Question

• Draw FBDs as required. Clearly state all assumptions made towards solving the problem . Please enter all answers on paper/tablet and convert to single pdf file and submit • Eram is for a total of 100 points 1) Member BCD is supported in the position shown using the in' rod AC and in diameter pins at B.C&D. The ultimate bearing stress for the pin at B is 5ksi, the unltimate shear stress for the pin at B is 7 ksi and the ultimate normal stress for the link AC is 12ksi. Assuming that the pins at A and Care strong and a factor of safety of 3 is desired, determine (a) The allowable force P that can be applied (b) Explain, with a logical reason, why the allowable value is a maximum/minimum? (25 points) 6 in. B с D 4 in. -4 in. Side view at B

Answer 1

1) Member BCD is supported in the position shown using the x in rod AC and in diameter pins at B, C&D. The ultimate bearing stress for the pin at B is 5ksi, the unltimate shear stress for the pin at B is 7 ksi and the ultimate normal stress for the link AC is 12ksi. Assuming that the pins at A and C are strong and a factor of safety of 3 is desired, determine (a) The allowable force P that can be applied (b) Explain, with a logical reason, why the allowable value is a maximum/minimum? (25 points) А 0-5631° tomo= 6/4 = ₂ 6 in. fresine 4С D Bx in. Side view at B By E Mat Bio PX8- Fac sin (56'31') xu co -4 in. P FAC= 2.404 P Bri= 240Uxcos ( 56:319 Ba= 1.333 p Efnzo, Efy=0, -P - Byt f 2p=0, By =p Recetion at B, R3 = √ B²+ By? J (1.333p)²+ p2 RB= 1.667 P O for AC ( Factor of safety F5= 3 = 4ksi こ Saccallow) = 3 The (cultimate) FS 2.404 P Fire A Lxlly SAC (allewesha - 19.2304 p ote= Stes 19.2804 P= 4X1000 P= 208.eb (Roz (6) . A = 16.98p pri pen at B shear stress on are plane 1.667 P TB- lų 7 Es Callowana) = ksi 2X [ X (4²)

For safely TBS to call) 16.98 p = to x lovo p= 137.42 eb (3) beame of B RB/ Ab 20:004 p psi U |-667P 1/3 x 4 OB = poltro Ab= dt o B Calloway) = 5/ kot BSB Callersalde) 20004P = X 1000 p=83.316 lb allewable lood p = min ( 83:32, 208, 137-41) p = 83.32 lb then minsam unp we choose minimum because if il 83.32, then the B will be lood greater failed in bearing