How to Design a Wood Shear Wall — A Complete Step-by-Step Guide (SDPWS 2021 & ASCE 7-22)

A beam tells you when it is undersized — it sags, it cracks a ceiling, somebody calls. A shear wall keeps its opinion to itself until the one afternoon the design earthquake or the design windstorm arrives, and then the whole lateral system gets graded at once: the sheathing nails, the hold-downs, the sill bolts, the little framing clips nobody drew. A shear wall is not a member; it is a chain of five connections wrapped around a piece of plywood, and the chain is graded by its weakest link.

This guide designs that whole chain, start to finish, on one real building — and like every honest design story, the first trial fails. The wall line everyone would sketch first comes out at a demand/capacity ratio of 196%, and the obvious fix (nail it harder) walks you straight into a code clause that changes the framing under the nails. You will see the three levers available — tighter nails, a thicker panel, more wall — what each one buys, what each one costs, and why the cheapest fix is usually the one drawn last.

Along the way this page carries the lesson most shear wall tutorials skip entirely: wind and seismic do not distribute to the same walls in the same proportions. On this building, the garage piers pass their seismic check at 88% — and fail their wind check at 103%. Size them from a single "share of V" and you ship that failure.

By the end you will have: the finished shear wall schedule for all three lines, every check with its substituted formula and code citation, the hold-down and end-post selections, the complete base anchorage (bolts on concrete, clips over framing), the collector (drag strut) force diagram with its strapping, the drift calculation, the construction mistakes that undo all of it on site, and a full hand-calculation verification table. Every number below was produced by the calculation engine that runs StructSuite's Wood-Frame Shear Wall module and locked by automated regression tests (the three drift terms of Step 12, which the code leaves with the engineer, are computed by hand and locked by the same suite); the hand-verification is at the end of the page.

#The building we are going to design

This shear wall design example runs on a single-story wood-framed house on an L-shaped plan: a 60 × 30 ft main body and a 30 × 20 ft wing, walls 9 ft tall, in Seismic Design Category D (S_DS = 1.0 g; redundancy ρ = 1.0 in the direction we design — a result shown in Step 1, not a default). The lateral analysis — which you can reproduce with our seismic base shear guide and wind load guide — delivered, per principal direction:

Behind that E sits a seismic weight of W = 162,500 lb — E/C_s with C_s = S_DS/(R/I_e) = 1.0/6.5 for light-frame wood shear walls — and for a one-story building that is also w_px, the weight tributary to the roof diaphragm, which Step 11 will need. The walls carry a light roof: D = 400 plf of dead load and L_r = 200 plf of roof live load land on top of every shear wall line. The east–west direction is the one we will design — three wall lines resist it:

• Line 1 — the back wall of the body (on the concrete stem wall);
• Line 2 — the wall at the step between body and wing;
• Line 3 — the front of the wing: two narrow piers beside the garage door, sitting on the framed floor of a split-level (this detail will matter twice).

Sheathing everywhere is the most ordinary product in the industry: 7/16″ wood structural panel (OSB) sheathing with 8d common nails — Table 4.3A's bread-and-butter row - on studs at 16″ o.c. The design questions are the ones every job has: how much wall does each line need, at what nail spacing, held down by what, bolted with what?

#How a shear wall works — the load path is the design

Lateral force enters at the roof. The roof deck — the diaphragm — spans horizontally like a wide, flat beam between the shear wall lines, and each line receives a share of the story force V as a horizontal reaction at its top. From there every wall in the line does the same four jobs at once:

1. Shear — the panel-and-nails assembly drags the force down to the base; its currency is unit shear, v = V/b (plf).
2. Overturning — V at the top of a wall of height h creates a moment V·h that tries to tip the wall; a hold-down at the uplift end and an end post at the compression end — the wall's tension and compression chords — resist the couple.
3. Base shear transfer — the sill or bottom plate must hand the same v to whatever is below: anchor bolts into concrete, or framing clips into floor framing.
4. Gravity — the same wall is usually also a bearing wall; the dead load helps overturning and punishes the compression post.

Each of those four jobs is a numbered step below, and each one has its own code home. That map is worth pinning to the wall:

One basis note before any numbers, because it silently controls everything: this design runs in ASD. ASCE 7-22 wind and seismic forces arrive at strength level; the allowable-stress combinations scale them — wind enters at 0.6W, seismic at 0.7E — and SDPWS hands you nominal sheathing capacities meant to be divided by 2.0 for wind and 2.8 for seismic (§4.1.4.1, §4.1.4.2, "no further increases shall be permitted"). Demand and capacity must land on the same side of that conversion. Every mismatched shear wall calc ever red-lined got this pairing wrong somewhere.

#Choosing the input parameters — where each one comes from

No number in a shear wall design is invented on this page, and none should be on yours:

#Step 1 — The story forces, and what "E" and "W" mean here

From the lateral analysis: E = 25,000 lb and W = 30,000 lb per direction, strength level, applied at the roof line. Three things ride along with them:

• ρ (redundancy): seismic combinations use E_h = ρ·E, and in Seismic Design Category D the standard starts at ρ = 1.3 — ASCE 7-22 §12.3.4.2 permits 1.0 only where one of two conditions is shown, direction by direction. This L-shaped plan fails the second condition outright: both plan projections beyond the reentrant corner exceed 20% of the plan dimension (30 of 60 ft, 20 of 50 ft), a Type 2 horizontal irregularity of Table 12.3-1. So ρ = 1.0 east–west rests on the first condition, the notional-removal check of Table 12.3-3, applied to the one story (which carries 100% of the base shear):
  • (a) at least two bays of seismic force-resisting framing on each side of the center of mass. Table 12.3-3 counts a shear wall bay as the wall length divided by the story height, rounded down. The center of mass sits at y = 21.25 ft: line 1's 22 ft of wall gives 22/9 → 2 bays south of it; line 2's 20 ft gives 2 bays north of it, and line 3's 4-ft piers add none. Two each side — exactly enough, and only on the reading this page uses: each line's sheathed length taken as the wall, on the intact system. Counted segment by segment (8/9, 8/9, 6/9 → 0) this plan would have no bay at all, and ρ would be 1.3 — the FAQ carries that case.
  • (b) removing any one wall or pier with h/b > 1.0, or its connections, reduces the story's lateral strength by no more than 35%. Counting strength as v_n·b at each line's final nailing (line 1 at 4″: 980 × 22 = 21,560; line 2 at 3″: 1,260 × 20 = 25,200; line 3's piers at their aspect-ratio factor: 1,260 × 0.889 × 8 = 8,960; total 55,720 plf·ft), the largest removable element is one of line 2's 8-ft walls at 10,080 — an 18% reduction. Line 1's 8-ft wall is 14%, its 6-ft wall 11%, a garage pier 8%. The 12-ft wall is not a candidate (h/b = 0.75).
  • (c) the system left after the removal has no Type 1 torsional irregularity with TIR > 1.4. §12.3.2.1.1 is explicit that the TIR "shall not apply to structures with flexible diaphragms", and a wood structural panel roof over light-frame walls is one. So ρ = 1.0 east–west is a demonstrated result. Had it come out 1.3, every seismic line shear on this page would scale by 1.3 — the chord forces by less, since the resisting moment does not move — line 1 would need 3″ nailing and line 2's 12 + 8 ft at 3″ would fail (the FAQ carries that case). StructSuite asks for ρ once per direction in Step 2 and withholds the seismic combinations until you choose, because 1.0 is a judgment the module cannot make for you.
• E_v (vertical seismic): ±0.2·S_DS·D accompanies E_h in the §2.4.5 combinations. It looks like a rounding detail until it reaches the hold-downs: in combination 10 it subtracts from the dead load that resists uplift.
• Wind is no afterthought here: at ASD level, 0.6W = 18,000 lb against 0.7E = 17,500 lb — near parity before distribution even starts. Which case governs each line depends on how the two distribute — that is Step 2, and it is the heart of this guide.

