# How to calculate the thickness of a circular plate centrally supported? [closed]

I wanted to calculate the required thickness of a circular plate of radius 0.06m made of A-36 steel. The plate is supported by a rod at its center welded together with the circular weld of radius 6mm. A uniform load of 981N is applied on the top of the plat. I wanted to calculate the minimum thickness of the plate so that the plate may not bend. Any help would be highly appreciated.

• Welcome to Engineering! This looks like a homework question. In order for such questions to be answered in this site, we need you to add details describing the precise problem you're having. What have you tried to solve this yourself? Also, the plate will always bend, even if you just consider it's own self-weight. The question is how much bending is allowed. Do you have a maximum allowable displacement or is the limit the thickness at which the plate won't collapse? Please edit your question to include this information. – Wasabi Oct 23 '16 at 19:44
• The plate will always bend. The question is how much is allowed. – ja72 Oct 24 '16 at 18:52

The governing equation for a circular ring is

$$\frac{{\rm d}}{{\rm d}r}\left[\frac{1}{r}\frac{{\rm d}}{{\rm d}r}\left(r\frac{{\rm d}y}{{\rm d}r}\right)\right]=\mbox{-}\frac{Q(r)}{D}$$

where $y$ is the axial deflection, $D=\frac{E\,t^{3}}{12(1-\nu^{2})}$ is a material constant with thickness $t$, and $Q(r)$ is the shear force rate at a particular radius. It is defined as $Q(r)=\frac{\mbox{load inside radius } {r}}{\mbox{perimeter at }{r}}$

For a uniform pressure $q = \frac{F}{\pi R^2}$ on the top surface the shear rate is thus $Q = \frac{F \;r}{2 \pi R^2}$

If only the center is fixed, then the deflection curve is solved by

$$y(r) = \frac{3F\,r^{2}(1-\nu)}{16\:\pi R^{2}\,E\,t^{3}}\left(2R^{2}(\nu+3)-r^{2}(\nu+1)\right)$$

Finally the deflection on the outer edge is $$\boxed{ \delta=\frac{3F\,R^{2}(1-\nu)(\nu+5)}{16\,\pi\,E\,t^{3}} }$$