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PROBLEM 7.26
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the x axis.
SOLUTION
First note that
123
2
2
2
[(5)(6) (4)(2) (4)(1)] in
(30 8 4) in
18 in
AA A A=−−
= −−
= −−
=
Now
123
() () ()
xx x x
II I I=−−
where
34
11
( ) (5 in.)(6 in.) 90 in
12
x
I= =
322
2
4
1
( ) (4 in.)(2 in.) (8 in )(2 in.)
12 2
34 in
3
x
I= +
=
2
32
3
4
13
( ) (4 in.)(1in.) (4 in ) in.
12 2
1
9 in
3
x
I
= +
=
Then
4
21
90 34 9 in
33
x
I
=−−
4
or 46.0 in
x
I=
and
4
24
46.0 in
18 in
x
xI
kA
= =
or 1.599 in.
x
k=
PROBLEM 7.27
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the x axis.
SOLUTION
22
3 4 22
(125)(250) (75) 22 414 mm
2
1(125)(250) (75) (75) (125)
3 82
x
A
I
π
ππ
= −=
= −+
64
500.56 10 mm
= ×
64
501 10 mm
x
I= ×
64
2
2
500.56 10 mm
22 414 mm
x
x
I
rA
×
= =
22
22 332 mm 149.4 mm
xx
rr= =
consent of McGraw-Hill Education.
PROBLEM 7.28
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the x axis.
SOLUTION
Portion
44 4
11
(6 ) 432 in.
33
x
Ia= = =
Portion
4
1 11
( for full circle)
4 44
xx
II r
π
= =
44
1(4 ) 16 50.3 in.
16
ππ
= = =
For entire area:
4
432 50.3 381.7 382 in.
xx
II=−= =
22
1
36 (4 ) 36 4 23.43 in.
4
A
ππ
=− =−=
22
381.7 16.29 in. 4.04 in.
23.43
x
xx
I
rr
A
= = = =
consent of McGraw-Hill Education.
PROBLEM 7.29
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the y axis.
SOLUTION
First note that
123
2
2
2
[(24 6) (8)(48) (48)(6)] mm
(144 384 288) mm
816 mm
AA A A=++
= ×+ +
= ++
=
Now
123
() () ()
yy y y
II I I=++
where
34
1
34
2
34
3
1
( ) (6 mm)(24 mm) 6912 mm
12
1
( ) (48 mm)(8 mm) 2048 mm
12
1
( ) (6 mm)(48 mm) 55,296 mm
12
y
y
y
I
I
I
= =
= =
= =
Then
44
(6912 2048 55,296) mm 64,256 mm
y
I= ++ =
or
34
64.3 10 mm
y
I= ×
and
4
22
64,256 mm
816 mm
y
y
I
kA
= =
or
8.87 mm
y
k=
PROBLEM 7.30
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the y axis.
SOLUTION
First note that
123
2
2
2
[(5)(6) (4)(2) (4)(1)] in
(30 8 4) in
18 in
AA A A
=−−
= −−
= −−
=
Now
123
() () ()
yy y y
II I I=−−
where
34
1
1
( ) (6 in.)(5 in.) 62.5 in
12
y
I= =
34
2
12
( ) (2 in.)(4 in.) 10 in
12 3
y
I= =
34
3
11
( ) (1 in.)(4 in.) 5 in
12 3
y
I= =
Then
4
21
62.5 10 5 in
33
y
I
= −−
4
or 46.5 in
y
I=
and
4
22
46.5 in
18 in
y
y
I
kA
= =
or 1.607 in.
y
k=
PROBLEM 7.31
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the y axis.
SOLUTION
22
34
(125)(250) (75) 22 414 mm
2
1(250)(125) (75)
38
y
A
I
π
π
= −=
= −
64
150.335 10 mm= ×
64
150.3 10 mm
y
I= ×
64
22
2
150.335 10 mm 6707.2 mm
22 414 mm
y
y
I
rA
×
= = =
81.9 mm
y
r=
PROBLEM 7.33
Determine the shaded area and its moment of inertia with respect to the
centroidal axis parallel to AA′ knowing that d1 = 25 mm and d2 = 10 mm
and that its moments of inertia with respect to AA′ and BB′ are 2.2 × 106
mm4 and 4 × 106 mm4, respectively.
SOLUTION
64 2
2.2 10 mm (25 mm)
AA
I IA
′=×=+
(1)
64 2
4 10 mm (35 mm)
BB
I IA
′=×=+
6 22
(4 2.2) 10 (35 25 )
BB AA
II A
′′
− =− ×= −
6
1.8 10 (600)A×=
2
3000 mmA=
Eq. (1):
62
2.2 10 (3000)(25)I×=+
34
325 10 mmI= ×
consent of McGraw-Hill Education.
PROBLEM 7.34
Knowing that the shaded area is equal to 6000 mm2 and that its moment of
inertia with respect to AA′ is 18 × 106 mm4, determine its moment of inertia
with respect to BB′ for d1 = 50 mm and d2 = 10 mm.
SOLUTION
Given:
2
6000 mmA=
2 62 2 2
1; 18 10 mm (6000 mm )(50 mm)
AA
I I Ad I
′=+ ×=+
64
3 10 mmI= ×
2 64 2 2
6 2 64
3 10 mm (6000 mm )(50 mm 10 mm)
3 10 6000(60) 24.6 10 mm
BB
I I Ad
′
=+=× + +
=×+ = ×
64
24.6 10 mm
BB
I
′
= ×
consent of McGraw-Hill Education.
