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# Ch 7 - Problem 12-1 An element in pmne stress is subjected...

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Unformatted text preview: Problem 12-1 An element in pmne stress is subjected to stresses 1,. or: = 6500 psi, 0'}, = 1300 psi, and 1'1}, = 1750 psi, as shown in the figure. as, = 11'00 psi Determine the stresses acting on an element oriented at Eltl angle 9 = 60° from the .1' axis, where the angle 5' is positive when counterclockwise. Show these stresses on a sketch of an element _ 0",. = 6500 p31 0 J.“ oriented tit the single 0. I“. = 27'50 psi 1,. or: = 6500 psi o", = 1300 psi 1'“, = 2750 psi ' s=sw ' ' 292B P31 “.1: + [T's ".1: _ [rs _ or,| = —2 ' +—ﬁ ' cos 2” + r_,._,. sin 2“ = 5180 psi 1'— ‘TJ; _ [T's _ 7_,.l_,.l — —T' sin 2H + 7,}. cos 2“ = —3=‘1-50 psi 1— rr_.Fl = rr_,. + tr}. — is“ = 2910 p51 1— Problem 12-14 Solve the preceding prohlent for ax = 31 MPa |_~.,~ and o", = —5£|I Iii-[Pa [see figure]. ' SCI MPa (3 /§ 32 MP4: —|— 0 .T a Solutlun 12-14 Biaitial stress a STEEss ELEMENT er n.“ =1] e = sass” /ﬂ( [1-H =:r_,.. +tr}. —r:'_1..| = — [8 MPa 4— ”" . — rr . a1 = 32 MPH '_.'|.'L't-'l = ——J ,3 J 51:12“ + s“. cosEIi o", = —51} MPa u ' a 1-5.: [I- = NIPa '— Find angles {1" for 0' = ELI .‘t‘ or = normal stress on plane a-a 15 NH (I: + “rs “'1'— “'1' , rrl-l = + cos 2“ + a}. sin 2“ = —‘-} + :1] coslliI [MPa] . 9 For rrl.l = I}, we ohtarn cos 2“ = 2H = 7132” and H = 33.6150 1— Problem 12-15 An element in plane stress from the frame of a racing car is oriented at a known angle 5' [see ﬁgure]. On this inclined element, the normal and shear stresses have the magnitudes and directions shown 4139 P51 2.360 psi ,r’i in the figure. I} = 30" Determine the nornta] and shear stresses acting on an element whose sides are parallel to the I}? axes; that is, determine o'x, 0",, and 1'“. Show the results on a sketch of an elentent oriented at H = ﬂ”; Problem 12-19 At a point in a structure subjected to plane stress, the stresses are or}, = —4ﬂﬂﬂ psi, 0". = 25oz: psi, and 1'“. = 2342"} psi [the sign convention for these stresses is shown in Fing-l }. A stress element located at the same point in the structure, but oriented at a counterclockwise angle H] with respect to the I axis, is subjected to the stresses shown in the figure iota, Tb, and Ellﬂ-EII psi). Assuming that the angle 9] is between zero and 913°, calculate the normal stress 0.5, the shear stress Tb, and the angle El]. Solution 12-19 Platte stress cl'I = —4ﬂ-DEI psi o", = 15GB psi Tm, = EEG-D psi ANGLE El, rr.+rr. tT.—U'. .1. J. .1. 1 FOR 9 = '5']: rr_1.|= T + cos 1“ + 7,}. sin 1“ "x. = 20”“ Psi “3-. = “in Ta,- = 1's soon psi = —?so — ssso cos so, + soon grass, Find ohm-b, and ﬁll or —65 cosEIEilJ + 56 sinErEi'1 — 55 = {I Solve numerically: STRESS Us se,= 89.12” e,= 44.5450 — o',, = or + o". — EDDIE psi = —35l2|ﬂ psi 4— ' SHEAR STRESS Tb I'FI. — (I 1. y . = —T sinEHl + 7,}. coslrl1 ' b ' .‘L'L'r'l 3'39"} psi 4— Prehlsm ?.3-1 An element in plane stress is subjected te stresses ax = esee psi, a", = lTDrIJ psi, and 1'”, = ETSD psi {see the figure fer F'rehlem "LE—l }. Determine the principal stresses and sheer them en a sketch ef a preperl'j.r erienterl element. Salutlen 13-1 Princile stresses cs, = 650'} Psi a, = tree psi en, = ease psi Therefere, er, = arse psi and a, = 24.44“ PRMCIPAL smsssss- I 0'2 = 451} psi and “P, = ] 14.44“ “ft— ... I 1}. tan En = = 1.1453 F' _ (II (1-,. 