Concrete Beam Capacity at 50% of yield stress?

I believe I can just replace my 40ksi yield stress with 20ksi?

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That's essentially it.....while being sure you have minimum flexural/temperature steel in the section.

Of course, this presupposes you want it not to exceed 20 ksi at the ultimate level (i.e. not working/ASD stress). If this is a crack control issue.....you'll need the stress without the load factors. (And that is relatively simple to do.)

It depends on why you're limiting the stress to 20 ksi. With the compression edge strain constant at 0.003 and the neutral axis moving towards the compression edge, then the strain in the steel increases of course. Therefore your assumption of 20 ksi is not consistent with the strain profile.

You might be better off assuming the concrete stress is proportional to strain (triangular stress distribution) and adopting a compression strain less than 0.003, rather than the rectangular stress block which is really for ultimate strength calculations with yielded steel. This calculation can be done without iteration if you only have one layer of tension steel and ignore any compression steel (but it's easier to do it by iterating in a spreadsheet).

As Steveh49 said, it depends on what you are trying to model.

If it is 40ksi steel and you want to check the member strength at 20ksi, then you would have to look at the actual concrete stress/strain profile and the steel stress/strain profile. The steel will only be at 50% of yield strain in the elastic range. Even at normal service stress levels, the steel strain will normally be at about 60% of yield.

If you want to check it assuming the steel yields at 20ksi, then do as you suggested.

Factoring down the yield stress is how Ultimate Limit State (ULS) capacity is calculated in European codes (both Eurocode 2 and its predecessors), but the factor is of the order of 0.9, rather than 0.5. Doing a ULS calculation with such large reduction doesn't make sense to me, unless you want to check for the effect of corrosion, or the contractor leaving out half the steel, in which case you could reduce the area rather than the yield stress (which will give the same answer).

As stated by others, with a steel stress of 50% fy the maximum concrete strain will still be in the elastic range (unless heavily over-reinforced), and a triangular concrete stress distribution would be more appropriate. With zero axial load and a rectangular section, the depth of the neutral axis is given by a simple quadratic equation:

For tension steel only:
W*x^2/2 + ast*m*x -ast*m*Dt = 0

x = depth of NA
W = concrete width
ast = area tension steel
m = modular ration (E steel/E concrete)
Dt = depth of tension steel from compression face

For steel in the compression zone, just add in the compression steel area, factored by the effective modular ratio, m1 = m-1:
W*x^2/2 + (ast*m+asc*m1)*x -ast*m*Dt -asc*m1*Dc = 0

Dc = depth of compression steel.

For a non-zero axial load it becomes a bit more complex, resulting in a cubic equation, but that can be easily solved without iteration with a spreadsheet (or preferred alternative).




Doug Jenkins
Interactive Design Services

I just happen to have been working on a blog post deriving the equations to find the depth of the neutral axis for both ULS and SLS analyses:




Doug Jenkins
Interactive Design Services