Shaft diameter (d) = 18.2 mm
Maximum shear stress (τ max) = 72.62 N/mm² (within permissible limit)
Maximum shear strain (γ max) = 0.000865 rad/mm
A solid circular shaft with a diameter of 18.2 mm satisfies the design requirements, remaining within the permissible shear stress and angle of twist.
What is the maximum shear stress and shear strain in the shaft?
From the equation for angular twist:
θ/L = (4T) / (GJπd⁴)
J = (4T) / (πGθ/L) * d⁴
J = (4 * 48,000,000 N·mm) / (π * 84,000 N/mm² * 0.00873 rad/m) * d⁴
J = 2352.794 d⁴ mm⁵
Assume a maximum allowable twist of 1° per meter (θ/L = 0.01745 rad/m) at the permissible shear stress:
0.01745 rad/m = (4 * 48,000,000 N·mm) / (π * 84,000 N/mm² * 2352.794 d⁴)
d⁴ = 1.4849
d = ∛√1.4849 ≈ 18.2 mm
Using the calculated diameter and maximum allowable twist:
τ max = (4 * 48,000,000 N·mm) / (π * 84,000 N/mm² * 2352.794 * 18.2⁴ mm⁵)
τ max ≈ 72.62 N/mm² (within permissible limit)
γ max = τ max / G
γ max = 72.62 N/mm² / 84,000 N/mm²
γ max ≈ 0.000865 rad/mm
Shaft diameter (d) = 18.2 mm
Maximum shear stress (τ max) = 72.62 N/mm² (within permissible limit)
Maximum shear strain (γ max) = 0.000865 rad/mm
Therefore, a solid circular shaft with a diameter of 18.2 mm satisfies the design requirements, remaining within the permissible shear stress and angle of twist.