Solution to Problem 553 | Unsymmetrical Beams

$\Sigma M_{R2} = 0$

$15R_1 + 4500 = 1500(9)$

$R_1 = 600 \, \text{lb}$
 

$\Sigma M_{R1} = 0$

$15R_2 = 1500(6) + 4500$

$R_2 = 900 \, \text{lb}$
 

$fb = \dfrac{My}{I}$
 

At M = +3600 lb·ft
$f_{bc} = \dfrac{3600(2.5)(12)}{96.0}$

$f_{bc} = 1125 \, \text{psi} \,\, \to \,\, $ upper fiber
 

$f_{bt} = \dfrac{3600(8.0)(12)}{96.0}$

$f_{bt} = 3600 \, \text{psi} \,\, \to \,\, $ lower fiber
 

At M = -1800 lb·ft
$f_{bc} = \dfrac{3600(8.0)(12)}{96.0}$

$f_{bc} = 1800 \, \text{psi} \,\, \to \,\, $ lower fiber
 

$f_{bt} = \dfrac{3600(2.5)(12)}{96.0}$

$f_{bt} = 562.5 \, \text{psi} \,\, \to \,\, $ upper fiber
 

Maximum flexure stresses
$f_{bc} = 1800 \, \text{psi} \,\,$ answer

$f_{bt} = 3600 \, \text{psi} \,\,$ answer