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# Equivalent Uniformly Distributed Loads

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## Equivalent Uniformly Distributed Loads

(OP)
Does anyone know if there is a reference that exists on how to approximate the equivalent uniform distributed loading on beams when there are concentrated loads on them. The uniform distributed load would provide the same or a bit more stress and deflection in the beams. I know there are many variables involved and there is no "one method fits all" ,but, I am looking for a approximate or empirical method to calculate  the equivalent uniform load for any concentrated load combination a beam.

Any help would be appreciated.

### RE: Equivalent Uniformly Distributed Loads

The AISC Steel Manual has some of this stuff in the Beams Part starting on Page 3-208.  To be honest, these never seemed very useful to me, other than to check the output from a program (very important).

### RE: Equivalent Uniformly Distributed Loads

Krus 1972, why do you want to go to the bother to create an artificially modified solution, when you could just simply solve the shears, moments and displacements by hand very quickly and simply for most problems??
Seems pointless.

### RE: Equivalent Uniformly Distributed Loads

For a stater, I would base it on the deflection curve, use the maximum deflection and find the uniform load using the deflection equation for uniform loading.

I wold also back check the max shear and moment to make sure there were no problems.  With all the simple to use beam programs, this should be a very easy process.

Cheers...

Mike McCann
McCann Engineering

### RE: Equivalent Uniformly Distributed Loads

We have an engineer in our department who always uses a large uniform live load in lieu of concentrated loads from equipment and piping supports. His rational is twofold.

First it is his contention that the final locations of concentrated loads will not be known until so late in the design development process that it will be impossible to do a precise analysis within the constraints of the project schedule.

Second he claims that a large uniform load provides an additional factor of safety to account for forces generated from operational loading.

The projects that we work on are typically cogeneration projects.

I am currently reviewing his design of a one storey turbine building for which his roof design he has used a live load of 100 psf and a piping load of 25 psf in addition to a roof snow load. He has no basis for these values other than "engineering judgement".

While I agree that these loads are more than adequate for the roof I disagree with this approach for numerous reasons. My biggest concern is that we will never know the actual state of stress in any member. This will be problematic in dealing with future structural issues.

There is also the issue of economy of design. My opinion is that engineering should produce a safe design at a reasonable cost. The results of the above has produced a very uneconomical roof.

I have reviewed this engineers work in the past and have had the same concerns. The last time I completely redid his RISA model to reflect the actual loading conditions (at quite the expense to our company) and of course his design was well within the allowables. The PM's opinion is that the design drawings are complete, the design is safe, so what's the problem?

I guess I have reached the same point as the original post. What I need is a justificaion and a basis for deterimining a design criteria that that correctly accounts for converting concentrated mechanical loads to equilivant uniform loads.

### RE: Equivalent Uniformly Distributed Loads

Steve1-

Wow! those are huge loads for mechanical equipment. At least if what we're talking about are typical HVAC units.

I have to agree with you on this one. It almost seems like cheating. Does this engineer have a concept of how much 100 PSF is?!

It's tremendously more economical to obtain a reasonable estimate of the actual mechanical loads. For your "engineering judgment" you could even bump those up by 50% or so. For worst case moment, place it at midspan and for worst case shear place it at the support.

If the building use changes at some point in the future, then it would fall to the new tenants or owner to assess the capacity of the roof and take appropriate measures. It seems unfair to make the current owner pay for what someone may do in the future. (Unless the owner specifically requests that the roof have adequate capacity for such a scenario.)

### RE: Equivalent Uniformly Distributed Loads

work out BM from point loads and work back converting into a UDL.Can make a conservative deflection calculation easier than using a number of point loads.

### RE: Equivalent Uniformly Distributed Loads

Steve1,

You stated that your project is a turbine building in a cogeneration project.  So it is an industrial structure, not a commercial or residential one.  Things are different with industrial structures.  100 psf live load for a roof does sound excessive, but perhaps the other engineer has his reasons which have not been adequately explained to him.  Who makes the ultimate decisions on this type of thing in your organisation?  As the reviewer, and if you can not agree with the designer, you need to talk to the managing engineer.

### RE: Equivalent Uniformly Distributed Loads

66,

The purpose of my post was to provide an example of what some of the problems are that can arise when uniform loads are used in lieu of actual concentrated loads.

For my particular case I am well aware, as department manager, of the types of facilities that we engineer. I have been EOR on numerous power projects with price tags in the tens of millions of \$.

The firm that currently employs me has expanded in the past few years from 15 to 40 employees. Unfortunately because our roots are in the mechanical and electrical disciplines the civil/structural department has no representative in the senior management staff. The person who is Director of Engineering is also the PM for the project that I mentioned above. Yes I will have a discussion with him concerning the applicable design criteria.

### RE: Equivalent Uniformly Distributed Loads

In terms of bending stress, the equivalent uniformly distributed load for a simple beam is P = 8M/L (kips), where M is the maximum moment (ft-kips), amd L is the span (ft).

### RE: Equivalent Uniformly Distributed Loads

steve1,

I was unaware of your experience and position when I posted.  As the structural department manager, you are the person who must make the ultimate decisions about these issues.  Sounds as if you have a training session ahead of you.  Good luck.  Old hands can be hard to change.

### RE: Equivalent Uniformly Distributed Loads

No response from krus1972 as to why he particularly wants to do this - I'm certainly intersted to know.

We often replace complex systems of loading with simpler loads.  Sometimes udl's sometimes point loads or combinations.  We do this empirically by looking at the actual loading and selecting a modified loading condition which has similar fundamental behaviour.

I have no problem with the strategy of the engineer working for Steve1.  To me, the question is can 100psf be justified?  There is a sub question - Can the actual loads be reasonably modelled as a udl?  If they can then the only arguement is that of load intensity.

We are now really talking about safety factors versus economy of design (including design fees).

In my area of work (process structures) it is not common to have accurate loads at the start.  We speak to the rest of the team to determine a level of confidence in the data available and we try to be more conservative where we feel it to be necessary.

It is not uncommon to have client organisations challenge us over utilisation ratios but I can onlt think of one case where we were callenged over our conservative assessment of loads.

### RE: Equivalent Uniformly Distributed Loads

personally, i'm really surprised by this approach.  a beam with a UDL deflects much less than a point-loaded beam (so to get the same deflection you'll apply an excessive load).

i agree with the earlier post that said in the early days of a project you're not sure where the load will be applied.  1st short-cut, put it in the middle of the span.  2nd short-cut, if this is excessive put a "no-go" zone on the middle 1/3 of the span and apply it at the edge of this.

### RE: Equivalent Uniformly Distributed Loads

I am suprised by your surprise, I have converted a BM from a series of point loads back to an equivilent UDL calculated deflection then run computer analysis on point loads and deflections are in approx agreement.

### RE: Equivalent Uniformly Distributed Loads

Herewegothen.

Confused.

Equation for deflection of a simply supported beam under a point load is:

(WL^3)/(48EI)= y1

Equation for deflection of a simply supported beam under a UDL load is:

(5Wa^4)/(384EI) = y2

Now, equating the above so that y1 = y2

W = Wa times L (or WaL simply put)

So if you have equivalent loads applied to the beam:

Y1 = Y2 can be seen as being :

(WaL^4)/48EI) = (5WaL^4)/(384EI)

Assuming identical material, 2nd moment etc etc, gives the ratios of load to equate equal deflection:

y1 = y2 shows:

(1/48) = (5/384) or simplified as (1/48) = (8/48)

Ergo, you can see that for equivalent centre point deflection, you require 8 times the total load when applied as a UDL against a point load.

Utter nonsence

### RE: Equivalent Uniformly Distributed Loads

40818 - Just a little correction:
(1/48)=(5/384) simplified is (1/48)=5/(8*48)
Therefore equivalent UDL is 8/5=1.6 times the total load.

### RE: Equivalent Uniformly Distributed Loads

Of course you wouldn't really model a single central point load as a udl.  That would be just plain silly.  You might model a predominantly uniformjly loaded beam with variation on load intensity and several small point loads at odd positions as a udl.  It's "engineering judgement" as to what to simplify the loading to.  You could model loads as a single point load if that seemed to be appropriate...

### RE: Equivalent Uniformly Distributed Loads

i wasn't commenting on simple centre load and udl. I was commenting on series of point load across the span. Say span on a girder of 20m. 8 point loads of high magnitude. You get a bending moment, rather than solve complex defelction formula in the 'old days' they would convert the bending moment back into a udl and use the udl deflection formula (i'm not willing to write it out here unless asked)to estimate the deflection. Why would i do that for a simple point load as the formula is just as simple.
Try it out. Computers didn't always exist.

### RE: Equivalent Uniformly Distributed Loads

i agree that the difference between a series of point loads and a UDL probably isn't great.  i think that a UDL = (sum of point loads)/L will probably underestimate the maximum moment in the beam (but probably not by much).

there seem to be some posts tlaking about matching the deflection of the point load case to the deflection of a UDL.  again, if there's a series of point loads the difference would small, but i think it'd be significant to replace a single point load with a UDL ... matching deflection would require a larger load (and i think needless math).

i was talking more in terms of how i read some of the posts(referring to what i took to be a single point load), particularly the post that seemed to propose using a UDL in lieu of a point load ('cause the location was uncertain).

### RE: Equivalent Uniformly Distributed Loads

2 point loads close to the ends would give the closest thing to uniform curvature. The BM between the loads is unofrm and tapers off to zero between the loads and the end.

Problem is that the loads may induce local buckling.

### RE: Equivalent Uniformly Distributed Loads

except that the bending moment for a UDL is parabolic ... three lines make a crappy parabola ... particularly if you don't want to underestimate the moment (ie pick tangents to the parabola)

### RE: Equivalent Uniformly Distributed Loads

You said that he doesn't know the exact location of the mechanical equipment.  I am assuming you do have a pretty good idea of the weight.  Why not just plop it at midspan and call it a day.  Then it doesn't matter where they put it since midspan is the worst case (unless shear controlls, which almost never happens, but if you are concerned about shear put one at midspan and one at the support and design for the worst).
There is no need to throw so much load up there and there is no way to take a load of known value and turn it into a uniform load - just put the point load at the worst place for design and let them put it where ever they want.

### RE: Equivalent Uniformly Distributed Loads

Quite often the design criteria for an industrial/power plant will include a uniform load allowance for for piping, conduit, and miscellaneous. I have seen these values range from 5 to 25 psf. The mechanical department at my firm does not do a flexiblity analysis on small bore and low temperature piping hence they provide no hanger loads for these systems. In the past they have informed me to use a concentrated load of 5 to 10 kips on each beam as a piping allowance. On a recent project that included four levels of steel over a plan area of 700 sf the summation of all of these 10 kip loads added up to almost one million pounds. So you can see that these things can get out of hand.

There really needs to be some guidance on setting up the design critria for industrial facilities. I have noticed that over the years building codes have greatly expanded the seismic criteria for mechanical and electrical equipment, but that the uniform design load for industrial facilities has remained light and heavy loading with no specific values based on type or use of facility.

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