How to Calculate MWFRS Wind Base Shear Using ASCE 7-22 Chapter 27 Directional Procedure (Gable Roof Example with Minimum Design Wind Loads and Figure 27.3-8)

This note is a stand-alone calculation for engineers checking horizontal shear at the base in two orthogonal plan directions and comparing the result to minimum design wind load in ASCE/SEI 7-22. The building is a rectangular, enclosed low-rise structure with a symmetric gable roof: the ridge runs parallel to the long plan side, so wind parallel to the ridge blows on the gable end and wind perpendicular to the ridge blows on the long wall-and-roof elevation (wall below the eaves plus sloping roof per Figure 27.3-1). It follows the Directional Procedure in ASCE 7-22 Chapter 27 (with supporting requirements in Chapter 26) and the MWFRS load cases in Figure 27.3-8. It does not replace a full project-specific wind study (topography, flexible buildings, torsion design, or local components).

Companion overview: ASCE 7-22 Wind Loads on MWFRS (StructSuite Wind Module). Seismic (ELF): StructSuite seismic loads — ELF wizard (ASCE 7-22).

1. Building and wind-assumption summary

Gust-effect factor (clarification): This example assumes the building is rigid with fundamental natural frequency n₁ > 1 Hz (the same intent as f₁ > 1 Hz when period is expressed as a frequency). G = 0.85 is then appropriate per Section 26.11.1. If the building were flexible (n₁ < 1 Hz), G_f would need to be calculated per ASCE 7-22 Section 26.11.2, not the tabulated rigid-building G.

Ridge orientation: Ridge is parallel to L (the x-axis). Therefore:

• Wind parallel to ridge ↔ wind along ±x (blows toward the 30 ft gable-end face).
• Wind perpendicular to ridge ↔ wind along ±y (blows toward the 50 ft long wall).

2. Geometry check (eave, ridge, mean height)

Half of the gable span in plan is B/2 = 15 ft. Vertical rise from eave to ridge along one slope:

Δh = (B/2) tan θ = 15 × 0.5 = 7.5 ft

• Ridge height above grade: h_e + Δh = 8 + 7.5 = 15.5 ft.
• Mean roof height (average of eave and ridge):
  h = h_e + Δh/2 = 8 + 3.75 = 11.75 ft (matches (8 + 15.5)/2).

All numerical steps below use L = 50 ft, B = 30 ft, h = 11.75 ft, h_e = 8 ft, θ = 26.565°.

3. Velocity pressure — Equation (26.10-1) and exact K_z at each height

Equation (26.10-1) gives velocity pressure q_z (lb/ft²); at mean roof height z = h, q_h uses K_h = K_z evaluated at h:

q_z = 0.00256 K_z K_zt K_e V²

K_d: Not part of Equation (26.10-1) for q; it enters Equation (27.3-1) with G and C_p (same as WindWizardApp Step 5 vs Step 7).

Exposure B, Chapter 27 Directional: use the numerical coefficients in Table 26.10-1 for each z. Do not apply the Chapter 28 / low-rise footnote that fixes K_z = 0.70 for Exposure B when z < 30 ft—that footnote is for the envelope procedure in Chapter 28, not for interpolating K_z at arbitrary z under Chapter 27.

Key elevations for this building (Section 2): eave z = 8.00 ft, mean roof z = h = 11.75 ft, ridge z = 15.5 ft.

q_z at these heights (K_zt = K_e = 1.0, V = 100 mph):

Mean roof height (hand check in Sections 4–6): use q_h = 14.592 lb/ft² (from K_z(h) = 0.57). Round for hand clarity in what follows: q_h ≈ 14.59 lb/ft² (or 14.6 lb/ft² if you round one step later).

Section 27.3.1 (Directional Procedure): the windward wall uses q_z at the height of each strip (z varies up the wall); leeward wall, sidewalls, and roof use q_h at h unless your criteria say otherwise. This example keeps a single-level wall strip using q = q_h on windward at z = h so p_net in Section 5 matches a uniform h-tall strip—but K_z values above are the exact table values at eave, mean, and ridge for documenting q_z variation (e.g. q_z = 14.72 lb/ft² at the ridge vs 14.59 lb/ft² at mean height).

