Distributed Forces: Centroids and Centers of Gravity

5 H P T E R Distributed Forces: entroids nd enters of Grvit 219

pter 5 Distributed Forces: entroids nd enters of Grvit 5.1 Introduction 5.2 enter of Grvit of Two- Dimensionl od 5.3 entroids of res nd Lines 5.4 First Moments of res nd Lines 5.5 omposite Pltes nd Wires 5.6 Determintion of entroids b Integrtion 5.7 Teorems of Pppus-Guldinus 5.8 Distributed Lods on ems 5.9 Forces on Submerged Surfces 5.10 enter of Grvit of Tree- Dimensionl od. entroid of Volume 5.11 omposite odies 5.12 Determintion of entroids of Volumes b Integrtion 5.1 INTRDUTIN We ve ssumed so fr tt te ttrction eerted b te ert on rigid bod could be represented b single force W. Tis force, clled te force of grvit or te weigt of te bod, ws to be pplied t te center of grvit of te bod (Sec. 3.2). ctull, te ert eerts force on ec of te prticles forming te bod. Te ction of te ert on rigid bod sould tus be represented b lrge number of smll forces distributed over te entire bod. You will lern in tis cpter, owever, tt ll of tese smll forces cn be replced b single equivlent force W. You will lso lern ow to determine te center of grvit, i.e., te point of ppliction of te resultnt W, for bodies of vrious spes. In te first prt of te cpter, two-dimensionl bodies, suc s flt pltes nd wires contined in given plne, re considered. Two concepts closel ssocited wit te determintion of te center of grvit of plte or wire re introduced: te concept of te centroid of n re or line nd te concept of te first moment of n re or line wit respect to given is. You will lso lern tt te computtion of te re of surfce of revolution or of te volume of bod of revolution is directl relted to te determintion of te centroid of te line or re used to generte tt surfce or bod of revolution (Teorems of Pppus-Guldinus). nd, s is sown in Secs. 5.8 nd 5.9, te determintion of te centroid of n re simplifies te nlsis of bems subjected to distributed lods nd te computtion of te forces eerted on submerged rectngulr surfces, suc s drulic gtes nd portions of dms. In te lst prt of te cpter, ou will lern ow to determine te center of grvit of tree-dimensionl bod s well s te centroid of volume nd te first moments of tt volume wit respect to te coordinte plnes. RES ND LINES Poto 5.1 Te precise blncing of te components of mobile requires n understnding of centers of grvit nd centroids, te min topics of tis cpter. 5.2 ENTER F GRVITY F TW-DIMENSINL DY Let us first consider flt oriontl plte (Fig. 5.1). We cn divide te plte into n smll elements. Te coordintes of te first element W G = ΔW 220 ΣM : W = Σ ΔW ΣM : W = Σ ΔW Fig. 5.1 enter of grvit of plte.

re denoted b 1 nd 1, tose of te second element b 2 nd 2, etc. Te forces eerted b te ert on te elements of plte will be denoted, respectivel, b DW 1, DW 2,..., DW n. Tese forces or weigts re directed towrd te center of te ert; owever, for ll prcticl purposes te cn be ssumed to be prllel. Teir resultnt is terefore single force in te sme direction. Te mgnitude W of tis force is obtined b dding te mgnitudes of te elementl weigts. 5.2 enter of Grvit of Two-Dimensionl od 221 of : W 5 DW 1 1 DW 2 1??? 1 DW n To obtin te coordintes nd of te point G were te resultnt W sould be pplied, we write tt te moments of W bout te nd es re equl to te sum of te corresponding moments of te elementl weigts, om : om : W 5 1 DW 1 1 2 DW 2 1??? 1 n DW n W 5 1 DW 1 1 2 DW 2 1??? 1 n DW n (5.1) If we now increse te number of elements into wic te plte is divided nd simultneousl decrese te sie of ec element, we obtin in te limit te following epressions: W 5 # dw W 5 # dw W 5 # dw (5.2) Tese equtions define te weigt W nd te coordintes nd of te center of grvit G of flt plte. Te sme equtions cn be derived for wire ling in te plne (Fig. 5.2). We note tt te center of grvit G of wire is usull not locted on te wire. W = ΔW G ΣM : W = Σ ΔW ΣM : W = Σ ΔW Fig. 5.2 enter of grvit of wire.

222 Distributed Forces: entroids nd enters of Grvit 5.3 ENTRIDS F RES ND LINES In te cse of flt omogeneous plte of uniform tickness, te mgnitude DW of te weigt of n element of te plte cn be epressed s DW 5 gt D were g 5 specific weigt (weigt per unit volume) of te mteril t 5 tickness of te plte D 5 re of te element Similrl, we cn epress te mgnitude W of te weigt of te entire plte s W 5 gt were is te totl re of te plte. If U.S. customr units re used, te specific weigt g sould be epressed in lb/ft 3, te tickness t in feet, nd te res D nd in squre feet. We observe tt DW nd W will ten be epressed in pounds. If SI units re used, g sould be epressed in N/m 3, t in meters, nd te res D nd in squre meters; te weigts DW nd W will ten be epressed in newtons. Substituting for DW nd W in te moment equtions (5.1) nd dividing trougout b gt, we obtin om : 5 1 D 1 1 2 D 2 1??? 1 n D n om : 5 1 D 1 1 2 D 2 1??? 1 n D n If we increse te number of elements into wic te re is divided nd simultneousl decrese te sie of ec element, we obtin in te limit 5 # d 5 # d (5.3) Tese equtions define te coordintes nd of te center of grvit of omogeneous plte. Te point wose coordintes re nd is lso known s te centroid of te re of te plte (Fig. 5.3). If te plte is not omogeneous, tese equtions cnnot be used to determine te center of grvit of te plte; te still define, owever, te centroid of te re. In te cse of omogeneous wire of uniform cross section, te mgnitude DW of te weigt of n element of wire cn be epressed s DW 5 g DL were g 5 specific weigt of te mteril 5 cross-sectionl re of te wire DL 5 lengt of te element It sould be noted tt in te SI sstem of units given mteril is generll crcteried b its densit r (mss per unit volume) rter tn b its specific weigt g. Te specific weigt of te mteril cn ten be obtined from te reltion g 5 rg were g 5 9.81 m/s 2. Since r is epressed in kg/m 3, we observe tt g will be epressed in (kg/m 3 )(m/s 2 ), tt is, in N/m 3.

5.4 First Moments of res nd Lines 223 = Δ L = Δ L ΣM : = Σ Δ ΣM : L = Σ Δ L ΣM : = Σ Δ ΣM : L = Σ Δ L Fig. 5.3 entroid of n re. Fig. 5.4 entroid of line. Te center of grvit of te wire ten coincides wit te centroid of te line L defining te spe of te wire (Fig. 5.4). Te coordintes nd of te centroid of te line L re obtined from te equtions L 5 # dl L 5 # dl (5.4) 5.4 FIRST MMENTS F RES ND LINES Te integrl e d in Eqs. (5.3) of te preceding section is known s te first moment of te re wit respect to te is nd is denoted b Q. Similrl, te integrl e d defines te first moment of wit respect to te is nd is denoted b Q. We write Q 5 # d Q 5 # d (5.5) ompring Eqs. (5.3) wit Eqs. (5.5), we note tt te first moments of te re cn be epressed s te products of te re nd te coordintes of its centroid: Q 5 Q 5 (5.6) It follows from Eqs. (5.6) tt te coordintes of te centroid of n re cn be obtined b dividing te first moments of tt re b te re itself. Te first moments of te re re lso useful in mecnics of mterils for determining te sering stresses in bems under trnsverse lodings. Finll, we observe from Eqs. (5.6) tt if te centroid of n re is locted on coordinte is, te first moment of te re wit respect to tt is is ero. onversel, if te first moment of n re wit respect to coordinte is is ero, ten te centroid of te re is locted on tt is. Reltions similr to Eqs. (5.5) nd (5.6) cn be used to define te first moments of line wit respect to te coordinte es nd

224 Distributed Forces: entroids nd enters of Grvit Fig. 5.5 P d' () (b) d ' P' to epress tese moments s te products of te lengt L of te line nd te coordintes nd of its centroid. n re is sid to be smmetric wit respect to n is 9 if for ever point P of te re tere eists point P9 of te sme re suc tt te line PP9 is perpendiculr to 9 nd is divided into two equl prts b tt is (Fig. 5.5). line L is sid to be smmetric wit respect to n is 9 if it stisfies similr conditions. Wen n re or line L possesses n is of smmetr 9, its first moment wit respect to 9 is ero, nd its centroid is locted on tt is. For emple, in te cse of te re of Fig. 5.5b, wic is smmetric wit respect to te is, we observe tt for ever element of re d of bsciss tere eists n element d9 of equl re nd wit bsciss 2. It follows tt te integrl in te first of Eqs. (5.5) is ero nd, tus, tt Q 5 0. It lso follows from te first of te reltions (5.3) tt 5 0. Tus, if n re or line L possesses n is of smmetr, its centroid is locted on tt is. We furter note tt if n re or line possesses two es of smmetr, its centroid must be locted t te intersection of te two es (Fig. 5.6). Tis propert enbles us to determine immeditel te centroid of res suc s circles, ellipses, squres, rectngles, equilterl tringles, or oter smmetric figures s well s te centroid of lines in te spe of te circumference of circle, te perimeter of squre, etc. D' D D' D ' ' () (b) Fig. 5.6 Fig. 5.7 d' d n re is sid to be smmetric wit respect to center if for ever element of re d of coordintes nd tere eists n element d9 of equl re wit coordintes 2 nd 2 (Fig. 5.7). It ten follows tt te integrls in Eqs. (5.5) re bot ero nd tt Q 5 Q 5 0. It lso follows from Eqs. (5.3) tt 5 5 0, tt is, tt te centroid of te re coincides wit its center of smmetr. Similrl, if line possesses center of smmetr, te centroid of te line will coincide wit te center. It sould be noted tt figure possessing center of smmetr does not necessril possess n is of smmetr (Fig. 5.7), wile figure possessing two es of smmetr does not necessril possess center of smmetr (Fig. 5.6). However, if figure possesses two es of smmetr t rigt ngle to ec oter, te point of intersection of tese es is center of smmetr (Fig. 5.6b). Determining te centroids of unsmmetricl res nd lines nd of res nd lines possessing onl one is of smmetr will be discussed in Secs. 5.6 nd 5.7. entroids of common spes of res nd lines re sown in Fig. 5.8 nd.

5.4 First Moments of res nd Lines 225 Spe re Tringulr re 3 b 2 b 2 b 2 Qurter-circulr re Semicirculr re r 4r 3 0 4r 3 4r 3 r 2 4 r 2 2 Qurter-ellipticl re Semiellipticl re b 4 3 0 4b 3 4b 3 b 4 b 2 Semiprbolic re Prbolic re 3 8 0 3 5 3 5 2 3 4 3 Prbolic spndrel = k 2 3 4 3 10 3 Generl spndrel = k n n + 1 n + 2 n + 1 4n + 2 n + 1 r irculr sector 2r sin α 3α 0 αr 2 Fig. 5.8 entroids of common spes of res.

