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First post so bear with me:

I am designing a frame for a press, which needs to be able to withstand 250 N. I am planning to make the frame an L-shaped beam, turned around. With a side view, it will look something like this (the press will be attached on the right side of the vertical beam, which will push downwards):

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l

l

l

l

The height of the vertical beam will probably be 160 mm and the length of the vertical beam will probably be 50 mm. The lower end of the vertical beam will be screwed on top of a table, so this will be a clamping.

For me, the question seemed simple at first, but I just can't seem to figure it out. I am almost sure I would need to use the E-modulus of the material (probably steel), but I can't find the correct formula. I have tried using the formula: $$\sigma = \frac{M * y}{I} $$ then I drew a cross-section of the beams, and made the assumption of using an 80 by 80mm beam.

Using this, I got M=12.5 Nmm, y= 40 mm and $I={1\over 12} * b * h^3$ (not sure how it is in English, probably 1/12wh^3). I could not figure this out further.

I am trying to find a suitable length and width for the horizontal and vertical beams, but this seems to be way harder than I thought... can anyone help me into the right direction?

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  • $\begingroup$ Hi, welcome to Engineering SE. You write The height of the vertical beam will probably be 160 mm and the length of the vertical beam will probably be 50 mm., maybe you mean than the horizontal beam length is 50[mm]? $\endgroup$
    – NMech
    Commented Oct 20, 2020 at 6:55
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    $\begingroup$ I think you need to provide a better diagram - your current description doesn't seem to add up, and teh diagram doesn't help at all. Can you add an image, even a simple drawn diagram would help. $\endgroup$
    – achrn
    Commented Oct 20, 2020 at 10:43
  • $\begingroup$ Please find a structural engineer to perform the necessary design calculation. The L shaped beam is a challenge even to the experienced engineer, as its odd geometry, tendency to twist, and prone to buckle. $\endgroup$
    – r13
    Commented Mar 20, 2021 at 2:05
  • $\begingroup$ A small component like this should be a weldment. Unless you can find beam stock that is the size you want , then cut off what is needed. Welding flat stock would also make micro alloyed/ TMCP steel possible with 3X the strength of cold rolled. $\endgroup$
    – blacksmith37
    Commented Mar 14, 2022 at 21:35
  • $\begingroup$ The equation will underestimate the stress which will occur in one of the corners. Also, the load will induce a torsion that you didn't account for. $\endgroup$
    – r13
    Commented Nov 10, 2022 at 0:41

2 Answers 2

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I concur with r13. The question seems to be referring to the loading of an angle. Also, the load does not pass through the shear center. Hence, torsion will exist.

Unless braced against lateral displacement, the member will not deflect parallel to the vertical load. A Lateral deflection of dL = (Ixy/Ix)dV will occur. This results in bi-axial moments about geometric axes or may be resolved into moments about the principle axes depending on the designers preference.

As such, use of the value (I) about the horizontal axis only is incorrect. Also, the moment of inertia equation used is for a vertical rectangular cross section and is not correct for an angle.

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  • $\begingroup$ Yup , if you only want vertical deflection you need to load it above the shear center. I imagine any Timoshenko book will cover this. You end up using Mohr's circle. $\endgroup$
    – Greg Locock
    Commented Jan 27 at 21:55
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First of all, the dimensions seem a bit problematic. Having a horizontal beam with 50[mm] length and a 80mm x 80mm cross-section doesn't make much sense to me.

The following is the procedure of determining the minimum square cross-section for the horizontal beam, if only strength is your concern then what you need to do is start from allowable stress.

If you are using steel then a good value (for a press) would be around $\sigma_{all} = 100[MPa]$$ (if you want to be safer use an even lower number).

Using this in the equation: $$\sigma_{all} \ge \sigma = \frac{M * y}{I}$$

substituting $I$:

$$\sigma_{all} \ge \frac{12 \cdot M \cdot y}{b h^3}$$

Assuming you use a rectangular cross-section $b=h$, and $y= {h\over 2}$.Then the above simplifies to:

$$\sigma_{all} \ge \frac{6 \cdot M }{h^3}\Rightarrow $$

$$h \ge \sqrt[3]{\frac{6 \cdot M }{\sigma_{all} }}$$

Using the values for M you have, then $h_{min} = 19.6mm$. So any square cross-section greater than 20x20 would be ok.

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  • $\begingroup$ the correct equation to use is $\sigma = M*c/I =M/(bh^2/6) = 6M/bh^2$. you are dividing by h^3. Also for these kinds of applications, they usually use a dynamic load factor of 3. so M=750*50mm. And one would consider the base attachment as a critical point. . $\endgroup$
    – kamran
    Commented Oct 20, 2020 at 16:19
  • $\begingroup$ But i am assuming b=h . $\endgroup$
    – NMech
    Commented Oct 20, 2020 at 17:01
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    $\begingroup$ I see it now, the "rectangular cross-section" confused me. you may want to change that to " square". $\endgroup$
    – kamran
    Commented Oct 20, 2020 at 17:19
  • $\begingroup$ I guess my Alzheimer is setting in. I was trying to find for a few minutes what the right word was and all I could come up was rectangular.... Thanks yet again. $\endgroup$
    – NMech
    Commented Oct 20, 2020 at 18:06
  • $\begingroup$ The shape is unsymmetrical, if not loaded through its shear center, the use of this equation is inappropriate. $\endgroup$
    – r13
    Commented Jul 18, 2021 at 16:16

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