Solution to Problem 554 | Unsymmetrical Beams

$\Sigma M_{R2} = 0$

$12R_1 = 1600(15) + 4000(6)$

$R_1 = 4000 \, \text{lb}$
 

$\Sigma M_{R1} = 0$

$12R_2 + 1600(3) = 4000(6)$

$R_2 = 1600 \, \text{lb}$
 

$f_b = \dfrac{My}{I}$
 

At M = -4800 lb·ft
$f_{bc} = \dfrac{4800(2)(12)}{84}$

$f_{bc} = 1371.43 \, \text{psi} \,\, \to \,\,$ lower fiber
 

$f_{bt} = \dfrac{4800(7)(12)}{84}$

$f_{bt} = 4800 \, \text{psi} \,\, \to \,\,$ upper fiber
 

At M = +9600 lb·ft
$f_{bc} = \dfrac{9600(7)(12)}{84}$

$f_{bc} = 9600 \, \text{psi} \,\, \to \,\,$ upper fiber
 

$f_{bt} = \dfrac{4800(2)(12)}{84}$

$f_{bt} = 2742.86 \, \text{psi} \,\, \to \,\,$ lower fiber
 

Maximum flexure stress:
$f_{bc} = 9600 \, \text{psi} \,\,$       answer
$f_{bt} = 4800 \, \text{psi} \,\,$       answer