Question 2 of 6: Roof Truss — Reactions and Member Forces

Question 2: Roof Truss — Reactions and Member Forces (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Bottom chord (1-8-7-6-5) is level; top chord (1-2-3-4-5) follows a 4:12 slope, giving node heights 0/1/2/1/0 m as tabulated. Verticals connect 2-8, 3-7, 4-6.

Find. The factored reactions $R_1$, $R_5$, and the factored axial forces in members 1-2, 2-3, 2-8 and 7-8.

Approach. Factor every joint load with the NBCC gravity combination $1.25D+1.5L$, find the two support reactions from global equilibrium, then solve the requested member forces by the method of joints, working outward from the pin support.

1. Factor the joint loads. $$P_f = 1.25D+1.5L$$ Node 1: $1.25(2)+1.5(5)=10.0$ kN. Node 2: $1.25(4)+1.5(10)=20.0$ kN. Node 3: $1.25(3)+1.5(8)=15.75$ kN. Node 4: $1.25(2)+1.5(6)=11.5$ kN. Node 5: $1.25(1)+1.5(3)=5.75$ kN. Total factored load $=63.0$ kN.
2. Reactions from global equilibrium. Taking moments about the pin at node 1 ($x=0$): $$\sum M_1=0:\quad R_5(12)=20.0(3)+15.75(6)+11.5(9)+5.75(12)=327.0\ \text{kN}\cdot\text{m}$$ $$\boxed{R_5=27.25\ \text{kN}}$$ Then $R_1=63.0-27.25=\boxed{35.75\ \text{kN}}$ (no horizontal load, so $R_{1x}=0$).
3. Joint 1 — members 1-2 and 1-8. Member 1-2 runs at slope 1:3 ($\sin\theta=1/\sqrt{10}$, $\cos\theta=3/\sqrt{10}$); member 1-8 is horizontal. With tension taken positive: $$\sum F_y=0:\ F_{12}\left(\tfrac{1}{\sqrt{10}}\right)+35.75-10.0=0 \ \Rightarrow\ F_{12}=-81.43\ \text{kN (C)}$$ $$\sum F_x=0:\ F_{12}\left(\tfrac{3}{\sqrt{10}}\right)+F_{18}=0 \ \Rightarrow\ F_{18}=+77.25\ \text{kN (T)}$$
4. Joint 2 — members 2-3 and 2-8. Member 2-3 has the same 1:3 slope as 1-2; member 2-8 is vertical. Using $F_{12}$ from Step 3 and the 20.0 kN joint load: $$\sum F_x=0:\ F_{23}=F_{12}=\boxed{-81.43\ \text{kN (C)}}$$ $$\sum F_y=0:\ F_{28}=-\left[20.0-\left(F_{12}-F_{23}\right)\tfrac{1}{\sqrt{10}}\right]=\boxed{-20.00\ \text{kN (C)}}$$
5. Joint 8 — member 7-8. Member 1-8 and 8-7 are both horizontal, so the vertical member 2-8 contributes no $x$-component here: $$\sum F_x=0:\ -F_{18}+F_{78}=0 \ \Rightarrow\ \boxed{F_{78}=+77.25\ \text{kN (T)}}$$