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SOLUTION Continued
Also
12
() ()
yy y
II I= −
where
32 2
1
4 64
1
( ) (60 mm)(108 mm) (6480 mm )[(56.54) mm]
12
(6,298,560 25,920) mm 6.324 10 mm
y
I= +
=+=×
32 2
2
4 64
1
( ) (36 mm)(72 mm) (1296 mm )[(56 46) mm]
36
(373,248 129,600) mm 0.502 10 mm
y
I= +−
=+=×
Then
64
(6.324 0.502)10 mm
y
I= −
or
64
5.82 10 mm
y
I= ×
consent of McGraw-Hill Education.
PROBLEM 7.56
Determine the moments of inertia
x
I
and
y
I
of the area shown with
respect to centroidal axes respectively parallel and perpendicular to
side AB.
SOLUTION
First locate C of the area:
Symmetry implies
12 mm.X=
2
, mmA
, mmy
3
, mmyA
SOLUTION Continued
Also
12
() ()
yy y
II I
= +
where
34
1
32
2
4
1
( ) (22 mm)(12 mm) 3168 mm
12
11
( ) 2 (18 mm)(12 mm) 18 mm 12 mm (4 mm)
36 2
5184 mm
y
y
I
I
= =
= +× ×
=
2
[( )
y
I
is obtained by dividing
2
A
into ]
Then
4
(3168 5184)mm
y
I= +
or
34
8.35 10 mm
y
I= ×
consent of McGraw-Hill Education.
PROBLEM 7.57
The shaded area is equal to
2
50 in .
Determine its centroidal moments
of inertia
x
I
and
,
y
I
knowing that
2
yx
II=
and that the polar moment
of inertia of the area about Point A is
4
2250 in .
A
J=
SOLUTION
Given:
2
50 inA=
4
2 , 2250 in
y xA
I IJ= =
2
(6 in.)
AC
JJA= +
4 22
2250 in (50 in )(6 in.)
C
J= +
4
450 in
C
J=
with 2
Cxy y x
J II I I=+=
4
450 in 2
xx
II= +
4
150.0 in
x
I=
4
2 300 in
yx
II= =
consent of McGraw-Hill Education.
PROBLEM 7.58
Determine the polar moment of inertia of the area shown with respect
to (a) Point O, (b) the centroid of the area.
SOLUTION
First locate centroid C of the figure.
SOLUTION Continued
4
2
64
22
3
64
4
4
64
( ) (42 mm)
4
2.44392 10 mm
( ) (54 mm)(27 mm)[(54 mm) (27 mm) ]
8
2.08696 10 mm
( ) (27 mm)
4
0.41739 10 mm
O
O
O
J
J
J
π
π
π
=
= ×
= +
= ×
=
= ×
Then
64
64
(12.21960 2.44392 2.08696 0.41739) 10 mm
12.15917 10 mm
O
J= +−− ×
= ×
or
64
12.16 10 mm
O
J= ×
(b)
2
OC
J J AX= +
or
64 2 2
12.15917 10 mm (4877.32 mm )( 22.3094 mm)
C
J= ×− −
or
6
9.73 10 mm
C
J
4
= ×
PROBLEM 7.59
Determine the polar moment of inertia of the area shown with respect to
(a) Point O, (b) the centroid of the area.
SOLUTION
Determination of centroid C of entire section:
Section Area, in2
, in.x
3
, inxA
2
ππ
16
PROBLEM 7.60
Two L6 × 4 ×
1
2
-in. angles are welded together to form the section
shown. Determine the moments of inertia and the radii of gyration
of the combined section with respect to the centroidal x and y axes.
SOLUTION
W section:
24 4
9.12 in 110 in 37.1in
xy
AI I= = =
Angle:
244
1.44 in 1.23 in 1.23in
xy
AI I= = =
total W A
2
4
9.12 4(1.44) 14.880 in
AAA
= +
=+=
Now
WA
() 4()
xx x
II I= +
where
A
2
A
42 2 4
()
1.23 in (1.44 in )(4.00 in. 0.836 in.) 34.907 in
xx
I I Ad= +
=+ +=
Then
44
(110 4 34.907) in 249.63 in
x
I= +× =
4
250 in
x
I=
and
4
22
total
249.63 in
14.880 in
x
xI
kA
= =
4.10 in.
x
k=
Also
WA
() 4()
yy y
II I= +
where
( )
( )
A
2
A
244
(1.23 1.44 5 0.836 ) in 26.198 in
yy
I I Ad= +
=+− =
44
(37.1 4 26.198) in 141.892 in
y
I= +× =
4
141.9 in
y
I=
4
22
total
141.892 in
14.880 in
y
y
I
kA
= =
3.09 in.
y
k=
consent of McGraw-Hill Education.
