# Fixing the Poisson equation to match the deformation of elastic sheet with experimental observation

I am working on the calculation of the deformation of a circular elastic sheet with radius $$R=1.2~m$$ when a plate with mass $$M$$ and radius $$r_0 = 4~cm$$ is sitting in the center of the sheet.

I used the wave equation without the $$z_{tt}$$ so that $$v_s^2 \Delta z = f(\mathbf{r})$$ where $$v_s$$ is the wave speed on the sheet. Here $$f=g= 9.81~m/s^2$$ and for the region the plate is sitting, I scaled the load as $$f=\tilde{g}=(1+(M/A)/\rho)g$$ where $$A$$ is the plate area $$A=\pi r_0^2$$ and $$\rho = 137~g/m^2$$ is the area density of the sheet.

Using the polar form of the Laplacian $$\Delta z = \frac{1}{r} \frac{\partial}{\partial r}(r \frac{\partial z}{\partial r}) + \frac{1}{r^2} \frac{\partial^2 z}{\partial \varphi^2}$$ and apply the axial symmetry, the general solution to $$v_s^2 \Delta z = g$$ is found to be $$z(r)=\frac{g}{4v_s^2}r^2 + C_1 \log{r} + C_2$$. I patched the turning point $$r=r_0$$ with

$$z(r_0^-)=z(r_0^+)$$ $$z'(r_0^-)=z'(r_0^+)$$ and used a boundary condition $$z(R)=0$$. This gives me the depression $$D \equiv 0-z(r_0)$$ as a function of the plate mass $$M$$: $$D(M) = \frac{g \log{(R/r_0)}}{2\pi \rho_m v_s^2} \cdot M + \frac{g}{4v_s^2} (R^2-r_0^2)$$ which is a very simple linear response.

The problem came after I plugged in the experimentally measured $$v_s^{expt} = 4.4~m/s$$. It turns out this value predicts a way large deformation $$D \sim 40~cm$$, while adopting $$v_s^{fit} = 8.0~m/s$$ gives a satisfactory prediction.

My doubt is did I missed a constant in the Poisson equation? Maybe it should be $$v_s^2 \Delta z = f/4$$ instead of $$v_s^2 \Delta z = f$$?

I'd really appreciate your kind help!

• Are you missing a $r \ln\left( r \right)$ term also? Commented Oct 8, 2018 at 22:47
• @ja72 Thanks for your attention! Do you mean the general solution should also include a $r \ln{(r)}$ term? Commented Oct 9, 2018 at 0:42
• Yes, the membrane solutions I have seen also have a term like that. Commented Oct 9, 2018 at 13:17
• @ja72, $r \ln(r)$ does not satisfy the differential equation. Commented May 28, 2019 at 21:02
• I would say that the equation is $T \nabla z = f$, where $T$ is the tension in the membrane. Also, for deflections of 40 cm the equation should be nonlinear. Commented May 28, 2019 at 21:05

Your derivation seems valid to me. Here is a derivation based on a more physical intuition yielding the same result. The weight P of the mass must be balanced at $$r=R$$, by a lineic reaction force $$F_{y}$$ such that $$2\pi R F_{y}=-P$$. This is actually true if we perform a virtual cut of the elastic sheet along any circumference at $$r \geq r_{0}$$, so if we introduce the tension of the elastic sheet $$T$$, for small deflections and neglecting the weight of the sheet for the time being: $$2\pi r T \frac{\partial w}{\partial r} = -P \tag{1}$$ Substituting $$T=\rho V_{s}^{2}$$ and integrating we find: $$w=\frac{-P}{2 \pi \rho V_{s}^{2}} \ln(r) + C_{1}$$ The boundary condition at $$r=R$$ yields at once: $$w=-\frac{P}{ 2 \pi \rho V_{s}^{2}} \ln (\frac{r}{R})$$ and finally: $$d=w(r_{0})= -\frac{mg}{2\pi\rho V_{s}^{2}} \ln \left(\frac{r_{0}}{R}\right)$$ which is exactly the first term of your solution, the second term accounts for the deflection of the sheet due to ist own weight and is an offset in the deflection with mass. To explain the difference with experimental data:
• the $$V_{s}$$ measurement you have might be inconsistent (how was $$V_{s}$$ measured? If you measured the frequency of a fundamental vibration mode, are you sure you associated this frequency with the correct mode? )
• Thank you so much! I actually am worried about the measurement of wave speed. I simply tapped at a spot on the membrane close to the edge of the circle and see the time $T_0$ it takes to traverse the diameter $2R$. It seems to me $T_0/(2R)$. It seems to me this makes sense. But there may be a problem. I will be very happy if you tell me this is wrong. I also wonder what will be a way to generate a fundamental vibration mode? Commented Apr 13, 2020 at 15:23
• I'm not sure I understand the way you measure the speed $V_{s}$. If you hit the elastic sheet in its center just a like drum, you will most likely excite the most fundamental mode. From the wiki page on membrane vibration, then $V_{s}\approx \omega_{01}(R/2.40483)$. Of course, this assumes you may somehow experimentally record the transient displacement signal and measure $\omega_{01}$ but I cannot think of better way to yield a consistent $V_{s}$. Commented Apr 13, 2020 at 16:13
• After I tapped a spot on the membrane (almost on the circumference), I see a single wavefront spreading out and record the time $T_0$ it takes the wavefront reaches the other end of the diameter $d=2R$. Then I assume $V_s \approx 2R/T_0$. Commented Apr 13, 2020 at 18:33
• I actually have thought about if the speed of the wavefront $2R/T_0$ is actually $V_s$ so I have simulated the wave propagation using the wave equation $V_s^2\Delta z= g + z_{,tt}$ and it turns out the speed of the wave front is indeed $V_s$. Commented Apr 13, 2020 at 20:18
• Thank you for accepting my answer, although your problem doesn't seem to resolve. The way you measured $V_{s}$ seems fine by me. Can you vary the tension on the membrane and measure the effect it has both on $V_{s}$ and static deflection? (i.e. does the discrepancy decrease when the tension is increased?) Commented Apr 17, 2020 at 15:13