Concrete f'c from Modulus of Rupture (ACI 318-19 19.2.3)
Back out the equivalent f'c from a flexural-beam (modulus-of-rupture) test.
Example
You enter
- Modulus of rupture fr (psi) 474.342
- Lightweight factor lambda (1.0 NW, 0.75 LW) 1
You get
- Fc (psi) 4000
Checked against published worked examples from: StructurePoint.
Details, formula, and sources
Scopef'c = (fr / (7.5 x lambda))^2. A 474 psi normalweight rupture strength implies ~4,000 psi. The code fr is a conservative lower bound, so the implied f'c is a lower-bound equivalent, not a cylinder-break value. A design aid; the engineer of record's stamped design governs.
Formula or tablef'c = (fr / (7.5 x lambda))^2, the ACI 318-19 19.2.3.1 relation fr = 7.5 x lambda x sqrt(f'c) solved for the compressive strength; lambda = 1.0 normalweight, 0.75 all-lightweight.
Source and editionThe ACI 318-19 19.2.3.1 modulus of rupture of concrete, solved for f'c, by name.
Free accessACI 318 is readable free through the ACI online reading room at concrete.org; the 19.2.3 rupture provisions are in the published code.
What governsEstimate. AHJ and licensed professional govern.
API field namesfr_psi, lambda, fc_psi
- Rupture stress inverted f'c = (fr / (7.5 x lambda))^2ACI 318-19 19.2.3.1
- Lightweight factor lambda = 1.0 normalweight, 0.75 all-lightweightACI 318-19 19.2.4
- Equivalent strength a lower-bound equivalent f'c, not a cylinder-break valuescope of this tile