Rectangular Stress Block

[rabi24]

Hello everyone,

I have seen in the SAFE 14 manual that they use an equation for the depth of the compression block in concrete according to AS 8.1.2.2, I could not find the equation in the code, and I tried to get at it without much luck. The equation is a = d - sqrt(d^2-(2M*/0.85f'c*phi*b))

I would love if you could point me in the right direction.

Thank you!

 

[asixth]

From equilibrium.

M*=0.85*f'c*phi*b*a*(d-a/2)

Got to solve for a.

 

[CELinOttawa]

aka Whitney stress block, and asixth has you sorted out... Most codes give you a = As * fy /(0.85 * fc' * b)

 

[msquared48]

Old WSD theory.

Mike McCann, PE, SE (WA)

 

[Ingenuity]

Old WSD theory.

I think you meant ULTIMATE STRENGTH DESIGN, not WORKING STRESS DESIGN?

 

[CELinOttawa]

Ingenuity: Yup. Working stress design formulae look more like you're working out the reinforcement ratio, because that's effectively what you did - And that then let you use the lever arm formulae to figure our what the stress was in your steel.

Working Stress Design would be to compute the area of steel required directly from the moment applied, and then check your concrete stress to see if you needed to add compressive steel to carry the full load.
Tensile Reinforcement: As = M/(fs*j*d) where j = 1-k/3 and k = [SQRT(2*n*p + (n*p)^2) - n*p] or, more commonly, k = 1/(1+fs/(n*fc))
Then the moment of compressive concrete stresses about the tensile reinforcement:
Mr = 1/2 fc*k*j*(bd^2)
if this exceeds your allowable (working) tensile stress, increase some parameters or add steel and try again.

And despite the "simplicity" [note: This should be seem as very sarcastic - I find the working stress, or any stress based formulation for concrete, very complex and onerous], this was nearly never done - Everyone used the tables of "canned" beams for flexural strengths... It was common to specifically design stirrups, however, much as we do today.

 

[hokie66]

But CEL, in WSD of flexural elements, we just didn't let the compressive stresses get that high, generally by controlling the percentage of reinforcement. In your first formula, just use .85 for j, and you are good to go.

 

[CELinOttawa]

Sorry Hokkie, you're absolutely right... I should have been more clear: That was ACI 317-1946, pretty much the earliest "modern" WSD code.

Later refinements corrected those gross assumptkons...