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I have a strip of stainless steel encastree'd at one end to which is applied a constant pressure on one side, and I need to know what the deflection equation y = f(x) is at equilibrium. If the deflection was small I would be able to use one of the very well known deflection formulas which assume vertical loads, but it isn't.

I am trying to work out the bending moment M=f(x) using integrals to use: $$y(x)=\int\int_{beam} \frac{M}{EI}dx^2$$ But I'm getting more and more confused. What is the right approach to solving this problem?

Here is a diagram:

enter image description here

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  • $\begingroup$ What's your deflection if you used the normal formulas? Small is typically 1/2 the depth of the item you are deflecting. $\endgroup$ – Mark Jul 10 '15 at 16:50
  • $\begingroup$ I haven't calculated it because it would be completely off - the deflection expected is around 5 to 10 times the depth since the strip is supposed to bend under pressure. $\endgroup$ – Mister Mystère Jul 10 '15 at 16:52
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    $\begingroup$ for the sake of joy, I would try an iterational approach, finding the deflection with a beam geometry, and then calculating the moments with the deflected geometry of step 1, and continue till convergence. it may even diverge if the material is too elastic (or the force too great). $\endgroup$ – Gürkan Çetin Jul 10 '15 at 18:21
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For large deformation, there are changes to the strain-displacement relation and vertical equilibrium, but horizontal equilibrium remains unchanged.

Let N be the axial force and M the bending moment, for finite rotations, these are defined as: $$ N=EA\epsilon^o \\ M=EI\kappa $$ where $$\epsilon^o = \frac{du}{dx} + \frac{1}{2}\left(\frac{dw}{dx}\right)^{2} $$ and $$ \kappa = -\frac{\frac{d^{2}w}{dx^{2}}}{\left[1 + \left(\frac{dw}{dx}\right)^2\right]^{3/2}} $$ so vertical equilibrium is given by: $$ \frac{d^2M}{dx^2} + N\frac{d^{2}w}{dx^2} + P = 0 $$ where $w$ is the vertical deflection The solution is then the solution to this differential equation dependent upon your set boundary conditions

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  • $\begingroup$ Thanks a lot, this looks useful. Could you remind me of what u is? I assume P is the uniform pressure applied? $\endgroup$ – Mister Mystère Jul 14 '15 at 15:11
  • $\begingroup$ Glad to be of help @MisterMystere, u is the deflection in the x direction (or essentially the extension along the length of the beam), w is the deflection in the y-direction and you are correct, P is the uniform applied pressure $\endgroup$ – bern Jul 14 '15 at 15:19
  • $\begingroup$ But isn't the displacement along the centreline zero, as the distributed forces are always perpendicular to the beam? $\endgroup$ – Mister Mystère Jul 14 '15 at 15:51
  • $\begingroup$ yes, for your case of a uniformly distributed load perpendicular to the beam, there should be no axial extension. This is a boundary condition that you apply to solve the differential equation. Therefore in the epsilon^o equation the du/dx term would be zero for your case as the beam is not displacing in the x-direction. Does this answer your question? $\endgroup$ – bern Jul 14 '15 at 15:56

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