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2.3-11 b

# 2.3-11 b - SECTION 2.3 Changes in Lengths under Nanuniform...

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Unformatted text preview: SECTION 2.3 Changes in Lengths under Nanuniform Conditions 115 (:3) IF x : 2L8 AND P12 n1“ JOINT 3 IS REPLACED or HP, 1 L8 + ”i3 3 PL_ FIND ,8 so THAT 5; = PL/EA 9 EA EA Ni=(l+,G)P N3=BP x=3— (“113) 9 l substitute in axial deformation expression above & B = "f <— solve for [5' B x 0.09] 2L 2L [(1 + l8)P]— {31° L — 7 . L 3 + = PL (o Draw AFD, ADD « see plots above tor x z 3 3 EA EA ‘ BIA . WMW Problem 23-12 A prismatic bar AB of length L, cross-scetioaai area A. modulus of elasticity E, ‘- -3 and weight W hangs vertically under its own weight (see ﬁgure). (a) Derive a formula for the downward displacement 5c of point C, located at distance 1'1 From the lower end of the bar. (b) What is the elongation 53 ot' the entire bar? (e) What is the ratio [3 of the elongation of the upper half of the bar to the elongation of the lower half of the but“? a ' Solution 2.3-12 Prismatic bar hanging vertically W = Weight of bar (b) ELONGATION OF BAR (Ir = O) (a) Dowmwnno DISPLACEMENT 5C 5a : WL _.__ Consitier an element at clis- BEA lance y trorn the lower end. to) RATIO OF ELONGATIONS Lt Elongation of upper hall” of bar (it = 3—) : 5 = on N(\') = I}: no = NOW}. 2 “(My upper SEA . ' 1'. EA EAL Elongation of lower half of bar: | . 11/}: t I 1 r J J 5“” = jifdﬁ = if 113:? = 21:114le w I") 5"“"L" = 5’3 _ 5W, 2 211:: _ 38:": = iii“: a W '1 1 bC=2EAL(LM-h~) (— ’3':(supper:§1§23 (— 5 lower 1/8 W ...
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