Problem 913 | Combined Axial and Bending

$\Sigma M_C = 0$

$36D_V + 12(\frac{3}{5} \times 6) + 6(\frac{4}{5} \times 6) = 18(\frac{4}{5} \times 10) + 6(\frac{3}{5} \times 10)$

$36D_V + 43.2 + 28.8 = 144 + 36$

$36DV = 108$

$D_V = 3^k$
 

 

$\Sigma F_H = 0$

$D_H = \frac{4}{5}(6) + \frac{3}{5}(10)$

$D_H = 10.8^k$
 

$M_{AB} = \Sigma M_{\text{to the right of } AB}$

$M_{AB} = 12D_V = 12(3)$

$M_{AB} = 36 ~ \text{kip}\cdot\text{in}$
 

 

$\sigma_a = \dfrac{D_H}{A_{AB}} = \dfrac{10.8(1000)}{2(6)}$

$\sigma_a = 900 ~ \text{psi}$
 

$\sigma_f = \dfrac{6M_{AB}}{bd^2} = \dfrac{6(36)(1000)}{2(6^2)}$

$\sigma_f = 3000 ~ \text{psi}$
 

$\sigma_A = \sigma_a - \sigma_f = 900 - 3000$

$\sigma_A = -2100 ~ \text{psi}$       answer
 

$\sigma_B = \sigma_a + \sigma_f = 900 + 3000$

$\sigma_B = 3900 ~ \text{psi}$       answer