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Design

With the internal forces from the structural analysis, SPX designs the circular reinforced concrete section in accordance with NBR 6118.

There are two checks: biaxial bending with axial force (longitudinal reinforcement) and shear (transverse reinforcement).

Biaxial bending with axial force

Design forces

The maxima are taken from the two models — X axis and Y axis —, and the resultant moment is combined vectorially:

\[ M_{d,1^a} = \gamma_f \sqrt{M_{kx}^2 + M_{ky}^2} \qquad N_d = \gamma_f N_k \]

with \(\gamma_f = 1.4\).

The vector combination is legitimate for a circular section, which has no preferred direction: any moment direction meets the same geometry and the same distribution of reinforcement.

Stress block parameters

According to the concrete class:

\[ \lambda = 0.8 \qquad \alpha_c = 0.85 \qquad \varepsilon_{cu} = 3.5\text{‰} \qquad \varepsilon_{c2} = 2.0\text{‰} \]
\[ \lambda = 0.8 - \frac{f_{ck}-50}{400} \qquad \alpha_c = 0.85\left(1 - \frac{f_{ck}-50}{200}\right) \]
\[ \varepsilon_{cu} = \frac{2.6 + 35\left(\frac{90-f_{ck}}{100}\right)^4}{1000} \qquad \varepsilon_{c2} = \frac{2.0 + 0.085\,(f_{ck}-50)^{0.53}}{1000} \]

The design strengths come from \(\gamma_c = 1.4\) and \(\gamma_s = 1.15\), with \(\sigma_{cd} = \alpha_c f_{cd}\).

Second-order effects

The slenderness ratio uses the radius of gyration of the circular section (\(i = D/4\)):

\[ \lambda = \frac{\ell_e}{D/4} \]

For \(40 < \lambda \le 140\), the second-order moment is added by the standard-column method with approximate curvature:

\[ M_{2d} = N_d \cdot \frac{\ell_e^2}{10} \cdot \frac{1}{r} \]

with the curvature limited by

\[ \frac{1}{r} = \min\left(\frac{0.005}{(\nu + 0.5)\,D},\; \frac{0.005}{D}\right) \qquad \nu = \frac{N_d}{A_c f_{cd}} \]

The total design moment is \(M_d = M_{d,1^a} + M_{2d}\).

Above λ = 140

NBR 6118 requires more rigorous methods for \(\lambda > 140\). The program does not add second-order effects in this range — it is up to the designer to check whether the slenderness is acceptable and, if so, to handle it with a specific analysis.

Reinforcement not required

Reinforcement can be omitted when the compressive stress is low enough:

\[ \sigma_{sd} = \frac{N_d}{A_c} \le 5\text{ MPa} \qquad\text{and}\qquad \sigma_{sd} \le 0.85\,f_{ck} \]

Even so, a minimum detailing reinforcement is usually adopted for other reasons — lifting, driving, tying into the cap.

Interaction diagram

The program builds the \(N\)–\(M\) interaction diagram of the section with the adopted reinforcement and checks whether the pair \((N_d, M_d)\) falls inside it. It is the most transparent check possible: you see the margin, not just the verdict.

The minimum number of bars is 6, as required by the code for a circular section.

Shear

Design force

The shear is combined vectorially from the two models:

\[ V_k = \sqrt{V_{kx}^2 + V_{ky}^2} \qquad V_d = \gamma_f V_k \]

Strut check

First, concrete crushing:

\[ \tau_{wd} = \frac{V_d}{b_w d} \qquad \tau_{wu} = \frac{0.27\,\alpha_v\,f_{ck}}{\gamma_c} \qquad \alpha_v = 1 - \frac{f_{ck}}{250} \]

If \(\tau_{wd} > \tau_{wu}\), the section is insufficient and the program refuses the design: no reinforcement can fix strut crushing, only increasing the diameter or the concrete strength.

For the circular section, \(b_w = D\) and \(d = D - c - \phi_t - \phi_\ell/2\) are adopted.

Transverse reinforcement

By Model I of NBR 6118 (struts at 45°):

\[ \tau_c = \frac{0.126\, f_{ck}^{2/3}}{\gamma_c} \quad (f_{ck} \le 50) \]
\[ \frac{A_{sw}}{s} = \frac{100\, b_w\, \cdot 1.11(\tau_{wd} - \tau_c)}{f_{yd}} \]

with minimum reinforcement

\[ \rho_{min} = \frac{0.2\, f_{ct,m}}{f_{ywk}} \qquad f_{ct,m} = 0.30\, f_{ck}^{2/3} \]

Adopted spacing

The spacing is the smallest of three criteria — and the program shows which one governed:

Criterion Limit
Theoretical From the calculated \(A_{sw}/s\)
Code (ULS) \(\min(0.6d;\,30\text{ cm})\), or \(\min(0.3d;\,20\text{ cm})\) if \(\tau_{wd} > 0.67\,\tau_{wu}\)
Detailing \(\min(20\text{ cm};\, D;\, 12\phi_\ell)\)

With a lower bound of 5 cm — below that the concrete cannot be placed.

Minimum stirrup diameter

\(\phi_t \ge \max(5\text{ mm};\, \phi_\ell/4)\). A 5.0 mm stirrup is calculated with \(f_{ywk} = 600\) MPa (CA-60); above that, 500 MPa.

Limits of validity

Range of application

  • The design covers a solid circular section. Hollow, steel or composite sections are not covered.
  • The vector combination of forces assumes that the X and Y maxima occur at the same section, which is conservative when they do not.
  • Second-order effects are only handled in the range \(40 < \lambda \le 140\).
  • The following are not checked: fatigue, cracking in service, transient driving or lifting situations, nor the anchorage of the reinforcement in the cap.
  • The buckling length \(\ell_e\) is an input. Determining it for a partially embedded pile requires judgement — the embedded length is restrained by the soil, but the stiffness of that restraint depends on \(K_h\) itself.