Question

The 0.5-in.-diameter rod CE and the 0.75 -in.-diameter rod DF are attached to the rigid bar ABCD as shown. Knowing that the rods are made of aluminium and using E = 10.6 X 10 psi, determine (a) the force in each rod caused by the loading shown, (b) the corresponding deflection of point A. 12 in., 8 in. 18 in. B C D 10 kips 24 in. 30 in. E F

Answer 1

Soms FBD of link ABCD А B VB lokips VD clearly, {fyco → VB-Ve-VD = 10 → Ug = 10 + vctus EMB=0 VeX 12 + 15X20 = 10x 18 [Clocknerse +ve 6Vct lov0 = go 3vct SVD=45 three so third But here. unkorowos but only two equations equation can be written form rugidity of ABCD. since, AB CD is a To Sc regid rod, so no deformation relation between sis. Sc So с SD Cot B 205 11 Be=360 . Estension is kod CE = $c = (CXⓇ) AE + Extension in hud DFS So = {Vox e CE DF & So= NOX 30 Sc = Vcx 24 7410.5)2 x E 700.ỷ xE Se= 3280 & Vex24 Vox30 (0.5² € form efa (1. 7 (0.25) XE CS

Vox 365 VOX 24" (Oos)? 0.75) 2 Vc = 0.555VD putting this value of vc in en 3Vc t5VD= 45 3x0.555VD+50=45 Vo = = 6075 keps 6. 665 & vc = 0.555 Vo = 3.72 Kep's putting these valves los esa : Vas 10 +Ve+VD = (0+ 6.75+ 3.72) = 20.47 Rups SAE Sc 782 122 → SA = ssc SA=158 8c 8c = AE 8.72 X 24 4 10.5 10.68106 4.299x105 0.04289 inch ..SA = 1'5 x 0.04289 = 0.064 335 inch Hence the forces is rods ct is Ve= 3.72 kips & in of is Vo = 6.75kips and Deflection of point A SA= 0.064335 inch
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