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# Circular Plates

Stresses and Strains in loaded Circular Plates and Rings.

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:

The radius of curvature in the plane is given by: (Approximately)

Thus from equation (1)

Note that, on a circle of radius and centre , lines such as form part of a cone with as the apex. Hence is the Centre of Curvature in the plane and: (Approximately)

If is the distance of any "fibre" from the neutral axis (which is assumed to be central) then proceeding as for "Pure Bending" in the planes and the linear Strains are:

And,

Where and are the Stresses in the directions and , is zero

Solving equations (4) and (5) for the Stresses and incorporating equations (2) and (3) gives:

The Bending Moment per unit length along is which is given by:

By substitution from equation (6)

Similarly if is the Bending Moment per unit length about then:

Using equation (7)

Note that:

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:

If is known as a function of , this equation can be integrated to determine and hence . Bending Moments and Stresses can then be calculated.

### Particular Case

A Plate Loaded with Uniformly distributed load of per unit Area and a Concentrated load at Centre of .

Per unit length of circumferentially (Except at )

Substituting in Equation (12) and Integrating:

But From Equation (1)

Example:
##### Example - A particular case
Problem
A Plate Loaded with Uniformly distributed load of w per unit Area and a Concentrated load at Center of P.
Workings

Per unit length of circumferentially (Except at x = 0)

Substituting in Equation (#19) and Integrating:

But From Equation (#1)

## Solid Circular Plate

Let the radius of the plate be and the thickness .

### 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 at

From 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.

At ,

i.e. from Equation (13),

Using Equation (14),Central Deflection =

Eliminating by using equation (9)

From Equation (6)

This Stress has its greatest numerical value when ( i.e. at the clamped edge), thus

From Equation (7)

From which, (At the Centre)

### Central Load P, Edge Freely Supported (w=0)

At , therefore from equation (13) and . From equation(14)

Note:

At , do from equation (8)

From which,

Thus, Central deflection

From 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

Hence, from Equation (13) . And from Equation (14),

Since and are not infinite at then , and since when and , then above equations reduce to: And,

• If and

From Equation (13),

And from Equation (14),

Equating the values of and at gives the following equations:

And,

at gives:

From Equations (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:

• For

Which has a maximum value at

Hence from equation (11)

Similarly:

## Annular Ring , Loaded Around The Inner Edge The ring is loaded with a total load around the inner edge and is freely supported around the outer edge. at and at .

And,

Subtracting and solving:
And then

Then,

The maximum Bending Moment is at , therefore