TDS Lec 2a-Dams.pdf

THEORY & DESIGN OF
STRUCTURES
C3 -05 (5S3 NVQ 2045)
Eng. Y.A.P.M Yahampath
B Sc. Eng (Hons), Dip Highway & Traffic Eng, AMIESL, AMECSL

Lecture Hours Allocated
Lecture / Tutorial Practical Demonstrations/
Industrial visits
Self Study Total
72 Hrs 00 Hrs 38 Hrs 110 Hrs
Unit Title Time (Hrs)
Combined Direct & Bending stresses 10
Principle stresses & strain 06
Structural design principles 06
Designing of Reinforced concrete
elements
26
Design of structural steel elements 18
Structural detailing 06
Total 72

AIM OF THE MODULE
To develop fundamental understanding of the behavior of structures with
particular reference to statically determinate civil engineering structures.
✓ To develop & understanding of the factors and constraints in determining
suitable structural components.
✓ To develop awareness of the economical, Engineering & esthetic aspect
in designing in selecting a particular structural component for given
condition.
✓ To introduce students to use codes of practice and design charts/ Tables
etc. for designing of structural elements.

Retaining Structures – Dams / Retaining walls
• Analysis of dams applying theory of combined stress
• Forces Applying on Dam
• Resultant force acting on base of a dam
• Stress variation diagram at the base of the dam

When we consider water at given depth
the pressure acting on a water particle
in any direction is equal.

Under water,
h
Water surface
σ
σ
σ
σ
σ = ϒ w .h
At given depth the pressure acting on a water
particle in any direction is equal.
Water
element

a = Top width of the dam
d = Bottom width of the dam
H = Height of the dam
h = Height of the water retain by the dam
ϒ w = Specific weight/ unit weight of the water (kN/ m3 ) (ϒ w = ρ wg)
Γ = Unit weight of the dam masonry (kN/ m3 )
Consider a unit length (b=1 m) of the dam,
Volume (m3) of unit length of the dam,
V = [(a+d)/2] x H X 1
Weight (KN) of unit length of the dam,
W = [(a+d)/2] x H X 1 X Γ
P = F/A (kN/ m2 )
ϒ = ρ g (g = 9.81 m/s-2)
ρ = M/V
ϒ = M/V X g
F = Ma = Mg
ϒ = F/V = kN/ m3
b

a
w1
w2
+
w1
w2
=
W
x
a
d
a/2
a+(d-a)/3
W.X = w1.a/2 + w2.[a+(d-a)/3]
X = {w1.a/2 + w2.[a+(d-a)/3]}/W
A B
MA
H
(d-a)/3
a/2

Water Pressure Equation
P = hρwg
• Hence the equation P = hρg represents the
pressure due to the weight of any fluid having an
average density ρ at any depth h below its
surface. Pressure varies with depth.
P = F/A
F = PA
dF = P.dA dA = 1.dy
dF = y.ρwg .1.dy
F = y.ρwg .1.dy = 1/2 ρwg y2 =1/2 ρwgh2
y
h
b=1 m
F
h
0

h
F
?
Water head
F = 1/2. (ρwg).h2 But, ϒw = ρwg
F = 1/2. ϒ w .h2 ϒ w = 1000 kg/m3 x 9.81 m/s-2 /1000 kN / m3
9.81 10 kN / m3
ϒ w = 1000 kg/m3 x 9.81 m/s-2 =10 kN/m3

The deeper the water, the more horizontal pressure it exerts on the dam. So at the surface of the
reservoir, the water is exerting no pressure and at the bottom of the reservoir, the water is
exerting maximum pressure. The force acts at the center of gravity of the triangle -- one-third of
the way up from the bottom.
h
F
h/3
Water head
h/3

