Computes the required area of reinforcing steel for a singly reinforced rectangular concrete cross section. Takes into account axial force and moment. Only checks strength requirements. Does not apply minimum reinforcement requirements. Uses ACI 318-19.
| Number | Name | Object Type | Description | Optional |
|---|---|---|---|---|
| 1 | Thickness | System.Double | Thickness of the concrete section, in the display unit system. | No |
| 2 | Width | System.Double | Width of the concrete section, in the display unit system. | No |
| 3 | Depth to Steel | System.Double | Distance from the extreme compression fiber to the centroid of the reinforcement, in the display unit system. | No |
| 4 | Comp Stress | System.Double | Specified concrete compressive strength, in the display unit system. | No |
| 5 | Yield Stress | System.Double | Yield stress of reinforcing, in the display unit system. | No |
| 6 | Moment | System.Double | Ultimate moment demand, in the display unit system. | No |
| 7 | Axial | System.Double | Axial force (compression = negative, tension = positive), in the display unit system. | No |
| 8 | Spirals | System.Boolean | Whether the section has conforming spiral reinforcement. Defaults to false. | Yes |
| 9 | Max Iter | System.Int32 | Maximum number of iterations. Defaults to 20. | Yes |
| Number | Name | Object Type | Description |
|---|---|---|---|
| 1 | Area | System.Double | The required area of reinforcing steel such that the factored nominal capacity exceeds the combined ultimate axial + moment demand. |
This component solves for the required area of steel based on the following:
$$ P=-0.85F_c^\prime a b +A_s F_s $$
$$ \frac{M}{\phi}=A_s F_s \left(d-\frac{a}{2} \right)-P \left(\frac{t}{2}-\frac{a}{2}\right) $$
where $P$ is negative for compression and positive for tension
and $F_s$ is the stress in the steel, which is equal to $min\left(\epsilon E_s, F_y\right)$
Solving the first equation for $a$ yields:
$$ a=\frac{A_s F_s - P}{0.85 F_c^\prime b} $$
Then substituting for $a$ in the second equation:
$$ \frac{M}{\phi}=A_s F_s \left(d-\frac{A_s F_s - P}{2 \left(0.85 F_c^\prime b\right)}\right)-P \left(\frac{t}{2}-\frac{A_s F_s - P}{2 \left(0.85 F_c^\prime b\right)}\right) $$
$$ \frac{M}{\phi}=A_s F_s d - \frac{A_s^2 F_s^2 -A_s F_s P}{1.7 F_c^\prime b}-\frac{P t}{2} + \frac{P A_s F_s - P^2}{1.7 F_c^\prime b} $$
$$ \frac{M}{\phi}=A_s F_s d - \frac{A_s^2 F_s^2}{1.7 F_c^\prime b} + \frac{A_s F_s P}{1.7 F_c^\prime b}-\frac{P t}{2} + \frac{P A_s F_s}{1.7 F_c^\prime b} - \frac{P^2}{1.7 F_c^\prime b} $$
Moving $\frac{M}{\phi}$ to the right side yields:
$$ 0=A_s F_s d - \frac{A_s^2 F_s^2}{1.7 F_c^\prime b}+ \frac{A_s F_s P}{1.7 F_c^\prime b}-\frac{P t}{2} + \frac{P A_s F_s}{1.7 F_c^\prime b} - \frac{P^2}{1.7 F_c^\prime b}-\frac{M}{\phi} $$