9.1
Determine PROBLEM by direct integration the moment of inertia of the shaded area with respect to the y axis.
SOLUTION
At x = a, y = b: b = ka 2
b a
5 2 5
5
or
a
2
k =
b a2
5
y =
x2
or
x=
b
2 5
y5
dI y =
=
1 3 x dy 3
1 a3 6 y 5 dy 3 b6 5 1 a3 b 6 y 5 dy 3 b6 0 5
b
Then
Iy =
1 5 a3 11 = y5 3 11 b 6 5
0
=
5 a3 11 b5 33 b 6 5 or I y = 5 3 ab 33
PROBLEM 9.2
Determine by direct integration the moment of inertia of the shaded area with respect to the y axis.
SOLUTION
At x = 0, y = 0: 0 = ka 2 + c k =- x = a, c a2 b=c
y =b: b a2
k = - y =- =-
Then
b 2 x - a) + b 2( a b 2 x - 2ax + a 2 + b a2
(
)
Now
2b 3 b dI y = x 2dA = x 2 ( ydx ) = - 2 x 4 + x - bx 2 + bx 2 dx a a 2b 3 b x dx = - 2 x4 + a a
Then
b 2b 3 a I y = dI y = 0 - 2 x 4 + x dx a a 1 x5 2 x 4 = b - 2 + a 4 0 a 5
a
a 3 2a 3 1 3 1 = b 5 + 4 = ba 2 - 5
Iy =
3a3b 10
PROBLEM 9.3
Determine by direct integration the moment of inertia of the shaded area with respect to the y axis.
SOLUTION
By observation
y =h-
h x b
x = h 1 - b Now
dI y = x 2dA = x 2 ( h - y ) dx
x = x 2 h - h 1 - dx b = hx3 dx b
b
Then
hx3 hx 4 I y = dI y = dx = 4b b
b 0
=
0
hb 4 4b Iy = b 3h 4
PROBLEM 9.4
Determine by direct integration the moment of inertia of the shaded area with respect to the y axis.
SOLUTION
Have At or At or y = kx 2 + c x = 0, y = b: b = k ( 0 ) + c
c=b
x = 2a, y = 0: 0 = k ( 2a ) + b
2
k =- y =- = I y = x 2dA, I y = a x 2dA = b 4a 2
2a
b 4a 2
Then
b 2 x +b 4a 2
b 4a 2 - x 2 2 4a
(
) ( )
Then
dA = ydx =
b 4a 2 - x 2 dx 4a 2
b 2a 2 x 4a 2 - x 2 dx 2 a 4a
2a
(
)
=
2 x3 x5 - 4a 3 5 a
=
b b 8a3 - a3 - 32a5 - a5 3 20a 2
7a3b 31a 3b - 3 20
(
)
(
)
=
Iy =
47 3 ab 60
PROBLEM 9.5
Determine by direct integration the moment of inertia of the shaded area with respect to the x axis.
SOLUTION
At x = a, y = b: b = ka 2 or k =
y = b a2 b 3 x2 a3 2
3
3
I x = y 2dA
dA = xdy
a 2 b = 0 y 2 2 y 5 dy b5
5 17 = 2 y5 b 5 17 a
b
0
5a b 5 = 17 b 2 5
17
or I x =
5 3 ab 17
PROBLEM 9.6
Determine by direct integration the moment of inertia of the shaded area with respect to the x axis.
SOLUTION
At
x = 0, y = 0: 0 = ka 2 + c k =- c a2 b=c
x = a, y = b k =- y =b- b a2
Then Now From above
b 2 x - a) 2( a
dI x = y 2dA = y 2 ( xdy )
( x - a )2
=
a2 (b - y ) b y b y +a b
Then
x - a = a2 1 -
and
x = a2 1 -
Then
y dI x = ay 2 1 + 1 - dy b y b I x = dI x = a 0 y 2 1 + 1 - dy b
and
y3 =a 3
b
0
y b + a 0 y 2 1 - dy 6
PROBLEM 9.6 CONTINUED
For the second integral use substitution
u =1- y 1 du = - dy, y = b(1 - u ) b b y = 0 u =1 y=b u =0
Now
b 2 b 2 2 0 y 1 - b dy = -0 b (1 - u ) u 2 du
3 u 1 - 2u 3 + u 5 du = -b3 2 u 2 - 4 u 5 + 2 u 7 2 2 2 2 2 5 7 1 3
y
1
=
0 -b3 1
0
2 4 2 70 - 84 + 30 16b3 = +b 3 - + = b 3 = 105 3 5 7 105
Then
Ix = a
b3 16ab3 51 3 + = ab 3 105 105 or I x = 17 3 ab 35
PROBLEM 9.7
Determine by direct integration the moment of inertia of the shaded area with respect to the x axis.
SOLUTION
By observation
y =h-
h x b
x = h 1 - b or Now y x = b 1 - h dI x = y 2dA = y 2 ( b - x ) dy by = y2 b - b + dy h =
h by 0
by 3 dy h
h
Then
Ix =
by 4 dy = h 4h
3
=
0
bh 4 4h or I x = bh3 4
PROBLEM 9.8
Determine by direct integration the moment of inertia of the shaded area with respect to the x axis.
SOLUTION
Have At or At or y = dI x = =
y = kx 2 + c x = 0, y = b: b = k (0) + c c=b x = 2a, y = 0: 0 = k (2a) 2 + b k =- b 4a 2
Then
b 4a 2 - x 2 4a 2
(
)
Now
1 3 y dx 3
3 1 b3 4a 2 - x 2 dx 6 3 64a
(
)
PROBLEM 9.8 CONTINUED
Then
I x = dI x
=
3 1 b3 2 a 4a 2 - x 2 dx 6 a 3 64a
(
)
=
b3 2 a 6 4 2 2 4 6 64a - 48a x + 12a x - x dx 192a 6 a
(
)
b3 = 192a 6 =
12 2 5 x 7 64a 6 x - 16a 4 x3 + a x - 5 7 a
2a
b3 64a 7( 2 - 1) - 16a 7 ( 8 - 1) 6 192a + 12 7 1 a ( 32 - 1) - (128 - 1) 5 7
=
ab3 372 127 3 - 64 - 112 + = 0.043006ab 192 5 7
I x = 0.0430ab3
PROBLEM 9.9
Determine by direct integration the moment of inertia of the shaded area with respect to the x axis.
SOLUTION
At or
x = a, y = b: b = ka 2
k = y = b a b 1 x2 a
3
1
Then
Now
dI x =
1 3 1 b 3 2 y dx = x dx 3 3 a
3
Then
3 a a1 b 2 I x = 20 dI x = 20 x dx 3 a
2 b 2 5 2 = x 3 a 5
3
a
=
0
4 b3 5 a2 3 15 a 2
Ix =
4 3 ab 15
PROBLEM 9.10
Determine by direct integration the moment of inertia of the shaded area with respect to the x axis.
SOLUTION
At or
x = 2a, y = b: k = 2a b3 x=
y =
2a = kb3
Then
2a 3 y b3
b
or
( 2a )
1 3
x3
1
Now
dI x =
1 3 1 b3 y dx = xdx 3 3 2a
2a
Then
1 b3 2a 1 b3 1 2 I x = dI x = xdx = x 3 2a a 6 a 2 a = b3 4a 2 - a 2 12a
(
)
Ix = 1 3 ab 4
PROBLEM 9.11
Determine by direct integration the moment of inertia of the shaded area with respect to the x axis.
SOLUTION
At Then Now
x = 0, y = b: b = ke0 = k y = be
dI x =
a
x -a
x 1 3 b 3 - a 3 y dx = e dx 3 3
Then
I x = dI x = 0
x b 3 - a 3 e dx 3
b3 - 3ax b3 -a - 3ax = e dx = e 3 3 3 =
a
=-
0
b3a -3 e - e0 9
(
)
ab3 ( 0.95021) = 0.10558ab3 9 or I x = 0.1056ab3
PROBLEM 9.12
Determine by direct integration the moment of inertia of the shaded area with respect to the y axis.
SOLUTION
At
x = a, y = b: b = ka 2
y =
dI y = x 2dA,
1
or
k =
b a
Then Now
b 1 x2 a
dA = ydx b 5 x 2 dx a
a 0
dI y = x 2 ydx =
a 0
Then
I y = dI y = 2 = 4 b 7 a2 7 a1 2
b 5 4 b 7 x 2 dx = x2 7 a a
or I y =
4 3 ab 7
PROBLEM 9.13
Determine by direct integration the moment of inertia of the shaded area with respect to the y axis.
SOLUTION
At Then x = 2a, y = b: 2a = kb3 x=
y =
or
s
2a 3 y b3
b
1 3
or
( 2a )
x3
1
Now Then
I y = x 2dA I y = a x 2 b
2a
dA = ydx b x 3 dx
1
( 2a )
1 3
=
( 2a )
b
1 3
2a a x 3 dx
5
2a
=
( 2a ) 3
1 3
1
3 10 x3 10
a
=
3b 10 ( 2a )
2a 10 - a 10 3 3 ( )
=
10 3ba3 10 2 3 - 13 1 10 ( 2 ) 3
= 2.1619a3b or I y = 2.16a3b
PROBLEM 9.14
Determine by direct integration the moment of inertia of the shaded area with respect to the y axis.
SOLUTION
At Then Now
x = 0, y = b: b = ke0 = k y = be
x -a
dI y = x 2dA = x 2 ydx
= x 2be a dx
-x
Then Use integration by parts
I y = dI y = 0 bx 2e a dx = b0 x 2e a dx
a
-x
a
-x
u = x2 du = 2 xdx
dv = e a dx v = -ae
a
0
x -a
-x
Then
-x -x a I y = 0 x 2e a dx = b -ax 2e a
-x a - 0 -ae a 2 xdx
-x a = b -a3e -1 + 2a 0 xe a dx
Again use integration by parts:
u = x du = dx dv = e a dx v = -ae
x -a
-x
PROBLEM 9.14 CONTINUED
Then
a 0 xe a dx = -axe
-x
x -a
a
0
- 0 -ae
a
x -a
x -a
dx
= -a 2e-1 - a 2e
a 0
= -a 2e-1 - a 2e-1 + a 2e0 = -2a 2e -1 + a 2
Finally,
I y = b -a 3e -1 + 2a -2a 2e-1 + a 2 = ba3 -5e-1 + 2 = 0.1606ba3
(
)
(
)
or I y = 0.1606ba3
PROBLEM 9.15
Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the x axis.
SOLUTION
At x = a, y1: b = ma y2: b = ka3 Then y1 = y2 = b x a b 3 x a3 or y1 = y2 = b or or x1 = or m= k = a y b 1 y3 b a b a3
a x2 = 1 3 b
Now
a 1 a dA = ( x2 - x1 ) dy = 1 y 3 - y dy 3 b b
a A = 2 dA = 2 y - 1 b b3
1 3
b 0
a
a 3 4 a 1 2 y dy = 2 1 y 3 - y b2 0 b3 4
b
=
3ab 1 - ab = ab 2 2
Then
a 1 a dI x = y 2dA = y 2 1 y 3 - y dy 3 b b 3 y 10 3 y3 y4 dy = 2a I x = 2 dI x = 2a - - 1 b 4b 10 b 1 3 b3 0
b 0
Now
y7 3
b
3 1 b3 3 = 2a b3 - = 2ab3 - 10 4 10 4
or I x = And
2 kx =
1 3 ab 10
Ix = A
1 ab3 10 1 ab 2
=
1 2 b 5 kx = b 5
PROBLEM 9.16
Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the x axis.
SOLUTION
At or
y1 = x1 =
x = a, y = b:
b = k1a 4
k1 = b a4 y2 = x2 =
b = k2a 4
k2 = b a
1 4
1
b a4
1
Then
b 4 x a4 a b
1 4
x4
1
and
y4
1
a 4 y b4
Now
1 x4 a x4 A = ( y2 - y1 ) dx = b0 1 - 4 dx a4 a 4 x 5 1 x5 4 = 3 ab = b - 1 4 5 5 a4 5 a 0
a
Then
I x = y 2dA Ix =
2 0 y
dA = ( x1 - x2 ) dy a b
1 4
b
y4 -
1
a 4 y dy b4
b
4 y 13 1 y 7 4 = a - 1 7 b4 13 b 4 0
1 4 = ab3 - 13 7 or I x = kx = Ix = A
15 ab3 91 3 ab 5
15 3 ab 91
Now
=
25 2 b = 0.52414b 91 or k x = 0.524b
PROBLEM 9.17
Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the y axis.
SOLUTION
At x = a, y1: b = ka3 y2: b = ma Then y1 = y2 = Now y1 = y2 = b or or b 3 x a3 b x a k = m= b a3 b a
b b dA = ( y2 - y1 ) dx = x - 3 x3 dx a a
a b a x3 b 1 1 A = dA = 2 0 x - 2 dx = 2 x 2 - 2 x 4 a a 2 a 4a 0
1 b a2 1 =2 - 2 a 4 = ab a 2 4a 2
Now
dI y = x 2dA =
b 3 x 5 x - 2 dx a a
Then
b a x5 a I y = 0 dI y = 2 0 x3 - 2 dx a a
a b 1 1 b a 4 1 a6 = 2 x 4 - 2 x6 = 2 - a 4 a 4 6 a2 6a 0
= And
1 3 ab 6
Iy A =
1 a 3b 6 1 ab 2
or I y =
= 1 2 a 3
1 3 ab 6
a 3
2 ky =
or k y =
PROBLEM 9.18
Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the y axis.
SOLUTION
At or
x = a, y = b:
b = k1a 4
k1 = b a4
b = k2a 4
k2 = y2 = b a4 b a
1 4 1
1
Then
y1 =
b 4 x a4
and
x4
1
Now
A = ( y2 - y1 ) dx = b
a
a 0
x1 4 a
1 4
-
x4 dx a4
4 x 5 1 x5 4 = 3 ab = b - 1 4 5 5 a4 5 a 0 Now
I y = x 2dA dA = ( y2 - y1 ) dx
Then
b 1 b a I y = 0 x 2 1 x 4 - 4 x 4 dx 4 a a 9 x6 a x4 = b 0 1 - 4 dx a4 a 4 x 13 1 x7 4 = b - 7 a4 13 a 1 4 0 1 4 = b a3 - a3 7 13 or I y = ky = Iy A =
15 a 3b 91 3 ab 5
a
15 3 ab 91
Now
=
25 a = 0.52414a 91 or k y = 0.524a
PROBLEMS 9.19 AND 9.20
P 9.19 Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the x axis. P 9.20 Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the y axis.
SOLUTION
First determine constants m, b, and c y1: at x = 2a, y =0
0 = m ( 2a ) + b At x = a, y =h
h = m(a) + b Solving yields Then m=- y1 = - h a b = 2h
h x + 2h a y =0 0 = c sin k ( 2a )
y2: at x = 2a,
At
x = a,
y = 2h
2h = c sin ka
Solving
c sin k ( 2a ) = 0
c0
sin k ( 2a ) = 0, k ( 2a ) = , k = Substitute k, 2h = c sin ka Then yields 2h = c sin
2a or c = 2h
2
y2 = 2h sin
2a
x
To calculate the area of shaded surface, a differential strip parallel to the y axis is chosen to be dA. h dA = ( y2 - y1 ) dx = 2h sin x - - x + 2h dx 2a a
PROBLEMS 9.19 AND 9.20 CONTINUED
h 2a A = dA = 0 2h sin x - 2hx + x dx 2a a
4a x2 x - 2x + = h - cos 2a 2a a
2a
4a ( 2a ) 2 = h - cos 2a - 2 ( 2a ) + 2a 2a 4a a2 - h - cos a - 2 ( a ) + 2a 2a
a 4a 4 1 = h - 4a + 2a - h -2a + = ah - 2 2 A = 0.77324ah PROBLEM 9.19 Moment of inertia where dI x = I x = a dI x dI x = 1 3 3 y2 - y1 dx 3
2a
(
)
Now
3 3 1 h x - 2h - x dx 2h sin 3 2a a
3 1 3 3 x 3 x - h 2 - dx = 8h sin 3 2a a
Then
Now
8h3 2a 3 h3 2 a x Ix = -a sin 2a xdx - 3 a 2 - a dx 3
3 2 sin 2a xdx = sin 2a x 1 - cos 2a x dx
3
= sin x dx - sin x cos 2 x dx 2a 2a 2a 2a 2a x+ x cos3 = - cos 2a 3 2a
Then
2a sin 3 a
2a 1 x dx = - cos x - cos3 x 2 2a 3 2a a
2a
=-
2a 1 4a -1 + = 3 3
PROBLEMS 9.19 AND 9.20 CONTINUED
And x a x - dx = - 2 - a a 4
2a 2 a 3 4 2a
a
4 4
a 2a a a a = - 2 - + 2 - = 4 a 4 a 4 Then
Ix = 8h3 4a h3 a h3a 32 1 - - = 3 3 3 4 3 3 4
I x = 1.0484ah3 and kx = Ix = A 1.0484ah3 = 1.1644h 0.77324ah
k x = 1.164h PROBLEM 9.20 I y = dI y
dI y = x 2dA dA = ( y2 - y1 ) dx
From Problem 9.19
y1 = 2h -
h x a
y2 = 2h sin
2a
x
Now
x3 dI y = 2hx 2 sin x - h 2x2 - dx a 2a
Then Now using integration by parts
x x3 2a 2a I y = a dI y = h a 2 x 2 sin - 2x2 + dx 2a a
u = x2 du = 2 xdx Then
dv = sin v=-
2a
xdx
2a
cos
2a 2a
x
2 2 x sin 2a xdx = x - cos 2a x - - cos 2a x 2 xdx 2a
PROBLEM 9.20 CONTINUED
Now let u = x du = dx Then dv = cos v= 4a 2a
2a sin
xdx
2a
2a
x 2a
2 2 x sin 2a xdx = - x cos 2a x + x sin 2a x - sin 2a x dx 2a
2a 2 8a 2 4a 2 2a 1 3 1 4 I y = 2h - x cos x + 2 x sin x+ 2 cos x - x + x 2a 2a 2a 3 8a a
3 2a a3 a 4 16a3 ( 2a ) 1 8a 2 2 = 2h ( 2a ) - 3 - + ( 2a )4 - 2 a + - = 1.5231a3h 3 8a 3 8a 2a
Finally,
I y = 1.523a3h
and
2 ky =
Iy A
=
1.5231a3h = 1.4035a 2 0.77324 k y = 1.404a
PROBLEM 9.31
Determine the moment of inertia and the radius of gyration of the shaded area with respect to the x axis.
SOLUTION
First note that
A = A1 + A2 + A3
= (1.2 in.)( 0.3 in.) + ( 2.4 in.)( 0.4 in.) + ( 2.4 in.)( 0.3 in.) = ( 0.36 + 0.96 + 0.72 ) in 2 = 2.04 in 2
Now where
I x = ( I x )1 + ( I x )2 + ( I x )3
( I x )1 = ( I x )2 ( I x )3
= =
1 (1.2 in.)( 0.3 in.)3 + 0.36 in 2 (1.36 in.)2 = 0.6588 in 4 12 1 ( 0.4 in.)( 2.4 in.)3 = 0.4608 in 4 12 1 ( 2.4 in.)( 0.3 in.)3 + 0.72 in 2 (1.35 in.)2 = 1.3176 in 4 12
(
)
(
)
Then
I x = 0.6588 in 4 + 0.4608 in 4 + 1.3176 in 4 = 2.4372 in 4 or I x = 2.44 in 4
and
2 kx =
Ix 2.4372 in 4 = = 1.1947 in 2 A 2.04 in 2 or k x = 1.093 in.
PROBLEM 9.32
Determine the moment of inertia and the radius of gyration of the shaded area with respect to the x axis.
SOLUTION
First note that
A = A1 - A2 - A3 = (100 mm )(120 mm ) - ( 80 mm )( 40 mm ) - ( 80 mm )( 20 mm ) = 7200 mm 2 = (12 000 - 3200 - 1600 ) mm 2 = 7200 mm
Now where
I x = ( I x )1 - ( I x )2 - ( I x )3
( I x )1 = ( I x )2 ( I x )3
= =
1 (100 mm )(120 mm )3 = 14.4 106 mm4 12 1 (80 mm )( 40 mm )3 + 3200 mm2 ( 40 mm )2 = 5.5467 106 mm4 12
(
)
1 (80 mm )( 20 mm )2 + 1600 mm2 ( 30 mm )2 = 1.4933 106 mm4 12
(
)
Then
I x = (14.4 - 5.5467 - 1.4933) 106 mm 4 = 7.36 106 mm 4
or I x = 7.36 106 mm 4 and
2 kx =
Ix 7.36 106 = = 1022.2 mm 2 7200 A
or k x = 32.0 mm
PROBLEM 9.33
Determine the moment of inertia and the radius of gyration of the shaded area with respect to the y axis.
SOLUTION
First note that
A = A1 + A2 + A3
= (1.2 in.)( 0.3 in.) + ( 2.4 in.)( 0.4 in.) + ( 2.4 in.)( 0.3 in.) = ( 0.36 + 0.96 + 0.72 ) in 2 = 2.04 in 2
Now Where:
Iy = Iy
( )1 + ( I y )2 + ( I y )3
1 ( I y )1 = 12 ( 0.3 in.)(1.2 in.)3 = 0.0432 in 4 1 ( I y )2 = 12 ( 2.4 in.)( 0.4 in.)3 = 0.0128 in 4 1 ( I y )3 = 12 ( 0.3 in.)( 2.4 in.)3 = 0.3456 in 4
Then
I y = ( 0.0432 + 0.0128 + 0.3456 ) in 4 = 0.4016 in 4
or I y = 0.402 in 4 And
2 ky =
Iy A
=
0.4016 = 0.19686 in 2 2.04 in 2 or k y = 0.444 in.
PROBLEM 9.34
Determine the moment of inertia and the radius of gyration of the shaded area with respect to the y axis.
SOLUTION
First note that
A = A1 - A2 - A3 = (100 mm )(120 mm ) - ( 80 mm )( 40 mm ) - ( 80 mm )( 20 mm ) = 7200 mm 2 = (12 000 - 3200 - 1600)mm 2 = 7200 mm 2
Now where
Iy = Iy
( )1 - ( I y )2 - ( I y )3
1 ( I y )1 = 12 (120 mm )(100 mm )3 = 10 106 mm4 1 ( I y )2 = 12 ( 40 mm )(80 mm )3 = 1.7067 106 mm4 1 ( I y )3 = 12 ( 20 mm )(80 mm )3 = 0.8533 106 mm4
Then
I y = (10 - 1.7067 - 0.8533) 106 mm 4 = 7.44 106 mm 4
or I y = 7.44 106 mm 4 And
2 ky =
Iy A
=
7.44 106 mm 4 = 1033.33 mm 2 2 7200 mm
k = 32.14550 mm
or k y = 32.1 mm
PROBLEM 9.35
Determine the moments of inertia of the shaded area shown with respect to the x and y axes.
SOLUTION
Have
I x = ( I x )1 + ( I x )2 + ( I x )3
3 3 1 1 = ( 2a )( 4a ) + ( a )( 3a ) 3 3
2 2 2 4a 2 4a 4 + a - a + a 3a + 4 3 4 3 16
4 9 4 4 128 4 27 4 a + a + = - + +2+ a 3 3 16 9 4 9 161 37 4 a = 60.9316a 4 = + 3 or I x = 60.9a 4 Also Iy = Iy
( )1 + ( I y )2 + ( I y )3
3 3 1 1 = ( 4a )( 2a ) + ( 3a )( a ) + a 4 3 3 16
32 = + 1 + a 4 = 11.8630a 4 16 3
or I y = 11.86a 4
PROBLEM 9.36
Determine the moments of inertia of the shaded area shown with respect to the x and y axes.
SOLUTION
Have I x = ( I x )1 - ( I x )2 - ( I x )3
3 1 = ( 3a )( 2a ) - a 4 - a 4 12 8 8
= 2 - - a4 = 2 - a4 8 8 4
or I x = 1.215a 4 Also Iy = Iy
( )1 - ( I y )2 - ( I y )3
2 1 3 a = ( 2a )( 3a ) + ( 3a )( 2a ) 2 12 2 2 4a 4a - a 4 - a 2 + a 2 2a - 2 3 2 3 8 2 2 2 4a 2 4a 4 - a - a + a a - 2 3 2 3 8
8 8 8 9 3 = + a4 - - + 2 - + 3 9 2 2 8 9
4 a
8 4 8 4 11 4 - - + - + a = 10 - a 8 9 2 3 9 4 = 1.3606a 4 or I y = 1.361a 4
PROBLEM 9.37
For the 6-in2 shaded area shown, determine the distance d 2 and the moment of inertia with respect to the centroidal axis parallel to AA knowing that the moments of inertia with respect to AA and BB are 30 in4 and 58 in4, respectively, and that d1 = 1.25 in.
SOLUTION
Have and subtracting or Solve for d 2
2 d 2 = 1.252 + 4.6667 in 2 = 6.2292 in 2
I AA = I + Ad12
2 I BB = I + Ad 2 2 I AA - I BB = A d12 - d 2
(
)
( 30 - 58) in 4
2 2 = 6 in 2 (1.25 in.) - d 2
(
)
(
)
Then
d 2 = 2.4958 in.
or d 2 = 2.50 in. and
I = I AA - Ad12 = 30 in 4 - 6 in 2 (1.25 in.) = 20.625 in 4
2
(
)
or I = 20.6 in 4
PROBLEM 9.38
Determine for the shaded region the area and the moment of inertia with respect to the centroidal axis parallel to BB knowing that d1 = 1.25 in. and d2 = 0.75 in. and that the moments of inertia with respect to AA and BB are 20 in4 and 15 in4, respectively.
SOLUTION
Have and subtracting
I AA = I + Ad12
2 I BB = I + Ad 2 2 I AA - I BB = A d12 - d 2
(
)
2 2 20 in 4 - 15 in 4 = A (1.25 ) - ( 0.75 ) in 2
5 in 4 = A [1.5625 - 0.5625] in 2 or A = 5 in 2 and
I = I AA - Ad 2 = 20 in 4 - 5 in 2 (1.25 in.) = 12.1875 in 4
2
(
)
or I = 12.19 in 4
PROBLEM 9.39
The centroidal polar moment of inertia J C of the 15.5 103 mm 2 shaded region is 250 106 mm 4 . Determine the polar moments of inertia J B and J D of the shaded region knowing that J D = 2 J B and d = 100 mm.
SOLUTION
Have and Now Then Now Substituting or a2 =
2 J B = J C + AdCB 2 J D = J C + AdCD
J D = 2J B
2 2 J C + AdCD = 2 J C + AdCB 2 dCB = a 2 + d 2
(
)
2
and
2 dCD = ( 2a ) + d 2
A 4a 2 + d 2 = J C + 2 A a 2 + d 2 1 JC + d2 2 A
(
)
(
)
=
1 250 106 mm 4 2 + (100 mm ) = 13064.5 mm 2 3 2 2 15.5 10 mm
or Then
a = 114.300 mm
2 2 J B = 250 106 mm 4 + 15.5 103 mm 2 (114.300 mm ) + (100 mm )
(
)
= 250 106 + 357.5 106 mm 4 = 607.5 106 mm 4 or J B = 608 106 mm 4 And
2 2 J D = 250 106 mm 4 + 15.5 103 mm 2 ( 228.60 mm ) + (100 mm )
(
)
(
)
= 250 106 + 964.99 10 mm 4 = 1214.99 106 mm 4 or J D = 1215 106 mm 4
(
)
PROBLEM 9.40
Determine the centroidal polar moment of inertia J C of the 10 103 mm 2 shaded area knowing that the polar moments of inertia of the area with respect to points A, B, and D are J A = 45 106 mm 4 , J B = 130 106 mm 4 , and J D = 252 106 mm 4 , respectively.
SOLUTION
Have Then Have Then Have Then Then Equation (3) - Equation(2): and Equation(4) - 3[ Equation(1)]: or JC = J A - 1 ( JD - JB ) 3 1 252 106 - 130 106 mm 4 = 4.3333 106 mm 4 3 or J C = 4.33 106 mm 4 Note and a = 63.77 mm d = 92.195 mm
2 J A = J C + AdCA
where
2 dCA = a 2
J A = J C + Aa 2
2 J B = J C + AdCB
(1)
2 dCB = a 2 + d 2
where
J B = JC + A a2 + d 2
2 J D = J C + AdCD
(
) )
= -3 J C
(2)
where
2 dCD = 4a 2 + d 2
J D = J C + A 4a 2 + d 2 J D - J B = 3 Aa 2
(
(3) (4)
( J D - J B ) - 3J A
= 45 106 mm 4 -
(
)
PROBLEM 9.41
Determine the moments of inertia I x and I y of the area shown with respect to centroidal axes respectively parallel and perpendicular to side AB.
SOLUTION
First calculate the centroid C of the area From symmetry To compute X use the equation
2 ( 3 5.4 ) in ( 2.7 in.) -
Y = 0.6 in. + 0.9 in. = 1.5 in.
XA = Ax
or
X =
1 2 2 (1.8 3.6 ) in ( 2.3 in.) 1 2 2 ( 3 5.4 ) in - (1.8 3.6 ) in 2
= 2.8 in. The moment of inertia of the composite area is obtained by subtracting the moment of inertia of the triangle from the moment of inertia of the rectangle I x = ( I x )1 - ( I x )2 where
( I x )1 =
( I x )2
1 ( 5.4 in.)( 3 in.)3 = 12.15 in 4 12
and
Then
3 1 = 2 ( 3.6 in.)( 0.9 in.) = 0.4374 in 4 12
I x = (12.15 - 0.4374 ) in 4 = 11.7126 in 4
or I x = 11.71 in 4 Similarly, where
Iy = Iy
( )1 - ( I y )2
1 ( I y )1 = 12 ( 3 in.)( 5.4 in.)3 + (3 5.4) in 2 ( 2.8 in. - 2.7 in.)2
= 39.582 in 4
PROBLEM 9.41 CONTINUED
and 1 ( I y )2 = 36 (1.8 in.)( 3.6 in.)3 + 1 (1.8)( 3.6) in 2 ( 2.8 in. - 2.3 in.)2 2
= 3.1428 in 4
Then
I y = ( 39.582 - 3.1428 ) in 4 = 36.4392 in 4
or I y = 36.4 in 4
PROBLEM 9.42
Determine the moments of inertia I x and I y of the area shown with respect to centroidal axes respectively parallel and perpendicular to side AB.
SOLUTION
By symmetry Have Where
AY = yA
X = 0.9 in.
A = (1.8 in.)(1.4 in.) +
1 (1.8 in.)( 2.1 in.) 2
= ( 2.52 + 1.89 ) in 2 = 4.41 in 2
Then
( 4.41 in ) Y
2
= ( 0.7 in.) 2.52 in 2 + ( 2.1 in.) 1.89 in 2 = 5.733 in 3
(
)
(
)
or Now where
Y = 1.3 in. I x = ( I x )1 + ( I x )2
( I x )1 =
1 (1.8 in.)(1.4 in.)3 12
+ 2.52 in 2 (1.3 in. - 0.7 in.) = 1.3188 in 4
2
(
)
And
( I x )2
=
1 (1.8 in.)( 2.1 in.)3 36 + 1.89 in 2 ( 2.1 in. - 1.3 in.) = 1.67265 in 4
2
(
)
Then
I x = (1.3188 + 1.67265 ) in 4 = 2.99145 in 4 or I x = 2.99 in 4
Also where
Iy = Iy
( )1 + ( I y )2
1 ( I y )1 = 12 (1.4 in.)(1.8 in.)3 = 0.6804 in 4
PROBLEM 9.42 CONTINUED
and 1 ( I y )2 = 2 36 ( 2.1 in.)( 0.9 in.)3
2 1 + 1.89 in 2 ( 0.3 in.) = 0.25515 in 4 2
Then
I y = ( 0.6804 + 0.25515 ) in 4
or I y = 0.936 in 4
PROBLEM 9.43
Determine the moments of inertia I x and I y of the area shown with respect to centroidal axes respectively parallel and perpendicular to side AB.
SOLUTION
A = A1 - A2 = (100 mm )(160 mm ) - ( 40 mm )(100 mm ) = (16 000 - 4000 ) mm 2 = 12 000 mm 2
First locate the centroid: Have or or And or or
2
AX = Ax
(12 000 mm ) X = (16 000)(50) - ( 4000)( 38) mm
X =
648 000 mm3 = 54 mm 12 000 mm 2
AY = Ay
3
= 648 000 mm3
(12 000 mm ) Y
2
= (16 000 )( 86 ) - ( 4000 )( 86 ) mm3 = 936 000 mm3 Y =
936 000 mm3 = 78 mm 12 000 mm 2
PROBLEM 9.43 CONTINUED
Now where
I x = ( I x )1 - ( I x )2
( I x )1 =
1 ( 40 mm )(100 mm )3 + 16 000 mm2 (80 mm - 78 mm )2 12
(
)
= 34.197 106 mm 4
( I x )2
=
1 ( 40 mm )(100 mm )3 + 4000 mm2 (80 mm - 78 mm )2 12
(
)
= 3.5893 106 mm 4
Then
I x = ( 34.197 - 3.5893) 106 mm 4
or I x = 30.6 106 mm 4
Also where
Iy = Iy
( )1 - ( I y )2
1 ( I y )1 = 12 (160 mm )(100 mm )3 + (16 000 mm2 ) ( 54 mm - 50 mm )2 = 13.589 106 mm4 1 ( I y )2 = 12 (100 mm )( 40 mm )3 + ( 4000 mm2 ) ( 54 mm - 38 mm )2 = 1.5573 106 mm4
Then
I y = (13.589 - 1.5573) 106 mm 4
or I y = 12.03 106 mm 4
PROBLEM 9.44
Determine the moments of inertia I x and I y of the area shown with respect to centroidal axes respectively parallel and perpendicular to side AB.
SOLUTION
First locate centroid
x1 = 160 mm
y1 = 60 mm
A1 = 320 mm 120 mm = 38 400 mm 2 x2 = -30 mm y2 = 40 mm
A2 = 60 mm 80 mm = 4800 mm 2 x3 = 280 mm y3 = -105 mm
A3 = 80 mm 210 mm = 16 800 mm 2 Then X = xA A
160 ( 38 400 ) - 30 ( 4800 ) + 280 (16 800 ) mm3 = ( 38 400 + 4800 + 16 800 ) mm2 = 178.4 mm And Y = yA A
60 ( 38 400 ) + 40 ( 4800 ) - 105 (16 800 ) mm3 = ( 38 400 + 4800 + 16 800 ) mm 2 = 12.20 mm
PROBLEM 9.44 CONTINUED
Then I x = ( I x )1 + ( I x )2 + ( I x )3
3 2 1 = ( 320 mm )(120 mm ) + 38 400 mm 2 ( 60 mm - 12.2 mm ) 12 3 2 1 + ( 60 mm )( 80 mm ) + 4800 mm 2 ( 40 mm - 12.2 mm ) 12 3 2 1 + ( 80 mm )( 210 mm ) + 16 800 mm 2 (105 mm + 12.2 mm ) 12
(
)
(
)
(
)
= ( 46.080 + 87.7379 ) + ( 2.5600 + 3.7096 ) + ( 61.7400 + 230.7621) 106 mm 4
= 432.5896 106 mm 4 or I x = 433 106 mm 4 And Iy = Iy
( )1 + ( I y )2 + ( I y )3
3 2 1 = (120 mm )( 320 mm ) + 38 400 mm 2 (178.4 mm - 160 mm ) 12 3 2 1 + ( 80 mm )( 60 mm ) + 4800 mm 2 ( 30 mm + 178.4 mm ) 12 3 2 1 + ( 210 mm )( 80 mm ) + 16 800 mm 2 ( 280 mm - 178.4 mm ) 12
(
)
(
)
(
)
= ( 327.6800 + 13.0007 ) + (1.4400 + 208.4667 ) + ( 8.9600 + 173.4190 ) 106 mm 4
= 732.9664 106 mm 4 or I y = 733 106 mm 4
PROBLEM 9.45
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 area A, in 2 1 2
y , in.
yA, in 3
2 -
( 2.7 )(1.8) = 7.6341
2
0.76394 0.38197
5.8319 0.4860 5.3460
( 0.9 )2
= -1.2723
6.3618
Then (a)
J O = ( J O )1 - ( J O )2 =
Y A = yA:
Y =
5.3460 in 2 = 0.84033 in. 6.3618 in 2
8
( 2.7 in.)(1.8 in.) ( 2.7 )2 + (1.8)2 in 2 - ( 0.9 in.)4
4 or J O = 19.58 in 4
= 19.5814 in 4
(b) or
JO = JC + A ( y )
2 2
J C = 19.5814 in 4 - 6.3618 in 2 ( 0.84033 in.) = 15.0890 in 4 J C = 15.09 in 4
(
)
PROBLEM 9.46
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 Symmetry implies
x1 = 0 Y =0 A1 = ( 20 in.)(12 in.) = ( 240 ) in 2 x2 =
4 ( 9 in.) = (12 ) in. 3
A2 = -
2
( 9 in.)2
= - ( 40.5 ) in 2
x3 = - A3 = -
xA X = = A
4 ( 6 in.) 8 = - in. 3 2
( 6 in.)2
= - (18 ) in 2
Then
( 0 ) ( 240 in 2 ) + (12
in.) -40.5 in 2 -
(
)
8 in.
