Problem 1006 | Width of fastened steel plate for balanced reinforcement

$\dfrac{M}{I} = \dfrac{f_b}{y}$
 

At the extreme wood fiber
$\left( \dfrac{M}{I} \right)_\text{wood} = \dfrac{1200}{\bar{y}}$
 

At the extreme steel fiber
$\left( \dfrac{M}{I} \right)_\text{steel} = \dfrac{18\,000 / 20}{10.5 - \bar{y}} = \dfrac{900}{10.5 - \bar{y}}$
 

$\left( \dfrac{M}{I} \right)_\text{wood} = \left( \dfrac{M}{I} \right)_\text{steel}$

$\dfrac{1200}{\bar{y}} = \dfrac{900}{10.5 - \bar{y}}$

$1200(10.5 - \bar{y}) = 900\bar{y}$

$12\,600 - 1200\bar{y} = 900\bar{y}$

$2100\bar{y} = 12\,600$

$\bar{y} = 6 ~ \text{in}$
 

You can also find $\bar{y}$ with the aid of stress diagram as follows:

$A_w = 6(10) = 60 ~ \text{in}^2$

$A_s = 0.5(20b) = 10b$

$A = A_w + A_s = 60 + 10b$
 

$A\bar{y} = A_w y_w + A_s y_s$

$(60 + 10b)(6) = 60(5) + 10b(10.25)$

$360 + 60b = 300 + 102.5b$

$42.5b = 60$

$b = \frac{24}{17} ~ \text{in} = 1.41 ~ \text{in}$           answer