226 Distributed Forces: entroids nd enters of Grvit Spe Lengt Qurter-circulr rc Semicirculr rc r 2r 0 2r 2r r 2 r r rc of circle r sin 0 2r Fig. 5.8 entroids of common spes of lines. 5.5 MPSITE PLTES ND WIRES In mn instnces, flt plte cn be divided into rectngles, tringles, or te oter common spes sown in Fig. 5.8. Te bsciss X of its center of grvit G cn be determined from te bscisss 1, 2,..., n of te centers of grvit of te vrious prts b epressing tt te moment of te weigt of te wole plte bout te is is equl to te sum of te moments of te weigts of te vrious prts bout te sme is (Fig. 5.9). Te ordinte Y of te center of grvit of te plte is found in similr w b equting moments bout te is. We write M : X(W 1 1 W 2 1... 1 W n ) 5 1 W 1 1 2 W 2 1... 1 n W n M : Y(W 1 1 W 2 1... 1 W n ) 5 1 W 1 1 2 W 2 1... 1 n W n ΣW = W 3 W 1 W2 G 3 X G Y G 1 G 2 ΣM : X Σ W = Σ W ΣM : Y Σ W = Σ W Fig. 5.9 enter of grvit of composite plte.

or, for sort, 5.5 omposite Pltes nd Wires 227 X W 5 W Y W 5 W (5.7) Tese equtions cn be solved for te coordintes X nd Y of te center of grvit of te plte. 3 X Σ = 1 3 2 Y 1 2 Q = X Σ = Σ Q = Y Σ = Σ Fig. 5.10 entroid of composite re. If te plte is omogeneous nd of uniform tickness, te center of grvit coincides wit te centroid of its re. Te bsciss X of te centroid of te re cn be determined b noting tt te first moment Q of te composite re wit respect to te is cn be epressed bot s te product of X nd te totl re nd s te sum of te first moments of te elementr res wit respect to te is (Fig. 5.10). Te ordinte Y of te centroid is found in similr w b considering te first moment Q of te composite re. We ve 1 W 1 W 2 2 W 3 Q 5 X( 1 1 2 1... 1 n ) 5 1 1 1 2 2 1... 1 n n Q 5 Y( 1 1 2 1... 1 n ) 5 1 1 1 2 2 1... 1 n n 3 or, for sort, Q 5 X 5 Q 5 Y 5 (5.8) Tese equtions ield te first moments of te composite re, or te cn be used to obtin te coordintes X nd Y of its centroid. re sould be tken to ssign te pproprite sign to te moment of ec re. First moments of res, like moments of forces, cn be positive or negtive. For emple, n re wose centroid is locted to te left of te is will ve negtive first moment wit respect to tt is. lso, te re of ole sould be ssigned negtive sign (Fig. 5.11). Similrl, it is possible in mn cses to determine te center of grvit of composite wire or te centroid of composite line b dividing te wire or line into simpler elements (see Smple Prob. 5.2). Fig. 5.11 1 1 2 3 2 3 1 Semicircle + 2 Full rectngle + + + 3 irculr ole +

80 mm 120 mm 60 mm 40 mm SMPLE PRLEM 5.1 For te plne re sown, determine () te first moments wit respect to te nd es, (b) te loction of te centroid. 60 mm SLUTIN omponents of re. Te re is obtined b dding rectngle, tringle, nd semicircle nd b ten subtrcting circle. Using te coordinte es sown, te re nd te coordintes of te centroid of ec of te component res re determined nd entered in te tble below. Te re of te circle is indicted s negtive, since it is to be subtrcted from te oter res. We note tt te coordinte of te centroid of te tringle is negtive for te es sown. Te first moments of te component res wit respect to te coordinte es re computed nd entered in te tble. 120 mm r 1 = 60 mm r 2 = 40 mm 60 mm = + + 40 mm 4r 1 = 25.46 mm 3 r 1 = 60 mm _ r 2 = 40 mm 80 mm 60 mm 80 mm 105.46 mm 80 mm 40 mm 20 mm 60 mm 60 mm omponent, mm 2, mm, mm, mm 3, mm 3 Rectngle (120)(80) 5 9.6 3 10 3 60 40 1576 3 10 3 1384 3 10 3 1 Tringle 2(120)(60) 5 3.6 3 10 3 40 220 1144 3 10 3 272 3 10 3 1 Semicircle 2p(60) 2 5 5.655 3 10 3 60 105.46 1339.3 3 10 3 1596.4 3 10 3 ircle 2p(40) 2 5 25.027 3 10 3 60 80 2301.6 3 10 3 2402.2 3 10 3 o 5 13.828 3 10 3 o 5 1757.7 3 10 3 o 5 1506.2 3 10 3 X = 54.8 mm Y = 36.6 mm. First Moments of te re. Using Eqs. (5.8), we write Q 5 5 506.2 3 10 3 mm 3 Q 5 506 3 10 3 mm 3 Q 5 5 757.7 3 10 3 mm 3 Q 5 758 3 10 3 mm 3 b. Loction of entroid. Substituting te vlues given in te tble into te equtions defining te centroid of composite re, we obtin X 5 : X(13.828 3 10 3 mm 2 ) 5 757.7 3 10 3 mm 3 X 5 54.8 mm Y 5 : Y(13.828 3 10 3 mm 2 ) 5 506.2 3 10 3 mm 3 Y 5 36.6 mm 228

10 in. 26 in. SMPLE PRLEM 5.2 Te figure sown is mde from piece of tin, omogeneous wire. Determine te loction of its center of grvit. 24 in. SLUTIN 10 in. 12 in. 26 in. 24 in. 5 in. Since te figure is formed of omogeneous wire, its center of grvit coincides wit te centroid of te corresponding line. Terefore, tt centroid will be determined. oosing te coordinte es sown, wit origin t, we determine te coordintes of te centroid of ec line segment nd compute te first moments wit respect to te coordinte es. Segment L, in., in., in. L, in 2 L, in 2 24 12 0 288 0 26 12 5 312 130 10 0 5 0 50 ol 5 60 L 5 600 L 5 180 Substituting te vlues obtined from te tble into te equtions defining te centroid of composite line, we obtin X L 5 L: X(60 in.) 5 600 in 2 X 5 10 in. Y L 5 L: Y(60 in.) 5 180 in 2 Y 5 3 in. 229

r SMPLE PRLEM 5.3 uniform semicirculr rod of weigt W nd rdius r is ttced to pin t nd rests ginst frictionless surfce t. Determine te rections t nd. SLUTIN 2r 2r G Free-od Digrm. free-bod digrm of te rod is drwn. Te forces cting on te rod re its weigt W, wic is pplied t te center of grvit G (wose position is obtined from Fig. 5.8); rection t, represented b its components nd ; nd oriontl rection t. Equilibrium Equtions 1l om 5 0: (2r) 2 W 2r p b 5 0 W 51 W p 1 F 5 0: 1 5 0 5 W p = W 52 52 W p 5 W p = W 1 F 5 0: 2 W 5 0 5 W dding te two components of te rection t : 5 c W 2 1 W 2 1/2 p b d 5 W 1 1 1 p 2b 1/2 tn 5 W W/p 5 p 5 tn21 p Te nswers cn lso be epressed s follows: 5 1.049W b72.3 5 0.318W 230

SLVING PRLEMS N YUR WN In tis lesson we developed te generl equtions for locting te centers of grvit of two-dimensionl bodies nd wires [Eqs. (5.2)] nd te centroids of plne res [Eqs. (5.3)] nd lines [Eqs. (5.4)]. In te following problems, ou will ve to locte te centroids of composite res nd lines or determine te first moments of te re for composite pltes [Eqs. (5.8)]. 1. Locting te centroids of composite res nd lines. Smple Problems 5.1 nd 5.2 illustrte te procedure ou sould follow wen solving problems of tis tpe. Tere re, owever, severl points tt sould be empsied.. Te first step in our solution sould be to decide ow to construct te given re or line from te common spes of Fig. 5.8. You sould recognie tt for plne res it is often possible to construct prticulr spe in more tn one w. lso, sowing te different components (s is done in Smple Prob. 5.1) will elp ou to correctl estblis teir centroids nd res or lengts. Do not forget tt ou cn subtrct res s well s dd tem to obtin desired spe. b. We strongl recommend tt for ec problem ou construct tble contining te res or lengts nd te respective coordintes of te centroids. It is essentil for ou to remember tt res wic re removed (for emple, oles) re treted s negtive. lso, te sign of negtive coordintes must be included. Terefore, ou sould lws crefull note te loction of te origin of te coordinte es. c. Wen possible, use smmetr [Sec. 5.4] to elp ou determine te loction of centroid. d. In te formuls for te circulr sector nd for te rc of circle in Fig. 5.8, te ngle must lws be epressed in rdins. 2. lculting te first moments of n re. Te procedures for locting te centroid of n re nd for determining te first moments of n re re similr; owever, for te ltter it is not necessr to compute te totl re. lso, s noted in Sec. 5.4, ou sould recognie tt te first moment of n re reltive to centroidl is is ero. 3. Solving problems involving te center of grvit. Te bodies considered in te following problems re omogeneous; tus, teir centers of grvit nd centroids coincide. In ddition, wen bod tt is suspended from single pin is in equilibrium, te pin nd te bod s center of grvit must lie on te sme verticl line. It m pper tt mn of te problems in tis lesson ve little to do wit te stud of mecnics. However, being ble to locte te centroid of composite spes will be essentil in severl topics tt ou will soon encounter. 231

PRLEMS 5.1 troug 5.9 Locte te centroid of te plne re sown. 30 mm 20 mm 30 mm 300 mm 36 mm 12 in. 21 in. 30 mm 240 mm 24 mm 15 in. Fig. P5.1 Fig. P5.2 Fig. P5.3 3 in. 6 in. 6 in. 6 in. 8 in. 8 in. 120 mm 6 in. 6 in. r = 4 in. 12 in. r = 75 mm Fig. P5.4 Fig. P5.5 Fig. P5.6 r = 15 in. 20 in. 60 mm 16 in. 20 in. r = 38 in. 30 in. 30 in. 60 mm Fig. P5.7 232 Fig. P5.8 Fig. P5.9

5.10 troug 5.15 Locte te centroid of te plne re sown. Problems 233 Semiellipse 47 mm 47 mm 26 mm 50 mm Verte Prbol 70 mm r 2 = 12 in. r 1 = 8 in. 15 mm 80 mm Fig. P5.10 Fig. P5.11 Fig. P5.12 Verte Prbol = k 2 60 mm r 20 mm 20 mm = k 2 30 mm 20 in. 20 in. = k 2 60 mm 75 mm Fig. P5.13 Fig. P5.14 Fig. P5.15 5.16 Determine te coordinte of te centroid of te sded re in terms of r 1, r 2, nd. α r 1 r 2 α Fig. P5.16 nd P5.17 b = k 2 5.17 Sow tt s r 1 pproces r 2, te loction of te centroid pproces tt for n rc of circle of rdius (r 1 1 r 2 )/2. 5.18 For te re sown, determine te rtio /b for wic 5. Fig. P5.18 5.19 For te seminnulr re of Prob. 5.11, determine te rtio r 2 /r 1 so tt 5 3r 1 /4.

234 Distributed Forces: entroids nd enters of Grvit 5.20 composite bem is constructed b bolting four pltes to four 60 3 60 3 12-mm ngles s sown. Te bolts re equll spced long te bem, nd te bem supports verticl lod. s proved in mecnics of mterils, te sering forces eerted on te bolts t nd re proportionl to te first moments wit respect to te centroidl is of te red sded res sown, respectivel, in prts nd b of te figure. Knowing tt te force eerted on te bolt t is 280 N, determine te force eerted on te bolt t. 300 mm 12 mm 60 mm 12 mm 450 mm 60 mm 12 mm 12 mm Fig. P5.20 () (b) 7.5 in. 1 2 4.5 in. Fig. P5.21 4.5 in. 5.21 nd 5.22 Te oriontl is is drwn troug te centroid of te re sown, nd it divides te re into two component res 1 nd 2. Determine te first moment of ec component re wit respect to te is, nd eplin te results obtined. 1.50 in. 0.75 in. 1.50 in. 2.00 in. 1 4.00 in. 1.50 in. 2 2.00 in. 0.75 in. Fig. P5.22 2.00 in.

5.23 Te first moment of te sded re wit respect to te is is denoted b Q. () Epress Q in terms of b, c, nd te distnce from te bse of te sded re to te is. (b) For wt vlue of is Q mimum, nd wt is tt mimum vlue? Problems 235 c c b Fig. P5.23 5.24 troug 5.27 tin, omogeneous wire is bent to form te perimeter of te figure indicted. Locte te center of grvit of te wire figure tus formed. 5.24 Fig. P5.1. 5.25 Fig. P5.2. 5.26 Fig. P5.3. 5.27 Fig. P5.7. r 5.28 uniform circulr rod of weigt 8 lb nd rdius 10 in. is ttced to pin t nd to te cble. Determine () te tension in te cble, (b) te rection t. 5.29 Member DE is component of mobile nd is formed from single piece of luminum tubing. Knowing tt te member is supported t nd tt l 5 2 m, determine te distnce d so tt portion D of te member is oriontl. d 1.50 m D Fig. P5.28 0.75 m 55 55 l E Fig. P5.29 nd P5.30 5.30 Member DE is component of mobile nd is formed from single piece of luminum tubing. Knowing tt te member is supported t nd tt d is 0.50 m, determine te lengt l of rm DE so tt tis portion of te member is oriontl.

