DAM ENGINEERING CONCRETE DAMS (HE-613/IE-613) Exercise Question 1: a) factors responsible for the stability of concrete dams are: b) Describe the stability of concrete dams due to seismic forces. Overturning Stability The forces responsible for over turning are: • The hydrostatic (water) force: this is horizontal force act on the upstream of the dam and its moment with respect to the toe of the dam is clock wise direction. • Slit force: this is a force due to the dead load of the dam (or deposition of the sediment upstream of the dam). The moment of this force is clock wise with respect to the toe of the dam. Hence it is the overturning force. • Uplift force: this is a force due to the seepage water under the bottom of the dam. The moment of this force is clock wise direction and it is overturning moment with respect to the toe of the dam.
Sliding stability:
The forces responsible for sliding are as follows: • Hydrostatic horizontal force: • Silt force: • Wind force: • Shear stress The forces responsible for resisting sliding are as follows: • Shear strength of concrete and rock : • Weight of the dam: b. Seismic forces affects the stability of dam: The dam section is given: Required: i) to determine the maximum stresses at the heel and toe of the dam ii) The factor of safety against sliding shear and overturning Given Data: Unit weight of concrete dam, γc = 2.4 t/m3 Coefficient of friction, = 0.7 Shear force = 200 t/m2 Horizontal Seismic Coefficient = 0.2g Vertical seismic coefficient = 0.1g Solution:
1
By considering all the forces exerted on the dam for extreme case: The forces of all kind, its center of action and moment about the toe of the dam are tabulated as follows. It is considered all extreme case and the Hydrodynamic forces may act in both directions, so it is necessary to consider both cases separately.
2
Type Forces
of Formula
Hydrostati c
Self weight b1=4.5m b2=10m b3=60m
Fv1= γ w * ∀
1545.075
Moment arms w.r.t toe of dam (m) 1/3*80 = 26.67 72.25
Fv2= γ w . y. sin θ .b1
993.25
72.252
+71764.3
γ c .h1.b1 2
2383.83
71.5
+170443845
W2= γ c .H .b2
18835.2
65
+1224288
52974
40
+2118960
2 Fh= γ w H
W1=
W3= Uplift force
Seismic Load
Vertical (KN)
31392 2
γ c. h3.b3 2
Moments Moment arm Moment w.r.t w.r.t toe of w.r.t to centroid of dam Dam (KN.m) centroid of dam -837224.64 26.67 -837224.64 +104406.67
7.4
+ 11433.555
6.65
+ 6605.1125
6.65
+ 15852.4695
0.15
+ 2825.28
-24.85 -1316403.9
Fu= ( H 1 + Hd ).B1 γw 2 H .B + γ w. d 2 2 Hd = 0.33H = 26.4 m B1 = 12.5 m B2 = 60 m Fses= +α h .Wtotal
-14293.17
Fses= +α v .Wtotal
+ 751.483
48.33
+ 15026.96
-69088.91
47.58
----
47.58
+ 714982.7568
-22.85
+ 17171.38655
+714982.8
Fewh=
summation
Horizontal (KN)
60814.82
0.66Ceα h Z 1γ w ( Z1 .Z max )
1
Fewv= +α v .Pwv
+253.8325
42
+31562.29
32
+1946074
32
+ 1946074.24
+18339.398
7.4
-1878.3605
2
63443.501* 61432.8695** 75726.0395*** 77736.6705****
72.25 107233.8*
175767908.9*
528851.8478****
77179.86**
174238139.9**
-870527.6138***
3
N.B. the mark of star (*) is for summation of all forces and moments (parameters) considering the seismic force in positive direction. But (**) is summation of all forces and moments (Parameters) considering seismic force in another direction, in negative direction. (***) is summation of forces and moments (parameters) excluding uplift force and considering negative seismic forces. (****) is summation of forces and moments (parameters) excluding uplift force and considering seismic force in positive direction.