240 in 2 - 40.5 in 2 - 18 in 2
( -18 in )
2
=
-486 in 3 + 144 in 3 -342 in 3 = = -0.59979 in. 181.5 in 2 181.5 in 2
PROBLEM 9.46 CONTINUED
(a) Have
J O = ( J O )1 - ( J O )2 - ( J O )3 =
4
( 20 in.)(12 in.) ( 20 in.)2 + (12 in.)2 - ( 9 in.)4 - ( 6 in.)4
4 4
= ( 32640 - 1640.25 - 324.00 ) in 4 = 96,371 in 4
or J O = 96.4 103 in 4 (b) Have Then
J O = J C + Ax 2 J C = 96,371 in 4 - 181.5 in 2 ( -0.59979 in.)
(
)
2
= 96,371 in 4 - 204.5629 in 4 = 96,166.4379 in 4
or J C = 96.2 103 in 4
PROBLEM 9.47
Determine the polar moment of inertia of the area shown with respect to (a) point O, (b) the centroid of the area.
SOLUTION
A, mm 2
y , mm
yA, mm3
1 2
2
-
(120 )2
= 22 619.5
50.9296 30
1.1520 106
-0.324 106
1 ( 240 )( 90 ) = -10 800 2 11 819.5
0.828 106
Now (a) where and
Y =
AY 0.828 106 mm3 = = 70.054 mm A 11819.5 mm 2 J O = ( J O )1 - ( J O )2
( J O )1 =
(120 mm ) = 162.86 10 4
4
6
mm 4
( J O )2 = ( I x ) 2 + ( I y ) 2
=
1 1 ( 240 mm )( 90 mm )3 + 2 ( 90 mm )(120 mm )3 12 12
= 40.5 106 mm 4
Then
J O = (162.86 - 40.5 ) 106 mm 4 = 122.36 106 mm 4
or J O = 122.4 106 mm 4
PROBLEM 9.47 CONTINUED
(b) or
J O = J C + Ay 2 J C = 122.36 106 mm 4 - 11 819.5 mm 2 ( 70.054 mm ) = (122.36 - 58.005 )106 mm 4
(
)
2
or J C = 64.4 106 mm 4
PROBLEM 9.48
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
x1 = 100 mm y1 = -28.75 mm
A1 = ( 400 mm )( 57.5 mm ) = 23 000 mm 2 x2 = 100 mm A2 = y2 = 80 mm
1 ( 300 mm )( 240 mm ) = 36 000 mm2 2
y3 = 40 mm
x3 = 50 mm A3 = -
1 (150 mm )(120 mm ) = -9000 mm2 2
Then
2 2 2 xA (100 mm ) 23 000 mm + (100 mm ) 36 000 mm + ( 50 mm ) -9000 mm = X = A 23 000 mm 2 + 36 000 mm 2 - 9000 mm 2
(
)
(
)
(
)
=
( 2.3 + 3.6 - 0.45) 106 mm3
50 103 mm 2
= 109.0 mm
And
2 2 2 yA ( -28.75 mm ) 23 000 mm + ( 80 mm ) 36 000 mm + ( 40 mm ) -9000 mm Y = = A 50 103 mm 2
(
)
(
)
(
)
=
( -661.25 + 2880 - 360 ) 103 mm3
50 103 mm 2
= 37.175 mm
PROBLEM 9.48 CONTINUED
(a) Now where 1 3
JO = I x + I y I x = ( I x )1 + ( I x )2 - ( I x )3
( I x )1 = ( 400 mm )( 57.5 mm )3 ( I x )2
= =
= 25.3479 106 mm 4
1 ( 300 mm )( 240 mm )3 = 345.6000 106 mm4 12 1 (150 mm )(120 mm )3 = 21.6000 106 mm4 12
( I x )3
Then
I x = ( 25.3479 + 345.6000 - 21.6000 ) 106 mm 4
= 349.348 106 mm 4 Also where
Iy = Iy
( )1 + ( I y )2 - ( I y )3
1 ( I y )1 = 12 ( 57.5 mm )( 400 mm )3 + ( 23 000 mm2 ) (100 mm )2 = ( 306.6667 + 230.0000 ) 106 mm 4 = 536.6667 106 mm 4 1 ( I y )2 = 12 ( 240 mm )( 300 mm )3 = 540.0000 106 mm4 1 ( I y )3 = 12 (150 mm )(120 mm )3 = 33.7500 106 mm4
Then Finally,
I y = ( 536.6667 + 540 - 33.75 ) 106 mm 4 = 1042.917 106 mm 4
J O = ( 349.348 + 1042.917 ) 106 mm 4 = 1392.265 106 mm 4
or J O = 1392 106 mm 4 (b) Have Then
J O = J C + Ad 2
where
d2 = X + Y
2
2
2 2 J C = 1392.265 106 mm 4 - 50 103 mm 2 (109.0 mm ) + ( 37.175 mm )
(
)
= (1392.265 - 594.050 - 69.099 ) 106 mm 4 = 729.1660 106 mm 4 or J C = 729 106 mm 4
PROBLEM 9.49
Two 1-in. steel plates are welded to a rolled S section as shown. Determine the moments of inertia and the radii of gyration of the section with respect to the centroidal x and y axes.
SOLUTION
S-section
A = 14.7 in 2 I x = 305 in 4
I y = 15.7 in 4
A = AS + 2 Aplate
= 14.7 in 2 + 2 ( 8 in.)(1 in.) = 30.7 in 2
Ix = Ix
( )S + 2 ( I x )plate
( 8 in.)(1 in.)3 2 = 305 in 4 + 2 + ( 8 in.)(1 in.)( 6.5 in.) 12
= ( 305 + 677.33) in 4 = 982.33 in 4 or and or Also
Iy = Iy
I x = 9.82 in 4 k x2 = Ix 982.33 in 4 = = 31.998 in 2 4 A 30.7 in k x = 5.66 in.
( )S
+ 2 Iy
( )plate
(1 in.)( 8 in.)3 = 15.7 in 4 + 2 12
= (15.7 + 85.333) in 4 = 101.03 in 4 or I y = 101.0 in 4 and k y2 = Iy A = 101.03 in 4 = 3.29098 in 2 30.7 in 2 or k y = 1.814 in.
PROBLEM 9.50
To form a reinforced box section, two rolled W sections and two plates are welded together. Determine the moments of inertia and the radii of gyration of the combined section with respect to the centroidal axes shown.
SOLUTION
W-section A = 7.08 in 2 I x = 18.3 in 4
I y = 82.8 in 4
A = 2 AW + 2 Aplate = 2 7.08 in 2 + ( 7.93 in.)( 0.3 in.) = 18.918 in 2 Now
Ix = 2 Ix
( )W
+ 2 Ix
( )plate
2 6.495 in. = 2 18.3 in 4 + 7.08 in 2 2
(
)
( 7.93 in.)( 0.3 in.)3 2 + 2 + ( 7.93 in.)( 0.3 in.) ( 6.495 in. + 0.15 in.) 12
= 2 92.967 in 4 + 2 105.07 in 4 = 396.07 in 4 or and or k x2 = Ix 396.07 in 4 = = 20.936 in 2 2 A 18.918 in
k x = 4.58 in.
I x = 396 in 4
PROBLEM 9.50 CONTINUED
Also
Iy = 2 Iy
( )W + 2 ( I y )plate
= 2 82.8 in
(
4
)
( 0.3 in.)( 7.93 in.)3 = (165.60 + 24.9339 ) in 4 + 2 12
= 190.53 in 4 or I y = 190.5 in 4 and k y2 = Iy A = 190.53 in 4 = 10.072 18.918 in 2 or k x = 3.17 in.
PROBLEM 9.51
Two C250 30 channels are welded to a 250 52 rolled S section as shown. Determine the moments of inertia and the radii of gyration of the combined section with respect to its centroidal x and y axes.
SOLUTION
Use Figure 9.13B (textbook) properties of rolled-steel shapes (SI units) to get the values for C250 and S250 shapes
S250 52 section:
A = 6670 mm 2 I x = 61.2 106 mm 4
I y = 3.59 106 mm 4
C250 30 section:
A = 3780 mm 2 I x = 32.6 106 mm 4
I y = 1.14 106 mm 4
How, for the combined section:
A = AS + 2 AC
= 6670 + 2 ( 3780 ) mm 2 = 14 230 mm 2
I x = ( I x ) S + 2 ( I x )C = 61.2 106 + 2 32.6 106 mm 4
(
)
Iy = Iy
where d is the distance from the centroid of the C section to the centroid C of the combined section Now
2 126 + 69 - 15.3 mm 2 I y = 3.59 106 mm 4 + 2 1.14 106 mm 4 + 3780 mm 2 2
( )S + 2 ( I y )C + AC d 2
or I x = 126.4 106 mm 4
(
) (
)
= ( 3.59 + 2.28 + 102.9588 ) 106 mm 4 or I y = 108.8 106 mm 4 Also kx =
=
Ix A
126.4 106 mm 4 14 230 mm 2 or k x = 94.2 mm
PROBLEM 9.51 CONTINUED
And
ky = =
Iy A
108.8 106 mm 4 14 230 mm 2 or k y = 87.5 mm
PROBLEM 9.52
Two channels are welded to a d 300-mm steel plate as shown. Determine the width d for which the ratio I x / I y of the centroidal moments of inertia of the section is 16.
SOLUTION
Channel:
A = 3780 mm 2 I x = 32.6 106 mm 4
I y = 1.14 106 mm 4
Now
Ix = 2 Ix
( )C + ( I x )plate
= 2 32.6 106 mm 4 +
(
)
d ( 300 mm )3 12
= 65.2 106 + 2.25d 106 mm 4 And
Iy = 2 Iy
(
)
( )channel + ( I y )plate
2 d = 2 1.14 106 mm 4 + 3780 mm 2 + 15.3 mm 2
(
)
( 300 mm )d 3 + 12
= 2.28 106 + 1890d + 115.668 103 d + 1.7697 106 + 25d 3 mm 4 = 25d 3 + 1890d 2 + 115.67 103 d + 4.0497 106 mm 4 Given Then 65.2 106 + 2.25d 106 = 16 25d 3 + 1890d 2 + 115.67 103 d + 4.0497 106 or 25d 3 + 1890d 2 - 24.955d - 25300 = 0 d = 12.2935 mm or d = 12.29 mm I x = 16I y
(
)
(
)
(
)
Solving numerically
PROBLEM 9.53
Two L3 3 1 -in. angles are welded to a C10 20 channel. Determine 4 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:
A = 1.44 in 2
I x = I y = 1.24 in 4
Channel:
A = 5.88 in 2
I x = 2.81 in 4 I y = 78.9 in 4
Locate the centroid X =0
Y = 2 1.44 in 2 ( 0.842 in.) + 5.88 in 2 ( -0.606 in.) Ay = 2 A 2 1.44 in + 5.88 in 2
(
)
(
)
(
)
=
Now
( 2.42496 - 3.5638) in 3
8.765 in 4
= -0.12995 in.
( I x ) = 2 ( I x )L + ( I x )C = 2 1.24 in 4 + (1.44 in 2 ) ( 0.842 in. + 0.12995 in.)2
2 + 2.81 in 4 + 5.88 in 2 ( 0.606 in. - 0.12995 in.)
(
)
= 2 ( 2.6003) in 4 + 4.1425 in 4 = 9.3431 in 4
or I x = 9.34 in 4 Also
( I y ) = 2 ( I y )L + ( I y )C = 2 2.14 in 4 + 1.44 in 2 ( 5 in. - 0.842 in.)2 + 7.89 in 4
= 2 ( 26.136 ) in 4 + 78.9 in 4 = 131.17 in 4
or I y = 131.2 in 4
PROBLEM 9.54
To form an unsymmetrical girder, two L3 3 1 -in. angles and two 4 L6 4 1 -in. angles are welded to a 0.8-in. steel plate as shown. 2 Determine the moments of inertia of the combined section with respect to its centroidal x and y axes.
SOLUTION
Angle:
L3 3
A = 1.44 in 2
1 : 4
I x = I y = 1.24 in 4
L6 4
1 : 2
A = 4.75 in 2
I x = 6.27 in 4
I y = 17.4 in 4
Plate:
A = ( 27 in.)( 0.8 in.) = 21.6 in 2 Ix = Iy =
1 ( 0.8 in.)( 27 in.)3 = 1312.2 in 4 12 1 ( 27 in.)( 0.8 in.)3 = 1.152 in 4 12
PROBLEM 9.54 CONTINUED
Centroid:
X =0 Y = Ay A
or
2 2 1.44 in 2 ( 27 in. - 0.84 in.) + 2 4.75 in 2 ( 0.987 in.) + 21.6 in 2 (13.5 in.) Y = 2 1.44 in 2 + 4.75 in 2 + 21.6 in 2
(
)
(
(
) )
(
)
=
Now
376.31 in 3 = 11.0745 in. 33.98 in 2
I x = 2 ( I x )1 + 2 ( I x )3 + ( I x )2
2 2 = 2 6.25 + 4.75 (11.075 - 0.987 ) in 4 + 2 1.24 + 1.44 ( 27 - 0.842 - 11.075) in 4 2 + 1312.2 + 21.6 (13.5 - 11.075) in 4
= 2 ( 489.67 ) in 4 + 2 ( 328.84 ) in 4 + 1439.22 in 4 = 3076.24 in 4
or I x = 3076 in 4 Also
( I y ) = 2 ( I y )1 + 2 ( I y )3 + ( I y )2
2 2 = 2 17.4 + 4.75 ( 0.4 + 1.99 ) in 4 + 2 1.24 + 1.44 ( 0.4 + 0.842 ) in 4 + 1.152 in 4
= 2 ( 44.532 ) in 4 + 2 ( 3.4613) in 4 + 1.152 in 4 = 97.139 in 4 or I y = 97.1 in 4
PROBLEM 9.55
Two L127 76 12.7-mm angles are welded to a 10-mm steel plate. Determine the distance b and the centroidal moments of inertia I x and I y of the combined section knowing that I y = 3I x .
SOLUTION
Angle:
A = 2420 mm 2 I x = 3.93 106 mm 4
I y = 1.074 106 mm 4
Plate:
A = ( 200 mm )(10 mm ) = 2000 mm 2 Ix = Iy =
1 ( 200 mm )(10 mm )3 = 0.01667 106 mm4 12 1 (10 mm )( 200 mm )3 = 6.6667 106 mm 4 12
Centroid
X =0 Y =
Ay A = 205.380 mm3 6840 mm 2
or
Y =
2 2420 mm 2 ( 44.5 mm ) + 2000 mm 2 ( -5 mm ) 2 ( 2420 ) + 2000 mm 2
(
)
= 30.026 mm Now
I x = 2 ( I x )angle + ( I x )plate
2 = 2 3.93 106 + ( 2420 )( 44.5 - 30.026 ) mm 4 2 + 0.01667 106 + ( 2000 )( 30.026 + 5 ) mm 4
= 2 4.43698 106 mm 4 + 2.4703 106 mm 4 = 11.344 106 mm 4 or I x = 11.34 106 mm 4
(
)
PROBLEM 9.55 CONTINUED
Also Where
Iy = 2 Iy
( )angle + ( I y )plate
( I y )angle = 1.074 106 mm4 + ( 2420 mm2 ) ( b - 19.05 mm )2
= 1.074 106 + ( 2420 ) b 2 - 38.1b + 362.9 mm 4 = 2420b 2 - 92202b + 1.9522 106 mm 4
(
)
and Now Then or
( I y )plate = 6.6667 106 mm4
I y = 3 Ix
( )
2 2420b 2 - 92202b + 1.9522 106 mm 4 + 6.6667 106 mm 4 = 3 11.34 106 mm 4 2420b 2 - 9.2202b + 1.9522 106 - 13.6767 106 = 0
(
)
b 2 - 38.1b - 4844.8 = 0 b = 91.2144 mm
or b = 91.2 mm
PROBLEM 9.56
A channel and an angle are welded to an a 20-mm steel plate. Knowing that the centroidal y axis is located as shown, determine (a) the width a, (b) the moments of inertia with respect to the centroidal x and y axes.
SOLUTION
(a) Using Figure 9.13B From the geometry of L152 152 19, C150 15.6, plate a 20 mm and how they are welded
x A = 44.9 mm AA = 5420 mm 2 xC = -12.5 mm AC = 1980 mm 2
a xP = - - 152 mm 2 From the condition AP = ( 20a ) mm 2 xA =0 A a - 152 mm 20a mm 2 = 0 2
X =
( 44.9 mm ) ( 5420 mm2 ) - (12.5 mm ) (1980 mm2 ) -
or a 2 - 304a - 21860.8 = 0
(
)
a = 364.05 mm or a = 364 mm
And
AP = ( 20 mm )( 364 mm ) = 7280 mm 2
PROBLEM 9.56 CONTINUED
(b) Locate the centroid Y = Ay A
=
(5420 mm ) ( 44.9 mm ) + (1980 mm ) 152 mm + ( 7280 mm ) ( -10 mm ) 2
2 2 2
( 5420 + 1980 + 7280 ) mm2
= 21.867 mm Now I x = ( I x ) A + ( I x )C + ( I x ) P
2 = 11.6 106 mm 4 + 5420 mm 2 ( 44.9 mm - 21.867 mm ) 2 + 6.21 106 mm 4 + 1980 mm 2 ( 76 mm - 21.867 mm ) 3 2 1 + ( 364.05 mm )( 20 mm ) + 7281 mm 2 (10 mm + 21.867 mm ) 12
(
)
(
)
(
)
= (11.6 + 2.8754 ) + ( 6.21 + 5.8022 ) + ( 0.2427 + 7.3939 ) 106 mm 4
= (14.4754 + 12.0122 + 7.6366 ) 106 mm 4 = 34.1242 106 mm 4 or I x = 34.1 106 mm 4 And Iy = Iy
( ) A + ( I y )C + ( I y )P
2 = 11.6 106 mm 4 + 5420 mm 2 ( 44.9 mm ) 2 + 0.347 106 mm 4 + 1980 mm 2 (12.5 mm ) 2 1 3 364.05 mm + ( 20 mm )( 364.05 mm ) + 7821 mm 2 - 152 mm 2 12
(
)
(
)
(
)
= (11.6 + 10.9268 ) 106 mm 4 + ( 0.347 + 0.3094 ) 106 mm 4 + ( 80.4140 + 6.5638 ) 106 mm 4 = ( 22.5268 + 0.6564 + 86.9778 ) 106 mm 4 = 110.161 10-6 mm 4 or I y = 110.2 106 mm 4
PROBLEM 9.57
The panel shown forms the end of a trough which is filled with water to the line AA . Referring to Sec. 9.2, determine the depth of the point of application of the resultant of the hydrostatic forces acting on the panel (the center of pressure).
SOLUTION
Using the equation developed on page 491 of the text have yP = I AA yA
For a quarter circle y =
I AA =
16
r4
and
4r , A = r2 3 4
Then
yP =
16 4r 2 r 3 4
r4
or yP =
3 r 16
PROBLEM 9.58
The panel shown forms the end of a trough which is filled with water to the line AA . Referring to Sec. 9.2, determine the depth of the point of application of the resultant of the hydrostatic forces acting on the panel (the center of pressure).
SOLUTION
Using the equation developed on page 491 of the text have yP = I AA yA
For a semiellipse y =
I AA =
8
ab3
4b , A = ab 3 2
Then
ab3 8 yP = 4b ab 3 2
or yP =
3 b 16
PROBLEM 9.59
The panel shown forms the end of a trough which is filled with water to the line AA. Referring to Sec. 9.2, determine the depth of the point of application of the resultant of the hydrostatic forces acting on the panel (the center of pressure).
SOLUTION
Using the equation developed on page 491 of the text Have Now yP =
YA = yA
I AA yA
=
h 4 1 ( 2b h ) + h 2b h 2 3 2 7 2 bh 3
= And where
I AA = ( I AA )1 + ( I AA )2
( I AA )1 = ( 2b )( h )3
1 3
=
2 3 bh 3
2
( I AA )2 = I x + Ad 2 =
= 11 3 bh 6 I AA =
1 1 4 ( 2b )( h )3 + 2b h h 36 2 3
Then
2 3 11 3 5 3 bh + bh = bh 3 6 2
5 3 bh yP = 2 7 2 bh 3
Finally,
or yP =
15 h 14
PROBLEM 9.60
The panel shown forms the end of a trough which is filled with water to the line AA. Referring to Sec. 9.2, determine the depth of the point of application of the resultant of the hydrostatic forces acting on the panel (the center of pressure).
SOLUTION
Using the equation developed on page 491 of the text Have where
yP = I AA = ( I AA )1 + ( I AA )2
I AA yA
2 2 4r 2 4r 3 1 = ( 2r )( r ) + r 4 - r 2 + r r + 3 8 2 3 2 3
=
2 4 8 4 9 4 5 4 r + - + + + r = 2 + r 3 2 3 8 8 8 9 4r 2 r YA = yA = ( 2r r ) + r + r 3 2 2
And
2 5 = 1 + + r 3 = + r 3 2 3 3 2
5 4 2 + r 8 = 1.2242r yP = 5 3 + r 3 2
Then
or yP = 1.224r
PROBLEM 9.61
The cover for a 10 22-in. access hole in an oil storage tank is attached to the outside of the tank with four bolts as shown. Knowing that the specific weight of the oil is 57.4 lb/ft3 and that the center of the cover is located 10 ft below the surface of the oil, determine the additional force on each bolt because of the pressure of the oil.
SOLUTION
Using the equation developed on page 491 of the text have yP = Then I AA yA R = yA 1 ft 2 144 in 2
R = 57.4 lb/ft 3 10 ft ( 22 10 ) in 2 = 876.94 lb
and
I AA =
1 ( 22 in.)(10 in.)3 + ( 22 in.)(10 in.) (120 in.)2 12
= 3.169833 106 in 4 = 152.8662 ft 4 and yA = 10 ft ( 22 10 ) in 2 yp = 1 ft 2 = 15.27778 ft 3 2 144 in
Then Now symmetry implies
152.8662 ft 4 = 10.00579 ft 15.27778 ft 2
FA = FB
and
FC = FD
Equilibrium M CD = 0:
(1 ft )( 2FAa ) - ( 0.5 - 0.00579 ) ft 876.94 lb = 0
FA = 216.70 lb or FA = FB = 217 lb
Also or
Fx = 0: -2 ( 216.70 ) + 876.94 - 2FC = 0 or FC = FD = 222 lb
PROBLEM 9.62
A vertical trapezoidal gate that is used as an automatic valve is held shut by two springs attached to hinges located along edge AB. Knowing that each spring exerts a couple of magnitude 8.50 kN m , determine the depth d of water for which the gate will open.
SOLUTION
d = ( h + 1.58 ) m From page 491 Now yp = I ss yA
= g
R = yA
yA = yA
1 1 = ( h + 0.34 m ) 2.4 m 1.02 m + ( h + 0.68 m ) 1.68 m 1.02 m 2 2
= ( 2.0808h + 0.99878 ) m3 Also,
R = 103 kg/m3 9.81 m/s 2 ( 2.0808h + 0.99878 ) m3 = 20 413 ( h + 0.480 ) N 1 3 2 1 And I ss = ( I ss )1 + ( I ss )2 = ( 2.4 m )(1.02 m ) + ( 2.4 m )(1.02 m ) ( h + 0.34 ) m 2 36 2 1 3 2 1 + (1.68 m )(1.08 m ) + (1.68 m )(1.02 m ) ( h + 0.68 ) m 2 2 36
2 2 = 0.07075 + 1.224 ( h + 0.34 ) + 0.04952 + 0.8568 ( h + 0.68 ) m 4
(
)(
)
= 0.12027 + 1.224 h 2 + 0.68h + 0.1156 + 0.8568 h 2 + 1.36h + 0.4624 m 4 = 2.0808h 2 + 1.9976h + 0.65795 m 4
(
)
(
)
(
)
PROBLEM 9.62 CONTINUED
Then
yp
( 2.0808h =
=
2
+ 1.9976h + 0.65795 m 4
3
( 2.0808h + 0.99878) m
)
h 2 + 0.960h + 0.3162 m h + 0.480
For gate to open
M AB = 0: M open - y p - h R = 0
(
)
h 2 + 0.960h + 0.3162 - h m ( 20 413)( h + 0.48 ) N = 0 2 ( 8500 N m ) - h + 0.480 or or 17 000 = 20 413 h 2 + 0.96h + 0.3162 - h 2 - 0.480h
(
)
0.48h - 0.5166 = 0 h = 1.0763 m
Now
d = h + 1.58 m = (1.0763 + 1.58 ) m = 2.6563 m or d = 2.66 m
PROBLEM 9.63
Determine the x coordinate of the centroid of the volume shown. (Hint: The height y of the volume is proportional to the x coordinate; consider an analogy between this height and the water pressure on a submerged surface.)
SOLUTION
Have where Now
x dV x = EL dV dV = ydA y = and xEL = x 60 1 x= x 300 5
Then
1 x = 1 5 x dA
x 5 x dA
( I z )A x 2dA = = xdA ( xA) A
where ( I z ) A is the moment of inertia of the area with respect to the z axis, and x is analogous to y p Now
( Iz )A
=
1 1 ( 240 mm )( 300 mm )3 + ( 240 mm )( 300 mm ) ( 200 mm )2 36 2
= 1.620 109 mm 4 and 1 xA = ( 200 mm ) ( 240 mm )( 300 mm ) = 7.20 106 mm3 2 x = 1.620 109 mm 4 7.20 106 mm3 or x = 225 mm
Then
PROBLEM 9.64
Determine the x coordinate of the centroid of the volume shown; this volume was obtained by intersecting an elliptic cylinder with an oblique plane. (Hint: The height y of the volume is proportional to the x coordinate; consider an analogy between this height and the water pressure on a submerged surface.)
SOLUTION
Have
x dV = xEL dV ,
y =
h x a
where xEL = x
And
h dV = ydA = x dA a
h ( Iz )A x 2dA = = x = h xdA ( xA) A a x dA
Now
x a x dA
For the given volume
( Iz )A
=
4
( 2 in.)( 3.5 in.)3 + ( 3.5 in.)( 2 in.) ( 3.5 in.)2
= ( 21.4375 + 85.7500 ) in 4 = 336.74 in 4 and Then
( xA) A
= 3.5 in. ( 3.5 in.)( 2 in.) = 76.969 in 3
x =
336.74 in 4 = 4.375 in. 76.969 in 3 or x = 4.38 in.
PROBLEM 9.65
Show that the system of hydrostatic forces acting on a submerged plane area A can be reduced to a force P at the centroid C of the area and two couples. The force P is perpendicular to the area and is of magnitude P = Ay sin , where is the specific weight of the liquid, and the couples are M x = ( I x sin ) i and M y = I xy sin j , where I xy = xy dA (see Sec. 9.8). Note that the couples are independent of the depth at which the area is submerged.
(
)
SOLUTION
The pressure p at an arbitrary depth ( y sin ) is p = ( y sin ) so that the hydrostatic force dF exerted on an infinitesimal area dA is dF = ( y sin ) dA Equivalence of the force P and the system of infinitesimal forces dF requires
F: P = dF = y sin dA = sin ydA
or P = Ay sin Equivalence of the force and couple requires M x: - yP - M x = ( - ydF ) Now - ydF = - y ( y sin ) dA = - sin y 2dA = - ( sin ) I x Then or - yP - M x = - ( sin ) I x M x = ( sin ) I x - y ( Ay sin ) = sin I x - Ay 2
( P, M x + M y ) and
the system of infinitesimal hydrostatic forces
(
)
or M x = I x sin
M y : xP + M y = xdF Now
xdF = x ( y sin ) dA = sin xydA
= ( sin ) I xy
( Equation 9.12 )
PROBLEM 9.65 CONTINUED
Then or xP + M y = ( sin ) I xy M y = ( sin ) I xy - x ( Ay sin ) = sin I xy - Ax y or, using Equation 9.13,
(
)
or M y = I xy sin
PROBLEM 9.66
Show that the resultant of the hydrostatic forces acting on a submerged plane area A is a force P perpendicular to the area and of magnitude P = Ay sin = pA , where is the specific weight of the liquid and p is the pressure at the centroid C of the area. Show that P is applied at a point C p , called the center of pressure, whose coordinates are x p = I xy / Ay and y p = I x / Ay , where I xy = xy dA (see Sec. 9.8). Show also that the difference of ordinates y p - y is equal to k x2 / y and thus depends upon the depth at which the area is submerged.
SOLUTION
The pressure p at an arbitrary depth ( y sin ) is p = ( y sin ) so that the hydrostatic force dP exerted on an infinitesimal area dA is dP = ( y sin ) dA The magnitude P of the resultant force acting on the plane area is then
P = dP = y sin dA = sin ydA
= sin ( yA ) Now p = y sin P = pA
Next observe that the resultant P is equivalent to the system of infinitesimal forces dP. Equivalence then requires
M x: - yP P = - ydP
Now
2 ydP = y ( y sin ) dA = sin y dA
= ( sin ) I x Then or yP P = ( sin ) I x yP =
( sin ) I x sin ( yA)
or yP = Ix Ay
M y : xP P = xdP Now
xdP = x ( y sin ) dA = sin xydA
= ( sin ) I xy
( Equation 9.12 )
PROBLEM 9.66 CONTINUED
Then or xP P = ( sin ) I xy
xP =
( sin ) I xy sin ( yA)
or xP =
I xy Ay
Now From above By definition Substituting Rearranging yields
I x = I x + Ay 2 I x = ( Ay ) yP I x = k x A
2
( Ay ) yP
= k x2 A + Ay 2
yP - y = k x2 y
Although k x is not a function of the depth of the area (it depends only on the shape of A), y is dependent on the depth. ( yP - y ) = f ( depth )
PROBLEM 9.67
Determine by direct integration the product of inertia of the given area with respect to the x and y axes.
SOLUTION
First note
y = a 1- =
x2 4a 2
1 4a 2 - x 2 2
Have where dI xy = 0
dI xy = dI xy + xEL yELdA
( symmetry )
1 1 y = 4a 2 - x 2 2 4
xEL = x
yEL =
dA = ydx = Then
1 4a 2 - x 2dx 2
2a 1 1 4a 2 - x 2 4a 2 - x 2 dx I xy = dI xy = 0 x 4 2
1 2a 1 1 = 0 4a 2 x - x3 dx = 2a 2 x 2 - x 4 8 8 4 0
= a4 8 1 2 4 2 ( 2) - 4 ( 2)
(
)
2a
or I xy =
1 4 a 2
PROBLEM 9.68
Determine by direct integration the product of inertia of the given area with respect to the x and y axes.
SOLUTION
At or
x = a, y = b: a = kb 2 k = x= a b2 a 2 y b2
Then Have where
dI xy = dI xy + xEL yELdA dI xy = 0 yEL = y
( symmetry )
dA = xdy =
xEL =
1 a x = 2 y2 2 2b
a 2 y dy b2
Then
b a a I xy = dI xy = 0 2 y 2 ( y ) 2 y 2dy 2b b
a2 b a2 1 = 4 0 y 5dy = 4 y 6 2b 2b 6 0
b
or I xy =
1 2 2 ab 12
PROBLEM 9.69
Determine by direct integration the product of inertia of the given area with respect to the x and y axes.
SOLUTION
First note that y =- Now where h x b
dI xy = dI xy + xEL yELdA dI xy = 0 ( symmetry ) xEL = x yEL = 1 1h y =- x 2 2b
h dA = ydx = - xdx b Then
0 1 h h I xy = dI xy = -b x - x - xdx 2 b b
=
1 h2 0 3 1 h2 1 4 x dx = x 2 b 2 -b 2 b 2 4 -b
0
1 or I xy = - b 2h 2 8
PROBLEM 9.70
Determine by direct integration the product of inertia of the given area with respect to the x and y axes.
SOLUTION
At
x = a, y = b: b = c sin a 2a y = b sin dI xy = 0
or
c=b
Now Have
2a
x
(symmetry)
Now or
dI xy
dA = ydx 0 = dI xy + xEL yELdA
1 a a 1 I xy = 0 x y ( ydx ) = 0 xb 2 sin 2 xdx 2 2a 2
b x 2 x sin x cos a a = - - 2 4 4 2a 8 2a
2
( )
( )
x 2 0
a
=
b 2 a 2 4a 2 4a 2 + 2 + 2 4 8 8 2
or I xy =
a 2b 2 4 + 2 8 2
(
)
The following table is provided for the convenience of the instructor, as many problems in this and the next lesson are related. Type of Problem Compute I x and I y Compute I xy I x, I y, I xy by equations Principal axes by equations
I x, I y, I xy by Mohr's circle
Fig. 9.12
Fig. 9.13B
Fig. 9.13A
9.67
9.72
9.73
9.74
9.75
9.78
9.79
9.80
9.81
9.83
9.82
9.84
9.85
9.86
9.87
9.89
9.88
9.90
9.91
9.92
9.93
9.95
9.94
9.96
Principal axes by Mohr's circle
9.97
9.98
9.100
9.101
9.103
9.106
PROBLEM 9.71
Using the parallel-axis theorem, determine the product of inertia of the area shown with respect to the centroidal x and y axes.
SOLUTION
Have Symmetry implies For the other rectangles Where symmetry implies A in 2 1 3
I xy = I xy
( )1 + ( I xy )2 + ( I xy )3 ( I xy )2 = 0
I xy = I xy + x yA I xy = 0 x , in. -2.75 2.75 y , in. 1.0 -1.0 Ax y in 4 -5.5 -5.5 -11.00
4 ( 0.5 ) = 2 4 ( 0.5 ) = 2
or I xy = -11.00 in 4
PROBLEM 9.72
Using the parallel-axis theorem, determine the product of inertia of the area shown with respect to the centroidal x and y axes.
SOLUTION
Note: Orientation of A3 corresponding to a 180 rotation of the axes. Equation 9.20 then yields
I xy = I xy
Symmetry implies
Using Sample Problem 9.6
X 2 = 9 in.
1 ( I xy )2 = - 72 ( 9 in.)2 ( 4.5 in.)2 = -22.78125 in 4 Y2 = 1.5 in. A2 = 1 ( 9 in.)( 4.5 in.) = 20.25 in 2 2
( I xy )1 = 0
and
Similarly, X 3 = -9 in.
1 ( I xy )3 = - 72 ( 9 in.)2 ( 4.5 in.)2 = -22.78125 in 4 Y2 = -1.5 in. A3 = 1 ( 9 in.)( 4.5 in.) = 20.25 in 2 2 with
and Then and
I xy = I xy
( )1 + ( I xy )2 + ( I xy )3
0
( I xy )2 = ( I xy )3
I xy = I xy + x y A
I xy = 2 -22.78125 + ( 9 )(1.5 )( 20.25 ) in 4
Therefore,
= 501.1875 in 4
or I xy = 501 in 4
PROBLEM 9.73
Using the parallel-axis theorem, determine the product of inertia of the area shown with respect to the centroidal x and y axes.
SOLUTION
Have For each semicircle Thus
A, mm 2
I xy = I xy
( )1 + ( I xy )2
and
I xy = 0 (symmetry)
I xy = I xy + x y A I xy = x y A
x , mm - 60
y , mm
Ax y , mm 4 69.12 106 69.12 106 138.24 106
or I xy = 138.2 106 mm 4
1
2
(120 )2 (120 )2
= 7200
-
160
2
2
= 7200
60
160
PROBLEM 9.74
Using the parallel-axis theorem, determine the product of inertia of the area shown with respect to the centroidal x and y axes.
SOLUTION
Have For each rectangle Thus
A, mm 2
I xy = I xy
( )1 + ( I xy )2
and I xy = 0 (symmetry)
I xy = I xy + Ax y I xy = x y A x , mm -12.9 21.9
y , mm
Ax y , mm 4 -58 980.86 -100 643.29 -159 624.15
or I xy = -0.1596 106 mm 4
1 2
76 ( 6.4 ) = 486.4 44.6 ( 6.4 ) = 285.44
9.4 -16.1
PROBLEM 9.75
Using the parallel-axis theorem, determine the product of inertia of the area shown with respect to the centroidal x and y axes.
SOLUTION
Have Now symmetry implies and for the other rectangles Thus
I xy = I xy
( )1 + ( I xy )2 + ( I xy )3 ( I xy )1 = 0
I xy = I xy + x y A
where
I xy = 0 (symmetry)
I xy = ( x y A)2 + ( x y ) A3
= ( -69 mm )( -25 mm ) (12 mm )( 38 mm ) + ( 69 mm )( 25 mm ) (12 mm )( 38 mm )
= ( 786 600 + 786 600 ) mm 4 = 1 573 200 mm 4
or I xy = 1.573 106 mm 4
PROBLEM 9.76
Using the parallel-axis theorem, determine the product of inertia of the area shown with respect to the centroidal x and y axes.
