EXERCICIOS DE RESISTENCIA DOS MATERIAIS 3°ed - Resoluções, Teses (TCC) de Engenharia Civil

Baixe EXERCICIOS DE RESISTENCIA DOS MATERIAIS 3°ed - Resoluções e outras Teses (TCC) em PDF para Engenharia Civil, somente na Docsity! WWW .icivil.ir Ulpis quuaimo g ylbgauvila gola JU p Uljac puaigo UBIIy Ulgjo q gut» cl Uljc joy Ullôs Usp q Úsjigy Uljac puigo ganái (elo Vasil Uljac punigo pon obiligjó  | CHAPTER 2 UI o 3 La CS = a O e O A — O cl toda ++ | Es o! 1 id DJ ta ca m | O TS iq) 3 PROBLEM 1.3 1.3 Two solid cylindrical vods AB and BC are welded together at B and loaded as shown. Knowing that d, = 1.25 in. and d, = 0.75 in, find the normal stress at the midpaint of (a) rod 48, (b) rod BC. SOLUTION (a vod ÀB P= Ia+rio = 22kips A= Fats Tas = Lazra im . Po. Rê - : «a Sis € A "TRA - 17.93 ks (o) rod BE P = I0 kips As Ta, = Elo) = om ia* A - Gr = É sega * 22.6 ku - PROBLEM L4 & 10 lips rod BL: Pe dº- LP 2 TS” 1,4 Two solid cylindrical rods 48 and BC are welded together at B and loaded as shown. Knowing that the normal stress must not exceed 25 ksi in either rod, determine the smaltest allowable values of the diameters d, and d;, SOLUTION rod AB: Pr Aros 2% kips Ga 3 RS ks! Aa Td Gas = A o Ana a Td, 2 - HP. (MA) NE d, “Tre TS LIZ0S in di= 1.057 m a dO lips Car âsks AncTdo No) ass E 05093 in! dE Otima  fa sq | vo 3 LI PROBLEM L5 200 K 1.5 A strain gage located at C on the surface of bone 48 indicates tha the average normal stress in the bone is 3.80 MPa when the bone is subjected to two 1200-N forces asshown. Assuming the cross section of the bone at ( to be annular and knowing that its outer diameter is 25 mm, determine the inner diameter of the bones cross section ato SOLUTION à 2 BA 2 2: d + -d, -Ee 2 3320 (4VUZ00) dy = Cassol) — Gtagoxior = 222.9 n10f m? d= 4.43 m10 dE AT mo 8  Los o o —a 3 on q o LI ] cal 3 Cl 1 | 1.6 Two steel plates are to be held together by means of 4-in.-diameter high- strength steel bolts Sitting snugly inside cylincrical brass spacers. Knowing that the average normal stress must not exceed 30 ksi in the botts and 13 ksi in the spacers, determine the outer diameter of the spacers which yields the most economical and safe design. SOLUTION At cach boft Joucatio He upper plate is pulled clown by +e tensite force Pi of he bolt. Ay lhe sume time He spacer pushes that plate upward with a compressive Force PB. In order ta maintain eguilibrivm Ps = P, For He boPt S = e = dh oo BETE = s - Ps m. 2. 2 For He Spacer Gs = E = TES o BR = Eóstds do) Equeting P, and Pe Id = Tela - A) dt= dia Sao (+ CDA 2 de= (+ SEMEF) = O. IC667 in ds; * 0.408. « é í J L. DJ a o ET TI —s Ls PROBLEM 1.9 1.9 Two horizontal 5-kip forces are applied to pin E of the assembly shown. Knowing that a pin of 0.8-in. diameter is used at each comection, determine the maximum value of the average normai stress (a) in ink AB, (5) ih link 80 SOLUTION Use joint B as Tree body. 10 Ksps Fe q sinto sin6o Sin 5º Ia kips Force +rianglde Fus = 7.3205 kips Fac = 8.958 kips Link AB is a tensioa member Minimum sectim at pra Pnad = U.8- a.8K6.5) = OS int (o) Stress in AB Ciz = e = =stes = I4+.64% ks «<a Link BC is q Tompression member Cross sectimal area is Àz BOSS) = 0.9 qn? (o) Sivescin BC So = Cm. D8I6SS | qacusi a À 0.4  PROBLEM 1.10 1.10 The frame shown consists of four wooden members. ABC, DEF, BE, and CF. Knowirg that each member has a 2x 4-n rectangular cross section and that each pin has a 3% » in. diameter, determine the imaximum value dl ihe average normal stress (a) in'member BE, (b) in member CF. AL - e SOLUTION ua ] Ada support veuchtos do figure am os shown, Using entre Frame as Free body ZM=o “o D, - (4S+30Y480) = o — De= q00 db. Dx Ust member DEF as Free body FF +* =o + AU 8D, -<D=0 =P a ce, 3 D,= &Dx = 1200 do. E O A E D E EMp=o AcoXÊ Be) - (30+15)D, = O Fre = - 2250 7/4 ZHM,=-o0 (so X& Ee) -053D, =o0 Fe* 750 dh. 4 te tg Stress in compressioa member BE / , Area Az Rinxtm = Bim? / (ay Seg fee = =2aso. = — 28 py — / Stress in tensrom membev CF Faso fl. Minimum section area Dccurs at pin, Dio 3 (QUHO-OS = TO in* mir imuma section LA bo aqu! cida dae cao) CL Ga, 3 PROBLEM LI1 2.31 For the Pratt bridge truss and loading shown, determine the average normal stress in member BF, knowing that the cross-sectional area of that member is 5.87 i in? B D F SOLUTION pá Use entire truss as Free body H EM=O (940) +(18 30) + (2780) - 36 À, = O or hp = 120 kips A C| E C Lan ak: 9k sakips sokips SO kips 8 E Use portion & truss to the dot E a section 8» cuthina members BD, BE, and CE. Ne 360 x 12 . . A c 2 x |ã0 - 80 - ja Fes * Fagz 50 kips ce = Es. «8 = 2 es; to lips BO lips Ser = A * 557 8.52 usi < ao ips A 1.12 Knowing that the average normai stress in member CE ol the Pratt bridge truss shown must not exceed 21 ksi for the given loading, determine the cross-sectional area of that member which will yield the most economical and safe design. Assume that both ends of the member will be adequately reinforced. PROBLEM 1,12 B D F SOLUTION Use entive trass as Pree body al SEM, (tango) + (13 K80) HT) — SEA, Ay = Io lips or 9% s sê sOkips SOkips Sokips Use portion e tross to the Lodt sta section 8 fas eubting members BD, BE, and CE , ç Os Mg =0 c a Ee -(Xno)= O A Fes = 70 kips A Lo» Fa Gus = Fer tro ps 30 Kips Ace H f r As = te - Je. WRP in? à, Ges I n -L vv) ca lo) Lv a! Lo IJ CA 23 (hd q PROBLEM L15 1:15 The wooden members 4 and E are to be joined by plywcod splice plates which will be fully glzed on the surfaces in contact As part of the design of the joint and knowing that :he clearance berween the ends of che members is to be 8 mm, determine the smallest allowable length Z if the average shearing stress in the glue is not to exceed 800 kPa. SOLUTION There are Fow separate areas & qlue. Each orea must tramsmit haofP of Ha 2484 Loud, ThereBe F=2 AN = tAxi0oÊN *€= 800 x/10* Pa pao? - Bo0w[0* Shearina stress in due v=E Ass Isso" m Let = eng & ql apea and w= width = 100 mn = 0] m “As lu a Ls 28 + quo A Ismiot A= wo “al = (aXiso)+ 8 = So atom = ISO mm 308 mm «tê PROBLEM 1.16 1.16 Determine the diameter of the largest circular hole which can be punched into à sheet OF polystyrene 6-mm thick, knowing that the force excrted by the punch is 45 kN and that à 55-MPa average shearing stress is required to cause the material to Fail. SOLUTION A= mat for cylindnical Parduvo guet ars Crencing stress t= f Às + Egu atira String Fr d: As qdt = nto - z AQUA a = d = FF" Tio.ooc FS Ns) 43.dwlo m d= 43.4 hmm a [LI ll =) É. 39 CI lã lo. L. Es ES E] 1 ES a PROBLEM 1,17 1.17 Two wooden planks, cach 7% - in. thick and 6 in. wide, are joined by the glued mortise joint shown. Kaowing that the joint will fail when the average shearing stress in (he glue reaches 120 psi, determine the smallest alicwabie tength « of the cuts if the joint is to withstand an axial load of magnitude P = 1200 lb. SOLUTION Seven surfaces carry the Jota? r Joud P= 200 fb. Area A=NG)d=- Bd co for ho E Gac! test, PROBLEM L.t8 1.18 Ajoad P is applicá to a steel rod supported as shown by an aluminum plate into which a 0,6-in, diameter hole has been drilled. Knowing that the shearing stress must not exceed 18 ksi in the steel rod and 10 ksi in the aluminum plate, determine the largest load P which may be applied to the rod. TR d4im. SOLETION - 4 | Fo see? Aj= mdt= T(paxos) 06in. = OSHO int ta E dt Pe AT = (ors4eÃm = 13.57 kips For aluninum Ass wdt - T(LO(0aS) - LASÇE in? t=R « PAT = (LaSGXIO & 12.97 kips Limeting valve dd Pis the smablervalçe * P=I2S? kips -ê  Lodo LA ES E E od El ES TS PROBLEM L,19 1.19 The axial force in the coluran supporting the timber beam showm is P = 75 kN. Determine the smailest alowable length Z of the bearing plate if the bearing stress in the timber is not to exceed 3.0 MPa SOLUTION 2. 