Solved examples of Dams

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Abdüsselam ALTUNKAYNAK www.altunkaynak.net
Research Assist. İsmail DABANLI www.akademi.itu.edu.tr/dabanli
PROBLEM 1 Analyze the stability of given gravity dam (Figure 1) for the following conditions: Friction coefficient between
concrete-foundation is 0.70, respectively. Allowable shear stress at the foundation level is 2200 kN/m2
, allowable compressive and
shear stresses in concrete are 2700 kN/m2
, and 2400 kN/m2
, respectively. Allowable compressive stress in foundation material is
2700 kN/m2
. Take specific weights of concrete and water as 24 kN/m3
, and 10 kN/m3
, respectively.
PROBLEM 2 Analyze the stability of given gravity dam (Figure 2) for the following conditions: The ice tickhness at the resevoir
surface is 50 cm with the increase in temperature of 5 o
C/h. Friction coefficients between concretes, and concrete-foundation are
0.75 and 0.65, respectively. Allowable shear stress at the foundation level is 2000 kN7m2
, allowable compressive and shear stress
in concrete are 2500 kN/m2
, and 2200 kN/m2
, respectively. Allowable compressive stress in foundation material is 2500 kN/m2
.
Relief drainage may reduce the uplift force by 50 %. The earthquake coefficient is 0.1. Take specific weights of concrete and water
as 25 kN/m3
, and 10 kN/m3
, respectively.
PROBLEM 3 Determine the total volume of an arch dam 120 m high to span a 300 m wide U-shaped valley. The crest width is 6
m. Take  = 10 kN/m3
,  = 120o
, all = 6200 kN/m2
. Ignore the variation of span width and a in the vertical direction. Consider
vertical upstream face.
PROBLEM 4 Determine the optimum central angle of an arch dam giving the minimum volume of rib.
Answer :a = 133o
.34’
PROBLEM 5 Elevation of an arch dam (Figure 3) is 75 m. Span (top width) of a U-shaped valley is 350 m. Central angle is 150o
.
Allowable compressive stress in concrete is 5500 kN/m2
. Ignoring the variation of , and span width with respect to depth,
determine:
a) The cross-section for a vertical upstream faced arch dam having 6 m thickness at the crest,
b) Thickness of arch at 30 m below the water surface,
c) Volume of arch per unit height assuming 1 m unit height at the bottom,
d) Total volume of arch dam
In the computations, take con = 24 kN/m3
, and w = 10 kN/m3
hx
6 m
tb
75 m
70 m
6 m 4 m 45 m
55 m
O
65 m
Figure 1
Figure 3
Figure 2
55 m
6 m 5 m 30 m
41 m
O
50 m
0.7 H : 1 V
1 H : 10 V
15 m

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Abdüsselam ALTUNKAYNAK www.altunkaynak.net
Research Assist. İsmail DABANLI www.akademi.itu.edu.tr/dabanli
SOLUTION 1
 Forces and loads acting the dam:
Fwh: Hydrostatic force produced by water in the reservoir and tail water in the downstream
Fwv: Water load produced by water weight
Fu : Uplift force produced by groundwater
W : The weight of the dam (W1, W2, W3)
 Free body diagram is shown in figure 4
 The value of the forces, total vertical and total horizontal forces, and moments:
W1 = 6 * ½ * 70 * 24 = 5040 kN XW1 = 1/3 * 6 + 4 + 45 = 51.00 m W1 * XW1 = 257040 kNm
W2 = 4 * 70 * 24 = 6720 kN XW2 = ½ * 4 + 45 = 47.00 m W2 * XW2 =315840 kNm
W3 = 45 * ½ * 70 * 24 = 37800 kN XW3 = 2/3 * 45 = 30.00 m W3 * XW3=1134000 kNm
Fwv = 6 * ½ * 65 * 10 = 1950 kN Xwv = 2/3 * 6 + 4 + 45 = 53.00 m Fwv * XFv =103350 kNm
Fwh = 65 * ½ * 65 * 10 = 21125 kN XFwh = 1/3 * 65 = 21.67 m Fwh * XFh =457779 kNm
Fu = 65 * ½ * 55 * 10 = 17875 Kn XFu = 2/3 * 55 = 36.67 m Fu * XFu =655476 kNm
 MO = 457779 + 655476 = 1113255 kNm/m
 Mr = 257040 + 315840 + 1134000 + 103350 = 1810230 kNm/m
 V = W1 + W2 + W3 + Fwv - Fu = 33635 kN/m
 H = Fwh = 21125 kN/m
FORCE (kN/m) MOMENT (kN/m/m)
MOMENT ARM ABOUT O(m)
70 m
6 m 4 m 45 m
55 m
O
65 m
W1 W2 W3
Fwv
Fwh
Fu
xm
Figure 4

