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Assembly Design

Step Four,Determine the length of insert based on shear strength of parent material Nominal Diameter 16.0 mm (selected bolt)Pitch 2.0 mm Pitch Diameter (min)17.299mm(refer taped hole data)Nominal Diameter 0.500" (selected bolt)TPI 13 Pitch Diameter (min)0.550"(refer taped hole data)

Tensile Strength of Bolt Tensile Strength of Bolt

L =

Shear Circumference Strength of Hole x Arbitrary Constant

L =

Shear Circumference Strength of Hole x Arbitrary Constant L = Required length of fitted insert L = Required length of fitted insert Arbitrary Constant = 0.5 (0.5 Based on shearing of the parent material occurring along the pitch diameter of the tapped hole) Arbitrary Constant

Conclusion:

= 0.5 (0.5 Based on shearing of the parent material occurring along the pitch diameter of the tapped hole) L =

For this application a 1/2" diameter bolt has been selected.Insert engagement of 1.09" was calculated. The suitable diameter of the insert can be determined by dividing the length of the insert by the diameter of the bolt.For example:L/dia= 1.09"/0.5"= 2.2 select next highest size Therefore use a 2.5D insert

L =
2 L = 20.1mm

1034 X (13.797 X

Ï€

/4)283 X 17.299

2

Ï€

181,000 X (0.407 X /4)25,000 X 0.550 X 0.5

L = 1.09"

π π

X 0.5 Conclusion: For this application a 16mm diameter bolt has been selected.Insert engagement of 20.1mm was calculated. The suitable diameter of the insert can be determined by dividing the length of the insert by the diameter of the bolt.For example:L/dia= 20.1mm/16mm= 1.26 select next highest size Therefore use a 1.5D insert

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