Conc. shall be proportioned to provide an average cylinder fc΄ and shall satisfy the durability criteria. Requirements for fc΄ shall be based on tests of cylinders molded and standard-cured in accordance with C 31 and tested in accordance with C 39. Cylinders' dimensions are 10*20 cm. or 15*30 cm. unless they are specified. fc΄ shall be based on 28-day tests.
2.1.8.3.5. Fcu for lightweight Conc. in (ACI -213R-14, 2014):
To account the fcu of lightweight Conc., as well as normal weight Conc., it shall be proportioned to provide an average cylinder fcu fc΄ and shall satisfy the durability criteria of ASTM C 3i and ASTM C 39. ACI code defines the minimum fcu for both Conc. type (NWC and …show more content…
can be determined using its fcu. The flexure strength does not correlate well with its fcu but rather with its square root. The (ECP 203-2007) gives an EXPression for estimating the Conc. modulus of rupture fctr as a function of its fcu as follows eq. (2.12): fctr = o.6 √fcu
2.1.8.4.1.2. Modulus of rupture for normal weight Conc. in (AC1-318R-14, 204):
MR is the basis for estimating flexural fatigue in Conc.. The ft of Conc. in flexure (Modulus of Rupture) is a more variable property than the fcu and is about io to i5 percent of the fcu. Literature published in the last decade suggested that the most important factor governing Conc. tensile capacity was the aggregate matrix bond, see eq. (2.i3). MR = o.62 √(fc' ) (2.13)
The modulus of rupture was performed according to ASTM C 78 and eCP 203-2007 with third-point loading. The modulus of rupture is calculated by the eq. (2.14): fr = …show more content…
The modulus of rupture is calculated by the eq. (2.15): fr = 3PL/2bd2
21.1.8.41.1.3. Modulus of rupture for lightweight Conc. in (ACI -318R-14, 2014):
(ACI -318R-05, 2005) presents the modulus of rupture for lightweight Conc. by using a modification factor λ as a multiplier for NWC eq. o.62 (fc ('o.5 fcu.
21.1.8.41.1.4. Modulus of rupture for normal weight Conc. in (EN -1992-1-4):
The ft refers to the highest σ reached under concentric tensile loading. The flexural ft shall be made to eq. (2.16). fctm = 0.30 x fck (2/3)