Solution to Problem 589 | Design for Flexure and Shear

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

$R_1 = R_2 = 3W$
 

$6w_o = 4W$

$w_o = \frac{2}{3}W$
 

$A = 3(6)(1)$

$A = 18 \, \text{ in}^2$
 

 

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

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

$\bar{y} = 2.17 \, \text{ in}$       (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