Homework Answers
Given,
Load at failure (P) = 91280 lb.
Diameter of cylinder = 4 in.
Therefore, gross cross sectional area of the cylinder
(Ag) = (
/4) *
42
= 12.5664 in2
Therefore, compressive strength of concrete (f
c) =
(P/Ag)
= 91280 / 12.5664
= 7263.815 PSI
Modulus of elasticity of concrete (Ec) = 57000 *
(fc)0.5
[As per ACI]
= 57000 * (7263.815)0.5
= 4857997 PSI
Modulus of rupture of concrete (fcr) =
4 * (fc)0.5
[As per ACI]
= 4 * (7263.815)0.5
= 340.912 PSI
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please solve completely and correctly, I will give thumbs
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