Seismic Loads and Lateral Force Distribution

Seismic Loads and Lateral Force Distribution

Seismic problems become manageable when each symbol is assigned one job. For this exam, use ASCE 7-16 with IBC 2018; use NDS 2018/SDPWS 2015 ASD only when the resulting wood diaphragm, wall, or connection is designed. Do not substitute ASCE 7-22 multi-period procedures.

Distinctions that must remain separate

A frequent wrong approach is to say “Site Class D means SDC D” or “R = 5 means Site Class 5.” Neither statement is meaningful. Site Class modifies hazard through site coefficients; SDC is assigned from the code criteria using risk category and design spectral values; R comes from the chosen lateral system; and Cs is calculated after those inputs are established.

Base-shear workflow

Follow the reference in this order:

1. Select risk category and seismic importance factor Ie.
2. Obtain mapped Ss and S1, then determine Site Class from the geotechnical information.
3. Use the ASCE 7-16 site coefficients Fa and Fv to calculate SMS = FaSs and SM1 = FvS1; then calculate SDS = (2/3)SMS and SD1 = (2/3)SM1. Apply the standard's site-specific requirements where triggered.
4. Assign SDC using the applicable ASCE 7-16/IBC 2018 criteria.
5. Select a permitted seismic-force-resisting system and obtain its R, overstrength factor, and deflection amplification factor. System selection must respect SDC and height limits.
6. Assemble effective seismic weight W. It includes dead load and the code-required portions of other loads, permanent equipment, partitions, storage loads, or snow where applicable. It is not simply “all live load.”
7. Determine the analytical period and calculate Cs. The familiar short-period expression is Cs = SDS/(R/Ie), but the ASCE 7-16 upper and lower bounds must also be checked.
8. Compute base shear V = CsW, distribute it vertically, and then distribute each level's force horizontally with the required torsional effects.

Worked base shear and vertical distribution

Consider a three-story building for which the ASCE 7-16 steps have produced SDS = 0.50, R = 5, and Ie = 1.0. Assume the period-dependent maximum and the code minimum have been checked and do not change the short-period result. Then

Cs = 0.50/(5/1.0) = 0.10.

Each floor contributes 1,000 kip to effective seismic weight, so W = 3,000 kip and

V = 0.10(3,000) = 300 kip.

For the given short-period case, take distribution exponent k = 1.0. Floor heights above the base are 12, 24, and 36 ft. With equal floor weights,

Cvx = wx hx^k / Σ(wi hi^k).

The denominator is 1,000(12 + 24 + 36) = 72,000 kip-ft. Therefore:

The forces sum to the 300-kip base shear. This arithmetic check is essential. Do not distribute shear equally merely because floor weights are equal; elevation and the exponent matter. Story shear at a level is the sum of the forces at and above that level, so story shear is not the same quantity as that floor's Fx.

Diaphragms, collectors, and foundations

At each level, inertia force begins at the distributed mass. The diaphragm carries in-plane shear and bending. Chords resist diaphragm bending tension and compression; collectors or drag struts gather force around openings or discontinuities and deliver it to frames or shear walls. Those vertical elements transfer shear and overturning through connections, hold-downs, base plates, anchor rods, and foundations into the ground. Stopping the sketch at a shear wall is an incomplete load path.

Do not declare a diaphragm rigid or flexible solely because it is concrete, steel deck, or wood. Use the ASCE 7-16 classification and deformation criteria applicable to the configuration. Large openings, offsets, reentrant corners, and discontinuous vertical elements can redirect forces and create collector demands.

Accidental torsion

Horizontal force acts through the center of mass, while resistance acts through the center of rigidity. Actual eccentricity creates torsion. ASCE 7-16 also requires accidental torsion by offsetting the center of mass by 5% of the building dimension perpendicular to the applied force, with additional torsional amplification where required.

If a level force is 100 kip and the perpendicular plan dimension is 60 ft, the accidental eccentricity is 0.05(60) = 3 ft, producing an added torsional moment of 100(3) = 300 kip-ft for that sign of offset. Evaluate the required positive and negative offsets. This moment changes how the diaphragm distributes force among vertical resisting lines; it is not another vertical load and does not simply add 300 kip-ft to base shear.

Final exam check

Confirm the chain: mapped hazard → Site Class/site coefficients → design spectrum → SDC → permitted system and R → W → bounded Cs → V → Fx → diaphragm/torsion → collectors and vertical elements → foundation. If two adjacent terms in that chain have been treated as synonyms, the solution is probably wrong.