(a) A 300 mm diameter concrete pile is to be driven into a medium dense to dense sand with an embedded length of 12 m. The soil conditions are shown in the figure. No ground water was encountered and the ground water table is not expected to rise dur

Question

(a) A 300 mm diameter concrete pile is to be driven into a medium dense to dense sand with an embedded length of 12 m. The soil conditions are shown in the figure. No ground water was encountered and the ground water table is not expected to rise during the life of the structure. Given: The coefficient lateral earth pressure (k) = 0·95, tan δ = 0·45 and for φ = 38° bearing capacity factor, N_q = 80. Determine the pile's axial capacity with a factor of safety of 2. Assume critical depth of the pile is equal to 20 times the diameter of the pile. (15 marks)

(b) A Pelton wheel develops 5520 kW power under a head of 240 m at an overall efficiency of 80% when revolving at a speed of 200 rpm. Find the unit discharge, unit power and unit speed. Assume peripheral coefficient as 0·46. If the head on the same turbine falls during summer season to 150 m, find the discharge, power and speed at this head. (15 marks)

(c) A 1 m long hollow shaft is to transmit a torque of 400 N-m. The outer diameter of the shaft must be 25 mm to fit existing attachments. The relative rotation of the two ends of the shaft is limited to 0·375 rad. The shaft can be made of either titanium alloy or aluminium. Using the data given in the table below, determine the maximum inner radius to the nearest millimeter of the lightest shaft that can be used for transmitting the torque. (20 marks)

| Material | Shear Modulus G (GPa) | Maximum Shear Stress τₐₗₗₒw (MPa) | γ (density) (Mg/m³) |
|----------|----------------------|-----------------------------------|---------------------|
| Titanium alloy | 36 | 450 | 4·4 |
| Aluminium | 28 | 150 | 2·8 |

UPSC Answer Check · Accepted Answer

(a)
Use the static pile-capacity formula by effective stress. Since no groundwater is present, effective stress equals total stress.
D = 300 mm = 0.30 m, L = 12 m, γ = 20.1 kN/m³, k = 0.95, tan δ = 0.45, N_q = 80.
Critical depth L_c = 20D = 20 × 0.30 = 6 m.

Base area:
A_b = πD²/4 = π(0.30)²/4 = 0.07069 m².
Perimeter:
p = πD = π × 0.30 = 0.94248 m.

Effective overburden at critical depth:
σ′_c = γL_c = 20.1 × 6 = 120.6 kPa.
Unit point resistance:
q_p = N_q σ′_c = 80 × 120.6 = 9648 kPa.
Point load:
Q_p = q_p A_b = 9648 × 0.07069 = 681.98 kN.

For skin friction, up to L_c:
f_s = k γ z tan δ = 0.95 × 20.1 × 0.45 z = 8.59275z kPa.
At z = 6 m, f_s = 51.56 kPa. Below critical depth it remains constant at 51.56 kPa.
Q_s = p [∫₀⁶ 8.59275z dz + ∫₆¹² 51.56 dz]
= 0.94248 [(8.59275 × 18) + (51.56 × 6)]
= 0.94248 [154.67 + 309.34]
= 437.33 kN.

Ultimate capacity:
Q_u = Q_p + Q_s = 681.98 + 437.33 = 1119.31 kN.
Allowable capacity with FS = 2:
Q_all = Q_u/2 = 1119.31/2 = 559.66 kN.

Final axial capacity = 560 kN (approximately).
Valid for a single homogeneous sand layer in compression, with effective stress capped at L_c = 6 m.

(b)
Use overall efficiency: P = η_o ρgQH.
Given P = 5520 kW, H = 240 m, η_o = 0.80, N = 200 rpm.

Actual discharge:
Q = P/(η_o ρgH)
= 5520 × 10³/(0.80 × 1000 × 9.81 × 240)
= 2.9307 m³/s.

Unit quantities:
Unit speed N_u = N/√H = 200/√240 = 12.91 rpm.
Unit discharge Q_u = Q/√H = 2.9307/√240 = 0.1892 m³/s.
Unit power P_u = P/H^(3/2) = 5520/240^(3/2) = 1.485 kW.

Unit speed = 12.91 rpm, unit discharge = 0.1892 m³/s, unit power = 1.485 kW.

For H₂ = 150 m, assuming same unit quantities and efficiency:
Q₂ = Q_u√H₂ = 0.1892 × √150 = 2.317 m³/s.
N₂ = N_u√H₂ = 12.91 × √150 = 158.1 rpm.
P₂ = P_u H₂^(3/2) = 1.485 × 150^(3/2) = 2727.5 kW.

At 150 m head: discharge = 2.317 m³/s, power = 2728 kW, speed = 158 rpm.
The peripheral coefficient φ = 0.46 ensures the same peripheral velocity ratio; it does not alter the unit-quantity calculations.

(c)
Use torsion formulae:
τ_max = T rₒ/J ≤ τ_allow, and θ = TL/(GJ) ≤ 0.375 rad.
Here T = 400 N·m = 400000 N·mm, L = 1000 mm, rₒ = 12.5 mm.
J = π/2 (rₒ⁴ − rᵢ⁴) = π/32 (Dₒ⁴ − Dᵢ⁴).

Required J from stress: J_s = T rₒ/τ_allow.
Required J from twist: J_t = TL/(Gθ).

For titanium alloy: G = 36000 N/mm², τ_allow = 450 N/mm².
J_s = 400000 × 12.5/450 = 11111 mm⁴.
J_t = 400000 × 1000/(36000 × 0.375) = 29630 mm⁴.
Twist governs: J_req = 29630 mm⁴.
rᵢ⁴ = rₒ⁴ − 2J_req/π = 12.5⁴ − 2 × 29630/π
= 24414.06 − 18862.7 = 5551.4 mm⁴.
rᵢ = 8.63 mm.

For aluminium: G = 28000 N/mm², τ_allow = 150 N/mm².
J_s = 400000 × 12.5/150 = 33333 mm⁴.
J_t = 400000 × 1000/(28000 × 0.375) = 38095 mm⁴.
Twist governs: J_req = 38095 mm⁴.
rᵢ⁴ = 24414.06 − 2 × 38095/π
= 24414.06 − 24251.6 = 162.5 mm⁴.
rᵢ = 3.57 mm.

Mass comparison per metre length:
Titanium: rᵢ = 8.63 mm, area = π(12.5² − 8.63²) = 256.9 mm².
Volume = 256900 mm³ = 2.569 × 10⁻⁴ m³.
Mass = 4.4 × 2.569 × 10⁻⁴ = 1.13 kg.

Aluminium: rᵢ = 3.57 mm, area = π(12.5² − 3.57²) = 450.8 mm².
Volume = 450800 mm³ = 4.508 × 10⁻⁴ m³.
Mass = 2.8 × 4.508 × 10⁻⁴ = 1.26 kg.

Titanium alloy shaft is lighter.

Lightest shaft = titanium alloy; maximum inner radius = 8.63 mm ≈ 9 mm to the nearest millimetre.
If only whole-millimetre sizes are permitted, the largest safe inner radius is 8 mm. Valid for elastic torsion of a homogeneous hollow shaft.

(a) A 300 mm diameter concrete pile is to be driven into a medium dense to dense sand with an embedded length of 12 m. The soil…

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More from Civil Engineering 2021 Paper I

पदार्थ	अपरूपण मापांक G (GPa)	अधिकतम अपरूपण प्रतिबल τ_अनुज्ञेय (MPa)	γ (घनत्व) (Mg/m³)
टाइटेनियम मिश्रधातु	36	450	4·4
एल्युमिनियम	28	150	2·8