Question

Calculate the forces in members BC, BE, and EF. Solve for each force from an equilibrium equation which contains that force as the only unknown. The forces are positive if in tension, negative if in compression. 11.5 kN 4.0 m 4.0 m с 4.0 m D 3.1 m 4.0 m 4.0 m E po G Answers: BC= i KN BE = i KN EF = i KN

Answer 1

In this problem, first Calculate the angle of inclined members (which is required for Calculation) and then use Method of section through the members BC, BE, and EF for Calculation of forces in these members by considering right sectias below:-

solution: - Free body diagram 11.5 kn In A B C E 4m 3 4m D > x = ton' ( 3 ) 3.1m 3.16 4m d = = 37.776 M Bt E 10.9m & In A EFM F 0.9 4.0 he Ba tan ( > For calculation of force in members B- 12.689 BC, BE & EF we method of sletion through these members (as thown in figure x-x) & consider right seation : Foc UEM FBE (Foe * 3.1) – (11.5 x4) = 0 Free = 11.5x4 tud FER 1.5 kwa B с Am --4 4m E 3.im 4.0m! 14.84 kl. E 3.1 (H) Emo0 -(11.5x B) - (FEFCB) (3:1) – (FEF SinB).(4-0) = 0 - 92 (FEF. Cob 19.68)(3-1) - (FEF Sin 1868) (4-0) =0 -92 3.0244. FEF 0.878. FEF 3.9024 FEF - FEF = -23.575 kn FEF 20 92

(lü E Fy=0. FBE. Sin 37.776 FBE. Sina FEF SinB - 11.5 = 0 FEF Sin 12.68 -11.5=0 (-23.575) (0.2195) – 11.5 = 0.6126 FBE 0.6126 FOE 11.5 - 5. 1748 6.3252 FBE 10.325 KN. Answer: - o Foca 14.84 ES FBE ka (Tension) 10.325 KN (Tension) 23.575 kn ( compression) () FEF