Internal pressure q_i: for an enclosed building, q_i = q_h evaluated at h (Section 26.13, used in Equation (27.3-1)).

4. Design wind pressure on MWFRS surfaces — Equation (27.3-1)

Equation (27.3-1) (general form used with Chapter 27):

p = q K_d G C_p − q_i K_d (GC_pi)

Define the shorthand (same q = q_h, K_d, G for all walls in this example), with q_h = 14.592 lb/ft² from K_z(h) = 0.57 (Section 3):

C_ref = q_h K_d G = 14.592 × 0.85 × 0.85 = 10.543 lb/ft²

Internal term magnitude (both signs of GC_pi use q_i = q_h here):

q_i K_d |GC_pi| = 14.592 × 0.85 × 0.18 = 2.233 lb/ft²

Windward wall (all directions): Figure 27.3-1, C_p = +0.8.

• With GC_pi = +0.18:
  p_W = 10.543×(0.8) − 2.233 = 6.20 lb/ft².
• With GC_pi = −0.18:
  p_W = 10.543×(0.8) + 2.233 = 10.67 lb/ft².

Leeward wall: Figure 27.3-1 uses L / B = (horizontal dimension parallel to wind) / (horizontal dimension normal to wind).

• Wind along x (parallel to ridge): L_parallel = 50 ft, B_perp = 30 ft → L/B = 50/30 = 1.667. Between 1 and 2, C_p interpolates from −0.50 to −0.30 → C_p ≈ −0.37.
• Wind along y (perpendicular to ridge): L_parallel = 30 ft, B_perp = 50 ft → L/B = 0.60 ≤ 1 → C_p = −0.50.

Leeward q = q_h = 14.592 lb/ft².

• GC_pi = +0.18: p_L = 10.543×C_p − 2.233.
• GC_pi = −0.18: p_L = 10.543×C_p + 2.233.

5. Along-wind net pressure for wall line load (Figure 27.3-8, Case 1)

For each orthogonal direction, p_W and p_L are the signed design pressures from Equation (27.3-1) (normal to each surface). For global horizontal base shear from the windward and leeward walls parallel to the wind, the along-wind resultant per unit area is p_net = p_W − p_L (lb/ft²): with p_L negative (suction on the leeward face), p_W − p_L is the vector sum in the wind direction (effective addition of windward push and leeward suction), not the algebraic sum p_W + p_L, which would understate the shear.

Same numbers in “push + pull” form: For the −GC_pi column in Section 4, parallel (x) gives p_W = 10.67 psf (push) and p_L = −1.67 psf (pull). The net lateral driving pressure is 10.67 + 1.67 = 12.34 psf, not 10.67 + (−1.67) = 9.00 psf. Perpendicular (y): 10.67 + 3.04 = 13.71 psf, not 10.67 + (−3.04) = 7.63 psf. To keep C_p signs straight with one expression, use p_net = q K_d G (C_p,windward − C_p,leeward) when the external velocity pressures and G match that form on both walls (internal GC_pi still cancels in the difference).

Note: For a given internal-pressure column, the q_i K_d (GC_pi) terms are the same on windward and leeward in Equation (27.3-1), so they cancel in p_W − p_L. Internal pressure still governs component design; it does not change global MWFRS base shear from opposing parallel walls in this combination. While GC_pi significantly impacts individual component design, it cancels out in p_net for this global shear check because it acts equally and oppositely on the windward and leeward surfaces in Equation (27.3-1).

Using Section 4 values (same GC_pi column on both walls):

Wind along x (parallel to ridge), B_normal = B = 30 ft:

Wind along y (perpendicular to ridge), B_normal = L = 50 ft:

Governing net pressures for Case 1 wall shear: p_net,x ≈ 12.33 lb/ft², p_net,y ≈ 13.71 lb/ft² (either internal-pressure column gives the same p_net within rounding).