#Step 2 — Distribute the story force to the wall lines (both cases, always)

For wood diaphragms the standard idealization is the flexible diaphragm: the roof spans like a simple beam between wall lines, and each line collects the load from the strip of building tributary to it — halfway to the neighboring line on each side. That much is familiar. What surprises engineers on irregular plans is that wind and seismic do not split the same way across the same walls, because they are different physics:

• Seismic force is inertia. It is generated wherever there is mass, and for a uniform roof the mass in a line's tributary strip is proportional to the strip's area.
• Wind force is pressure on the projected face of the building. Along the span it arrives per foot of building depth, so a line's share is proportional to its tributary length of the span — the plan being locally wide adds mass, but it adds no wind.

On a clean rectangle the two measures agree and nobody notices. On our L-shape they diverge, and one line will fail because of it. The tributary boundaries fall midway between lines: at y = 15 ft and y = 40 ft.

Seismic — share by tributary area (uniform roof mass):

Wind — share by tributary length of the 50-ft span (pressure on the projected face arrives per foot of depth; local plan width is irrelevant):

Same walls, same story — two different splits. Look at line 3: 12.5% of the seismic force but 20% of the wind — a 60% larger share of the wind case, because the wing is shallow in plan: little mass, but its full slice of the projected face. Line 1 leans the other way. Hold that thought; it detonates in Step 6.

Both distributions above (and every share in this guide) come from StructSuite's distribution engine, which always computes both cases; the within-line and per-case arithmetic is locked by the same regression pins as this page. In the module this is Step 1: a grid at every line of resistance, corner and wall end (only the main grids named), the footprint checked bay by bay on the plan, and the shear walls checked on the plan as the grid segments between wall ends. Two or more lines in a direction and the story force distributes exactly as above; a single line takes 100% of it, with no footprint asked for. Each line is then designed in turn from the Line dropdown — its gravity loads and sheathing — and each wall's hold-down, anchor bolts and end post from Step 4's Shear wall dropdown, either copied to every line or every wall in one click.

#Step 3 — Choose the walls on each line, and respect the aspect ratio law

A wall line is rarely one long wall — it is the segments left over between windows and doors. Each full-height segment (SDPWS §4.3.2.1) works as its own shear wall of length b and height h, and the first legality check is pure geometry, from SDPWS 2021 Table 4.3.3 — Maximum Shear Wall Aspect Ratios:

with the table's own footnote 1 — attached to its gypsum wallboard and portland cement plaster rows: "Walls having aspect ratios exceeding 1.5:1 shall be blocked shear walls." At h = 9 ft, a blocked WSP segment may be as narrow as 9/3.5 = 2 ft 7 in. — but "may" is doing heavy lifting, because narrowness is taxed twice:

• §4.3.3.2 — for WSP walls with h/b > 2:1, multiply the nominal capacity by the Aspect Ratio Factor = 1.25 − 0.125·h/b;
• §4.3.5.5.1, Exception 1 — when segments of unequal stiffness share a line by the capacity-proportional simplification, a wall with h/b > 2:1 contributes with a 2b/h reduction instead. The engine computes both and takes whichever governs.

Our first-pass elevations, straight off the architect's plan:

Line 3's piers at h/b = 9/4 = 2.25 carry both penalties: Aspect Ratio Factor = 1.25 − 0.125(2.25) = 0.969, and Exception 1's 2b/h = 2(4)/9 = 0.889 — the smaller one, 0.889, governs. Every plf of tabulated capacity on those piers is worth 89 cents on the dollar before we start.

#Step 4 — Sheathing capacity: reading Table 4.3A like the engine does

SDPWS Table 4.3A tabulates nominal unit shear capacities v_n for wood-based panels, by panel type, thickness, nail, and edge-nail spacing (6″, 4″, 3″, 2″). Our row — 7/16″ WSP Sheathing, 8d common (2½″ × 0.131″) nails, 1⅜″ minimum nail bearing length in the framing (the table's ℓ_m column):

Two conversions stand between that row and a check:

Nominal → ASD (§4.1.4). Divide by 2.0 for wind (§4.1.4.2) or 2.8 for seismic (§4.1.4.1) — "no further increases shall be permitted", which retires the old habit of a 1.33 or 1.4 duration bump. At 4″ spacing: v_ASD,seismic = 980/2.8 = 350 plf, v_ASD,wind = 980/2.0 = 490 plf. Notice wind gets 40% more capacity from the same nails — remember that asymmetry when the two demands are close.

Per-wall, then per-line. Each segment's ASD capacity is (v_n/Ω)·(aspect factor)·b, and the line's capacity is the sum. The engine distributes the line force to segments in proportion to their design capacities (the §4.3.5.5.1 Exception 1 simplification) — so equal-height, equal-construction segments all run at the same unit shear, and a factored narrow pier automatically takes less.

Three footnotes on Table 4.3A earn their keep so often the module carries each as its own control in Step 3 — footnotes 2 and 10 as checkboxes that are live only when the chosen row carries that superscript, footnote 3 as a checkbox that opens the framing species dropdown that sets G — and names every one it applied in the step summary (footnote-apply standard):

• Footnote 2: the 3/8″ and 7/16″ rows may claim the 15/32″ row's values with the same nailing, provided studs are at a maximum of 16″ o.c. or the panels run long-dimension across the studs. Our wall qualifies at 16″ studs — the upgrade (980 → 1,065 plf at 4″, the 15/32″ Sheathing row with the same 8d nails) stays deliberately unclaimed, as margin.
• Footnote 3: framing other than Douglas Fir-Larch or Southern Pine multiplies v_n by the Specific Gravity Adjustment Factor = 1 − (0.5 − G) ≤ 1.0 — SPF framing (G = 0.42) knocks 8% off every number in this guide.
• Footnote 10: 10d rows take a 0.92 factor when the hold-down sits on the inside face of the end post.

We claim none of them here — DF framing keeps footnote 3 away, and footnote 2's upgrade stays in the back pocket — which is exactly why the trial that follows is as clean a failure as you will ever get.

#Sheathing both sides: the capacity doubles, and nothing else does

A wall can be sheathed on opposing sides of common framing — the same studs carry a panel on the outside face and another on the inside face. SDPWS §4.3.5.4.1 gives the combined capacity of similar sheathing systems as two equations, and this is the place where engineering intuition is reliably wrong:

G_a(c) = G_a(1) + G_a(2)    (4.3-3)

v_n(c) = K_min · G_a(c)    (4.3-4)    where K_min is the smaller of v_n(1)/G_a(1) and v_n(2)/G_a(2)

The stiffnesses add. The capacity is then governed by whichever side is less efficient at converting stiffness into strength. Work it through for two identical sides — our 7/16″ WSP at 6″ on both faces, v_n = 670 plf and G_a = 15 kips/in each:

K_min = min(670/15, 670/15) = 44.67  →  v_n(c) = 44.67 × 30 = 1,340 plf = 2 × 670

G_a cancels, and the answer is exactly twice one side. That is the only case where adding the two sides is correct. Put 8d @ 6″ on one face (670 plf, G_a = 15) and 8d @ 4″ on the other (980 plf, G_a = 22) and the sum would say 1,650 plf, while Eq. (4.3-4) says K_min = min(44.67, 44.55) = 44.55, G_a(c) = 37, v_n(c) = 1,648 plf — close here, but the gap opens as the two sides diverge, and it opens on the unconservative side. Mismatch the systems rather than the spacing and you leave §4.3.5.4.1 altogether for §4.3.5.4.2, where the answer is the greater of twice the smaller side and the larger side alone — a 730 plf panel against 200 plf gypsum is worth 730 plf for seismic, not 930. (For wind only, §4.3.5.4.2 has an exception that does permit wood structural panels, hardboard siding or structural fiberboard on one side to be summed with gypsum on the other. Taking that sum for seismic is the single most common two-sided mistake.)

