Use of trusses; cantilever truss member forces

Use of trusses in engineering

A truss is a light, stiff framework of straight members arranged in triangles, so every member carries only tension or compression and the material is used efficiently over long spans. Trusses are used for roofs of halls and factories, bridges (railway and road), transmission and telecom towers, crane jibs and gantries, floor and wall frames, aircraft and ship structures, and temporary stages and scaffolding.

Data

Wall supports at A(0,6) and G(0,0) (pin-type, 4 reaction components). Bottom chord G(0,0), F(4,0), E(8,0), D(12,0); top chord A(0,6), B(4,4), C(8,2), D(12,0) in one straight line (slope 6 m in 12 m); verticals BF and CE; diagonals AF and BE. Loads: 8 kN ↓ at F, 6 kN ↓ at E, 4 kN ↓ at D. (m = 10, r = 4, j = 7: m + r = 2j, determinate.) Lengths: AB = BC = CD = √(4² + 2²) = 4.472 m (cos 0.8944, sin 0.4472); BE = 5.657 m; AF = 7.211 m.

Method of joints, starting at the free end (no need for reactions)

Joint D: ΣFy: −4 + 0.4472F_CD = 0, so F_CD = +8.94 kN (T). ΣFx: −0.8944(8.94) − F_ED = 0, so F_ED = −8.00 kN (C).

Joint C: CB and CD are collinear and CE is the only other member (no load at C), so F_CE = 0 and F_BC = F_CD = +8.94 kN (T).

Joint E: ΣFy: −6 + 0.7071F_BE = 0, so F_BE = +8.49 kN (T). ΣFx: −8.00 − 0.7071(8.49) − F_EF = 0, so F_EF = −14.00 kN (C).

Joint B (BC = 8.94 T, BE = 8.49 T): ΣFx: 0.8944(8.94) + 0.7071(8.49) − 0.8944F_AB = 0, so 8.00 + 6.00 = 0.8944F_AB and F_AB = +15.65 kN (T). ΣFy: −0.4472(8.94) + 0.4472(15.65) − F_BF − 0.7071(8.49) = 0, so −4.00 + 7.00 − 6.00 = F_BF and F_BF = −3.00 kN (C).

Joint F: ΣFy: −8 − 3.00 + 0.8321F_AF = 0, so F_AF = +13.22 kN (T).

Result

Also CD = 8.94 T, DE = 8.00 C, CE = 0 and GF = 21.33 C (from ΣFx at F). Reactions: GF is the only member at G and it is horizontal, so the wall at G gives only a horizontal reaction of 21.33 kN →; A gives 21.33 kN ← and 18 kN ↑ (the total load). Check ΣM about G: 4(8) + 8(6) + 12(4) = 128 = 6 × 21.33 ✓.