7,. =48.E~9° and EP=E=L=14° as . q as} erase“ and sp= 114.44“ 4'GP“ TIE!) psi rrl. + {1'}. rr_,.. — rrv re“ — ,3 + ,3 ' ces EH + 7,}. sin 2H Fer 1HP= 48.89”: er_,.l = Ti‘ﬁﬂ psi Fer EHP= 223.39”: rrxl = 451] psi Prehlsm 1.3-2 An element in plane stress is subjected te stresses 0', = SI} MPa, e". = 52 MPa, and 1'“, = 48 MPa [see the figure fer Prehlent "LE—2}. ' ' Determine the principal stresses and shew them en a sketch ef a preperly eriented element. 1.} Problems 13-11 through 13-15 An e1emettt in plane stress {see ﬁgure} is subjected to stresses err, 0",, and 1'“. [st] Determine the principal stresses and show them on a sketch of EL properly.r oriented element. [11 J Determine the moximutii shear stresses and associated normal Tn- a). stresses and show them on a sketeh of El properl],r oriented eleirtent. 0 2: Data tor 13-11 or}: = 3512"] psi, o3. = 1121} psi, i'M = —12ll-|J psi SDIIJIIIJII 13-11 Platte. stress or: = 351er psi o", = ] 121] psi Tn, = —12Ill-|I|I psi Therefore, ' ' titJ = 41301] psi and “pl = —22.61° h ta] PRINCIPAL STRESSES I.‘J'_._. = 62!] psi” and tipa = 45133” 27A}. taitﬂtiP = = — ] .ﬂﬂSﬁ (TI — rt"). 23? = —45.24° and HP = —12.62° 29? = 134.7%” and Hp = 6133” “rs; + “T's fr.1.'_ [rs- 2 ' + ' cos 2“ + .3}. stnEH For 23F = —45.E4°: re“ =4£|Iﬂll psi For 23? = 1351.16”: :rxl =ﬁ2ll psi rrl.| = Salutlun 13-19 Plane stress HI = 65m psi 1-1}, = —EEI}EI psi 0'}, = ? lCline principal stress = T301} psi (tension) [a] STRESS 0", Because crI is smaller than the given principal stress, we knew that the given stress is the larger principal stress. 0'] = T301} psi rr_1..+ tr}. til". — :il' I r 2 2 rr1 = 1 + 2 + TI}. Substitute numerical values and solve for a"; a", = —25EHI|I psi 4— 1'} {b} PRINCIPAL STRESSES 27A}. tan EN}, = = —ﬂ.62222 rrl. — If}. 23 = —3l.391° and HP = —15.945° seﬁ = 143.1139“ and tip = “H.053” rr.+r,1'. :r.—rr. (r11 = A '1 + A J cos 1“ + Tl}. sin 1“ “1| ‘4 For 23? = —31.39]°: [rm = TED-El psi For 23? = 143.109“: crxl = —33{IE| psi Therefore, -:JI'J = TECH} psi anti“?| = —15.95° tr2 = —33ﬂﬂ psi and upI = 7435” 3300 psi Problem 14-1 An element in mimic? stress is subjected to tensile stresses a1 = 14,513] psi, as shown in the figure. Using Mohr's circle, determine {a} the stresses acting on all 145m] P51 eleiitent oriented at a counterclockwise angle H = 24° from the .‘r .‘r axis and [h] the maximum shear stresses and associated normal stresses. Show all results on sketches of properlyr oriented elements. Solutlnn 1.4-1 Ultimtisl stress err = 14,501} psi at,“ = {I 1'”. = {II Point 5': rrIl = R + R cos2t1= 12,100 psi rm = R sihEIr = —53‘;HJ psi [rt] ELEMENT AT 5' = “4° [All stresses in psi] point 5-: rrl.l = R — R cos EN = All-l] psi 33 = 43° = “L ° R = T251] psi 7W = 539G psi Point ('3': ct“ = T251] psi {h} MAKE-:IUM SHEAR STRESSES Point .5]: EHEI = —+§.Iﬂ° MEI = —45° Tmﬂ = R = T251] psi Point .52: ENE: = 911° us: = 45° ijn = —R = —?15ﬂpsi ﬂ'mr = R = 7251] psi TE 50 psi Problem "L445 An elernertt in bi'oxr'at‘ stress is subjected to stresses y err = EDEN} psi and o", = —15E|ﬂ psi, as shown in the ﬁgure. Using Mohr's circle, determine {a} the stresses acting on an element 150:] psi oriented at a counterclockwise angle [1" = 6|)" front the I axis and {b1 the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elernents. 