PROBLEM 5.1

Locate the centroid of the plane area shown.

SOLUTION

1'(

I CD (j) TI~O-

f

ee,Lie _ ;500.......

I · ~2'2t""'...

-L I I~ .. 2"0 J fOO"'-' tS'o......~ ~

A,nun2 x,nun y,nun x:4,nun3 yA,nun3

1 200 x 150= 30000 -100 250 -30000000 6750000

2 400 x 300 = 120000 200 150 24000000 18000000

1: 150000 21000000 24750000

- UA 21000000 -Then X =- = nun or X = 140.0 nun <11III

1:.4 150000

Y = 1:jA= 24750000nnn or Y = 165.0nnn <11III1:.4 150000

/';';JJ

PROPRIETARY MATERlA.L IC 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed. reproducedor distributed in any 101711or by any means. without the prior written pennission of the publisher. or used beyond the limited distribution to teachers andeducators pennitted by McGraw-Hili for their individual course preparation. If you are a student using this Manual. you are using it without permission.

611

PROBLEM 5.13

Locate the centroid of the plane area shown.

SOLUTION

:!>..., )I

Then x = ~ =-810.00 in.1:.4 255.82

Y = :tjiA = 171.00 in.1:.4 255.82

or X =-3.17 in. ~

or Y=0.668in. ~

PROPRIETARY MATERIAL. 0 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed. reproducedor distributed in a~vform or by any means, without theprior writtenpemlission of thepublisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hili for their individual course preparation. If you are a student using this Manual. you are using it without permission.

624

A.2 X,in. y,in. XA,in3 yA,in3,m

1 (18)(8)=144 -3 4 -432 576

2 !(6)(9) = 27 2 -3 54 -812

3 1r(12)(9) = 84.823 -5.0930 -3.8197 -432.00 - 324.004

1: 255.82 -810.00 171.00

yPROBLEM 5.18

t.

140

r100

LLti-I 20 f-

The horizontal x axis is drawn through the centroid C of the area shownand divides the area into two component areas ~ and A2. Determine thefirst moment of each component area with respect to the x axis, andexplain the results obtained.x

Dimensions in mm

..I

i ,i

PROPRIETARY MATERIAL. ibi 2007 The McGraw-Hill Companies. Inc. All rights reserved. No part of this Manual may be displayed. reproducedor distributed in any f01711or by any means. withollt the prior written pennission of the puhlisher. or IIsed beyond the limited distribution to teachers and

educators permitted by McGraw-Hillfor their individual course preparation. If you art' a swdell1 u.~ing (hisManual. you are u.~illgit withoutpermission.

630

SOLUTION

'fl

I 'I ·I

· i'10W\M

.tLC'L4-0""''''\

i-- 'I

A nun2 y,nun yA, nun3,

1 (80)(20) = 1600 90 144000

2 (20)(80) = 1600 40 64000

I: 3200 208000

Then - I:j1A 208000 =65.000nuny---- -LA 3200

Now, for the first moments about the x-axis:

Area I

VI

I(l)

Il::: )c

QJ = I:j1A = 25(80 x 20) + 7.5(20 xiS) = 42 250 nun3, or QJ = 42.3 X 103 mm3 <11III

PROBLEM 5.18 CONTINUED

Area II

~I

-

QTI = !:yA = -32.5(20 x 65) = 42250 rom3,

Note that ~area) =QJ + Q/I =0 which is expectedas y = 0 and ~area) = yA sincex is a centroidalaxis.

PROPRIETARY MATERIAL (Q 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be di~played. reprodllcedor distribllted in any form or by any means, witholltthe prior written permission of the publisher. or IIsed beyond the limited distriblltion to teachers andeducators permitled hy McGraw-Hili for their individual cOllrse preparation. If YOIlare a studellt using this Manllal. you are IL~illgit without permission.

