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PROBLEM 7.41
Two channels are welded to a rolled W section as 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:
2
4
4
9.12 in
110 in
37.1in
x
y
A
I
I
=
=
=
Channel:
2
4
4
3.37 in
1.31in
32.5 in
x
y
A
I
I
=
=
=
total W chan
2
2
9.12 2(3.37) 15.86 in
AAA= +
=+=
Now
W chan
() 2()
xx x
II I= +
where
chan
2
chan
42 2 4
()
1.31in (3.37 in )(4.572 in.) 71.754 in
xx
I I Ad= +
=+=
Then
44
(110 2 71.754) in 253.51in
x
I= +× =
4
254 in
x
I=
and
4
22
total
253.51in
15.86 in
x
xI
kA
= =
4.00 in.
x
k=
Also
W chan
() 2()
yy y
II I= +
44
(37.1 2 32.5) in 102.1in= +× =
4
102.1in
y
I=
and
4
22
total
102.1in
15.86 in
y
y
I
kA
= =
2.54 in.
y
k=
consent of McGraw-Hill Education.
PROBLEM 7.42
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
From Figure 9.13A
PROBLEM 7.43
Two channels and two plates are used to form the column
section shown. For b = 200 mm, determine the moments of
inertia and the radii of gyration of the combined section with
respect to the centroidal x and y axes.
SOLUTION
For C250 × 22.8
2
64
64
2890 mm
28.0 10 mm
0.945 10 mm
x
y
A
I
I
=
= ×
= ×
Total area
4 32
2[2890 mm (10 mm)(375 mm)] 13.28 10 mmA=+=×
Given b= 200 mm:
64 3 2
1
2[28.0 10 mm ] 2 (375 mm)(10 mm) (375 mm)(10 mm)(132 mm)
12
x
I
=×+ +
6
186.743 10= ×
64
186.7 10 mm
x
I= ×
6
2 32
3
186.743 10 14.0620 10 mm
13.28 10
x
x
I
kA
×
= = = ×
×
118.6 mm
x
k=
From Figure 9.13B
consent of McGraw-Hill Education.
SOLUTION Continued
PROBLEM 7.44
Two 20-mm steel plates are welded to a rolled S section as 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
S section:
2
64
64
6010 mm
90.3 10 mm
3.88 10 mm
x
y
A
I
I
=
= ×
= ×
Note:
total S plate
2
2
2
6010 mm 2(160 mm)(20 mm)
12,410 mm
A AA= +
= +
=
Now
S plate
() 2()
xx x
II I= +
where
plate
2
plate
32 2
64
()
1(160 mm)(20 mm) (3200 mm )[(152.5 10) mm]
12
84.6067 10 mm
xx
I I Ad= +
= ++
= ×
Then
64
(90.3 2 84.6067) 10 mm
x
I= +× ×
64
259.5134 10 mm= ×
or
64
260 10 mm
x
I= ×
and
64
22
total
259.5134 10 mm
12410 mm
x
x
I
kA
×
= =
or
144.6 mm
x
k=
Also
S plate
() 2()
yy y
II I= +
64 3
64
1
3.88 10 mm 2 (20 mm)(160 mm)
12
17.5333 10 mm
=×+
= ×
or
6
17.53 10 mm
y
I
4
= ×
and
64
22
total
17.5333 10 mm
12,410 mm
y
y
I
kA
×
= =
or
37.6 mm
y
k=
consent of McGraw-Hill Education.
PROBLEM 7.45
Two L4 × 4 ×
1
2
-in. angles are welded to a steel plate as shown. Determine
the moments of inertia of the combined section with respect to centroidal
axes respectively parallel and perpendicular to the plate.
SOLUTION
For
1
4 4 -in.
2
××
angle:
24
23
3.75 in , 5.52 in
(12.5 in ) 33.85 in
2.708 in.
xy
A II
YA y A
Y
Y
= = =
= Σ
=
=
Section Area, in2
in.y
3
, inyA
Plate
(0.5)(10) 5=
5 25
Two angles
2(3.75) 7.5=
1.18 8.85
Σ
12.5 33.85
consent of McGraw-Hill Education.
