5 Distributed Forces: 5.I INTRODUCTIONCentroids ond Centers of Grovity5.1 Introduction5,2 Center of Grovity of q Two-

Dimensionol Bodv5.3 Centroids of Aruo, qnd Lines5.4 First Moments of Areos qnd Lines5.5 Composite Plotes ond Wires5.6 Determinotion of Ceniroids

by Integrotion5.7 Theorems of Poppus-Guldinus5.8 Distributed Loods on Beoms5.9 Forces on Submerged Surfoces5.10 Center of Grovity of o Three-

Dimensionol Body. Centroid ofo Volume

5.1I Composite Bodies5.12 Determinotion of Centroids of

Volumes by Integrotion

we have assumed so far that the attraction exefted bv the earth o' algld body could be represented by a single force W 'ihis force, calledthe lorce ol gravity or t]e weight of the 6ody. was to be applied at thecenter of grauity of the body (Sic. 3.2). Actually. the earth

"*"rt, u fo.""

on each of the particles forming the body. The action of the earth on arigrdbody should thus be represented bya large number of small forcesdistributed over tre entire body. you wiil lear"tiin this chapter, however,that all of these small forces can_be replaced by a single eq:uivalent forceW You will also leam how to determine the centeiof giavig,, i.e., thepoint of application of the resultant W, for bodies of .,"io.,, ,h"o",

^ In the first part of the chapter, two-dimensional bodiesl suchas flat plates and wires contaitted in a given plane, are considered.Two concepts closely associated with thsdetermination of the centerof gravity of a plate or a wire are introduced: the concept of thecentroid of an area or a line and the concept of thefrsr iwment ofan area or a line with respect to a given axis.

You will also learn that the computation of the area of a sur{aceof revolution or of the v^ol-ume of a body of revolution is directly relatedto the determination of the centroid of the line or area used io n"rr"r_ate that surface or body of revolution (Theorems of pappus-Gul&nus).And, as is shown in secs. 5.8 and 5.9, the determinatio" of tt

" centroid

9f g arga simpliffes the analysis of beams subjected to distributedloads and the computation of

'the forces exerted on submerged rect-

angular surfaces, such as hydraulic gates and portions of daJs._ In the last part of the chapter, you will learn how to determine

the center of gravity of a three-dimensional body as well as the cen-troid of a volume and the first moments of that volume with respectto the coordinate planes.

5.2 CENTER OF GRAVIW OF ATWO.DIMENSIONAL BODY

Let us first consider a flat horizontal plate (Fig. 5.1). we can dividethe plate into n small elements. The coordinat6s of the first element

Phoro 5.1 The precise boloncing of ihecomponents of o mobile requires on understondingof ceniers of grovity ond centroids, the moin topicsof this chopter.

ZMo: iW=Lx LW

2 M , : y W = Z y L W

Fig. 5.1 Center of grovity of o plote.

220

9 c''6:r 'oJr/v\ o ro i!r,ro.rO 1o .re1ua3

,14y fr.7:,yk :xytr7

,t4y x7= 1nx ,fryntr7

'err,t.r. eql uo peltcol lou lllensn sr erl\. B Jo 11 frn"r8 Jo reluecetll ]Bql olou e \'(A'g'3t,{) errc1d kx eW uI Su1l1 err.an € roJ pe uepeq u€c suonenba erues eql 'ep1d pg B Jo 1; fir,rer8 Jo reluec arp Jo4 pu" t saluurprooc aql pue A l{31" eq} augep suonsnba eseql

(z's)^o o I,: ̂O ^0, I: /a! nn l: n

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(r's)"A\V"fr + ... + 61y\Vzk + IzttV rn : A4,4 :yy7",MV"*+ ... + ulyyu*+ InVrx - lyyx,frW<

'qq8ge.,rn Flueuiele

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'slq8gerr,l

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IIB roJ ta.teznoq :rlpee eq+ Jo roluac arll Pre^\ol PaperIP e.re qqErezlrro secroJ eser{J ''aLv ' "' '6arv

"r!tv lq lle,rncedser 'pe}ouep

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lZ7, iJu-olru.*16-o^g o 1o ,!r,ror9 Jo roru€J 2.9 'zk prrc 6x dq ]ueurelo puoces eq] Jo asoq 'rn pue rr dq pelouep ere

222 Distributed Forces: Centroids ond CentersoI \rrovtty 5.3 CENTROIDS OF AREAS AND LINES

In the case of a flat homogeneous prate of uniform thickness, themagnitude Aw of the weight of an element of the plate can beexpressed as

AW : yt AA

where 7 : specific weight (weight per unit volume) of the materiarf : thickness of the olate

M : area of the elemint

Similarly, we can express the magnitude Iv of the weight of the entireplate as

1ry : yA

where A is the total area of the plate.If U.S. customary units are used, the specific weight 7 should

be expressed in lb/ft3, ihe thickness t in feet, and the are"as M andAin square feet. we observe that Aw and w will then be expressedin pounds. If SI units are used, 7 shourd be erpressed rn N/in3, t inmeters, and the areas M and A in square, -"i"rr; the weights AWand W will then be expressed in newttns.i

,. . _.Srb:atauting for AW and W in the moment equations (5.1) anddr\4drng throughout by yl , we obtain

?M4 iA : x1M, + x2M2 + . . . + x ,M,2 M , : i A : y r M r - t y 2 M 2 + . . . + i , M ,If we increase the number of elements into which the area A isdivided and simultaneously decrease the size of each

"l"-"rrt, -"

obtain in the limit

(5.3)

These equations define the coordinates r and y of the center ofgravity of a homogeneous plate. The point *hosi coordinates are .rard y is also known as the centroid c of the area A of the plate(Fig. 5.3). If,the plate is not homog"'"orr, these equatior.,

"J,,r,oabe used to determine the center. of gravity of the plater they stilrdefine, however, the centroid of the aiea.

In the case of a homogeneous wire of ,niform cross section, themagnitude AW of the weight of an element of wire can be expressed as

A W : y a A L

where 7 = specific weight of the materiala : cross-sectional area of the wire

AL : Iength of the element

llt should be noted that in the SI system ofunits a given material is generally charac-terized by its densiry p (mass per unit volume) rathei tran by its speific *"igr, f. ir,specific weight of the materlal can then be obtained from the relaiion

y = p g

yh:.",g =, ?:t l,

-i l'"::

p "'.

expressed in kg/m3. we observe that 7 will be expressed in(kgm")(nr/s'). that is, in N/m,.

, e : [ * a e i e : l a a e

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/c'c)

O's)

'ourl o Jo plor+ue3 t.g .Ell

lV [rZ=.Ik "INg.

7y xg=.1x ,67tg7

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orol: do* I: ^d

,enJ:t4,p*[:rz

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suonenba er1+ tuo{ pou1e]qo are T aufl a{+ Jo plo4uec eql Jo 4 puer selauTprooc aql '(L'g 'ftg) erynr eg+ 3o adeqs egt Surugap I aun atfi

to 3 ppo.tquac aq+ Wur soplculoc ua{+ arlr\ egl 3o fir,ter8 Jo ra}uoc ar{I

'Dsro uo Jo plortual C.g .6!l

VV fiZ=V fi ,"WK

VV xK=Vx tfrWK

eZZ saull puo soerv Jo stuor.uow tsrll f.g

2243i'!:TiJ

Forces: Cenrroids ond centers

Fig. 5.5

to express these mom_ents_as the products of the length L of the lineand the coordinates i and y of iis centroid.

An area A is said to be sErurnntric usith respect to an axis BB,if for every point f of the

"t""" th"." exists a point p' of the same

area such that the line PP' is pelpendicular to B'8, and,is divided intotwo equal parts by that axis Fig. s.Ea), A line L is said to be sran-metric with respect to an ans BB' if it satisffes similar corrditilr,r.When an area A or a line L possesses an axis of slrnmetry BB', itsfirst moment with respect to

-BB' is zero, and its centroid is located

on that a:<is. For example, in the case of the area A of Fig. 5.5b, whichis s)rmmetri" *i+ respect to the. g axis, we observe fiat fo.

".r"ryelement of area dA of abscissa r there exists an element dA' of equararea and with abscissa -x.rt follows that the integral in the ffrst ofEqs. (5.5) is zero and, thus, that p, : 0. It also foll-ows from the {irstof the relations (5.3) that t : 0. Thus, if an area A or a line L pos_sesses an aris of symmetry its centroid c is located on that axis,'

we further note that.if ?n area- or line possesses two axes of syrn-metry', its centroid c must be located at the intersection of the trvo axes(Fig; 5:6). This property enables us to determine immediately the cen_troid ofareas such as circles, ellipses, squares, rectangles, equilateral tri-Tgles, or other s)rrnmetric ftgurbs as well as the ceniroid oi lines in theshape of the circumference of a circle, the perimeter of a square, etc.

(b)

(b)\a)

Fig. 5.6

An area A is said to be sym.m.etric with respect to a center o iffor every element of area dA of coordinates x and, a there exists anelement dA' of equal area with coordinates -r and"-rr (Fis. 5.7). Itthen follows that the integrals in Eqs. (5,5) are both zero"and thatQ,:.Qr: 0. It also- follows from Eqs. (5.3) that i : i :0, that is,that the centroid of the area coincideJ with its center of srrmmetrv O.similarly, if_a line p_ossesses a center of symmetry o, the centroid ofthe line will coincide with the center O.

It should be noted that a figure possessing a center of s)rynme_try does not ndcessarily possess u" *ir of symmetry (Fig. 5.7[ whilea {igure possesfiing two axes of symmetry does not necesiarily possessa center of symmetry (Fig. 5.6a). However, if a figure possesies twoaxes o{ symmetry at a right angle to each other, the point of intersec-tion of these a6es is a center of syrnmetry (fig. F.Ob).

- Determining the centroids of unsymmetrical areas and linesand of areas and lines possessing only one axis of symmetry will bediscussed in Secs. 5.6 and 5.7. centroids of common shapes of areasand lines are shown in Fig. 5.BA and B.Fis. 5.7

'so3lD Jo sedoqs uoululoc Jo splo4ue) vg.g .6!l

6Jn0q

p tll.s J'6rolces Je[ncJrc

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?aJv0edeqg

9ZZ seull pub sberv lo stueurow lsrll y.9

226 3i.!lTiJ Forces: centroids ond cenfers

Quarter-circulararc

IIv C

rsinr,a

Fig. 5.8B Centroids of common shopes of lines.

5.5 COMPOSITE PTATES AND WIRESIn manyinstances, a flat plate can be divided into rectangles, triangles,or tjre other common shapes shown in Fig. 5.8A. The abicissa x o? it,center of gravity G can be determined from the abscjssas i.t,iz, . . . ,inof the centers of gravity of the various parts by

"*pr"rrirrf tt"t it"

moment of tle weight of the whole plate about the 4 axi"s is equalto the sum of the moments of the weiglqs of the variorls parts abtutthe same axi; (f1S. 5.9). The ordinatJl of the center o?gravity ofthe plate is found in a similar way by equating moments about'ther axis. We write

Z M i X ( W , + W z - l . . . + W , ) : i t W r + i 2 W 2 + . . . + i . W n2 M * : Y ( W r + W z * . . . + W . ) : y t W r + i r W , + . . . + i , W *

LMo: XZW=DrW

2 M , : f > W = L y W

Fig. 5.9 Cenier of grovity of o composite plote.

++

+

T

eloq ftlncrlc cv

e18ue1cer 1ng 6y

elcrnuueg Iy

v!v

l1'9 '6!t

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(8's)

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v4<: v<t

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vIK: VKX: 6d

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3o 'j tueuroru lsrs aqt Suuaprsuoc lq

xec eql Jo tr eleurpro "qJ

(Of 'g 'blg),{-reluauale er1+ Jo queurour }sJS eql Jo , ̂reluauela al[+ Jo queulour }sJs eql Jo

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ofl l r"-o*eq u"c BerB erll Jo Plo4uac eq]

C plo4uec el[+ $r^a saplculoc ,$rner8 yopue snoeueSouoq $ aield eq131

'Dero a+rsoduoc o Jo prorlueJ Ol.g .5ll

=vlL ='d

=vzx =od

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sAtvtplE pRoB[EM 5.1

':,:,:;::l:

For the plane area shown, determine (a) the first moments with respect rothe r and y axes, (b) the location of the centroid.

r '

:tt:lltitt:), ;..tt:t ::):.::l:t :,, l:l

r t = 6 0 m m

r z = 4 0 m m

c'r;;"*.;;;";";;;-';;;:;-;";;;r,*;;;-^il;;"angle, and a semicircle and by then subtracting'a circlel Usins th?