StructSuite offers the two-sided option only with the same Table 4.3A row at the same nail spacing on both faces — the case above where G_a cancels. That restriction is not squeamishness about the arithmetic: dissimilar sides are a §4.3.5.4.2 wall, and §4.3.5.5.1 Exception 1 (distributing a line's shear in proportion to capacity, which every line in this guide relies on) names §4.3.5.4.1 only. A §4.3.5.4.2 wall has to be distributed by equal calculated deflection instead.

Two consequences worth carrying into Step 5.

Table 4.3A footnote 6 turns on by itself. "Where panels are applied on both faces of a shear wall and nail spacing is less than 6″ on center on either side, panel joints shall be offset to fall on different framing members. Alternatively, the width of the nailed face of framing members shall be 3″ nominal or greater at adjoining panel edges and nails at all panel edges shall be staggered." At 6″ o.c. it does not apply. Tighten either face to 4″ and it does — and then a two-sided wall carries the same 3x-or-offset-joints detailing that Step 5 treats as the hidden cost of nailing harder. The module raises the note the moment both conditions are met.

The 980 plf threshold does not read the combined value. §4.3.7.1(5)(c) is written on the nominal unit shear capacity "on either side of the shear wall". Two 670 plf faces give v_n(c) = 1,340 plf and still do not trip it, because neither side exceeds 980. Items (5)(a) and (5)(b) read nail spacing and nail size, which the second face does not change either. Sheathing both sides is the one capacity lever in this chapter that can leave the framing alone. The module tests the per-side value for exactly this reason, so checking the two-sided box never invents a 3x-framing requirement that the code does not ask for.

#Step 5 — Design line 1: the trial that fails, and the three levers

Line 1's governing ASD demand is seismic — combination 8. D + 0.7E_v + 0.7E_h (ASCE 7-22 §2.4.5):

V_ASD = 0.7 × 9,375 = 6,562.5 lb, spread over Σb = 14 ft → v = 468.8 plf of demand.

Capacity at each nail spacing is (v_n/2.8) × Σb, and here is the whole trial in one table — the demand never moves; only the nails do:

The 6″ row is the one everyone sketches first, and it is not even close: 196%. Tightening nails looks like it works — 2″ spacing passes at 80% — but read the fine print before celebrating. SDPWS §4.3.7.1(5) requires the framing at adjoining panel edges to be 3x nominal (or two 2x stitched together) with staggered nailing when any of these is true: (a) edge nailing at 2″ o.c., (b) 10d common nails at 3″ o.c. or closer, or (c) nominal unit shear capacity on either side of the wall over 980 plf in Seismic Design Category D, E or F. Our 2″ trial trips (a) and (c). So does the "upgrade the hardware" lever: 15/32″ panels with 10d nails at 3″ o.c. reaches v_n = 1,680 → 600 plf ASD → D/C 78% — and trips (b) and (c). In SDC D, every nail-side path out of this failure converts the wall's studs at panel joints from 2x to 3x, with the plate splices, the special-inspection attention and the subcontractor phone calls that follow.

The lever study. Three ways out, priced honestly:

Lever D is the one that surprises people, and the reason it is not the answer here is instructive. At 6″ on both faces the line capacity is (1,340/2.8) × 14 = 478.6 × 14 = 6,700 lb, a D/C of 98% — it passes, on ordinary 2x framing, without touching the floor plan. What it does not do is reduce the demand. Capacity is what doubles; the applied shear is still 6,562.5 lb over 14 ft, so the unit shear stays at 468.8 plf and the chord tension at each segment end stays at 468.8 × 9 = 4,219 lb gross — against 2,685 lb for Lever C's 22 ft line. Lever D buys sheathing capacity and pays for it at the hold-downs, the end posts and the base anchorage. That is the general shape of it: two-sided sheathing solves a sheathing problem, and only a sheathing problem.

Lever C wins, and not just on cost: at 4″ spacing v_n = 980 plf sits exactly at the 980 threshold — "exceeds 980" it does not — so the whole line stays on ordinary 2x framing. That is not luck; it is why the 4″/980 row is the workhorse of high-seismic residential design. The cheapest shear wall fix is almost always length, not hardware. Fight for wall in the floor plan before you fight physics with nails.

Final line 1: three segments, 8 + 8 + 6 ft (Σb = 22 ft), 7/16″ WSP, 8d @ 4″/12″.

(The 12″ field spacing leans on SDPWS §4.3.7.1(3)'s exception: intermediate-framing nailing may relax from 6″ to 12″ o.c. only where panels are thicker than 7/16″ or studs are closer than 24″ o.c. — our 16″ studs qualify; a 24″-o.c. wall would not.)

• Capacity: (980/2.8) × 22 = 350 × 22 = 7,700 lb → D/C = 6,562.5/7,700 = 85% ✓ (seismic governs; wind runs at 50%)
• Demand unit shear: v = 6,562.5/22 = 298.3 plf

#Step 6 — Line 3: the wind trap, sprung

Now the wing. Its seismic share was small — 12.5%, an ASD demand of 0.7 × 3,125 = 2,187.5 lb. Its wind share was not — 20%, an ASD demand of 0.6 × 6,000 = 3,600 lb. Run the two 4-ft piers (with their 0.889 aspect factor) through both cases at the spacing the seismic number suggests:

At 4″ spacing — capacity per case, Σ(v_n/Ω)·0.889·b over both piers:

• Seismic: (980/2.8) × (8/9) × 8 = 2,488.9 lb → D/C = 2,187.5/2,488.9 = 88% ✓ passes
• Wind: (980/2.0) × (8/9) × 8 = 3,484.4 lb → D/C = 3,600/3,484.4 = 103% ✗ FAILS

There it is. Sized by its seismic share alone, the garage front passes at 88% and ships. The wind share fails it at 103%. Nothing exotic happened — the wing is shallow (small mass, small E) but stands its full slice of the projected face (full W), and even wind's friendlier Ω = 2.0 cannot cover a 65% larger force on 11% less capacity. Any workflow that distributes one "share of V" — a spreadsheet with one tributary column, a rigid habit of "seismic always governs in SDC D" — ships this wall.

At 3″ spacing both cases clear comfortably:

• Wind: (1,260/2.0) × (8/9) × 8 = 4,480 lb → D/C = 3,600/4,480 = 80% ✓
• Seismic: (1,260/2.8) × (8/9) × 8 = 3,200 lb → D/C = 2,187.5/3,200 = 68% ✓

So line 3 ships at 8d @ 3″ — and pays two taxes that belong in the drawing notes: v_n = 1,260 > 980 plf in SDC D triggers §4.3.7.1(5c) 3x framing at adjoining panel edges, with nailing staggered at all panel edges (the abutting stud between panels and the blocked horizontal joint at the 8-ft panel break), and the narrow piers keep their 0.889 aspect-ratio haircut forever.

