Member forces in truss with 20, 10, 12, 50 kN

Data and assumptions

A(0,0) is a pin, D(10,0) a roller, E(5,0) lies on the bottom chord; the top joints are B(2.5,4) and C(7.5,4) (taken at the middle of AE and ED, so BC = 5 m). Members: AB, BC, CD, AE, ED, BE, CE (m = 7, j = 5, r = 3, so m + r = 2j: determinate and stable). Loads: 20 kN ↓ at B; 10 kN ↓ and 12 kN → at C; 50 kN ↓ at E. Length AB = BE = CE = CD = √(2.5² + 4²) = 4.717 m, so sin = 4/4.717 = 0.848 and cos = 2.5/4.717 = 0.530.

Reactions

• ΣM_A = 0: 10D_y − 20(2.5) − 10(7.5) − 50(5) − 12(4) = 0, so D_y = 42.3 kN ↑.
• ΣFy: A_y = 80 − 42.3 = 37.7 kN ↑; ΣFx: A_x = 12 kN ←.

Method of joints (tension +, compression −)

Joint A: ΣFy: 37.7 + 0.848F_AB = 0, so F_AB = −44.46 kN (C). ΣFx: −12 + 0.530(−44.46) + F_AE = 0, so F_AE = +35.56 kN (T).

Joint D: ΣFy: 42.3 + 0.848F_CD = 0, so F_CD = −49.88 kN (C). ΣFx: −0.530(−49.88) − F_ED = 0, so F_ED = +26.44 kN (T).

Joint B (AB pushes B with 23.56 kN → and 37.70 kN ↑):

• ΣFy: 37.70 − 20 − 0.848F_BE = 0, so F_BE = +20.87 kN (T).
• ΣFx: 23.56 + F_BC + 0.530(20.87) = 0, so F_BC = −34.62 kN (C).

Joint C: BC is in compression and pushes C with 34.62 kN to the right; CD is in compression and pushes C with 26.44 kN to the left and 42.30 kN upward; the loads are 12 kN → and 10 kN ↓. CE goes down-left from C.

• ΣFx: 34.62 − 26.44 + 12 − 0.530F_CE = 0 (tension in CE pulls C towards E, i.e. down-left), so F_CE = +38.09 kN (T).
• ΣFy (check): 42.30 − 10 − 0.848(38.09) = 0 ✓.

Result