# Solution to Problem 586 | Design for Flexure and Shear

**Problem 586**

The distributed load shown in Fig. P-586 is supported by a box beam having the same cross-section as that in Prob. 585. Determine the maximum value of w_{o} that will not exceed a flexural stress of 10 MPa or a shearing stress of 1.0 MPa.

**Solution 586**

## Click here to expand or collapse this section

$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 mm^{4}

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 mm^{3}

I = 334 375 000 mm^{4}

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*

- Log in to post comments