"oordi_nate axes shown, the area and the coordinates o?the centroid oTeach ofthecomponent areas are determined and entered in the table below The areaof the circle is indicated as negative, since it is to be subtracted fro* ih"other areas. we note thatthe coordinate y of the centroid of the trianqle rsnegative for the axes shown. The first moments of the component areaJwithrespect to the coordinate axes are computed and entered in the table.

a4 f t- '= t5 46 - -Jlf

+ v _r r = 6 0 m m

80 mm 105.46 mm

I _t__-20 mm 60 mm 60 mm

A, mmz i, mm , m m -A, mm3 [A, mm3RectangleTiiangleSemicircleCircle

(120X80) :9 .6 x 103;(12ox6o) :3.6 x lo3

ln(60)2 = 5.655 x 103- n ( 4 0 ) 2 : - 5 . 0 2 7 x 1 0 3

60406060

40-20

r05.4680

+576 x 103+144 x 103

+339.3 x 10s-301.6 x 103

+394 x 103-72 x 103

+596.4 x 103-402.2 x I03

) A = 1 3 . 8 2 8 x 1 0 3 ) i A : + 7 5 7 . 7 x r c 3 > V A : + 5 0 6 . 2 x 1 0 3

c, Firct fi4rmentg ef the Ares. Using Eqs. (5.8), we write

X = 54.8 mm

Y = 36.6mm

Q. : >iA: 506.2 x 103 mms \), - 6t)6 x 3{)'r 1tttt" ,,erii

Qo : 2iA : 787.7 X 103 mm3 e,, * TSg x l{}ii mrnl .,:d

b. location of centrcid. substituting the values given in the tab]e intothe equations defining the centroid of a composite irea, we obtain

X>a : >4, X(IS.AZS x 103 mm2) : 757.7 X lOs mm3X : 54.8 mn "{i

f(te.szs x 103 mm2) : 506.2 x to3 mmgY : 36,6 ntm "*.1

228

r2A:2 iA :

6Z(,

W'ulE :4

W 'uI0I : X

6ur 0BI : ('ut 09)I :1hg: 1gLzut 009 : ('ut 09)] tjlK: IKX

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08I :74K009 = 7{K09:7<oq

08I0

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q

c(.1

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A uniform semicircular rod of weight w and radius r is attached to a ninat A and rests against a frictionless surface at B. Determine the reacti6nsat A and B.

Free-Body Dicgrom. A free-body diagram of the rod is drawn. The forcesaciing on the rod are its weight w, which is applied at the center of gravityG (whose.position is

"bq1"a froy

{ig. b.8Bi; reaction at A, repres"entel

by its components A* and Ar; and a hJrizontal reactjon at B. r

ffi:;T -*(+):,

:;: "T:j, 'r u : # * q

N''* :Y*\ = w I

edding the two components of the reaction at A:

,t : tu(, *;!)"'

t a n a : w / n : n

The answers can also be expressed as follows:

A : 1.049\Iz L72.3'" : * , : :

230

Itz

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(se1oq 'eldurexe ro3) ..pe^ouer,, eJ€ qcrq,,rA s€ere leql requaural o1 no,,( roJ lelluessesr ]I 'sprorluec eql Jo selBurprooc e,trlcadser eql pu€ sqfuel ro s€orB eql Sululet-uoc elqel € lJnrlsuoc nod urelqord qcee roJ l€ql pueururocer ,{18uo.qs e A ,q

_ 'edeqs parrsep E urelqo o] ureq] pp€ s€ IIe \ se s€eru +curlqns uec nol 1eq1

1e8ro; +ou oCJ 'sq18uel ro s€or€ puu sprorluoc rlar{} qsrrq€}sa {1}carroc o1 no,{ dlaqru^\ (I'g 'qor4 aldrues ur auop sr su) queuoduroc quere33Tp er{l Sur.ttoqs 'os1y de,nauo utr{}_eJoru ur edeqs relncnrud € }crulsuoc o1 alqrssod uego sr }r s€eJe eueld ro3]uq1 azluBocer plnoqs no 'B'g '8tg

Jo sedeqs uoururoc eql tuor3 aurl ro eere ue,L13eq1 lcrulsuoc o+ ,,\\oq eplcop ol eq ppoqs uoqnlos .rnol ur dels 1s.rg e{J "n

'pezrseqdure aq pporis leq] slugod leroles te,te,ttoq 'ere areq; 'adft

slql Jo surelqord Surzrlos ueq.^a ̂\olloJ plnoqs nol arnpecord aql olurlsnl[ 6'9 pueI'g sualqor4 aldurug .ssurl puD ipaJo sllsoduros to spro4ua' aW Su1;B)ol "t

'l(g'g) 'sbg] sale1d elrsoduloc roJ Ber€ eql Jo sluauourlsJrJ aql eurruJalop ro sauq pu€ sBoJB alrsoduoc Jo sploJluac aql a+Ecol ol e^Er{IIIm nol 'sruelqord 3upro11o3 aql uI '[(t'g) 'sbg] seurT pu€ t(e'g)

'sbgl seere eueld

Jo sprorluec eql pue [(f'g) 'sbg] serr,r pue seroog leuorsueurp-omt Jo ,Qprer8y

Jo sroluec eql Supecol ro3 suoqenba preua8 eql padole,rep e,{\ uossel sql uI

5"I through 5,9 Locate the centroid of the plane area shown.

,o--*] F

t[**-]

'1,r,,*-2rin --*r

lt-----,1Fig. P5.3

30mm

L

Fig. P5.l

II

300 mm

III

i--

U

T-3 in.tAII

6 in.IIt

| , " / r , - - /'ra-.--(l-Fig. P5.6

L--.120 in.

Fig. P5.7

232

x

Fig. P5.9

.Vltrt=fteqtost-t/6-t o\lBt eql euprrelap ,II.g .qotd

Jo 'eer€ relnuuerues eq] rod 6l.g

8l'gd '6lJ '4 : I qcq.{\ roJ q/o o\er arlt eulurretep <u,^ oqs BerE eq+ rod gl.g

I '6/(t

+ I;) snrpur Jo alcrrc Jo cr€ uB roJ ]eql saqceorddeI plorluec aql Jo uortecol eqf 'zr saqceordde rr se +Bqt ̂ orls ll"gII

ll.gd puD 91.9d .6!t

.i? pu€^ ,6-r. ,I,r Jo srurel

ur eerE pepeqs eql Jo plorluec aql Jo eleurprooc /, eql eunrralo(l 9L.g

9t'9d '6!ttt'gd'6!tet'9d '6!J

r-'**--l

11'9d'5!lo1'9d '6!J

t,----f--

ultu 9uesdqelne5

wt,

I'r402

_1

zt'gd '6!j

08 --*l

t,

et[ suiatqord'u.4aoqs ee.re eueld eqt Jo pJorluec eqt etecoT g1.g q$norql Ol.S

Distributed Forces: Cenfroids ond CentersOI \rfovlry

234

60 mm

(b)\a )

/D rn .

f-1.50 itr.I

0,1.50 in

Ht ll l

L--t2.00 in.

I

,5.20 A composite beam is constructed by bolting four plates to four60 X 60 X l2-mm angles as shown. The bolts are equally spacecalong the beam, and the beam gupports a vertical load. Ai provecin mechanics of materials, the shearing forces exerted on the boltsat A and B are proportional to the fir;t moments with respect tothe centroidal r axis of the red shaded areas shown. ,espectiuely,in parts a and b of the figure. Knowing that the forc6 exerte.don the bolt at A is 280 N, determine the force exerted on the boltat B.

450 mm

III

_ t-T12 mm

Fig. P5.2O

5.2t snd 5.22 The horizontal r axis is drawn throush the centroidC of the area shown, and it dlvides the area into iwo componentareas ,41 and A2. Determine the first moment of each comtponentarea with respect to the r axis, and explain the results obtiined.

0.75 in.

Fig. P5.22

T-

Fig. P5.21

^ .leluozrrorl sr raquaw er{t Jo uoruod srql }eq1 os go

IrIr€ Jo I r{}tsue1 eq} aqurro}ep '!u 0S.0 sI p t€q} puu D tE pepoddns

sr requtetu erp l€rp bqmoul 'bqqn1 ynu$mp J0 ocotd eliup t

Irrog peurJoJ sJ pup alrqou e 7o lueuoduoc e sr J{CIJA7 re<Iuory

'l€luozrroq sr reqweu oql Jo (ICg uorlrodleql os p ecuetsrp eql eururelep ,u

6 : I l€ql pu? p +e peyoddnssr roqruoru oql l€ql Sur,rouy '3u1qn1 runururnle yo ecard el8urs euro{ pauroJ sr pue olrqoul e3o;uauoduoc E sr IOJ7V rrqurol{ 6Z.5

'J le uollcBer rq] (g) 'dlqer arllur uorsue] eql (r) eururela6l .gV elqer erl] o] pu€ 3 te u1d e o]peqcelle q'q 0I snlpBr pu€ ql 8 fqbla \Jo por rBlncrrc ruroJrun V gr.g

'|sa'lr'7 1''' .r.gir .!r.{ qr.g'6.9d'tsr,{ ga.g.I.gd .tsrd

- ?(.9 'peturoJ snql ernilJ err. A erll5o ,,(lraer8 Jo roluoc eql el€coT 'peleclpur ern8g eql 3o .relaurr.radaqt ruroJ ol luoq sr err,^A snoeue;ouroq ,uql

V IU'* q6no,tql ?A.g

o0'gOe.gd puD 67.gd.6H

ez'gd'6tJq

ienle^ tuntulx€ru leql sI tBr{A\ pue 'urnurrxeur "d sl f yo en1e,l

lpr{.r rod (g) 'sue r eq} ot €ere pep€qs eql Jo aseq aql urorS fecu€lslp aql pu? '!'qp sturet ut 'fl sserdxg (o) ''d Aqpelouap srsrxe r eql ol lceoser qlvl\ eer€ pep€qs eql Jo lueulotu lsrrJ aql tc.g

8Z'9d '6:l

ge[ surelqord

236 Dishibuted Forces: Centroids ond Centersof Grovity

5.3r The homogeneous wire ABC is bent into a semicircular arc and astraight section as shown and is attached to a hinge at A. Deter-mine the value of d for which the wire is in equilibrium for theindicated position.

Determine the distance h for which the centroid of the shadedarea is as far above Tine BB' as possible when (a) k : 0.10,(b) k = 0.80.

5.32

h

L!:1-] B '

Fig. P5.32 ond P5.33

5.33 Knowing that the distance h has been selected to maximize thedistance y ftomline BB' to the centroid of the shaded area, showthat y : 21113.

5.6 DETfRiAINATION Of CTNTROIDSBY INTTGRATION

The centroid of an area bounded by analwical curves (i,e.. curvesdefined by algebraic equations) is usually determined by evaluatingthe integrals in Eqs. (5.3) of Sec. 5.3:

r e : [ . a e (5.3), t d : l , a a e" J "

If the element of area dA is a small rectangle of sides dr and dq,the evaluation of each of these integrals .eqrires a double i,ntegr"a-tion with respect to r and y. A double integration is also necesioryif polar coordinates are used for which dA is a small element ofsides dr and r d0.

In most cases, however, it is possible to determine the coordi-nates of the centroid of an area by performing a single integration.This is achieved by choosing dA to be a thin rectangle or strip or athin sector or pie-shaped element (Fig. 5.12); the centroid of thethin rectangle is located at ils center, and the centroid of the thinsector is located at a distance fr from its vertex (as it is for a triangle).The coordinates of the centroid of the area under consideration arethen obtained by expressing that the first moment of the entire areawith respect to each of the coordinate axes is equal to the sum (orintegral) of the corresponding moments of the elements of area.

