Solution to Problem 586 | Design for Flexure and Shear

$3R_1 = 4w_o (1) $

$R_1 = \frac{4}{3}w_o$
 

$\Sigma M_{R1} = 0$

$3R_2 = 4w_o (2)$

$R_2 = \frac{8}{3}w_o$
 

From shear diagram
$\dfrac{x}{\frac{4}{3}w_o} = \dfrac{3 - x}{\frac{5}{3}w_o}$

$\frac{5}{3}x = 4 - \frac{4}{3}x$

$x = \frac{4}{3} \, \text{ m}$
 

Based on allowable bending stress
$f_b = \dfrac{Mc}{I}$

Where (From Solution 585):
c = 125 mm
I = 334 375 000 mm⁴
 

Thus,
$10 = \dfrac{\frac{8}{9}w_o (1000^2)(125)}{334\,375\,000}$

$w_o = 30.09 \, \text{ kN/m}$
 

Based on allowable shear stress
$f_v = \dfrac{VQ}{Ib}$

Where (From Solution 585):
Q = 1 781 250 mm³
I = 334 375 000 mm⁴
b = 100 mm
 

Thus,
$1 = \dfrac{\frac{5}{3}w_o(1000)(1\,781\,250)}{334\,375\,000(100)}$

$w_o = 11.26 \, \text{ kN/m}$
 

For safe value of w_o, use w_o = 11.26 kN/m           answer