SOLUTION Continued
Also
12
() ()
yy y
II I= −
where
32 2
1
4 64
1
( ) (60 mm)(108 mm) (6480 mm )[(56.54) mm]
12
(6,298,560 25,920) mm 6.324 10 mm
y
I= +
=+=×
32 2
2
4 64
1
( ) (36 mm)(72 mm) (1296 mm )[(56 46) mm]
36
(373,248 129,600) mm 0.502 10 mm
y
I= +−
=+=×
Then
64
(6.324 0.502)10 mm
y
I= −
or
64
5.82 10 mm
y
I= ×
consent of McGraw-Hill Education.
PROBLEM 7.56
Determine the moments of inertia
x
I
and
y
I
of the area shown with
respect to centroidal axes respectively parallel and perpendicular to
side AB.
SOLUTION
First locate C of the area:
Symmetry implies
12 mm.X=
2
, mmA
, mmy
3
, mmyA
SOLUTION Continued
Also
12
() ()
yy y
II I
= +
where
34
1
32
2
4
1
( ) (22 mm)(12 mm) 3168 mm
12
11
( ) 2 (18 mm)(12 mm) 18 mm 12 mm (4 mm)
36 2
5184 mm
y
y
I
I
= =
= +× ×
=
2
[( )
y
I
is obtained by dividing
2
A
into ]
Then
4
(3168 5184)mm
y
I= +
or
34
8.35 10 mm
y
I= ×
consent of McGraw-Hill Education.
PROBLEM 7.57
The shaded area is equal to
2
50 in .
Determine its centroidal moments
of inertia
x
I
and
,
y
I
knowing that
2
yx
II=
and that the polar moment
of inertia of the area about Point A is
4
2250 in .
A
J=
SOLUTION
Given:
2
50 inA=
4
2 , 2250 in
y xA
I IJ= =
2
(6 in.)
AC
JJA= +
4 22
2250 in (50 in )(6 in.)
C
J= +
4
450 in
C
J=
with 2
Cxy y x
J II I I=+=
4
450 in 2
xx
II= +
4
150.0 in
x
I=
4
2 300 in
yx
II= =
consent of McGraw-Hill Education.
PROBLEM 7.58
Determine the polar moment of inertia of the area shown with respect
to (a) Point O, (b) the centroid of the area.
SOLUTION
First locate centroid C of the figure.
SOLUTION Continued
4
2
64
22
3
64
4
4
64
( ) (42 mm)
4
2.44392 10 mm
( ) (54 mm)(27 mm)[(54 mm) (27 mm) ]
8
2.08696 10 mm
( ) (27 mm)
4
0.41739 10 mm
O
O
O
J
J
J
π
π
π
=
= ×
= +
= ×
=
= ×
Then
64
64
(12.21960 2.44392 2.08696 0.41739) 10 mm
12.15917 10 mm
O
J= +−− ×
= ×
or
64
12.16 10 mm
O
J= ×
(b)
2
OC
J J AX= +
or
64 2 2
12.15917 10 mm (4877.32 mm )( 22.3094 mm)
C
J= ×− −
or
6
9.73 10 mm
C
J
4
= ×
PROBLEM 7.59
Determine the polar moment of inertia of the area shown with respect to
(a) Point O, (b) the centroid of the area.
SOLUTION
Determination of centroid C of entire section:
Section Area, in2
, in.x
3
, inxA
2
ππ
16
PROBLEM 7.60
Two L6 × 4 ×
1
2
-in. angles are welded together to form the section
shown. Determine the moments of inertia and the radii of gyration
of the combined section with respect to the centroidal x and y axes.
SOLUTION
W section:
24 4
9.12 in 110 in 37.1in
xy
AI I= = =
Angle:
244
1.44 in 1.23 in 1.23in
xy
AI I= = =
total W A
2
4
9.12 4(1.44) 14.880 in
AAA
= +
=+=
Now
WA
() 4()
xx x
II I= +
where
A
2
A
42 2 4
()
1.23 in (1.44 in )(4.00 in. 0.836 in.) 34.907 in
xx
I I Ad= +
=+ +=
Then
44
(110 4 34.907) in 249.63 in
x
I= +× =
4
250 in
x
I=
and
4
22
total
249.63 in
14.880 in
x
xI
kA
= =
4.10 in.
x
k=
Also
WA
() 4()
yy y
II I= +
where
( )
( )
A
2
A
244
(1.23 1.44 5 0.836 ) in 26.198 in
yy
I I Ad= +
=+− =
44
(37.1 4 26.198) in 141.892 in
y
I= +× =
4
141.9 in
y
I=
4
22
total
141.892 in
14.880 in
y
y
I
kA
= =
3.09 in.
y
k=
consent of McGraw-Hill Education.
