Question 9.2: The figure below (Fig. 9.15) is a proposed weir floor with t......

(A) UPSTREAM PILING 1

Figure 9.6 gives the pressures (from \phi_{\mathrm{D}}  and  \phi_{\mathrm{{E}}} curves) at D and E of the extreme downstream pile – see the definition sketch on this figure – as a first estimate which are then corrected for the floor thickness.

Theoretical d (from the floor level) = 100.00 – 93.00 = 7 m
Total floor length, b = 50.5 m
\therefore 1 / \alpha=d / b=7 / 50.5=0.139 \text {. }

From Fig. 9.6 P_{\scriptscriptstyle\mathrm{D}}/H = 23 %, if the piling had been located at the downstream end of the floor. Since the piling 1 under question is located at the upstream end, the law of reversibility gives

P_{ D 1} / H=100-23=77 \% (see Fig. 9.15)

Similarly from the chart we can get P_{ E } / H(\text { at } 1 / \alpha=0.139)=33 \% and hence (point E of the downstream pile corresponds to C_{1} of the pile at upstream end) by the law of reversibility,

Correction for floor thickness

Total pressure drop on the d/s face of the piling = 77 – 67 = 10 % over a height of 7  m; since the floor is 0.75 m thick (100 – 99.25) the corresponding correction

CORRECTION FOR THE MUTUAL INTERFERENCE OF PILES (SEE FIG.9.1 5 )

Interference of pile 2 on pile 1

True depths of the piles: D = d = 99.25 – 93.00 = 6.25 m and the distance between them, b_{1}= 15 m

\therefore the correction, C (equation (9.10))

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

Therefore the corrected pressure behind (downstream side) the first piling

( B ) INTERMEDIATE PILING – PILE 2 ( SEE FIG.9.1 5)

Theoretical depth of piling, d = 7 m (pile 1) and hence α = 50.5 / 7 = 7.21 and the distance b’ from the piling to the upstream end of the floor = 15.75 m.

From the pressure chart (left side diagram of Fig. 9.7) the pressure at C – see the notes on the chart for computations – corresponding to 1-b^{\prime} / b=0.688 \text { on }  x \text {-axis, } P_{ C } / H=31 \% \text { which gives } P_{ E } / H=100-31=69 \% .

Also, P_{ D } / H (from the right-hand diagram of Fig. 9.7) = 36 % which gives the actual

For P_C / H direct use of left-hand diagram of Fig. 9.7 gives (for α = 7.21 and b’/b = 0.312)

Correction for floor thickness

For P_E / H=(69-64) 0.75 / 7=-0.54 \%
For P_C / H=(64-53) 0.75 / 7=+1.18 \%.

Correction for mutual interference

INFLUENCE OF PILE 1 ON PILE 2 – ON P_E / H

As D = d, i.e. the piles 1 and 2 are of the same length, this effect is the same as the effect of pile 2 on pile 1. However, this correction is subtractive and  = – 3.03%.

INFLUENCE OF PILE 3 ON P _{ c } / H (SEE FIG.9.15)

Now D (measured from the top level of pile 2) = 99.25 – 91.00 = 8.25 m. Also, d = 99.25 – 93.00 = 6.25 m.

Correction for sloping floor for P _{ C } / H

The slope being \frac{1}{4} the theoretical correction is around 3.3% (see Section 9.1.5 a (iii)) and the corrected slope with the horizontal length of the sloping floor, b_2 (100.50 – 99.25) being 5 m

= 3.3 (5/34) = – 0.48 %.

\therefore The corrected pressure behind (E) the pile 2 is

and ahead (C) of the pile,

P_C / H=53.00+1.18+2.69-0.48=56.39 \approx 56 \% .

(C) DOWNSTREAM END PILING – PILE 3 (SEE FIG.9.15)

Theoretical pile depth, d = 98.50 – 91.00 = 7.50 m

1/α = d/b = 7.5/50.50 = 0.148 and

P_E / H (directly from pressure chart – see Fig. 9.5) = 36%.

Similarly, P_D / H =  26%.

CORRECTION FOR FLOOR THICKNESS, † (98.50 – 97.25) = 1.25 m

= (36 – 26) 1.25 / 7.5 = – 1.67 %.

Correction for interference of pile 2 on pile 3 (see Fig. 9.15)

Depth of pile 2, D = 97.25 – 93.00= 4.25 m
Depth of pile 3, d = 97.25 – 91.00 = 6.25 m.

\therefore The correction, C = 19 \sqrt{4.25 / 34} (4.25 + 6.25)/50.50 = – 0.073%.

Note: If the intermediate pile (pile 2, D) is shallower than the end pile (pile 3, d), i.e. D < d and b_1 \geq 2d ( = 12.50 m) the mutual interference correction is negligible. The corrected pressures behind (E) of the pile 3,

The actual pressure distributions are summarized in the following table:

Table 1

Bligh’s theory suggests linear pressure distribution with a safe exit gradient (see equation (9.1)),

G_{ e }=H_{ s } / L                     (9.1)

The actual exit gradient (after Khosla et al. – see equation (9.8)) = 1 in 7.75.

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