Determine the minimum diamter required for the shaft of a simple torsion-bar.

Question

A simple torsion-bar spring is shown. The shear stress in the steel [G = 11,500 ksi] shaft is not to exceed 11000 psi, and the vertical deflection of joint D is not to exceed 0.650 in. when a load of P = 3000 lb is applied. Neglect the bending of the shaft and assume that the bearing at C allows the shaft to rotate freely. Determine the minimum diameter required for the shaft. Use dimensions of a = 76 in., b = 40 in., and c = 30 in.

Answer 1

This requires several calculations in order to fully appreciate the outcomes.

Step #1 - Gather All the Given Data

1. G = 11,500 ksi
2. τ = 11,000 psi
3. D = 0.45" (In the provided image it states that the vertical deflection should not exceed 0.45," not
4. 0.65 as in the initial question.)
5. L_a = 78"
6. L_b = 26"
7. c = 20"
8. P = 2,900 lb

Step #2 - Calculate the Total Rotational Angle in Radians

1. θ=2arcsin(c/(2R))=2arcsin(0.45/(2*20))=1.289º
2. Radians = πθ/180º = π*1.289º/180º = 0.0225 rad

Step #3 - Calculate Radius for the Longest Shaft

Using the following equation to determine the radius: τ/r = T/J = Gθ/L

Transposing the last two equations to find the radius(r), one is gets: J = TL/Gθ. The polar inertia equation for this question is r⁴π/2. Knowing this and plugging it into the prior equation, we get the following:

r = [2F*d*L_a/(Gθπ)]^(1/4) = [2*2900*20*78/(11.5*10⁶*0.0225π)]^(1/4) = 1.83" ∴ D = 3.66"

Step #4 - Check to Make Sure the Shaft Diameter Does Not Exceed Shear Stress Limitations

Using the following equation to determine shear stress due to the shaft diameter: τ/r = T/J.

Transposing the prior equation: τ = [2F*d/(r³π)]^(1/3) = [2*2900*20/(1.83³π)^(1/3) = 18.2 psi.

18.2 psi < 11.5*10⁶ psi

Answer: D = 3.66"