Step-by-step solved example
f'c=4000 psi, NWC λ=1. fr and Mcr for Ig=12000 in⁴, yt=12 in?
1. fr
7.5√4000=7.5×63.25=474 psi.
2. Mcr
fr Ig/yt=474×12000/12=474,000 lb·in=39.5 kip·ft.
Answer: fr ≈ 474 psi, Mcr ≈ 39.5 kip·ft
f'c=4000 psi, NWC λ=1. fr and Mcr for Ig=12000 in⁴, yt=12 in?
1. fr
7.5√4000=7.5×63.25=474 psi.
2. Mcr
fr Ig/yt=474×12000/12=474,000 lb·in=39.5 kip·ft.
Answer: fr ≈ 474 psi, Mcr ≈ 39.5 kip·ft
1.√f'c with f'c=3600 psi is
2.Mcr uses the
3.λ=0.75 for
4.Direct tension of concrete is even lower than fr; design usually takes
5.SI: 0.62√f'c for f'c=25 MPa is nearest
6.Ig of a rectangle bh³/12 for 12×24 in is
7.Ec=w_c^{1.5} 33 √f'c is
8.Cracked-section analysis starts when
9.fr=7.5√f'c is in
10.Higher f'c increases fr only with