SOLUTION
Symmetry implies
( I xy )1 = 0
( )
Using Sample Problem 9.6 and Equation 9.20, note that the orientation of A2 corresponds to a 90 rotation of 1 2 2 the axes; thus I xy = b h 2 72 Also, the orientation of A3 corresponds to a 270 rotation of the axes; thus Then x2 = 6 in., 1 ( I xy )2 = 72 ( 9 in.)2 ( 6 in.)2 = 40.5 in 4 y2 = -2 in., A2 = 1 ( 9 in.)( 6 in.) = 27 in 2 2 1 ( I xy )3 = 72 b2h2
and Also and Now Then
( I xy )3 = ( I xy )2 = 40.5 in 4
x3 = -6 in., I xy = I xy 0 y3 = 2 in., and A3 = A2 = 27 in 2 I xy = I xy + x y A
( )1 - ( I xy )2 - ( I xy )3
( I xy )2 = ( I xy )3
I xy = -2 40.5 in 4 + ( 6 in.)( -2 in.) 27 in 2 = -2 ( 40.5 - 324 ) in 4
or I xy = 567 in 4
(
)
PROBLEM 9.77
Using the parallel-axis theorem, determine the product of inertia of the area shown with respect to the centroidal x and y axes.
SOLUTION
Have Where I xy = 0 for each rectangle Then
I xy = I xy + x y A
I xy = I xy
= x yA Now
( )1 + ( I xy )2 + ( I xy )3
x1 = - (178.4 mm - 160 mm ) = -18.4 mm y1 = 60 mm - 12.2 mm = 47.8 mm A1 = 320 mm 120 mm = 38400 mm 2
and
x2 = - (178.4 mm + 30 mm ) = -208.4 mm y2 = 40 mm - 12.2 mm = 27.8 mm A2 = 60 mm 80 mm = 4800 mm 2
and
x3 = ( 320 mm - 178.4 mm ) - 40 mm = 101.6 mm y3 = - (12.2 mm + 105 mm ) = -117.2 mm A3 = ( 80 mm 210 mm ) = 16800 mm 2
Then
I xy = ( -18.4 mm )( 47.8 mm ) 38400 mm 2 + ( -208.4 mm )( 27.8 mm ) 4800 mm 2 + (101.6 mm )( -117.2 mm ) 16800 mm 2 = - ( 33.7736 + 27.8089 + 200.0463) 106 mm 4 = -261.6288 106 mm 4
or I xy = -262 106 mm 4
(
)
(
)
(
)
PROBLEM 9.78
Using the parallel-axis theorem, determine the product of inertia of the area shown with respect to the centroidal x and y axes.
SOLUTION
Have For each rectangle I xy = I xy + x yA Then
I xy = I xy
( )1 + ( I xy )2
I xy = 0 (symmetry)
and
I xy = x yA = ( -0.75 in.)( -1.5 in.) ( 3 in.)( 0.5 in.)
+ ( 0.5 in.)(1.00 in.) ( 4.5 in.)( 0.5 in.)
= (1.6875 + 1.125 ) in 4 = 2.8125 in 4
or I xy = 2.81 in 4
PROBLEM 9.79
Determine for the quarter ellipse of Problem 9.67 the moments of inertia and the product of inertia with respect to new axes obtained by rotating the x and y axes about O (a) through 45o counterclockwise, (b) through 30o clockwise.
SOLUTION
From Figure 9.12: Ix = = Iy = = From Problem 9.67: First note
16
( 2a )( a )3
8
a4
16
( 2a )3 ( a )
a4 1 4 a 2
2
I xy =
1 1 5 I x + I y = a4 + a4 = a4 2 28 2 16
(
)
1 1 3 I x - I y = a4 - a4 = - a4 2 2 8 2 16
(
)
Now use Equations (9.18), (9.19), and (9.20). Equation (9.18): I x = = Equation (9.19): I y = = Equation (9.20): I xy = 1 1 Ix + I y + I x - I y cos 2 - I xy sin 2 2 2 5 3 1 a 4 - a 4 cos 2 - a 4 sin 2 16 16 2 1 1 Ix + I y - I x - I y cos 2 + I xy sin 2 2 2 5 3 1 a 4 + a 4 cos 2 + a 4 sin 2 16 16 2 1 I x - I y sin 2 + I xy cos 2 2 3 1 a 4 sin 2 + a 4 cos 2 16 2
(
)
(
)
(
)
(
)
(
)
=-
PROBLEM 9.79 CONTINUED
(a)
= +45:
I x =
5 3 1 a 4 - a 4 cos90 - a 4 sin 90 16 16 2 or I x = 0.482a 4
I y =
5 3 1 + a 4 cos 90 + a 4 16 16 2
or I y = 1.482a 4
I xy = -
3 1 a 4 sin 90 + a 4 cos 90 16 2
or I xy = -0.589a 4
(b) = -30 : I x = 5 3 1 a 4 - a 4 cos ( -60 ) - a 4 sin ( -60 ) 16 16 2 or I x = 1.120a 4 I y = 5 3 1 a 4 + a 4 cos ( -60 ) + a 4 sin ( -60 ) 16 16 2
or I y = 0.843a 4
I xy = -
3 1 a 4 sin ( -60 ) + a 4 cos ( -60 ) 16 2
or I xy = 0.760a 4
PROBLEM 9.80
Determine the moments of inertia and the product of inertia of the area of Problem 9.72 with respect to new centroidal axes obtained by rotating the x and y axes 45 clockwise.
SOLUTION
From the solution to Problem 9.72
I xy = 501.1875 in 4
A2 = A3 = 20.25 in 2 First compute the moment of inertia
I x = ( I x )1 + ( I x )2 + ( I x )3
with
( I x )2 = ( I x )3
3 3 1 1 = (12 in.)( 9 in.) + 2 ( 9 in.)( 4.5in.) 12 12
= ( 729 + 136.6875 ) in 4 = 865.6875 in 4 and
Iy = Iy
( )1 + ( I y )2 + ( I y )3
with
( I y )2 = ( I y )3
3 3 2 1 1 = ( 9 in.)(12 in.) + 2 ( 4.5 in.)( 9 in.) + 20.25 in 2 ( 9 in.) 12 36
(
)
= (1296 + 182.25 + 3280.5 ) in 4 = 4758.75 in 4 From Equation 9.18
I x = Ix + I y
2
+
Ix - I y
2
cos 2 - I xy sin 2
=
865.6875 in 4 + 4758.75 in 4 865.6875 in 4 - 4758.75 in 4 cos 2 ( -45 ) + 2 2
-501.1875 in 4 sin 2 ( -45 )
= ( 2812.21875 + 501.1875 ) in 4 = 3313.4063 in 4 or I x = 3.31 103 in 4
PROBLEM 9.80 CONTINUED
Similarly
I y = Ix + I y
2
-
Ix - I y
2
cos 2 + I xy sin 2
= ( 2812.21875 - 501.1875 ) in 4 = 2311.0313 in 4
or I y = 2.31 103 in 4
and
I xy =
Ix - I y
2
sin 2 + I xy cos 2
=
865.6875 in 4 - 4758.75 in 4 sin 2 ( -45 ) 2
+ 501.1875cos 2 ( -45 )
= ( -1946.53125 )( -1) in 4 = 1946.53125 in 4
or I xy = 1.947 103 in 4
PROBLEM 9.81
Determine the moments of inertia and the product of inertia of the area of Problem 9.73 with respect to new centroidal axes obtained by rotating the x and y axes through 30o clockwise.
SOLUTION
From Problem 9.73,
I xy = 138.24 106 mm 4
I x = ( I x )1 + ( I x )2
4 = 2 (120 mm ) 8
( I x )1 = ( I x )2
= 51.84 106 mm 4
Iy = Iy
( )1 + ( I y )2
( I y )1 = ( I y )2
4 2 2 = 2 (120 mm ) + (120 mm ) ( 60 mm ) 2 8
= 103.68 106 mm 4 Have and Then
I x = 2 25.92 106 = 51.84 106 mm 4 I y = 2 51.84 106 = 103.68 106 mm 4
(
)
(
)
1 I x + I y = 77.76 106 mm 4 2 1 I x - I y = -25.92 106 mm 4 2
(
)
and
(
)
PROBLEM 9.81 CONTINUED
Now, from Equations 9.18, 9.19, and 9.20 Equation 9.18: I x = 1 1 Ix + I y + I x - I y cos 2 - I xy sin 2 2 2
(
)
(
)
= 77.76 106 - 25.92 106 cos ( -60 ) - 138.24 106 sin ( -60 ) mm 4 = 323.29 106 mm 4 or I x = 323 106 mm 4 Equation 9.19: I y = 1 1 Ix + I y - I x - I y cos 2 + I xy sin 2 2 2
(
)
(
)
= 77.76 106 + 25.92 106 cos ( -60 ) + 138.24 106 sin ( -60 ) mm 4 = 165.29 106 mm 4
or I y = 165.29 106 mm 4
Equation 9.20:
I xy =
1 I x - I y sin 2 + I xy cos 2 2
(
)
= -25.92 106 sin ( -60 ) + 138.24 106 cos ( -60 ) mm 4 = 139.64 106 mm 4
or I xy = 139.6 104 mm 4
PROBLEM 9.82
Determine the moments of inertia and the product of inertia of the area of Problem 9.75 with respect to new centroidal axes obtained by rotating the x and y axes through 60o counterclockwise.
SOLUTION
From Problem 9.75 Now where and
I xy = 1.5732 106 mm 4
I x = ( I x )1 + ( I x )2 + ( I x )3
( I x )1 = ( I x )2 = ( I x )3
=
1 (150 mm )(12 mm )3 = 21 600 mm4 12
1 (12 mm )( 38 mm )3 + (12 mm )( 38 mm ) ( 25 mm )2 12
= 339 872 mm 4 Then Also where and
I x = 21 600 + 2 ( 339 872 ) mm 4 = 701 344 mm 4 = 0.70134 106 mm 4
Iy = Iy
( )1 + ( I y )2 + ( I y )3
1 ( I y )1 = 12 (12 mm )(150 mm )3 = 3.375 106 mm4 1 ( I y )2 = ( I y )3 = 12 ( 38 mm )(12 mm )3 + (12 mm )( 38 mm ) ( 69 mm )2 = 2.1765 106 mm 4
Then Now and
I y = ( 3.375 + 2 ( 2.1765 ) 106 mm 4 = 7.728 106 mm 4
1 I x + I y = 4.2146 106 mm 4 2 1 I x - I y = -3.5133 106 mm 4 2
( (
)
)
PROBLEM 9.82 CONTINUED
Using Equations 9.18, 9.19, and 9.20 From Equation 9.18: I x = Ix + I y 2 + Ix - I y 2 cos 2 - I xy sin 2
= 4.2147 106 + -3.5133 106 cos (120 ) - 1.5732 106 sin (120 ) mm 4 = 4.6089 106 mm 4 or I x = 4.61 106 mm 4 From Equation 9.19: I y = Ix + I y 2 - Ix - I y 2 cos 2 + I xy sin 2
(
)
= 4.2147 106 - -3.5133 106 cos (120 ) + 1.5732 106 sin (120 ) mm 4 = 3.8205 106 mm 4
or I y = 3.82 106 mm 4
(
)
From Equation 9.20: I xy = Ix - I y 2 sin 2 + I xy cos 2
= -3.5133 106 sin (120 ) + 1.5732 106 cos (120 ) mm 4 = -3.8292 106 mm 4
or I xy = -3.83 106 mm 4
PROBLEM 9.83
Determine the moments of inertia and the product of inertia of the L76 51 6.4-mm angle cross section of Problem 9.74 with respect to new centroidal axes obtained by rotating the x and y axes through 45o clockwise.
SOLUTION
From Problem 9.74
I xy = -0.1596 106 mm 4
From Figure 9.13 I x = 0.166 106 mm 4
I y = 0.453 106 mm 4
Now
1 I x + I y = 0.3095 106 mm 4 2 1 I x - I y = -0.1453 106 mm 4 2
(
)
(
)
Using Equations (9.18), (9.19), and (9.20) Equation (9.18): I x = Ix + I y 2 + Ix - I y 2 cos 2 - I xy sin 2
= 0.3095 106 + -0.1435 106 cos ( -90 ) - -0.1596 106 sin ( -90 ) mm 4 = 0.1499 106 mm 4 or I x = 0.1499 106 mm 4
(
)
(
)
PROBLEM 9.83 CONTINUED
Equation (9.19): I y = Ix + I y 2 - Ix - I y 2 cos 2 + I xy sin 2
= 0.3095 106 - -0.1435 106 cos ( -90 ) + -0.1596 106 sin ( -90 ) mm 4 = 0.4691 106 mm 4
or I y = 0.469 106 mm 4
(
)
(
)
Equation (9.20): I xy = Ix - I y 2 sin 2 + I xy cos 2
= -0.1435 106 sin ( -90 ) + 0.1596 106 cos ( -90 ) mm 4 = 0.1435 106 mm 4
or I xy = 0.1435 106 mm 4
PROBLEM 9.84
Determine the moments of inertia and the product of inertia of the L5 3 1 -in. angle cross section of Problem 9.78 with respect to new 2 centroidal axes obtained by rotating the x and y axes through 30o counterclockwise.
SOLUTION
From Problem 9.78
I xy = 2.8125 in 4
From Figure 9.13
I x = 9.45 in 4 , I y = 2.58 in 4
Now
1 I x + I y = 6.015 in 4 2 1 I x - I y = 3.435 in 4 2
(
)
(
)
Using Equations (9.18), (9.19), and (9.20) Equation (9.18): I x = Ix + I y 2 + Ix - I y 2 cos 2 - I xy sin 2
= 6.015 + 3.435cos ( 60 ) - 2.8125sin ( 60 ) in 4 = 5.2968 in 4
or I x = 5.30 in 4 Equation (9.19): I y = Ix + I y 2 - Ix - I y 2 cos 2 + I xy sin 2
= 6.015 - 3.435cos ( 60 ) + 2.8125sin ( 60 ) in 4 = 6.7332 in 4 or I y = 6.73 in 4
Equation (9.20):
I xy =
Ix - I y 2
sin 2 + I xy cos 2
or I xy = 4.38 in 4
= 3.435sin ( 60 ) + 2.8125cos ( 60 ) in 4 = 4.3810 in 4
PROBLEM 9.85
For the quarter ellipse of Problem 9.67, determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia.
SOLUTION
From Problem 9.79:
Ix =
8
a4
Iy = 1 4 a 2
2
a4
Problem 9.67:
I xy =
Now, Equation (9.25):
tan 2 m = -
2 I xy Ix - I y
=-
8
1 2 a4 2 a4 -
2
a4
= Then
8 = 0.84883 3
2 m = 40.326
and
220.326
or m = 20.2 and 110.2 Also, Equation (9.27):
I max, min = Ix + I y 2 Ix - I y 2 + I xy 2
2
=
1 4 4 a + a 2 8 2
1 1 a4 - a4 + a4 2 2 2 8
2 2
= ( 0.981748 0.772644 ) a 4 or I max = 1.754a 4 and I min = 0.209a 4 By inspection, the a axis corresponds to Imin and the b axis corresponds to Imax.
PROBLEM 9.86
For the area indicated, determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia. Area of Problem 9.72
SOLUTION
From the solutions to Problem 9.72 and 9.80 I xy = 501.1875 in 4 1 I x + I y = 2812.21875 in 4 2
(
)
1 I x - I y = -1946.53125 in 4 2 Then Equation (9.25): or
tan 2 m = - 2 I xy Ix - I y =- 501.1875 = 0.257477 -1946.53125
(
)
2 m = 14.4387
and
194.4387 or m = 7.22 and 97.2
Equation (9.27):
I max, min =
Ix + I y 2
Ix - I y 2 + I xy 2
2
= 2812.21875
( -1946.53125)2 + ( 501.1875)2
= ( 2812.21875 2010.0181) in 4 or I max = 4.82 103 in 4 and I min = 802 in 4
By inspection, the a axis corresponds to I min and the b axis corresponds to I max .
PROBLEM 9.87
For the area indicated, determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia. Area of Problem 9.73
SOLUTION
From Problems 9.73 and 9.81 I x = 51.84 106 mm 4
I y = 103.68 106 mm 4 I xy = 138.24 106 mm 4
Equation (9.25):
tan 2 m = -
2 I xy Ix - I y
=-
2 138.24 106
6
(
)
51.84 10 - 103.68 106
= 1.69765 2 m = 59.500 and 239.500 or m = 29.7 and 119.7
Ix - I y 1 2 = Ix + I y + I xy 2 2
Then
I max, min
(
)
2
=
( 51.84 + 103.68) 106
2
( 51.84 - 103.68 ) 106 6 + 138.24 10 2
2
(
)
2
= ( 244.29 160.44 ) 106 mm 4 or I max = 405 106 mm 4 and I min = 83.9 106 mm 4
Note: By inspection the a axis corresponds to I min and the b axis corresponds to I max .
PROBLEM 9.88
For the area indicated, determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia. Area of Problem 9.75
SOLUTION
From Problems 9.75 and 9.82
I x = 0.70134 106 mm 4
I y = 7.728 106 mm 4 I xy = 1.5732 106 mm 4
Then
1 I x + I y = 4.2147 106 mm 4 2 1 I x - I y = -3.5133 106 mm 4 2
(
)
(
)
Equation (9.25):
tan 2 = -
2 I xy Ix - I y
=-
2 1.5732 106
6
(
)
0.70134 10 - 7.728 106
= 0.44778 Then 2 m = 24.12 and 204.12 or m = 12.06 and 102.1 Also, Equation (9.27):
I max, min = Ix + I y 2 Ix - I y 2 I 2 xy
2
= 4.2147 106
( -3.5133 10 ) + (1.5732 10 )
6 2 6
2
= ( 4.2147 3.8494 ) 106 mm 4 or I max = 8.06 106 mm 4 and I min = 0.365 106 mm 4
By inspection, the a axis corresponds to I min and the b axis corresponds to I max .
PROBLEM 9.89
For the angle cross section indicated, determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia. The L76 51 6.4-mm angle cross section of Problem 9.74
SOLUTION
From Problems 9.74 and 9.83
I x = 0.166 106 mm 4
I y = 0.453 106 mm 4 I xy = -0.1596 106 mm 4
Then
1 I x + I y = 0.3095 106 mm 4 2 1 I x - I y = -0.1435 106 mm 4 2
(
)
(
)
Equation (9.25): Then or Also, Equation (9.27):
tan 2 m = -
2 I xy Ix - I y
=-
2 -0.1596 106
( 0.166 - 0.453) 106
and 131.96
(
)
= -1.1122
2 m = -48.041
m = -24.0 and 66.0
I max, min =
( Ix + I y )
2
Ix - I y 2 + I xy 2
2
= 0.3095 106
( -0.1435 10 ) + ( -0.1596 10 )
6 2 6
2
= ( 0.3095 0.21463) 106 mm 4 or
I max = 0.524 106 mm 4 I min = 0.0949 106 mm 4
By inspection, the a axis corresponds to I min and the b axis corresponds to I max .
PROBLEM 9.90
For the angle cross section indicated, determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia. The L5 3 1 -in. angle cross section of Problem 9.78 2
SOLUTION
From Problems 9.78 and 9.84
I xy = 2.81 in 4
I x = 9.45 in 4
I y = 2.58 in 4
Then
1 I x + I y = 6.015 in 4 2 1 I x - I y = 3.435 in 4 2
(
)
(
)
Equation (9.25): Then
tan 2 m = -
2 I xy Ix - I y
=-
2 ( 2.81) 9.45 - 2.58
= -0.8180
2 m = -39.2849
and
140.7151 or m = -19.64 and 70.36
Also, Equation (9.27):
I max, min
( Ix + I y ) =
2
Ix - I y 2 + I xy 2
2
= 6.015 3.4352 - 2.812 = ( 6.015 4.438 ) in 4 or I max = 10.45 in 4 and I min = 1.577 in 4
Note: By inspection, the a axis corresponds to I max and the b axis corresponds to I min .
PROBLEM 9.91
Using Mohr's circle, determine for the quarter ellipse of Problem 9.67 the moments of inertia and the product of inertia with respect to new axes obtained by rotating the x and y axes about O (a) through 45o counterclockwise, (b) through 30o clockwise.
SOLUTION
From Problem 9.79:
Ix = Iy =
8
a4 a4
2
Problem 9.67:
I xy =
1 4 a 2 1 Y a4, - a4 2 2
The Mohr's circle is defined by the diameter XY, where 1 X a4, a4 8 2 Now
I ave =
and
1 1 5 I x + I y = a 4 + a 4 = a 4 = 0.98175a 4 2 2 8 2 16
(
)
and
R=
Ix - I y 2 + I xy = 2
2
1 4 4 1 4 2 8 a - 2 a + 2 a
2
2
= 0.77264a 4 The Mohr's circle is then drawn as shown.
tan 2 m = - 2 I xy Ix - I y
1 2 a4 2 a4 -
=-
8
2
a4
= 0.84883 or 2 m = 40.326
PROBLEM 9.91 CONTINUED
Then
= 90 - 40.326
= 49.674
= 180 - ( 40.326 + 60 )
= 79.674 (a) I x = I ave - R cos = 0.98175a 4 - 0.77264a 4 cos 49.674 or I x = 0.482a 4
I y = I ave + R cos = 0.98175a 4 + 0.77264a 4 cos 49.674 or I y = 1.482a 4 I xy = - R sin = -0.77264a 4 sin 49.674 or I xy = -0.589a 4
(b)
I x = I ave + R cos = 0.98175a 4 + 0.77264a 4 cos 79.674 or I x = 1.120a 4
I y = I ave - R cos = 0.98175a 4 - 0.77264a 4 cos 79.674 or I y = 0.843a 4 I xy = R sin = 0.77264a 4 sin 79.674 or I xy = 0.760a 4
PROBLEM 9.92
Using Mohr's circle, determine the moments of inertia and the product of inertia of the area of Problem 9.72 with respect to new centroidal axes obtained by rotating the x and y axes 45 clockwise.
SOLUTION
From the solution to Problem 9.72: Problem 9.80:
I xy = 501.1875 in 4
I x = 865.6875 in 4
I y = 4758.75 in 4
Now
1 I x + I y = 2812.21875 in 4 2 1 I x - I y = -1946.53125 in 4 2
(
)
(
)
The Mohr's circle is defined by the points X and Y where X: Now
( I x , I xy )
I ave =
2
Y:
( I y , -I xy )
1 I x + I y = 2812.2 in 4 2
(
)
and
R=
Ix - I y 2 + I xy = 2
( -1946.53125)2 + 501.18752
in 4
= 2010.0 in 4 Also,
tan 2 m = I xy Ix - I y 2 or Then 2 m = 14.4387 = 501.1875 = 0.2575 1946.53125
= 180 - (14.4387 + 90 ) = 75.561
PROBLEM 9.92 CONTINUED
Then I x , I y = I ave R cos = 2812.2 2010.0cos 75.561 or I x = 3.31 103 in 4
and I y = 2.31 103 in 4
and
I xy = R sin = 2010.0sin 75.561
or I xy = 1.947 103 in 4
PROBLEM 9.93
Using Mohr's circle, determine the moments of inertia and the product of inertia of the area of Problem 9.73 with respect to new centroidal axes obtained by rotating the x and y axes through 30o clockwise.
SOLUTION
From Problems 9.73 and 9.81
I xy = 138.24 106 mm 4
I x = 51.84 106 mm 4 = 162.86 106 mm 4
I y = 103.68 106 mm 4
= 325.72 106 mm 4 Now I ave = 1 Ix + I y 2
(
)
= 244.29 106 mm 4
R= Ix - I y 2 + I xy 2
2
= 160.4405 106 mm 4 From Problem 9.87 Then Then 2 m = 59.5
= 180 - 60 - 2 m = 60.5
I x = I ave + R cos = 244.29 + 160.4405cos 60.5
= 323.29 106 mm 4 or I x = 323 106 mm 4 I y = I ave - R cos = 244.24 - 160.4405cos 60.5 = 165.29 106 mm 4
or I y = 165.3 106 mm 4 I xy = R sin = 160.44sin 60.5 = 139.6 106 mm 4
PROBLEM 9.94
Using Mohr's circle, determine the moments of inertia and the product of inertia of the area of Problem 9.75 with respect to new centroidal axes obtained by rotating the x and y axes through 60o counterclockwise.
SOLUTION
From Problems 9.75 and 9.82 I x = 0.70134 106 mm 4
I y = 7.728 106 mm 4 I xy = 1.5732 106 mm 4
Now
I ave =
1 I x + I y = 4.2147 106 mm 4 2
Ix - I y 2 + I xy = 3.8494 106 mm 4 2
2
(
)
and
R=
Then and Then
-2 (1.5732 ) 2 m = tan -1 = 24.12 0.70134 - 7.728
= 120 - 24.12 - 90 = 5.88
I x = I ave + R sin = ( 4.2147 + 3.8494sin 5.88 ) 106 mm 4 = 4.6091 106 mm 4 or I x = 4.61 106 mm 4
I y = I ave - R sin = ( 4.2147 - 3.8494sin 5.88 ) 106 mm 4
= 3.8203 106 mm 4
or I y = 3.82 106 mm 4 I xy = - R cos = -3.8494 cos 5.88 = -3.8291 106 mm 4 or I xy = -3.83 106 mm 4
PROBLEM 9.95
Using Mohr's circle, determine the moments of inertia and the product of inertia of the L76 51 6.4-mm angle cross section of Problem 9.74 with respect to new centroidal axes obtained by rotating the x and y axes through 45o clockwise.
SOLUTION
From Problems 9.74 and 9.83 I x = 0.166 106 mm 4
I y = 0.453 106 mm 4 I xy = -0.1596 106 mm 4
Now
I ave =
1 I x + I y = 0.3095 106 mm 4 2
2
(
)
and
R=
Ix - I y 2 + I xy 2
= 0.21463 106 mm 4 Then and Then I x = I ave - R sin = ( 0.3095 - 0.21463sin 48.04 ) 106 mm 4 = 0.14989 106 mm 4 or I x = 0.1499 106 mm 4 and
I y = I ave + R sin = ( 0.3095 + 0.21463sin 48.04 ) 106 mm 4
-2 ( -0.1596 ) 2 m = tan -1 = -48.04 0.166 - 0.453
+ 90 - 2 = 90; = 2 m
= 0.46910 106 mm 4
or I y = 0.4690 106 mm 4
and
I xy = R cos = 0.21463cos 48.04 = 0.1435 106 mm 4 or I xy = 0.1435 106 mm 4
PROBLEM 9.96
Using Mohr's circle, determine the moments of inertia and the product of inertia of the L5 3 1 -in. angle cross section of Problem 9.78 with 2 respect to new centroidal axes obtained by rotating the x and y axes through 30o counterclockwise.
SOLUTION
Have I x = 9.45 in 4
I y = 2.58 in 4
From Problem 9.78 Now
I xy = 2.8125 in 4
I ave =
2
Ix + I y 2
2
= 6.015 in 4
and
R=
Ix - I y + I xy 2
( )
= 4.43952 in 4 Then
-2 ( 2.8125 ) 2 m = tan -1 = -39.31 9.45 - 2.58
2 m + 60 + = 180, Then
= 80.69
I x = I ave - R cos = 6.015 in 4 - ( 4.43952 in 4 ) cos80.69 = 5.29679 in 4 or I x = 5.30 in 4 I y = I ave + R cos = 6.015 in 4 + ( 4.43952 in 4 ) cos80.69 = 6.73321 in 4
or I y = 6.73 in 4
I xy = R sin = ( 4.43952 in 4 ) sin 80.69 = 4.38104 in 4
or I xy = 4.38 in 4
PROBLEM 9.97
For the quarter ellipse of Problem 9.67, use Mohr's circle to determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia.
SOLUTION
From Problem 9.79: Ix = I xy =
8
a4
Iy =
2
a4
Problem 9.67:
1 4 a 2 1 Y a4, - a4 2 2
The Mohr's circle is defined by the diameter XY, where 1 X a4, a4 8 2 Now and R= 1 2 2 I x - I y + I xy = I ave = and
1 1 I x + I y = a 4 + a 4 = 0.98175a 4 2 2 8 2
(
)
(
)
2
1 4 4 1 4 a - a + a 2 2 2 8
2
2
= 0.77264a 4 The Mohr's circle is then drawn as shown. tan 2 m = - 2 I xy Ix - I y
1 2 a4 2
=-
8
a4 -
2
a4
= 0.84883 or and 2 m = 40.326
m = 20.2
PROBLEM 9.97 CONTINUED
The principal axes are obtained by rotating the xy axes through 20.2 counterclockwise About O. Now
I max, min = I ave R = 0.98175a 4 0.77264a 4
or I max = 1.754a 4 and I min = 0.209a 4 From the Mohr's circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max .
PROBLEM 9.98
Using Mohr's circle, determine for the area indicated the orientation of the principal centroidal axes and the corresponding values of the moments of inertia. Area of Problem 9.72
SOLUTION
From the solution to Problem 9.72:
I xy = 501.1875 in 4
From the solution to Problem 9.80:
I x = 865.6875 in 4
I y = 4758.75 in 4
1 I x + I y = 2812.21875 in 2 2 1 I x - I y = -1946.53125 in 4 2 The Mohr's circle is defined by the point Now
2
(
)
(
)
X: I ave =
( I x , I xy ) ,
Y:
( I y,
- I xy
)
1 I x + I y = 2812.2 in 4 2
(
)
and
R=
Ix - I y + I xy = 2
( -1946.53125)2 + 501.18752
= 2010.0 in 4
PROBLEM 9.98 CONTINUED
tan 2 m = -
I xy
Ix - I y 2
=-
501.1875 = 0.2575, -1946.53125
2 m = 14.4387
or m = 7.22 counterclockwise Then
I max, min = I ave R = ( 2812.2 2010.0 ) in 4
or I max = 4.82 103 in 4 and I min = 802 in 4
Note: From the Mohr's circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max .
PROBLEM 9.99
Using Mohr's circle, determine for the area indicated the orientation of the principal centroidal axes and the corresponding values of the moments of inertia. Area of Problem 9.76
SOLUTION
From the solution to Problem 9.76 Now
I xy = 576 in 4
I x = ( I x )1 - ( I x )2 - ( I x )3 , where
=
( I x )2 = ( I x )3
4
(15 in.)4 - 2 ( 9 in.)( 6 in.)3 = ( 39761 - 324 ) in 4
1 12
= 39, 437 in 4
and
Iy = Iy =
( )1 - ( I y )2 - ( I y )3 ,
4 1 36
where
( I y )2 = ( I y )3
1 2
(15 in.)4 - 2 ( 6 in.)( 9 in.)3 + ( 9 in.)( 6 in.)( 6 in.)2
= ( 39, 761 - 243 - 1944 ) in 4 = 37,574 in 4 The Mohr's circle is defined by the point (X, Y) where X: Now I ave =
R=
( I x , I xy )
)
Y:
( I y , -I xy )
1 1 I x + I y = ( 39, 437 + 37,574 ) in 4 = 38,506 in 4 2 2
2 + I xy =
(
and
Ix - I y 2
1 2 4 2 ( 39, 437 - 37,574 ) + 567 = 1090.5 in
2
PROBLEM 9.99 CONTINUED
tan 2 m = - I xy Ix - I y 2 = -567 = -0.6087 1 ( 39, 437 - 37,574 ) 2 or m = -15.66 clockwise Then
I max, min = I ave R = ( 38,506 1090.50 ) in 4
or I max = 39.6 103 in 4 and I min = 37.4 103 in 4
Note: From the Mohr's circle it is seen that the a axis corresponds to the I max and the b axis corresponds to
I min .
PROBLEM 9.100
Using Mohr's circle, determine for the area indicated the orientation of the principal centroidal axes and the corresponding values of the moments of inertia. Area of Problem 9.73
SOLUTION
From Problems 9.73 and 9.81
I x = 162.86 106 mm 4
I y = 325.72 106 mm 4 I xy = 138.24 106 mm 4
Define points
X (162.86,138.24 ) 106 mm 4
I ave =
Y ( 325.72, -138.24 ) 106 mm 4
Now
1 1 I x + I y = (162.86 + 325.72 ) 106 mm 4 2 2
(
)
= 244.29 106 mm 4 and
R= Ix - I y 2 + I xy = 2
2
(162.86 - 325.72 ) 106 + 138.24 106 2
2
(
)
2
= 160.44 106 mm 4 and
-2 (138.24 ) 106 2 m = tan -1 = 59.4999 6 (162.86 - 325.72 ) 10 or m = 29.7 counterclockwise Then
I max, min = I ave R = 244.29 106 160.44 106 mm 4 or I max = 405 106 mm 4 and I min = 83.9 106 mm 4
(
)
Note: From the Mohr's circle it is seen that the a axis corresponds to I min and the b axis corresponds to
I max .
PROBLEM 9.101
Using Mohr's circle, determine for the area indicated the orientation of the principal centroidal axes and the corresponding values of the moments of inertia. Area of Problem 9.74
SOLUTION
From Problems 9.74 and 9.83
I x = 0.166 106 mm 4 , I y = 0.453 106 mm 4 , I xy = -0.1596 106 mm 4
Define points Now
X ( 0.166, -0.1596 ) 106 mm 4 I ave =
and
Y ( 0.453, -0.1596 ) 106 mm 4
1 1 I x + I y = ( 0.166 + 0.453) 106 mm 4 2 2
(
)
= 0.3095 106 mm 4 and
R=
Ix - I y 2 + I xy = 2
2
( 0.166 - 0.453)106 6 + -0.1596 10 2
2
(
)
2
= 0.21463 106 mm 4 Also -2 I xy 2 m = tan -1 Ix - I y -2 ( -0.1596 ) = tan -1 = -48.04 0.166 - 0.453
m = -24.02
or = -24.0 clockwise Then
I max,
min
= I ave R = ( 0.3095 0.21463) 106 mm 4 or I max = 0.524 106 mm 4 and I min = 0.0949 106 mm 4
Note: From the Mohr's circle it is seen that the a axis corresponds to I min and the b axis corresponds to
I max .
PROBLEM 9.102
Using Mohr's circle, determine for the area indicated the orientation of the principal centroidal axes and the corresponding values of the moments of inertia. Area of Problem 9.75
SOLUTION
From Problems 9.75 and 9.82
I x = 0.70134 106 mm 4 , I y = 7.728 106 mm 4 , I xy = 1.5732 106 mm 4
Now
I ave =
1 1 I x + I y = ( 0.70134 + 7.728 ) 106 mm 4 = 4.2147 106 mm 4 2 2
2
(
)
and
R=
Ix - I y 2 + I xy = 2
( 0.70134 - 7.728 ) 106 6 + 1.5732 10 2
2
(
)
2
= 3.8495 106 mm 4 Define points
X ( 0.70134, 15732 ) 106 mm Y ( 7.728, - 1.5732 ) 106 mm
Also
-2 (1.5732 ) 2 m = tan -1 = 24.122, m = 12.06 0.70134 - 7.728
or m = 12.06 counterclockwise Then
I max,
min
= I ave R = ( 4.2147 3.8495 ) 106 mm 4 or I max = 8.06 106 mm 4 and I min = 0.365 106 mm 4
Note: From the Mohr's circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max .
PROBLEM 9.103
Using Mohr's circle, determine for the area indicated the orientation of the principal centroidal axes and the corresponding values of the moments of inertia. Area of Problem 9.71
SOLUTION
From Problem 9.71 Compute I x and I y for area of Problem 9.71
Ix = 5 in. ( 0.5 in.) 12
3
I xy = -11.0 in 4
( 0.5 in.)( 4 in.)3 2 + 2 + ( 4 in. 0.5 in.)(1.0 in.) 12
= 9.38542 in 4
( 0.5 in.)3 ( 4 in.) 0.5 in. ( 5 in.)3 2 Iy = 2 + ( 4 in. 0.5 in.)( 2.75 in.) + 12 12
= 35.54167 in 4 Define points Now
I ave =
X ( 9.38542, - 11) ,
and
Y ( 35.54167, 11)
Ix + I y 2
2
=
9.38542 in 4 + 35.54167 in 4 = 22.46354 in 4 2
2
and
R=
Ix - I y + I xy 2
( )
=
2 9.38542 - 35.54167 + (11.0 ) 2
2
= 17.08910 in 4 Also Then
-2 ( -11.0 ) 2 m = tan -1 = -40.067 9.38542 - 35.54167
or m = -20.033 clockwise
I max, min = I ave R = 22.46354 17.08910
= 39.55264, 5.37444
or I max = 39.55 in 4 I min = 5.37 in 4
Note: The a axis corresponds to I min and b axis corresponds to I max .
PROBLEM 9.104
Using Mohr's circle, determine for the area indicated the orientation of the principal centroidal axes and the corresponding values of the moments of inertia. Area of Problem 9.77
SOLUTION
From Problems 9.44 and 9.77
I x = 432.59 106 mm 4 , I y = 732.97 106 mm 4 , I xy = -261.63 106 mm 4
Define points
X ( 432.59, - 261.63) 106 mm 4 Y ( 732.97, 261.63) 106 mm 4
Now
I ave =
1 1 I x + I y = ( 432.59 + 732.97 ) 106 = 582.78 106 mm 4 2 2
2
(
)
and
R=
2 Ix - I y 1 432.59 - 732.97 2 6 + I xy = 10 + -261.63 106 2 2 2
(
)
2
= 301.67 106 mm 4 Also or Then or and I max,
min
tan 2 m = -
2 I xy
Ix - I y
=
( 432.59 - 732.97 ) 106
-2 ( -261.63) 106
= -60.14
m = -30.1 clockwise
= I ave R = ( 582.78 301.67 ) 106 mm 4 I max = 884 106 mm 4 I min = 281 106 mm 4
Note: From the Mohr's circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max .
