# StaticsSMChap09

DisciplinaEstática Tecnica18 materiais233 seguidores
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```CCHHAAPPTTEERR 99
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1377
PROBLEM 9.1
Determine by direct integration the moment of inertia of the shaded area with
respect to the y axis.
SOLUTION
By observation
h
y x
b
=
Now 2 2
2
[( ) ]
1
ydI x dA x h y dx
x
hx dx
b
= = \u2212
\ufffd \ufffd= \u2212\ufffd \ufffd\ufffd \ufffd
Then 2
0
1
b
y y
x
I dI hx dx
b
\ufffd \ufffd= = \u2212\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd
4
3
0
1
3 4
b
x
h x
b
\ufffd \ufffd
= \u2212\ufffd \ufffd\ufffd \ufffd
31or
12y
I b h= \ufffd\ufffd
1375
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1377
PROBLEM 9.1
Determine by direct integration the moment of inertia of the shaded area with
respect to the y axis.
SOLUTION
By observation
h
y x
b
=
Now 2 2
2
[( ) ]
1
ydI x dA x h y dx
x
hx dx
b
= = \u2212
\ufffd \ufffd= \u2212\ufffd \ufffd\ufffd \ufffd
Then 2
0
1
b
y y
x
I dI hx dx
b
\ufffd \ufffd= = \u2212\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd
4
3
0
1
3 4
b
x
h x
b
\ufffd \ufffd
= \u2212\ufffd \ufffd\ufffd \ufffd
31or
12y
I b h= \ufffd\ufffd
1376
reproduced or distributed in any form form for or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for raw-Hill for raw-Hill their individual course preparation. If you If you If are a student using this Manual,
you are using it without permiyou are using it without permiyou are using it without per ssion.
PROBLEM 9.2
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
SOLUTION
At , :x a y, :a y, :a y, :a y, :, :a, :, := =, := =a y= =a y, :a y, := =, :a y, :
k
a
a
= 2or k ak a=k a
Then
2a
y
x
=
Now 2 2
2 2
( )ydIydIy x d
2 2x d2 2A x2 2A x2 2x dA xx d2 2x d2 2A x2 2x d2 2 y d( )y d( )( )x( )( )y d( )x( )y d( )
x a x dx
= =x d= =x dA x= =A xx dA xx d= =x dA xx d
\ufffd \ufffd2\ufffd \ufffd22 2\ufffd \ufffd2 2a\ufffd \ufffda2 2a2 2\ufffd \ufffd2 2a2 2= =x= =x2 2\ufffd \ufffd2 22 2a2 2\ufffd \ufffd2 2a2 22 2dx2 2\ufffd \ufffd2 2dx2 22 2\ufffd \ufffd2 2\ufffd \ufffd2 2\ufffd \ufffd2 22 2a2 2\ufffd \ufffd2 2a2 2\ufffd \ufffd2 2a2 2\ufffd \ufffd2 2a2 22 2dx2 2\ufffd \ufffd2 2dx2 2\ufffd \ufffd2 2dx2 2\ufffd \ufffd2 2dx2 2\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd2 2\ufffd \ufffd2 2\ufffd \ufffd2 2\ufffd \ufffd2 2dx\ufffd \ufffddx\ufffd \ufffddx\ufffd \ufffddx2 2dx2 2\ufffd \ufffd2 2dx2 2\ufffd \ufffd2 2dx2 2\ufffd \ufffd2 2dx2 2\ufffd \ufffd\ufffd \ufffd\ufffd \ufffddx\ufffd \ufffddx\ufffd \ufffddx\ufffd \ufffddx= =\ufffd \ufffd= =\ufffd \ufffd= =\ufffd \ufffd= == =\ufffd \ufffd= =\ufffd \ufffd= =\ufffd \ufffd= =dx= =dx\ufffd \ufffddx= =dx\ufffd \ufffddx= =dx\ufffd \ufffddx= =dx\ufffd \ufffdx\ufffd \ufffdx\ufffd \ufffd\ufffd \ufffd\ufffd \ufffdx\ufffd \ufffdx\ufffd \ufffdx\ufffd \ufffdx\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd
Then
2 22
2 2 2 2 2[(2 2[(2 22 )2 22 )2 2 ( ) ]
2 2
a
a
y y
a
a2 2a2 2
a2 2a2 2Iy yIy y a x dx
2 2dx2 2a x a a2 )a a2 ) ( )a a( )\ufffd \ufffd2 2\ufffd \ufffd2 21\ufffd \ufffd1= = = =a x= =a x a a\u2212a a\ufffd \ufffd2 2\ufffd \ufffd2 2a x\ufffd \ufffda x\ufffd \ufffda x\ufffd \ufffda x\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd2 2\ufffd \ufffd2 2\ufffd \ufffd2 2\ufffd \ufffd2 2= =\ufffd \ufffd= =a x= =a x\ufffd \ufffda x= =a xa x= =a x\ufffd \ufffda x= =a x\ufffd \ufffd2 2\ufffd \ufffd2 2\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd2 2\ufffd \ufffd2 2\ufffd \ufffd2 2\ufffd \ufffd2 2a x\ufffd \ufffda x\ufffd \ufffda x\ufffd \ufffda x\ufffd \ufffdy y\ufffd \ufffdy y a\ufffd \ufffdadI\ufffd \ufffddIy ydIy y\ufffd \ufffdy ydIy y= =\ufffd \ufffd= =dI= =dI\ufffd \ufffddI= =dI
43or
2y
I yI y a= \ufffd\ufffd
1377
reproduced or distributed in any form form for or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for raw-Hill for raw-Hill their individual course preparation. If you If you If are a student using this Manual,
you are using it without permiyou are using it without permiyou are using it without per ssion.
PROBLEM 9.3
Determine by direct integration the moment of inertia of the shaded area with
respect to the y axis.
SOLUTION
2y kxy kxy ky k=y k
For :x a= 2b kab kab kb k=b k
2
b
k
a
=
Thus: 2
2
b
y x
a
=
( )dA b y( )b y( )dx= \u2212( )= \u2212( )( )b y( )= \u2212( )b y( )
2 2 2 2
3 3
0
( )
1 13 31 13 3
3 5
y
a
y y
dIydIy x d
2 2x d2 2A x2 2A x2 2x dA xx d2 2x d2 2A x2 2x d2 2 b y( )b y( )dx x b2 2x x b2 2dx x bd x dxx dxx d
Iy yIy y x dx ax ax dx ax d b ab a
3 3b a3 33 3
1 13 3b a3 3
1 13 3b
\ufffd \ufffd2 2\ufffd \ufffd2 2b\ufffd \ufffdb2 2b2 2\ufffd \ufffd2 2b2 22 2x x b2 2\ufffd \ufffd2 2x x b2 2x d\ufffd \ufffdx d2 2x d2 2\ufffd \ufffd2 2x d2 2= =x d= =x dA x= =A xx dA xx d= =x dA xx d \u2212 =( )\u2212 =( )( )b y( )\u2212 =( )b y( )d\u2212 =dx x b\u2212 =x x bdx x bd\u2212 =dx x bd \ufffd \ufffd\ufffd \ufffd2 2\ufffd \ufffd2 22\ufffd \ufffd2x x b\ufffd \ufffdx x b2 2x x b2 2\ufffd \ufffd2 2x x b2 2x d\ufffd \ufffdx d2 2x d2 2\ufffd \ufffd2 2x d2 22 2\ufffd \ufffd2 2\ufffd \ufffd2 2\ufffd \ufffd2 22 2
b2 2\ufffd \ufffd2 2b2 2\ufffd \ufffd2 2b2 2\ufffd \ufffd2 2b2 22 2x x b2 2\ufffd \ufffd2 2x x b2 2\ufffd \ufffd2 2x x b2 2\ufffd \ufffd2 2x x b2 2x d\ufffd \ufffdx d\ufffd \ufffdx d\ufffd \ufffdx d2 2x d2 2\ufffd \ufffd2 2x d2 2\ufffd \ufffd2 2x d2 2\ufffd \ufffd2 2x d2 2\u2212\ufffd \ufffd\u2212\ufffd \ufffd2\ufffd \ufffd2a\ufffd \ufffda\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd2\ufffd \ufffd2\ufffd \ufffd2\ufffd \ufffd2x x b\ufffd \ufffdx x b\ufffd \ufffdx x b\ufffd \ufffdx x b x d\ufffd \ufffdx d\ufffd \ufffdx d\ufffd \ufffdx d