6. Base shear from Figure 27.3-8 — Case 1 (full pressure)

Figure 27.3-8 defines MWFRS cases applied with pressures from Section 27.3.1. Case 1 uses 100% of the design wall pressures.

Wall-driven shear (hand steps 6.1–6.2 below) uses the uniform line-load strip over h:

V_walls = (p_W − p_L) B_normal h = p_net B_normal h

Important: Section 27.3.1 assigns q_z varying up the windward wall; using a single q at h is a hand simplification. StructSuite Chapter 27 Step 8 reports V = V_walls + V_{roof,horizontal} for sloped gable/hip and θ ≥ 10°: V_y includes (p_rw − p_rl) A_half sin θ with A_half = L (B/2) / cos θ, q_h, and windward primary roof C_p (Figure 27.3-1); V_x,roof = 0 (symmetric slopes cancel for wind parallel to ridge). Flat roofs or θ < 10° add no roof horizontal term in that summary. Roof uplift still follows Note 2 (|p| × L×B). Sections 6.1–6.2 keep wall-only numbers so you can check p_net without mixing in roof tributary geometry.

6.1 Wind parallel to ridge (along x)

Use p_net = 12.33 lb/ft², B_normal = B = 30 ft, h = 11.75 ft:

V_x (Case 1) = 12.33 × 30 × 11.75 = 4,346 lb

6.2 Wind perpendicular to ridge (along y)

Use p_net = 13.71 lb/ft², B_normal = L = 50 ft, h = 11.75 ft:

V_y (Case 1) = 13.71 × 50 × 11.75 = 8,055 lb

So perpendicular-to-ridge wind controls Case 1 wall-only shear in this numeric set (8,055 lb > 4,346 lb).

6.3 Roof horizontal shear — wind perpendicular to ridge (θ ≥ 10°)

When wind is perpendicular to the ridge (along ±y), net pressure on the windward and leeward roof halves has a component along the wind direction proportional to sin θ. For a symmetric gable, StructSuite (and this hand check) use one sloped half-area and primary windward roof C_p from Figure 27.3-1 (localized windward zones are not subdivided for global shear):

A_half = L (B/2) / cos θ

V_y,roof = (p_rw − p_rl) A_half sin θ

Roof pressures use q_h and Equation (27.3-1). For a given internal-pressure case, q_i K_d (GC_pi) is the same on both roof surfaces, so it cancels in (p_rw − p_rl):

p_rw − p_rl = q_h K_d G (C_p,rw − C_p,rl)

Figure 27.3-1 inputs for this direction: plan dimension parallel to wind = B = 30 ft, so h / L = 11.75 / 30 ≈ 0.392. Roof pitch θ = 26.565°. Interpolating the table (between h/L rows and θ columns) gives windward roof primary C_p,rw ≈ −0.239 and leeward roof C_p,rl = −0.60 (same convention as StructSuite Chapter 27 Step 8).

Using q_h K_d G = 10.543 lb/ft²:

p_rw − p_rl = 10.543 × (−0.239 − (−0.60)) = 10.543 × 0.361 = **3.81 lb/ft²**

Geometry:

• A_half = 50 × 15 / cos 26.565° = 750 / 0.89443 = 838.5 ft²
• sin θ = sin 26.565° = 0.4472

V_y,roof = 3.81 × 838.5 × 0.4472 ≈ **1,427 lb**

(Rounding C_p and areas gives ~1,427 lb; V_y for Case 1 with walls + roof ≈ 8,055 + 1,427 ≈ 9,482 lb.) Wind parallel to the ridge produces V_x,roof = 0 for this symmetric gable (opposing slopes cancel).

6.4 Other Figure 27.3-8 cases (brief)

Cases 2, 3, and 4 apply fractions of these loads (0.75, 0.75, and 0.563) and add eccentricity for torsion (Cases 2 and 4). You must check all cases and both GC_pi signs for governing shear and moment. For example, Case 2 shear magnitude in a given direction is 0.75 times the Case 1 value if only the factor changes:

V_x (Case 2) ≈ 0.75 × 4,346 = 3,260 lb

Always use the full Figure 27.3-8 rules from the standard for the final design.

