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

$\mathrm{\Pi }[y(x)]=\frac{1}{2}\underset{0}{\overset{L}{\int }}EI(y^{{\text{'}}^{\prime }}{)}^2-\underset{0}{\overset{L}{\int }}pydx+M_0{y}^{\text{'}}(0)-V_{0}y(0)-M_L{y}^{\text{'}}(L)+V_{L}y(L)$

where $x$ is the position along the length of the beam and $y(x)$ is the beam's deflection curve.

File:Beam bending final.png 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 $\mathrm{\Pi }$, we have

$\delta \mathrm{\Pi }=\underset{0}{\overset{L}{\int }}EI{y}^{{\text{'}}^{\prime }}\delta y^{{\text{'}}^{\prime }}dx-\underset{0}{\overset{L}{\int }}p\delta ydx+M_0\delta y^{\text{'}}(0)-V_0\delta y(0)-M_L\delta y^{\text{'}}(L)+V_L\delta y(L)$

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

$\delta \mathrm{\Pi }={(EI{y}^{{\text{'}}^{\prime }}\delta y^{\text{'}})|}_0^L-\underset{0}{\overset{L}{\int }}(EI{y}^{{\text{'}}^{\prime }}{)}^{\text{'}}\delta y^{\text{'}}dx-\underset{0}{\overset{L}{\int }}p\delta ydx+M_0\delta y^{\text{'}}(0)-V_0\delta y(0)-M_L\delta y^{\text{'}}(L)+V_L\delta y(L)$

Integrating by parts again,

$\delta \mathrm{\Pi }={(EI{y}^{{\text{'}}^{\prime }}\delta y^{\text{'}})|}_0^L-{(EI{y}^{{\text{'}}^{\prime }}{)}^{\text{'}}\delta y|}_0^L+\underset{0}{\overset{L}{\int }}(EI{y}^{{\text{'}}^{\prime }}{)}^{{\text{'}}^{\prime }}\delta ydx-\underset{0}{\overset{L}{\int }}p\delta ydx+M_0\delta y^{\text{'}}(0)-V_0\delta y(0)-M_L\delta y^{\text{'}}(L)+V_L\delta y(L)$

Expanding out,

$\begin{array}{cc}\delta \mathrm{\Pi }=& EI{y}^{{\text{'}}^{\prime }}(L)\delta y^{\text{'}}(L)-EI{y}^{{\text{'}}^{\prime }}(0)\delta y^{\text{'}}(0)-(EI{y}^{{\text{'}}^{\prime }}{)}^{\text{'}}(L)\delta y(L)+(EI{y}^{{\text{'}}^{\prime }}{)}^{\text{'}}(0)\delta y(0)\\ & +\underset{0}{\overset{L}{\int }}(EI{y}^{{\text{'}}^{\prime }}{)}^{{\text{'}}^{\prime }}\delta ydx-\underset{0}{\overset{L}{\int }}p\delta ydx+M_0\delta y^{\text{'}}(0)-V_0\delta y(0)-M_L\delta y^{\text{'}}(L)+V_L\delta y(L)\end{array}$

Rearranging,

$\begin{array}{cc}\delta \mathrm{\Pi }=& \underset{0}{\overset{L}{\int }}[(EI{y}^{{\text{'}}^{\prime }}{)}^{{\text{'}}^{\prime }}-p]\delta ydx+[M_0-EI{y}^{{\text{'}}^{\prime }}(0)]\delta y^{\text{'}}(0)+[EI{y}^{{\text{'}}^{\prime }}(L)-M_L]\delta y^{\text{'}}(L)\\ & +[(EI{y}^{{\text{'}}^{\prime }}{)}^{\text{'}}(0)-V_0]\delta y(0)+[V_L-(EI{y}^{{\text{'}}^{\prime }}{)}^{\text{'}}(L)]\delta y(L)\end{array}$

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

$\begin{array}{cc}0=& \underset{0}{\overset{L}{\int }}[(EI{y}^{{\text{'}}^{\prime }}{)}^{{\text{'}}^{\prime }}-p]\delta ydx+[M_0-EI{y}^{{\text{'}}^{\prime }}(0)]\delta y^{\text{'}}(0)+[EI{y}^{{\text{'}}^{\prime }}(L)-M_L]\delta y^{\text{'}}(L)\\ & +[(EI{y}^{{\text{'}}^{\prime }}{)}^{\text{'}}(0)-V_0]\delta y(0)+[V_L-(EI{y}^{{\text{'}}^{\prime }}{)}^{\text{'}}(L)]\delta y(L)\end{array}$

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

$(EI{y}^{{\text{'}}^{\prime }}{)}^{{\text{'}}^{\prime }}-p=0$

and the associated boundary conditions are

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

and

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