27 March 2020
- Last edited 27 March 2020
Please delete the duplicate posts and leave just one.
Your model is very different from my first example. In my first example, there is only a shaft and nothing else. The ends of the shaft are Fixed. The point of that was to show the relationship between the requested preload and the resultant adjustment when the clamped part was infinitely stiff and did not deform at all because it was modeled as two fixed supports.
Your model is very different from my second example. In my second example, a spring represents the clamped part. You have a U shaped part as the clamped part. The point of my second example was to show the relationship between the deformation of the part, modeled as a spring, under the preload and to add that to the stretch of the bolt then do a hand calculation on the Adjustment.
Your model has the peak stress in the base of the two arms that rise up from the base. If you double the thickness of the two arms, the stress would go down a lot. If you fill in the center of the U-shaped part so it is a solid block with a hole in it, the stress would go down a lot more.
I have not explained my second example sufficiently for you to understand that Adjustment is the sum of two deformations: Bolt Stretch and Part Compression. These two components each have their own stiffness values: Kbolt and Kpart. These two springs are in series. The equivalent spring rate for springs in series is 1/Keq = 1/Kbolt + 1/Kpart. My first example shows how to calculate Kbolt (shown as k in that example). In my second example, I created the part by using a spring and was able to type in Kpart. In your model, you would have to divide the Pretension force, F = 25kN, by the change in the distance between the holes on the outside of the U-shaped part to calculate Kpart to use with Kbolt to calculate Keq.
Then you will find that the Adjustment = F/Keq