Question

As shown, a truss is loaded by the forces P1 = 490 lb and P2 = 192 lb and has the dimension a = 9.00 ft. y P X G H P2 E B > a с D > 7 a/2 a/2 Determine Fcg and Fgh, the magnitudes of the forces in members CG and GH, respectively, using the method of sections. Assume for your calculations that each member is in tension, and include in your response the sign of each force that you obtain by applying this assumption. Express your answers numerically in pounds to three significant figures separated by a comma.

Answer 1

Calculations for finding out force in the members CG and GH are done below.

Note:

• BG= DH= AB= DE= a= 9 ft
• BC= CD= a/2= 9/2 ft
• P1= 490 lb and P2= 192 lb

Calculations:

# Supports A and E are replaced by support reactions Rax & Ray in support a and RE in support B. Ľ 490 lb х 192 lb I ft RAX E B C D TRAY RE 9 ft gebot q ft eft GEMat A =0 490 (AB) = 0 → RE (AE) + 192 (DH) o RE (27) + 192 (9) RE 99.333 lb 490 (9) = 0 + EF, -o → * Ray +RE = 490 Ray + 99. 333 = 490 390.667 Lb RAY +, EFy-o → RAX – 192 =0 | RAX = 192 lb # Take a section QQ' as shown in below figure. 49066 Lox H 192 lb 19 ft RAX A TRAY k 9 ft 9 ft.

# Take left part of section QQ' and do the calculations to find Fea and FGH : 490 lb ㄴ Х FGH Ech 9 ft. RAX FBC tan" (Besa) O- Ray tant/ ķ in " (1943) 9 ft * 2 ft o = 26.565° G E Mat C =0 490 (BC) – FGH (BG) - Ray (AC) = 0 (9) 390.667 (9+ 2) = 0 490 ( 2 ) - FGH → FGH = -341.0005 Lb → FOH = - 341.001.16 GΣΜ at B → - Ray (AB) (AB) - EGH - GH (BG) – Fcg sino.(BG) = 0 -390.667(9) - (-341.001)(9) Fag sin(26.565) (9) = 0 F = -1/1.057 Lb 'CG