Flexural Design Example

Example Problem

Determine the LRFD flexural design strength for a W10x12 beam with an unbraced length of 2 ft.

1. Determine if Section is Compact

\begin{aligned} \lambda_{flange} &= 9.43\;\text{ }(\text{AISC Table 1-1}) \\[10pt] \lambda_{web} &= 46.6\;\text{ }(\text{AISC Table 1-1}) \end{aligned}

2. Determine the limiting ratios (AISC Table B4.1b)

Check Flange

\begin{aligned} E &= 29000\;\mathrm{ksi} \\[10pt] F_{y} &= 50\;\mathrm{ksi} \\[10pt] \lambda_{pf} &= 0.38 \cdot \sqrt{\frac{E}{F_{y}}} = 0.38 \cdot \sqrt{\frac{29000\;\mathrm{ksi}}{50\;\mathrm{ksi}}} = 9.15\;\text{ }(\text{Case 10}) \\[10pt] \lambda_{r} &= 1 \cdot \sqrt{\frac{E}{F_{y}}} = 1 \cdot \sqrt{\frac{29000\;\mathrm{ksi}}{50\;\mathrm{ksi}}} = 24.08 \end{aligned}

\\[20pt] \left( \lambda_{pf} < \lambda_{flange} < \lambda_{r} \right) = \left( 9.15 < 9.43 < 24.08 \right) = 1 \;\;\; \therefore \text{{\color{blue} \ Noncompact Flange}}

Check Web

\begin{aligned} \lambda_{pw} &= 3.76 \cdot \sqrt{\frac{E}{F_{y}}} = 3.76 \cdot \sqrt{\frac{29000\;\mathrm{ksi}}{50\;\mathrm{ksi}}} = 90.55\;\text{ }(\text{Case 10}) \end{aligned}

\\[20pt] \left( \lambda_{web} < \lambda_{pw} \right) = \left( 46.6 < 90.55 \right) = 1 \;\;\; \therefore \text{{\color{blue} \ Compact Web}}

3. Calculate the LB strength with AISC Spec F3

\begin{aligned} S_{x} &= 10.9\;\mathrm{inch}^{3}\;\text{ }(\text{AISC Table 1-1}) \\[10pt] Z_{x} &= 12.6\;\mathrm{inch}^{3}\;\text{ }(\text{AISC Table 1-1}) \end{aligned}

\begin{aligned} M_{p} &= F_{y} \cdot Z_{x} = 50\;\mathrm{ksi} \cdot 12.6\;\mathrm{inch}^{3} = 52.5\;\mathrm{ft}\,\mathrm{kip} \end{aligned}

\begin{aligned} M_{nLB} &= M_{p} - \left( M_{p} - 0.7 \cdot F_{y} \cdot S_{x} \right) \cdot \frac{\lambda_{flange} - \lambda_{pf}}{\lambda_{r} - \lambda_{pf}} \\[10pt] &= 52.5\;\mathrm{ft}\,\mathrm{kip} - \left( 52.5\;\mathrm{ft}\,\mathrm{kip} - 0.7 \cdot 50\;\mathrm{ksi} \cdot 10.9\;\mathrm{inch}^{3} \right) \cdot \frac{9.43 - 9.15}{24.08 - 9.15} \\[10pt] &= 52.11\;\mathrm{ft}\,\mathrm{kip} \end{aligned}

4. Calculate LTB strength with AISC spec F2.2

\begin{aligned} L_{b} &= 2\;\mathrm{ft} \\[10pt] r_{y} &= 0.78\;\mathrm{inch} \end{aligned}

\begin{aligned} L_{p} &= 1.76 \cdot r_{y} \cdot \sqrt{\frac{E}{F_{y}}} = 1.76 \cdot 0.78\;\mathrm{inch} \cdot \sqrt{\frac{29000\;\mathrm{ksi}}{50\;\mathrm{ksi}}} = 2.77\;\mathrm{ft} \end{aligned}

\\[20pt] \left( L_{p} > L_{b} \right) = \left( 2.77\;\mathrm{ft} > 2\;\mathrm{ft} \right) = 1 \;\;\; \therefore \text{{\color{blue} \ Full Plastic Behavior}}

\begin{aligned} M_{nLTB} &= M_{p} = 52.5\;\mathrm{ft}\,\mathrm{kip} \end{aligned}

5. Design Strength

\begin{aligned} M_{n} &= \mathrm{min}\left( M_{nLTB}, M_{nLB} \right) = \mathrm{min}\left( 52.5\;\mathrm{ft}\,\mathrm{kip}, 52.11\;\mathrm{ft}\,\mathrm{kip} \right) = 52.11\;\mathrm{ft}\,\mathrm{kip} \\[10pt] \phi_{b} &= 0.9 \\[10pt] {\phi}M_{n} &= \phi_{b} \cdot M_{n} = 0.9 \cdot 52.11\;\mathrm{ft}\,\mathrm{kip} = 46.9\;\mathrm{ft}\,\mathrm{kip} \end{aligned}