consent of McGraw–Hill Education.
SOLUTION Continued
Then
4
(213.73 22.43) in
x
I= −
or
4
191.3 in
x
I=
Also
12
() ()
yy y
II I= −
where
32 2
1
44
32 2
2
44
1
( ) (8 in.)(5 in.) (40 in )[(2.7 2.5) in.]
12
(83.333 1.6) in 84.933 in
1
( ) (5 in.)(2 in.) (10 in )[(2.7 1.9) in.]
12
(3.333 6.4) in 9.733 in
y
y
I
I
= +−
=+=
= +−
=+=
Then
4
(84.933 9.733) in
y
I= −
or
4
75.2 in
y
I=
consent of McGraw–Hill Education.
PROBLEM 7.36
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
18 mm.X=
2
, mmA
, mmy
3
, mmyA
1
36 28 1008×=
14 14,112
2
136 42 756
2××=
42 31,752
Σ
1764 45,864
Then
23
: (1764 mm ) 45,864 mmY A yA YΣ=Σ =
or
26.0 mmY=
Now
12
() ()
xx x
II I= +
where
32 2
1
4 34
32 2
2
4 34
1
( ) (36 mm)(28 mm) (1008 mm )[(26 14)mm]
12
(65,856 145,152)mm 211.008 10 mm
1
( ) (36 mm)(42 mm) (756 mm )[(42 26)mm]
36
(74,088 193,536)mm 267.624 10 mm
x
x
I
I
= +−
=+=×
= +−
=+=×
Then
34
(211.008 267.624) 10 mm
x
I=+×
or
34
479 10 mm
x
I= ×
Dimensions in mm
consent of McGraw–Hill Education.
SOLUTION Continued
Also
12
() ()
yy y
II I= +
where
3 34
1
32
2
4 34
1
( ) (28 mm)(36 mm) 108.864 10 mm
12
11
( ) 2 (42 mm)(18 mm) 18 mm 42 mm (6 mm)
36 2
2(6804 13,608)mm 40.824 10 mm
y
y
I
I
= = ×
= +× ×
=+=×
2
[( )
y
I
is obtained by dividing
2
A
into ]
Then
34
(108.864 40.824 10 mm
y
I= +×
or
34
149.7 10 mm
y
I= ×
consent of McGraw–Hill Education.
PROBLEM 7.37
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 centroid C of the area.
2
, inA
, in.x
, in.y
3
, inxA
3
, inyA
1
3.6 0.5 1.8
×=
1.8 0.25 3.24 0.45
2
0.5 3.8 1.9×=
0.25 2.4 0.475 4.56
3
1.3 1 1.3
×=
0.65 4.8 0.845 6.24
Σ
5.0 4.560 11.25
Then
23
: (5 in ) 4.560 inX A xA XΣ=Σ =
or
0.912 in.X=
and
23
: (5 in ) 11.25 inY A yA YΣ=Σ =
or
2.25 in.Y=
consent of McGraw–Hill Education.
SOLUTION Continued
Now
123
() () ()
xx x x
II I I=++
where
32 2
1
44
32 2
2
44
1
( ) (3.6 in.)(0.5 in.) (1.8 in )[(2.25 0.25) in.]
12
(0.0375 7.20) in 7.2375 in
1
( ) (0.5 in.)(3.8 in.) (1.9 in )[(2.4 2.25) in.]
12
(2.2863 0.0428) in 2.3291in
x
x
I
I
= +−
=+=
= +−
=+=
32 2
3
44
1
( ) (1.3 in.)(1in.) (1.3 in )[(4.8 2.25 in.)]
12
(0.1083 8.4533) in 8.5616 in
x
I= +−
=+=
Then
44
(7.2375 2.3291 8.5616) in 18.1282 in
x
I= ++ =
or
4
18.13 in
x
I=
Also
123
() () ()
yy y y
II I I=++
where
32 2
1
44
1
( ) (0.5 in.)(3.6 in.) (1.8 in )[(1.8 0.912) in.]
12
(1.9440 1.4194) in 3.3634 in
y
I= +−
=+=
32 2
2
44
32 2
3
44
1
( ) (3.8 in.)(0.5 in.) (1.9 in )[(0.912 0.25) in.]
12
(0.0396 0.8327) in 0.8723 in
1
( ) (1in.)(1.3 in.) (1.3 in )[(0.912 0.65) in.]
12
(0.1831 0.0892) in 0.2723 in
y
y
I
I
= +−
=+=
= +−
=+=
Then
44
(3.3634 0.8723 0.2723)in 4.5080 in
y
I= ++ =
or
4
4.51in
y
I=
consent of McGraw–Hill Education.
