HS1DEGD - Design examples of Gravity Dam PDF

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Design examples of Gravity Dam...

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Design Example Gravity dam Design the non- overflow section of a gravity dam with the following data for an extreme loading condition: I. R. L. of deepest foundation level 100 m II. R. L. of roadway at the top of dam 161 m III. Full reservoir level 152 m IV. Design flood level 155 m V. Maximum tail water level 106.9 m VI. Minimum tail water level 100 m VII. Maximum deposited silt level on the upstream 105.3 m. VIII. Location of centre of drainage gallery form u/s face of dam 7 m IX. Roadway width at the top 6.1 m X. Downstream vertical face up to El. 154.28 m XI. Upstream face of dam vertical XII. Downstream face of dam 0.9 horizontal to 1.0 vertical XIII. Dry density of silt 1360 kg/m3 XIV. Unit weight of concrete 25 KN/m3 XV. Safe bearing capacity 1672 KN/m2 XVI. Shearing resistance of concrete (average) 2060 KN/m2 XVII. Shearing resistance of foundation (average) 1300 KN/m2 XVIII. Minimum allowable shear friction factor under a. Normal loading = 5.0 b. Extreme loading = 4.0 XIX. Maximum coefficient of sliding under Normal loading = 0.75 XX. Horizontal seismic coefficient = 0.2 XXI. Vertical seismic coefficient = 0.1 XXII. Fetch distance 18 km XXIII. Average wind speed 25 m/s Design 6.1

161 154.28

152 0.9

W1

1 W2

105.3

106.9 100

7

A. Selection of Dam Profile The base width as provided for the given section:

B

B = 0.9*(54.28) + 6.1 = 54.95 m The angle of the downstream face from the vertical is: tan-1(0.9/1) = 41.987o Preliminary section: Depth of water at the deepest foundation level (h): 152-100 = 52m Specific gravity of the section material (G) = 25/9.81 = 2.55 The bottom width of the elementary profile will be: For no overturning condition, B = 52/(2.55-1)0.5 = 41.77 m For no sliding condition, B = 52/(0.7*2.55) = 29.13 m Thus the minimum bottom width to be provided will be the maximum of the above results B = 41.77 m. This confirms that the provided (54.95 m) is safe and can be adopted. B. Steps in the Stability Analysis 1. Determination of forces and their moments All the force acting on the dam should be written first. For earthquake, the direction of acceleration should be mentioned, so as to determine the force correctly.  Dead load: The weight of the dam section can be found by dividing the section into simpler geometrical shapes, like triangle, rectangle, etc. The area of each part can be easily calculated and multiplying by the unit weight of concrete (25 KN/m3), the weight of each part can be calculated. The weight W 1 in the dam section is the weight of the rectangular section with width 6.1 m and height 61 m ; The weight W 2 in the dam section is the weight of the triangular section with width 48.85 m (0.9*54.28) and height 54.28 m. taking the thickness of the dam section perpendicular to this page as 1 m. The total dead load will be: (6.1*61 + 0.5*54.28*48.85)*1*25 = 42448.582 KN Moment arm length of the dead load from the toe of the dam will be: (6.1*61*(48.85 + 6.1/2) + 0.5*54.28*48.85*(2*48.85/3))/(6.1*61 + 0.5*54.28*48.85) = 36.8 m Moment of the dead load about the toe positive will be: 42448.582*36.8 = 1562107.82KNm We consider all clockwise moments about toe as negative all anticlockwise moments as positive. Thus the moment due to the dead load is positive.  Water Thrust The water pressure that acts horizontally on the upstream face is given by ½ϒhu2. In this case, water thrust = ½*52*52*9.81 = (-) 13263.12 KN. It acts at a distance of h/3 i.e. 52/3 m from toe. Hence moment is negative and equals to 13263.12*52 /3 = (-) 229894.08 KNm. The water pressure that acts horizontally on the downstream face is given by ½ϒhd2. In this case, water thrust = ½*6.9*6.9*9.81 = (+)233.527 KN. It acts at a distance of h/3 i.e. 6.9/3 m from toe. Hence moment is positive and equals to 233.527*6.9 /3 = (+) 537.112 KNm. The tail water weight that lies over the downstream face is given by: 9.81*½*6.9*(0.9*6.9) = (+) 210.174 KN. It acts at a distance of (0.9*6.9) /3 i.e. 2.07 m from toe. Hence moment is positive and equals to 210.174*2.07 = (+) 435.06 KNm.