2 SS =4 = ly 4 . . à Pedi] Solving for Li L Sw (Goro Yo.yo) = 78.610 m L= 178.6 mm <q 1.20 An axial load P is supported by a short W250 x.67 colusm of cross-sectional area À = 8580 mm and is distributed to à concrete foundaton by a square plate as shown. Knowing that the average normal stress in the column must not exceed 150 MPa and that the bearing stress ort the concrete foundation must not exceed 12.5 MPa, determine the side a of the plate which will provide the most economical and safe design PROBLEM 1.20 SOLUTION Area SF codomn: À = 8580mm = ESDO m/0 om Normal stress in colgmn? O = ISOxio! Pa E = £ . P= AS: (gssowoYUso vo?) = Las7x)0* N Bearing plate : 64 = + end À OÉ For square plate. “ ” E aguas — aero! or 82) mm mt  s— Ls Lo =] co 1.8 Each of the four vertical links has an 8 x 36- mm uniform rectangular cross PROBLEM 1.24 section and each of the tour pins has a 16- nun diameter. 1,24 For the assembly and loadisyg of Prob. 1.8, determine (a) the average shearing stress in the pin at C, (b) 1he average bearing stress at in link CE, (e) the average bearing stress at ( in member 4BC, knowing that tus member has à 10 x 50-mm | uniform rectangular cross section. SOLUTION Use bar ABC as a Pree body EMp=o -(o04)F, - (ooaslzoxo!)= o Eles - 1208 Aa=Fdt= Eloo6)= ao!.v6 mo w - Fe 12.5 x103 - tz o 2 - 2 (2X 201.04 n10º*) 3i.ixijo 311 MPa <a (a) Shear in pin af € Double sheam (b) Bearing in dink CE at G Az dt= (o. o)fo.0os)= Izamo* m” Si: Ee - Ses ifizs to?) = U8.2010º uLS HP a te) Bearing in ARC at O A= dt: tpomkocio): Icone" m Fe. gia! S= RE = Tessfo-e o 781 x10º 78.1 MPa at = . . =otum nss int é unica cm cmeeneaa maias re Ee entrega ego uti a mto infere 1 — U. [| L — [a | (| m, oa a PROBLEM L25 1.9 Two horizontal 5-kip forces are applied to pin B of the assembly shown. Knowing that a pin of O 8-in. diameter is used at each connection, determine the maximum value of the average normai stress (a) ia link 42, (6) in ink BO 05in. 1.25 For the assembly and loading of Prob. 1 9, determine (a) the average shearing stress in the pin at 4, (5) the average bearing stress at A in member 48. SOLUTION Jse Qoiwt B as Free body. to uips Fug Fac Fa Eu to ki: ps Law ob Sines Force triang fe Fag - Foc a lo º . . Sinus O Sto” Sn as Pag * 1.3208 kips ta) Shearing stress mm pin ot À p= ua “ph eva has Fa - Elo.g)= 0.5026 in* . TÃO . . Tese 128 7.48 bes; a (0) Beanina stress at À in member AB bp: tds (osÃoas)= 0.4 mt 6, = Fe - Los . jg.30 18.30 ks; «at As 0.4  [| PROBLEM 1.26 1.9Ywo horizontal S-kip forces are applied to pin B of the assembly shown Knowing that a pin of 0.8-in. diameter is used at each connection, determine the maximum value of the average normal stress (4) in link AB, (Jin tink 8€ 1.26 For the assembly and loading of Prob. 1.9, determine (a) the average shearing stressin the pin at €, (6) the average bearing stress at ( in member BC, (c) the average bearing stress at B in member 8€. 0Sin. SOLUTION Use joint Bos Free body IO Kips E PE ? a8 Fa 10 Kkips E Law of Sines Force rinnade no ro - pa = sã Fe = 8.9658 kips n tm) Shearing stress in pr at € T= e N Ap= Fato Elos) = o.s0m in* Uo- Lt 3688 - 3.42 8.92 des; - into. sore) mm El (bl Bearing sbress od € in member BC G,= + [ Az td = (osXos ot" n G= ESSE . 22.4 aus : €» Bearing stress at B in member BC Sp e l | Asztd -2l0.50.8) = 0.8 in* = ES. ai Walk a ca ll Da. E Ú [A ll 1 L Io t E] Lo) ll CI L LA | o o PROBLEM 1,29 1.29 The 6- kN load P is supported by two wooden members of 75 x 125- mm uniform rectangular cross section which are joined by the simple glued scarf splice shown. Dexermine te normal and shearing stresses in the glued splice SOLUTION P= exi N 8 = 9%) joº = 20º Ap= (oo Kotasy= q875m10 mt LEUIOSS cum" 20º . 3 aafevios 965 MID g = É cos'6 = G = 56% kPa « mo Po « lext0!) sin Mo? g = za sin 20 cayadignios) * 206M0 Y= 206 kPa ai, PROBLEM 1.30 125 mu Sum 25 do” 1.30 Two wooden members of 75 x 125- mm uniform rectangular cross section are joined by the simple glued scarf splice shown. Knowing that the maximum allowable tensite stress in the glued splice is 500 kPa, determine (a) the largest load E which can be safely supported, (b) the corresponding shearing stress in the splice. SOLUTION A = (on Modas) = ABS to ml 8 = g0º-7o” = 20º 6 = Soo xi! Pa GS = esto o -s p- ME = (ag7sx10* sos x 18) S.308S% jo? cos? 20º ta) P= 5.3 kN -— . PsinZO (5 308smio! ) am dO o TE Rà Casemod) EU (6) T= 82.0 kPa a  5 PROBLEM (31 ta) - Proto b & AO 1.31 Two wooden members of 3 x 6- in. uniform rectangular cross section are joined by the simple glued scarf splice shown. Knowing thar the maximum allowable shearing stress in the glued splice is 90 psy, determine (a) the largest load P which can be safely » applied, (b) the corresponding tensite stress in the splice SOLUTION O= 4%, -g0º = 50º A, = (3U6) = 18 inf T= 7 sin2o = BRE - AXU )(IO) "da * micos * 32% P= 329% th. 4 3240 cost 50º 18 = 18.5 S=75Sps PROBLEM 1.32 1.32 Two wooden members of 3 * 6- in. uniform rectangular cross section are joined by the simple glued svarf splice shown. Knowing that 2 = 2400 lb, determine the normal and shearing stresses in the glued splice, SOLUTION 6 =9%º-4o" = So0º P= 2400 dh, Aos (BO) = Ig int 6 = Bento - (2i00)csis0" . 95,1 G= S&lps a a Pe = (2900)s0a Jo0º 1 T 3a sra 20 CAJU) 6s.7 €<= 657 psi -—s  IJ cd a Lo [nm Ss ) od ll L. tl í DO a O PROBLEM 1.33 1.33 A centricload Pis applied to the granite block shown. Knowirg that the resulting º maximum value of'the shearing stress in the block is 2 $ Ksi determine (a) the magnitude of P, (5) the orientarion of the surface on which the maximum shearing stress occurs, p (e) the normal stress exerted on that surface, (dd) the maximum value ofthe normal stress | in the block. SOLUTION AG 7 (EMC) E 36 in? Toon RS si 8= 4º for plane SP Tur (083 Co IE: PIE GA, Toe = QASEXAS) 2Ao A - 180 Kips «a b) sind0=) 28=90º Bags a LP costusta É . cito 2. ; te Bs A,“ 45 ZA * TRIGO) 2.sksi «a - I80 , (d) Cuz da Ss -Shksi — PROBLEM 1.34 2.34 A 240- Kip load P is applied to the granite block shown. Determine the resulting maximum value of (4) the normal stress, (b) the shearing stress. Specify the orientation of the plane on which each of these maximum values occurs P | SOLUTION A > (6X 36 in” S= foco = Rà css tO = 6.67 ento ta) mor tensife strese = O of O = 70º ag. compreasive stress = 6.67 ksr E ct 0= 0º 2. Po. ato - R (b) Lu 2a Cala 338 ks E) ad G=W5º a  3 a q L. —— o cos f o + rs Lodo tá E Li i — lo =3 — 3 Co o Ca ca [ Dera, 5 1.39 Member 4BC, which is supported by & pin and bracket at € and a cable BD, was designed to support the 4-Kip load Pas shown Knowing thar the ultimate load for cable BD is 25 kips, determine the factor of safety with respect to cable failure. 