In day-to-day structural design, we often focus on gravity loads — but it is the lateral loads (wind and earthquake) that truly test a column’s design. When a building sways under lateral forces, each story undergoes a horizontal displacement known as story drift (Δ). This drift forces columns to resist bending moments — and the magnitude of those moments depends entirely on how the column ends are restrained.
This post gives you a ready-to-use reference for the moment formulas in columns under lateral drift, broken down by end conditions. Whether you’re doing a quick hand check or verifying ETABS output, this guide will save you time.
Imagine two identical columns — same cross-section, same material, same height — but one is fixed at both ends and the other is pinned at both ends. Under the same story drift, the fixed-fixed column will attract 100% of the moment, while the pinned-pinned column attracts zero moment. This is the fundamental reason why modeling boundary conditions correctly in software like ETABS or SAP2000 is critical — a wrong assumption here directly affects reinforcement design.
(All moments resist the drift direction; CW = Clockwise, CCW = Counter-Clockwise)
| End Conditions | Moment at Top (Mtop) | Moment at Bottom (Mbot) |
|---|---|---|
| Both Ends Fixed | (6 × E × I × Δ) / L² (CW) | (6 × E × I × Δ) / L² (CCW) |
| Top Pinned, Bottom Fixed | 0 | (3 × E × I × Δ) / L² (CCW) |
| Top Fixed, Bottom Pinned | (3 × E × I × Δ) / L² (CW) | 0 |
| Both Ends Pinned | 0 | 0 |
| Top Guided, Bottom Fixed | (3 × E × I × Δ) / L² (CW) | (3 × E × I × Δ) / L² (CCW) |
Consider a concrete column (M25 grade) in a multi-storey building with the following properties:
Substituting into the formulas (M = k × EI × Δ / L², where k = 6 for Fixed-Fixed, k = 3 for others):
| End Condition | Formula | M_top (kNm) | M_bot (kNm) | Design Implication |
|---|---|---|---|---|
| Fixed-Fixed | 6 × 1250 × 0.02 / 9 | 16.67 (CW) | 16.67 (CCW) | Highest demand — check both ends for flexure + axial interaction |
| Pinned-Fixed | 3 × 1250 × 0.02 / 9 | 0 | 8.33 (CCW) | Design base for combined bending; top connection is simple |
| Fixed-Pinned | 3 × 1250 × 0.02 / 9 | 8.33 (CW) | 0 | Design top for bending; base acts as a pin — verify foundation |
| Guided-Fixed | 3 × 1250 × 0.02 / 9 | 8.33 (CW) | 8.33 (CCW) | Equal moments top and bottom — design both ends; uncommon in practice |
| Pinned-Pinned | — | 0 | 0 | No moment from drift — ensure lateral system carries all shear |
Quick Calculation Check (Fixed-Fixed):
EI = 25×10⁶ kN/m² × 5×10⁻⁵ m⁴ = 1,250 kN·m²
M = (6 × 1,250 × 0.02) / (3)² = 150 / 9 = 16.67 kNm ✓
The moment induced in a column by story drift is not just a textbook concept — it directly affects how you design and detail column reinforcement, beam-column connections, and lateral load resisting systems. The stiffer the boundary conditions, the greater the moment demand. Use the formulas and tables above as a first-principles check alongside your structural analysis software to ensure your designs are both safe and efficient.