#Step 7 — Line 2, and the finished shear wall schedule

Line 2 carries the largest forces of all (E_line = 12,500, W_line = 15,000 lb) on 20 ft of wall. At 4″ the seismic case fails (125%); at 3″ it works: capacity 450 × 20 = 9,000 lb against 0.7 × 12,500 = 8,750 lb → D/C 97% — tight, deliberate, and legal. (Wind: 71%.) Being over v_n = 980, line 2 also joins the 3x-framing club.

The line-by-line result — this is the table that becomes the drawing's shear wall schedule: the nailing schedule, the governing case, and the special-framing flags for all three lines in one place:

Within each line, the engine splits the line force among segments in proportion to design capacity (§4.3.5.5.1 Exc. 1): on line 1 that is 2,386 lb to each 8-ft wall and 1,790 lb to the 6-ft wall — all three at the same 298.3 plf, which is what "capacity-proportional" buys you. Those per-wall shears are the inputs to everything below the sheathing: hold-downs, posts, and the base.

#Step 8 — Hold-down design: overturning, and where 0.6D and E_v earn their keep

Each wall's shear V acts at the top; V·h wants to rotate the wall out of its own footprint. What resists is the wall's tributary dead load — and the code deliberately shrinks it. The uplift check runs under combination 10: 0.6D − 0.7E_v + 0.7E_h (ASCE 7-22 §2.4.5): only 60% of the dead load may help, and the vertical seismic component E_v = 0.2·S_DS·D takes another bite — at S_DS = 1.0 the net multiplier is 0.6 − 0.7(0.2)(1.0) = 0.46D.

One thing the engine does that the textbook sketch skips: the moment arm. SDPWS 2021 §4.3.6.1.2 writes the chord forces as T = C = V·h/b_eff (Eq. 4.3-7), with b_eff the "effective length (e.g. moment arm) for calculation of T and C force" — not the wall length. The tension acts at the hold-down's anchor rod and the compression at the centroid of the end post, and both sit inside the wall ends. With (2) 2×4 chords (3″ along the wall) and a device whose catalog C_L dimension puts the rod C_L from the post face, b_eff = b − (3″ + C_L) − 1½″. On a 6-ft wall with an HDUE3-SDS3 (C_L = 1⅜″) that is 6 − 4.375/12 − 0.125 = 5.510 ft, 8% shorter than b — and the resisting dead load, acting at mid-length, is taken about the compression centroid, 3 − 0.125 = 2.875 ft away.

Worked for line 1's 6-ft wall (its 298.3 plf × 6 ft = 1,789.8 lb share):

• Overturning: M_OT = V·h = 1,789.8 × 9 ≈ 16,108 lb-ft
• Resisting: the wall's dead load W_D = 400 × 6 = 2,400 lb at mid-length → M_R = 0.46 × 2,400 × 2.875 = 3,174 lb-ft
• Hold-down tension: T = (M_OT − M_R)/b_eff = (16,108 − 3,174)/5.510 = 2,347.2 lb

The compression end is checked under the heaviest vertical case — combination 8 (D + 0.7E_v + 0.7E_h, net 1.14D) — taking moments about the rod, so the dead load's arm to the rod is 3 − 0.365 = 2.635 ft:

• C = M_OT/b_eff + 1.14·W_D·2.635/b_eff = 2,923.2 + 1,308.5 = 4,231.7 lb

Across the line, the 8-ft walls come out at T = 2,100 lb and C = 4,625 lb over their 7.51-ft arm (their bigger V, but also more helping dead load). The worst hold-down demand on the line is the 6-ft wall's 2,347 lb — narrow walls punch above their weight on uplift, which is why hold-down schedules are set by the shortest segment, not the longest.

Selecting the hardware. Hold-down capacities are manufacturer allowables, by species column — and because the device sets where its rod sits, the demand is recomputed for each candidate's C_L. The lightest 3×-member row, a DTT2Z (2,145 lb DF/SP, (8) ¼″ × 1½″ S_DS screws), puts its rod 13/16″ from the post: b_eff = 5.56 ft, T = 2,327 lb, D/C 109% — it does not fit, and what sinks it is the moment arm, not the load: the textbook T over the full 6 ft would read 2,132.7 lb and a legal-looking 99%. One step up, an HDUE3-SDS3 — 3,790 lb (DF/SP), (7) ¼″ × 3″ S_DS screws on a 5/8″ anchor bolt — carries 2,347 lb at 62%, with room for the day the architect widens that window. Note what the catalog is telling you in the fine print: the rated deflection at the allowable load (0.127″ for the HDUE3-SDS3) is not trivia — it feeds Δ_a in the drift equation of Step 12, and hold-down stretch is routinely the largest single term in wood shear wall drift.

Every segment end gets a hold-down in this design — SDPWS §4.3.6.4.2 requires uplift anchorage wherever the dead-load stabilizing moment cannot cover the overturning, and at 0.46D almost nothing residential covers it.

#Step 9 — The end post (compression chord): the boring check that fails first on heavy walls

The compression end of each wall stacks C = M_OT/b_eff + the gravity reaction at the chord into a stud pack. For line 1 we try the default: (2) 2×4 DF No. 2 — F_c = 1,350 psi, C_F = 1.15, E_min = 580,000 psi, F_c⊥ = 625 psi (NDS Supplement Table 4A), A = 2 × 5.25 = 10.5 in² — braced by sheathing in-plane, buckling out-of-plane over l_e = 9 ft − plates = 103.5″, l_e/d = 29.6:

• Column capacity (NDS 3.7, C_P = 0.208 at this slenderness): P_allow = 5,438 lb
• Plate bearing (NDS 3.10, F_c⊥, no C_D): 6,563 lb
• Demand: worst C = 4,625 lb (the 8-ft walls under combination 8, over b_eff = 7.51 ft) → D/C = 85% ✓ — column buckling governs, as it usually does at 9-ft walls.

Two habits worth stealing: check bearing and buckling every time (they trade places as h changes — bearing wins on short heavy walls, buckling on tall ones), and remember C_D = 1.6 applies to F_c but never to F_c⊥ (NDS Table 2.3.2). The full stability math — C_P, F_cE, the works — is the same machinery as our wood column guide; in the module it is Substep 4C, checked per wall in Step 6.

#If the wall does not continue below, the member under it is a different design

Every wall in this example runs to the foundation, so its end post lands on concrete and the check above is the whole story. Change one thing — let a shear wall stop at the floor below, over a garage opening or offset from the wall line beneath it — and the post or header carrying it leaves the scope of Step 9 entirely.

ASCE 7-22 §12.3.3.4 is the provision: "Structural elements supporting discontinuous walls or frames of structures that have horizontal irregularity Type 4 of Table 12.3-1 or vertical irregularity Type 3 of Table 12.3-2 shall be designed to resist the seismic load effects, including overstrength of Section 12.4.3." The supporting member is designed for E_mh = Ω₀Q_E (Eq. 12.4-7) rather than Q_E. With Ω₀ = 3 for light-frame wood shear walls (Table 12.2-1) — 2½ where the diaphragm is flexible, by the table's footnote b — the seismic part of that member's demand roughly triples — and it is almost always what governs it.

