5 – Analysis of Gravity Dams, Slides of Fluid Mechanics

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ENGR. BON RYAN ANIBAN

A dam is subjected to a hydrostatic forces due to water which is raised on its upstream side. These forces cause the dam to

slide horizontally on its foundation and overturn about its downstream edge or toe. These tendencies are resisted by friction

on the base of the dam and gravitational forces which causes a moment opposite to the overturning moment. The dam may

also be prevented from sliding by keying its base.

Typical section of gravity dam

Upstream Downstream

Heel Toe

Upstream

x 1

Heel Toe

Steps in Analysis:

W 1

W 2

W 3

W 4

p = γh

h

U

Vertical

projection

of the

submerged

face of dam

F

1

IV. Moment About the toe

R

R y

R x

A. Righting Moment, RM

x 2

x 3

x 4 RM = 𝑊 1

1

2

2

3

3

4

4

B. Overturning Moment, OM

OM = Fy + Uz

y

z

V. Location of R y

( 𝑥ҧ)

xҧ

𝑥^ ҧ =

𝑦

Factor of Safety

Factor of Safety against sliding, FSs ;

𝑦

𝑥

Factor of Safety against overturning, FSs ;

𝐹𝑆o =

Where ;

𝜇 = cofficient of friction between the base of the dam and the foundation

Heel Toe

Foundation Pressure

R y

e

xҧ

e

cg

For e ≤ 𝐵/ 6

1 m

B

B/ 2 B/ 2

From combines axial and

bending stress formula :

q

t

q

H

𝑦

𝑦

3

𝑦

𝑦

3

H

𝑦

T

𝑦

B/ 6

6 m

8 m

L = 1

Sg concrete = 2.

Neglect Uplift

𝜇 = 0. 8

𝐹𝑆𝑠 =?

𝐹𝑆𝑜 =?

𝑞 𝑇

=?

𝑞 𝐻

=?

W 1

Steps in Analysis:

I. Consider 1 m length of dam (perpendicular to the sketch)

II. Determine all the forces acting:

A. Vertical Forces

1. Weight of the dam

2. Weight of the water in the upstream side (if any)

3. Hydrostatic Uplift

B. Horizontal Force

1. Total Hydrostatic Force acting at the vertical projection of

the submerged portion of the dam

III. Solve for the reaction

A. Vertical Reaction, R y

B. Vertical Reaction, R x

W

1

= γ c

V

1

F = γ

ℎA

= 565. 056 kN

F

R

y

= 565. 056 kN

R

x

= 313.92 kN

= 9. 81 ( 4 )( 8 )( 1 ) = 313. 92 kN

6 m

8 m

L = 1

Sg concrete = 2.

Neglect Uplift

𝜇 = 0. 8

𝐹𝑆𝑠 =?

𝐹𝑆𝑜 =?

𝑞 𝑇

=?

𝑞 𝐻

=?

W 1

Steps in Analysis:

III. Solve for the reaction

B. Vertical Reaction, R x

F

R

y

= 565. 056 kN

R

x

= 313. 92 kN

IV. Moment About the toe

A. Righting Moment, RM

RM =

B. Overturning Moment, OM

OM = 313. 92 kN( 8 / 3 )

𝑥^ ҧ =

𝑦

V. Location of R y

( 𝑥ҧ)

4 m

( 565. 056 )( 4 ) = 2260. 224 kN−m

8/

= 837. 125 kN−m

2260. 224 − 837. 125 kN−m

= 2. 519 m

Factor of Safety

𝑦

𝑥

𝐹𝑆o =

Foundation Pressure

R y

cg

ҧ 𝑒 𝑥

1. 481

H

𝑦

T

𝑦

For e ≤ 𝐵/ 6

A. Vertical Reaction, R y

H

= − 48. 877 kPa

T

= − 139. 475 kPa

4 m

2 m

W 1

Steps in Analysis:

A. Vertical Forces

B. Horizontal Force

III. Solve for the reaction

W

1

F

= 376 kN

F

R

y

R

x

= 176. 58 kN

= 176. 58 kN

L = 1

γ c

= 2 3.5 kN/m

3

Consider Uplift

𝜇 = 0. 6

𝐹𝑆𝑠 =? 𝐹𝑆𝑜^ =?

𝑞 𝑇

=? 𝑞 𝐻

=?

6 m

2 m

W 2

W

2

= 188 kN

γh

58.86 kPa

U

U =^117.^72 kN

R y

R x

IV. Moment About the toe

A. Righting Moment, RM

RM =

B. Overturning Moment, OM

𝑥 ҧ =

𝑦

V. Location of R y

( 𝑥ҧ)

( 376 ) = 1378. 667 kN−m

Factor of Safety

𝑦

𝑥

𝐹𝑆o =

3 m

4 / 3

2

8 / 3

OM = ( 176. 58 )( 2 )+ (117.72) ( 8 / 3 )= 667. 08 kN−m

𝑥 ҧ

= 1. 594 m

Foundation Pressure

cg

ҧ 𝑒 𝑥

1. 406

= 0. 667 For e^ ≤^ 𝐵/^6

H

𝑦

T

𝑦

H

= − 43. 624 kPa

T

= − 179. 516 kPa