Solution to Problem 564 | Unsymmetrical Beams

Force acting on the shaded region
$F = ( \, f_b \, / \, c \, ) \, A' \, y'$
 

where
$f_b = 1200 \, \text{ psi}$

$c = 6 \, \text{ in.}$

$A' = 10(2) = 20 \, \text{ in.}^2$

$y' = 5 \, \text{ in.}$

 

Thus,
$F = (1200/5)(20)(5)$

$F = 24\,000 lb$

$F = 24 \, \text{kips}$           answer
 

Moment of force F about NA
$M = f_b \, I' \, / \, c$
 

where
$I' = I + Ad^2$

$I' = \dfrac{10(2^3)}{12} + (10 \times 2)(5^2)$

$I' = 506.67 \, \text{in}^4$

 

Thus,
$M = 1200(506.67)/6$

$M = 101\,333.33 \, \text{lb}\cdot\text{in}$

$M = 8\,444.44 \, \text{lb}\cdot\text{ft}$