Download Fluids Assignment 5 – Gravity Dams and more Quizzes Fluid Mechanics in PDF only on Docsity!
PROBLEM 1
THE SECTION OF THE MASONRY DAM IS AS SHOWN.THE SPECIFIC WEIGHT OF WATER IS 9.
kN/m³ AND THAT OF CONCRETE IS 23.54 kN/m³. ASSUMING UPLIFT PRESSURE VARIES
LINEARLY FROM MAXIMUM HYDROSTATIC PRESSURE AT THE HEEL TO ZERO AT THE LOCATION
OF THE DRAIN, DETERMINE THE
(A) LOCATION OF THE RESULTANT FORCE,
(B) FACTOR OF SAFETY AGAINST SLIDING IF COEFFICIENT OF FRICTION IS 0.75 ,
(C) FACTOR OF SAFETY AGAINST OVERTURNING,
(D) THE PRESSURE AT THE HEEL AND AT THE TOE.
SOLUTION :
VERTICAL FORCES
WEIGHT OF THE CONCRETE DAM
W
1
= γ
c
V
1
- 54 kN/m
3
1
2
∙ 5. 2 m ∙ 52 m ∙ 1 m) = 3182. 608 kN
W
2
= γ
c
V
2
= ( 23. 54 kN/m
3
)( 7 m ∙ 52 m ∙ 1 m) = 8568. 560 kN
W
3
= γ
c
V
3
- 54 kN/m
3
1
2
∙ 26 m ∙ 52 m ∙ 1 m) = 15913. 040 kN
WEIGHT OF THE WATER IN THE UPSTREAM SIDE
W
4
= γ
W
V
4
= ( 9. 81 kN/m
3
1
2
∙ 5 m ∙ 50 m ∙ 1 m) = 1226. 250 kN
HYDROSTATIC UPLIFT
U = V = (
1
2
∙ 9. 81 kN/m
3
∙ 50 m ∙ 23. 2 m ∙ 1 m) = 5689. 800 kN
HORIZONTAL FORCES
TOTAL HYDROSTATIC FORCE
F = γh
A =
- 81 kN/m
3
25 m
( 50 m ∙ 1 ) = 12262. 500 kN
REACTION
R
y
= W
1
+ W
2
+ W
3
+ W
4
− U
R
y
= 3182. 608 kN + 8568. 560 kN + 15913. 040 kN + 1226. 250 kN − 5689. 800 kN
R
y
= 23200. 658 kN
R
x
= F = 12262. 500 kN
MOMENT ABOUT THE TOE
RIGHTING MOMENT
RM = W
1
(x
1
) + W
2
(x
2
) + W
3
(x
3
) + W
4
(x
4
RM = 3182. 608 kN ( 26 m + 7 m +
- 2
3
m) + 8568. 560 kN
26 m + 3. 5 m
26 ∙ 2
3
m) + 1226. 250 kN ( 26 m + 7 m + 5. 2 m −
5
3
m)
RM = 683940. 131 kN − m
OVERTURNING MOMENT
OM = F(y) + U(z)
OM = 12262. 500 kN (
50
3
m) + 5689. 800 kN ( 26 m + 7 m + 5. 2 m −
- 2
3
m)
OM = 377724. 240 kN − m
LOCATION OF THE RESULTANT FORCE
x̅ =
RM−OM
R
y
x̅ =
131 kN−m− 377724. 240 kN−m
658 kN
x̅ = 13. 199 m
FACTOR OF SAFETY
FOR SLIDING
FS
S
μR y
R
x
FS
S
( 0. 75 )( 23200. 658 kN)
- 500 kN
FS
S
= 1. 419 > 1 ; SAFE
FOR OVERTURNING
FS
O
RM
OM
FS
O
131 kN−m
240 kN−m
FS
O
= 1. 811 > 1 ; SAFE
PROBLEM 2
A DAM IS TRIANGULAR IN CROSS SECTION WITH THE UPSTREAM VERTICAL. WATER IS FLUSH
AT THE TOP. THE DAM IS 8 m HIGH AND 6 m WIDE AT THE BOTTOM, AND WEIGHS 23.5 kN/m³.
THE COEFFICIENT OF FRICTION BETWEEN THE BASE AND THE FOUNDATION IS 0..
HYDROSTATIC UPLIFT VARIES FROM FULL AT THE HEEL TO ZERO AT THE TOE.
(A) DETERMINE THE FACTOR OF SAFETY AGAINST OVERTURNING
(B) DETERMINE THE FACTOR OF SAFETY AGAINST SLIDING
(C) DETERMINE THE MAXIMUM PRESSURE AT THE BASE OF THE DAM
SOLUTION :
VERTICAL FORCES
WEIGHT OF THE CONCRETE DAM
W
1
= γ
c
V
1
- 5 kN/m
3
1
2
∙ 8 m ∙ 6 m ∙ 1 m) = 564. 000 kN
HYDROSTATIC UPLIFT
U = V = (
1
2
∙ 9. 81 kN/m
3
∙ 8 m ∙ 6 m ∙ 1 m) = 235. 440 kN
HORIZONTAL FORCES
TOTAL HYDROSTATIC FORCE
F = γh
A = ( 9. 81 kN/m
3
)( 4 m)( 8 m ∙ 1 ) = 313. 920 kN
REACTION
R
y
= W
1
− U
R
y
= 564. 000 kN − 235. 440 kN
R
y
= 328. 560 kN
R
x
= F = 313. 920 kN
MOMENT ABOUT THE TOE
RIGHTING MOMENT
RM = W
1
(x
1
RM = 564. 000 kN( 4 m)
RM = 2256 kN − m
OVERTURNING MOMENT
OM = F(y) + U(z)
OM = 313. 920 kN (
8
3
𝑚) + 235. 440 kN( 4 m)
OM = 1178. 880 kN − m
LOCATION OF THE RESULTANT FORCE
x̅ =
RM−OM
R y
x̅ =
2256 kN−m− 1178. 880 kN−m
- 560 kN
x̅ = 1. 452 m
FACTOR OF SAFETY
FOR SLIDING
FS
S
μR y
R
x
FS
S
( 0. 80 )( 328. 560 kN)
- 920 kN
FS
S
= 0. 837 < 1 ; NOT SAFE
FOR OVERTURNING
FS
O
RM
OM
FS
O
2256 kN−m
- 880 kN−m
FS
O
= 1. 268 > 1 ; SAFE
e =
6
2
− x̅
e = 1. 548
B
6
THUS e > B/
q
T
2R y
3 x̅
2 ( 328. 560 kN)
3 ( 1. 452 m)
q
T
= − 150. 838 kPa
