(a) A uniform beam has a mass of 50 kg per metre length. Compute the reactions at the support O. The force loads shown lie in a…

(a) Reactions at the support O of the bent cantilever (10 marks)

Take O as the origin, x to the right and y upward. Weight per metre = 50 × 9.81 = 490.5 N/m = 0.4905 kN/m. The horizontal part is 1.8 + 0.6 = 2.4 m long, and the inclined part is 0.6 + 0.6 = 1.2 m long at 30° to the horizontal. The 1.4 kN load at C is taken perpendicular to the inclined member (pointing down and to the right), and the 4 kN·m couple at B is counter-clockwise, as drawn.

Loads.

• Self-weight of the horizontal part: 2.4 × 0.4905 = 1.177 kN at x = 1.2 m.
• Self-weight of the inclined part: 1.2 × 0.4905 = 0.589 kN at its middle, B (2.920, 0.300) m.
• 3 kN downward at A, x = 1.8 m.
• 1.4 kN at C (3.439, 0.600) m, with components 1.4 sin 30° = 0.700 kN to the right and 1.4 cos 30° = 1.212 kN downward.
• Couple 4 kN·m counter-clockwise.

Force balance.

ΣFx = 0: O_x + 0.700 = 0, so O_x = −0.70 kN, i.e. 0.70 kN acting to the left.

ΣFy = 0: O_y − 3 − 1.177 − 0.589 − 1.212 = 0, so O_y = 5.98 kN upward.

Moment balance about O (counter-clockwise positive).

M_O − 3(1.8) − 1.177(1.2) − 0.589(2.920) − [1.212(3.439) + 0.700(0.600)] + 4 = 0

M_O − 5.400 − 1.413 − 1.718 − 4.590 + 4 = 0, so M_O = 9.12 kN·m counter-clockwise.

The reactions at O are therefore a horizontal force of 0.70 kN to the left, a vertical force of 5.98 kN upward (resultant 6.02 kN) and a fixing couple of 9.12 kN·m counter-clockwise.

(b) Fixed-ended stepped bar (10 marks)

Data: A₁ = 840 mm², L₁ = 200 mm (segments AB and CD), A₂ = 1260 mm², L₂ = 250 mm (segment BC), P_B = 25.5 kN acting towards C at B, and P_C = 17.0 kN acting towards B at C, so P_B and P_C compress the middle segment.

Equilibrium. Let x be the compressive force in AB. Then F_BC = x + P_B and F_CD = x + P_B − P_C. Equilibrium alone cannot fix x, so the bar is statically indeterminate to the first degree.

Compatibility. The total change in length between the two fixed ends is zero, and E is common:

x L₁/A₁ + (x + P_B) L₂/A₂ + (x + P_B − P_C) L₁/A₁ = 0

L₁/A₁ = 200/840 = 0.2381 and L₂/A₂ = 250/1260 = 0.1984 (mm per mm²), so

x (2 × 0.2381 + 0.1984) + 25.5 (0.1984) + 8.5 (0.2381) = 0

0.6746 x + 5.060 + 2.024 = 0, so x = −10.5 kN.

The negative sign means AB is in tension of 10.5 kN.

(i) Reactions. R_A = 10.5 kN, acting away from the bar (a pull towards the left wall). F_CD = −10.5 + 25.5 − 17.0 = −2.0 kN, so CD is in tension of 2.0 kN and R_D = 2.0 kN, also pulling on the bar towards the right wall. Check: 25.5 − 17.0 − 10.5 + 2.0 = 0.

(ii) Middle segment. F_BC = x + P_B = −10.5 + 25.5 = 15.0 kN (compressive).

(c) Grain size: importance, estimation, ASTM number (10 marks)

Importance. Grain boundaries block the movement of dislocations, so strength rises as grain size falls (Hall–Petch relation σ_y = σ₀ + k d^(−1/2)). Fine grains also improve toughness and lower the ductile-to-brittle transition temperature, and give better fatigue strength and a smoother surface after forming. Coarse grains, on the other hand, resist creep at high temperature, and grain size also affects corrosion, machinability and magnetic and electrical behaviour. Grain size is controlled by cooling rate, alloying and heat treatment.

