Double shear pin for the snow blower

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The discussion revolves around the calculations for the shear pin used in the Stanley SS30 dual-stage snow blower. The maximum output torque is calculated to be 258.72 Nm, while the shear force for the pin is determined to be 42072 N, leading to a moment of 1851.168 Nm. There is confusion regarding why the shear pin's calculated strength exceeds the maximum output torque, suggesting it should break at a lower value. The shear strength of the material is specified as 186 MPa, and the pin diameter is 12 mm. Clarifications and additional details about the pin's installation and dimensions are requested to better understand the calculations.
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Snow blower Stanley SS30
Type :Dual Stage
Maximum RPM: 3600
Torque : 21.56 Nm
Impeller Diameter :16" inch
Auger Diameter:16" inch
CC: 302
Tire Size : 16" * 6.5
I do not know the gear ratio : if assume its 12:1
Gear output torque will be 258.72 Nm

Homework Equations


Assume the Shaft diameter = 44 mm=0.044m

shear force=(pi *d^2/2)∗ shear strength of the material ( for double shear)

shear force = (pi * 12^2/2)∗186
shear force = 42072 N

M = shear force * L

M= 42072* 0.044

M = 1851.168 Nm

why i am getting so value for the shear pin because the maximum output torque is just 258.72 so the shear pin should break before this value isn't it?

The Attempt at a Solution


Assume the Shaft diameter = 44 mm=0.044m

shear force=(pi *d^2/2)∗ shear strength of the material ( for double shear)

shear force = (pi * 12^2/2)∗186
shear force = 42072 N

M = shear force * L

M= 42072* 0.044

M = 1851.168 Nm

why i am getting so value for the shear pin because the maximum output torque is just 258.72 so the shear pin should break before this value isn't it?

 
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bal said:
Snow blower Stanley SS30
Type :Dual Stage
Maximum RPM: 3600
Torque : 21.56 Nm
Impeller Diameter :16" inch
Auger Diameter:16" inch
CC: 302
Tire Size : 16" * 6.5
I do not know the gear ratio : if assume its 12:1
Gear output torque will be 258.72 Nm

Homework Equations


Assume the Shaft diameter = 44 mm=0.044m

shear force=(pi *d^2/2)∗ shear strength of the material ( for double shear)

shear force = (pi * 12^2/2)∗186
shear force = 42072 N

M = shear force * L

M= 42072* 0.044

M = 1851.168 Nm

why i am getting so value for the shear pin because the maximum output torque is just 258.72 so the shear pin should break before this value isn't it?
[/B]

The Attempt at a Solution


Assume the Shaft diameter = 44 mm=0.044m

shear force=(pi *d^2/2)∗ shear strength of the material ( for double shear)

shear force = (pi * 12^2/2)∗186
shear force = 42072 N

M = shear force * L

M= 42072* 0.044

M = 1851.168 Nm

why i am getting so value for the shear pin because the maximum output torque is just 258.72 so the shear pin should break before this value isn't it?
[/B]

It's hard to follow these calculations. Units appear only for certain quantities, and numbers show up without reference.

What is 186? What units does it have? What is 12?

Can you provide some sort of sketch of this pin and what it connects to? You want to see if the pin fails, but no dimensions are provided. How whack is that? o_O
 
Sorry for not providing the information properly and thank you for the reply.
shear strength of the material=186 MPa
d=12 mm ( diameter of the pin)
M = moment
L = distance between two notch
upload_2015-4-16_13-13-22.jpeg

In the above figure,two yellow pointed arrow is where the shear pin will be installed

I am planning to use the similar type of shear pin with two notch as shown in figure below
shearbolt.jpg
 
the pin will be connect to the auger shaft of the snow blower
 
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