Elasticity/Sample final 5

Sample Final Exam Problem 5
Assuming that plane sections remain plane, it can be shown that the potential energy functional for a beam in bending is expressible as

\Pi[y(x)] = \frac{1}{2}\int_0^L EI(y^{''})^2 - \int_0^L p~y~ dx + M_0~ y^{'}(0) - V_0~ y(0) - M_L~ y^{'}(L) + V_L~ y(L) $$ where $$x$$ is the position along the length of the beam and $$y(x)$$ is the beam's deflection curve.
 * (a) Find the Euler equation for the beam using the principle of minimum potential energy.
 * (b) Find the associated boundary conditions at $$x = 0$$ and $$x = L$$.

Solution:
Taking the first variation of the functional $$\Pi$$, we have

\delta\Pi = \int_0^L EI~y^{}\delta y^{}~dx - \int_0^L p~\delta y~dx + M_0~ \delta y^{'}(0) - V_0~ \delta y(0) - M_L~ \delta y^{'}(L) + V_L~ \delta y(L) $$ Integrating the first terms of the above expression by parts, we have,

\delta\Pi = \left.(EI~y^{''}\delta y^{'})\right|_0^L - \int_0^L (EI~y^{''})^{'}\delta y^{'}~dx - \int_0^L p~\delta y~dx + M_0~ \delta y^{'}(0) - V_0~ \delta y(0) - M_L~ \delta y^{'}(L) + V_L~ \delta y(L) $$ Integrating by parts again,

\delta\Pi = \left.(EI~y^{''}\delta y^{'})\right|_0^L - \left.(EI~y^{''})^{'}\delta y\right|_0^L + \int_0^L (EI~y^{})^{}\delta y~dx - \int_0^L p~\delta y~dx + M_0~ \delta y^{'}(0) - V_0~ \delta y(0) - M_L~ \delta y^{'}(L) + V_L~ \delta y(L) $$ Expanding out,
 * $$\begin{align}

\delta\Pi = & EI~y^{}(L)\delta y^{'}(L) - EI~y^{}(0)\delta y^{'}(0) - (EI~y^{})^{'}(L)\delta y(L) + (EI~y^{})^{'}(0)\delta y(0) \\ & + \int_0^L (EI~y^{})^{}\delta y~dx - \int_0^L p~\delta y~dx + M_0~ \delta y^{'}(0) - V_0~ \delta y(0) - M_L~ \delta y^{'}(L) + V_L~ \delta y(L) \end{align}$$ Rearranging,
 * $$\begin{align}

\delta\Pi = &\int_0^L \left[(EI~y^{})^{} - p\right]\delta y~dx + \left[M_0 - EI~y^{''}(0)\right] \delta y^{'}(0) + \left[EI~y^{''}(L)- M_L\right] \delta y^{'}(L) \\ & + \left[(EI~y^{''})^{'}(0) - V_0\right]~ \delta y(0) + \left[V_L - (EI~y^{''})^{'}(L)\right]~ \delta y(L) \end{align}$$ Using the principle of minimum potential energy, for the functional $$\Pi$$ to have a minimum, we must have $$\delta\Pi = 0$$. Therefore, we have
 * $$\begin{align}

0 = &\int_0^L \left[(EI~y^{})^{} - p\right]\delta y~dx + \left[M_0 - EI~y^{''}(0)\right] \delta y^{'}(0) + \left[EI~y^{''}(L)- M_L\right] \delta y^{'}(L) \\ & + \left[(EI~y^{''})^{'}(0) - V_0\right]~ \delta y(0) + \left[V_L - (EI~y^{''})^{'}(L)\right]~ \delta y(L) \end{align}$$ Since $$\delta y$$ and $$\delta y^{'}$$ are arbitrary, the Euler equation for this problem is

{(EI~y^{})^{} - p = 0} $$ and the associated boundary conditions are

{ \text{at}~x = 0; EI~y^{''} - M_0 = 0 \text{and} (EI~y^{''})^{'} - V_0 = 0 } $$ and

{ \text{at}~x = L; EI~y^{''} - M_L = 0 \text{and} (EI~y^{''})^{'} - V_L = 0 } $$