PROBLEM 9.105
The moments and product of inertia for an L102 76 6.4-mm angle cross section with respect to two rectangular axes x and y through C are, respectively, I x = 0.166 106 mm 4 , I y = 0.453 106 mm 4 , and I xy < 0 , with the minimum value of the moment of inertia of the area with respect to any axis through C being I min = 0.051 106 mm 4 . Using Mohr's circle, determine (a) the product of inertia I xy of the area, (b) the orientation of the principal axes, (c) the value of I max .
SOLUTION
Given:
I x = 0.166 106 mm 4 , I y = 0.453 106 mm 4 and I xy < 0
Note: A review of a table of rolled-steel shapes reveals that the given values of I x and I y are obtained when the 102 mm leg of the angle is parallel to the x axis. For I xy < 0 the angle must be oriented as shown. (a) Now I ave = 1 1 I x + I y = ( 0.166 + 0.453) 106 mm 4 2 2
(
)
= 0.3095 106 mm 4 Now Then
I min = I ave - R
or
R = I ave - I min
R = ( 0.3095 - 0.051) 106 mm 4 = 0.2585 106 mm 4
From
Ix - I y + I xy R = 2
2
2
( )
2
I xy =
2 2 0.166 - 0.453 6 4 ( 0.2585 ) - 10 mm 2
I xy = 0.21501 106 mm 4
Since
I xy < 0,
I xy = -0.21501 106 mm 4
or I xy = -0.215 106 mm 4
PROBLEM 9.105 CONTINUED
(b)
-2 ( -0.21501) 2 m = tan -1 = -56.28 0.166 - 0.453
or m = -28.1 clockwise (c) I max = I ave + R = ( 0.3095 + 0.2585 ) 106 mm 4 or I max = 0.568 106 mm 4
PROBLEM 9.106
Using Mohr's circle, determine for the cross section of the rolled-steel angle shown the orientation of the principal centroidal axes and the corresponding values of the moments of inertia. (Properties of the cross sections are given in Fig. 9.13.)
SOLUTION
From Figure 9.13
I x = 9.45 in 4
I y = 2.58 in 4
From Problem 9.78
I xy = -2.81 in 4
The Mohr's circle is defined by the diameter XY where X ( 9.45, - 2.81) in 4 Y ( 2.58, 2.81) in 4 Now I ave = 1 1 Ix + I y = 9.45 in 4 + 2.58 in 4 2 2 = 6.015 in 4 and R= = 1 2 2 I x - I y + I xy
(
)
(
)
(
)
2
1 9.45 in 4 - 2.58 in 4 2
(
) + ( 2.81 in )
2 4
2
= 5.612 in 4
tan 2 m = -2 I xy Ix - I y = -2 -2.81 in 4
4
(
)
9.45 in - 2.58 in 4
= 0.81805
or
2 m = 32.285 or m = 19.643 counterclockwise
About C. Now I max,
min
= I ave R = ( 6.015 5.612 ) in 4 or I max = 11.63 in 4 and I min = 0.403 in 4
From the Mohr's circle it is seen that the a axis corresponds to I max and the b axis corresponds to I min .
PROBLEM 9.107
Using Mohr's circle, determine for the cross section of the rolled-steel angle shown the orientation of the principal centroidal axes and the corresponding values of the moments of inertia. (Properties of the cross sections are given in Fig. 9.13.)
SOLUTION
From Figure 9.13B: Have Now I xy = I xy x1 =
I x = 7.20 106 mm 4 ,
I y = 2.64 106 mm 4
( )1 + ( I xy )2 ,
where
I xy = I xy + x yA y1 = 50.3 -
and
I xy = 0
102 - 25.3 = 25.7 mm, 2
12.7 = 43.95 mm 2
A1 = 102 12.7 = 1295.4 mm 2 x2 = -25.3 - 12.7 = -18.95 mm 2 1 y2 = - (152 - 12.7 ) - ( 50.3 - 12.7 ) = 32.05 mm 2
A2 = (12.7 )(152 - 12.7 ) = 1769.11 mm 2 Then
I xy = ( 25.7 mm )( 43.95 mm ) 1295.4 mm 2 + ( -18.95 mm )( -32.05 mm ) 1769.11 mm 2 106
{
(
)
(
)}
= (1.46317 + 1.07446 ) 106 mm 4 = 2.5376 106 mm 4 The Mohr's circle is defined by points X and Y, where X I x , I xy , Y I y , - I xy Now I ave =
(
) (
)
1 1 I x + I y = ( 7.20 + 2.64 ) 106 mm 4 = 4.92 106 mm 4 2 2
(
)
PROBLEM 9.107 CONTINUED
and R= Ix - I y 1 2 2 + I xy = ( 7.20 - 2.64 ) + 2.53762 106 mm 4 2 2
= 3.4114 106 mm 4
tan m = - 2 I xy Ix - I y =-
( 7.20 - 2.64 )
2 ( 2.5376 )
= -1.11298,
2 = -48.0607
or Now or and
I max, min = I ave R = ( 4.92 3.4114 ) 106 mm 4
= -24.0 clockwise
I max = 8.33 106 mm 4 I min = 1.509 106 mm 4
Note: From the Mohr's circle it is seen that the a axis corresponds to I max and the b axis corresponds to I min .
PROBLEM 9.108
For a given area the moments of inertia with respect to two rectangular centroidal x and y axes are I x = 640 in 4 and I y = 280 in 4 , respectively. Knowing that after rotating the x and y axes about the centroid 60 clockwise the product of inertia relative to the rotated axes is -180 in 4 , use Mohr's circle to determine (a) the orientation of the principal axes, (b) the centroidal principal moments of inertia.
SOLUTION
Have I ave = 1 1 Ix + I y = 640 in 4 + 280 in 4 = 460 in 4 2 2
(
)
(
)
1 1 Ix - I y = 640 in 4 - 280 in 4 = 180 in 4 2 2 Also have
I xy = -180 in 4 , 2 = -120, Ix > I y
(
)
(
)
Letting the points I x , I xy and I x , I xy be denoted by X an X , respectively, three possible Mohr's circles can be constructed
(
)
(
)
Assume the first case applies Then Also Ix - I y 2 = R cos 2 m or or R cos 2 m = 180 in 4 R cos = 180 in 4
I xy = R cos
= 2 m Also have = -2 m Note 120 = 2 m + ( 90 - ) and 2 ( 2 m ) = 30 or or 2 m - = 30 2 m = = 15
2 m > 0 implies case 2 applies < 0
PROBLEM 9.108 CONTINUED
(a) Therefore,
m = 7.5 clockwise
(b) Have Then or and
R cos15 = 180 I max,
min
or
R = 186.35 in 4
= I ave R = 460 186.350
I max = 646 in 4 I min = 274 in 4
Note: From the Mohr's circle it is seen that the a axis corresponds to I max and the b axis corresponds to I min .
PROBLEM 9.109
It is known that for a given area I y = 300 in 4 and I xy = -125 in 4 , where the x and y axes are rectangular centroidal axes. If the axis corresponding to the maximum product of inertia is obtained by rotating the x axis 67.5o counterclockwise about C, use Mohr's circle to determine (a) the moment of inertia I x of the area, (b) the principal centroidal moments of inertia.
SOLUTION
First assume
Ix > I y
(Note: Assuming I x < I y is not consistent with the requirement that the axis corresponding to the I xy obtained by rotating the x axis through 67.5 counterclockwise) From Mohr's circle have (a) From 2 m = 2 ( 67.5 ) - 90 = 45 tan 2 m = 2 I xy
( )max is
Ix - I y
125 in 4 = 550 in 4 tan 45 or I x = 550 in 4
Have
Ix = I y + 2
I xy
tan 2 m
= 300 in 4 + 2
(b) Now
I ave =
1 550 + 300 4 Ix + I y = in = 425 in 4 2 2
(
)
and
R= I max,
I xy
sin 2 m
=
125 in 4 = 176.78 in 4 sin 45
Then
min
= I ave R = ( 425 176.76 ) in 4 = ( 601.78, 248.22 ) in 4 or I max = 602 in 4 and I min = 248 in 4
PROBLEM 9.110
Using Mohr's circle, show that for any regular polygon (such as a pentagon) (a) the moment of inertia with respect to every axis through the centroid is the same, (b) the product of inertia with respect to every pair of rectangular axes through the centroid is zero.
SOLUTION
Consider the regular pentagon shown, with centroidal axes x and y. Because the y axis is an axis of symmetry, it follows that I xy = 0. Since
I xy = 0, the x and y axes must be principal axes. Assuming I x = I max and I y = I min , the Mohr's circle is then drawn as shown. Now rotate the coordinate axes through an angle as shown; the resulting moments of inertia, I x and I y , and product of inertia, I xy , are indicated on the Mohr's circle. However, the x axis is an axis of symmetry, which implies I xy = 0. For this to be possible on the Mohr's circle, the radius R must be equal to zero (thus, the circle degenerates into a point). With R = 0, it immediately follows that (a)
I x = I y = I x = I y = I ave (for all moments of
inertia with respect to an axis through C)
(b)
I xy = I xy = 0 (for all products of inertia with respect to
all pairs of rectangular axes with origin at C)
PROBLEM 9.111
Using Mohr's circle, prove that the expression
2 I x I y - I xy
is
independent of the orientation of the x and y axes, where I x , I y , and
I xy represent the moments and product of inertia, respectively, of a
given area with respect to a pair of rectangular axes x and y through a given point O. Also show that the given expression is equal to the square of the length of the tangent drawn from the origin of the coordinate system to Mohr's circle.
SOLUTION
First observe that for a given area A and origin O of a rectangular coordinate system, the values of I ave and R are the same for all orientations of the coordinate axes. Shown below is a Mohr's circle, with the moments of inertia, I x and I y , and the product of inertia. I xy , having been computed for an arbitrary orientation of the xy axes. From the Mohr's circle
I x = I ave + R cos 2 I y = I ave - R cos 2 I xy = R sin 2 Then, forming the expression
2 I x I y - I xy 2
2 I x I y - I xy = ( I ave + R cos 2 )( I ave - R cos 2 ) - ( R sin 2 )
2 = I ave - R 2 cos 2 2 - R 2 sin 2 2 2 = I ave - R 2 which is a constant
(
) (
)
2 I x I y - I xy is independent of the orientation of the coordinate axes
Q.E.D. Shown is a Mohr's circle, with line OA, of length L, the required tangent. Noting that OAC is a right angle, it follows that
2 L2 = I ave - R 2 2 or L2 = I x I y - I xy
Q.E.D
PROBLEM 9.112
Using the invariance property established in the preceding problem, express the product of inertia I xy of an area A with respect to a pair of rectangular axes through O in terms of the moments of inertia I x and I y of A and the principal moments of inertia I min and I max of A about O. Use the formula obtained to calculate the product of inertia I xy of the 76 51 6.4-mm angle cross section shown in Fig. 9.13B knowing that its maximum moment of inertia is 524 103 mm 4.
SOLUTION
From Problem 9.111 have, I x I y - I xy = constant Now consider the following two cases Case 1: Case 2: Then or From Figure 9.13B: Now
With then Finally
I x = 453 103 mm 4
I x = I x , I x = I max ,
I y = I y , I y = I min ,
I xy = I xy I xy = 0
2 I x I y - I xy = I max I min
I xy = I x I y - I max I min
I y = 166 103 mm 4
I ave =
1 1 Ix + I y = I max + I min 2 2
(
)
(
)
I max = 524 103 mm 4 I min = ( 453 + 166 - 524 ) 103 mm 4 = 95.0 103 mm 4
I xy =
( 453)(166 ) - ( 524 )( 95.0 ) 103 mm4
= 159.43 103 mm 4
(a)
The two roots corresponding to the following orientations of the cross section: (a) I xy = -159.4 103 mm 4
(b)
(b) I xy = 159.4 103 mm 4
PROBLEM 9.113
The quarter ring shown has a mass m and was cut from a thin, uniform plate. Knowing that r1 = 1 r2 , determine the mass moment of inertia of 2 the quarter ring with respect to (a) axis AA, (b) the centroidal axis CC that is perpendicular to the plane of the quarter ring.
SOLUTION
First note mass = m = V = tA = t Also
4
(r
2 2
- r12
)
I mass = tI area
=
4
m
(
r22 - r12
)
I area
(a) Using Figure 9.12, Then I AA, mass =
I AA, area =
(r 16
4 2
- r14
)
m r22 - r12
( 4
)
16
(r
4 2
- r14
)
= =
m 2 r2 + r12 4
(
)
2
m 2 1 r2 + r2 4 2
m5 = r22 44 or I AA = 5 mr22 16
(b) Symmetry implies Then,
I BB, mass = I AA, mass
I OO = I AA + I BB
5 = 2 mr22 16 = 5 2 mr2 8
PROBLEM 9.113 CONTINUED
Now locate centroid C.. X A = xA or
4r 4r X r22 - r12 = 2 r22 - 1 r12 4 4 3 4 3 4
X = r = X 2
1 - r2 4 2 2 = 14 2 r = 2 3 2 1 2 9 r2 - r2 2 r23
3
or Now
4 r23 - r13 3 r22 - r12
Finally
I OO = I CC + mr 2
14 2 5 2 mr2 = I CC + m r 9 2 8
2
or
or I CC = 0.1347mr22
PROBLEM 9.114
A thin, semielliptical plate has a mass m. Determine the mass moment of inertia of the plate with respect to (a) the centroidal axis BB, (b) the centroidal axis CC that is perpendicular to the plate.
SOLUTION
First note
mass = m = V = tA = t ab 2
Also
I mass = tI area = 2m I ab area
(a) Have or
I x, area = I BB, area + Ay 2
I BB, area = I BB, area = I BB, mass = = 2m
8
4b ab3 - ab 8 2 3 64 ab3 1 - 9 2 64 ab3 1 - 9 2
2
or Then
8
ab
1 2 64 mb 1 - 4 9 2 I AA, area = I AA, mass = = 2m
or I BB = 0.0699mb 2
(b) Using Figure 9.12, Then
8
a3b a3b
8
ab
1 2 ma 4
Finally,
I CC, mass = I AA, mass + I BB, mass = 1 2 1 2 64 ma + mb 1 - 4 4 9 2 or I CC = 1 m a 2 + 0.279b 2 4
(
)
PROBLEM 9.115
The elliptical ring shown was cut from a thin, uniform plate. Denoting the mass of the ring by m, determine its moment of inertia with respect to (a) the centroidal axis BB, (b) the centroidal axis CC that is perpendicular to the plane of the ring.
SOLUTION
First note mass = m = V = tA
= t ( 2a )( 2b ) - ab
= 3 tab Also I mass = tI area = m I area 3 ab
(a) Using Figure 9.12,
I BB, area = = I BB, mass =
3 3 ( 2a )( 2b ) - ab
4 15 ab3 4 m 15 ab3 3 ab 4 or I BB = 5 2 mb 4
Then
(b) Using Figure 9.12 and symmetry, can conclude that Now I CC, mass = I AA, mass + I BB, mass = 5 2 5 2 ma + mb 4 4
I AA, mass =
5 2 ma 4
or I CC =
5 m a 2 + b2 4
(
)
PROBLEM 9.116
The machine component shown was cut from a thin, uniform plate. Denoting the mass of the component by m, determine its mass moment of inertia with respect to (a) the axis BB, (b) the centroidal axis CC that is perpendicular to the plane of the component.
SOLUTION
First note or (a) Now
I BB, mass = tI BB, area m = tA A = ( 2a ) - a 2 = 3a 2
4
t =
m 3a 2
1 m 16 1 4 3 3 1 = t ( 2a )( 2a ) - ( a )( a ) = 2 - a 3 3 3a 3 3 or I BB, mass = (b) First locate centroid a 2 2 7 yA a 4a + 2 -a Y = = = a 2 2 6 A 4a - a Symmetry implies Also Now and Then X =Y = I AA, mass = I BB, mass = 7 a 6 5 2 ma 3 10 2 ma 3 d2 = X 2 + Y 2 5 2 ma 3
( )
( )
I O, mass = I BB, mass + I AA, mass = I O, mass = I CC , mass + md 2 I CC, mass where
7 2 7 2 10 2 ma - m a + a = 3 6 6 = 22 2 ma 36 or I CC , mass = 11 2 ma 18
PROBLEM 9.117
The rhombus shown has a mass m and was cut from a thin, uniform plate. Determine the mass moment of inertia of the rhombus with respect to (a) the x axis, (b) the y axis.
SOLUTION
First note and Now 1 A = 2 2a b = 2ab 2 m = tA or
t =
m 2ab
3 1 I DD, area = 2 ( 2a )( b ) 12
= Then
1 3 ab 3
I DD, mass = tI DD, area m 1 3 1 2 = ab = mb 2ab 3 6
(a) Then
1 3 1 I x, area = 2 ( 2b )( a ) = a3b 12 3 m 1 3 I x, mass = tI x, area = a b 2ab 3 or I x, mass = 1 2 ma 6
(b) Have
I CC, mass = I x, mass + I DD, mass = = 1 m a 2 + b2 6
1 2 1 2 ma + mb 6 6
(
)
where d 2 = b2
Now
I y, mass = I CC , mass + md 2
=
1 m a 2 + b 2 + mb 2 6 or I y, mass = 1 m a 2 + 7b 2 6
(
)
(
)
PROBLEM 9.118
The rhombus shown has a mass m and was cut from a thin, uniform plate. Knowing that the AA and BB axes are parallel to the z axis and lie in a plane parallel to and at a distance a above the zx plane, determine the mass moment of inertia of the rhombus with respect to (a) the axis AA, (b) the axis BB.
SOLUTION
From Problem 9.117 (a) Have
I DD, mass = I AA, mass = I DD, mass + md 2 =
1 2 mb 6 where
d 2 = a2
1 2 2 mb + m ( a ) 6 or
I AA, mass = d 2 = a 2 + b2
1 m 6a 2 + b 2 6
(
)
(b) Have
I BB, mass = I DD, mass + md 2 =
where
1 2 mb + m a 2 + b 2 6
(
)
or
I BB, mass =
1 m 6a 2 + 7b 2 6
(
)
PROBLEM 9.119
A thin plate of mass m has the trapezoidal shape shown. Determine the mass moment of inertia of the plate with respect to (a) the x axis, (b) the y axis.
SOLUTION
First note mass = m = V = tA 1 = t ( 2a )( a ) + ( 2a )( a ) 2
= 3 ta 2
Also
I mass = tI area = m I area 3a 2
(a) Now
I x, area = ( I x )1, area + ( I x )2, area
= =
1 1 ( 2a )( a )3 + ( 2a )( a )3 3 12 5 4 a 6
m 5 a4 2 6 3a
Then
I x, mass =
or I x, mass = (b) Have
I z , area = ( I z )1, area + ( I z )2, area
5 ma 2 18
2 1 1 1 3 3 1 = ( a )( 2a ) + ( a )( 2a ) + ( 2a )( a ) 2a + 2a 2 3 3 36
= 10a 4
PROBLEM 9.119 CONTINUED
Then
I z , mass = = m 10a 4 3a 2
10 2 ma 3
Finally,
I y, mass = I x, mass + I z , mass = =
5 10 2 ma 2 + ma 18 3 65 2 ma 18
or I y, mass = 3.61ma 2
PROBLEM 9.120
A thin plate of mass m has the trapezoidal shape shown. Determine the mass moment of inertia of the plate with respect to (a) the centroidal axis CC that is perpendicular to the plate, (b) the axis AA which is parallel to the x axis and is located at a distance 1.5a from the plate.
SOLUTION
First locate the centroid C
X A = xA: or
1 X 2a 2 + a 2 = a 2a 2 + 2a + 2a a 2 3
(
)
( )
( )
X = Z A = zA:
14 a 9
1 1 Z 2a 2 + a 2 = a 2a 2 + a a 2 2 3 Z = 4 a 9
(
)
( )
( )
or (a) Have
I y,mass = I CC ,mass + m X 2 + Z 2
(
)
From the solution to Problem 9.117
I y,mass =
Then
65 2 ma 18
I cc,mass =
14 2 4 2 65 2 ma - m a + a 18 9 9
or I cc = 0.994ma 2 (b) Have and Then
I x,mass = I BB,mass + m Z
( )
2
I AA,mass = I BB,mass + m (1.5a )
2
I AA,mass = I x,mass
2 2 4 + m (1.5a ) - a 9
PROBLEM 9.120 CONTINUED
From the solution to Problem 9.117
I x,mass =
Then
5 ma 2 18
2 5 2 4 2 = ma + m (1.5a ) - a 18 9
I AA,mass
or I AA = 2.33ma 2
PROBLEM 9.121
The parabolic spandrel shown is revolved about the x axis to form a homogeneous solid of revolution of mass m. Using direct integration, express the moment of inertia of the solid with respect to the x axis in terms of m and b.
SOLUTION
At Then Now
x = a, y = b: y =
b = ka 2 b 2 x a2
or
k =
b a2
dm = r 2 dx
b = 2 x 2 dx a
2
( )
Then
m =
b2 a 4 x dx a4 0
a
1 b2 = 4 x5 5 a 0
=
1 ab 2 5
or
=
5m ab 2
Now
2 2 1 b 1 b d I x = r 2 dm = 2 x 2 2 x 2 dx 2 a 2 a
=
5m 1 b 2 4 b 2 4 5 b2 x 4 x dx = m 9 x8dx 2 a ab 2 2 a 4 a
a
Then..
5 b2 a 5 b2 1 I x = m 9 0 x8dx = m 9 x9 2 a 2 a 9 0
or I x =
5 mb 2 18
PROBLEM 9.122
Determine by direct integration the moment of inertia with respect to the z axis of the right circular cylinder shown assuming that it has a uniform density and a mass m.
SOLUTION
For the cylinder For the element shown
m = V = a 2 L dm = a 2dx = m dx L
and
dI z = dI z + x 2dm =
L 0
1 2 a dm + x 2dm 4
L
Then
1 2 1 m m 1 I z = dI z = a + x 2 dx = a 2 x + x3 3 0 4 L L 4
=
m1 2 1 3 a L+ L 3 L4
or I z =
1 m 3a 2 + 4 L2 12
(
)
PROBLEM 9.123
The area shown is revolved about the x axis to form a homogeneous solid of revolution of mass m. Determine by direct integration the moment of inertia of the solid with respect to (a) the x axis, (b) the y axis. Express your answers in terms of m and a.
SOLUTION
At Then Now
x = 2a
2a = k ( 2a )
y =
3
or
k =
1 4a 2
1 3 x 4a 2
dm = r 2dx
(
)
2
6 1 = 2 x3 dx = x dx 16a 4 4a
Then m=
=
16a 4
4
2a 6 a x dx
1
16a 7 112m 127a3
2a
x
7 a
=
112a
127 7 7 2a ) - ( a ) = a3 4 (
112
or =
(a) Now
1 1 1 6 d I x = r 2 dm = 2 x3 x dx 4 2 2 4a 16a = x6 1 6 112m 7m x dx = x12dx 4 3 4 32a 127a 16a 4064a11
2a
2
Then
Ix = = 7m 7m 1 13 2 a 12 x dx = x 11 a 4064a 4064a11 13 a
7m 57337 2 13 13 2 2a ) - ( a ) = 11 ( 52832 ma = 1.0853ma 52832a or I x = 1.085ma 2
PROBLEM 9.123 CONTINUED
Have
2 1 1 6 1 dI y = r 2 + x 2 dm = 2 x3 + x 2 x dx 4 4 4 4a 16a
=
1 112m 1 12 x + x8 dx 4 16a 127a3 64a 4
2a
Then
Iy = = =
7m 2a 1 12 7m 1 1 x + x8 dx = x13 + x8 7 a 4 7 4 9 a 127a 127a 832a 64a 7m 1 1 1 1 13 9 13 9 2a ) + ( 2a ) - a - (a) 7 4( 4( ) 9 9 127a 832a 832a 7m 8191 9 511 9 a + a = 3.67211ma 2 9 127a 7 832
or I y = 3.67ma 2
PROBLEM 9.124
Determine by direct integration the moment of inertia with respect to the x axis of the tetrahedron shown assuming that it has a uniform density and a mass m.
SOLUTION
Have and x= z = a y y + a = a 1 + h h b y y + b = b 1 + h h
2
For the element shown
y 1 1 dm = xzdy = ab 1 + dy h 2 2 1 y ab 1 + dy m = dm = 2 h
0 -h 3 1 h y = ab 1 + 2 3 h -h 0 2
Then
= =
1 3 3 abh (1) - (1 - 1) 6 1 abh 6 I AA,area 1 3 1 3 y = xz = ab 1 + 36 36 h
4
Now, for the element
Then
4 1 y dI AA, mass = tI AA, area = ( dy ) ab3 1 + h 3
PROBLEM 9.124 CONTINUED
Now dI x = dI AA,mass 2 1 2 + y + z dm 3
4
1 y = ab3 1 + dy 36 h
2 2 1 y 1 y + y 2 + b 1 + ab 1 + dy h 2 h 3 4 1 y 1 y3 y 4 + 2 dy ab3 1 + dy + ab y 2 + 2 12 h 2 h h
=
Now
m=
1 abh 6
Then and
4 1 b2 y 3m 2 y3 y 4 + 2 dy dI x = m 1 + + y + 2 h h h h 2 h
I x = dI x = -h
0
4 m 2 y y3 y 4 + 2 dy b 1 + + 6 y 2 + 2 2h h h h 0
5 1 m 2 h y 1 y4 y5 = + 2 b 1 + + 6 y 3 + 3 2h 5 h 2 h 5h -h
=
m 1 2 1 1 5 3 1 ( -h )4 + 2 ( -h )5 b h (1) - 6 ( -h ) + 2h 5 2h 5h 3 or I x = 1 m b2 + h2 10
(
)
PROBLEM 9.125
Determine by direct integration the moment of inertia with respect to the y axis of the tetrahedron shown assuming that it has a uniform density and a mass m.
SOLUTION
Have and x= z = a y y + a = a 1 + h h b y y + b = b 1 + h h
2
For the element shown
y 1 1 dm = xzdy = ab 1 + dy h 2 2 1 y ab 1 + dy m = dm = 2 h
0 -h 3 1 h y = ab 1 + 2 3 h -h 0 2
Then
= = Also Then, using
1 3 3 abh (1) - (1 - 1) 6 1 abh 6 1 3 xz 12 I DD,area = 1 3 zx 12
I BB,area =
I mass = tI area have 1 dI DD,mass = ( dy ) zx3 12
1 dI BB,mass = ( dy ) xz 3 12
PROBLEM 9.125 CONTINUED
Now dI y = dI BB,mass + dI DD,mass = 1 xz x 2 + z 2 dy 12
2 1 y ab 1 + a 2 + b 2 12 h
(
)
=
(
)
2 y 1 + dy h
Have Then
m=
1 m 2 abh dI y = a + b2 6 2h
(
)
y 1 + dy h
4
m 2 y I y = dI y = a + b 2 1 + dy 2h h
0 -h 5 m 2 h y 2 = a + b 1 + 2h 5 h -h
(
)
4
(
)
0
=
m 2 5 5 a + b 2 (1) - (1 - 1) 10 or I y = 1 m a 2 + b2 10
(
)
(
)
PROBLEM 9.126
Determine by direct integration the moment of inertia with respect to the z axis of the semiellipsoid shown assuming that it has a uniform density and a mass m.
SOLUTION
First note that when
x2 2 y = b 1 - 2 a x2 2 z = c 1 - 2 a x2 dm = ( yzdx ) = bc 1 - 2 dx a m = dm =
1 1
z = 0:
y = 0:
For the element shown
Then
a bc 1 0
-
a
x2 dx a2
1 2 = bc x - 2 x3 = abc 3 3a 0
For the element Then Now
I AA,area =
4
zy 3
dI AA,mass = tI AA,area = ( dx ) zy 3 4
dI z = dI AA,mass + x 2dm =
4
b3c 1 -
x2 x2 dx + x 2 bc 1 - 2 dx 2 a a
2
=
x2 x4 2 x4 3m b 2 1 - 2 2 + 4 + x - 2 dx 2a 4 a a a
PROBLEM 9.126 CONTINUED
Finally
I z = dI z
=
3m a b 2 x2 x4 2 x4 0 4 1 - 2 a 2 + a 4 + x - a 2 dx 2a
a
3m b 2 2 x3 1 x5 1 3 1 x5 = + x - + x - 2a 4 3 a2 5 a4 3 5 a2 0
=
3 b2 2 1 1 1 m 1 - + + a 2 - 2 4 3 5 3 5
or I z =
1 m a 2 + b2 5
(
)
PROBLEM 9.127
A thin steel wire is bent into the shape shown. Denoting the mass per unit length of the wire by m , determine by direct integration the moment of inertia of the wire with respect to each of the coordinate axes.
SOLUTION
First note
2 2 2 dy -1 = -x 3 a 3 - x 3 dx 1
Then
2 2 -2 dy 1 + = 1 + x 3 a3 - x3 dx
2
a 3 = x
2
For the element shown
dy dm = mdL = m 1 + dx dx
a 3 = m dx x
1
2
Then
m = dm =
a 0
m
a3 x
1 3
1
dx =
1 2 a 3 3 ma 3 x 3 = ma 0 2 2
Now
I x = y dm =
2
a 2 a3 0
1 3 m a 3 dx -x x1 3 2 3
5 2 1 4 1 2 a a = ma 3 0 1 - 3a 3 x 3 + 3a 3 x - x 3 dx 3 x 1 3 2 9 4 4 3 2 3 8 = ma 3 a 2 x 3 - a 3 x 3 + a 3 x 2 - x 3 4 2 8 0 2
a
3 9 3 3 3 = ma3 - + - = ma3 2 4 2 8 8 or I x = Symmetry implies Iy = 1 2 ma 4 1 2 ma 4
PROBLEM 9.127 CONTINUED
Alternative Solution I y = x dm =
1
2
a 0
x
2
m
a3
5 1 a dx = ma 3 0 x 3 dx 1 x3
1
= ma 3 = Also 1 2 ma 4
3 8 a 3 x 3 = ma3 8 0 8
I z = x 2 + y 2 dm = I y + I x or I z = 1 2 ma 2
(
)
PROBLEM 9.128
A thin triangular plate of mass m is welded along its base AB to a block as shown. Knowing that the plate forms an angle with the y axis, determine by direct integration the mass moment of inertia of the plate with respect to (a) the x axis, (b) the y axis, (c) the z axis.
SOLUTION
For line BC
=-
h x+h a 2
= Also
h ( a - 2x ) a
1 m = V = t ah 2 = 1 tah 2
1 2 dm + dm 12 2 1 2 dm 3
2
(a) Have
dI x = =
where Then
dm = t dx
a 1 2 I x = dI x = 2 0 2 ( t dx ) 3
a h 2 2 t 0 ( a - 2 x ) dx 3 a
3
=
2 h3 1 1 4 = t 3 - ( a - 2 x ) 2 0 3 a 4 2
=- =
a
1 h3 4 4 t 3 ( a - a ) - ( a ) 12 a
1 tah3 12 or I x = 1 2 mh 6
PROBLEM 9.128 CONTINUED
Now and
I = x 2dm
2 I = x 2dm = 2 0 x 2 ( t dx ) a
a h 2 = 2 t 0 x 2 ( a - 2 x ) dx a
h a 1 2 = 2 t x3 - x 4 a 3 4 0
= 2t
a
h a a 1a - a 3 2 4 2
3
4
=
1 1 ma 2 ta3h = 48 24
(b) Have
2 I y = ry2dm = x 2 + ( sin ) dm
= x 2dm + sin 2 2dm
Now
I x = 2dm I y = I + I x sin 2
=
1 1 ma 2 + mh 2 sin 2 24 6 m 2 a + 4h 2 sin 2 24
or I y = (c) Have
(
)
I z = rz2dm = x 2 + y 2 dm
2 = x 2 + ( cos ) dm
(
)
= x 2dm + cos 2 2dm
= I + I x cos 2
=
1 1 ma 2 + mh 2 cos 2 24 6 or I z =
m 2 a + 4h 2 cos 2 24
(
)
PROBLEM 9.129
Shown is the cross section of a molded flat-belt pulley. Determine its mass moment of inertia and its radius of gyration with respect to the axis AA. (The specific weight of brass is 0.306 lb/in3 and the specific weight of the fiber-reinforced polycarbonate used is 0.0433 lb/in3.)
SOLUTION
First note for the cylindrical ring shown that m = V = t and, using Figure 9.28, that I AA = = = = = 1 d2 1 d m2 - m1 1 2 2 2 2
2 2
4
(d
2 2
- d12
)
1 2 2 2 2 t 4 d 2 d 2 - t 4 d1 d1 8 1 4 4 t d 2 - d1 8 4
(
) )( )
1 2 2 2 2 t d 2 - d1 d 2 + d1 8 4
(
1 2 m d12 + d 2 8
(
)
Now treat the pulley as four concentric rings and, working from the brass outward, have
m=
0.306 lb/in 3
4 32.2 ft/s 2
( 0.875 in.) ( 0.55 in.)2 - ( 0.25 in.)2
+
1.0433 lb/in 3
4 32.2 ft/s
2
{(0.875 in.) (0.85 in.)
2
2 - ( 0.55 in.)
2 2 + ( 0.10 in.) (1.1 in.) - ( 0.85 in.)
2 2 + ( 0.475 in.) (1.4 in.) - (1.1 in.)
}
=
128.8
( 0.06426 + 0.01593 + 0.00211 + 0.015426 ) lb s 2/ft
PROBLEM 9.129 CONTINUED
Now m = 1567.38 10-6 + 388.553 10-6 + 51.465 10-6 + 376.259 10-6 lb s 2 /ft = 2383.657 10-6 lb s 2 /ft
Then
I AA = 0.25 2 0.55 2 1 1567.38 10-6 lb s 2 /ft ft + ft 8 12 12
(
)
0.55 2 0.85 2 ft + ft + 388.553 10-6 lb s 2 /ft 12 12 0.85 2 1.1 2 ft + ft + 51.465 10-6 lb s 2 /ft 12 12
+ 376.259 10
-6
1.1 2 1.4 2 lb s /ft ft + ft 12 12
2
=
1 ( 3.9728 + 2.7657 + 0.69067 + 8.2829 ) 10-6 lb ft s2 8
= 1.96401 10-6 lb ft s 2
or I AA = 1.964 10-6 lb ft s 2 and
2 k AA =
I AA 1.96401 10-6 lb ft s 2 = m 2383.657 10-6 lb s 2 /ft
= 8.23947 10- 4 ft 2
k AA = 2.87044 10- 2 ft = 0.34445 in.
or k AA = 0.344 in.
PROBLEM 9.130
Shown is the cross section of an idler roller. Determine its moment of inertia and its radius of gyration with respect to the axis AA. (The density of bronze is 8580 kg/m3; of aluminum, 2770 kg/m3; and of neoprene, 1250 kg/m3.)
SOLUTION
First note for the cylindrical ring shown that m = V = t and, using Figure 9.28, that I AA 1 d 1 d = m2 2 - m1 1 2 2 2 2 = = = =
2 2
(d 4
2 2
- d12 =
)
4
2 t d 2 - d12
(
)
1 2 2 2 2 t 4 d 2 d 2 - t 4 d1 d1 8 1 4 4 t d 2 - d1 8 4
( (
) )( )
1 2 2 2 2 t d 2 - d1 d 2 + d1 8 4 1 2 m d12 + d 2 8
(
)
2 2 - ( 0.006 m )
Now treat the roller as three concentric rings and, working from bronze the outward, have Have m=
(8580 kg/m ) ( 0.0195 m ) ( 0.009 m ) 4{
3
2 2 + 2770 kg/m3 ( 0.0165 m ) ( 0.012 m ) - ( 0.009 m ) 2 2 + 1250 kg/m3 ( 0.0165 m ) ( 0.027 m ) - ( 0.012 m )
(
)
(
)
}
[7.52895 + 2.87942 + 12.06563] 10-3 kg 4 = 5.9132 10-3 kg + 2.26149 10-3 kg
= + 9.47632 10-3 kg = 17.6510 10-3 kg
PROBLEM 9.130 CONTINUED
And I AA = 1 2 2 5.9132 10-3 kg ( 0.006 ) + ( 0.009 ) m 2 8
2 2 + 2.26149 10-3 kg ( 0.009 ) + ( 0.012 ) m 2 2 2 + 9.47632 10-3 kg ( 0.012 ) + ( 0.027 ) m 2
{(
(
)
)
(
)
}
=
1 ( 691.844 + 508.835 + 8272.827 )10-9 kg m 2 8
= 1.18419 10-6 kg m 2 or I AA = 1.184 10-6 kg m 2 Now
2 k AA =
I AA 1.18419 10-6 kg m 2 = m 17.6510 10-3 kg
= 67.08902 10-6 m 2 k AA = 8.19079 10-3 m or k AA = 8.19 mm
PROBLEM 9.131
Given the dimensions and the mass m of the thin conical shell shown, determine the moment of inertia and the radius of gyration of the shell with respect to the x axis. (Hint: Assume that the shell was formed by removing a cone with a circular base of radius a from a cone with a circular base of radius a + t . In the resulting expressions, neglect terms containing t2, t3, etc. Do not forget to account for the difference in the heights of the two cones.)