236 Distributed Forces: entroids nd enters of Grvit 5.31 Te omogeneous wire is bent into semicirculr rc nd strigt section s sown nd is ttced to inge t. Determine te vlue of u for wic te wire is in equilibrium for te indicted position. q r 5.32 Determine te distnce for wic te centroid of te sded re is s fr bove line 9 s possible wen () k 5 0.10, (b) k 5 0.80. r Fig. P5.31 kb ' b Fig. P5.32 nd P5.33 5.33 Knowing tt te distnce s been selected to mimie te distnce from line 9 to te centroid of te sded re, sow tt 5 2/3. 5.6 DETERMINTIN F ENTRIDS Y INTEGRTIN Te centroid of n re bounded b nlticl curves (i.e., curves defined b lgebric equtions) is usull determined b evluting te integrls in Eqs. (5.3) of Sec. 5.3: 5 # d 5 # d (5.3) If te element of re d is smll rectngle of sides d nd d, te evlution of ec of tese integrls requires double integrtion wit respect to nd. double integrtion is lso necessr if polr coordintes re used for wic d is smll element of sides dr nd r du. In most cses, owever, it is possible to determine te coordintes of te centroid of n re b performing single integrtion. Tis is cieved b coosing d to be tin rectngle or strip or tin sector or pie-sped element (Fig. 5.12); te centroid of te tin rectngle is locted t its center, nd te centroid of te tin sector is locted t distnce 2 3 r from its verte (s it is for tringle). Te coordintes of te centroid of te re under considertion re ten obtined b epressing tt te first moment of te entire re wit respect to ec of te coordinte es is equl to te sum (or integrl) of te corresponding moments of te elements of re.

Denoting b el nd el te coordintes of te centroid of te element d, we write 5.6 Determintion of entroids b Integrtion 237 Q 5 5 # el d Q 5 5 # el d (5.9) If te re is not lred known, it cn lso be computed from tese elements. Te coordintes el nd el of te centroid of te element of re d sould be epressed in terms of te coordintes of point locted on te curve bounding te re under considertion. lso, te re of te element d sould be epressed in terms of te coordintes of tt point nd te pproprite differentils. Tis s been done in Fig. 5.12 for tree common tpes of elements; te pie-sped element of prt c sould be used wen te eqution of te curve bounding te re is given in polr coordintes. Te pproprite epressions sould be substituted into formuls (5.9), nd te eqution of te bounding curve sould be used to epress one of te coordintes in terms of te oter. Te integrtion is tus reduced to single integrtion. nce te re s been determined nd te integrls in Eqs. (5.9) ve been evluted, tese equtions cn be solved for te coordintes nd of te centroid of te re. Wen line is defined b n lgebric eqution, its centroid cn be determined b evluting te integrls in Eqs. (5.4) of Sec. 5.3: L 5 # dl L 5 # dl (5.4) P(, ) P(, ) el d el el = el = /2 el = + 2 el = d = d el d = ( ) d () (b) Fig. 5.12 entroids nd res of differentil elements. el d r 2r 3 θ el el el = 2r cosθ 3 el = 2r sinθ 3 d = 1 r 2 2 dθ (c) P( θ, r)

238 Distributed Forces: entroids nd enters of Grvit Te differentil lengt dl sould be replced b one of te following epressions, depending upon wic coordinte,,, or u, is cosen s te independent vrible in te eqution used to define te line (tese epressions cn be derived using te Ptgoren teorem): dl 5 1 1 d d b 2 d dl 5 1 1 d d b 2 d dl 5 r 2 1 dr du b 2 du fter te eqution of te line s been used to epress one of te coordintes in terms of te oter, te integrtion cn be performed, nd Eqs. (5.4) cn be solved for te coordintes nd of te centroid of te line. 5.7 THEREMS F PPPUS-GULDINUS Tese teorems, wic were first formulted b te Greek geometer Pppus during te tird centur.d. nd lter restted b te Swiss mtemticin Guldinus, or Guldin, (1577 1643) del wit surfces nd bodies of revolution. surfce of revolution is surfce wic cn be generted b rotting plne curve bout fied is. For emple (Fig. 5.13), te Poto 5.2 Te storge tnks sown re ll bodies of revolution. Tus, teir surfce res nd volumes cn be determined using te teorems of Pppus-Guldinus. Spere Fig. 5.13 one Torus surfce of spere cn be obtined b rotting semicirculr rc bout te dimeter, te surfce of cone cn be pro duced b rotting strigt line bout n is, nd te surfce of torus or ring cn be generted b rotting te circumference of circle bout nonintersecting is. bod of revolution is bod wic cn be generted b rotting plne re bout fied is. s sown in Fig. 5.14, spere, cone, nd torus cn ec be generted b rotting te pproprite spe bout te indicted is. Spere one Torus Fig. 5.14 THEREM I. Te re of surfce of revolution is equl to te lengt of te generting curve times te distnce trveled b te centroid of te curve wile te surfce is being generted. Proof. onsider n element dl of te line L (Fig. 5.15), wic is revolved bout te is. Te re d generted b te element

dl L 5.7 Teorems of Pppus-Guldinus 239 d Fig. 5.15 2 dl is equl to 2p dl. Tus, te entire re generted b L is 5 e 2p dl. Reclling tt we found in Sec. 5.3 tt te integrl e dl is equl to L, we terefore ve 5 2pL (5.10) were 2p is te distnce trveled b te centroid of L (Fig. 5.15). It sould be noted tt te generting curve must not cross te is bout wic it is rotted; if it did, te two sections on eiter side of te is would generte res ving opposite signs, nd te teorem would not ppl. THEREM II. Te volume of bod of revolution is equl to te generting re times te distnce trveled b te centroid of te re wile te bod is being generted. Proof. onsider n element d of te re wic is revolved bout te is (Fig. 5.16). Te volume dv generted b te element d is equl to 2p d. Tus, te entire volume generted b is V 5 e 2p d, nd since te integrl e d is equl to (Sec. 5.3), we ve V 5 2p (5.11) d dv 2 Fig. 5.16 were 2p is te distnce trveled b te centroid of. gin, it sould be noted tt te teorem does not ppl if te is of rottion intersects te generting re. Te teorems of Pppus-Guldinus offer simple w to compute te res of surfces of revolution nd te volumes of bodies of revolution. onversel, te cn lso be used to determine te centroid of plne curve wen te re of te surfce generted b te curve is known or to determine te centroid of plne re wen te volume of te bod generted b te re is known (see Smple Prob. 5.8).

= k 2 b SMPLE PRLEM 5.4 Determine b direct integrtion te loction of te centroid of prbolic spndrel. SLUTIN Determintion of te onstnt k. Te vlue of k is determined b substituting 5 nd 5 b into te given eqution. We ve b 5 k 2 or k 5 b/ 2. Te eqution of te curve is tus d = d 5 b 2 2 or 5 1/2 1/2 b Verticl Differentil Element. We coose te differentil element sown nd find te totl re of te figure. el = 2 el = 5 # d 5 # d 5 # 0 b 2 2 d 5 c b 3 2 3 d 0 5 b 3 Te first moment of te differentil element wit respect to te is is el d; ence, te first moment of te entire re wit respect to tis is is Q 5 # el d 5 # d 5 # Since Q 5, we ve 5 # el d 0 b 2 2 b d 5 c b 2 4 4 d 0 5 2 b 4 b 3 5 2 b 4 5 3 4 Likewise, te first moment of te differentil element wit respect to te is is el d, nd te first moment of te entire re is Q 5 # el d 5 # 2 d 5 # 1 2 b 2 2 b 2d 5 c b2 5 2 4 5 d 5 b2 10 Since Q 5, we ve 0 0 5 # el d b 3 5 b2 10 5 3 10 b d = ( ) d el = + 2 b el = Horiontl Differentil Element. Te sme results cn be obtined b considering oriontl element. Te first moments of te re re Q 5 # el d 5 # 1 b ( 2 ) d 5 2 # 0 5 1 2 # b 0 2 2 2 b b d 5 2 b 4 Q 5 # el d 5 # ( 2 ) d 5 # 2 b 5 # 2 b 1/2 3/2 b d 5 b2 10 0 2 2 2 d 2 b 1/2 1/2 b d To determine nd, te epressions obtined re gin substituted into te equtions defining te centroid of te re. 240

SMPLE PRLEM 5.5 r Determine te loction of te centroid of te rc of circle sown. α α SLUTIN Since te rc is smmetricl wit respect to te is, 5 0. differentil element is cosen s sown, nd te lengt of te rc is determined b integrtion. r dθ θ = α dl = r dθ L 5 # dl 5 # 2 r du 5 r # 2 du 5 2r Te first moment of te rc wit respect to te is is Q 5 # dl 5 # 2 5 r 2 3sin u4 2 5 2r 2 sin (r cos u)(r du) 5 r 2 # 2 cos u du θ = r cosθ Since Q 5 L, we write (2r) 5 2r 2 sin 5 r sin θ = α SMPLE PRLEM 5.6 r Determine te re of te surfce of revolution sown, wic is obtined b rotting qurter-circulr rc bout verticl is. 2r SLUTIN ccording to Teorem I of Pppus-Guldinus, te re generted is equl to te product of te lengt of te rc nd te distnce trveled b its centroid. Referring to Fig. 5.8, we ve 2r 2r 5 2r 2 2r p 5 2r 1 2 1 p b 5 2pL 5 2p c 2r 1 2 1 pr bd p 2 b 5 2pr 2 (p 2 1) 241

100 mm 20 mm SMPLE PRLEM 5.7 30 mm 400 mm 60 mm 20 mm 20 mm Te outside dimeter of pulle is 0.8 m, nd te cross section of its rim is s sown. Knowing tt te pulle is mde of steel nd tt te densit of steel is r 5 7.85 3 10 3 kg/m 3, determine te mss nd te weigt of te rim. 100 mm 60 mm I 50 mm I 30 mm II _ II 375 mm 365 mm SLUTIN Te volume of te rim cn be found b ppling Teorem II of Pppus- Guldinus, wic sttes tt te volume equls te product of te given cross-sectionl re nd te distnce trveled b its centroid in one complete revolution. However, te volume cn be more esil determined if we observe tt te cross section cn be formed from rectngle I, wose re is positive, nd rectngle II, wose re is negtive. Distnce Trveled re, mm 2, mm b, mm Volume, mm 3 I 15000 375 2p(375) 5 2356 (5000)(2356) 5 11.78 3 10 6 II 21800 365 2p(365) 5 2293 (21800)(2293) 5 24.13 3 10 6 Volume of rim 5 7.65 3 10 6 Since 1 mm 5 10 23 m, we ve 1 mm 3 5 (10 23 m) 3 5 10 29 m 3, nd we ob - tin V 5 7.65 3 10 6 mm 3 5 (7.65 3 10 6 )(10 29 m 3 ) 5 7.65 3 10 23 m 3. m 5 rv 5 (7.85 3 10 3 kg/m 3 )(7.65 3 10 23 m 3 ) W 5 mg 5 (60.0 kg)(9.81 m/s 2 ) 5 589 kg? m/s 2 m 5 60.0 kg W 5 589 N SMPLE PRLEM 5.8 Using te teorems of Pppus-Guldinus, determine () te centroid of semicirculr re, (b) te centroid of semicirculr rc. We recll tt te volume nd te surfce re of spere re 4 3pr 3 nd 4pr 2, respectivel. r = r 2 2 L = r SLUTIN Te volume of spere is equl to te product of te re of semicircle nd te distnce trveled b te centroid of te semicircle in one revolution bout te is. 4 V 5 2p 3pr 3 5 2p( 1 2pr 2 ) 5 4r 3p Likewise, te re of spere is equl to te product of te lengt of te generting semicircle nd te distnce trveled b its centroid in one revolution. r 5 2pL 4pr 2 5 2p(pr) 5 2r p 242