631

PROBLEM 5.28

E Knowing that the figure shown is fonned of a thin homogeneous wire,detennine the length 1of portion CE of the wire for which the center of

I--J gravityof the figureis locatedat point C when (a) 0 '= 15°, (b) 0 = 60°.B

PROPRIETARY MATERIAL iO 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed. reproducedor distributed in any form or by any means, without the prior written permission of the publisher. or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hili for their individual course preparation.lfyou are a student using this Manual. you are using it without permission.

641

SOLUTION

The centroid coincides with the center of gravity because the wire is homogeneous.

<." "tL x xL

Ic1.,

I 1'2I' I''" .LA

, I

-- --2 2

2 201' rsinO -21'2sinO . ,---C,9Y c.!

'"0

\\ /3 1 1 12- - ,2 2 9

Then- !:XLX =- = 0 ::::} !:XL= 0 and

U

1'2 12-- - 2,.2sinO + - = 0, or 1 = r.JI + 4sinO

2 2

(a) 0 = 15°:

1 = r.JI + 4sin 15° or 1 = 1.4271' .....

(b) 0 = 60°:

1 = r.Jl + 4sin 60° or/=2.llr.....

SOLUTION

PROBLEM 5.36

Determine by direct integration the centroid of the area shown.

x

First note that symmetry implies

For the element(EL)shown

y = Rcos8, x = Rsin8

dx = Rcos8 d8

dA = ydx = Ifcos28d8

X Hence

J Ja..1 2 "..1(8 Sin28

)l

a 1 2(.

)A = dA =2 0 lr cos 8d8 =L.lC "2+~ 0 ="2 R 2asm2a

JYELdA = 2J; ~ cos8(R2cos28d8) =.JiI(~cos2 8sin8 + ~Sin8 )1:

= ~3 (cos2asina + 2sina)

_ JiI (cos2asina + 2sina)Y _ 3

-~-(2a + sin2a)2

or _ 2 . (cos2a + 2)y =-Rsma3 (2a + sin2a)

Alternatively,- 2

R . 3- sin2a ....y =- sma .....3 2a + sin2a

PROPRIETARY MATERIAL. to 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed. reproduced

or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hili for their individual course preparation. If you are a student using this Manual. you are ILvingit without permission.

651

PROBLEM 5.42 I1tI

y

A homogeneous wire is bent into the shape shown. Detennme by directintegration the x coordinate of its centroid

r

I

SOLUTIONFirst note that because the wire is homogeneous, its center of gravitycoincides with the centroid of the corresponding line

Now XEL= rcos8 and dL = rd8

Then f f7TC14

[ ]7TC14 3

L = dL = rd8 = r 8 = - rrrTCl4 TCl4 2

'I- and

2[

'

]7TC14 2(1 1

) 2 {;;2= r sm8 rr/4 = r -.J2 - .J2 = -r ,,~

Thus

PROPRIETARYMATERIAL.1::12007 The McGraw-Hili Companies, Inc. All rights reserved. No parl of this Manual may be displayed. reproducedor distributed in any fo/'m or by any means, wirhalll the prior written permission of the publisher. or used beyond the limited distribution to teachers andeducarors permirred by McGraw-Hili for rheir individual course preparation. If you are a stlldent using rhis Manllal. you are IIsing it withour permission.

659

'~"'~T;'- IIl ~. 0]1

PROBLEM 5.47

Detennine the volume and the surface area of the solid obtained byrotatingthe area ofProb. 5.2 about (a) the x axis, (b) the lines x = 19 in.

SOLUTION

I ~

-It(9

From the solution to problem 5.2:

IXA = 1102 in3,

and from the solution of problem 5.22

L = 56 in., and "£XL = 488 in2

>Applying the theorems of Pappus-Guldinus, we have

(a) Rotation about the x-axis:

Volume =2nYareaA =21ttyA =2n(536 in3) =3367.8in3 or v = 1.949ft3 <III!

Area = 2nYIengthL =21ttyL = 2n[ 6(15) + 10(4) + 8(10) + 4(18)] =1520.53 in2

or A =10.56 ft2 <III!

