ASCE 7-22 Seismic Load Worked Example — ELF Procedure

Complete worked example for calculating seismic design forces using the Equivalent Lateral Force (ELF) procedure per ASCE 7-22 Chapter 11-12. This example covers a 5-story steel special moment frame building in a high-seismicity region (Ss = 1.5, S1 = 0.6) on Site Class D. All calculations follow ASCE 7-22 strength-level procedures with the two-stage risk-adjusted maximum considered earthquake (MCER) spectral accelerations.

Related pages: Seismic Design Basics | US Load Combinations | Steel Seismic Design | Seismic Load Calculator | Wind Load Calculator

Problem Statement

A 5-story steel office building with the following characteristics:

Step 1 — Determine Seismic Design Category (Sections 11.4-11.6)

1a. Site Coefficients Fa and Fv (Tables 11.4-1 and 11.4-2)

For Site Class D:

Fa (short-period site coefficient): For Ss = 1.5, Site Class D: [ F_a = 1.0 \text{ (Table 11.4-1: Ss >= 1.25, Site Class D)} ]

Fv (long-period site coefficient): For S1 = 0.6, Site Class D: [ F_v = 1.7 \text{ (Table 11.4-2: S1 >= 0.5, Site Class D; Fv requires site-specific geotechnical investigation by Section 11.4.8)} ]

Note: For S1 >= 0.5 and Site Class D, ASCE 7-22 Section 11.4.8 requires a site-specific ground motion hazard analysis. We proceed with the mapped values and note that site-specific analysis may reduce Fv from 1.7 toward 1.5 per ASCE 7-22 Section 21.2.

1b. Adjusted MCER Spectral Accelerations (Section 11.4.4)

[ S_{MS} = F_a \times S_s = 1.0 \times 1.5 = 1.50 ]

[ S_{M1} = F_v \times S_1 = 1.7 \times 0.6 = 1.02 ]

1c. Design Spectral Accelerations (Section 11.4.5)

[ S*{DS} = \frac{2}{3} \times S*{MS} = \frac{2}{3} \times 1.50 = 1.00 ]

[ S*{D1} = \frac{2}{3} \times S*{M1} = \frac{2}{3} \times 1.02 = 0.68 ]

1d. Seismic Design Category (Tables 11.6-1 and 11.6-2)

Based on SDS = 1.00 and SD1 = 0.68, for Risk Category II:

• Short-period SDC: SDS = 1.00 >= 0.50 → SDC D (Table 11.6-1)
• Long-period SDC: SD1 = 0.68 >= 0.20 → SDC D (Table 11.6-2)
• Governing: SDC D

SDC D triggers numerous seismic design and detailing requirements per AISC 341 (Seismic Provisions for Structural Steel Buildings).

Step 2 — Selecting the Analysis Procedure (Section 12.6)

For SDC D buildings, Section 12.6 permits the Equivalent Lateral Force (ELF) procedure when the building is regular (< 5 irregularities) and height <= 160 ft. Our 65 ft building qualifies.

The design spectral response acceleration for ELF:

[ Cs = \frac{S{DS}}{\left(\frac{R}{I_e}\right)} = \frac{1.00}{\left(\frac{8}{1.0}\right)} = 0.125 ]

But Cs must satisfy minimum and maximum bounds per Section 12.8.1.1:

Step 3 — Fundamental Period T (Section 12.8.2)

Two values are required: the approximate fundamental period Ta and the calculated period T from analysis (which must not exceed Cu x Ta for determining base shear).

3a. Approximate Fundamental Period Ta

For steel special moment frames, ASCE 7-22 Table 12.8-2:

[ T_a = C_t \times h_n^x = 0.028 \times (65)^{0.8} ]

[ T_a = 0.028 \times 65^{0.8} = 0.028 \times 28.3 = 0.79 \text{ s} ]

Where: Ct = 0.028, x = 0.8 for steel moment-resisting frames.

3b. Coefficient for Upper Limit on Calculated Period Cu

For SD1 = 0.68, from Table 12.8-1: [ C_u = 1.4 \text{ (SD1 >= 0.4)} ]

Upper limit on period for base shear calculation: [ T_{max} = C_u \times T_a = 1.4 \times 0.79 = 1.11 \text{ s} ]

If a computer model calculates a longer period, the base shear must be based on T <= 1.11 s per Section 12.8.2. We use T = Ta = 0.79 s for this hand calculation.

