A W18x50 beam has $F_y=50\text{ ksi}$, $d=18.0\text{ in}$, $t_w=0.355\text{ in}$, and $h/t_w=46.6$. Service loads are $D=0.8\text{ kip/ft}$ and $L=1.2\text{ kip/ft}$ over 24 ft. For this example use $1.2D+1.6L$.

A built-up girder has $h=60\text{ in}$, $t_w=0.50\text{ in}$, $F_y=50\text{ ksi}$, $E=29,000\text{ ksi}$ and $k_v=5.34$.

For the W18x50, use $I_x=800\text{ in}^4$, span 24 ft, service live load $1.2\text{ kip/ft}$, and an explicitly selected $L/360$ live-load criterion.

For the same beam, total service load is $0.8+1.2=2.0\text{ kip/ft}$. Suppose the project criterion for this check is $L/240$.

A W24x68 has $F_{yw}=50\text{ ksi}$, $t_w=0.415\text{ in}$, $k=1.18\text{ in}$, and bearing length $l_b=6\text{ in}$.

For the same W24x68 with $l_b=4\text{ in}$, use the near-end expression applicable to the example.

For the W24x68 under the interior bearing load, let $t_f=0.585\text{ in}$, $d=23.7\text{ in}$, $t_w=0.415\text{ in}$, $l_b=6\text{ in}$, and $F_{yw}=50\text{ ksi}$. Use the stated example expression:

$R_n=0.80t_w^2\left[1+3\left(\frac{l_b}{d}\right)\left(\frac{t_w}{t_f}\right)^{1.5}\right]\left(\frac{EF_{yw}t_f}{t_w}\right)^{1/2}.$

A reaction $R_u=120\text{ kips}$ acts near the end of a W16x31 with $F_{yw}=50\text{ ksi}$, $t_w=0.275\text{ in}$, and $k=0.827\text{ in}$. For the stated example, take the applicable design factor as 1.00.