 Uplift forces The uplift intensity at the upstream heel end is ϒh i.e. 52*9.81= 510.12 KN/m2 and on the downstream toe it will be ϒhd i.e. 6.9*9.81 = 67.689 KN/m2. 54.95m U2 442.43

67.689

U1

For extreme load combination design of a gravity dam the drains are assumed to be choked thus the uplift pressure force is computed as: ½ (510.12 + 67.689)*54.95 = (-) 15875.302 KN Its line of action from the toe will be 54.95*(2*52 + 6.9)/(3*(52 + 6.9) = 34.49 m. Thus the moment about the toe is negative and equals (-) 547500.722 KNm.  Silt load The silt pressure that acts horizontally on the upstream face is given by ½( ϒs - ϒw)hs2. In this case, silt thrust = ½*5.3*5.3*(13.34 - 9.81) = (-)49.601 KN. It acts at a distance of hs/3 i.e. 5.3/3 m from toe. Hence moment is negative and equals to 49.601*5.3 /3 = (-) 87.629 KNm.  Inertial forces due to earthquake Horizontal inertial force on the dam is computed as: αh*(W 1 + W 2) = 0.2*42448.582= (-) 8489.72 KN. Its line of action is through the centroid of the dam i.e. (6.1*61*(61/2) + ½*54.28*48.85*(54.28/3))/(6.1*61 + 0.5*54.28*48.85) = 20.8 m above the toe of the dam. The least stable structure of the dam exists if this horizontal force acts in the downstream direction (note the earth quake acts upstream in such cases). Hence moment is negative and equals to 8489.72*20.8 = (-) 176586.10 KNm. Vertical inertial force on the dam is computed as: αv*(W 1 + W 2) = 0.1*42448.582 = (-) 4244.86 KN. Its line of action is through the centroid of the dam i.e. 36.8 m from the toe of the dam. The least stable structure of the dam exists if this vertical force acts in the upward direction (note the earth quake acts in the direction of gravity in such cases). Hence moment is negative and equals to 4244.86*36.8 = (-) 156210.8 KNm. For upstream impounded water the vertical inertial force on the impounded water is computed as: αv*(½ϒhu2) = 0.1*13263.12 = (+)1326.312 KN. For downstream water αv*(½ϒhd2) = 0.1*233.527 = (-)23.353 KN. (Note the vertical inertial force tries to increase or decrease the net acceleration of the impounded water i.e. the unit weight of the water. Altering the unit weight has an impact on the horizontal water pressure). Their line of action is at 52/3 and 6.9/3 from the toe of the dam for the upstream and downstream water, respectively. The least stable structure of the dam exists if this vertical force acts in the upward direction (note the earth quake acts in the direction of gravity in such cases). Thus this force should be reduced from the respective water pressure forces of the upstream and downstream. For the upstream water the moment is positive and equals to 1326.312*17.33 = (+)22989.408 KNm. However the moment is negative and equals to (-) 53.711.

