EXAMPLE
EXAMPLE # 1:
A plastic cube of side L and sp. gravity of 0.82 is placed vertically in water. Is the cube stable?
Solution:
Note: The body is stable if M is above G.
For draft:
sp . gr . of cube
0.82
L=
L
sp . gr . of water
1
¿ 0.82 L
D=
1
1
L B3
( L )( L )3
I
12
12
M Bo = =
=
=0.102 L
VD
VD
( L ) ( L )( 0.82 L )
L D L 0.82 L
G Bo = − = −
=0.09 L
2 2 2
2
Since M Bo >G B o
The body is STABLE
, M is above G
EXAMPLE # 2:
A solid wood cylinder of specific gravity 0.6 is 600 mm in diameter and 1200 mm high. If placed vertically in oil (sp. gr. =
0.85), would it be stable?
Solution:
For Draft:
D=
sp . gr . of wood
0.60
H=
( 1200 )=847.06 mm
sp . gr . oil
0.85
Distance from M and Bo :
π 2
r
I
4
M Bo = =
=
VD VD
Distance from G and Bo:
G Bo =600 mm−
π
(300 mm )4
4
=26.56 mm
300 mm 2
(π)
( 847.06 mm )
1
(
)
D
847.06 mm
=600 mm−
2
2
¿ 176.5 mm

Since M Bo <G B o , The Metacenter is below G
Therefore, the body is UNSTABLE.
EXAMPLE # 3:
A rectangular scow 9 m wide, 15 m long, and 3.6 m high has a draft in sea water of 24 m. Its center of gravity is 2.7 m
above the bottom of the scow. Determine the following:
a) The initial metacentric height
b) The righting or overturning moment when the scow tilts until one side is just at the point of submergence.
Solution:
a) Initial Position
Distance of Metacenter to the center of buoyancy:
[
]
B2
tan 2 θ
1+
whereθ=0 °
12 D
2
2
2
9m
tan 0 °
¿
1+
2
12 ( 2.4 m )
45
¿ ∨2.81 m
16
M Bo =
[
]
Distance from G to the center of buoyancy:
G Bo =2.7 m−1.2 m=1.5 m
Initial Metacentric Height:
M G=M Bo +G B o
¿ 2.81 m−1.5 m
¿ 1.31m
b)
For Tilting Position:
Angle of tilting:
θ=arctan
1.2m
=14.93 °
4.5 m
Distance of Metacenter to the center of buoyancy:
[
B2
tan 2 θ
M Bo =
1+
12 D
2
[
1.2 m
2
4.5m
9m
¿
1+
2
12 ( 2.4 m )
¿ 2.91 m
(
]
2
)
]
Metacentric Height:
MG=M B o +G Bo
¿ 2.91 m−1.5 m
¿ 1.41m
 since
MG> M B o , the moment is righting moment
Weight of an object:
W =BF=γ V D
KN
¿ ( 1.03 ) 9.81 3 ( 9 m ) (15 m ) ( 2.4 m )
m
¿ 3273.79 kN
(
)
Righting Moment:
RM =W ( MG sinθ )
¿ 3273.79 kN ( 1.41m ) sin14.93 °
EXAMPLE # 4:
A barge floating in fresh water has the form a parallelpiped having a dimension of 10m x 30m by 3 m. It weights 4500
kN when loaded with center of gravity along its vertical axis 4 m from the bottom. Find the metacentric height about its longest
and shortest centerline, and determine wether or not the barge is stable.
Solution:
To solve for draft D:
[ BF=W ]
γ V D =W
KN
9.81 3 [ ( 10 m )( 30 m ) D ] =4500 kN
m
D=1.53 m
(
)
Distance from G to the center of buoyancy:
G Bo =4 m−
D
1.53 m
=4 m−
2
2
¿ 3.24 m
Along longitudinal axis (rolling)
B = 10 m
2
[
Along traverse direction (pitching):
B = 30 m
2
]
2
B
tan θ
M Bo =
1+
where θ=0 °
12 D
2
( 10 m)2
¿
( 1+ 0 )
12 ( 1.53 m )
¿ 5.45 m
[
2
]
B
tan θ
1+
where θ=0 °
12 D
2
( 30 m )2
¿
( 1+ 0 )
12 ( 1.53 m )
¿ 49.02 m
M Bo =
MG=5.45 m−3.24 m
(the barge is stable in rolling)
¿ 2.21 m
MG=49.02 m−3.24 m
( the barge is stable in
¿ 45.78 m
pitching)
PROBLEM # 5:
A crane barge, 20 m long, 8 m wide, and 2 meters high loaded at its center with a road roller weighing 20 short tons,
floats on fresh water with a draft of 1.20 meters and has its center of gravity located along its vertical axis at a point 1.50 m
meters above its bottom. compute the horizontal distance out to one side from the centerline of the barge through which the
crane would swing the 20-ton load which it had lifted from the center of the deck, and tip the barge with the 20-meter edge just
touching the water surface?
