Problem 8'1 8.1Two solid cylindricalrods l8 and BCareweldedtogethertt B andtoaded as shown. Ifurowingthat d; = 50 mm md d2= 30 mm, find the avcfiage norrnslstressat thcmidretim of (a) rodlf, (6) rodBC. = 70 3@ mm ILU = 7D ,t lD= N (So)t = t.q#s*1o3-,,'l= t-ct635' /o"'^' '>f*fl Di ' g= 3 5 - 7 x l o oP o 6e = 35't 250 mm MPa I = 3a* lo3 N , (A)" = 706 -86 n,.' = lo?-96x lti' -" loxlo3 - = +Z-4ylo"pa > 6,gGr l o' e Probrem 82 :"' Oe. " +2.4 MPa. 4 3*ff'tri'ff1'',tffi:;*#3#"fi#:,Tlti::?:'ls.'1flffi$if# rod, determinc the smallest allowable values of d1 and d2. = 10 klJ = '?O >tto3 hJ 300 mm '- = 'lP )," Ifd,^ = 2^q A o2,xlo-3r' -4 -@ 1m d, = 2S'2 250 mm h,.|4 'J\ = 3 o * 1 o sN P-=-tr- trd] r/TF..@..f d.=/#*" Id TTA;.- vh = rc.sTxto" ^n d " =1 6 - 5 Z pR.opRIETAR.y *|ATERIAL O 20il The McGraw-Hill comprnicl lnc. All nghts rcscrved.Ho prt of thir Mrnual.ryv-F dbpla]rcd' rcpmduccd, the limited dlcfrution to tcechcn rd or dictributcd in rny form or by any ncrni, without the prior r".itlo pcrmimion ofihe publisher,or usedbeyond it without p€rmislbn' e&lcatorr pcnnitrd uy u"ct *-ttitt for thcir individual courseprcpontion. Strdcntsusingthis manualareusing < --F PrcbremB.3f###trf#fl#ffi,T#'l*.s*:,ti#iffi# BC. 8.3 Two solid cylindrical rodsAB andBC are weldedtogetherat B and fo"A"O* shown. Determinethe averagenormalstressat themidsectionof (a) rodl8, (b) rod BC. (@) R"J AB. P = {o hps (*e.n:io". ) A."={4= rl rlal-=3.1.11 -^n T OAn o.e=fi-effiZ = t?.'7 (\) R.l Bclc;ps.. O lcips F . &lo- C?Xgo) = - 2O i.e\.e. 20 iora ion .. 2o k: lJps oa,rpress ?s cc olrzrfv^eSs n({':7-oG 86ina d e t -r -r 6 - -- v-T7- Oe 86 irn" A.< o."-=E Aa. Pmblgm ^ AAB = .3 .f = -2o -?-81 ksi ?-83 ks,' 4 E.4 In Ptob. Prob. 8.3, tra 8.3, d€tnninc determine thc the magnihde magninrde of the the force fonccP which the P for for which the tensile tcnsilc !tscss sfrcss rod lB has in rod,l.8 has the the sune samemrgaitudc magnitude asthc the compressive stressin rod rod rC, BC. compressivcstr6s 8.4 l. ,-\r ) : t'(Z a loedcdrs and loadod together at B and weldcdtogcdrGr-rt, 8C are are welded and BC rods l8 and Two lolid solid cylindrical cylinddcalrods,t, r !33 Two (b) tod ro (a) rod,{A, rod AB, (D) of (a) midsection of at the the mirlscction stress at normal st€ss average nonul the averigp rown. Daennine Daermine the fiovn. 3'l\16in- BC. c' P - fi-:l- = = 7-oea6 3 in. 3 in' = O-3tB3t P A."= S (a)' ' 7-oese in" - (2\(3")- P r-v1 A^" = f? , _=! 7 - 6 eA g E1',i\^? -- O?-18;g -- s-lr.iaz +" Pp l.{ l +? o-r.r 3 6a. l. 6i".y ,) O.3r83r P . 8 -'9t 888s33-- o . lrtyltq. {?? P p= = tf S 8 -. 4 6( k ,: o?,s < P No part of this Manuelmly bc dirpUpC, rcproducc4 PROPRIETAR.YMATERIAL. O 2011The McGraw-Hill Companies,Inc. All rightsreserved. or dirtributcd in rny fom or by any means,without the prior written permissionof the publisher,or usedbeyondthe limitcd dbhibution b tcrch€rt ild odupatonpcrtnittcdby McGnw-Hill for their individualcoursepreparation. Studentsusingthis manualareusingit withoutpcnnirbn. E J Link 8O con*i*c of a ainglebar 30 mrn wide ad 12 mm thick. Ifuowing thrt each pin hs! a lO-ilet diraldor, deterrnincthe rmximurn value of thc sy€ragG norurelstncsg in link BD if (a) d= 0o,(b) g= 90" Problem8.5 Use bor A Be as fr'ee boill , 2o Rn DEMI = oi ( a) es O ( o . l S o sin 30' )(|ortor ) - ( 0.300 cos 3o') FcD = F"o = (b) e a ?oo 1 1. g Q x p 3 N, i-e. l"*eio,t ( o.gso c.,s 3o'Xaorros) - (o.3oo c'rs 3oo ) Ftr E F6o : - 3o r ros N J o \"e. ca\P1e'ssium Areos (al *e"sin, I"");ntr: fr= ( o , 0 3 o - o - o t o ) ( 0 . o t a . \i (b) cor'rprcrsio.,ri ff: (o.ogo Xo.ot?,) = 24or/d' 86o r/dG h4r -g'lne$ses (o\ 6- E G) 6= 1 7 . 32 x l o g 2tlo x fo'3 - 3 o Fr o s 3go r I gl-6 ' 12.2 t lo3 = - I3.grfo6 7?.2Mfu - 8A.?MP+< problem 9.6 t.6 Iftrowing t the cenfial portion of the link 8D has a uniform cross-smtiond 1 areaof t00 mm' determinethe magnitudeof the load P for which the norrnolsfress, BD is 50 MPa. b MPa- = Soxlo'?a t Eoo ,.rnl = 8oo ,( lo- vt' iroAuo = (t5ox,os)(too*lo') = 4o r,o* U boil-1 ),nr4wt ,F bAf Ats D . +)M.= o: Lo.+so)(# tr") - o.rsgP = o P = l-scg| F"o ''') P=_(t.::::li:r P = 6?.7 kU -{ Problem 8.7 8.2Link,{chasr uniform rectangular cross section Vein. thick and I in. widc. Detennine the nor nal stressin the central portion of the link. 'o*e *.oqe-*lne,. ,u.!lejs'or 4 N"ie fl'at le",s iorn li lZoolL - in covpl e +. act ora *he bol t- f;" b>r'n^ = <>: z ._ r tB - (12++ )( r^" ..,s 3o' ) + ( lo )( Fo. sir, 3o') F - - _Ql ? 6 ? I Ac Aweo. "f f6 c a s 3 o ._ l o s i n 3 oo l;. k AC: Strcss i" /;nk AC i l TOO = O - t 35. 50 ^pb. A = li,,"'' } ; . = a - les - 6o-.{' = -t.. A : - Cj.t1*S r L .r'r IOS{ P', = l,Ot# ksi { 8,t Two horizontal 5-kip forces are applied to pin I of the assemblyshown.. Knowing that a pin of 0,8-in. diameteris usedat eachconnection,determinethe ; morimumvalueof the averagenormalstress(a) in lnkAB, (b) in link BC. Problem8.8 Use joi,^f B d,s {ne" b"J1. F*u lc k'P. c tnian Xl. Fornc lo 5in 75' sin 6d tr^" : Fu. = 8.9658 kips. 1.j2os kips L i nk A B is a- *gnsiora t'r €nn te^, t An"t = ( t - g - o . 8 X o - S ) = o . s i n Mit irarJrtr s ec+iov4qt p ia i (*) Stress i, AB, 6^*= ffi F- = -a{ ZIM = l+.G+ ksi is a. ooh4p Fess io"r tn €r* L en. (1.8)(Ct-s) C y ^ c r sS s golict^e,) 4Fe4 is A= L;^ k BC (b) St'ressin tsC: -Eg (s^r= -1-\)Bc A -8'76s! = O.q = in" 1.gG ksi 8.9 For the Pratt bridgc truss and loading shown, determinethe averagenmnral i areaof that mcmbcris 5.87 stnees in mcmbcr8E, knowing that the cross-sectional in2. Problem8.9 Use enfire, *n.. a.sf,r.. b*ly' ZM* = o: (qXto)+(texao)+(ztl(ao)- 36 At = o Ar = l2o kiTs stt kip"- Sttkips g fit) kips Use p"*i"^ Fco of *r^.rrs *o *he /€)+ c,rl-h'na m€mbe.^s BD, BE, c+nd af a. s ee*io*t cF. +f Zry = o: l2o - 8o #n, = o F"c ,SttHps S{.}kips Feo <{ 8.10 Knowing that the average normal stress in mernber CE of the Pratt bridge truss i shown must not srceed 2l ksi for the given loading, determine thc croes-tGctional , area of thet member that will yield the most economical and safe design. Assume . that,both qrds of the me,mber will be adequately rEinforced. rsft Sttkips = 8'r;2 ksi 6es=ff= # Problem 8.10 ; F." -- 5o k,i'= Use er,*ire *r".rss 4s $n** bo*, DZMH = o'(q\(ao)+ (r8X8o) +(azXao) sC 4 Ay -- l?o kips s o Dsa po*ion .tF *russ h *l.te -A"fr of o seu{:ian cultinl rr€h bens BD, BE, enJ C F CZMc=o: le Eo :" (q[rao) = o :. F.e = 1o k'pt 10 hiy* i +.261 int { 8'11 A goupleM of magnitude1500N . m is appliedto the crankof an engine. For the positionshown,determine(a) the force P requireato hold the engineJystemin equilibrium,.