Fig. P5.3I

(r)

'slusurels lDrluererjlp Jo sDsrD puo spro4usJ zt.g .6!l

tq /D\

7oPzr i=vPA.o$ L =pk

r6

4soJg-lar

fr.p(x-o)=yp

n=Pn

rpfr.:yp

zth=P4

._12-

o-TIh

@'s)

(o's)

elu 4, e1\4.'yp]uetrrele eq+ Jo proJluec erl+ Jo seleurprooc e\t 1" 4 pue lux Aqfiunoue C

-tpol:, ,p*l:,:g'g 'ces p ft'g)'sbg ur spr8alur eq18unen1e.rc lq pautrue+ap aq

rrec prorluec s1r 'uoeenba cprqaSle ue lq peugep $ euq ? ueq A.BarB aq+ Jo

plorluec eqlJo 4p* t sepurprooc aql roJ pe los eq u€c s,rott"n6eeseql 'pel€nle e ueeq e,req (6'9) .sbg ur qer8elur eqt pue paulu-relap ue_eq seq eere eql acuo .uorler8alur e18urs e ol pecnper snq]sr uoperBalul eql xeqlo aq+ Jo sture+ ur se]eurpJooc aql Jo auosserdxa ol pesn eq ppor{s e,trnc Surpunoq eq};o-uonenbe aq} pue '(6'9) sepuroJ olut pelnlqsqns aq ppoqs suorsserdxa epgdorddeeql 'seleurp-rooc relod ur ue.l,r8 sr BarB eq] Surpunoq e^rnc eq]3o uonenba eql uerl \ posn eq pForls c yed Jo tuetuele pedeqs-erda{} jsluaruala

3o sad,(t uoruruoc aarq} roJ 6I'9 '8Id ur euop ueeq

seq srql 's[EpuareJJrp alegdordde eq] pue lurod leql yo seleuro-rooceql Jo suue] ur pesse-rdxe eq ppor{s yp }uauele eq} Jo eere e{+ 'oslv_'uoll"Japlsuoc repun €aJB aql Surpunoq e^rnJ erp uo pel€collurod e Jo seleulprooc aq+ Jo srurol ur passerdxa aq pporls Vp EarBJo +uatuala arll Jo plorluoc orll Jo I" fi pue l'r seleurprooc aql

'sluatuele aserFruor; palndruoc eq os[B rrec ]r '<u,&\orDl fpee;p ]ou $ V eaJ€ arp JI

v?FiJ:v&:"d

veP! I: v!: nd

(h'x)a

rcZ uouo.rOetul ,(q sp;olue] Jo uorrouluJerec 9'g

238 Distributed Forces: Centroids ond Centersof Grovity

Photo 5.2 The storoge tonks shown ore ollbodies of revolul ion. Thus, their surfoce oreosond volumes con be determined using thetheorems of Poppus-Guldinus.

The differential length dL should be replaced by one of the followingexpressions, depending upon which coordinate, x: U, 2r ̂ 0, is chosen

as the inde'oendent variable in the equation used to define the line(these e"prissions can be derived using the Pythagorean theorem):

d L : [.e,J',,FE,J,-d L :

d L :

After the equation of the line has been used to express one- of thecoordinates in terms of the other, the integration can be performed,and Eqs. (5.4) can be solved for the coordinates i and y of the cen-troid of the line.

5.7 THEOREMS OF PAPPUS.GULDINUSThese theorems, which were first formulated by the Greek geometerPappus during the third century e.o. and later restated by-the Swissmathematiciun G,tldittrrs, or Guldin, (7577-1643) deal with surfacesand bodies of revolution.

A surface of reaolution is a surface which can be generated byrotating a plane curve about a fixed axis. For example (Fig. 5.f3), the

,2" | _ _ _ _

Sphere

Fis. 5.13

C '.}A C '.-jA CCone Torus

surface of a sphere can be obtained by rotating a semicircular arc ABCabout the diameter AC, the surface of a cone can be produced byrotating a straight line AB about an axis AC, and the surface of a torusor ring can be generated by rotating the circumference of a circleabout a nonintersecting axis. Abody of reoolution is a body which canbe generated by rotating a plane area about a ffxed axis. As shown inFig. 5.f4, a sphere, a cone, and a torus can each be generated byrotating the appropriate shape about the indicated axis.

A f f idq\ -4 1

i_ ffifl"\ _ , Mi 1,r \ ) , - )

Sphere Cone Tonrs

Fig. 5.14

T'HsORHM t. The area of a surface of reaolution i's equal to thelength of the generating curae times the distance traueled by thecentroid of the curae while the surface is bei'ng generated.

Proof. Consider an element dt of the line L (Fig. 5.15), whichis revolved about the r axis. The area dA generated by the element

(#)'*

'(g'q 'qora eldures ees) urvrorq sI perp arp lq palr:a'cg ,l''o.l -.r, r-etunlol orl+ uory\{ eare aueld € Jo plor}uoc erp outtura}ep ol Jo rr \orDIsI o^Jnc e{+ lq pelenaa? eceynJ aq+ Jo Ber€ aT+ o"q ,-." ,.r"1deJo pro4uec aw euluua+ap ol pasn eq osF rlr.,c ,{eql .flesranuo3 .uou-nlo er Jo se.poqJo_ saumlo orB pus uoqnlo oJ Jo socBJJns J0 SBOJB erDelnduroc ol lezv' elduns € re#o im4ppg-snddedJo^sureroerl+ aql

'eare buperauaE aql slcesJeluT uoll-elor J: strB aq] I lldde lou seop {rreroerf+ erp leql pe+ou eq ppoqslI 'uIBBv 'v

Jo plorlusc eql .,(q pelezterl ecuelsrp ",{r

q ku6'ir"rq

(rls)vfruT: A, el"q e \ ,(g.g .ces)

VA ot pnba sr yp n I pr8elur oql aculs prra 'vp ku,6 I : ,ysI V lq pepreua8 aumlo^ orrlue erl+ 'srr-q.; .Vp fru6'o1 1".r1" rtlueurela eql lq pelereue8 Ap eurnlo eql .(gl'g '3rg) sure r eql ln

'palotaua? Bu1,aq v hpoq aq+ a1lqmaql to p?oquac aq+ hq palaaDr+ aiuolw aq+ s"utxt oato 3up1o-

pe^lo er sl qcFI.,lA v Eere eql Jo yp luaurele ue replsuoD looJd

dldde 1ou p1norAuarooql aql pu€ 'su8rs ellsoddo Surz'uq seere alereua8 ppozn spre elrllJo aPIS rerl}Ie uo suoncas 01\ + eql 'prp ll

JI :pelElor sr I qJrq \ ]noqBsDre arp ssoJc 1ou lsnru a.trnc Suuereue8 erp leq] palou eq ppoqs ]I '(gf 'S 'Bf.{)

TJo prorluec aql lq pelo €r} ecu€}srp arn, sr hu6 ereq^

aq+ o+ Tonba n uo?tnloaar {o fipoq o to auryoa aq1 .ll

(0r'q)74sz : v

el"q eroJareql eM'I& o1 pnba q :fp n IprFelul eql teql g'g 'ces uI punoJ e \ lBrF SuqJeceg 'W fruZ t : VsI Z lq peleraue8 BerB erqua eql 'snqJ '.Ip nL6 o1 pnba sr

9t'9'6H

s1'9'6H

6eZ snulp;ngrsnddo; Jo suaroaql l'9

Determine by direct integration the location of the centroid of a parabolicspandrel.

k$ermilttion d the Consfoat k fhe value of /c is determined by sub-stituting^r : a artdy : b lnto tlle grven equation. We have b ='kaz ork : b/a'. The equatiorr of the curve is tlus

, - b - , o . u zA : 1*' or *: fuA''"Varticcl Oiffurerdiol Element We choose the differential element shor.nand.find the total area of the ftgure.

d:[ ae:l , o.:[5r - :13:1",: +The ffrst moment of the differential element with respect to the y axis isi"1 ilAi hence, the ffrst moment of the entire area with respect to this axis is

240

rv(,

\r I )161116 : 7xrt6 : v/)

\r, - L I'b 16: r/ '6

o^eq o,\\ ,Sg.S .Bl.{ o1 Euu"reSag .plorl-ua: sll ,{q pa1a,te4 acuetsrp eqt pup cre arll 3o qfual

"q1 3o 1".pora ".p

o1pnbe sr pale-reue8 Bare eqt ,snurpp3-snddea

3b 1 -"rb"q1 o1 B.rroro""y

tu 11 * niltt7 : V

( z\l('\r"/L\l

-

('-\I

.--_"..1

'srxe Iecruo e lnoq€ cr€ relncrrc_rapenb e Buqelor .(q

pJur€lqo sr rlJrr{1!\ rw\\orls uolln[o^Jt Jo JJe#ns Jw Jo eerB Jrl] JulrurJ]ec

o uys ,-t6 : (nr.6)x.

'rr \or{s olcrrc Jo cre ar{l Jo plol}uec orl} Jo uoqecol eq} eulurJelec

?"8 W3?808d 3?dtfifvs

elrra a,tr '7r : hfl ecurg

n uls z16 : ";lg urs]u-r :o-r o-r

r "- 0p 0soc | "t

-_ tep"i.)(psoc'l) | : lp, | : o0oJ oJ J

sr srn 6 eql ol lcedser {}-r.ll. c.re orp Jo lueurour lsrL} aql

-'-r '-[ | nr6:0p l":ep" l:Wl:t oJ pJ J

'uope,r8elur.{q peuru;alep sr crt aqt 3o qrBual eq} pu€ 'u,\\oqs sB uasoqc sr }uaurolaIelluereJJrp V

'0 : 6 'srr"e r eql o1 lcadser qll.^ IBJulru[u,{s sr cre arll JruIS

sistt{?1{}s

;t;,1';:;:;;.;,;,.)::;;1L;;;111;11);1,;:1a:1;:':.;:,.::,::::::t:jili

x

op"=-r&

$"9 $t:nTgsud 3?dwvtl'l.ll:ll!:,l':,::.:::;:'::l;:l::::...'::::!:|.)|::.::.::|;.|:.:|;:i|.:;

n- =0

The outside diameter of a pulley is 0.8 m, and the cross section of its rim is as

shown. Knowins that the pulley is made of steel and that the density of steel

is p : 7.35 x io3 kgm3, ieteimine the mass and the weight of the rim'

l-*l* oo ---f-]20 rnm 20 mm

400 m

II

__LSOLTJTION

60 mm

r -*l too'"*

l-roT"'ir.;---'=l-T-- |' l

I:' 375 mm

The volume of the rim can be found by applFng Theorem II of Pappus-

Guldinus, which states that the volume equals the product of the given

cross-sectional area and the distance traveled by its centroid in one complete

revolution. However, the volume can be more easily determined if we

observe that the cross section can be formed from rectangle I, whose area

is positive, and rectangle II, whose area is negative'

365 mm

I

Ittl

Area, mme

sA't'TPLE

Distance Traveledby C, mm Volume, mm3

(5000)(2356) = 11.78 x 106(-1800x229s) : -4.13 x 106

Volume of rirn : 7.65 x 106

Since 1 mm : 10-3 m, we have I mm3 : (10-3 m)3 : 10-e m3, and rye ob-

,5 ' 'v- : i .os x 106mm3 : (7.65 x 106)(10-emi) :7 .65 x l0-3m3.

m : pv: (7.85 x 103 kg/m3)(7,65 x 10-3 m3) m : 60,0 kg 4

w : mg: (60.0 kgXo.SirrVs'9) : 589 kg ' m/s2 trtr/ : 5iJ9 N 'q

, : - i '

{.::i.' PROBLEf{I 5.8

Using the theorems of Pappus-Guldinus, determine (a) the centroid of a

semi;ircular area, (b) the centroid of a semicircular arc. We recall that the

volume and the surface area of a sphere ate tnra and 4nf , respectively'

- , . : . , . j . ) . . . . . . : : . : : . : t : ; : 1 1 , : .

. . . :

soruiloNThe volume of a sphere is equal to the -product of the area of a semicircle

and the distance traveled by the centroid of the semicircle in one revolution

about the r axis.

V : 2 r i A 3 r r r 3 : 2 r r y r l n r 2 ) i : * #" ,7Tl

Likewise, the area of a sphere is equal to the product of the length of the gen-

eratinq semicircle and the distance traveled by its centroid in one revolution.

