Concentrated Load Equivalents 3

[vmirat]

I have a project that includes OWSJ's at 4' o.c. supported on a 30 foot W section. The layout is seven joists, evenly spaced. Load per joist is the same, so it's symetrical. My concern is the design of the W section beam.
On small stuff, I like to do the calcs by hand, however the deflection on this one gets a little hairy, so I used STAAD. I know about Table 3-22a in the Steel Manual, but that only goes to four loads per span.
Has anyone developed a rule of thumb where they assume uniform loading if there are so many concentrated loads? The difference between concentrated and uniform diminishes as the number of concentrated loads per span goes up.

 

[msquared48]

The steel joist table should list the maximum resisting moment for any span for the joist. Just figure out the moment and make sure it is less than that. I assume the joist manufacturer can supply the joist to suppport the five or seven interior panel point loading configurations (Is it possible that two of the joists are over columns at the ends of the girder?)

Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask

 

[hokie66]

I think the assumption of uniform load is universal. It is overkill to use a program like STAAD to check the deflection of a simple span beam.

 

[nutte]

Agree with hokie on both points.

Although if you wanted to check the deflection with your seven point loads, you could outsmart the manual and its limited tables by using superposition. Calculate the centerline deflection under each of your seven load and add them all up. I bet this is pretty close to your uniform load deflection.

 

[JAE]

We almost never use concentrated loads on WF beams to model joist loads. The difference is very negligible in the resulting designs.

 

[csd72]

The moments and deflection go down to only a few percent above udl at about 6 or so point loads. I used to have a good chart that mapped these and I will see if I can find it.

Remember that in these situations even if you had a central point load you would usually only have half the udl in that point load.

If this is the case P=wL/2 and M=(wL/2)*L/4 = wL^2/8

As deflection is proportionate to the area under the moment diagram then the deflection in this case would be less than for the udl.

2 to 6 point loads are a bit more complex but also have a reduced load which helps keep the moment down closer to the udl.

 

[paddingtongreen]

I would absolutely use the UDL on the beam, but the OP leads to some wool gathering on my part. From my archives, the following general cases:

Beam divided into n parts, for n-1 loads P. There are different formulas for even and odd divisions. The first part is the same for both:

CF=PL^3/192EI

When n is odd; dmax=CF.[1-1/n][3-(1-1/n^2)/2]

When n is even; dmax=CF.n[3-(1+4/n^2)/2]



Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[BAretired]

The moment diagram for the concentrated loads touches the uniform load parabola at every joist. Using uniform load for moment calculation contains no error. It is theoretically accurate, even for a single joist at midspan.

Deflection is overestimated when "smearing" the point loads to uniform. The moment curve for the uniform load case is outside the point load moment curve between adjacent loads. This means using uniform load is always conservative.

The error can easily be found using moment-area principles. Between adjacent joists, a uniform load would produce a moment of wa²/8 where a is the joist spacing. The area under that moment curve is:
A = 2/3(wa²/8)a = wa³/12

If beam is divided into n parts, then if n is even, there are n/2 areas of A each side of midspan.

Error in using UDL for deflection = (n/2)(A/EI)(L/4) = wL⁴/(96n²EI)

If n = 8, Error = 0.0125 or 1.25%
n = 6, Error = 0.0222 or 2.22%
n = 4, Error = 0.05 or 5.0%

The case of odd n is not included here.

Using a uniform load instead of point loads results in no error for moment calculation, but a slight overestimate for deflection calcs.

BA

 

[msquared48]

OK Guys:

Take a 12 foot span beam with five 1K loads 2 feet apart on the beam.

Considering the point loads, the moment is 9K'.
1(2 + 4) + 1(12)/4 = 9 K'

But considering the 5K load spread over the 12 foot span, the moment is 7.5 K'. .4167(12)^2/8 = 7.5 K'

That's a 20% difference. You can't make the blanket statement that a uniform load would be the same. It's not necessarily true.
Maybe for longer spans wqith a lot of point loads, but not short ones with a few loads.

Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask

 

[trainguy]

If you're going to be that precise, shouldn't you spread the 5k load over 10 feet and then compare?

I personally use UDL over the full length, but the UDL is calculated using the girder's full tributary area width. I know it's a little overkill, but I've almost always done it this way.

tg

 

[nutte]

msquared, you should be comparing your 9 kip-ft moment to that produced by 6 point loads (half of each joist reaction at the columns). (0.5 k/ft)(12 ft)^2/8 = 9 kip-ft

 

[BAretired]

I agree with nutte. 1000# at 2' o/c is equivalent to w = 500#/'.

BA

 

[msquared48]

OK. I thought the load was being applied a different way.

The moment and shear will be conservative with the uniform load at .5 klf applied across the entire length of the beam.

Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask