Introduction to 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.

Beam bending problem

• (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

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