(b) Rotation about x = 19 in.:

I.i . =9072.9in3 or v = 5.25 ft3 ~

Area = 2n(19 - Xiine)L= 2n(19L - "£XL)= 2n[(19 in.)(56 in.) - 488 in2]

=3619.1in2

or A =25.1ft2~

~. ,.!Ir"

I. n

r Iif

PROPRIETARY MATERIAL iO 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproducedor distributed in any form or by any means, withuut the prior Ii?'itten pelwission ufthe publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hill for their individual course preparation. lfyou are a student using this Manual, you are using it without permission.

666

I,

i

;, I" I.

PROBLEM 5.55

Determine the volume and total surface area of the body shown.

160mm

20mm ~

SOLUTION I ,

Volume:

The volume can be obtainedby rotating the triangulararea shownthrough n radiansabout they axis.

The area of the triangleis:

A =.!.(52)(6O) = 1560 mm22

Applyingthe theoremsof Pappus-Guldinus.we have

V = nXA = n(52 mm)(156Omm2) or V = 255 X 103 mm3 .....

The surface area can be obtained by rotating the triangle shown through an angle of n radians about the y axis.

y

&

Considering each line BD. DE. and BE separately:

Line BD: £. = ~222 + 602 =63.906 mm Xl=20+ 22 =31mm2

Line DE: ~ = 52 mm x2=20+22+26=68mm

Xi= 20 + 74 = 57 mm2Line BE: L:J= J742 + 602 = 95.268 mm

CQntinuetJ

PROPRIETARY MATERIAL 0 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed. reproducedor distributed in allYform or by any means, without the prior writtenpermission of the publisher, or used beyond the limited distribution to teachers andeducatorspermitted by McGraw-Hilifor their individualcoursepreparation. If you are a student using this Manual.you are using it withoutpermission.

675

fPROBLEM 5.55 CONTINUED

Then applying the theorems of Pappus-Guldinus for the part of the surface area generated by the lines:

AL =n:IXA =H[(3l)(63.906) + (68)(52)+ (57)(95.268)] =H[10947.6] = 34.392 x 103mm2

The area of the "end triangles":

Total surface area is therefore:

A =AL + AE = (34.392 + 3.12) x 103 mm2 or A =37.5 X 103 nun2 ~

,..

"

~t

n

PROPRIETARY MATERIAL () 2007 The McGraw-HilI Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproducedor distributed in anyform or by any means, without the prior writtenpermission of the publisher, or used beyond the limiteddistribution to teachers andeducators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual. you are using it without permission.

676

L

2-ION/IOB

4.2m---l

PROBLEM 5.61

For the beam and loading shown, detennine (a) the magnitude andlocation of the resultant of the distributedload, (b) the reactions at thebeam supports.

A

SOLUTION

~,

(a) Note that in the free-body diagram:

1Rt = -(4.2 m)(600 N/m) = 1260N,2

1and R2= -(4.2 m)(240N/m)= 504N2

Then for the equivalence of the systems of forces:

~y: R = Rt + R2 = 1260 + 504 = 1764 N R =1764N ~

!MA: -x(1764 N) = [(2 +~4.2)m ](1260 N)+[(2 +%4.2)m ](504 N) = 3.8000m

or X =3.80 m ~

(b) Equilibrium:

r.F'x =0: Ax =0

Ay - 1764=0

A =1764N f~

MA - (3.80 m)(1764 N) = 0

MA =6.70 kN.m ) ~

PROPRIETARY MATERIAL. IC 2007 The McGraw-Hili 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 andeducators permitted by McGraw-Hili for their individual course preparation. If you are a student using this Manual. you are using it without permission.

683

I:PROBLEM 5.74

A grade beam AB supports three concentrated loads and rests on soil andB the top of a large rock.The soil exerts an upward distributedload, and theu" rock exerts a concentrated load RR as shown. Knowing that

WB = O.4WA,detennine (a) the largest value ofP for which the beam isin equilibriwn, (b) the corresponding value of WA'

SOLUTION

fI'I

t, "1 1'I I

~k

III~"I

Rt = (3.6 ft)(WAkipslft) = 3.6wA kips

R2= .!.(5.4ft)(0.6WAkipslft) = 1.62wA kips2

R3 = (5.4 ft)(O.4wAkipslft) = 2.16wA kips

(a) Equilibriwn:

