Question

Module -9. Test Q. 1 Determine the resultant forces at pins B and C on member ABC of the four- member frame. 5 8 Ax to 411 E F9 ΣΜΕ εο Ax (4) -150 (1) (-1) = 0 ..Ax = 918.052b FED 'ABC' 15 Fa FEE 3-14 FBE = 1531.2526 F BE = 1531.25 Ib (c) Fco = 35026 (T) MAO (0)-(Fra) ***120)(2)- -. Feo - 36026

Answer 1

Sol Equivalent force of the distributed load (P) = WL P= 150 lbx 7 ft = 1050 lb ft It acts at Inalta 3.5 ft from point A Free body diagram of the truss. 3.5 ft K P= 1050 lb. AX A Ау. u ft. BX B BY

(5) Z mg=0 » (Ax xy) - (Px 3:5)=0 4 Ax - (1050 X3.5)=0 Ax = 918.75 1b.

sol Free body diagram of the member ABC P=150x7 1050 lb. 3.5 ft K 2 B BC. eft k 44 FB. Fc. K 5th FB x 5 (0) Z Mg = 0. → (Fcx 7) + 5)-o (-1050X 3.5) - 3675 -7 Fc + 4FB =0 UFB-TFC = 3675 (+) ŽEx=0 - Axt 3 FB =0 918-75+ 3 FB=0 5 COMPRESSION FB = 1531.25 lb From 4 x 1531.25 – 7 FC = 3675 7FC = 2450 Fc = 350 lb (TENSION)