Seismic Load Worked Example — ELF Procedure per ASCE 7-22

Problem: Determine the seismic base shear and distribution of lateral forces for a 4-story steel moment frame office building located in Los Angeles, California. Use the Equivalent Lateral Force (ELF) procedure per ASCE 7-22 Chapter 12. Building is 60 ft × 120 ft in plan, 52 ft tall total, with 13 ft typical story heights.

Step 1: Building Parameters

Step 2: Seismic Design Category (SDC)

From ASCE 7-22 Figures 22-1 through 22-7, for Los Angeles:

S_S (short period mapped acceleration) = 1.50g S_1 (1-second mapped acceleration) = 0.60g

Site Class D — Site coefficients (Table 11.4-1 and 11.4-2):

F_a = 1.0 (for S_S ≥ 1.25 and Site Class D) F_v = 1.5 (for S_1 ≥ 0.50 and Site Class D)

Adjusted MCER spectral accelerations:

S_MS = F_a × S_S = 1.0 × 1.50 = 1.50g S_M1 = F_v × S_1 = 1.5 × 0.60 = 0.90g

Design spectral accelerations (ASCE 7-22 Eq 11.4-1 and 11.4-2):

S_DS = (2/3) × S_MS = (2/3) × 1.50 = 1.00g S_D1 = (2/3) × S_M1 = (2/3) × 0.90 = 0.60g

Seismic Design Category (Tables 11.6-1 and 11.6-2):

From Table 11.6-1 (S_DS = 1.00g): Risk Category II, S_DS ≥ 0.50 → SDC = D From Table 11.6-2 (S_D1 = 0.60g): Risk Category II, S_D1 ≥ 0.20 → SDC = D

Result: SDC = D (both criteria give D)

Step 3: System Parameters (Table 12.2-1)

For Special Steel Moment Frame:

• Response modification coefficient (R) = 8.0
• Overstrength factor (Ω₀) = 3.0
• Deflection amplification factor (Cd) = 5.5
• Allowable height limit for SDC D = No limit (SMF, Seismic Design Category D)

Step 4: Fundamental Period (Section 12.8.2)

Approximate period (Ta) per Table 12.8-2:

For steel moment frames: Ct = 0.028, x = 0.80

Ta = Ct × hn^x = 0.028 × (52)^0.80

ln(52) = 3.951 0.80 × 3.951 = 3.161 52^0.80 = e^3.161 = 23.6

Ta = 0.028 × 23.6 = 0.66 seconds

Upper limit for period (Section 12.8.2.1):

Cu = 1.4 (for S_D1 = 0.60g ≥ 0.30g, from Table 12.8-1)

T_max = Cu × Ta = 1.4 × 0.66 = 0.92 seconds

For a steel moment frame, a more accurate period from structural analysis is typically in the 0.9-1.2 second range. We will use T = 0.92 seconds (the code maximum for ELF).

Step 5: Seismic Response Coefficient (Cs)

ASCE 7-22 Equation 12.8-2:

Cs = S_DS / (R/I_e)

Where I_e = 1.0 (Risk Category II, Table 1.5-2)

Cs = 1.00 / (8.0 / 1.0) = 0.125

Check upper limit (Equation 12.8-3): Cs_max = S_D1 / (T × (R/I_e)) Cs_max = 0.60 / (0.92 × 8.0) = 0.082

Check lower limit (Equation 12.8-5): For S_1 ≥ 0.60g: Cs_min = 0.5 × S_1 / (R/I_e) = 0.5 × 0.60 / 8.0 = 0.038

Check absolute minimum (Equation 12.8-6): Cs_min_abs = 0.044 × S_DS × I_e = 0.044 × 1.00 × 1.0 = 0.044

Governing Cs = min(0.125, 0.082) = 0.082, max(0.038, 0.044) = 0.044

Cs = 0.082 (the period-based upper limit governs)

Step 6: Base Shear (ASCE 7-22 Equation 12.8-1)

