# Solution to Problem 589 | Design for Flexure and Shear

**Problem 589**

A channel section carries a concentrated loads W and a total distributed load of 4W as shown in Fig. P-589. Verify that the NA is 2.17 in. above the bottom and that I_{NA} = 62 in^{4}. Use these values to determine the maximum value of W that will not exceed allowable stresses in tension of 6,000 psi, in compression of 10,000 psi, or in shear of 8,000 psi.

**Solution 589**

$R_1 = R_2 = \frac{1}{2}(W + 4W +W)$

$R_1 = R_2 = 3W$

$6w_o = 4W$

$w_o = \frac{2}{3}W$

$A = 3(6)(1)$

$A = 18 \, \text{ in}^2$

$A\bar{y} = \Sigma ay$

$18\bar{y} = 2 [ \, 6(1)(3) \, ] + 6(1)(0.5)$

$\bar{y} = 2.17 \, \text{ in}$ (*okay!*)

By transfer formula for moment of inertia

$I_{NA} = 2\left[ \dfrac{1(6^3)}{12} + 6(0.83^2) \right] + \left[ \dfrac{6(1^3)}{12} + 6(1.67^2) \right]$

$I_{NA} = 61.5002 \, \text{ in}^4$ (*okay!*)

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

**For M = -2W lb·ft**

Top fiber in tension

$6000 = \dfrac{2W(12)(3.83)}{62}$

$W = 4045 \, \text{ lb}$

Bottom fiber in compression

$10\,000 = \dfrac{2W(12)(2.17)}{62}$

$W = 11\,905 \, \text{ lb}$

**For M = W lb·ft**

Top fiber in compression

$10\,000 = \dfrac{W(12)(3.83)}{62}$

$W = 13\,490 \, \text{ lb}$

Bottom fiber in tension

$6000 = \dfrac{W(12)(2.17)}{62}$

$W = 14\,286 \, \text{ lb}$

#### Based on allowable shear stress:

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

Where

$V = 2W$

$Q_{NA} = 2 [ \, 3.83(1)(3.83/2) \, ] = 14.6689 \, \text{ in}^3$

$I = 62 \, \text{ in}^4$

$b = 2 \, \text{ in}$

Thus,

$8000 = \dfrac{2W(14.6689)}{62(2)}$

$W = 33\,813 \, \text{ lb}$

For safe value of W, use **W = 4045 lb** *answer*