F
W
G
h/3
R
J K
N
L
M
x
J K
N
L
M
h/3
x
F
W
X/F = h/3/W
X = F.h/3W
R
R2 = √ W2+F2
Φ
✓ W
✓ F
✓ R
✓ h/3
✓ Φ = W/F
X = ?
Using force diagram,
X = The
horizontal
distance
between
the center
of gravity of
the dam &
point where
the
resultant
force ( R )
pass
through the
dam base

F
Water head
h/3
h
b =1m
H
a
A B
w1
W
w2
R
Toe
XX-Axis
x
ZZ- Axis of
symmetry
G
G2
G1
a/2
d/2
(d-a)/3
e
X
K
L
YY-
Axis
(d-a)

Trapezoidal
Dam with
Water Face
Vertical
h
H
F
a
d/2
h/3
A B
Water head
w1
W w2
R
Heel Toe
L
x x
d/2
e
e = [L – d/2]
σmin = W/d { 1 – 6e/d } σmax = W/d { 1 + 6e/d }
Stress Distribution at base of the dam,
Axis of
symmetry
L = Liver arm
L = X+X
G
G2
G1
C+ C+
C+
T-
Heel Toe

Derivation of Middle third Rule for Dam Base
• The minimum stress (σMin) must be greater or equal to zero for no tensile stress
at any point along the width of the column.
Z = I/eMax
I = bd
3
/12
b = 1 m & d = d m,
I = 1.d3/12
eMax = d/2
Z = d
2
/6
A = b.d
But, b = 1 m
A = 1.d
σmin = W/A - M/Z
σmin > 0
W/A - M/Z > 0
M = We
W/d – We/ (d
2
/6) > 0
W/d - 6We/d
2
> 0
W/d { 1 – 6e/d } > 0
W/d = 0,
So,
{ 1 – 6e/d } > 0
d/6 > e
σ = W/A ± M/Z
Assume Bending around x-x,

Derivation of
Middle third
Rule for Dam
Base
d
1m
d/3 d/3 d/3
d/6 d/6
d/2
A B
A B
K
e
L
e = [L – d/2]
x
x
W
W
Y Y
z
z
Plan view of Base
L = Liver arm
L = X+X
Axis of symmetry Z-Z,

F
f
Friction Effect on Dam
Frictional force (f),
f = µR
considering equilibrium of vertical forces,
W = R
f = µ.W
If,
f > F
Hence no sliding effect.
If,
f < F
Failure due to sliding effect.
Safety factor against water pressure = f/F
f/F >1 - no sliding effect
W
G
R
A

Methods of failure
Tension in the base
If, d/3 < L < 2d/3 – No
failure( Resultant force
pass through the middle
third section of the base)
If, d/3 > L > 2d/3 –
Failure
Over turning
If, 0< L < d – No
failure( Resultant
force pass within the
section of the base)
If, 0 > L > d – Failure
( Resultant force pass
outside the section
of the base)
Sliding
If, f > F - No failure
If, f < F - Failure

• Check for Failure Due to Tension
in the base
• Check for Failure Due to
overturning
F
W
G
F
W
G
R R
d/3 d

h =7m
F = ?
a =1.2m
h/3 =7/3m
Water head
W=?
H =8m
w1
w2
G2
G1
d =4.8 m
10kN/m3
24kN/m3
x
A B
Up stream

Considering a 1 m length of dam,
• F = 1/2. ϒ w .h2
• F = ½ X 10 kN/m3 X 72 m2
• F = 245 kN/m
• W = (1.2+4.8)/2 X 8 X 24 X 1
• W = 576kN/m
• X = F.h/3W
• X = (245 x 7)/(3 x 576)
• X = 0.99 m

To find location of the center of gravity of the dam section (X) ,
Consider the dam section only & split in to two shapes
(rectangle + triangle) with known center of gravity
Considering moments around A,
W.X = w1.x1 + w2.x2
576 x X = (8x1.2)x 24 x (1.2/2) + ½ x 8 x 3.6 x 24 x(1.2+3.6/3 )
X = 1.68 m
L = X+X
L = 1.68 + 0.99 = 2.67 m
w1
w2
G
G2
G1
W
X = ?
a/2 =0.6m
a =1.2m
d =4.8m
1.2+(3.6/3)= 2.4m
A