(D) the avercgenormalstressin thi connectingrodBC,which has a z+)u-rnmunrtormcrosssection. Problem 8.11 Use pisforar rodr an) crank (o.a8orn)t1 - l5oo f,l-pr = () H s .5.35?lxlos N 60 mm 'AY Urg f;itq, 1!on,"o.,\" 2: = ?oE .81 Frh^ P: l?.86xfo3 P- l-7,gc lN lv Fg.= lg.€+s xlos N Rod BC b cor.aprae.gst'o'ri.'gr.^ben. q. Fn = Tfs qlaa. is +-fo p,r,rz= ?f,o xlOt - / S- G + 3 xl o 3 = - l l . L +x r o c ? q . +so .,;:-" 6u.= -+l-+ MPa- PROPRTETARYMATERIAL' o 2011The McGraw-Hill Companies,Inc..All rights reserved.No part of this Manualmay be displayed,reproduced, or distributedin any form or by any means,withoutthe prior writien peimissionof the publisher, or usedbeyondth€ limiteddistributionto teachercand pcrmittedby McGraw-Hillfor their individualcoursepreparation.Students educators usingthis manualareusingit withoutpermission. ml E.12 Two hydraulic cylindws are used to control the position of the robotic rrn ABC. Knowing that the control rods attached at A and D each have a 20-mm ; disrneter and happen to be parallel in the position shown, dctermine the avcrrge t normal stressin (a) memberlE, (b) member DG. Prublem 8,12 150mm EOIIN c use men/lp^Age a.s S rea bo)y. E 200mm (0.rso) + Fo. DzMB?o; Areo, df ,,,e*. Le" A E iq (O. 6ooxgoo) = O A = {d"= = St+ t4 x tdL rn"- {(no*6")' r= = Ft. = +x1c3 = "FE A 3lT. 16xlo-- Ee= eFrlogN 12.1?xlo' ?<- 6rp = 12.73 MPa Use co-r b;ncJ yutpn6e,^sABC o-) BFD 4s fnee hoJ,y, (o,tso[#F_)- (o.eoox$ F*) ( t.osa - O.3.So)(6ao) = 11 Av e-^ in fs,i D G Slncss i'. noJ DC I rs Fo+ !i r l5oo N fl= fi )' = {(a"rto*)' = gtl,t6*ti' rnt q" a +t= A -r5oo - = -+.-tixlo^ pa 3. lt{l6xlce (b) 5o* = 1.17 MPa '- PROFR'If,TARY MATERIAL- @ 20ll The McGraw-Hilt Corryenicr, IBc. All rights reserved.No part of thi,r Manurl nuy bc dirphyrd, rGprp61lcc4 or dirtributed in rny form or by eny means,without the prir wriucn pcrmissionofihe publisher,or usedbcfrord the limitod distributioi to tcachen md educatorrpamittod by McGnw-Hill for their individual counGprcprrttion. Studentsusingthis manualareuting it without pcnniglm, E /3 The woodenmembersA andB arc to bejoined by plywoodspliceplateswhich I will be fully gluedon the surfacesin contact. As part of the designof thejoint, and i knowing that the clearancebetweenthe ends of the membersis to be 8 mm, r determinethe smallestallowablelengthZ if the avera1e shearingstressin the glue is I not to exceed800kPa. 8.13 Problem 4 L Thene. clne Su^ se,pa.rals 4rtao-;s oi} Xlve. ry onea- hnus+frnansr,"i| ha-l| ,rf +l'.c--2+hN -!oJ, ) T-h"*cfare lfi) mm tr = le "&U : t/.x tos N Sheo.""X gfress ia . {v -t t = a- xlo3 Soor(I O F Le* tI" A '/o t = area vJl = , u i l h h a'nJ = 8 a ( , ^ /t o s F q | 5 >t lo'" loo F1h : o.l /sY to-3 = A I So v rrj3 o,jW ( a ) ( t s o ) + I = 3og ^nr^ y.n = h1 )Yl l5d2 p,q { 8.14 Determine the diameter of the largest circular hole that can be punched into a r of polystyrene 6 mm thick, knowing that the force exerted by the punch is 45 ; -sfeet kN and that a 55-MPa average shearing sffess is required to cause the material to fail. Problem 8.14 SOLUTION A = TTdt f"^ .y/i nJri .^l ; S h" orl nfl s+Y"€ss ETud'n? A's, S'fvintrf.n C, S^,'/r,ro S t,rl q.q- - . tvl FP rTdt = e= -g +3,1x/en d = +3-1 wn, Problem 8.15 8./5 Two wooden planks, each'A in. thick and 6 in. wide, are joined by the glued i mortise joint shown. Knowing that the joint will fail when the average shearing , stress in the glue reaches 120 psi, determine the smallest allowable length d of the r cuts if the joint is to withstand an axial load of magnitude P : 1200 lb. Seven sor faces 6arY^/ *he t"*.P 1""A f = t?oa lb,, A = cz)G)d = fr3 :- A= A P L Pa= 2, r uin.-l Fo'^ Tt= A" = rrJt Fon o -er* r h U t n 'tz=fi L i'.,'/ i" X j. v n.lu, d = 1.G43in. .{ 8.16 A load P is applied to a steel rod supported as shown by un aluminum plate into ' which a 0.6-in.-diameter hole has been drilled. Knowing that the shearing stress : must not exceed l8 ksi in the steel rod and l0 ksi in the aluminum plate, determine : the largest load P that may be applied to the rod. Problem 8.16 f- -P stee^0 A,= r Jt = Tf(o.oXo.q) P. 4" O.1g+O i" p = A,t, s (o,Zs+eY fg) : .p = Art" = (t.2566xto-; F "f P is th. L sua+.llen'rtJve. 13.57 k ips l?. f,? I(ips P = t2.5'7klas 8.17 An axial load P is supported by a shon W250 x .67 column of cross-sectionalarea I : 8580 mm2 and is distributed to a concrete foundatio:r by a square plate as shown. Knowing that the average normal stress in the column must not excced 150 MPa and that the bearing stresson th" concrete foundation must not excred 12.5 MPa, dctcnnine the side a of the plate that will provide the most economical and safe design. Problem 8.17 A,reo "F colo^n i A = &,:son,i = &fgo * 164v4 'sfness in c o l o r . , . r , r i S = t S ? C Y ! O 6P a " Nor rnc,l g= F .'. p= A g' : g s a o i n Qg I J e : Ab= b= t'= JS-g _E_n " C Au l.?87 ,( log l ? . . sx I 0 6 0-T Problem.8.18 = (gsaoxro"XlSo 'tCIa) t.787 x l}e N ? Ab--i fo" s1oa-'.e- p./d.*e lo&D x 1d3 a lL=get o, gQ-l ln Yu,ry 4 t.!t The axial force in the columnsupportingthe timberbeamshownis P = 75 tfrI. Determinethe smallestallowablelengthI of thebearingplateif thebearingstressin t thetimberis not to excecd3.0 MPa. 6t=fr = _E_ Lw Solvintrfo* L., I t: A trlm r m 118.6 xlO t = 17 8 .6 tn'u1 PROPRIETARY MATERIAL. O 2011The McGraw-Hill Companies,Inc. All rightsreserved,No part of this Manual may be displayed,reproduced, or disfributedin any form or by anymeans,withoutthe prior written permissionof the publisher,or usedbeyondthe limited distributionto teachersand permittedby McGraw-Hillfor their individualcoursepreparation. educators usingthis manualareusingit without permission. Students Problem8.19 E,I9 Tltree woode,n planls arc fastened together by a series of bolts to form a column. The diamcter of each bolt is Vz in. andthe inner diameter of each washcr ic Vt in., which is slightly largcr than the diameter of the holes in the plankr. Determine thd smallcst allowable outer diametq d of the washers, knowing that the av€,ragenormal stress in the bolts is 5 ksi and that the bearing sfess betweeir the wa$ers and the planks must not exceed 1.2 ksi. BoWt AuJt= Sdr" = +(*)' -- o. ti63f iu' t;te fr^". in bolf, p =,6oA -,(5 )( o.'t?63s)= o.1sl?5.kip Q = * :- Te,n i.srJe J;or^e-len= d; = # ir., oufsiileJ;,"",*.