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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f9835e9a66a640b1f70008c78c94ca52e96ffa62)
where
is the position along the length of the beam and
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
and
.
Taking the first variation of the functional
, 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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d4d0701f8d69eacfec62ba5ebd755852ebb178f)
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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/97572fed76bf80a8a10aa3b56193e2e53088fb57)
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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ef031e7ea225f3a1957e99c2127ba0b34f5f2470)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e3f9965092fb5a00da59cfa864533107a5ef928)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ade4c831fe06da023044af5b700ca286d759b192)
Using the principle of minimum potential energy, for the functional
to have a minimum, we must have
. 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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/912a6e7b1dd8b8127a8dfcfc5ffa6e9a446d5247)
Since
and
are arbitrary, the Euler equation
for this problem is
![{\displaystyle {(EI~y^{''})^{''}-p=0}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b15f0e71c5e90ea213a19412c76e3501eeac9ded)
and the associated boundary conditions are
![{\displaystyle {{\text{at}}~x=0~~;~~EI~y^{''}-M_{0}=0~~{\text{and}}~~(EI~y^{''})^{'}-V_{0}=0}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f1b9b70b3e2c98d644e827a7e6f6af8e97ab9d7a)
and
![{\displaystyle {{\text{at}}~x=L~~;~~EI~y^{''}-M_{L}=0~~{\text{and}}~~(EI~y^{''})^{'}-V_{L}=0}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/07352083733177769feddaecad4578fa106f3e0e)