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Point Load on One-Way Slab 5

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vincentpa

Structural
Nov 9, 2005
223
1. I have two point loads on a one way slab. The span is 14'. Does anyone have an idea how much of an effective width, "b" I can use in resisting these loads? I think the usual b = 12" wide for distributed load is a lot conservative. There is no mention in ACI dealing with point loads on a one-way slab.

2. The point loads are pumps by the way. Does anyone have any idea how to determine the natrual frequency of a one-way slab?

3. There is a hatch in the slab 3'x4' at the edge of the slab. I would like to frame it with beams. My boss just wants to add more rebar. I think this is risky since the slab is supporting heavy loads.
 
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Is the slab:

formed and poured w/o metal deck
formed and poured w/ metal form deck
formed and poured w/ composite metal deck
precast plank
or another type?

The type of construction will make a difference in determining the appropriate effective width. Please note that if the loads are near to one another, you would need to look at overlapping stresses.
 
Are the pumps heavy enough to be considered concentrated loads or can they be looked at as part of the uniform live load? I assume they have a housekeeping pad underneath to help spread out the load. What is the uniform load under the contact area of the pad?
 
There is an AASHTO equation for effective widths from concentrated loads that you can use.

Also, I have done this before:

1. Draw four lines extending out from your point load, 30 degrees from a line that is parallel to the span direction. You would have four lines going northwest, northeast, southwest, and southeast.
2. Extend these lines to the supports.
3. From a point halfway between the point load and the support, draw lines perpendicular to the span direction.
4. The distance between points of intersction between your half-way lines and your diagonal lines can be assumed as an effective width.

Keeping in mind that the above is not supported by lab tests, it at least feels right.

The other way would be to use 4 x slab thickness or some other empirical measure. If you model the slab with finite elements, you would get variations of moments but would still need to assume an effective spread (weighted average) of the moments over some discrete width.

 
The two pumps with impact weigh 5.9 kips each. the weight is 2533 pounds and the thrust is 2273 pounds. I used a factor of 1.5 for impact. They operate at 1800 rpm. They are over an 18' deep concrete tank. The tank is 30 feet long. the tank contains acid mine drainage with pH of 2.5 sulphuric acid. I would like to form the bottom of the tank and paint it. I don' know if it is a good idea to use form deck with that environment.
 
See thread507-112794 and thread592-142932.
 
The best reference for the distribution of concentrated loads on formed slabs is the one I describesd in
Conservatively, the effective width for a single concentrated load on a one way slab is 0.6 times the span length. This assumes that the load is not near a free edge. Note for a single concentrated load at the center of the span M = PL/4/0.6L = 0.42 P. You must add to this the dead load and any other live load stresses.

You should also have more than just temperature & shrinkage steel at right angles since this helps spread out the load to that effective width.

If the distance between concentrated loads is less than the effective width then you will have overlapping stresses and must reduce the effective width appropriately.
 
Should the effective width for slab design be different for positive bending and negative bending (if any, at support)?

Nowadays, more and more I used a simple shell element model to get the maximum bending moment (in k.ft per foot). The reinforcement (for concrete slab) is determined right away, no hassle of "effective width". And when the slab is not a very simple one-way slab, there is a thickening portion, computer modeling shows its advantage. FEA sounds complicated and overkill, but we just use the tool in the structural analysis software. It is only a matter of a few minutes for modeling. This can save a lot of search and justification.
 
There is no effective width for negative moment. As a concentrated load approaches a slab support such as a beam, the negative moment approaches P/(2xPI). If the support is a fixed condition, the negative moment approaches P/(PI). Note that a lift truck straddling a beam (with one wheel on each side) can develop the P/(PI) negative moment.

Don't forget to add in the DL negative moment and any other LL negative moment.

Although I often use FEA computer modeling for many things, I don't generally use this for concentrated loads on slabs because of the difficulty of modeling the concentrated load over a finite area rather than a point load. This can result in unrealistic hot spots for the moments.
 
jike
Can you explain 'There is no effective width for negative moment', and what is meant by P/(PI)?

Thanks
 
PI = 3.14157

The studies that were done by Westergaard in the 1930's proved that as a single concentrated approaches a fixed support, the negative moment approaches P/3.14157. The study could not relate this to an effective width like that for positive moment. As P approaches the fixed support of a "beam", the negative moment approaches zero, however on a one way "slab" the negative moment approaches P/3.14157. Based upon that info. what is the effective width for negative moment? Eff. Width = M beam / M slab = zero? This is the same calculation that we use to determine the efective width for the positive moment but it doesn't work for negative moment.

Note that effective width is just a theorectical concept that helps us understand the load distribution. The concept doesn't work in the case of negative moment.
 
close enough......us old guys get forgetful!
 
Modeling a "point" load should be easy enough in FEA if you know what the real point is. For example, if it is a truck wheel, just use a FE to be the contact area, then apply area load on the FE.
 
vincentpa forget the slab point load idea.

My experience with pumps like this is to run a wide concrete beam, or two beams, one on each side of the shaft hole opening (if it is large). Mass is needed to dampen the vibration.

The vibration consideration is real. If this pump starts dancing it will wear out bearings very quickly. The fix later is not easy! Look for some reference on simple structural dynamics to size the beam.

I'll see what I can find.
 
SlideRuleEra has some info. on his web site slideruleera.net on machine foundations. Most deal with mass foundations on soil not beam supports. However, the natural frequency calcs are there and maybe helpful.

Pumps that you are dealing with have rotation parallel to the slab and are less problematic than machinery such as compressors that rotate perpindicular to the slab. Still adding mass under the pump is a good idea even if you cannot calculate it.
 
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