SOLUTION
First note or
h h = a+t a h = h (a + t ) a
For a cone of height H whose base has a radius r, have where Ix = 3 mr 2 10
m = V = Then Ix = = Now, following the hint have mshell = mouter - minner = =
3
r 2H
3 2 2 r H r 10 3
10
r 4H
3
( a + t ) h - a 2h
2
3
( a + t )
2
h ( a + t ) - a 2h a
=
3
a 2 h 1 +
3 t t 2 - 1 = a h 1 + 3 + ... - 1 a 3 a
Neglecting the t2 and t3 terms obtain mshell aht
PROBLEM 9.131 CONTINUED
Also
( I x )shell = ( I x )outer - ( I x )inner
= =
10
( a + t ) h - a 4h
4
10
( a + t )
4
h ( a + t ) - a 4h a
=
10
a 4h 1 +
5 t - 1 a
=
10
a 4h 1 + 5
t + ... - 1 a
Neglecting t2 and higher order terms, obtain
( I x )shell
2
a3ht
or I x = 1 2 ma 2
Now
1 2 ma Ix 2 kx = = 2 m m
or k x =
a 2
PROBLEM 9.132
A portion of an 8-in.-long steel rod of diameter 1.50 in. is turned to form the conical section shown. Knowing that the turning process reduces the moment of inertia of the rod with respect to the x axis by 20 percent, determine the height h of the cone.
SOLUTION
8-in. rod and Therefore,
( I x )0 = 1 m0a 2 2
a = 0.75 in.
where and
m0 = V0 = a 2 L L = 8 in.
(
)
( I x )0 = 1 a4 L 2
Ix = Ix
Rod and cone where and
3 ( )cyl + ( I x )cone = 1 mcyla2 + 10 mconea 2 2
mcyl = Vcyl = a 2 ( L - h ) 1 mcone = Vcone = a 2h 3 Ix =
Then Given Then
1 1 a 4 ( L - h ) + a 4h 2 10
I x = 0.8 I x
( )0
1 1 1 a 4 ( L - h ) + a 4h = 0.8 a 4 L 2 10 2 5 5 1 4 L- h+ h= L 10 10 10 10
h=
or
or
1 1 L = ( 8.00 in.) = 2.00 in. 4 4 or h = 2.00 in.
PROBLEM 9.133
The steel machine component shown is formed by machining a hemisphere into the base of a truncated cone. Knowing that the density of steel is 7850 kg/m3, determine the mass moment of inertia of the component with respect to the y axis.
SOLUTION
First note and Now
L L - 240 = 120 80 m = stV
or
L = 720 mm
2 1 2 m1 = st a1 h1 = 7850 kg/m3 ( 0.120 m ) ( 0.720 m ) 3 3
= 85.230 kg
2 2 2 2 m2 = st a2 = 7850 kg/m3 ( 0.090 m ) = 11.9855 kg 3 3 2 1 2 m3 = st a3 h3 = 7850 kg/m3 ( 0.080 m ) ( 0.720 - 0.240 ) m 3 3
= 25.253 kg
Now
Iy = Iy
( )1 - ( I y )2 - ( I y )3
PROBLEM 9.133 CONTINUED
where (using Figure 9.28) 3 3 ( I y )1 = 10 m1a12 = 10 (85.230 kg )( 0.120 m )2 = 0.36819 kg m2 2 ( I y )2 = 1 ( I y )sphere = 1 5 mspherea22 2 2
=
where
msphere = 2mhemisphere
12 2 2 11.9855 kg ( 0.090 m ) 25
= 0.038833 kg m 2
3 3 ( I y )3 = 10 m3a32 = 10 ( 25.253 kg )( 0.080 m )2 = 0.048486 kg m2 Then
I y = ( 0.36819 - 0.038833 - 0.048486 ) kg m 2 = 0.28087 kg m 2
= 0.281 kg m 2
or I y = 281 10-3 kg m 2
PROBLEM 9.134
After a period of use, one of the blades of a shredder has been worn to the shape shown and is of weight 0.4 lb. Knowing that the moments of inertia of the blade with respect to the AA and BB axes are 0.6 10-3 lb ft s 2 and 1.26 10-3 lb ft s 2 , respectively, determine (a) the location of the centroidal axis GG, (b) the radius of gyration with respect to axis GG.
SOLUTION
Have (a) and
dB =
4 - d A = ( 0.33333 - d A ) ft 12
2 I AA = I GG + md A 2 I BB = I GG + md B
Then
2 2 I BB - I AA = m d B - d A
(
)
)
2 2 = m ( 0.33333 - d A ) -d A
= m ( 0.11111 - 0.66666d A )
Then
(1.26 - 0.6 ) 10-3 lb ft s2
= 0.40 lb ( 0.11111 - 0.66666d A ) ft 2 32.2 ft/s 2
or or (b) or
d A = 0.08697 ft d A = 1.044 in.
2 I AA = I GG + md A
I GG = 0.6 10-3 lb ft s 2 - 0.4 lb ( 0.08697 ft )2 32.2 ft /s 2
= 0.50604 10-3 lb ft s 2 Then
2 kGG =
I GG 0.50604 10-3 lb ft s 2 = 0.4 lb m 32.2 ft/s 2
= 0.04074 ft 2
kGG = 0.20183 ft = 2.4219 in. or kGG = 2.42 in.
PROBLEM 9.135
The cups and the arms of an anemometer are fabricated from a material of density . Knowing that the moment of inertia of a thin, hemispherical shell of mass m and thickness t with respect to its centroidal axis GG, is 5ma2/12, determine (a) the moment of inertia of the anemometer with respect to the axis AA, (b) the ratio of a to l for which the centroidal moment of inertia of the cups is equal to 1 percent of the moment of inertia of the cups with respect to the axis AA.
SOLUTION
(a) First note and marm = Varm = dmcup = dVcup
4
d 2l
= ( 2 a cos )( t )( ad )
Then
2 mcup = dmcup = 0 2 a 2t cos d 2 = 2 a 2t [sin ]0
= 2 a 2t Now
( I AA )anem. = ( I AA )cups + ( I AA )arms
Using the parallel-axis theorem and assuming the arms are slender rods, have
( I AA )anem.
2 = 3 ( I GG )cup + mcup d AG
+ 3 I arm + marm d AGarm
2 5 2 a = 3 mcup a 2 + mcup ( l + a ) + 12 2 2 1 l 2 +3 marml + marm 2 2
5 = 3mcup a 2 + 2la + l 2 + marml 2 3 5 = 3 2 a 2t a 2 + 2la + l 2 + d 2l l 2 3 4
(
)
( )
or
( I AA )anem
5 a2 d 2l a = l 2 6a 2t 2 + 2 + 1 + 3 l l 4
PROBLEM 9.135 CONTINUED
(b) Have
( IGG )cup ( I AA )cup
= 0.01
or
5 5 mcup a 2 = 0.01mcup a 2 + 2la + l 2 12 3 a l 5 5 2 = 0.12 2 + 2 + 1 3 40 2 - 2 - 1 = 0
( From Part a )
Now let = Then or Then or
=
2
( -2 )2 - 4 ( 40 )( -1) 2 ( 40 )
and
= 0.1851
= - 0.1351
a = 0.1851 l
PROBLEM 9.136
A square hole is centered in and extends through the aluminum machine component shown. Determine (a) the value of a for which the mass moment of inertia of the component with respect to the axis AA, which bisects the top surface of the hole, is maximum, (b) the corresponding values of the mass moment of inertia and the radius of gyration with respect to the axis AA. (The density of aluminum is 2800 kg / m3 .)
SOLUTION
First note And m1 = V1 = b 2 L m2 = V2 = a 2 L
I AA = ( I AA )1 - ( I AA )2
2 1 a = m1 b 2 + b 2 + m1 2 12
(a) Using Figure 9.28 and the parallel-axis theorem, have
(
)
2 1 a - m2 a 2 + a 2 + m2 2 12
(
)
1 1 5 = b2 L b2 + a 2 - a 2 L a 2 4 6 12
(
)
(
)
= Then or Also.. Now, for a = 0,
L
( 2b 12
4
+ 3b 2a 2 - 5a 4
) )
dI AA L = 6b 2a - 20a3 = 0 da 12
(
a=0
and
a=b
3 10
d 2 I AA L 1 6b 2 - 60a 2 = L b 2 - 10a 2 = 2 12 2 da
(
)
(
)
d 2 I AA >0 da 2
and for
a=b a=b
3 d 2 I AA , <0 10 da 2
( I AA )max
occurs when
a = 84
3 10
3 = 46.009 mm 10
a = 46.0 mm
or
PROBLEM 9.136 CONTINUED
(b) From part (a)
( I AA )mass
2 4 3 3 = 2b + 3b b - 5 b 12 10 10 4 2
L
=
49 49 4 2800 kg/m3 ( 0.3 m )( 0.4 m ) Lb 4 = 240 240
(
)
= 8.5385 10-3 kg m 2
or
( I AA )mass
m
= 8.54 10-3 kg m 2
and where
2 k AA =
( I AA )mass
m = m1 - m2 = L b - a
(
2
2
)
2 3 7 2 2 = L b - b 10 = 10 Lb
Then
2 k AA
49 Lb 4 7 2 7 240 = = b = (84 mm )2 = 2058 mm2 7 24 24 2 Lb 10
k AA = 45.3652 mm or k AA = 45.4 mm
To the instructor: The following formulas for the mass moment of inertia of thin plates and a half cylindrical shell are derived at this time for use in the solutions of Problems 9.1379.142 Thin rectangular plate
( I x )m
= I x =
( )m + md 2
b 2 h 2 1 m b 2 + h 2 + m + 12 2 2
(
)
=
1 m b2 + h2 3
(
)
2
( I y )m = ( I y )m + md 2
= = 1 b mb 2 + m 12 2 1 2 mb 3
I z = I z = =
( )m + md 2
1 h mh 2 + m 12 2 1 2 mh 3
2
Thin triangular plate Have and Then 1 m = V = bht 2 I z , area = 1 3 bh 36
I z , mass = tI z , area = t = 1 3 bh 36
1 mh 2 18
Similarly, Now Thin semicircular plate Have And Then m = V = a 2t 2 I y, area = I z, area =
I y, mass = I x, mass = I y, mass
1 mb 2 18 1 + I z , mass = m b2 + h2 18
(
)
8
a4
I y, mass = I z , mass = tI y, area = t =
8
a4
1 2 ma 4 I x, mass = I y, mass + I z , mass = 1 2 ma 2
Now Also And
I x, mass = I x, mass + my 2 I z , mass = I z, mass + my 2
or or
16 2 1 I x, mass = m - a 2 2 9 16 2 1 I z, mass = m - a 2 4 9 4a 3
Thin Quarter-Circular Plate Have and Then m = V = a 2t 4 I y, area = I z, area =
y = z =
16
a4
I y, mass = I z , mass = tI y, area = t =
16
a4
1 2 ma 4
Now Also or and or
I x, mass = I y, mass + I z , mass =
1 2 ma 2
I x, mass = I x, mass + m y 2 + z 2 32 2 1 I x, mass = m - a 2 2 9
I y, mass = I y, mass + mz 2
(
)
16 2 1 I y, mass = m - a 2 4 9
PROBLEM 9.137
A 0.1-in-thick piece of sheet metal is cut and bent into the machine component shown. Knowing that the specific weight of steel is 0.284 lb / in 3 , determine the moment of inertia of the component with respect to each of the coordinate axes.
SOLUTION
m1 = V = =
g
tA
0.284 lb/in 3 ( 0.10 in.)( 38 in.)( 24 in.) 32.2 ft/s 2
= 0.80438 lb s 2 /ft m2 = 1 0.284 ( 0.1 in.) ( 38 in.)( 24 in.) 32.2 2
= 0.402186 lb s 2 /ft m3 = 0.284 ( 0.1 in.) ( 24 in.)2 = 0.39900 lb s2/ft 32.2 4
Using Fig. 9.28 for component 1 and the equations derived for components 2 and 3, have I x = ( I x )1 + ( I x )2 + ( I x )3 1 24 2 24 2 2 2 = 0.80438 lb s 2 /ft + ft + 0.402186 lb s /ft 12 12 2 12
(
)
(
)
1 24 2 24 2 2 + ft 3 12 18 12
1 24 2 + 0.39900 lb s 2 /ft ft 2 2 12
(
)
= (1.0725 + 0.26812 + 0.7980 ) lb ft s 2 = 2.13862 lb ft s 2 or I x = 2.14 lb ft s 2
PROBLEM 9.137 CONTINUED
Also Iy = Iy
( )1 + ( I y )2 + ( I y )3
1 38 2 24 2 38 2 24 2 2 = 0.80438 lb s 2 /ft + ft 2 + + ft 12 12 12 2 12 2 12
(
)
1 38 2 38 2 2 + 0.402186 lb s 2 /ft + ft 3 12 18 12
(
)
1 24 2 + 0.39900 lb s 2 /ft ft 2 4 12
(
)
= ( 3.76122 + 0.672172 + 0.3990 ) lb ft s 2 = 4.83234 lb ft s 2 or I y = 4.83 lb ft s 2 I z = ( I z )1 + ( I z )2 + ( I z )3 1 38 2 38 2 2 = 0.80438 lb s 2 /ft + ft 2 12 12 12
(
)
1 38 2 24 2 38 2 24 2 2 + 0.401286 lb s 2 /ft + + + ft 12 3 12 3 12 18 12
(
)
1 24 2 + 0.3990 lb s 2 /ft ft 2 4 12
(
)
= ( 2.68871 + 0.940296 + 0.3990 ) lb ft s 2 = 4.0280 lb ft s 2 or I z = 4.03 lb ft s 2
PROBLEM 9.138
A 3-mm-thick piece of sheet metal is cut and bent into the machine component shown. Knowing that the density of steel is 7850 kg / m3 , determine the moment of inertia of the component with respect to each of the coordinate axes.
SOLUTION
First compute the mass of each component. Have m = stV = st tA m1 = 7850 kg/m3 ( 0.003 m )( 0.70 m )( 0.780 m ) = 12.858 kg 2 m2 = 7850 kg/m3 ( 0.003 m ) ( 0.39 m ) = 5.6265 kg 2
(
)
(
)
1 m3 = 7850 kg/m3 ( 0.003 m ) ( 0.780 m )( 0.3 m ) = 2.7554 kg 2
(
)
Using Fig. 9-28 for component 1 and the equations derived above for components 2 and 3 have
I x = ( I x )1 + ( I x )2 + ( I x )3
2 1 2 0.78 2 = (12.858 kg ) ( 0.78 ) + m 2 12
4 0.39 2 1 16 2 + ( 5.6265 kg ) - + ( 0.39 ) m 2 ( 0.39 )2 + 2 3 2 9
2 2 2 2 0.78 1 0.30 2 + ( 2.7554 kg ) ( 0.78 ) + ( 0.30 ) + + m 3 18 3
= ( 2.6076 + 1.2836 + 0.3207 ) kg m 2 = 4.2119 kg m 2
or I x = 4.21 kg m 2
PROBLEM 9.138 CONTINUED
And
Iy = Iy
( )1 + ( I y )2 + ( I y )3
1 2 2 2 2 1 = (12.858 kg ) ( 0.7 ) + ( 0.78 ) + ( 0.7 ) + ( 0.78 ) m 2 4 12
2 2 1 + ( 5.6265 kg ) ( 0.39 ) + ( 0.39 ) m 2 4 2 1 2 2 0.78 4 + ( 2.7554 kg ) ( 0.78 ) + ( 0.7 ) + m 3 18
= ( 4.7077 + 1.0697 + 1.6295 ) kg m 2 = 7.4069 kg m 2
or I y = 7.41 kg m 2 And
I z = ( I z )1 + ( I z )2 + ( I z )3 1 2 1 = (12.858 kg ) ( 0.7 ) + m2 12 4
2 1 + ( 5.6265 kg ) ( 0.39 ) m 2 4 2 1 2 2 0.30 2 + ( 2.7554 kg ) ( 0.3) + ( 0.70 ) + m 18 3
= ( 2.1001 + 0.21395 + 1.39145 ) kg m 2 = 3.7055 kg m 2
or I z = 3.71 kg m 2
PROBLEM 9.139
The cover of an electronic device is formed from sheet aluminum that is 2 mm thick. Determine the mass moment of inertia of the cover with respect to each of the coordinate axes. (The density of aluminum is 2770 kg / m3 .)
SOLUTION
Have Now
m = V
m1 = 2770 kg/m3 ( 0.002 m )( 0.048 m )( 0.06 m ) = 0.015955 kg m2 = 2770 kg/m3 ( 0.002 m )( 0.06 m )( 0.124 m ) = 0.041218 kg m3 = 2770 kg/m3 ( 0.002 m )( 0.048 m )( 0.124 m ) = 0.032974 kg
(
)
(
)
(
)
Using Fig. 9.28 and the parallel axis theorem, have
I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x )4
and
( I x )3 = ( I x ) 4
1 1 2 1 2 1 = ( 0.015955 kg ) ( 0.048 ) + m 2 + ( 0.041218 kg ) ( 0.124 ) + m2 12 4 12 4 1 2 2 2 1 + 2 ( 0.032974 kg ) ( 0.048 ) + ( 0.124 ) + m 12 4 = 0.012253 10-3 + 0.211256 10-3 + 0.388654 10-3 kg m 2 = 0.61216 10-3 kg m 2 or I x = 0.612 10-3 kg m 2
(
)
PROBLEM 9.139 CONTINUED
Iy = Iy
( )1 + ( I y )2 + ( I y )3 + ( I y )4
2 2 2 1 1 = ( 0.015955 kg ) ( 0.06 ) m 2 + ( 0.041218 kg ) ( 0.06 ) + ( 0.124 ) m 2 3 3 1 2 1 2 2 2 1 + ( 0.032974 kg ) ( 0.124 ) m 2 + ( 0.032974 kg ) ( 0.124 ) + ( 0.06 ) + ( 0.124 ) m 2 3 12 4 = ( 0.019146 + 0.260718 + 0.16900 + 0.287709 ) 10-3 kg m 2 = 0.73657 10-3 kg m 2 or I y = 0.737 10-3 kg m 2 I z = ( I z )1 + ( I z )2 + ( I z )3 + ( I z )4 2 2 1 2 1 = ( 0.015955 kg ) ( 0.06 ) + ( 0.048 ) m 2 + ( 0.041218 kg ) ( 0.06 ) m 2 3 3 1 2 1 2 2 + ( 0.032974 kg ) ( 0.048 ) m 2 + ( 0.032974 kg ) ( 0.048 ) + ( 0.06 ) m 2 3 3 = ( 0.031399 + 0.049462 + 0.025324 + 0.14403) 10-3 kg m 2 = 0.250215 103 kg m 2 or I z = 0.250 10-3 kg m 2
PROBLEM 9.140
A framing anchor is formed of 2-mm-thick galvanized steel. Determine the mass moment of inertia of the anchor with respect to each of the coordinate exes. (The density of galvanized steel is 7530 kg/m3.)
SOLUTION
First compute the mass of each component Have Now m = V = At m1 = 7530 kg/m3 ( 0.002 m )( 0.045 m )( 0.070 m ) = 0.047439 kg m2
3
( ) = ( 7530 kg/m ) ( 0.002 m )( 0.045 m )( 0.020 m ) = 0.013554 kg
( )
1 ( 0.04 m )( 0.095 m ) = 0.028614 kg 2
m3 = 7530 kg/m3 ( 0.002 m )
Using Fig. 9.28 for components 1 and 2 and the equations derived above for components 3, have I x = ( I x )1 + ( I x )2 + ( I x )3
2 2 2 2 1 1 = ( 0.047439 kg ) ( 0.07 ) m 2 + ( 0.013554 kg ) ( 0.020 ) + ( 0.07 ) + ( 0.01) m 2 3 12
1 2 2 2 2 1 + ( 0.028614 kg ) ( 0.095 ) + ( 0.04 ) + ( 2 0.095 ) + ( 0.040 ) m 2 9 18 = ( 0.077484 + 0.068222 + 0.136751) 10-3 kg m 2 = 0.282457 10-3 kg m 2 or I x = 0.2825 10-3 kg m 2
PROBLEM 9.140 CONTINUED
Iy = Iy
( )1 + ( I y )2 + ( I y )3
2 2 2 1 1 = ( 0.047439 kg ) ( 0.045 ) m 2 + ( 0.013554 kg ) ( 0.045 ) + ( 0.02 ) m 2 3 3 1 2 2 2 1 + ( 0.028614 kg ) ( 0.04 ) ( 0.045 ) + ( 0.04 ) m 2 9 18 = ( 0.03202 + 0.010956 + 0.065574 ) 10-3 kg m 2 = 0.10855 103 kg m 2 or I y = 0.1086 103 kg m 2
I z = ( I z )1 + ( I z )2 + ( I z )3
2 2 1 21 2 = ( 0.047439 kg ) ( 0.045 ) + ( 0.070 ) m 2 + ( 0.013554 kg ) ( 0.045 ) + ( 0.070 ) m 2 3 3
2 1 2 2 2 2 + ( 0.028614 kg ) ( 0.095 ) + 0.045 + 0.095 m 3 18
= ( 0.0109505 + 0.075564 + 0.187064 ) 10-3 kg m3 = 0.37213 10-3 kg m 2 or I z = 0.372 10-3 kg m 2
PROBLEM 9.141
A 2-mm-thick piece of sheet steel is cut and bent into the machine component shown. Knowing that the density of steel is 7850 kg/m3, determine the moment of inertia of the component with respect to each of the coordinate axes.
SOLUTION
Have Then
m = stV = sttA
m1 = 7850 kg/m3 0.002 m ( 0.320 0.240 ) m 2
= 1.20576 kg
m2 = 7850 kg/m3 0.002 m ( 0.320 0.180 ) m 2
= 0.90432 kg
1 m3 = 7850 kg/m3 0.002 m 0.180 0.240 m 2 2 = 0.33912 kg
2 m4 = 7850 kg/m3 0.002 m ( 0.100 m ) 2
= 0.24662 kg Using Fig. 9-2B for components 1 and 2 and the equations derived above for components 3 and 4, have Now where I x = ( I x )1 + ( I x )2 + ( I x )3 - ( I x )4
( I x )1 = (1.20576 kg )( 0.240 m )2
= 23.151 10-3 kg m 2
1 3
( I x )2
=
1 ( 0.90432 kg )( 0.180 m )2 3
= 9.7667 10-3 kg m 2
( I x )3
=
1 ( 0.33912 kg ) ( 0.180 )2 + ( 0.240 )2 m2 6
= 5.0868 10-3 kg m 2
PROBLEM 9.141 CONTINUED
( I x )4
=
1 ( 0.24662 kg )( 0.100 m )2 4
= 0.61655 10-3 kg m 2 Then
I x = ( 23.151 + 9.7667 + 5.0868 - 0.61655 ) 10-3 kg m 2 or I x = 37.4 10-3 kg m 2
And where
Iy = Iy
( )1 + ( I y )2 + ( I y )3 - ( I y )4
( I y )1 = 1 (1.20576 kg ) ( 0.320)2 + ( 0.240 )2 m2 3
= 64.307 10-3 kg m 2
( I y )2 = 1 ( 0.90432 kg )( 0.320 m )2 3
= 30.867 10-3 kg m 2
( I y )3 = 1 ( 0.33912 kg )( 0.240 m )2 6
= 3.2556 10-3 kg m 2
( I y )4 = ( 0.24662 kg ) 1 - 9162 ( 0.100 m )2 2
2 2 4 0.100 2 + ( 0.160 ) + m 3
= 7.5466 10-3 kg m 2 Then I y = ( 64.307 + 30.067 + 3.2556 - 7.5466 )10-3 kg m 2 or I y = 90.9 10-3 kg m 2 And where 1 3
I z = ( I z )1 + ( I z )2 + ( I z )3 - ( I z )4
( I z )1 = (1.20576 kg )( 0.320 m )2 ( I z )2
=
= 41.157 10-3 kg m 2
1 ( 0.90432 kg ) ( 0.320 )2 + ( 0.180 )2 m2 3
= 40.634 10-3 kg m 2
( I z )3
=
1 ( 0.33912 kg )( 0.180 m )2 = 1.83125 10-3 kg m2 6
PROBLEM 9.141 CONTINUED
( I z )4 = ( 0.24662 kg ) ( 0.100 m )2 + ( 0.160 m )2
= 6.9300 10-3 kg m 2 Then
1 4
( I z )z
= ( 41.157 + 40.634 + 1.83125 - 6.9300 ) 10-3 kg m 2 or I z = 76.7 10-3 kg m 2
PROBLEM 9.142
The piece of roof flashing shown is formed from sheet copper that is 0.032 in. thick. Knowing that the specific weight of copper is 558 lb/ft3, determine the mass moment of inertia of the flashing with respect to each of the coordinate axes.
SOLUTION
Have
m = copperV
=
copper
g
tA
Then
m1 = m2
=
558 lb/ft 3 1 ft 0.032 in. ( 48 6 ) in 2 2 32.2 ft/s 12 in.
3
= 0.092422 lb s 2 /ft Using Fig. 9-2B for components 1 and 2, have Now Then
I x = ( I x )1 + ( I x )2
and
( I x )1 = ( I x )2
2 1 6 ft = 1.54037 lb ft s 2 I x = 2 0.092422 lb s 2 /ft 12 3
(
)
or I x = 1.54 10-3 lb ft s 2
Iy = Iy
( )1 + ( I y )2
2 2 6 + ft 2 12
where
48 ( I y )1 = 1 ( 0.092422 lb s2/ft ) 12 3 = 500.62 10-3 lb ft s 2
and where
( I y )2 = ry2dm
ry2 = x 2 + z 2 = x 2 + ( cos30 )
2
PROBLEM 9.142 CONTINUED
and
dm = dV =
copper
g
td dx
Then
( I y )2 =
=
copper
g
L a t 0 0 x 2 + 2 cos 2 30 d dx
(
)
copper
g
1 L t 0 ax 2 + a3 cos 2 30 dx 3
=
1 copper t aL3 + a3 L cos 2 30 3 g
(
)
A = aL
=
1 m2 L2 + a 2 cos 2 30 3 1 0.092422 lb s 2 /ft 3
(
)
2
=
(
) ( 4 )
1 + cos 30 2
2
ft 2
= 498.69 10-3 lb ft s 2 Finally,
I y = ( 500.62 + 498.69 ) 10-3 lb ft s 2 or I y = 999 10-3 lb ft s 2
I z = ( I z )1 + ( I z )2
where
( I z )1 =
1 48 0.092422 lb s 2/ft ft 3 12
(
)
2
= 492.92 10-3 lb ft s 2 and where and Then
( I z )2
= rz2dm
rz2 = x 2 + y 2 y = sin 30
( I z )2
2 = x 2 + ( sin 30 ) dm
PROBLEM 9.142 CONTINUED
Similarly, as I y
( )2
= 1 m2 L2 + a 2 sin 2 30 3
( I z )2
(
)
1 = 0.092422 lb s 2 /ft 3
(
)
2 1 2 2 ( 4 ) + sin 30 ft 2 2
= 494.84 10-3 lb ft s 2 Then I x = ( 492.92 + 494.84 ) 10-3 lb ft s 2 or I z = 988 10-3 lb ft s 2
PROBLEM 9.143
The machine element shown is fabricated from steel. Determine the mass moment of inertia of the assembly with respect to (a) the x axis, (b) the y axis, (c) the z axis. (The specific weight of steel is 0.284 lb / in 3. )
SOLUTION
Have m = V =
g
V =
0.284 lb/in 3 V 32.2 ft/s 2
= 0.0088199 lb s 2 /ft in 3 V
2 Then m1 = 0.0088199 lb s 2 /ft in 3 ( 4 ) ( 2 ) in 3 = 0.88667 lb s 2 /ft 2 m2 = 0.0088199 lb s 2 /ft in 3 (1) ( 3) in 3 = 0.083126 lb s 2 /ft 2 m3 = 0.0088199 lb s 2 /ft in 3 (1) ( 2 ) in 3 = 0.055417 lb s 2 /ft
(
)
(
)
(
)
(
)
Using Fig. 9-28 and the parallel theorem, have (a) I x = ( I x )1 + ( I x )2 - ( I x )3
2 2 2 1 1 ft = 0.88667 lb s 2 /ft 3 ( 4 ) + ( 2 ) + (1) in 2 12 12 in.
(
)
2
2 2 2 1 1 ft + 0.083126 lb s 2 /ft 3 (1) + ( 3) + (1.5 ) in 2 12 12 in. 2 2 2 1 1 ft - 0.055417 lb s 2 /ft 3 (1) + ( 2 ) + (1) in. 12 12 in.
(
)
2
(
)
2
= 0.034106 lb ft s 2 or I x = 0.0341 lb ft s 2
PROBLEM 9.143 CONTINUED
(b) Iy = Iy
( )1 + ( I y )2 - ( I y )3
2 1 1 ft = 0.88667 lb s 2 /ft ( 4 ) in 2 2 12 in.
(
)
2
2 1 2 1 ft + 0.083126 lb s 2 /ft (1) + ( 2 ) in 2 2 12 in.
(
)
2
-
( 0.055417 lb s /ft ) 1 (1) 2
2
2
2 1 ft + ( 2 ) in 2 12 in.
2
= 5.0125 10-2 lb ft s 2
or I y = 0.0501 lb ft s 2
(c)
I z = ( I z )1 + ( I z )2 - ( I z )3
2 2 2 1 1 ft = 0.88667 lb s /ft 3 ( 4 ) + ( 2 ) + (1) in 2 12 12 in. 2
(
)
2
2 2 2 2 1 1 ft + 0.083126 lb s /ft 3 (1) + ( 3) + ( 2 ) + (1.5) in 2 12 12 in. 2 2 2 2 2 1 1 ft - 0.055417 lb s /ft 3 (1) + ( 2 ) + ( 2 ) + (1) in 2 12 12 in. 2
(
)
2
(
)
2
= 0.034876 lb ft s 2 or I z = 0.0349 lb ft s 2
To the Instructor: The following formulas for the mass of inertia of a semicylinder are derived at this time for use in the solutions of Problems 9.1449.147. From Figure 9.28 Cylinder:
( I x ) cyl
=
1 mcyla 2 2
1 ( I y )cyl = ( I z )cyl = 12 mcyl ( 3a2 + L2 ) Symmetry and the definition of the mass moment of inertia I = r 2dm imply
(
)
( I )semicylinder
=
1 (I ) 2 cylinder
( I x )sc
=
11 2 mcyla 2 2
and
1 ( I y )sc = ( I z )sc = 1 12 mcyl ( 3a2 + L2 ) 2 msc = 1 mcyl 2 = 1 msca 2 2
However,
Thus,
( I x )sc
and
1 ( I y )sc = ( I z )sc = 12 msc (3a2 + L2 )
Also, using the parallel axis theorem find 16 2 1 I x = msc - a 2 9 2 1 16 2 1 2 I z = msc - a + L 2 12 4 9 where x and z are centroidal axes.
PROBLEM 9.144
Determine the mass moment of inertia of the steel machine element shown with respect to the y axis. (The density of steel is 7850 kg/m3.)
SOLUTION
Have Then
m = steel V m1 = 7850 kg/m3 ( 0.200 0.120 0.600 ) m3 = 113.040 kg m2 = 7850 kg/m3 ( 0.200 0.080 0.360 ) m3 = 45.216 kg
2 m3 = 7850 kg/m3 ( 0.100 ) ( 0.120 ) m3 2
= 14.7969 kg
2 m4 = 7850 kg/m3 ( 0.050 ) ( 0.120 ) m3
= 7.3985 kg
Using Figure 9.28 for components 1 and 2 and the equations derived above for components 3 and 4, have Now
Iy = Iy
( )1 + ( I y )2 + ( I y )3
2
where
1 ( I y )1 = (113.040 kg ) 12 ( 0.600 )2 + ( 0.200)2 + 0.600 2
0.200 + 2
2
m2
= 15.0720 kg m 2
PROBLEM 9.144 CONTINUED
( )2
Iy
1 0.360 2 0.200 2 2 2 2 = ( 45.216 kg ) ( 0.360 ) + ( 0.200 ) + + m 2 12 2
= 2.5562 kg m 2
( )3
Iy
2 16 4 0.100 2 1 = (14.7969 kg ) - ( 0.100 )2 + ( 0.100 )2 + 0.600 + m 2 3 2 9
= 6.3024 kg m 2
( I y )4 = ( 7.3985 kg ) 1 ( 0.050)2 + ( 0.100 )2 + ( 0.600)2 m2 2
= 2.7467 kg m 2
Then
I y = (15.0720 + 2.5562 + 6.3024 - 2.7467 ) kg m 2
= 21.1839 kg m 2
or I y = 21.2 kg m 2
PROBLEM 9.145
Determine the mass moment of inertia of the steel machine element shown with respect to the z axis. (The density of steel is 7850 kg/m3.)
SOLUTION
See machine elements shown in Problem 9.145 Also note
m1 = 113.040 kg m2 = 45.216 kg m3 = 14.7969 kg m4 = 7.3985 kg
Using Fig. 9.28 for components 1 and 2 and the equations derived above for components 3 and 4, have Now where
I z = ( I z )1 + ( I z )2 + ( I z )3 - ( I z )4
( I z )1 = (113.040 kg )
= 2.0498 kg m 2
2 2 1 ( 0.200 )2 + ( 0.120 )2 + 0.200 + 0.120 m 2 2 2 12
( I z )2
2 2 2 2 1 = ( 45.216 kg ) ( 0.200 ) + ( 0.080 ) + ( 0.100 ) + ( 0.160 ) m 2 12
= 1.78453 kg m 2
( I z )3 = (14.7969 kg )
1 2 2 2 2 2 3 ( 0.100 ) + ( 0.120 ) + ( 0.100 ) + ( 0.060 ) m 12
= 0.25599 kg m 2
( I z )4 = ( 7.3985 kg )
1 2 2 2 2 2 3 ( 0.050 ) + ( 0.120 ) + ( 0.100 ) + ( 0.060 ) m 12
= 0.114122 kg m 2 Then I z = ( 2.0498 + 1.78453 + 0.25599 - 0.114122 ) kg m 2 = 3.97629 kg m 2
or I z = 3.98 kg m 2
PROBLEM 9.146
An aluminum casting has the shape shown. Knowing that the specific weight of aluminum is 0.100 lb / in 3 , determine the moment of inertia of the casting with respect to the z axis.
SOLUTION
Have
m = V =
g
V =
0.10 lb/in 3 V 32.2 ft/s 2
= 0.0031056 lb s 2 /ft in 3 V Then m1 = 0.0031056 lb s 2 /ft in 3 (12 in.)( 3 in.)( 4.8 in.) = 0.53665 lb s 2 /ft m2 = 0.0031056 lb s 2 /ft in 3 (1.5 in.)( 4.8 in.)( 3 in.) = 0.06708 lb s 2 /ft
2 m3 = 0.0031056 lb s 2 /ft in 3 ( 2.4 in.) (1.5 in.) 2
(
)
(
)
(
)
(
)
= 0.042148 lb s 2 /ft
2 m4 = m5 = 0.0031056 lb s 2 /ft in 3 ( 2.75 in.) (1.0 in.) 2
(
)
= 0.036892 lb s 2 /ft
PROBLEM 9.146 CONTINUED
Using Fig. 9.28 for components 1 and 2 and the equations derived above for components 3, 4, and 5, have I z = ( I z )1 + ( I z )2 + ( I z )3 + ( I z )4 + ( I z )5 where
( I z )4
= ( I z )5
2 2 2 1 2 2 12 in. 3 in. 1 ft I z = 0.53665 lb s 2 /ft (12 in.) + ( 3 in.) + + 2 2 12 in. 12
(
)
2 2 2 2 1 ft 1 + 0.06708 lb s 2 /ft ( 3 in.) + (1.5 in.) + (1.5 in.) + ( 4.5 in. - 0.75 in.) 12 12 in.
(
)
2
1 16 1 2 2 + 0.042148 lb s 2 /ft - ( 2.4 in.) + (1.5 in.) 4 4 12
(
)
2 2 4 2.4 in. 2 1 ft + 3 in. + + ( 4.5 in. - 0.75 in.) 3 12 in. 2 2 2 2 1 ft 1 + 2 0.036892 lb s 2 /ft 3 ( 2.75 in.) + (1.0 in.) + (12 in - 2.75 in ) + ( 0.5 in ) 12 in. 12
(
)
2
= 0.252096 lb ft s 2 or I z = 0.252 lb ft s 2
PROBLEM 9.147
Determine the moment of inertia of the steel machine element shown with respect to (a) the x axis, (b) the y axis, (c) the z axis. (The specific weight of steel is 490 lb/ft3.)
SOLUTION
Have
m = ST V =
ST
g
V
3
Then
1 ft 490 lb/ft 3 ( 3 1 4 ) in 3 m1 = 2 32.2 ft/s 12 in. = 105.676 10-3 lb s 2 /ft
m2 =
1 ft 490 lb/ft 3 2 -3 (1.5 1 2 ) in 3 = 26.419 10 lb s /ft 2 12 in. 32.2 ft/s
3
3
490 lb/ft 3 2 1 ft 3 2 m3 = ( 0.5 ) 1.5 in 3 = 5.1874 10 lb s /ft 2 32.2 ft/s 2 12 in.
m4 =
1 ft 490 lb/ft 3 2 (1.4 ) 0.4 in 3 2 32.2 ft/s 2 12 in.
