Question 4 of 6: Lateral Force Resisting Systems and Shearwall Design

Question 4: Lateral Force Resisting Systems and Shearwall Design (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.

4a) Three lateral force resisting systems

1) Plywood/OSB shearwalls with a horizontal diaphragm. The roof (or a horizontal bracing plane) acts as a deep, thin horizontal beam — a "diaphragm" — spanning between the two end walls, collecting wind pressure on the long side walls and delivering it as in-plane shear into vertical sheathed shearwall panels at each end, which then carry it down to the foundation. Magnitude: the wind pressure on the long-wall tributary area (NBCC Cl. 4.1.7, external pressure coefficient $C_p$ times reference velocity pressure $q$, factored) is summed over the barn's tributary half-height and half-length to get a total diaphragm shear, then split between the two end shearwalls in proportion to their stiffness/tributary length — this is exactly the design load used in Question 4b.

2) Braced frames (diagonal X- or K-bracing). Diagonal steel rods, cables or timber braces are added within selected structural bays (e.g. between two columns of Figure 1's side-wall frames), converting the lateral load path into pure axial tension/compression in the braces rather than relying on sheathing shear capacity. Magnitude: the same tributary wind pressure is collected by the wall girts and columns into a point load at each brace-bay eave, and the brace axial force follows directly from the geometry, $F_{brace}=V/\cos\theta$, where $\theta$ is the brace angle from horizontal and $V$ is the storey shear carried by that bay.

3) Moment-resisting (rigid) frames. The side-wall columns and roof trusses (Figure 1's Q3/Q4 elements) are connected with moment-resisting (rigid) connections at the knee and ridge, so the frame resists lateral load through flexure and rigid-joint continuity rather than through sheathing or bracing — useful where door/window openings prevent continuous shearwall or braced-bay coverage. Magnitude: the total wind shear on the frame's tributary width is applied as a lateral point load at eave height, and each frame's share is found by distributing the total storey shear among the parallel frames in proportion to relative lateral stiffness (a portal-frame or matrix stiffness analysis).

4b) End-wall shearwall design

Given.

Find. The required unit shear resistance and the resulting hold-down (chord) tension/compression force at each end of the shearwall.

Approach. Treat the full 12 m end wall as one shearwall segment, find the unit (per-metre) shear demand on the sheathing, select a standard nailed sheathing schedule with adequate factored unit shear resistance, then find the overturning chord force from the wall's height-to-length aspect ratio.

1. Unit shear demand. $$v_f=\frac{V_f}{L}=\frac{16.0}{12.0}=\boxed{1.33\ \text{kN/m}}$$
2. Select sheathing/nailing. This demand is far below the factored unit shear resistance of even a light standard nailing schedule (CSA O86 Table 9.5.2 gives roughly 5–7 kN/m for 9.5 mm OSB/plywood with 8d common nails at 150 mm o.c. panel edges, 300 mm o.c. field, applied single-sided) — specify that minimum standard schedule along the full 12 m wall length; it provides roughly 4–5 times the required capacity.
3. Overturning (chord) force. Treating the wall as a vertical cantilever shear panel, the couple resisted by tension/compression chord posts at each end is $$T=C=\frac{V_f\,h}{L}=\frac{16.0(3.6)}{12.0}=\boxed{4.8\ \text{kN}}$$ Provide a mechanical hold-down connector at each end-wall chord post sized for 4.8 kN tension, anchored through to the foundation.