Circular Plates Symmetrically Loaded.Consider a Diametral Section through a plate of thickness . is the centre of the plate and and are the principal axes in the plane of the diagram. The axis is perpendicular to the screen.
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Uniformly Loaded, Edge Freely Supported
- and since and can not be infinite at the centre, then from Equation (14) at , , and therefore from equation (15).
Using equations (9) and (14). At , therefore
Thus,Central Deflection = at . Thus,Eliminating by substitution from Equation (10)From Equation (7)And atFrom Equation (8)As above when , . Therefore Note: the Maximum Stresses occur at the centre.
Uniformly Loaded With The Edge Clamped
- As in the last case, and at , . Therefore from equation(15)
Central Load P, Edge Freely Supported (w=0)
- At , therefore from equation (14) and . From equation(15)
At , do from equation (9)
From which, Thus, Central deflectionFrom Equation (7)Note: And from equation (8),These Stresses appear to become infinite at the centre, but it must be realised that the load can not be applied at a point but must extend over a finite area. If this area can be estimated then the maximum Stresses can be obtained.
Loaded Round A Circle, Edge Freely Supported
- Let a total load be distributed around a circle of radius .
It is necessary to divide the plate into two regions, one for and the other for . At the values of , and must be the same for both regions.
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- If , and
Equating the values of and at gives the following equations:And,at gives: 16) to (17) the constants are found to be:The Central Deflection is given by the value of at and by substitution equation (15) reduces to:
- If and
Which has a maximum value at Hence from equation (12)Similarly:
Annular Ring , Loaded Around The Inner Edge
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