Large deflection of a cantilever beam with distributed normal load

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:

• What's your deflection if you used the normal formulas? Small is typically 1/2 the depth of the item you are deflecting.
– Mark
Commented Jul 10, 2015 at 16:50
• 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. Commented Jul 10, 2015 at 16:52
• 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). Commented Jul 10, 2015 at 18:21

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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