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Uplift at inside corners, 50 m floor weir
Short answer
Khosla's theory: α=6.25, λ=3.665, basic pressures u/s inside corner 65.01%, d/s 34.99%. After interference (+2.43% for u/s,–2.43% d/s) and thickness corrections (+1.36%,–2.73%), final u/s corner uplift 68.8% (3.44m), d/s corner 29.8% (1.49m). Exit gradient 0.104.
A weir has a solid horizontal floor length of 50 m with two lines of cutoff of 8 m depth below the river bed at its two ends. The floor thickness is 1 m at upstream end and 2 m at downstream end, with its upper level being in flush with river bed. For an effective head of 5 m over the weir, calculate the uplift pressure at the two inside corner points and also the exit gradient.
Given: floor length m, cutoffs at both ends, depth m below the river bed (the floor top is flush with the bed), head m, floor thickness m (u/s end) and m (d/s end). Khosla's theory is used. The inside corners are the junctions of the floor with the u/s pile () and with the d/s pile ().
Khosla's constants
Pressures from the simple profile (thin floor)
| Point | Formula | Pressure (%) |
|---|---|---|
| (top of u/s pile) | 100 | 100 |
| (tip of u/s pile) | 75.91 | |
| (u/s inside corner) | 65.01 | |
| (d/s inside corner) | 34.99 | |
| (tip of d/s pile) | 24.09 | |
| (d/s end) | 0 | 0 |
Correction for mutual interference of piles
Both piles are 8 m deep ( m), spaced m apart, total floor length m:
It is positive at (upstream of the interfering pile) and negative at (downstream of it).
Correction for floor thickness
The uplift acts on the underside of the floor, a distance below the key point, so the pressure is interpolated along the pile:
Corrected pressures and uplift heads
Exit gradient
Answer: Uplift at the u/s inside corner of (3.44 m); at the d/s inside corner (1.49 m); exit gradient , which is less than the safe gradient of about 1/6 (0.167) for fine sand, so the structure is safe against piping.
✓ Answer checked by Claude Opus 5.5 (high effort)
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