I am trying to solve for the deflection $w$ of a rectangular membrane with tension $T$ acting on two opposite edges. These two edges are fixed i.e. $w = 0$. The other two opposite edges are free. The size of the membrane is $L$x$L$. There is a pressure $P$ acting on an area $\frac{L}{3}$x$\frac{L}{3}$ at the centre of the membrane.

A schematic is presented below:

enter image description here

If the tension was acting on four edges and they were fixed. The equations and boundary conditions could be written as: $$T\frac{\partial^2w}{\partial x^2}+T\frac{\partial^2w}{\partial y^2} = p$$ with boundary conditions: $$w(x=0) = w(x=L) = w(y=0) = w(y=L) = 0$$

I do not know how to modify these equations and boundary conditions for my problem.


2 Answers 2


This answer expands upon the approach suggested by @setun-90 and shows that the algebra gets quite complicated. I'd suggest a Green's function approach instead.


  • isotropic
  • deflection of membrane is small
  • applied tension large so that its variation due to deflection is small

Governing equation

Let the tension in the $x, y$-directions be $T_x, T_y$ per unit length, respectively. Let $w$ be the lateral deflection. The resultant in the $w$-direction due to the tension $T_x$ on the $dy$-edge of a small element of area $(dx \times dy)$ is $$ R_x := (T_x dy) \left(\theta_x + \frac{\partial \theta_x}{\partial x}\, dx\right) - (T_x dy)\, \theta_x = T_x\, \frac{\partial \theta_x}{\partial x}\, dx dy $$ where we have used $\sin\theta_x \approx \theta_x$ for small $\theta_x$ and $ \theta_x := \frac{\partial w}{\partial x} \,. $ Similarly, the resultant in the $w$-direction due to $T_y$ is $$ R_y := (T_y dx) \left(\theta_y + \frac{\partial \theta_y}{\partial y}\, dy\right) - (T_y dx)\, \theta_y = T_y\, \frac{\partial \theta_y}{\partial y}\, dx dy $$ where $ \theta_y := \frac{\partial w}{\partial y} \,. $ Therefore, the total lateral force due to $T_x$ and $T_y$ is $$ R = R_x + R_y = \left(T_x\, \frac{\partial \theta_x}{\partial x} + T_y\, \frac{\partial \theta_y}{\partial y}\right) dx dy \,. $$ This force is balanced by the applied lateral pressure, $p(x,y)$, and we have $$ p(x, y)\, dx dy + \left(T_x\, \frac{\partial \theta_x}{\partial x} + T_y\, \frac{\partial \theta_y}{\partial y}\right) dx dy = 0 \,. $$ Plugging in the definitions of $\theta_x$ and $\theta_y$, and eliminating $dx dy$, we get the governing equation get $$ \boxed{ p(x, y) + T_x\, \frac{\partial^2 w}{\partial x^2} + T_y\, \frac{\partial^2 w}{\partial y^2} = 0 \,.} $$

Case where $T_x = T$ and $T_y = 0$

In this case, the governing equation reduces to $$ \frac{\partial^2 w}{\partial x^2} = -\frac{p(x,y)}{T} \,. $$ Integrating once, we have $$ \frac{\partial w}{\partial x} = -\int\frac{p(x,y)}{T}\,dx + F(y) + F_c $$ where $F(y)$ is an integration function and $F_c$ is an integration constant.
Integrating again $$\boxed{ w(x, y) = \int\left[-\int\frac{p(x,y)}{T}\,dx + F(y) + F_c\right] dx + G(y) + G_c } $$ where $G(y)$ is another integration function and $G_c$ is an integration constant.

Case where $p(x,y)$ is a discontinuous function

Consider the case where $$ p(x,y) = \begin{cases} 0 & \text{for} \quad 0 \le x < \tfrac{L}{3}, 0 \le y \le L \\ 0 & \text{for} \quad \tfrac{L}{3} \le x \le \tfrac{2L}{3}, 0 \le y < \tfrac{L}{3} \\ P & \text{for} \quad \tfrac{L}{3} \le x \le \tfrac{2L}{3} , \tfrac{L}{3} \le y \le \tfrac{2L}{3}\\ 0 & \text{for} \quad \tfrac{L}{3} \le x \le \tfrac{2L}{3}, \tfrac{2L}{3} < y \le L \\ 0 & \text{for} \quad \tfrac{2L}{3} < x \le L, 0 \le y \le L \end{cases} $$ Then, $$ \int \frac{p(x,y)}{T}\,dx = \begin{cases} 0 & \text{for} \quad 0 \le x < \tfrac{L}{3}, 0 \le y \le L \\ 0 & \text{for} \quad \tfrac{L}{3} \le x \le \tfrac{2L}{3}, 0 \le y < \tfrac{L}{3} \\ \tfrac{Px}{T} & \text{for} \quad \tfrac{L}{3} \le x \le \tfrac{2L}{3}, \tfrac{L}{3} \le y \le \tfrac{2L}{3}\\ 0 & \text{for} \quad \tfrac{L}{3} \le x \le \tfrac{2L}{3}, \tfrac{2L}{3} < y \le L \\ 0 & \text{for} \quad \tfrac{2L}{3} < x \le L , 0 \le y \le L \end{cases} $$ This leads to $$ \frac{\partial w}{\partial x} = \begin{cases} F_1(y) + F_{c1} & \text{for} \quad 0 \le x < \tfrac{L}{3}, 0 \le y \le L \\ F_2(y) + F_{c2} & \text{for} \quad \tfrac{L}{3} \le x \le \tfrac{2L}{3}, 0 \le y < \tfrac{L}{3} \\ -\tfrac{Px}{T} + F_3(y) + F_{c3} & \text{for} \quad \tfrac{L}{3} \le x \le \tfrac{2L}{3}, \tfrac{L}{3} \le y \le \tfrac{2L}{3}\\ F_4(y) + F_{c4} & \text{for} \quad \tfrac{L}{3} \le x \le \tfrac{2L}{3}, \tfrac{2L}{3} < y \le L \\ F_5(y) + F_{c5} & \text{for} \quad \tfrac{2L}{3} < x \le L , 0 \le y \le L \end{cases} $$ From the second integration: $$\boxed{ w(x,y) = \begin{cases} [F_1(y) + F_{c1}]\,x + G_1(y) + G_{1c} & \text{for} \quad 0 \le x < \tfrac{L}{3}, 0 \le y \le L \\ [F_2(y) + F_{c2}]\,x + G_2(y) + G_{2c}& \text{for} \quad \tfrac{L}{3} \le x \le \tfrac{2L}{3}, 0 \le y < \tfrac{L}{3} \\ -\frac{Px^2}{2T} + [F_3(y) + F_{c3}]\,x + G_3(y) + G_{3c} & \text{for} \quad \tfrac{L}{3} \le x \le \tfrac{2L}{3}, \tfrac{L}{3} \le y \le \tfrac{2L}{3}\\ [F_4(y) + F_{c4}]\,x + G_4(y) + G_{4c}& \text{for} \quad \tfrac{L}{3} \le x \le \tfrac{2L}{3}, \tfrac{2L}{3} < y \le L \\ [F_5(y) + F_{c5}]\,x + G_5(y) + G_{5c}& \text{for} \quad \tfrac{2L}{3} < x \le L , 0 \le y \le L \end{cases}} $$ The 20 unknown quantities in this solution need 20 boundary conditions, making the algebra quite complex.