PROBLEM 7.26
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the x axis.
SOLUTION
First note that
123
2
2
2
[(5)(6) (4)(2) (4)(1)] in
(30 8 4) in
18 in
AA A A=−−
= −−
= −−
=
Now
123
() () ()
xx x x
II I I=−−
where
34
11
( ) (5 in.)(6 in.) 90 in
12
x
I= =
322
2
4
1
( ) (4 in.)(2 in.) (8 in )(2 in.)
12 2
34 in
3
x
I= +
=
2
32
3
4
13
( ) (4 in.)(1in.) (4 in ) in.
12 2
1
9 in
3
x
I
= +
=
Then
4
21
90 34 9 in
33
x
I
=−−
4
or 46.0 in
x
I=
and
4
24
46.0 in
18 in
x
xI
kA
= =
or 1.599 in.
x
k=
PROBLEM 7.27
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the x axis.
SOLUTION
22
3 4 22
(125)(250) (75) 22 414 mm
2
1(125)(250) (75) (75) (125)
3 82
x
A
I
π
ππ
= −=
= −+
64
500.56 10 mm
= ×
64
501 10 mm
x
I= ×
64
2
2
500.56 10 mm
22 414 mm
x
x
I
rA
×
= =
22
22 332 mm 149.4 mm
xx
rr= =
consent of McGraw-Hill Education.
PROBLEM 7.28
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the x axis.
SOLUTION
Portion
44 4
11
(6 ) 432 in.
33
x
Ia= = =
Portion
4
1 11
( for full circle)
4 44
xx
II r
π
= =
44
1(4 ) 16 50.3 in.
16
ππ
= = =
For entire area:
4
432 50.3 381.7 382 in.
xx
II=−= =
22
1
36 (4 ) 36 4 23.43 in.
4
A
ππ
=− =−=
22
381.7 16.29 in. 4.04 in.
23.43
x
xx
I
rr
A
= = = =
consent of McGraw-Hill Education.
PROBLEM 7.29
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the y axis.
SOLUTION
First note that
123
2
2
2
[(24 6) (8)(48) (48)(6)] mm
(144 384 288) mm
816 mm
AA A A=++
= ×+ +
= ++
=
Now
123
() () ()
yy y y
II I I=++
where
34
1
34
2
34
3
1
( ) (6 mm)(24 mm) 6912 mm
12
1
( ) (48 mm)(8 mm) 2048 mm
12
1
( ) (6 mm)(48 mm) 55,296 mm
12
y
y
y
I
I
I
= =
= =
= =
Then
44
(6912 2048 55,296) mm 64,256 mm
y
I= ++ =
or
34
64.3 10 mm
y
I= ×
and
4
22
64,256 mm
816 mm
y
y
I
kA
= =
or
8.87 mm
y
k=
PROBLEM 7.30
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the y axis.
SOLUTION
First note that
123
2
2
2
[(5)(6) (4)(2) (4)(1)] in
(30 8 4) in
18 in
AA A A
=−−
= −−
= −−
=
Now
123
() () ()
yy y y
II I I=−−
where
34
1
1
( ) (6 in.)(5 in.) 62.5 in
12
y
I= =
34
2
12
( ) (2 in.)(4 in.) 10 in
12 3
y
I= =
34
3
11
( ) (1 in.)(4 in.) 5 in
12 3
y
I= =
Then
4
21
62.5 10 5 in
33
y
I
= −−
4
or 46.5 in
y
I=
and
4
22
46.5 in
18 in
y
y
I
kA
= =
or 1.607 in.
y
k=
PROBLEM 7.31
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the y axis.
SOLUTION
22
34
(125)(250) (75) 22 414 mm
2
1(250)(125) (75)
38
y
A
I
π
π
= −=
= −
64
150.335 10 mm= ×
64
150.3 10 mm
y
I= ×
64
22
2
150.335 10 mm 6707.2 mm
22 414 mm
y
y
I
rA
×
= = =
81.9 mm
y
r=
PROBLEM 7.33
Determine the shaded area and its moment of inertia with respect to the
centroidal axis parallel to AA′ knowing that d1 = 25 mm and d2 = 10 mm
and that its moments of inertia with respect to AA′ and BB′ are 2.2 × 106
mm4 and 4 × 106 mm4, respectively.
SOLUTION
64 2
2.2 10 mm (25 mm)
AA
I IA
′=×=+
(1)
64 2
4 10 mm (35 mm)
BB
I IA
′=×=+
6 22
(4 2.2) 10 (35 25 )
BB AA
II A
′′
− =− ×= −
6
1.8 10 (600)A×=
2
3000 mmA=
Eq. (1):
62
2.2 10 (3000)(25)I×=+
34
325 10 mmI= ×
consent of McGraw-Hill Education.
PROBLEM 7.34
Knowing that the shaded area is equal to 6000 mm2 and that its moment of
inertia with respect to AA′ is 18 × 106 mm4, determine its moment of inertia
with respect to BB′ for d1 = 50 mm and d2 = 10 mm.
SOLUTION
Given:
2
6000 mmA=
2 62 2 2
1; 18 10 mm (6000 mm )(50 mm)
AA
I I Ad I
′=+ ×=+
64
3 10 mmI= ×
2 64 2 2
6 2 64
3 10 mm (6000 mm )(50 mm 10 mm)
3 10 6000(60) 24.6 10 mm
BB
I I Ad
′
=+=× + +
=×+ = ×
64
24.6 10 mm
BB
I
′
= ×
consent of McGraw-Hill Education.