To find the eccentricity,
• e = [L – d/2]
• e = [2.67 – 4.8/2] = 0.27 m
X =1.68m
d =4.8m
1.2+(3.6/3)= 2.4m
a =1.2m
e = 0.27m
N/A
σmin = W/d { 1 – 6e/d }
σmin = 576/4.8 x { 1 – 6 x 0.27/4.8 }
σmin = 79.5 kN/m2
σmax = W/d { 1 + 6e/d }
σmax = 576/4.8 x { 1 + 6 x 0.27/4.8 }
σmax = 160.5 kN/m2

σmin = 79.5 kN/m2
σmax = 160.5 kN/m2
Stress ( kN/m2 )
Distance ( m )
4.8 m
Toe
Heel
STRESS DISTRIBUTION DIAGRAM
0

• Check the middle third rule,
d =4.8 m
d/3=1.6m
d/6 =0.8m
d/2 =2.4m
A B
A B
K
e=0.27m
L=2.67m
R
d/3=1.6m d/3=1.6m
N/A
d/6 =0.8m
d/6 =0.8m > e=0.27m
Middle third rule satisfied.
1.6 < L=2.67m < 3.2

• Check the stability of the dam for Failure Due to Tension & overturning
• Check for Failure Due to Tension in
the base
Check if,
d/3 < L < 2d/3 – No failure
L = 2.67 m
d/3 = 4.8/3 = 1.6m
2d/3 = 2 x 4.8/3 = 3.2m
So,
1.6m < 2.67 m < 3.2m
Resultant force pass through the
middle third section of the base.
So, No failure due to tension
• Check for Failure Due to overturning
Check if,
0< L < d – No failure
L = 2.67 m
d = 4.8 m
So,
0 < 2.67m < 4.8
Resultant force pass within the section of
the base.
So, No failure due to over turning.

• If the coefficient of friction between soil and dam base is 0.6, Check the stability of the
dam
R = W
R = 576 KN
f = µR = µw = 0.6 x 576
f = 345.6 kN
But, F = 245 kN/m
So, 345.6 kN > 245 kN/m
If,
f > F
Safety factor against water pressure = f/F = 345.6/245 = 1.4
F
f
W
G
R

Q2.
A concrete dam has its upstream face vertical & a top width of 3m. Its downstream face
has a uniform batter. It stores water to a depth of 15m with a freeboard of 2 m.
Calculate the minimum dam width(d) at the bottom for no tension in concrete.
ϒw = Specific weight/ unit weight of the water (10kN/ m3 ) (ϒw = ρ wg)
Γ = Unit weight of the dam masonry (25kN/ m3 )
Neglect uplift force.

h =15m
F = ?
a =3m
h/3 =5m
Water head
W=?
H =17m
w1
w2
G2
G1
d =? m
10kN/m3
25kN/m3
x
A B
Up stream
2m
x
L

• F = 1/2. ϒ w .h2
• F = ½ X 10 kN/m3 X 152 m2
• F = 1125 kN/m
• W = [(3+d)/2 X 17 X1]X 25
• W = 212.5(3+d)kN/m
• X = F.h/3W
• X = (1125 x 15)/[3 x 212.5(3+d)]
• X = 26.47/(3+d) m

w1
w2
G
G2
G1
W
X = ?
X1= a/2 =1.5m
a =3m
d = ? m
X2=3+[(d-3)/3]= [2+d/3]m
A (d-3)m

To find location of the center of gravity of the dam section (X) , Consider the dam section only &
split in to two shapes (rectangle + triangle) with known center of gravity
W.X = w1.x1 + w2.x2
212.5(3+d)X = (17x3)x 25 x (3/2) + ½ x 17 x (d-3) x 25 x [2+d/3]
1912.5X + 637.5dX = 5737.5 + 212.5d2 + 1275d - 637.5d - 3825
X = [d2 +3d + 9 ]/3(3+d)
L = X+X
L = 26.47/(3+d) + [d2 +3d + 9 ]/3(3+d)
L = [88.41 +d2 + 3d]/ 3(3+d)

