Lintel Load and Moment Calculator
Wall load above the opening, moment and steel estimate.
Inputs
Result
Formula
| Effective height = min(actual, span/3) | 45° spread |
| q = thickness × effective height × density + self weight | kN/m |
| M = q·l²/8, V = q·l/2 | simply supported |
| As ≈ M / (0.9·fy·h0) | rough estimate |
1.5 m clear span, 240 mm wall, 19 kN/m³ masonry, 3 m of wall above, 240×200 lintel: effective height min(3.0, 0.5) = 0.5 m gives a wall load of 2.28 kN/m plus 1.20 kN/m self weight, q = 3.48 kN/m. M = 0.98 kN·m, V = 2.61 kN, h0 = 165 mm, about 18 mm² of steel - nominal bars suffice.
FAQ
Why count only 0.5 m of a 3 m wall?
Masonry arches. The wall above the opening forms an arch and carries its weight to the piers either side instead of loading the lintel fully. Tests and codes accept the 45° spread, hence span/3. If a concrete beam or slab bears directly on the lintel, use the actual load - the simplification no longer applies.
How much bearing does a lintel need?
At least 240 mm each end and no less than the wall thickness. Shorter bearing crushes the masonry locally, showing up as corner cracking and settlement. Seismic regions require at least 360 mm and a reliable connection to ring beams or tie columns.
If the calculated steel is tiny, can it be omitted?
No. A small As means strength is not the issue, but minimum ratios and detailing still apply - typically not less than 2Φ10, or 2Φ12 per the catalogue - and the lintel must resist handling and shrinkage. It also frames the opening and ties to the wall either side.
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Results are estimating aids only and are not a design basis. Construction and final accounts must follow the drawings, the applicable standards and an approved method statement. Anything affecting structural safety, shoring or falsework must be checked by a suitably qualified engineer.