Stresses and Strains in loaded Circular Plates and Rings.
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. Let be the Centre of Curvature of a section at a distance from . Then if the deflection is small: 1) 4) and (5) for the Stresses and incorporating equations (2) and (3) gives:
Using equation (7)
The diagram shows the Forces and Moments per unit length acting on an element which subtends an angle at the centre. is the Shearing Force per unit length in the direction of . Now consider the equilibrium of the Couples in the Central Radial Plane. i.e., Which in the Limit reduces to:
Substituting from Equations (8) and (10) gives:
This can be written as:
Uniformly Loaded, Edge Freely Supported
- and since and can not be infinite at the centre, then from Equation (13) at , , and therefore from equation (14).
Using equations (8) and (13). At , therefore
Thus,Central Deflection = at . Thus,Eliminating by substitution from Equation (9)From Equation (6)And atFrom Equation (7)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(14)
Structural loads or actions are forces, deformations or accelerations applied to a structure or its components.Loads cause stresses, deformations and displacements in structures.
Central Load P, Edge Freely Supported (w=0)
- At , therefore from equation (13) and . From equation(14)
At , do from equation (8)
From which, Thus, Central deflectionFrom Equation (6)Note: And from equation (7),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.
- If , and
Equating the values of and at gives the following equations:And,at gives: 15) to (16) the constants are found to be:The Central Deflection is given by the value of at and by substitution equation (14) reduces to:
- If and
Which has a maximum value at Hence from equation (11)Similarly: