Concrete Pad - Stress Distribution

[Redacted]

Hi there,

I need to design a house-keeping pad for an air handling unit.

The mechanical engineer stated that the depth of the pad can be 4" thick.

The floor load where we will be putting these pads has a design live load of 100psf, so I am trying to make sure that the pressure from the pad and AHU do not go above this.

The AHU is around 1650 lbs. I applied a dead load safety factor to the AHU and pad load. In order to make the factored pressure less than 100 psf, I needed to increase the house keeping pad plan area to 1'4" around the AHU.

However, I am not sure if I can consider this full plan areas as spreading the load? Is there a point where the additional plan area contribution will not spread the load?

I drew a 45 degree angle line in red from the edge of the AHU (assuming this is the way the load will spread from the AHU). Which of the three lengths can I assume to spread the load? See below sketches :

First sketch shows the plan area :

Second sketch shows the section :

I believe I will run into some issues if I can't consider the full pad width, as the factored pressure with this load is ~ 100 psf.

Any help would be appreciated.

 

[driftLimiter]

You should just spread the load 45 degrees through the housekeeping pad to calculate the PSF. The housekeeping pad is somewhere in the middle of flexible and rigid but I think adding more slab in an attempt to lower the weight is a bit backwards.

The AHU is around 1650 lbs. I applied a dead load safety factor to the AHU and pad load.

What exactly do you mean. If I take the weight of the units floor area I get 51 psf. Then add the 4" of concrete I get 101 psf.

If your trying to keep the weight low, adding more housekeeping pad seems like your shooting yourself in the foot.

 

[Redacted]

Hi driftLimiter, thanks for this information. So for my understanding. Does this mean that the contribution from the pad to spread the load can only be a max of 4" around the AHU?

If I reduce the slab plan area to reflect the 4", it would result in a new plan area of 6'11"x5'10".

I normally work in metric and convert to imperial at the end so apologies if this is confusing but :

AHU load = 7.2 kN
Plan area = 3.75m^2

Factored pressure = (7.2*1.35)/3.75 = 2.592 kPa
Unfactored pressure = 7.2/3.75 = 1.92 kPa

4" pad load = 24*0.1016*3.75 = 9.144kN
Plan area = 3.75 m^2

Factored pressure = (9.144*1.35)/3.75 = 3.29 kPa
Unfactored pressure = 9.144/3.75 = 2.44 kPa

Total factored pressure (Pad and AHU) = 5.88 kPa (122.8 psf)
Total unfactored pressure (Pad and AHU) = 4.36 kPa (91.2 psf)

It would only work when using unfactored loads. I'm not sure if this would be acceptable or not for an indoor house keeping pad.

The original structural drawings for the building state that the design live load for the floor is 100 psf. I'm not sure if this is unfactored or if the original designer applied an additional live load safety factor to this.

 

[jayrod12]

Where I practice, the specified loads on the drawing are just that specified. Therefore you compared the actual loads to the specified loads. If I were to design a floor beam for a specified load of 100 psf, I'd use 150 PSF factored. You're factoring the loads of the pad and unit, and then comparing them to the specified unfactored load. That's unnecessary in my opinion.

 

[Redacted]

Thanks jayrod12, Thanks for this advice. I want to make sure I am not doubling up on the safety factors.

It would make sense for the load shown on the drawings to be un-factored and it would save a lot of headaches with the pad sizing.

This is what was included in the drawings:

 

[driftLimiter]

I agree with Jayrod you can look at the actual weight and compare that directly to 100 psf live. If I were designing the slab I would assume 100 psf service level load then factor it according to the design method selected for that element.

 

[le99]

Note the loads indicated on the design drawing are "working loads", so it is simpler to use the ASD method to figure out the footprint, then use USD to check the concrete strength against bearing and shear stresses.

4" concrete pad weights 50 psf, the net additional load available = 100 - 50 = 50 psf. The equipment weights 1650 lbs, so 1650/A < 50, or A[sub]req[/sub] = 1650/50 = 33 SF.

Try 6'-11" x 5'-10" pad size, A[sub]prov[/sub] = 40.3 SF > A[sub]req[/sub], ok.

Now you can apply load factors to the equipment and the concrete pad to check the impact on the existing slab if so desired.