(a) The consolidated undrained (CU) tests were performed on the four over-consolidated clay samples obtained from a site. The pre-consolidation pressure was 650 kN/m². The results of triaxial test in CU condition are as follows:
| Test sample | Cell

Question

(a) The consolidated undrained (CU) tests were performed on the four over-consolidated clay samples obtained from a site. The pre-consolidation pressure was 650 kN/m². The results of triaxial test in CU condition are as follows:
| Test sample | Cell pressure kN/m² | Deviator stress kN/m² | Pore pressure kN/m² | Remark |
|-------------|---------------------|----------------------|---------------------|--------|
| 1 | 100 | 290 | – 40 | All the tests were performed in CU condition. The deviator stress and pore pressure were at failure. |
| 2 | 200 | 430 | – 20 | |
| 3 | 400 | 600 | 50 | |
| 4 | 600 | 840 | 110 | |
Determine the effective shear strength parameters. Draw the variation of pore pressure parameter 'A' (at failure) with over-consolidation ratio.

(b) A 250 mm diameter concrete pile 8 m long was driven by a double-acting hammer. The driving was carried out by a short dolly and cushion. The average penetration recorded in the last five blows was 3.0 mm per blow. Determine the safe pile load. As per IS 2911 (Part I) – 1979, the coefficient of restitution of the materials under impact for double-acting hammer striking on steel anvil and driving RCC pile is 0.5. The rated energy of hammer is 16.5 kJ and mass of hammer is 22 kN. Assume that only 90% of the rated energy is consumed. The density of RCC pile may be considered as 25 kN/m³. Assume the factor of safety as 2.5.

(c) (i) In a horizontal, rectangular channel, the sluice gate is opened. A hydraulic jump is formed downstream of the sluice gate.
The depth of water before jump is 0.8 m and specific energy before jump is 12.0 m.
Find the sequent depth of the jump and energy lost in the jump.
What is the initial Froude number ? Classify the jump based on the results obtained in the problem.

(ii) A runoff river plant is proposed to generate hydroelectric power. The net head available is 30 m. The river carries a sustainable flow of 30 m³/s in dry weather. Determine the maximum generating capacity. Daily load pattern indicates 20 hrs of average load and 4 hrs of peak load. Estimate the volume of pondage to be provided to supply the daily demand.
Assume load factor = 85%, Efficiency = 80%.

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(a)

Effective stresses: σ₃′ = σ₃ − u, σ₁′ = σ₁ − u = σ₃ + Δσ − u.
Then p′ = (σ₁′ + σ₃′)/2, q = (σ₁′ − σ₃′)/2.
Compute:
1: σ₃′ = 100 − (−40) = 140 kN/m², σ₁′ = 100+290+40 = 430 kN/m², p′ = 285 kN/m², q = 145 kN/m².
2: σ₃′ = 200 − (−20) = 220 kN/m², σ₁′ = 200+430+20 = 650 kN/m², p′ = 435 kN/m², q = 215 kN/m².
3: σ₃′ = 400 − 50 = 350 kN/m², σ₁′ = 400+600−50 = 950 kN/m², p′ = 650 kN/m², q = 300 kN/m².
4: σ₃′ = 600 − 110 = 490 kN/m², σ₁′ = 600+840−110 = 1330 kN/m², p′ = 910 kN/m², q = 420 kN/m².

Use Mohr-Coulomb criterion in triaxial invariants: q = c′ cos φ′ + p′ sin φ′.
Least-squares fit of q on p′ gives slope m = 0.4355 and intercept b = 21.74 kN/m².
Thus sin φ′ = 0.4355 → φ′ = 25.8°.
c′ = b/cos φ′ = 21.74/0.900 = 24.1 kN/m².
Effective shear strength parameters: c′ ≈ 24 kN/m², φ′ ≈ 25.8°.

Pore pressure parameter at failure: A = u/Δσ.
OCR = σ_c′/σ₃ = 650/σ₃.
Pairs (OCR, A):
(6.5, −40/290 = −0.138),
(3.25, −20/430 = −0.047),
(1.625, 50/600 = 0.083),
(1.083, 110/840 = 0.131).
Plot A against OCR: A rises from negative values at high OCR to positive values near OCR = 1.

(b)

Use Hiley’s formula as per IS 2911 (Part I):
Q_u = ηE/(S + C/2) × (W + e²W_p)/(W + W_p).
Given: ηE = 0.9 × 16.5 = 14.85 kN m, e = 0.5, W = 22 kN.
Pile area = π(0.25)²/4 = 0.04909 m².
Volume = 0.04909 × 8 = 0.3927 m³.
W_p = 0.3927 × 25 = 9.82 kN.
For RCC pile with short dolly and cushion, take C = 25 mm = 0.025 m.
S = 3 mm = 0.003 m, C/2 = 0.0125 m.
(W + e²W_p)/(W + W_p) = (22 + 0.25×9.82)/(22 + 9.82) = 24.45/31.82 = 0.7686.
Q_u = 14.85/0.0155 × 0.7686 = 736.4 kN.
Safe load = Q_u/FS = 736.4/2.5 = 294.6 kN.
Safe pile load ≈ 294.6 kN.

(c)(i)

Specific energy before jump: E₁ = y₁ + V₁²/(2g).
12.0 = 0.8 + V₁²/(2×9.81) → V₁²/(2g) = 11.2 m → V₁² = 219.744 → V₁ = 14.82 m/s.
Fr₁ = V₁/√(g y₁) = 14.82/√(9.81×0.8) = 14.82/2.801 = 5.29 = 2√7.

Sequent depth formula for rectangular channel:
y₂/y₁ = ½(√(1 + 8Fr₁²) − 1) = ½(√(1 + 8×28) − 1) = ½(15 − 1) = 7.
y₂ = 7 × 0.8 = 5.6 m.
Energy lost: ΔE = (y₂ − y₁)³/(4 y₁ y₂) = (4.8)³/(4×0.8×5.6) = 110.592/17.92 = 6.17 m = 216/35 m.
Sequent depth = 5.6 m; energy lost = 6.17 m; initial Froude number = 5.29. Since 4.5 < Fr₁ < 9.0, it is a steady jump.

(c)(ii)

Power from available flow:
P = 9.81 Q H η = 9.81 × 30 × 30 × 0.80 = 7063.2 kW.
This is the 24-hour average firm power. Load factor = average load / peak load = 0.85.
Peak load = maximum generating capacity = 7063.2/0.85 = 8309.6 kW ≈ 8.31 MW.

Peak flow required = Q_peak = 30/0.85 = 35.29 m³/s.
River flow = 30 m³/s.
Deficit during 4-hour peak = 35.29 − 30 = 5.29 m³/s.
Pondage volume = 5.29 × 4 × 3600 = 76,235 m³.
Maximum generating capacity ≈ 8.31 MW; pondage volume ≈ 7.62×10⁴ m³.

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परीक्षण नमूना	कोष्ठिका दाब kN/m²	विचलक प्रतिबल kN/m²	रंध्र दाब kN/m²	टिप्पणी
1	100	290	– 40	सभी परीक्षण संघनित अनपवाहित अवस्था में किए गए । विचलक प्रतिबल और रंध्र दाब विफलन पर थे।
2	200	430	– 20
3	400	600	50
4	600	840	110