<^= 4o W*sher: P , ^ - r ,a Beori^3q.*eqiA* = +( 4: - d;. ) ? E7u#;''3,+U: - di') = s't do'=,4'++& = €)' + t' TT *^J A* = & o . 1 8 _ : - /J5 _ l . + g 2 3 i n z t? .5) do= Problem8.20 t l-lq7 in- E.20Link lB, of width b = 2 in. andthicknesst = Ynin., is usedto supportthe endof a horizontalbcarn. Knowing that the averagenormal stressin the link is -20 ksi and that the ever.Ee*rwing stressin eachof the two pins is 12 ksi, determine(a) thc diameterd of ilre pins, (b) the averagebearingstressin the link. Root A B is in Corr"Pr€5s r'Dlf, A= b t u , r l r e r c b = 2 i " , . o n o l l = P= Pin : (o.) d=ffi = oXaxt)= +in. to kips o,"d A? = +d' = l.O3O in. I PnO?R.IETARY MATERIAL O 2011The McGraw-Hill Companies,Inc. All riShb rcrcrvcd,No part of this Manualmay be displayed,rcprdm€d, or distributcd in rny forn or by rny means,without the prior writtcn permissionof thc pblirhar, or usedbeyondthe limited distribution to bechsn rnd Gfttcstorfpqtnitt€d by McCraw-Hillfor their individualcourseprrepamtion. Studentc ruing thir manualut uring it withoutpermission. E t Two horizontal 5-kip forces are applied to pin B of the assembly shown. Knowing that a pin of 0.8-in. diarneter is used at each connection, determine the maximum value of th_eaveragenormal stress(a) in linklB, (b) in linkBC. Problem8.21 0.5 in. t.21 For the assembly and loading of Prob. 8.8, determine (a) the average shearing strss in the pin at A, (b) the averagebearing stressat A in mernber,4B. B c.s^a$ree b oJy. U,.. j oi'* e l . ;) KtPs ;) Klps J l - 0.5in. IO IIJF Larv "l lo l;ps tnra nTfe Fornce S i', es Ee F 1.3?af J,,P, 'AB- si"r {5' (a-) Shear,'n st nessin pi*r a+ A . f= ? rrrhe+aA.' tA^ = *to-8)"= 0.5026t.4z 2Ar -7^8 ksi (b) 9eo^J"r # n s: a-l A Au: 6b= tA F tfirg- Au l" -,r,,e*,ben AB, (o-sXo,8) = O-+ 7'-32.o{ = 0-+ | g.3o .1 l,'l l?.3o ks; < PROPRIETARY MATDRIAL. O 20ll The McGraw-Hill Companies,Inc. All rightsreserved.No part of this Manualmay be displaycd,reproduced, or distibutod in any form or by any rcaru, withoutthe prior writtenpermissionof thepublisher,or usedbeyondthc limiteddistributionto teachcrsard permittcdby McGraw-Hillfor their individualcoursepreparation. educators Students usingthis manualureoiing it withoutpermission. l Problem8.22 t.22 The hydrauliccylinder CF, which partially controlsthe positionof rod DE, has gb is% in. thickand is conncctcdto the @ lgckedin the positionshown. Member -vclticalrod by a %-in'-diaurctsbolt. Determine(c) the averageshearingsbcrs in the bolt, (r) thebearingstressat C inmember.gD. bo&t., a-J Use rrnevaber BCD a-s o Se. vrole. th of AB is a. *uo {or.e ,;L*,bef', 4oo t{oo cs75" . $tn <!'C /S cr 8r- -t_ E.2 (Qro.lu")(&fi")- DzMe=oi (+siq?"o"\(#n") - (1c.rs{oo)(+uo \2fr=o'- -'* C* = -t-r- +t z\ [* + Cx + + Cl r\ uy , , ^qq 'a) - - Sheq,.i* tloo + +oo sia 750 = lcr" + C, c',-'75') = 4r.ocosTso 'Ioo c,e.s ?So = W = o: C: sia 7.1*) - ( 7 si-r lo"Xtoo 7?97.,35 = o 3-3e€78 F^, :i . Ae 7f.-3'I lL. r +aQ srr 75- = Q = ll?+.cs^gL. ltq7.7 ,lb 6.,^Pf ssff rcss r c s s ii"q ff,'h te b "!tt 5 P = ll17-? IL A ={a" = +(t)' ' o. troqsin' t = * = #tr# ) B-- e a , ^ i n q s l L r . e s s# C Au= dL= q: s- A, : lo.8+"ro3p,i i^ -e,-'A€^BCD.: (*Xf): o.Ls+s?.fin' = llq?'?- = ',tlxlosps, o.23rt37s lo.8{ ksf < p= IIq?.L !lr, = 5- lf ks, < Problem8.23 t'.tJ Knowing that & = 40" and P = 9 kN, determine(a) the smallestallowcblc $lg""f of thc pin at d if thc averageshearingstressin the pin is to not exceedl?0 MPa, (e) the correspondingaverqgebearing stressin mernberAB at B, (c) the correspondingaveragebearingstressin eachof the supportbracketsat.B. 16 mm{ Geor'n"fv"y: Trr'ar,Xle ABC rs c..'r isose,les fuianSle wi]ln e ant3les Shorun h€re . Use jo;nt A c-g o -fr.* bily Lqw of Sines npfl ;"4 lei +" foncc *nr'r.,a1 ? Fanc e tu.iarrXle ]P-sinl2" l=oo = Flq sinIlO"-- si"tfOo F; I s \ (a) Affo,.ro,Llepi', J;o*. rie., = & : 2Ar -+ 2Td' nd" (3Xa'|.zs[tgE). = A--- 1?et= f= + 1lt, n(taortD" ) wLr,.*F*e=a'J,7Jr, lo3lJ |gt.fgxto-c,' d = l f . 9 5 , 1 o - 3u ll.95 r',lrn bl EeqrinT slne"s i" AB o+ A, rr/o'3 Al= t ol = (0.otgX,t,l.qs ) = lA3.A6 ,lo-" h^a Sr: +t: Ar -1'l'!3xt'o3-= /33-?(, lo-6 tgrl.ivloc { 131.7P1Pq. c) A= fir= 1 3 ? . 9v l o - " n n qo.Oxlo' ?O,oPlP+{ E.24 Determine the largest load P that may be applied at I when 0 = 6C , knowing that the averageshearing stressin the l0-mm-diameter pin at B must not exceed 120 MPa and that the averagebearing stressin member AB and in the bracket at I must not exceed90 MPa. Problem8.24 P 16mmAs ' Gcorn.fry Tr,'qTJ" ABC is al.r i sose--Cestnia n1)e wi*ll ongles shou,rv, h etne. l2O" goi Usej"int A as $.. " b"Jy . l.au,r,rf s Ines ap f, h oJ {* lonle tnia n1)e i P P l 20 " ? sin 3oo Forre fuianl la F; I$ shec.inX sf rsss i', pi. .t B F^. =Fte'= P= ?= is cri |r'co,l.l sin 30" Sin f?g' Fo" =tl 3=o = g. s7735 F^. srn l2Oo t*ftff" = E. xrDt r' |, = $d" = T(o.oro)a= ?8.Sq (aX78,s',rto'c)(taovlot) = t8.85a*los N Fre= "A't,: + !eaninl slness in -e- ber AB "* brr.aJre*a* A is €rr't;.^i, ( 0 , o 1 6 ) ( o . o t o ) - r 1 6 o x l o ' 6r ' Ar= tJ: F*u: If Ab6t = (teoxto^cXqoxlo6)=l.l.ToxlLf A/ ber^in3 shncrs in *[,e bnaolcef a-] ts ,'s cni]i .4, Au= 2t,d = Rl;(o.ot2{o.oto)r ?qoylro-6;' F r e = A r 6 o ; ( 2 1 o * 1 6 c , ) ( q o" , o " ) = 2 l - € - * / o ' N A.0!""t.$Je Then, F4e is tl,e snq/J*t '..F,,ra,^, S4#f "s Prr*" t.e. lrl .tlo x ,,esM (0. -sz?gg )( l't.*ox lo3) = 8 - 3 l , ( l o?' A / 83t AN < Problem8.25 E.25Two woodenmernbersof 3 x 6-in. uniformrectangularcrosssectionarejoined by the simplcglued scarfsplice shown. Knowing that P = 2400 lb, determinethe in thegluedsplice. normalandshearingstrcsses = 5o' O = ?oo-4oo Ao = (sXG) = P: Z t l o oj L l8 i',r (aroo) er3d o 5 = #.