Unformatted text preview: PROBLEM 7.23 7.23. The steel pip? A33_has a 102rnm cute} diameter _and a Glynnwall 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 SIMv: 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 Nm
Mi = —(10xlo‘)(l$ouo" ) = 4500 Nm
I . u j w, T = M : 2000 w. I c : ry = Slxlo“ m JLv
K a ")_ ‘ ta: ___ __ (2000 5'! to 24.3.7on Pa. ,1 J ‘ '4.I3$5'xlo" " TwanSVe¢se Shear l _ .1
Few semcmﬁe A: Er J: Hr
Q= A” = ﬁr” 51‘ Qo‘ Q; = :az'rogén’ = 27.68“!wa “.3
= 27.88'4xlo“ m3
V = R = laulo'i N t smug,” t Rm, I2xlo'3m “A
g 2.9., roxuosnzlcsquo") n ‘P m
T“ h It (zmznro ) untos)  ”02’” “ l “x BEHJEHQ : Pofn+ H ﬂies 0" Hedi?!” axis. 63 = 0 13h? 5+Msses 42“ Paid H t 6; ; :9J 63 :0
2:: = mavuo‘ + ILDQHO‘ = 35.3%:0‘ PE (in =J§((>“,+6_,)= 0 R: 5%)H’zzf = (55.39 an)" ’94, 6...", : 6M+R = 35.3?RIO‘PQ. = 35A! MP4, 4
6m  Gm —R — —3$.301*10"Pa = 45.4 MP1 4
’11....“ = 1? = 35.4 mpg. 4 ...
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