A I.4.4

2n(375) : 23562n(365) : 2293

L = r r

242

A : 2nAL 4nra : 2rg(trr)

tvz'EJJP SurlEtJuJB dql Jo d,\ln,)

Suqereue8 eql So.q+Bual eql al€lnrlec o] e^eq lou op nor( :yf rc 7li'".rr.ur"1"pA1uo peou noA 'acteqs uoururoc auo ueql erou Jo s+srsrroc EoJE Jo auu Buqereuagaqt qclq.tt roJ suralqord.esoqt roJ .snqJ die,rqcadsar ,1yl\ vete uE puu (?4) au{E Jo s+ua[Iolll lsrlJ a_q+ .{ldruts ary qrlq,{\ 'satlrlurnb esoq} Jo slcnpord eq} riieluojl(tt'g) Pry (oi'g) 'sbg] suopenbe Suqpsar aq] 'Bar€ Surjr:"raua8 eqt ot ro e^rncSuqereua8 eql Jo llBue1 eqt o] pu€ prortuac eip Iq pele^Ert acu€lsrp aqt ol raJarsureroaq] e{r.q8noql1y sarrnloA put seer€yo ,tonelr-,doior aql ol tplorl.t"ryo

"iip" -i^\ou>l rno{,(dde o1 no(.{\olle srreroaq} InJasn ,(ra.t 1a,{ ,eldurrs as-eqt .g.g q;"oiqrg'g'sqord aldureS ur u^\oqs sv'$nu,ptfiff-rndd*6 *s swar*aqt aqisurr{y}dv .g

'sp"r8aqur llncrilrp alunlt^a ol porllaurlsalsuJ aql s1 sp"r8alur Jo alq€l e Sursn ,es-rnoc

a6 .sped ,{q uoqe-rEa1u, ro uoqnlqs

-qns crJleurouo8ul se r1,rns 'sanbruqJal paJu€^pB erour esn ol ,(ressecau aq (eur 1rsetur] ]B 'p.re,trropqSlerls ore uossel sFil q suoqer8elul eql Jo lsour q8noqlly .p

ilrletuLll,\s J€lncJrc sEq EOJE uB ro euq B ueq \salEurprooJ relod asn o1 snoeSelue^pp eq,{1ensn IIi.u }t'os1y 'seuo pcrue^Jo p€e}sursluatrlale repbuelcer lEluozuori esn o+ elq€JeJerd aq,{uur qg serlrqallros 'aldurexa rog'asn

IFA no^ ]Erl] ]ueurrle IErluJr+JIp aql Sulugep eroJaq aurl ro earu ua,tr8 eql 3oadeqs aql eurrrexe s.{e,n1u 'suoqelrlduoc jno,{ eziurrurur -ro,{Srtdurrs.{1q1ssod o1 "r

"

'1utod luepodurr srql puulsrepun.{1ry no,'( 1pun 6T.g

,Brg .{pn1s i11ryamc p1norlsnotr uorleJaprsuoc repun EaJB eql burcunoq aarnc eql uo pelEcol lurod e Jo salEu-lpJooc eq] ol IEnDe ]orr oJts Vp Jo proJluoc aql Jo salBurpJooc aql }rq} azluboce;r o1lu€uodrlrr q ]r '7p plE vp slrreurale IErluerelJlp eLIl Jo plor+Ltac aq+ {o sa+DLtlprooceql luase;dar suoqenba e^oqc eql ur f aqt pue rt aq+ 'g.g .res ur peureldxe sy :q'ploJluJJ slr ol sJ.)uplsrp Jql pul, ?f Jo yp JoJ dJrorlJ.rno,{;rrr1 ro l2rrc rro.nr8 aq+

Jo qcle{s -rno.{ uo ,r\orls o+ no,{ a8e.rnocue r(13uor1s eA\ .sulnurroJ pl8alur elqecudcleaql ul LLIrel LIcEe Eurururralap ro Surur3ep ,{gn3arec ,{q uoqnlos rno,{ ur8ag ,n

'slurod 3ur,to11o3 aql ol uoquol]€ -re1ncr1.red .{ed plnoqs no,i .uorlrppe u1'plorluec arllJo seluulprooc eq] roJ (l'g) ro (Cg) 'sbg r^los prru .ourl eql ro EarBaq] lo sluauroLu lsrrJ aql eururrelap '-l ro V elnduroc :g'g puu L'g

.sqord alduregur rlr\oqs uoilnlos Jo poqleur rqt ^\olloJ plnoqs no,{ 'edt slq} Jo surelqo.id 8uu.1osuaq6 's*url pup swar* ** sfw.t&u* ar..i{ {rsl{sJffie1l'fi a*ax.p &Effian*zttsta,;*ff "t

Jo,secp,+rns ro s.er€ "., ".'o.,"!$:ffjJ:fii1"'l'i;;3T':H",".ul"offij:""tllij:; .{1dde osp il1.{\ no1 dle,trlcadse-r 'selrrl pil€ sr:e,ru aurld Jo sprorluor aql aleJol o}

,pol:,

suorlenba erl] asn prn r-ro,{ 'uossel s}q} roJ surelclord aqr rrg

i.._.....:.._...:.:;,_-:_::.., .,.. ".* K, g,{ ,"&

,,*U * * ; ; * #

..... .. ..:

4 :4 {,€^{"a bT{ s{ a g 3 a{ *

,p, l: u

,1r, [

: ,l ,pn[:v!

\v'9)

(c'c)

w|{.{K\ffi{*?{ # {rt q6 tr\{effi

5.34 threugh 5,36the area shown.

Determine by direct integration the centroid ofExpress your answer in terms of a and h.

.r/

,/

l l=kxz

Fig. P5.35 Fig. P5.36

5.37 fhrough 5"Sg Determine by direct integration the centroid ofthe area shown.

IIh.*L

F_A_______tl

Fig. P5.34

t " , l

Fig. P5.37

5,4O crnd 5.4Iarea shown

Fig. P5.39

Determine by direct integration the centroid of theExpress your answer in terms of a andb.

' l

li " , - l . t * ^ \ t

I

q = m x

;:,:

tZa + r

a2 b2

Fig. P5.41

6r'9d '6!t

zr'gd'6tJ

zr'gd '6!l

f- ?--*l- 7--

grsa'6u

q9,7,'Ir-r-]

'rz-SOJD=hxlI

'ulAoqs eoJBeIIl Jo plorluoc eql uorlerSolur ]cerrp lq euruueleq 6?'g* pus g?.g*

', JO StrrJAJ Ur Je,r\Sue

rno.,( sserdxg 'plorluec slr Jo o]€urprooc r eql uoqer8elur lcelplq aururreleq 'u. Aoqs adeqs eql olul ]ueq sr orr.&\ snoeueSouroq y /t.g*

9r'9d '6:t9r'9d'6!J

'plorluoc str Jo el€urprooc r eql uorle,r8elur narrp lq oururreleq'u,^Aoqs adeqs eql olur ]ueq sr arr.4\ snoaueSoruoq y gf.g puo g?.g

I_1 ui

r

llt,

er'gd'6H

6,6, -+--Dl0l

_""- A 6-"t - '"

'qpve o Jo sruJal uI ro,^ sue .rno.{ sse.rdxg 'u^\oqs eereaqt Jo plorluec eqt uorl€r8alur lcarrp lq euruualaq vtr'g ?rtrl g?.g

'u.t\or{s €arp eql Jo plorluec aql uoper8e}ul lcerrp ,,{q eurur;eleq gj'g

Qsrllso=ft9gso) o = x

,.-, I t"-" 1

$VC suetqord

246 Distributed Forces: Centroids ond Cenlersof Grovify

Fig. P5.50 ond P5.5I

Fig. P5.58

5.50 Determine the centroid of the area shown when a : 2 in.

5.51 Determine the value of a for which the ratio ili is 9.

5.52 Determine the volume and the surface area of the solid obtainedby rotating the area of Prob. 5.1 about (a) the line x : 240 mm,(b) the y axis.

5.53 Determine the volume and the surface area of the solid obtainedby rotating the area of Prob. 5.2 about (a) the line y : 60 mm,(b) the y axis.

5.54 Determine the volume and the surface area of the solid obtained byrotating the area of Prob. 5.8 about (a) the x axis, (b) the y axis.

5.55 Determine the volume of the solid generated by rotating the para-bolic area shown about (a) the r axis, (b) the axis AA'.

5.56 Determine the volume and the surface area of the chain linkshown, which is made from a 6-mm-diameter bar, if R : 10 mrna n d L : 3 0 m m .

Fig. P5.56

5.57 Verify that the expressions for the volumes of the first four shapesin Fig. 5.21 on page 260 are correct.

5.58 A f-in.-diameter hole is drilled in a piece of 1-in.-thick steel; thehole is then countersunk as shown. Determine the volume of steelremoved during the countersinking process.

5,59 Determine the capacity, in liters, of the punch bowl shown ifR : 250 mm.

l-.

Fig. P5.59

g9'9d'5H'u^\or{s uoqcos ssorc crloqered eql seq I leq} 8uI^\oDI .epuqs

oqlJo eplslno oqlJo eore aceJrns eqt aulturaleq 'crlse1d Juacnlsuer]Jo laoqs uql € Luo$ paurroJ sl tqd{ p3}unour-llB \ B roJ op€qs arlJ gg.f *

'els?,r\ seluocaq lBql lo^\op eql

Jo erunlo^ Frtrq aq] 3o a8rluacred oql aurruroloq .3uo1 .ur 7 puB

relelllerp q 'ul I Ie. Aop e urorJ peurnl sr u,^^oqs 8ed uepoo,nr, arlJ ?g.s

lured so sl"oc o.^A1 ""'s'yfr?l^'?j3'.ilJ:tJi,TH:fr" in fT:?Q plnoqs lured 3o suolle8 ,{ueu .,ra,oq aulruro}ecl 'u,r\oqs edeqs eq]Suraeq s8ed uepoo,n 000'06 acnpord o1 Suruueld sr rarnlc€Jnueur V Cg.S

'uoaqclncsa

oq] Jo ssetu er{l aurrurelep 'eu/8{ g/fg sr sserq 3o lgrsueparll leql iur.4\ou) 'sserq uro$ lsEc sr u,ro,oqs (1p.rn € rrrou slrxaedrd eql ereq.r,r adrd e uo peceld e1e1d e.trlerocep e) uoeqclncsa eqJ A9.g

19'9d '6!t

urru 8

t9:9d puD e9.9d .6!t

-@--l wl w_lwrl --ff uuqz I

ffitlH-l(- # tutu

K3i'epeqs eql Jo ss€tu oq] eururelep '"u/3{

00gZ sr urnu-rurnle Jo llrsuep oq] leq] Sup,rou;1 urur I Jo sienlcgql urroJrun es€q u,&oqs durel llrsuelutq8q purs eqt roJ apeqs trrnururnp er{I tg.g

\q)

ffil

&G-r ffiJ 1

*r@: "i"

,,riM I -f7\-80'0 / \ /\\ ^r-t1 owv

'uI c.,

'uI

(c)

&€IH 'uie

AI 'w--r'uI960=r

o9'9d'6tt

'u8rsep qcee ro3 le11ndaI{} puE iloq oqf uoo,4Aleq oaro lcoluoc oq} ouuralap ,4,e11nd slr 3oecuereJuncrrc eql Jo Jlsr{-euo qlla }celuoc se{eur lleq rlcee alurlue,u8 lue l€JI 'palpn$ aq o] ere selSord tloq a^rrp tuareJJrp oarqJ

'ut 0g'0

ut91.81 0="r

IVZ suretqord09'9

248 Distributed Forces: Cenhoids ond Centersof Grovity

Fig. 5.17

*5.8 DISTRIBUTED IOADS ON BEAMS

The concept of the centroid of an area can be used to solve other

problems besides those dealing *ith the weights oJ fl?t plaGs. con-'sider,

for example, a beam t.rppottittg a distributed load; this load may

consist of the *"igftt of materials sJpported directly or indirectly by

the beam, or it rn"ay be caused by wrnd or hydrostatic pressure. The

distributed load can be represented by plotting the load u, supportedper unit leneth (Fie. 5.17); this load is-expressed in N/m or in lb/ft''the

masnit;de of tf,e force exerted on an element of beam of length

d.x is dfr : u dx, and the total load supported by the beam is

B x

(b)

w : l '(a )

w : l d A : A

. I

_ ;- J o

We observe that the product u; dx is equal i1 magnitude to the ele-

ment of area dA sho*n in Fig. 5.I7a.the load W is thus equal in

magnitude to the total area A under the load curye:

We now determine where a single concentrated loadW, of trre

same magnitude W as the total distributed load, should be applied

on the bJam if it is to produce the same reactions at the supports(Flg. 5.I7b). However, tlis concentrated load W, which represents,the

resrll1xnt of the given distributed loa&ng, is equivalent to the loading

only when considering the free-body diagram of the entire beam. Thepoint of upplication P of the equivalent concentrated load W is obtained

fy "*pt"rtitg

that the moment of W about P91"t O is^equal to the

rnrn of the ioments of the elemental loads dW about O:

t

( o P ) w : J x d w

or, since dW : w dx : dA andW : A,

u; dx

(oP)A x d A (5.r2)

Photo 5.3 The roofs of the bui ldings shownmust be oble to support not only the totol weight

of the snow but olso lhe nonsymmefric disir ibutedloods result ing from dri f t ing of lhe snow.

Since the integral represents the first moment with respect to- the u;

axis of the area ,rnder the load curve, it can be replaced by theoroduct rA. We therefore have OP : r, where r is the distanceirom the r,rl axis to the centroid C of the area A (thls is nof the cen-troid of the beam).

A distributed load on a beam can thus be replaced by a con-centrated load; the magni,tude of thl's si'ngle load is equal to the area

und.er the load curne, and lts li,ne of action passes through the cen-

troid of that area.It should be noted, however, that the concentratedload is equivalent to the given loading only as far as external forces

"r" .on""ired. It can be used to determine reactions but should not

be used to compute internal forces and deflections.

