Question

1 1. (25 pts.) A system of bars is loaded by axial loads as shown. Determine: a) the normal stress in each bar b) the deflection of point D with respect to point A. Loads: F1 = 35 (kips), F2 = 24 (kips), F3 = 16 (kips) 2 02:55:52 The normal stress: BAR AB (ksi) BAR BC = (ksi) BAR CD = (ksi) The axial deflection: BAR AB = (in.) BAR BC= (in.) BAR CD = (in.) The deflection of point D with respect to point A = (in.) F A B F2 F3 Outside Diam. = 2.75 in Inside Diam. = 2.5 in. Outside Diam. = 2.0 in Outside D Diam. - 1.375 in E = 10000 ksi F E = 10000 ksi E = 30000 ksi 48 in. 30 in. 18 in

Answer 1

Normal stress JCD si A B 24 kips >35kips outside dia 2.oin outside dia = 275in inside Dia = 2.5in outside 16 kips -> - 1.375 24 kips >35 kips 18 in -48in. 30in Bar CD D с outside Sia=1375 16 kips L 16 kips -16 18in TED = Ecd = 100co ks; *1395 4 (Tensile) JCD = 10.775 kse in CD Deflection DED JED X LED ELD Aco 10.775. :) 10000 X 18

Ә) ДcD О.019Ч gn ВС Bar B 32 k? 24 kips outside Эја = 2.0 B 24 kips 32kirk 16 kips >> Скря 24 kips 29kips EBC= 30in = 1000 ks 32 Увс exc2)2 (Compresive) Ова = - 10. 106 ksi Deflection Две TBC x LBC ЕВс Дъс - - 10-180 x 30 Jo000 ДВС — О.0306 in

Bar AB B A → 35 tips 35 kips outside sia = 2.75in 32 kips 32 kips inside Dia = 2.5 in > 35 kips 35 kips 48in EAB = 30000ksi A B 38 kips 38 kips JAB = 38 2.75 -2.53) 4 36.863 Ksi JAB 36.863 X 48 DAB 30000 0.059 in IDAB

with respect Deflection of point & to point A AART A Bit ALD AD = 0.059 - 0.0306 +0.0194 - AD = 0.0478 en stress; taking tensile stress (+) The Normal BAR AB = 36.863 ksi = -10.186 ksi 10.775 ksi BAR BC BAR CD The axial Deflection BAR AB = 0.059 in BAR BC = -0.0306 in BAR CD = 0.0194 in AD 0.0478 in