Determine the moment of resistance of the beam section given below.

Take;

f_{ck }= 20MPa

f_{yk} = 460MPa

Table of Contents

**Calculation**

Strain

Stress

Assume the steel has yielded.

Then, the tension force in steel can be found from:

T = F_{st}= A_{s}f_{yd}

T=3*π(20/2)^{2} mm^{2} * 400Nmm^{-2} = 376.99kN

Now, let’s find out the concrete compressive force C or F_{cc} by considering the rectangular stress distribution provided in BS EN 1992-1-1: 2004, Clause 3.1.7.3, Figure 3.5.

By using the rectangular stress block,

where,

From BS EN 1992-1-1: 2004, Clause 3.1.7.3, Equation 3.19, for f_{ck}≤50MPa;

λ = 0.8

From BS EN 1992-1-1: 2004, Clause 3.1.7.3, Equation 3.21, for f_{ck}≤50MPa;

η = 1.0

Hence,

C = (0.8*11.333) * (1*x) * 200

Now let’s find the neutral axis depth (“x”) considering the balance of the section.

T = C

376.99*10^{3} = (0.8*11.333*) * (1*x) * (200)

x = 207.91mm

Now let’s find the actual strain in steel by considering the strain distribution provided in Fig 4.

Hence, steel has yielded, and our assumption is correct.

This can also be confirmed by the BS EN 1992-1-1: 2004, Clause 3.1.7.1.

According to BS EN 1992-1-1: 2004, Clause 3.1.7.1; ε_{cu}=0.0035 from BS EN 1992-1-1: 2004, Table 3.1.

The condition that tensile reinforcement has yielded when the concrete strain is 0.0035 is:

Hence, steel has yielded.

EuroCode2 recommends that (x/d) be less than 0.45; this will give enough warning before failure.

Now let’s find the lever arm (“z”).

where; 0.95d =0.95*350 = 332.5mm

Hence, ok.

Now let’s find the moment of resistance (“M”).