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Eccentricity is given by, e =
∑ M * 528851.85 = = 6.8* 77736.671 ∑V *
For Seismic in another direction, e = -11.5**
For the Reservoir full state: The Vertical Normal stress at the heel of the dam is,
σ zheel =
∑ V 6e 1 − = 385.2162 KN/m2* B B
or
=1588.328 KN/m2**
σ ztoe =
The vertical Normal stress at the toe of the dam is
∑V B
6e 1 + = 1317.965 KN/m2* B
Or
= 60.87668 KN/m2**
Horizontal shear stresses are: ii) The factor of safety against sliding shear and overturning Taking c= 200t/m2 = 1962KN/m2 tanΦ = 0.7 c. Ah + ∑ V . tan φ Factor of safety against sliding, Fsf = = 44411.81379* ∑H OR
= 43004.90252
Factor of safety against overturning,
Fo =
∑ M + ve = 176674222.5/906313.55 =194.9372 ∑ M −ve
OR by considering opposite direction of seismic load
Fo =
∑ M + ve = 175909338/1671198.038 =105.26 ∑ M −ve
For the reservoir empty state: ΣV =74944.513*
or
ΣH = 15026.96*
or
ΣM = 174533638.1*
73441.547**
-15026.96**
or 173040547.9**
(Moment w.r.t toe of the dam)
ΣM = -599914.7802 * or -1995537.521** (Moment w.r.t centroid) Eccentricity, e’ = -8.00478589* or e’ = -27.1717795 ** The Vertical Normal stress at the heel of the dam is, OR
σ zheel =
∑V B
6e ' 1 − = 1654.493924* B = 3143.033265**
5
The vertical Normal stress at the toe of the dam is
σ ztoe =
∑ V 6e 1 + = 357.44* B B
OR
= -1171.44811**
From this we can see that maximum stress at the heel is 3143.033265 KN/m2 for full reservoir case for seismic load acting in negative direction. Similarly the maximum stress at the toe of the dam is 1317.965 KN/m2 for full reservoir case when the seismic load acts in positive direction. Horizontal shear stresses are: τ u = ( Pw − σ zu ) * tan φu = -1.15 KN/m2 Pw = 373.7143KN/m2 σzu = 385.2162 tan Φu = 0.1
τ d = σ zd * tan φ d = 446.8 KN/m2 σzd = 357.44 tan Φd = 75/60 = 1.25
Horizontal Normal Stresses
σ yu = Pw + (σ zu − Pw ) * tan 2 φu = 373.714 + (385.216 – 373.714) *(0.1)2 =373.83 KN/m2 σ yd = σ zd tan 2 φ d = 357.44 *(1.25)2 =558.5 KN/m2 ii) The factor of safety against sliding shear and overturning Taking c= 200t/m2 = 1962KN/m2 tanΦ = 0.7 c. Ah + ∑ V . tan φ Factor of safety against sliding, Fsf = = 52470.88622* ∑H OR
= 51399.35578**
Factor of safety against overturning,
Fo =
∑ M + ve = 174533638.1/0 = infinity. ∑ M −ve
OR by considering opposite direction of seismic load
Fo =
∑ M + ve = 173787093/746545.09 = 194.94 ∑ M −ve
c) Determine shear friction factor at section X-X when the X-X makes an angle α with horizontal anti clock wise.
6
ANSWER: For the dam base plane inclined with angle α with horizontal in anti clock wise direction, the shear friction factor is given by: ( S + Pp ) c. A Fsf = where Pp = + Ww tan(φ + α ) ∑H cos α (1 − tan φ tan α ) Ww is weight of the passive wedge at the base. d) Determine principal stresses at the heel and the toe.
σ1 =
σ z +σ y 2
2 2 + τ max = σ zu ((1 + tan φu ) − p w tan φu = 385.33 KN/m2
σ 1u = σ zu ((1 + tan 2 φu ) − p w tan 2 φu = 385.33 KN/m2 σ 1d = σ zd ((1 + tan 2 φ d ) − p w tan 2 φ d = Pw =373.714KN/m2 Pw = 373.7143KN/m2 σzu = 385.2162 tan Φu = 0.1
σ3 =
σ z +σ y
− τ max 2 = Pw = 373.714 KN / m 2 = 0 for no tail water downstream.
σ 3u σ 3d τ max
σ z −σ y = + τ 2 2
1
2
1
τ max u / s
σ zu − σ yu 2 = + τ u 2
τ max d / s
σ zd − σ yd 2 = + τ d 2
2
1
2
1654.5 − 373.83 = + (−1.15 2 ) 2
1
1317.97 − 558.5 = + (446.8) 2 2
2
1
2
= 25.33KN / m 2
= 447.22 KN / m 2
Question 2: a) Enumerate the various forces acting on the gravity dam: Describe in detail the force and moment of seepage and wind Answer:
7
General: In the design of concrete gravity dams, it is essential to determine the loads required in the stability and stress analysis. The following forces may affect the design: (1) Dead load (2) Headwater and tail water pressures. (3) Uplift. (4) Temperature. (5) Earth and silt pressures. (6) Ice pressure. (7) Earthquake forces. (8) Wind pressure. (9) Sub atmospheric pressure. (10) Wave pressure. (11) Reaction of foundation. Describe in detail the force and moment of seepage and wind on the dam stability
2 2 Prove σ 1 = σ sec φ − σ 3 tan φ τ = ( σ 1 − σ 3 ) tan φ
B
σ1 A
Solution: Let a triangle of sides AB = dr, the stress normal to it is σ 1 Side BC = ds, the stress normal to it σ 3 Side AC = dy, the stress normal to it is σ Resolving forces in vertical direction, ∑V = 0. σ .dy = σ 3 ds. sin φ + σ 1 .dr. cos φ
σ = σ 3 . sin 2 φ + σ 1 cos 2 φ σ1 = σ.
