Tubular shaft undergoes torsion

[Ry122]

For a system where a tubular shaft undergoes torsion, the maximum shearing stress in the steel shaft must not exceed 70MPa. The outside diameter of the tubular shaft is 50mm and the inside is 25mm
How do i determine the maximum torque that can be applied to the shaft?

 

[edziura]

The max shear stress occurs on the cross-section at the outer wall. The general torque equation is

$$\tau$$ = $$\frac{Tc}{J}$$

where $$\tau$$ in this case is the max shear stress, T is the max torque, and c is the outside radius.

J is the polar moment of inertia: J = $$\frac{\pi}{2}$$ (co⁴ - ci⁴).

co =outside radius
ci = inside radius

 

[piercas]

To determine the maximum torque that can be applied to the shaft, we can use the equation for maximum shearing stress, which is τmax = Tc/J, where T is the applied torque, c is the distance from the center to the outermost point of the cross section, and J is the polar moment of inertia of the cross section.

In this case, c = 25mm and J = π/2 * (50mm)^4 * (25mm)^2 = 1.25 x 10^7 mm^4.

Substituting these values into the equation for maximum shearing stress, we get:

70MPa = T * 25mm / 1.25 x 10^7 mm^4

Solving for T, we get:

T = 70MPa * 1.25 x 10^7 mm^4 / 25mm = 3.5 x 10^9 Nmm = 3.5 kNm.

Therefore, the maximum torque that can be applied to the shaft without exceeding the maximum shearing stress limit is 3.5 kNm.