Solution to Problem 552 | Unsymmetrical Beams

$M = 5(8) - 30$

$M = 10 \, \text{kip}\cdot\text{ft}$
 

$f_b = \dfrac{My}{I}$
 

At M = +10 kip·ft of moment diagram
$f_{bc} = \dfrac{10(6)(12)}{90} = 8 \, \text{ksi} \,\, \to \,\, $ upper fiber

$f_{bt} = \dfrac{10(2)(12)}{90} = 2.67 \, \text{ksi} \,\, \to \,\, $ lower fiber
 

At M = -20 kip·ft of moment diagram
$f_{bc} = \dfrac{20(2)(12)}{90} = 5.33 \, \text{ksi} \,\, \to \,\, $ lower fiber

$f_{bt} = \dfrac{20(6)(12)}{90} = 16 \, \text{ksi} \,\, \to \,\, $ upper fiber
 

Maximum bending stresses:
$f_{bc} = 8 \, \text{ksi} \,\,$ answer

$f_{bt} = 16 \, \text{ksi} \,\,$ answer