+ r.MA=0: -(1.8 ft)[(3.6wA)kips ] + (3.6 ft)RR

+(5.4 ft)[(1.62wA) kips] + (6.3 ft)[(2.16 wA)kips] - (1.5 ft)(6 kips)

-( 6 ft)(4.5 kips) - (7.2 ft)p = 0

or 28.836wA + 3.6RR -7.2P - 36 = 0

RR + 3.6wA + 1.62wA + 2. 16wA - 6 - 4.5 - P = 0

(1)

+ I:Fy.v = 0:

or 7.38wA + RR - P -10,5 = 0

(28.836)Eq.(2) - (7.38)Eq.(I) = 0 gives

2.268RR - 37.098 + 24,3P = 0

(2)

f~!i

PROPRIETARY MATERIAL iC 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part oJthis Manual may be displayed, reproducedor distributed in any Jorm or by any means, without the prior written permission oj the publisher. or used beyond the limited distribution to teachers andeducators pemritted by McGraw-Hili Jor their individual course preparation. If you are a sflldent using this Manual. you are using it without permission.

698

PROBLEM 5.74 CONTINUED

Since RR ~ 0, the maximwn acceptable value of P is that for which RR = 0, and

P = 1.52667 kips

(b) Now, ftom (2):

7.38wA - 1.52667 - lO.5 = 0

P = 1.527 kips ~or

or WA = 1.630kipslft~

PROPRIETARY MATERIAL. "2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproducedor distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hili for their individual course preparation. If you are a student using this Manual, you are using it without permission.

699

'.,.

-; 'iII (-

J

J

J .'.

..

;J.L

t;t."'i

r

If, "!f

'Ii

\'1

,.

',I'

g.

I

PROBLEM 5.99

For the machine elementshown, locate the y coordinate of the center ofgravity.

SOLUTION

tAssume that the machine element is homogeneous so that its center of gravity coincides with the centroid ofthe volume.

'I I

'!'I !~ I I.

, I

I

. 41't 1~I'

1

2

. . 3

4u"

.. ~ I

I.

In\II

, 1I

Then f = I:yV_ 36.309.I:V - 37.217In.

or f =0.976 in. ~

PROPRIETARY MATERIAL 02007 The McGraw-Hili Companies, Inc. All rights reserved. No part l?(this Manual may be displayed. reproducedor distributed in an.vform or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-HilIfor their individual course preparation. If you are a student using this Manual. you are using it without permission.

732

IiI..

-

V . 3

Iy, in. I yv, in'

,m

(8)(0.9)(2.7)=19.44 0.45 8.748

1 1.6 I 27.216

2(2.1)(6)(2.7) =17.01

n( 1.35)'(0.9)=2.5765 5 I \.15943

-1r(0.8)2(0.9) =-1.80956 /o:4s I -0.81430-

37.217 I I 36.309

i.

PROBLEM 5.102

Locate the center of gravity of the sheet-metal fonn shown.

SOLUTION.'

Assume that the body is homogeneous so that its center of gravity coincides with the centroid of the area.II

'III

Ii

I

I

~,

.II

First note that by symmetry:

For 1: y = 180 + 96 + 4(150) = 339.7 mm31r

x =150mm ...

Z=O

For 2: y = 180+ 2(96) = 241.1mm1r

2(96) = 61.11 mmz=-1r

For 3: Length DE = J(180)2+ (96)2 = 204 mm

y=90mm, z=48mm

,~

\"'

PROPRIETARY MATERlAL. C>2007 The McGraw-Hili Companies, Inc. All rights reserved. No part ~rthis Manual may be displayed. reproducedor distributed in any foml or by any means, without the prior written pennission of the publisher. or used be}'Ond the limited distribution to teachers andeducators pennitted by McGraw-Hill for their individual course preparation.lf)'Ou are a student using this Manual. you are using it without pennission.

736

.

