Stress Transformation Problem Solutions

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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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““ ...
View Full Document

## This note was uploaded on 12/12/2011 for the course EMCH 260 at USC.

### Page1 / 3

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online