Solution to Problem 507 | Flexure Formula

$\dfrac{x}{60 \times 10^{-3}} = \dfrac{120 - x}{100 \times 10^{-3}}$

$x = 0.6(120 - x)$

$x + 0.6x = 0.6(120)$

$1.6x = 72$

$x = 45 \, \text{mm}$
 

 

$\dfrac{\delta_{top}}{x + 30} = \dfrac{60 \times 10^{-3}}{x}$

$\delta_{top} = \dfrac{60 \times 10^{-3}}{45}(45 30)$

$\delta_{top} = 0.1 \, \text{mm}$   lengthening
 

$\dfrac{\delta_{bottom}}{195 - x} = \dfrac{100 \times 10^{-3}}{120 - x}$

$\delta_{bottom} = \dfrac{100 \times 10^{-3}}{120 - 45}(195 45)$

$\delta_{bottom} = 0.2 \, \text{ mm}$   shortening
 

From Hooke's Law
$f_b = E\varepsilon$

$f_b = \dfrac{E\delta}{L}$
 

$( \, f_b \, )_{top} = \dfrac{70\,000(0.1)}{200} = 35 \, \text{ MPa}$   tension           answer

$( \, f_b \, )_{bottom} = \dfrac{70\,000(0.2)}{200} = 70 \, \text{ MPa}$   compression           answer