Question

25 QUESTION 15 Each of the links AB and CD is made of aluminum (E = 10.9 x 106 psi) and has a cross-sectional area of 0.3 in. All pins have a diameter of 0.05 in. D 16 in R 14 in Knowing that the links support the rigid member BC, determine the following: Force in link AB kip Force in link CD = kip Deflection AB x 10^-3 in Deflection CD- X 10^-3 in Deflection of point E- x 10^-3 in Stress in link AB psi Make sure to show all the detailed calculations in the PDF Submittal

Answer 1

Solution P = 2kip. I 16 in -26 in Min let force in AB= P and force in CD= P.. Pitp 2 let elongation in AB= 8, elongation in and CD = d.

Now, moment ment abour B=0 Bx (26614). = Px14 P₂= 2x1u 0.7 kip, 40 P = 2- Pa 2-0.72 13 Kip

Now, deflection in AB= 8, = P, L = (1.3 X1000) x 16 0.3 X10.9x106 AE 8, = 6:36*10-3 in deflection in CD = 8 = PLL 0.7 X1000 x 16 AE 0:3x10.9x106 82= 3.425x10-3 in Now, let deflection at de. 14 in 26in B E. 162 S , S. BS Si-dz Now, by similar triangle; S.-82 6 26 S = 26 Чо (6.36-3.425 )*10-3 Х 40 > 8 = 1.90826 X10-3 in. SE = 8+2=(1-90826 + 3.425 )x10-3 in. SE 5.33 X103 in

Now, let stress in AB T a 어 ), = 1:3x1000 0.3 T = 4333.33 p.si