PROBLEM 5.102 CONTINUED

Then

f = :tyA = 28.420 x 106nun1:.4 141.78x 103

Z = m = 5.702x 106 nun1:.4 141.78x 103

or f =200 mm .....

or f =40.2 nun .....

.:J

I'..

.r:

,:,.

PROPRIETARY MATERIAL. 02007 The McGmw-HiII Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproducedor distributed in any form or by any means, without the prior wrinen permission of the publisher. or used beyond the limited distribution to teachers and

educators permitted by McGraw-Hilifor their individualcoursepreparation. Ifyou are a student using thisManual,you are using it withoutpermission.

737

A,nun2 y,nun "i,nun y.4, nun3 L4,nun3

11C(150)2 = 35.34 x 103

339.7 0 12.005 x 106 0

2

21C(96)(300) =45.24x 103

244.1 61.11 10.907 x 106 2.765 X 1062

3(204)(300) = 61.2x 103- 2.25

90 48 5.508x 106 2.938 X 106

1: 141.78 x 103 28.420 x 106 5.702 X 106

PROBLEM 5.111

The ftame of a portable equipment cover is fabricated ftom steel pipe ofunifonn diameter. Locate the center of gravity of the ftame.

First note by symmetry:

To simplify the calculations replace:

(a) The two rectangular sides with an element oflength

~a) =2[2(7 ft) + 2(5 ft)] =48 ft

and center of gravity at (3.5 ft, 2.5 ft, 3 ft)

(b) The two semicircular members with an element oflength

Lb = 2[ n(3 ft)] = 6n ft

and with center of gravity at (2ft, 5 + 2: 3 ft, 3 ft) = (2 ft, 6.9099 ft, 3 ft)

(c) The cross members 1 and 2 with an element oflength

4: = 2(6 ft) = 12ft

and with center of gravity at (2 ft, 5 ft, 3 ft)

This leaves a single straight piece of pipe, labeled (d) in the figure.Now for the centroid of the ftame:

,

SOLUTION

I'.I,"Ii

II

I

' "

I

1\

f

I

,:!II

..:!!

iii"'Ii

1~ j!

Z =3.00ft ~

PROPRIETARY MATERIAL Q 2007 The McGraw-HilI Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproducedor distributed in any form or by any means. without the prior written permission of the publisher. or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual. you are using it without permission.

748

L,ft x,ft y,ft xL,ft2 -£ft2y,

(a) 48 3.5 2.5 168 120

(b) 6n =18.8496 2 6.9099 37.699 130.249

(c) 12 2 5 24 60

(d) 6 7 5 42 30

,I: 85.850 271.70 340.25

Then X= m = 271.70 or X =3.20ftI:L 84.850

Y = 'E.JL =340.25 or y =4.01ftI:L 84.850

I

II

PROBLEM 5.121

Locate the centroid of the volume obtained by rotating the shaded areaabout the line x =a.

SOLUTION

I,

,.

First, by symmetry:x=a....

Y=O....

\

Next determine the constants kin y =kxV]:b

At x =a, b =kaV]or k =-v3a

Therefore, y = ~ xV],or x = ~ y3a bChoosing horizontal disks of thickness dy for volume elements (dV in the figure above)

V = f: n[tl- (a - x)2]

=nf:(2ax-~)dy

=nf:(2ax ;1- ~ i )dy

=n tl (2 x ! y 4 _.!. X ! y7

~b =~ntlb

b3 4 b3 7 14o

Now Y = 1. jYELdV, orV

- 14 rb[ (

tl..3 tl 6

) ]Y = 5na2b Joy n 2 b3 y - b6 Y dy

= 14(2 x yS _.!. y8

)

b

5b4 5 b3 8o

II...

~

!,

I

or - - Iib....y - 100

PROPRIETARY MATERIAL. (') 2007 The McGraw-Hill Companies, Inc. All rights reserved. No pan of this Manual m~v be displayed. reproducedor distributed in any form or by any means, without the prior written pelmission of the publisher. or used beyond the limited distribution 10teuchr:~ unaeducators permitted by McGraw-Hill for their individual coursepreparation. 1/you are a studellt using this Manual. you are u.ring it without permission.

764

---------