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## Main Question or Discussion Point

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...

Due to the fact that we use an M6 bolt and machine it down to get to the right diameter, it is very difficult to keep this peace from deforming as you machine. The solution is to machine it down to a 4mm diameter and drill a hole through the pin! To check the diameter of the hole I used the following equations:

Ssu=VQ/(It)

with V= T/r/2;

Q= y*A;

y=(4/3π)*(So²+SoSi+Si²)/(So+Si);

A= π*(So²-Si²);

I=¼π*(So4-Si4);

t=(So-Si)

for a input of: So (Outside diameter) = 4mm

Si (inside diameter) = 1mm

it gives: Ssu = 297 MPa

which is more or less the Ssu of the material...

I also realise 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, but I am not o worried about this.

Any thoughts or feedback on this problem would be appreciated! Am I over thinking it?

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...

Due to the fact that we use an M6 bolt and machine it down to get to the right diameter, it is very difficult to keep this peace from deforming as you machine. The solution is to machine it down to a 4mm diameter and drill a hole through the pin! To check the diameter of the hole I used the following equations:

Ssu=VQ/(It)

with V= T/r/2;

Q= y*A;

y=(4/3π)*(So²+SoSi+Si²)/(So+Si);

A= π*(So²-Si²);

I=¼π*(So4-Si4);

t=(So-Si)

for a input of: So (Outside diameter) = 4mm

Si (inside diameter) = 1mm

it gives: Ssu = 297 MPa

which is more or less the Ssu of the material...

I also realise 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, but I am not o worried about this.

Any thoughts or feedback on this problem would be appreciated! Am I over thinking it?