
$\Sigma M_{R2} = 0$
$6R_1 = 4w_o(2)$
$R_1 = \frac{4}{3}w_o \, \text{ N} $
$\Sigma M_{R1} = 0$
$6R_2 = 4w_o(4)$
$R_2 = \frac{8}{3}w_o \, \text{ N} $
From the shear diagram
$\dfrac{x}{\frac{4}{3}w_o} = \dfrac{4 - x}{\frac{8}{3}w_o}$
$2x = 4 - x$
$x = \frac{4}{3} \, \text{ m}$
Maximum moment = sum of area in Shear Diagram at the left of point of zero shear
$M_{max} = 2(\frac{4}{3}w_o) + \frac{1}{2}(\frac{4}{3})(\frac{4}{3}w_o)$
$M_{max} = \frac{32}{9}w_o \, \text{ N}\cdot\text{m}$
Based on allowable flexural stress
$f_b = \dfrac{Mc}{I}$
Where
c = 150 mm
I = 200(3003)/12 - 175(2503)/12
I = 222 135 416.67 mm4
Thus,
$10 = \dfrac{\frac{32}{9}w_o(1000)(150)}{222\,135\,416.67}$
$w_o = 4165.04 \, \text{ N/m}$
Based on allowable shear stress
$f_v = \dfrac{VQ}{Ib}$
Where
Q = 200(25)(137.5) + 125(25)(62.5)
Q = 882 812.5 mm3
I = 222 135 416.67 mm4
b = 25 mm
Thus,
$1.0 = \dfrac{\frac{8}{3}w_o(882 812.5)}{222\,135\,416.67(25)}$
$w_o = 2358.96 \, \text{ N/m}$
For safe value of wo, use wo = 2358.96 N/m. answer
Comments
bakit po hindi niyo sinama
hindi po ba dapat 2358.960 N/m yung safe load? di niyo po ata naisama yung base na 25 sa pagcalculate
Tnx. Na edit ko na.
Tnx. Na edit ko na.