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780_Mechanics Homework Mechanics of Materials Solution

780_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: PROBLEM 7.23 7.23. The steel pip? A33_has a 102-rnm cute}- diameter _and a Glynn-wall thickness. Knomgthat arm CD 13 ngtdlyattached tothe pnpe, detemunc the pnncxpa! straws and the maximum shearing stress at point H. SOLUTION 51 _ do _ I02 _ — _ _ mm (‘0.— "E- T- SIM-v: fic—I", t— 45mm J = ‘E(V‘:-Y‘;)= ‘1‘. ISSS x106 mm‘ = H. Igssxzo'“ m" I = %J-- 2.0927 "4* Force—‘Coupje adskm 44+ cen‘hw a“? +9)“, in 'Hne Pfiane com+amih3 Pofn‘fs H anal K. F, 2 10 31103 N x M = (Iowo‘flzoouU’) = 2000 N-m Mi = —(10xlo‘)(l$ouo" ) = 4500 N-m I . u j w, T = M -: 2000 w. I c :- ry = Slxlo“ m J-Lv K a ")_ ‘ ta: ___ __ (2000 5'! to 24.3.7on Pa. ,1 J ‘ '4.I3$5'xlo" " TwanSVe¢se Shear l _ .1 Few semcmfie A: Er J: Hr Q= A” = fir” 51‘ Qo‘ Q; = :az'rog-én’ = 27.68“!wa “.3 = 27.88'4xlo“ m3 V = R = laulo'i N t smug,” t Rm, I2xlo'3m “A g 2.9., roxuosnzlcsquo") n ‘P m T“ h It (zmznro- ) unto-s) - ”-02’” “- l “x BEHJEHQ : Pofn+ H flies 0" Hedi?!” axis. 63 = 0 13h? 5+Msses 42“ Paid H t 6; ; :9J 63 :0 2:: = mavuo‘ + ILDQHO‘ = 35.3%:0‘ PE (in =J§((>“,+6_,)= 0 R: 5%)H’zzf = (55.39 an)" ’94, 6...", :- 6M+R = 35.3?RIO‘PQ. = 35A! MP4, 4 6m - Gm —R -— —3$.301*10"Pa = 45.4 MP1 4 ’11....“ =- 1? = 35.4 mpg. 4 ...
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