
A short concrete post is reinforced with six symmetrically placed steel bars so that it can carry a load, P, of 1,000 kN. The concrete post is 200 mm by 200 mm and the crosssectional area of steel bars is circular. If the allowable stresses of the steel, σ_{st}, and concrete, σ_{co}, are 120 MPa and 6 MPa, respectively, what is the minimum diameter of the steel bars? Assume the modulus of elasticity of the steel, E_{st}, and and concrete, E_{co}, are 14 GPa and 200 GPa, respectively.

Static Equilibrium and Deformation
of Steel and Concrete


The applied load, P, will be distributed between the steel and the concrete. From equilibrium at an arbitrary section aa, these forces can be correlated as
ΣF = 0
P_{st} + P_{co} = P
P_{st} + P_{co} = 1,000 kN
σ_{st}A_{st} + σ_{co}A_{co} = 1,000 kN
Since no other equations of static equilibrium are available to indicate in what proportion of the load is distributed to each material, the elastic deformation of the structure has to be considered. It is evident from symmetry that the bearing plate causes the steel and concrete to deform equally.
δ_{st} = δ_{co}
σ_{st}L_{st}/E_{st} = σ_{co}L_{co}/E_{co}
Noting that steel and concrete are of equal length and substituting the actual modulus gives
σ_{st}/200×10^{6} = σ_{co}/14×10^{6}
σ_{st} = 200 σ_{co}/14
σ_{st} = 14.29 σ_{co}
To determine the correct amount of steel required, first assume the concrete will reach it allowable stress limit of 6 MPa. This gives,
σ_{st} = 14.29 (6)
σ_{st} = 88.74 MPa
This indicates that the steel can not be stressed to its limit of 120 MPa without overstressing the concrete. Thus, the assumption that the concrete will fail before the steel is correct. The actual working stresses will be σ_{co} = 6 MPa and σ_{co} = 88.74 MPa. These values are substituted into the previous static equilibrium equation to obtain
88,740 A_{st} + 6,000 (0.2)^{2} = 1,000
88,740 A_{st} + 240 = 1,000
88,740 A_{st} = 760
A_{st} = 8.564×10^{3} m^{2}
A_{st} = 8,564 mm^{2}
The total crosssectional area, A_{st}, is the total of the six steel bars. Thus, the area of a single steel bar will be
A = A_{st}/6
A = 8,564/6 mm^{2}
A = 1,427 mm^{2}
πd^{2}/4 = 1,427 mm^{2}
d^{2} = 5,708/π mm^{2}
d = 42.63 mm 