^t"t = 5'f. t Pri r=#sial.e- = 6.t.7 fl Problem8.26 4 joined E.26Two woodenmernbersof 3 x 6-in. unifonn rectangularcrosssectionare allowable by the simple glued scarf splice shown. Knowing that the maximum rtt"*ing stressiti ttt" gluedspliceis 90 psi, determine(a) the largestload P that can tensilestressin the splice. bc safelyapplied,(b) thecorresponding o=.9 0 0 - { o o =500 A o = (3XG) = b PI = 18 f'"t fi'Areinz€ sin Ze = sLTo = (axrsJi:e\ sin lr)c)' P - 321tt lb. (,a) ff= < 3?9o c.'sesd t8 = 75-S fl = 7€.F psi PROPRIETARY MATERIAL. O 2011The McGraw-Hill Companies, No part of this Manualmay be displayed,reproduced, Inc. All righs reserved. or distributedin anyform or by any means,withoutthe prior written permissionof thepublisher,or usedbeyondthe limited distributionto teachersand educators permittedby McGraw-Hillfor their individualcoursepreparation. Studensusingthis manualareusingit without permission. < 4' E.27 The 6-kN load P is supportedby two woodenmembersof 75 x 125-mm ,uniform cross section that are joined by the simple glued scarf splice shown. in the gluedsplice. Determinethe normalandshearingstresses Problem8.27 P 3 6 > 1b 3 l J Ao = (o.o7SXa, l2S) P _6 = fre"sre e = ?oo- 1oo = 20" = 1,375r(lo-3hn' { qxlgl) c,re"eo' _ 565. x to3 $ = 56-f (g*163) sia too = 20 6Y/os s i n 2 O = (a\(q.3?fx,o-s ) t-- t I28 Problem hPq = 2od kPe- E.2ETwo wooden mernbersof 75 x 125-mm uniform cross section are joined by the simple scarf splice shown. Knowing that the marcimum allowable tensile stress in the glued splice is 500 kPa, determine (a) the largest load P that can be safely supported,(D) the corresponding shearing stressin the splice. A. = /o, o75-)(o.lzs) = 9.-g?g .td' e = 7oo-7oo = ?O' 6 (a) L P s i ^ 2e = XAt + (b) rua I P= 5-s, kru (E3oa5rtlo3 ) si" 9o- : lgl.?q xroe t = f 8?-o kPo- PROPRIETARY MATERIAL. O 2011The McGraw-Hill Companies,Inc. All rightsreserved. No part of this Manualmay be displayed,reproduced, or distributedin any form or by any means,withoutthe prior written permissionof thepublisher,or usedbeyondthe limiteddistributionto teachersand pcrmittedby McGraw-Hillfor theirindividualcoursepreparation.Students educators usingthis manualareusingit withoutpermission. < PfOblem 8.29 8.29 A 240-kip load P is applied to the granite block shown. Determine the resulting maximum value of (c) the normal stress, (E) the shearing stress. Sprciff the orientation of the plane on which each of thesemaximum values occurs. (6)( E) = sE in' ?' +A o cos e- e'e tvtc*, t nsile. ={r*.s *+ e = ?oo t' O Yao.,,/r.co*-f |.45s i u€, s*rc sS *t e = c" t,o = #" = Affi1 = 3-33ksi ** probrem B.B0 < O = t+S" :#,**'#;f*l;#'lr':1i:j};,ffiJiiii:'f;ffr,ffH#i:l:; ttre magnitude of P, (D) the orientation of the surface on which the marimum shearing stres occurs, (c) the normal stress exerted on the zurface, (d) ttre maximum alue of the normal stressin the block. (6)(6 ) = 3a i"r" t,^o-n= 2-f ks; / sinTe=t (c\ 5+s= 2A--10" *, ".'"" (d) 6n-tt=#=# +5"= ,t \.q(-= ({ Xga )(r-s ) (ips 1 e={So < = = -2.{ksi< PROPRIETARY MATERIAL. @20ll The McGraw-Hill Companies, Inc, All rightsreserved.No part of this Manual may be displayed,reproduced, or distibuted in anyform or by any means,withoutthe prior written permissionof the publisher,or usedbeyondthe limited distribuiionto teachersand educators permittedby McGraw-Hillfor their individualcoursepreparation. usingthis manualareusingit without permission. Students Problem8.31 ao ni f" 3 *)" o.3oo vr,, = D' t$o t.l (>- t = 0. tro- O.ooG = O.l.14.Yor A. T(yL'- h") = fl (o.,so* - o. l+{t,) = s.St+ x, o-3 l'rt e= n50 c.,sr&5o 6:= + coste = ?go:F f4o =-37. I x loo t = & $- ='- 37-l HP6- a srr ?e = =-11.28x lo' ! = l7-?z |4fu 4 8.32 Astecl pipe of 300-mm outer diamcter is fabricated from 6-mm-thick plate by , we.lding along a helix that forms an angle of 25o with a plane perpardicular to the i axis ofthe pipe. Knowing that the ma:cimum allowable normal and shearing stressesL in the directions respectively normal and tangential to the weld are d= SOtvtpu and 1 = 30 MP4 determine the magnitude P of the largest axial force that can be ap'plied to the pipe. Problem8.42 tr\Sor^a f!=*J"' n= t - t = O . l S o - O . c n $ = O . , + t /h ^ A.'= T ( rot- nr') = r ( o- ,soa- o. ,9+1) do= 0.Wn Weld ?5" = T- S! r to-s w1z 0=25' MPa: 6= f, c,s'O = (Sgt xro-=)(soxl.a6)= gA7x los e',T MFa.: Z, = #-si'r 2e = (ZXs-s{ *f o-3X3o * toe) = +g4x lo3 srh 5o' SJfen Vafue is *le ^.0I"-J0e. vo.l,* "f P. E P= 337 KN ttSG i gff Link,4B is to be rd' of r *d for whichffit ultiffi mormalstrwr it r$ll of cefcy &sor fu sffi AB fu ih" ,ror.sc!fiid rlca for ] llF.;;;ir" thatthelink w,illlbt @telv reiahrd ffiDd &c pinslt,4 nd I il-lSo. il;; 1t Probhn 8.33 P = (t.?.)(e) = ?.d &tl tr^" t)znt"=Dl - (o,gXF^"s,',.3s1 + +(o-'t) (eo1 Fre = 3"1.€n kM Es: a "rrg Fre _ = ( E,s..) 6i* = t68-l vfo' w ? Ao*' t68-[ *r*t I Problem8.34 t 83{ MembcrABC, which is supportedby a pin and bracketat C urda cable.BD, wasdecignd to zuppofithe 4-kip load P as shown. Knowingthat theultimatclod for cableBD is 25 kips, determinethe factorof safetywith respectto cablefailure. Ose ]n€rnber ABC as 4. Sree bady. ort{ n ole fhaf ne#ber €D is a, {wo - Son"e rr€wr b erl8 in. ,n".J EM" = Q: (F e"s 4o') (gi,) + (P 40'Xrs i"-) - lfu cosgo"Xrs;".) "i' -(F?o sin 3o'Xt2 i.) = o F.o= +i# ? =# = G.87ts h;p, = F.g = trnt. = =S 6'87b 3'6+ '' R" Fa.at. of' s*Fe+' f. c*[le tsD: Problem8.35 E's-sKngwing Sratthc ultimate load for cableBD is 25 kips and that a factor of safetyof 3.2 rrith respectto cablefailure is required,determinethemagninrdeof the largestforceP thu canbe safelyappliedasshownto membcrABC. A TI Dse Fe€n Le. ABC as a- +*" il;- tS in a'^) no** A that v"oehber B D rs a. two - frrd"se lrrCrqber*, 18 in. (P"* ,,".-J ZM. = oi 9o")(3o ;.) + ( P si",{o"X ls ;") - (Fro cosJooXlsi.) = ,tt:'u.?"xta': 3r.62e FBD A!,!"*A,0" !"*) A)!"onil* IrJ =o O.SgZt6 F"o 'I" F* = Frr rrtl''be, tsD 3.Q 7.8125k.ps P = (o.fsetc\0.8 ta5) = +.ss kr'ps PROPRIETARY MATERIAL. O 201I The McGraw-Hill Companies,Inc. All rightsrercrvod.No port of this Manual may be displayed,reproduced, or dirtributed in any frtm or by my reans, without the prior writien permissionof the publirpq or urcd bcyondthe limited distribution to teachersand c&lcrtorspermittedby McGnr'-Hill for their individualcoutr" pt p"*tion. Students usingthi" menurlrrc ruing it withoutpermission. Problem8.36 t'tf Mcmbss AB andlC of thc trussshownconsist of barcof squarecrosssgtion is ofthesame alloywas *:i*:,y'^:PI._.It T:,*, r|u,12!-1m+uuoi* tcsM to failureandthatanultimate road l of 120kN ** ;;rd;."? saf€ty of 3'2 is t'o be achieved for both bars, "-#il of "r determinc ttre required dimensions th_ecross *Tion of (a) bar AB, (b)bar AC. I en3*!' L.4 ItI lL D \z z'M .e 0.85 y'n Dse enlrne |russ ag a Srve bdyt.aAx -(AlSXeg) = e Ar=15,(lr/ 6t 'l - ot +tIz zF, r! = o.?Seio.,t' .l^o I m g , , . , L g rA g : "f I Al =?8 ltN' Ay- l' ' us3le jro i n,+A a,s4^a. Au $ EF'= 17 kN Fp" _1 -t[ry=6t A,-F*-$IFre=o F^.