WeStatiX Documentation EN

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RC member design for bending and axial force – symmetrical reinforcement

In this example we use WeStatiX to design a RC cross-section with symmetrical reinforcement and we verify the results. We consider the cantilever beam represented in the following figure.

RC design symmetrical reinfocement

You can find the beam model in WestatiX ready for the calculation, while the loading conditions are summarized in the following table.

Bending moment\(M_{Ed}\)\(\)280kNm
Axial force\(N_{Ed}\)\(\)1875kN

The cross section characteristics are listed in the table below

Overall width of a cross-section\(b\)\(\)300mm
concrete cover\(d_1\)\(\)50mm
ratio for interaction diagram choice\(d_1/h\)\(\)0,10

And finally, here are the material parameters.

Characteristic compressive cylinder
strength of concrete at 28 days
Characteristic yield strength
of reinforcement
Coefficient taking account
of long term effects
Partial factor for concrete\(\gamma_c\)\(\)1,50
Partial factor for
reinforcing steel
Design value of concrete
compressive strength
\(f_{cd}\)\(\alpha_{cc} f_{ck}/\gamma_c\)17.000,00kPa
Design value for yield
strength of reinforcement

In the following pictures you can see how the section and material parameters have been selected in WeStatiX.

Concrete parameters
Reinforcement steel parameters
Nodal loads

Once you created the model as shown above, you can start the calculation. You will obtain the results shown in the following pictures.

Normal force
Bending moment Y
ULS reinforcement top Z. \(A_s1=11,52cm^2 \)
ULS reinforcement bottom Z. \(A_s2=11,52cm^2 \)

The computed reinforcement is symmetric and the total cross section reinforcement area is equal to \( A_{s,tot}=23,04 cm^2 \).

RC design reinforcement area
ULS reinforcement sum. \( A_{s,tot}=23,04 cm^2 \)

We can verify the results using the interaction diagrams for symmetric reinforcement.[1]

The first step consists in the calculation of the following parameters

Parameterized axial force\(\nu\)\(N_d/b \cdot h \cdot f_{cd}\)0,74
Parameterized bending moment\(\mu\)\(M_d/b\cdot h^2 \cdot f_{cd}\)0,22

Then, we can choose the interaction diagram to design a RC section with symmetrical reinforcement and reach the ratio \( \frac{A_s f_{yd}}{bhf_{cd}}\) as follows.

RC design interaction diagram

Finally, knowing the ratio you can calculate the total reinforcement area as

Coefficient from interaction diagram \( \frac{ A_s \cdot f_{yd} }{ b \cdot h \cdot f_{cd}} \) 0,40
Total reinforcement area\(A_{s,tot}\)\(\)23,45cm^2

So you can compare the two solutions as

\( \epsilon = 1-\frac{23,04}{23,45} = 1,79\% \)

Therefore, we can conclude that the design solution calculated by WeStatiX is verified.

[1] A.W.. BEEBY and R.S: NARAYANAN – Designers’ guide to Eurocode 2: design of concrete structures. – Designers’ guide to EN1992-1-1 and EN1992-1-2 Eurocode 2: design of concrete structures. General rules and rules for buildings and structural fire design.