SLVING PRLEMS N YUR WN In te problems for tis lesson, ou will use te equtions 5 # d 5 # d (5.3) L 5 # dl L 5 # dl (5.4) to locte te centroids of plne res nd lines, respectivel. You will lso ppl te teorems of Pppus-Guldinus (Sec. 5.7) to determine te res of surfces of revolution nd te volumes of bodies of revolution. 1. Determining b direct integrtion te centroids of res nd lines. Wen solving problems of tis tpe, ou sould follow te metod of solution sown in Smple Probs. 5.4 nd 5.5: compute or L, determine te first moments of te re or te line, nd solve Eqs. (5.3) or (5.4) for te coordintes of te centroid. In ddition, ou sould p prticulr ttention to te following points.. egin our solution b crefull defining or determining ec term in te pplicble integrl formuls. We strongl encourge ou to sow on our sketc of te given re or line our coice for d or dl nd te distnces to its centroid. b. s eplined in Sec. 5.6, te nd te in te bove equtions represent te coordintes of te centroid of te differentil elements d nd dl. It is importnt to recognie tt te coordintes of te centroid of d re not equl to te coordintes of point locted on te curve bounding te re under considertion. You sould crefull stud Fig. 5.12 until ou full understnd tis importnt point. c. To possibl simplif or minimie our computtions, lws emine te spe of te given re or line before defining te differentil element tt ou will use. For emple, sometimes it m be preferble to use oriontl rectngulr elements insted of verticl ones. lso, it will usull be dvntgeous to use polr coordintes wen line or n re s circulr smmetr. d. ltoug most of te integrtions in tis lesson re strigtforwrd, t times it m be necessr to use more dvnced tecniques, suc s trigonometric substitution or integrtion b prts. f course, using tble of integrls is te fstest metod to evlute difficult integrls. 2. ppling te teorems of Pppus-Guldinus. s sown in Smple Probs. 5.6 troug 5.8, tese simple, et ver useful teorems llow ou to ppl our knowledge of centroids to te computtion of res nd volumes. ltoug te teorems refer to te distnce trveled b te centroid nd to te lengt of te generting curve or to te generting re, te resulting equtions [Eqs. (5.10) nd (5.11)] contin te products of tese quntities, wic re simpl te first moments of line ( L) nd n re ( ), respectivel. Tus, for tose problems for wic te generting line or re consists of more tn one common spe, ou need onl determine L or ; ou do not ve to clculte te lengt of te generting curve or te generting re. 243

PRLEMS 5.34 troug 5.36 Determine b direct integrtion te centroid of te re sown. Epress our nswer in terms of nd. = m = k 2 = k 3 Fig. P5.34 Fig. P5.35 Fig. P5.36 5.37 troug 5.39 Determine b direct integrtion te centroid of te re sown. 2 b 2 + 2 = 1 2 b 2 r 1 r 2 2 Fig. P5.37 Fig. P5.38 Fig. P5.39 5.40 nd 5.41 Determine b direct integrtion te centroid of te re sown. Epress our nswer in terms of nd b. b = k( ) 2 1 = k 1 2 b 2 = k 2 4 244 Fig. P5.40 Fig. P5.41

5.42 Determine b direct integrtion te centroid of te re sown. Problems 245 5.43 nd 5.44 Determine b direct integrtion te centroid of te re sown. Epress our nswer in terms of nd b. ( = 1 L + 2 L ) 2 = k 2 b 2 = k 2 b Fig. P5.42 L L 2 2 b 2 b Fig. P5.43 Fig. P5.44 5.45 nd 5.46 omogeneous wire is bent into te spe sown. Determine b direct integrtion te coordinte of its centroid. = cos 3 θ = sin 3 θ π 0 θ 2 r 45 45 Fig. P5.45 Fig. P5.46 3 2 = k *5.47 omogeneous wire is bent into te spe sown. Determine b direct integrtion te coordinte of its centroid. Epress our nswer in terms of. *5.48 nd *5.49 Determine b direct integrtion te centroid of te re sown. Fig. P5.47 = cos p 2L r = e q L 2 L 2 q Fig. P5.48 Fig. P5.49

246 Distributed Forces: entroids nd enters of Grvit = (1 1 ) 5.50 Determine te centroid of te re sown wen 5 2 in. 5.51 Determine te vlue of for wic te rtio is 9. 5.52 Determine te volume nd te surfce re of te solid obtined b rotting te re of Prob. 5.1 bout () te line 5 240 mm, (b) te is. 1 in. Fig. P5.50 nd P5.51 5.53 Determine te volume nd te surfce re of te solid obtined b rotting te re of Prob. 5.2 bout () te line 5 60 mm, (b) te is. 5.54 Determine te volume nd te surfce re of te solid obtined b rotting te re of Prob. 5.8 bout () te is, (b) te is. ' 5.55 Determine te volume of te solid generted b rotting te prbolic re sown bout () te is, (b) te is 9. 5.56 Determine te volume nd te surfce re of te cin link sown, wic is mde from 6-mm-dimeter br, if R 5 10 mm nd L 5 30 mm. Fig. P5.55 R 90 1 1 4 in. R Fig. P5.56 L 1 in. 5.57 Verif tt te epressions for te volumes of te first four spes in Fig. 5.21 on pge 260 re correct. Fig. P5.58 3 4 in. 5.58 3 4-in.-dimeter ole is drilled in piece of 1-in.-tick steel; te ole is ten countersunk s sown. Determine te volume of steel removed during te countersinking process. 5.59 Determine te cpcit, in liters, of te punc bowl sown if R 5 250 mm. R R Fig. P5.59

5.60 Tree different drive belt profiles re to be studied. If t n given time ec belt mkes contct wit one-lf of te circumference of its pulle, determine te contct re between te belt nd te pulle for ec design. Problems 247 40 0.625 in. 40 0.08 in. r = 0.25 in. 0.125 in. 0.375 in. 3 in. 3 in. 3 in. Fig. P5.60 () (b) (c) 5.61 Te luminum sde for te smll ig-intensit lmp sown s uniform tickness of 1 mm. Knowing tt te densit of luminum is 2800 kg/m 3, determine te mss of te sde. 56 mm 32 mm 26 mm 66 mm 75 mm 25 mm 75 mm 26 26 Fig. P5.62 Fig. P5.61 32 mm 28 mm 8 mm 0.50 in. 0.625 in. r = 0.875 in. r = 0.1875 in. 1.00 in. 5.62 Te escutceon ( decortive plte plced on pipe were te pipe eits from wll) sown is cst from brss. Knowing tt te densit of brss is 8470 kg/m 3, determine te mss of te escutceon. 5.63 mnufcturer is plnning to produce 20,000 wooden pegs ving te spe sown. Determine ow mn gllons of pint sould be ordered, knowing tt ec peg will be given two cots of pint nd tt one gllon of pint covers 100 ft 2. 5.64 Te wooden peg sown is turned from dowel 1 in. in dimeter nd 4 in. long. Determine te percentge of te initil volume of te dowel tt becomes wste. *5.65 Te sde for wll-mounted ligt is formed from tin seet of trnslucent plstic. Determine te surfce re of te outside of te sde, knowing tt it s te prbolic cross section sown. 3.00 in. 0.50 in. Fig. P5.63 nd P5.64 100 mm 250 mm = k 2 Fig. P5.65

248 Distributed Forces: entroids nd enters of Grvit = w w L d W w d () (b) dw = d P Fig. 5.17 L W W = *5.8 DISTRIUTED LDS N EMS Te concept of te centroid of n re cn be used to solve oter problems besides tose deling wit te weigts of flt pltes. onsider, for emple, bem supporting distributed lod; tis lod m consist of te weigt of mterils supported directl or indirectl b te bem, or it m be cused b wind or drosttic pressure. Te distributed lod cn be represented b plotting te lod w supported per unit lengt (Fig. 5.17); tis lod is epressed in N/m or in lb/ft. Te mgnitude of te force eerted on n element of bem of lengt d is dw 5 w d, nd te totl lod supported b te bem is L W 5 # w d 0 We observe tt te product w d is equl in mgnitude to te element of re d sown in Fig. 5.17. Te lod W is tus equl in mgnitude to te totl re under te lod curve: W 5 # d 5 We now determine were single concentrted lod W, of te sme mgnitude W s te totl distributed lod, sould be pplied on te bem if it is to produce te sme rections t te supports (Fig. 5.17b). However, tis concentrted lod W, wic represents te resultnt of te given distributed loding, is equivlent to te loding onl wen considering te free-bod digrm of te entire bem. Te point of ppliction P of te equivlent concentrted lod W is obtined b epressing tt te moment of W bout point is equl to te sum of te moments of te elementl lods dw bout : (P)W 5 # dw or, since dw 5 w d 5 d nd W 5, L (P) 5 # d (5.12) 0 Poto 5.3 Te roofs of te buildings sown must be ble to support not onl te totl weigt of te snow but lso te nonsmmetric distributed lods resulting from drifting of te snow. Since te integrl represents te first moment wit respect to te w is of te re under te lod curve, it cn be replced b te product. We terefore ve P 5, were is te distnce from te w is to te centroid of te re (tis is not te centroid of te bem). distributed lod on bem cn tus be replced b concentrted lod; te mgnitude of tis single lod is equl to te re under te lod curve, nd its line of ction psses troug te centroid of tt re. It sould be noted, owever, tt te concentrted lod is equivlent to te given loding onl s fr s eternl forces re concerned. It cn be used to determine rections but sould not be used to compute internl forces nd deflections.

*5.9 FRES N SUMERGED SURFES Te pproc used in te preceding section cn be used to determine te resultnt of te drosttic pressure forces eerted on rectngulr surfce submerged in liquid. onsider te rectngulr plte sown in Fig. 5.18, wic is of lengt L nd widt b, were b is mesured perpendiculr to te plne of te figure. s noted in Sec. 5.8, te lod eerted on n element of te plte of lengt d is w d, were w is te lod per unit lengt. However, tis lod cn lso be epressed s p d 5 pb d, were p is te gge pressure in te liquid nd b is te widt of te plte; tus, w 5 bp. Since te gge pressure in liquid is p 5 g, were g is te specific weigt of te liquid nd is te verticl distnce from te free surfce, it follows tt w 5 bp 5 bg (5.13) wic sows tt te lod per unit lengt w is proportionl to nd, tus, vries linerl wit. Reclling te results of Sec. 5.8, we observe tt te resultnt R of te drosttic forces eerted on one side of te plte is equl in mgnitude to te trpeoidl re under te lod curve nd tt its line of ction psses troug te centroid of tt re. Te point P of te plte were R is pplied is known s te center of pressure. Net, we consider te forces eerted b liquid on curved surfce of constnt widt (Fig. 5.19). Since te determintion of te resultnt R of tese forces b direct integrtion would not be es, we consider te free bod obtined b detcing te volume of liquid D bounded b te curved surfce nd b te two plne surfces D nd D sown in Fig. 5.19b. Te forces cting on te free bod D re te weigt W of te detced volume of liquid, te resultnt R 1 of te forces eerted on D, te resultnt R 2 of te forces eerted on D, nd te resultnt 2R of te forces eerted b te curved surfce on te liquid. Te resultnt 2R is equl nd opposite to, nd s te sme line of ction s, te resultnt R of te forces eerted b te liquid on te curved surfce. Te forces W, R 1, nd R 2 cn be determined b stndrd metods; fter teir vlues ve been found, te force 2R is obtined b solving te equtions of equilibrium for te free bod of Fig. 5.19b. Te resultnt R of te drosttic forces eerted on te curved surfce is ten obtined b reversing te sense of 2R. Te metods outlined in tis section cn be used to determine te resultnt of te drosttic forces eerted on te surfces of dms nd rectngulr gtes nd vnes. Te resultnts of forces on submerged surfces of vrible widt will be determined in p. 9. Te pressure p, wic represents lod per unit re, is epressed in N/m 2 or in lb/ft 2. Te derived SI unit N/m 2 is clled pscl (P). Noting tt te re under te lod curve is equl to w E L, were w E is te lod per unit lengt t te center E of te plte, nd reclling Eq. (5.13), we cn write R 5 w E L 5 (bp E )L 5 p E (bl) 5 p E were denotes te re of te plte. Tus, te mgnitude of R cn be obtined b multipling te re of te plte b te pressure t its center E. Te resultnt R, owever, sould be pplied t P, not t E. d L Fig. 5.18 R () R 1 R W (b) Fig. 5.19 5.9 Forces on Submerged Surfces R w E P D R 2 249

w = 1500 N/m L = 6 m w = 4500 N/m SMPLE PRLEM 5.9 bem supports distributed lod s sown. () Determine te equivlent concentrted lod. (b) Determine te rections t te supports. = 4 m 1.5 kn/m I = 2 m 6 m II 4.5 kn/m SLUTIN. Equivlent oncentrted Lod. Te mgnitude of te resultnt of te lod is equl to te re under te lod curve, nd te line of ction of te resultnt psses troug te centroid of te sme re. We divide te re under te lod curve into two tringles nd construct te tble below. To simplif te computtions nd tbultion, te given lods per unit lengt ve been converted into kn/m. omponent, kn, m, kn? m Tringle I 4.5 2 9 Tringle II 13.5 4 54 o 5 18.0 o 5 63 18 kn X = 3.5 m Tus, X 5 : X(18 kn) 5 63 kn? m X 5 3.5 m Te equivlent concentrted lod is nd its line of ction is locted t distnce W 5 18 knw X 5 3.5 m to te rigt of 13.5 kn 4.5 kn b. Rections. Te rection t is verticl nd is denoted b ; te rection t is represented b its components nd. Te given lod cn be considered to be te sum of two tringulr lods s sown. Te resultnt of ec tringulr lod is equl to te re of te tringle nd cts t its centroid. We write te following equilibrium equtions for te free bod sown: 2 m 1 F 5 0: 5 0 1l om 5 0: 2(4.5 kn)(2 m) 2 (13.5 kn)(4 m) 1 (6 m) 5 0 4 m 6 m 5 10.5 kn 1l om 5 0: 1(4.5 kn)(4 m) 1 (13.5 kn)(2 m) 2 (6 m) 5 0 5 7.5 kn lterntive Solution. Te given distributed lod cn be replced b its resultnt, wic ws found in prt. Te rections cn be determined b writing te equilibrium equtions of 5 0, om 5 0, nd om 5 0. We gin obtin 5 0 5 10.5 kn 5 7.5 kn 250