Three things that catch people:

• ρ does not also apply. §12.3.4.1 item 6 permits ρ = 1.0 for members where the overstrength effects are required, so ρ and Ω₀ are alternatives, never factors that stack. The transfer post takes 3.0, not 3.0 × 1.3.
• Ω₀ belongs to the system being supported. Read it from the Table 12.2-1 row for the wall above, not for whatever frames into the post from the side.
• The section number moved. This is §12.3.3.4 in ASCE 7-22; it was §12.3.3.3 in ASCE 7-16, and §12.3.3.3 in 7-22 is now Extreme Weak Stories. Citing the old number on a 7-22 job points a plan checker at the wrong rule.

In StructSuite this is a Step 2 option in the beam and column modules — "supports a discontinuous wall or frame" — with Ω₀ entered below it once ticked. It is off unless you assert it, because nothing in a single member's geometry reveals whether the wall above lands on the wall below; and Ω₀ is never assumed, because Table 12.2-1 gives it by system — 3 for these walls, 2½ with the flexible-diaphragm footnote, other values for other systems. The shear wall module itself is unaffected: the wall above is designed exactly as this guide designs it. What changes is the member underneath.

#Step 10 — Base shear transfer: the check most drawings skip

Everything so far moved the force to the bottom plate. It still has to leave the wall. SDPWS §4.3.6.4.1 says it plainly: "Connections shall be provided to transfer the unit shear force induced by the design load, v, into and out of each shear wall." Two substrates, two solutions:

Line 1 sits on concrete — sill anchor bolts. The wood side of a bolt in a DF sill comes from NDS 2024 Table 12E (single shear, sawn lumber to concrete, 6″ embedment assumed, f′_c ≥ 2,500 psi): a 5/8″ bolt in a 1½″ sill gives Z_∥ = 930 lb. Adjusted (NDS Table 11.3.1): Z′ = Z_∥ × C_D × C_M = 930 × 1.6 × 1.0 = 1,488 lb per bolt (C_g = 1.0 — each bolt carries its own tributary length of a distributed shear; C_Δ = 1.0, which the module takes and which assumes the full NDS end distance for a bolt bearing toward the end of the sill, 7D = 4⅜″ for 5/8″ — a bolt set 4″ from the end of the sill would take C_Δ = 4/4.375 = 0.91, so hold the end bolts at 4⅜″ or more and the number stands).

At 48″ o.c.: demand per bolt = v × s = 298.3 × 48/12 = 1,193.2 lb → D/C = 80% ✓ (spacing could legally stretch to 59.9″, but 48″ ≤ the IBC 6-ft maximum with margin, and even spacing lays out cleanly on 8-ft walls). The prescriptive floor rides along regardless of calculation — IBC 2024 §2308.7.1: ½″ minimum diameter, 7″ embedment, two bolts per plate piece, one 4–12″ from each end — and SDPWS §4.3.6.4.3 adds the seismic-country signature: a 0.229″ × 3″ × 3″ steel plate washer under every nut, extending to within ½″ of the sheathed edge. Those washers are not bureaucracy; they are what stops the bottom plate from splitting in cross-grain bending before the bolts ever yield.

Line 3 sits on floor framing — framing clips. No toe-nail shortcut is available even if you wanted it: the piers' seismic unit shear is 273.4 plf, and SDPWS §4.1.10 bars toe-nailed connections above 150 plf (ASD) in SDC D, E and F. A mechanical clip is the code path. The governing unit shear is wind's 450 plf, and a Simpson A35 in its plates-to-rim configuration carries 650 lb in the catalog's (160) column — one column that serves wind and seismic alike, "no further increase allowed". At 16″ o.c.: demand per clip = 450 × 16/12 = 600 lb → D/C = 92% ✓ (required spacing ≤ 17.3″).

Behind these checks sit the fully digitized tables — NDS Table 12E cell-verified against the book, and the C-C-2026 clip rows with their footnotes verbatim. The base-transfer step built on them runs in development preview while its interface is finished; in the production module today, Step 4's hold-down and anchor-tension checks carry the same verbatim-source discipline.

#Step 11 — The collector (drag strut): the element nobody draws

Look again at line 1's real elevation. The diaphragm hands its shear to the whole 60 ft of that wall line, but only 22 ft of it is sheathed wall — the rest is 15 ft of glazing at each end and two 4 ft windows between the segments. (Step 5's elevation drew the sheathed segments; this is the entire line.) Something has to carry the shear from where the diaphragm delivers it to where the walls can take it. That element is the collector — a drag strut when it drags force into a wall — and in a wood building it is almost always the double top plate, which is why it is so easy to design by accident.

The code is blunt about it. ASCE 7-22 §12.10.2: "Collector elements shall be provided that are capable of transferring the seismic forces originating in other portions of the structure to the element providing the resistance to those forces." Not "consider" — provided.

#First, the force the collector is designed for — and it is not the line shear

The wall checks above ran on line 1's share of the story force, 0.7 × 9,375 = 6,562.5 lb at ASD. The collector does not. For a structure braced entirely by wood light-frame shear walls, §12.10.2.1's exception (quoted below) sends the collector to the diaphragm design force of §12.10.1.1, and that force has a floor the story force usually sits under. Eq. (12.10-1) gives F_px = (ΣF_i/Σw_i)·w_px, which for one story is simply the story force, 25,000 lb. Eq. (12.10-2) then requires F_px ≥ 0.2·S_DS·I_e·w_px = 0.2 × 1.0 × 1.0 × 162,500 = 32,500 lb, and Eq. (12.10-3) caps it at 0.4·S_DS·I_e·w_px = 65,000 lb. The floor governs — as it does for every one-story wood-structural-panel light-frame building (R = 6.5) on the plateau of the spectrum, where C_s = S_DS/6.5 = 0.154 can never reach 0.2·S_DS. Three more things ride on that number:

• ρ = 1.0 here regardless of Step 1. §12.3.4.1 item 7 permits ρ = 1.0 for diaphragm forces determined by Eq. (12.10-1), including its limits.
• Line 1 takes its tributary share, the same 37.5% the story force split by in Step 2: 0.375 × 32,500 = 12,187.5 lb.
• §12.3.3.5 adds 25%. In SDC D–F, a structure with a horizontal irregularity of Type 1, 2, 3 or 4 has its §12.10.1.1 forces "increased 25%" for collectors, their connections, and the diaphragm-to-vertical-element connections. The reentrant corner that cost us the regular-plan route in Step 1 is a Type 2 irregularity, so it applies: 12,187.5 × 1.25 = 15,234.4 lb at strength level. (Forces computed with overstrength are exempt — that exception matters below.)

At ASD, the §2.4.5 combinations carry 0.7 of that: V_collector = 0.7 × 15,234.4 = 10,664.1 lb — 1.6 times the line shear the walls were designed for. The module derives all of it in Step 5 from three inputs it asks for there — F_px by Eq. (12.10-1), w_px and I_e — and reads the Type 2 corner off the plan itself.

#The force is the running difference of two unit shears

The diaphragm delivers that collector shear uniformly at v_diaph = V/L = 10,664.1/60 = 177.7 plf. The walls resist at v_wall = V/Σb = 10,664.1/22 = 484.7 plf. The collector force at any point is the running integral of the difference:

F(x) = ∫₀^x (v_diaph − v_wall·[x on a wall]) dx

Over glazing or an opening only the diaphragm term acts, so F climbs at 177.7 lb per foot. Over a wall segment the wall pulls harder than the diaphragm delivers, so F falls at 484.7 − 177.7 = 307.0 lb per foot. It starts at zero at one diaphragm edge and must return to zero at the other — that closure is the equilibrium check.

