Problem 653 | Beam Deflection by Conjugate Beam Method

By symmetry,
$R_1 = R_2 = 2(600)$

$R_1 = R_2 = 1200 \, \text{ N}$
 

 

 

The loads of conjugate beam are symmetrical, thus,
$F_1 = F_2 = \frac{1}{2} [ \, \frac{1}{2}(5)(3000) + \frac{1}{3}(1)(75) - \frac{1}{3}(5)(1875) \, ]$

$F_1 = F_2 = 2200 \, \text{ N}\cdot\text{m}^2$
 

For this beam, the maximum deflection will occur at the midspan.
 

 

$M_{midspan} = \frac{1}{2}(2.5)(3000)[ \frac{1}{3}(2.5) ] + \frac{1}{3}(0.5)(75)[ \frac{1}{4}(0.5) ] - \frac{1}{3}(2.5)(1875)[ \frac{1}{4}(2.5) ] - 2200(2.5)$

$M_{midspan} = -3350 \, \text{ N}\cdot\text{m}^3$
 

Therefore, the maximum deflection is
$EI~\delta_{max} = M_{midspan}$

$EI~\delta_{max} = -3350 \, \text{ N}\cdot\text{m}^3$

$EI~\delta_{max} = 3350 \, \text{ N}\cdot\text{m}^3$   below the neutral axis           answer

(Conjugate beam method using the actual moment diagram)
 

 

By symmetry of conjugate beam
$F_1 = F_2 = \frac{1}{2} [ \, \frac{2}{3}(2)(1200) + 1200(1) + \frac{2}{3}(2)(1200) \, ]$

$F_1 = F_2 = 2200 \, \text{ N}\cdot\text{m}^2$
 

The maximum deflection will occur at the midspan of this beam
 

 

$M_{midspan} = \frac{2}{3}(2)(1200)(1.25) + 0.5(1200)(0.25) - 2.5(2200)$

$M_{midspan} = -3350 \, \text{ N}\cdot\text{m}^3$
 

Therefore, the maximum deflection is
$EI~\delta_{max} = M_{midspan}$

$EI~\delta_{max} = -3350 \, \text{ N}\cdot\text{m}^3$

$EI~\delta_{max} = 3350 \, \text{ N}\cdot\text{m}^3$   below the neutral axis           okay!