Estimation (ASTM E112). The specimen is polished and etched to reveal boundaries. Three methods are used:

• Comparison method: the microstructure at 100× is matched against standard charts of numbered grain sizes.
• Planimetric (Jeffries) method: the grains completely inside a circle of known area, plus half the grains cut by the circle, are counted, and multiplied by a factor to give grains per unit area.
• Intercept (Heyn) method: the number of grain boundaries crossed by test lines of known length gives the mean intercept length ℓ, and G = −6.6457 log₁₀ ℓ − 3.298 (ℓ in mm).

(i) ASTM number. At 100×, N = 2^(n−1) grains per square inch:

75 = 2^(n−1), so n − 1 = log₂ 75 = 6.229, and n ≈ 7.2.

(ii) Grains per square inch at 60×. The grains per unit area seen scale as 1/M², so N₆₀ (60)² = N₁₀₀ (100)²:

N₆₀ = 75 × (100/60)² = 75 × 2.778 = 208 grains/in² (about 200).

(d) Minimum pinion teeth and path of contact for stub teeth (10 marks)

cos φ = 0.94 gives φ = 19.95° ≈ 20°, and sin²φ = 1 − 0.94² = 0.1164. The wheel addendum is 0.85 m. Let G = T_w/T_p.

Minimum teeth on the pinion (to avoid interference).

T_min = 2 a_w / [ √(1 + (1/G)(1/G + 2) sin²φ) − 1 ], with a_w = 0.85 m in module units, so 2a_w = 1.7.

(i) G = 1: √(1 + 3 × 0.1164) = √1.3492 = 1.1616, so T_min = 1.7/0.1616 = 10.5, and hence 11 teeth.

(ii) G = 2.5: (1/G)(1/G + 2) = 0.4 × 2.4 = 0.96, √(1 + 0.96 × 0.1164) = √1.1117 = 1.0544, so T_min = 1.7/0.0544 = 31.3, and hence 32 teeth.

Length of path of contact. Path of approach = √(R_a² − R² cos²φ) − R sin φ, and path of recess = √(r_a² − r² cos²φ) − r sin φ, with r_a = r + 0.85 m and R_a = R + 0.85 m. Here sin φ = 0.3412.

(i) G = 1, T = 11: r = R = 5.5 m and r_a = 6.35 m. r cos φ = 5.17 m and r sin φ = 1.876 m, so each of approach and recess is √(40.32 − 26.73) − 1.876 = 3.687 − 1.876 = 1.810 m. The path of contact is 2 × 1.810 = 3.62 m.

(ii) G = 2.5, T_p = 32: r = 16 m, R = 40 m, r_a = 16.85 m, R_a = 40.85 m.

• Approach: √(40.85² − 37.6²) − 40 × 0.3412 = √254.96 − 13.647 = 15.968 − 13.647 = 2.32 m.
• Recess: √(16.85² − 15.04²) − 16 × 0.3412 = √57.72 − 5.459 = 7.597 − 5.459 = 2.14 m.
• Path of contact = 2.32 + 2.14 = 4.46 m.

(e) Largest permissible thrust on a shaft under thrust and torque (10 marks)

Data. d = 60 mm, T = 2 kN·m = 2 × 10⁶ N·mm and σ_all = 220 MPa.

Section properties. A = π/4 × 60² = 2827.4 mm². The torsional shear stress at the surface is

τ = 16T/(πd³) = 16 × 2 × 10⁶/(π × 60³) = 47.16 MPa.

The axial stress is σ = P/A, and the surface element has σ and τ with no other stress.

Principal stress. The larger principal stress in magnitude is

σ_max = σ/2 + √((σ/2)² + τ²)

Setting σ_max = 220 MPa: √(σ²/4 + τ²) = 220 − σ/2. Squaring gives σ²/4 + τ² = 220² − 220σ + σ²/4, so

σ = (220² − τ²)/220 = (48400 − 2224)/220 = 209.9 MPa.

Load. P = σ A = 209.9 × 2827.4 = 5.93 × 10⁵ N, i.e. P ≈ 593 kN.

Check. With σ = 209.9 MPa and τ = 47.16 MPa: σ/2 = 104.95, √(104.95² + 47.16²) = 115.05, and the sum is 220.0 MPa. The other principal stress is 104.95 − 115.05 = −10.1 MPa, and the maximum shear stress is 115 MPa. The result is the same whether P is tensile or compressive, since the largest principal stress in magnitude is what is limited.