SOLUTION Continued
Entire section:
23 2 2
1
( ) (0.5)(10) (0.5)(10)(2.292) 2[5.52 (3.75)(1.528) ]
12
xx
I I Ad
′
=Σ+ = + + +
4
41.667 26.266 1604 17.511 96.48 in= + ++ =
4
96.5 in
x
I=
32
1(10)(0.5) 2[5.52 (3.75)(1.43) ]
12
y
I= ++
4
0.104 11.04 15.367 26.51in=++ =
4
26.5 in
y
I=
consent of McGraw-Hill Education.
PROBLEM 7.46
A channel and a plate are welded together as shown to form a
section that is symmetrical with respect to the y axis.
Determine the moments of inertia of the combined section
with respect to its centroidal x and y axes.
SOLUTION
2
PROBLEM 7.47
Two L76 × 76 × 6.4-mm angles are welded to a C250 × 22.8 channel.
Determine the moments of inertia of the combined section with respect to
centroidal axes respectively parallel and perpendicular to the web of the
channel.
SOLUTION
Angle:
2
64
929 mm
0.512 10 mm
xy
A
II
=
= = ×
Channel:
2
64
64
2890 mm
0.945 10 mm
28.0 10 mm
x
y
A
I
I
=
= ×
= ×
First locate centroid C of the section
2
, mm
A
, mmy
3
, mmyA
Angle
2(929) 1858=
21.2 39,389.6
Channel 2890 −16.1 −46,529
Σ
4748 −7139.4
Then
23
: (4748 mm ) 7139.4 mmY A yA YΣ=Σ =−
or
1.50366 mmY= −
Now
2() ()
x xL xC
II I= +
where
64 2 2
64
2 64 2 2
64
( ) 0.512 10 mm (929 mm )[(21.2 1.50366) mm]
0.990859 10 mm
( ) 0.949 10 mm (2890 mm )[(16.1 1.50366) mm]
1.56472 10 mm
xL x
xC x
I I Ad
I I Ad
2
=+= × + +
= ×
=+= × + −
= ×
Then
64
[2(0.990859) 1.56472 10 ] mm
x
I= +×
or
64
3.55 10 mm
x
I= ×
consent of McGraw-Hill Education.
PROBLEM 7.41
Two channels are welded to a rolled W section as 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:
2
4
4
9.12 in
110 in
37.1in
x
y
A
I
I
=
=
=
Channel:
2
4
4
3.37 in
1.31in
32.5 in
x
y
A
I
I
=
=
=
total W chan
2
2
9.12 2(3.37) 15.86 in
AAA= +
=+=
Now
W chan
() 2()
xx x
II I= +
where
chan
2
chan
42 2 4
()
1.31in (3.37 in )(4.572 in.) 71.754 in
xx
I I Ad= +
=+=
Then
44
(110 2 71.754) in 253.51in
x
I= +× =
4
254 in
x
I=
and
4
22
total
253.51in
15.86 in
x
xI
kA
= =
4.00 in.
x
k=
Also
W chan
() 2()
yy y
II I= +
44
(37.1 2 32.5) in 102.1in= +× =
4
102.1in
y
I=
and
4
22
total
102.1in
15.86 in
y
y
I
kA
= =
2.54 in.
y
k=
consent of McGraw-Hill Education.
PROBLEM 7.42
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
From Figure 9.13A
PROBLEM 7.43
Two channels and two plates are used to form the column
section shown. For b = 200 mm, determine the moments of
inertia and the radii of gyration of the combined section with
respect to the centroidal x and y axes.
SOLUTION
For C250 × 22.8
2
64
64
2890 mm
28.0 10 mm
0.945 10 mm
x
y
A
I
I
=
= ×
= ×
Total area
4 32
2[2890 mm (10 mm)(375 mm)] 13.28 10 mmA=+=×
Given b= 200 mm:
64 3 2
1
2[28.0 10 mm ] 2 (375 mm)(10 mm) (375 mm)(10 mm)(132 mm)
12
x
I
=×+ +
6
186.743 10= ×
64
186.7 10 mm
x
I= ×
6
2 32
3
186.743 10 14.0620 10 mm
13.28 10
x
x
I
kA
×
= = = ×
×
118.6 mm
x
k=
From Figure 9.13B
consent of McGraw-Hill Education.