Step 4 — Seismic Response Coefficient Cs (Section 12.8.1.1)

4a. Cs Equation (12.8-2) — Short-Period Limit

[ Cs = \frac{S{DS}}{\left(\frac{R}{I_e}\right)} = \frac{1.00}{8 / 1.0} = \frac{1.00}{8} = 0.125 ]

4b. Cs Equation (12.8-3) — Long-Period Limit (must not exceed)

For T = 0.79 s <= TL (TL = 8 s for California, ASCE 7-22 Figure 22-12):

[ Cs = \frac{S{D1}}{T \times \left(\frac{R}{I_e}\right)} = \frac{0.68}{0.79 \times (8 / 1.0)} = \frac{0.68}{6.32} = 0.108 ]

4c. Cs Minimum (12.8-5)

[ C*{s,min} = 0.044 \times S*{DS} \times I_e = 0.044 \times 1.00 \times 1.0 = 0.044 ]

For S1 >= 0.6, also check (12.8-6): [ C_{s,min} = \frac{0.5 \times S_1}{\left(\frac{R}{I_e}\right)} = \frac{0.5 \times 0.6}{8/1.0} = \frac{0.30}{8} = 0.0375 ]

4d. Governing Cs

The long-period limit governs because T = 0.79 s exceeds Ts = SD1 / SDS = 0.68 / 1.00 = 0.68 s, placing the building on the descending branch of the design spectrum.

Step 5 — Seismic Base Shear V (Section 12.8.1)

5a. Effective Seismic Weight W (Section 12.7.2)

The effective seismic weight includes total dead load and applicable portions of other loads:

Floor levels (Levels 2-5): Area per floor = 120 ft x 90 ft = 10,800 ft^2 Floor dead load = 85 psf W_floor_DL = 10,800 x 85 = 918,000 lb = 918 kips per floor

Floor live load contribution (Section 12.7.2): For office occupancies in storage/warehouse areas with heavy live load — but for standard office, no live load need be included in W. Per exception, for office occupancy: minimum 0 psf inclusion.

However, for partition loads where floor live load < 80 psf: include 10 psf partition allowance per Section 4.3.2. Since the 85 psf dead load likely already includes partition allowance, we use dead load only.

Total per-floor weight = 918 kips

Cladding weight per floor: Perimeter = 2 x (120 + 90) = 420 ft Story height = 13 ft, cladding = 15 psf W_clad = 420 x 13 x 15 = 81,900 lb = 81.9 kips per floor

Roof level: Roof area = 10,800 ft^2 Roof dead load = 20 psf W_roof_DL = 10,800 x 20 = 216,000 lb = 216 kips Roof live load (less than 20 psf): not included in W per Section 12.7.2 exception. Roof cladding: assume parapet only, 3 ft x 420 ft x 15 psf = 18,900 lb = 19 kips

Effective seismic weight summary:

5b. Base Shear

[ V = C_s \times W = 0.108 \times 4,235 = 457 \text{ kips} ]

This represents approximately 10.8% of the building's seismic weight. For SDC D with special moment frames, this is consistent with the high seismicity (Ss = 1.5) but the ductility reduction (R = 8) limits the elastic demand.

Step 6 — Vertical Distribution of Seismic Forces (Section 12.8.3)

The base shear is distributed vertically as:

[ Fx = C{vx} \times V ]

[ C_{vx} = \frac{w_x \times h_x^k}{\sum w_i \times h_i^k} ]

Where k is the exponent related to building period:

• T <= 0.5 s: k = 1.0 (triangular distribution)
• T >= 2.5 s: k = 2.0 (parabolic distribution)
• T = 0.79 s: interpolate, k = 1.0 + (0.79 - 0.5) x (2.0 - 1.0) / (2.5 - 0.5) = 1.0 + 0.29 x 0.5 = 1.145

Use k = 1.15 (rounded to 2 decimal places per Section 12.8.3).

*The sum of Fx slightly exceeds the base shear of 457 kips due to k-exponent rounding and numerical approximation. Adjust proportionally to match V=457 kips.

**Cvx sum > 1.0 due to rounding. Normalize all forces by factor 457/492.9 = 0.927.

Adjusted vertical distribution (normalized):

Check: Sum Fx = 457 kips = V (matches).

Step 7 — Horizontal Distribution and Torsion (Section 12.8.4)

7a. Accidental Torsion

Per Section 12.8.4.2, the center of mass at each level is displaced 5% of the building dimension perpendicular to the direction of force:

For seismic force in the E-W direction (120 ft dimension): [ e_{acc} = 0.05 \times 120 = 6.0 \text{ ft} ]

For N-S direction: [ e_{acc} = 0.05 \times 90 = 4.5 \text{ ft} ]

7b. Torsional Amplification (Section 12.8.4.3)

For buildings in SDC C-F with torsional irregularity Type 1a or 1b, the accidental torsion must be multiplied by Ax. Assuming a regular building without torsional irregularity (verified by analysis or by checking drift ratios at extreme frames):

Ax = 1.0 (no torsional irregularity assumed for this preliminary calculation).

