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CHAPTER 7 i . ê a r . - | k YDocorãõioõirgaãioeo oo OT Dto cao cs ca end * o o e — - E] —) DRE o “TE tlaromgho 7,4 For tho given state Of stress, determine the normal aod shemins PROBLEM 7.1 stresses exeried on the oblique face of the shaded triangular element shomri. Use é method of analysis based on the equilibrium of thai ciement, as wes done in the decivations of Sec, 7.2. 4ksi a 1 SOLUTION - Ny R A & BA cos SA q ÂÃ Poe o 1 < 3Acos70 * Agindo “IA singo” Stresses qts Areas Forces | ynsingo Bysr ty» EF =o GA - FA costD'cos 20" - SAcos 20's:nZ0" - BÁsinZ0%Os DO! = UA sia sin" =O E = Ecoslzo”s 3coszo"sin 20" +3 sindo cos do" 4 Gsm'Zo” = Leg ks «x AzFE-o TA 4 BAcos20'sins20' - SAcos 0 'cosão?+ SA sinZ0'sim 20º «VÁ sin O%cos do! = O Tr Prosas Blcostzo"- sint20º) + UsinZ0'cos7o” = 1013 ls; À —a 7.1 through 74 For the given stato of stress, determino the normal and shearing PROBLEM 7.2 stresses exeried om the oblique face of the shaded triangular element shown. Use à method of analysis based on the equilibrium of that element, as was done in the derivationa of Sec. 7.2. 10 ksi SOLETION 10 kesi 10 À costgº Es EA cost] gr gua Atas 1 <— GA siniS a A A [> i sinis* + 6 4 bes; À Asin sh 4 Asinis* Peas Stregses «ancas Forces *+/ZF-o0 GA + HAcosIS'cinIS!s loAcoiS cos is” = CA Bins siniso 4 É AsinlS ess ISO = O G=- GeoslSsniSº - |O costs? + 6 sintiSe- smlS cos ist = 10.93ks «mf += Ff >o TA + UAcosIS cos IS? — OA cus ISSA ISO — CAsmIS cos IS? - 4AsmiSGnIS = O P=-ilcostsº -smtist) + Clo+e )cosiSO sn iSO = O.S36 ks» um a | PROBLEM 7,5 che pa 7.8 For the given state of stress, determine (0) he principal planes, (5) SOLUTION 6, = IB kei S,=-12 ki Ty E si GE. (2x8) O tun 28, = E6 "TB+ia * 05858 28p = 28.07º Op = ito4º totogyº - (3 Sopnin - SS 4 (ES tp - Vartz = Bla (esa ro ay = 38%17 ksr Coros 7 RO Ux ca Sin FIM des = PROBLEM 7,6 7.8 thromgh 7,9. For the given sato of stress, determine (a) the principal pimes, (5) the principal stresses. SOLUTION 67 Aksi Syr to ks Ty"-3ksi ES 73 (8) fem 20, = ie «deal . 0,710 28, = 36.87º Op 7 18.48", 108,93" (6) Sun = BS + EE 7 ada (o = GaS ks [E = ksr [EE * | kai í J [| [| PROBLEM 7,7 73 through 7,3 For the given state of stress, determine (a) the principal planes, (b) J ll O a ca | the principal stresses. 40 MPa SOLUTION TT 3sMpa Gy * -60 MPa Gr -'do MPa T)=35 MPa CM 6) dan 20,» 282. . AXE) (gs al CS, -gorto -—[— 20, = -r80s* Gee -3%02º s297º a 0) Guga = ES + SS, fo = =Se- de + [EE = -S0 & 36,4 MPa Cam FT “IS.GO MPa ti Cam T — B6HM MPa = PROBLEM 7.8 7.5 through 7,8 Par the given state of stress, determino (a) the principal planes, (5) thé principal stressem. SOLUTION 6y= 6 MPa 6,” -48 MPa ty T-60 MPa a 4 (2Y-80) to) tan 29, DS “Tosas — —I87S 29p” - 61,93º B- - 30.96", S%oy” -—a (3 Cos, mim = Ser&y 4 Heçãe yr + Ty” = JE-48 + CEPIS = Epa 4 [(1e248) +(-€0) --I6 + 68 Come * 5% MPa Gean , = -84 MPa À f— E rs PROBLEM 7.11 40 