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Abdüsselam ALTUNKAYNAK www.altunkaynak.net
Research Assist. İsmail DABANLI www.akademi.itu.edu.tr/dabanli
 Stability Check For the Whole Dam:
1. Overturning (F.S0): The dam must be safe against overturning for all loading conditions. F.S0 should
be greater than 2.0 for usual loading, and than1.5 for unusual or severe loading.
F.S0 =


O
r
M
M
=
1113255
1810230
= 1.626 > 1.5 O.K.
2. Sliding (F.SS): The dam must be safe against sliding over any horizontal plane. F.SS should be greater
than 1.5 for usual loading and than 1.0 for unusual or severe loading.
F.SS =



H
V
f
=
21125
33635
7
.
0 
= 1.11 > 1.0 O.K
3. Shear and sliding together (F.SSS): The dam must be also checked for shear and sliding together. F.SSS
should be greater 5.0 for usual loading and 3.0 unusual and severe loading.
F.SSS =

 



H
A
V
f S

5
.
0
=
21125
2200
55
5
.
0
33635
7
.
0 



= 3.98 > 3.0 O.K.
4. Stress (max/min): The contact stress between the foundation and the dam must be greater than zero and
all points or the dam will be unsafe against overturning. Maximum base pressure (max) should be
less than the allowable compressive stress and minimum base pressure (min) should be greater than
zero.

 


V
M
M
X
O
r
=
33635
1113255
1810230 
= 20.72 m
e = 72
.
20
2
55
 = 6.78 (eccentricity)
M =  V * e = 33635 * 6.78 = 228045.3 kNm/m (the net moment about the centerline of the base)
C =
2
55
= 27.5 m
I =
12
1
553

= 13864.58 m3
( the moment of interia)
I
C
M
B
V 



min
max/
 =
58
.
13864
5
.
27
3
.
228045
55
33635 

max = 1063.87 kN/m2
< 2700 kN/m2
O.K.
max = 159.22 kN/m2
> 0.00 kN/m2
O.K.

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Abdüsselam ALTUNKAYNAK www.altunkaynak.net
Research Assist. İsmail DABANLI www.akademi.itu.edu.tr/dabanli
SOLUTION 2
 Forces and loads acting the dam:
Fi : Ice Load (for cold climates and Fi1, and Fi2 for reservoir and tail water in the downstream,
respectively)
Fw: Water force produced by earthquake (Fw1, and Fw2 for reservoir and tail water in the
downstream, respectively)
Fwh: Hydrostatic force produced by water in the reservoir and tail water in the downstream (Fwh1,
and Fwh2)
Fwv: Water load produced by water weight (Fwv1, and Fwv2 for reservoir and tail water in the
downstream, respectively)
Fu : Uplift force produced by groundwater (since the tail water in the downstream, the diagram of
uplift force will be in trapezoidal shape)
W : The weight of the dam (W1, W2, W3…Wn)
Fd : Earthquake forces (Fdh1 and Fdv1: horizontally and vertically)
 Free body diagram is shown in figure 4
 Shear and sliding check due to ice force at the reservoir surface:
Average shear stress:
5
90
= 18 kN/m2
<< 2200 O.K.
Sliding check:
90
25
5
5
75
.
0 


= 5.21 >> 1.00 O.K.
55 m
6 m 5 m 30 m
41 m
50 m
W1 W2 W3
Fwv1
Fwh1
Fu
xm
Fw1
Fi1
Fdv
Fdh
Figure 5
O
15 m
Fwv2
Fwh2
Fw2
Fi2

