I have a small machine shop and need to size a beam for a gantry crane. The design is a 4 post fixed design with 2 horizontal beams running parallel across the shop ends with 1 beam to be mounted perpendicular on trolleys on top of the two parallels making a crane that can traverse x and y travel with an electric hoist rasing and lowering loads.

The span is approximately 28 feet and the biggest hoist I have is a mere 1 ton but would like to size it so that I have 3000-4000lbs load capacity. Also consider that the x and y axis trolleys are not motorized so a small dynamic/torsional load will be imparted in the x and y when the payload is pulled around.

I'm thinking a wide flange 12*45 would be a decent size. I'm unsure though if in beam calculators they account for the weight of the beam or if it is just the payload. With a centered load and 4000lbs the deflection shows .311" over 28 feet which seems reasonable enough if I understand what I'm looking at.

I'm thinking I'm looking for a moment of inertia in the x axis if the beam is standing in the traditional direction. I'm also not too clear on deflection limitations on steel.

Thanks for the help.

  • 1
    $\begingroup$ 3000 - 4000 ounces, pounds, stones, kilograms or tons ? $\endgroup$
    – Solar Mike
    Sep 14, 2020 at 6:37
  • $\begingroup$ Pounds forgot to add that $\endgroup$
    – Joe howell
    Sep 14, 2020 at 6:57
  • $\begingroup$ That is a significant beam. From industrial gantrys I have seen , you should probably be thinking -truss. $\endgroup$ Sep 16, 2020 at 18:45
  • $\begingroup$ I had all the beams and steel needed for everything just needed an appropriately sized spanner beam. It will look nearly identical to this one ton here cdn.shopify.com/s/files/1/0051/2291/6416/products/… I don't mind it being oversized or less than efficient I have the ceiling space for it all. I priced a w14 * 43 and that would still come in under 4k lbs over head. 2500 for load and 1200 for beam weight $\endgroup$
    – Joe howell
    Sep 17, 2020 at 0:58

1 Answer 1


Although this is probably a question best answered by @kamran (I'm not familiar with imperial units or with safety factors in US) I'll give it a shot (just to improve my game).

Assuming the wide flange properties I found are correct

  • $I_{xx}= 350 [in^4]$
  • $h = 12.06 [in]$ depth of beam
  • $m_{beam}= 45 \frac{lb}{ft} $

Although the mass of the beam is distributed, I am assuming that it is located in the center to make a more conservative estimation. the total load should be:

$$P_{tot}[lbf] = m_{beam}g L + P_{Load} $$


  • $ m_{beam}\cdot L \cdot g = 1260 [lbf]$
  • $ P_{beam}\cdot L \cdot g = 4000 [lbf]$

The maximum bending moment is at the center :

  • $ M_b = P_{tot}*\frac{L}{2}= 73640 [lbf\cdot ft]$

Then the maximum stress is $$\sigma_b = \frac{M_m}{I_xx}\cdot \frac{h}{2} = 15.2245 [ksi] = 105[MPa]$$

$$\delta = \frac{P \cdot L^3}{48\cdot E \;I_xx}= 0.4 [in]$$

For the static calculation, this is marginally ok. The safety factor seems to be about 2-3. However since you mention dynamic loads, I would feel prefer it to be slightly more beefy.

In any case, I would wait for @kamran's verdict.

  • $\begingroup$ Yeah the beam I had on hand to use was around a w12 and about 80 lb ft and the weight of it was more than what I wanted to use. Plus it was welded in the center and not a solid beam. I've heard before that load bearing cranes and what not have a safety factor of 10x in commercial settings. This isn't a shop where anyone but myself works in and if it has a safety factor of 2-3 on a 4k lb load I'd be comfortable hanging ~2500lbs off of it. My electric hoist is 200lbs plus 50lbs chain, another 50 for trolley and the hoist will lift a tad over 1 ton. I appreciate the input. $\endgroup$
    – Joe howell
    Sep 14, 2020 at 11:17
  • $\begingroup$ That's why I suggested in the first place to wait for kamran. He usually responds to this type of answers and his reponses are usually a treasure trove of succinct knowledge and experience. $\endgroup$
    – NMech
    Sep 14, 2020 at 11:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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