Factor of Safety Calculation 4

[DamsInc]

Hello,

I have a question regarding the calculation of safety factors. This is a pretty basic question but is causing me some confusion.

When calculating the safety factor against overturning, how do you arrange the forces in the equation (FS=resisting moments/overturning moments) if the force is acting on an inclined face of (for example) a rigid retaining wall. The way I see it, there are two methods:

1) Always resolve an inclined force into a resisting and overturning moment from its horizontal and vertical components,
2) Use the moment caused by the original force and define it as either resisting or overturning as appropriate.

See the attached file for an example of what I’m getting at. Obviously this is trivial when calculating stresses on the base, but it does have an effect on the acceptance criteria. For this case it is straight forward, but other cases with multiple forces also beg the question if a force’s components cause a “negative overturning moment” or a resisting moment.

It seems that it is correct to break the force into its resisting and overturning components first, but why is the second approach incorrect?

 

[AELLC]

see attached.

The definition of a structural engineer: overdesign by a factor of 1.999, instead of the usual 2.

 

[JoshPlumSE]

I've looked at this issue before and I don't know that there is an inherently correct answer. And, I think a good engineer could come up with either solution and be comfortable with his rationale.

That being said, I tend to hang my hat on the term "factor of safety". In that sense, I would take a look at the effect of each load. Per your sketch. I would calculate the total effect of W and come up with reduced stabilizing component (due to the eccentricity. Then I would calculate the total effect of F and come up with a reduced De-stabilizing component (due to the angle of the applied force).

The idea is that if I've got a Safety factor of 2.0, then my de-stabilizing force (F) should be multiplied by exactly 2.0 for the calculation to show instability.

 

[rb1957]

surely 1) and 2) are the saem thing ... whether you calc a moment from force components or from the vector the result is the same.

as josh says, FS*applied = zero stability. In this case, the moment due to the dead weight (Mrestoring?) is your "allowable" and the net overturning moment due to the applied loads (Moverturing?) is your applied. i don't think you should separate all the restoring moments into one number (whether they are from the dead weight or the applied loads) and all the overturning into another. oh, i see (i think) ... is this what you meant? but then you can't readily derive the FS (as applied forces are present in both restroing and overturning numbers).



Quando Omni Flunkus Moritati

 

[DamsInc]

I've looked at this issue before and I don't know that there is an inherently correct answer. And, I think a good engineer could come up with either solution and be comfortable with his rationale.

That being said, I tend to hang my hat on the term "factor of safety". In that sense, I would take a look at the effect of each load. Per your sketch. I would calculate the total effect of W and come up with reduced stabilizing component (due to the eccentricity. Then I would calculate the total effect of F and come up with a reduced De-stabilizing component (due to the angle of the applied force).

The idea is that if I've got a Safety factor of 2.0, then my de-stabilizing force (F) should be multiplied by exactly 2.0 for the calculation to show instability.

I'm glad to hear that I'm not the only one that has raised this question. I'm not sure I completely understand your second paragraph. In this case, F is a stabilizing force. I think you are indicating to use method 1) which would be to consider the entire effect of F itself (and thus an undefined FS in my example).

surely 1) and 2) are the saem thing ... whether you calc a moment from force components or from the vector the result is the same.

as josh says, FS*applied = zero stability. In this case, the moment due to the dead weight (Mrestoring?) is your "allowable" and the net overturning moment due to the applied loads (Moverturing?) is your applied. i don't think you should separate all the restoring moments into one number (whether they are from the dead weight or the applied loads) and all the overturning into another. oh, i see (i think) ... is this what you meant? but then you can't readily derive the FS (as applied forces are present in both restroing and overturning numbers).

Yes, from a mechanics standpoint 1) and 2) are the same thing. The problem comes when figuring out which to put in the numerator and which in the denominator of the FS calculation.

I think I used a poor example to ask my question. Will try again after some rest!

 

[rb1957]

FS = allowable/applied. in your case the allowable is the restoring moment due to the dead weight, and the applied is the net overturning moment.

it does get tricky to work out if there are multiple loads particularly if there is a non-linear combination of loads. but always FS is a factor applied to all of the applied loads to create a failure situation.

Quando Omni Flunkus Moritati

 

[kingnero]

The second FS in your original is not correct. you arbitrarily add values together, making your denominator = 0. you actually have an overturning moment (coming from Fhor), yet you say it's = 0

If F existed with only a vertical component (so, F is vertically downwards, and the force would be led into the body in a vertical fashion (welded connection, ...), then your FS would be undefind, as then, there is NO overturning moment.

 

[DamsInc]

The second FS in your original is not correct. you arbitrarily add values together, making your denominator = 0. you actually have an overturning moment (coming from Fhor), yet you say it's = 0

The values are not arbitrarily added together. That force of 16.8 kNm is the actual moment caused by the force F (the moment from F is the algebraic sum of it's component force's moments). I could have also calculated it by finding the perpendicular distance from it's line of action to the point of rotation.

F is the real force. Fhor and Fver do not exist in actuality, they are just the components of the force which we use to conveniently describe the force in terms of our (arbitrary) x and y axis. These components have the same effect as F in terms of their action on the rigid body. Yet they change the calculation of the factor of safety which is my issue.

If F existed with only a vertical component (so, F is vertically downwards, and the force would be led into the body in a vertical fashion (welded connection, ...), then your FS would be undefined, as then, there is NO overturning moment.

Which is method 2) above. Again, the force F produces a stabilizing moment. Whether it is vertically downwards, or at an inclination doesn't matter.