6l'9'6H

'u luel*sar aqr .s reruec su re ernsse;d i?;"^;:r'{^i#'#,:3"t{iffiil:fi"Jdq peuielqo eq u€r U Jo epqluSeiu eq+ 'snql 'a+o1d arp jo Eoft eql selouap V erarla

Vzd = 17q1tfl : 1sd4) = 77o, : td

ollr.^A irec a^\ '(tI g) 'bg Surqeca: pue 'e1e1d eq] Jo A raluoc orll le ql8ua1 lrun.red peol erp s1 aor ercqr 'Is(n o1 pnbe sI e^rnc p€ol o{r repun EorE eqt teqi 8ur-lo5l

'(eA) locs0d e peflEc s] zrulN tlun IS pa,ruep eqJ, ."rJlcli

ur ro zrulN u1 pesse;dxe sI '€ere llun .red puol r s1ulsarder qc1q,r,r, 'd a.rnsserd 5q11

'O 'deqC ur peurrrretep eq ilI^a qlpp,r elqeFs^ Jo seJ€Fns pe8raru-qns uo secroJ Jo sluellnseJ aqJ 'seu€A pue sale8 rqn8uepe; puesurBpJo secelns eql uo Peuexe secJoJ cnelsorp^q eqlJo luEllnsor erFeuruJelep 01 pesn oq uBc uoqces sql ur peullno spoqleru eqJ

'U- Jo esues eql burs.re.rer Lq peuretqo uer{+ $ ec€3rns pe^Jnc

eq] uo peuexe secroJ crJelsorp&{ erp Jo U }uellnser eql '?6I'9 '8ld

3o ,(poq aa4 eql ro3 runuqmnbe 3o suonenbe aql Surqos ,{q paurelqosr U- ecroJ eq+ 'punoJ uaeq e Bq senlBn rreq] relp :spoqleru preprrelslq paururralep eq uBc zU pue 'IU 'Al secroJ eqJ 'am{-ms paurn aq+uo p?nbn aq+ nq peuoxe secroJ ern Jo U lu€]Fser aq+ 'se uooc€ Jo eurrerrres eq+ s€q pue 'o1 elrsoddo pue pnbe sr U- luel1nsar er{I 'p?nb7 aUluo aco{-tns paemc aq+ fr,q pouoxo sacroJ oq} Jo U- }u€}Fsar aq+ prre'Oguo pepexa sacJoJ e{+Jo 6g }ue}pser aW'OV uo pouoxa sacroJ eq}

3o tg luelpser eqi 'p1nbllJo erunlol peqc€tep eW Jo A\ 1q31ezr,r oW ereOgV trpog, ae{ er1 uo Supce sarroJ eql 'q6l'g '8ld ul u aorls gO pulCV seceJuns eueld oz'r1 eq] {q pve €IV ec€Fns pa rnc eq} Xq papunoqCgyprnbrlJo eulnlol er{+ SuFIcB}ep lq peuprqo.{poq eeg eqt reprsuocen 'Asee aq lou ppo.4 , uorl€Jbalur perro-{q secroJ asarfl Jo U }tre{nsereql Jo uon€ulrurelep eqt ecurs '(ogt'g'bt.{) rpp}e }u€}suoc Jo ace#nspe^rnc € uo pFbll u lq pepaxe socroJ erp reprsuoc e.,ur '1xeN

l'atnssa-td to raluac er{+ sE u,r\ou>l sr paudde sr U areq \ e1e1d eql 3od +ulod eqJ '€or€ 1€rl1 Jo C plortuec eqr q8norql sessed uoncB Jo eurrsll lerll pue e rnc peol aql rapun €eff pplozade4 eW ol eprq1u8eruur pnba sr e1e1d erp Jo epls euo uo peuexa secroJ c!€lsorplq aqt 3oU luelpser er{r }Br{} e^resqo e,lr 'B'g 'ces

Jo s}Fser,eql BurnecaH'r

{}la AIJ€OuIT seuB^ 'snq}'pve

U o1 puonrodord sr ar qrFual lrun red pBol eq] teql s.AAoqs qclq \

(8r's)tlL7: dq : m

leql s.^'\olloJlr 'ec€3rns eo$ er{} ruog ecuelsrc l€crue^ eqr sI ? pu€ pFbII er{} JolqSlezrr clylceds eql sl /t arar{,AA 'qL : d sr prnbu € ur arnssard eBeBorp ecrns 'dq : m'snq1 :a1e1d eql Jo rppl \ eql sr q pu€ +p1nb{ eq}ur arnsserd aBeB eql sr d araqzn 'x? qd : Vp d se pesserdxa eq ospu€c p€ol srql ta,rezvrog 'qBue1 lrun rad p€ol aql sr m e:elqnt'xp msr rp qfuel So e1e1d eql Jo lueurele u€ uo peuexe p€ol eql 'g'g 'ces

ur pelou sy 'ern8g aql Jo eueld aql ol relncrpuedrad pernseeu sr

? erer{ \'? qtpl \pue ? {fualJo sr qcrq^\'8T'g'8l,tl ur ur'r.oqs a1e1drep3uelcer eql replsuoC 'ppbrr e ur pa8reruqns acn{tns nynBuopa"t.€ uo pauexa secroJ eJnssaJd crlElsoJP^q aql Jo ]u€lFSoJ ar{} eulur-re+ap o] pesn aq uec uoqcas Surpaca"rd eyl ul pasn qceo,rdde aq;

8l'9'6H

6Vl, secol.rng peO.retuqns uo se3rol 6 gst)vJuns ql9urwgns No s])uoJ 6'9*

; , - i i ' n , \ , S A ' $ P L E P R S B L T & I 5 . ?r f \ = 1 5 0 0 N / i l l

; - / - ssruT;oruc. Squivalent Co*centrafed Lcad" The magnitude of the resultant of theload is equal to the area under the load cuwe, and the line of action of theresuJtant-passes through the centroid of the same area. We divide the areaunder the load curve into two triangles and construct the table below. Tosimplif' the computations and tabulation. the given loads per unit lengthhave been converted inlo kN/m.

A beam supports a distributed load as shown. (a) Determine the equivalentconcentrated load. (b) Determine the reactions at the supporls.

x : : . s *

W : 1fi ]<;'r" J

4.5 kN/m

I

4.5 kNtf8 - -- - - w

T - - - - - q -

13.5 iiN*E - - '- - wY

A, KN f , m r A , k N ' m

Triangle ITriangle II

4.5r3.5

2+

I

)A : 18.0 ZiA : 63

Thus, X)a :ZiA: X(tA t N) : 63 kN ' m

The equivalent concentrated load is

and its line of action is located at a distance

7 : :].8 rn ta the riS-* at A .i

b, Reccfions. The reaction af A is r ertical and is denoted by A: the reacfionat B is represented by its components B, and Br. Ihe given load can beconsidered to be the sum of two triangular loads as shown. The resultant ofeach triangular load is equal to the area-ofthe triangle and acts at its centroid,We write the following equilibrium equations for the free body shown:

i ) r " : 0 : s ' : 0 ?

+\ )Ma : 0: -(4.5 kNX2 m) - (13.5 kNXa m) + By(6 m) : 0

3" : l8'5 i<N i'

+\ xM, : s. +(4.5 kNX4 m) + (13.5 kNX2 m) - 4(6 m) : 0

A : 7 . 5 k N ?

A.lt*rszre*ic.e Stsluti,*sEz" The given distributed load can be replaced by itsresultant, which was found in part a. The reactions can be determined bywriting the equilibrium equations )f, : O, 2Ma: 0, and )Ms : 0. Wea{ain obtain

L8 k\

250

&, =,' tJ Sry : l{1.5 kN'l A : 7 . 5 k N t . 4

t9(

7"::,,- ,,X'ir:A {i[ Q!i9'6i * X

ll + tl 0ll'0t : *{0 : 91 60I'0I - 1l

:elrsoddo pue pnbe sl JA ar"J aql uo rele.,n eql lq x B :

pauaxa sacJoJ aqlJo U luellnsar aqJ .p_aururratop ars H_ yo uorlca.,rp p,re, -

.l-l{--------*-7 ,:lspnlluiEtu eq] q_cF{,r\ uro$ u.e\erp sr a13ueu1 ecroJ v.d pu€ rzt{;o p uoa qt,gg.zr="

*rY". ,

--.# ,,t, -Jrsrr.)ur 10 lurod eqt q8no.rq+ sessed g- ';urrrn""ro, ,q lrn,u irrrn3

"rrqr "";;;=;

T* q= n- ,i

,,JJurS d JJroJ rll] pue 'tAI lq81e,t drl] 'rd.le,{\ rq} uo ulep Jrll ,{q fiepa*o ,,, 0,,,,,, =r'

-] .i I

seJroJaqlJou-lustpsaraq]aru,po^loqsacroJer{Jdpoqaa4usBuasorlJ $=:Jry".,$- : i i sr o)s rale.^AJo uopces cnoqered eql ,1xci,a{#A& f* ff *un:|n*sa q

.l,:il, gW

"JJil /, I

{680r:..qlJ!*:p H "t--i

i, u ' qt 096'o6s I tiurrl^\ .VJo lq8p eq1 ol p rJuelsrp n

1e Suqce acroS el8urs e .,(q paurelqo -"1rlr'"1d',o'"-.i":.J di;id;;;; "1a?; ! 1i " {ll {i9ti'{}El : }ai

0: tN + (I g)(ql 60I'01) + (U 06xqt 8sr/) - (+J /lxqt0006) -(u s rTxqt 00e'er) - (rr s)(q 098'7r)- ,0 : vruK !+

'- ; til {}f! -l * .X

0: ql 88VL - qI0006 - ql009'9T - qt0g8'7I - A ,s:'g3J+

:0 :'cKasrr,*y$)nt';{ ww*qVym&t,;g

qt 60r'0r : (ru/qt r'6e)(}J 8r)(rJ r)(U Srx : aqt 88rr : (eUlQI i'6e) ($ i ) (U 8T ) (+t oi )€ : u,.ltql 0006 : (ei.I/91 oeI ) (U I ) (U 8I )(U 0I ): : e,4tqt 00s'er : (cJ/qtOsr)(u i)(u 66)(u g) : u,.n

qr 0eB'71 : (eu/qr Oer )(u r )(u 66) (u 6)t : r,,!r

(cU/ql ' us)(I 8IXr.I I) =

,U t'

rT--

. "^erl e a

.og uopcos Jo lqiF aqt ot to]€.{\ oql^q Gg uoqcas uo po}Iaxe secroJ arnsserd arll Jo ;[ lu?]lnsar oq] pu" :r^tre]€.4 erl] Jo tq5ge,r eql :e,11 pup '.61t 'I41 slueuoduroc s1r jo slq8ra.r,r aqt lqpaluesarda.r 'tuep oql Jo tq81a.ln oql erp lpoq aa4 eqt uo Burt.le i""to; raq+1;'y +e urelsls eldnoc-ecroy luep,tnbe ue ,{q peluesa;dar e-re gy as€cl aq} uc)puno.r8 eql lq palrexr se.)roJ uooreer ;qJ .rrlu,tr pun urr?p erl] Jo ge)dVVuorlJds )tJlrfi-U-l aql lpoq rerl p se rsooqJ r1.1 .uotl)rla5 punoJg.s

dleapcedsa.r '$/qIv'zg pue eU/gl 0gr er€ rrte.{\ pu€ o}orcuoc 3o slqFre,racgrcads ar{J 'urep eqiJo Jg aceJ erp uo re}€.4A aq+ {q pegexe secroJ ernsserder{t Jo }u€}Fsor orF (q) 'ttrep oq+ Jo gy eseq rr{} uo punorB aqt dq pe}raxesocroJ uorlceer eql Jo tutllnsar erll (r) aurrrrrolrp pup 'urep aq+ Jo uorl-JJS lJrr.l.l-u-l e rJprsuo]'r!\^orls se sr ruep JlJlJuoc EJo uoLlJJs ssorJ J1IJ

dr1= ot

ol's wITSoUd EldWVg

Cr .- .i' ! L! -:.1

tlT9

$6

T|" problems.in

fis ]esson involve two_common and very important g,pes ot'

I loading: distributed loads on beams and forces on s.tb*eiged^surfaces tf "orr-stant width. As we discussed in Secs. 5.8 and 5.9 and illustrated In Sample probs- 5.g

iid 5 19, determining the single equivalent force for each of these load'ings r"q"rr",

a knowledge ol' centroids.

l. Anclyzing heoms rubiected |o dis*ibuied locds. In sec. 5.8, we showedthat a distributed load on a beam can be replaced by a single equivalent force.The magnitude of this force is equal to the aiea ,rndei the d[tributed load curveand its line of action passes through the centroid of that area. Thus, yo" ,r.o"rJbegin your solution by replacing th'e various distributed loads on a grven beam bytheir.respective single eqtti"aletit forces. The reactions at the supports of the beamcan then be determined by using the methods of Chap. 4.

fvhen possible, complex distributed loads should be divided into the common-shape areas shown in Fig. 5.BA fsample prob. 5.9]. Each of these areas can thenbe replaced by a single-equivalent force. If required, the system of equivalentforces can be reduced fu.rtter to a single equivient force. As you study SampleProb. 5.9, note how we have used the"analogy betrveen force'und

"r"u "rrd t'h"

techniques for locating the centroid of a com"posite area to analyze a beam sub-jected to a distributed load.