8,” SOLUTION E er Use member ABC as a TIN Fyee body and note T that member BD is a TN y SI EM,=0 two -Foree mem bar. (P cos dot) (80im) 4 (P Sinto ES e Y (Fan cos SOS im) (Fã sin BOAIZ IN) = — 3.623 32.023 (4); : Fgo = Joao Ts iescenca) = E.87$ kips Factor of saFeky Br calle BD Es = rãs g" 8 e à 1,40 Knowing that the ultimate load for cable BD is 25 kips and that a factor of safety 0f 3.2 with respect to cable failure is required, determine the magnitude of the largest force P which can be safely applied as shown to member ABC. PROBLEM 1.40 SOLUTION Use member ABC asa Free body and note that member BD is à A dg iwo-force member, a 2M. o (Pcos 40" )(30 im) + (P sin do"M IS ix) = (Fin cos 30) Sin) - (Fa sm 30º JOR mm) o p= JE Fo = O.Sg2E Fa Jatas "e e Abfowahte Poa For member BD is Fo = Rg = se 28125 kps AMowabde had P=(osSEUCÁTERT) = 4SS kips = ma Lo í ra PROBLEM LAt prtm IA ISkN 1.4] Members AB and AC of the truss shown consist of bars of square cross section made of the same alloy. [ris known that a 20- mm- square bar of the same alloy was tested to failure and that an ultimate load of 120 kN was recorded. Ifbar 48 has à 15- mm- square cross section, determine (a) the factor of safety for bar 4B, (b) the dimensions of the cross section of bar AC ifit is to have the same factor of safety as bar AB. SOLUTION Length À member AB Sua = joas* +04" = 0.8Sm Use entire fruss as a Free body As fAr DEM zo uAs-Wosza)= 6 As SEN fEf=o A-28=0 Ay: 28 UN é Use joint À as Free body zum A Ah 22-60 CE fe-A=o ce Ee Fis = dis). 17 kN Fe dZf-o A Fu-SE Faso For the test bar For He meterial (o) For bar AB (Cb) For bar AC a (ESSE, q = 8 - õ, = Fes às -Lostm « 20 kN Az(o.ozo)” = uooxD mm = izoxio!N 3 Sy = ae - J2oxel . aooxo! Pa Hogwto* É Gu À B0oxio* Ho.og)! Es.= e - SeA L3ooxio* No.01g) a = Fa 7x0 = 3.97 a E -. SA qa Ra Fe Fi - Fac - (3 mlaomiot) À cm7nios 300 x/0$ a: 6x0” m 16.27 mm at E : ca IE CI Cd o — | 1.42 Members 4B and AC of the truss shown consist of bars of square cross section made of the same alloy ft is known that a 20-mm-square bar of the same ailoy was tested to failure and that an ultimate load of 120 kN was recorded. Ifa factor of safety of 3.2 is to be achieved for both bars, determine the required dimensions of the cross section of (a) bar AB, (5) bar AC, PROBLEM 1.42 SOLUTION Length of member AB Ja = JON TrOM * 0.80m A £ Use entire tross as a Free body r DEM ze LUA, (OS A2) =o Av=iSkN e «1ZE, = 0 Ap- 23=0 Ay: Z84N Use joint À as free body askn Ay SER=0 SF -As=o € Ax Y € | Fa "a = SEQU » IT kN Fe Ego A-Fe-Ztm=o Fez 28 — LONA. = 20 kN For the test bar A = (0.020 = uooxto* m* PR = 120x40* N 3 For He matevia? Gy = te = Reiia = B00x10º Pa tai For member AB ES.= É - - SA. Soa =. (ES) Fa - BIA rod = q3).83n]0 mt o Soo oo x jot a = 138.47 xi0m 13.87 mm y uy 6,b* (b) Fow member AC ES, -E = GA 2 Te > - (Elm = fame faze xo!) a 2) - maggio im” o p= Ieglro*m 1LElm at  a [| aa] — " ] ido oia aos 1 1 Es la — 4.47 A load Pis supported as shown by a steel pin which has been inserted in a short wooden member hanging from the ceiling. The ultimate strength of the wood used is 60 MPa in tension and 7.5 MPa in shear, while the ultimate strength of the steelis 150 MPa in shear. Knowing that the diameter of the pin is d = 16 mm and that the magnitude of the load is? = 20 kN, determine (a) the factor of safety for the pin, (b) the required values of b and e if the factor of safety for the wooden member is to be the same as that found in part a for the pia. SOLUTION “o P=20kN= gor N | aa to) Pin: As Fats E(o0!6) = 20).06x0 m Double skear Tz t= Pe BP, = 2AT = (QMaontexo*Xiso mos) = €0.89x]0" N - Po cos! FS.= 5-= Goxios - 3.02 a (b) Tension in wood P,= 60.819 »/0! N for same ES. -Bo o. bel Sa A wlb-e) - Polo. -Go.SMnjo jo = Ulxio!m b= did—= CO + CODE) viere w= 40mm= O.Ó4Om b- URL mm -— Sbear in woo ad PD, = 60.819 HO" N for Same ES. Deuble shear; each orea is Az we Po to = ZA awe Po - tesao 2 joo.s no? C Two GXONI.Sx0º) ” C=100.5 mm a O E, a, LI Lo. + CI ld -] r i [a Ls 1.47 A load Pis supported as shown by a steel pin which has been inserted in a short wooden member hanging fom the ceiling The ultimate strength of the wood used is 60 MPa in tension and 7.5 MPa in shear, while the ultimate strength of the steel is 150 MPa in shear. 1.48 For the support of Prob. 1.47, knowing that 5 = 40 mm, c = 5$ mm and d'= 12 mm, determine she aliowable load P if an overall factor of safety of 3.2 is desired PROBLEM 148 SOLUTION Based on double shear in pin P= 2A%= 2Td'r, EU. om Misono!) = 33.98 w15* “ H Based on tension in wood PB: AS,= wlb-d)s, =(o.oy0)l0,040- oo )Ã(6ox 10º) > 67.2 x10ºN Based on double shear im the wood Pos 247%, = 2weT = (QNo.oso (0.08) 7. sx108) 33.0 xj0* M Use smallest P, = 380 x0! N Alowahdo Po Ee = $$: xjo? - to.30! 4 10.3) kN am “ao Ca C5 ta 4 CI ca [| Ca q Cl ra PROBLEM 149 1.49 Tn the structure shown, an 8- mm- diameter pin is used at 4, and 12- mm- diameter pins are used at E and D. Knowing that the ultimate shearing stress is 100 MPa at all connections and thal the ultimate normal stress is 250 MPa in each of'the two links joining 2 and D, determine the allowabie load P if an overall factor of safety of Top vie 3.0 is desired. pe cep mto mm SOLUTION a tme<L YO States: lise ABC os Free body. Ê e F 5 doi 4 e TEA Fao ZM, =6o o20 Fa-O!SP=o o . 12 rom +] E P= e Fa Front view Side view ZEMa=o Owhp-osP=a ta Based on double shear ju pin A P= ja A=54º = F(o.008) = So.gée x1o*w -G E = ZA, EL looxjor)( SO. 266 rj ) e B8stajo! N p= RE = saxo! N Based ow double sheam in pins ad B and D é z A = PP To ow) = l3.j0 x" m E. = ABA RNioo xj0º MB. loxio” €> - E] ES. 2.0 TS4m/0*N o P= EF 397x0?N Bosed on compression in Baks BD For one Pink As (0020 Y0.008) = I6Ox/0Um Fio = ASA . AaM2so mos Migas) ag 7mo"N "ES. 3.0 P= fo = Mox N AlBovabde valve oP Pis smallest à Po 372210ºN 3.72 kW at  (O) PROBLEM 1.52 1.51 Each of'the steel links 48 and CD is connected to a support and to member BCE by | -in.- diameter steel pins acting in single shear. Knowing that the ultimate shearing din. e stress is 24 ksi for the steel used in the pins and thar the ultimate normal stress is 60 ksi ; p for the steel used in the links, determine the allowable load P ifan overall factor of safety of 3.2 is desired. (Note thut the links are not reinforced around the pin holes A ia. N 1,52 An alternative design is being considered to support member AC ofProb. 1.51 in which link CD will be replaced by two links, each of 1 * 1-in. cross section, causing the pins at C and D to be in double shear. Assuming that all other specifications remain unchanged, determine the allowable load P if an overall factor of safety of 3.2 is a desired tosa Iein, SOLUTION | Es p Use member BCE as Free body 8 ZM, *0 8Ep-20P =o pr ãE, c E E o . 2 ha ZM-=0 gSEa-llP=o P-$ Hs Based on pin À in single shear As Edto EHk"- otoasin* BLA = (ao mess)= 4.712 keips Based on tension im Link AB As(b-d)t=(-EYG= Otas in? FR = GA = Goloias) = 250kips Ultimate Road For Dink AB'is smallet, te, Ha = NAU kips Conresponding vPkimate Joad Por strochurei Re ZE = 2.146 kips Based om pins at Cand Din doubhe shear A= Td" = Eh) = o.lgess im* Foz 2%6A = (Quo. 19635) = 7.4248 kips Based on tension in Pinks BC A=(b-ad)t o U- LHE) 06rS im" Cone Pink) = 264 = (Aleo)(o.0629)= 7.50 bips Coral, both Pinkts) Ultimate Poast for Minks BC is smalhest, re. = 7.50 kips Correspondina ultimate Load For structure Bs =Ê = 3.00 Vips. Actval ultimate Poa is smadfest ie P, = 800 kips Abfowable Jonol for stroclore Pe Ê = te = 0.938 kip P- 938 Sh. ca  PROBLEM 1,53 1.53 Each of the two vertical links Ci” comecting the two horizontal members AD and FG has a 10 x 40- mm uniform rectangular cross section and is made of'a steel with sa an ultimate strength in tension of 400 MPa, while each of the pins at € and E has a 20- mm mm diameter and is made of a steel with an ultimate strength in shear of 150 MPa. EA Determine the overall factor of safety for the links CF and the pins connecting them to l the horizontal members. A SOLUTION Use member EFG as free body. Fes