Truss indeterminacy; forces in BC, BG, HG, GD

Degree of indeterminacy

Joints: j = 8 (A, B, C, D, E, F, G, H). Members: AB, BC, CD, DE (top chord and end rafters), AH, HG, GF, FE (bottom chord), BH, CG, DF (verticals), BG, GD (diagonals): m = 13. Reactions: pin at A (2) + roller at E (1): r = 3.

• External indeterminacy: D_e = r − 3 = 3 − 3 = 0.
• Internal indeterminacy: D_i = m − (2j − 3) = 13 − 13 = 0.
• Total: m + r − 2j = 13 + 3 − 16 = 0. The truss is statically determinate and stable (all panels triangulated).

Data

A(0,0), H(4,0), G(8,0), F(12,0), E(16,0) on the bottom (4 m panels); B(4,4), C(8,4), D(12,4) on the top (height 4 m). Loads: 10 kN → at D, 5 kN ↓ at H, 10 kN ↓ at G, 15 kN ↓ at F.

Reactions

• ΣM_A = 0: 16E_y − 5(4) − 10(8) − 15(12) − 10(4) = 0, so E_y = 20 kN ↑.
• ΣFy: A_y = 30 − 20 = 10 kN ↑; ΣFx: A_x = 10 kN ←.

Method of joints

Joint A (AB at 45°): ΣFy: 10 + 0.7071F_AB = 0, so F_AB = −14.14 kN (C). ΣFx: −10 + 0.7071(−14.14) + F_AH = 0, so F_AH = +20 kN (T).

Joint H: ΣFy: −5 + F_BH = 0, so F_BH = +5 kN (T). ΣFx: −20 + F_HG = 0, so F_HG = +20.00 kN (T).

Joint B (AB pushes B with 10 kN → and 10 kN ↑; BH pulls down 5 kN):

• ΣFy: 10 − 5 − 0.7071F_BG = 0, so F_BG = +7.07 kN (T).
• ΣFx: 10 + F_BC + 0.7071(7.07) = 0, so F_BC = −15.00 kN (C).

Joint C: BC and CD are collinear and no load acts at C, so F_CG = 0 and F_CD = −15 kN (C).

Joint G (HG = 20 T pulls left, BG pulls up-left, GD goes up-right):

• ΣFy: −10 + 0.7071(7.07) + 0.7071F_GD + F_CG(0) = 0, so F_GD = +7.07 kN (T).
• ΣFx: −20 + F_GF − 0.7071(7.07) + 0.7071(7.07) = 0, so F_GF = +20 kN (T).

Result

Remaining members (check at D and E): DF = +15 T, DE = −28.28 C, EF = +20 T, CD = −15 C.