510 Circular bar used as beam

Problem 510
A 50-mm diameter bar is used as a simply supported beam 3 m long. Determine the largest uniformly distributed load that can be applied over the right two-thirds of the beam if the flexural stress is limited to 50 MPa.

Solution 510
$\Sigma M_{R1} = 0$
$3R_2 = 2w(2)$
$R_2 = \frac{4}{3} w$

$\Sigma M_{R2} = 0$
$3R_1 = 2w(1)$
$R_1 = \frac{2}{3} w$

$(f_b)_{\text{max}} = \dfrac{Mc}{I}$
Where:
$f_b = 20 \text{ MPa}$
$M = \frac{8}{9} w$
$c = 25 \text{ mm}$
$I = \dfrac{\pi r^4}{4}$

$I = \dfrac{\pi (25^4)}{4} = 97656.25\pi \text{ mm}^4$

Thus,
$50 = \dfrac{\frac{8}{9}w(1000)(25)}{97656.25\pi}$
$w = 690.29 \text{ N/m}$