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Abdüsselam ALTUNKAYNAK www.altunkaynak.net
Research Assist. İsmail DABANLI www.akademi.itu.edu.tr/dabanli
 The value of the forces, total vertical and total horizontal forces, and moments:
W1 = 50 * 6.0 * 0.5 *25 = 3750 kN 37.00 m 138750 kNm
W2 = 55 * 5.0 * 1.0 *25 = 6875 kN 32.50 m 223438 kNm
W3 = 50 * 30 * 0.5 * 25 = 18750 kN 20.00 m 375000 kNm
Fwv1 = 50 * 6 * 0.5 * 10 = 1500 kN 39.00 m 58500 kNm
Fwv2 = 15 * 9 * 0.5 * 10 = 675 kN 3.00 m 2025 kNm
Fwh1 = 55 * 55 * 0.5*10 = 15125 kN 17.00 m 257125 kNm
Fwh2 = 15 * 15 * 0.5*10 = 1125 kN 5.00 m 5625 kNm
Fu =[(50+15)/2]*41*10*0.5= 6663 kN 23.30 m 168295 kNm
FW1 = 0.726*C*k**h1
2
FW1 = 0.726*0.65*0.1*10*502
= 1180 kN 17.00 m 20056 kNm
FW2 = 0.726*0.46*0.1*10*152
= 75 kN 5.00 m 375 kNm
Fi1 = 90 kN 50.00 m 4500 kNm
Fi2 = 90 kN 15.00 m 1350 kNm
Fdv = k *W=0.1* 29375 =2937.5 kN 30.10 m 88419 kNm
Fdh = k * W=0.1*29375 =2937.5 kN 20.64 m 60630 kNm
 Stability Check For the Whole Dam (When vertical Fd (Fdv) is considered):
 MO = 257125+168295+20056+4500+88419 = 538395 kNm/m
 Mr = 138750+223438+375000+58500+2025+5625+375+1350 = 805063 kNm/m
 V = W1 + W2 + W3 + Fwv1 + Fwv2- Fu -Fdv = 21950 kN/m
 H = Fwh1- Fwh2 + FW1 - FW2 + Fi1 - Fi2 = 15105 kN/m
1. Overturning (F.S0): The dam must be safe against overturning for all loading conditions. F.S0 should
be greater than 2.0 for usual loading, and than1.5 for unusual or severe loading.
F.S0 =


O
r
M
M
=
538395
805063
= 1.495 < 1.5 NOT O.K.
2. Sliding (F.SS): The dam must be safe against sliding over any horizontal plane. F.SS should be greater
than 1.5 for usual loading and than 1.0 for unusual or severe loading.
F.SS =



H
V
f
=
15105
21950
65
.
0 
= 0.94 < 1.0 NOT O.K.
FORCE (kN/m) MOMENT (kN/m/m)
MOMENT ARM ABOUT O(m)

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Abdüsselam ALTUNKAYNAK www.altunkaynak.net
Research Assist. İsmail DABANLI www.akademi.itu.edu.tr/dabanli
3. Shear and sliding together (F.SSS): The dam must be also checked for shear and sliding together. F.SSS
should be greater 5.0 for usual loading and 3.0 unusual and severe loading.
F.SSS =

 



H
A
V
f S

5
.
0
=
15105
2200
41
5
.
0
21950
65
.
0 



= 3.93 > 3.0 O.K.
4. Stress (max/min): The contact stress between the foundation and the dam must be greater than zero and
all points or the dam will be unsafe against overturning. Maximum base pressure (max) should be
less than the allowable compressive stress and minimum base pressure (min) should be greater than
zero.

 


V
M
M
X
O
r
=
21950
538395
805063 
= 12.15 m
e = 15
.
12
2
41
 = 8.35 (eccentricity)
M =  V * e = 21950 * 8.35 = 183307 kNm/m (the net moment about the centerline of the base)
C =
2
41
= 20.5 m
I =
12
1
413

= 5743.42 m4
( the moment of interia)
I
C
M
B
V 



min
max/
 =
42
.
5743
5
.
20
183307
41
21950 

max = 1189.64 kN/m2
< 2700 kN/m2
O.K.
max = -118.91 kN/m2
< 0.00 kN/m2
NOT O.K.
 Stability Check For the Whole Dam (When horizontal Fd (Fdh) is considered):
 MO = 257125+168295+20056+4500+60630 = 510606.0 kNm/m
 Mr = 138750+223438+375000+58500+2025+5625+375+1350 = 805063.0 kNm/m
 V = W1 + W2 + W3 + Fwv1 + Fwv2 - Fu = 19012.5 kN/m
 H = Fwh1- Fwh2 + FW1 - FW2 + Fi1 - Fi2 + Fdh = 18042.5 kN/m

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Abdüsselam ALTUNKAYNAK www.altunkaynak.net
Research Assist. İsmail DABANLI www.akademi.itu.edu.tr/dabanli
1. Overturning (F.S0): The dam must be safe against overturning for all loading conditions. F.S0 should
be greater than 2.0 for usual loading, and than1.5 for unusual or severe loading.
F.S0 =


O
r
M
M
=
510606
805063
= 1.58 > 1.5 O.K.
2. Sliding (F.SS): The dam must be safe against sliding over any horizontal plane. F.SS should be greater
than 1.5 for usual loading and than 1.0 for unusual or severe loading.
F.SS =



H
V
f
=
5
.
18042
5
.
19012
65
.
0 
= 0.65 < 1.0 NOT O.K.
3. Shear and sliding together (F.SSS): The dam must be also checked for shear and sliding together. F.SSS
should be greater 5.0 for usual loading and 3.0 unusual and severe loading.
F.SSS =

 



H
A
V
f S

5
.
0
=
5
.
18042
2200
41
5
.
0
5
.
19012
65
.
0 



= 3.19 > 3.0 O.K.
4. Stress (max/min): The contact stress between the foundation and the dam must be greater than zero and
all points or the dam will be unsafe against overturning. Maximum base pressure (max) should be
less than the allowable compressive stress and minimum base pressure (min) should be greater than
zero.