= 2g - I+tJ!il. o.Es t r-i For tI" tet* Fa. tle bq, r,n-tet*,0 A = (o.ozo)' = g" - & vu = A 4oo r/d6 P, = t2o x los = ,l?o"{ol, = soo>ltae Pa, *ooxlo-6 o F. S. = -!-L = Fra, (q.\ h4' = 20 kN G,A = 614," Fre, Fro (3,1Xtzx/os) = / 8 1 - & 3 x l o - ' , q . 3o x rog 1 3 .1 ? x / o - 3 m Ec | .Y. P" =J9: rk |--. 3oo r loL t4-G, xro-s rn 13.47rr,rr,a< q,A vuf't l-* 6, b' =-}1!l.3 Fn" t 2 l3 .33 )./c>- r' 19.6 I ra*, 8.37 Problem steelbolts areto be usedto attachthe steelplate shown 8,37 Threot/, -in.-diameter the to a woodenbeam. Knowing that the plate will supporta}4't'tp load and that for safety of factor the determine ksi, ultimateshearingstressfor G steelusedis 52 this design. = o'4\tsin' Fo"ec,,,lb"Jt fi=fat=*(ff P" = Ato = (o.4+ts)(s:)= 2?-q7l*'Ps Pe"bo|t; p: t= 8k:Fs 8.38 Two plates,each 3 mm thick, are used to splice a plastic strip-as shown' Knowing that the ultimate shearingstressofthe bondingbetweenthe surfacesis 900 kpa"d*ennine the factorof safetywith respectto shearwhenP: 1500N' 8.38 Problem Bar/ a.n,a.: ( See f;g r^. ) [= ](a")(a")+(ls)(a.\ = l SOo -'," = l!;Do rloe m- LD *-_ ,5 r Pu= 2 A,yb u : Q\(tfioox td)(qoo 'fos ) = 2zoo 1., : 1,'c,o, = Es. = tu P 8.39 Two wooden mernbers of 3.5 x 5.5-in. uniform rectangular cross section are joined by the simple -rtr"rrglued scarf splice shown. Knowing that the maximum in the glued splice is ?5 psi, determine the largest axial shearing allowabti load P that can be safely applied. A o = ( 3 . 5 x S -s ) - - | q , 2 5 i n O = ?o"- 20' f = * P= si"6coso = ,fr*''"'2o @ s in l{o" = 4417 Ib 4.47 V,r,, PROPRIETARY MATERHL. @2011The McGraw-Hill Companies,Inc. All rightsreserved. No part of this Manual may be displayed,reproduced, or distributedin anyform or by any means,withoutthe prior written permissionof thepublisher,or usedbeyondthe limited distributionto teachersand educators permittedby McGraw-Hillfor their individualgoursepreparation. usingthis manualareusingit without permission. Students Problem8.40 t.40 A load P is supported as shown by a steel pin that has been inserted in a short woodcn mernber hanging from the ceiling. The ultimate strength of the wood usod is 60 MPa in tension and 7.5 MPa in shear, while the ultimate strength of the stccl is 150 MPa in shear. Knowing that the diameter of the pin is d = 16 mm and that the magnitude of the load is P:20 kN, determine (a) the factor of safety for the pin, (D) the required values of 6 and c if the factor of safety for the wooden member is to be the same as that found in part a for the pin. ?o t(N = F: -fi = + d;-= + ('' 0'6)l 2ol'06"l6q^ (a) Pi", : rl b lg Do,.r s I e4ra "t,=f; tu=ft- x/03) ?Ato = (2X?ot,t6xt66Xrso R W= 4Or+r., ? P'O.I*O 6 ?Ov IDNN 5 60,gt?x los N Fu - 6o.3t?xros= 3 - O 2 AO trlog F 6 0 . 3 1 1 v l l s N $ o " Sara € F, S. w|ene b-- C + = = O.O{D r',r U = 4O w'l''r- = +l-lxlog ('-016-1 'l/. l' tuln b = Pu = €O.gq Do.rLle sAea.j 4P, Lu g ao[r a-f?a- is " lDs N A = $o,* s4.h^e ES. WC 2A ? c -- Pu = - 6o.9!1 x lD3 - = (iXo"ovo)(7.sY/o') lOO.5 x/ds rn C = loo.{ mn FROPRIETARY MATERHL. O 20lI The McGraw-Hill Companies,Inc. AII rightsreserved.No part of this Mrnual mey bc disph),td, repruducod, or dirtibutod in my form or by eny rrcNns,without the prior written permissionof the publisher,or usedbeyord thc limitcd distribution to terchen rnd e&rcaton pcrnittcd by McGnw-Hill for thcir individual coursepreparation.Studentsusingthis manualare usingit witbut pcrmisbn Problem8.41 5 t.4l A steelplate TI in. thick is embeddedin a horizontalconcreteslabandir usGd to anchora high-shengthvertical cableas shown. Thc diameterof the hole in thc plate is '/e in., the ultimate $trengthof the stccl used is 36 ksi, and the ultimatc bondingstressbetwcenplate and concreteis 300 psi. Knowing that a factor of ssfetyof 3.60is desiredwhenP = 2.5 kips, determine(a) therequiredwidth a of thc plalc,(6) the minimumd€pth6 to which a plateof thatwidth shouldbe embcddedin the concreteslab. (Neglcct the normal stre$cesbetwecnthe concreteand the lowcr endof theplate.) 2.5 lrJpt B ,s e"l ov1*ens ia^ i n p!**e', A E (A-J) L P, = 6'4 E,S- = Pu = F So.0u'ni {or fl-= d + Bosed or, * h e.rr Q-) ( ns.)P 6r 1s.5o)(as) LseXf ) (a) o- bet-ee.^ p,0r"ie anJ eonc,t,v..*e s.loL ". A = pe^ no*er * Jep*h= B= T"A = 2f"(a+t)b Solv drr3, f"," b, b= GI P 2(a+a) b { t, = 0.3oo }zsi P" FiS. = P (3.G) ( z,5 ) (2 X/.sro*f,Xo. 3oo) b = 8- oS ,tn. r<t Pn'0PRIETARY I{ATERIAL' o 2011 The McGnw'Hill Congrd*, Im. All rfihts rcaenrud.No part of this Manurl mey bc dirplrrr4 meroar*sd, or dfutibutGdin any fwm or by any means,withou! -nt ptl* *titlo pcrmirsion or-trrcprutirtrcr, or usedbeyondthe lirnitod dirtrfrution to tsrchon rd educrtorr pormittcoby McGnw-Hill for their indivirtrnl couracpnfrrtto. Shrdcntruring thi* manualareusing it withut pcrmirrbn 5 is uscdto t.41 A steelplatelc in. ttrick is embeddedin a horizontalconcreteslaband plrtc it in the hole the of diamefcr Thc anchora high-strengthvertical cablc as shown. strcss bonding ultimate the and ksi, is 36 used stcel the of % in., the ultimateJtrength OT3n plateandconcreteis:00 nsi. Problem8.42 t.fi Detcrminethe factor of safetyfor the cableanchorin Prob. 8.41 whenP = 3 kips,knowingthata= 2in. andb = 7.5 in. 3 kfps B osed on *ens,'ovt i'a PIa*e i A = (2' + X#) = 0.3106in' Pu= 6rA ltips = B6 Xo.3?06)= \q.OG F.S.= F= *# B o.seJ oh sheor be*weem pla[e and -- +.Gg concreJe sla.E3 A= penimelerr Jept[, = ?(a-+ t) b = 2(a+#)(z,s) Tu = O . 3 o o l * i A= 3+ . Gq int = lo.+l kips P"= tuA = (o,?oo)(g+"r-?') F. S. = 3r- = P Ac*lrol ln in lo' til- ?= 3'+7 o| so|+ly is *Ae snal.ler ur,P.re, ES.= 3.17{ PR'OPRIETARY MATEruAL. O 201I The McGraw-HillConmnis, Im. All rightsreserved. No part of tqisManul Imy bc diryhyl{ Fproducd, or dbtrbutad in rny form or by any means,without the prir rriucn pcrmirrion ofihe publisher,or uied beyondthe limibd amrituti.rn to terchsn rnd c&lortor! pcldtd by Mcfnw-Hill for their individual courrc prqrntion Studentsusingthis manualareusing it without permiubn. Problem8.43 E dJ In the structure shown, an 8-mm-diametcr pin is used at14 and l2-mm-dirmcfcr pins are used at B andD. Knowing that the ultimate shearing stress is 100 MPe af all connections and the ultimate normal stress is 250 MPa in each of the two links joining B and D, determine the allowable load I if an overall factor of safety of 3.0 Top view 8mm{ l_ --\- | Stc'ir'cs : Use ABC r F^ 8mm Z M " = o> : D 12mml F"o 8mm p= l-- Side view = \? ft = 4e" etul _ (edtoox,lo.)