The diagram peaks where the last of the unbraced strip has been collected and the first wall is about to take it: F_max = 177.7 plf × 15 ft = 2,666 lb, at the outer end of each end wall (x = 15 ft and x = 45 ft). That is line 1's design drag force — and note where it is: not at the openings everyone worries about, but at the two ends of the sheathed group.

#What §12.10.2.1 requires — and the exception that saves this building

In Seismic Design Category C through F, collectors are normally an overstrength element. ASCE 7-22 §12.10.2.1 requires collector elements "and their connections, including connections to vertical elements" to be designed for the maximum of three cases — (a) and (b) both use "the seismic load effects including overstrength of Section 12.4.3", and (c) uses the §2.3.6 combinations with the Eq. (12.10-2) diaphragm force. For light-frame wood walls sheathed with wood structural panels, Table 12.2-1 (row 16) gives Ω₀ = 3 — and on this path cases (a) and (b) coincide for one story and govern — Ω₀ on the line's share of the story force: 3 × 6,562.5 = 19,687.5 lb of collector shear, exempt from the §12.3.3.5 increase, a drag peak of 4,921.9 lb through that plate and its straps.

Then comes the sentence worth knowing by heart:

EXCEPTION: In structures or portions thereof braced entirely by wood light-frame shear walls, collector elements and their connections, including connections to vertical elements, need only be designed to resist forces using the load combinations of Section 2.3.6 with seismic forces determined in accordance with Section 12.10.1.1.

Our building is braced entirely by wood light-frame shear walls, so the exception applies: no Ω₀, the §12.10.1.1 force with its 25% increase instead, and the design force is 2,666 lb. Two cautions decide whether you may actually use it:

• "Entirely" means entirely. One steel moment frame at the garage opening, one masonry stair core, one cantilever column — and that portion is no longer braced entirely by wood light-frame shear walls. Mixed systems lose the exception where they mix.
• It is an exception to the overstrength requirement, not to the collector requirement. §12.10.2 still demands a collector that works; the exception only sets the force level.

This is the Step-5 choice StructSuite refuses to guess: pick Structure braced entirely by wood light-frame shear walls or Apply overstrength (then enter Ω₀), and the drag check follows. It stays unselected — and red — until you decide, because the two answers differ by nearly a factor of two here (2,666 against 4,922 lb) and by about 2.3 on a plan with no irregularity increase (Ω₀ = 3 against the 1.3× F_px floor).

#Practical notes — straps, splices, and the details that actually fail

A collector is only as real as its continuity. A double top plate is a spliced member: it comes in lengths, and every splice is a break in the drag path unless something carries the force across it.

• Strap the splices that fall inside the drag zone, not wherever the splice happens to land. Read the design force at the splice location off the diagram: on line 1 a splice at x = 15 ft carries the full 2,666 lb, one at x = 23 ft carries 210 lb, one out at x = 55 ft carries about 890 lb. Same plate, twelve-fold difference.
• Match the strap family to the force. Light flat straps (LSTA/MSTA-type) cover the low hundreds of pounds; heavier flat straps (MST-type) and coil straps (CS/CMST-type) cover the kilopound range and the long laps. Take the allowable from the manufacturer's table at the fastener schedule and species you will actually install — a strap's rating assumes every listed nail is driven into the specified member.
• Decide the strap face in the detail, not on the roof deck. A collector strap drawn on the sheathed face becomes a shim under the panel; strap the inside face of the plate, or use a plate-top strap installed before the roof framing lands.
• The drag force does not stop at the wall — it enters it. §12.10.2.1's "connections to vertical elements" is that connection: the strap must lap onto the shear wall's own top plate, where the wall's edge nailing can receive it. A strap that dies 6 in short of the wall has transferred nothing.
• Lateral load reverses, so the collector is a tension and compression member. The plate takes compression by bearing — as long as the splice is tight and the joint is not open. Detail both: strap for tension, snug bearing for compression.
• Do not notch the collector. Plumbers and electricians treat the top plate as a highway. A drilled or notched plate in the peak-drag zone is a broken collector; protect it in the notes and check it at framing inspection.
• Headers can be collectors, but only if you say so. Where the drag path crosses a large opening, the header is the continuous element — and it needs the strap connection at both ends plus a path into the wall below. That is a detail, not an assumption.

One habit worth keeping: sketch the drag diagram, even roughly, before choosing strap locations. The peak is a geometric fact of the elevation — long unbraced ends make big drags — and it usually lands somewhere nobody would have strapped by instinct.

#Step 12 — Drift: the check nobody asks for until the windows crack

Strength is not the only acceptance criterion — SDPWS §4.3.4.1 gives the deflection of a full-height segment (Equation 4.3-1):

δ_sw = 8vh³/(EAb) + vh/(1000·G_a) + h·Δ_a/b

— bending of the wall as a cantilever (chords), shear racking of the panel-and-nails (that is what G_a measures), and rigid-body rotation from anchorage stretch. Drift is evaluated at the strength-level shear — the unreduced E, so for the 6-ft wall v = 1,789.8/0.7/6 = 426.1 plf — with (2) 2×4 DF chords (E = 1.6×10⁶ psi, A = 10.5 in²), G_a = 22 kips/in, and Δ_a on the same basis as v. The HDUE3-SDS3's catalog deflection, 0.127″, is measured at its 3,790-lb allowable; at this wall's strength-level tension — T = (23,011 − 0.7 × 2,400 × 2.875)/5.510 = 3,299 lb under the §2.3.6 combination 0.9D − 0.2S_DSD + E_h — the device stretches in proportion, 0.127 × 3,299/3,790 = 0.111″. That is the device alone: §4.3.4.1 defines Δ_a as the total vertical elongation of the anchorage system, so post crushing under the device and any shrinkage gap between rod and plate are added from the detail, not from the catalog.

Read the proportions, not just the total: the chords barely matter; nail slip and hold-down stretch are ~93% of the drift. That is why tightening nail spacing stiffens a wall as much as it strengthens it (G_a jumped from 15 to 22 when we left 6″ behind), and why a soft hold-down on a narrow wall is a drift problem before it is a strength problem. Amplify δ_sw per the ASCE 7-22 Chapter 12 drift provisions (C_d/I_e for your system row) and compare against the story-drift limit for your building — the module reports the strength chain; the drift comparison stays with you and your Table 12.2-1 row.

#Building it: what actually goes wrong on site

Every number above assumes the wall in the drawing is the wall that gets built. These are the failures inspectors actually find, in rough order of frequency:

• Over-driven nails. A gun set hot enough to bury heads through the face veneer of OSB can quietly delete a third of the wall's capacity — the head punches through instead of clamping. Specify flush-driven, and expect the inspector's fingernail test.
• Edge distance. Nails at 3/8″ minimum from panel edges; a nail split into the edge is a nail that is not there. On 3x framing at abutting edges, stagger the two panels' rows — that is what §4.3.7.1(5) is buying.
• Missing blocking. Table 4.3A is a blocked-wall table. An unblocked horizontal joint at the 8-ft panel line converts your 350 plf wall into something the table never promised.
• The washer that stayed on the truck. Standard cut washers under sill nuts in SDC D are a plan-check classic; the 0.229″ × 3″ × 3″ plates of §4.3.6.4.3 are the difference between bolt bearing and plate splitting.
• Hold-down stack-up. Devices installed over uncompressed shims, rods left finger-tight, or green lumber shrinking half a season later — every 1/16″ shows up as Δ_a, and Step 12 showed anchorage stretch is roughly half the drift. Retighten after dry-in on shrinkage-critical jobs.
• The clip that vanished at framing. Base transfer hardware (our A35s) never makes it from the calc package to the framer unless it is on the schedule with a count and a spacing. Draw it like you mean it.

