Hi..
I think it is
qmax = P/(B*L) + 6Mx/(B*L^2)+ 6My/(L*B^2)
qmin = P/(B*L) - 6Mx/(B*L^2) - 6My/(L*B^2)
Which L= length of footing // to X axis
B = length of footing // to Y axis
Alright?
That looks right so long as the resultant is in that diamond shape in the middle of the raft which puts it in the middle third on either of the axes i.e.6x/L + 6y/B <1. In other words, the whole of the raft has a positive bearing pressure on the ground. If the resultant is outside the diamond, then I think it gets a whole lot more complicated!
R
AASHTO Standard Specification for Highway Bridges (16 Edition) has Figure 4.4.7.1.1C p. 53 for Contact Pressure for Footing Loaded Eccentrically About Two Axes.
Similar info may be found in AREA Manual for Railway Engineering.
Send me your e-mail address and I will try to send you this info.
MGinsburg@leoadaly.com
Thanks Chaps however i may not have given you the whole story. POP2514's equation is correct as long as the eccentricity falls within the kern (full marks to richard M). I was looking at the more complex area where, as WHYMRG points out, it seems to get tricky. I have always used tables for this region (Teng) which are based on US AREA docs.
If you study these closely, bi eccentricity introduces a more or less constant 0,84 factor to POP2514's equation. I just cant figure out why or how that balances out. Obviously I dont want to resort to graphical methods for rectangular pads. Someone must know!
Do you have any further advice ---TIA
Your problem just requires a bit of programming and you have it solved by FEM and ready for examination in my sheets Foot_rect.mcd and Foot_circ.mcd in the Mathsoft collaboratory (rectangular and circular footings subject to eccentrical in x and y compression).
By using compression only springs and adequate model with plates you can get also an answer for a footing of any shape through any modern FEM structural analysis having such elements.
