Problem 908 | Combined Axial and Bending

 

 

$M_{1-1} = 28P ~ \text{lb}\cdot\text{ft}$
 

 

$\sigma_A = \sigma_a + \sigma_f$

$\sigma_A = \dfrac{P}{120} + \dfrac{28P(8)}{4000}$

$\sigma_A = \dfrac{1}{120}P + \dfrac{7}{125}P$

$\sigma_A = \dfrac{193}{3000}P$
 

$\sigma_B = \sigma_a - \sigma_f$

$\sigma_B = \dfrac{P}{A} - \dfrac{Mc}{I}$

$\sigma_B = \dfrac{P}{120} - \dfrac{28P(12)}{4000}$

$\sigma_B = \dfrac{1}{120}P - \dfrac{21}{250}P$

$\sigma_B = -\dfrac{277}{3000}P$
 

Based on fibers at A

$18 = \dfrac{193}{3000}P$

$P = 279.79 ~ \text{kips}$

 

Based on fibers at B

$-18 = -\dfrac{277}{3000}P$

$P = 194.94 ~ \text{kips}$

 

Use $P = 194.94 ~ \text{kips}$       answer