I have a 2 mm thick steel plate which is 300 mm long and 30 mm wide, supported at either end. It supports a weight-bearing wheel that can roll along the plate. It currently supports the maximum weight that I expect it to support when the wheel is in the middle, but it flexes a little bit too much. Would making it wider help to support the weight and increase its stiffness, or do I need to make it thicker?

Also is there a way to calculate how the stiffness will change with the thickness (or width if that would affect it)?


3 Answers 3


Short answer: make it thicker.

Long answer: The moment of inertia affects the beam's ability to resist flexing.

Use one of the many, free, online moment of inertia calculators (like this one) to see how increasing the height of the beam will have an exponential effect on increasing the stiffness of the beam.

And this site helps provide a pictorial view of the load(s) upon a beam depending upon differing configurations, such as where the supports are and where the load is applied. It also provides a calculator to determine the forces involved.

Wikipedia has a decent article for area moments of inertia. In your particular case, you're asking about a filled rectangular area and Ix = bh3/12. The height has an exponential factor of 3 whereas increasing the base does not have an exponential factor. So for the same amount of material, increasing the height stiffens the beam better.

To be clear, you can make the beam sag less by increasing the width of the plate. It's just more effective to make the plate thicker.

Current moment: Ix = 30 * 23 / 12 = 20 mm4
Increase width by 1mm: Ix = 31 * 23 / 12 = 20.6 mm4
Increase height by 1mm: Ix = 30 * 33 / 12 = 67.5 mm4

And if for some reason you can't easily increase the thickness of the plate, you can consider a different beam structure. Currently, your beam is a simple rectangle. You can easily use a T-beam or an I-beam in order to stiffen the plate instead.

Again, while I've provided some suggested links to online calculators feel free to search for and use others that you may prefer.

  • 2
    $\begingroup$ Wow, this is amazing thanks. Despite theory/maths being way over my head, I used this to calculate that by making it 1mm thicker, the displacement will go from about 5mm down to about 1.5mm, and increasing the thickness by 2mm takes me down to 0.6mm displacement. Spot on! :) $\endgroup$
    – jhabbott
    Commented Jan 21, 2015 at 3:46
  • $\begingroup$ This answer is good and probably the best advice, but I'm going to add another answer explaining exactly how making the plate wider could also help, depending on the situation. $\endgroup$
    – Rick
    Commented Jan 21, 2015 at 16:05
  • 1
    $\begingroup$ This is an excellent answer. It's written in a way that concisely answers the question asked, it has links for further reading but avoids creating an "answer in another castle" situation, it clearly explains the concepts involved and shows their application through example, and it is well written and formatted in an easy to digest manner. Answers like this are the reason the stack exchange value proposition works, and it's the reason I keep coming back. Just wanted to say thanks to GlenH7 for spending the time to make this answer great. $\endgroup$
    – CBRF23
    Commented Apr 27, 2016 at 15:09

The stiffness of a rectangular cross section, be it steel, concrete, wood, or any other material, is related almost entirely to it's modulus of elasticity, $E$, and it's moment of inertia about the axis of bending, $I$.

Since you already have your material set, steel, you cannot change $ E $. What you can change is your $ I $.

The moment of inertia for a rectangular cross section about its neutral axis is $ \frac{b \cdot d^3}{12} $. You will increase your stiffness exponentially by increasing the depth of your plate.

It's important to note that you can increase your I in other ways as well. For example, if you welded another plate to your existing plate to make a T shape in cross section, you will significantly increase your $I$ and greatly stiffen your member.

  • 1
    $\begingroup$ What this answer implies but doesn't say outright is that in many cases you will find it more cost-effective to add stiffeners than to buy a thicker plate. It can even be as simple as slapping on a piece of angle iron, for small projects, with bolts or epoxy. $\endgroup$
    – Air
    Commented Mar 7, 2016 at 19:57

As mentioned in other answers, what controls the deflection is the second moment of area, and the easiest way to increase it is increase the thickness. Doubling the thickness would increase the 16 fold the second moment of area and it would increase the weight by 100%

However below I am outlining, another way to improve the second moment of area is to change the cross-section, which is more efficient. I am presenting the base line example (30 mm X 2 mm), then I am giving two additional configurations which increase the weight by 2/3 (or 66%) and I am also presenting the increase in second moment of area which is significantly greater)

More specifically

Cross-section Description $I_{xx}$ Increase
enter image description here Flat sheet $$20 mm^4$$ -
enter image description here H- section (adding 5mm flanges at either side) $$593 mm^4$$ ~ x30
enter image description here C- section (adding 10mm flanges) $$1217 mm^4$$ ~60 times

From the above three configurations you can see the vast difference the relatively small addition of flanges can make on the stiffness of the beam ( I won't go through the calculation because its in most textbooks - and the question has already an accepted answer).

As a final point, it is noteworthy that the configurations H and C have the weight, yet the C section will have half the deflection. (So its really a matter of intelligently using the cross-section).


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