(str.tge tr - -F.s. I$ rA = s.o P= +F^ = 3.72,(lotN '.). ll oo q )i o: ZMl= Based on do.rtle s[,eo* in pin A: o.2D Fn- o.rt P = e Fe O.20 F " o - O . 3 8 F = Q lo F"o lct = 3-sstylog N BaseJ on Jo,rU. she.in ir,r Pi.s tuJ 8 c,!al D; A= tCt= Fro -== - ? M - q(o.olt)r = ll3. lo x/o'6 ^' - ( a X t o ox t o ' ) ( t r g - t o t l o : 6) F.S" P = jr" Fro r 3-o s.q1 x r o 3 N ks B oseJ oh GDrra P Fessiovr ir, ,f ;" Fon Eo on? l;^lc ? 6, A F-s. = T.$rf x/Oj N 8D : A=(o.oeoXo,ooE) = (a)( ?$b '/0. X tGoxlo-c) 2A J. L' P = i+-|1o= lt.o+vfosN AlJo.ro,[,0uva.lue oF P is t-oll,qs+. p= g,72"roEN P E ?-jZ ki., Problem8.44 Top view f.ff fn the stnrcturcshown,an 8-mm-diameter pin is usedat A andl2-mm-diamcter pins areusedat.B andD. Knowing that the ultimateshearingstressis 100MPa st all conncctionsandfrreultimatenormal stressis 250 MPa in eachof the two links joining I andD, determinethe allowableloadP if an overallfretor of safetyof 3.0 is dcsfud. 8.44In an altemativedeign for the structureof Prob.8.43,a pin of 10-mm-diamaer is to be used at l. Aszuming that all other spocificationsremain unchangd, dctcnninethe allowableloadP if an overallfactorof safetyof 3.0 is desircd. Stah'cs i lJse ABC a,s 9"c. boJy. A B l=A Feo ZM" = o 3 p= f p^ u mm-'l Z M6 = O : Bc,sed oh do.rUe slreqr inapi. A l = \ fi = 4 et : S (o.ofe)r J ff,-,rt. xtd6 ,,,n'\ i-A -T.s. o.ZDFn - O.rg P = o E O.2oF"o- O.38P:g o FL" 3.o P = .f FA = 5.[12 iltDt N BaseJ on Jorbl* sLec"nit'r Pirs LJ 8 c'nJ D: A = + C ' = T ( o . o t ? ) r s l l 3 . l o N I D ' 6t n L (aXr"o*to')(rrs.foxlo-g) F r s-o _= _ ? f r A = = zs+ yto"N 3-o F P = jfr' Fro r s.q7 x/os N BoseJ oh corvrpFess iou ir, ,e;r,ks 8D ; For onc l;"t Eo= # A=(aoloXo,ooa) \(tgo'ro.' ) = (4)?.*..r/o' l6o x/o-t *r' ?G-7x/ot f.l P = i+Eo = l,f.o.lxfosN AlJ"-",[.0u vol,,reoF P is "-ol]es+, p= g.q7 r/o$ N P = g-qT RN Problem8.45 E.15Link AC is madeof a steel with a 65-ksi ultimatenormal sfressand has a % x %-in. uniform rectangulmctoss section. It is connectedto a supportat I and to mcmberBCD at C by %-in-diameterpins, while mernberBCD is connectedto its E supportat B by a ft-in.-diameter pin; all of the pins aremadeof a steelwith a 25-kei ultimateshearingstressandarein singleshear.Knowingthat a factorof safetyof 3.25 is desired,determinethe largcstload P that canbe applicdat D. Note that link lC is not rcinforcedaroundthepin holes. l--u U s e S " " b " l y B C D+ J M e = o : ( 6 ) [ € F ^ " )- l o p - o ".-l--n,n p = O.tA F*. B"-frtr^.= o 't'zE'= o: 6? lo'he Bx= -z p 3r t.as'* (€)' P r I f = l-2,5? .- 68, -ttP +) Mc =' O ' . BJ= cr) r.€. = o B= ePI P = o-705t8 l.tf ILL-t P B Sheo. in pi^s ,-+ A o,.) C. h.= t Ap,,= *ffi11=e#+xsf +ffiJ HtxE)= O-9+a51 T.^a ' ot4 ot-1 nr* s"4io.n "-f A o-rJ C . Fo.= 6-A; = g A".r=€*IdXi-?) ; S**1!/e. vJu" F.o*.(r) p= df Fn. i5 lc;p" O.6Z5 k,p. O.6eS k,p,, (o-'ls)f o-62f) = o-3ooLJps SL"o. in pi^ ,J B. =(#ifxTxf)' , B = t Ae;^: * Te'. F^"^ (a) P= (o-zosaa{cl.s8?{q) = o.+le Ailu* J,0" vJuu J P ,'s fL.- "-&'n v*!oe. 0.5 81qq l;p* l<:ps p = O - g o ok J p s or P= %o.th PR'OPRIETAR'YMATERIAL. O 20ll The McGraw-Hill Companies,Inc. All rightsreserved.No part of this Mrnuel nreyhc dr6ph)€4 rqo6rc4 or dirtributod in any forra ot by my rnstns, without the prior writien peimissionof the publisher,or usedbeyord th limitcd dfutrftutbn to tcrcbn rnd adrcebn pcrmitbd by McGnw-Hill for their individualioursepreparation. Students usingthis manualur" uiiog it witho't pormi*bn. (z\ 5 Problem8.46 -t to useF-i"'|.ff Solveprob. 8.45,assumingthat the structurehasbeenrodceigned madc' hes been changc no other that and as at.B dimreterpins at A arriC aswcll 8.d5 Link AC is made of a stecl with a 65-ksi ultimate normal stress and has a tA x %-in. uniform rectangular cross section. It is connected to a support at I and to monbcr BCD at C by %-in-diamcter pins, while mernber BCD is connected to its 5 support at B by a E-in.diameter pin; all of the pins are made of a steel with a 25-ksi ultimate shearing sffess and arc in single shear. Knowing that a factor of safety of 3.25 is desired, determine the largest load P that can be applied at D. Note that link ,dC is not rEinforced around the pin holes. -o in.+l+4 in- BC o Use f.-o. hJt +) ZMB = o: (6)GtF*) - lo ? = o (r) f = O.t+g F*. B,,- 3q.= c +tZFr=o: B"= Etr*. = -GBt +)Z trle = o S, = -l8.rt * Sh.on l 3 y= - Z " =O '1.2s'*(€)'P''. = A in pi ns -+f t.7€ ? - ine Br=3e F' D.7o583B l.qtefJP onJ C. k:p" = *, {/" =(;ftX+)€f : o-58?7q Fk = l-Ap,^ s ea l i a . Tcnti on on n.l FA"= 64,'r' S,. nflr. F^. p= ? o'qs?slc,'ps, is o. SEqqq L;p. ' O.?83 lc,'ps B. g,= dApr*= # At!.-"Jol" C. P = (O.t3XO.St?dt?) = S[.co. in Pi^ -I Fr.r*. (a), A o^l *A^*=(f=X+)(*-F) vo)u,e of F,o.^ (r1.1 A 4e" =(#XFX#f . o-syqqllc;ps ( O . T o E t a X o - S E q i q ) - O . t 1 f dk i p s vn!,te J P i. lL" s^J.!e, uo.,fr"- p= c.-293 lc;ps P= 2#lh pRoFnIETARy MATEHAL. c) 2011rh McGnw-Hill coEFDbr, Irc. All rightsreserved.No partof thir Yry"l.ryy.bc dirphpd, mrcaucaq bcyondthc limitcd dirtrbution to tcrchcrt ffid or dbributcd in rny forrn or by any means,without the prinr wrien pctmiarion of-the publisher,or used win3 it without pcrmi$ion' are manual using this ptrpli.tion strdents counc e&rcetorrFffnitcd by Mccnw-Hill for their individr,rel Q\ Problem8.47 8.4? Eachof the two vertical tinks CF connectingthe two horizontal membss lD crosssectionand is madeof a steel andEG hasa l0 x 40-mmuniform rectangutar. with an ultimatestrengthin tensiono14ffi MPA while eachof the pins at C andF has a 20-mm diameterand is made of a steel with an ultimate strengthin shesrof 150 MPa. Determinethe overall factor of safety for the links CF and the pins connectingthemto the horizontalmembets. Use ncv,tberEFC 4s fuee bJl. \ Ozmr = o: e o o.qoF." -(o,g'SXe+vros) 24kN F"" : 3? rt lor N BnseJ on lensiEn in /;^ ks CF: = (o.o+o -o.ol\(o-oto)= A= (6-J)t F, = 26u A = (Z{+o ottat) (Zoo r,o-5 ) = Basex oh doublu'shea.*in A = q)" 20orlo-6rt1\ (ane*^f) l6cl.cl x los N pinsJ = Xb.ozol' = 3tq-,6xlD-' Y'4' F u = 2 t , A = ( 2 \ ( t € o x 1 s e [ S t +l .€ y l o - c ) = ? + . | L t 8x / o s N q+.2't1 *lo3 N Aa{- oe.I Fntig sna,!,ferVahe, i.q F; = F*{- -"} s^{o\ 71L1g llos = 2.+? 