5 ft 9 ft 10 ft SMPLE PRLEM 5.10 22 ft Verte Prbol 18 ft Te cross section of concrete dm is s sown. onsider 1-ft-tick section of te dm, nd determine () te resultnt of te rection forces eerted b te ground on te bse of te dm, (b) te resultnt of te pressure forces eerted b te wter on te fce of te dm. Te specific weigts of concrete nd wter re 150 lb/ft 3 nd 62.4 lb/ft 3, respectivel. SLUTIN H 2.5 ft. Ground Rection. We coose s free bod te 1-ft-tick section 4 ft EFD of te dm nd wter. Te rection forces eerted b te ground 9 ft 6 ft E F on te bse re represented b n equivlent force-couple sstem t. ter forces cting on te free bod re te weigt of te dm, represented 6 ft D b te weigts of its components W 1, W 2, nd W 3 ; te weigt of te wter W 4 ; nd te resultnt P of te pressure forces eerted on section D b 22 ft te wter to te rigt of section D. We ve W 4 18 ft W P W 1 5 1 2 2(9 ft)(22 ft)(1 ft)(150 lb/ft 3 ) 5 14,850 lb W 1 W 3 6 ft W 2 5 (5 ft)(22 ft)(1 ft)(150 lb/ft 3 ) 5 16,500 lb W 3 5 1 3(10 ft)(18 ft)(1 ft)(150 lb/ft 3 ) 5 9000 lb M 14 ft W 4 5 2 3(10 ft)(18 ft)(1 ft)(62.4 lb/ft 3 ) 5 7488 lb 3 ft w = bp V = (1 ft)(18 ft)(62.4 lb/ft 3 ) P 5 1 2(18 ft)(1 ft)(18 ft)(62.4 lb/ft 3 ) 5 10,109 lb Equilibrium Equtions 4 ft D R W 4 R G P = 10,109 lb P = 36.5 R = 12,580 lb 6 ft W 4 = 7488 lb 1 F 5 0: H 2 10,109 lb 5 0 H 5 10,110 lb 1oF 5 0: V 2 14,850 lb 2 16,500 lb 2 9000 lb 2 7488 lb 5 0 V 5 47,840 lb 1l om 5 0: 2(14,850 lb)(6 ft) 2 (16,500 lb)(11.5 ft) 2 (9000 lb)(17 ft) 2 (7488 lb)(20 ft) 1 (10,109 lb)(6 ft) 1 M 5 0 M 5 520,960 lb? ft l We cn replce te force-couple sstem obtined b single force cting t distnce d to te rigt of, were 520,960 lb? ft d 5 5 10.89 ft 47,840 lb b. Resultnt R of Wter Forces. Te prbolic section of wter D is cosen s free bod. Te forces involved re te resultnt 2R of te forces eerted b te dm on te wter, te weigt W 4, nd te force P. Since tese forces must be concurrent, 2R psses troug te point of intersection G of W 4 nd P. force tringle is drwn from wic te mgnitude nd direction of 2R re determined. Te resultnt R of te forces eerted b te wter on te fce is equl nd opposite: R 5 12,580 lb d36.5 251

SLVING PRLEMS N YUR WN Te problems in tis lesson involve two common nd ver importnt tpes of loding: distributed lods on bems nd forces on submerged surfces of constnt widt. s we discussed in Secs. 5.8 nd 5.9 nd illustrted in Smple Probs. 5.9 nd 5.10, determining te single equivlent force for ec of tese lodings requires knowledge of centroids. 1. nling bems subjected to distributed lods. In Sec. 5.8, we sowed tt distributed lod on bem cn be replced b single equivlent force. Te mgnitude of tis force is equl to te re under te distributed lod curve nd its line of ction psses troug te centroid of tt re. Tus, ou sould begin our solution b replcing te vrious distributed lods on given bem b teir respective single equivlent forces. Te rections t te supports of te bem cn ten be determined b using te metods of p. 4. Wen possible, comple distributed lods sould be divided into te commonspe res sown in Fig. 5.8 [Smple Prob. 5.9]. Ec of tese res cn ten be replced b single equivlent force. If required, te sstem of equivlent forces cn be reduced furter to single equivlent force. s ou stud Smple Prob. 5.9, note ow we ve used te nlog between force nd re nd te tecniques for locting te centroid of composite re to nle bem subjected to distributed lod. 2. Solving problems involving forces on submerged bodies. Te following points nd tecniques sould be remembered wen solving problems of tis tpe.. Te pressure p t dept below te free surfce of liquid is equl to g or rg, were g nd r re te specific weigt nd te densit of te liquid, respectivel. Te lod per unit lengt w cting on submerged surfce of constnt widt b is ten w 5 bp 5 bg 5 brg b. Te line of ction of te resultnt force R cting on submerged plne surfce is perpendiculr to te surfce. c. For verticl or inclined plne rectngulr surfce of widt b, te loding on te surfce cn be represented b linerl distributed lod wic is trpeoidl in spe (Fig. 5.18). Furter, te mgnitude of R is given b R 5 g E were E is te verticl distnce to te center of te surfce nd is te re of te surfce. 252

d. Te lod curve will be tringulr (rter tn trpeoidl) wen te top edge of plne rectngulr surfce coincides wit te free surfce of te liquid, since te pressure of te liquid t te free surfce is ero. For tis cse, te line of ction of R is esil determined, for it psses troug te centroid of tringulr distributed lod. e. For te generl cse, rter tn nling trpeoid, we suggest tt ou use te metod indicted in prt b of Smple Prob. 5.9. First divide te trpeoidl distributed lod into two tringles, nd ten compute te mgnitude of te resultnt of ec tringulr lod. (Te mgnitude is equl to te re of te tringle times te widt of te plte.) Note tt te line of ction of ec resultnt force psses troug te centroid of te corresponding tringle nd tt te sum of tese forces is equivlent to R. Tus, rter tn using R, ou cn use te two equivlent resultnt forces, wose points of ppliction re esil clculted. f course, te eqution given for R in prgrp c sould be used wen onl te mgnitude of R is needed. f. Wen te submerged surfce of constnt widt is curved, te resultnt force cting on te surfce is obtined b considering te equilibrium of te volume of liquid bounded b te curved surfce nd b oriontl nd verticl plnes (Fig. 5.19). bserve tt te force R 1 of Fig. 5.19 is equl to te weigt of te liquid ling bove te plne D. Te metod of solution for problems involving curved surfces is sown in prt b of Smple Prob. 5.10. In subsequent mecnics courses (in prticulr, mecnics of mterils nd fluid mecnics), ou will ve mple opportunit to use te ides introduced in tis lesson. 253

PRLEMS 5.66 nd 5.67 For te bem nd loding sown, determine () te mgnitude nd loction of te resultnt of te distributed lod, (b) te rections t te bem supports. Prbol Verte 150 lb/ft 120 lb/ft 200 N/m 800 N/m 9 ft 4 m Fig. P5.66 Fig. P5.67 5.68 troug 5.73 Determine te rections t te bem supports for te given loding. 6 kn/m 480 lb/ft 600 lb/ft 2 kn/m D Fig. P5.68 6 m 4 m 3 ft Fig. P5.69 6 ft 2 ft 200 lb/ft 1000 N/m 1200 N/m 6 ft 9 ft 6 ft 3.6 m Fig. P5.70 Fig. P5.71 Verte 200 lb/ft 12 ft Prbols 6 ft 100 lb/ft Prbol 300 N/m 6 m 900 N/m Fig. P5.72 254 Fig. P5.73

5.74 Determine () te distnce so tt te verticl rections t supports nd re equl, (b) te corresponding rections t te supports. 5.75 Determine () te distnce so tt te rection t support is minimum, (b) te corresponding rections t te supports. 5.76 Determine te rections t te bem supports for te given loding wen w 0 5 150 lb/ft. 5.77 Determine () te distributed lod w 0 t te end D of te bem D for wic te rection t is ero, (b) te corresponding rection t. 1800 N/m 4 m Fig. P5.74 nd P5.75 Problems 600 N/m 255 5.78 Te bem supports two concentrted lods nd rests on soil tt eerts linerl distributed upwrd lod s sown. Determine te vlues of w nd w corresponding to equilibrium. 5.79 For te bem nd loding of Prob. 5.78, determine () te distnce for wic w 5 20 kn/m, (b) te corresponding vlue of w. 24 kn 30 kn = 0.6 m 0.3 m 450 lb/ft 44.1 kip ft 4 ft 12 ft Fig. P5.76 nd P5.77 2 ft w 0 D w w 1.8 m Fig. P5.78 In te following problems, use g 5 62.4 lb/ft 3 for te specific weigt of fres wter nd g c 5 150 lb/ft 3 for te specific weigt of concrete if U.S. customr units re used. Wit SI units, use r 5 10 3 kg/m 3 for te densit of fres wter nd r c 5 2.40 3 10 3 kg/m 3 for te densit of concrete. (See te footnote on pge 222 for ow to determine te specific weigt of mteril given its densit.) 5.80 Te cross section of concrete dm is s sown. For 1-m-wide dm section determine () te resultnt of te rection forces eerted b te ground on te bse of te dm, (b) te point of ppliction of te resultnt of prt, (c) te resultnt of te pressure forces eerted b te wter on te fce of te dm. 7.2 m 4.8 m Verte Prbol 2.4 m 5.81 Te cross section of concrete dm is s sown. For 1-ft-wide dm section determine () te resultnt of te rection forces eerted b te ground on te bse of te dm, (b) te point of ppliction of te resultnt of prt, (c) te resultnt of te pressure forces eerted b te wter on te fce of te dm. Fig. P5.80 8 ft r = 21 ft r = 21 ft Fig. P5.81

256 Distributed Forces: entroids nd enters of Grvit d T 3 m Fig. P5.82 nd P5.83 T 5.82 Te 3 3 4-m side of tnk is inged t its bottom nd is eld in plce b tin rod. Te mimum tensile force te rod cn witstnd witout breking is 200 kn, nd te design specifictions require te force in te rod not to eceed 20 percent of tis vlue. If te tnk is slowl filled wit wter, determine te mimum llowble dept of wter d in te tnk. 5.83 Te 3 3 4-m side of n open tnk is inged t its bottom nd is eld in plce b tin rod. Te tnk is to be filled wit glcerine, wose densit is 1263 kg/m 3. Determine te force T in te rod nd te rections t te inge fter te tnk is filled to dept of 2.9 m. 5.84 Te friction force between 6 3 6-ft squre sluice gte nd its guides is equl to 10 percent of te resultnt of te pressure forces eerted b te wter on te fce of te gte. Determine te initil force needed to lift te gte if it weigs 1000 lb. Mrs = 6 ft 15 ft Fig. P5.84 cen d 3 ft 6 ft 5.85 freswter mrs is drined to te ocen troug n utomtic tide gte tt is 4 ft wide nd 3 ft ig. Te gte is eld b inges locted long its top edge t nd bers on sill t. If te wter level in te mrs is 5 6 ft, determine te ocen level d for wic te gte will open. (Specific weigt of slt wter 5 64 lb/ft 3.) 5.86 Te dm for lke is designed to witstnd te dditionl force cused b silt tt s settled on te lke bottom. ssuming tt silt is equivlent to liquid of densit r s 5 1.76 3 10 3 kg/m 3 nd considering 1-m-wide section of dm, determine te percentge increse in te force cting on te dm fce for silt ccumultion of dept 2 m. Wter 6.6 m Silt Fig. P5.85 Fig. P5.86 nd P5.87 D T 0.27 m 0.45 m 0.48 m 0.64 m Fig. P5.88 nd P5.89 5.87 Te bse of dm for lke is designed to resist up to 120 percent of te oriontl force of te wter. fter construction, it is found tt silt (tt is equivlent to liquid of densit r s 5 1.76 3 10 3 kg/m 3 ) is settling on te lke bottom t te rte of 12 mm/er. onsidering 1-m-wide section of dm, determine te number of ers until te dm becomes unsfe. 5.88 0.5 3 0.8-m gte is locted t te bottom of tnk filled wit wter. Te gte is inged long its top edge nd rests on frictionless stop t. Determine te rections t nd wen cble D is slck. 5.89 0.5 3 0.8-m gte is locted t te bottom of tnk filled wit wter. Te gte is inged long its top edge nd rests on frictionless stop t. Determine te minimum tension required in cble D to open te gte.