b / c sin φ =
σ3 C
σ
ds dr , cos φ = dy dy
1 sin 2 φ − σ . 3 cos 2 φ cos 2 φ
σ 1 = σ . sec 2 φ − σ 3 tan 2 φ To determine shear stress at the base, resolving all forces in the horizontal direction, ∑ H = 0. τ .dy = σ 3 . cos φ .ds − σ 1 sin φ .dr ds dr − σ 1 sin φ . dr dy Substituting the value of σ 1 from equation above:
τ = σ 3 . cos φ .
8
τ = σ 3 cos φ sin φ − (σ sec 2 φ − σ 3 tan φ ) sin φ cos φ = cos φ sin φ (σ 3 sec 2 φ − σ sec 2 φ )
τ = (σ 3 − σ ) tan φ b) short notes on : I) RCC dam II) Elementary profile III) Drainage galleries II) Elementary profile: Answer: Density of concrete, γ c = 24 KN / m3 Density of water, γ w = 10 KN/m3. Let the dam shape is vertical upstream or slight slope u/s and 1 to m downstream. Assume there is water upstream and may not be water downstream. Self-weight of the dam due to concrete material Assume the water level is up to crest of the dam for maximum flood level. The shape of the dam is triangular. The forces exerted on the dam are: Hydrostatic u/s = ½ γ w .h
2
h γ wh3 = 3 6 γ h2 1 Weight of the dam = γ c h.mh. = c 2 2 γ ch2 2 1 Moments about the toe= * (mh) = γ c .m 2 .h 2 3 2 3 If only these two forces are exerted, for stable dam profile Restoring moment due to weight of the dam should be greater or equal to the overturning moment about the toe of the dam. 1 γ w h3 2 2 Hence, >= γ c .m .h 3 6 SIMPLIFYING THIS: m>= 0.456 (d/s slope of the dam). m= b/h, Therefore, bottom side of the dam is b >= 0.456h. Moment about a toe: ½. γ w h 2 .
•
For Seepage or uplift pressure from bottom of the dam is considered in addition to the above two forces: γ w .mh 2 1 Fu = γ mh.h = 2 w 2
9
γ w .mh 2 2 γ w .m 2 h 3 Moment about the toe of the dam is given by * mh = 2 3 6 γ c .m 2 h 3 γ h3 γ w .m 2 h 3 >0 − w − 3 6 6 Rearranging the equations m>= 0.597 m= b/h b >= 0.597h • The base of the dam is assumed is given in the last equation greater than 0.597 h, to withstand the wave action and maximum flood level the height of the dam is increased. Free board of 1.5m to 2m is added. • The top width is varying from 6m to 10m for installation of d/nt machinery, for vehicle etc. • The d/s slope of the dam vary from 0.7 to 0.9 • If height of the dam is increased and it needs further increase of the base width, then u/s also provided with steep slope for stability of dam. (i.e 1:10).