3
= 10.8491 10-3 lb s 2 /ft (a) Using Fig. 9.28 for components 1 and 2 and the equations derived above for components 3 and 4, have
I x = ( I x )1 + ( I x )2 + ( I x )3 - ( I x )4
where
( I x )1 = (105.676 10
-3
1 1 2 4 2 2 1 ft 2 2 2 lb s /ft (1) + ( 4 ) + + in 2 2 12 in. 12
2
)
= 4.1585 10-3 lb ft s 2
( I x )2
1 ft 2 2 2 2 1 = 26.419 10-3 lb s 2 /ft (1) + ( 2 ) + ( 0.5 ) + ( 5 ) in 2 12 12 in.
(
)
2
= 4.7089 10-3 lb ft s 2
PROBLEM 9.147 CONTINUED
2 2 16 4 0.5 2 1 ft 2 2 1 0.5 ) + ( 0.5 ) + 6 + I x )3 = 5.1874 10-3 lb s 2 /ft - in ( 2 ( 3 12 in. 2 9
(
)
= 1.40209 10-3 lb ft s 2
( I x )4
= 10.8451 10
(
-3
2 2 1 4 0.5 2 1 ft 2 2 lb s /ft 3 (1.4 ) + ( 0.4 ) + 6 in 3 12 in. 12 2
)
= 0.38736 10-3 lb ft s 2 Then
I x = ( 4.1585 + 4.7089 + 1.40209 - 0.38736 ) 10-3 lb ft s 2
= 9.8821 10-3 lb ft s 2 or I x = 9.88 10-3 lb ft s 2
(b) Have where
Iy = Iy
( )1 + ( I y )2 + ( I y )3 - ( I y )4
2
( I y )1 = (105.676 10
-3
3 2 4 2 2 1 ft 2 2 1 2 lb s /ft ( 3) + ( 4 ) + + in 2 2 12 in. 12
)
= 6.1155 10-3 lb ft s 2
( )2 (
Iy
1 ft 2 2 2 2 1 = 26.419 10-3 lb s 2 /ft (1.5 ) + ( 2 ) + ( 0.75) + ( 5 ) in 2 12 12 in.
)
2
= 4.7854 10-5 lb ft s 2
( )3 (
Iy
2 2 1 16 1 4.05 2 1 ft = 5.1874 10-3 lb s 2 /ft - in ( 0.5)2 + (1.5)2 + ( 0.75)2 + 6 + 2 12 3 12 in. 4 9
)
= 1.4178 s 10-3 lb ft s 2
( I y )4 = (10.8451 10
-3
2 2 1 ft 16 4 1.4 2 2 2 1 lb s /ft - 1.4 ) + 3 - + ( 2 ) in 2 ( 3 12 in. 2 9 2
)
= 0.78438 10-3 lb ft s 2
PROBLEM 9.147 CONTINUED
Then
I y = (16.1155 + 4.7854 + 1.41785 - 0.78438 ) 10-3 lb ft s 2
= 11.5344 10-3 lb ft s 2 or I y = 11.53 10-3 lb ft s 2
(c) Have where
I z = ( I z )1 + ( I z )2 + ( I z )3 - ( I z )4
( I z )1 = (105.676 10
-3
1 3 2 1 2 2 1 ft 2 2 2 lb s /ft ( 3) + (1) + + in 2 2 12 in. 12
2
)
= 2.4462 10-3 lb ft s 2
1.5 2 1 2 2 1 ft 2 2 2 1 I z )2 = 26.419 10-3 lb s 2 /ft (1.5 ) + (1) + ( + in 2 2 12 in. 12
(
)
= 0.198754 10-3 lb ft s 2
( I z )3
= 5.1874 10
(
-3
1 ft 2 2 2 2 1 lb s /ft 3 ( 0.5 ) + (1.5 ) + ( 0.75 ) + ( 0.5 ) in 2 12 12 in.
2
)
2
= 0.038275 10-3 lb ft s 2
( I z )4
= 10.8451 10
(
-3
2 2 1 1 ft 16 1 4 1.4 2 2 2 2 lb s /ft - 1.4 ) + ( 0.4 ) + 3 - + ( 0.8 ) in 2 ( 12 3 12 in. 4 9 2
)
= 0.49543 10-3 lb ft s 2 Then
I z = ( 2.4462 + 0.198754 + 0.038275 - 0.49543) 10-3 lb ft s 2
= 2.1878 10-3 lb ft s 2 or I z = 2.19 10-3 lb ft s 2
To the instructor: The following formulas for the mass moment of inertia of wires are derived or summarized at this time for use in the solutions of problems 9.1489.150
Slender Rod
Ix = 0 I y = Iz = Circle Have Now And symmetry implies Ix = Iz = Semicircle Following the above arguments for a circle, have I x = Iz = 1 2 ma 2 I y = ma 2 I y = r 2dm = ma 2 Iy = Ix + Iz Ix = Iz 1 2 ma 2 I y = I z = 1 mL2 (Fig. 9.28) 12
1 2 mL (Sample Problem 9.9) 3
Using the parallel-axis theorem I z = I z + mx 2 or x= 2a
4 1 I z = m - 2 a 2 2
PROBLEM 9.148
Aluminum wire with a mass per unit length of 0.049 kg/m is used to form the circle and the straight members of the figure shown. Determine the mass moment of inertia of the assembly with respect to each of the coordinate axes.
SOLUTION
First compute the mass of each component. Have Then m = L m1 = ( 0.049 kg/m ) 2 ( 0.32 m ) = 0.09852 kg m2 = m3 = m4 = m5
= ( 0.049 kg/m )( 0.160 m ) = 0.00784 kg
Using the equation given above and the parallel axis theorem, have I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x )5 1 1 2 2 = ( 0.09852 kg ) ( 0.32 m ) + ( 0.00784 kg ) ( 0.160 m ) 2 3
2 + ( 0.00784 kg ) 0 + ( 0.160 m )
1 2 2 2 + ( 0.00784 kg ) ( 0.16 m ) + ( 0.08 m ) + ( 0.32 m ) 12
2 2 2 1 + ( 0.00784 kg ) ( 0.16 m ) + ( 0.16 m ) + ( 0.32 m - 0.08m ) 12
= ( 5.0442 + 0.06690 + 0.2007 + 0.86972 + 0.66901) 10-3 kg m 2 = 6.8505 10-3 kg m 2
or I x = 6.85 10-3 kg m 2
PROBLEM 9.148 CONTINUED
Have where Then Iy = (Iy ) + (Iy ) + (Iy ) + (Iy ) + (Iy )
1 2 3 4 5
(I ) = (I )
y 2
y 4
and
(I ) = (I )
y 3
y 5
2 2 I y = ( 0.09852 kg ) ( 0.32 m ) + 2 ( 0.00784 kg ) 0 + ( 0.32 m )
2 2 1 + 2 ( 0.00784 kg ) ( 0.16 m ) + ( 0.24 m ) 12
= 10.088 + 2 ( 0.80282 ) + 2 ( 0.46831) 10-3 kg m 2
= 12.6303 10-3 kg m 2
or I y = 12.63 10-3 kg m 2 By symmetry Iz = Ix or I z = 6.85 10-3 kg m 2
PROBLEM 9.149
The figure shown is formed of 3-mm-diameter steel wire. Knowing that the density of the steel is 7850 kg/m3 , determine the mass moment of inertia of the wire with respect to each of the coordinate axes.
SOLUTION
Have Then
2 m1 = m2 = 7850 kg/m3 ( 0.0015 m ) ( 0.36 m )
m = V = AL
(
)
m2 = m1 = 0.062756 kg
2 m3 = m4 = 7850 kg/m3 ( 0.0015 m ) ( 0.36 m )
(
)
= 0.019976 kg Using the equations given above and the parallel axis theorem, have I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x )4 where Then
2 2 2 1 1 I x = ( 0.062756 kg ) ( 0.36 m ) + ( 0.062756 ) ( 0.36 m ) + ( 0.36 m ) 2 2 2 2 2 1 + 2 ( 0.019976 kg ) ( 0.36 m ) + ( 0.18 m ) + ( 0.36 m ) 12
( I x )3 = ( I x ) 4
= 4.06659 + 12.19977 + 2 ( 3.45185 ) 10-3 kg m 2
= 23.1701 10-3 kg m 2 or I x = 23.2 10-3 kg m 2 Have where Iy = Iy
( )1 + ( I y )2 + ( I y )3 + ( I y )4
and
( I y )1 = ( I y )2
( I y )3 = ( I y ) 4
`
PROBLEM 9.149 CONTINUED
Then
2 2 I y = 2 ( 0.062756 kg ) ( 0.36 m ) + 2 ( 0.019976 kg ) 0 + ( 0.36 m )
= 2 8.13318 10-3 kg m 2 + 2 2.58889 10-3 kg m 2 = 21.44414 10-3 kg m 2
(
) (
)
or I y = 21.4 10-3 kg m 2 Have I z = ( I z )1 + ( I z )2 + ( I z )3 + ( I z )4 where Then
2 1 I z = ( 0.062756 kg ) ( 0.36 m ) 2
( I z )3 = ( I z )4
2 1 4 2 2 0.36 m 2 + ( 0.062756 kg ) - 2 ( 0.36 m ) + + ( 0.36 m ) 2 2 1 + 2 ( 0.019976 kg ) ( 0.36 m ) 3
= 4.06659 + 12.1998 + 2 ( 0.86296 ) 10-3 kg m 2
= 17.9923 10-3 kg m 2 or I z = 17.99 10-3 kg m 2
PROBLEM 9.150
A homogeneous wire with a weight per unit length of 0.041 lb/ft is used to form the figure shown. Determine the moment of inertia of the wire with respect to each of the coordinate axes.
SOLUTION
First compute the mass of each component. Mass of each component is indentical
m=
( m/L ) L
g
Have
=
( 0.041 lb/ft )(1.5 ft )
32.2 ft/s 2
= 0.00190994 lb s 2 /ft Using the equations given above and the parallel axis theorem, have
( I x )1 = ( I x )3 + ( I x )4 = ( I x )6
Then
and
( I x ) 2 = ( I x )5
I x = 4 ( I x )1 + 2 ( I x )2
2 1 I x = 4 0.00190994 lb s 2 /ft ( 5 ft ) 3 2 + 2 0.00190994 lb s 2 /ft 0 + (1.5 ft )
(
)
(
)
= 0.0143246 lb ft s 2 or I x = 14.32 10-3 lb ft s 2 Now
( I y )1 = 0 ( I y )2 = ( I y )6
( I y )4 = ( I y )5
`
PROBLEM 9.150 CONTINUED
Then
Iy = 2 Iy
( )2 + ( I y )3 + 2 ( I y )4
1 2 2 = 0.0019094 lb s 2 /ft 2 (1.5 ft ) + 0 + (1.5 ft ) 3
(
)
2 2 2 1 + 2 (1.5 ft ) + (1.5 ft ) + ( 0.75 ft ) 12
= 0.0019094 (1.5 + 2.25 + 6 ) lb ft s 2 = 0.0186219 lb ft s 2
I y = 18.62 10-3 lb ft s 2
By symmetry
Iz = I y I z = 18.62 10-3 lb ft s 2
PROBLEM 9.151
Determine the products of inertia Ixy, Iyz, and Izx of the steel machine element shown. (The specific weight of steel is 490 lb/ft3.)
SOLUTION
From the solution to Problem 9.147 m1 = 105.676 10-3 lb s 2 /ft m2 = 26.419 10-3 lb s 2 /ft m3 = 5.1874 10-3 lb s 2 /ft m4 = 10.8451 10-3 lb s 2 /ft
First note that symmetry implies I xy = I yz = I z x = 0 for each component
Now so that Then
I uv = I uv + mu v = mu v
0
( Iuv )body
= mu v
1.5 0.5 0.75 0.5 I xy = mx y = 105.676 10-3 lb s 2 /ft ft ft + 26.419 10-3 lb s 2 /ft ft ft 12 12 12 12 0.75 0.5 ft ft + 5.1874 10-3 lb s 2/ft 12 12
4 1.4 in. 1 ft 0.8 ft - 10.8451 10-3 lb s 2 /ft 3 in. - 3 12 in. 12
(
)
(
)
(
)
(
)
= ( 550.40 + 68.799 + 13.5089 - 144.952 ) 10-6 lb ft s 2 = 487.76 10-6 lb ft s 2 or I xy = 0.488 10-3 lb ft s 2
PROBLEM 9.151 CONTINUED
0.5 2 0.5 5 I yz = my z = 105.676 10-3 lb s 2 /ft ft ft + 26.419 10-3 lb s 2 /ft ft ft 12 12 12 12
4 0.5 in. 1 ft 0.5 + 5.1874 10-3 lb s 2 /ft ft 6 in. + 3 12 12 in.
(
)
(
)
(
)
0.8 2 ft ft - 10.8451 10-3 lb s 2 /ft 12 12 = ( 733.86 + 458.66 + 111.893 - 120.501) 10-6 lb ft s 2 = 1183.91 10-6 lb ft s 2 or I yz = 1.184 10-3 lb ft s 2
(
)
2 1.5 5 0.75 I zx = m z x = 105.676 10-3 lb s 2 /ft ft ft + 26.419 10-3 lb s 2 /ft ft ft 12 12 12 12 4 0.5 1 ft 0.75 ft + 5.1874 10-3 lb s 2 /ft 6 + in. 3 12 in. 12
(
)
(
)
(
)
1 ft 4 1.4 2 - 10.8451 10-3 lb s 2 /ft ft 3 - in. 3 12 12 in.
(
)
= ( 2201.6 + 687.99 + 167.840 - 362.38 ) 10-6 lb ft s 2 = 2695.1 10-6 lb ft s 2 or I zx = 2.70 10-3 lb ft s 2
PROBLEM 9.152
Determine the products of inertia I xy , I yz , and I zx of the steel machine element shown. (The specific weight of steel is 0.284 lb / in 3. )
SOLUTION
First compute the mass of each component m1 = m2 = m3 =
m=
g
V
Then
0.284 lb/in 3 5 in. 4.5 in. 0.9 in.) = 0.1786 lb s 2 /ft 2 ( 32.2 ft/s 0.284 lb/in 3 3 in. 2.5 in. 0.8 in.) = 0.05292 lb s 2 /ft 2 ( 32.2 ft/s 0.284 lb/in 3 2 ( 0.6 in.) 0.5 in. = 0.0049875 lb s 2 /ft 32.2 ft/s 2
Now observe that the centroidal products of inertia, I xy , I yz , and I z x , of each component are zero because of symmetry. Now I uv = I uv + muv so that ( I uv )body = mu v . m, lb s 2 /ft x , ft 2.5 0.20833 y , ft 0.45 0.0375 z , ft 2.25 0.1875 mx y lb ft s 2 0.20093 1.3953110-3 my z lb ft s 2 0.18083 1.25578 10-3 mz x lb ft s 2
0
1
0.1786
1.0046
6.97656 10-3
2 3
0.05292 0.0049875
4.6 0.38333 5.25 0.4375
2.40 0.20 2.70 0.225
2.25 0.1875 2.25 0.1875
0.58424 4.0572 10-3 0.07069 0.49095 10-3 0.85586 5.94347 10-3
0.28577 1.9845110-3 0.03030 0.2104110-3 0.4969 3.45069 10-3
0.54772 3.80362 10-3 0.05891 0.40913 10-3 1.61123 11.18909 10-3
PROBLEM 9.152 CONTINUED
Then or I xy = 5.94 10-3 lb ft s 2 or I yz = 3.45 10 -3 lb ft s 2 or I zx = 11.19 10-3 lb ft s 2
PROBLEM 9.153
Determine the mass products of inertia Ixy, Iyz, and Izx of the cast aluminum machine component shown. (The density of aluminum is 2700 kg/m3.)
SOLUTION
Have Then
m = alV
kg m1 = 2700 3 ( 0.350 0.100 0.030 ) m3 m
= 2.8350 kg
kg m2 = 2700 3 ( 0.200 0.100 0.050 ) m3 m
= 2.7000 kg
kg 2 m3 = 2700 3 ( 0.025 ) 0.100 m3 m
= 0.53014 kg
First note that symmetry implies I xy = I yz = I z x = 0 for each component Now where
I uv = I u v + mu v I u v = 0 I xy = mx y = ( 2.8350 kg )( 0.175m )( 0.050 m )
+ ( 2.7000 kg )( 0.100 m )( 0.050 m ) - ( 0.53014 kg )( 0.080 m )( 0.050 m )
= ( 24.806 + 13.500 - 2.1206 ) 10-3 kg m 2 = 36.1854 103 kg m 2
or I xy = 36.2 10-3 kg m 2
I yz = my z = ( 2.8350 kg )( 0.050 m )( 0.015m )
+ ( 2.7000 kg )( 0.050 m )( 0.055 m ) - ( 0.53014 kg )( 0.050 m )( 0.040 m )
= ( 2.1263 + 7.4250 - 1.06028 ) 10-3 kg m 2 = 8.49102 10-3 kg m 2
or I yz = 8.49 10-3 kg m 2
PROBLEM 9.153 CONTINUED
I zx = mz x = ( 2.8350 kg )( 0.015 m )( 0.175 m ) + ( 2.7000 kg )( 0.055 m )( 0.100 m ) - ( 0.53014 kg )( 0.040 m )( 0.080 m )
= ( 7.4419 + 14.850 - 1.69645 )10-3 kg m 2 = 20.59545 10-3 kg m 2
or I zx = 20.6 10-3 kg m 2
PROBLEM 9.154
Determine the mass products of inertia Ixy, Iyz, and Izx of the cast aluminum machine component shown. (The density of aluminum is 2700 kg/m3.)
SOLUTION
Have Then
m = V
m1 = 2700 kg/m3 ( 0.118 0.036 0.044 ) m3 = 0.50466 kg
2 m2 = 2700 kg/m3 ( 0.022 ) 0.036 m3 = 0.07389 kg 2
(
)
(
)
m3 = 2700 kg/m3 ( 0.028 0.022 0.024 ) m3 = 0.03992 kg Now observe that I xy , I yz and I z x are zero because of symmetry Now 4 0.023 x2 = - 0.118 + m = -0.12734 m 3 0.062 y3 = 0.036 - m = 0.025 m 2 m, kg 1 2 3 0.50466 0.07389 0.03992
x, m y, m z, m
(
)
mx y kg m 2
my z kg m 2
mz x kg m 2
-0.059 -0.12734 -0.041
0.018 0.018 0.025
0.022 0.022 0.026
-0.53595 10-3 -0.16932 10-3 -0.01397 10-3
0.19985 10-3 0.02926 10-3 0.02594 10-3
-0.65505 10-3
-0.20695 10-3 -0.01453 10-3
PROBLEM 9.154 CONTINUED
And I xy = I xy + mx y I yz = I yz + my z
0
( (
0
) )
0
I zx = ( I zx + mx z ) Finally I xy = I xy
( )1 + ( I xy )2 - ( I xy )3 = -0.6913 10-3 kg m2
or I xy = -0.691 10-3 kg m 2
I yz = I yz
( )1 + ( I yz )2 - ( I yz )3 = 0.20317 10-3 kg m2
or I yz = 0.203 10-3 kg m 2
I zx = ( I zx )1 + ( I zx )2 - ( I zx )3 = -0.84747 10-3 kg m 2
or I zx = -0.848 10-3 kg m 2
PROBLEM 9.155
A section of sheet steel 3 mm thick is cut and bent into the machine component shown. Knowing that the density of the steel is 7860 kg/m3 , determine the mass products of inertia I xy , I yz , and I zx of the component.
SOLUTION
Have Then
m = v = st tA
m1 = ( 7860 kg/m3 ) ( 0.003 0.420 0.360 ) m3 = 3.5653kg
2 m2 = ( 7860 kg/m3 ) ( 0.003 m ) ( 0.210 m ) = 1.6334 kg 2
x2 = Now observe that m, kg 1 2
4 ( 0.210 m ) 3
= 0.089127 m
I xy = I yz = I z x = 0
x, m
0 0.089127
y, m
z,m
0.21 0.21
mx y , kg m 2
my z , kg m 2
mz x , kg m 2
3.5653 1.6334
0.8 0.36
0 52.409 10-3 52.409 10-3
134.768 10-3 123.485 10-3 258.253 10-3
0 30.572 10-3 30.572 10-3
Then
I xy = I xy + mx y I yz = I yz + my z
0
( (
0
) )
or I xy = 52.4 10-3 kg m 2 or I yz = 258 10-3 kg m 2 or I zx = 30.6 10-3 kg m 2
0
I zx = ( I zx + mz x )
PROBLEM 9.156
A section of sheet steel 3 mm thick is cut and bent into the machine component shown. Knowing that the density of the steel is 7860 kg/m3 , determine the mass products of inertia I xy , I yz , and I zx of the component.
SOLUTION
First compute the mass of each component Have Then
m = sT V = sT tA
m1 = ( 7860 kg/m3 ) ( 0.003)( 0.08 )( 0.09 ) m3
= 0.169776 kg 1 m2 = 7860 kg/m3 ( 0.003) 0.09 0.036 m3 2 = 0.03820 kg Now observe that
(I ) = (I ) = (I )
x y 1 y z 1 z x
1
=0
(I ) = (I )
y z 2 z x
2
=0
From Sample Problem 9.6
(I )
x y
2,area
2,area
=-
1 2 2 b2 h2 72
Then
(I )
x y 2
= sT t ( I xy )
1 1 2 = sT t - b22 h2 = - m2 b2 h2 36 72
Also
x1 = y1 = z2 = 0
0.09 x2 = -0.045 + m = -0.015 m 3
PROBLEM 9.156 CONTINUED
Finally..
1 I xy = I xy + mx y = ( 0 + 0 ) + - ( 0.03820 kg )( 0.09 m )( 0.036 m ) 36 0.036 m + ( 0.03820 kg )( -0.015m ) 3
= ( -3.4379 10-6 - 6.876 10-6 ) kg m 2 = -10.3139 10-6 kg m 2
(
)
or I xy = -10.31 10-6 kg m 2
And I yz = I yz + m y z = ( 0 + 0 ) + ( 0 + 0 ) = 0 I zx = ( I zx + m z x ) = ( 0 + 0 ) + ( 0 + 0 ) = 0
(
)
or I yz = 0 or I zx = 0
PROBLEM 9.157
A section of sheet steel 3 mm thick is cut and bent into the machine component shown. Knowing that the density of the steel is 7860 kg/m3 , determine the mass products of inertia I xy , I yz , and I zx of the component.
SOLUTION
First compute the mass of each component Have.. Then m = sT V = sT tA
m1 = ( 7860 kg/m3 ) ( 0.003)( 0.7 )( 0.78 ) m3 = 12.785 kg
m2 = ( 7860 kg/m3 ) ( 0.003) 0.392 m3 = 5.6337 kg 2 1 m3 = ( 7860 kg/m3 ) ( 0.003) 0.78 0.3 m3 = 2.7589 kg 2 Now observe that because of symmetry the centroidal products of inertia of components 1 and 2 are zero and I xy = I zx = 0
Also
( )3 ( )3 ( I yz )3,mass = sT t ( I yz )3,area
1 ( I yz )3,area = 72 b32h32
Using the results of Sample Problem 9.6 and noting that the orientation of the axes corresponds to a 90 rotation, have
Then 1 1 ( I yz )3 = sT t 72 b32h32 = 36 m3b3h3 Also Finally
y1 = x2 = 0 y2 =
4 0.39 m = 0.16552 m 3
I xy = I xy + mx y = ( 0 + 0 ) + ( 0 + 0 )
(
)
0.3m 2 + 0 + ( 2.7589 kg ) ( 0.7 m ) = -0.19312 kg m 2 or I xy = -0.1931 kg m 2
PROBLEM 9.157 CONTINUED
I yz = I yz + my z
(
)
= ( 0 + 0 ) + 0 + ( 5.6337 kg ) ( 0.16552 m )( 0.39 m )
1 -0.3 m 0.78 m + ( 2.7589 kg ) ( 0.78 m )( 0.3 m ) + 3 3 36 = ( 0.36367 + 0.017933 - 0.07173) kg m 2 = 0.30987 kg m 2 or I yz = 0.310 kg m 2
I zx = I zx + mz x = 0 + (12.875 kg ) ( 0.35 m )( 0.39 m )
(
)
0.78 m + ( 0 + 0 ) + 0 + ( 2.7589 kg ) ( 0.7 m ) 2 = (1.75744 + 0.50212 ) kg m 2 = 2.25956 kg m 2 or I zx = 2.26 kg m 2
PROBLEM 9.158
A section of sheet steel 0.08 in. thick is cut and bent into the machine component shown. Knowing that the specific weight of steel is 490 lb/ft3, determine the mass products of inertia I xy , I yz , and I zx of the component.
SOLUTION
First note
m = sT V =
sT
g
tA
3
Then
490 lb/ft 3 1ft m1 = ( 0.08 in.) ( 6 3.6 ) in 2 2 32.2 ft/s 12in.
= 15.2174 10-3 lb s 2 /ft 490 lb/ft 3 2 1 ft m2 = ( 0.08 in.) (1.8 in.) 2 2 12 in. 32.2 ft/s = 3.5855 10-3 lb s 2 /ft 490 lb/ft 3 2 1 ft m3 = ( 0.08 in.) ( 3.6 in.) 2 4 12 in. 32.2 ft/s = 7.1710 10-3 lb s 2 /ft
3 3
Note that symmetry implies
(I )
x y
x y
1,2
= I yz
y z
( )
3
1,2
= I z x
( )
1,2
=0
(I ) = (I )
3
=0
Now
I uv = I uv + mu v
PROBLEM 9.158 CONTINUED
Thus
I xy = mx y
0
= m1x1 y1 - m2 x2 y2 + m3 x3 y3
0.6 1.8 = (15.2174 10-3 lb s 2 /ft ) - ft ft 12 12 4 1.8 1 ft 1.8 ft - ( 3.5855 10-3 lb s 2 /ft ) 2.4 - in. 3 12 in. 12 = ( -114.131 - 73.326 ) 10-6 lb ft s 2 Now Also
0 0 0 I yz = my z = m1 y1 z1 - m2 y2 z2 + m3 y3 z3
or I xy = -187.5 10-6 lb ft s 2 or I yz = 0
I zx = ( I zx )1 - ( I zx )2 + ( I zx )3
0 0 = m1 z1x1 - m2 z2 x2 + ( I zx )3
Now determine ( I zx )3 Have
( dI zx )3 = ( dI zx )3 + z x dm
x = ( z ) - sT t x dz 2 g
0
=-
1 sT 2 tz a3 - z 2 dz 2 g
(
)
Now
m3 =
sT
g t =
2 t a3 4
or
sT
g
4m3 2 a3
a
Therefore,
( I zx )3
2m3 a 2 2m3 1 2 1 a3 z - z 3 dz = - 2 a3 z 2 - z 4 = 2 0 4 0 a3 a3 2 =- 1 2 m3a3 2
(
)
Finally
I zx
1 3.6 7.1710 10-3 lb s 2 /ft ft =- 2 12
(
)
2
or I zx = -102.7 10-6 lb ft s 2
PROBLEM 9.159
Brass wire with a weight per unit length w is used to form the figure shown. Determine the products of inertia I xy , I yz , and I zx of the wire figure.
SOLUTION
First compute the mass of each component. Have m= Then.. m1 = m2 = m3 = W 1 = wL g g
w 3 3 w a = a 2 2 g g w w ( 3a ) = 3 a g g w w ( a ) = a g g
Now observe that the centroidal products of inertia, I xy , I yz , and I z x , of each component are zero because of symmetry.
m
1 2 3 3 w a 2 g 3 -
x
23 a 2 0 2
y
z
mx y
my z
mz x
2a 1 a 2
1 a 2 2a
-9
w 3 a g
0
3 w 3 a 2 g 3
-
9w 3 a 4g 0
w a g w a g
w 3 a g w 3 a g
2
(a)
-a
a
-2
w 3 a g w 3 a g
-
w 3 a g
-11
w 3 + 3 a g2
-
1w 3 a 4g
PROBLEM 9.159 CONTINUED
Then
I xy
0 = I xy + mx y
(
) )
or I xy = -11 or I yz =
w 3 a g
0 I yz = I yz + my z
(
1w 3 a ( + b ) 2g 1w 3 a 4g
I zx
0 = I zx + mz x
(
)
or I zx = -
PROBLEM 9.160
Brass wire with a weight per unit length w is used to form the figure shown. Determine the products of inertia I xy , I yz , and I zx of the wire figure.
SOLUTION
First compute the mass of each component. Have
m=
Then
W 1 = wL g g w w ( 2 a ) = 2 a g g w w (a) = a g g w w ( 2a ) = 2 a g g w 3 w 2 a = 3 a 2 g g
m1 = m2 = m3 = m4 =
Now observe that the centroidal products of inertia, I xy , I yz , and I z x , of each component are zero because of symmetry.
PROBLEM 9.160 CONTINUED
m
1 2 3 4 2
x
2a 2a 2a 2a
y
z
mx y
my z
mz x
w a g
a
1 a 2 0 3 - a 2
-a
4
w 3 a g
-2
w 3 a g
0 0
-4
w 3 a g
0
w a g
2 3
0
w 3 a g
0
w a g w a g
a
4
w 3 a g w 3 a g
2a
-9
w 3 a g
-9 -11
w 3 a g w 3 a g
4
12
w (1 - 5 ) a3 g
0 = I xy + mx y
w (1 + 2 ) a3 g w 3 a (1 - 5 ) g w 3 a g
Then
I xy
(
)
or I xy =
0 I yz = I yz + my z
(
)
or I yz = -11 or I zx = 4
0 I zx = I zx + mz x
(
)
w 3 a (1 + 2 ) g
PROBLEM 9.161
The figure shown is formed of 0.075-in.-diameter aluminum wire. Knowing that the specific weight of aluminum is 0.10 lb/in 3 , determine the products of inertia I xy , I yz , and I zx of the wire figure.
SOLUTION
First note
m = V =
g
V =
g
AL
Specific weight of aluminium = 0.10 lb/in 3 = 172.8 lb/ft 3 Then
172.8 lb/ft 3 ( 0.075 ft ) m= 12 32.2 ft/s 2 4
2
L
= 0.16464 10-3 L lb s 2 /ft 2
Now
(
)
L1 = L4 = 12.5 in. = 1.04167 ft m1 = m4 = 0.17150 10-3 lb s 2 /ft L2 = L5 = 9 in. = 0.75 ft m2 = m5 = 0.12348 10-3 lb s 2 /ft L3 = L6 = 15 in. = 1.25 ft m3 = m6 = 0.20580 10-3 lb s 2 /ft
and
I xy = I yz = I z x = 0
PROBLEM 9.161 CONTINUED
m, lb s 2 /ft
x , ft
0.75 0.375 0 0 0.375 0.75
y , ft
0.5208 1.04167 1.04167 0.5208 0 0
z , ft
0 0 0.625 1.25 1.25 0.625
mx y, lbft s2
0.06699 10-3 0.04823 10-3 0 0 0 0 0.11522 10-3
my z , lb ft s 2
0 0 0.13398 10-3 0.111646 10-3 0 0 0.24563 10-3
m z x , lb ft s 2
1 2 3 4 5 6
0.17150 10-3 0.12348 10-3 0.20580 10-3 0.17150 10-3 0.12348 10-3 0.20580 10-3
0.05788 10-3 0.09647 10-3 0.15435 10-3
0 I xy = ( I xy + mx y ) = 0.115222 10-3 lb ft s 2 0 I yz = ( I yz + my z ) = 0.24563 10-3 lb ft s 2 0 I zx = ( I z x + mz x ) = 0.15435 10-3 lb ft s 2 or I xy = 0.1152 10-3 lb ft s 2
or I yz = 0.246 10-3 lb ft s 2
or I zx = 0.1543 10-3 lb ft s 2
PROBLEM 9.162
Thin aluminum wire of uniform diameter is used to form the figure shown. Denoting by m the mass per unit length of wire, determine the products of inertia I xy , I yz , and I zx of the wire figure.
SOLUTION
First compute the mass of each component. Have
m m = L = mL L
Then
m1 = m5 = m R1 = mR1 2 2
m2 = m4 = m ( R2 - R1 )
m3 = m R2 = mR2 2 2
Now observe that because of symmetry the centroidal products of inertia, I xy , I yz , and I z x , of components 2 and 4 are zero and
( I xy )1 = ( I zx )1 = 0
Also
( I xy )3 = ( I yz )3 = 0
z4 = z5 = 0
( I yz )5 = ( I zx )5 = 0
x1 = x2 = 0 y2 = y3 = y4 = 0
Using the parallel-axis theorem [Equations (9.47)], it follows that
I xy = I yz = I zx for components 2 and 4.
To determine I uv for one quarter of a circular arc have
where
dI uv = uvdm
u = a cos
v = a sin
and
dm = dV = A ( ad )
where A is the cross-sectional area of the wire. Now
m = m a = A a 2 2
so that and
dm = mad dI uv = ( a cos )( a sin )( mad )
= ma3 sin cos d
PROBLEM 9.162 CONTINUED
Then
I uv = dI uv = 02 ma3 sin cos d
1 2 = ma sin = ma 3 2 2 0
2 3 1
Thus,
( I yz )1 = 1 mR13 2
( I zx )3
1 3 = - mR2 2
and
( I xy )5 = - 1 mR13 2
0
Because of 90 rotations of the coordinate axes. Finally,
I xy = I xy = I xy
( )
( )1 + m1x1 y1 + ( I xy )3 + m3x3 y3 + ( I xy )5
1 3 or I xy = - mR1 2 0 0 + I yz + m3 y3 z3 + I yz + m5 y5 z5 3 5
0
I yz = I yz = I yz
( ) ( )1 ( )
0
( )
or I yz =
1 3 mR1 2
I zx = ( I zx )
= I z x
( )1
+ m1z1x1 + ( I zx )3 + I zx
( )5 + m5 z5 x5
1 3 or I zx = - mR2 2
0
PROBLEM 9.163
Complete the derivation of Eqs. (9.47), which express the parallel-axis theorem for mass products of inertia.
SOLUTION
Have and Consider Substituting for x and for y
I xy = xydm x = x + x
I yz = yzdm y = y + y I xy = xydm
I zx = zxdm z = z + z
(9.45) (9.31)
I xy = ( x + x )( y + y ) dm
= xydm + y xdm + x ydm + x y dm
By definition and
I xy = xydm
xdm = mx ydm = my
However, the origin of the primed coordinate system coincides with the mass center G, so that
x = y = 0
I xy = I xy + mx y Q.E.D.
The expressions for I yz and I zx are obtained in a similar manner.
PROBLEM 9.164
For the homogeneous tetrahedron of mass m shown, (a) determine by direct integration the product of inertia I zx , (b) deduce I yz and I xy from the results obtained in part a.
SOLUTION
(a) First divide the tetrahedron into a series of thin vertical slices of thickness dz as shown. Now and
x=-
a z z + a = a 1 - c c
b z y = - z + b = b 1 - c c
The mass dm of the slab is
z 1 1 dm = dV = xydz = ab 1 - dz c 2 2
Then
2 3 c 1 z 1 z m = dm = ab 1 - dz = ab - 1 - 2 c c 3 02 0 c c
2
=
Now where and
1 abc 6
dI zx = dI z x + z EL xEL dm
dI zx = 0 z EL = z xEL =
( symmetry )
1 1 z x = a 1 - 3 3 c
Then
2 c 1 z 1 z I zx = dI zx = z a 1 - ab 1 - dz c 2 c 0 3
=
1 2 c z2 z3 z 4 a b 0 z - 3 + 3 2 - 3 dz c 6 c c
c
m 1 z3 3 z 4 1 z5 = a z2 - + - 4 c 2 5 c3 0 c 2 c
or I zx = 1 mac 20
PROBLEM 9.164 CONTINUED
(b) Because of the symmetry of the body, I xy and I yz can be deduced by considering the circular permutation of Thus
( x, y, z )
and
( a, b, c ) .
1 mab 20 1 mbc 20
I xy = I yz =
Alternative solution for part a First divide the tetrahedron into a series of thin horizontal slices of thickness dy as shown. Now and
x=- z =-
a y y + a = a 1 - b b c y y + c = c 1 - b b
The mass dm of the slab is
y 1 1 dm = dV = xzdy = ac 1 - dy b 2 2
Now where and dI zx, Area =
dI zx = tdI zx, Area
2
t = dy
1 2 2 x z from the results of Sample Problem 9.6 24
Then
2 2 1 y y dIzx = ( dy ) a 1 - c 1 - b b 24
=
1 y 1m y ac 1 - dy a 2c 2 1 - dy = 24 4 b b b
b
4
4
Finally
I zx = dI zx = 0
1m y ac 1 - dy 4 b b
b
4
5 1 m b y ac - 1 - = 4 b 5 b 0
or I zx =
1 mac 20
PROBLEM 9.164 CONTINUED
Alternative solution for part a The equation of the included face of the tetrahedron is
x y z + + =1 a b c
so that
x z y = b 1 - - a c
For an infinitesimal element of sides dx, dy, and dz
dm = dV = dydxdz
From part a Now
I zx = zxdm =
z x = a 1 - c
x ( ) b(1- a - cz ) zx ( dydxdz )
0 0
z c a 1- c
0
= 0 0 ( = b
z c a 1- c
) zx b (1 -
2
x a
-
z c
) dxdz
z a 1- c
c 1 z 0
1 x3 1 z 2 - x - x 3 a 2 c 0 2
( )
dz
2 3 2 z 1 3 z 1 z 2 z c 1 a 1 - - a 1 - dz = b 0 z a 2 1 - - 3a 2c c c c 2
= b 0
=
c
1 2 z a z 1 - dz 6 c
3
1 2 c z2 z3 z 4 a b 0 z - 3 + 3 2 - 3 dz 6 c c c
c
m 1 z3 3 z 4 1 z5 = a z2 - + - 4 c 2 5 c3 0 c 2 c
or I zx = 1 mac 20
PROBLEM 9.165
The homogeneous circular cylinder shown has a mass m. Determine the moment of inertia of the cylinder with respect to the line joining the origin O and point A which is located on the perimeter of the top surface of the cylinder.
