Pré-visualização50 páginas

CCHHAAPPTTEERR 99 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, 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 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, 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 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may may ma be displayed, 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 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may may ma be displayed, 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 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, 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 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may may ma be displayed, 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