5.90 long troug is supported b continuous inge long its lower edge nd b series of oriontl cbles ttced to its upper edge. Determine te tension in ec of te cbles, t time wen te troug is completel full of wter. Problems 257 20 in. 20 in. 20 in. r = 24 in. Fig. P5.90 5.91 4 3 2-ft gte is inged t nd is eld in position b rod D. End D rests ginst spring wose constnt is 828 lb/ft. Te spring is undeformed wen te gte is verticl. ssuming tt te force eerted b rod D on te gte remins oriontl, determine te minimum dept of wter d for wic te bottom of te gte will move to te end of te clindricl portion of te floor. d 4 ft Fig. P5.91 3 ft 2 ft D 5.92 Solve Prob. 5.91 if te gte weigs 1000 lb. 5.93 prismticll sped gte plced t te end of freswter cnnel is supported b pin nd brcket t nd rests on frictionless support t. Te pin is locted t distnce 5 0.10 m below te center of grvit of te gte. Determine te dept of wter d for wic te gte will open. 5.94 prismticll sped gte plced t te end of freswter cnnel is supported b pin nd brcket t nd rests on frictionless support t. Determine te distnce if te gte is to open wen d 5 0.75 m. d 0.75 m 0.40 m Fig. P5.93 nd P5.94 5.95 55-gllon 23-in.-dimeter drum is plced on its side to ct s dm in 30-in.-wide freswter cnnel. Knowing tt te drum is ncored to te sides of te cnnel, determine te resultnt of te pressure forces cting on te drum. 23 in. 11.5 in. Fig. P5.95

258 Distributed Forces: entroids nd enters of Grvit Poto 5.4 To predict te fligt crcteristics of te modified oeing 747 wen used to trnsport spce suttle, te center of grvit of ec crft d to be determined. VLUMES 5.10 ENTER F GRVITY F THREE-DIMENSINL DY. ENTRID F VLUME Te center of grvit G of tree-dimensionl bod is obtined b dividing te bod into smll elements nd b ten epressing tt te weigt W of te bod cting t G is equivlent to te sstem of distributed forces DW representing te weigts of te smll elements. oosing te is to be verticl wit positive sense upwrd (Fig. 5.20) nd denoting b r te position vector of G, we write tt r G W = Wj = r ΔW = ΔWj Fig. 5.20 W is equl to te sum of te elementl weigts DW nd tt its moment bout is equl to te sum of te moments bout of te elementl weigts: of: 2Wj 5 o(2dwj) (5.14) om : r 3 (2Wj) 5 o[r 3 (2DWj)] Rewriting te lst eqution in te form rw 3 (2j) 5 (or DW) 3 (2j) (5.15) we observe tt te weigt W of te bod is equivlent to te sstem of te elementl weigts DW if te following conditions re stisfied: W 5 o DW rw 5 or DW Incresing te number of elements nd simultneousl decresing te sie of ec element, we obtin in te limit ΔW W 5 # dw r W 5 # r dw (5.16) We note tt te reltions obtined re independent of te orienttion of te bod. For emple, if te bod nd te coordinte es were rotted so tt te is pointed upwrd, te unit vector 2j would be replced b 2k in Eqs. (5.14) nd (5.15), but te reltions (5.16) would remin uncnged. Resolving te vectors r nd r into rectngulr components, we note tt te second of te reltions (5.16) is equivlent to te tree sclr equtions W 5 # dw W 5 # dw W 5 # dw (5.17)

If te bod is mde of omogeneous mteril of specific weigt g, te mgnitude dw of te weigt of n infinitesiml element cn be epressed in terms of te volume dv of te element, nd te mgnitude W of te totl weigt cn be epressed in terms of te totl volume V. We write 5.10 enter of Grvit of Tree-Dimensionl od. entroid of Volume 259 dw 5 g dv W 5 gv Substituting for dw nd W in te second of te reltions (5.16), we write or, in sclr form, r V 5 # r dv (5.18) V 5 # dv V 5 # dv V 5 # dv (5.19) Te point wose coordintes re,, is lso known s te centroid of te volume V of te bod. If te bod is not omogeneous, Eqs. (5.19) cnnot be used to determine te center of grvit of te bod; owever, Eqs. (5.19) still define te centroid of te volume. Te integrl e dv is known s te first moment of te volume wit respect to te plne. Similrl, te integrls e dv nd e dv define te first moments of te volume wit respect to te plne nd te plne, respectivel. It is seen from Eqs. (5.19) tt if te centroid of volume is locted in coordinte plne, te first moment of te volume wit respect to tt plne is ero. volume is sid to be smmetricl wit respect to given plne if for ever point P of te volume tere eists point P9 of te sme volume, suc tt te line PP9 is perpendiculr to te given plne nd is bisected b tt plne. Te plne is sid to be plne of smmetr for te given volume. Wen volume V possesses plne of smmetr, te first moment of V wit respect to tt plne is ero, nd te centroid of te volume is locted in te plne of smmetr. Wen volume possesses two plnes of smmetr, te centroid of te volume is locted on te line of intersection of te two plnes. Finll, wen volume possesses tree plnes of smmetr wic intersect t well-defined point (i.e., not long common line), te point of intersection of te tree plnes coincides wit te centroid of te volume. Tis propert enbles us to determine immeditel te loctions of te centroids of speres, ellipsoids, cubes, rectngulr prllelepipeds, etc. Te centroids of unsmmetricl volumes or of volumes possessing onl one or two plnes of smmetr sould be determined b integrtion (Sec. 5.12). Te centroids of severl common volumes re sown in Fig. 5.21. It sould be observed tt in generl te centroid of volume of revolution does not coincide wit te centroid of its cross section. Tus, te centroid of emispere is different from tt of semicirculr re, nd te centroid of cone is different from tt of tringle.

Spe Volume Hemispere 3 8 2 3 3 Semiellipsoid of revolution 3 8 2 2 3 Prboloid of revolution 3 1 2 2 one 4 1 2 3 Prmid b 4 1 3 b Fig. 5.21 entroids of common spes nd volumes. 260

5.11 MPSITE DIES If bod cn be divided into severl of te common spes sown in Fig. 5.21, its center of grvit G cn be determined b epressing tt te moment bout of its totl weigt is equl to te sum of te moments bout of te weigts of te vrious component prts. Proceeding s in Sec. 5.10, we obtin te following equtions defining te coordintes X, Y, Z of te center of grvit G. X W 5 W Y W 5 W Z W 5 W (5.20) 5.12 Determintion of entroids of Volumes b Integrtion 261 If te bod is mde of omogeneous mteril, its center of grvit coincides wit te centroid of its volume, nd we obtin: X V 5 V Y V 5 V Z V 5 V (5.21) P(,,) 5.12 DETERMINTIN F ENTRIDS F VLUMES Y INTEGRTIN Te centroid of volume bounded b nlticl surfces cn be determined b evluting te integrls given in Sec. 5.10: V 5 # dv V 5 # dv V 5 # dv (5.22) If te element of volume dv is cosen to be equl to smll cube of sides d, d, nd d, te evlution of ec of tese integrls requires triple integrtion. However, it is possible to determine te coordintes of te centroid of most volumes b double integrtion if dv is cosen to be equl to te volume of tin filment (Fig. 5.22). Te coordintes of te centroid of te volume re ten obtined b rewriting Eqs. (5.22) s el d el d el =, el =, el = dv = d d Fig. 5.22 Determintion of te centroid of volume b double integrtion. 2 el V 5 # el dv V 5 # el dv V 5 # el dv (5.23) nd b ten substituting te epressions given in Fig. 5.22 for te volume dv nd te coordintes el, el, el. using te eqution of te surfce to epress in terms of nd, te integrtion is reduced to double integrtion in nd. If te volume under considertion possesses two plnes of smmetr, its centroid must be locted on te line of intersection of te two plnes. oosing te is to lie long tis line, we ve 5 5 0 nd te onl coordinte to determine is. Tis cn be done wit single integrtion b dividing te given volume into tin slbs prllel to te plne nd epressing dv in terms of nd d in te eqution V 5 # el dv (5.24) For bod of revolution, te slbs re circulr nd teir volume is given in Fig. 5.23. d el r el = dv = r 2 d Fig. 5.23 Determintion of te centroid of bod of revolution.

SMPLE PRLEM 5.11 100 mm 60 mm Determine te loction of te center of grvit of te omogeneous bod of revolution sown, wic ws obtined b joining emispere nd clinder nd crving out cone. 60 mm SLUTIN ecuse of smmetr, te center of grvit lies on te is. s sown in te figure below, te bod cn be obtined b dding emispere to clinder nd ten subtrcting cone. Te volume nd te bsciss of te centroid of ec of tese components re obtined from Fig. 5.21 nd re entered in te tble below. Te totl volume of te bod nd te first moment of its volume wit respect to te plne re ten determined. 60 mm + 3 (60 mm) = 22.5 mm 50 mm 3 (100 mm) = 75 mm 8 4 omponent Volume, mm 3, mm V, mm 4 Hemispere 1 4p 2 3 (60)3 5 0.4524 3 10 6 222.5 210.18 3 10 6 linder p(60) 2 (100) 5 1.1310 3 10 6 150 156.55 3 10 6 one 2 p 3 (60)2 (100) 520.3770 3 10 6 175 228.28 3 10 6 ov 5 1.206 3 10 6 ov 5 118.09 3 10 6 Tus, XoV 5 ov: X(1.206 3 10 6 mm 3 ) 5 18.09 3 10 6 mm 4 X 5 15 mm 262

SMPLE PRLEM 5.12 4.5 in. 2.5 in. Locte te center of grvit of te steel mcine element sown. Te dimeter of ec ole is 1 in. 2 in. 0.5 in. 0.5 in. 1 in. 1 in. 2 in. 1 in. SLUTIN 4.5 in. 2 in. I + 2 in. II _ 1 in. dim. _ III IV Te mcine element cn be obtined b dding rectngulr prllelepiped (I) to qurter clinder (II) nd ten subtrcting two 1-in.-dimeter clinders (III nd IV). Te volume nd te coordintes of te centroid of ec component re determined nd re entered in te tble below. Using te dt in te tble, we ten determine te totl volume nd te moments of te volume wit respect to ec of te coordinte plnes. 0.5 in. 4r 4(2) = = 0.8488 in. 3 3 1 in. 1 in. III I 2.25 in. IV 0.25 in. 8 in. 3 I, III, IV II 0.5 in. II 0.25 in. 2 in. 1.5 in. V, in 3, in., in., in. V, in 4 V, in 4 V, in 4 I (4.5)(2)(0.5) 5 4.5 0.25 21 2.25 1.125 24.5 10.125 1 II 4p(2) 2 (0.5) 5 1.571 1.3488 20.8488 0.25 2.119 21.333 0.393 III 2p(0.5) 2 (0.5) 5 20.3927 0.25 21 3.5 20.098 0.393 21.374 IV 2p(0.5) 2 (0.5) 5 20.3927 0.25 21 1.5 20.098 0.393 20.589 ov 5 5.286 ov 5 3.048 ov 5 25.047 ov 5 8.555 Tus, XoV 5 ov: X(5.286 in 3 ) 5 3.048 in 4 X 5 0.577 in. YoV 5 ov: Y(5.286 in 3 ) 5 25.047 in 4 Y 5 20.955 in. ZoV 5 ov: Z(5.286 in 3 ) 5 8.555 in 4 Z 5 1.618 in. 263

