Question 9.1: This example considers the design of a glacis-type weir. The......

The régime width (equation (9.9)) of the upstream waterway,

B=4.75 Q^{1 / 2}                     (9.9)

B =  4.75\sqrt{1800}\simeq200\,\mathrm{m}. . Adopting the gross length of the weir as 200m and assuming 20 spans of 10m each centre-to-centre of piers, and a pier thickness of 1.5 m,

clear waterway = 200 19 × 1.5 = 171.5 m.

Neglecting the pier and abutment contractions and assuming the weir to be broad crested ( Q = 1.7\, b H^{3/2} – modular flow; Chapters 4 and 8),

total head over the crest,  H= (1800/1.7\times171.5)^{2/3}=3.36\,\mathrm{m},

velocity of approach, V=1800/200 \times7=1.3\,\mathrm{m} \,\mathrm{s} ^{-1}.

approach velocity head \simeq0.08\,\mathrm{m},

therefore

upstream total energy level (TEL) = 300.00 + 0.08 = 300.08 m AOD

where n is the number of piers (= 19), k_{\mathrm{P}} is the pier coefficient ( = 0.01 for semicircular piers), and k_{\mathrm{a}} is the abutment coefficient (= 0.1 for 45° wing walls). Therefore,

and the actual energy head,

Hence the RL of weir crest = 300.083.39 = 296.69 m AOD. Since the allowable afflux is 1 m,

downstream HFL = 300.00 – 1.00 = 299.00 m AOD.

Check whether the flow is free (modular) or submerged (non-modular). For the flow to be modular, i.e. not affected by submergence, the ratio H_2 / H_1 \text {, where } H_1 \text { and } H_2  are the upstream and downstream heads above the weir crest, is less than 0.75 (BSI, 1969; Bos, 1976):

upstream head, \textstyle H_{1} = 300.00 – 296.69 = 3.31 m

and

downstream head, H_2=299.00-296.69=2.31 \, m

Therefore, the submergence ratio,

Hence the flow is modular, and the weir discharges the design flow with the desired upstream and downstream water levels.

Note that if a structure is submerged its discharge capacity is reduced. The submerged flow, Q_{s} , may be estimated by using the modular flow (Q_{m}) equation with a correction factor, f (i.e. Q_{s} = f Q_{m}). The correction factor depends on the type of structure, the submergence ratio H_{2}/H_{1} , and the ratio P_{2}/H_{1} , where P_{2} is the crest height above the downstream channel bed (Table 9.1).

The régime scour depth (equation (9.10)),

R_{ s }=0.475(Q / f)^{1 / 3}               (9.10)

Provide cut-offs for (1) upstream scour depth = 1.75 × 5.78 = 10.11 m and (2) downstream scour depth = 2.00×5.78 = 11.96 m. The RL of the bottom of the upstream cut-off = 300.0010.11 = 289.89 m AOD and the RL of the bottom of the downstream cut-off = 299.00-11.96 = 287.04 m AOD. Therefore the depth of upstream cut-off pile, d_{1} = 293.00-289.89  \simeq 3.0 m, and the depth of the downstream cut-off pile, d_{2} =293.00-287.00 =6.0 m. The pool level, i.e. the upstream storage level or canal FSL upstream of the head regulator= 299.0 m AOD. Therefore, the maximum seepage head (assuming tailwater depth= 0),

The exit gradient, G_{\mathrm{e}}=1 / 6=H_{\mathrm{s}} / \pi d_2 \lambda^{1 / 2} (equation (9.8)), which gives

G_{ e }=H_{ s } / \pi d \sqrt{\lambda}                     (9.8)

and, from \lambda=\frac{1}{2}\left[1+\left(1+\alpha^2\right)^{1 / 2}\right] ,

therefore b = 6.22 × 6 = 37.32 m.
Providing an upstream glacis of 2 in 1, 5m of horizontal apron (safety factor for additional creep length), a downstream glacis of 3 in 1, and an 18 m long stilling basin (USBR type III basin – Chapter 5, Fig. 5.6 – length \simeq 3 times sequent depth), the total floor length,

(Fig. 9.10). The total creep length (Bligh) = 2×3+42.45+2×6 = 60.45 m. Therefore the rate of head loss (gradient)= H/L = 6/60.45 = 0.0992. The head loss in the upstream cut-off = 0.0992×6 = 0.595 m, and in the downstream cut- ff = 0.0992×12 = 1.19 m. The total head loss to the gate (point 1) = 0.595 + 0.0992 × 12.88 = 1.87 m. The total head loss up to end of the downstream glacis (point 2)  = 0.595 + 0.0992 × 24.45 = 3.02 m. The total head loss up to the mid-point of the stilling basin (point 3) = 0.595 + 0.0992 × 33.45 = 3.91 m. The RL of the HGL at

point 1 = 299.00 – 1.87 = 297.13 m AOD,

point 2 = 299.00 – 3.02 = 295.98 m AOD,

point 3 = 299.00 – 3.91 = 295.01 m AOD.

The uplift pressure head above the top surface of the structure at

point 1, h = 297.13 – 296.69 = 0.44 m,

point 2, h = 295.98 – 293.00 = 2.98 m,

point 3, h = 295.01 – 293.00 = 2.01 m.

The thickness of the concrete at

point 1 (crest) \simeq 1.2 h \simeq 0.5 \, m .

point 2 (upstream of stilling basin floor) \simeq 1.2 h \simeq 3.5 \, m ,

point 3 (centre of stilling basin floor) \simeq 1.2 h \simeq 2.5 m .

Adopt a stepped floor thickness downstream of the downstream glacis as shown in Fig. 9.10 and a nominal thickness of 0.3 m for the upstream glacis and the horizontal floor slab (uplift pressures are less than the weight of water above).