No tail water is given in the figure. What head H is used in the uplift check?

H = 230 - 210 = 20 m, with no water downstream

H = 230 - 213 = 17 m (upstream floor)

H = 230 - 216 = 14 m (crest level)

H = 230 - 205 = 25 m (pile bottom)

How is the total creep length of 88 m obtained?

Horizontal floor lengths plus twice each pile depth plus the 2 m of vertical step creep

Each pile depth counted once plus floor lengths

H multiplied by the creep coefficient

Sum of floor lengths only: 15 + 25 + 20 = 60 m

How is the residual head at a point found?

H x (creep to the point) / L_total

2071 Magh · 12 marks

Khosla: is the section safe at A and B

Short answer

For worst-case head 20 m and G=2.24, Bligh creep length 88 m gives residual heads above tailwater: B 7.27 m, A 2.27 m; unbalanced above floor top B 5.27 m, A 2.27 m. Required thickness t=h/(G-1): B 4.25 m, A 1.83 m, so A safe, B not safe.

Find whether the section provided is safe against uplift at A and B.

[Figure: water level 230 m; crest 216 m; upstream floor 213 m and 212 m with 4:1 upstream slope; upstream pile bottom 208 m; intermediate pile bottom 209 m; downstream floor levels 212 m (point B), 211 m, 210 m (point A); downstream pile bottom 205 m; distances 15 m, 25 m and 20 m]

The structure is checked for uplift by finding the residual seepage head under the floor at A and B (creep theory), converting it to the floor thickness needed to resist it, and comparing this with the floor provided. The figure gives no tail water, so the worst case is taken: no water on the downstream side, which gives the maximum static head. Missing values are assumed as stated.

Assumptions

Upstream water level 230 m; downstream water level at the lowest floor level (point A), 210 m. $H = 230 - 210 = 20$ m.
Upstream pile depth $d_1 = 213 - 208 = 5$ m; intermediate pile $d_2 = 212 - 209 = 3$ m; downstream pile $d_3 = 210 - 205 = 5$ m.
Floor lengths: 15 m upstream floor, 25 m under the crest, 20 m downstream floor (A at the end of the downstream floor, B at the end of the crest floor). The two steps of the downstream floor (212 to 210 m) add 2 m of vertical creep.
Specific gravity of concrete $G = 2.24$ .

Creep path (Bligh)

\begin{aligned} L_{total} &= 2d_1 + 15 + 2d_2 + 25 + 20 + 2 + 2d_3 \\ &= 10 + 15 + 6 + 25 + 20 + 2 + 10 = 88\ \text{m} \end{aligned}

The creep length from the upstream end to B is $56$ m and to A is $78$ m. Residual head above the downstream water level $= H\,(L_{total} - L_x)/L_{total}$ .

Point	Creep to point (m)	Residual head above 210 m (m)	Unbalanced head above floor top (m)	Thickness needed $t = h/(G-1)$ (m)
B (212 m)	56	7.27	5.27 (7.27 less 2 m, since B is 2 m above 210 m)	4.25
A (210 m)	78	2.27	2.27	1.83

For the floor to be safe, the submerged weight of the floor must at least balance the uplift: $t(G-1) \ge h$ .

Conclusion

At A, only 2.27 m of head remains, which needs a floor about 1.83 m thick (2.44 m with the usual extra 33 %). A floor of 2 m or more is safe.
At B, the head still to be dissipated is 5.27 m, which needs 4.25 m of floor (5.67 m with the extra 33 %). A normal 1.5 to 2.5 m floor is not safe here against the full static head.
Remedies: allow for the actual tail water (it reduces $H$ sharply), lengthen the floor, deepen the intermediate pile, or provide pressure-relief weep holes with an inverted filter under the crest floor.

Answer: With the worst-case head of 20 m, the section is safe at A (needs 1.83 m) but not safe at B (needs 4.25 m) unless more floor thickness, creep length or tail water is provided.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Khosla: is the section safe at A and B

Assumptions

Creep path (Bligh)

Conclusion

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