SOLUTION Continued
PROBLEM 7.44
Two 20-mm steel plates are welded to a rolled S section as 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
S section:
2
64
64
6010 mm
90.3 10 mm
3.88 10 mm
x
y
A
I
I
=
= ×
= ×
Note:
total S plate
2
2
2
6010 mm 2(160 mm)(20 mm)
12,410 mm
A AA= +
= +
=
Now
S plate
() 2()
xx x
II I= +
where
plate
2
plate
32 2
64
()
1(160 mm)(20 mm) (3200 mm )[(152.5 10) mm]
12
84.6067 10 mm
xx
I I Ad= +
= ++
= ×
Then
64
(90.3 2 84.6067) 10 mm
x
I= +× ×
64
259.5134 10 mm= ×
or
64
260 10 mm
x
I= ×
and
64
22
total
259.5134 10 mm
12410 mm
x
x
I
kA
×
= =
or
144.6 mm
x
k=
Also
S plate
() 2()
yy y
II I= +
64 3
64
1
3.88 10 mm 2 (20 mm)(160 mm)
12
17.5333 10 mm
=×+
= ×
or
6
17.53 10 mm
y
I
4
= ×
and
64
22
total
17.5333 10 mm
12,410 mm
y
y
I
kA
×
= =
or
37.6 mm
y
k=
consent of McGraw-Hill Education.
PROBLEM 7.45
Two L4 × 4 ×
1
2
-in. angles are welded to a steel plate as shown. Determine
the moments of inertia of the combined section with respect to centroidal
axes respectively parallel and perpendicular to the plate.
SOLUTION
For
1
4 4 -in.
2
××
angle:
24
23
3.75 in , 5.52 in
(12.5 in ) 33.85 in
2.708 in.
xy
A II
YA y A
Y
Y
= = =
= Σ
=
=
Section Area, in2
in.y
3
, inyA
Plate
(0.5)(10) 5=
5 25
Two angles
2(3.75) 7.5=
1.18 8.85
Σ
12.5 33.85
consent of McGraw-Hill Education.
SOLUTION Continued
Entire section:
23 2 2
1
( ) (0.5)(10) (0.5)(10)(2.292) 2[5.52 (3.75)(1.528) ]
12
xx
I I Ad
′
=Σ+ = + + +
4
41.667 26.266 1604 17.511 96.48 in= + ++ =
4
96.5 in
x
I=
32
1(10)(0.5) 2[5.52 (3.75)(1.43) ]
12
y
I= ++
4
0.104 11.04 15.367 26.51in=++ =
4
26.5 in
y
I=
consent of McGraw-Hill Education.
PROBLEM 7.46
A channel and a plate are welded together as shown to form a
section that is symmetrical with respect to the y axis.
Determine the moments of inertia of the combined section
with respect to its centroidal x and y axes.
SOLUTION
2
PROBLEM 7.47
Two L76 × 76 × 6.4-mm angles are welded to a C250 × 22.8 channel.
Determine the moments of inertia of the combined section with respect to
centroidal axes respectively parallel and perpendicular to the web of the
channel.
SOLUTION
Angle:
2
64
929 mm
0.512 10 mm
xy
A
II
=
= = ×
Channel:
2
64
64
2890 mm
0.945 10 mm
28.0 10 mm
x
y
A
I
I
=
= ×
= ×
First locate centroid C of the section
2
, mm
A
, mmy
3
, mmyA
Angle
2(929) 1858=
21.2 39,389.6
Channel 2890 −16.1 −46,529
Σ
4748 −7139.4
Then
23
: (4748 mm ) 7139.4 mmY A yA YΣ=Σ =−
or
1.50366 mmY= −
Now
2() ()
x xL xC
II I= +
where
64 2 2
64
2 64 2 2
64
( ) 0.512 10 mm (929 mm )[(21.2 1.50366) mm]
0.990859 10 mm
( ) 0.949 10 mm (2890 mm )[(16.1 1.50366) mm]
1.56472 10 mm
xL x
xC x
I I Ad
I I Ad
2
=+= × + +
= ×
=+= × + −
= ×
Then
64
[2(0.990859) 1.56472 10 ] mm
x
I= +×
or
64
3.55 10 mm
x
I= ×
consent of McGraw-Hill Education.
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