Fer [ED Dl TED O.to le-gas ! y 2FkN DzMezo 0% Fr (DESVa4 elo!) e O aum Fez 37x10" N Based on fensisa jn Pinks CF A=C(b-d)t = (0.040 - O o)(v.oto)= 200 x10€ m* (one Pink) F, =28,4= (2(no0x10)g00 x10') = go. O x]0º N Based on doulide sheamin pins A = Ed" T(o.o20) = 814.16 x10! m* " Fr= LUA =tQYisoxo Bué x0)= quasgro! N Aetoa) E, is smaller valve, Le Fy= 94248 *t0* E. tua rio? Factor of sotety Es, = Fe = Ba Kia = 2.42 = — | Us 1 ed Loi+ [a É mom ca ll 1.53 Each of the wo vertical links CF connecting the two horizantal members AD and EG hasa 10 * 40- mm uniform rectangular cross section and is made ofa steel with an ultimate strength in tension of 4C0 MPs, while each of the pins at C and E has a 20- mra diameter and is made of a steel with an ultimate strength in shear of 150 MPa. ão mm Determine the overall factor of safety for the links CF and the pins connecting hem to the horizontal members a 1.54 Solve Prob. 1.53, assuming that the pins at €” and & have been replaced by pins with à 30- mm diameter. Tama SOLUTION e Use member EFG as free body. For Fe kk g4o —slooas DIM -o dn OMO Fer -(0.65 Yay=15) =0 Fe = 89x/0! N MEN Based on tension im Jnks CF A =th-ajt = (o.odo- O 080)(0.010) " 100 0197 mº Cone Piale) 80.0x10º N E: 26,4 = GUuooxio ivo xio'f) Based on dooble shear in pins A=Tadt= E(ooso)! = J06.86 x10% m* = 2%A = QMisonio! Aoc 8eviD!)= 212.06 x IO! N Actual Fis smalfer vadue ie. F, = 80.0] N Factor ob saley ps. E = SOSH0 . 205 =  PROBLEM 1.57 *1.57 A 40-kg platform is attached to the end 3 ofa 50-kg wooden beam 42, which is supported as shown by a pin at 4 and by a slender steel rod BC witha 12-kN ultimate load. (a) Using the Load and Resistance Factor Design method with a resistance factor &= 0.90 and load factors pp = 1.25 and y, = 1.6, determine the largest load which can be safely placed on the platform. () What is the corresponding conventional factor of safety for rod BC? SOLUTION DzM = o auge 244 cha 2 PoEW EM For dead hoadina W, = (WoW) =- 392.4 N = (GoXE)- 4I%o.SN Po = (EX392.9) (É U90.S) = 10623 x Io* N For Pive Poa eling W mg W so PL - e mg From — mhich m = É A Design emterion FP + NA = PR p = PR-NB . (pgo)iawiot)- (has )thocaaxio!) YA = 1.6 = S.920x10* N AP ou adbe Load mes & Sjtzio Egânio, * 362 kg «a Con ventionad Factor sabehy PP: BrP: LoGaêmo! + E 920x0* = EasaNio* N Es. DB dameo P eaBanos + 718 A  a E Do 1 L LA (9 ca va Do QB o ' IS Lis PROBLEM 1,58 P=B+P - Pa Se To — E E *1.58 The Load and Resistance Factor Design method is to be used to select the two cables which will raise and lower a platform supporting two window washers. The platform weighs 160 Tb and each of the window washers is assumed to weigh 195 lb with his equipment. Since these workers are free to move on the platform, 75% of their total weight and of the weight of their equipment «will be used as the design live load of cach cable. (a) Assuming a resistance factor &= 0.85 and load factors 4p = 1.2 and y. = 1.5, determine the required minimum ultimate load of one cable (b) What is the conventional factor Df safety for the selected cables? SOLUTION LATA = PR P, = BANP P . UANAL ICO) AUGI) Conventiona? factor of safety 0.85 = 623 fb. <a = Ju80 + ON XAxias = gs dh = Ser = 1.esa a  PROBLEM 1.59 A É SOLUTION TRADE f dam Using method o qoints te find member forces C| ID 150kN E Joint B: AB aud BD ave zero Sovee members, 225m Joint À * das E /34AST = 3.75 m ER. ME go [a NS 130 tn 3m —— Fe Fae By isimihas trianghes t Fio Force Po e For Rs db Trangte 375 (compression ) int Jem D Es By similar trrang Pes q he Fog 228 Eo * 2.25 | 3.75 he = 125 kN (comp) ar = I85x/0º N Area: Ág= 2500 um = 2500m]0 m" , IBSiO . “ = HO MP a Stress: Op = “ao ros * - Siro" Pa SHO MPa 2.59 For the truss and loading shown, determine the average normal stress in member DF, knowing that the cross-sectional area of that member is 2500 mm?.  a CI Cl PROBLEM 1.64 1.64 A 4-in- diameter steel rod AB is fitted to à round hole near end Cofthe wooden member CD. For the loading shown, determine (a) the maximum average normal stress in the wood, (5) the distance b for which (he average shearing stress is 90 psi on the surfaces indicated by the dashed lines, (c) the average bearing stress on the wood. SOLUTION (4) Mazimom norma? stress in the wood Ant o S(B-A 189 int - PR . Jeso . . Câm" res * SS pu - = l000 - . ZE Gado Nim A Po P !o98 0. . GA E Ga) O 287 pr ” 65 Two plates, each 3-mm (hick, arc used to splice a plastic strip as shown. Knowing that the ultimate shearing stress of the bonding between the surfaces is 900 kPa, determine the factor of safety with respect to shear when P =: 1500 N. PROBLEM 1.65 15 mm Pp Ro SOLUTION Ea Bond area: (See Fig vre) r P 2 do As Slate É = 1500 mm” = |506r710ºm ” P; = RAL, = (QUIsooxS MJ0o=10º) = 2700 N enilimedors Es. - E. 422 . 1.800 <a PROBLEM 1.66 1.66 Two wooden members ví 3.5 x 5,5-in, uniform rectangular cross section are joined by the simple glued scarf splice shown. Knowing that the maximum allowable shearing stress in the glued splice is 75 psi, determine the largest axial load P wbich can be safely applied SOLUTION Aa (ES ASS) 19.25 da ag: 9º-20º = 70º T= P sueco = a sin 26 po. 2ã - QUA uso pp Sinjo Sin [os = 0.49 léps -—  1.67 A steel loop ABCD of length 1.2 m and of 10-mm diameter is placéd as shown PROBLEM 1.67 around a 24-mm-diameter aluminum rod AC. Cables BE end DF, each of 12-mm diameter, are used tó apply the load Q. Knowing that the ultimate strength of the aluminum used for the rod is 260 MPa and that the ultimate strength of the steel used for the loop and the cables is 480 MPa, determine the largest load Q which can be applied if an overall factor of safety of 3 is desired. SOLUTION a Q Usina ja, E cttbree doda ond conside ima sym 4 EE 2dhe-Qro ES Fm Q= É Fa pinga a Dr, 2-&m-Fe-o Far 2FQ-Felo : qu dE Based on strenghh o? cable BE Qu= SÃ=- GEd"= imoxo)F(007= suzauo N Based on shresghth ob steel Loop Q * Éh: ÉGA- SGT = Elysono) Elo oro) = us aymo* N Based on strength of rod AC QE - ÉGA = GE - Elacomot)E(oo2v)" = sg,q2x10º N Actos? vltimade Pod Oo is Fe smadfest = Gu= 4S.Z440 N AMovabhe Jon Q= & = 45241”, IS.08 aj M = 18.08 kN «a  1.68 Link AC bas à uniform 4x +- in. uniform rectangular cross section and is made of a steel with a 60-ksi ultimate normal stress. Tt is connected to a support at 4 and to member BCD at Cby ê -in.-diameter pins, while member BCD is connected to asupportatSbya + “jn.-diameter pin. All of the pins are in single shear and are made of a steel with a 25-ksi ultimate shearing stress. Knowing that an overall factor of safety 0F 3.25 is desired, determine the largest load P which can be safely applied at D. Note that link AC is not reinfarced around the pin holes. PROBLEM 1.68 SOLUTION DEM, ro GXéR)-10 Po Fa = 2.0838 P P= Osso Fa E bzFfro Be-$R.