Question

Using the method of joints, determine the force in each member of the truss shown. The load P = 390 lb.

Using the method of joints, determine the force in each member of the truss shown. The load P= 390 lb. 20 in. 48 in. 15 in. The force in member AB (FAB) is 1800 The force in member BC (FBC) is 1950 The force in member AC (FAC) is 3000 lb. (Tension) lb. (Compression) lb. (Compression)

Answer 1

Sol! P: 390 Lb p=39016 P-39026 RAMA Тес, EMAIO I tve 390(48+15) – Ry (15)=0 his 390 (63) = 15 Rcy l Rey = 163800 Lb ] {fy=0 Itve - Ray + Rey -390=0 1638-3912480 B Tray = 124826 / тузе [Елу 20 ] Jointa Ay I to os tan ( ) 0953-13) at 1248 To AB DAC 1. 1248 V ABC cose Efy = 0 Acsino 1248 + Ac sin 53.1320 Jaca - 1560.00 lb

EFXzo AB + Ac CSO 20 ABS -/-1560) cos(53.13) AB = 9366 390 Joint B 1 48 BC sirge AB J Bc sing N /20 TØ = 22.610 {Fx30 - BCCS $ - AB =0 - BC COS (22.6)) -93670 - 1013.92 kb Check Efyzo -390 – Bc sin (22.61)=0 -390 + 1013.92 (5192261)= 0 -390 + 389.89 90 100] Answer AB = 936 26 (Tension) BC = 1013.92 (compression) AC = 1560 Compression)