Question

The composite shaft, consisting of aluminum, copper, and steel sections, is subjected to the loading shown. Determine the displacement of end A with respect to end D and the normal stress in each section. The cross-sectional Aluminum Соpper Steel E=10(10') ksi AAB 0.09 in2 Eg=18(10') ksi ARc 0.12 in2 E-29(10) ksi ACD 0.06 in 1.75 kip 3.50 kip 1.50 kip 2.00 kip area and modulus of elasticity for each section are shown in the figure. Neglect the size of the collars at B and C. D A B 3.50 kip 1.75 kip 16 in- 18 in-

Answer 1

Aluminum I copper - Steel Eal = 10 X 10% ksi Ecu- 188109ksi Ex= 29x109 ksi A AB=0.09 in² ABC = 0.12 in2 ACD=0:06 in? 2.00 kip to 3.50 kip 175KIP V to 1.75 KIP 15kip D 1.50 kip B 3.SOKIP - 16 in. 18 in 12 in. FiB.DE TO - T10 T2 B DT3 B o - C 4 From eauilibrium of forces W At A: (i) Ate Ti = 2:00 Kip (ii) At B VV Equilibrium at T-T3 = -1.7582 Tz+1.5=-3.5, T2=-5 kip. - 13=-1.50 kip = T-T2=3.542 2-T2-3.5x2 T2=-5 kip.

W normal stresses (a) in section AB JAB = TILAB - - AAB - 0.0g AAB 0.00 - 22.22 kiplinz (Tensile) (b)(a) in section Bc DC = T2 ar .-5 7 =-41.667 Kiplina . . ABC OBC=-41.667 Kiplin2 * compressive. (c) in section CD JCD = T3.-1.5 A CD 0.06 = 25 kiplina i.o. compressive (2) Deflection Clie.displacements) (a) in section AB DAB - JABLAB 22.22x 18 = 0.04 33849 DAB = Ear Tox 103 Ebanoin.) CAB=0.04 in.) ang - 0 0 338 in. (b) in section BC ABC = TBC LBC to - 41.667812 E eu - 0.02778 ABC

(C) in section CD o aco. O Coleo O Est 3 = -0.01379 in. 29x103 | ALD =-0.01379 in.) - -25 16 Deflection Of A Weret. D ie. LAD DAB+ ABC+OLD 7 000338 50.04 -0.02778-0.01379 LAAD=5574153 in.) Hence deflection of a worit. D is -0.00157in. Normal stresses | Nature I ( ksi) 22.22 Tensile AB. 41.667 Bc compressive compressive 25.