SMPLE PRLEM 5.13 Determine te loction of te centroid of te lf rigt circulr cone sown. SLUTIN el = Since te plne is plne of smmetr, te centroid lies in tis plne nd 5 0. slb of tickness d is cosen s differentil element. Te volume of tis element is dv 5 1 2 pr 2 d el r Te coordintes el nd el of te centroid of te element re obtined from Fig. 5.8 (semicirculr re). el 5 el 5 4r 3p We observe tt r is proportionl to nd write r 5 r 5 Te volume of te bod is V 5 # dv 5 # 0 1 2 pr 2 d 5 # 1 2 p 2 b d 5 p2 6 Te moment of te differentil element wit respect to te plne is el dv; te totl moment of te bod wit respect to tis plne is 0 Tus, # el dv 5 # 0 ( 1 2 pr 2 ) d 5 # 0 ( 1 2 p) b 2 d 5 p2 2 8 V 5 # el dv p 2 5 p2 2 5 3 6 8 4 Likewise, te moment of te differentil element wit respect to te plne is el dv; te totl moment is Tus, # el dv 5 # 0 4r 3p (1 2 pr 2 )d 5 2 3 # 3 b d 5 3 6 0 V 5 # el dv p 2 5 3 5 6 6 p 264

SLVING PRLEMS N YUR WN In te problems for tis lesson, ou will be sked to locte te centers of grvit of tree-dimensionl bodies or te centroids of teir volumes. ll of te tecniques we previousl discussed for two-dimensionl bodies using smmetr, dividing te bod into common spes, coosing te most efficient differentil element, etc. m lso be pplied to te generl tree-dimensionl cse. 1. Locting te centers of grvit of composite bodies. In generl, Eqs. (5.20) must be used: XwoW 5 oww YwoW 5 oww ZwoW 5 oww (5.20) However, for te cse of omogeneous bod, te center of grvit of te bod coincides wit te centroid of its volume. Terefore, for tis specil cse, te center of grvit of te bod cn lso be locted using Eqs. (5.21): XwoV 5 ow V YwoV 5 owv ZwoV 5 owv (5.21) You sould relie tt tese equtions re simpl n etension of te equtions used for te two-dimensionl problems considered erlier in te cpter. s te solutions of Smple Probs. 5.11 nd 5.12 illustrte, te metods of solution for two- nd tree-dimensionl problems re identicl. Tus, we once gin strongl encourge ou to construct pproprite digrms nd tbles wen nling composite bodies. lso, s ou stud Smple Prob. 5.12, observe ow te nd coordintes of te centroid of te qurter clinder were obtined using te equtions for te centroid of qurter circle. We note tt two specil cses of interest occur wen te given bod consists of eiter uniform wires or uniform pltes mde of te sme mteril.. For bod mde of severl wire elements of te sme uniform cross section, te cross-sectionl re of te wire elements will fctor out of Eqs. (5.21) wen V is replced wit te product L, were L is te lengt of given element. Equtions (5.21) tus reduce in tis cse to XwoL 5 owl YwoL 5 owl ZwoL 5 owl b. For bod mde of severl pltes of te sme uniform tickness, te tickness t of te pltes will fctor out of Eqs. (5.21) wen V is replced wit te product t, were is te re of given plte. Equtions (5.21) tus reduce in tis cse to Xwo 5 ow Ywo 5 ow Zwo 5 ow 2. Locting te centroids of volumes b direct integrtion. s eplined in Sec. 5.12, evluting te integrls of Eqs. (5.22) cn be simplified b coosing eiter tin filment (Fig. 5.22) or tin slb (Fig. 5.23) for te element of volume d V. Tus, ou sould begin our solution b identifing, if possible, te d V wic produces te single or double integrls tt re te esiest to compute. For bodies of revolution, tis m be tin slb (s in Smple Prob. 5.13) or tin clindricl sell. However, it is importnt to remember tt te reltionsip tt ou estblis mong te vribles (like te reltionsip between r nd in Smple Prob. 5.13) will directl ffect te compleit of te integrls ou will ve to compute. Finll, we gin remind ou tt el, el, nd el in Eqs. (5.23) re te coordintes of te centroid of dv. 265

PRLEMS 5.96 Determine te loction of te centroid of te composite bod sown wen () 5 2b, (b) 5 2.5b. b Fig. P5.96 5.97 Determine te coordinte of te centroid of te bod sown. b 2 Fig. P5.97 nd P5.98 5.98 Determine te coordinte of te centroid of te bod sown. (Hint: Use te result of Smple Prob. 5.13.) 5.99 Te composite bod sown is formed b removing semiellipsoid of revolution of semimjor is nd semiminor is /2 from emispere of rdius. Determine () te coordinte of te centroid wen 5 /2, (b) te rtio / for wic 5 20.4. 2 266 Fig. P5.99

5.100 For te stop brcket sown, locte te coordinte of te center of grvit. Problems 267 5.101 For te stop brcket sown, locte te coordinte of te center of grvit. 5.102 nd 5.103 For te mcine element sown, locte te coordinte of te center of grvit. 62 mm 12 mm 100 mm 12 mm 51 mm 88 mm 40 mm 28 mm 16 mm 60 mm 24 mm 10 mm 10 mm 55 mm 34 mm 45 mm Fig. P5.100 nd P5.101 18 mm r = 13 mm r = 12 mm 100 mm 20 mm 25 mm 25 mm 20 mm Fig. P5.102 nd P5.105 3 in. 2 in. 2 in. 1 in. r = 1.25 in. r = 1.25 in. 2 in. 0.75 in. 5.104 For te mcine element sown, locte te coordinte of te center of grvit. 5.105 For te mcine element sown, locte te coordinte of te center of grvit. 2 in. 2 in. Fig. P5.103 nd P5.104 125 mm 80 mm r = 1.8 m 1.2 m 150 mm 0.8 m 1.5 m Fig. P5.106 250 mm Fig. P5.107 5.106 nd 5.107 Locte te center of grvit of te seet-metl form sown.

268 Distributed Forces: entroids nd enters of Grvit 5.108 wstebsket, designed to fit in te corner of room, is 16 in. ig nd s bse in te spe of qurter circle of rdius 10 in. Locte te center of grvit of te wstebsket, knowing tt it is mde of seet metl of uniform tickness. 10 in. 10 in. 16 in. Fig. P5.108 5.109 mounting brcket for electronic components is formed from seet metl of uniform tickness. Locte te center of grvit of te brcket. r = 0.625 in. 1 in. 2.5 in. 0.75 in. 3 in. 1.25 in. 0.75 in. Fig. P5.109 5.110 tin seet of plstic of uniform tickness is bent to form desk orgnier. Locte te center of grvit of te orgnier. 6 in. 60 mm r = 5 mm 74 mm 69 mm 75 mm 30 mm r = 6 mm r = 6 mm Fig. P5.110 r = 6 mm

5.111 window wning is fbricted from seet metl of uniform tickness. Locte te center of grvit of te wning. Problems 269 34 in. r = 25 in. 4 in. Fig. P5.111 5.112 n elbow for te duct of ventilting sstem is mde of seet metl of uniform tickness. Locte te center of grvit of te elbow. r = 400 mm r = 200 mm 76 mm Fig. P5.112 100 mm 5.113 n 8-in.-dimeter clindricl duct nd 4 3 8-in. rectngulr duct re to be joined s indicted. Knowing tt te ducts were fbricted from te sme seet metl, wic is of uniform tickness, locte te center of grvit of te ssembl. 12 in. 12 in. 8 in. 4 in. Fig. P5.113

270 Distributed Forces: entroids nd enters of Grvit 5.114 tin steel wire of uniform cross section is bent into te spe sown. Locte its center of grvit. 0.6 m 0.6 m D 0.8 m 60 Fig. P5.114 5.115 nd 5.116 Locte te center of grvit of te figure sown, knowing tt it is mde of tin brss rods of uniform dimeter. D 1.5 m 30 in. 1 m 0.6 m D r = 16 in. E Fig. P5.115 Fig. P5.116 5.117 Te frme of greenouse is constructed from uniform luminum cnnels. Locte te center of grvit of te portion of te frme sown. r 5.118 scrtc wl s plstic ndle nd steel blde nd snk. Knowing tt te densit of plstic is 1030 kg/m 3 nd of steel is 7860 kg/m 3, locte te center of grvit of te wl. 5 ft 50 mm 90 mm 10 mm 3 ft Fig. P5.117 2 ft 25 mm r Fig. P5.118 80 mm 3.5 mm

5.119 brone busing is mounted inside steel sleeve. Knowing tt te specific weigt of brone is 0.318 lb/in 3 nd of steel is 0.284 lb/in 3, determine te loction of te center of grvit of te ssembl. 5.120 brss collr, of lengt 2.5 in., is mounted on n luminum rod of lengt 4 in. Locte te center of grvit of te composite bod. (Specific weigts: brss 5 0.306 lb/in 3, luminum 5 0.101 lb/in 3.) 1.125 in. 0.5 in. Problems 271 1.00 in. 1.6 in. 0.40 in. 0.75 in. 1.80 in. 4 in. 2.5 in. Fig. P5.119 3 in. Fig. P5.120 5.121 Te tree legs of smll glss-topped tble re equll spced nd re mde of steel tubing, wic s n outside dimeter of 24 mm nd cross-sectionl re of 150 mm 2. Te dimeter nd te tickness of te tble top re 600 mm nd 10 mm, respectivel. Knowing tt te densit of steel is 7860 kg/m 3 nd of glss is 2190 kg/m 3, locte te center of grvit of te tble. 5.122 troug 5.124 Determine b direct integrtion te vlues of for te two volumes obtined b pssing verticl cutting plne troug te given spe of Fig. 5.21. Te cutting plne is prllel to te bse of te given spe nd divides te spe into two volumes of equl eigt. r = 280 mm r = 180 mm 5.122 emispere Fig. P5.121 5.123 semiellipsoid of revolution 5.124 prboloid of revolution. 5.125 nd 5.126 Locte te centroid of te volume obtined b rotting te sded re bout te is. = k( ) 2 = k 1/3 Fig. P5.125 Fig. P5.126

272 Distributed Forces: entroids nd enters of Grvit 5.127 Locte te centroid of te volume obtined b rotting te sded re bout te line 5. 2 2 2 + 2 = 1 Fig. P5.127 *5.128 Locte te centroid of te volume generted b revolving te portion of te sine curve sown bout te is. = b sin π 2 b Fig. P5.128 nd P5.129 *5.129 Locte te centroid of te volume generted b revolving te portion of te sine curve sown bout te is. (Hint: Use tin clindricl sell of rdius r nd tickness dr s te element of volume.) *5.130 Sow tt for regulr prmid of eigt nd n sides (n 5 3, 4,... ) te centroid of te volume of te prmid is locted t distnce /4 bove te bse. Fig. P5.131 R R 5.131 Determine b direct integrtion te loction of te centroid of one-lf of tin, uniform emispericl sell of rdius R. 5.132 Te sides nd te bse of punc bowl re of uniform tickness t. If t V R nd R 5 250 mm, determine te loction of te center of grvit of () te bowl, (b) te punc. R R Fig. P5.132

5.133 fter grding lot, builder plces four stkes to designte te corners of te slb for ouse. To provide firm, level bse for te slb, te builder plces minimum of 3 in. of grvel benet te slb. Determine te volume of grvel needed nd te coordinte of te centroid of te volume of te grvel. (Hint: Te bottom surfce of te grvel is n oblique plne, wic cn be represented b te eqution 5 1 b 1 c.) Problems 273 50 ft 3 in. 30 ft 5 in. 6 in. 8 in. Fig. P5.133 5.134 Determine b direct integrtion te loction of te centroid of te volume between te plne nd te portion sown of te surfce 5 16( 2 2 )(b 2 2 )/ 2 b 2. 3 Fig. P5.135 b Fig. P5.134 5.135 Locte te centroid of te section sown, wic ws cut from tin circulr pipe b two oblique plnes. *5.136 Locte te centroid of te section sown, wic ws cut from n ellipticl clinder b n oblique plne. Fig. P5.136 b b