3q rlor pR.OpnIETAny MATEruAL. @ 20ll The McGraw-Hill Companks,Inc. All rightsreserved.No part of this Manurl mey bc dnphycq npme*d' or diatrbupd in my forrn or by any means,witlout the prior written pcrmision of the publisher,or usedbeyondthc limibd dirtributbn b brchot' rnd c&lqtbt| pcrnitfi by Mccnw-Hill for their individual coursepreprntbn. Stndentlruing this manualareusing it without pcenilrion < memberr lD and !.47 Each of the trvo vertical linlcs cF connecting the two horizontal of a stecl with an made EG has a l0 H 40-mm uniform rcctangular cross scction and is a 20-mm andFhas pins at C the of each while ultimate sficngth in tension of 400 MPa, MPa" 150 of. shear in str$gth ultimate an with steel a of diamctcr end is made them to pins connecting Detcrminc the overall factor of safety for the links CF and the Problem8.48 the horizontal mernbers. replaced by pina 8.48 Solve Prob. 8.47, assuming that the pins at C and Fhave been with a 30-mm diameter. Use fttember Efo t- a6 fre.boJ,1. . ,t. o.?g o^,1o DZY, = o3 oAo F.. - (o.esXe+ x/o3) = 6 ltf kN F., = 31 *to3 N B o'tuJ orf Iensian A = (t - d)t l-u ? F irn ,l;n k. C tr : - O.oso)(o.olo) = lOo*lO'"^" (o^, ^!,'^k\ = (O.O,lo 26rA = izlt4oo*tot)(l0axlo-c) = 1o.ox,os N Bose) oh {ooLle. shea,r in pins: A -- ':t S of" = T (o.o3a)a = 1oG. Br, lD'6rn' F; = 2 t, A = (a,X\.5orf 03 XZoA.86*ra't ) = ?t? -oGx lo3 Acl'tol Fu is sj+tallec v aloe, i-e. Fu = Fo"lo, ol sale].v ES.=E - E= go.o >t/o3N = -8W-:lo" = Z.as xtDs 3q pnOpRIETARy MATERIAL. O 20ll The McGraw-HiltConprnicc, Inc. All rightsreserved.No part of this Manualmry bc di!pl|)'od' t€ProdtH4 or dirtributod ia any form or by my reanr, without the prior nitLn pcrminion of the publisher,or usedbeyondthe limited disFfoutionto Erchcrs rd edrrcrton pr3rgdttcdby McGnw-Hill for thcir individual coursepmprntion. Snrdcnsusingthis manualareusingit without pcrmi$hn -4 Problem8-49 E.49 A 40-kN axial load is applied to a short wooden post that is supportcd by a concrete footing resting on undisturbed soil. Determine (a) the morimum bearing stress on the concrete footing, (b) the size of the footing for which the average bearing stressin the soil is 145 kPa. (a) Beatt",g s*^ess o"'' corc.ste f.,o* i"'e. 1o l4N = \o*lot N 3 A = (toaXla@)= 12* ,D3**a = l?, 16 ,n' p= g= -lgi-tg: * = o-?-1s8e,,,',' A= b' sl uafl ) o. 27 586 MPa- d g= lr{5. [.P<= t]Sxlo= ?q 't i : t s.s3xtD6 ?o, a33 Llo x /o3 N = *"t,tl.#sx,os t = O.5?S v-, b + 525 hu I PROPRIETARY MATERIAL. O 201I The McGraw-HitlConrpanies, Irtc. All rightsreserved.No part of this Manurl mey bc displayed,reproduced, or dirtributcd in eny form or by any msans,without the prior writtcn permissionof the publisher,or usedbeyondthe limitc.ddirtribution to tcrclrcrr ud educatorspermittcdby McGraw-Hillfor their individualcourseprcparation.Studentsusingthis manualareusingit withoutpcrmission. Problem8.50 E 50 The frameshor*nconsistsof four woodenmembers,ABC, DEF, BE,and CF. Knowingrthateachmcmberhas a 2 x 4-in.rectangularcrosssectionand ttru eacl pin hasat/z'in. diameter,detenninethe maximum of the averagenormel strcss (c) in memberBE,(b) in mernberCF. "ilrr" ls in'-1- w h'-lc A)e suppo.t vea,.Atov,s +r' fig,r,n a-S sh aurn. Dsi"'1 entiF{. Fra*,e as $r'e< btl- Zl+ = o: rsin.*|I- 4o D, - LrlS+ So[+eo) = o Dx = loo )L' soin.-Jr Usb ms,\be* DE F a.6f"** bdf . IF t-' =e : FD/-+Dx=o Dy = *O" = ttoo )b- D {sc14Efr') EH= = O i 7H-= = O: Fee=-225o lL. ( S o + r g ) D f= o (eo)(S F"") -(ts)Pt =o F..g = 1So lL. ynaep, in Cor,rpne.ssica,r LAn BF Ay"ea, A = ?in,* 1;n. z 8ina (4) S*ress , 6es= .F,e"= - -32so-= g A (b) Siress "zlion 64r€a- OCCuv\S A.i,, r- (a)(eo -o-5) \ rynit irnu*t see*io. 4 i.a tensron rner,nbe,^CF M i n in, ,.rp', se 7golb, -4alpsi 6 -=F"--7'So Q"": ;; = tr = 7.o i',' "t = tol-l ps; P in. Problem8.51 A+ €aaLr boll t.Sl Two steelplatcc are to be held togetherby meansof %-in.-diamdcrhighrtrcngth gteelbolts fitting snugly inside cylindricd brassspoccrs.Knowing tr* thc wcregc normalstressmustnot orceed30 ksi in the bolts and 18ksi in the opl€ffi, dctcnninethe outerdiarnctcrof the spacersthat yieldsthc mosteconomicalandsafc design. J*olio", *\ a uppqn p,lo.*e is polle/ donrn by *lt *usih A+ *[,e scrh^e*;rne *le spa.c.er p,,rslres Force R of *l,e b"!t. +h^+ glale upwarJ wi*Ji a. c or,rpr€ssir/e fo*"e rn a.i", * oin P" . In ud e n *o e1.li /,'btiuly\, Ps=e F* +[. E"I*, F-t f 5 rl Ite sPa,ce"r E 1r,*ing Pu Au E = q R = A* r+R Td' +R or 9q f b - T €, drr" or R =+G(do'-J.') frffiIfT il (dot-otr') Pr arhJ P-s, +sbdr' = Ft"(d"' d.t = drt .r * ;; d==(f + #X+)' - rlf ) ( r+$) dr' 0. ,c€e7 ds 3 O. tlO8 i',- tlprods4 rnoFnlETARy MATERIAL. O 20ll The McGraw-Hill coryaicr, Inc. All rightsreserved.No part of this Manual.mgr-pediwhyo4 pcmieion ofihe publisher,or usedbeyondthe limited distribution to terchon ud or dirtributed h rny form or by any means,without the prior *tlipermission. drc.1ort eG-.tt61i by McGraw-Hillfor their individualcounc pnqdrtioo- Sudcntsusingthis manualareusingit without E 5? Whenthe force P reached8 kN, the woodenspecimenshownfailed in shcar along the surfaceindicatedby the dashedline. Determinethe averageshearing stressalongthat surfaceat the time of failure. Are". being sh een eJ : f i = Q O n n q s rl S n * = l35o ^.t N Fo*" =g Problem8.53 = t3f,o xl(JG wl P S-qs*[o3 Pa=5.?3 ]aPa-< A 8.53Knowingthat link DE is I in. wide and % in. thick, determinethe normalshess in thecentralportionof thatlink when(a) 0= 0", (b) 0:90". Usa hne'^te* CEF "'s ct f'""it bJY' DEM.= ol = o - l" Fo, - GYeo sino) (tel(eo .^,s6 )' Foe = - \O sin I D aose A o " = (r)C*\ = o- l " S b"Jo /;'qta:1 Ree -: c* 'lV ir'" 5ot = (al O= 6Lt (b) e= Ge - go .flL. - 6Io P=i E- = -9o.,0L' = - Af,o psi PROPRIETARY MATERIAL. @2011The McGraw-Hill Companies, No part of this Manualmay be displayed,reproduced, Inc. All rightsreserved. or diskibutedin anyform or by any means,without the prior written permissionof the publisher,or usedbeyondthe limited distributionto teachersand permittedby McGraw-Hillfor their individualcoursepreparation. educators usingthis manualareusingit without permission. Students / i .I t Probhn A.** ctneeus, eot\ Six bainl A$f Trp iloo&n planks,each l2 mm 6idr nd225 mm wide, rt joind hf fu &1 mfitiF joint town. Knowin3 Ss thGuruodusedshearsoff along ib 1l*r Sr t MPr, determinethe magnitudeI of lh nil , Stc rrurfe drming stressre# hd 6J will causethejoint to ftil. 16 rr"n * 12 nn aFe sf,eq.rcC. A,'r*: A . (e\( lgXr*) = I l.f,?