For this activity, you will be examining a repair on an aircraft structure with a fastener hole. Your analysis of the repair will be done in two parts: calculating the stress for the pristine structures as well as the structure that has material loss as a result of corrosion and providing a written synopsis explaining your calculations.

Calculations

The 1/4 inch plates shown above are manufactured from 2024-T4 and are 2 inches wide.

• Calculate tension stress, bearing stress, and tear-out stress. Use MIL-HDBK 5 or MMPDS for material properties.
• For the tear-out stress, calculate the margin of safety.
• Corrosion has reduced the cross-sectional thickness by 10%. Calculate the new tear-out stress and the new margin of safety.
• How much can the cross-section be reduced before the margin of safety is less than 1? Show your answer in terms of cross-sectional thickness.

My answer:

The tension stress is the stress in the plate due to an applied tensile force. The formula to calculate the tension stress is:

tension stress = applied force / cross-sectional area

The cross-sectional area of the plate is:

area = width x thickness

Before corrosion:

area = 2 in x 0.25 in = 0.5 in^2

After corrosion:

area = 2 in x 0.225 in = 0.45 in^2

The tension stress for the original thickness is:

tension stress = applied force / area = F / 0.5 in^2

The tension stress for the corroded thickness is:

tension stress = applied force / area = F / 0.45 in^2

Bearing Stress:

The bearing stress is the stress in the plate due to an applied compressive force. The formula to calculate the bearing stress is:

bearing stress = applied force / (diameter x thickness)

Assuming the applied force is perpendicular to the plate and acts on a circular area with a diameter of 1 inch, the bearing stress for the original thickness is:

bearing stress = F / (1 in x 0.25 in)

The bearing stress for the corroded thickness is:

bearing stress = F / (1 in x 0.225 in)

Tear-out Stress:

The tear-out stress is the stress in the plate due to an applied tensile force that could cause the plate to tear out from its fasteners. The formula to calculate the tear-out stress is:

tear-out stress = 2F / (hole diameter x thickness)

Assuming the applied force is perpendicular to the plate and acts on a circular area with a diameter of 1 inch, the tear-out stress for the original thickness is:

tear-out stress = 2F / (1 in x 0.25 in)

The tear-out stress for the corroded thickness is:

tear-out stress = 2F / (1 in x 0.225 in)

To calculate the margin of safety, we need to compare the stresses to the material's yield strength:

margin of safety = yield strength / stress

For each stress, we'll calculate the margin of safety for the original and corroded thicknesses.

Margin of safety for tension stress:

original thickness: MOS_tension = YS / (F / 0.5 in^2)

corroded thickness: MOS_tension = YS / (F / 0.45 in^2)

Margin of safety for bearing stress:

original thickness: MOS_bearing = YS / (F / (1 in x 0.25 in))

corroded thickness: MOS_bearing = YS / (F / (1 in x 0.225 in))

Margin of safety for tear-out stress:

original thickness: MOS_tear-out = YS / (2F / (1 in x 0.25 in))

corroded thickness: MOS_tear-out = YS / (2F / (1 in x 0.225 in))