-0 à B, = &F. -(EXaosszP)= 1.25 P a fee fipro Brinctro 8, ! B,: P- glromsP)« - 0.66667 P B:[8+B" = LWa7P, ProsseB Based on strength of Sink AC: 64= 60 bes Art = (AE-É)- ocosmsin) Fagu = SÃut >(60YDOBIT)= L8IS he, Po = (Og80 )l.87) = 0.900 kip, Based on strength pin ate: Ano Tato FB) = O MONS int T=25k Fio” UA = (25MO.llOSS) = 2761 kip P = (Duzo)2.761)= 1.825 ip, Based cu strength E pia at Bo Apos Td! = FGF)! = 007670 int Bs = Ap = (28)0.07670) = |.9475 hip. = (OvosssXi.97s) = 1.3585 kip Acducl Po is the smallest: Par 0.900 hp ABowoA e value For P: P- É = <Sidoe = 0.277 ip 2477 bh ai  PROBLEM 1.C2 1.C2 A 20-kN force is applied as shown to the horizontal member ABC. — Member ABC has a 10 x 50-mm.uniform rectangular cross section and is sup- ported by four vertical links, each of 8 X 36-mm uniform rectangular cross section. Each of the four pins at A, 8, €, and D has the same diameter d and is in double shear, (a) Write a computer program to calculate for values of d from 10 to 30 mm, using 1-mm increments, (1) the maximum value of the av- erage normal stress in the links connecting píns 8 and », (2) the average nor- mal stress in the links connecting pias C and E, (3) the average shearing stress in pin B, (4) the average shearing stress in pin C, (5) the average bearing stress at Bin member ABC, (6) the average bearing stress at € in member ABC, (b) Check your program by comparing the values obtained for d = 16 mm with the answers given for Probs. 1.8, 1.23, and 1,24, (c) Use this program to find the permissible values of the diameter d of the pins, knowing that the allow- able valnes of the normal, shearing, and bearing stresses for the steel used are, 1espectively, 150 MPa, 90 MPa, and 230 MPa. (d) Solve part c, assuming that the thickness of member ABC has been reduced from 10 to 8 mm. SOLUTION FORCES IN LINKS P=20kN "a BDircenth OF ABC M=0: 2 Fap(BO-P(AC)=0 Fop= P(AC)/2(BC) (TENSION) Fer= PlAB/Z(B Cc) (comP) (1) Link BD (2) LINK CE |'B» Thicknees= ty, E: Thickness = t Agp=É, (uj-d) j Ao DO Cop=4 Ep/A ES ce = ba up) ? Bp/ ED ED ce =— Fez Ace (3) Et B o (4) PIN Za = ep / (Meda) T= Ee /(Td/a) c (5) BEARNG STRESS ATB | SHEARINE STRESS INARC Thickness ef Member AL=taç UNDER PIN B : 26 Fa Dic tac (Nac /2) SigBea B= Fan /(d tio) plgeipege ll (5) BEARING STRESS ATC “or? . & - 28 “SsBearc- Fes/ CA tac) “e T Tac Us (CONTINUED)  PROBLEM 1.€2 CONTINUED PROGRAM OU IPUTS INPUT DATA FOR PARTS (a), (b), (c): P= 20kN, AB=025m, BC=040m, AC=0.65m, TL = 8mm, WL = 36 mm, TAC = 10 mm, WAC = 50 mm d Sigma BD Sigma CE Tau B Tau C SigBear B SigBear € 10.60 78.13 -21.70 79.58 PE 125.00 11.00 81.25 -21.70 65.77 113.64 12.00 84.64 -21.70 55.26 3] 104.17 13.00 88.32 -21.70 27.09 asi, 96.15 14.00 92.33 -21.70 40.60 4 89.29 15.00 96.73 -21.70 35.37 - 83.33 b) 16.00 101.56 -21.70 - 31.08 203.12 78,13 de ( I7TTO 106. 3T =2T.7U T-57 15 TST. TS Ts 18.00 112.85 -21.70 63.86 24.56 180.56 69.44 19.00 119.49 -21.70 57.31 22.04 171.05 65.79 20.00 126.95 -21.70 51.73 19.89 162.50 62.50 22.00 135.42 -21.70 46.92 18,04 154.76 59.52 22.00 42,75 16.44 147.73 56.82 23.00 39.11 15.04 142.30 54.35 24.00 35.92 13.82 135.42 52.08 25.00 33.10 12.73 130.00 50.00 26.00 30.61 11.77 125.00 48.08 27.00 28.38 10.92 120.37 46.30 28.00 26.39 10.15 116.07 44.64 29.00 24.60 9.46 112.07 43.10 30.00 22.99 8.84 108.33 41.67 (0) ANSWER: lómm<d<22mm (O) CHECK: For d= 22 mm, Tau AC =65 MPa <90MPa OK INPUT DATA FOR PART (d): P=20kN,AB=025m BC=0.40m, AC =0.65m, TL=8 mm, WL =36 mm, TAC = 8 um, WAC = 50 mm d Sigma BD Sigma CE Tau B Tau C SigBear B SigBear € 10.00 78.13 21.70 650) 79.58 156.25 11.00 81.25 21.70 , 65.77 142.05 12.00 84.64 21.70 [43% 55.26 130.21 13.00 88.32 21.70 S1 47.09 120.19 12.00 92.33 21.70 40.60 111.61 15.00 96.73 21.70 E] 35.37 104.17 16.00 101.56 -21.70 Bogz 31.08 97.66 17.00 106.91 21.70 71.59 27.54 92.91 18.00 112.85 -21.70 63.86 24.56 B6.81 19.00 119.49 -21.70 57.31 22.04 213,82 82/24 20.00 126.95 -21.70 51.73 19.89 203,12 98.13 21.00 135.42 -21.70 46.92 18.04 19345 4.40 22.00 145.09 21.70 42.75 16.44 184.66 71.02 23.00 [Tê -21.70 39.11 15.04 196.63 67,93 24.00 -21.70 35,92 13.82 169.27 65.10 25.00 -21.00 33.10 12.73 162.50 62.50 26.00 -21.70 30.61 11.77 156.25 60.10 27.00 -21.70 28,38 10.92 150.46 57.87 28.00 -21.70 26.39 10.15 145.09 55.80 29.00 -21.70 24.60 9.46 140.09 53.88 30.00 :-21.70 22.99 8.84 135.42 52.08 (d) ANSWER : 18 mm <d<22mm me G CHECK: For d=22 mm, Tau AC = 81.25 MPa < 90 MPa OK. d)  PROBLEM 1.3 1.C3 “Two horizontal 5-kip forces are applied to pin B of the assembly 05im shown, Each of the three pins at A, B, and C has the same diameter 4 and is in double shear. (a) Write a computer program to calculate for values of d' from BR 0.50 to 1.50 in., using 0.05-in. increments, (1) the maximum value of the av- erage normal stress in member AB, (2) the average normal stress in member BC, (3) the average shearing stress in pin A, (4) the average shearing stress in pin C, (5) the average bearing stress at A in member 48, (6) the average bear- ing stress at C in member BC, (7) the average bearing stress at B in member BC. (b) Check your program by comparing the values obtained for d = 0.8in. with the answers given for Probs. 1.9, 1.25, and 1.26. (c) Use this program to find the permissible values of the diameter d of the pins, knowing that the al- lowable values of the normal, shearing, and bearing siresses for the steel used are, respectively, 22 ksi, 13 ksi, and 36 ksi. (d) Solve part c, assuming that a new design is being investigated, in which the thickness and width of the two members are changed, respectively, from 0.5 to 0.3 in. and from 1.8 im. to 24in. SOLUTION FORCES IN MEMBERS BE AND BC FREE Boprs PINB Fed FORCE TRIANGLE! + Fa F 2P = > 2P * PB 86 o dr É / “Ss fa s Sin AS T gindo” sim7s te, ES . ) 5 , e. e cg P=2PÓSin45ySim 75) Copo fae=2P(Sinto/sin75) (MAX, AVE, STRESS IN AB (2) AVE, STRESS JN BE e Widih= ur É sa (SE =É Bc Age = Wl Ee nel Fae/Ane Es Gac? Fec/Age CG) PINA (4) PIN C Ca = (Pas /0) TAVA) CoPi) (C)BEBRING STRESS ATA | (6) BEMRINE STRESS ATC Sig Bear A = Fas/dt SigBear C= Re/dt (7) RERRINC STRESS ATE INMEMBÉER Bl Sig Bear B= For /24I (CONTINUED)  PROBLEM 1.€5 1.€5 A load P is supported as shown by two waoden members of uni- form rectangular cross section which are joined by a simple glued scarí spice. (a) Denoting by o and Te, respectively, the ultimate strength of the joint in tension and in shear, write a computer program which, for given values of €,b, P, o and Ty, expressed in either SI or U.S. customary units, and for val- ues of a from 5 to 85º at 5º intervals, can be used to calculate (1) the normal stress in the joint, (2) the shearing stress in the joint, (3) the factor of safety relative to failure in tension, (4) the factor of safety relative to failure in shear, (5) the overall factor of safety for the glued joint. (b) Apply this program, us- ing the dimensions and loading of the members of Probs. 1.29 and 1.32, know- ing that o; = 1.26 MPaand 74 = 1.50 MPa for the glue used in Prob. 1.29, and that cy = 150 psiand Ty = 214 psi for the glue used in Prob. 1.32. (c) Verify in each of these two cases that the shearing stress is maximum for m=45º, SOLUTION (I)ana(2) Draw the FB diagram of lower member: V VE dZE=or -VtPox=O O V=Pesw EE=o: F-Psinx=o F=Psina x * Areaz ab/sinwy E. Normal stress: P 7 E E = (F/ab) Sinto Área Shearing stress: EM = (P/ab) Sinx cos (3) Es for fensien ÔNormal e es) FsN = 04/02 (4) FS for shear; Fss 2 CG/O E) OvERALL FS. FS = The smaller ot FsSN and FSS (CONTINUED)  PROBLEM L.€5 CONTINUED PROGROM OUTPU TS For Pol. ledo Pa okN A RI mn, b=750m, X= 70, G, 1,26 MPa, “47 L5UMPa, ALPHA SIG(MPa) TAU(MPa) FSN Fss FS 5.0000 0.0049 0.0556 259.1782 26.9942 26.9942 10.0000 0.0193 0.1094 €65.2905 13.7053 13.7053 15.0000 0.0429 0.1600 29.3899 9.3750 9.3750 :.0000 0.0749 0.2057 16.8301 7.2925 7.2925 25.0000 0.1143 0.2451 11.0229 6.119] 6.1191 30.0000 0.1600 0.277 7.8750 5.4127 5,4127 35.0000 0.2106 0.3007 5.9842 4.9883 4.9883 40.0000 0.2644 0.315] 4.7649 4.7598 4,7598 45.0000 0.3200 0.3200 3.9375 4.6875 3.9375 «ali (e) 50.0000 0.3756 0.3151 3.3549 4.7598 3.3549 55.0000 0.4294 0.3007 2.9340 4.9883 2.9340 60.0000 0.4800 0.277] 2.6250 5.4127 2.6250 65.0000 0.5257 0.245] 2.3968 6.1191 2.3968 72.0090 0.5651 0.2057 2.2296 7.2925 * 2.2296 «df (b) 75.0000 0.597] 0.1600 2.1101 9/3750 “II01 80.0000 0.6207 0.1094 2.0300 13.7053 2.0300 85.0000 0.6351 0.0556 1.9838 26.9942 1.9838 For Prob Li2! Po otools Aa z6Gin,b=3in, X= 405, C= 150 poi, TZ =2t poi U U , ALPHA sIG(psi) TAU(psi) FSN FSs Fs 5.0000 1.0128 11.5765 148.1018 18.4857 18.4857 10.0000 4.0205 22.8013 27.3089 9.3854 9,3854 "0000 8.9316 33.3333 16.7942 6.4200 6.4200 .9.0000 15.5970 42.8525 9.6172 4.9939 4.9939 25.0000 23.8142 51.0696 6.2988 4.1904 4.1904 30.0000 33.3333 57.7350 4.5000 3.7066 3.7066 35.0000 43.8653 62.6462 3.4196 3.4160 3.4160 20 0000. 