2" sclring proLlerns jnr:l"i*s forces on- suhmerged bedies. The followingpoints and techniques should be remembered when rJ,ritrg problems of this tlpe:

, o. The pressure p ^t,^ depth h below the free surface of a liquid is equar to

yh or pgh, where 7 and p are the speci{ic weight and the density of the iiquid,respectively. The load per unit length-to acting oi a submerged surface of

"onr'tuniwidth b is then

u : b p : b y h : b p g h

b" the line of action of the resultant force R acting on a submerged planesurface is pelpendicular to the surface.

c. For a vertical or inclined plane rectangular surface of width b,the loadingon the surface can be rep:esented by a linearly distributed load which is trapezoildal in shape (Fig. 5.18). Further, the magnitude of R is given by

n : lh"ewhere hu is the vertical distance to the center of the surface and A is the area ofthe surface.

2,s2

egt,

'uossel slql ulpacnporlur seapl eql asn ol .{lrunpoddo aldruu e^€rl il$\ no,,( '(scrueqcau pln6pu€ sFIJal€u Jo sJrutrlcaur telncrlred ur) sasJnoc scrueqcau luanbasqns u1

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'31,{ 3o tg acro3 er{+ +€q} a^resqo '(Ot

S '31.{)

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5"66 o'nd 5,67 For the beam and loading shown, determine (a) themagnitude and location of the result*nt oi th" distributed load.(b) the reactions at the beam supports.

Fig. P5.66

Fis. P5.68

Parabola 7"'

-rf1a1\ott\,1,f00\'n,_+_++ _j"-dL- rfu

l__+_ ____!Fig. P5.67

5.68 tlrrough 5.73 Determine the reactions at the beam supportsfor the given loading.

'1E0 ]b/fi

f\ -*fi-:f-i.i(x)lr,/rt{ {\,-t-t I { { I iA [in - "* ire* : _:*: D

I u * c s ' It t t lf.-rr,_-l.-6ft_#

Fig. P5.69

1200 N/rn

Fig. P5.7O Fig. P5.71

Fig. P5.72

254

200 lb/lt

Fig. P5.73

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u'dH t w

ly(il 09r

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Jo buroeol pue urEeq eq] ro.{ 6l'5

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'slroddns erl] l€ suopceer Surpuodsarroc eql (q) 'urnurrurur

sr g poddns Je uorlceer eql ler{} os 2 acuelsrp aql (a) eurure]ee g1.g

'sl.roddns

eql lB suollcee.r Surpuodserroc aq] (q) 'pnbe ere g pue y sgod-dns 1e suollceer l€cruol or{l lerll os 2 ecu€lslp eql (a) eururre1ee Vl'S

u u-]yl -r

sw)

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-i --l

[,GuiiN 008I

256 Distributed Forces: Cenhoids ond Centersot (,rovrry

T C

Fig. P5.82 ond P5.83

Fig. P5.84

The 3 x 4-m side AB of a tank is hinged at its bottom A and isheld in place by a thin rod BC. The maximum tensile force therod can'withstand without breaking is 200 kN, and the designspecifications require the force in the rod not to exceed 20 percentol this value. If the tank is slowly {illed with water, determine themaximum allowable depth of water d in the tank.

The 3 x 4-m side of an open tank is hinged at its bottom A andis held in place by a thin rod BC. The tank is to be filled withglycerine, whose density is 1263 kg/m'. Determine the force T inthe rod and the reactions at the hinge after the tank is filled to adepth of 2.9 m.

The friction force between a 6 X 6-ft square sluice gate AB andits guides is equal to l0 percent of the resultant of the pressureforces exerted by the water on the face of the gate. Determine theinitial force needed to Iift the gate if it weighs 1000 Ib.

5.82

s.83

5.84

Fig. P5.85

5.85 A freshwater marsh is drained to the ocean through an auto-matic tide gate that is 4 ft wide and 3 ft high. the gate is heldby hinges located along its top edge at A and bears on a sill atB. If the water level in the marsh is h : 6 ft, determine theocean Ievel d for which the gate will open. (Specific weight ofsalt water : 64 lb/ft'.)

5.86 The dam for a lake is designed to withstand the additional forcecaused by silt that has settled on the lake bottom. Assuming thatsilt is equivalent to a liquid of density p, : 1.76 X I0" kg/m' andconsidering a 1-m-wide section of dam, determine the percentageincrease in the force acting on the dam face for a silt accumulationof depth 2 m.

lig. P5.86 and P5.87

5.87 'the base of a dam for a lake is designed to resist up to 120 percentof the horizontal force of the water. After construction, it is foundthat silt (that is equivalent to a liquid of density p, : 116 x 103 kg/m3)is settling on the lake bottom at the rate of 12 mm/year. Consider-ing a l-m-wide section of dam, determine the number of yearsuntil the dam becomes unsafe.

5.88 A 0.5 X 0.8-m gate AB is located at the bottom of a tank filledwith water. The gate is hinged along its top edge A and rests on africtionless stoD at B. Determine the reactions at A and B whencable BCD is slack.

5.89 A 0.5 x 0.8-m gate AB is located at the bottom of a tank filledwith water. The gate is hinged along its top edge A and rests on africtionless stop at B. Determine the minimum tension requiredin cable BCD to open the gate.

0.27

I0.45

A l-T0.48

Fig. P5.88 qnd P5.89

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, ur 07'0 ,

l6'sd '6!l

96'9d'6H

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'qord o los e6.g

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'role,^A Jo 1ry .{1e1e1duroc sr qBno4

aql uar{.4A erurl B }B 'selqBc eql Jo qcEe ul uolsuel eql eulturel_oc

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T-I--lrtt- It

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III

l9(, sualqord05'g

258 3i.3|o1i,i

Forces: Centroids ond Centers

Photo 5.4 To predict the f l ight chorocferist icsof the modified Boeing 747 when used totronsport o spoce shuti le. the center of grovity ofeoch croft hod to be determined.

The,center of graoity G ol a three-dimensional body is obtained bydivlding the body into small elements and by then expressing thatthe weight W of the body acting at G is equivalent to the system ofdistributed forces AW representing the weights of the small ele-ments. Choosing the y aris to be vertical with positive sense upward(Fig. 5.20) and denoting by i the position vector of G, we write that

5.I O CENTER OF GRAVITY OF A THREE-DIMENSIONALBODY. CENTROID OI A VOLUME

v t !

A1'\ - Alllj

II'J7 z

Fig. 5.2O

W is equal to the sum of the elemental weights AW and that itsmoment about O is equal to the sum of the mornents about O ofthe elemental weights:

)F: -wj : >(-Arvj)) M e : F x ( - W j ) : ) [ r x ( - A W j ) ]

Rewriting the last equation in the formiW x (-jl : ()r Aw) x (-j)

(5.14)

(5.r5)we observe that the weight W of the body is equivalent to the systemof the elemental weights AW if the following conditions are satisfied:

W : > L W i W : ) r A W

Increasing the number of elements and simultaneously decreasingthe size of each element, we obtain in the limit

(5,16)

We note that the reiations obtained are independent of the orienta-tion of the body. For example, if the body and the coordinate axeswere rotated so that the e axis pointed upward, the unit vector -jwould be replaced by -k in Eqs. (5.14) and (5.15), but the relations(5.16) would remain unchanged. Resolving the vectors i and r intorectangular components, we note that the second of the relations(5.16) is equivalent to the three scalar equations

w : I a w i w : l , o *

( l= ) a d w z w - - l z d wi w : l . o * i w (5.17)

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(8r'9)Lpt

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Shape x Volume

Hemisphere 3aB

2 ^

iooo

Semiellipsoidof revolution

shB

2 " ,ifieLn

Paraboloidof revolution

- -h

L3

I o r

tTa-n

Cone

r-o

L/ !m2h

$e'amid L4 !"un

Fig. 5.21 Cenhoids of common shopes ond volumes.

260

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- : : - " " " - " ; ; : ' : ' :. a ' '

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j { O O m - t = 2 2 . 5 m m

SATYIPIE PROBLEfr,i 5.I I

Determine the location of_the center of gravity of the homogeneous bodyof revolution shown, which was obtained by joining a hemiiphere and acylinder and carving out a cone.

SOLUTIOI-l

Becalse of symmetry tle center of gravity lies on the r axis. As shown inthe ffgure below, the body can be obtained by adding a hemisphere to acylinder and then subtracting a cone. The volume and the absdssa of thecentroid of each of these components are obtained from Fig. 5.21 and areentered in the table below. The total volume of the bodv and the firstmoment of its volume with respect to the y.:, plane are then determined,.

50 mm

Thus,

X>v: >;v X(I.ZOO x 106 mm3) - 18.09 X 106 mma

262

, l , l

Co Volume, mm3 i, mm iV, mma

Hemisphere

Cylinder

Cone

I /,+

i ; r o o r 3 : 0 . 4 5 2 4 x 1 0 6

?-(60)'z(r00) : 1.1310x 106- , (0or - r roor : -0 .3770 x t06

- Y . v 5

+50

-10.18 x 106

+56.55 x 106

-28.28 x 106

) V = 1 . 2 0 6 x 1 0 6 ) i V : + 1 8 . 0 9 x 1 0 6

X - 1 5 r n n i 4

e94

'r1I 8I9"I

'ilI CtCG"at-

'ut ltg'0

*L

*;_;" -_ ,\

rut 999'8 : (€ui 986'9)2 :A.ZK: LKZ_

vvr L?0'9- : (err ggZ'g),t ,A4K: AKL

rut 8i0'e : (€q 9BG'9)X :A{K : AKX.snt[J

irr s6 o- -

Arc,rlrc.rc-T'u]

II

'seuElct eleurprooc erF Jo rlcEe 01 lceoser qll^\ eunlo^ eql Josluaruou aql puE alunlo lBlol oql oururralap uaql a^\ 'olq€} aql uI €l€p eqlEuIsO '.t\oleq elqBl eq] uI parelue eff pup paulturelep eft lueuoclwoc qc€e

Jo plorluec aq] Jo seteurprooc aql pue eunlo^ eql '(1I pus III) s.repurtlcreleu€rp-'ur-1 o,Lr1 Suncertqns ueql put (II) repuulc ralrenb e o1 (1) ped-ide1a1e"red reln8uelca"r e Suppe lq peurulqo eq u€c ]ueruala eulqc€ur erlJ

uIl

t7

+

.".ffi,NOltnl0s

uI 9'0

rz

'uI I q alor{ r{cBa Jo ral

-dur€p or{J 'u \oqs }uauldlJ autllJeul [ar]s aI{} Jo 'Q1nur8 3o raluer aq} el€ro'l

Yl'g wrr80ud rldwvs'r :."

ss9'8 : AlK1fr0's- : L-K870'e : AIK9BZ'9 : AK

AIIIIIII

68g'0-VLT'I-868'096I'0I

t6t'086e '0

888'I -

860'0-860'0-6II'696I'I

gT

cc

92',0q6',Z

I-I-

8878'0-I-

q7o.

9ZO887t'I

96'0

L66t'o-: (SO)u{SO;.',-L66t'0-: (90)r(90)a-

r/9'r : (s'\rlar\9'r : (9'0x6xg'7)

#I,AZrul 'A

vu'l'Ax'ut'z'ur'UI .J

rul'^

- - j : : : : : - . . . . . : . : . . . j . . : . - : : : " : . . : : . . : : : . . . : . . ' . . . : . . . j . . . . . : : . . : . . . : . : . . : : . . . . - - .'i

s{}*,1rT*#*{

g**tPlf png**-Ffvl 5"13

Determine the location of the centroid of ihe half right circular conesnown.

Since the xg plane is a plane of symmetry the centroid lies in this planeand a : 0. A slab of thickness dr is chosen as a differential element. Thevolume of this element is

4y : )rr2 dx

The coordinates i,1_and -y"i

ol the centroid ol 'the element are obtained fromFig. 5.8 (semicircular area).

4rrot : x !/"t : ^- 3 f i

We obserye that r is proportional to x and write

r a ar : -; : i ' : i *

The volume of tJre body is

| , h r h / ' i

v : I dv -- I ' f , , ' t r : l " i , (?r\ or : TJ J r

' J o ' \ h / 6

The moment of the differential element with respect to the ya plane isia d.V; the total moment of the bodywith respect to this plane is

I l h l h / o \ r , o o r h tI x" tdv : I x ( ! r r2) dr : I r ( jz ) ( 7x I dx : .J J o J o \ n / d

Thus.

