(a) A 20 mm diameter shaft is subjected to a torque of 80 Nm and a downward bending moment of 100 Nm at the centre. Draw the state of stress on the bottom surface of the shaft at the centre and find principal stresses and shear stress at the centre o

Question

(a) A 20 mm diameter shaft is subjected to a torque of 80 Nm and a downward bending moment of 100 Nm at the centre. Draw the state of stress on the bottom surface of the shaft at the centre and find principal stresses and shear stress at the centre of the bottom surface. What is the angle of shear plane? (15 marks)

(b) What is Atomic Packing Factor of a crystal structure? Calculate the atomic packing factor of aluminium assuming atoms to be of spherical shape with atomic radius 'R'. (15 marks)

(c) The cranks of a three-cylinder single acting engine are set equally at 120°. The engine speed is 540 rpm. The turning moment diagram for each cylinder is a triangle for the power stroke with a maximum torque of 100 Nm at 60° after dead-centre of the corresponding crank. On the return stroke, the torque is sensibly zero.

Determine the:
(i) power developed by the engine
(ii) coefficient of fluctuation of speed if the flywheel has a mass of 7·5 kg with a radius of gyration of 65 mm
(iii) coefficient of fluctuation of energy
(iv) maximum angular acceleration of the flywheel (20 marks)

UPSC Answer Check · Accepted Answer

(a) 
For the shaft, d = 20 mm = 0.02 m, bending moment M = 100 Nm, torque T = 80 Nm.
At the bottom surface, taking the downward bending moment as sagging, the bending stress is tensile along the shaft axis.
Bending stress: σx = 32M/(πd³) = 32×100/(π×0.02³) = 1.27324×10⁸ Pa = 127.324 MPa.
Torsional shear stress: τ = 16T/(πd³) = 16×80/(π×0.02³) = 5.09296×10⁷ Pa = 50.930 MPa.
On the free surface, σy = 0. Draw the element with σx acting outward on the axial faces and τ acting on the same faces in the circumferential direction.

Principal stresses by the plane-stress principal stress formula:
σ1,2 = σx/2 ± √((σx/2)² + τ²)
= 63.662 ± √(63.662² + 50.930²) MPa
= 63.662 ± 81.527 MPa.
Hence, σ1 = 145.189 MPa and σ2 = −17.865 MPa.
Maximum shear stress τmax = (σ1 − σ2)/2 = 81.527 MPa.

Angle of principal plane: tan 2θp = 2τ/σx = 2×50.930/127.324 = 0.800.
Thus θp = 19.33° from the shaft axis.
The shear planes are at θp ± 45°, i.e. 64.33° and −25.67° from the shaft axis, equivalently 45° from the principal planes.

(b) 
Atomic Packing Factor is the fraction of unit-cell volume occupied by atoms:
APF = volume of atoms in one unit cell / volume of the unit cell.
Aluminium has an FCC crystal structure. For FCC, number of atoms per unit cell n = 4. Atoms touch along the face diagonal:
4R = a√2, so a = 2√2 R.
Volume of atoms = 4 × (4/3)πR³ = 16πR³/3.
Unit-cell volume = a³ = (2√2 R)³ = 16√2 R³.
Therefore,
APF = (16πR³/3)/(16√2 R³) = π/(3√2) = π√2/6 ≈ 0.74048 = 74.05%.
This assumes hard spherical atoms touching along the FCC face diagonal.

(c) 
Each cylinder gives a triangular turning-moment pulse. Base = 180° = π rad, maximum torque = 100 Nm.
Work per cylinder per revolution = area of triangle = (1/2)×π×100 = 50π J.
Three cylinders give total work per revolution = 150π J.
Speed N = 540 rpm = 9 rev/s.

(i) 
Power developed = total work per second
= 150π × 9 = 1350π W = 4241.15 W ≈ 4.241 kW.
Mean torque = total work per revolution/(2π) = 150π/(2π) = 75 Nm.

(ii) 
With cranks at 120° and a 180° power stroke, the total torque diagram repeats every 120°. It varies linearly from 50 Nm to 100 Nm and back to 50 Nm.
Maximum fluctuation of energy ΔE is the area above the mean torque 75 Nm. This is a triangle from 30° to 90°:
base = 60° = π/3 rad, height = 100 − 75 = 25 Nm.
ΔE = (1/2)×(π/3)×25 = 25π/6 J = 13.09 J.
Flywheel moment of inertia:
I = mk² = 7.5×(0.065)² = 0.0316875 kg m².
Angular speed: ω = 2πN/60 = 18π rad/s.
Coefficient of fluctuation of speed:
Cs = ΔE/(Iω²) = (25π/6)/(0.0316875×(18π)²)
= 0.1292.

(iii) 
Coefficient of fluctuation of energy:
Ce = ΔE/(work done per cycle).
Work done per cycle = work per revolution = 150π J.
Ce = (25π/6)/(150π) = 25/900 = 1/36 = 0.02778.

(iv) 
Maximum angular acceleration occurs at maximum excess torque over mean torque:
αmax = (Tmax − Tmean)/I = (100 − 75)/0.0316875
= 788.95 rad/s² ≈ 789 rad/s².
The maximum retardation has the same magnitude.

(a) A 20 mm diameter shaft is subjected to a torque of 80 Nm and a downward bending moment of 100 Nm at the centre. Draw the…

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