Problem 657 | Beam Deflection by Conjugate Beam Method

From the load diagram
$\Sigma M_{R1} = 0$

$6R_2 = \frac{1}{2}(4)(600)[\frac{1}{3}(4)]$

$R_2 = 266.67 \, \text{ N}$
 

$\dfrac{y}{1} = \dfrac{600}{4}$

$y = 150 \, \text{ N/m}$
 

 

From the moment diagram
$a = 3R_2 = 3(266.67)$

$a = 800 \, \text{ N}\cdot\text{m}$
 

$b = -\frac{1}{2}(1)(y)[\frac{1}{3}(1)]$

$b = -\frac{1}{6}y = -\frac{1}{6}(150)$

$b = -25 \, \text{ N}\cdot\text{m}$
 

From the conjugate beam
$\Sigma M_{F1} = 0$

$6F_2 + \frac{1}{4}(4)(1600)[\frac{1}{5}(4)] = \frac{1}{2}(6)(1600)[\frac{1}{3}(6)]$

$F_2 = 1386.67 \, \text{ N}\cdot\text{m}^2$
 

 

$M_{midspan} = \frac{1}{2}(3a)[\frac{1}{3}(3)] - \frac{1}{4}(1b)[\frac{1}{5}(1)] - 3F_2$

$M_{midspan} = \frac{1}{2}(3)(800)[\frac{1}{3}(3)] - \frac{1}{4}(1)(25)[\frac{1}{5}(1)] - 3(1386.67)$

$M_{midspan} = -2961.25 \, \text{ N}\cdot\text{m}^3$
 

Thus, the deflection at the midspan is
$EI \, \delta_m = M_{midspan}$

$EI \, \delta_m = -2961.25 \, \text{ N}\cdot\text{m}^3$

$EI \, \delta_m = 2961.25 \, \text{ N}\cdot\text{m}^3$   below the neutral axis           answer