• e = [L – d/2]
• e = [88.41 +d2 + 3d]/ 3(3+d) – d/2
• e = [176.82 - d2 – 3d] / (18 + 6d)
• Stress Distribution at base of the dam,
• σmin = W/d { 1 – 6e/d }
• The minimum stress (σMin) must be greater or equal to zero for no tensile stress
at any point along the width of the column.
• σmin >= 0
• W/d { 1 – 6e/d } >= 0
• d/6 >= e
• Minimum d value at,
• d/6 = e
• d = 6e

• d/6 = e = [176.82 - d2 – 3d] / (18 + 6d)
• d.(18 + 6d) = [176.82 - d2 – 3d] x 6
• d2 + 3d - 88.41 = 0
• d = 8.21m or d = -11. 021m
• d can’t be a negative value. So,
• d = 8.21m

Q3.
A trapezoidal shape concrete dam has a top width of 1m and base width of 7m. It
stores water to a depth of 10m without a freeboard.
i. List out assumptions clearly
ii. Find the total force due to upstream water and point of action
iii. Find the weight of the dam and point of action
iv. Calculate the ecstaticity of the resultant force at the base
v. Check the stability of the dam for Failure Due to Tension & overturning
vi. Find the factor of safety against sliding at the base
vii. Find the maximum and minimum stress at the base
viii. Draw the stress distribution diagram
µ = Friction coefficient between dam base and soil is 0.5
ϒ w = Specific weight/ unit weight of the water (10kN/ m3 ) (ϒ w = ρ wg)
Γ = Unit weight of the dam masonry (20kN/ m3 )
Neglect uplift force.

a =1m
h/3 =10/3m
Water head
W=?
h=H =10m
w1
w2
G2
G1
d =7 m
10kN/m3 20kN/m3
A B
Up stream
w3
G3
w4
G4
F = ?
x
R
L x e f
E D C
x1
x2
x3
x4
1m 5m

• F = 1/2. ϒ w .h2
• F = ½ X 10 kN/m3 X 102 m2
• F = 500 kN/m
• W = Weight of dam + weight of the AED water mass
• W = [(1+7)/2x 10 x 1x 20] + [½ x 10 x 1x1 x 10]
• W = 850kN/m
• X = F.h/3W
• X = (500 x 10)/[3 x 850]
• X = 1.96 m
Assumptions
Consider 1 m length of dam.
Neglect up thrust.
Water pressure force at h
depth is equal to any direction.
So consider water force acts
perpendicular to AE line.
Weight of AED section due to
water mass also added to total
weight of the dam.

split in to four shapes (water trinangle+trinangle+rectangle + triangle) with known center of
gravity.
W.X = w1.x1 + w2.x2 + w3.x3 + w4.x4
850X = (½ x1x10x1)x 10 x (1/3) + (½ x 1x10x1) x20x(2/3) + (1x10x1)x20x(1+0.5) + (½ x1x10x1) x20x(2+5/3)
850X = 16.667 +66.667 + 300 +1833.333
850X = 2216.667
X = 2.607 m
L = X+X
L = 1.96 + 2.607
L = 4.567 m

• e = [L – d/2]
• e = [ 4.567 – 7/2 ]
• e = 1.067m
• To satisfy middle third rule to avoid tension at the dam base,
• d/6 >= e
• But, d/6 = 7/6 = 1.1667m
• 1.1667m > 1.067m
• Dam base satisfy middle third rule, so no tension at the dam base.

