Question

P5.027 An aluminum alloy [E = 8700 ksi) pipe with a cross-sectional area of A1 = 4.22 in.2 is connected at flange B to a steel [E = 26000 ksi) pipe with a cross- sectional area of A2 = 2.64 in.2. The assembly is connected to rigid supports at A and C. Assume L1=152 in., L2=236 in., and P=31 kips. For the loading shown, determine: (a) the normal stresses in aluminum pipe (1) and steel pipe (2) ( 01, 02 use positive if tensile, negative if compressive). (b) the deflection UB (use positive if it moves to the right and negative if it moves to the left) of flange B. RI L J L2 Answers: (a) 01= ksi, 02 = ksi, (b) Up =

Answer 1

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Overall deformation will be zero as both ends are fixed.

Using this condition the forces for individual sections are calculated.

At AC cbe RAORC Let Reaction Respectively. RAS & Pero In parts: Rae RA T ERC ~ par RA-2P Ra-ap RA-2p - Rc-o LRA-2p=Rc As end A & cane fixed. Deformation=0 CAIC - DAIB +DBIC As ends fixed overall Deflection Edeformation A= PL AE O RAXIS? + CRA-2p) x236 4.22* 0900 2.64 x 26000 RAXIS2 x 2.64 x26000 = 2p- RA 4.22 070 236 1.RRA-2P-RA 2.2RA = 20 RA 2x31/2.2 = 20.18 ksi

Hence 20.10 the force :- 2018 TE 33.82 RA-2p = 28.18-2x31 33.82 KÉP Jis ? - RA - 2010 - 667 kipslin? A AAB 4.22 6.67 Ksi T2- RAZP AAC -33.82 2.64 -12.81 kips lin? =-12.81 ksi Movement of Bi DAB = PL = 28.10X 152 4.22 x 2700 AE =0.116 inches. Hence B will moves toward Right by 0.16 inches