1010
Ans:
Ix=
ab3
3(3n
+
1)
1018
10–9.
SOLUTION
Moment of Inertia:Performing the integration, we have
Ans.
(b) Differential Element:Here,The area of the differential
element parallel to xaxis is
Moment of Inertia:Applying Eq. 10–1 and performing the integration, we have
Ans.=23.8 ft4
=2c–
y2
15 125 –10y23
2–
2y
375 125 –10y23
2–
2
13125 125 –10y27
2d
`
0
2.5 ft
=2L2.5 ft
0
y2225 –10ydy
Ix=L
A
y2dA
dA =2xdy =2225 –10ydy.
x=225 –10y.
=23.8 ft4
=1
3a–0.001
7x7+0.075
5x5–1.875
3x3+15.625xb
`
–5 ft
5 ft
Ix=LdIx=1
3L5 ft
–5 ft 1–0.001x6+0.075x4–1.875x2+15.6252dx
=1
31–0.001x6+0.075x4–1.875x2+15.6252dx
=1
312.5 –0.1x223dx
=1
12 1dx2y3+ydxay
2b2
dIx=dIx¿+dAy
‘2
Determine the moment of inertia of the area about the x
axis. Solve the problem in two ways,using rectangular
differential elements: (a) having a thickness dx and
(b) having a thickness of dy.
y
x
y=2.5 – 0.1x2
5ft
2.5 ft
Ans:
I
x=
23.8
ft
4
1019
10–10.
Determine the moment of inertia for the shaded area about
the xaxis.
SOLUTION
Ans.
Also,
Ans.=
2
15 bh3
=cb
3y3–
b
5h2y5dh
0
=Lh
0
y2(b–
b
h2y2)dy
Ix=Ly2dA
dA =(b–x)dy =(b–
b
h2y2)dy
=
2
15 bh3
=
1
3ah2
bb3>2
a2
5bx5>2 ]b
0
=Lb
0
y3
3dx =Lb
0
1
3ah2
bb3>2
x3>2 dx
Ix=LdI
x
dI
x=
1
3y3dx
b
x
y
y
2
—x
h
h
2
b
15
1022
8
1023
2
*10–16.
Determine the moment of inertia for the shaded area about
the y axis.
SOLUTION
y
x
16 in.
4 in.
y2 x
1026
10–17.
Determine the moment of inertia for the shaded area about
the x axis.
SOLUTION
30
y
x
h
b
y x3
h
b3
30
10–18.
Determine the moment of inertia for the shaded area about
the y axis.
6
y
x
h
b
y x3
h
b3
10–19.
Determine the moment of inertia for the shaded area about
the x axis.
SOLUTION
3
1
15
y
y2 1 x
x
1 m
1 m
1 m
*10–20.
Determine the moment of inertia for the shaded area about
the y axis.
SOLUTION
3
3
3
105
y
y2 1 x
x
1 m
1 m
1 m