#Summary — the finished lateral design, east–west

#Verification — hand calculation against StructSuite

Every engine number on this page was computed twice: once by the StructSuite engine (locked by the V-SWG regression suite) and once by hand, long-form (V-SWH). The Δ column is the difference. Rows marked hand-only are exactly that — the module does not compute drift, and ρ is the engineer's input.

#Distribution and line shear

#Overturning, hold-down, end post (the chord forces)

#Base transfer and drift

Agreement of arithmetic is what this proves — not judgment. The choices (three segments instead of two, 48″ bolts instead of 59″, a hold-down at 62% rather than a lighter one that no longer fits its own moment arm) remain engineering, and they remain yours.

#Seven traps that sink shear wall designs

1. One split for two forces. Distributing a single "share of V" to walls that see both wind and seismic. This page's garage piers: 88% pass by seismic, 103% fail by wind.
2. The duration-factor double-dip. Applying C_D = 1.6 (or the old 1.33) on top of Table 4.3A values. §4.1.4's Ω = 2.0/2.8 already own that ground: "no further increases shall be permitted."
3. Full dead load resisting uplift. Let 1.0D resist instead of 0.6D − 0.7E_v and the resisting moment is overstated 2.2×; on these walls the computed hold-down demand comes out roughly 30–40% low (the 6-ft wall's 2,347 lb shrinks to a fictitious 1,671 lb, the 8-ft walls' 2,100 lb to 1,209 lb), and the error grows with wall length.
4. Blocked table, unblocked wall. Table 4.3A assumes every panel edge lands on framing or blocking. One unblocked joint and the tabulated value no longer describes the wall you built.
5. No path out of the wall. Sheathing and hold-downs designed, base transfer left to toe-nails — which §4.1.10 bars above 150 plf in SDC D–F. The v must land in bolts or clips, on the drawing, with a spacing.
6. No collector, or a collector with a splice in it. The diaphragm delivers over the whole line; the walls occupy part of it. Everything between is a drag path — and an unstrapped top-plate splice inside the peak-drag zone is a collector that ends there. §12.10.2 requires the element; §12.10.2.1 sets its force.
7. The 980-plf surprise. Specifying 3″ or 2″ nailing in SDC D without noticing §4.3.7.1(5)'s 3x-framing requirement — discovered at rough framing inspection, fixed at time-and-materials. The software catches this one — and traps 1, 2 and 3 as well: both cases always run, the 2.0/2.8 factors are built into the table, and 0.6D − 0.7E_v is the only resisting case it offers. Step 3 evaluates (a), (b) and (c) against the row, the spacing and the Seismic Design Category, and prints the requirement the moment one fires.

#Frequently Asked Questions

#Why is the seismic reduction factor 2.8 but wind only 2.0?

Because the two factors were set to give back the historical allowable values, not derived from one another. The SDPWS commentary (C2.2) explains that the tabulated nominal capacities for wood structural panel walls were built by multiplying the historical ASD seismic values by 2.8 — a factor long taken as the target minimum safety factor for these walls under seismic load — and that nominal strengths were factored up so that dividing by the specified reduction factor returns the prior codes' design value; on the wind side those prior values stood 1.4 times higher, which is where 2.0 (= 2.8/1.4) comes from. The practical consequence: the same wall has 40% more ASD capacity against wind, so wind can only govern where its distributed demand is enough larger — which is exactly what happens on shallow wings, garage fronts and stair towers.

#Why is ρ = 1.0 on an L-shaped plan — and what if it were 1.3?

Because §12.3.4.2's first condition is satisfied, not its second. The reentrant corner makes the plan irregular (Type 2), which rules out the regular-plan route; what remains is the notional-removal check of Table 12.3-3, worked in Step 1: two wall bays on each side of the center of mass, no single removable wall taking more than 18% of the story strength against a 35% limit, and no torsional irregularity to evaluate on a flexible diaphragm. If any of those had failed, ρ = 1.3 would multiply every seismic line shear on this page: line 1's ASD demand becomes 8,531 lb against 7,700 at 4″ (111%) and needs 3″ nailing (9,900 lb, 86%); line 2 at 3″ goes from 97% to 126% and needs 2″ nailing (11,714 lb, 97%) or more wall; line 3's piers at 3″ still pass, but their governing case flips from wind (80%) to seismic (89%). That is why the module asks for ρ per direction and refuses to assume it.

#Do I really have to check both wind and seismic on every line, even in SDC D?

Yes — and not because of the cases' totals, but because of their splits. Seismic follows tributary mass (area); wind follows tributary length of the projected face. On any non-rectangular plan those differ line by line, and the governing case can flip between neighboring wall lines of the same building — on the L-shaped example above, seismic governs the body lines while wind governs the wing.

#Can I sheathe both sides? Does the capacity just double?

You can, and it doubles only when the two sides are identical. SDPWS §4.3.5.4.1 gives G_a(c) = G_a(1) + G_a(2) (Eq. 4.3-3) and v_n(c) = K_min·G_a(c) (Eq. 4.3-4), where K_min is the smaller of the two sides' v_n/G_a ratios. Same row, same spacing on both faces and G_a cancels: v_n(c) = 2·v_n. Different sides and K_min bites, so the combination is worth less than the sum — and if the two sheathing systems differ (panel one side, gypsum the other), you are in §4.3.5.4.2, where the answer is the greater of twice the smaller side and the larger side alone. The exception that permits summing panel-plus-gypsum is written for wind only; using it for seismic is the classic error. StructSuite offers the two-sided option only for identical sides, because dissimilar sides also leave the §4.3.5.5.1 Exception 1 distribution the module uses.

#If I sheathe both sides, do my hold-downs get smaller?

No. Sheathing is a capacity question; hold-downs are a demand question. The line still carries the same shear over the same length of wall, so the unit shear v is unchanged, and the chord tension T — which is v·h less the dead-load resisting moment — is unchanged with it. End posts, anchor bolts, base shear transfer and the collector are all unchanged too. The only things the second face moves are the sheathing capacity and the wall's stiffness. If the hold-down is what is failing, the lever is more wall length or a bigger hold-down, never a second layer of panels.

#Does the second face trip the 3x framing rule?

Not by itself. §4.3.7.1(5)(c) reads the nominal unit shear capacity "on either side of the shear wall", so two 670 plf faces give a combined 1,340 plf and still sit under the 980 plf SDC D-F threshold. Items (5)(a) and (5)(b) read nail spacing and nail size, which a second face does not change. What does turn on is Table 4.3A footnote 6: panels on both faces with nail spacing under 6″ on either side require panel joints offset onto different framing members, or 3″ nominal framing at adjoining panel edges with staggered nailing. At 6″ o.c. both faces, nothing changes in the framing.

#When does a shear wall have to be blocked?

Table 4.3.3 lists blocked and unblocked wood structural panel walls as separate rows with separate limits (3.5:1 blocked, 2:1 unblocked); its footnote 1 — walls over 1.5:1 must be blocked — is attached to the gypsum wallboard and portland cement plaster rows, not to panel walls. For a panel wall the question is which table you are reading: the Table 4.3A capacities used in this guide are blocked-wall values, so every panel edge must land on framing or blocking; an unblocked panel wall is designed with the Unblocked Shear Wall Adjustment Factor C_ub of §4.3.5.3 instead. For engineered work the honest summary is still: block the edges, nail the edges, and the table means what it says.

