(a) The resistance force F of a ship is a function of its length L, velocity V, acceleration due to gravity g and fluid properties like density ρ and viscosity μ. Write this relationship in a dimensionless form. (15 marks)

(b) The stream function fo

Question

(a) The resistance force F of a ship is a function of its length L, velocity V, acceleration due to gravity g and fluid properties like density ρ and viscosity μ. Write this relationship in a dimensionless form. (15 marks)

(b) The stream function for a two-dimensional flow is given by ψ = 2xy. Calculate the velocity and velocity potential at point P(2, 3). (15 marks)

(c) A group of nine friction piles is driven through 5 m of clay with unconfined compressive strength of 60 kN/m² followed by 10 m of clay with unconfined compressive strength of 100 kN/m². The piles are in 3 rows and will be 1·00 m centres in a row and the rows will be 750 mm on centres. Each pile has a diameter of 300 mm. If a factor of safety of 2·5 is required, determine the maximum load that can be carried by the group. Take Nc = 9 and unit weight of clay as 16·4 kN/m³. (20 marks)

UPSC Answer Check · Accepted Answer

(a) By the Buckingham Pi theorem, the variables are F, L, V, g, ρ and μ, so n = 6. Their dimensions are:
F = MLT⁻², L = L, V = LT⁻¹, g = LT⁻², ρ = ML⁻³, μ = ML⁻¹T⁻¹.
With three basic dimensions M, L, T, the number of independent Π terms is n − m = 6 − 3 = 3.

Choose ρ, V and L as repeating variables. Then:
Π₁ = F ρᵃ Vᵇ Lᶜ. Equating powers gives a = −1, b = −2, c = −2, so
Π₁ = F/(ρ V² L²).

Π₂ = g ρᵃ Vᵇ Lᶜ. Equating powers gives a = 0, b = −2, c = 1, so
Π₂ = gL/V² = 1/Fr², where Fr = V/√(gL).

Π₃ = μ ρᵃ Vᵇ Lᶜ. Equating powers gives a = −1, b = −1, c = −1, so
Π₃ = μ/(ρ V L) = 1/Re, where Re = ρVL/μ.

Therefore the dimensionless relationship is:
F/(ρ V² L²) = f(gL/V², μ/(ρVL))
or, equivalently,
F/(ρ V² L²) = f(V/√(gL), ρVL/μ).
This is valid for geometrically similar ships. Complete similarity requires equal Froude and Reynolds numbers; if viscous effects are negligible, the function reduces to the Froude-number form.

(b) For a two-dimensional flow, using the standard convention
u = ∂ψ/∂y, v = −∂ψ/∂x.
Given ψ = 2xy:
u = ∂(2xy)/∂y = 2x,
v = −∂(2xy)/∂x = −2y.

At P(2, 3):
u = 2(2) = 4 m/s,
v = −2(3) = −6 m/s.
Thus V = 4i − 6j m/s.

Magnitude:
|V| = √(4² + (−6)²) = √52 = 2√13 m/s ≈ 7·21 m/s.

Check irrotationality:
ω_z = ∂v/∂x − ∂u/∂y = 0 − 0 = 0.
Hence a velocity potential φ exists.

Using u = ∂φ/∂x and v = ∂φ/∂y:
∂φ/∂x = 2x ⇒ φ = x² + f(y).
Then ∂φ/∂y = f′(y) = −2y ⇒ f(y) = −y² + C.
So φ = x² − y² + C.

At P(2, 3), taking C = 0:
φ = 2² − 3² = 4 − 9 = −5 m²/s.
Thus velocity potential at P = −5 m²/s (apart from an arbitrary constant).

(c) Unconfined compressive strength qu = 2c_u, so:
Layer 1: c_u1 = 60/2 = 30 kN/m².
Layer 2: c_u2 = 100/2 = 50 kN/m².

Pile diameter d = 0·30 m.
Since the piles are friction piles, end bearing of an individual pile is neglected. The perimeter of one pile is
p = πd = 0·30π m.

Individual pile capacity
Skin friction in layer 1:
Q_s1 = c_u1 p L1 = 30 × 0·30π × 5 = 45π kN ≈ 141·37 kN.

Skin friction in layer 2:
Q_s2 = c_u2 p L2 = 50 × 0·30π × 10 = 150π kN ≈ 471·24 kN.

Ultimate capacity of one friction pile:
Q_u1 = 45π + 150π = 195π kN ≈ 612·61 kN.

For 9 piles:
Q_ug,individual = 9 × 195π = 1755π kN ≈ 5513·50 kN.

Safe load by individual action:
Q_safe,individual = 5513·50/2·5 = 2205·40 kN.

Block failure check
The group has 3 piles per row at 1·00 m centres and 3 rows at 0·75 m centres.
Overall length = (3 − 1)(1·00) + 0·30 = 2·30 m.
Overall width = (3 − 1)(0·75) + 0·30 = 1·80 m.

Perimeter of group, P_g = 2(2·30 + 1·80) = 8·20 m.
Base area of group, A_g = 2·30 × 1·80 = 4·14 m².

Side friction on block in layer 1:
Q_sg1 = 30 × 8·20 × 5 = 1230 kN.

Side friction on block in layer 2:
Q_sg2 = 50 × 8·20 × 10 = 4100 kN.

Net base resistance of block:
Q_bg = N_c c_u2 A_g = 9 × 50 × 4·14 = 1863 kN.

Ultimate block capacity using net base resistance:
Q_ug,block = 1230 + 4100 + 1863 = 7193 kN.

Check with unit weight: gross base pressure = c_u2 N_c + γD = 450 + 16·4 × 15 = 696 kN/m²; gross base force = 696 × 4·14 = 2881·44 kN; weight of block = 16·4 × 4·14 × 15 = 1018·44 kN. Then Q_ug,block = 5330 + 2881·44 − 1018·44 = 7193 kN. The unit weight cancels in the net capacity.

Safe load by block action:
Q_safe,block = 7193/2·5 = 2877·20 kN.

The lesser of the two safe loads governs:
Q_safe = min(2205·40, 2877·20) = 2205·40 kN.

Maximum load carried by the group = 2205·4 kN ≈ 2·21 MN (2205 kN).

The analysis assumes adhesion equal to undrained cohesion (α = 1), as no adhesion factor is specified. The unit weight γ = 16·4 kN/m³ cancels in the net block capacity; if gross base resistance is used, the weight of the displaced block must be subtracted.

(a) The resistance force F of a ship is a function of its length L, velocity V, acceleration due to gravity g and fluid…

Model answer

What "Solve" is asking you to do

Structure that answers it

Where marks are lost

How this answer will be evaluated

Approach

Key points expected

Evaluation rubric

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Practice this exact question

More from Civil Engineering 2024 Paper I