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MECHANICS - EXAMPLE


Reinforced Concrete Post
  Example

 

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 cross-sectional area of steel bars is circular. If the allowable stresses of the steel, σst, and concrete, σco, are 120 MPa and 6 MPa, respectively, what is the minimum diameter of the steel bars? Assume the modulus of elasticity of the steel, Est, and and concrete, Eco, are 14 GPa and 200 GPa, respectively.

   
  Solution


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 a-a, these forces can be correlated as

      ΣF = 0
      Pst + Pco = P
      Pst + Pco = 1,000 kN
      σstAst + σcoAco = 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.

     δst = δco
      σstLst/Est = σcoLco/Eco

Noting that steel and concrete are of equal length and substituting the actual modulus gives

     σst/200×106 = σco/14×106
     σst = 200 σco/14
     σst = 14.29 σco

To determine the correct amount of steel required, first assume the concrete will reach it allowable stress limit of 6 MPa. This gives,

     σst = 14.29 (6)
     σst = 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 σco = 6 MPa and σco = 88.74 MPa. These values are substituted into the previous static equilibrium equation to obtain

     88,740 Ast + 6,000 (0.2)2 = 1,000
     88,740 Ast + 240 = 1,000
     88,740 Ast = 760
     Ast = 8.564×10-3 m2
     Ast = 8,564 mm2

The total cross-sectional area, Ast, is the total of the six steel bars. Thus, the area of a single steel bar will be
     A = Ast/6
     A = 8,564/6 mm2
     A = 1,427 mm2
     πd2/4 = 1,427 mm2
     d2 = 5,708/π mm2
     d = 42.63 mm

     
   
 
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