Designing a shear pin for a shaft?

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SUMMARY

The discussion focuses on designing a shear pin for a gearbox in a gate valve application, specifically to prevent excessive torque transmission. The calculated output torque of the gearbox is 763 N.m, with a gearbox efficiency of 85% and a spindle diameter of 0.05 m. Two shear stress calculations were performed: average direct shear stress, resulting in a pin diameter of 2.5 mm, and transverse shear stress, suggesting a pin diameter of approximately 3 mm. The preference for transverse shear stress is based on the pin's potential bending due to a gap between the hand wheel and spindle.

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WillemBouwer
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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
 
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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...
 

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