This document discusses the different modes of failure for gravity dams: 1. Overturning or rotation about the toe if the resultant force passes outside the toe. 2. Compression or crushing if compressive stresses exceed allowable stresses in the dam material. 3. Development of tension, which concrete dams are designed to avoid as concrete cannot withstand sustained tension. 4. Shear failure or sliding if the net horizontal force exceeds the frictional resistance between surfaces of the dam. The factor of safety against sliding must be greater than 1.

1. Lecturer: Prof. Dr. Thamer Ahmad Mohammad
Lecture 6/ 30 December 2020
Dam Engineering
4th Year
Department of Water Resources Engineering
College of Engineering
University of Baghdad
COLLEGE OF
ENGINEERING

2. Modes of Failure and Criteria for Structural Stability
of Gravity Dams
A gravity dam may fail in the following ways:
1. By overturning or rotation about toe
2. By compression or crushing
3. By development of tension causing ultimate failure by
crushing
4. By shear failure called sliding
The failure may occur at the foundation plane (i.e at the base
of the dam ) or at any other plane at higher elevation

3. 1. By overturning or rotation about toe
If the resultant of all forces acting on the dam at
any of its sections, passes outside the toe, the dam
shall rotate and overturn about the toe. Practically,
such a condition shall not arise, as the dam will fail
much earlier by compression.
Moment
Clockwise
Moment
Clockwise
Anti
g
Overturnin
Against
Safety
of
Factor 
The value of the factor of safety is generally varies between 2 to 3.

4. 2. By Compression or Crushing
A dam may fail by the failure of its material, i.e the compressive
stress produced may exceed the allowable stresses and the dam
material may get crushed. The vertical direct stress distribution at
the base is given by the following equation:
P = Direct stress + Bending stress
y
I
M
B
F
P v



1
*
min
max
B
1 m
12
1
*
3
B
I 

5. Neutral
Axis
y=B/2
1 m























B
e
B
F
P
B
e
F
B
F
P
B
e
F
B
B
e
F
y
I
M
v
v
v
v
v
6
1
6
*
6
*
2
12
*
min
max
2
min
max
2
3









B
e
B
F
P v 6
1
max 








B
e
B
F
P v 6
1
min
E=Eccentricity of the resultant force from
the center of the base
Fv= Total vertical force
B= width of dam base

6. Toe
Heel
Fv
R=Resultant Force
FH
R
Pmax
Pmin
Pmax
Pmin
e
Resultant
nearer to the
toe which
produced a
maximum
compressive
stress at the toe
when the
reservoir is full

7. Toe
Heel
Fv
R=Resultant Force
FH
R
Pmax
Pmin
Pmax
Pmin
e
Resultant
nearer to the
heel which
produced a
maximum
compressive
stress at the
heel when the
reservoir is
empty

8. 3. By development of tension causing ultimate
failure by crushing
Concrete gravity dams are usually designed in such a
way that no tension is developed anywhere because
these materials cannot withstand sustained tensile
stress. If subjected to such stresses, these materials
may finally cracks. However, for achieving economy in
designs of very high gravity dams, certain amount of
tension may be permitted because of the fact that
such worst loading conditions shall occur only
momentarily for short time and would neither last
long nor occur frequently. In order to ensure that no
tension is developed anywhere, Pmin should equal to
zero.

9. 6
0
6
1
0
6
1
0
6
1
min
min
B
e
B
e
B
e
B
F
P
B
e
B
F
P
v
v






















The maximum value of the eccentricity that can be
permitted on the other side of the center is equal to
B/6 which leads to the famous statement (the
resultant must lie within the middle third).

10. Toe
Heel
Fv
R=Resultant Force
FH
R
Pmax
Pmin=0
e=B/6

11. 4. By shear failure called sliding
Sliding or shear failure will occur when the net horizontal
force above any plane in the dam or at the base of the dam
exceeds the frictional resistance developed at that level. The
friction developed between the two surfaces is equal to
Fv where Fv is the algebraic sum of all vertical forces
whether upward or downward and  is the coefficient of
friction between two surfaces. In order that no sliding takes
place, the external horizontal force, FH must be less than
the shear resistance Fv.
Fv/FH represents nothing but the factor of safety
against sliding which must be greater than unity

12. FH
Fv
H
v
F
F
sliding
against
safety
of
Factor




Factor of safety against sliding 1

13. FH =5022.72 kN
Fv =7978.25 kN
Calculate the followings:
1. The resultant of all horizontal
forces
2. The resultant of all vertical
forces
3. The shear force if the
coefficient of friction between
the dam and its foundation is
0.70
4. The factor of safety against
sliding
5. The specific weights of water
and concrete are 9.87 and 23.5
kN/m3
6. F.S=1.6
Exercise for Stuents Practice
7 m
30 m
25 m
32 m
5 m
Fv =11397.5 kN