Truss member forces BC, BD, BE, DE, DG, EG

Data and assumptions

The top chord A–B–D–F–H is horizontal with 4 m panels: A(0,0), B(4,0), D(8,0), F(12,0), H(16,0). The lower chord runs in a straight line from A down to G, so C(4,−1.5), E(8,−3), G(12,−4.5) (depth 4.5 m under F). Members: AB, BD, DF, FH, AC, CE, EG (chords), BC, DE, FG (verticals), BE, DG (diagonals) and HG. The figure shows a vertical 48 kN load at A and supports at G and H. (m = 13, r = 3, j = 8: determinate.) Whichever of G and H is taken as the pin, the horizontal reaction is zero and the member forces are the same.

Reactions

• ΣM_H = 0: 48(16) = G_y(4), so G_y = 192 kN ↑; H_y = 144 kN ↓ (the support at H holds the truss down). The horizontal reactions are zero.

Method of joints

Direction of the lower chord: AC = CE = EG = √(4² + 1.5²) = 4.272 m, cos = 0.9363, sin = 0.3511.

Joint A (48 kN ↓): ΣFy: −48 − 0.3511F_AC = 0, so F_AC = −136.70 kN (C). ΣFx: F_AB + 0.9363(−136.70) = 0, so F_AB = +128.00 kN (T).

Joint C: AC and CE lie on one straight line and BC is the only other member, with no load at C, so F_BC = 0 and F_CE = F_AC = −136.70 kN (C).

Joint B: members AB, BD (collinear), BC = 0 and BE. Since BC = 0, ΣFy gives −0.6F_BE = 0, so F_BE = 0 and F_BD = F_AB = +128.00 kN (T).

Joint E: CE and EG are collinear and BE = 0, so F_DE = 0 and F_EG = F_CE = −136.70 kN (C).

Joint D: BD = 128 kN (T) and DE = 0, so DF = 128 kN (T), and ΣFy gives −0.7474F_DG = 0, so F_DG = 0.

Joint G (check): EG pushes G with (128 →, 48 ↓); HG (compression 192.67 kN) pushes G with (128 ←, 144 ↓); G_y = 192 kN ↑ balances 48 + 144 = 192 ✓.

Result

BC, BE, DE and DG are zero-force members: at each unloaded joint (C, B, E, D in turn), once the zero members already found are removed, the member is the only one not collinear with the others.