chapter vii dams

7/24/2019 Chapter Vii Dams

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Dams are structures built for the purpose of impounding water.

The dam is subjected to hydrostatic forces due to water which is raised on its upstream side. These

forces causes the dam to slide horizontally on its foundation and overturn it 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.

Purpose of Dam

a. Water supply

b. Irrigation

c. Power supply


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P

y

Headwater, HW

Upstream Side Downstream Side

Tailwater, TW

Vertical

Projection

of the

submerged

face of damh

1m

U2

Toe

x4

x1

x2

x3

W4 W1

W3

W2

Uplift Pressure Diagram

U1

Heel

z1

z2

Ry

Rx

R

Analysis of Gravity Dam


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Steps of Solution

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

II. Determine all the forces acting:

A. Vertical forces

1. Weight of the dam

1 = 12 = 23 = 3

2. Weight of the water

4 = 43. Weight or permanent structures on the dam

4. Hydrostatic uplift

1 = 12 = 2

B. Horizontal Force

1.

Total Hydrostatic Force acting at the vertical projection of the submerged portion ofthe dam,

= 2. Wind pressure

3. Wave Action

4. Floating Bodies

5. Earthquake Load


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III. Solve for the reaction

A. Vertical reaction, Ry

= = 1 + 2 + 3 + 4 1 2

B. Horizontal Reaction, Rx

=

= IV. Moment about the Toe

a. Righting Moment, RM (rotation towards the upstream side)

= 11 + 22 + 33 + 44

b. Overturning Moment, OM (rotation towards the downstream side)

= + 11 + 22


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V. Location of Ry () = =

where: = = 9.81 3 =

= 2.4 = 2.4

Factor of SafetyA. Against sliding, FSs:

= > 1.50

=coefficient of friction between the masonry and the foundation

B. Against Overturning, FSo

= > 2.0


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B/2 B/2

B

heel toe

c.g.

RY

Soil Pressure Diagram

Pmax

1m

= 1

= ()

PRESSURE DISTRIBUTION AT THE BASE

1.

Resultant at the middle of the base


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B/3

Middle Third

B/3 B/3

RY

heel toe

Pmax

=

3

=+ 0

2(1)

=

2()

2.

Resultant at the middle thirds nearer the toe.


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B/3

Middle Third

B/3 B/3

RY

heel toe

Pmax

3

= (3)2

= 23

3. Resultant outside the middle-thirds


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B/3 B/3 B/3

Middle Third

heel toe

c.g.

RY

Soil Pressure Diagram

fh

ft

e

1m

4. Resultant within the middle thirds

From combine axial and bending stress formula:

=

where : = = 1 = = = 1

3

12

= 2


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Substitute to the interaction formula

= (2)

(1)312

: = 2

= 62

=

(1 6

)


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4m

18.8m

1.20m

Masonry

Dam

(sg=2.40)

2.40m

20m

Example:

For the gravity dam with cross section shown, determine:

1.

Factor of safety against sliding

2.

Factor of safety against overturning3. Stress of foundation at the toe of dam

4. Stress of foundation at the heel of dam

Assume the uplift to vary linearly from full hydrostatic pressure at the toe to full hydrostatic

pressure at the heel, coefficient of friction in siding is 0.65.


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4m

18.8m

1.20m

Masonry

Dam

(sg=2.40)

2.40m

20m

F1F2

U1

U2

P1=18.8w P2=2.40w

(P1-P2)

heel toe

W1

W3W2

Solution:

Forces acting on dams


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Considering 1m length of wall

1 =1

211 =

1

2(18.8w1

2 = 1221 = 1

2(2.4w1

Forces Magnitude (kN)Moment Arm

about toe (m)

Resisting

Moment

(kN.m)

Overturnin

g Moment

(kN.m)

W1 =4(20)(1)(2.4x9.81)=1,883.52 18 33,903.36W2 =1/2 (16)(20)(1)(2.4x 9.81)=3,767.04 32/3 40,181.76

W3 =1/2(2.4)(1.92)(1)(9.81)=22.6 1.92/3 14.46

F1 =1/2(18.8x9.81)(18.8)(1)=1,733.62 18.8/3 10,864.02

F2 =1/2(2.4x9.81)(2.4)(1)=28.25 2.4/3 22.6

U1 =1/2(18.8-2.4)(9.81)(20)(1)=1,608.84 40/3 21,451.2

U2 =(2.4x9.81)(20)(1)=470.88 10 4,708.8

TOTAL 74,122.18 37,024.02

= 1 + 2 + 3 1 2 = 1,883.52 + 3,767.04 + 22.6 1,608.84 470.88 = 3,593.44 = 1 2 = 1,733.63 28.25 = 1,705.37


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1. Factor of safety against sliding

= =0.65(3,593.44)

1,705.37= 1.37 < 1.5 ,

2.

Factor of safety against overturning

= =74,122.18

37,024.02= 2.6 > 2.0 ()

3. Location of Ry at base of dam from toe

=

= 74,122.18 37,024.023,593.44

= 10.32 > 3

chapter vii dams

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