Solution to Problem 688 | Beam Deflection by Method of Superposition

From the figure below, the total deformation at the end of overhang is

$\delta = 2\theta + \delta_1$
 

 

The rotation θ at the left support is combination of Case No. 12 and by integration of Case No. 7.
 

$\theta_L = \dfrac{ML}{3EI}$	$\theta_L = \dfrac{Pb(L^2 - b^2)}{6EIL}$

 

Solving for θ

EIθ = EIθ due to 800 N·m moment at left support - EIθ due to 400 N/m uniform load

$EI \, \theta = \dfrac{800(4)}{3} - \displaystyle \int_0^2 \dfrac{400 dx(4 - x) [ \, 4^2 - (4 - x)^2 \, ]}{6(4)}$

$EI \, \theta = \dfrac{3200}{3} - \displaystyle \dfrac{50}{3}\int_0^2 [ \,16(4 - x) - (4 - x)^3 \, ]dx$

$EI \, \theta = \dfrac{3200}{3} - \dfrac{50}{3} \left[ -8(4 - x)^2 + \dfrac{(4 - x)^4}{4} \right]_0^2$

$EI \, \theta = \dfrac{3200}{3} - \dfrac{50}{3} \left[ -8(2^2) + \dfrac{2^4}{4} \right] + \dfrac{50}{3} \left[ -8(4^2) + \dfrac{4^4}{4} \right]$

$EI \, \theta = \dfrac{3200}{3} + \dfrac{1400}{3} - \dfrac{3200}{3}$

$EI \, \theta = \dfrac{1400}{3} \, \text{ N}\cdot\text{m}^2$
 

Apply Case No. 3 for solving δ₁. From Case No. 3:

$\delta = \dfrac{w_oL^4}{8EI}$
 

Solving for δ₁:

$EI \, \delta_1 = \dfrac{400(2^4)}{8}$

$EI \, \delta_1 = 800 \, \text{ N}\cdot\text{m}^3$
 

Total deflection at the free end
$EI \, \delta = 2 EI \, \theta + EI \, \delta_1$

$EI \, \delta = 2 \left( \dfrac{1400}{3} \right) + 800$

$EI \, \delta = \dfrac{5200}{3} \, \text{ N}\cdot\text{m}^3$           answer

This problem can be done with less effort by area-moment method.
 

 

$EI \, t_{A/B} = (Area_{AB}) \, \bar{X}_A$

$EI \, t_{A/B} = -\frac{1}{3}(2)(800)(\frac{3}{2})$

$EI \, t_{A/B} = -800 \, \text{ N}\cdot\text{m}^3$
 

$EI \, t_{D/B} = (Area_{AB}) \, \bar{X}_A$

$EI \, t_{D/B} = -\frac{1}{3}(2)(800)(\frac{7}{2})$

$EI \, t_{D/B} = -\frac{5600}{3} \, \text{ N}\cdot\text{m}^3$
 

The negative sign above indicates that the elastic curve is below the tangent line.
 

$\dfrac{y_A}{2} = \dfrac{t_{D/B}}{4}$

$y_A = \frac{1}{2}t_{D/B}$

$EI \, y_A = \frac{1}{2}(EI \, t_{D/B})$

$EI \, y_A = \frac{1}{2}(\frac{5600}{3})$

$EI \, y_A = \frac{2800}{3} \, \text{ N}\cdot\text{m}^3$
 

$EI \, \delta = EI \, y_A + EI \, t_{A/B}$

$EI \, \delta = \frac{2800}{3} + 800$

$EI \, \delta = \frac{5200}{3} \, \text{ N}\cdot\text{m}^3$

$EI \, \delta = \dfrac{5200}{3} \, \text{ N}\cdot\text{m}^3$       (okay!)