PROBLEM 7.38
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.
2
, inA
, in.y
3
, inyA
1
2
(6) 56.5487
2
π
=
2.5465
π
8=
144
2
1(12)(4.5) 27
2
−=−
1.5 –40.5
Σ
29.5487 103.5
Then
23
: (29.5487 in ) 103.5 inY A yA YΣ=Σ =
or
3.5027 in.Y=
(a)
12
()()
OO O
JJ J= −
where
44
1
222
( ) (6 in.) 107.876 in
4
() () ()
O
Ox y
J
JII
π
′′
= =
= +
Now
34
21
( ) (12 in.)(4.5 in.) 91.125 in
12
x
I′= =
and
34
2
1
( ) 2 (4.5 in.)(6 in.) 162.0 in
12
y
I
′
= =
[Note:
2
()
y
I′
is obtained using ]
SOLUTION Continued
consent of McGraw–Hill Education.
Then
4
2
4
( ) (91.125 162.0) in
253.125 in
O
J= +
=
Finally
44
(1017.876 253.125) in 764.751in
O
J=−=
or
4
765 in
O
J=
(b)
2
OC
J J AY= +
or
4 22
764.751in. (29.5487 in )(3.5027 in.)
C
J= −
or
4
402 in
C
J=
consent of McGraw–Hill Education.
PROBLEM 7.39
Determine the polar moment of inertia of the area shown with respect
to (a) Point O, (b) the centroid of the area.
SOLUTION
Dimensions in mm Symmetry:
0X=
SOLUTION Continued
(b) Section :
3 64
1(80)(60) 1.44 10 mm
12
x
I= = ×
For Iy consider the following two triangles
3 64
6 64
1
2 (60)(40) 0.640 10 mm
12
(1.44 0.640)10 2.08 10 mm
y
Oxy
I
J II
= = ×
=+= + = ×
Entire section
66
12
( ) ( ) 13.653 10 2.08 10
OO O
JJ J
= − = ×− ×
64
11.573 10 mm
O
J= ×
64
11.57 10 mm
O
J= ×
(c)
Polar moment of inertia of intire area
O
J
2
62
64
11.573 10 (4000)(30.667)
7.811 10 mm
OC
C
C
J J AY
J
J
= +
×=+
= ×
64
7.81 10 mm
C
J= ×
consent of McGraw–Hill Education.
SOLUTION Continued
Then
4
(213.73 22.43) in
x
I= −
or
4
191.3 in
x
I=
Also
12
() ()
yy y
II I= −
where
32 2
1
44
32 2
2
44
1
( ) (8 in.)(5 in.) (40 in )[(2.7 2.5) in.]
12
(83.333 1.6) in 84.933 in
1
( ) (5 in.)(2 in.) (10 in )[(2.7 1.9) in.]
12
(3.333 6.4) in 9.733 in
y
y
I
I
= +−
=+=
= +−
=+=
Then
4
(84.933 9.733) in
y
I= −
or
4
75.2 in
y
I=
consent of McGraw–Hill Education.
PROBLEM 7.36
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
18 mm.X=
2
, mmA
, mmy
3
, mmyA
1
36 28 1008×=
14 14,112
2
136 42 756
2××=
42 31,752
Σ
1764 45,864
Then
23
: (1764 mm ) 45,864 mmY A yA YΣ=Σ =
or
26.0 mmY=
Now
12
() ()
xx x
II I= +
where
32 2
1
4 34
32 2
2
4 34
1
( ) (36 mm)(28 mm) (1008 mm )[(26 14)mm]
12
(65,856 145,152)mm 211.008 10 mm
1
( ) (36 mm)(42 mm) (756 mm )[(42 26)mm]
36
(74,088 193,536)mm 267.624 10 mm
x
x
I
I
= +−
=+=×
= +−
=+=×
Then
34
(211.008 267.624) 10 mm
x
I=+×
or
34
479 10 mm
x
I= ×
Dimensions in mm
consent of McGraw–Hill Education.
SOLUTION Continued
Also
12
() ()
yy y
II I= +
where
3 34
1
32
2
4 34
1
( ) (28 mm)(36 mm) 108.864 10 mm
12
11
( ) 2 (42 mm)(18 mm) 18 mm 42 mm (6 mm)
36 2
2(6804 13,608)mm 40.824 10 mm
y
y
I
I
= = ×
= +× ×
=+=×
2
[( )
y
I
is obtained by dividing
2
A
into ]
Then
34
(108.864 40.824 10 mm
y
I= +×
or
34
149.7 10 mm
y
I= ×
consent of McGraw–Hill Education.
PROBLEM 7.37
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 centroid C of the area.
