Solution to Problem 503 | Flexure Formula

$M = F(\frac{1}{3}x)$
 

$\dfrac{y}{x} = \dfrac{1000}{6}$

$y = \frac{500}{3}x$
 

$F = \frac{1}{2}xy$

$F = \frac{1}{2}x(\frac{500}{3}x)$

$F = \frac{250}{3}x^2$
 

Thus,
$M = (\frac{250}{3}x^2)(\frac{1}{3}x)$

$M = \frac{250}{9}x^3$
 

Part (a):
The maximum moment occurs at the support (the wall) or at x = 6 m.

$M = \frac{250}{9}x^3 = \frac{250}{9}(6^3)$

$M = 6000 \, \text{N}\cdot\text{m}$
 

$(f_b)_{max} = \dfrac{Mc}{I} = \dfrac{Mc}{\dfrac{bh^3}{12}}$

$(f_b)_{max} = \dfrac{Mc}{I} = \dfrac{6000(1000)(75)}{\dfrac{50(150^3)}{12}}$

$(f_b)_{max} = 32 \, \text{MPa}$           answer
 

Part (b):
At a section 2 m from the free end or at x = 2 m at fiber 20 mm from the top of the beam:

$M = \frac{250}{9}x^3 = \frac{250}{9}(2^3)$

$M = \frac{2000}{9} \, \text{N}\cdot\text{m}$
 

$f_b = \dfrac{My}{I} = \dfrac{\frac{2000}{9}(1000)(55)}{\dfrac{50(150^3)}{12}}$

$f_b = 0.8691 \, \text{MPa} = 869.1 \, \text{kPa tension}$           answer