Hw8 Solution - Problem 7.15 7.13 through 7.16 For the given...

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Unformatted text preview: Problem 7.15 7.13 through 7.16 For the given state of stress, detemfine the normal and shearing su-osses after the element shown has been rotated through (a) 25° clockwise, (b) 10" counterclockwise. 6; =‘ 8 ksi 6d: -12 ksf 7x3: -€ ksf : 63—126” 2 ~2 ks: 462—6 2 lo ks.“ “ SK’ 3 “SJ-:96; + a—ggi (3:529 -I- ’1’“ SM 26 6.3-6. ~ 'I fly: ': ‘- "—2-1‘ 5”! 29 + TV!) C95 26 i' w to.) e = - 25° 29: «50° " 5;. = —2 -+ to “Sc—5‘0") — c s.'n(-So') - 9.02 ksn‘ "" 7" 113' =- - '0 65m ('50,) - G (-05 (‘59.) T 3-30 ksl.‘ A 53,. = -2 - lo CO3 (-50') + c. 5.». (-50“) ? — I3.02- km‘ 4 1 l m e = 10° 29 = 20"}1 6,. e - :2 + to may) - a who“) = 5.34 ks; 4 ‘ [my 2 _" '0 5:“ (10.) "' G 605 (20.3 = ’Q—OG k5" ‘ ‘ 63: -.~ -2 — IO ca; (2.0. ) + G SM (20‘) 7 — 9.3"} ‘6: 4 Problem 3A, 5 8.45 1111739. forces are applied to the bar shown. Determine the nonnalsand shearing stresses at (a) point a, (1:) point I), (c) point c. A = (4:0st) =‘ 1220 wt 1 = 16120110" m 12 : [gamma = mafiiloim” = leagwto" m" if fiCSR.)(€o\3'- 5'76 vtq’ mu = 576130flw‘ AJr H‘e 53%.‘0-1 can+mintn1 Pam‘i‘s g. I: M239:- ) - J P = Io m V7: 750M J v: =500 N M: = (|3outo‘9)(-15:>\ = {MN-"M ‘ V ' M3: (210210")(500) 1" '= NO N-m I ,__ w: EL » T: o 220 mm‘—* SOON 6‘: E +. m2! .. [VJ-:2 '2: = gt— 1 J = 1c. mm 2: -l5mm 4 Q : A2: 61)(153(22.53= Jazz/0‘ W. ' g; 5. _ mite" + (uasxlexlo's) _(no)§—/5xto’5) ' lfiZDxlo“ 163-8‘fxl0“ 5743340" .—L , = 2:3 MPa. 4: —6 ’t“ (Snazglmsxm ) _ 0.295 MPq, .- ‘ (374x ro")(32><w") _ Pfgble ‘6' 8.69 Twoforces mappli'ed tothe bar shown. At points, damsel; m 8 9 stresses and principal planes, (lathe maximum shearing stresses .60 kips A'} +1.: SEC'HOH gon+ainfinfl Pam-f 3' an“; bi : ~ V = \O k‘IPS ‘P: 60 HF: (COMPR-‘asiw M = (8)00) ‘—' 30 kip-[Vt _‘ :- A = (23(3) : <2; in" 1 = M23835? 4.5- mm. I 1 . - _ 13 ago . M POIH+ a. 63 '— —E- 3—;— — .. '0 k5! ”is 3.3% = 5:11-93) = 2.5 ks: 6;.=o ...
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