V = Cs × W_total

V = 0.082 × 4,500 kips

V = 369 kips

Step 7: Vertical Distribution of Seismic Forces (Section 12.8.3)

The lateral force at each level (Fx) is:

Fx = Cvx × V

Where:

Cvx = (wx × hx^k) / Σ(wi × hi^k)

And k = 1.0 for T ≤ 0.5 sec, k = 2.0 for T ≥ 2.5 sec. For T = 0.92 sec, k = 1.0 + (0.92 - 0.50) × (2.0 - 1.0)/(2.5 - 0.5) = 1.0 + 0.42 × 0.5 = 1.21

*Note: Small rounding difference from the 369 kip total. Always use the exact computed distribution sum.

Step 8: Story Drift Check (Section 12.8.6)

For a preliminary check, assume the moment frame provides a lateral stiffness that limits elastic drift to Δ_e:

Using a lateral analysis of the frame (performed with the seismic calculator):

Estimated elastic drifts:

Drift check at roof: 3.19 in > 3.12 in → Drift ratio exceeds 0.020h_sx limit. The moment frame section sizes need to be increased (approximately 5-10% stiffness increase) to bring roof drift under 3.12 in.

Step 9: Diaphragm Forces (Section 12.10.1)

Design force for diaphragm at each level:

F_px = Σ(F_i from level i to roof) × w_px / w_total_above

At the 2nd floor: F_p2 = (129 + 122 + 77) × 1,200 / (900 + 1,200 + 1,200) F_p2 = 328 × 1,200 / 3,300 = 119 kips

Minimum: 0.2 × S_DS × I_e × w_px = 0.2 × 1.00 × 1.0 × 1,200 = 240 kips ← governs

Maximum: 0.4 × S_DS × I_e × w_px = 0.4 × 1.00 × 1.0 × 1,200 = 480 kips

Design diaphragm force at 2nd floor = 240 kips

This force must be distributed to the collectors and drag struts at each end of the diaphragm.

Step 10: Redundancy Factor (Section 12.3.4)

For SDC D, the redundancy factor ρ = 1.3 unless the structure meets either:

• Condition a: Each story has at least 2 bays of perimeter SFRS (moment frame) on each side, OR
• Condition b: Removing a single frame does not reduce story strength by more than 33%

For a 60 ft × 120 ft building with moment frames on all sides, at least 2 bays of moment frame on each side is likely → ρ = 1.0.

Summary of Results

Try the Calculator

Use the Seismic Load Calculator to compute base shear and lateral forces per ASCE 7-22 for your own building parameters and location.

Frequently Asked Questions

When is the Equivalent Lateral Force (ELF) procedure not permitted? ELF is not permitted for structures with significant torsional irregularity, extreme soft story irregularity, or nonparallel lateral force systems in SDC D-F. It is also not permitted for structures exceeding certain height limits depending on the SFRS type. For these cases, Modal Response Spectrum Analysis (Ch 12.9) or Nonlinear Response History Analysis (Ch 16) is required.

What is the difference between seismic base shear from ELF and Response Spectrum Analysis? ELF tends to be conservative for regular structures because it assumes the fundamental mode dominates. Response Spectrum Analysis captures higher mode effects and typically gives lower base shear (10-30% less) for long-period structures. However, RSA requires a full 3D structural model and modal analysis.

How do I determine the seismic weight of a floor? Seismic weight includes dead load plus applicable portions of other loads (25% of floor live load for storage, snow load where applicable, partition weight). For office buildings, include: self-weight of structure (steel frame, slab, deck), superimposed dead loads (MEP, ceiling, flooring, partitions), and the weight of any permanent equipment.

What does the deflection amplification factor Cd mean? Cd amplifies the elastic displacement from the design base shear to estimate the maximum inelastic displacement. Cd = 5.5 for special steel moment frames means the actual roof drift during a design-level earthquake is estimated to be 5.5 times the elastic drift computed from the reduced (R-factor) seismic forces.

See Also

• Seismic Load Calculator
• Seismic Load Worked Example
• Seismic Design Basics