# Designing a shear pin for a shaft?

• WillemBouwer
In summary: The average direct shear stress should be used because it is a more accurate representation of the shearing stress that the pin would experience. The Transverse shear stress should be used because it takes into account the fact that the pin will not rupture linearly.
WillemBouwer
Background: To prevent too much torque having to be transmitted through the gearbox on a gate valve, a shear pin is to be fitted between the hand wheel and the gearbox spindle. When a certain torque is reached the pin should rupture and no torque transmitted. See attachment for reference diagram.

Solution needed: What diameter pin to use at certain torque figures?

Calculations: There are 2 kinds of shear stress I am considering.
Average direct shear stress and Transverse shear stress.

Inputs: Output torque of the gearbox: To = 763 N.m (15% SF included)
Gearbox ratio: rg = 12
Gearbox efficiency: η = 85%
Spindle diameter: ∅D = 0.05 m (Note the spindle would not be threaded as in the picture)
Material: SS 304 thus Su = 505 MPa and Ssu = 0.6*505 = 303MPa

Average direct shear stress:
Ssu = V/A
with Ssu the ultimate shear stress in MPa
V = Normal load on pin due to torque in N
A = Cross sectional area of pin in m^2

V = [(To/(rg*η))/(D/2)]/2 because we have double shear
V = 1496.08 N

A = ∏p^2 where p is the radius of the pin

Thus 303MPa = 1496.08N/∏p^2
so p = 1.25 mm and ∅P = 2.5 mm

Transverse shear stress:
Ssu = VQ/It = (4/3)*(V/A)
303MPa = (4/3)*(1496.08/∏p^2)
so p = 1.45 mm and ∅P = 2.9 mm≈ 3 mm

The reason I think the Transverse shear stress should be used is because the pin would not be rupturing linearly, the hand wheel hole diameter would also be larger than the spindle diameter leaving a gap which would cause a bend hence not rupturing the pin due to normal shearing...

I also realize that the material is strain hardening which could cause the material to get even harder during normal working torque so this could affect the Ssu of the material...

Any thoughts or feedback on this problem would be appreciated!

Thanks guys/girls

It would be quick to set up a cheap test fixture and see how various pins behave. Or you could do a complicated analysis and never be sure of the results because of uncertainies in your input data.

Thanks Pkruse

We currently have a design in progress, but I want my theoretical calculations to be a good reflection on my background studies, hence the reason for asking which equation would be better suited for the pin design...

## 1. What is the purpose of a shear pin in a shaft design?

A shear pin is designed to break under excessive force or torque, preventing damage to the shaft and other components. It acts as a safety mechanism to protect against equipment failure or potential hazards.

## 2. How do you determine the appropriate material for a shear pin?

The material for a shear pin should be chosen based on its strength and ability to withstand shear forces. Common materials used include steel, aluminum, and brass. Factors such as the application, operating conditions, and budget should also be considered when selecting a material.

## 3. What factors should be considered when designing the dimensions of a shear pin?

The diameter and length of a shear pin should be determined based on the expected load and the shear strength of the material. Other factors such as the shaft diameter and the location of the pin on the shaft should also be taken into account.

## 4. How do you properly install a shear pin?

The shear pin should be installed in a clean and properly aligned hole in the shaft. It should be inserted and secured tightly, ensuring that it can withstand the expected forces. Proper installation is crucial for the shear pin to function effectively.

## 5. Can a shear pin be reused after it breaks?

No, a shear pin should not be reused after it has broken. It is designed to break under excessive force, and using it again could compromise the safety and functionality of the shaft. A new shear pin should be installed in its place.

• Mechanical Engineering
Replies
11
Views
17K
• Mechanical Engineering
Replies
2
Views
2K
• Mechanical Engineering
Replies
1
Views
2K
• Mechanical Engineering
Replies
1
Views
4K
• Mechanical Engineering
Replies
19
Views
4K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Mechanical Engineering
Replies
13
Views
19K
• Mechanical Engineering
Replies
3
Views
6K
• Introductory Physics Homework Help
Replies
3
Views
2K
• Mechanical Engineering
Replies
5
Views
10K