# Solution to Problem 585 | Design for Flexure and Shear

**Problem 585**

A simply supported beam of length L carries a uniformly distributed load of 6000 N/m and has the cross section shown in Fig. P-585. Find L to cause a maximum flexural stress of 16 MPa. What maximum shearing stress is then developed?

**Solution 585**

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

Where

f_{b} = 16 MPa

M = 1/8 w_{o}L^{2} = 1/8 (6000)L^{2} = 750L^{2} N·m

c = ½(250) = 125 mm

I = 300(250^{3})/12 - 200(150^{3})/12 = 334 375 000 mm^{4}

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_{o}L = ½(6000)(7.55) = 22 650 N

Q = 10 000(100) + 2(6 250)(62.5)

Q = 1 781 250 mm^{3}

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*