**' ?.25mm -a> = lrge *lo'o r'r\ t'= .P, A p= tA i (dxlalXtf-fexlo'.)= q.2Ayto"tt = 7.nkh} ry tr *fhl O Sll Th ]le{hr-|fl b. All rights reserved.No part of this Md l.lOtl|3TAfT XtT[ltlL Ceryir+ fttdt of the publisheaor usedbeyord tb: hu c trrH i ry lrr * ly ry n, r{ld tte tr|t r'di.pnilon tr r|lri h#rifurl m *ffi fy Xdlr-l* ltrdents usingthis manualareusing it riH Pn*f F*a rulbr rr|rfu.1 d f * F Problem8.55 Us" one S"" t< as a. f"". boo,\./, +)Z Me = c); 71 E - (zolllSrro) E = ,?5o .!L. --> |^5oo!6 LZF* = o: E+Br = O +t (al & = t25o .lL * zt=o: B *: 8f * 8; _She".!h1 st*.r fSoo = o (t) B Ap;., BJ = lioo -lL l?Soa + fsoo in pin A Ar,.=+Ar:={hf d( - , \= Brr=-E B- = o.t q C Z S tqs?.sc in' 7.qq*loz O.lq635 f)t q-?.+k s , B ea.,'^3 s4 r ess a,-[ Bg- B-= dt 1152:-56 (+x+) = 6,-?5*1o3y=, 6-e5 PROPRIETARY MATERIAL. O 20ll The McGraw-Hill Companies,Inc. All rightsreserved.No part of this Manualmay be disphycd reproduccd, or distributcd in rny forn or by any rcaru, without the prior written permissionofihe publisher,or usedbeyondthc limitcd distribution to tcaihers a116 pcrmittedby McGraw-Hillfor tbeir individualcoursepreparation.Students educatoru usingthis manual uiing it withoutpcrmirrion "r" 4 l/"-in-rliompf: L t/z-in.-diarnetrr d-l ffif /R ie fitted Gtr.oA to tn t t t'tf A is # sd lB r rurnd lrolc near end C 0f ffrG ttoodn memberCD, Fs Srplolding shown,detanrninc(e) ec maximumfifi*r1n naill rhess in the umo{ $) SG distance6 for whi& trc mmge shearingf; if 90 pd m the surfacesiilfi*Gd hf dre dashedlinel {c} thc lrrunge bearingstu m ProbHn8.S : L# fursd. (A) Pl**iF u*r not"y^a.I s*ress An"t = 3 P (t) t b (ct 6, P A P 2t,t $= A..t P in J4re woo/,- l. 8?S itn' 533 P $ , < ?bf Iooo (?X+Xqo) 7-41 in. ' 1 - - 4 1i n , b = '/ooo (*)(+) 26C7 F"i O 201I The McGraw-Hill C,ort:*., hc. AI rib ruerved. No part of this Manud ry h fihrf' ?]ftlffTAfY HATIIIAL ef*r of ir p-*her, or usedbeyondthe lH or 1||'ha L ry frr r ty ry means,without thc prior writtc f* rrfil {ti manualareusingit without pil*r. ly Ha0n-ilf;n for their individual courseprepalilin lth *q5lr tda < prU' b * d is placedT-tlo*" 8.57A steelloop ABCD of length 1.2 mand of lO-mmdiameter eachof l2-mm DF, and BE Cables aluminum tod AC. arounda 24-mm-diameter s(ength of the ultimate the that Knowing load the Q. diameter,arc usedto uppfy for the steel strength is 260 l,tpa anOthat the ultimate aluminumused tor ih;il thst cenb€ load largest the Q determine MPa" usedfor the loop andthe cablesis 480 desired' is of 3 safety of wiratl factor applied Problem8.57 ,l 1an Usinq ioi*t B osq$n.c- b"J1 anolocJr,t,clcria1 sy *-"in1T ?-t'fio - a = o e= 8n. [)si'.n ioi^] A as q-S". bnlu o.J4.1. siol*h3 sy"nr'.a*""t; F^. = o Fo.=o Z- $ri. E'{A "' .8":;el ovrs.l**3+lr # c",Lle tsF; rf\ *l Q, = q,A = qr\ol ")= (+ao/166) + (o-oti-)'= Stl.?qxtd Bose J on =t^e', X++J Jee/ -!"., y: tr^q = 36"4 = 3q,ffid) 3 r.*" e - ^f\ 6 -6 t1 tq{u - = .JA -c- \-,rr A n F = f; t +Boxto') + (o.oto)" = 45.L+*lo' N Bose) oA sfnr" {an "F noJ AC I c:$tr".,E+qA = +gr(+d-) = tr (u" vlDn) + (o.oa't)"= Ar*o"! uf+;^te A!,lruJ.!" ) rJ ,l"J O, is *he s^o.lfert, -'fl A r^' = JIa trs. 45'?txf ?33 g g . 1 . 2x l o 3 M Q,, = +-5. vtd ",t lS. of x loo lJ = l,S.Og k iJ FROFRIETARY MATERIAL. O 201I The McGraw-Hill Coryldr+ lrp. Att righu reserved.No part of this Manual may bc dirpleytd, tpprudud, or dirufr@ ia rny fonn or by rny rmans,without the prior wrinm prmirion of th publisher,or usedbeyondthe limited dirffiutioo b brch€rr rad cdnrlofi paruittd by McGn*-Hill for their individual coursepilfrsion. Shdcffi uring thb manualareusingit withoutperni$bn, joid bf 6c rinryle glued joirt *rorur. Knowinj S* P - 3.6 kt{ and trr ilr *b ril!6 of the gluo ir l.l MPr in tensiond l.{ }tPt ia shear,dcadr hbofrfuy. f A = Io'- 6-5' ! f;S' fr= 3-6 Lu = -A.=(zs)Ctzs) Tentioq o4 qlue- -- q-v-.a - s: P "u{a A' 3.G * lo3 N 7.g?5 ' ki r# = 7.975 * fdo *n (g-e"tol)(ses*s"\o 3 9- 3?5- * fo- P"" " E S - = C6- 3.+7 (g-€ >"ld X*t" 5o') (A\t1.gzs' to-s) l + ? . o 8 t x f o 3 Fc, trs. 1 . 1x t o ' l{7- o8f r [o3 q.57 i{ Tle s"'.o.I.l=n A**o. J =*{"*7 3ouet'Y\s. E s. = 3.'+1 l$fEfAlt il*TIINAL O t6ll Tte McGraw-Hill Companies,Inc. Atl rights reserved.No part of this hdrd cfrlhtltryhrtyqn;witlroutthepriorwrittenpermissionofthepublisher,orusedbeyondtbHrhlebrfud .-fir 17 lChr.lfl fu rr.ir individual courscpr€paration.Shrdentsusingthiu manualare usingit tlh tr*a ry h t;tcA pr*n ryrfu* -1, Problem8.59 r-600 8.J9 Link 8C is 6 mm thick, hasa width w = 25 mm, and is madeof a steelwith a 480-MPaultimate strengthin tension. What was the safety factor usd if the structureshownwasdesignedto supporta l6-kN loadP? _l mm Use bo* AC D as a $ne. Fe. V : " d y , a n J n o f e tl- ^.1 mer*"'herBD r's c L l w o fol'i,e rt e*-Lel . ZMo = o i (+eo1Fea - (6oo) F- 6oo Fec=E;r ^ = = 2O t, /Os |vt rn€raryLey.tsC : )( o.o06)(o.025) F.S.= + = rTe pROpnIETAR.y MATERIAL. A zfill The McGraw-Hill Companies,Inc. All rightsreserved.No part of this Manusl mey be dirplayd rrprofixr4 or distributcd in any form or by any means,without the prior writtcn permissionof the publisher,or usedbeyondthe limited distnlbutionto tcachcruand usingthis manualareusingit withoutpermission cdpcatorspcrmittcdby McGraw-Hillfor their individualcouneprcparation.Students { 8.60 Problem t.60 The two portions of membcz_AB are glued together along a plane forming an angle d with the horizontal. Knowing that the ultimate stress for the glued joint is 2.5 l$i in te,nsionand 1.3 ksi in shear, determine the range of values of 0 for which the factor of safetv of the members is at least 3.0. rl I Ao= (l,oJ(t.ts) ? 2.So,i^L P= 8.+ krps B nsel evtte'.r i le v^* 5a= Jl fSosel g ev, sh eanin3 slncss ; " A'frsi', 20 = -', r) Hg"..- ?l-3; g6g'@ = q& = (n-s.l(a-s?-) = 0-868 J'u z zl-3 0 7- e ? e : 3 ? .B o olo Or n = = sinOc-rs@ tr= &. A,L : (zXe-so){l.3f _?_ 2 g = 6 + -S ? ' s*res: : jl0 c.osee 6t>s9 = O.13t€q R = (FJ)P = 7-2.k:y, e i - ^20 sin o. qo?7 g g -< 32.3o o PROPRIETARY MATERIAI. O 201| Tb McGraw-Hill Companies,Inc. All rightsreserved.No part of this Manualmay be displayc4reprcduced, or dktributedin any frrm or by any tmens,withouttheprior writtenpermissionof thepublisher,or usedbeyondthe limited distributionto teachersand cducatorspcrmittcdby McGnr'-Hitl fot thcir individual coursepreparation.Studentsusingthis manualareusingit without pcrmi$$ion. €