SOLUTION
From Figure 9.28 and using the parallel-axis theorem
Iy =
1 2 ma 2
Ix = Iz =
Symmetry implies
1 1 h m 3a 2 + h 2 + m = m 3a 2 + 4h 2 12 2 12
(
)
2
(
)
I xy = I yz = I zx = 0
For convenience, let point A lie in the yz plane. Then
OA = 1 h + a2
2
( hj + ak )
With the mass products of inertia equal to zero, Equation (9.46) reduces to 0 2 2 I OA = I xx + I y y + I z z2
1 h = ma 2 h2 + a 2 2 1 m 3a 2 + 4h 2 + 12
2
(
)
a h2 + a 2
2
or I OA =
1 10h 2 + 3a 2 ma 2 2 12 h + a2
Note: For point A located at an arbitrary point on the perimeter of the top surface, OA is given by
OA = 1 h + a2
2
( a cos i + hj + a sin k )
which results in the same expression for I OA
PROBLEM 9.166
The homogeneous circular cone shown has a mass m. Determine the moment of inertia of the cone with respect to the line joining the origin O and point A.
SOLUTION
First note that
dOA =
9 2 2 3 a + ( -3a ) + ( 3a ) = a 2 2
2
Then
OA =
1 3 1 ai - 3aj + 3ak = ( i - 2 j + 2k ) 9 2 3 a 2
For a rectangular coordinate system with origin at point A and axes aligned with the given x, y, z axes, have (using Figure 9.28)
Ix = Iz =
3 1 2 2 m a + ( 3a ) 5 4
Iy =
3 ma 2 10
= Also, symmetry implies
111 2 ma 20 I xy = I yz = I zx = 0
With the mass products of inertia equal to zero, Equation (9.46) reduces to
2 2 I OA = I xx + I y y + I z z2
=
111 2 1 3 111 2 2 2 ma + ma 2 - + ma 20 3 10 3 20 3 193 2 ma 60
2
2
2
=
or I OA = 3.22ma 2
PROBLEM 9.167
Shown is the machine element of Problem 9.143. Determine its moment of inertia with respect to the line joining the origin O and point A.
SOLUTION
First compute the mass of each component Have Then
2 m1 = 0.008819 lb s 2 /ft in 3 ( 4 in.) ( 2 in.) = 0.88667 lb s 2 /ft
m = sT V =
0.284 lb/in 3 V = ( 0.008819 lb s 2 /ft in 3 )V 32.2 ft/s 2
m2 = 0.008819 lb s 2 /ft in 3 (1 in.) ( 3 in.) = 0.083125 lb s 2 /ft
2 2 m3 = 0.008819 lb s 2 /ft in 3 (1 in.) ( 2 in.) = 0.055417 lb s 2 /ft
Symmetry implies
I yz = I zx = 0
( I xy )1 = 0
and Now
( I xy )2 = ( I xy )3 = 0
( )
I xy = I xy + mx y = m2 x2 y2 - m3 x3 y3 1 ft 2 = ( 0.083125 lb s 2 /ft ) ( 2 in.)(1.5 in.) 144 in 2 1 ft 2 - ( 0.055417 lb s 2 /ft ) ( -2 in.)( -1 in.) 144 in 2 = 0.96209 10-3 lb ft s 2 From the solution to Problem 9.143: I x = 34.106 10-3 lb ft s 2
I y = 50.125 10-3 lb ft s 2
I z = 34.876 10 -3 lb ft s 2
PROBLEM 9.167 CONTINUED
By observation Then
I OA =
2 I xx
OA =
1 ( 2i + 3j) 13
+
2 I y y
0 0 + I z z2 - 2 I xy x y - 2 I yz yz - 2I zxz x
-3
0
= 34.106 10
(
2 lb ft s 13
2
)
2
+ 50.125 10
(
-3
3 lb ft s 13
2
)
2
2 3 - 2 0.96209 10-3 lb ft s 2 13 13
(
)
= (10.4942 + 34.7019 - 0.8881) 10-3 lb ft s 2 = 44.308 10-3 lb ft s 2 or I OA = 44.3 10-3 lb ft s 2
PROBLEM 9.168
Determine the moment of inertia of the steel machine element of Problems 9.147 and 9.151 with respect to the axis through the origin which forms equal angles with the x, y, and z axes.
SOLUTION
From Problem 9.147:
I x = 9.8821 10-3 lb ft s 2
I y = 11.5344 10-3 lb ft s 2
I z = 2.1878 10-3 lb ft s 2
Problem 9.151:
I xy = 0.48776 10-3 lb ft s 2 I yz = 1.18391 10-3 lb ft s 2
I zx = 2.6951 10-3 lb ft s 2
Now and Therefore, or Equation 9.46
x = y = z
2 2 x + y + z2 = 1 2 3x = 1
x = y = z =
1 3
2 2 I OL = I xx + I y y + I z z2 - 2 I xy x y - 2 I yz y z - 2I zxz x 2 2 2 1 1 1 1 1 = 9.8821 + 11.5344 + 2.1878 - 2 ( 0.48776 ) 3 3 3 3 3
- 2 (1.18391)
1 1 1 1 -3 2 - 2 ( 2.6951) 10 lb ft s 3 3 3 3
= 4.95692 10-3 lb ft s 2 or I OL = 4.96 10-3 lb ft s 2
PROBLEM 9.169
The thin bent plate shown is of uniform density and weight W. Determine its mass moment of inertia with respect to the line joining the origin O and point A.
SOLUTION
First note that And that
m1 = m2 =
OA =
1W 2 g
1 (i + j + k ) 3
Using Figure 9.28 and the parallel-axis theorem have I x = ( I x )1 + ( I x )2 1 1W = 12 2 g 2 1W a + 2 g
2 a 2
1 1W 2 1W 2 + a +a + 2 g 12 2 g
(
)
a 2 a 2 + 2 2
=
1 W 1 1 1 1 1W 2 + a2 + + a2 = a 2 g 12 4 6 2 2 g
Iy = Iy
( )1 + ( I y )2
1 1W 1W 2 2 = a +a + 2 g 12 2 g
(
)
a 2 a 2 + 2 2
1 1 W 2 1 W 2 a 2 + ( a ) + a + 2 g 2 12 2 g
=
1 W 1 1 2 1 5 2 W 2 6 + 2 a + 12 + 4 a = g a 2 g
PROBLEM 9.169 CONTINUED
I z = ( I z )1 + ( I z )2 1 1W = 12 2 g 2 1W a + 2 g a 2
2
1 1W 2 1W + a + 2 g 12 2 g
2 a 2 ( a ) + 2
=
1 W 1 1 2 1 5 2 5 W 2 12 + 4 a + 12 + 4 a = 6 g a 2 g
Now observe that the centroidal products of inertia, I xy , I yz , and I z x , of both components are zero because of symmetry. Also, y1 = 0
Then
0 1W a 1W 2 I xy = I xy + mx y = m2 x2 y2 = a (a) = 2 g 2 4 g
(
)
0 1 W a a 1 W 2 I yz = I yz + my z = m2 y2z2 = a = 2 g 2 2 8 g 0 I zx = I z x + mz x = m1z1x1 + m2 z2 x2
(
)
(
)
=
1W 2 g
3W 2 a a 1 W a a + (a) = 2 2 2 g 2 8 g
Substituting into Equation (9.46)
2 2 I OA = I xx + I y y + I z z2 - 2 I xy x y - 2 I yz yz - 2I zxz x
Noting that
2 2 x = y = z2 = x y = y z = zx =
1 3
Have I OA = 1 1 W 2 W 2 5 W 2 a + a + a g 3 2 g 6 g
1 W 2 1 W 2 3 W 2 - 2 a + a + a 8 g 8 g 4 g
=
1 14 3 W - 2 a2 3 6 4 g or I OA = 5 W 2 a 18 g
PROBLEM 9.170
A piece of sheet metal of thickness t and density is cut and bent into the shape shown. Determine its mass moment of inertia with respect to a line joining points A and B.
SOLUTION
Have Then
m = V = tA m1 = ta 2 m2 = 1 ta 2 2
Compute moments and moments of inertia with respect to point A Now I x = ( I x )1 + ( I x )2
1 = ta 2 a 2 + 2 1 1 2 2 a 2 2 + ( a ) + ta 2 a 2 + a 2 2 18 3
=
19 ta 4 12
I y = I y
( )1 + ( I y )2
2
1 a 2 2 = ta a 2 + + ( a ) 12 2
2 2 2 1 2 1 2 2 b + 2 + ta a + a + 3 3 2 18
=
5 ta 4 3
I z = ( I z )1 + ( I z )2 1 1 1 = ta 2 a 2 + a 2 + ta 2 a 2 3 2 6 = 3 ta 4 4
(
)
Now note symmetry implies
( I xy )1 = ( I yz )1 = ( I zx )1 = 0 ( I xy )2 = ( I yz )2 = 0
PROBLEM 9.170 CONTINUED
Now Therefore I uv = I uv + mu v 0 a a 1 I xy = m1x1 y1 + m2 x y = ta 2 = ta 4 2 2 2 2 4
0 1 a I yz = m1 y z + m2 y z = ta 2 ( -a ) = - ta 4 1 1 2 2 2 2 I z x = m1z x + I zx 1 1 From Sample Problem 9.6 Then Then
( )2 + m2 z 2 x 2
1 ( I z x ) = - a4 2 area 72 =- 1 ta 4 72
( I zx )2 = t ( I zx )2 area
I z x = 1 2 a a ( -a ) 2 2
1 1 1 2 1 + - ta 4 + ta 2 - a a 2 72 2 3 3 5 = - ta 4 8 By observation Now, Equation 9.46
2 2 I AB = I xx + I y y + I zz2 - 2 I xyxy - 2I yz yz - 2 I zxzx
AB =
1 (i + j - k ) 3
19 1 2 5 1 2 3 1 2 = ta 4 + + - 3 3 4 3 12 3
1 1 1 1 1 1 - 2 - 2 - - 3 4 3 3 2 3
5 1 1 - 2 - - 3 3 8 or I AB = 5 ta 4 12
PROBLEM 9.171
Determine the mass moment of inertia of the machine component of Problems 9.138 and 9.157 with respect to the axis through the origin characterized by the unit vector = ( -4i + 8j + k ) / 9 .
SOLUTION
From Problem 9.138: I x = 4.212 kg m 2
I y = 7.407 kg m 2
I z = 3.7055 kg m 2 From Problem 9.157:
I xy = -0.19312 kg m 2 I yz = 0.30987 kg m 2
I zx = 2.25956 kg m 2 Now Eq. (9.46): OL = 1 ( -4i + 8j + k ) 9
2 2 I OL = I xx + I y y + I z z2 - 2 I xy x y - 2 I yz y z - 2I zxz x 2 2 2 4 8 1 = 4.212 - + 7.407 + 3.7055 9 9 9
4 8 8 1 - 2 ( -0.19312 ) - - 2 ( 0.3098 ) 9 9 9 9 1 4 - 2 ( 2.25956 ) - kg m 2 9 9 = ( 0.832 + 5.85244 + 0.04575 - 0.15259 - 0.061195 + 0.22317 ) kg m 2 = 6.73957 kg m 2 I OL = 6.74 kg m 2
PROBLEM 9.172
For the wire figure of the problem indicated, determine the mass moment of inertia of the figure with respect to the axis through the origin characterized by the unit vector = ( -3i - 6 j + 2k ) / 7. Problem 9.150
SOLUTION
Mass of each leg is identical: W /L m= L g = Also, I xy = I yz = I zx = 0 for each leg, and Now x1 = x6 = 0 y4 = y5 = y6 = 0 z1 = z2 = z3 = 0 0 I xy = I xy + mx y = m2 x2 y2 + m3x3 y3 0.041 lb/ft (1.5 ft ) 32.2 ft/s 2 = 0.00190994 lb s 2 /ft
(
)
= 0.00190994 lb s 2 /ft ( 0.75 )(1.5 ) + (1.5 )( 0.75 ) ft 2 = 0.0042974 lb ft s 2 = 4.2974 10-3 lb ft s 2 I yz = 0 I zx = ( I zx + mz x ) = m4 z4 x4 + m5z5 x5 = 0.00190994 lb s 2 /ft ( 0.75 )(1.5 ) + (1.5 )( 0.75 ) ft 2 = 0.0042974 lb ft s 2 = 4.2974 10-3 lb ft s 2
(
)
(
)
PROBLEM 9.172 CONTINUED
From Problem 9.150 I x = 14.32 10-3 lb ft s 2
I y = I z = 18.62 10-3 lb ft s 2
Now
I OL
1 ( -3i - 6 j + 2k ) and then 7 0 2 2 = I xx + I y y + I z z2 - 2 I xy x y - 2 I yz y z - 2I zxz x OL =
2 -3 3 = 14.32 10 - + 18.62 10-3 7
Eq. ( 9.46 )
(
)
(
) -76
2
2 + 7
2
3 -6 - 2 4.2974 10-3 - 7 7
(
)
2 -3 - 2 4.2974 10-3 lb ft s 2 7 7
(
)
= 2.6302 10-3 + 15.20 10-3 - 3.1573 10-3 + 1.05242 10-3 lb ft s 2 = 15.725 10-3 lb ft s 2 or I OL = 15.73 lb ft s 2
(
)
PROBLEM 9.173
For the wire figure of the problem indicated, determine the mass moment of inertia of the figure with respect to the axis through the origin characterized by the unit vector = ( -3i - 6 j + 2k ) / 7. Problem 9.149
SOLUTION
First compute the mass of each component Have m = stV = mAL
2 = 7850 kg/m3 ( 0.0015 m ) L
(
)
= ( 0.055488L ) kg/m Then m1 = m2 = 0.055488 kg/m ( 0.36 m ) = 0.062756 kg m3 = m4 = 0.055488 kg/m ( 0.36 m ) = 0.019976 kg Now observe that the centroidal products of inertia I xy = I yz
= I z x = 0 for each component.
Also Then
x3 = x4 = 0,
y1 = 0,
z1 = z2 = 0
0 I xy = I xy + mx y = m2 x2 y2
(
)
2 0.36 m -3 2 = ( 0.062756 kg ) - ( 0.36 m ) = -5.1777 10 kg m 0 I yz = I yz + my z = m3 y3 z3 + m4 y4 z4
(
)
where
m3 = m4 ,
y3 = y4 , z4 = - z3 ,
so that
I yz = 0
I zx = ( I zx + mz x ) = m1z1x1 + m2 z2 x2 = 0
PROBLEM 9.173 CONTINUED
From the solution to Problem 9.149 I x = 23.170 10-3 kg m 2
I y = 21.444 10-3 kg m 2
I z = 17.992 10-3 kg m 2 Now Have
I OL =
2 I xx
OL =
1 ( -3i - 6 j + 2k ) 7 0 0
+
2 I y y
+
I z z2
- 2 I xy x y - 2 I yz y z - 2 I zx z x Eq. ( 9.46 )
2 2 2 3 6 2 = 23.170 - + 21.444 - + 17.992 7 7 7
3 6 - 2 ( -5.1777 ) - - 10-3 kg m 2 2 7 = ( 4.2557 + 15.755 + 1.4687 + 3.8040 ) 10-3 kg m = 25.283 10-3 kg m 2 or I OL = 25.3 10-3 kg m 2
PROBLEM 9.174
For the wire figure of the problem indicated, determine the mass moment of inertia of the figure with respect to the axis through the origin characterized by the unit vector = ( -3i - 6 j + 2k ) / 7. Problem 9.148
SOLUTION
First compute the mass of each component. Have m = ( m/L ) L = ( 0.049 kg/m ) L Then m1 = ( 0.049 kg/m )( 2 0.32 m ) = 0.09852 kg m2 = m3 = m4 = m5 = ( 0.049 kg )( 0.160 m ) = 0.00784 kg Now observe that Also, Then 0 I xy = I xy + mx y = m2 x2 y2 + m3x3 y3 I xy = I yz = I z x = 0 for each component. y1 = 0, z1 = z2 = z3 = 0
x1 = x4 = x5 = 0,
(
)
= ( 0.00784 kg ) ( 0.32 m )( 0.08 m ) + ( 0.24 m )( 0.16 m )
= 0.50176 10-3 kg m 2 By symmetry Now
I yz = I xy = 0.50176 10-3 kg m 2
I zx = ( I zx + mz x ) = 0
From the solution to Problem 9.148 I x = I z = 6.8505 10-3 kg m 2
I y = 12.630 10-3 kg m 2
PROBLEM 9.174 CONTINUED
Now Have
I OL =
2 I xx
OL =
1 ( -3i - 6 j + 2k ) 7 0
+
2 I y y
+
I z z2
- 2 I xy x y - 2 I yz y z - 2I zxz x Eq. ( 9.46 )
3 2 6 2 = ( 6.8505 ) - + - 10-3 kg m 2 7 7
2 2 - (12.63) 10-3 kg m 2 7
3 6 6 2 -2 ( 0.50176 ) - - + - 10-3 kg m 2 7 7 3 7 = ( 6.29128 + 1.03102 - 0.12288 ) 10-3 kg m 2 = 0.719942 kg m 2 or I OL = 7.2 10-3 kg m 2
PROBLEM 9.175
For the rectangular prism shown, determine the values of the ratios b/a and c/a so that the ellipsoid of inertia of the prism is a sphere when computed (a) at point A, (b) at point B.
SOLUTION
(a) Using Figure 9.28 and the parallel-axis theorem have at point A.. I x = I y = I z = 1 m b2 + c2 12
(
)
)
2
1 1 a m a2 + c2 + m = m 4a 2 + c 2 12 2 12 1 1 a m a 2 + b2 + m = m 4a 2 + b 2 12 2 12 I xy = I yz = I z x = 0
(
(
) )
(
)
2
(
Now observe that symmetry implies
Using Equation (9.48), the equation of the ellipsoid of inertia is then
I x x 2 + I y y 2 + I z z 2 = 1
or
1 1 1 m b2 + c2 x2 + m 4a 2 + c 2 y 2 + m 4a 2 + b 2 z 2 = 1 12 12 12
(
)
(
)
(
)
For the ellipsoid to be a sphere, the coefficients must be equal. Therefore, 1 1 1 m b2 + c 2 = m 4a 2 + c 2 = m 4a 2 + b 2 12 12 12 Then and b 2 + c 2 = 4a 2 + c 2 b 2 + c 2 = 4a 2 + b 2
(
)
(
)
(
)
b =2 a c =2 a
or or
(b) Using Figure 9.28 and the parallel-axis theorem, we have at point B I x = I y = I z = 1 1 c m b2 + c2 + m = m b 2 + 4c 2 12 12 2 1 1 c m a2 + c2 + m = m a 2 + 4c 2 12 12 2 1 m a 2 + b2 12
(
)
2
(
)
)
(
)
2
(
(
)
PROBLEM 9.175 CONTINUED
Now observe that symmetry implies I xy = I yz = I z x = 0 From part a it then immediately follows that 1 1 1 m b 2 + 4c 2 = m a 2 + 4c 2 = m a 2 + b2 12 12 12 Then b 2 + 4c 2 = a 2 + 4c 2 b 2 + 4c 2 = a 2 + b 2
(
)
(
)
(
)
b =1 a c 1 = a 2
or
and
or
PROBLEM 9.176
For the right circular cone of Sample Prob. 9.11, determine the value of the ratio a/h for which the ellipsoid of inertia of the cone is a sphere when computed (a) at the apex of the cone, (b) at the center of the base of the cone.
SOLUTION
(a) From sample Problem 9.11, we have at the apex A
Ix =
3 ma 2 10
3 1 2 m a + h2 5 4
I y = Iz =
Now observe that symmetry implies I xy = I yz = I zx = 0 Using Equation (9.48), the equation of the ellipsoid of inertia is then
I x x2 + I y y 2 + I z z 2 = 1
or
3 3 1 3 1 ma 2 x 2 + m a 2 + h 2 y 2 + m a 2 + h 2 z 2 = 1 10 5 4 5 4
For the ellipsoid to be a sphere, the coefficients must be equal. Therefore, 3 3 1 ma 2 = m a 2 + h 2 10 5 4 (b) From Sample Problem 9.11, we have I x = and at the centroid C 3 ma 2 10 I y = 3 2 1 2 m a + h 20 4
2
or
a =2 h
Then
I y = I z = =
3 2 1 2 h m a + h + m 20 4 4
1 m 3a 2 + 2h 2 20
(
)
Now observe that symmetry implies I xy = I yz = I zx = 0 From part a it then immediately follows that 3 1 ma 2 = m 3a 2 + 2h 2 10 20
(
)
or
a = h
2 3
PROBLEM 9.177
For the homogeneous circular cylinder shown, of radius a and length L, determine the value of the ration a/L for which the ellipsoid of inertia of the cylinder is a sphere when computed (a) at the centroid of the cylinder, (b) at point A.
SOLUTION
(a) From Figure 9.28 Ix = 1 2 ma 2 I y = Iz = 1 m 3a 2 + L2 12
(
)
Now observe that symmetry implies I xy = I yz = I zx = 0 Using Equation (9.48), the equation of the ellipsoid of inertia is then I x x 2 + I y y 2 + I z z 2 = 1: 1 2 2 1 1 ma x + m 3a 2 + L2 y 2 + m 3a 2 + L2 = 1 2 12 12
(
)
(
)
For the ellipsoid to be a sphere, the coefficients must be equal. Therefore, 1 2 1 ma = m 3a 2 + L2 2 12 (b) Using Fig. 9.28 and the parallel-axis theorem Have I x = 1 2 ma 2 1 7 2 L 1 m 3a 2 + L2 + m = m a 2 + L 12 48 4 4 I xy = I yz = I z x = 0 From Part a it then immediately follows that 1 2 7 2 1 ma = m a 2 + L 2 4 48 or
(
)
or
a 1 = L 3
I y = I z =
(
)
2
Now observe that symmetry implies
a 7 = L 12
PROBLEM 9.178
Given an arbitrary body and three rectangular axes x, y, and z, prove that the moment of inertia of the body with respect to any one of the three axes cannot be larger than the sum of the moments of inertia of the body with respect to the other two axes. That is, prove that the inequality I x I y + I z and the two similar inequalities are satisfied. Further, prove that I y 1 Ix 2 if the body is a homogeneous solid of revolution, where x is the axis of revolution and y is a transverse axis.
SOLUTION
(i) To prove By definition I y + Iz Ix
I y = z 2 + x 2 dm
(
)
I z = x 2 + y 2 dm
(
)
Then
I y + I z = z 2 + x 2 dm + x 2 + y 2 dm = y 2 + z 2 dm + 2 x 2dm
(
)
(
)
(
)
Now..
2 2 ( y + z ) dm = I x
and Q.E.D.
2 x dm 0
I y + Iz Ix
The proofs of the other two inequalities follow similar steps. (ii) If the x axis is the axis of revolution, then I y = Iz and from part (i) or or Iy I y + Iz Ix 2I y I x 1 Ix 2 Q.E.D.
PROBLEM 9.179
Consider a cube of mass m and side a. (a) Show that the ellipsoid of inertia at the center of the cube is a sphere, and use this property to determine the moment of inertia of the cube with respect to one of its diagonals. (b) Show that the ellipsoid of inertia at one of the corners of the cube is an ellipsoid of revolution, and determine the principal moments of inertia of the cube at that point.
SOLUTION
(a) At the center of the cube have (using Figure 9.28) Ix = I y = Iz = 1 1 m a 2 + a 2 = ma 2 12 6 I xy = I yz = I zx = 0
(
)
Now observe that symmetry implies
Using Equation (9.48), the equation of the ellipsoid of inertia is 1 2 2 1 2 2 1 2 2 ma x + ma y + ma z = 1 6 6 6 or which is the equation of a sphere. Since the ellipsoid of inertia is a sphere, the moment of inertia with respect to any axis OL through the center O of the cube must always 1 be the same R = . I OL
I OL = 1 2 ma 6
x2 + y 2 + z 2 =
6 = R2 2 ma
(
)
(b) The above sketch of the cube is the view seen if the line of sight is along the diagonal that passes through corner A. For a rectangular coordinate system at A and with one of the coordinate axes aligned with the diagonal, an ellipsoid of inertia at A could be constructed. If the cube is then rotated 120 about the diagonal, the mass distribution will remain unchanged. Thus, the ellipsoid will also remain unchanged after it is rotated. As noted at the end of section 9.17, this is possible only if the ellipsoid is an ellipsoid of revolution, where the diagonal is both the axis of revolution and a principal axis.
It then follows that
I x = I OL =
1 2 ma 6
PROBLEM 9.179 CONTINUED
In addition, for an ellipsoid of revolution, the two transverse principal moments of inertia are equal and any axis perpendicular to the axis of revolution is a principal axis. Then, applying the parallelaxis theorem between the center of the cube and corner A for any perpendicular axis
I y = I z = 3 1 2 ma + m 2 a 6
2
or I y = I z =
11 2 ma 12
Note: Part b can also be solved using the method of Section 9.18. First note that at corner A
Ix = I y = Iz =
2 2 ma 3
I xy = I yz = I zx =
1 2 ma 4
Substituting into Equation (9.56) yields
k 3 - 2ma 2k 2 +
55 2 6 121 3 9 m a k- ma =0 48 864 11 2 ma 12
For which the roots are
k1 =
1 2 ma 6
k2 = k3 =
PROBLEM 9.180
Given a homogeneous body of mass m and of arbitrary shape and three rectangular axes x, y, and z with origin at O, prove that the sum I x + I y + I z of the moments of inertia of the body cannot be smaller than the similar sum computed for a sphere of the same mass and the same material centered at O. Further, using the results of Prob. 9.178, prove that if the body is a solid of revolution, where x is the axis of revolution, its moment of inertia I y about a transverse axis y cannot be smaller than 3ma 2 , where a is the radius 10 of the sphere of the same mass and the same material.
SOLUTION
(i) Using Equation (9.30), we have
I x + I y + I z = y 2 + z 2 dm + z 2 + x 2 dm + x 2 + y 2 dm
(
)
(
)
(
)
= 2 x 2 + y 2 + z 2 dm
= 2 r 2dm where r is the distance from the origin O to the element of mass dm. Now assume that the given body can be formed by adding and subtracting appropriate volumes V1 and V2 from a sphere of mass m and radius a which is centered at O; it then follows that m1 = m2 mbody = msphere = m .
(
)
(
)
Then
( I x + I y + I z )body = ( I x + I y + I z )sphere + ( I x + I y + I z )V
- Ix + I y + Iz or
1
(
)V
2
( I x + I y + I z )body = ( I x + I y + I z )sphere + 2m r 2dm - 2m r 2dm
1 2
Now, m1 = m2 and r1 r2 for all elements of mass dm in volumes 1 and 2.
m r 2dm - m r 2dm 0 1 2
so that
( I x + I y + I z )body ( I x + I y + I z )sphere
Q.E.D.
PROBLEM 9.180 CONTINUED
(ii) First note from Figure 9.28 that for a sphere
Ix = I y = Iz =
Thus,
2 2 ma 5
( I x + I y + I z )sphere = 6 ma 2 5
I y = Iz
For a solid of revolution, where the x axis is the axis of revolution, have
Then, using the results of part i
( I x + 2I y )body 6 ma2 5
From Problem 9.178 have or
Iy
1 Ix 2
( 2I y - I x )body 0
Adding the last two inequalities yields
( 4I y )body 6 ma 2 5
or 3 ( I y )body 10 ma2 Q.E.D.
PROBLEM 9.181
The homogeneous circular cylinder shown has a mass m, and the diameter OB of its top surface forms 45 angles with the x and z axes. (a) Determine the principal moments of inertia of the cylinder at the origin O. (b) Compute the angles that the principal axes of inertia at O form with the coordinate axes. (c) Sketch the cylinder, and show the orientation of the principal axes of inertia relative to the x, y, and z axes.
SOLUTION
(a) First compute the moments of inertia using Figure 9.28 and the parallel-axis theorem.
a 2 a 2 13 1 2 m 3a 2 + a 2 + m + 2 = 12 ma 12 2 1 3 2 I y = ma 2 + m ( a ) = ma 2 2 2 Ix = Iz =
(
)
Next observe that the centroidal products of inertia are zero because of symmetry. Then 0 a a I xy = I xy + mx y = m - = 2 2 0 a a I yz = I yz + my z = m - = 2 2 0 a a I zx = I z x + mz x = m = 2 2 Substituting into Equation (9.56)
- - 1 ma 2 2 2 1 2 2 ma 2
1 2 ma 2
13 3 13 2 2 K3 - + + ma K 12 2 12 13 3 3 13 13 13 1 1 1 + + + - - - - - 2 2 2 12 2 2 12 12 12 2 2
2 2 2
2 2 2
ma 2
(
)
2
K
3 13 3 13 13 1 3 1 13 1 1 1 1 2 - - - - 2 2 - 12 - - 2- - 2 ma = 0 2 2 2 2 12 2 12 12 2 2 2 2
(
)
Simplifying and letting K = ma 2 yields
3 -
11 2 565 95 + - =0 3 144 96
PROBLEM 9.181 CONTINUED
Solving yields
1 = 0.363383
2 =
19 12
3 = 1.71995
The principal moments of inertia are then
K1 = 0.363ma 2 K 2 = 1.583ma 2 K3 = 1.720ma 2
(b) To determine the direction cosines x , y , z of each principal axis, we use two of the equations of Equations (9.54) and Equation (9.57). Thus (9.54a) ( I x - K ) x - I xy y - I zxz = 0
- I zxx - I yz y + ( I z - K ) z = 0
2 2 x + y + z2 = 1
(9.54c) (9.57)
Note: Since I xy = I yz , Equations (9.54a) and (9.54c) were chosen to simplify the "elimination" of y during the solution process. Substituting for the moments and products of inertia in Equations (9.54a) and (9.54c)
1 13 2 1 ma 2 y - ma 2 z = 0 ma - K x - - 12 2 2 2 1 1 13 - ma 2 x - - ma 2 y + ma 2 - K z = 0 2 12 2 2
or and
1 1 13 - x + y - z = 0 2 2 2 12 1 1 13 - x + y + - 2 2 2 12
(i) (ii)
z = 0
Observe that these equations will be identical, so that one will need to be replaced, if 13 1 - = - 12 2 or
=
19 12
Thus, a third independent equation will be needed when the direction cosines associated with K 2 are determined. Then for K1 and K3
PROBLEM 9.181 CONTINUED
13 1 13 1 Eq.(i) Eq.(ii) - - - x + - - - z = 0 12 2 2 12
or Substituting into Eq.(i) or
z = x
1 1 13 - x + y - x = 0 12 2 2 2
y = 2 2 -
7 x 12
Substituting into Equation (9.57)
2 x + 2 2 - x + ( x ) = 1 12 2 2 7 2 2 + 8 - x = 1 12
7
2
or
(iii)
K1 : Substituting the value of 1 into Eq.(iii)
2 7 2 2 + 8 0.363383 - ( x )1 = 1 12
or and then
( x )1 = ( z )1 = 0.647249
( y )1 = 2
7 2 0.363383 - ( 0.647249 ) 12
= -0.402662
( x )1 = ( z )1 = 49.7
( y )1 = 113.7
K 3 : Substituting the value of 3 into Eq.(iii)
2 7 2 2 + 8 1.71995 - ( x )3 = 1 12
or and then
( x )3 = ( z )3
= 0.284726
( y )3 = 2
7 2 1.71995 - ( 0.284726 ) 12
= 0.915348
( x )3 = ( z )3
= 73.5
( y )3 = 23.7
PROBLEM 9.181 CONTINUED
K 2 : For this case, the set of equations to be solved consists of Equations (9.54a), (9.54b), and (9.57). Now - I xy x + I y - K y - I yz z = 0 Substituting for the moments and products of inertia. 1 1 3 - - ma 2 x + ma 2 - K y - - ma 2 z = 0 2 2 2 2 2 or 1 1 3 x + - y + z = 0 2 2 2 2 2 (iv)
(
)
(9.54b)
Substituting the value of 2 into Eqs.(i) and (iv) 1 13 19 - ( x )2 + y 2 2 12 12 1 3 19 ( x )2 + - y 2 2 2 12 or and Adding yields and then Substituting into Equation (9.57) 0
2
( )2 - 1 ( z )2 = 0 2
( )2 + 2 1 2 ( z )2 = 0
- ( x )2 +
1 y 2
( )2 - ( z )2 = 0
( x )2 -
2 y 6
( )2 + ( z )2 = 0
( y )2 = 0
( z )2
= - ( x )2
( x ) 2 + ( y ) 2 + ( - x ) 2 2 2
= 1 2 and
=1 =- 1 2
or
( x )2
( z )2
( x )2
= 45.0 y
( )2 = 90.0 ( z )2 = 135.0
(c) Principal axes 1 and 3 lie in the vertical plane of symmetry passing through points O and B. Principal axis 2 lies in the xz plane.
PROBLEM 9.182
Prob. 9.167 For the component described in the problem indicated, determine (a) the principal moments of inertia at the origin, (b) the principal axes of inertia at the origin. Sketch the body and show the orientation of the principal axes of inertia relative to the x, y, and z axes.
SOLUTION
From the solution to Problem 9.143 and 9.167 I x = 34.106 10-3 lb ft s 2 I y = 50.125 10-3 lb ft s 2 I z = 34.876 10-3 lb ft s 2 I xy = 0.96211 10-3 lb ft s 2 I yz = I zx = 0 (a) From Equation 9.55
Ix - K I xy 0 I xy Iy - K 0 =0 0 0 Iz - K or or Then Now
( Ix - K )( I y - K )( Iz
( Iz
2 - K ) - ( I z - K ) I xy = 0
2 - K ) ( I x - K ) I y - K - I xy = 0
(
)
2 I z - K = 0 and I x I y - I x + I y K + K 2 - I xy = 0
(
)
K1 = I z = 34.876 10-3 lb ft s 2 or K1 = 34.9 10-3 lb ft s 2
and
(34.876 10 )(50.125 10 ) - (34.106 10 + K - ( 0.96211 10 ) = 0
-3 -3
2
-3
+ 50.125 10-3 K
)
-3
2
or Solving yields
1.70864 10-3 - 84.231 10-3 K + K 2 = 0 K 2 = 34.0486 10-3 K3 = 50.1824 103 or K 2 = 34.0 10-3 lb ft s 2 and K3 = 50.2 10-3 lb ft s 2
PROBLEM 9.182 CONTINUED
(b) To determine the directions cosines x , y and z of each principal axis use two of the Equations 9.54 and Equation 9.57
K1 : Using Equation 9.54(a) and Equation 9.54(b) with I yz = I zx = 0 ,
we have
( I x - K1 )( x )1 - I xy ( y )1 = 0
- I xy ( x )1 + I y - K1 y Substituting
(
)( )1 = 0
( )1 = 0
1
(34.106 10
or
-3
- 34.876 10-3 ( x )1 - 0.96211 10-3 y
)
-0.96211 10-3 ( x )1 + 50.125 10-3 - 34.876 10-3 y
(
)( )
=0
-0.770 10-3 ( x )1 - 0.96211 10-3 y
( )1 = 0 ( )1 = 0
or
-0.96211 10-3 ( x )1 + 15.249 10-3 y Solving yields 0 From Equation 9.57 and
( x )1 = ( y )1 = 0
+ y
( )
2 x 1
( )1 + ( z )12 = 1
2
0
( z )1 = 1
( x )1 = 90.0,
( y )1 = 90.0, )( )2 = 0
( z )1 = 0
K 2 : Using Equation 9.54(b) and Equation 9.54(c) with I yz = I zx = 0
- I xz ( x )2 + I y - k2 y
(
( Iz
Now Substituting
- K 2 )( z )2 = 0
I z K 2 ( z )2 = 0
-0.96211 10-3 ( x )2 + 50.125 10-3 - 34.0486 10-3 y
(
)( )
2
=0
or
( y )2 = 0.05985 ( x )2
PROBLEM 9.182 CONTINUED
Then
( )
2 x 2
0 2 0.05985 ( x ) + ( z ) = 1 + 2 2
( x )2 ( x )2 ( Iz
Now Substituting
= 0.99821
( y )2 = 0.05974
and = 3.43,
( y )2 = 86.6, ( z )2 = 90.0 )( )3 = 0
K3 :
- I xy ( x )3 + I y - K3 y
(
- K3 ) ( z )3 = 0
I z K3 ( z )3 = 0
-0.96211 10-3 ( x )3 + 50.125 10-3 - 50.1824 10-3 y
(
)( )
3
=0
-0.96211 10-3 ( x )3 - 0.0574 10-3 y or Have yields and
( )3 = 0
0
2 z 3
( y )3 = -16.7615 ( x )3
( )
2 x 3
+ -16.7615 ( x )3 + ( and
2
)
=1
( x )3
= -0.059555
( y )3 = 0.998231
z = 90.0
( x )3
= 93.4,
( y )3 = 3.41,
(c) Principal axis 1 coincides with the z axis, while the principal axes 2 and 3 lie in the xy plane
PROBLEM 9.183
Prob. 9.147 and 9.151 For the component described in the problem indicated, determine (a) the principal moments of inertia at the origin, (b) the principal axes of inertia at the origin. Sketch the body and show the orientation of the principal axes of inertia relative to the x, y, and z axes.
