Pin-Jointed Roof Truss Under Asymmetric Snow Load
A symmetrical pin-jointed roof truss has a 6.0 m span with five equal panels of 1.2 m along the bottom chord. The apex is 1.5 m above the supports. Under a partial snow-drift case, the left half of the roof carries 4.0 kN per node at the two upper nodes on the left slope, while the right half carries only 1.5 kN per node at the two corresponding right-side nodes. The truss is supported on a pin at the left and a roller at the right. A student analysing the truss assumes the loading is symmetrical, calculates the reactions as half the total load each, and reports the maximum member force as 9.8 kN. A second student checks the horizontal reaction at the pin and finds it is zero.
1. Explain why the second student's finding of zero horizontal reaction is correct even though the loading is asymmetric, and state what this implies about the truss analysis.
Read the worked answer
All applied loads are vertical (gravity snow load), and the right support is a roller that can supply only a vertical reaction. Taking moments about the left pin eliminates both the vertical and horizontal reactions at that pin; the only remaining unknown is the roller reaction, and the moment equation contains no horizontal force term. Summing horizontal forces then gives H_left = 0 directly, because there is no other horizontal force in the system.
The implication is important: asymmetry in the vertical load distribution changes the magnitudes of the vertical reactions and the member forces, but it does not create a horizontal reaction. The horizontal reaction would only become non-zero if a horizontal load existed - wind, a horizontal component of load at a node, or a support settlement or temperature effect in a statically indeterminate structure. This truss is statically determinate (m + r = 2j), so no horizontal thrust can develop.
A common error is to assume asymmetry implies a horizontal reaction; it does not. The correct check is to confirm the direction of every applied load and the type of every support before writing equilibrium equations.
2. Calculate the correct vertical reactions and explain what the first student's error does to the reported maximum member force.
Read the worked answer
Total load. Left side: 2 nodes × 4.0 kN = 8.0 kN. Right side: 2 nodes × 1.5 kN = 3.0 kN. Total = 11.0 kN.
Reactions. Take moments about the left support (span 6.0 m). The four loaded upper nodes are at horizontal positions determined by the 1.2 m panels. Using the left-slope nodes at 1.2 m and 2.4 m from the left, and the right-slope nodes at 3.6 m and 4.8 m: R_right × 6.0 = 4.0 × 1.2 + 4.0 × 2.4 + 1.5 × 3.6 + 1.5 × 4.8 = 4.8 + 9.6 + 5.4 + 7.2 = 27.0 kN·m R_right = 4.5 kN, so R_left = 11.0 − 4.5 = 6.5 kN.
Error. The first student assumed R_left = R_right = 5.5 kN. The true left reaction is 18% higher than assumed and the right is 18% lower. Because member forces are linear in the applied loads and reactions, the forces in members near the heavily loaded left side are underestimated and those near the right side overestimated. The reported 9.8 kN maximum is therefore not conservative - the true governing force will exceed it, and a design based on it would be unsafe. The correct procedure is to solve the asymmetric case explicitly, or to use the envelope of symmetric and asymmetric load cases.