2.4.3 Determining the Statically Required Reinforcement

Determining the Statically Required Reinforcement

The stress-strain diagrams for concrete and reinforcing steel described in chapter 2.4.2 together with the allowable range of strain distributions (limit strains) represent the basis for determining the required longitudinal reinforcement for the previously determined design moments. This process is also documented in the design details.

The first subentries for the required longitudinal reinforcement are the top and bottom side of the plate. There are main entries for the Bottom surface (+z) and Top surface (-z) that contain further details for each reinforcement direction.

Figure 2.39 shows that the reinforcement directions 2 and 3 require very little to no reinforcement at the plate's bottom.

Reinforcement Direction 1 is to be designed for the design bending moment m_end,+z,φ1 = 35.89 kNm/m. The strains provide information about the determination of the longitudinal reinforcement.

The example shown in Figure 2.39 is checked for a dimensionless design procedure by means of a design table. The following input parameters are given:

• Cross-section [cm]: rectangle w/h/d = 100/20/17
• Materials: concrete C20/25 B 500 S (A)
• Design internal forces: M_Eds = ns_end,+z,φ1 ⋅ z_+z,φ1 = 240.005 ⋅ 0.161 = 38.64 kNm/m
  N_Ed = 0.00 kNm/m

$f_{cd} = \frac{\alpha · f_{ck}}{{\gamma }_c} = \frac{0.85 · 2.0}{1.5} = 1.13\mathrm{kN}/{\mathrm{cm}}^2$

${\mu }_{Eds} =\hspace{0.17em}\frac{M_{Eds}}{b · d^2 · f_{cd}} = \frac{3864}{100 · {17}^2 · 1.13} =\hspace{0.17em}0.1183$

For μ_Eds = 0.1183, the following values can be interpolated from the design tables (e.g. [3] Annex A4):

${\omega }_1 =\hspace{0.17em}0.1170 + \frac{\left(0.1285 - 0.1170\right) · \left(0.1183 - 0.11\right)}{0.12 - 0.11} = 0.1265$

${\sigma }_{sd} =\hspace{0.17em}45.24 + \frac{\left(45.40 - 45.24\right) · \left(0.1183 - 0.11\right)}{0.12 - 0.11} = 45.37 \mathrm{kN}/{\mathrm{cm}}^2$

With these values, the required longitudinal reinforcement can be determined:

$A_{s1} =\hspace{0.17em}\frac{{\omega }_1 · b · d · f_{cd} + N_{Ed}}{{\sigma }_{sd}} = \frac{0.1265 · 100 · 17 · 1.13 + 0}{45.37} =\hspace{0.17em}5.36 {\mathrm{cm}}^2/\mathrm{m}$