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Consider an arm of length $D$ supporting $W$ weight. The arm is welded to the plate with bolts $d$ inches apart. How do I determine the thickness of the steel plate 6 in wide by a length supported every $D_1$ inches that the arm is bolted to?

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$W$ is nominally 300 lbs.

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  • $\begingroup$ What are the plate supports like? Those things D1 distance apart? $\endgroup$ – John Alexiou May 15 '16 at 14:47
  • $\begingroup$ Please assume that the remaining structure can be considered relatively stiff, and that the plate's connection to its supports (around D1) is hinged. – Homer yesterday So the solution that I am looking for and would understand I think would be something like T=(WD)(dx)(D1y) with some factor that would consider hot rolled vs cold rolled steel. – Homer yesterday $\endgroup$ – Homer May 17 '16 at 1:32
  • $\begingroup$ Please assume that the remaining structure can be considered relatively stiff, what I am looking for is the thickness required of the 6 in wide steel to which I can mount an arm. The more I increase d the less steel I would need, The plate is supported every D1 distance apart. I do not care if the plate flexes only if it is deformed – Assume all other parts adequate So the solution that I am looking for and would understand I think would be something like Thickness = (WD)(dx)(D1y) with some factor that would consider hot rolled vs cold rolled steel. $\endgroup$ – Homer May 17 '16 at 1:47
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This will depend greatly on the stiffness of the remaining structure (the supports indicated around the $D_1$ span, whatever those rest on, etc).

Assuming that the remaining structure can be considered relatively stiff, but that the plate's connection to its supports (around $D_1$) is hinged, then the structure is isostatic and can be trivially solved.

Here's the bending moment, for example (this obviously depends on the actual values of $D$ and $D_1$) (ignore the units, they're irrelevant).

enter image description here

On the other hand, if the supports can be considered fixed (not allowing rotations, then the bending moment diagram is as follows:

enter image description here

As it happens, regardless of the chosen condition, the maximum bending moment on the plate in this example is 150 kNm (or whatever unit you desire). The maximum stress applied on the plate will be equal to

$$\sigma = \dfrac{My}{I}$$

where $M$ is the moment, $y$ is the distance to the fiber farthest from the centroid (in this case, $y = \frac{t}{2}$, where $t$ is the thickness of the plate) and $I$ is the section's inertia (in this case, $I = \dfrac{bt^3}{12}$).

Adding appropriate safety factors to both the applied load and the maximum allowable stress in your steel, you can therefore reverse-engineer the necessary thickness of your plate. This ignores the bending moment due to the structure's self-weight, which is obviously a function of the plate's thickness, but it works as a very decent first approximation (and, actually, in this layout, the plate's thickness shouldn't change the bending moment anyway).

There are other verifications such as for shear strength, but these follow a similar concept as shown for bending moment. You must also check the connection itself, but that requires more details regarding the connection.

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  • $\begingroup$ Please assume that the remaining structure can be considered relatively stiff, and that the plate's connection to its supports (around D1) is hinged. $\endgroup$ – Homer May 15 '16 at 15:54
  • $\begingroup$ So the solution that I am looking for and would understand I think would be something like T=(WD)(dx)(D1y) with some factor that would consider hot rolled vs cold rolled steel. $\endgroup$ – Homer May 15 '16 at 16:04
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To come up with the internal forces/moments you can use Free Body Diagrams, or you can use AutoDesk ForceEffect which is free.

fbd

res

with full results for each element in the structure.

table

I suspect that the bolt connection to the beam is the weak spot so I would check the shear forces on the plate near the connections and I would make sure there is sufficient area to spread the support loads.

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