MPa 35 MPa 60 MPa 7.9 threngh 7.12 For the given state of stress, determine (a) the orientation of the planes of masimum in-plane shearing stross, (5) the maximum in-plane shearing stress. (c) the corresponding normal stress. SOLUTION S = -60 MPa S,=- 40 MPa ay E 35 MPa (a) San too - Sade o as - O.2859 28,7 15.95º = TP, GRITO a 0 Tomo (EZS + (BH ea = 364 MPa a co 6'- Gu - Stôr , c60-%0 (soma a PROBLEM 7.12 48 MPa 7.9 through 7.12 For the given state of stress, determine (9) the oriemtation of the planes of maximum in-plane shearing stress, (b) the maximum in-piane shearing stress, (c) the corresponding normal stress. SOLUTION Se= 16 MPa 67-48 MPa, Ty -coMPa 8-6 16 +48 9. =-400y 2. Jetiê - ós333 tm) tan 28. TT TRIER) s33 20,= 28.07º = 40º, tod, 0 at 6) Em (ES + tj « (1eps + Ce) = cêMPa «a e» 6'= Gu = St8r, 16-VE. ieMpa cm PROBLEM 7,13 7.13 through 7.16 For the given state of stress, determine the normal and shearing stresses after the element shown has been rotated through (4) 25º clockwise, (b) 10º counterclockwise, SOLUTION Gu = - to MPa Sy= 60 MPa Ty = RO MPa Er 1% = 10 MPa SS = -S0 MPa Su = G+8, + Ber 8 cos 20 + Ty Sinão Ty * - 84 sin 20 + Ty cos 26 St = Set Sy - SS cos ÃO — Ty sin 28 to) O =-a5º Ra = - soº 6w 7 IO - S0 cosliso?) + ZJOsin(-S0) = -375 MPa «a Ty = +50 sin(-$) + 20 cos(-SO”) = -28.4 MPa nd 6 = Jo +50 coslso)- 20 sin(-So') = 57.5 MPa «a td 0=140º Re = 20º GS = to -50 costa) + go sinZo = - got MPa = Ty +50 sinl40) + 20 colo = Bs. MPa < Gt fo +50 cos(tZo”) - Zo sixlzo)= 50.1 MPa - PROBLEM 7.15 7.13 throagh 7.16 For the given state of stress, determine the normal and shearing stresses after the element shown has been rotated through (a) 25º clockwise, (b) 10º counterclockwise. 18 ksi a SOLUTION aki Ss, =8 ks S,= 2h Tg 3 -6 ks: “AM G+6 = 2h S-S lots —A eks 2 z 6 : &+O + SS cos 28 + Ty Sin 20 Ty - Ses sin Zo + Ty cos 20 Gy” Su r8y — BS co520 — By Sin RO to) = -2S* 28= -S0º Gu = -2 +10 cos(so) — G sin(-Sot) - Q02Z ks! — t Lys -jo sms) - Gesso) = 3.0 uy a GG 7:-2 — IO cos (iso) + Gsi (SS) -12,02 Us! a to O = do 20 = 20º Gu -A + IOcos(20) - Esinlro) = sad si am Gy - |O sm(zor) - E cos(a0! 1 - 7,06 ksi À 6 *-A — IO cos (20) + G sin (2º) = - 9.84 ks — 1 / q Li AM PROBLEM 7.16 16ksi to) 0 =-asº 7.13 through 7.16 For the given state of stress, determine the normal and shearing stresses after the element shown has been rotated through (4) 25º clockwise, (b) 10º counterolockwise. SOLUTION S4=0 8 =16 ks Ty = |Oksr Sr. guy SS. gts Sw - Et + Be Sa cos 20 + Cy Sin 28 Ty = — Be Or sim 20 + Ly cos 20 Gy = Bu rOs - Sou tos 20 — Try sin 26 26: - Soº Sur B -Beos(s0)+ IOsintso) = -weo ks —m ey E 8 sinfiso!) 