Eccentricity in an isolated footing

[Buleeek]

Hello,

I have an isolated footing and need to calculate maximum soil bearing pressure. The footing has a column offset as shown in the attached picture. In addition there is a moment. My question is, what "e" should I take into consideration? Should I just combine "e" from M/P and "e" from column offset?

And then, what is the formula for qmax assuming that e > B/6 (see attached)?

Please advise. Thank you

 

[phamENG]

Link

 

[Buleeek]

phamENG,

Thank you, that is very interesting. However that chart/method has a limitation that forces occur at center of the footing. I am looking for a solution where column is located off-center and moment is applied to it.
Thank you,

 

[phamENG]

Sorry about that. I'd look at it as concentric with a moment to resolve the e from the offset, and then combine it with the other moment.

 

[BAretired]

See below. Your 0.5' dimension is off scale, but the effective eccentricity is 1.1' which exceeds B/6, so most of the footing will have zero pressure on it.

Usually the load on a column has a maximum and a minimum value. You show only one load, 5000#. If the minimum load is less than 5000#, the effective footing width will be even less than 2.1'.

BA

 

[Buleeek]

BA,
Thank you for explaining this. It makes sense. If the moment was in other direction I would subtract one from another.

 

[BAretired]

If the moment was in other direction I would subtract one from another.

Yes, the net eccentricity would be e = 0.1' if the moment was reversed. That would make the entire 5'x5' footing effective and the soil pressure would be close to 200 psf throughout.

BA

 

[Buleeek]

BA,

According to my calculation the length of "active" part of footing is 4.2, not 2.1' as you pointed out. Please advise.

 

[Celt83]

4.2 sounds accurate

**Assuming a weightless footing or your loads already account for the footing weight**
Sum Moments about bottom left (clockwise positive) = 5000 (2.0 ft) - 3000 ft-lbs = 7,000 ft-lbs
statically equivalent system puts your 5000 lbs force at 7000/5000= 1.4 ft from the bottom left of the ftg.

Limit of the Kern zone is 5ft/6 = 0.833 ft from the footing center or 1.67 ft from the left edge, so you are outside the kern.

Bearing pressure total load = [1/2 * B' * Q,max] * L
For static moment equilibrium the resultant of the bearing pressure needs to align with the static equivalent location of P:
centroid of a triangle is 1/3*base from the tall end
1/3 * B' = 1.4 --> B' = 3*1.4 = 4.2 ft = bearing length

From sum of vertical forces = 0, P = [1/2 * B' * Q,max] * L
5000 = 1/2 * 4.2 * Q,max * L --> 5000 = 10.5 Q,max --> Q,max = 476.19 psf

My Personal Open Source Structural Applications:

https://github.com/buddyd16/Structural-Engineering

Open Source Structural GitHub Group:

https://github.com/open-struct-engineer

 

[BAretired]

According to my calculation the length of "active" part of footing is 4.2, not 2.1' as you pointed out. Please advise.

You are absolutely correct! Looks like I was asleep when I responded before.

BA