Vertical inertial force on the tail water weight is computed as: αv*(Wtw) = 0.1*210.174 = ()21.017 KN. Its line of action is through the centroid of the tail water section i.e. 0.9*6.9/3 = 2.07 m from the toe of the dam. The least stable structure of the dam exists if this vertical force acts in the upward direction (note the earth quake acts in the direction of gravity in such cases). Hence moment is negative and equals to 21.017*2.07 = (-) 43.505 KNm. Vertical inertial force on the silt accumulated is computed as: αv(½(ϒs - ϒw)hs2) = 0.1*49.601 = 4.96 KN. (Note the vertical inertial force tries to increase or decrease the net acceleration of the accumulated silt i.e. the unit weight of the silt and water. Altering the unit weight has an impact on the horizontal silt pressure). Its line of action is at 5.3/3 from the toe of the dam. The least stable structure of the dam exists if this vertical force acts in the upward direction (note the earth quake acts in the direction of gravity in such cases). Thus this force should be reduced from the silt pressure force and/or the moment could be taken as positive and equals to 4.96*5.3/3 = 8.763 KNm.  Hydrodynamic Pressure due to earthquake The horizontal earthquake will produce a force given by: Fe = 0.726 P eh, where pe = c αhϒ h; c = cm (at the base where h = z); cm = 0.73*90/90 = 0.73; (Since αh = 0.20 and upstream face is vertical) pe = 0.73*0.2*9.81*52 = 74.48 KN/m2 Fe = 0.726*74.48*52 = (-)2811.68 KN The moment about heel is given by 0.3peh2 = -0.3*74.48*52*52= (-) 60418.18 KNm. As the line of action of the hydrodynamic force equal from both toe and heel, this momentum is also acts with the same magnitude about the toe.  Wave pressure The wind speed in km/hr will be 25*60*60/1000 = 90 km/hr. Since the fetch distance (18 km) is less than 32 km, the wave height is computed as: hw = 0.032*(90*18)1/2 + 0.763 – 0.271(18)1/4 = 1.5 m The total pressure due to waves is given by 2ϒhw2. Fw = 2*9.81*1.52 = (-)44.145 KN. Since the pressure due to waves acts at height of (3hw/8) = 0.5625 m from reservoir surface, lever arm of the pressure about the toe = 52 + 0.56 = 52.56 m. The moment about toe is negative and is equal to 44.145*52.56 = (-) 2320.37 KNm.

2. Tabulation of Forces and Moments: Having determined forces and moments, they can be tabulated and summed up. No. 1

Item 2

VF

HF

MAL

M (+)

M(-)

KN

KN

m

KNm

KNm

4

5

6

7

8

36.8

1562107.82

Dead Load 1

(W1+W 2)

42448.582 Water thrust

3

PFu

-13263.12

17.33

229894.08

4

PFd

5

Wtw

233.527 210.174

2.3

537.112

2.07

435.06

Silt load 3

Fs

-49.601

1.77

87.629

Inertial force due to earth quake 3

Fv (W 1+W 2)

5

Fh (W1+W2)

7

36.8

156210.85

-8489.72

20.8

176586.10

Fh(PFu)

1326.312

17.33

8

Fh(PFd)

-23.353

2.3

53.711

9

Fv (Wtw)

2.07

43.506

10

Fh(F s)

11

Fe

-4244.86

-21.02 4.96

1.77

22989.408

8.763

Hydrodynamic force due to earth quake -2811.68

60418.18

Uplift forces 12 Σ

U1+U2

-15875.30 22517.58

34.49 -23072.67

547500.72 1586078.16

1170794.78

Dividing total moments (algebraic addition of positive and negative moments) by the total vertical force, the point of application of the resultant on the dam bottom line from heel is determined. Subtracting half bottom width from the above, the eccentricity of the resultant force is found. The eccentricity e will be: e = 54.95/2 - (1586078.16 - 1170794.78)/ 22517.58 = 9.03 m < 54.95/6 The total stress can be computed as: Total stress at the toe: 22517.58/54.95 + 6*9.03*22517.58/54.952 = 813.82 KN/m2 Total stress at the heel: 22517.58/54.95 - 6*9.03*22517.58/54.952 = 5.74 KN/m2 There is compression at both toe and heel. The maximum major principal stress at the toe occurs when there is no tail water load and is computed as: 819.57*Sec2(41.987) = 1483.57 KN/m2. By comparing the major principal stress at the toe with the bearing strength of the foundation (1672 KN/m2) it can be seen that there will not be any failure due to overstress. Coefficient of friction = 23072.67/22517.58 = 1.02 > 0.75, check for shear friction factor: SFF = (0.7*22517.58 + 0.5*1300*54.95)/23072.67 = 2.23 (For extreme loading condition the recommendation is that SFF shall be >1.3 for foundation rock. Thus the structure is safe against sliding) Factor of safety against overturning: FSo = 1586078.16/1170794.78 = 1.35 (For extreme loading condition the FSo shall be greater than 1.25. Thus the structure is safe against overturning)...