Design¶
The sixth tab. It designs the circular reinforced concrete section for the internal forces from the analysis, in accordance with NBR 6118.
Inputs¶
| Field | Unit | Note |
|---|---|---|
| \(f_{ck}\) | MPa | Concrete strength |
| \(f_{yk}\) | MPa | Longitudinal steel, usually 500 (CA-50) |
| \(E_s\) | MPa | Steel modulus, default 210 000 |
| Longitudinal bar diameter \(\phi_\ell\) | mm | |
| Number of bars | — | Leave it on automatic for the program to search |
| Stirrup diameter \(\phi_t\) | mm | Minimum \(\max(5;\, \phi_\ell/4)\) |
| Cover | cm | Comes from the exposure class, but can be edited |
| Buckling length \(\ell_e\) | cm | See the note below |
The buckling length is your judgement
\(\ell_e\) is an input, not a calculated value. 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. A pile fully embedded in competent soil rarely has a buckling problem; one with an exposed length does.
Longitudinal reinforcement¶
The program checks the section for biaxial bending with axial force, combining the moments of the two axes vectorially and adding the second-order effect when \(40 < \lambda \le 140\).
The result is shown as an \(N\)–\(M\) interaction diagram: the boundary of the section with the adopted reinforcement, and the design point marked on it. The reading is immediate — you see the margin, not just the verdict.
| Situation | Reading |
|---|---|
| Point well inside the curve | Generous section; consider reducing the reinforcement or the diameter |
| Point close to the boundary | Tight design, but valid |
| Point outside | Insufficient section — increase the reinforcement, \(f_{ck}\) or the diameter |
Minimum of 6 bars
A circular section requires at least six longitudinal bars, by code requirement. Even when the calculation does not require reinforcement, this detailing minimum usually prevails — for lifting, driving and tying into the cap.
Reinforcement not required¶
When \(\sigma_{sd} = N_d/A_c \le 5\) MPa and \(\sigma_{sd} \le 0.85 f_{ck}\), the calculation does not require reinforcement. The program flags the condition, but the decision to take advantage of it is the designer's.
Transverse reinforcement¶
Stirrups are designed by Model I of NBR 6118.
The first check is strut crushing. If \(\tau_{wd} > \tau_{wu}\), the program refuses the design with an explicit message: no reinforcement can fix strut crushing; the pile diameter or the concrete strength must be increased.
The result shows:
| Output | Content |
|---|---|
| \(V_d\) | Design shear, combining the two axes |
| Code minimum area | \(A_{sw,min}\) in cm²/m |
| Effective area adopted | The larger of the calculated and the minimum |
| Stirrup diameter | As entered |
| Adopted spacing \(s\) | The smallest of theoretical, code and detailing |
| Stirrups per metre | \(100/s\) |
Which criterion governed the spacing
The adopted spacing is the smallest of three limits — the theoretical one from the calculation, the ULS code limit and the detailing limit (\(\min(20\text{ cm};\, D;\, 12\phi_\ell)\)) — with a lower bound of 5 cm.
In piles, the detailing criterion often governs: the shear is usually modest, and what tightens the stirrups is the geometric limit.
See shear formulation.
Error messages¶
\"Invalid stirrup diameter. Minimum required: X mm\"
The stirrup must be at least \(\max(5\text{ mm};\, \phi_\ell/4)\). Increase \(\phi_t\) or reduce \(\phi_\ell\).
\"Strut crushing\"
The section is insufficient for the shear. Increase the pile diameter or \(f_{ck}\) — reinforcement does not solve it.
The point falls outside the interaction diagram
The section cannot resist the \(N\)–\(M\) combination. Increase the reinforcement, \(f_{ck}\) or the diameter. If \(\lambda\) is high, much of the moment may be second-order — in that case, increasing the diameter is more effective than adding reinforcement.