Torsion Spring Design Techniques: Logic Diagram and Example

The logic diagram below is intended as a tool for torsion spring design. Hopefully it will be a guide and provide some inspiration for the designer. It is a template for one of many torsion spring design techniques, outlining the step-by-step procedure. No single logic diagram is the answer for all torsion spring problems. Hopefully this will help get you going in the right direction.

This logic diagram shows the basic steps and procedures which will determine if the torsion spring can be made to meet the specifications. The design that is derived is only an initial design and may not be the most economical one. What also needs to be considered are the manufacturing techniques required, the quantities involved, the tolerances needed, etc.

torsion spring design

LINK 1: This page has the chart on minimum tensile strength (TS) of spring wires
LINK 2: This page has the chart "Design Stresses for Static Service" to help you determine S_max

Torsion Spring Design Example

The example we will now review follows the logic diagram above and will show you how to use it properly.

PROBLEM: Need a cabinet door hinge torsion spring to hold the door closed, exerting a torque M₁=0.5 in. lb. at

=90 degree between legs, each 0.750 in. long. When the door is fully opened the spring deflects through additional overtravel 120 degrees or

or 120/360 = 1/3 turn.

Maximum spring length is 0.5 in. and the spring operates over a 0.25 in. diameter arbor D_s.

Use oil-tempered wire.

PROCEDURE: torsion spring design

Now we can proceed...

Assuming a clearance between the arbor and the ID of the spring to be, say, 20% of arbor diameter, calculate mean diameter.
D = 1.2 x D_s + d = 1.2 x 0.250 + 0.037 = 0.337 in. dia.
Calculate number of coils N.
N = (Ed⁴)/(10.8Dk) = (30 x 10⁶ x 0.037⁴)/(10.8 x 0.337 x 1.5) = 10.2
Adjust N so that partial coil agrees with the desired position of arms when spring is unloaded. Because 90^o + 120^o is greater than 180^o, the desired partial coil is
(540 - (90 + 120))/360 = 0.9
Therefore, N₁ = 9.9 and the adjusted mean diameter is
D₁ = (30 x 10⁶ x 0.037⁴)/(10.8 x 9.9 x 1.5) = 0.350 in.
Calculate length of spring in loaded position.
L = (N₁ + ₂ + 1)d = (9.9 + 0.66 + 1) x 0.037 = 0.428 in.
0.428<0.9 x 0.5 = 0.45 in. O.K.

Calculate clearance D.
D = ((N₁D₁)/(N₁ + ₂)) - d - D_s = ((9.9 x 0.35)/(9.9 + 0.66)) - 0.037 - 0.250 = 0.041 in.

Check whether the clearance is safe.
0.041 in. > 0.1 x 0.250 = 0.025 in.
and
0.041 in. < 0.2 x 0.250 = 0.050 in.
O.K.

Determine that S₂ ≥ 75% TS. Tensile strength of 0.037 in. wire is 268,000 psi. (This is taken from the chart on this page.)

S₂ = 200,000/268,000 x 100 is approx 75%TS

END OF DESIGN

Special Note: Calculated length of wire in the spring body is

DN = 3.14 x 0.350 x 9.9 = 11 in. which is much more than the length of the wire in the spring ends. Therefore, bending of the spring ends contributes little to the over-all spring deflection.

Now that you have completed this page it is time to look at the design from a different perspective:

Design for Manufacture and Assembly.

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