HGE Part 2 Seta Solutions
HGE Part 2 Seta Solutions
HGE Part 2 Seta Solutions
Solution:
Given
R = 18m L = 40m
kN
w = 9.79
Given 3
m
F = w
D = 1.2m W = 900N Vblock = 0.5m
3 R
R L = 63439.200 kN answer
kN
#27 2
c = 24 W block = c Vblock = 12 kN
m
3
#28 ( 2 )
W = 0.25 R L w = 9.965 10 kN
4
kN
SGsea = 1.03 water = 9.81 2 2
3 RF = F + W = 118129.874 kN answer
m
Consider the diagram of the block to get the tension,T R−a R
in the cable #29 Guess: a = 2m
W RF
( 2 )
BF2( h ) = SGsea water 0.25 D ( 0.3m + h )
A. 31
Solution:
B. 43 C. 62 D. 86
Given
W+T BF2( h ) Guess: h = 1m kN
R = 100mm water = 9.81
Find ( h ) = 0.387 m answer 3
m
Situation 1 - The canal shown in cross section in Fig. HYD 16.27 runs 40 h = R = 0.1 m
m into the paper. 2 2
27. Determine the horizontal hydrostatic force (kN). Use unit weight = A = R = 0.031 m
9.79 kN/m3.
A. 61127 B. 58786 C. 65112 D. 63439
28. Determine the magnitude of the hydrostatic force (kN).
F = water h A = 30.819 N answer
A. 175002 B. 118130 C. 154207 D. 131284
29. Find the vertical location of the center of pressure from A.
A. 8.33 m B. 15.18 m C. 2.82 m D. 9.67 m
1
Prepared by: ENGR. JOBERT S. DE LA CRUZ, CE
University of Cebu – CE Review HGE PART 2 Set A
MIDTERM EXAM SOLUTIONS
( )
Find Vcrown = 9.174 10
−5 3
m
W air kN
crown = = 128.624
Vcrown 3
m
crown
SGcrown = = 13.112 answer
water
2
Prepared by: ENGR. JOBERT S. DE LA CRUZ, CE
University of Cebu – CE Review HGE PART 2 Set A
MIDTERM EXAM SOLUTIONS
38. What is the depth of water in the tank when brought to rest after Solution:
spilling 2.6 m3 of water? Given
A. 2.685 m B. 2.524 m C. 2.793 m D. 1.586 m kN
Solution: a = 200mm h = 400mm water = 9.81
2 3 3
Given Vblock = a h = 0.016 m m
d = 4m H = 4m
W water = 50N BF = water Vblock = 156.96 N
h = 3m D = H − h = 1 m r = 0.5d = 2 m
W water W air − BF
#36 For no spillage, y 1 = 2D = 2 m
#39
2 2
r rad W air = W water + BF = 206.96 N
y1 Guess: = 3.132 answer
2g s
#40 W air
rad SGblock = = 1.319 answer
Find ( ) = 3.132 water Vblock
s
m #41
v = r = 6.264 answer
s W air kN
block = = 12.935 answer
Vblock 3
2 3 3 m
#37 Vairi = 0.25 d D = 12.566 m Vspill = 2.6m
2 2
2 r
y2 y 2 2g
2g w2 = = 32.852 rpm answer
2
r
2 3
#38 Vwateri = 0.25 d h = 37.699 m
( ) 2
Vwaterf h f = 0.25 d h f
2
0.25 d h Vwateri − Vspill
Vwateri − Vspill
h = = 2.793 m answer
2
0.25 d
3
Prepared by: ENGR. JOBERT S. DE LA CRUZ, CE
University of Cebu – CE Review HGE PART 2 Set A
MIDTERM EXAM SOLUTIONS
Given Given
kN
s = 0.6m w = 9.81 = 30deg kN
3 SGo = 0.68 w = 9.81
m 3
m
h = 1m + 0.5s cos ( ) = 1.26 m
#42
2 2
( )
W o Vo = SGo w Vo
A = s = 0.36 m
BF( Vs ) = w Vs
FH = w h A = 4449.137 N answer
#45 Wo BFo
4
s 4
#43 I = = 0.011 m
12 Vs
= SGo = 0.68 answer
h Vo
y = = 1.455 m
cos ( )
3
I #46 Ve = 0.36m
e = = 0.021 m
Ay
Vs Vo − Ve
location of the force from the bottom of the gate,z
Vo − Ve SGo Vo
z = 0.5s − e = 0.279 m answer
Ve 3
Vo = = 1.125 m answer
#44 Sum up moment about hinge: 1 − SGo
s F − FH ( 0.5s + e) 0 Guess: F = 100kN #47
W o = SGo w Vo = 7.505 kN answer
Find ( F) = 2377.491 N answer
Situation 7 - A cylindrical buoy 0.60 m in diameter and 1.8 m high weighs
Situation 6 - An object having a specific gravity of 0.68 is floating in water. 205 kg. It is moored to a salt water to a 12 m length of chain weighing
12 kg/m of length. At high tide, the height of the buoy protruding
45. What is the ratio of its volume submerged to its total volume? above water is 0.84 m. Density of steel and salt water are 7790 kg/m3
A. 0.32 B. 0.68 C. 0.58 D. 0.72 and 1030 kg/m3.
46. If its volume exposed is 0.36 m3, what is the total volume of the object 48. What is the depth of water during high tide?
in m3? A. 11 m B. 8 m C. 9 m D. 10 m
A. 1.075 B. 0.875 C. 1.125 D. 1.225 49. What could be the height of protrusion of the buoy if the tide dropped
47. What is the weight of the object in kN? 2.1 m?
A. 7.505 B. 7.625 C. 7.425 D. 7.875 A. 0.91 m B. 1.03 m C. 1.25 m D. 0.63 m
Solution: 50. How far is the bottom of the buoy from the ground at low tide?
A. 4.78 m B. 7.88 m C. 6.11 m D. 5.13 m
Solution:
4
Prepared by: ENGR. JOBERT S. DE LA CRUZ, CE
University of Cebu – CE Review HGE PART 2 Set A
MIDTERM EXAM SOLUTIONS
Given
D = 0.6m H = 1.8m mbuoy = 205kg
kg
Lchain = 12m wchain = 12
m
kg kg
st = 7790 w = 1030
3 3
m m
wchain −3 2
A st = = 1.54 10 m
st
BFst ( y ) = w A st y
2
BFbuoy = w 0.25 D ( H − h ) = 279.577 kg
Find ( y ) = 7.16 m
therefore the depth of water at hightide
d ht = y + ( H − h ) = 8.12 m use d = 8 m answer #48
(
mbuoy + wchain d lt − x ) BF'st ( x) + BF'buoy ( x) solve → 0.887 m
5
Prepared by: ENGR. JOBERT S. DE LA CRUZ, CE
University of Cebu – CE Review HGE PART 2 Set A
MIDTERM EXAM SOLUTIONS
6
Prepared by: ENGR. JOBERT S. DE LA CRUZ, CE