Question 15.S-P.2: For the beam and loading shown, determine (a) the equation o...

Differential Equation of the Elastic Curve. From Eq. (15.32),

\frac{d^{4}y}{dx^{4}} = – \frac{w(x)}{EI}                 (15.32)

EI \frac{d^{4}y}{dx^{4}} = -w(x) = -w_{0}  sin \frac{πx}{L}                                      (1)

Integrate Eq. (1) twice:

EI \frac{d^{3}y}{dx^{3}} = V = +w_{0} \frac{L}{π} cos  \frac{πx}{L} + C_{1}                         (2)

EI \frac{d^{2}y}{dx^{2}} = M = +w_{0} \frac{L^{2}}{π^{2}}  sin  \frac{πx}{L} + C_{1}x + C_{2}                             (3)

Boundary Conditions:

[x = 0, M = 0]: From Eq. (3), we find        C_{2} = 0
[x = L, M = 0]: Again using Eq. (3), we write

0 = w_{0} \frac{L^{2}}{π^{2}} sin π + C_{1}L                   C_{1} = 0

Thus:

EI \frac{d^{2}y}{dx^{2}} = +w_{0} \frac{L^{2}}{π^{2}}   sin  \frac{πx}{L}                      (4)

Integrate Eq. (4) twice:

EI \frac{dy}{dx} = EI  θ = -w_{0} \frac{L^{3}}{π^{3}}  cos  \frac{πx}{L} + C_{3}                             (5)

EI  y = -w_{0} \frac{L^{4}}{π^{4}}  sin  \frac{πx}{L} + C_{3}x + C_{4}                           (6)

Boundary Conditions:

[x = 0, y = 0]: Using Eq. (6), we find    C_{4} = 0
[x = L, y = 0]: Again using Eq. (6), we find     C_{3} = 0

a. Equation of Elastic Curve                                        EIy = -w_{0} \frac{L^{4}}{π^{4}}  sin  \frac{πx}{L}

b. Slope at End A. For x = 0, we have

EI  θ_{A} = -w_{0} \frac{L^{3}}{π^{3}}   cos  0              θ_{A} = \frac{w_{0}L^{3}}{π^{3}EI} ⦪

c. Maximum Deflection. For x = \frac{1}{2} L

ELy_{max} = -w_{0} \frac{L^{4}}{π^{4}}  sin  \frac{π}{2}                               y_{max} = \frac{w_{0}L^{4}}{π^{4}EI} ↓