Clause 7.6.3 specifies how the total design base shear VB is distributed as lateral forces at each floor level. The distribution assumes a parabolic (approximately first-mode) deflection shape, with forces increasing quadratically with height. Each floor receives a lateral force proportional to the product of its seismic weight and the square of its height from the base.
Key Requirements
•The design lateral force at floor i: Qi = VB × (Wi × hi²) / Σ(Wj × hj²)
•Wi is the seismic weight lumped at floor i
•hi is the height of floor i from the base of the building
•The summation in the denominator is over all floors
•This distribution applies to the equivalent static method (Cl. 7.6)
•For dynamic analysis (response spectrum or time history), the distribution comes from the modal analysis
Formulas
Qi = VB × (Wi × hi²) / Σ(Wj × hj²)
Design lateral force at floor level i
Qi = Design lateral force at floor i in kNVB = Total design base shear in kNWi = Seismic weight at floor i in kNhi = Height of floor i from the base in metresWj, hj = Seismic weight and height at any floor j
Practical Notes
✓The top floors receive the largest lateral forces due to the hi² term — this governs column and beam design at upper levels
✓For a uniform building, approximately 50% of the base shear goes to the top third of the building
✓This distribution is conservative for regular buildings and may underestimate forces for irregular buildings — use dynamic analysis for irregular cases
✓The base shear check must be satisfied: sum of all Qi must equal VB
Common Mistakes
⚠Using linear (hi instead of hi²) distribution — IS 1893 uses a parabolic distribution
⚠Assigning seismic weight at the wrong floor level — Wi should be at the floor slab level, not at mid-storey
⚠Not including roof-level weight (water tanks, parapets, etc.) at the topmost level
⚠Applying the equivalent static distribution to irregular buildings where dynamic analysis is required by Cl. 7.7