 


V
M
M
X
O
r
=
5
.
19012
510606
805063 
= 15.49 m
e = 49
.
15
2
41
 = 5.0 (eccentricity)
M =  V * e = 19012.5 * 5.0 = 95299.25 kNm/m (the net moment about the centerline of the base)
C =
2
41
= 20.5 m
I =
12
1
413

= 5743.42 m4
( the moment of interia)
I
C
M
B
V 



min
max/
 =
42
.
5743
5
.
20
25
.
95299
41
5
.
19012 

max = 803.87 kN/m2
< 2700 kN/m2
O.K.
max = 123.57 kN/m2
> 0.00 kN/m2
O.K.

8. C
Ci
iv
vi
il
l E
En
ng
gi
in
ne
ee
er
ri
in
ng
g F
Fa
ac
cu
ul
lt
ty
y W
WA
AT
TE
ER
R R
RE
ES
SO
OU
UR
RC
CE
ES
S E
EN
NG
GI
IN
NE
EE
ER
RI
IN
NG
G
C
Ci
iv
vi
il
l E
En
ng
gi
in
ne
ee
er
ri
in
ng
g D
De
ep
pa
ar
rt
tm
me
en
nt
t
H
Hy
yd
dr
ra
au
ul
li
ic
cs
s 2
20
01
11
1-
-F
FA
AL
LL
L
E
EX
XE
ER
RC
CI
IS
SE
ES
S 2
2-
- D
DA
AM
MS
S -
- A
AC
CT
TI
IN
NG
G F
FO
OR
RC
CE
ES
S
Abdüsselam ALTUNKAYNAK www.altunkaynak.net
Research Assist. İsmail DABANLI www.akademi.itu.edu.tr/dabanli
SOLUTION 3
2
2

Sin
B
r

 =
2
120
2
300
Sin

= 173 m
Thickness at the base:
tb =
all
r
h

 

=
6200
173
120
10 

= 33.5 m
173
10
6200
6




h = 21.5 m
L = r *  = 120
180
173 


= 362 m
VT =   L
h
h
t
b
h
b b











 

2
=   362
5
.
21
120
2
5
.
33
6
5
.
21
6 










 = 750924 m3
SOLUTION 4
The optimum central angle  for a minimum volume of arch rib can be determined by differentiating the equation written below
with respect to  and equating to zero.
V = 





all
r
h 2
 V’ = Continue
120o
6 m
300 m
120o
6 m
300 m
h*=21.5 m
6 m
tb = 33.5 m
120 m
Figure 6
Figure 7

9. C
Ci
iv
vi
il
l E
En
ng
gi
in
ne
ee
er
ri
in
ng
g F
Fa
ac
cu
ul
lt
ty
y W
WA
AT
TE
ER
R R
RE
ES
SO
OU
UR
RC
CE
ES
S E
EN
NG
GI
IN
NE
EE
ER
RI
IN
NG
G
C
Ci
iv
vi
il
l E
En
ng
gi
in
ne
ee
er
ri
in
ng
g D
De
ep
pa
ar
rt
tm
me
en
nt
t
H
Hy
yd
dr
ra
au
ul
li
ic
cs
s 2
20
01
11
1-
-F
FA
AL
LL
L
E
EX
XE
ER
RC
CI
IS
SE
ES
S 2
2-
- D
DA
AM
MS
S -
- A
AC
CT
TI
IN
NG
G F
FO
OR
RC
CE
ES
S
Abdüsselam ALTUNKAYNAK www.altunkaynak.net
Research Assist. İsmail DABANLI www.akademi.itu.edu.tr/dabanli
SOLUTION 5
a)
2
2

Sin
B
r

 =
2
150
2
350
Sin

= 181 m
Thickness at the base:
tb =
all
r
h

 

=
5500
181
75
10 

= 24.7 m
181
10
5500
6




h = 18.23 m
b)
t30 =
5500
181
30
10 

= 9.87 m
c)
V = 





all
r
h 2
= 150
180
5500
181
75
10 2



 
= 11695.65 m3
/m
d)
L = r *  = 150
180
181 


= 474 m
VT =   L
h
h
t
b
h
b b











 

2
=   474
23
.
18
75
2
7
.
24
6
23
.
18
6 










 = 464899 m3
hx=18.23 m
6 m
tb=24.7 m
75 m
Figure 8