# Problem 1006 | Width of fastened steel plate for balanced reinforcement

**Problem 1006**

Determine the width b of the 1/2-in. steel plate fastened to the bottom of the beam in Problem 1005 that will simultaneously stress the wood and the steel to their permissible limits of 1200 psi and 18 ksi, respectively.

**Solution 1006**

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$\dfrac{M}{I} = \dfrac{f_b}{y}$

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:

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$\dfrac{\bar{y}}{1200} = \dfrac{10.5 - \bar{y}}{18\,000 / 20}$

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

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

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

$\bar{y} = 6 ~ \text{in}$ (*okay!*)

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

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

$A = A_w + A_s = 60 + 10b$

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

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

$360 + 60b = 300 + 102.5b$

$42.5b = 60$

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

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