Member forces in given truss (2074 Asoj)

Data and assumptions

Joints: A(0,0), C(5,0), E(10,0), G(15,0), H(20,0) on the bottom chord; B(5,3), D(10,5), F(15,3) on the top. The four panels of 5 m make 20 m. A is a pin and H a roller. Loads: 5 kN at D at 30° to the horizontal, pointing down-right (4.33 kN → and 2.50 kN ↓), 15 kN ↓ at F and 10 kN ↓ at G. The members marked in the figure are taken as BD, DF, DE, BE and CE. (m = 13, r = 3, j = 8: determinate.)

Reactions

• ΣM_A = 0: 20H_y − 10(15) − 15(15) − 2.5(10) − 4.33(5) = 0, so 20H_y = 421.65 and H_y = 21.08 kN ↑.
• ΣFy: A_y = 27.5 − 21.08 = 6.42 kN ↑; ΣFx: A_x = 4.33 kN ←.

Section 1–1 between B–C and D–E (cuts BD, BE, CE; left part)

Lengths: BD = √(5² + 2²) = 5.385 m, BE = √(5² + 3²) = 5.831 m.

• CE: ΣM_B = 0: 3F_CE − 6.42(5) − 4.33(3) = 0 → F_CE = +15.03 kN (T).
• BD: ΣM_E = 0 (clockwise +): 6.42(10) + F_BD(0.3714×5 + 0.9285×3) = 0 → 64.17 + 4.643F_BD = 0 → F_BD = −13.82 kN (C).
• BE: ΣFy: 6.42 + 0.3714F_BD − 0.5145F_BE = 0 → 6.42 − 5.13 = 0.5145F_BE → F_BE = +2.49 kN (T).

Joint D (cos = 5/5.385 = 0.9285, sin = 2/5.385 = 0.3714)

• BD is in compression and pushes D with 13.82 × 0.9285 = 12.84 kN to the right and 13.82 × 0.3714 = 5.13 kN upward.
• ΣFx: 12.84 + 0.9285F_DF + 4.33 = 0 → F_DF = −18.49 kN (C).
• ΣFy: 5.13 − 0.3714F_DF − F_DE − 2.5 = 0 → 5.13 + 6.87 − 2.5 = F_DE → F_DE = +9.50 kN (T).

Result (indicated members)

Other members (for completeness): AB −12.47 C, FH −40.98 C, AC = CE = 15.03 T, EG = GH = 35.14 T, EF −20.96 C, FG = 10.00 T (it carries the load at G), BC = 0.