the dam. It also shows calculations for checking the stability of the dam against sliding, overturning, tension and sufficient shear resistance.

1. Swami Keshvanand Institute of Technology
Management & Gramothan Jaipur
1
SUBJECT: WATER RESOURCE ENGINEERING (WRE)
SITARAM SAINI
ASSISTANT PROFESSOR
DEPARTMENT OF CIVIL ENGINEERING

2. Introduction
1. What is the gravity dam.
2. Need of gravity dam.
3. Forces acting on gravity dam.
4. Stability analysis of gravity dam.
5. Numerical problems on gravity dam.
2

3. Numerical problems on gravity dam.
Q.1. Determine the maximum and minimum vertical stresses to which the
foundation of the dam will be subjected from the following data:
1. Total overturning moment about toe(ΣMo ) = 1.2× 106 kN-M
2. Total resisting moment about toe(ΣMR ) = 2.5× 106 kN-M
3. Total vertical force above base (Σ V) = 6× 104kN
4. base width of the dam = 55m
No tail water depth at d/s. and d/s slope = 0.8:1
Answer :
_
𝑥
=
𝑀
𝑉
=
𝑀𝑅− 𝑀𝑂
𝑉
3

4. _
𝑥
=
2.5×106 −1.2×106
6×104 = 21.67m
e =
𝑏
2
- _
𝑥
= 55
2
− 21.67= 5.83m
Maximum vertical stress at toe:
fyd =
𝑉
𝑏
(1 +
6𝑒
𝑏
) =
6×104
55
(1 +
6×5.83
55
) = 1784.73
𝑘𝑁
𝑀2
Minimum vertical stress at heel:
fyu =
𝑉
𝑏
(1 −
6𝑒
𝑏
) =
6×104
55
(1 −
6×5.83
55
) = 397.09
𝑘𝑁
𝑀2
4

5. Q.2. Check the stability of the gravity dam shown in fig. for the reservoir empty
and reservoir full conditions. Also find the principal stresses and shear stresses at
toe and heel of the dam . Assume µ=0.75.consider
1. weight of dam.
2.water pressure.
3. uplift pressure.
Take the average shear strength (q)=1400kN/m2
Unit weight of concrete = 24kN/M3
Slope of d/s face = 0.8:1 (40/50) and Slope of u/s face = 0.1:1 (3/30)
5

6. S.
N
O
ITEM DESCRIPTION AND
DIMENSION
FORCE(KN) LEVER
ARM
FROM
TOE (M)
MOMENT ABOUT
TOE(KN-M)
VERTICAL HORIZONTAL (+)IVE (-)IVE
1 W1 .5X3X30X24 1080 48 51840
2 W2 7X65X24 10920 43.50 475020
3 W3 .5X40X50X24 24000 .67X40 640000
4 PH 9.81X62X62X.5 18854.82 .33X62 -389666.2
5 PV1 3X32X9.81 941.76 48.50 45675.36
6 PV2 .5X3X30X9.81 441.45 49 21631.05
7 U1 .33X62X9.81X8 -1621.9 46 -74608.32
8 U2 .5X.67X62X9.81X8 -1621.9 142X.33 -76770.88
9 U3 .5X.33X62X9.81X42 -4257.5 28 -119211.1
SUM OF FORCES&MOMENTS; 1-3 36000 1166860 0.0
1-6 37383.2 18854.82 1234166 -389666.2
1-9 29881.8 18854.82 1234166 -660256.6
6

7. 1.0 WEIGHT OF THE DAM:
7M
3M
15M
32M
40M
62M 65M
30M
8M 42M
50M
Drainage gallery
𝛾𝐻 Uplift pressure diagram.
7
W2
W1 W3

8. 2.0 WATER PRESSURE:
PV1
PH
62M =H PV2
H/3
9.81X62
8

9. 3.0 UPLIFT-PRESSURE:
U1,U2,U3
U= 9.81(H`+1/3(H-H`))
BUT H`=0 ; U=9.81XH
9

10. 1.) Check the stability of the gravity dam for the reservoir empty :
Forces 1 TO 3.
_
𝑥
=
𝑀
𝑉
=
1166860
36000
= 32.41 m
e =
𝑏
2
- _
𝑥
=
50
2
− 32.41= - 7.41m
𝑏
6
=
50
6
= 8.33 , e <
𝑏
6
Vertical stress at toe:
fyd =
𝑉
𝑏
(1 +
6𝑒
𝑏
) =
36000
50
(1 −
6×7.41
50
) = 79.78
𝑘𝑁
𝑀2
Vertical stress at heel:
fyu =
𝑉
𝑏
(1 −
6𝑒
𝑏
) =
36000
50
(1 +
6×7.41
50
) = 1360
𝑘𝑁
𝑀2
10

11. Principal stress at toe :
𝜎d = fyd 𝑠𝑒𝑐2
∅ = 79.78 (1+𝑡𝑎𝑛2
∅) = 79.78 (1+.82
) = 130.84
𝑘𝑁
𝑀2
shear stress at toe :
𝜏d = fyd tan∅ = 79.78× (.8/1) = 63.82
𝑘𝑁
𝑀2
Principal stress at heel :
𝜎𝑢 = fyu 𝑠𝑒𝑐2∅ = 1360.22 (1+𝑡𝑎𝑛2∅) =1360.22 (1+.12) = 1373.82
𝑘𝑁
𝑀2
shear stress at toe :
𝜏u = - fyu tan∅ = -1360.22 × (.1/1) = - 136.02
𝑘𝑁
𝑀2
Check for tension = no tension
Check for sliding = no sliding
Check for overturning = no overturning
11

12. 2.) Check the stability of the gravity dam for reservoir full conditions:
_
𝑥
=
𝑀
𝑉
=
1234166.4−660256.60
29881.8
= 19.21 m
e =
𝑏
2
- _
𝑥
=
50
2
− 19.21= 5.79 m
𝑏
6
=
50
6
= 8.33 , e <
𝑏
6
Vertical stress at toe:
fyd =
𝑉
𝑏
(1 +
6𝑒
𝑏
) =
29881.83
50
(1 +
6 × 5.79
50
) =1012.87
𝑘𝑁
𝑀2
Vertical stress at heel:
fyu =
𝑉
𝑏
(1 −
6𝑒
𝑏
) =
29881.83
50
(1 −
6 × 5.79
50
) = 169.49
𝑘𝑁
𝑀2
12

13. Principal stress at toe :
𝜎d = fyd 𝑠𝑒𝑐2
∅ = 1012.87 (1+𝑡𝑎𝑛2
∅) = 1012.87 (1+.82
) = 1661.11
𝑘𝑁
𝑀2
shear stress at toe :
𝜏d = fyd tan∅ = 1012.87 × (.8/1) = 810.29
𝑘𝑁
𝑀2
Principal stress at heel :
𝜎𝑢 = fyu 𝑠𝑒𝑐2∅ = 169.49 (1+𝑡𝑎𝑛2∅) =169.49 (1+.12) = 171.18
𝑘𝑁
𝑀2
shear stress at toe :
𝜏u = - fyu tan∅ = -169.49 × (.1/1) = - 16.95
𝑘𝑁
𝑀2
Check for tension = no tension
Check for sliding =
F.S=
𝜇 𝑉
𝐻
=
.75 ×29881.83
18854.82
= 1.19 > 1
13

14. Check for overturning =
FO =
1234166.44
660256
= 1.87 > 1.50
Shear friction factor =
S.F.F =
𝜇 𝑉+𝑏𝑞
𝐻
=
.75 ×29881.83+50 ×1400
18854.82
= 4.90
14

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