Truss forces in CE, BE, BD, DE; stability

Data

A(0,0) pin and H(20,0) roller; bottom joints C(5,0), E(10,0), G(15,0); top joints B(5,3), D(10,5), F(15,3). The figure shows four equal panels between A and H (the label 5@4 m = 20 m gives the same 20 m span). Loads: 5 kN ↓ at B; 10 kN at D at 30° to the horizontal pointing down-left (8.66 kN ←, 5 kN ↓); 20 kN ↓ at E. (m = 13, r = 3, j = 8: determinate.)

Reactions

• ΣM_A = 0: 20H_y − 5(5) − 5(10) − 20(10) + 8.66(5) = 0, so H_y = 11.58 kN ↑.
• ΣFy: A_y = 30 − 11.58 = 18.42 kN ↑; ΣFx: A_x = 8.66 kN →.

Section through BD, BE and CE (left part: A, B, C)

BD = √(5² + 2²) = 5.385 m (cos 0.9285, sin 0.3714); BE = √(5² + 3²) = 5.831 m (cos 0.8575, sin 0.5145).

• CE: ΣM_B = 0: 3F_CE − 18.42(5) + 8.66(3) = 0, so F_CE = +22.03 kN (T).
• BD: ΣM_E = 0 (anticlockwise +): −18.42(10) + 5(5) − 4.643F_BD = 0, so F_BD = −34.28 kN (C).
• BE: ΣFy: 18.42 − 5 + 0.3714(−34.28) − 0.5145F_BE = 0, so F_BE = +1.33 kN (T).

Joint D (for DE)

BD (compression) pushes D with (31.83 →, 12.73 ↑); DF goes down-right (0.9285, −0.3714); DE is vertical.

• ΣFx: 31.83 + 0.9285F_DF − 8.66 = 0, so F_DF = −24.96 kN (C).
• ΣFy: 12.73 + 0.3714(24.96) − F_DE − 5 = 0, so F_DE = +17.00 kN (T).

Result

Stability and determinacy

• Stability: a structure is stable if it keeps its shape and position under any load (supports prevent rigid-body motion and the members form a rigid arrangement of triangles). Example: the triangular truss above on a pin and a roller. A truss with a square panel without a diagonal, or supports whose reactions are all parallel or concurrent, is unstable.
• Determinacy: externally r − 3 and for a truss m + r − 2j (= 0 determinate, > 0 indeterminate, < 0 unstable). The truss above: 13 + 3 − 16 = 0, so it is determinate. If a diagonal is added to a panel already braced, m + r − 2j = 1 and the truss becomes indeterminate to the first degree.