55.0901 65.6538 27228 3.2595 2.7228 af tb) 45.0000 66.6667 66.6667 — 2-2500 — 3.ZIUU TESUT «h(C) 50.0000 78.2432 E5.6538 1,917] 3.2595 1.9171 55/0000 89.4680 62.6462 1.6766 3.4160 1.6766 60.0000 100.0000 57.7350 1.5000 3.7066 1.5000 65. 0000 109.5192 51.0696 1.369 4.1904 1.3696 70.0000 117.7363 42.8525 1.2740 4.9939 1.2740 75. 0000 124.4017 33.3333 1.2058 6.4200 1.2058 800000 129.3128 22.8013 1.1600 9.3854 1.1600 85 0000 132.3205 11.5765 1.1336 18.4857 1.1336  PROBLEM 1.C6 1.C6 Member 4BC is supported by a pin and bracket at 4 and by two links wihich are pin-connected to the member at B and to a fixed support at D. (a) Write a computer program to calculate the allowable load P,y for any given values of (1) the diameter à; of the pin at 4, (2) the common diameter «, of the pins at B and D, (3) the ultimate normal stress q, in each of tho two links, (4) the ultimate shearing mm stress t, in each ofthe three pins, (5) the desired overall factor of safety F.S. Your Topview program should also indicate which of the following three stresses is critical: the 200 mn -e-180 uu normal stress in the links, the shearing stress in the pin at A, or the shearing stress Ie | e mim in the pins at Band D. (b and c) Check your program by using the data of Probs. 1.49and 1.50, respectively, and comparing the answers obtained for P, with those given inthe text. (dd) Use your program to determine the allowable load P,y, as well as which of the stresses is critical, when «, = d, = 15 mm, q, = 110 MPa for aluminum links, 7,,= 100 MPa for steel pins, and F.8. = 3.2. SOLUTION PT” (0) EB DIDERAM OF ABC: E A = 0 Eesm A N > Ma ESA th Y 200 mm | jEon] EMps0: Side view K Fep P po >00 =jgo Ja () Foraiwmea di sé pint rs (o, des (iairo), P dh, (2) or given cof pins BandD' p= 20, FO? |), Bo 200 E, 63 for ultimate stress jo links BD: Pap 2(T/FSO 020.008), p= fp É, shearinpctresc in ainso Ro to) Tor UE sfencinp tres “o inês Fy is the smaller of P and É, (E) For desired ovecall ES! Po is the smaller pf E amd P, If FB <P, stress ig critical in links g Bs ch, and P< PB stress is Critical tm pin À Py <B and B< Po shrese is critical in pis Band D PROGRAM OuTPUTS (5) Lob. 1.47, DATA: d,= Bm, d,=12mun,0;,=250 MPa Cj = 100MPa ES=3,0 Fate 3,72 kN. Stress in pin A is critical «x C) Probt.50.DATA: dj= 0a dy =|2 mm Oz 250MPa, 2, IbUMPa, FS = 20) FenT GI97kN. Stress in pins Band Dis critical «4 &) DATA: d=d,>150m,G,=HOMPa, T=/00MPa, ES.= 3.2 Pu SIIkN, Stress in links is critical «ad  ra t— a a 1 ces ES 2.5 A 9-m length of 6-mn-diameter steel wire is to be used in a hanger. It is noted PROBLEM 2.5 that the wire stretches 18 mm when a tensile force P is applied. Knowing that E=200 ra determine (a) the magnitude of the force P, (b) the corresponding normal stress SOLUTION in the wire. - 2 - ta) A=qul= Bloco) = 28.q74x107! mt g= Ph. po MES, (28 474x10€ X 200210 Mi3n103) “AE UV 4 = Biot No = M.3 kN a bb) &s f = = uooxio* Pa = 400 MPa mesa 2,6 A 4,5-t, aluminum pipe should not stretch more than 0,05 in. when it is subjected PROBLEM 2.6 to a tensile load, Knowing that 1 = 10.1 x 10º psi and that the allowable tensile strength is 14 ksi., determine (a) the maximum allowable length of the pipe, (b) the SOLUTION required area of the pipe if jne tensile load is 127.5 kips. - PL «| « EAS . ES Gozo) 0.08) , tw Ss RE “L 2 pass 86.1) in a = : - Po, Tso 2 6) c=-D a A- To ZiSo. qui im et PROBLEM 2.7 2.7 A nylon thread is subjected to a 8.5-N tension forve. Knowing that E = 3.3 GPa ” and that the length of the thread increases by 1.1 %, determine (a) the diameter of SOLUTION the thread, (5) the stress in lhe thread. (a) 2 = pa a F = 90.904 - PL PL (8.5)(40907) 5. = o 2 =. TSE — , X/o SE À: E ES riDd 23416 x10” m AE: ds E O. ssExiOÊ Mm = 0.546 mm = bt) 6 = É no - Be3xi0º Fa = 33 MPa 2.8 A cast-iron tube is used to support a compressive load, Knowing that E = 10 x PROBLEM 2.8 10º psi and that the maximum allowable change in length is 0.025 percent, determine (a) the maximum normal stress in the tube, (5) the minimum wa!l thickness for a load SOLUTION af 1600 Ib if the outside diameter of the tube is 2.0 in. 03 Be = 040025 S = Es = (Ox 0tAn uvas) = 2.5 molps; = 2.5 ks aa P J606 Ja - “ == -— .eto & S&S Ê Â = LEIO 0.6 im “ . A = Td - di) 2 ( J 1 . x d= do - e = qt. bt set = 3Bstintã de= i7847 in te lldo-A) A(AO- 17847) = OO7 im <a  2.9 A block of 10-in. length and 1.8x1,6 in. cross section is to support a centric compressive load P. The material to be used is a bronze for which E =14 x 10º psi. Determine the largest load which can be applied, knowing that the normal stress must SOLUTION not exceed 18 ksi and that the decrease in length of the block should be at most 0,12 percent of its original length. À PROBLEM 2.9 comida ollowalode alvess GC =I8hs = 8708 psi = ML) = 288” GÊ 2 = GA z=(3H0M2SB)2 S1.8 mio! dk Considerina affowab fe debormedia Dez LÊ - ooo! L Togo 5= FE « P5ÃEÊ = (assiyxiotlo.o0r)= 43.410! Mb Smalher value gevems P=ugimo! MH = us. kips - PROBLEM 2.10 2.10 A 9-kN tensile load will be applicd to a 50-m length of steel wire with E' = 200 GPa. Determine the smallest diameter wire which can be used, knowing that the normal stress must not exceed 150 MPa and that the increase in the length of the wire SOLUTION should be at most 25 mm. Com sidering alhowabhe stress S=I40x0ºPa p Po. Ie ea G=: 2» As = G0%0" m A FT isoxios . -3 Con sidering alhow able ebonguti om S=2A5vio"m E PL. LIxloê)iso) ca = - = = DwWIO m 8 A= ES e O asmos) go mt Largo area goverus = go wx/0'* m A = Ea d «JE FA . [ng ti(goxto rt) . 10-10 40º m = |fo.7JOo mm CJ CI 1 C 3 Ca 7 / qa O 1 Ta [o À Es - a 2.41 The 4mm-diameter cable BC is made of a steel with E =200 GPa. Knowi PROBLEM 241 tar the maximum stress in the cable must not exceed 190 MPa and that the ne elongation of the cable must not exceed 6 mm, find the mazcimum load P that cam be applied as shown. SOLUTION La=) 64h = FR m ; w Use bo AR as a Prec body Ê Aê A. 1 Losoml JM =0 3asP- (6 Xsém Fro)= o Pe o.ys0% Fe As Consideru alfomeb he stress S= 190 * Pa A, “ As Bata Elucos)t = Ia.s66 r/otM G= Eis Fer GA= 0% motduascexo = 2.888*410! N Considering alPow ole edongution 8: € xjo? m 5 « Gula o po DES, Mto tMan Mo! K ENE). 20450? Smaler value qeueras Re Reogixio? N P: o.9s04 Re = loasorÃa oaad')= 198810 N € [.788 K) a ) ss 242 Rod BD is made of steel (1 = 29 x 10º psi) and is used to brace the axially compressed member 48€. The maximum force that can be developed in member BD is 0.02P. If the stress must not execed 18 ksi and the maximum change in length of P= 130 kips BD must not exceed 0.001 times the lengih of ABC, determine the smallest diameter rod that can be uscd for member BD. PROBLEM 2.12 SOLUTION Fo = C02P = (or) = acho = 2.6 x10º 4h Consideriny sivress S= 13 si = 1800" pai Fão co as BE. og tm Considera, defornation SrlOooN(MD = O. HE im. - Eos. - Fole — (Zemos EH) a s-*E% À= ÉS * Geo )to duo) 0-03862 in Larger area goveras AS O UU no AsEdt a d-[E - [toped = 0.429 in. a  2.17 The rod 4BC is made of an aluminum for which E = 70 GPa. Knowing that P | | ' i & PROBLEM 217 =6kN and Q = 42 kN, determine the deflection of (4) point A, (b) point B. r SOLUTION . 