I'm an electrical engineer, so please bear with me :). I came across this situation and I tried to find some equations to solve it, but all I could find was some Finite Element Theory, and I was wondering, if there is an easier way to solve the following situation: Shells

A thin shell is supported in the middle and a force is applied on the four corners (by a screw or a bolt for instance), or (I believe it is equivalent) a thin shell is supported in one corner and a force acts on the other corner. I am interested in finding the resulting bending (assuming a linear material) or finding the necessary strength of the material to avoid breaking. Can someone point me to a textbook with a (perhaps simplified) solution, or is it necessary to use a Finite Element software?

Thanks in advance!

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    $\begingroup$ It's perhaps simpler to turn it around, a thin shelf supported at the corners with a point load in the middle. $\endgroup$ Commented Jun 20, 2017 at 15:03
  • $\begingroup$ not sure if this applies or not and may be dependent on your software. In finite element analysis you have plate elements and shell elements. If I recall correctly, plate elements will calculate bending forces, where a shell element will not. $\endgroup$
    – Forward Ed
    Commented Sep 28, 2017 at 23:33

1 Answer 1


You'd do well with minimum potential energy here. It's a wild ride, hang on.

The potential energy associated with a set of stresses and strains in a flat panel of thickness t is:

$$\frac{1}{2}\int\int_A\int^{t}_{0}(\sigma_x\epsilon_x + \sigma_y\epsilon_y + \sigma_{xz}2\epsilon_{xz}+\sigma_{yz}2\epsilon_{yz}+\sigma_{xy}2\epsilon_{xy}dzdA - \int\int_Ap(x,y)w(x,y)dA$$

Converting to displacements and defining the following functions $w(x,y)$ as the displacement of the beam, $u_0(x,y)$ as the x-ward displacement of the shell, midplane, $v_0(x,y)$ as the y-yard displacement of the shell, and assigning some constants based off the modulus, E, and the poisson's ratio, $\nu$ $$A = \frac{Et}{1-\nu^2} $$ $$G = \frac{Et}{2(1+\nu)} $$ $$D = \frac{Et^3}{1-\nu^2} $$ $$H = \frac{Et^3}{2(1+\nu)} $$

Then the potential energy is simply

$$V = \frac{1}{2}\iint_A\left[ A \left((\frac{\partial u_0}{\partial x})^2 + 2\nu\frac{\partial u_0}{\partial x}\frac{\partial v_0}{\partial y} + (\frac{\partial v_0}{\partial y})^2 \right) + D\left((\frac{\partial^2 w}{\partial x^2})^2 + 2\nu\frac{\partial^2 w}{\partial x^2}\frac{\partial^2 w}{\partial y^2} + \frac{\partial^2 w}{\partial y^2}^2 \right) + G\left((\frac{\partial u_0}{\partial y})^2 + 2\nu\frac{\partial u_0}{\partial y}\frac{\partial v_0}{\partial x} + (\frac{\partial v_0}{\partial x})^2 \right) + H\left((\frac{\partial^2 w}{\partial y^2})^2 + 2\nu\frac{\partial^2 w}{\partial y^2}\frac{\partial^2 w}{\partial x^2} + \frac{\partial^2 w}{\partial x^2}^2 \right)\right]dydx - Fw(\frac{L}{2},\frac{W}{2}) - Fw(\frac{-L}{2},\frac{-W}{2}) - Fw(\frac{-L}{2},\frac{W}{2}) - Fw(\frac{L}{2},\frac{-W}{2})+ 4Fw(0,0)$$

Then the procedure is to guess! Trying a bunch of different polynomials, one that is bound to fit with the boundary conditions (or you relax a boundary condition) to find one that fits. My suggestion is polynomials, but trigonometric functions have been used before as well. The goal is to find a simple function that satisfies the conditions, then perform variance with the arbitrary coefficients to find the best coefficients to minimize the potential energy.

As a step, I assume the following, which satisfy many of the boundary conditions: $w(x,y)=Kx^2y^2$, $u_0(x,y)=B_1x+B_2y$, $v_0(x,y)=C_1x+C_2y$

Plugging in above, we achieve a function for the potential function. Focusing only on the w(x,y) function, we find

$$V = \frac{K^2(D+H)}{1440}(9L^5W +9LW^5+10\nu L^3W^3)-FKLW + other_stuff$$

Differentiating V with respect to the arbitrary constant K, we find that:

$$K = \frac{720F}{(D+H)(9L^4+9W^4+10\nu L^2W^2)}$$

Similar work with the other functions will find that $B_1 = 0, B_2 = 0, C_1 = 0, C_2 = 0$. So, we know the deflection of the panel. From there, it's simple enough to follow appropriate formulas to derive the stresses.

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    $\begingroup$ So I'm reading this equation as "Just use finite element method". That looks spooky. $\endgroup$
    – JMac
    Commented Jun 20, 2017 at 19:01
  • $\begingroup$ Perhaps when I play around a bit more, I can come up with a simple demonstration. It's a long answer, but it can be a good one at the end. $\endgroup$
    – Mark
    Commented Jun 20, 2017 at 20:13
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    $\begingroup$ There is no easy way around it... for a good classic text on the matter look for "Leissa - Vibration of Shells", I think you can find free copies on the web $\endgroup$
    – hdrz
    Commented Jun 21, 2017 at 6:58
  • $\begingroup$ I'm thinking K|x||y| may be a better fit. But yeah, not a guaranteed match either. $\endgroup$
    – Mark
    Commented Jun 21, 2017 at 7:26
  • $\begingroup$ I'd go with 'just use finite element method'. In the good old days there were standard charts and solutions you could look up. I expect 'Roark' has one (more formally: Roark's Formulas for Stress and Strain, a book of equations about stress and strain) though I'm temporarily out of reach of my copy. These days, it's probably quicker and easier to run an FE analysis, even than applying a standard result, certainly than doing a double integration algebraically. $\endgroup$
    – achrn
    Commented Jun 21, 2017 at 10:31

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