Uplift Shear resistance footings

[slickdeals]

Folks,
A colleague of mine is working on a project where a footing is embedded in rock and in order to negate the effect of uplift, the geotechnical engineer provided a perimeter shear resistance value of 2 ksf.

The footing is not rectangular and he is modeling it in a computer program (SAFE). A 350 pci modulus of subgrade reaction is used for sizing the footing.

I was thinking of using a line-spring around the perimeter to model the effects of shear resistance but I am unsure how a 2 ksf value of shear resistance can be converted to a line-spring value (psi * height in contact)

 

[ishvaaag]

To develop the allowable shear strength some shear distortion will be required. This is the data you need, as in elastomeric pads in shear. You need also hypothesize the distance of the surrounding rock mobilized in shear to equilibrate tension. A distortion means a shearing force reaction. If you get to have such value you will get reasonable values of the displacement, for you will be able to adjudicate vertical spring reactions for any computed vertical displacements at the node in shear.

then, thinking just 2D

F=G·gamma=G·atan(y/d)

where G is is the shear modulus, y the vertical movement of the footing, and d the width in plan of the soil (or rock mobilized in shear). F is, by the way, your allowable 2 ksf friction-adhesion force, assumed to happen still in some elastic realm.

For sound rock, not much distortion should be required to develop the top shear strength; say, that even allowing this allowable shear strength is but a reduced by a SF situation at ultimate strength, by 0.5% distortion the characterized allowable (and maximum allowable) shear strength will be developed. Hence, taking the angle in radians by its tangent value you will be using

G= 2/.005 = 400 ksf. Since this is a very modest value for a shear modulus, either it is one consistent with a cracked rock status, or else we have overestimated the need of deformation to develop full allowable shear strength. Tryng to bracket behaviour we will be also examining 5 per thousandt distorsion, or G=4000 ksf, still a modest value but directly showing that not much vertical movement will be required in either case to develop the proposed allowable shear strength.

Just to situate where we are, we correlate G with E for the soils. Poisson's = 0.3 average (Bowles) for cohesionless, hence E=G*2*(1+0.3)=2.6 times G

hence for our lower distorsion case E=1040 ksf (50 MPa) and for our lesser distorsion case E=10400 ksf (500 MPa). The lower value is about what dense sand would give, and the upper value better than the best gravel would do. Surely the upper value is a better guess of the behaviour than the lower, for the contact area won't be converted to gravel but partly part of the rock at the interface, and more considered after concrete is poured, so we will be using G=4000 ksf as the value to determine our vertical spring constants, procedent of the allowable strength in shear and standing 0.0005 distortion.

Now we will be considering the case in which the shear is mobilized by distorting 1 inch of soil around the footing. Hence, an upwards movement of just 0.0005 inches will be causing the required distortion of 0.0005 and hence mobilizing f=2 ksf.

Hence a vertical ballast modulus for shear contact (on vertical area elements) of 2 ksf/.0005 in = 27777 pci would result. This is an staggering ballast modulus, around 100 times the vertical ballast modulus for a good soil, indicating us that far more than 1 inch around the footing will be involved taking the forces in shear, say it is 100 inches. 100 inches deforming at .0005 distortion move one end .05 inches upward, hence

2 ksf/.05 in = 277.78 pci

this, seeing you have been allowed only 350 pci in compression seems still a somewhat high value for the ballast modulus in shear action at vertical faces; probably bracketing the behaviour between 300 and 100 pci will give you better insight on what to expect.

For linear spring, just take the tributary area for the ballast modulus taken.

 

[ishvaaag]

There are other interesting aspects in the problem above, such the volume of soil that would take in energy of deformation in shear some given tensile action (the FEM program solves in another way for the equilibrium, as per our input) or, perhaps more importantly for your practical case, if the surroundings of the footing and its depth allow for an area to deform in SHEAR for, say, 100 inches. This means a somewhat deep footing, and that no other footings will be in the range of, say, 20ft requiring akin equilibrium. If one has 2 ft of footing embedment one shouldn't think he has 100 inches in shear, other mechanisms would be involved. And the other mechanisms require unrealistic.

Also, although we dismissed the lower G value as above determined, the staggering final ballast first derived (having then to amplify to a very wide shear zone of 100 inches to then gain normalcy in the ballast value), this leaves us with no other option than to think that (once more) the f=2 ksf given is very conservative and not at all consistent with the behaviour of even altered rock; hence the lower value would have provided 10 times less at 1 inch wide area of shear deformation, then

2777 pci (very high modulus)

and we would have hence only to extend from 1 to 10 inches our shear zone to get the proposed (high) value

277 pci.

That is, geotechs have more in mind some loose sand (for calculation in shear almost degraded to silty sand) deforming in shear in say a 10 inches band around the footing. Again, as usual, overly safe.

 

[BAretired]

I would have thought that the resistance to uplift would be perimeter*height*2 ksf.

BA

 

[slickdeals]

Yes, you are correct. But in order to not input a downward force for each and every load combination, is there a way to model frictional resistance as a spring such that the downward resistance is automatically computed. I realize that frictional resistance offered will most likely be the same (approximately straight line) until the resistance is overcome and not linear elastic.

 

[ishvaaag]

Well you are trying to model a non linear behaviour using a linear model, some tradeoffs are always needed when not relying in an entirely consistent environment.

On the other hand as long the friction-adhesion is required to counteract forces, I don't think will pass from 0 to allowed value instantly, it will graduate its response to the applied force, only that the response will be irregular and difficult tho characterize without tests.

The procedure above involves of course guess but also shows quite splendorously the lack of consistence between what can be expected from different parameters known to be true or at least apply by ordinary assumptions to the structures we design. Here it is apparent, but custom occults that on quite many other times, for example that when using conventional push forces on walls we are making exactly the same, the formulations do not comply with the best that is known about the true behaviour, are not the best portrait of equilibrium.

Geotechnical people live the hard variation of the soils and need to conjugate their ouput both with safety and reality and with engineered structures of properties of 10 and 100 times more consistent behaviour. So it is not uncommon that even when they expect average settlement of 1 they may need to predict a worse case of 20 just to ensure safety at the same percentiles than the structures above.
This is of course source of an interesting debate for even if wee need to know that might be a settlement of 20 could develop, it is patent we are far more interested, I would say, 20 times more, in knowing what they expect to be the average behaviour, for that is truly what we and they are probabilistically expecting (and needing) to see. Special structures, nuclear, marine, in watery environments may need to apply the 20, but most on ordinary soils with ordinary buildings just the 1, and someone will have to put some order on this.

 

[BAretired]

I would regard a rock socket as unmoving. It is a tiedown for uplift. It is not a spring and there is no spring constant. Uplift may be determined for various load combinations and compared to the tiedown capacity.

BA

 

[ishvaaag]

BAretired, I think most wouldn't model any spring there, just check the tensile reaction. However I told on how to get estimates of what wanted from the data, just an exercise of thought.

 

[csd72]

This is a classic example of where things get more complicated by using a computer program.

It is rock, pretty solid and unyielding, so using a modulus of subgrade reaction is really over analysing it. Just use classic footing theory.

Same applies for the uplift, calculate the uplift resistance including all friction and dead weight and compare this to the worst uplift. Spring is also not really appropriate here as the friction will apply with almost zero strain.