1Design of concrete members using Eurocode 2(EN1992):

1.1Criterion to be used (Ales course project)

1.2Summary of code provision

Figure 1 shows a section of a rectangular reinforced concrete beam. Figure 1(b) shows the cracked section with concrete and steel at yield this is based on the recommendations in Eurocode 2. The simplified rectangular stress block of Figure 1(c) is also allowed in Eurocode and is normally used. Presumably the concrete stress block of (c) is statically equivalent to that of (b).

Definitions

  • d the effective depth of the section i.e. the distance from the top of the beam to the centre of area of the reinforcement.
  • h the total depth of the beam
  • As the area of the tensile reinforcement
  • fck the characteristic cylinder strength of the concrete
  • 0.85 fck the 0.85 factor is to allow for the fact that the strength of the concrete in a beam is different from that in a cylinder test
  • x the distance from the top of the beam to the neutral axis
  • fyk the characteristic yield strength of the reinforcement
  • z the lever arm for the moment
  • s the depth of the stress block = 0.85x
  • M is the applied moment

Moment equilibrium for the section

Figure 2 shows the forces on the section

The resultant compression in the concrete is:

C = 0.85bs (fck/c) (1)

where c is the partial safety factor for the concrete

The resultant tensile force in the steel is:

T = Ast(fyk/ s)(2)

where s is the partial safety factor for the concrete

Horizontal equilibrium is C = T (3)

Taking moments about the centre of compression:

M = Tz = Ast(fyk/ s)z = Cz = 0.85 bs(fck/c) (4)

z = d – s/2(5)

Manipulating (4) and (5) and using c = 1.5 and s = 1.15 (from Eurocode 0) gives the expression:

(z/d)2 - (z/d) + K/1.134 = 0.0(6)

where K = M/bd2fck

Solving the quadratic equation of (6) gives:

(7)

Using Equation (4) Ast = M/(0.87fykz)(8)

Equations (7) and (8) are used to calculate the required area of tension steel.

Under-reinforced section

A standard rule in reinforce concrete beam design is that, in make the failure more ductile, steel failure should precede concrete failure. A section for which steel and concrete yield at the same time is called a balanced section. It can be shown that this condition occurs when

K = M/bd2fck = 0.167(9)

Steel fails first when K 0.167. Such a section is called under-reinforced.

If K> 0.167 then compression reinforcement is needed.

2Shear strength

2.1Criterion to be used (Ales course project)

2.2Summary of code provision

Section 6.2.2 of Eurocode 2 gives the following empirical expression for the design shear resistance of reinforced concrete sections

VRd,c = [ 0.12k(1001fck)1/3]bwd

where:

VRd,c is the design shear resistance of a reinforced concrete section.

k = [1 + (100/d)1/2]  2.0 with d in mm

1 = Asl /(bwd)  0.02

Asl is the area of the tensile reinforcement at the section at which the shear is define. This area of steel needs to extend for a full anchorage length plus the effective depth of the section beyond the section.

bw is the width of the section at the level of the tension reinforcement

d is the effective depth

fck is in MPa (N/mm2)

The criterion for shear strength without shear reinforcement is:

VEd < VRd,c

where VEd is the shear force at the limit state.

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I MacLeod