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| | | Assembly Design | | |
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| | | Determine the length of insert based on shear strength of parent material | | |
| | | Step Four, Nominal Diameter Pitch Pitch Diameter (min) | | |
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| | | 16.0 mm (selected bolt) 2.0 mm 17.299mm (refer taped hole data) Tensile Strength of Bolt | | Nominal Diameter TPI Pitch Diameter (min) | | 0.500" (selected bolt) 13 0.550" (refer taped hole data) | | |
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| | | Tensile Strength of Bolt | | |
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| | | L = - | | L =- | | |
| | | Shear Circumference Strength of Hole x Arbitrary Constant | | | Shear Circumference Strength of Hole x Arbitrary Constant | | |
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| | | 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) | | |
| | | 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) L = 181,000 X (0.4072 X K/4) 25,000 X 0.550 K X 0.5 L = 1.09" Conclusion: 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 | | |
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| | L | | 1034 X (13.7972 X K/4)
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| | 283 X 17.299 K X 0.5 L = 20.1mm | | |
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| | | 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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| | | Calculated Load of Assembly (pounds) | | Tensile Strength - Bolt | | |
| | | Tensile Strength - Bolt | | |
| | | Calculated Load of Assembly (kN) | | |
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| | | 200 | | |
| | | 50,000 | | |
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| | | 1250 Mpa bolt | | |
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| | | 40,000 | | |
| | | 160 | | |
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| | | 1034 Mpa bolt | | |
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| | | 700 Mpa bolt | | |
| | | 220,000psi bolt | | |
| | | 30,000 | | |
| | | 120 | | |
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| | | 180,000psi bolt | | |
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| | | 20,000 | | |
| | | 80 | | |
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| | | 120,000psi bolt | | |
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| | | 10,000 | | |
| | | 40 | | |
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| | | 60,000psi bolt | | |
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| | | 5,000 | | 15,000 25,000 35,000 45,000 Shear Strength - Parent Material (psi) | | |
| | | 20 | | 100 180 260 Shear Strength - Parent Material (Mpa) | | |
| | | 340 | | |
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| | | The shaded area in the graph indicates the region in which bolt failure will occur. | | The shaded area in the graph indicates the region in which bolt failure will occur. | | |
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| | | Note: Inserts are available in standard lengths which are multiples of the diameter. For example an insert with a length of 1.5D will measure one and a half times as long as the diameter when installed. Note: The example above is an indication only. Professional engineering advice must be sought when exact design calculations are required. | | |
| | | 1.5D = 1 1/2 times length of insert | | |
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| | | 0> E b | | |
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| | | 37 | | |
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