# 6.4: Design Concept for Plates

The plates loaded in the transverse direction can be design for:

• Stiffness
• Strength(yielding or plastic collapse)
• fracture

Plastic collapse and fracture of ductile materials will be covered in separate lectures. Stiffness is a global property of the plate and is the ratio of force to displacement. For a uniformly loaded plate the stiffness is defined as

$K = \frac{\pi R^2 p_o}{w_o}$

For the clamped plate with $$\nu = 0.3$$

$K = 18.5 \equiv \frac{h^3}{R^2} \left[\frac{N}{m}\right]$

Stiffness can be controlled by choosing a suitable material $$(E)$$, thickness $$(h)$$ and distance between support $$(R)$$. The boundary conditions enter through the numerical coefficient. The concept of optimum design includes the weight and cost of a given structure. Leaving the complex issue of cost, the wight can be easily included by calculating stiffness per unit weight. The wight of the circular plate $$W = \pi R^2 \rho$$, so the stiffness per unit weight is

$\bar{K} = \frac{K}{W} = \frac{\pi R^2 p_o}{\pi R^2 \rho w_o} = \frac{p_o}{\rho w_o}$

In the case of a clamped plate

$\bar{K} = 4.8 \frac{E}{\rho}\frac{h^2}{R^4} \left[\frac{N}{mKg}\right]$

The dependance of $$K$$ and $$\bar{K}$$ on $$h$$ and $$R$$ is different. While the stiffness favors thicker plates, the stiffness per unit weight increases faster with a large radius. The effect of the ratio $$E/\rho$$ can be shown on the example of steel and aluminum plates, see Table ($$\PageIndex{1}$$).

 E[GPa] $$\rho[g/cm^3]$$ $$E/\rho$$ Steel 2.1 7.8 3.7 Al 0.8 2.8 3.5

Aluminum alloys seem much lighter but they lose elasticity modulus in the same proportion. It is seen that there are not much gain in the stiffness per unit weight by replacing steel by aluminum. So, what else could be done to increase plate stiffness? The answer is:

• Sandwich plates
• Stiffened plates

Each of the above concept is studied separately.