0L 630'] P51 I Salutlun 14-5 Blaxial stress ax=EDEHJpsi o",=—15[HII-psi 1' =1} 1'}- [a] ELEMENTATH = 61]” [All stresses in psi} 2E3 = 121]” E = 61]” ER = 'i‘ﬁﬂﬂ psi R = STE!) psi Point C: rrII = 2151} psi Point D: ail = 125i] — R c0561)“ = 37')" psi [11] MAKIMUM SHEAR STREESES TIM = ‘R Sinﬁﬂo = ‘3343 P5” Point 31: 2n.” = —-;ui° “5' = 45'? pm'm Dr: (Tl-l = 1251] + R 60560” = 4125 psi Tmﬂ = R = 3351} psi Tim = R sinéﬂ” = 3243. psi Point 5'3: 2:13: = 91]: HS: = :159 _ rmin = _—R = —3T3ﬂ p51 (Tamer = 225D p51 3: S: 2250 psi I 4125 psi 2250 pm 600 as: = 45° I I 3250 psi 31-20 psi EDIIIHDH 2.4-3 Purl: shear EI=U Jv=ﬂ le=—iﬁMPa I,“ {a} ELEMENTATH = 20° 103 MP3 5. [All stresses in MPa‘J D. / “13 MPa 2ﬂ=4ﬂ° H=2EI° R=16MPEL \ Origin 0 is at center of circle. Point D: rrxl = —R sin 2F! = —i|II.2E MPa Tm.I = —R cos 2f.I = —12.26 MPEL Point 0': = R sin 2” = 10.23 Mia;l mm = R ms 2F.I = 12.26 I'm-{Pa Problems 1445 TI'IIDIJIJII LII-23 An element in plans stress is subjected to stresses crx, or? and TI}. {see figure}. Using Mohr's circle, determine {it} the principal stresses and lib} the maximum shear stresses and associated normal stresses. Show all results on sketches of properly.r oriented elements. ‘1}. [11113 1D! Til-11] 01. = D. a}, = —22.£1- MPa, TI? = —6.6 MP3. Eulutiun 14-13 Principal slrtssts arr = 0 Cry = —22.4 MPa _‘1' 1'“ = —E.EIMPﬂ {15111 stresses in MPa‘J P2 24.2 MP3 {b1 MAKE-[UM SHEAR 5111135 3135 21’!5| = —1'J.' — 90° = — 120.51D H;I = —6|I|.EIS° 315: = 90° — a: = 59.49" N5: = 2934-” 13"I:1111'11.11'l:1:11amI = — 11.2 MPs rm“ = R = 13.3 MPﬂ 13"I:1111'11.11'2:1:11amI = — 11.2 MPs Tm" = —13.0 MP1: R = 1’111.2J2+[6.632=13.0 MP3. 6.6 = 3121.51“ 11-2 11.2 MB: ﬂ! = ﬂl'CtﬂTL 1:11 PRWCIPAL STRESES 211?. = —a = —30.51° 11m: —15.ss° 211“: = 130° — a: = 149.119" 1113 = MJ4° lF"u::111'1t.F'1:-::Irl = R — 11.2 = 1.3 MPﬂ 1:11:111'11.33'2:-::Ir2 = —11.2— R = —24.2MPa s 1‘1 39.?4° thlem 15-? Solve the preceding problem for an aluminum plate with cl:r = 12,33!) psi {tension}, cry = —3,ﬂﬂﬂ psi fccmpressicnlt dimensicns 2:} X 3|} X 115 in., E'= 10.5 X 1135 psi, and 1' = 1133. Ecluticn 15"." Binxinl slress‘ err = 110001351 cry = —3,Elﬂﬂ psi (b; CMNGE IN TI-EEC‘ICNESS E = “3-5 X “35135-1 F = ‘133 Eq. tT-B‘icl: s;= —'—.;.rx+ urn = —EE~.2.9>< 10-5 Dimensions of Plate: 20 in. X 30 in. X {3.5 in. E ' ShearMcclulus (Eq. 7-38]: hr = Pa! = — 141 X 10—5 in. 1— E [Decrease in thickness] 6 = — = 3.941% x let psi EH + J'l {c} CHANGE MVOLWJE _ 2,. fa; WWW JIM-PLANE SHEAR mm me E‘l- (7'43: 11" = VUC E JWx'l' ".3 Principal stresses: 0'1 = 12 DEED ps1 Vt. = [Emlmlfﬂ-Sl = 39'] 111-3 0'2 = —3.0E|ﬂpsi 1—21' _6 Ir] _ “2 Also.( )trrx + rry} = 291.4 X 1'3 E . res]: .—m = = 1504:: si _ q l _ 2 P sv= tans m.3lf291.4 >< 10-53 Eq‘ H-351]: Tm“ = =1~gm x 1.3—6 ‘— : EDEFM in} 1— flncrease in vclu me] ...
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Ch 7 - Problem 12-1 An element in pmne stress is subjected...

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