Question

A torsion-bar spring consists of a prismatic bar, usually of round cross section, that is twisted at one end and held fast at the other to form a stiff spring. An engineer needs a stiffer one than usual and so considers building in both ends and applying the torque somewhere in the central portion of the span, as shown in the figure. This effectively creates two springs in parallel.

The engineer is forced by geometric considerations to apply the torque on the spring at the location x = 0.37l. For a uniform-diameter spring, this would cause one leg of the span to be underutilized when both legs have the same diameter. For optimal design the diameter of each leg should be designed such that the maximum shear stress in each leg is the same Using x = 0.37l, l=10 in, T=1,500 lbf.in, and G = 10.4 Mpsi, design the spring such that the maximum shear stresses in each leg are equal and the spring will have the same spring rate (angle of twist) as a uniform-diameter bar with T=1,500 lbf in, x=1/2-5 in, and d1=d2=0.5
Find the required diameters of each leg:
d1
d2

Answer 1

Answer:

d₁ = 0.29 in

d₂ = 0.505 in

Explanation:

Given:

T = 1500 lbf in

L = 10 in

x = 0.5 L = 5 in

$T_{1} =\frac{T(L-x)}{L} =\frac{1500*(10-5)}{10} =750lbfin$

First case: T = T₁ + T₂

T₂ = T - T₁ = 1500 - 750 = 750 lbf in

If the shafts are in series:

θ = θ₁ + θ₂

θ = ((T₁ * L₁)/GJ) + ((T₂ * L₂)/GJ)

Second case: If d₁ ≠ d₂

θ = ((T₁ * L₁)/GJ₁) + ((T₂ * L₂)/GJ₂) = 0 (eq. 1)

t₁ = t₂

$\frac{16T_{1} }{\pi d_{1}^{3} } =\frac{16T_{2} }{\pi d_{2}^{3} }$ (eq. 2)

T₁ + T₂ = 1500 (eq. 3)

θ₁ first case = θ₁ second case

Replacing:

$\frac{750*5}{G(\frac{\pi }{32})*0.5^{4} } =\frac{T_{1}*3.7 }{G(\frac{\pi }{32})*d_{1} ^{4} }\\T_{1} =16216d_{1} ^{4}$

The same way to θ₂:

$\frac{750*5}{G(\frac{\pi }{32})*0.5^{4} } =\frac{T_{2}*6.3 }{G(\frac{\pi }{32})*d_{2} ^{4} } \\T_{2} =9523.8d_{2} ^{4}$

From equation 2, we have:

d₁ = 0.587 * d₂

From equation 3, we have:

d₂ = 0.505 in

d₁ = 0.29 in