Shaft AB has a 30-mm diameter and is made of a steel with an allowable shearing stress of 100 MPa; shaft BC has a 50-mm diameter and is made of an aluminum alloy with an allowable shearing stress of 60 MPa. Neglecting the effect of stress concentrations, determine the largest torque T that can be applied at A.

The largest torque that can be applied at A is the smaller of both

Tᵇc and Tᵃᵇ

which is (Tᵃᵇ = 530.14376 Nm)

Explanation:

First we take a look at Torque formula

T = M × Shear stress

T = π/2 × (d/2)³ × shear stress

{ where M = π/2 × (d/2)³ = π/2 × d³/8

For Shaft AB

d = 30mm = 30 × 10⁻³m

shaft stress = 100Mpa = 100 × 10⁶ N/m²

now Torque at A due to AB

Tᵃᵇ = π/2 × (d/2)³ × shear stress

Tᵃᵇ = π/2 × (30 × 10⁻³)³/2³ × 100 × 10⁶

Tᵃᵇ =  π/2 × (0.000027 / 8) × 100 × 10⁶

Tᵃᵇ = 530.14376 Nm

For Shaft BC

the value for M is changed

M = π/2 × { (d₂/2)⁴ - (d₁/2)⁴}

d₁ = 30mm = 30 × 10⁻³m

d₂ = 50mm = 50 × 10⁻³m

M = π/2 × { (50 × 10⁻³/2)⁴ - (30 × 10⁻³/2)⁴}

M = 5.34 × 10⁻⁷

Torque formula is also changed which is

T = M × share stress / (d₂/2)

shear stress = 60Mpa = 60 × 10⁶ N/m²

so Torque A due to BC is

Tᵇc =  (5.34 × 10⁻⁷ × 60 × 10⁶) / ( 50 × 10⁻³ / 2

Tᵇc  = 1281.6 Nm

Therefore the largest torque that can be applied at A is the smaller of both

Tᵇc and Tᵃᵇ

which is (Tᵃᵇ = 530.14376 Nm)

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