• Check for Over turning
• If, 0< L < d – No failure( Resultant force pass within the section of the base)
• 0 < 4.567 m < 7m
• So resultant force pass within the section of base, no failure due to overturning

Check for Sliding,
R = W
R = 850 KN
f = µR = µw = 0.6 x 850
f = 510 kN
But, F = 500 kN/m
So, 510 kN > 500 kN/m
If,
f > F
Safety factor against water pressure = f/F = 510/500 = 1.4

σmin = W/d { 1 – 6e/d }
σmin = 850/7 x { 1 – 6 x 1.067/7 }
σmin = 10.37 kN/m2
σmax = W/d { 1 + 6e/d }
σmax = 850/7 x { 1 + 6 x 1.067/7 }
σmax = 232.48 kN/m2

σmin = 10.37 kN/m2
σmax = 232.48 kN/m2
Stress ( kN/m2 )
Distance ( m )
7 m
Toe
Heel
0

Q4. Calculate the stresses of dam base for water level at D-C
a =0.8m
Water head
W=?
0.5m
w1
w2
G2
G1
d =1.4 m
10kN/m3
20kN/m3
Up stream
F = ?
R
L
f
x1
x2
B
D C
A
0.6m
0.5m
H=h=1m
x
w3
G3
w4
G4
x3
x4
0.2m
0.2m
x
e
E
I J
K
M

• F = 1/2. ϒ w .h2
• F = ½ X 10 kN/m3 X 12 m2
• F = 5kN/m
• W = Weight of dam + weight of the AEDIJM water mass
• W = [0.8x 1 x 1x 20]+[0.6x0.5x1x20] + [ 0.2x
1x1 x 10] + [0.6x0.5x1x10]
• W = (16+6+2+3) = 27kN/m
• X = F.h/3W
• X = (5x 1)/[3 x 27]
• X = 0.0617 m
Assumptions
Consider 1 m length of dam.
Neglect up thrust.
Water pressure force at h
depth is equal to any direction.
So consider water force acts
perpendicular to dam.
Weight of AEDIJM section due
to water mass also added to
total weight of the dam.

split in to four shapes (water trinangle+trinangle+rectangle + triangle) with known center of
gravity.
W.X = w1.x1 + w2.x2 + w3.x3 + w4.x4
27X = (0.2x1x1)x 10 x (0.2/2) + (0.6x0.5x1) x10x[0.2+(0.6/2)] + (0.6x0.5x1)x20x[0.2+(0.6/2)] + (1x0.8x1)
x20x[0.2+0.6+(0.8/2)]
27X = 0.2 + 4.5 + 3 + 19.2
27X = 26.9
X = 26.9/27
X = 0.996 m
L = X+X
L = 0.0617 + 0.996
L = 1.057 m

• e = [L – d/2]
• e = [ 1.057 – 1.4/2 ]
• e = 0.3579m
• To satisfy middle third rule to avoid tension at the dam base,
• d/6 >= e
• But, d/6 = 7/6 = 0.233m
• 0.233m < 0.3579m
• Dam base satisfy doesn't middle third rule, so tension will built up at the dam
base.

• Check for Over turning
• If, 0< L < d – No failure( Resultant force pass within the section of the base)
• 0 < 1.057 m < 1.4m
• So resultant force pass within the section of base, no failure due to overturning

Check for Sliding,
R = W
R = 27KN
f = µR = µw = 0.6 x 27
f = 16.2 kN
But, F = 5 kN/m
So, 16.2 kN > 5 kN/m
If,
f > F
Safety factor against water pressure = f/F =16.2/5 = 3.24

σmin = W/d { 1 – 6e/d }
σmin = 27/71.4 x { 1 – 6 x 0.3579/1.4 }
σmin = -10.29 kN/m2
σmax = W/d { 1 + 6e/d }
σmax = 27/1.4 x { 1 + 6 x0.3579 /1.4 }
σmax = 48.86 kN/m2

σmin = -10.29 kN/m2
σmax = 48.86 kN/m2
Stress ( kN/m2 )
Distance ( m )
1.4 m
Toe
Heel
0

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