PROBLEM 5.12
SOLUTION
Area mm2
, mmx
, mmy
3
, mmxA
3
, mmyA
3
1(200)(480) 32 10
6
11.52 10×
6
1.92 10
×
PROBLEM 5.13
The horizontal x–axis is drawn through the centroid C of the area shown,
and it divides the area into two component areas A1 and A2. Determine
the first moment of each component area with respect to the x–axis, and
explain the results obtained.
SOLUTION
Area above x–axis (Area A1):
1
33
(25)(20 80) (7.5)(15 20)
40 10 2.25 10
Q yA=Σ= ×+ ×
=×+ ×
33
1
42.3 10 mmQ= ×
Area below x–axis (Area A2):
2( 32.5)(65 20)Q yA=Σ=− ×
33
2
42.3 10 mmQ=−×
Dimensions in mm
PROBLEM 5.14
The horizontal x–axis is drawn through the centroid C of the area shown,
and it divides the area into two component areas A1 and A2. Determine the
first moment of each component area with respect to the x–axis, and
explain the results obtained.
PROBLEM 5.16
A composite beam is constructed by bolting
four plates to four 60 × 60 × 12-mm angles as
shown. The bolts are equally spaced along the
beam, and the beam supports a vertical load.
As proved in mechanics of materials, the
shearing forces exerted on the bolts at A and B
are proportional to the first moments with
respect to the centroidal x–axis of the red–
shaded areas shown, respectively, in parts a
and b of the figure. Knowing that the force
exerted on the bolt at A is 280 N, determine
the force exerted on the bolt at B.
SOLUTION
x
Therefore,
()
, or
() () ()
xB
AB BA
xA xB xA
Q
FF FF
QQ Q
= =
For the first moments,
3
3
12
( ) 225 (300 12)
2
831,600 mm
12
( ) ( ) 2 225 (48 12) 2(225 30)(12 60)
2
1,364,688 mm
xA
xB x
Q
Q QA
=+×
=
= + − ×+ − ×
=
Then
1,364,688 (280 N)
831,600
B
F=
or
459 N
B
F=
consent of McGraw–Hill Education.
PROBLEM 5.17
A thin, homogeneous wire is bent to form the perimeter of the figure indicated.
Locate the center of gravity of the wire figure thus formed.
SOLUTION
, mmL
, mmx
, mmy
2
,mmxL
2
,mmyL
PROBLEM 5.18
A thin, homogeneous wire is bent to form the perimeter of the figure indicated.
Locate the center of gravity of the wire figure thus formed.
SOLUTION
L, in.
, in.x
, in.y
2
, inxL
2
, inyL
1 4 0 2 0 8
2
22
3 6 6.7082+=
3 5.5 20.125 36.894
3 7 6 3.5 42 24.5
4 6 3 0 18 0
Σ
23.708 80.125 69.394
Then
(23.708) 80.125X L xL XΣ=Σ =
3.38 in.X=
(23.708) 69.394Y L yL YΣ=Σ =
2.93 in.Y=
consent of McGraw–Hill Education.
PROBLEM 5.19
A thin, homogeneous wire is bent to form the perimeter of the figure
indicated. Locate the center of gravity of the wire figure thus formed.
SOLUTION
Perimeter of Figure P5.4
L
x
y
2
, mmxL
2
, mmyL
I 30 0 15 0
3
0.45 10×
II 210 105 30
3
22.05 10×
3
6.3 10×
III 270 210 165
3
56.7 10×
3
44.55 10×
IV 30 225 300
3
6.75 10
×
3
9 10×
V 300 240 150
3
72 10×
3
45 10×
VII 240 120 0
3
28.8 10×
0
Σ
1080
3
186.3 10×
3
105.3 10×
32
(1080 mm) 186.3 10 mm
X L xL
X
Σ=Σ
= ×
172.5 mmX=
32
(1080 mm) 105.3 10 mm
Y L yL
Y
Σ=Σ
= ×
97.5 mmY=
Dimensions in mm
consent of McGraw–Hill Education.
PROBLEM 5.20
A thin, homogeneous wire is bent to form the perimeter of the figure
indicated. Locate the center of gravity of the wire figure thus formed.
SOLUTION
By symmetry,
.XY=
L, in.
, in.x
2
, inyL
1
1(10) 15.7080
2
π
=
2(10) 6.3662
π
=
100
2 5 0 0
3 5 2.5 12.5
4 5 5 25
5 5 7.5 37.5
Σ
35.708 175
Then
(35.708) 175X L xL XΣ=Σ =
4.90 in.XY= =
consent of McGraw–Hill Education.
