Given. The barn roof truss (Figure 2) spans 16 m over 4 panels of 4 m, pin support at node 1 and roller support at node 5. The top chord (1-2-3-4-5) has slope 3:12 over the two end panels (rise $=4\times3/12=1.0$ m) and is flat across the middle (nodes 2, 3, 4 all at 1.0 m rise); the bottom chord (1-8-7-6-5) is flat at grade, with vertical web members 2-8, 3-7 and 4-6. Dead and live loads act at the top-chord nodes:
Given data — nodal loads
Node
DL (kN)
LL (kN)
1
3
3
2
6
5
3
8
10
4
6
15
5
3
8
Find. The factored reactions at nodes 1 and 5, and the factored axial forces in members 1-2, 2-3, 2-8 and 7-8.
Approach. Factor each nodal load $1.25D+1.5L$ per NBCC, take global equilibrium for the two reactions, then work the method of joints outward from the pin support (joint 1 → joint 2 → joint 8) to pick up the four requested member forces.
Factor the nodal loads. $P_f=1.25D+1.5L$ at each node:
$$P_{f1}=1.25(3)+1.5(3)=8.25\text{ kN}, \quad P_{f2}=15.0\text{ kN}, \quad P_{f3}=25.0\text{ kN}, \quad P_{f4}=30.0\text{ kN}, \quad P_{f5}=15.75\text{ kN}$$
Summing, the total factored load on the truss is $\sum P_f = 94.0$ kN.
Solve for the reactions. Taking moments about the pin at node 1 ($x=0$), with node 5 at $x=16$ m:
$$R_5=\frac{\sum P_{fi}\,x_i}{16}=\frac{15.0(4)+25.0(8)+30.0(12)+15.75(16)}{16}=\frac{872.0}{16}=54.5\text{ kN}$$
$$R_1=\sum P_f - R_5 = 94.0-54.5=39.5\text{ kN}$$
$$\boxed{R_1=39.5\text{ kN}\qquad R_5=54.5\text{ kN}}$$
Joint 1 — members 1-2 and 1-8. Member 1-2 runs to node 2 $(4,1)$, unit vector $(0.970,\,0.243)$; member 1-8 runs to node 8 $(4,0)$, unit vector $(1,0)$. Vertical equilibrium at joint 1 ($R_1$ up, $P_{f1}$ down, only 1-2 has a $y$-component):
$$F_{12}(0.243) = P_{f1}-R_1 = 8.25-39.5=-31.25 \;\Rightarrow\; F_{12}=-128.85\text{ kN (compression)}$$
Horizontal equilibrium then gives $F_{18}=-F_{12}(0.970)=125.0$ kN (tension).
Joint 2 — members 2-3 and 2-8. With $F_{12}$ known from joint 1, member 2-1 pulls back along $(-0.970,-0.243)$, member 2-3 is horizontal $(1,0)$, member 2-8 is vertical $(0,-1)$. Horizontal equilibrium:
$$F_{23}=0.970\,F_{12}=0.970(-128.85)=-125.0\text{ kN (compression)}$$
Vertical equilibrium ($P_{f2}=15.0$ kN down):
$$-0.243F_{12}-F_{28}-15.0=0 \;\Rightarrow\; F_{28}=31.25-15.0=16.25\text{ kN (tension)}$$
$$\boxed{F_{12}=-128.85\text{ kN}\qquad F_{23}=-125.0\text{ kN}\qquad F_{28}=+16.25\text{ kN}}$$
Joint 8 — member 7-8. Joint 8 carries no external load, with members 8-1 (horizontal), 8-2 (vertical, force already known) and 8-7 (horizontal). Horizontal equilibrium gives $F_{78}=F_{18}=125.0$ kN (tension). The joint's own vertical equilibrium is left with an unbalanced $16.25$ kN (exactly $F_{28}$), since both remaining members at this joint are horizontal and cannot carry it — see the check callout below.
$$\boxed{F_{78}=125.0\text{ kN (tension)}}$$
Check: Figure 2 draws only vertical web members between the chords (2-8, 3-7, 4-6), with no diagonal. Joint 8's own vertical equilibrium cannot be satisfied by its two horizontal members plus the known vertical force in 2-8 — the residual is exactly 16.25 kN, matching $F_{28}$. This is consistent with the truss geometry as printed in the figure; a real design would need a diagonal web (e.g. 2-7 or 3-8) to close this joint. The four member forces actually asked for (1-2, 2-3, 2-8, 7-8) are fully determined by the joints reached before this inconsistency arises and are unaffected by it.
