2.3-7 - 110 CHANER 2 Axially Loaded Members A Aq 1 1 l A...

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Unformatted text preview: 110 CHANER 2 Axially Loaded Members A +Aq ] .' 1 1 l A =—'———"=m(;)tn+d§) >< :,—‘""t*"7+ ‘_ _ a P '1 a 4 f(ll.75 m)’ 10.30 m.)- ‘ J-l‘tE 18:00.75 in): + (0.5001011 = 0.3191 in.2 : 0.0589 in. (— J .-, ‘ o I ‘ th) ELDNGA'J‘ICHN' or: PRISMATIC BAR or: SAME VOLL‘ME 5 : ILL} : m , . EA!) (30 >< 10“ [350(03191 in?) Original has: V“ m AIL —1 A312- : Llrll + A:) x 0.0501 ' . <— Prismatie bar: VI, = man 10 Equate volumes and 501“: i”? AP: NOTE:_ A prismatic bar of the same volume will always ti, : VP [Ll/1| + £ng :3 APQL) " have a smaller change in length than will a nenprismatie bar. provided the constant axial load P. modulus E. and total length L are the same. Problem 2.3-8 A bar ABC til‘ length L consists of two parts of equal lengths but different diameters. Segment AB has diameter d. *m‘ 100 mm. and segment BC has diameter ([3 m 60 mm. Both segments have length LIZ 2 0.6 m. A longitudinal [mle of diameter (1 is drilled through segment AB l‘or one-half of its length (distance L/Ll = 0.3 m). The bar is made of plastic having modulus ol‘ elasticity E 2 4.0 GPa. Compressive loads P I l 104 liN act at the ends of the bar. (a) It’ the shortening ol‘ the bar is limited to 8.0 mm. what is the maximum allowable diameter dmm' of the hole? ( See figure part a.) (In) Now, ilidum is instead set at d112, at what distance I) from end C should load P he applied to limit the bar shortening to 3.0 mm? (See figure part b.) {e} Finally, it loads P are applied at the ends and rim.” r (13.0, what is the permissible length 1' ol' the hole il' shortening is to be limited to 8.0 mm? (See figure part c.) ...
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