Steel Plate Check

[RG4]

I have a trench that spans 7 ft. and a steel plate is needed for a load of 16 kips plus impact. I need to check if a 1.5 thick plate works. Also how to get the stresses? Any suggestion? The plate is free at two ends and simply supported at the other two. fy = 36 ksi.

Thank you

 

[canwesteng]

Yield line analysis probably

 

[Lomarandil]

16 kips plus impact -- sounds like we're talking about public traffic loading, correct?

I've yet to find a good reference for yield line analysis, so maybe that helps you out enough to make it work. Do you know of one CanWest?

But my gut feel (without yield line) is that 7' is too long a span for 1-1/2" A36.

 

[CELinOttawa]

http://www.gratingpacific.com/images/algrip_carbon_steel_floor_plate_load_table.gif

 

[CELinOttawa]

Analytically you can use yield line, or ready tables. There are suitable tables of pre-solved moments in the Structural Engineering Formulas by Ilya Mikhelson.

 

[WARose]

I ran it myself on STAAD (i.e. a 7'x14'x1.5" A36 steel plate; simply supported on the long sides, free on the short sides). I spread it (i.e. a single 32 kip load, I used a factor of 2 for impact) over 144 in^2 at mid-span/center of the plate. This assumes we are talking about a tire here. Results: Max. Deflection= 0.423". Max shear stress= 289 psi. Max. Principal stress= 19.9 ksi.

Back checking the stress with a simple-span model, it's about what I would expect to get if about 7' of the plate was engaged to resist the (developed) moment.

A critical condition occurs when the plate is loaded near a free edge (@ mid-span). In that case, you have stresses more than twice what I quoted. Not sure if that can happen in your case.....but wanted to advise.

 

[BAretired]

I ran it myself on STAAD (i.e. a 7'x14'x1.5" A36 steel plate; simply supported on the short ends, free on the long sides). I spread it (i.e. a single 32 kip load, I used a factor of 2 for impact) over 144 in^2 at mid-span/center of the plate.

Did you mean to say simply supported on the long sides, free on the short ends?

BA

 

[WARose]

Correct BA.....good catch. I altered my post to fix that (and add some more info).

 

[dik]

Can you use Z=b*d^2/4 for the plastic section? Advantage with plastic design as well as strength is isotropic, ie. moment resistance same about any axis.

Dik

 

[daranguiz]

Timoshenko or yield analysis.

[URL unfurl="true"]http://wsmurti.lecture.ub.ac.id/files/2012/10/Perencanaan-Praktis-Garis-Leleh1.pdf[/url]

 

[canwesteng]

daranguiz beat me to it. That publication is for concrete but the same principles apply for steel. For this plate, consider a yield line parallel to the supports at the location of the load (failure as if it were a 1D beam), diagonal yield lines from the point load to the corners, then diagonal yield lines going from the location of the load to any point along the unsupported edge. I do this in Mathcad and vary the dimensions until I find something critical.

 

[BAretired]

The dimension of the plate parallel to the trench has not been given. The answer to the problem will be much different for a square plate (7' x 7') than it is for a very long plate.

The nature of the impact has not been given; the impact factor cannot be determined with precision.





BA

 

[RG4]

Thanks for all the help. The plate is 15x8. 15ft spans the 7 ft trench while 8ft is parallel to the trench.

Yes public traffic loading.

IM = 1.3

 

[canwesteng]

Consider grating as well, something like this

http://www.amico-grating.com/heavyduty.htm

 

[BAretired]

The plate is free at two ends and simply supported at the other two.

The plate is 15x8. 15ft spans the 7 ft trench while 8ft is parallel to the trench.

How can this be? If true, the span of the plate is 15'-0" which does not make sense.

Are there additional supports at each edge of the trench? In that case, the plate is not a simple span.

Perhaps a sketch showing the 15x8 plate and the location of the supports would be helpful.



BA

 

[jayrod12]

BA, If I'm envisioning this correctly, He's covering an approximate 7' wide trench opening with a giant piece of steel plate that overhangs the trench 4' on each side. Which seems unrealistic. It would weigh over 6000lbs. Seems kind of unlikely that is manageable to move around.

 

[BAretired]

It doesn't make sense to me either, jayrod. I think we need more information from RG4.

BA

 

[rowingengineer]

Often in these large span grate situtations you have small columns that reduce the spans down but dont affect the flow to much.

http://www.nceng.com.au/

"Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the Universe trying to produce bigger and better idiots. So far, the Universe is winning."

 

[RG4]

BA, If I'm envisioning this correctly, He's covering an approximate 7' wide trench opening with a giant piece of steel plate that overhangs the trench 4' on each side. Which seems unrealistic. It would weigh over 6000lbs. Seems kind of unlikely that is manageable to move around.

You're correct jayrod. The trench is 7ft by 18ft and the trench plate is being placed in a way that 15ft spans over the trench so 4 feet is pass the trench on each side. Couple plates are needed to cover the trench. Hope that helps.

 

[jayrod12]

Ok but why a 15 ft long piece of plate. The 4 ft overlap seems wasteful