Elasticity/Energy methods example 3

Suppose that the cross-section of the cylinder is bounded by the curves:

${\displaystyle {\begin{aligned}x_{1}&=c\\x_{1}&=d\\x_{2}&=g(x_{1})\\x_{2}&=-g(x_{1})\end{aligned}}}$

A statically admissible Prandtl stress function for this cross-section is

${\displaystyle \phi =[x_{2}^{2}-g^{2}(x_{1})]f(x_{1})}$

with the restrictions that ${\displaystyle f(x_{1})\,}$ is twice continuously differentiable and ${\displaystyle f(c)=f(d)=0\,}$.

We seek to derive the best approximate Prandtl stress function of this form by minimizing the complementary energy.

You can show that the complementary energy per unit length of the cylinder can be expressed as

${\displaystyle \Pi ^{c}[\phi (x_{1},x_{2})]=\int _{\mathcal {S}}\left[{\frac {1}{2\mu }}(\phi _{1}^{2}+\phi _{,2}^{2})-2\alpha \phi \right]~dA}$

Plugging in the given form of ${\displaystyle \phi }$ and after some algebra, we get

${\displaystyle \Pi ^{c}[f(x_{1})]={\frac {15}{16\mu }}\int _{c}^{d}\left[{\frac {1}{2}}g^{5}(f^{'})^{2}+{\frac {5}{2}}g^{4}g^{'}ff^{'}+{\frac {15}{4}}g^{3}(g^{'}f)^{2}+{\frac {5}{4}}g^{3}f^{2}+{\frac {5}{2}}\mu \alpha g^{3}f\right]~dx_{1}}$

Taking the variation of ${\displaystyle \Pi }$, and after considerable manipulation, we get

${\displaystyle \delta \Pi ^{c}={\frac {15}{16\mu }}\int _{c}^{d}{\frac {g^{5}}{4}}\left[f^{''}-{\frac {5}{g^{2}}}(g^{2}f)^{''}+{\frac {10}{g^{2}}}f+{\frac {10\mu \alpha }{g^{2}}}\right]\delta f~dx_{1}}$

Now, if we consider the cross-section to be rectangular, then we have ${\displaystyle c=-a}$, ${\displaystyle d=a}$, and ${\displaystyle g=b}$. Therefore, the above equation reduces to

${\displaystyle \delta \Pi ^{c}={\frac {15}{16\mu }}\int _{-a}^{a}{\frac {b^{5}}{4}}\left[f^{''}-5f^{''}+{\frac {10}{b^{2}}}f+{\frac {10\mu \alpha }{b^{2}}}\right]\delta f~dx_{1}}$

Therefore, the function ${\displaystyle f}$ that minimizes ${\displaystyle \Pi ^{c}}$ satisfies the equation

${\displaystyle -4f^{''}+{\frac {10}{b^{2}}}f+{\frac {10\mu \alpha }{b^{2}}}=0}$

or,

${\displaystyle f^{''}-{\frac {5}{2b^{2}}}f-{\frac {5\mu \alpha }{2b^{2}}}=0}$

with the static admissibility conditions ${\displaystyle f(-a)=f(a)=0\,}$.

The general solution of the above equation is

${\displaystyle f=A\sinh \left({\sqrt {2.5}}~{\frac {x_{1}}{b}}\right)+B\cosh \left({\sqrt {2.5}}~{\frac {x_{1}}{b}}\right)-\mu \alpha }$

Therefore, from the BCs,

${\displaystyle A=0~~;~~B={\frac {\mu \alpha }{\cosh {{\sqrt {2.5}}~a/b}}}}$

We can substitute back into ${\displaystyle \phi }$ to get the approximate Prandtl stress function for this problem. The error between the exact and this approximate solution is generally less than 1%.