\ufffd \ufffd2 4\ufffd \ufffd2 4b\ufffd \ufffdb2 4b2 4\ufffd \ufffd2 4b2 4bx\ufffd \ufffdbx x d\ufffd \ufffdx d2 4x d2 4\ufffd \ufffd2 4x d2 4= = x d\u2212 =x dx a\u2212 =x ax dx ax d\u2212 =x dx ax d b a\u2212b a\ufffd \ufffd2 4\ufffd \ufffd2 42\ufffd \ufffd2bx\ufffd \ufffdbx x d\ufffd \ufffdx d2 4x d2 4\ufffd \ufffd2 4x d2 4\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd2 4\ufffd \ufffd2 4\ufffd \ufffd2 4\ufffd \ufffd2 42 4
b2 4\ufffd \ufffd2 4b2 4\ufffd \ufffd2 4b2 4\ufffd \ufffd2 4b2 4bx\ufffd \ufffdbx\ufffd \ufffdbx\ufffd \ufffdbx x d\ufffd \ufffdx d\ufffd \ufffdx d\ufffd \ufffdx d2 4x d2 4\ufffd \ufffd2 4x d2 4\ufffd \ufffd2 4x d2 4\ufffd \ufffd2 4x d2 4\u2212 =\ufffd \ufffd\u2212 =\u2212 =\ufffd \ufffd\u2212 =x d\u2212 =x d\ufffd \ufffdx d\u2212 =x d\ufffd \ufffd2\ufffd \ufffd2a\ufffd \ufffda\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd2\ufffd \ufffd2\ufffd \ufffd2\ufffd \ufffd2bx\ufffd \ufffdbx\ufffd \ufffdbx\ufffd \ufffdbx x d\ufffd \ufffdx d\ufffd \ufffdx d\ufffd \ufffdx d\ufffd \ufffd0\ufffd \ufffd0y y\ufffd \ufffdy ydI\ufffd \ufffddIy ydIy y\ufffd \ufffdy ydIy y= =\ufffd \ufffd= =dI= =dI\ufffd \ufffddI= =dI 3215yI yI y a b3a b3= \ufffd\ufffd
1378
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1380
PROBLEM 9.4
Determine by direct integration the moment of inertia of the shaded
area with respect to the y axis.
SOLUTION
2
2
4
x x
y h
a a
\ufffd \ufffd
= \u2212\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd
2
2 2
2
3 4
20
4
4
y
a
y
dA y dx
x x
dI x dA hx dx
a a
x x
I h dx
a a
=
\ufffd \ufffd
= = \u2212\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd
\ufffd \ufffd
= \u2212\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd
4 5 3 3
2
0
4 4
4 4 55
a
y
x x a a
I h h
a a
\ufffd \ufffd \ufffd \ufffd
= \u2212 = \u2212\ufffd \ufffd\ufffd \ufffd \ufffd \ufffd\ufffd \ufffd \ufffd \ufffd
31
5y
I ha= \ufffd\ufffd
1379
reproduced or distributed in any form form for or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for raw-Hill for raw-Hill their individual course preparation. If you If you If are a student using this Manual,
you are using it without permiyou are using it without permiyou are using it without per ssion.
PROBLEM 9.5
Determine by direct integration the moment of inertia of the shaded
area with respect to the x axis.
SOLUTION
By observBy observBy ation
h
y xy x
b
y x
b
y xy x=y x
or
b
x y
h
=
Now 2 2
2
3
( )xdIxdIx y d
2 2y d2 2A y2 2A y2 2y dA yy d2 2y d2 2A y2 2y d2 2 x d( )x d( )y( )y( )( )x d( )y( )x d( )
y y
b
y d3y d3 yy dyy d
h
= =y d= =y dA y= =A yy dA yy d= =y dA yy d
\ufffd \ufffdb\ufffd \ufffdb
dy
\ufffd \ufffd
dy= \ufffd \ufffdy y\ufffd \ufffdy yy y\ufffd \ufffdy y dy\ufffd \ufffddy\ufffd \ufffd\ufffd \ufffd\ufffd \ufffdb\ufffd \ufffdb\ufffd \ufffdb\ufffd \ufffdb dy\ufffd \ufffddy\ufffd \ufffddy\ufffd \ufffddy\ufffd \ufffdh\ufffd \ufffdh\ufffd \ufffd\ufffd \ufffd\ufffd```