I am trying to design a lifting mechanism and I would love some insight/opinions as to how I can shed some weight off it.

The current weight I have for my mechanism is 187.5kg, however, this is supposed to be something that is "easily transportable"

I am using stainless steel. Originally we had 20.3m of steel with a cross section of 40mm x 40mm and 10mm thick. Seeing as 187.5kg is quite high I am looking to reduce the weight. I will admit the material and cross section were chosen relatively arbitrarily, but we wanted to be sure that our structure could hold any load without actually worrying about the calculations for that.

The main stress that will be exerted on the beam is bending, as the structure is built like a table, and will have a weight suspended from the middle. I looked up that the yield stress for stainless steel is 193 MPa (I took the lowest value I could find for safety reasons) and I have applied various different cross sections for the steel.

For my original dimensions 40 by 40 by 10mm thick:

$$\sigma\text{ (bending)} = \dfrac{My}{EI}$$

Assuming my table top is completely made of a hollow rectangle of stainless steel:

$$\begin{gather} M = 200\cdot9.81\cdot500\text{ (lifting 200kgs 50cm away from it)} = 98100\text{ Nmm} \\ I = \dfrac{40^4 - 20^4}{3} = 800000\text{ mm}^4 \\ y = 20\text{ mm} \\ E = 193\text{ MPa} \\ \therefore \sigma_b = 0.0127\text{ MPa} \end{gather}$$

I worked in mm because 1 kg/(mm*s2) = 1 MPa

A reduced, solid cross section of 20mm gave a bending stress of:

$$\begin{align} M &= 98100\text{ Nmm} \\ I &= \dfrac{20^4}{3} = 53333\text{ mm}^4\\ y &= 10\text{ mm}\\ E &= 193\text{ MPa}\\ \therefore \sigma_b &= 0.0953\text{ MPa} \end{align}$$

I know stainless steel is a very strong metal, which is why I leaned towards using it originally. I would just really like to know if the logic behind all of my calculations make sense? Am I on the right track for this or have I done something horribly wrong that I am unaware of? I am aware I did not include self weight, but did not see the need for it.

Thank you for reading, I would appreciate any thoughts on this.

Edit: this is for a university project, I am mainly looking for pointers on my conceptual design and work.

  • $\begingroup$ Buy one that will be certified and, therefore covered by insurance... $\endgroup$
    – Solar Mike
    Mar 26, 2018 at 7:38
  • $\begingroup$ Hi this is for a university project. $\endgroup$
    – gunter
    Mar 26, 2018 at 7:48
  • $\begingroup$ Then you should make tgat clear at the start of your question... $\endgroup$
    – Solar Mike
    Mar 26, 2018 at 7:56
  • 2
    $\begingroup$ Can you provide a diagram? I read "...design a ... mechanism...", saw there was no diagram, and stopped reading... $\endgroup$ Mar 26, 2018 at 8:33
  • $\begingroup$ Does your weight move around or is it fixed in one spot? $\endgroup$ Mar 26, 2018 at 15:46

2 Answers 2


Your equation for bending is wrong. Force x distance is appropriate for a cantilever. For a point load applied to a "table" you are either talking simply supported (in which case Mmax = Force x span / 4) or fixed-ended (in which case Mmax = Force x span / 8). Note that you have not provided enough info to show whether a calc this simple is appropriate or not.

Your equation for second moment of area is wrong. For a square section it's b^4/12.

Your equation for stress is wrong. Stress = My/I. No E. E is Young's modulus, not yield stress. You just need to calculate Stress = My/I and make sure that the stress is less than the yield stress.

As you've said, you've neglected self-weight. What's your justification for this? You have a self-weight of 187.5kg and a live load of 200kg; self-weight is a significant proportion of your load!

You haven't explicitly included any safety factors on your load or on your yield. Are there any? Do you need any? (I can't answer this for you, I work in bridge design not small mechanical products, so I have no idea of applicable standards / codes of practice. But I suspect you'll need to make allowance with some sort of factor.)

  • $\begingroup$ Thank you so much for your response AndyT this is exactly the kind of feedback I needed :) $\endgroup$
    – gunter
    Mar 26, 2018 at 8:48

Structural shapes are your friend with this. Look at I-beams, channels, and struts...

enter image description here

You won't believe how much more weight this piece can carry than a table that is much heavier.

Stainless will add cost to your project with no structural advantage over carbon steel. The additional elements in stainless prevent corrosion, which you haven't mentioned as a design issue.

Aluminum comes in a zillion flavors, but a stronger aluminum alloy is pound-for-pound much stronger than typical stainless or carbon steel. Aluminum resists corrosion like stainless steel, too... it won't rust outside.

A long horizontal span of a table increases the stress in the top without increasing the capacity. Is there any way to reduce the horizontal span? Also, if you adopt a truss or A-frame design, rather than a cantilever beam, you will achieve much greater strength per weight.

Casters! Put wheels on the thing and let it roll around.

Quick assembly/disassembly. Can you use toggle clamps or quick-release fasteners, like the wheels on a bike, to disassemble it? Then you transport it in pieces rather than the whole.

Don't rule out wood. Use 4 X 4 or 4 X 6 douglas fir or redwood... wood is really light compared with metal for many applications.

Do the statics on your design by making a free-body diagram, and remember the basic truss premise: a truss shape is light and strong.


  • $\begingroup$ Thanks so much Elliot. That was super helpful! Much appreciated. $\endgroup$
    – gunter
    Mar 27, 2018 at 19:01

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