HW13

# HW13 - 3.53 The composite shaﬁ shown is to be twisted by...

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Unformatted text preview: 3.53 The composite shaﬁ shown is to be twisted by applying a torque T at end A. Knowing that the modulus of rigidity is 77 GPa for the steel and 27 GPa for the aluminum, determine the largest angle through which end-A may be rotated if the following allowable stresses are not to be exceeded I'M. = 60 MPa and rm”, 45 MPa. PROBLEM 3.53 SOLUTION Tm: : 1% 25 5%., = V 2’ \$0., CGOL‘ “dew‘h'i Steel core /\,_/ T - L. G Ch'ﬁ Aluminum jacket l/ Steel) Core: 214 = 60x106 PA} C~%::.'{d :' 0.02.7mJ G 2: 7731/07 P4 £4 = Goxlo‘ L (77x/o")(o.027‘) = 28.860f10-3 rad/m 2:4.- 4s'xlo‘Pq) cw=ﬁd=o.036m) G=27xlo* Pq = 46.274; vzo's rad/m Smaﬂﬂev Unpue jot/eff): (pf/4 = 28.860’00-3 V‘aaI/M Aﬁﬂowdﬂe anjpe o‘P +wis+ Q)“ : L %’ : (2-5)(28.860x(o'5) = 72.15 >404qu = 4-Is° 14/). 3.56 Two solid steel shaﬁs are ﬁtted with ﬂanges which are then connected by ﬁtted bolts so that there is no relative rotation between the ﬂanges. Knowing that G = 77 GPa, determine the maximum shearing stress in each shaft when a 500 N-m torque is applied to ﬂange B. PROBLEM 3.56 SOLUTION Sha‘Fi’ AB T: 113) L“: 0.6M) C=ia= 0.0lfM I“ = 350*: 130.05? = 7152 Ho“ m = TAB LA ‘39” Gala Ll _ {77xlo‘)(7‘1.\$2v!6’) ‘ 0.5 ‘93 Jen : 194.8% on‘7 m‘ — G , " 14439440“ “3.3% -(letglLﬂcpc z magma (pa * _ 'Cb Co 0,? Male/king V‘O‘i’tv‘l't‘ou at} 44c rjanjes (p6: (Pc : Q Tb‘l'ap +or1ve an ‘Fjamjes T: Thai-12° -"- 500 N-M 500 = (10.205 x103 4 H.108XI03) 90 g) = 20.565wo“3 m4 TM = (10.20918 )(2o.\$c5 Ho's) 20 9. 87 N-m 720 = (H. Ioaxlo’X20565 x10") 2 290,.»3 N.“ Mmimum shearins Sire.” in _. TMC _ (209.87)go.o;s) _ c P 4 T“ J» ‘ mane-a ’ 3154"” P“ 3"" M " Mmimum sideline) 541%: lv‘ CD It»: 13.29.. = W = 3!,é7xlO‘ Pa 3].? MP0. «I J; 194.3% do“? ...
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## This note was uploaded on 12/09/2009 for the course CE 3400 taught by Professor Moorthy during the Fall '09 term at LSU.

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HW13 - 3.53 The composite shaﬁ shown is to be twisted by...

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