0) Amas Edo = FHo.020)! = 84. 1e io wo ; À me damer Ag = Foo” = Wlo.oco)' = 28274 xo? mt Paz Pr GH N | | Pac =P-Q= Gxot-Nxio'- “34 N vem | som dancer bas 7 Om Lee 0.5 m | sus Palm. (emiot No) o | [A Breton Goxio?) - 109.135 xi07* mm a. | Se Peclm. (serto 08) “AE sinris Yo rio?) = - 0.47 x/0" m E Sa7 Das + Sge 7 lO7 I80x[6!- Go JY7RiO Mm = IB IGujo! mm fr | = 0.084 mm a | Cl! (bl Sg= Sw 2 -PIriytm = - 0.090 mm = | A mm E [| CI la E, 3H TT] 2.18 The 36-mm-diamcier steel rod ABC and a brass rod CD of'the same diameter are Joined at point C'to form the 7.5-m rod ABCD. For the loading shown, and neglecting the weight of the rod, determine the deflection of (a) point €, (b) point D. PROBLEM 2.18 SOLUTION “o A=Edts Eloom)= Logus mr Stesk E = 200 CP: e º Portioa] É Lz E | PL;/ÃE SN m AB |i50 kN | 2m | 200 GP: | LUI xo! m BC jjoun | 3m |2006Pa| WA Io m Et CD | I90kN | 2.5m|105Gfa| 23344102 m Brass: E -- 105 CPa têm (o) DS SagtSu = LIPIedo + LATO” = R.9948ri0tm = 2.95 mm «am () So D+ So MIJRO! 4 2.989 108 JOD EN S2sPrio tm = SAP mm «at 2.19 The brass tube 4B (E = 15 x 10º psi) has a cross-sectional area of 0,22 in? and is fitted with a plug at 4. The tube is attached at B to a rigid plate which is itself attached at Cto the bottom of an aluminum cylinder (E = 10.4 x 10º psi) with a cross-sectianal area or 0.40 irê. The cylinder is then hung from a support at D. In order to close the cylinder, the plug must move down through &; in. Determine the force P that must be upplicd to the cylinder. Shovkening of brass tube AB Le Std fBoMTin Ag O2in" Es I5x1D* par Pla Pus£esm) - . Elas = 4554740 iba “me Echo Too) * tóstmio Lengthenina E abuminum eyfinder cD PROBLEM 2.19 Leo tiSim, Agr OHO in” o - Plo Pts) - é" EsÃo QodnSIodo) Er to Meo psi + 3.6058 mo Total dePhecÃtou Sa Tag à So 3 (usas! sessao NP 2 Po spp uto! dh 64 = 5.74 ips et  2,28 A 1,2-m section of aluminum pipe of cross-sectional area 1100 mm? rests on à fixed support at 4. The 15mm-diameter steel rod BC hangs ftom a rigid bar that tests onthe top of the pipe at 2. Knowing that the modulus of elasticity is 200 GPa for steel and 72 GPa for aluminurm, determine the deflection of point C when a 60 kN force is applied at C. PROBLEM 2.20 B SOLUTION Rod BC les Ul ms Es. = 200 410" Pa A Sus Pla - (Gono! Can) de Eh (RODO NINE IDE) = 3.565x10"* m Td" = Elo.0s)" - mesmo mt +” Pipe AB: LagT 2m, Em NAXIO! Pa, Ang TIDO mei 1OOIS Cm! - Pla . oxio')(.a) - ca Som Tha Gai Xircosioe) TOTIXID m Seo Semi Sat POLI BESC E MAPS a EM Tom ma 2.21 The steel frame (E = 200 GPa) shown has a diagonal brace BD with an area of PROBLEM 2.21 1920 mm?. Determine the largest allowable load P if the change in length ofmember Pp BD is not to exceed 1.6 mm. Ê e SOLUTION “= -o So cloriom, Amos |920 mm = [920 H0* mm” sm LuzyS"+6'= 780m, Egzâvono! fa so - Pele | de Erhm . il = Le 7.81 em = 78.67x/0? N Use joint B as a Free body: +ER> o mofe-P=o P Ss . QUE O) Pã Fe = Elo Fac = 50.4xi0 N = 40.4 kN) em  UI Ca o DS DS. o 2.24 Members 4B and CD are 1 4 «in.-diameter steel rods, and members BC and AD are t -in.-diameter steel rods. When the turbuckls is tightened, the diagonal member . AC'is putin tension. Knowing that E = 29 x 10º psi and » = 4 ft, determine the largest B e allowable tension in AC so that the deformations in members AB and CD do not exceed 0.04 in. PROBLEM 2,25 ] 2.25 For the structure in Prob. of 2.24, determine (a) the distance k so that lhe h deformations in members 43, BC, CD and AD are all equal to 0.04 in., (b) the corresponding tension in member 4€. a p SOLUTION Lo (0) Statics? Vie quint B as a Tree body E Evo similar triangdes F, F Loo Fis Fe Fo Fo eo Fl Nº hM h bo la À N [A + Faso = h Coe Force Triana fe Geometry For. equal detormatious ' bA mom BRO me fder Equaling express tous For Fio be cb As bh Am - Edo . da bach fo fu O Re fd” dai todo. db. s b= 3H = BE in h=Fb = E(3)= 2.86 Fh = 46.8 in mi (b Sefting Seg = See * Q.0f (n, % qo - Fab o p= EfuSe (roxo!) E(G) Coou) CC Che se b 3€ " 17.376 40º Pb, Fas ? t Fa Fagen )= ar azxo! A From the torce Inangde Fer f JRR = aneno d -  2.26 Members ABC and DEF gre joined with steel links (E = 200 GPa). Eachofthe PROBLEM 2,26 links is made of a pair of 25x35-mm plates. Determine the change in length of (a) member BE, (b) member CF, 180mm SOLUTION c E. Use member ABC as a Free. body 260 mm cr | ellos DEM-o E fesmram=o Fe (o.2go)ús mio?) - (0.136)Fe = O A For = (odte ig nto!) . Rex N [BAN » ZM.> o (o.s4o Miê > 10º) + (0.180) Fez = O her Dink mod Ávea ink made Far (esse Miami) qyujot M SE fo ples À = (20.025 %0.085)2 1260! mt Ê - Po. = do do be E aanitos va x 3 « - 06) Se = Fe er = o ay 17. 88910 “m = O ÓNRBS mm cad  2.27 Bach ofthe links 48 and CD is made of aluminum (E=75 GPa) and has a cross- PROBLEM 2,27 sectional area of 125 mm. Knowing that they support lhe rigid member BC, determine the deflection of point E. SOLUTION Paim Fa Fo Use member BC as « 1 o Free body tispat om é Sujo? no Dem co (0.6) Fa tlON SED O Fas T 3.4375x0º N DzmM=o (0.68) Fi, - (0 20KGxo*)- 0 Fo * LSGAS AN For Pinks AB ad CD A=I2S mm = IAGO im" E Fuel — (3H do! 0,36) - 182.00 jo” 5 EA CsmvIKUAS XISES 2 te m e se Sof, (SACOS, qm it 1 8 E e Shope O: dec 8 - Alceu “NAS eo pod Se= S + 16 DeSorma bom olinmpam = Eooorio* + (0.44 K2.5 mio) = |orS=otm = 0.1045 mm «é  2.31 The volume of a tensile specimen is essentially constant while plastic PROBLEM 2.31 defurmation occurs. If the initial diameter of the specimen is dl, show that when the diameter is d, the true strain is 6 = 2 Ind, /d). SOLUTION 1F the volome is constant Fair = Edi, L - ; E Se -(g) = ME - MS) = am E) 2.32 Denoting by ethe “engineering strain” in a tensile specimen, show that the true PROBLEM 2.52 straimis G=In(! + 6). SOLUTION E Ma E - Ea = MUS Ê) = datise) Thos E, s Pais) -—  Fa e e o O Doo oOoOoROo y 2,33 An axial force of 60 kN is applied to the assembly shown by means of rigidend PROBLEM 2,33 plates. Determine (4) the normal stress in the brass sheil, (b) the corresponding deformation of the assembly. E SOM som 20mm 3 sum “es eds mm SOLUTION o Let Ps portion of axial force compied by bress shedp Steel core ne . E=20CPa R= portion ob axial Ponce camred by steef come Brass shell Se AL Po: Er AS E= 105GPa 280 mm A Es L BL P,= EA S AsE L PO BARBA SA Do / 2.e = L Es A, 3 Es As As = (o 0z000.020)= gooxlo* m? As= (0.030 X0.080) - taozoyo.020) = 100 rD mm e 60 x10$ .E S=£= = 4$2.88 x/0 U (las 10º Y500x10*) 4 (doo=10º Moo xIO E) (0) 6: Ez = (losmo'Anszesxo OD urso! Pa = 4.5 Mp — (b) 8 = Le = (QsoxmoNisisameo)- [I8.2x/0%m = o. 1132 »/0* m = O ISA mm -t  2.34 The length of the assembly decreases by 0.15 mm when an axial force is applied PROBLEM 2.34 by means of rigid end plates. Determine (a) the magnitude of the applied force, (b) the corresponding stress in the steel core. 5 mm, SOLUTION . Let Ps portion E axial Force comel by brass shedg. Steelcore f * portion oF axial force corritd by stee) core. Pl . EAS Ee amo fr Ss = RL A = EA O hs Es l P= RAP (EA LEA) É . Às = lo020X0.020) = Jos x(D"* m À = Co.o30Y%0.030)-C0.0200.020) = S00w0*m” 37] DIS? to Pr Itosmo! (sooxio!) + (200 mio! Kyoo to) ] 255 =10* = 7.5 wo* M = 75.9 kW as 0) q- ne - Sê, (GoorplMo fuel . go miot Pa u RO Ma ame) [| ld DD cc a i 2.3? For the composite block shown in Prob. 2.36, determine (4) the value of k if the portion of the load carried by the aluminum plates is haif the portion of the load carried by the brass core, (b) the total load if the stress in the brass is 80 MPa. PROBLEM 2.37 Brass core (E = 105 GPa) SOLUTION Lt & a portion HF axiul Forea cormied by bross core = porbia camied ly The tuo omingm plates RL EA, Ss S= Po m Ebhy EA h L s. BL p= EiMaS EA, L (23 Given R>4R Elas = | EhAyS Arco nto m? om = QMeo)Jh 15 mm -. Pr = AS =(Mooxo Xgo mio) = Iggxto8 N PAP Re N P= BR+P = gramolny = 88 4 -a  2.35 The 4.5-fi concrete post is reinforced with six steel bars, each with a Him. diameter. Knowing that E, = 29 x 10º psi and E, = 4.2 x 10º psi, determine the normal stresses in the steel and in the concrete wben a 350-kip axial centric force P is applied to the post. 2.38 For the post of Prob. 2.35, determine the maximum centric force which may be. applied if the allowabie normal stress is 20 ksi in the stecl and 2.4 ksi in the concrete. PROBLEM 2.38 CS LS m | SOLUTION 7 Determne allovabde strasa in gaeh matencl . « - O aoxot -€ . Sreed: Es E “muige * 6920 L; Cononte: E.