REVIEW ND SUMMRY Tis cpter ws devoted ciefl to te determintion of te center of grvit of rigid bod, i.e., to te determintion of te point G were single force W, clled te weigt of te bod, cn be pplied to represent te effect of te ert s ttrction on te bod. enter of grvit of two-dimensionl bod In te first prt of te cpter, we considered two-dimensionl bodies, suc s flt pltes nd wires contined in te plne. dding force components in te verticl direction nd moments bout te oriontl nd es [Sec. 5.2], we derived te reltions W 5 # dw W 5 # dw W 5 # dw (5.2) wic define te weigt of te bod nd te coordintes nd of its center of grvit. entroid of n re or line In te cse of omogeneous flt plte of uniform tickness [Sec. 5.3], te center of grvit G of te plte coincides wit te centroid of te re of te plte, te coordintes of wic re defined b te reltions 5 # d 5 # d (5.3) Similrl, te determintion of te center of grvit of omogeneous wire of uniform cross section contined in plne reduces to te determintion of te centroid of te line L representing te wire; we ve L 5 # dl L 5 # dl (5.4) First moments Te integrls in Eqs. (5.3) re referred to s te first moments of te re wit respect to te nd es nd re denoted b Q nd Q, respectivel [Sec. 5.4]. We ve Q 5 Q 5 (5.6) Te first moments of line cn be defined in similr w. 274 Properties of smmetr Te determintion of te centroid of n re or line is simplified wen te re or line possesses certin properties of smmetr. If te re or line is smmetric wit respect to n is, its centroid

lies on tt is; if it is smmetric wit respect to two es, is locted t te intersection of te two es; if it is smmetric wit respect to center, coincides wit. Review nd Summr 275 Te res nd te centroids of vrious common spes re tbulted in Fig. 5.8. Wen flt plte cn be divided into severl of tese spes, te coordintes X nd Y of its center of grvit G cn be determined from te coordintes 1, 2,... nd 1, 2,... of te centers of grvit G 1, G 2,... of te vrious prts [Sec. 5.5]. Equting moments bout te nd es, respectivel (Fig. 5.24), we ve XwoW 5 o w W YwoW 5 o w W (5.7) enter of grvit of composite bod ΣW = W 3 W 1 W 2 G 3 X G Y G 1 G 2 Fig. 5.24 If te plte is omogeneous nd of uniform tickness, its center of grvit coincides wit te centroid of te re of te plte, nd Eqs. (5.7) reduce to Q 5 Xwo 5 ow Q 5 Ywo 5 ow (5.8) Tese equtions ield te first moments of te composite re, or te cn be solved for te coordintes X nd Y of its centroid [Smple Prob. 5.1]. Te determintion of te center of grvit of composite wire is crried out in similr fsion [Smple Prob. 5.2]. Wen n re is bounded b nlticl curves, te coordintes of its centroid cn be determined b integrtion [Sec. 5.6]. Tis cn be done b evluting eiter te double integrls in Eqs. (5.3) or single integrl wic uses one of te tin rectngulr or pie-sped elements of re sown in Fig. 5.12. Denoting b el nd el te coordintes of te centroid of te element d, we ve Determintion of centroid b integrtion Q 5 5 # el d Q 5 5 # el d (5.9) It is dvntgeous to use te sme element of re to compute bot of te first moments Q nd Q ; te sme element cn lso be used to determine te re [Smple Prob. 5.4].

276 Distributed Forces: entroids nd enters of Grvit 2 Teorems of Pppus-Guldinus L () Fig. 5.25 2 (b) Distributed lods Te teorems of Pppus-Guldinus relte te determintion of te re of surfce of revolution or te volume of bod of revolution to te determintion of te centroid of te generting curve or re [Sec. 5.7]. Te re of te surfce generted b rotting curve of lengt L bout fied is (Fig. 5.25) is 5 2pL (5.10) were represents te distnce from te centroid of te curve to te fied is. Similrl, te volume V of te bod generted b rotting n re bout fied is (Fig. 5.25b) is V 5 2p (5.11) were represents te distnce from te centroid of te re to te fied is. Te concept of centroid of n re cn lso be used to solve problems oter tn tose deling wit te weigt of flt pltes. For emple, to determine te rections t te supports of bem [Sec. 5.8], we cn replce distributed lod w b concentrted lod W equl in mgnitude to te re under te lod curve nd pssing troug te centroid of tt re (Fig. 5.26). Te sme pproc cn be used to determine te resultnt of te drosttic forces eerted on rectngulr plte submerged in liquid [Sec. 5.9]. w d W w dw = d = w W W = L d L P Fig. 5.26 enter of grvit of treedimensionl bod Te lst prt of te cpter ws devoted to te determintion of te center of grvit G of tree-dimensionl bod. Te coordintes,, of G were defined b te reltions W 5 # dw W 5 # dw W 5 # dw (5.17) entroid of volume In te cse of omogeneous bod, te center of grvit G coincides wit te centroid of te volume V of te bod; te coordintes of re defined b te reltions V 5 # dv V 5 # dv V 5 # dv (5.19) If te volume possesses plne of smmetr, its centroid will lie in tt plne; if it possesses two plnes of smmetr, will be locted on te line of intersection of te two plnes; if it possesses tree plnes of smmetr wic intersect t onl one point, will coincide wit tt point [Sec. 5.10].

Te volumes nd centroids of vrious common tree-dimensionl spes re tbulted in Fig. 5.21. Wen bod cn be divided into severl of tese spes, te coordintes X, Y, Z of its center of grvit G cn be determined from te corresponding coordintes of te centers of grvit of its vrious prts [Sec. 5.11]. We ve XwoW 5 o w W YwoW 5 o w W Zw ow 5 o w W (5.20) If te bod is mde of omogeneous mteril, its center of grvit coincides wit te centroid of its volume, nd we write [Smple Probs. 5.11 nd 5.12] XwoV 5 o w V YwoV 5 o w V Zw ov 5 o w V (5.21) Wen volume is bounded b nlticl surfces, te coordintes of its centroid cn be determined b integrtion [Sec. 5.12]. To void te computtion of te triple integrls in Eqs. (5.19), we cn use elements of volume in te spe of tin filments, s sown in Fig. 5.27. Review nd Summr enter of grvit of composite bod Determintion of centroid b integrtion 277 P(,,) el el el d d Fig. 5.27 el =, el =, el = dv = d d 2 Denoting b el, el, nd el te coordintes of te centroid of te element dv, we rewrite Eqs. (5.19) s V 5 # el dv V 5 # el dv V 5 # el dv (5.23) wic involve onl double integrls. If te volume possesses two plnes of smmetr, its centroid is locted on teir line of intersection. oosing te is to lie long tt line nd dividing te volume into tin slbs prllel to te plne, we cn determine from te reltion V 5 # el dv (5.24) wit single integrtion [Smple Prob. 5.13]. For bod of revolution, tese slbs re circulr nd teir volume is given in Fig. 5.28. d Fig. 5.28 el r el = dv = r 2 d

REVIEW PRLEMS 5.137 nd 5.138 Locte te centroid of te plne re sown. 30 mm 54 mm = k 2 48 mm b = 4 in. Fig. P5.137 54 mm 72 mm Fig. P5.138 = 8 in. 5.139 Te frme for sign is fbricted from tin, flt steel br stock of mss per unit lengt 4.73 kg/m. Te frme is supported b pin t nd b cble. Determine () te tension in te cble, (b) te rection t. 0.6 m 0.8 m 0.75 m R 0.2 m Fig. P5.139 1.35 m 5.140 Determine b direct integrtion te centroid of te re sown. Epress our nswer in terms of nd. = k(1 c 2 ) = m + b 278 Fig. P5.140

5.141 Determine b direct integrtion te centroid of te re sown. Epress our nswer in terms of nd b. Review Problems 279 = 2b c 2 2b b = k 2 4 in. Fig. P5.141 4 in. 5.142 Knowing tt two equl cps ve been removed from 10-in.- dimeter wooden spere, determine te totl surfce re of te remining portion. Fig. P5.142 10 in. 5.143 Determine te rections t te bem supports for te given loding. 900 N/m Fig. P5.143 3 m 1 m 5.144 bem is subjected to linerl distributed downwrd lod nd rests on two wide supports nd DE, wic eert uniforml distributed upwrd lods s sown. Determine te vlues of w nd w DE corresponding to equilibrium wen w 5 600 N/m. w D E w 1200 N/m F 0.6 m 0.8 m 3.1 m 6 m w DE 1.0 m Fig. P5.144 d 5.145 Te squre gte is eld in te position sown b inges long its top edge nd b ser pin t. For dept of wter d 5 3.5 ft, determine te force eerted on te gte b te ser pin. 1.8 ft Fig. P5.145 30

280 Distributed Forces: entroids nd enters of Grvit 5.146 onsider te composite bod sown. Determine () te vlue of wen 5 L/2, (b) te rtio /L for wic 5 L. b L b 2 Fig. P5.146 5.147 Locte te center of grvit of te seet-metl form sown. 0.12 m r = 0.18 m 0.16 m 0.05 m 0.1 m 0.2 m Fig. P5.147 5.148 Locte te centroid of te volume obtined b rotting te sded re bout te is. = (1 1 ) 1 m 3 m Fig. P5.148

MPUTER PRLEMS 5.1 bem is to crr series of uniform nd uniforml vring distributed lods s sown in prt of te figure. Divide te re under ec portion of te lod curve into two tringles (see Smple Prob. 5.9), nd ten write computer progrm tt cn be used to clculte te rections t nd. Use tis progrm to clculte te rections t te supports for te bems sown in prts b nd c of te figure. w 0 w 1 w 2 w n+ 1 300 lb/ft 400 lb/ft 240 lb/ft 150 lb/ft 420 lb/ft L 01 L 12 L 3 ft 2 ft 5 ft 4.5 ft 4 ft 3 ft 3.5 ft Fig. P5.1 () (b) (c) 5.2 Te tree-dimensionl structure sown is fbricted from five tin steel rods of equl dimeter. Write computer progrm tt cn be used to clculte te coordintes of te center of grvit of te structure. Use tis progrm to locte te center of grvit wen () 5 12 m, R 5 5 m, 5 90 ; (b) 5 570 mm, R 5 760 mm, 5 30 ; (c) 5 21 m, R 5 20 m, 5 135. R α Fig. P5.2 d 5.3 n open tnk is to be slowl filled wit wter. (Te densit of wter is 10 3 kg/m 3.) Write computer progrm tt cn be used to determine te resultnt of te pressure forces eerted b te wter on 1-m-wide section of side of te tnk. Determine te resultnt of te pressure forces for vlues of d from 0 to 3 m using 0.25-m increments. Fig. P5.3 2.1 m 60 281

282 Distributed Forces: entroids nd enters of Grvit 5.4 pproimte te curve sown using 10 strigt-line segments, nd ten write computer progrm tt cn be used to determine te loction of te centroid of te line. Use tis progrm to determine te loction of te centroid wen () 5 1 in., L 5 11 in., 5 2 in.; (b) 5 2 in., L 5 17 in., 5 4 in.; (c) 5 5 in., L 5 12 in., 5 1 in. = (1 ) L 10 Fig. P5.4 L 5.5 pproimte te generl spndrel sown using series of n rectngles, ec of widt D nd of te form bcc9b9, nd ten write computer progrm tt cn be used to clculte te coordintes of te centroid of te re. Use tis progrm to locte te centroid wen () m 5 2, 5 80 mm, 5 80 mm; (b) m 5 2, 5 80 mm, 5 500 mm; (c) m 5 5, 5 80 mm, 5 80 mm; (d) m 5 5, 5 80 mm, 5 500 mm. In ec cse, compre te nswers obtined to te ect vlues of nd computed from te formuls given in Fig. 5.8 nd determine te percentge error. Δ 2 = k m d c d c b Δ b Fig. P5.5 5.6 Solve Prob. 5.5, using rectngles of te form bdd9b9. *5.7 frmer sks group of engineering students to determine te volume of wter in smll pond. Using cord, te students first estblis 2 3 2-ft grid cross te pond nd ten record te dept of te wter, in feet, t ec intersection point of te grid (see te ccompning tble). Write computer progrm tt cn be used to determine () te volume

of wter in te pond, (b) te loction of te center of grvit of te wter. pproimte te dept of ec 2 3 2-ft element of wter using te verge of te wter depts t te four corners of te element. omputer Problems 283 ord ord 1 2 3 4 5 6 7 8 9 10 1............ 0 0 0......... 2...... 0 0 0 1 0 0 0... 3... 0 0 1 3 3 3 1 0 0 4 0 0 1 3 6 6 6 3 1 0 5 0 1 3 6 8 8 6 3 1 0 6 0 1 3 6 8 7 7 3 0 0 7 0 3 4 6 6 6 4 1 0... 8 0 3 3 3 3 3 1 0 0... 9 0 0 0 1 1 0 0 0...... 10...... 0 0 0 0............

Trusses, suc s tis Prtt-stle cntilever rc bridge in New York Stte, provide bot prcticl nd n economicl solution to mn engineering problems. 284