Step-by-step solved example
f'c = 4000 psi. Approximate modulus of rupture fr = 7.5 √f'c (normalweight, λ = 1)?
1. Root
√4000 = 63.25.
2. fr
fr = 7.5×63.25 = 474 psi.
Answer: 474 psi
f'c = 4000 psi. Approximate modulus of rupture fr = 7.5 √f'c (normalweight, λ = 1)?
1. Root
√4000 = 63.25.
2. fr
fr = 7.5×63.25 = 474 psi.
Answer: 474 psi
1.US standard cylinder is commonly
2.f'c is usually specified at
3.Higher w/c generally
4.fr ≈ 7.5 √f'c is in
5.A cube strength is typically
6.4000 psi is about
7.λ for lightweight concrete is
8.f'c is used in Whitney as
9.Air entrainment typically
10.A 6×12 cylinder fails at 118 kip. f'c nearest