# Analysis of Under Reinforced Concrete Beam Section in Flexure According to Euro Codes

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

Take;

fck = 20MPa

fyk = 460MPa

## Calculation

Strain

Stress

Assume the steel has yielded.

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

T = Fst= Asfyd

T=3*π(20/2)2 mm2 * 400Nmm-2 = 376.99kN

Now, let’s find out the concrete compressive force C or Fcc 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 fck≤50MPa;

λ = 0.8

From BS EN 1992-1-1: 2004, Clause 3.1.7.3, Equation 3.21, for fck≤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*103 = (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; εcu=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”). I am Kaumadi Ganiarachchi, a dedicated civil engineer with expertise in construction, structural analysis, project management, and structural design. With hands-on experience and a deep understanding of the engineering domain, I bring practical insights to my writings, aiming to inform, inspire, and guide readers in the ever-evolving world of civil engineering. Here is my LinkedIn Profile and About Us Page