#How narrow can a shear wall segment be?

Blocked WSP: h/b ≤ 3.5:1 (Table 4.3.3) — at 9 ft, about 2′-7″. But above 2:1 the §4.3.3.2 factor (1.25 − 0.125·h/b) and the §4.3.5.5.1 Exception 1 factor (2b/h) shave the capacity, and the drift equation's h·Δ_a/b term grows. Narrow piers are legal; they are just expensive per plf.

#When do I need 3x framing at panel edges?

SDPWS §4.3.7.1(5): edge nailing at 2″ o.c.; 10d commons at 3″ o.c. or less; or nominal capacity on either side over 980 plf in SDC D, E or F. That third trigger catches every 3″ row of the 7/16″ table in high-seismic work — plan for it before pricing, not after inspection. Three details are easy to miss:

• "On either side" is per side. A wall sheathed both faces under §4.3.5.4.1 has a combined v_n(c) that can sit well over 980 plf while neither face does. The combined value is not what the threshold reads.
• "Exceeds 980 plf" is strict. The 4″ row at exactly 980 does not trip it, which is most of why that row is the workhorse of high-seismic residential design.
• The list is not the same for every material. §4.3.7.3(5) (particleboard) carries (a) and (b) with no 980 plf condition — particleboard is not permitted in SDC D–F at all — and §4.3.7.4 (structural fiberboard) has no 3″ rule whatsoever; its item 4 asks for 2″ nominal at adjoining panel edges and stops there. Generalising §4.3.7.1 across Table 4.3A is the mistake here.

StructSuite evaluates all three conditions from the row, the spacing and the Seismic Design Category, and states the requirement in Step 3 when one fires. It is reported as a construction requirement, never as a D/C: it changes no number in the design, it changes the drawings.

#My shear wall sits on a beam or post, not on the foundation. Does anything change?

Not for the wall — it is designed exactly as in this guide. For the member underneath, a great deal: ASCE 7-22 §12.3.3.4 makes it an element supporting a discontinuous wall, designed for the seismic load effects including overstrength, E_mh = Ω₀Q_E. At Ω₀ = 3 (2½ with the flexible-diaphragm footnote b of Table 12.2-1) the seismic part of its demand triples, and that normally governs. ρ is not applied on top (§12.3.4.1 item 6 permits ρ = 1.0 there), and Ω₀ comes from the Table 12.2-1 row for the wall above. See Step 9 for the full note.

#Does a shear wall need a hold-down at every segment end?

Wherever the reduced dead load (0.6D − 0.7E_v) cannot cover the overturning — SDPWS §4.3.6.4.2. At S_DS = 1.0 that net 0.46D covers almost nothing residential; assume a device at both ends of every segment until a calculation says otherwise.

#What is a collector (drag strut), and when does a shear wall line need one?

A collector is the element that carries diaphragm shear along a wall line to the wall segments that can actually resist it — in wood framing, almost always the double top plate. You need one whenever the sheathed wall length is shorter than the diaphragm length delivering to that line, which is nearly always. Its force is the running difference between the diaphragm unit shear (V/L) and the wall unit shear (V/Σb), peaking at the outer ends of the sheathed group — but the V is the line's share of the §12.10.1.1 diaphragm force, not the story force: on this page's line 1 that is 10,664 lb at ASD (the Eq. 12.10-2 floor, 0.2·S_DS·I_e·w_px, raised 25% by §12.3.3.5 for the reentrant-corner plan), and 177.7 plf × 15 ft of unbraced end = 2,666 lb. ASCE 7-22 §12.10.2 requires the element; §12.10.2.1 puts collectors in SDC C–F on the overstrength load combinations — with an exception for structures braced entirely by wood light-frame shear walls, which keeps the force at the §12.10.1.1 level (no Ω₀).

#How do you size a shear wall hold-down?

Compute the wall's overturning moment M_OT = V·h, subtract the resisting moment of the reduced dead load (0.6D − 0.7E_v, acting at mid-length, taken about the compression chord), divide the remainder by the effective length b_eff of SDPWS Eq. 4.3-7 — hold-down rod to end-post centroid, not the wall length — and select a device whose published allowable, in the correct species column, meets that tension: T = (M_OT − M_R)/b_eff. Because the device sets where its rod sits, the demand is recomputed for each candidate. On this page's 6-ft wall that is (16,108 − 3,174)/5.510 = 2,347.2 lb, met by an HDUE3-SDS3 on a 3× post at 3,790 lb (62%); the DTT2Z everyone reaches for first comes out at 109% over its own 5.56-ft arm.

#Can toe-nails transfer shear wall base shear?

Not for seismic shear above 150 plf (ASD) in SDC D–F — §4.1.10 limits the seismic force toe-nails may transfer from diaphragms to shear walls or collectors, and says nothing about wind. Below that, and for wind, they still deserve suspicion: the clip's few dollars buy a tested, listed, inspectable connection.

#What if my shear wall framing is SPF instead of DF?

Table 4.3A footnote 3: multiply v_n by 1 − (0.5 − G). SPF at G = 0.42 gives 0.92 — every capacity in this guide drops 8%, and line 2's 97% stops passing. Species is not a footnote; it is a design input.

#What about openings — is there a method that skips hold-downs at every pier?

Yes: the perforated shear wall method (SDPWS §4.3.5.6, the C_o factor), which trades hardware for capacity. This guide uses the segmented method throughout because it is the transparent one — every pier's numbers are visible — and it is what the module implements.

#Where does the IBC come into this?

IBC 2024 §2305.1 delegates engineered wood lateral design to SDPWS, and keeps for itself the prescriptive minimums this page cited where they bind (sill bolt diameter, spacing, embedment — §2308.7.1 — and the SDC D/E washer and bolt upgrades).

#Design this building in StructSuite

Everything on this page is the Wood-Frame Shear Wall module doing its ordinary work: enter the plan in Step 1 — the grid, the footprint and the shear walls checked on it; enter the building's story forces and ρ per direction, S_DS and the Seismic Design Category once in Step 2; set each line's gravity loads and sheathing (Steps 2–3), each wall's hold-down and end post (its chords) and anchor bolts (Step 4) and the collector (Step 5, with its own building inputs: F_px by Eq. (12.10-1), w_px and I_e) — both optional steps — copying any of them to every line or every wall in one click; Step 3 reproduces Table 4.3A row for row with its footnotes as labelled checkboxes; and Step 6 checks every line: both aspect-ratio factors computed side by side, per-wall overturning with 0.6D − 0.7E_v over Eq. 4.3-7's b_eff, hold-down and anchor-tension checks against the catalog columns, end posts per NDS 3.7/3.10, and every check stated with its substituted numbers, with Step 7 drawing the diagrams. The distribution that produced this page's line forces is that Step 1; the bolt-and-clip base-transfer step runs in development preview while its interface is finished — its engine and verification pins are already in the build. The sample project runs the same module on its own single-story L-shaped building — a different set of walls, loads, sheathing and hardware, so expect different numbers — which you can open without an account, and the method behind every number here is documented in How StructSuite Works.

Companion guides: seismic base shear (ASCE 7-22 §12.8) · wind loads, Chapter 27 · wood column design · NDS Supplement tables.