SOLUTION
From Problem 9.147: I x = 9.8821 10-3 lb ft s 2
I y = 11.5344 10-3 lb ft s 2
I z = 2.1878 10-3 lb ft s 2 From Problem 9.151:
I xy = 0.48776 10-3 lb ft s 2 I yz = 1.18391 10-3 lb ft s 2
I zx = 2.6951 10-3 lb ft s 2 (a) From Equation 9.56
2 2 2 K 3 - I x + I y + I z K 2 + I x I y + I y I z + I z I x - I xy - I yz - I zx K 2 2 2 - I x I y I z - I x I yz - I y I zx - I z I xy - 2 I xy I yz I zx = 0
(
)
(
)
(
)
Substituting K 3 - ( 9.8821 + 11.5344 + 2.1878 ) 10-3 K 2 + ( 9.8821)(11.5344 ) + (11.5344 )( 2.1878 )
2 2 2 + ( 2.1878 )( 9.8821) - ( 0.48776 ) - (1.18391) - ( 2.6951) 10-6 K 2 2 - ( 9.8821)(11.5344 )( 2.1878 ) - ( 9.8821)(1.18391) - (11.5344 )( 2.6951) 2 - ( 2.1878 )( 0.48776 ) - 2 ( 0.48776 )(1.18391)( 2.6951) 10-9 = 0
{
}
or
K 3 - 23.6043 10-3 K 2 + 151.9360 10-6 K - 148.1092 10-9 = 0
(
)
(
)
Solving numerically K1 = 1.180481 10-3 lb ft s 2 K 2 = 10.72017 10-3 lb ft s2 K3 = 11.70365 10-3 lb ft s2 or or or K1 = 1.180 10-3 lb ft s2 K 2 = 10.72 10-3 lb ft s2 K3 = 11.70 10-3 lb ft s2
PROBLEM 9.183 CONTINUED
Solving numerically K1 = 1.180481 10-3 lb ft s 2 K 2 = 10.72017 10-3 lb ft s2 K3 = 11.70365 10-3 lb ft s2 (b) From Equations 9.54(a) and 9.54(b) or or or K1 = 1.180 10-3 lb ft s2 K 2 = 10.72 10-3 lb ft s2 K3 = 11.70 10-3 lb ft s2
( I x - K )( x ) - I xy ( y ) - I zx ( z ) = 0
- I xy ( x ) + I y - K1 y - I yz ( z ) = 0
(
)( )
K1: Substitute K1 and solve for to get ( x ) , xy and y
( 9.8821 - 1.180481) ( ) - 0.48776 x 1 y
( )
( )3 .
( )1 - 2.6951( z )1 10-3 = 0
( )1 - 1.18391( z )1 10-3 = 0
-0.48776 ( ) + (11.5344 - 1.180481) x 1 y
or
17.83996 ( x )1 - y
( )1 - 5.52546 ( z )1 = 0
-0.0471( x )1 + y1 Then and Equation 9.57: Substituting
( )1 - 0.11434 ( z )1 = 0
( z )1 = 3.1549 ( x )1
( y )1 = 0.40769 ( x )1
2 ( x )12 + ( y )1 + ( z )1 2
=1
2
( x )12 + 0.40769 ( x )1
or and
2
+ 3.1549 ( x )1 = 1 then then then
( x )1 = 0.29989
( x )1 = 72.5
( y )1 = 0.122262
( z )1 = 0.94612
( y )1 = 83.0
( z )1 = 18.89
K 2: Substitute K 2 and solve for .
( 9.8821 - 10.72017 ) ( ) - 0.48776 x 2 y
( )2 - 2.6951( z )2 10-3 = 0 ( )2 - 1.18391( z )2 10-3 = 0
-0.48776 ( ) + (11.5344 - 10.72017 ) x 2 y
PROBLEM 9.183 CONTINUED
or -1.718202 ( x )2 - y -0.599045 ( x )2 + y Then and Then
( )2 - 5.52546 ( z )2 = 0
( )2 - 1.45402 ( z )2 = 0
= -0.33201( x )2
( z )2
( y )2 = 0.116306 ( x )2
( x )2 + 0.116306 ( x )2 2 or And
2
+ -0.33201( x )2 = 1 = 0.94333 then then then
2
( x )2
( z )2
( x )2
( z )2
= 19.38
( y )2 = 0.109715
= -0.31320
( y )2 = 83.7
= 108.3
K 3: Substitute K3 and solve for .
( 9.8821 - 11.70365 ) ( ) - 0.48776 x 3 y
( )3 - 2.6951( z )3 10-3 = 0 ( )3 - 1.18391( z )3 10-3 = 0
-0.48776 ( ) + (11.5344 - 11.70365 ) x 3 y
or
-3.73452 ( x )3 - y 2.88189 ( x )3 + y
( )3 - 5.52546 ( z )3 = 0
( )3 + 6.99504 ( z )3 = 0
( z )3
= 0.58019 ( x )3
Then and Then
( y )3 = -6.9403( y )3
2 ( x )3 + -0.69403( x )3 2
+ 0.58019 ( x )3 = 1
= -0.142128* then
2
or
( x )3
( x )3
( z )3
= 98.2
( y )3 = 0.98641
( z )3
*
then then
( y )3 = 9.46
= 94.7
= -0.082461
Note: the negative root of ( x )3 is taken so that axes 1, 2, 3 form a right-handed set.
PROBLEM 9.184
Prob. 9.169 For the component described in the problem indicated, determine (a) the principal moments of inertia at the origin, (b) the principal axes of inertia at the origin. Sketch the body and show the orientation of the principal axes of inertia relative to the x, y, and z axes.
SOLUTION
(a) From the solution of Problem 9.169 have Ix = 1W 2 a 2 g W 2 a g 5W 2 a 6 g I xy = I yz = I zx = 1W 2 a 4 g 1W 2 a 8 g 3W 2 a 8 g
Iy = Iz = Substituting into Equation (9.56)
2 2 2 2 1 1 5 W 5 5 1 1 1 3 W K 3 - + 1 + a 2 K 2 + (1) + (1) + - - - a 2 K 6 g 6 6 2 4 8 8 g 2 2 3 2 2 1 5 1 1 2 3 5 1 1 1 3 W 2 - (1) - - (1) - - 2 a = 0 8 6 4 4 8 8 g 2 6 2 8
Simplifying and letting K =
W 2 a yields g
3 - 2.33333 2 + 1.53125 - 0.192708 = 0
Solving yields
1 = 0.163917
The principal moments of inertia are then
2 = 1.05402
3 = 1.11539
W 2 a g W 2 a g W 2 a g
K1 = 0.1639 K 2 = 1.054 K3 = 1.115
PROBLEM 9.184 CONTINUED
(b) To determine the direction cosines x , y , z of each principal axis, use two of the equations of Equations (9.54) and (9.57). Then
K1 : Begin with Equations (9.54a) and (9.54b).
( I x - K1 )( x )1 - I xy ( y )1 - I zx ( z )1 = 0
- I xy ( x )1 + I y - K 2 y Substituting
1 1W 2 W 2 a y - 0.163917 a ( x )1 - g 4 g 2 3 ( )1 - 8 W a2 ( z )1 = 0 g
(
)( )1 - I yz ( z )1 = 0
1W 2 W - a ( x )1 + (1 - 0.163917 ) a 2 y 4 g g
( )1 - 1 W a 2 ( z )1 = 0 8 g
Simplifying yields 1.34433( x )1 - y -0.299013 ( x )1 + y Adding and solving for ( z )1
( )1 - 1.5 ( z )1 = 0
( )1 - 0.149507 ( z )1 = 0
( z )1 = 0.633715 ( x )1
and then
( y )1 = 1.34433 - 1.5 ( 0.633715) ( x )1
= 0.393758 ( x )1
Now substitute into Equation (9.57) ( x )12 + 0.393758 ( x )1 or and
2
+ 0.633715 ( x )1 = 1
2
( x )1 = 0.801504
( y )1 = 0.315599
( z )1 = 0.507925
( x )1 = 36.7
( y )1 = 71.6
=0
( z )1 = 59.5
K 2 : Begin with Equations (9.54a) and (9.54b).
( I x - K 2 )( x )2 - I xy ( y )2 - I zx ( z )2
- I xy ( x )2 + I y - K 2 y
(
)( )2 - I yz ( z )2 = 0
PROBLEM 9.184 CONTINUED
Substituting
1 1W 2 W 2 a y - 1.05402 a ( x )2 - g 4 g 2 3 ( )2 - 8 W a 2 ( z )2 = 0 g
1W 2 W - a ( x )2 + (1 - 1.05402 ) a 2 y 4 g g
( ) 2 - 1 W a 2 ( z )2 = 0 8 g
Simplifying yields -2.21608 ( x )2 - y 4.62792 ( x )2 + y Adding and solving for ( z )2
( )2 - 1.5 ( z )2 = 0
( )2 + 2.31396 ( z )2 = 0
( z )2
and then
= -2.96309 ( x )2
( y )2 = -2.21608 - 1.5 ( -2.96309) ( x )2
= 2.22856 ( x )2
Now substitute into Equation (9.57)
( x )2 + 2.22856 ( x )2 2
or and
2
+ -2.96309 ( x )2 = 1 = 0.260410
2
( x )2
( y )2 = 0.580339
( z )2
= 74.9
= -0.771618
( x )2
( y )2 = 54.5
=0
( z )2
= 140.5
K 3 : Begin with Equations (9.54a) and (9.54b).
( I x - K3 )( x )3 - I xy ( y )3 - I zx ( z )3
- I xy ( x )3 + I y - K3 y Substituting
1 1W 2 W 2 a y - 1.11539 a ( x )3 - g 4 g 2
(
)( )3 - I yz ( z )3 = 0
3 ( )3 - 8 W a 2 ( z )3 = 0 g 1 ( )3 - 8 W a2 ( z )3 = 0 g
1W 2 W - a ( x )3 + (1 - 1.11539 ) a 2 y 4 g g
PROBLEM 9.184 CONTINUED
Simplifying yields -2.46156 ( x )3 - y 2.16657 ( x )3 + y Adding and solving for ( z )3
( )3 - 1.5 ( z )3 = 0
( )3 + 1.08328 ( z )3 = 0
( z )3
and then
= -0.707885 ( x )3
( y )3 = -2.46156 - 1.5 ( -0.707885) ( x )3
= -1.39973 ( x )3
Now substitute into Equation (9.57)
2 ( x )3 + -1.39973( x )3 2
+ -0.707885 ( x )3 = 1 = 0.537577
2
(i)
or and
( x )3
( y )3 = -0.752463
( z )3
= 57.5
= -0.380543
( x )3
( y )3 = 138.8
( z )3
= 112.4
(c) Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain the direction cosines corresponding to the labeled axis, the negative root of Equation (i) must be chosen; that is, ( x )3 = -0.537577 Then
( x )3
= 122.5
( y )3 = 41.2
( z )3
= 67.6
PROBLEM 9.185
Prob. 9.170 For the component described in the problem indicated, determine (a) the principal moments of inertia at the origin, (b) the principal axes of inertia at the origin. Sketch the body and show the orientation of the principal axes of inertia relative to the x, y, and z axes.
SOLUTION
From Problem 9.170 I x = ( I x )1 + ( I x )2 = Iy = Iy
m1 = ta 2
m2 =
1 ta 2 2
Now
1 11 5 ta 2 a 2 + ta 2 a 2 = ta 4 3 6 2 12
(
)
( )1 + ( I y )2 = 1 ( ta 2 ) a2 + 1 1 ta 2 ( a2 + a 2 ) = 1 ta4 3 6 2 2
1 11 3 ta 2 a 2 + a 2 + ta 2 a 2 = ta 4 3 6 2 4
I z = ( I z )1 + ( I z )2 = Now note that symmetry implies
(
)(
)
( I xy )1 = ( I yz )1 = ( I zx )1 = 0 ( I xy )2 = ( I yz )2 = 0
Have
I uv = I uv + mu v
PROBLEM 9.185 CONTINUED
Then 0 a a 1 I xy = m1x1 y1 + m2 x2 y2 = ta 2 = ta 4 2 2 4 0 0 I yx = m1 y1z1 + m2 y2 z2 0 I zx = m1z1x1 + I z x From Problem 9.170 I zx = -
( )2 + m2 z2 x2
1 ( I zx )2 = - 72 ta 4
Then (a) Equation 9.56
1 1 1 1 1 ta 4 + ta 2 a a = ta 4 72 2 3 3 24
2 2 2 K 3 - I x + I y + I z K 2 + I x I y + I y I z + I z I x - I xy - I yz - I zx K 2 2 2 - I x I y I z - I x I yz - I y I zx - I z I xy - 2 I xy I yz I zx = 0
(
)
(
)
)
(
Substituting
2 5 1 1 3 3 5 1 2 5 1 3 1 K 3 - + + ta 4 K 2 + + + - - 0 - ta 4 24 12 2 4 12 2 2 4 4 12 4
(
) K
2
2 2 5 1 3 1 1 3 1 - - 0 - - - 0 ta 4 2 24 4 4 12 2 4
(
)=0
3
Simplifying and letting yields Solving numerically...
K = ta 4
3 =
5 2 479 125 + - =0 3 576 1152
1 = 0.203032 2 = 0.698281 3 = 0.765354
or or or
K1 = 0.203 ta 4 K 2 = 0.698 ta 4 K3 = 0.765 ta 4
(b) Equations 9.54a and 9.54b
( I x - K )( x ) - I xy ( y ) - I zx ( z ) = 0
- I xy ( x ) + I y - K y - I yz ( z ) = 0
(
)( )
PROBLEM 9.185 CONTINUED
Substituting K1 5 1 12 - 0.203032 ( x )1 - 4 y 1 ( )1 - 24 ( z )1 ta 4 = 0
1 1 - 4 ( x )1 + 2 - 0.203032 y or and Equation 9.57 Substituting
( )1 - 0 ta 4 = 0
( y )1 = 0.841842 ( x )1
( z )1 = 0.0761800 ( x )1
2 ( x )12 + ( y )1 + ( z )1 2
=1
2
( x )12 + 0.841842 ( x )1
or
2
+ 0.0761800 ( x )1 = 1 then then then
( x )1 = 0.763715
( x )1 = 40.2
( y )1 = 0.642927
( z )1 = 0.0581798
Substituting K 2 5 1 12 - 0.698281 ( x )2 - 4 y 1 ( )2 - 24 ( z )2 ta4 = 0
( y )1 = 50.0
( z )1 = 86.7
1 1 - 4 ( x )2 + 2 - 0.698281 y or and Then
( )2 - 0 ta 4 = 0
( y )2 = -1.260837 ( x )2
( z )2
= 0.806278 ( x )2
2
( x )2 + -1.260837 ( x )2 2
or
+ 0.806278 ( x )2 = 1 = 0.555573 then then then
2
( x )2
( x )2
= 56.2
( y )2 = -0.700487
( z )2
= 0.447946
( y )2 = 134.5
( z )2
= 63.4
PROBLEM 9.185 CONTINUED
Substituting K3 5 1 12 - 0.765354 ( x )3 - 4 y 1 ( )3 - 24 ( z )3 ta4 = 0
1 1 - 4 ( x )3 + 2 - 0.765354 y or And Then
( )3 - 0 ta 4 = 0
( y )3 = -0.942138 ( x )3
( z )3
= -2.71567 ( x )3
2 2 ( x )3 + -0.942138 ( x )3
+ -2.71567 ( x )3 = 1 = 0.328576 then then then
2
or
( x )3
( x )3
= 70.8
( y )3 = -0.309564
( z )3
(c) = -0.892304
( y )3 = 108.0
( z )3
= 153.2
PROBLEM 9.186
For the component described in the problem indicated, determine (a) the principal moments of inertia at the origin, (b) the principal axes of inertia at the origin. Sketch the body and show the orientation of the principal axes of inertia relative to the x, y, and z axes. Problem 9.150 and 9.172
SOLUTION
(a) From the solutions to Problem 9.150 I x = 14.32 10-3 lb ft s 2
I y = I z = 18.62 10-3 lb ft s 2
From Problem 9.172
I xy = I zx = 4.297 10-3 lb ft s 2 , I xz = 0
Substituting into Eq. (9.56) and using
I y = Iz,
I xy = I zx ,
I yz = 0
2 2 2 2 K 3 - I x + 2I y K 2 + I x 2 I y + I y - 2 I xy K - I x I y - 2I y I xy = 0
(
)
( )
( )
(
)
K 3 - 14.32 10-3 + 2 18.62 10-2 K 2 + 14.32 10-3 ( 2 ) 18.62 10-3 + 18.62 10-3
2 - 2 4.297 10-3 K - 14.32 10-3 18.62 10-3
(
)
(
) (
) (
)
2
(
)
(
)(
)
2
2 - 2 18.62 10-3 4.297 10-3 = 0
(
)(
)
or
K 3 - 51.56 10-3 K 2 + 0.84305 10-3 K - 0.004277 10-3 = 0 K1 = 0.010022 lb ft s 2 K 2 = 0.018624 lb ft s 2 K3 = 0.022914 lb ft s 2 or K1 = 10.02 10-3 lb ft s 2 or K 2 = 18.62 10-3 lb ft s 2 or K3 = 22.9 10-3 lb ft s 2
Solving:
(b) To determine the direction cosines x , y , z of each principal axis, use two of the equations of Equations (9.54) and Equation (9.57). Then
K1 : Begin with Equations (9.54b) and (9.54c):
0
- I xy (x )1 + ( I y - k1)( y )1 - I yz (z ) = 0 - I zx (x )1 - I yz ( y )1 + ( I z - K1) (z ) = 0 or
-4.297 10-3 (x )1 + 18.62 10-3 - 10.02 10-3 ( y )1 = 0 -4.297 10-3 (x )1
-3
( + (18.62 10
- 10.02 10-3
) ) ( )
z 1
=0
PROBLEM 9.186 CONTINUED
( y )1 = ( z )1 = 0.49965 ( x )1
2 ( x )1 + 2 0.49965 ( x )1 2
=1
( x )1 = 0.81669
( y )1 = ( z )1 = 0.40806
( x )1 = 35.2;
K2: Begin with Equations (9.54a) and (9.54b):
( y )1 = ( z )1 = 65.9
( I x - K 2 )( x )2 - I xy ( y )2 - I zx (z )2
- I xy ( x )2 + I y - K 2 y
=0
(
)( )2 - I yz ( z )2 = 0
(i) (ii)
0
Substituting:
(14.32 10
-3
- 18.62 10-3 ( x )2 - 4.297 10-3 ( y )2 - (z ) 2 = 0
-4.297 10-3 ( x )2 + 18.62 10-3 - 18.62 10-3 y
(
)
)( )
From (ii) From (i) Substituting:
( y )2 = - ( z )2 2 2 ( x )2 + ( y )2 + - ( z )2 = 1 2
( y )2 =
1 2
( x )2
2
=0
=0
( x )2
K3: Begin with Equations (9.54b) and (9.54c)
= 90.0, y
( )2 = 45.0, ( z )2 = 135.0
- I xy ( x )3 + I y - K3 y
(
- I zx ( x )3 - I yz y + ( I z - K3 ) ( z )3 = 0 Substituting: -4.297 10-3 ( x )3 + 18.62 10-3 - 22.9 10-3 ( z )3 = 0 -4.297 10 Simplifying:
-3
( )
0
)( )3 + I yz ( z )3 = 0
( x )3
( + (18.62 10
1 3
-3
- 22.9 10
-3
) ) ( )
z 3
=0
( y )3 = ( z )3 = - ( x )3
2
( x )2 + 2 - ( x )3 3
= 1 ( x )3 =
and
( y )3 = ( z )3 = -
( x )3
= 54.7, y
1 3
( )3 = ( z )3 = 125.3
PROBLEM 9.186 CONTINUED
(c) Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set to obtain the direction cosines corresponding to the labeled axis, the negative root of Equation (i) must be chosen; that is:
( x )3
Then:
=-
1 3
( x )3
= 125.3
( y )3 = ( z )3 = 54.7
PROBLEM 9.187
Determine by direct integration the moment of inertia of the shaded area with respect to the y axis.
SOLUTION
At or Then Now
x = a, y = b :
k = y = b a3 b 3 x a3
b = ka3
dI y = x 2dA = x 2 ( b - y ) dx
x3 = bx 2 1 - 3 dx a
Then
x3 a I y = dI y = 0 bx 2 1 - 3 dx a
1 x6 = b x3 - 3 6a 0 3
a
or
Iy =
1 3 ab 6
PROBLEM 9.188
Determine the moments of inertia I x and I y of the area shown with respect to centroidal axes respectively parallel and perpendicular to side AB.
SOLUTION
First locate centroid C of the area. A, in 2 1 2 3 x , in. y , in. xA, in 3 yA, in 3
3.6 0.5 = 1.8 0.5 3.8 = 1.9 1.3 1 = 1.3 5.0
1.8 0.25 0.65
0.25 2.4 4.8
3.24 0.475 0.845 4.560
0.45 4.56 6.24 11.25
Then or And or Now where
X A = xA:
X (5 in 2 ) = 4.560 in 3
X = 0.912 in. Y A = yA: Y (5 in 2 ) = 11.25 in 3
Y = 2.25 in.
I x = ( I x )1 + ( I x )2 + ( I x )3
( I x )1 =
2 1 ( 3.6 in.)( 0.5 in.)3 + 1.8 in 2 ( 2.25 - 0.25) in. 12
(
)
= ( 0.0375 + 7.20 ) in 4 = 7.2375 in 4
( I x )2
=
2 1 ( 0.5 in.)( 3.8 in.)3 + 1.9 in 2 ( 2.4 - 2.25) in. 12
(
)
= ( 2.2863 + 0.0428 ) in 4 = 2.3291 in 4
( I x )3
=
2 1 (1.3 in.)(1 in.)3 + 1.3 in 2 ( 4.8 - 2.25 in.) 12
(
)
= ( 0.1083 + 8.4533) in 4 = 8.5616 in 4 Then
I x = ( 7.2375 + 2.3291 + 8.5616 ) in 4 = 18.1282 in 4
or I x = 18.13 in 4
PROBLEM 9.188 CONTINUED
Also where
Iy = Iy
( )1 + ( I y )2 + ( I y )3
1 ( I y )1 = 12 ( 0.5 in.)( 3.6 in.)3 + (1.8 in 2 ) (1.8 - 0.912 ) in. 2 = (1.9440 + 1.4194 ) in 4 = 3.3634 in 4 1 ( I y )2 = 12 ( 3.8 in.)( 0.5 in.)3 + (1.9 in 2 ) ( 0.912 - 0.25) in. 2 = ( 0.0396 + 0.8327 ) in 4 = 0.8723 in 4 1 ( I y )3 = 12 (1 in.)(1.3 in.)3 + (1.3 in 2 ) ( 0.912 - 0.65) in. 2 = ( 0.1831 + 0.0892 ) in 4 = 0.2723 in 4
Then
I y = ( 3.3634 + 0.8723 + 0.2723) in 4 = 4.5080 in 4
or I y = 4.51 in 4
PROBLEM 9.189
Using the parallel-axis theorem, determine the product of inertia of the area shown with respect to the centroidal x and y axes.
SOLUTION
Have
I xy = I xy
( )1 + ( I xy )2 + ( I xy )3
Now, symmetry implies
( I xy )1 = 0
and for each triangle
I xy = I xy + x yA
where, using the results of Sample Problem 9.6, I xy = - triangles. Note that the sign of I xy
1 2 2 b h for both 72 is unchanged because the angles of
rotation are 0 and 180 for triangles 2 and 3, respectively. Now
A, in 2
x , in.
y , in.
x yA, in 4
2 3
1 ( 9 )(15) = 67.5 2 1 ( 9 )(15) = 67.5 2
-9 9
7 -7
-4252.5 -4252.5 -8505
Then
2 2 1 I xy = 2 - ( 9 in.) (15 in.) - 8505 in 4 72
= -9011.25 in 4 or I xy = -9010 in 4
PROBLEM 9.190
A piece of thin, uniform sheet metal is cut to form the machine component shown. Denoting the mass of the component by m, determine its moment of inertia with respect to (a) the x axis, (b) the y axis.
SOLUTION
First note
mass = m = V = tA 1 a = t ( 2a )( a ) - ( 2a ) 2 2 = 3 ta 2 2
Also
I mass = tI area = 2m I area 3a 2
(a) Now I x, area = ( I x )1, area - 2 ( I x )2, area = = Then 1 a 1 ( a )( 2a )3 - 2 ( a )3 12 12 2 7 4 a 12 I x, mass = 2m 7 a4 3a 2 12 or I x = (b) Have
I z , area = ( I z )1, area - 2 ( I z )2, area
7 ma 2 18
=
3 1 1 a ( 2a )( a )3 - 2 ( a ) 3 2 12
= Then
31 4 a 48 2m 31 4 a 3a 2 48 31 2 ma 72
I z , mass = =
PROBLEM 9.190 CONTINUED
Finally, I y, mass = I x, mass + I z, mass = = 7 31 2 ma 2 + ma 18 72 59 2 ma 72 or I y = 0.819 ma 2
PROBLEM 9.191
A piece of thin, uniform sheet metal is cut to form the machine component shown. Denoting the mass of the component by m, determine its moment of inertia with respect to (a) the axis AA (b) the axis BB where the AA and BB axes are parallel to the x axis and lie in a plane parallel to and at a distance a above the zx plane.
SOLUTION
First note that the x axis is a centroidal axis so that
I = I x, mass + md 2
and that from the solution to Problem 9.115,
I x, mass = (a) Have I AA, mass =
7 ma 2 18
7 2 ma 2 + m ( a ) 18 or I AA = 1.389ma 2
(b) Have
I BB, mass =
7 ma 2 + m a 2 18
(
)
2
or I BB = 2.39ma 2
PROBLEM 9.192
For the 5 3 1 -in. angle cross section shown, use Mohr's circle to 2 determine (a) the moments of inertia and the product of inertia with respect to new centroidal axes obtained by rotating the x and y axes 45 counterclockwise, (b) the orientation of new centroidal axes for which I x = 2 in 4 and I xy < 0.
SOLUTION
From Figure 9.13A
I x = 9.45 in 4
I y = 2.58 in 4
From Problem 9.106 Now
I ave = =
I xy = -2.8125 in 4
1 Ix + I y 2
(
)
1 ( 9.45 + 2.58) in 4 = 6.015 in 4 2
2
R=
Ix - I y 2 + I xy 2
2
2 9.45 - 2.58 = + ( -2.8125 ) 2
= 4.4395 in 4
tan 2 m = - 2I xy Ix - I y =- 2 ( -2.8125 ) in 4 = 0.81877
( 9.45 - 2.58) in 4
2 m = tan -1 ( 0.81877 ) = 39.310
PROBLEM 9.192 CONTINUED
(a) First note Then
= 90 - 39.310 = 50.690
I x , I y = I ave R cos = ( 6.015 4.4395cos ) in 4
or I x = 8.83 in 4 and I y = 3.20 in 4 Also
I xy = R sin = 4.4395sin
or I xy = 3.43 in 4 (b) Have or Then or
I x = I ave - R cos
2 = 6.015 - 4.4395cos
or
= 25.260
2 = 180 - ( 39.310 + 25.260 ) = 115.430
= 57.715
Rotate the centroidal axis 57.7 clockwise.
PROBLEM 9.193
Four 3 3 1 -in. angles are welded to a rolled W section as shown. 4 Determine the moments of inertia and the radii of gyration of the combined section with respect to its centroidal x and y axes.
SOLUTION
W section:
A = 9.13 in 2 I x = 110 in 4
I y = 37.1 in 4
Angle:
A = 1.44 in 2
I x = I y = 1.24 in 4
Note:
Atotal = AW + 4 Aangle
= 9.13 + 4 (1.44 ) in 2
= 14.89 in 2 Now
where Ix = Ix
( )W + 4 ( I x )angle
( I x )angle
= I x + Ad 2
= 1.24 in 4 + 1.44 in 2 ( 4.00 + 0.842 ) in.
(
)
2
= 35.0007 in 4 Then
I x = 110 + 4 ( 35.0007 ) in 4 = 250.0028 in 4
or I x = 250 in 4 and k x2 = Ix 250.0028 in 4 = Atotal 14.89 in 2
or k x = 4.10 in.
also
Iy = Iy
( )W + 4 ( I y )angle
PROBLEM 9.193 CONTINUED
where
( I y )angle = I y + Ad 2
= 1.24 in 4 + 1.44 in 2 ( 5 - 0.842 ) in.
(
)
2
= 26.1361 in 4 Then
I y = 37.1 + 4 ( 26.1361) in 4 = 141.6444 in 4 or I y = 141.6 in 4
and
k y2 =
Iy Atotal
=
141.6444 in 4 14.89 in 2 or k y = 3.08 in.
PROBLEM 9.194
For the 2-kg connecting rod shown, it has been experimentally determined that the mass moments of inertia of the rod with respect to the center-line axes of the bearings AA and BB are, respectively, I AA = 78 gm 2 and I BB = 41 gm 2 . Knowing that ra + rb = 290 mm , determine (a) the location of the centroidal axis GG (b) the radius of gyration with respect to axis GG
SOLUTION
(a) Have
I AA = I GG + mra2
ra + rb = 290 mm
= 0.29 m
and Subtracting
I BB = I GG + mrb2 I BB - I AA = m rb2 - ra2
(
)
( 41 - 78) g m2
or or now so that
= (2000 g) ( rb + ra )( rb - ra )
-37 = (2000) (0.29) ( rb - ra )
ra - rb = 63.793 10-3 m ra + rb = 0.29 m
2ra = 0.35379 m
ra = 0.17689 m
or ra = 176.9 mm (b) Have Then
I AA = I GG + mra2 I GG = 78 g m 2 - (2000 g) (0.17689 m)2
= 15.420 g m 2
Finally,
2 kGG =
I GG 15.420 g m 2 = = 0.007710 m 2 m 2000 g kGG = 0.08781 m kGG = 87.8 mm
PROBLEM 9.195
Determine the mass moment of inertia of the 0.9-lb machine component shown with respect to the axis AA .
SOLUTION
First note that the given shape can be formed adding a small cone to a cylinder and then removing a larger cone as indicated.
Now The weight of the body is given by
h h + 2.4 = 0.4 1.2
or
h = 1.2 in.
W = mg = g ( m1 + m2 - m3 ) = g (V1 + V2 - V3 )
or
2 2 2 1 ft 0.9 lb = 32.2 ft/s 2 ( 0.8 ) ( 2.4 ) + ( 0.2 ) (1.2 ) - ( 0.6 ) ( 3.6 ) in 3 3 3 12 in.
= 32.2 ft/s 2 ( 2.79253 + 0.02909 - 0.78540 ) 10-3 ft 3
3
or Then
= 13.7266 lb s 2 /ft 4
m1 = (13.7266 ) 2.79253 10-3 = 0.038332 lb s 2 /ft m2 m3
-3 2
( ) = (13.7266 ) ( 0.02909 10 ) = 0.000399 lb s /ft = (13.7266 ) ( 0.78540 10 ) = 0.010781 lb s /ft
-3 2
Finally, using Figure 9.28 have,
I AA = ( I AA )1 + ( I AA )2 - ( I AA )3
= 1 3 3 2 2 2 m1a1 + m2a2 - m3a3 2 10 10
2
3 3 2 2 2 1 1 ft = ( 0.038332 )( 0.8 ) + ( 0.000399 )( 0.2 ) - ( 0.010781)( 0.6 ) (lb s 2 /ft) in 2 10 10 2 12 in. = ( 85.1822 + 0.0333 - 8.0858 ) 10-6 lb ft s 2
or I AA = 77.1 10-6 lb ft s 2
PROBLEM 9.196
Determine the moments of inertia and the radii of gyration of the steel machine element shown with respect to the x and y axes. (The density of steel is 7850 kg/m3.)
SOLUTION
First compute the mass of each component. Have
m = stV
Then
m1 = 7850 kg/m3 ( 0.24 0.04 0.14 ) m3
= 10.5504 kg
2 m2 = m3 = 7850 kg/m3 ( 0.07 ) 0.04 m3 = 2.41683 kg 2
(
)
(
)
2 m4 = m5 = 7850 kg/m3 ( 0.044 ) ( 0.04 ) m3 = 1.90979 kg
(
)
Using Figure 9.28 for components 1, 4, and 5 and the equations derived above (before the solution to Problem 9.144) for a semicylinder, have I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x )4 - ( I x )5 where
( I x )2 = ( I x )3
2 2 1 1 = (10.5504 kg ) 0.042 + 0.142 m 2 + 2 ( 2.41683 kg ) 3 ( 0.07 m ) + ( 0.04 m ) 12 12 2 2 2 1 + (1.90979 kg ) 3 ( 0.044 m ) + ( 0.04 m ) + (1.90979 kg )( 0.04 m ) 12 2 2 1 - (1.90979 kg ) 3 ( 0.044 m ) + ( 0.04 m ) 12
(
)
= ( 0.0186390 ) + 2 ( 0.0032829 ) + ( 0.0011790 + 0.0030557 ) - ( 0.0011790 ) kg m 2
= 0.0282605 kg m 2 or I x = 28.3 10-3 kg m 2
PROBLEM 9.196 CONTINUED
Iy = Iy where Then
( )1 + ( I y )2 + ( I y )3 + ( I y )4 - ( I y )5
( I y ) 2 = ( I y )3 ( I y )4 = ( I y )5
1 I y = (10.5504 kg ) 0.242 + 0.142 m 2 12
2 16 4 0.07 2 1 2 0.07 m + ( 2.41683 kg ) 0.12 + + 2 ( 2.41683 kg ) - m 2 3 2 9
(
)
(
)
= ( 0.0678742 ) + 2 ( 0.0037881 + 0.0541678 ) kg m 2
= 0.1837860 kg m 2
or I y = 183.8 10-3 kg m 2
Also
m = m1 + m2 + m3 + m4 - m5 where m2 = m3 , m4 = m5 = (10.5504 + 2 2.41683) kg = 15.38406 kg
Then
2 kx =
Ix 0.0282605 kg m 2 = 15.38406 kg m
or k x = 42.9 mm and
2 ky =
Iy m
=
0.1837860 kg m 2 15.38406 kg
or k y = 109.3 mm
PROBLEM 9.197
Determine the moments of inertia of the shaded area shown with respect to the x and y axes.
SOLUTION
Have where Now and or
I x = ( I x )1 + ( I x )2 - ( I x )3
( I x )1 =
1 ( 2a )( 2a )3 = 4 a 4 12 3 = ( I BB )3 =
( I AA )2
( I AA )2
2 3
8
a4
= ( I xz )2 + Ad 2
2
4a ( I x )2 = ( I x )3 = a 4 - a 2 3 8 2 8 4 = - a 8 9
Then
( I x ) = I x2
2
( )
2
8 4 2 4a 2 + Ad 2 = - a + a a + 8 9 2 3 4 a
2
4 5 = + 8 3
( I x )3
= I x3
( )
3
8 4 2 4a 2 + Ad3 = - a + a a - 3 2 8 9
2
4 5 4 = - + a 8 3 Finally
Ix = 4 4 4 5 4 4 5 4 a + + a - - + a 3 8 3 8 3
or I x = 4a 4 Also where
Iy = (Iy ) + (Iy ) - (Iy )
1 2 3
(Iy )
=
1
1 ( 2a )( 2a )3 + ( 2a )2 ( a )2 = 16 a 4 12 3
4 2 2 = a + a ( a ) 2 8
(Iy )
2
= (Iy )
3
Iy =
16 4 a 3
PROBLEM 9.198
A 20-mm-diameter hole is bored in a 32-mm-diameter rod as shown. Determine the depth d of the hole so that the ratio of the moments of inertia of the rod with and without the hole with respect to the axis AA is 0.96.
SOLUTION
First note Cylinder: I AA = = I AA = = 1 2 ma 2 1 a 4 L 2 3 ma 2 10 1 a 4h 10 m = V = a 2h 3 m = V = a 2 L
(
)
Cone:
Now or or
( I AA )bored ( I AA )solid
= 0.96
0.96 ( I AA )solid = ( I AA )solid - ( I AA )hole
0.04 ( I AA )solid = ( I AA )cylinder + ( I AA )cone hole
Then
1 1 1 4 4 4 0.04 arod Lrod = aholed + aholehcone 2 2 10 a 1 d = 0.04 rod Lrod - hcone 5 ahole 1 16 mm = 0.04 ( 70 mm ) - ( 5.8 mm ) 5 10 mm or d = 17.19 mm
4 4
or