# 6.3: Equivalence of Square and Circular Plates

In the section of Chapter 6 on stiffened plates, the analogy between the response of circular and square plates was exploit to demonstrate the effectiveness of stiffeners. It was stated that stiffness of these two types of plates are similar if the arial surface was identical. We are now in the position to assess accuracy of the the earlier assertion.

Consider a clamped square plate $$2a \times 2a$$, uniformly loaded by the pressure $$p_o$$. The total potential energy of the system $$\prod$$ is

$\prod = \frac{D}{2} \int_{S} [(\kappa_x^2 + \kappa_y^2) + 2(1 − \nu)\kappa_G] ds − \int_{S} −q_ow ds$

It can be shown (Shames & Dign 1985) that for the fully clamped boundary conditions, the integral of the Gaussian curvature $$\kappa_G$$ vanishes. The expression for $$\prod$$ simplifies to

$\prod = \frac{D}{2} \int_{0}^{a} \int_{0}^{a} \left[ \frac{d^2w}{dx^2} + \frac{d^2w}{dy^2} \right] dx dy − a^2 q \int_{0}^{a} \int_{0}^{a} w dx dy \label{7.37}$

For simplicity only one quarter of the plate is considered with the origin at the plate center. As a trial deflection shape, we assume

$w(x, y) = C(x^2 − a^2 )^2 (y^2 − a^2 )^2 \label{7.38a}$

$\frac{dw}{dx} = C2(x^2 − a^2 )2x(y^2 − a^2 )^2$

$\frac{dw}{dy} = C(x^2 − a^2 )^2 2(y^2 − a^2 )2y \label{7.38c}$

It is seen that both the deflections and slopes are zero at the clamped boundary. Furthermore, the slopes at the plate center $$x = y = 0$$ vanishes, as they should due to symmetry. The maximum amplitude is at center and is equal to $$Ca^8 = w_o$$. Thus, the kinematic boundary conditions are satisfied identically for any value of the unknown constant $$C$$. Substituting the expression \ref{7.38a} - \ref{7.38c} into Equation \ref{7.37} and performing integration yields

$\prod = 9a^4DC^2 − 0.384q_oC$

According to the Ritz method, equilibrium is maintained if

$\delta \prod = \frac{\partial \prod}{\partial C} \delta C = 0$

This means that for a given load intensity and the assumed normalized shape function, the true deflection amplitude is chosen by the condition

$\frac{\partial \prod}{\partial C} = 0 \quad \text{ or } \quad 18a^4DC − 0.383q_o = 0$

Having found the amplitude $$C$$, the load-displacement relation of the square plate becomes

$w_o = \frac{p_oa^4}{47D} \label{7.42}$

The corresponding solution for the clamped circular plate is

$w_o = \frac{p_oR^4}{64D}$

The stiffnesses of both plates are identical if $$\frac{R^4}{64} = \frac{a^4}{47}$$ or if $$a = 0.92R$$. The area equivalence $$4a^2 = \pi R^2$$ gives a similar result $$a = 0.88R$$. For simplicity in the qualitative analysis throughout the present lecture notes one can approximately assume $$a = R$$. The difference between the exact and approximate solution from the area and stiffness equivalence does exist, but it is small. It is interesting that the approximate solution obtained by the Ritz method is very close to the exact series solution where the coefficient 47 in Equation \ref{7.42} should be replaced by 49.5.