4 lo cos (50) = O.3o ks mt Sy: 2 48 calço) - losm(-50") = Roso ks; a (b) 6 =Ioº 26 = 20º Gu = B - Gestal + losnlo) = 3.90 ks a Ty * & sin(ao) 4 lOcos (20) = I2t3 ksi e 6 =P +Bes(Ao) — Decos lo) = Ia. ksy a 7.49 The centric force P is applied to a short post as shown, Knawing that the stresses PROBLEM 7.19 on plane a-a are o=-15 ksi and 7=5 ksi, determine (a) the angle fihai plane a-a forms with thc horizontal, (b) the maximum compressive stress in the post. SOLUTION 3 FA 6,=0 Te &- Suco = - É — * TA p E tamf thscop ) A eh id 2; 8 PIN = Asma tl RE 's - e Cal = $ “Eca Areas Foree (a) = area & Triangle = 18,49º -—- Forces tamna, ) = 7 - . W Pr CGA sepXoa) frog migo Eri = PROBLE 7.20 Two members of uniform cross section 50 x 80 mm are glued together along M 7.20 plane a-a, which forms an angle of 25” with the horizontal, Knowing thar the allowable stresses for the glueú joint are «= 800 kPa and = 600 kPa, determine the iargest axial load P that can be applied. SOLUTION “ S RA tra / For plane aa 0-6" — X [A Vox ss? - -L 0 To S- à G = Cestb+ SysinO +Ty sin0 cs O = 04 Femestso P- AS - (sono Msoxio* Hgooxnio!) = 3.40n10º N snes O EE . T=(S- Sy) sa0cosO + Ty (esto -sinO) = E sm6S Costão pp. AT = (eoxio go nto! Ygoo mit) sc 27410) N ih 65 cosEso Sin6sº cos 45” Abowahhe valve oP Pis the smaller, P-s90wo'N-39%0kM PROBLEM 7.21 721 Two steel pintes of uniform cross section 10 x 80 mm are welded together as shown. Knowing that centric 100-N forces are applied to the welded plates and that B= 25”, determine (a) the in-plane shearii stress parallel to the weld; (5) the normal Seres perpendicuto io die nada ne º veis (b) SOLUTION / Area oF uol) jno kN k A= Leno Asomor*) damn) 457 vo Cos 25? F “mn = 28207 |"! m* Forces te) EF -o E - 100 sin25º = o F; = 42.26 ky 1 2 TE. AIRE 810) Lo 7 Ro” TEBaTmlos * 4Gxtio'Pa = 4LI MPa — bt) ZH= o Fu - 100 cos 25º= O 7 70.63 kM = fa - Goganio! s - — Sus do” BRAGA ioE é 102. 7wl0* Pa = 102.7 MPa SOLUTION Area TP welf - -s vor | ÁF ho fomotNsonet) | <—g cos . - do fo" E E - Ea m Forces , e ZF-o E - 100 sn8= O F= 100 sin KM = JoONO sin M “E c. 100 x 10º sinB Jo! si Tu E A Sonio Cgooalot fo A = IQSH0 sinficosA x e > simfBcesf = om 28 = úsrios” O240 fartáaç a bi ZK=o F.- [00 cosf= O Fa = 100 cos IU3= 96:85 kN - Boongt ... Av Coe guao * FAS.THO m . E. Fsêrio? e . 6 D Fasmamçe” 7.gxotfa = rs Ma PROBLEM 7,22 7.22 Two steel plates of uniform cross section 10 x 80 mm are welded together as shown, Knowing that centric 100-kN forces are applied to the welded plates and thnt the in-plane shearing stress parallel to the weld is 30 MPa, determine (a) lhe angle 8, (b) the corresponding normal stress perpendicular to the weld. Pp 7.24 The steel pipe 4B has a 102-mm vuter diameter and a 6-mm wall thickness. ROBLEM 7.24 Knowing that arm CD is rigidly attached to the pipe, determine the principal stresses and the medmum shearing stress at point K. SOLUTION od MR. s) - - fo = GE = Slmm te ft: 4Smm E(rtint) qisss nO mm" = ULIBSS jo m* 4IT = aotarro* m* Force - couple system as center of tube in the plane containing poreis H aud K Fer J0kN = loxto! My My (bm! KRoonto 3) = 2000 Mm Mis «(oo Ysork!)