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I am wondering how to determine screw size for a system.

Here is the attached pictures of an example system. enter image description here enter image description here

Lets say the mass of object, m = 10kg
distance, x = 0.5m, x2=0.5m

with moment = 0,
0 = F*1.0m - 100N*0.5m
F = 50N as the screw clamping force

is the calculation to determine required clamping force correct?
If its correct, how to determine the required size and number of screw? (FYI, socket head cap screw is used)

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  • $\begingroup$ ask yourself does the thickness of the vertical plate matter? $\endgroup$ – agentp Dec 12 '16 at 12:29
  • $\begingroup$ What assumptions are you making about the mode of failure? Do you have reason to ignore tear-out? $\endgroup$ – Air Dec 12 '16 at 16:27
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I believe your equillibrium is very wrong. This is like a rigid body and we have to consider all forces applied to it. The base shouldn't be part of the rigid body, since the screw force would be then an internal one, with no way to find it. So, an apporopriate equillibrium would be as shown in the figure below:

rigid body equiilibrium

The forces are:

  • Weight F
  • Reaction from base at rotation point R
  • Screw tension T

Assuming small displacements, the equillibrium equations should be: $$ F + T - R = 0 \\ F\cdot 0.5b - T \cdot 0.5d = 0 $$

From the second one we get: $$ T = F\frac{b}{d}$$

If b/d is let's say 5, then T = 5F !!! That means the screws are loaded with 5x the weight F. It can only get worse with a more slender post.

How to improve it

Add some stiffeners between the base and the posts. The rotation point is moved away from the screws, offering a larger lever arm for T, and simultaneously reducing the lever arm of the weight F.

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