= E atra? = SI ABajO* E. maxios r Smalher value govems £+ 2 = 57.48 x10'* Let R.= potou 5 Poud carried by comerete Fã = potim carried by six steel vods s = EA P- EAE $ Ee? Pç EA, É = EA E P=P+B = (EA+rEAJE A, 6 Td CE(riaÊ= gia ri” Ac = Edo A = E(e)- say = 248.5 in* E Ê e + co ca Op r> " H 9 Cl P= [tuanotX 248.5)+ (2a mio! AGA) CS 74310) = 695 x/0º 4 ” 695 kips o] - Í I Co Ca 3 E 1 (— o o o! E 1 lh) 2.39 Threc steel rods (E = 200 GPa) support a 36-kN load P. Each of the rods AB and PROBLEM 2,39 CU has a 200-mm? cross-sectional area and od EF has a 625- mam? cross-sectional area, Determine (he (4) lhe change in lengih of rod EF, (6) the stress in cach rod. SOLUTION Use member BED as a Free body he Co By sqnmetoy, or by EMpro Ro * Ps ZR:o PastPao tt -Pea P- 2Fa + Per Pis Las Posleo — Pal Ce = Chao Se ER o Ser FRe Since Lgclo ond Agr Ao, a T Seo Since ponta AC, and E ave fixed So = das, Ep So Ses Ser e Since mewber BED is rig Se = Sa: S Pas Lia Peel Agel 200 = e bas . . Ae. Le - . HQ E Asa E Age 2 Pa Às» Lam Per = Gas So Fe = 0.256 Per P= 2 + Per ROLO Pa 4 Pro SIR Po - Po - Bexios s Per = isa * tea - 28.80 mo! N Pe: Po: toaseVas mono) = eors xo! N (23.840 m03W 400 war) -€ Sr Tioómos leis) ” 7e.2x0" m 3 0.0H2 mm at « Le oerxo! Ms00 mt08) “ Sm "Tioxio zoo rio) * 78 AXO m J eo ” “o ” Sa Gu + fe. Como! . so sunt Ph = 30.5 MPa a “o Apa 200 rio 2. .ABBIOXIDI So - Ser = AE Ceasa = -38. xi fa = 38IMPa CI ta 1 f LI - Lu mw o | ; UI [a 2.42 A steel mbe (E = 200 GPa) with a 12-mm outer diameter end a 4-mm thickness is placed in a vise that is adjusted so thut its jaws just touch the ends of the tube without exerting any pressure on them. The two forces shown arc then applied to lhe tube, After these forces are applied, the vise is adjusted to decrease the distance SOLUTION betweca its jaws by 0.2 mm. Determine (a) the forces exerted by the vise on the tube at A and D, (5) the chanpe in length of the portion BC af the tube. PROBLEM 2.42 Ra É 8 c D E ds E HA kN Jokb For the tube dizdç-2t =32 (24) = 2%mm A=E(d -di)= (32! 20") = 351.86 mm = 3SLBGuio mi AB: Pr R,L=:O080m FL Ra (0.080) = igeg eo? R Sua? EA * Goowoskast RENDA + BC: P= Ri+ 4Axio*, Le: 0.080 Sa é EE. ati) c pigga no! RA + 47 mio "Fà (2o0x01 X351Z6 x0€) Co: P=Rtizco?, L: 0.080 = (Rar IQxi0! 0.086) - PL s 2 re TEREM = LIZGRO R+ 13.442x16 des EA —Goono Xasi.8exio) 8 8 ê Total: So - Set SutSo + 34IO4nIO PR, + ci. 388x/0 Civen jam movement So *-ORem = Quo" w (=D lo"= Egbyrio "A, +cLsgao! . Rir -7EC ON A =-76.6 kN ham Ro: Rj + IQuIos Ro = -64exioê MN - - 646 UN hn =34.4 w40"*m -0.0398 mm a &) Se : (LiZegxo Veree mo) + qr 746 nto  ( co Es ca PROBLEM 2.43 2.42 A steel tube (E = 200 GPa) with a 32-mm outer diameter and a 4-mm thickness is placed in a vise that is adjusted so thát its jaws just touch the ends of the tube wilhout cxcrting any pressure on them. The two forces shown arc then applicd to the SOLUTION tube, After these forces are applied, the vise is adjusted to decrease Lhe distance - between its jaws by 0,2 mm, Determine (a) the forces exerted by the vise on the tube at À and D, (b) the change in length of the portion BC vf the tube. 243 Solve Prob, 2.42, assuming that after the forces have been applied, the vise is adjusted to decrease the distance betwcen its jaws by 0.1 mm. 80mm | Sômm mm RA 8 c D p, 4kN 3oky For the tube di=d,-2t =32 (204) = 24mm 2 - Ela diy= Tata) SSL mm = 351.86 010 m* > u “ AR: Pr RM, L-0080m Sig * El . Falooso) ad PP EA * Qocuokesamos igeg uol Ry BC: P= Ra+ 4axio? L=0.080m = PL ARa+ ago!) (o) 1 : -. == AMO MIO! = LI368X10 7 x Be CEA CCaoonios ESTE 50) HI3 Re + 47 IO P=Ratriaxio?, L=Oogo S. = BL. (Mariano! Mo.oso) e EA Taoovio' KEsIBenio*) en <D: LIES IG RA I3G4aniot Total: So Sat SutSo * I404nIO RP, + el.38au/0'E Due do He movement of +e jaws Sao = — Olmm = - Ob m (o -Otxio = sylognio "RA + Gl. 388m0* fy = - Hrezado! n =- 472.3 4 me Ro = Ras tamo! --g5. 2ºr"10'N =-B8.83 kN mai) (b) Set(iizesno KIT ETA E) + 47. THE md = — Gar Cm = —0,00408mm E , oca r o [Io Cs PROBLEM 2.44 2.44 Three wires are used to suspend the plate shown. Aluminum vires are used at A and É with a diameter of 3 in. and a steel wire is used at C with a diameter of TF in. Knowing that the allowable stress tor alurainum (E = 10.4 x 10º psi) is 14 Ksi and that the ellowabls stress for steel (E = 29 x 10º psi) is 16 ksi, determine the maximum load P-that may be applied. -- SOLUTION By symmetry Paz Po, and Sa= Sa Abdo, S.= 84=S * $ Strain in each wire As, EsÊ E Determine ab fable straim A:B as Sa. lino 3462 nO! Ea Joctaige — Ec =ê re eo a € Ee = E = ade = 0.G2o7xi0* LA E Es = Se - 9.3103 » 19º... Abfovable straca Br wire € governs à 6. = i8x10? pai GH Re ABe HEfGano Nose) = 276! Ab fa = jsmer SerE£ O PAS = EGhyUsmo)= 18.17 Md For eguil briom &? He plate Pe PPP = 77 a  Ls PROBLEM 2.47 2.47 The rigid rod ABCD is suspended from three wires of the same material. The cross-sectional area of the wire at B is equal to half of the cross-sectional area of the wires 4 and €. Determine the tension in each wire caused by the load P. SOLUTION e Ler À BE THE LENETH OF THE WIRES BA Agp te Ê à ao Fa ) dE Fl E -É & —— à Bl A (êr-dh) “ LA Fon THE pereRimem DlrsBAM , 3 -5=3-6 e OR des (Batã) £ Ela) FA SB=7º3' 5º B= 0,zoop BP Hb)-He £= sz p= Er HE) go? prarsp “a CEC 3 BR rare noc6 Fº, Ok  248 The rigid bar ABCD is suspended from four identical wires. Determine the tension in each wire caused by the load P. PROBLEM 2,48 SOLUTION Let O be the shope &f bar ABCD aflen dePormetion Sec S4+ LB Sd = S4+ 2L0 So = 4 +3L0 p= is, Pos EA S + Es s, + Edo Poo Êo = Eis, + 2886 Pr EA G - ER s, + SEAL O P dzr-o ReBAR+Po-P-o ho, + SEA jo = P ns +G Lo = E nm DTM -o LP +2LP. + GL - QALP=O GEA 5, + JEAL | 6 - UP ES + mio - SÉ (2) Solving (1) and (3) aims )tuneovsd; Le = sê + E Pas se - 5” - e Ei SAE = dr - o Psdp Aedo a - DR o -  2.49 A steel railroad track (E = 200 GPa, a = 11.7 x 10%ºC) was laid out at a | . PROBLEM 2.49 temperature of 6ºC. Determine the normal stress in the rails when the temperature o C " reaches 48 ºC, assuming that the rails (a) are welded td form a continuous track, (b) [ í SOLUTION are 10 m long with 3-mm gaps between them, ta) Sr AMDL= (To A4S-CN1o) = parpxjd"m 7 EL Le tloy€ 514 ] Se: AE E “Soro s0x0º € T S= +18 - Egyro' +soxo"G = o « Ga-98,38m0 ="48.3 MPa «8 (b) Se 8 +S = 4ayalo'+ Sono "G = axo” - Into - pommo? ag 8 not > 383 MA «a S Soro 38.3 »jo* Pa fa o 9] O Lo LT