Roof truss member forces (1-11, 1-10 etc.)

Data and assumptions

Bottom joints 12, 1, 2, 3, 4, 5, 6 are 4 m apart (12 at the pin, 6 at the roller). The left rafter rises at 30°, so the top joints are at heights 11: 2.309 m, 10: 4.619 m, 9 (apex): 6.928 m. Members 1-11, 2-10 and 3-9 are vertical. The loads of 20 kN (at 11), 15 kN (at 10) and 20 kN (at 9) act normal to the left rafter, i.e. at 30° to the vertical, pointing down-right. Components: 20 kN → (10.00 →, 17.32 ↓), 15 kN → (7.50 →, 12.99 ↓).

Reactions

• Resultant horizontal load = 10 + 7.5 + 10 = 27.5 kN, so H₁₂ = 27.5 kN ←.
• ΣM_12 = 0: 24R₆ = 17.32(4) + 10(2.309) + 12.99(8) + 7.5(4.619) + 17.32(12) + 10(6.928) = 508.1, so R₆ = 21.17 kN ↑.
• Total vertical load = 17.32 + 12.99 + 17.32 = 47.63 kN, so V₁₂ = 47.63 − 21.17 = 26.46 kN ↑.

Members: top chord 12-11-10-9-8-7-6 (straight rafters); bottom chord 12-1-2-3-4-5-6; verticals 1-11, 2-10, 3-9, 4-8, 5-7; diagonals 1-10, 2-9, 4-9, 5-8 (m = 21, j = 12, r = 3, so determinate).

Method of joints (axes along and normal to the rafter)

Joint 12 (rafter 12-11 at 30°): ΣFy: 26.46 + 0.5F₁₂₋₁₁ = 0, so F₁₂₋₁₁ = −52.92 kN (C). ΣFx: −27.5 + 0.866(−52.92) + F₁₂₋₁ = 0, so F₁₂₋₁ = +73.33 kN (T).

Joint 11: members 11-12 and 11-10 lie along the rafter, and the load acts normal to it. Resolving normal to the rafter: −20 − 0.866F₁₋₁₁ = 0, so F₁₋₁₁ = −23.09 kN (C). Along the rafter: F₁₁₋₁₀ − F₁₁₋₁₂ + (−0.5)F₁₋₁₁ = 0, so F₁₁₋₁₀ = −52.92 − 11.55 = −64.47 kN (C) (this is member 10-11).

Joint 1 (member 1-10 has direction (4, 4.619)/6.110 = (0.6547, 0.7559)):

• ΣFy: −23.09 + 0.7559F₁₋₁₀ = 0, so F₁₋₁₀ = +30.55 kN (T).
• ΣFx: −73.33 + F₁₋₂ + 0.6547(30.55) = 0, so F₁₋₂ = +53.33 kN (T).

Joint 10 (normal-to-rafter equation): −15 − 0.866F₂₋₁₀ − 0.3273(30.55) = 0, so F₂₋₁₀ = −28.87 kN (C).

Result

(If the loads were taken as vertical instead of normal to the rafter, the values would change. The normal direction is the usual wind-type loading on a roof truss and is what the arrows in the figure show.)