Solution to Problem 696-697 | Beam Deflection by Method of Superposition

Apply Case No. 8 and Case No. 11 to find the slope at the right support.
 

 

$\theta = \dfrac{w_oL^3}{24EI} - \dfrac{ML}{3EI}$

$\theta = \dfrac{80(9^3)}{24EI} - \dfrac{3P(9)}{3EI}$

$\theta = \dfrac{2430}{EI} - \dfrac{9P}{EI}$

$3\theta = \dfrac{7290}{EI} - \dfrac{27P}{EI}$
 

Use Case No. 1 for the deflection at the free end due to concentrated load P.

$\delta = \dfrac{PL^3}{3EI}$

$\delta = \dfrac{P(3^3)}{3EI}$

$\delta = \dfrac{9P}{EI}$
 

$\delta = 3\theta$

$\dfrac{36P}{EI} = \dfrac{7290}{EI}$

$P = 202.5 \, \text{ lb}$           answer

From Solution 696,

$\theta = \dfrac{2430}{EI} - \dfrac{9P}{EI}$

$0 = \dfrac{2430}{EI} - \dfrac{9P}{EI}$

$\dfrac{9P}{EI} = \dfrac{2430}{EI}$

$P = 270 \, \text{ lb}$           answer