Stress Transformation Problem Solutions

Stress Transformation Problem Solutions - Q—l The state...

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Unformatted text preview: Q—l The state of stress at £1 point in a member is shuwn t‘Jll the element. Determine the 5111355 eenipenenu; aelingI en the inclined plane AH. Selve the problem using the tnetherl of equilibrium described in flee. 9.1. k+EFr=D AFr+mmwmw—{smwmm°—aumwmw+ {mwmsmw Mean Amati Afir-[mmflmflw-{Sfilfin wiwsr+fiem 40°th+ GIMME WWW“ -'|J I AF}. mum-1M I miu' m \ “5:53! flr-M‘J*I%-4.Mm All! ‘mi hF” % F J ' a}, I“; ‘ 9,. =1imud “if =4.m hi Au “m EM'W' 4.. Jaime“ mmswmqusmoiw men.- mung-mi: mMMHoIFflD 9—7. Solve Pl‘uh. 9—2 using, the stress-trailnsl't'n'matien equaliens developed in Sec 9.2. Gx=5hi 03=3k5| 171,. =8ksl'. B=13D“ 5+1: 6-H 53.: ‘ 7+__..._..._" ' we 29+!”sin 29 ————me260“+85in260“=—4.Mhi Ana 2 i12+5-3 2 The ungalim sign imitates 01-15 a cemwnssive stress. 1',-,.- = -Ei:2£lsh 29+ enema 29 =-(:;3}aii1260°+8¢w260”=-D.404hi Am Tbmglfiuaignhflicmrfi-ishflw —y'dimction_ 9—33. Solve Prob. 9—32 for the normal stress acting perpendicular to the seam. 10 N ION P to a: _ = -——----v---v-—--- = A 30.031 F 0.028!) ' i , ‘1'. Hp-‘ml—ZaL-l-Eiflmfifli-rnstH It" x = ._1m'7: "' 0 + ——lm‘?6 * 0 custfiD") + o: 313 km; All: 9—66. Determine the equivalent slate of stress if an element is oriented 2'0" clockwise from the element shown. Show the result on the element. Construction DJ" the Cinch: In accordance wilh lhc sign canmticn. ct, = 3 hi. 6, = —-2 Itsi and. 1” = -4 hi. Hence. J = a,+cr,. 2 3+t—2: “I 2 2 =u.5m m 111-: coordinates for refcrem‘c poinls A and C are M3. '4} ("01500. III] The tadius; Hi the circle is R = J:3—t1.5u£n’ +41 = 4.711 kg Sum cm The Rotated Element: The numui and shes: stress c11mpun¢nl¢£firl.:md fir} unrepresented by lhc coordinate uI' putnt P an the circle. a).- can he deleminud by calculating the coordinates uf 'FIGIIIJ'II Qrm II'u: cutie. Q_ I or. ; t].5m+4.?]?cus [1.990 :43» kg Ans " “‘5' fh. =-4.T]?§h11?,99"=-I.46k5i Ans 23‘ 0‘,- =U.SDU—-1.?li'ccs 1?.99"=—3.99 ksi An: I I - 1% t3- 9”“ 9-?4. DL‘1L'l'flilil1i.‘ [11} 111:: principal SLL'L'SSCS and [b] 111:; Si] MPQ LImXimum ill-plane shear stress and average Imrmal slrcs‘s. Specify the orientalij “I lheelemenl in each case. 45 MPa {"30 MPH AI45.-5fl] “30.50:! {21375.13} fl=m=CB=f151+502=lfl55 a] U1 53T3+50j5=38J MP3 All! m=313*50.561-13.IHP1 All]. W123, =¥ 1&,=fll.41“ fl,= —4D.?‘-‘ In} u-mfi rim:- = R = 50.6 MP3; Am a". :55 “Pa Am J.. 2n, = 90-29, a. = 4.11” Au 5““ ...
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Stress Transformation Problem Solutions - Q—l The state...

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