Coefficient of Friction 2

[ChiEngr]

Hello,

I was given a soils report for a project I am working on which stated that for sliding checks of footings, the friction factor given should be multiplied by the dead load to obtain my sliding resistance. Does anybody else use only the dead load when checking sliding resistance? It seems to me that if you have other vertical downward loads acting on the foundation, that you should be able to include those forces in the sliding resistance check unless I am grossly mistaken. Does anything have any thoughts on this?

 

[Tomfh]

Don't start discounting G just for the sake of it

It's what the code says to do. Where a load, e.g. G, causes a destablisizing effect, you factor it up. Where a load, e.g. G, causes a stabilizing effect, you factor it down.

It uses the Factored-Load <= Factored-Resistance, inequality

Yes and it does so to ensure that there's an overwhelming probability that the real load is less than the real resistance. That is the condition the code is trying to avoid - real load equalling (or exceeding) real resistance. If however you multiply Load and Resistance by the same factor (say 1.0), on the basis that you know them perfectly, you subvert this aim, as your real load equals your real resistance to begin with. Your real load and your real resistance, are, to use your phrase, "the same".

 

[human909]

It's what the code says to do. Where a load, e.g. G, causes a destablisizing effect, you factor it up. Where a load, e.g. G, causes a stabilizing effect, you factor it down.

That is a pretty simplistic argument, and doesn't show suitable engineer judgment. Most codes also allow suitable engineering judgment. It makes ZERO rational sense to factor the same load up and down in the same combination. That is a logical impossibility and using 'the codes says that' shouldn't preclude you from thinking for yourself.

For example when I'm calculating wind load the stability of silos I use 0.9G and 0.0Q. When I'm calculating seismic I use 1.0G, 1.0Q and 1.0Eu even though there is no where in my code that to use 1.0Q for seismic mass. I choose a value that is sensible for the use pattern of the structure I'm analysing.

If however you multiply Load and Resistance by the same factor (say 1.0), on the basis that you know them perfectly, you subvert this aim, as your real load equals your real resistance to begin with. Your real load and your real resistance, are, to use your phrase, "the same".

No. That is NOT what I'm suggesting. I explained that in my last post. The destabilising load isn't simply 'G'. And the stabilising resistance isn't simply 'G' either. G is a mass not a force** (the post got deleted but yes, in these combinations 'G' is a force', but the input is a mass and that can't be two things at the same time). Your load and resistance factors come into the equations where G is inputted, be it seismic or friction.