Accurate and quick way to determine tension reinforcing area in a concrete beam 1

[jspritzer]

thread507-215879

Known: Mu (in kips), b (in), d (in), f'c (ksi) and fy (ksi).

Cu = β₁*f'c*b*d/fy
(Remember: β₁ is the depth factor for an equivalent compression stress block)

Equation 1a:
Required Steel Area (exact solution), As = Cu(1-(1-2Mu/Øfy d Cu)^0.5)
Ø = 0.9 with the assumption the beam is a tension controlled member

Equation 1b:
If the factored moment is in units of kip-ft, simply change the equation to As = Cu(1-(1-24Mu/Øfy d Cu)^0.5)

NOTE: If (1-2Mu/Øfy d Cu) is negative, then the section, b*d, is too small and needs to be increased in size. Increasing f'c may be another solution as well.

Since ∅ is assumed to equal 0.9, check to confirm section is in fact tension controlled, IE, ε_t ≥ 0.005 or c/d ≤ 0.375 (c=a/β₁)

Don't forget to check minimum reinforcing requirements of ACI 318 as well: the lesser of equations (a) or (b) of section 9.6.1.2 or meet requirements of section 9.6.1.3. (2014 or 2019 code)

 

[MIStructE_IRE]

For some reason I’m very reluctant to download that file...

 

[jspritzer]

The file is a pdf stored in Eng Tips. It's the algebraic derivation for the equations above from the equation Mn = Asfy(d-a/2). Without the derivation (not choosing to down load the file) it's simple enough to verify the given equation is an exact solution of As by taking the result, As (exact solution above), calculating a (which equals Asfy/(.85F'c b)) and inserting this a and As into ⇒ Mn = Asfy(d-a/2). ØMn will equal Mu used above.

 

[DaveAtkins]

I generally use As = Mu/(4*d) to get me in the right ballpark for area of steel, then calculate ΦMn with the actual area of steel I intend to use.

DaveAtkins

 

[jspritzer]

Certainly simple. For those who are wondering wondering where As ≅ Mu/4d comes from . . . using our well know equation ⇒ Mu ≤ 0.9Asfy(d-a/2).

Providing Mu in k-ft units, assuming the term (d-a/2) is equal to 8/9*d, rearranging the terms to As=Mu/[(0.9fy*8d/9)/12] and using a steel yield of fy=60ksi the typical single reinforced concrete beam equation reduces to As = Mu/4d. Note the division by 12 is needed to obtain units of sq-in in the final result.

If fy=75ksi the quick approximation is as simple as As ≅ Mu/5d.