PROBLEM 5.12
SOLUTION
Area mm2
, mmx
, mmy
3
, mmxA
3
, mmyA
3
1(200)(480) 32 10
6
11.52 10×
6
1.92 10
×
PROBLEM 5.13
The horizontal x–axis is drawn through the centroid C of the area shown,
and it divides the area into two component areas A1 and A2. Determine
the first moment of each component area with respect to the x–axis, and
explain the results obtained.
SOLUTION
Area above x–axis (Area A1):
1
33
(25)(20 80) (7.5)(15 20)
40 10 2.25 10
Q yA=Σ= ×+ ×
=×+ ×
33
1
42.3 10 mmQ= ×
Area below x–axis (Area A2):
2( 32.5)(65 20)Q yA=Σ=− ×
33
2
42.3 10 mmQ=−×
Dimensions in mm
PROBLEM 5.14
The horizontal x–axis is drawn through the centroid C of the area shown,
and it divides the area into two component areas A1 and A2. Determine the
first moment of each component area with respect to the x–axis, and
explain the results obtained.
PROBLEM 5.16
A composite beam is constructed by bolting
four plates to four 60 × 60 × 12-mm angles as
shown. The bolts are equally spaced along the
beam, and the beam supports a vertical load.
As proved in mechanics of materials, the
shearing forces exerted on the bolts at A and B
are proportional to the first moments with
respect to the centroidal x–axis of the red–
shaded areas shown, respectively, in parts a
and b of the figure. Knowing that the force
exerted on the bolt at A is 280 N, determine
the force exerted on the bolt at B.
SOLUTION
x
Therefore,
()
, or
() () ()
xB
AB BA
xA xB xA
Q
FF FF
QQ Q
= =
For the first moments,
3
3
12
( ) 225 (300 12)
2
831,600 mm
12
( ) ( ) 2 225 (48 12) 2(225 30)(12 60)
2
1,364,688 mm
xA
xB x
Q
Q QA
=+×
=
= + − ×+ − ×
=
Then
1,364,688 (280 N)
831,600
B
F=
or
459 N
B
F=
consent of McGraw–Hill Education.
PROBLEM 5.17
A thin, homogeneous wire is bent to form the perimeter of the figure indicated.
Locate the center of gravity of the wire figure thus formed.
SOLUTION
, mmL
, mmx
, mmy
2
,mmxL
2
,mmyL
PROBLEM 5.18
A thin, homogeneous wire is bent to form the perimeter of the figure indicated.
Locate the center of gravity of the wire figure thus formed.
SOLUTION
L, in.
, in.x
, in.y
2
, inxL
2
, inyL
1 4 0 2 0 8
2
22
3 6 6.7082+=
3 5.5 20.125 36.894
3 7 6 3.5 42 24.5
4 6 3 0 18 0
Σ
23.708 80.125 69.394
Then
(23.708) 80.125X L xL XΣ=Σ =
3.38 in.X=
(23.708) 69.394Y L yL YΣ=Σ =
2.93 in.Y=
consent of McGraw–Hill Education.
PROBLEM 5.19
A thin, homogeneous wire is bent to form the perimeter of the figure
indicated. Locate the center of gravity of the wire figure thus formed.
SOLUTION
Perimeter of Figure P5.4
L
x
y
2
, mmxL
2
, mmyL
I 30 0 15 0
3
0.45 10×
II 210 105 30
3
22.05 10×
3
6.3 10×
III 270 210 165
3
56.7 10×
3
44.55 10×
IV 30 225 300
3
6.75 10
×
3
9 10×
V 300 240 150
3
72 10×
3
45 10×
VII 240 120 0
3
28.8 10×
0
Σ
1080
3
186.3 10×
3
105.3 10×
32
(1080 mm) 186.3 10 mm
X L xL
X
Σ=Σ
= ×
172.5 mmX=
32
(1080 mm) 105.3 10 mm
Y L yL
Y
Σ=Σ
= ×
97.5 mmY=
Dimensions in mm
consent of McGraw–Hill Education.
PROBLEM 5.20
A thin, homogeneous wire is bent to form the perimeter of the figure
indicated. Locate the center of gravity of the wire figure thus formed.
SOLUTION
By symmetry,
.XY=
L, in.
, in.x
2
, inyL
1
1(10) 15.7080
2
π
=
2(10) 6.3662
π
=
100
2 5 0 0
3 5 2.5 12.5
4 5 5 25
5 5 7.5 37.5
Σ
35.708 175
Then
(35.708) 175X L xL XΣ=Σ =
4.90 in.XY= =
consent of McGraw–Hill Education.