Beam Flexural Design & Ductility Screening

Design checks

Calculation

1. 1Effective depth
   d = h - \text{cover}
   d = 24.00 - 3.10 = 20.90
   d=20.90 in
2. 2Coefficient of resistance, assuming tension-controlled (φ = 0.9)ACI318 10.2
   R_n = \frac{M_u}{\phi b d^2}
   R_n = \frac{188.0 \times 12}{0.9 \times 12.00 \times 20.90^2} = 0.478
   R_n=0.4782 ksi
3. 3Required steel ratio and areaACI318 10.2
   \rho = \frac{0.85f'_c}{f_y}\left[1 - \sqrt{1 - \frac{2R_n}{0.85f'_c}}\right], \quad A_s = \rho b d
   \rho = 0.007, \quad A_s = 1.760
   A_{s,req}=1.760 in²
4. 4Minimum flexural reinforcementACI318 10.5.1
   A_{s,min} = \max\!\left(\frac{3\sqrt{f'_c}}{f_y}, \frac{200}{f_y}\right) b d
   A_{s,min} = 0.734
   A_{s,min}=0.734 in²
5. 5Governing tension steel
   A_{s,gov} = \max(A_{s,req}, A_{s,min})
   A_{s,gov} = 1.760
   A_{s,gov}=1.760 in²
6. 6Bar count and clear spacing in one layer
   n = \left\lceil \frac{A_{s,gov}}{A_b} \right\rceil, \quad s_{clear} = \frac{(b - 2\,\text{cover}) - n\,d_b}{n-1}
   n = 3 \times \#8 \Rightarrow A_{s,prov} = 2.370, \quad s_{clear} = 1.40
   A_{s,prov}=2.370 in²
7. 7Equivalent stress-block depth and neutral axisACI318 10.3
   a = \frac{A_s f_y - A'_s f_y}{0.85 f'_c b}, \quad c = \frac{a}{\beta_1}
   \beta_1 = 0.800, \quad a = 6.482, \quad c = 8.103
   c=8.103 in
8. 8Net tensile strain and yield strainACI318 10.3
   \varepsilon_t = \frac{0.003(d-c)}{c}, \quad \varepsilon_y = \frac{f_y}{E_s}
   \varepsilon_t = 0.005, \quad \varepsilon_y = 0.003
   \varepsilon_t=0.00474
9. 9Strength reduction factor — transition zone, interpolatedACI318 9.3.2
   \phi = \begin{cases}0.90 & \varepsilon_t \ge 0.005\\0.65 & \varepsilon_t \le \varepsilon_y\\0.65+(\varepsilon_t-\varepsilon_y)\dfrac{0.25}{0.005-\varepsilon_y} & \text{otherwise}\end{cases}
   \phi = 0.874
   \phi=0.874
10. 10Nominal and design flexural strengthACI318 10.3
    M_n = A_s f_y\!\left(d-\frac{a}{2}\right) + A'_s f_y\!\left(\frac{a}{2}-d'\right), \quad \phi M_n
    M_n = 486.5, \quad \phi M_n = 425.1
    \phi M_n=425.1 kip·ft

Assumptions

• Singly or doubly reinforced rectangular section, tension steel in one layer.
• Compression steel, where present, is assumed to yield at fy in the classification step — conservative, not exact, for a doubly reinforced section far from balanced.
• Section A assumes a tension-controlled section (φ = 0.9) to solve for the required steel; Section B then confirms that assumption from the steel actually provided.
• The Chapter 21 screen in Section C checks only the flexural reinforcement, geometry and joint-moment limits that follow from the section and moment data already entered.