# 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

${\displaystyle \Pi [y(x)]={\frac {1}{2}}\int _{0}^{L}EI(y^{''})^{2}-\int _{0}^{L}p~y~dx+M_{0}~y^{'}(0)-V_{0}~y(0)-M_{L}~y^{'}(L)+V_{L}~y(L)}$

where ${\displaystyle x}$ is the position along the length of the beam and ${\displaystyle y(x)}$ 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 ${\displaystyle x=0}$ and ${\displaystyle x=L}$.

## Solution:

Taking the first variation of the functional ${\displaystyle \Pi }$, we have

${\displaystyle \delta \Pi =\int _{0}^{L}EI~y^{''}\delta y^{''}~dx-\int _{0}^{L}p~\delta y~dx+M_{0}~\delta y^{'}(0)-V_{0}~\delta y(0)-M_{L}~\delta y^{'}(L)+V_{L}~\delta y(L)}$

Integrating the first terms of the above expression by parts, we have,

${\displaystyle \delta \Pi =\left.(EI~y^{''}\delta y^{'})\right|_{0}^{L}-\int _{0}^{L}(EI~y^{''})^{'}\delta y^{'}~dx-\int _{0}^{L}p~\delta y~dx+M_{0}~\delta y^{'}(0)-V_{0}~\delta y(0)-M_{L}~\delta y^{'}(L)+V_{L}~\delta y(L)}$

Integrating by parts again,

${\displaystyle \delta \Pi =\left.(EI~y^{''}\delta y^{'})\right|_{0}^{L}-\left.(EI~y^{''})^{'}\delta y\right|_{0}^{L}+\int _{0}^{L}(EI~y^{''})^{''}\delta y~dx-\int _{0}^{L}p~\delta y~dx+M_{0}~\delta y^{'}(0)-V_{0}~\delta y(0)-M_{L}~\delta y^{'}(L)+V_{L}~\delta y(L)}$

Expanding out,

{\displaystyle {\begin{aligned}\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{aligned}}}

Rearranging,

{\displaystyle {\begin{aligned}\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{aligned}}}

Using the principle of minimum potential energy, for the functional ${\displaystyle \Pi }$ to have a minimum, we must have ${\displaystyle \delta \Pi =0}$. Therefore, we have

{\displaystyle {\begin{aligned}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{aligned}}}

Since ${\displaystyle \delta y}$ and ${\displaystyle \delta y^{'}}$ are arbitrary, the Euler equation for this problem is

${\displaystyle {(EI~y^{''})^{''}-p=0}}$

and the associated boundary conditions are

${\displaystyle {{\text{at}}~x=0~~;~~EI~y^{''}-M_{0}=0~~{\text{and}}~~(EI~y^{''})^{'}-V_{0}=0}}$

and

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