Unformatted text preview: 3.133 A 1.25mlong steel angle has an L 127x 76 x 6.4 cross section. From
Appendix C we ﬁnd that the thickness of the section is 6.4 mm and that its area is 1252 mm2 . Knowing that IE" = 60 MPa, G = 77 GPa, and ignoring the effect of stress concentrations, determine (a) the largest torque T which may be applied, (b) the
corresponding angle of twist. PROBLEM 3.133 SOLUTION b = GJI m». = 0.0064 in
”75.6; mm = 0.1956 In H .= ca: 3L("O'é3°%) = 0.3265 _ T . _ 3
TM”... EraE: .. T  Clab 211a? ca.) T40.3265)(0.l756)(0.006‘fl1(60xlo‘)= IJéﬁrrtog Nm
= 1510 kNm «2 (b) (P: TL. _ gab me_ crave; _ ’r L. .L gals‘6" czabsG " (:sz ” be AM _ 2. A = 7 ., «a
GP (0006‘fK77x10“)  152.10! no m 8. 2 3.134 A 3000 lbin. torque is applied to a (Suﬁlong steel angle with a L 4 x 4 x g
PROBLEM 3.134 cross section. From Appendix C we ﬁnd that the thickness of the section is 2 in. and that its area is 2.86 in2 . Knowing that G = 1 1.2 X 10" psi, determine (a) the maximum L4>< 4 x% SOLUTION shearing stress along line aa, (b) the angle of twist.
a t . .84 .
a A =2.86 [.an b=§’m=0.37§mJ Q=%=§€§? =7.627m.
% = "32:35? : 20.3% a, = C2 : g0 — 0.43:: £1) = 0.3230
_ _.._I_. _. 3°°° _ s . .
(0“) 214:)! A C. OLE?" ' (0.3230)(7627)(0.375‘11 ‘ 8.£6 >40 PSI =' 8.6g k5] 4
= __ﬂ__ :: ”MW—H : wayswad“;
0"] {P CLQE'G (0.3230)L7.5273(o.37§)3m.2no") 4
— 3.51 " ' ...
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