*v :l *",av ,+: + c * j* .,1Likewise. the moment of the differential element with respect to the arplane is g"1 dV: the lolal moment is

f - - l h 4 r , e \ , 2 l h / o \ ' . o 3 hl i rdv

: I o ' - r2 r r ' ) t l x :

- | ( ; x ld r : - -

L Ln 37T '

3Jo \ 1 , / 6

Thus,

iv : Ii",av

264

na'h oth {i .' 6 6 ' , ? i

99L

'Ap Jo plorluec arpJo saleurprooD aq+ ara (t6'g'sbg ur /': pue 'P4'p.y1ur11 noi

pulurer. ure8u a.tt ,{ltn.rt.q 'alnduroc ol o^r?rl fltu no,{ spr8alur atp 5o ,Qxaldrnoc aq]1ce33e .{pcarip ru \ (Sf g 'qor4 eldures ur r pue r rroe^qeq dn1suoile1al al+ e{r[) solq€-uEA arl] Suoute qsrrqelse no.{ 1eq} dlqstroqeler eq} ltrll -roqusureJ o1 lueyoduri sr 1rle.te,tro11 'flerls

Ierupu{.{r urq] r: ro (g1'g 'qor4 aldureg ur se) qels uq} E eq.'(eru srq}'uoqnio^er

Jo sapoq rog 'alnduroc ol +sersua aql are lrr11 spr8alur elqnop .io a18urseqt sacnpord rlrll.\\ Ap aqt 'alqrssod y 'Sud3puapr .{q uoqnlos .rnoi ur8eq pporlsno.{ 'snq1 'Ap allrnlo^ Jo tualuale rqt roJ (gZ'q 'Bl.{) qEIS uFIt e rc (76'9 31g) lrreur-EIIJ ulql e rerlllo Sursooqc,{q pagqciurls aq uec (Af'q) 'rbS jo sp;Betur eqt Supenp.ta'6I'g 'ces nr peurelclre sy "**a4**fu4zltry&*t*Xq saruergee *e €ffesals> eWfuWwZ "&

VZK: V<Z Y4< : Y<,t V{K : VKXot asec slql q ecnpe.r snq (16'9) suoqunbg 'e1u1d ue.tr8 p Jo eaie aqt q v ererp\

'y1 lcnpold aqt qli \ pccrlda.r sr A rreq \ (tZ'g) 'sbS

Jo tno rotceJ II.rr!\ s.llu1d eql go/ sseDlJrql eLp,'ssau:p!LL+ uuo{nm ?nns aLp p sa1oyd llraaas Jo apeur ,{poq e rod 'q

TZK : TKZ il< : I<I 7{K : 7KXot esEc slql ut ecnpe-r snql (IZ'g) suorlenbg

'tueurele_ ue,rr8 r 3o ql8ue1 eql sl T a.raq,\\ '7y lcnpord aq+ rllp\ pecelder sr A uoqa

(tf'g) 'sbS Jo lno rolcp,J III \ s+uourole orl\\ eql Jo y r?ore lurrollces-ssorc oLIl'uo!+

4as' ssoJl uuottun alLtos aql Jo slLLauap alroT lDranas'Jo ap€Lu ,{poq e -rod "*'lelrel€Lrr arues aql Jo apr?ur sa1e1d urro,ygurl ro saJI.\\ uIJo.]Iun .Iaqlla

Jo slsrsuoJ fpoq ue,rr8 eql uerl \ ,rncco +se"reluL +o sasr)e Tonads o3r? 1tstl] elou e'11

'alcrrJ rJ+l?rrh e jo illorluec eq] roJ srroq-enba eql Sursn parrrelqo ala,{\ .rapur1.,{c ralrenb arl} Jo pioJlual eql Jo sa}BurpJooDf pur: r aql ^\oq a.lrasqo '6I'9 'qord aldureg .,(pn1s no( s€ 'oslv 'salpoq alrsod-uroc Surzl,pue ueq,r\ solqe+ Ilue suru-r8urp eler"rdorddr lcnrlslror o1 rro.,( aFerrtocua.{13uor1s uru8e ecuo a.u 'snrlJ

lecr]uepl o,re suralqord leuorstreurlp-corq] pu€ -o,\\l

-ro.I uoqnlos Jo spoqlaur arp 'altrlsnllI ZI'g prrr? [I'g

'sqord aldureg Jo suor]nloseq+ sV 'reldeqc oql ul rorlrea pe-reprsuoc sruelqord 1luolsuetulp-o,\\l aq] ro1 posnsuorlenbe oq] lo Lrorsualxa ue ,Lldurrs e.re suorlenba asaql +Eq] ezlleor plnoqs notr

(r6'9)A."7 - A7'7 AII- - A7 I A\'7 - A'7V1r:.\ _ 1r.\z 111-\ _ ./r-\-\ 1r::\ _ lr\^

:(16'9) 'sbg Sursn pelecol aq oslu utc {poq aq} Jo {1rtr:"r8 Jo rotuocaql 'asec prcads slrlt roJ 'oroJe"raql 'aunloo s4t .{o pro4uoo oql rllp\ seplculor,{poq eql 3o ,Qr,ter8 Jo relrier eqt 'l)poq snoaua7outorl e Jo rsec aq} ro3 le.te,trog

A\4K :,4\KI ,4\IK : A\KX:Pesn eq

qsnur (96'9) 'sbg'prauaB uI '6*irry-&a*&*&e3 g* &Az'r"x& &*xa*41*e3*r"q*xg*,w3 "g

'esuc l€uorsueurp-eerq+ p.reua8 eql o] peqdde eq oslt .{eur- t1a '}uaurlJ

IBriuareJJIp llrercgJr tsour eql Sursooqc 'sedeqs uotuuror olur '(poq et11 3u1ppr1p'.(r1aurur.{s Sursn-sarpoq luuorsuaulp-o^q .IoJ passnJslp .{lsnoltard e.u senbru-qce] oql Jo IIv

'settlnlo^ rreql Jo splo"rluor eql ro selpoq lEuolsuerurp-rrrtll Jox-ilr,ter8Jo s-reluec eql elucol ol pe{se eq ilpl\ no,{'uossel slq} roJ suLelqord erll u$

(06'9) A\ZK: X\KZ

%ttrf .{tr }ffi{*:r{4 ffi{ f 6 f ,3ff*%

5.96 Determine the location of the centroid of the composite bodvshown when (a) h : 2b, (b) h : 2.5b.

5.97 Determine the y coordinate of the centroid of the body shown.

' t , \ l

,4*_[t\o=_

Fig. P5.96

Fig. p5.97 ond p5.98

5"98 Determine the a coordinate of the centroid of the bodv shown.(Hint: IJse the result of Sample prob. 5.13.)

5.99 The composite body shown is formed by removing a semieilipsoidof revoiution of semimajor axis h and semiminoiaxis a/2 fim ahemisphere of radius a. Determine (a) the r7 coordinate of thecentroid when h : a/2, (b) the ratio h/a for -hi"h 7 : -0.4a.

Fig. P5.99

'u^\oqs

urroJ l€taur-leeqs or{t Jo llr.rer8 Jo ratuec eql etecoT tOL,g pug ggl.g

rol'gd '6!r

lOt'gd puo gg1'94 '6;1

erll Jo alBurpJooc , oql elecol

erll Jo el€urprooc z eql elEcol

'llr,ter8 Jo reluec

'u.^Aorls ]ualuale eulqcBru arll rod gal,g

'llr,ter8 Jo roluoc

'u,^^oqs luetuele eurqcBrr aql rod t01.9

uJrrr 9zuJtJJn,z /

fir,ter8 Jo reluec eql Jo a]€up-rcoc fi oql attcol'u^\oqs ]uourolo aurrlcuur oql rod e01"9 puo lAl'9

_ ,{1r,te.r8 yoreluec eql Jo aleurprooc z eql elBcol 'u.^eoqs le{cBrq do}s eql lo.f 1ol'g

.,(1r,rer8 3oroluoc oqlJo eleurprooc I arll ot€col'u1(oqs le{cerq dop aq} ro,f 001'9

901'9d'6!r

tOt'gd puo ggl'96 '6;1

(;

l*-16 (lrur 09

l9Z suatqord

268 31.3|of# Forces: Centroids ond centers 5"!Og A wastebasket, designed to fit in the corner of a room, is 16 in.

high and has a base in the shape of a quarter circle of radius l0 in.Locate the center of gravity of the wastebasket, knowing that it ismade of sheet metal of uniform thickness.

Fig.

$-lQg A mountingsheet metalthe bracket.

P5.r08

bracket for electronic components is formed fromof uniform thickness. Locate the center of gravity of

5"?lC A thin sheet ofplastic ofuniform thickness is bent to form a deskorganizer. Locate the center of gravity of the organizer.

, leO ̂rf,

(

(' ';i:'')t;,J,,-.-r\ x

3 / 3 0 m m \ r = 6 m m

r = 6 m m r = 6 m m

Fig . P5. l I0

",rffio. ruK. . .

'sseu{ciql urroJlun Jo-IJqeJ eJo,{\ slcnp oqllcnp r€ln8u€lce|ur-g

etr'gd '6t!

dlquresse eql;o llr,ter8 Jo raluac eql elr?colsl rlclq.{\ 'lBlelu

leerls elues eql uorJ peler1eq1 Sugatoul 'pel€crpur se paurol aq o] ol€

X p e pue lcnp lecrrpurl,{J rJtrruprp- uf g uy Sl t,E

zlr'gd'6!l

.,{\oqlo arl] Jo AI^€ri Jo ralu€c aql olEcoT 'ssau)lcrql urroJrun JolE]etrl ]eeqs Jo epr?ul sr ue]s^s tsu4€lrlue E Jo pnP er{l roJ .{\oqla uv g i [.9

llt'gd'6!t

uIC7=J

'8uru.Lre erl] Jo .{lr^€r8 Jo raluec aql elecoT 'sseu

-{clql urroJlun Jo lelaur }aoqs luo41 paleclrqeJ sr Suru.tte ̂\opurl\ V 111'g 692 suetqoid

270 Distributed Forces: Centroids ond Centersor (,rovrry

Fig . P5. l l5

5.114 A thin steel wire of uniform cross sectionshown. Locate its center of gravity.

is bent into the shape

Fig. P5.lt4

5.1t5 ond 5,116 Locate the center of gravity of the figure shown,knowing that it is made of thin brass rods of uniform diameter.

30 in.

I"\ ,:

z

Fig. P5.l

, lffiru,w#^1 7 x

Fig. P5.tt6

5.117 The frame of a greenhouse is constructed from uniform aluminumchannels. Locate the center of gravity of the portion of the frameshown.

5.118 A scratch awl has a plastic handle and a steel blade and shank.Knowing that the density of plastic is 1030 kg/m3 and of steel is7860 kg/m", locate the center of gravity of the awl.

- l [ 'o*-

{.-_ 80 mm ____J

, l ' l^

(I

5ft

3.5 mm

Fis. P5.ll8

97t'gd'6lJ

r-.t___-__l xl I

IW'slx€ x oql lnoq€ €ar€ popeqs aql EUI

-tetor lq pourElclo olunlo^ orllJo plorluac oql alecoT 921"9 PUP 921'9

'uorlnlo^ar Jo ploloqered v tu t's

uorlnlo^er Jo plosd{lelrrlos v gg1'E

areqdsuleq Y A7,Z'g'rq8t"q

lenbe 3o saurn

10^ o.^ l olur edeqs aq+ sap1.tlp pue adeqs ue.r,13 eql 3^o asuq^oql ol

lelpred sr eueld Surpnc ̂eq1, 'II'F '8r,q

3o adeqs ue,u8 aql q8norqreueld Surpnc I€crue^ e Burssed ,{q paurelqo seurnlol o,tl} eql roJ r

Jo sanle^ or{} uolter8e}ul tcerrp lq eurturelaq ttrl'g q5no*q* eet'S

'alqel oq] 3o l1rcr8 Jo raluec orll olecol

j-l8l OO1Z sr sse13 Jo pup c./3{ OgSl st lrr}s 3o ,{lrsuap rq} }Ptllbur,,vrouy 'l,1a,rr;;edsa"r'uur 0l puB ulru 009 Jre dol rlqEl rtllJo ssJu-{crql oq} Pu€ roleul€Ip eql '6urul

09I Jo Eo-re l€uol}ces-ssorc e pueLrLLr VZ p relaru€Ip aplslno uB seq qcrq,t\ 'tsu1qn1

Iools Jo epetu er€p.r" peceds llenbe er€ olqet peddol-ssu18 IIEurs eJo s8el aerql er{J 121"9

921'9d '6H

rutu 086 = r.

lZl'gd '6!t

6lr'gd'5!J

ozt'gd '6tt

[-ril:;rI T- uI0t0t

II('"rtlql tOt'O : rrnurunle 'rul/ql g0t'0 : ss€rq :slqBra.tt cSlceds)

$oq elsodruoc aq+ 3o .,{1r.ner8 Jo ro}uac aql_ ot€coT u1 7 qr8ual

Jo por runururnle uB uo palunout sl "ul 9'6 qFualSo 'rBIIoc sserq Y QEE'9

.,(lquresse eql 3o ,{1r,,re-r8 Jo reluac arl} Jo uol}Ecol eI{} eulluJa}ep' ,ylnt vgz'O sI Iaels Jo pu€ eul/ql 8It'0 q azuorq 1o tqBle.u c151cadsiql w r Sur.ttou;tr 'enaels

laals € oprsul polunorll sr Burqsnq azuorq Y 61['9

l

Ll(, suotqord

00 II

II

272 Dishibuted Forces: Cenlroids ond Centersof Grovity

5.127 Locate the centroid of the volume obtained by rotating the shadedarea about the line K : h.

tig. P5.127

*5.128 Locate the centroid of the volume generated by revolving the por-tion of the sine curve shown about the r axis.

u=bsnF"// za

r{-l;ik I

ffi_i_F-a------r-"--4

Fig. P5.128 qnd P5.129

'5.129 Locate the centroid ofthe volume generated by revolving the por-tion of the sine curve shown about the y axis. (Hint: Use a thincylindrical shell of radius r and thickness dr as the element ofvolume.)