For a building known to have a torsional irregularity, Ax would be computed from the ratio of maximum story drift at one end to average story drift.

7c. Distribution to Moment Frames

For the E-W direction with 4 perimeter moment frames (assume 2 on each of the north and south faces):

Per-frame shear at Level 5: V5,frame = 171.1 / 4 = 42.8 kips

Per-frame shear including 5% accidental torsion (frame located at distance d = 45 ft from center):

Torsional moment: Mt = V x e_acc = 171.1 x 6.0 = 1,027 kip-ft

Torsional force at frame (stiffness proportional, assuming equal stiffness for 4 frames separated by 30 ft typical bay spacing):

Ft = Mt x (K_frame x d_frame) / sum(K_i x d_i^2)

With equal frame stiffness and symmetric layout: Ft = 1,027 x 45 / (2 x (15^2 + 45^2)) = 1,027 x 45 / (2 x (225 + 2,025)) = 1,027 x 45 / 4,500 = 10.3 kips

Total frame force at Level 5: F_total = 42.8 + 10.3 = 53.1 kips (worst frame, amplified side).

Step 8 — Story Drift Determination (Section 12.8.6)

The design story drift is the difference in deflections at the center of mass of the top and bottom of the story, multiplied by the deflection amplification factor Cd.

For steel special moment frames: Cd = 5.5 (ASCE 7-22 Table 12.2-1).

8a. Elastic Deflection (from analysis)

Without running a full structural analysis model, a simplified estimate based on seismic weight and approximate stiffness:

For a 4-bay moment frame 120 ft wide with W24 sections: Approximate stiffness per frame at each level: k ≈ 100 kips/inch (typical for medium-span steel moment frames)

Elastic deflection at roof: [ \delta*{xe,roof} = V*{roof,frame} / k_effective = 52.1/4 / 100 = 13.0/100 = 0.13 \text{ in} ]

8b. Amplified Deflection

[ \deltax = \frac{C_d \times \delta{xe}}{I_e} = \frac{5.5 \times 0.13}{1.0} = 0.72 \text{ in} ]

8c. Interstory Drift Ratio

Roof displacement amplified = 0.72 in Level 5 displacement amplified ≈ 0.58 in (scaled proportionally)

Interstory drift (Roof to Level 5): [ \Delta = 0.72 - 0.58 = 0.14 \text{ in} ] [ \text{Drift ratio} = \frac{0.14}{13 \times 12} = \frac{0.14}{156} = 0.00090 = 0.09% ]

Check against allowable drift per Table 12.12-1: For Risk Category II, all other structures: allowable drift = 0.020 x hsx = 0.020 x (13 x 12) = 3.12 in.

0.14 in << 3.12 in → drift is acceptable by a large margin, typical for steel moment frames controlled by strength rather than drift at this height.

Step 9 — Overturning Moment and P-Delta Check (Section 12.8.7)

9a. Overturning Moment at Base

[ M_{OT} = \sum F_x \times h_x ]

[ M_{OT} = 19,502 \text{ kip-ft} ]

9b. Stability Coefficient P-Delta Check (Section 12.8.7)

The stability coefficient theta per Section 12.8.7:

[ \theta = \frac{Px \times \Delta \times I_e}{V_x \times h{sx} \times C_d} ]

At Level 2 (maximum gravity load, maximum theta): Px = total unfactored vertical load at and above Level 2 ≈ 4,235 kips Δ = amplified interstory drift (Level 2 to Level 3) Vx = story shear at Level 2 = 456.9 kips hsx = story height = 13 ft = 156 in Cd = 5.5

Estimated elastic drift at Level 2: similar analysis gives Δ_elastic ≈ 0.08 in Amplified drift: Δ = 5.5 x 0.08 / 1.0 = 0.44 in

[ \theta = \frac{4,235 \times 0.44 \times 1.0}{457 \times 156 \times 5.5} = \frac{1,863}{392,106} = 0.0047 ]

Theta_max = 0.5 / (beta x Cd) = 0.5 / (1.0 x 5.5) = 0.091

Since theta = 0.0047 < 0.10, no P-Delta effects need be considered per Section 12.8.7 (theta < 0.10 satisfactory).

