22-Agric-B2 Structural Design for Agricultural, Biosystems, and Food Industries · May 2018

Question 4 of 6: Lateral Force Resisting Systems and End-Wall Brace

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 4: Lateral Force Resisting Systems and End-Wall Brace (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.

a) Three lateral force resisting systems for the barn. (1) Diaphragm roof to braced gable end walls, the system this exam adopts for part (b): wind pressure on the long side walls is collected as a uniformly distributed load on the roof sheathing, which spans as a deep horizontal beam between the two gable end walls and delivers a single concentrated reaction to each end. The magnitude at each end wall is found from simple tributary-area statics: total wind pressure (from NBCC wind pressure $q$ times exposure/gust factors) times the barn's side-wall height and half its length, split between the two end-wall reactions. (2) Braced (knee-braced) post-frame system: instead of relying on the roof diaphragm, each interior post-to-truss connection is stiffened with diagonal knee braces so the post-and-truss bent itself resists sidesway as a rigid frame, distributing wind load along the barn's length rather than concentrating it at the ends. Here the lateral load per bent is found by tributary width (post spacing, 3.6 m in Figure 1) times wind pressure times wall height, and each braced bent must independently resist its own share. (3) Continuous shearwall (sheathed) end walls or side walls: plywood or OSB sheathing nailed to the wall framing acts as a vertical cantilever shear panel, with unit shear $v_f=V_f/L_{wall}$ checked against the sheathing's nailing schedule and chord (tension/compression boundary member) forces $=V_f h/L_{wall}$ checked at the wall ends. The design lateral load for a shearwall segment is again its tributary share of the total wind reaction, but the load path runs directly down through the sheathed wall rather than through knee braces or a diaphragm-to-brace path. All three systems ultimately trace back to the same source load (NBCC wind pressure on the exposed wall/roof area) but differ in how that load is collected (diaphragm vs. distributed bents) and resisted (braced frame vs. shear panel vs. diagonal brace).

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. Barn end wall (Figure 4), 16 m wide, vertical wall height 3.6 m below a gambrel roof; total factored wind lateral load $V_f=25$ kN delivered by the roof diaphragm at eave level to one top corner of the end wall.

Find. The axial force in a diagonal wood brace spanning the end wall, and a member size to carry it.

Approach. Model the diagonal brace as a two-force axial member running from the loaded top corner to the opposite bottom corner of the wall's dimensioned rectangle (Figure 4 gives no separate roof-slope dimensions for the gambrel's lower plane, so the wall's own 16 m × 3.6 m rectangle — the only fully dimensioned plane in the figure — is used per the exam's own Note 1 to state assumptions where doubt exists). Resolve the 25 kN horizontal reaction into the brace axis, then size an SPF tension member using Table 6.3.1A.

1. Brace geometry and axial force. The brace spans the full wall rectangle: $$L_{brace}=\sqrt{16.0^2+3.6^2}=16.4\text{ m}, \qquad \cos\theta=\frac{16.0}{16.4}=0.976$$ For a strut/tie carrying the entire 25 kN horizontal reaction along its own axis: $$\boxed{F_{brace}=\frac{V_f}{\cos\theta}=\frac{25.0}{0.976}=25.6\text{ kN}}$$
2. Size the brace member (tension). Try a single-ply 38 × 184 SPF No.1/No.2 ($f_t=5.8$ MPa, Table 6.3.1A), net of a 3/4" bolted end connection (21 mm net width deduction): $$A_n = 38(184)-38(21)=6194\text{ mm}^2, \qquad T_r=\phi_t f_t A_n=0.9(5.8)(6194)\times10^{-3}=32.3\text{ kN}$$ $$T_r=32.3\text{ kN} \ge F_{brace}=25.6\text{ kN}\quad\checkmark\ (\text{27\% reserve})$$ A smaller 38 × 140 ply gives only $T_r=23.6$ kN, which would not be adequate — 38 × 184 is the minimum standard size that works.