7. Minimum design wind load (enclosed building) — Section 27.1.5

Section 27.1.5 requires that MWFRS wind loads be not less than specified minimums. For enclosed and partially enclosed buildings, the horizontal minimum used in this example follows the vertical projection rule aligned with Figure C27.1-1 (commentary) and Section 28.3.6 (minimum design wind load expression repeated for low-rise MWFRS checks):

V_min = 16 A_wall,proj + 8 A_roof,proj

Open buildings and other classifications follow Section 27.1.5 differently (e.g. 16 psf on total projected area); this example stays enclosed.

7.1 Wind perpendicular to ridge (wind toward the L = 50 ft face)

• Wall projection (Figure C27.1-1 style rectangle using mean roof height for this minimum check, consistent with common low-rise MWFRS minimum application):
  A_wall,proj = L × h = 50 × 11.75 = 587.5 ft².
  (If your office standard uses only h_e for the wall strip and adds roof projection separately for envelope methods, document that; the 16/8 minimum uses the commentary vertical projection split.)
• Roof projection (one sloped half, vertical “silhouette”):
  A_roof,proj = L × (B/2) tan θ = 50 × 7.5 = 375 ft².

V_{min, perp} = 16 × 587.5 + 8 × 375 = 9,400 + 3,000 = 12,400 lb

7.2 Wind parallel to ridge (wind toward the B = 30 ft gable end)

• Wall projection (rectangle to eave plus gable triangle):
  A_wall,proj = B × h_e + ½ B × ((B/2) tan θ) = 30×8 + ½×30×7.5 = 240 + 112.5 = 352.5 ft².
• Roof projection (triangular vertical silhouette for one roof half when wind is parallel to ridge in this geometry):
  A_roof,proj = ½ B × ((B/2) tan θ) = 112.5 ft².

V_{min, par} = 16 × 352.5 + 8 × 112.5 = 5,640 + 900 = 6,540 lb

8. Governing horizontal shear (this example)

Here the Section 27.1.5 / 28.3.6 minimum still controls both directions even after adding V_y,roof from Section 6.3. Your project may differ once full Figure 27.3-8 cases (fractions and torsion), roof contributions to overturning, and q_z variation along the windward wall are included.

9. Relation to Chapter 28 Envelope (low-rise)

This note uses Chapter 27 only. For qualifying low-rise buildings, Chapter 28 applies design pressure normal to each surface from Figure 28.3-1 zones and Equation (28.3-1); horizontal base shear is obtained by summing along-wind components of those normal forces on walls and sloped roof zones (including F sin θ type resolution on roof halves), which captures roof contribution without adding a separate roof shear term the way Section 8 describes for StructSuite’s Chapter 27 summary. Use one governing MWFRS procedure per direction on the calculation package—do not combine Chapter 27 C_p strips with Chapter 28 GC_pf zones for the same check.

10. References (ASCE/SEI 7-22)

• Equation (26.10-1) — velocity pressure.
• Section 27.3.1 — q and q_i for surfaces; Equation (27.3-1).
• Figure 27.3-1 — C_p for walls and roof.
• Figure 27.3-8 — MWFRS cases (1–4) for the Directional Procedure.
• Section 27.1.5 — minimum design wind loads.
• Section 28.3.6 — minimum design wind load (16 psf / 8 psf on projections; cross-check with Section 27.1.5).
• Figure C27.1-1 (commentary) — illustration of wall and roof vertical projections for minimum load.

This technical note is educational. Confirm all coefficients, K_z from Table 26.10-1 at each relevant height z (Chapter 27 uses interpolated table values; do not substitute the Chapter 28 Exposure B footnote for K_z when tabulating q_z), G, and minimum-load applicability with the governing edition of ASCE 7 and the authority having jurisdiction.