1
$\begingroup$

Let's say I'm designing a table surface which has to be stiff, rigid, but at the same time, not too heavy. It is rectangular in shape and ratio of length:width is quite large. It is only supported on both ends of its length. If I use a 50 mm slab of gray cast iron for the surface of the table, during maximum loading the table will bend and the Z-axis displacement at the center would be 0.04 mm. This table will be used as a machining table, and the maximum tolerance of the workpiece will be 0.005 in (0.127 mm), so I think the 0.04 mm deflection of the table is already good because it will give room for tolerance from other component of the machine, such as cutting tools, spindle, etc. Problem is, with 50 mm steel slab, the maximum weight criterion is exceeded. The whole table is designed to be able to tilt on A-axis (it's a trunnion table), driven by a motor. The heavier the table is, the bigger and exponentially more expensive the motor would be. I have to reduce the weight of this structure, but I have to keep the stiffness.

What if instead a simple flat 50 mm slab I make it like 20 mm slab with extra stiffener profile on its underside? Like honeycomb pattern, or any other stiffener configuration. which is possible because it's a cast iron anyway, just have to create the mould.. by doing that I could reduce the weight by almost 40kg

The question is: Is it theoretically possible? 50 mm slab gives 0.04 mm bending displacement, now I want to make it like 20 mm slab plus some stiffener to reduce weight, but I would like to keep same performance with 0.04 mm maximum displacement. The maximum load is 1200 N, concentrated in small area in the middle of the table.

EDIT: previously I wrote the requirement as 0.005 mm deflection, but it is actually 0.005 in (0.127 mm). However, this +- 0.127 mm figure is basically the machining tolerance of the workpiece that will be made on top of this table, so I guess the tolerance of table itself should be much lower than that, I don't know the exact number yet but it just won't go as low as 0.005 mm that's probably too extreme. I'm still in the early stage of the design so a lot of design requirements is still open to revision..

$\endgroup$
  • $\begingroup$ Holy Jesus, that's a tight tolerance! 5 micrometers! $\endgroup$ – Wasabi Sep 7 '17 at 9:09
  • $\begingroup$ Some more info please: what's the loading pattern? Is it distributed over the entire table span and width, is it a concentrated load, or is it something else entirely. Also, is there a thickness limit to the table? How "tall" can the stiffeners be? $\endgroup$ – Wasabi Sep 7 '17 at 9:10
  • $\begingroup$ Uhm.. yes it's tight, to be honest I'm not entirely sure the 5 micron requirements is right. Basically it will be a milling table.. And I want the milling machine to be able to produce parts with 5 micron tolerances. $\endgroup$ – JohnDoe Sep 7 '17 at 9:16
  • 1
    $\begingroup$ @Wasabi just now realized that I've made a stupid mistake when deciding the tolerance for my machine.. I took that 0.005 figure from various machinist forums, generally in any topics that talk about machining tolerances, they say 0.005 is a good tolerance that is still "relatively easy to do".. That's why I decide that my milling machine should have at least 0.005 mm tolerances, because it's good enough and still easy to do.. I visited those forums again to reassure that 0.005 mm is "still relatively easy to do", just to realize that they're talking about inches, not millimeters. *big facepalm $\endgroup$ – JohnDoe Sep 7 '17 at 10:06
  • 1
    $\begingroup$ @Wasabi Alright, I thought it was needed to keep track of what being edited. Quite new to stackexchange here so I don't know the norms yet :) $\endgroup$ – JohnDoe Sep 8 '17 at 0:34
0
$\begingroup$

The formula for central displacement shows that the displacement is inversely proportional to both the elastic modulus, E, and the second moment of area, I.

You need to increase either E or I, eight times. Increasing E is not allowed because it is an intrinsic property of the material that you are using.

I is proportional to the cube of height, so doubling height would directly solve your problem. However, the increase in mass would not be allowed.

What you can do is make a hollow rectangular section and thus increase the height of the cross section as much as possible. This will take some calculations ( by hand or by using a spreadsheet program).

It seems quite doable.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you! This is the kind of answer that I'm looking for. I was looking for a theoretical basis to modify my design. However it appears that the increase in area moment of inertia wouldn't need to be eight times. As I added in the comment above, I was wrongly took the maximum deflection requirement as 0.005 mm while it was actually 0.005 in (0.127 mm). Actually I must improve the requirement a little because +- 0.127 mm is actually the tolerance of the final machining products. If the table alone already gives +- 0.127 mm then there's no room for error in cutting tools, spindle, etc.. $\endgroup$ – JohnDoe Sep 7 '17 at 17:30
  • $\begingroup$ Pleased to hear that you liked my answer. You can choose to select this as the answer to your question. (Or wait for another, better one). $\endgroup$ – Gürkan Çetin Sep 7 '17 at 17:32
  • $\begingroup$ I have a little more question sir, rectangular cross section is actually a simplification of the real design, because just like any other machining table, there would be a lot of details on the surface like T-slots for example, also by adding stiffener under the table it would make the cross section stray even more from a rectangular shape.. I imagine the determining the area moment of inertia would be difficult.. Is this still something that can be done with hand calculation or do I have to use finite element software? $\endgroup$ – JohnDoe Sep 7 '17 at 17:55
  • $\begingroup$ You can use hand calculation methods, and if it gets too complicated you can use software , not necessarily FEM software. $\endgroup$ – Gürkan Çetin Sep 8 '17 at 5:46
1
$\begingroup$

Wooden tables have been doing that for a long time using an apron. In fact a steel clad wooden surface may be enough. But if you seal the top with a slab of steel you will also want to seal the underside to avoid moisture related warping.

Add a T-profile beam on the underside connecting the supports and your table will be much stiffer.

If necessary you can also add a cross shape or some beams connecting the other two beams to add some support in the center.

| improve this answer | |
$\endgroup$
  • $\begingroup$ In this case I can't use wood, because there would be some machining going on on top of that table. What makes feel uneasy is that even with a thick slab of cast iron, the deformation is already quite high. And now I must reduce the thickness of the slab and at the same time increasing stiffness.. Like those big bulky 80 kg table is not stiff enough, and here I come with a 40 kg table expecting it to be stiffer than before.. It makes me doubt myself a little bit there.. $\endgroup$ – JohnDoe Sep 7 '17 at 8:30
  • 1
    $\begingroup$ @JohnDoe that's why you add support beams, even a centimeter wide vertical member will have a lot of resistance against a vertical load. I suggested the T profile so you can attach the wide flat bit to the table and let the vertical member provide stiffness. $\endgroup$ – ratchet freak Sep 7 '17 at 8:53
  • $\begingroup$ I agree with this. Do not just put plain sheet as table, instead create an effective framing for your table. This is similar to the framing of trucks. $\endgroup$ – Jem Eripol Sep 7 '17 at 9:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.