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Background/Purpose

The shear pin is a small metal/nylon pin used to hold together the steering column that is designed to break when the car crashes. When the pin breaks, the steering column will be allowed to collapse into the dash, thus reducing impact forces on the driver’s chest. The objective is to find a suitable pin diameter for this purpose.

Force Calculations

The pin will undergo double shear, the formula for double shear is:

τallow= F/2A

To solve for the pin diameter with a safety factor of F/2 (safety factor of two, but designed for the pin to break, therefore the force is divided by 2):

D=sqrt(F/π*τallow)

From a research paper found on google books (reference), the force required to cause damage to a human chest was 8800N, therefore, a pin diameter was found where F=8800N with the materials nylon, steel and aluminum. Shear strength reference steel and aluminum, shear strength reference for nylon.

Material

τallow (MPa)

Diameter (mm)

Nylon

70

6.33

Aluminum (6061-T6)

207

5.20

Steel (ASTM - A36)

400

2.65

Operational forces on the pin were found for a lower bound on the pin diameters. The operational force was found by using 600N transferred to the pinion from the steering system (see FEA of rack extension), the torque transferred to the steering column would be 7500 N*mm with a pinion radius of 12.5mm. The inner diameter of the steering column is 19.05mm, therefore the operational force on the pin would be 787 N.

The minimum diameter for the pin to operate in each material using a safety factor of 2, and F=787N:

Material

τallow (MPa)

Diameter (mm)

Nylon

70

3.78

Aluminum (6061-T6)

207

2.20

Steel (ASTM - A36)

400

1.58

Previous on MSXII, the shear pin used was (believed) to be 4-5mm nylon pin, therefore, it would be suitable for MSXIV to also use a nylon pin, from calculations above, the nylon pin should be between 2.68 to 6.33mm in diameter.

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