Step 10 — Seismic Load Combinations (Section 12.4)

The seismic load effect E combines horizontal and vertical effects:

[ E = E_h \pm E_v ]

Where: [ Eh = \rho \times Q_E \text{ (horizontal)} ] [ E_v = 0.2 \times S{DS} \times D = 0.2 \times 1.00 \times D = 0.20 D \text{ (vertical, for LRFD)} ]

Redundancy factor rho: for SDC D with special moment frames, rho = 1.0 per Section 12.3.4.2 (each story has > 2 bays of moment frame on each perimeter face, satisfying the redundancy requirement without penalty).

Basic LRFD load combinations with seismic per ASCE 7-22 Section 2.3.6:

Combination 5: (1.2 + 0.2 SDS) D + 1.0 E + L + 0.2 S = (1.2 + 0.20) D + 1.0 QE + 0.5 L = 1.40 D + 1.0 QE + 0.5 L

Combination 7: (0.9 — 0.2 SDS) D + 1.0 E = (0.9 – 0.20) D + 1.0 QE = 0.70 D + 1.0 QE

Key Cross-Code Comparisons

Frequently Asked Questions

Why does Site Class D increase the design forces so significantly compared to Site Class C?

Site Class D (stiff soil) amplifies ground motion, particularly at longer periods, because softer soils have lower shear wave velocities (600–1,200 ft/s) that trap and amplify seismic energy. For S1 = 0.6, Fa rises from 1.0 (Class C) to 1.0 (Class D) — not much change at short periods — but the critical difference is Fv: 1.5 for Class C vs 1.7 for Class D, a 13% increase that directly scales the long-period design spectrum and base shear when T > Ts. For sites with S1 in the 0.4–0.6 range, Site Class D almost always requires a site-specific ground motion analysis per ASCE 7-22 Section 11.4.8, and the results often moderate Fv toward 1.5, recovering some design efficiency.

When can I use the Equivalent Lateral Force method instead of modal response spectrum analysis?

The ELF procedure is permitted by ASCE 7-22 Section 12.6 when: (a) the building is in SDC B or C, or (b) in SDC D-F with no structural irregularities of Types 1a, 1b, 4, or 5 and building height <= 160 ft. For SDC D buildings, the ELF method is still permitted when the building is regular, but modal response spectrum analysis is increasingly standard practice for buildings over 3 stories. Our 5-story regular building technically qualifies for ELF, but a modal analysis would verify that higher-mode effects do not significantly alter the story shear distribution, especially at upper stories.

How do I handle the vertical distribution when k > 1.0?

The exponent k accounts for higher-mode effects that concentrate seismic force at upper stories. When T > 0.5 s, the upper-story forces increase relative to the triangular distribution (k = 1.0). This is because longer-period structures respond with significant contributions from second and third modes, which push force upward. For T = 0.79 s, k = 1.15 represents only a modest 15% shift from triangular. At T = 2.5 s, k = 2.0 shifts nearly 40% of the base shear into the top quarter of the building — significant for tall steel frames.

What is the practical effect of the redundancy factor rho?

The redundancy factor rho penalizes structural systems with too few lateral-force-resisting elements by increasing the seismic design force. For SDC D-F, the default rho = 1.3 unless the building demonstrates adequate redundancy per Section 12.3.4.2. A value of rho = 1.3 adds 30% to the seismic lateral force, which cascades through the entire design. Achieving rho = 1.0 requires: at least 2 bays of seismic-force-resisting perimeter framing on each side of the building in each direction, or that removal of any single element does not create an extreme torsional irregularity. Most well-configured steel moment-frame buildings meet the rho = 1.0 criteria. Always document the redundancy check in the Basis of Design.

Why does ASCE 7-22 require site-specific analysis for S1 >= 0.5 on Site Class D?

The requirement exists because the standard mapped spectral values and site coefficients become unconservative for soft soils in very high seismicity. The Fv = 1.7 value for Site Class D was calibrated against a limited set of sites, and for S1 >= 0.5, the ground-motion prediction equations show greater variability in long-period amplification than the single deterministic coefficient captures. A site-specific probabilistic seismic hazard analysis (PSHA) develops a site-specific uniform hazard spectrum that reflects the actual soil profile, basin effects, and near-source characteristics. The result is often a reduction in Fv (from 1.7 toward 1.5) but the analysis costs $15,000–$30,000 — on a large project, the steel-weight savings usually justify the geotechnical investment.

Reference only. Verify all values against the current edition of ASCE/SEI 7-22 Minimum Design Loads and Associated Criteria for Buildings and Other Structures. This worked example does not constitute professional engineering advice and must be independently verified by a licensed Professional Engineer for the specific project conditions, site characteristics, and governing building code edition.