h
mh
γ w .h
III) Drainage Galleries: Question 3: a) A remote steep sided and narrow valley in a given water course for Hydropower generation Catchments area 60 Km2, Flood 180-200 m3/s Thin superficial mantle, frequent intensive rock exposure, hydrological data minimal Access to valley is difficult from ground survey options are: Site A, Dam height = 50m nominal height and 450m length b/n steep valley and sides A deep deposit of fills overlying rock on the valley floor Site B, Located at 300m d/s of the site A with crest 320m b/n steep rocky abutments but requires a dam of 60m height. Valley floor has competent rock at the shallow depth i) Determine site and justify, the type of dam, likely to prove most appropriate on each site A and B. ii) Define & explain a programme of further investigations design to confirm the suitability of each site to the type of dam proposed under (i). Question 4:
10
a) Design a non overflow concrete dam by using the following data: The depth of water = 80m Free board = 3m Tail water = 3m Ultimate shear resistance of concrete = 56kg/cm2 Velocity of wind = 80 miles/hour Fetch 4 miles αh = 0.1 Depth of silt = 6m Assume top width = 6m Uplift area factor = 0.4 Design a dam by cantilever analysis considering a single block under reservoir empty and full condition. Answer:
11
Type of Formula Forces
Hydrostati c Fhu= Fvd=
Fhd=
Self weight
W1=
γw H2
2
γw H 2
Horizontal (KN)
Horizontal
Moment Moments arms w.r.t toe of toe of dam (KN.m) (m)
31392
31392
26.67
-837224.64
-837224.64
0
0.8
28.2528
28.2528
0
1
44.145
44.145
0
B+3
(11952B+35856) 908322
908322
2/3 B
600B2
3197400
3197400
2
-122.28
-122.28
-122.28
0.65456B
106.6B2 - 639.5B
-614754.9
0
32
2880B+38246.4
248486.4
248486.4
35.316
γ w *∀
-44.145
-44.145
w.r.t Moment Moment Dam w.r.t toe for w.r.t toe full for empty
2
γ c .h1.b1
W2= Silt force
Vertical (KN)
11952
γ c .H .b2 /900B 2
Fs=1/2KaGs'.hs
61.14
61.14
2
Fu= 162.846B+ ή.Tγw(Z1+Z2)/2 977.076 Uplift force Seismic Load
90B+1195.2 7765.2 Fses= +α
h
.Wtotal
Fewh= Wave pressure
0.66Ceα h Z1γ w ( Z1 .Z max )
Rw = 2.0 γw.hw2
1
2
3038.9214
3038.9214
6.806635802
32
-97245.48448
-97245.48448
0
80.220876
-546.0342849
546.034284 9
0
-860963.641 14192.5B 494B2
4354280.798 -1549893.339
4354208.4 -122.28
2804387.459
4354086.1
hw=0.032(V.F)1/ 2 +0.7630.271(F)3/4 Full Reservoir hw Empty Reservoir
12971.19864 1062.846B 0.589001981
sum
42213.1164 7826.34
11952 900B
ΣM
ΣV
ΣH
e
Xcentroid
2804417.459
90558.9566
42213.1164
5.532135934
30.96786407
when the reservoir is full
4354086.12
89539.758 7826.34 Empty Reservoir
-12.1274055
48.62740549
Empty
B Trial for B
73
Full Reservoir Vertical stress σ
Normal U/s D/s U/s D/s 676.47 1804.6 2449.19 3.96 All normal stresses are less than available shear strength of the material. i.e less than 28MN/m2
Factor of safety against overturning
2.81
35608.5
Friction Factor of safety against sliding(tan =0.75)
11.89
64.03
Question 5: a) What is the significance of survey work in the planning of a concrete dams and extent of survey work to be carried out? b) What are the various stages of investigation? c) Explain magnitude of investigation required for the construction of a concrete dam of height 100m and above. Question 6: a) Enumerates the type of geophysical investigation. What are the objectives of such investigations? Explain any one of them which most relevant according to you as an in charge of design of concrete dams. b) The distance to velocity change point is 200m, velocity of seismic wave near ground is 240m/s and below the ground at the expected rock level is 600m/s. calculate the depth of rock. Answer: Given: V1 =240m/s and V2= 600m/s Distance, x=200m Required: The depth of rock, d Solution: x v2 − V 1 (600 − 240) d= = 100 * = 65.465m of rock depth. 2 V 2 + V1 600 + 240 Question 7: a) How sand density test is carried out at the site to confirm the density of the foundation material. Answer: The sand follows:
density test at the site is carried out to confirm the foundation material. This is done as Four pit is dug on the horizontal surface having 20cm diameter and 30 cm depth; The material excavated is preserved and weighted. The excavated material dried and again weighted; The difference between initial weight and final weight gives the weight of the moisture. To know the volume standard stand 20 Kg is used which is completely dry. The pit is filled with the sand and removed thereafter to know the volume (cm3). weight of the moisture ( grams) Density = Volume
b) A permeability test was conducted at the bore hole of dia. 60 cm over a length of 110 cm. The injected pressure was 10 Kg/cm2. The rate of water injected was 45 m3/s. Determine coefficient of permeability. Answer: Given: Q= 45 m3/s, r= radius of bore hole = 0.3m R = radius of influence in feet (0.5L to 1.0L) = 1.10 m H= pressure head (m) = p/ρ= (10 *10000 Kg/m2)/ (100 kg/m3) = 100 m L = 1.10m Required: Permeability of the foundation material Solution; Ke =
Q. ln( R / r ) = 0.084638 m/s 2πLH
c) A rock at the site was encountered. How do you carry out the block shear test at the site in situ? Answer: In situ determination of shear strength parameters may also be necessary, using plate loading tests in trial pits or adits, or dilatometer or pressure meter testing conducted within boreholes. The latter techniques are particularly suitable in softer rocks containing very fine and closely spaced fissures. d)
What is the purpose served by drilling drifts, trenches and pits?
