Problem 655 | Beam Deflection by Conjugate Beam Method

$\Sigma M_{D} = 0$

$8R_1 = 200(5) + 400(1)$

$R_1 = 175 \, \text{ lb}$
 

$\Sigma M_{A} = 0$

$8R_2 = 200(3) + 400(7)$

$R_2 = 425 \, \text{ lb}$
 

 

By ratio and proportion
$\dfrac{y_C}{1} = \dfrac{2125}{5}$

$y_C = 425 \, \text{ lb}\cdot\text{ft}$
 

From the conjugate beam
$\Sigma M_D = 0$

$8F_1 + \frac{1}{2}(4)(1600)[1 + \frac{2}{3}(4)] = \frac{1}{2}(3)(525)[5 + \frac{1}{3}(3)] + \frac{1}{2}(5)(2125)[\frac{2}{3}(5)]$

$F_1 = 1337.5 \, \text{ lb}\cdot\text{ft}^2$
 

 

$\Sigma M_A = 0$

$8F_2 + \frac{1}{2}(4)(1600)[3 + \frac{1}{3}(4)] = \frac{1}{2}(3)(525)[\frac{2}{3}(3)] + \frac{1}{2}(5)(2125)[3 + \frac{1}{3}(5)]$

$F_2 = 1562.5 \, \text{ lb}\cdot\text{ft}^2$
 

Consider the section to the left of B in conjugate beam
$M_B = \frac{1}{2}(3)(525)[\frac{1}{3}(3)] - 3F_1$

$MB = 787.5 - 3(1337.5)$

$M_B = -3225 \, \text{ lb}\cdot\text{ft}^3$
 

Thus, the deflection at B is
$EI ~ \delta_B = M_B$

$EI ~ \delta_B = 3225 \, \text{ lb}\cdot\text{ft}^3$           answer
 

Consider the section to the right of C in conjugate beam
$M_C = \frac{1}{2}(1)(y_C)[\frac{1}{3}(1)] - 1F_2$

$M_C = \frac{1}{2}(1)(425)[\frac{1}{3}(1)] - 1(1562.5)$

$M_C = -1491.67 \, \text{ lb}\cdot\text{ft}^3$
 

Thus, the deflection at C is
$EI ~ \delta_C = M_C$

$EI ~ \delta_C = -1491.67 \, \text{ lb}\cdot\text{ft}^3$

$EI ~ \delta_C = 1491.67 \, \text{ lb}\cdot\text{ft}^3$   downward           answer