 

[kingnero]

The values are not arbitrarily added together. That force of 16.8 kNm is the actual moment caused by the force F (the moment from F is the algebraic sum of it's component force's moments). I could have also calculated it by finding the perpendicular distance from it's line of action to the point of rotation.

yes they are. why haven't you also added the moment coming from its own weight? If you did, you have the net moment around point A, and while everything upto this point is done correctly, you are calculating something else but not the FS.

for the second point , you need to read my case again. Maybe I don't explain well enough, because I'm not native english speaking. The only way to have a FS that is undefined, would be when you have no component that is horizontal. When everything is as stable as it gets, there is no risk for overturning, hence your FS is indefinitely high.

 

[rb1957]

i think both your FS are wrong.

from your calcs (sorry, before i saw only the sketch) i think FS = 119.9/(70.5-53.7)
= 119.9/16.8 = 7.13.

ie if you have 7x the applied load, F, you'll be close to failing (overturning) the structure.

Quando Omni Flunkus Moritati

 

[DamsInc]

yes they are. why haven't you also added the moment coming from its own weight? If you did, you have the net moment around point A, and while everything upto this point is done correctly, you are calculating something else but not the FS.

I did, it is W (forgot to add the dimensions in the sketch). It produces a clockwise (resisting) moment of 119.9 kNm.

i think both your FS are wrong.

from your calcs (sorry, before i saw only the sketch) i think FS = 119.9/(70.5-53.7)
= 119.9/16.8 = 7.13.

ie if you have 7x the applied load, F, you'll be close to failing (overturning) the structure.

This is actually how I was originally calculating the factor of safety! Although if I'm not mistaken, the denominator would be FS= 119.9/(53.7-70.5)= -7.1. My boss took one look at a negative factor of safety and told me to use the components because it made no sense.

 

[rb1957]

mea culpa ...

i mistook your moments arrows.

your structure is stable because F has a net restoring moment (= 16.8)

so the conclusion of 2) is right, but i don't like the way you express it. I think you should have all the terms that include the "load" (in your case F) together. i can see the logic of the approach, but it seems to be specific to this type of application, and i don't think your calc produces a true FS (can you factor the applied load by this value and get failure ?)
i'd rather have FS = allowable/applied (in this case 119.9/-16.8). if the force components were flipped then my previous calc (FS = 7.13) would apply.

Quando Omni Flunkus Moritati

 

[DamsInc]

Thanks rb1957. I came to similar conclusions as you.

This issue came about as I am developing a somewhat lengthy and complex mathcad sheet for the analysis of retaining walls and gravity dams. With the dams, there are multiple loads acting in different locations on both sides of the structure in various directions. The principle of multiplying the overturning forces by the factor of safety to cause instability begins to break down/get complex (would it still apply?) This example is my attempt to distill the fundamental issue into a simple example. A similar issue arises with sliding FS.

My original approach was to do as you say and take the component moments of the inclined force, add them together to get one overturning or stabilizing moment (or alternatively, multiply the magnitude of the force by the distance of the line of action from the point of rotation, same result). This makes sense to me because it is the effect of one force. But it is also the method which results in a negative factor of safety.

So, what is the factor of safety of the example?

1) FS=3.5
2) FS=undefined
3) FS=-7.1
4) Other?

 

[rb1957]

the true FS is infinite because the basic loading is stable. i'd express it that way ...
if the load has a net restoring moment (ie is inherently stable), FS = stable
if the load has a net overturning moment thenit has a finite FS.

of course this is only one of many failure scenarios you're considering, at some stage the load will cause the body of the dam to shear.

Quando Omni Flunkus Moritati

 

[JoshPlumSE]

There is another similar situation that I have dealt with in the past. Let's say you've got an eccentric dead load. The classic example for me is a piece of equipment that hangs off the side of my column. You can multiple that load by 1000x and it will still be stable because the eccentricity of the load is constant. But, is it really reasonable to say the safety factor is infinite?

In those cases, I prefer use the 2nd method / suggestion and break that dead load into two components. The concentric stabilizing component and the de-stabilizing component due to eccentricity. In this case, I am no longer calculating a factor of safety vs the loading, but a factor of safety vs the eccentricity of the load. I can increase that eccentricity by what factor before the footing becomes unstable?

In your case, it would be something akin to a factor of safety vs the inclination of the load.

 

[rb1957]

i don't think at an eccentric load on a column is inherently stable. as you increase the load you increase the stress in the column, approaching your allowable ... no?

in the example given the loading is stable, you cannot ever, no matter how large a load, have the overturning moment > the restoring moment ... yes?

Quando Omni Flunkus Moritati

 

[DamsInc]

Correct. I don't quite understand the example JoshPlum gave either.

 

[CELinOttawa]

Josh's example is true if the eccentricity does not exceed the pertinent dimension of the column.

This is a study in what I have come to call the "Three Ss": Strength, Stability, Stiffness.

The load will NEVER cause the problem to be unstable, as it is always within the width of the column and in effect results in a "returning" or "stabilising" moment.

That is *not* to say that the problem will always result in a safe condition, with the load eventually causing a failure of STRENGTH.

 

[AELLC]

DamsInc,

You mentioned you are automating the design for retaining walls and dams.

I do have an Excel which does CMU and concrete retaining walls, which may give you ideas on how to set it up - however NOT to be used for actual design - it has bugs that I am trying to correct - meanwhile, I am using RetainPro canned software.

The definition of a structural engineer: overdesign by a factor of 1.999, instead of the usual 2.

 

[DamsInc]

AELLC,

I would be very interested in taking a look at your spreadsheet. Of course it will be for my own learning purposes only.