Solution to Problem 585 | Design for Flexure and Shear

Flexural Stress
$f_b = \dfrac{Mc}{I}$

Where
f_b = 16 MPa
M = 1/8 w_oL² = 1/8 (6000)L² = 750L² N·m
c = ½(250) = 125 mm
I = 300(250³)/12 - 200(150³)/12 = 334 375 000 mm⁴
 

Thus,
$16 = \dfrac{750L^2(1000)(125)}{334\,375\,000}$

$L = 7.55 \, \text{ m}$           answer
 

Shearing Stress
$f_v = \dfrac{VQ}{Ib}$

Where
V = ½ w_oL = ½(6000)(7.55) = 22 650 N
Q = 10 000(100) + 2(6 250)(62.5)
Q = 1 781 250 mm³
b = 2(50) = 100 mm
 

Thus,
$f_v = \dfrac{22\,650(1\,781\,250)}{334\,375\,000(100)}$

$f_v = 1.21 \, \text{ MPa}$           answer