=-I$0a Nm y . Torsion > Bt point k place Doca X-ase im (a) w + * Hs negative global 2- divecktian K [x T= M,= 2000 N.m C=h= silo? m mk - Te . (QuooXsmiol) to Pi + Ty 3" Csemioso * 2.S7xlo Pa = 24,37 MPa Transverse Sheaw 2 Siress due te transverse shemm V= Fe is zera e? pk K. Bending: 16,1= IME, Lipo) ter.) = 36.56 nO! Pa - 36.56 HA Potnt K Pies on compression sine 55 neutval axis: & - 3e.so MPa Total stresses af point K Go 6,7-36.56 MP, Ty 524.87 MPa Came = HAS) E -12.28 MPa Re ASGS ray = 20.4 mpa Cm 7 Cau tR E 18.284 30.46 = + 12.18 MPa a Cuiu T Gu R= -18.28-S0,46 = -U8.74 MPa a Tuna R = 30.46 MPa a r L q L = Li— PROBLEM 7.25 Shaft crss sectica. Torsion: Benaling: Transverse shedr '| « - Je JT - Me sz 4.om ksi X RA.NU6 ki 7.25 A 400-Ib vertical force is applied at D to a gear attached to the solid one-inch diameter shaft 48, Determine the principal stresses and the maximum shearing stress at point H located as shown an top of the shafl. SOLUTION Equivalemt Force- couple system alcenter o) shaPl in seckion af perl H. V= 400 )b, M=(4oole)= avoo fhiu T= (Goo): 00 4biim. dm c-id=0.Sim J=Fe' zo os int I=fJ= o 04087 in? « X8001(0.5) So - : = A ILiIs BOTO par = 4 oN bes: - (eavoYo.s) 0. 049087 E) Qu dedo psi 7 AME ksi Stress ad point H is zevo. 6x = RIM ks: S;= o Ty = 4.07f ker Sou? H(G,AS,)T 2223 kesi Re (ES + = duras fimo = 2.88% kw Gu Gu + R = AS.107 les att GG Gu-R= - 0.66] ks — Toa o RT 1a ks: -— PROBLEM 7.27 8, o (ki & = Ga-Av- IR Tt 7.27 For the state of plane stress shown, determine the largest value of 9, for which the maximum in-plane shearing stress is equal to or less than 15 ksi. SOLUTION SIR ka, S= 2, Tys tsi Lt us SS &* 6x2 R=4 us ca = 485 ks DE a/Rº- Tl ca fist Ut me I4.957 hai 2U ENS) = HO.7 kesi, 16.9! ks: Largest valve Por EM is required, Sy = 40.7 des a PROBLEM 7,2% &ksi 7.28 For the state of plane stress shown, determine (a) the largest value of 5, for which the maximum in-plane shearing stress is equal to or less than 12 ksi, (b) the corresponding principal stresses. SOLUTION Gu=to ka, 2-8 ksi, Gy? Teo Re AS BD ay = V4t4 Ty = 18 ks: to Ly ÁIRPOGT = TAM ts — b) Gue= HGASDE ts 6: Cut R GS = Gue- RO ] 13 desv | +12 ] I- 12 -JE bs PROBLEM 7,29 2.29 Determine the range of values of 9, for which the maximum in-planc shcaring stress is equal to or less than 50 MPa. 3 MPa SOLUTION TP aU MPa 4-2, G=7 MPa, Ty" 4 MPa a Sx Let v + SeSr 6, r Sy +au nn R= JU ty = E * So MP DER Ty =a/S0l-Ho* = +30 MPa G= Syrâu - 75 +(2X30) = 35 MPa, I5 MPa Abbowahte range IsMPa < 6, € 135 MPa ——a PROBLEM 7.30 7.30 For the state of plane stress.shown, determine (2) the valus of £,, for which the in-plane shearing stress parallel to the weld is zero, (5) the corresponding principal stresses. 2MPa —> e tg SOLUTION | S4* 12 MPa, Gy ZM, Lys? 12 MPa Jos Since Toy =0, x. direchou as is a principal dirtedron. = Op = —15º x tam 20p= Ea no (03 Tg =A(G-G) dan2o, = QRO tan( 3) - 287% MPa cam tb) R «(Sms + Lo = Ste zsat e 6,7735 MPa Cia 466) TM Cuz Cant R = 7+5.739 = 12.77 MPa h E) So Cu RE 7-57735 = Lage MPa 
