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Sol HW10

Sol HW10 - PrOblem 7-55 7.55 through 7.58 Determine the...

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Unformatted text preview: PrOblem 7-55 7.55 through 7.58 Determine the principal planes and the principal stresses for qr]! ‘Z 57 the state of plane stress resulting from the superposition of the two states of stress ' shown. ' 50 MPa I ' 3/ 80 MM \ 7OMPa 7‘ Molw‘s circle in.» Znal s-lvess slag-lg + ’5'} Y 6}“: 9+0 +40 (.95 éo" = 60 MP6. 63 = 40—40025 éO' le— : 213 MPa 71:, -‘ ‘1‘0 am 60° = @1864 MPa. Resug‘l‘qul’ slrcsscs 6',= 70+ 60 = :30 MPa 6‘= 50. +20 T YDMPa 2’3]: 0 +3H,é'+ : 34.6% HPa. 5am = *HEMGQ) = aL(I\$D+7o)= loot/1P4 1:} ("823 . _ 22¢ _ (2)8459), ’ Jam 29?. ﬁg? - ——————BD_7O : /./547 29,= H‘IJI" 90‘: 29.4“ 95: IH_6° an R = 1’(§12‘—5i)‘+ 219" b #518. MP4. 5.. : Gnu + R 6;? "A 61,: 6a.}: “ Q 6L, = Problem 761 7.61 For the state of stress shown, determine the range of values of 6 for which the it,” 7 52? normal stress axris equal to or less than 20 ksi‘ (Ty: w...» .12 ksi 5;. s ‘20 k5} 43w s‘l’q‘les a; s‘l‘wess Corresponalfn +0 arc. H‘BK a? Mohr‘s air: a. From “Me ct‘vxcje R “5 QSP= 20 - a! ‘=' 1mg; 05 29 = = 0.73333 250 = 42.233” g> = 2: 417° 9H 2 905‘ S? r — 26-5é5°+ gun-7° : _ 51,59; 26K = 29,, +3éo'— 450 = —Io.2a7° +340“ - gasegﬂ 20MB?” 9“ = I32.02' Perm'ssihe ﬂange JP 9 is 9» 5 9 5 9k —5’.l5” s e s 132.02° 4 G AQso, was? s ,, w——————-—-——x‘ . L P rob, em 71 4 5 7.145. The strams detexmmed by the use of the rosette shown during the test ofa ; machme element are ‘ ll,” Ib/ Z el=+600,u 62=+450p €3=—75,u y ' Detennine (a) the in-plane pn'ncipal strains, (17) the in—plane maximum shearing strain. ‘ [m 3 1 ._x Q: 3 30° 2 I 92 = 150’ 93 = ‘10 x Ex eagle, + 9y 513416, + Y‘qSinQCose, = E, 0.75 E, + 0.25 SJ + 0.4330) ny : 600A 0) Ex C0319! + 5M2 92 + 7/20 513092 60392 '5 81 0,75 s, + 0.2:: a; — 0.4330: m = 450/4 <2) 8% £05193 + E! \$51193 4: Y‘xj SI‘AQJ Co: 93 ‘-' £3 0 -+ 2:! O = “75/4 (3) Sopw‘nﬁ (£7) (23) «Ml (33 .Sx‘wwxg‘l’amCousjjJ ex = 725/4) sj: 45;” ﬂy: I73.zl,u l d A 4 7.158 The grain of a wooden member forms an angle of 15° with the vertical ‘i “1 PrOblem 7'1 58 For the state of stress shown, determine (a) the in-plane shearing stress parallel to the grain, (b) the normal stress perpendicular to the grain. 6; = A25 MP4. 6"}: 4.5 MP9. 11., > 0 e = .. 15° 29 = v30° / 2.5MPa (an) 5.34 + ’ﬁyC/OS/Ze ,: __ — 2,514.57 5.-.,,(-3o°) + o = — 0450 MP4 <3 do) 6')”.- LS—‘ngi 4‘6—1—i-5-é-ca529-l— 215 57" 25 ____.__L.————~2'§+ 4‘5) 4r .“__,____<.___).2'5.“ "-5 cos(_go°j + O "S “' 2.43 MP4. 4 2 2. 7.158 The grain of a wooden member forms an angle of 15 ° with the vertical. For the state of stress shown, determine (a) the in-plane shearing stress parallel to the grain, (b) the normal stress perpendicular to the grain. 6x:-3MPQ Gd: [8 Mch 28:0 Suﬁ—5.42193— : ~24MPo. PJ’oHeAL oCWfs Meier‘s CJWcjei TD (MPal X: (Sm—T‘s) : (-SMPQJO) - Y: (63) ’1’»: )= (~ 1.8 MP5 0) CI (GM) 0 ) = (—29 M9430) 6mm 9 = ‘ l? 29 = ~30° a = 0.6 MP4. 12 : 0.6 MPG. (a) 213,! = ~Ex’sin 50 = — I? 5:» 30" =- -O.6 5m 30’ 11y:-O.3°MPQ «a m 6;- = 6‘... - EX'co: 3o° = ~24 - 0.4 Co: 30" Sw-ZWWQ .4 H ...
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Sol HW10 - PrOblem 7-55 7.55 through 7.58 Determine the...

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