Green's function approach

An alternative is to use the Green's function approach in which we start with an assumed sinusoidal lateral load $$ p(x,y) = \sum_{m=1}^\infty \sum_{n=1}^\infty p_0 \sin \frac{m\pi x}{a} \sin \frac{n\pi y}{b} $$ where $a$ and $b$ are the dimensions of the membrane. We can then specialize the solution to the case where the applied load is a point load at the location $(\xi,\eta)$. Let $K(x,y,\xi,\eta)$ be this solution for a point load. Then the total deflection can be computed using $$ w(x,y) = \iint_A p(\chi,\eta) K(x, y, \chi, \eta) \,d\chi d\eta \,. $$ Details of this approach can be found in Section 29 of "Theory of Plates and Shells" by Timoshenko and Woinowsky-Krieger.

  • $\begingroup$ I think you still need to split the domain into two regions (with $P$ and without) when using Green's function. $\endgroup$ Jun 2, 2016 at 18:25
  • $\begingroup$ @ja72: That's correct. The $(\xi,\eta)$ coordinates determine that region. However, the complicated matching of boundary conditions at the edges and corners can be avoided because you are just integrating over a square. $\endgroup$ Jun 2, 2016 at 20:07

For the sake of a solution:
$w$ can be considered a function only of $x$ since two sides are bound. Therefore ${\partial w\over\partial y}=0$ therefore ${\partial^2w\over\partial y^2}=0$.

Then you have three regions ($[0;{L\over3}],[{L\over3};{2L\over3}],[{2L\over3};L]$, called $R_1$,$R_2$ and $R_3$):

  • $x\in R_1\implies P=0 \\ \implies {\partial^2w_1\over\partial x^2}(x)=0\implies{\partial w_1\over\partial x}(x)=A_1\\ \implies w_1(x)=A_1x+B_1$
  • $x\in R_2\implies P\ne 0 \\ \implies {\partial^2w_2\over\partial x^2}(x)={P\over T}\implies{\partial w_2\over\partial x}(x)={P\over T}x+B_2\\ \implies w_2(x)={P\over2T}x^2+B_2x+C$
  • $x\in R_3\implies P=0 \\ \implies {\partial^2w_3\over\partial x^2}(x)=0\implies{\partial w_3\over\partial x}(x)=A_3\\ \implies w_3(x)=A_3x+B_3$

Finally, you need to join the functions together (continuity, derivatives and function itself), which will take an enormous amount of time in this case (you're basically dealing with a spline here).

  • $\begingroup$ I wasn't sure about $$\frac{\partial^2w}{\partial y^2} = 0$$ but I think you are right. Other than that since P is only applied in a fraction of the membrane, we need to break down the problem into two parts: one where P is non zero and other where it is zero. Thanks for the answer. $\endgroup$
    – user163917
    May 26, 2016 at 15:14
  • $\begingroup$ Yes I know about the split region. I'm working on it, but it's taking a long time (continuity & boundary conditions). But the method for this problem basically starts with noting dw/dy = 0. $\endgroup$
    – setun-90
    May 26, 2016 at 15:19
  • $\begingroup$ I don't think $\dfrac{\partial^2w}{\partial y^2}=0$ is valid in this case since the load isn't applied along all of $y$. The deflection at $x=0, y=\dfrac{L}{2}$ will be different than that at $x=y=\dfrac{L}{2}$. $\endgroup$
    – Wasabi
    May 26, 2016 at 22:19
  • $\begingroup$ ...I'm sorry, I failed to understand your illustration. Did you mean $x={L\over2},y=0$ or $x=y=0$ (since we are talking about y direction)? In any case, I unfortunately can only offer this solution, as multivariable differential equations escape my knowledge (strangely, we haven't yet seen any examples in second year). $\endgroup$
    – setun-90
    May 26, 2016 at 22:30
  • $\begingroup$ @Wasabi Do you think $$\frac{\partial^2w}{\partial y^2} = 0$$ is valid for region where pressure is not acting? $\endgroup$
    – user163917
    May 27, 2016 at 16:11

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