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206_Mechanics Homework Mechanics of Materials Solution

# 206_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: L___VJ (“J l_mJ [j I . . 1' ' 'alrodBC fie hL=24'.' u beam“: PROBLEM 3.50 35”““5’ “‘er “1“!er o ngt 1:: 18a ac rigid lever A3 of length a = 15 in. and to the support at C. When a 1004b force 1’ is applied at A. design speciﬁcations require that the displacement of A not exceed 1 in. 1.. when a lOO-Ib force P is applied at A. For the material indicated determine the required diameter ofthc rod. 3.50 Steel: 7..“ 15 ksi, G = 11.2 x lo‘psi. SOLUTION Al" ‘Hne. anoulaJnie {1013+ amalc 51'” g) = g : «é : 043%“; 3.82%” = 0.066714 mat. T - Pa. «.05 30 = (/00)(15)Cos 3.8226“ = H‘MJ It.- in . L TL ._ 2T1. . r a 2T1. Based on +wts‘l‘ q)— (:1: -— 17-5-6, -- C WG-9 (szeenxzu) -3. . TTZI1.2xlo‘)(o.ogg7;5) : 30-503’“O 6— 0-418 m. Based on .Shess ”E. '= 5—5“ = 11%;; 3- C3 = % (7::[5000’030 as: 204%.?) = Gasmﬂodin, ”(LS-000) (2": C. = 0.3% in. Use [Parser value For design C. :- O. 397 1-91 d=..2C =‘ 0.337 in. -" 3.50 and3.51 The solid cylindrical rod BC oflength L = 24 in. is attached to the rigid leverABoflengtha =15 in. and to the support at C. Whena lOO-lb three P is applied at A. design speciﬁcations require that the displacement OfA not exceed 1 in_ when a 100-113 force P is applied at A. For the material indicated determine the required diameter of the rod. 3.51 Aluminum: rm= 10 ksi, G = 3.9 x 1o°psi. SOLUTION A+ +1.3 momma +ms+ mare . 5.». go = 137: Kg: 0.06667 3.312s” _— 0.066116 rut. *T = Po. case? = (100 )(Isloos 3.82%" = weevils-in W M +~~+ =1“ til- = 17% 1% 1;. momenta“ .. x "3 w .— 0.544' Cm " “'8“ ’° ' C '"‘ Based ow stress ’[1‘ E9 =I 1?; -— (Zr-10000 psi) as: W = 95‘. 233 x10" if c--- o.‘I\$71'n Use ﬂea/1e» Val): '92» 010.333" C. -‘ 0.544 in air-2c: 1.089 in an. ...
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