Solution:
If the Load is along the Center of Gravity:
Weight of the road roller:
W R=20 short tons
W R=20 short tons
lb
0.4536 kg
m
9.81
( 1 2000
short ton )( 1 lb ) (
s )
¿ 177992.64 N∨177.993 kN
Buoyant Force:
BF=γ V D
kN
¿ 9.81 3 ( 1.2 m )( 8 m ) ( 20 m )
m
¿ 1883.52kN
(
)
Total weight of the barge:
W T =BF
W T =1883.52 kN
Weight of the Barge:
W B=BF−W R =1883.52 kN−177.993 kN
¿ 1705.53 kN
For Tilted Position: (Load is at distance L from the Centerline)
New Position of G:
tan θ=
0.80m
4.0 m
θ=11.31 °
W T ( 0.5 m )=W B ( d )
( 1883.52 kN ) ( 0.5 m) =( 1705.53 kN ) d
2
d=0.55 m
z=d sin θ=( 0.55 m ) sin ( 11.31° )
z=0.11 m
Distance from G to the center of buoyancy:
G Bo =1.4−d=1.4 m−0.55 m
¿ 0.85 m
Distance of Metacenter to the center of buoyancy:
( 8 m )2
B2
tan 2 θ
tan 2 11.31 °
M Bo =
1+
=
1+
12 D
2
2
12 (1.2 m )
¿ 4.53 m
[
]
[
]
Metacentric Height:
MG=M B o−G Bo=4.53 m−0.85 m
¿ 3.68 m
Horizontal Distance of Bo from the new center of gravity:
x=MG sin θ=( 3.68 m ) sin 11.31°
¿ 0.72 m
Horizontal distance of road roller from the center of the deck:
[ ∑ M G=0 ]
( BF ) x =W R ( L+ z )
( 1883.52 kN ) ( 0.72m )=( 177.99 kN )( L+0.11 m )
L=7.51 m
PROBLEM # 6:
A wooden barge of rectangular cross section is 8 m wide, 4 m high, and 16 m long. It is transporting in seawater (s =
1.03) a total load of 1,500 kN including its own weight and cargo. If a weight of 75 kN (included in the 1,500-kN) is shifted a
distance of 2.5 m to one side, it will cause the barge to go down 450 mm in the wedge of immersion and also rise 450 mm in the
corresponding wedge of immersion. The barge floats vertically (on an even kneel) before the shifting of the weight. Compute how
far above the waterline is the center of gravity of the loaded barge.
Solution:
To solve for the draft, D:
BF=W
(
( 1.03 ) 9.81
kN
[ ( 8 m )( 16 m ) ( D ) ] =1500 kN
m3
)
D=1.16 m
For the Tilted Position:
Angle of tilting:
0.45m
4m
0.45m
θ=arctan
4m
θ=6.42°
tan θ=
Distance of Metacenter to the center of buoyancy:
2
[
2
]
2
[
2
(8 m)
B
t an θ
tan 6.42 °
M Bo =
1+
=
1+
12 D
2
2
12 (1.16 m )
¿ 4.63 m
Inclined Distance of Bo from the Metacentric Height:
]
c=M B o sin θ=( 4.63 m ) sin 6.42°
¿ 0.52 m
Inclined Distance from Bo to Cargo:
a=( 3.42m ) sin 6.42° + ( 2.5 m ) cos 6.42°
¿ 2.87 m
Inclined Distance from Bo to the center of Gravity, G:
b=( h+0.58 m ) sin 6.42 °
Height of the center of gravity from the water surface, h:
[∑ M
=0 ]
1425 ( b ) +75 ( a )=BF ( c )
( 1425 kN ) [ ( h+0.58 m ) sin 6.42 ° ] +75 ( 2.87 m) =( 1500 kN ) ( 0.52m )
h=2.96 m
Bo
PROBLEM # 7:
The water line section of a 1500 kN barge is as shown. Its center of gravity is 1.5 m above the center of buoyancy.
Compute the initial metacentric height against rolling.
Solution:
Total moment of inertia of an object, Itotal:
T TOTAL =I rectangle + I triangle + I semicircle
1
1
π 4
3
3
¿ ( b ) ( h ) + ( b )( h ) ( 2 ) + r
12
12
8
1
1
π
3
¿ (12 m )( 8 m) + ( 6 m )( 4 m )3 ( 2 ) + ( 4 m) 4
12
12
8
4
¿ 676.53 m
Volume of displacement, VD :
[ BF=W ]
(9.81 kNm ) V =1500 kN
3
D
V D =152.91 m3
Distance of Metacenter to the center of buoyancy:
M Bo =
I
676.53 m4
=
=4.42 m
V D 152.91 m3
Initial Metacentric Height:
MG=M B o−G Bo=4.42 m−1.5 m
MG=2.92m
BOARDWORK
PROBLEM # 1:
A cylindrical buoy weighing 20 KN is to float in sea water above density is 1025 kg/m3. The
buoy has a diameter of 2 m and 2.5 m high. Is the buoy stable or unstable?
PROBLEM # 2:
a)
If the wood floats in the seawater (s = 1.03) with the
smaller base at the bottom, would it be stable?
b)
If the wood floats in the seawater (s = 1.03) with the
larger base at the bottom, would it be stable?
PROBLEM # 3:
The center of gravity is located at a point 3.5 m from the bottom.
Determine the following:
a) Initial metacentric height
b) Metacentric height when the scow tilts until one side is
just at a point of submergence
c) Righting moment
d) the maximum distance from the center in which the cargo
should be shifted until one side ia at a point of submergence.
PROBLEM # 4:
Would a wooden cylinder (s = 0.61), 660 mm in diameter and and 1.3 m long be stable if placed vertically in oil (s = 0.85)?
PROBLEM # 5:
Determine:
a) metacentric height
b) is it stable or not?