2
, inA
, in.x
, in.y
3
, inxA
3
, inyA
1
3.6 0.5 1.8
×=
1.8 0.25 3.24 0.45
2
0.5 3.8 1.9×=
0.25 2.4 0.475 4.56
3
1.3 1 1.3
×=
0.65 4.8 0.845 6.24
Σ
5.0 4.560 11.25
Then
23
: (5 in ) 4.560 inX A xA XΣ=Σ =
or
0.912 in.X=
and
23
: (5 in ) 11.25 inY A yA YΣ=Σ =
or
2.25 in.Y=
consent of McGraw–Hill Education.
SOLUTION Continued
Now
123
() () ()
xx x x
II I I=++
where
32 2
1
44
32 2
2
44
1
( ) (3.6 in.)(0.5 in.) (1.8 in )[(2.25 0.25) in.]
12
(0.0375 7.20) in 7.2375 in
1
( ) (0.5 in.)(3.8 in.) (1.9 in )[(2.4 2.25) in.]
12
(2.2863 0.0428) in 2.3291in
x
x
I
I
= +−
=+=
= +−
=+=
32 2
3
44
1
( ) (1.3 in.)(1in.) (1.3 in )[(4.8 2.25 in.)]
12
(0.1083 8.4533) in 8.5616 in
x
I= +−
=+=
Then
44
(7.2375 2.3291 8.5616) in 18.1282 in
x
I= ++ =
or
4
18.13 in
x
I=
Also
123
() () ()
yy y y
II I I=++
where
32 2
1
44
1
( ) (0.5 in.)(3.6 in.) (1.8 in )[(1.8 0.912) in.]
12
(1.9440 1.4194) in 3.3634 in
y
I= +−
=+=
32 2
2
44
32 2
3
44
1
( ) (3.8 in.)(0.5 in.) (1.9 in )[(0.912 0.25) in.]
12
(0.0396 0.8327) in 0.8723 in
1
( ) (1in.)(1.3 in.) (1.3 in )[(0.912 0.65) in.]
12
(0.1831 0.0892) in 0.2723 in
y
y
I
I
= +−
=+=
= +−
=+=
Then
44
(3.3634 0.8723 0.2723)in 4.5080 in
y
I= ++ =
or
4
4.51in
y
I=
consent of McGraw–Hill Education.
PROBLEM 7.38
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.
2
, inA
, in.y
3
, inyA
1
2
(6) 56.5487
2
π
=
2.5465
π
8=
144
2
1(12)(4.5) 27
2
−=−
1.5 –40.5
Σ
29.5487 103.5
Then
23
: (29.5487 in ) 103.5 inY A yA YΣ=Σ =
or
3.5027 in.Y=
(a)
12
()()
OO O
JJ J= −
where
44
1
222
( ) (6 in.) 107.876 in
4
() () ()
O
Ox y
J
JII
π
′′
= =
= +
Now
34
21
( ) (12 in.)(4.5 in.) 91.125 in
12
x
I′= =
and
34
2
1
( ) 2 (4.5 in.)(6 in.) 162.0 in
12
y
I
′
= =
[Note:
2
()
y
I′
is obtained using ]
SOLUTION Continued
consent of McGraw–Hill Education.
Then
4
2
4
( ) (91.125 162.0) in
253.125 in
O
J= +
=
Finally
44
(1017.876 253.125) in 764.751in
O
J=−=
or
4
765 in
O
J=
(b)
2
OC
J J AY= +
or
4 22
764.751in. (29.5487 in )(3.5027 in.)
C
J= −
or
4
402 in
C
J=
consent of McGraw–Hill Education.
PROBLEM 7.39
Determine the polar moment of inertia of the area shown with respect
to (a) Point O, (b) the centroid of the area.
SOLUTION
Dimensions in mm Symmetry:
0X=
SOLUTION Continued
(b) Section :
3 64
1(80)(60) 1.44 10 mm
12
x
I= = ×
For Iy consider the following two triangles
3 64
6 64
1
2 (60)(40) 0.640 10 mm
12
(1.44 0.640)10 2.08 10 mm
y
Oxy
I
J II
= = ×
=+= + = ×
Entire section
66
12
( ) ( ) 13.653 10 2.08 10
OO O
JJ J
= − = ×− ×
64
11.573 10 mm
O
J= ×
64
11.57 10 mm
O
J= ×
(c)
Polar moment of inertia of intire area
O
J
2
62
64
11.573 10 (4000)(30.667)
7.811 10 mm
OC
C
C
J J AY
J
J
= +
×=+
= ×
64
7.81 10 mm
C
J= ×
consent of McGraw–Hill Education.