Question

P2 An aluminum alloy [E = 10,000 ksi; oy 40 ksi] pipe (1) is connected to a bronze [E = 16,000 ksi; oy 45 ksi] pipe at flange B, as shown in the Figure. The pipes are attached to rigid supports at A and C. Pipe (1) has an outside diameter of 2.375 in., a wall thickness of 0.203 in., and a length of L1 = 6 ft. Pipe (2) has an outside diameter of 4.50 in., a wall thickness of 0.226 in., and a length of L2 = 10 ft. If a minimum factor of safety of 1.67 is required for each pipe, determine: (a) the maximum load P that may be applied at flange B. (b) the deflection of flange B at the load determined in part (a) (2) (I) В L1 L2 FIGURE P2

Answer 1

E, = 1000 ksi ty = 40kst Eq=16000 ksi Tye - 45 ksi out side dia =4.5 in call thickness - 0.226. outside diameter = 2.375 in wall thickness= 0.203 in Inside diameter = 1.969 in L = 6.inft = 6*12= 72 Inside diameter=4.048 in L2=10ft elox 12 = 120 in. 1.969 P+P =P Compatability Equation AAB + ABC on -P, 4 + (P-P.) L2 - 0 ÃE, A2 E2 - > Pix 72 od to wool - EXO (P-P.)120 42-3754 m2 <10000 .5-2404%)&160cm * -P, x 4.0824x163 +(P-P.) 109411x153.0 P = 0.322P Ty = 40 ksi 0.322 Pro I (2-375202032) (P = 546.31 ksito

1.67 Pe=P-P, = 0.678 P. Ty2 = 45 ksi - 40.678P = 45 (4.5² 4.0482) P=201.41**pl - from ② & ③ P= 3D 201.41kip factored load (P) = 201.41 = 120.6 kip. 1) max. load P that can applied at flange Bei = 120.6 kip 2 deflection of flange Bu ABAR = (P-P,) 120 I x4.6324.0482) * 16000 = 0-678x 1806 x 120 * *(4-54-018?) * 16000 = 0.202 inch.