*5,130 Show that for a regular pyramid of height h and n sides (n =

3, 4, . . . ) the centroid of the volume of the pyramid is located ata distance h/4 above the base.

5.131 Determine by direct integration the location of the centroid ofone-half of a thin, uniform hemispherical shell of radius R.

5.132 The sides and the base of a punch bowl are of uniform thickness t.If t << R and R : 250 mm, determine the location of the centerof gravity of (a) the bowl, (b) the punch.Fig. P5.t3l

Fig. P5.132

)xtr..t\ q

get'ga '6ll 'eue1d anbllqo ue lqrepur/c pcqdlla

z ue ruo$ lnc s€,^A r{cil{,4^ 'u.4Aorls uopces eq} Jo plorluec eql o}BcoT ggl.g*

'seueld enbllqo o.,r,r1 lq edrd .relncrrc urqlB urog lnc s€,^ qctq \ 'uaoqs uopcas aql Jo plo4uec er{} el€coT gtt.g

rgrsa'6tr

, 'zq|D/(zz,- zq)(ic - xn)r4g : fi

ac€Jrns er{l Jo u.4Aor{s uoFroo eql pu? euslcl zI eql uae \leq elunlo^eIIl Jo plorluoc eI{} Jo uollecol eql uoper8elul lcerrp lq eurur"releq tgl'g

eel'gd '6H

(zc_y xg a o : k uollenbe eql lq pelueserdereq uec qclqll 'euu1d enblqo ue s1 lezrer8 eql Jo eceJrns tuol-foq or{I :tuxH) 'le^etd er{l Jo^eurnlo^ orll Jo plo4uoc ol{l Jo olEulp-Jooc r eql pu€ pepeeu le Brts Jo eunlo^ eql eulrurelec 'q€Is eq]qpauoq 1e,re.r8 5o

'ul t Jo ulntulullu e seceld replnq el{r 'q€ls aql

roJ es€q Ie el 'ur{I e eppord oJ 'asnoq e roJ qels eql Jo sreuroceql eru8gsep ot se{Bts rno; seceld rep[nq P 'pl B Super3 raUY gg['g

9gt'9d '6!l

elZ surotqord

eenter of gravity nf ctwo-dimemsioncl b*dy

This chapter was devoted chiefly to the determination of the centerof grar:i,ty of a rigid body, i.e., to the determination of the point Gwhere a single force W, called the weight of the body, can be appliedto represent the effect of the earth's attraction on the body.

In the first part of the chapter, we considered ttao-dimensionalb2!les, such as flat plates and wires contained in the xy plane. Byadding force components in the vertical z direction and momentsabout the horizontal y and r axes [Sec. 5.2], we derived therelations

w : I a w

Similarly, the determination of the center of gravity of a homnge-neous u;ire of unifunn cross section contained in a plane reduces tothe determination of the centroicJ C of the line L iepresenting thewire; we have

which define the weight of the body and the coordinates i and y ofits center of gravity.

f;enfrold o$ nn #res sr l;ne In the case of ahomogeneous flat plate of unifurm thickness fSec. 5.3],the center of gravity G of the plate coincides with the centroid C ofthe area A of the plate, the coordinates of which are defined bv therelations

(5.3)

iw : l .o* iw : l ro* (s .2 )

. e : [ . a e a e : l a a e

.r : f.ar or: la ar (5.4)

Finst ffimmenfs The integrals in Eqs. (5.3) are referred to as the first mamznts ofthe area A with respect to the y an! x axes and are denoted by Qoand Q,, respectively [Sec. 5.4]. We have

Q r : i A Q " : Y A (5.6)

The ftrst moments of a line can be defined in a similar wav.

Fropertier cf symm*fry The determination of the centroid C of an area or line is simplifiedwhen the area or line possesses certain properties of symmetry. lfthe area or line is symmetric with respect to an axis, its centroid C

274

'ftr'g 'qora apueg] V eere eq] eulurretep o1pasn oq osl€ uuc ]uotualo oruus eql :"fl pue hfl slueurour 1srrJ arp Jorpoq elndruoc ol €eJe Jo lueurele errres ar1 esn ol snoe8eluerrpe sr 11

G's)

eABq e/\{ 'vP lueluele erlr Jo plorluec eql Jo sel€urproocarql P4prre lu!.{q Sunoueq '6I'9 '8t.f ur rr ̂oqs ear€ Jo sluetuolepadeqs-ard ro rep8uelcer ur-r[+ er[+ Jo euo sesn qclry\\ yo-fiaryt a73-u!,s E ro (g'S) 'sbg ur spr8alur elqnop erll raple Sunenp.te lq auopeq uec sFU '[g'g 'ceg] uogqu.Zap.g Aq peurrure+ep aq uec plortuecsI Jo seluurpJooc erfl 'seAJnJ puArcue Aq pepunoq $ €ere u€ uerl1\

'[tr'g 'qora elduug] uoqsuJ relrrum B ur lno peursc sT erl \ elrsod-ruoc B 3o filuter8 Jo roluac aW Jo uorleurruralep eql '[I'g 'qo.r4 e1d-ureg] pgortuac sIJo,t pue X sel€urprooc eIF roJ pe^los aq uec deql.ro 'eere elrsoduroc aql Jo sluauour lsrlJ eql plarl suonunbe eser{J

vpt"li l:

w:,o v.p.! l:

vz: no

uolpr$qry ,QPpJ|ue3 ro uopsu!ililst3{

{poq eysoduroro to ,t!^or6 p ralua3

(e'q) vfrK: vKz : "dWK : VK4: nd

tl\IK: tl\KX

ol ecnper (Z'g) 'sbg

puu 'ep1d erp Jo EarB eql Jo C plorluec eql ql.r^\ seppuloc firu'u.r8Jo reluac slr 'ssarDlcrq+ ruroJlun Jo pue snoeueSoruoq sr a1e1d aql y1

?u'9'6!t

(t's)IY\LK: tvt<l

elsr{ e \ 'fte'S'^L) llenncadser 'sexe r pue f aq+ +noqeslueruour Suuenbg '[g'g'ceS] qred snope,r etlrJo' ' ' '6'f)]C fir,re-r8Jo sroluac el[+ "}o

' %&'rh PtJc- ' ' ' '6!'rr saleulProoc oql uro{

peulurrelep eq u"c p fir,mr8 5o rsluec srl Jo tr pIrB X sel€urprooc etp'sedeqs eserllJo lure^es orq pep1,r1p eq uec alqd pgn uer{z\A.'s'g '8tg

ur pelelnqe] arc sadnqs uanmnc movna{o sp?oqun a#pw noJD arl1-

'o qll a seplculoc p '6 raluec e o1 pedsa;rl}1 A clJleuruls sr 1l fl :serru o q etp Jo uoqcesrelq eql 13 pe]€Jolsr C

'sexe o ^+ ol pedser rFI/\ cpleuur{s sr U JI :strE l€{l uo sett !LZ, fuoru'rng puo /,.or^€d

276 ]'[|ofri Forces: cenhoids ond centers

Theorennc of Pcppus-suidinus

t--*

,rA=l \ l-\-l-

z"i \,,i(a)

Fig. 5.25

Fig. 5.26

Cenler o* grcvity of s three- The last part of the chapter was devoted to the determination of thedirnensionol bodv center of graai,ty G of a three-di,m,ensi,onal body. The coordinates r,

a, Z of G were defined by the relations

In the case of a homngeneous body , the center of gravity G coincidesCenlroid nf s volunne with the centroid C ol the oolumi V of the body;1he coor&nates of

C are defined by the relations

zv: Izd,v (5 .1e)

If the volume possesses a plane of sym:rnntry, its centroid C will liein that plane; if it possesses two planes of symmetry C \l'ill be locatedon the line of intersection of the two planes; if it possesses threeplanes of symmetry which intersect at only one point, C will coincidewith that point [Sec. 5.10].

(b)

Dirtribuled losds

The theorems of Pappus-Culdlrus relate the determination of thearea of a surface of revolution or the volume of a body of revolutionto the determination of the centroid of the generating curve 0I AI0A[Sec. 5.7]. The area A of tb.e surface 6-tt.it"d by 'i,tatirtg a cuNe

of length L about a fixed axrs (Fig. 5,250) rsA : 2 r i L (5.r0)

where y represents the distance from the centroid C ofthe curve tothe fixed axis. Similarly, the volume V of the body generated byrotating an area A about a fixed axis (Fig. 5.25b) is

V :2 r r yA (5.1r)

where y represents the distance from the centroid C of the area tothe fixed axis.

The concept of centroid of an area can also be used to solve problemsother than those dealing with the weight of flat plates. For example,to determine the reactions at the supports of a beam [Sec. 5,8], wecan replace a distributed Load w by a concentrated load W equal inmagnitude to the area A under the load curve and passing throughthe centroid C of that area (Fig. 5.26). The same approach can beused to determine the resultant of the hydrostatic forces exerted on arectangular plate submerged in a liquid [Sec. 5.9].

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5.137 ond 5.138 Locate the centroid of the plane area shown.

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5.139 The frame for a sign is fabricated from thin, flat steel bar stock ofmass per unit length 4.73 kg/m. The frame is supported by a pinat C and by a cable AB. Determine (a) the tension in the ca6le(b) the reaction at C.

48 mm

54 mm 72mm

Fig. P5.137 Fig. P5.138

Fig. P5.t39

5.140 Determine by direct integration the centroid of the area shown.Express your answer in terms of a and h.

Fig. P5.l4O

0.6 m

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280 Distributed Forces: Centroids ond Centersof Grovity

5.t46 Consider the composite bodv shown. Determine (a) the value of rwhen h = LtZ, (b) tk ratioh/L for whrch t = L.

Fig. P5.146

5.147 Locate the center of gravity of the sheet-metal form shown.

5.148 Locate the centroid of the volume obtained by rotating the shadedarea about the r axis.

3 m

Fig. P5.147

Fig. P5.148

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292 Distributed Forces: Centroids ond CeniersOI \rrovlry

5.c4 Appronmate the curve shown using to srraightrine segments, andthen write a computer program that can be used to

-determine"the location

of the centroid of the line. use this program to determine the ]ocation of thec e n t r o i d w h e n ( a ) a : I i n . , L : I I i n . , h : 2 i n . ; ( b ) a : 2 i n . . L : 1 7 i n . .h = 4 in; (c) a : 5 in, L : 12 in., h : I in.

Fig. P5.C4

5.C5 Approximate the general spandrel shown using a series of n rectan-gles, each of width Aa and of the form bcc'b', and then write a compurerprogram that can be used to calculate the coordinates of the centroid o-f thearea. Use this program to locate the centroid when (a) m : 2, a : 80 mrn,h : 8 0 m m ; ( b ) m : 2 , a : 8 0 m m , h : 5 0 0 m m ; ( c ) m : 5 , a : 8 0 m r n ,h : 80mm; (d) m : 5, a : 80mm, h : 500 mm. In each case. comDarethe answers obtained to the exact values of - and I computed fromtheformulas given in Fig, 5.8A and determine the percent"g" 6rror.

Fig. P5.C5

5.C6 Solve Prob. 5.C5, using rectangles of the form bdrJ'b' .

*5.C7 A farmer asks a group of engineering students to determine thevolume of water in a small pond. using cord, the students first establish a2 x. z-ft grid across the pond and then record the depth of the water, infeet, at each interseciion point of the grid (see the accompanying table).Write a computer program that can be used to determine (a) the volume

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