- 88 Views
- Uploaded on
- Presentation posted in: General

Read Chapter 1

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Basic Elasticity

- Equilibrium Equations

- Plane Stresses

- Principal Stresses

- Mohr’s Circle

- Stress Strain Relationships

- Equilibrium under external forces
- Continuous and deformable material
- Forces are transmitted throughout its volume

Resultant force at point O δP

Force must be in equilibrium: Equal and opposite force δP on the particle

Divide the particle along plane nn containing O

δP can be considered as uniformly distributed over a small area δA

Normal or direct stress

Shear stress

First two terms Taylor Series Expansion

Plane Stresses

Body Forces per unit mass times density

- Taking sum of the moments about an axis through the centre line of the element parallel to the z axis it would be found that:

Body forces ignored (second order term)

δx and δy are small stress distribution is assumed uniform

Sum of the Forces perpendicular to ED:

Sum of the Forces parallel to ED:

Cantilever beam of solid cross-section

Compressive load of 50 KN, 1.5 mm below horizontal diameter plane

Torque of 1200 Nm

Calculate direct and shear stresses on a plane 600 to the axis of the cantilever beam on a point located the lower edge of the vertical plane of symmetry

θ = 300

σy = 0

Principal stresses determine the maximum or minimum value given a

loading stress, σn

Starting from the plane stress equation, derivating with respect to θ and

equating to zero, we can obtain an expression of the maximum and minimum

stresses:

Two solutions are obtained:

θ and θ + π/2

These planes correspond to those on which there is no shear stresses

The direct stresses on these planes (principal planes) are called principal stresses

- Derivation on page 16 Megson.

Maximum principal stress

Minimum principal stress

If negative (compressive) could be numerically larger that σI

- Similarly, we can find the maximum and minimum shear stresses (principal):

Two solutions are obtained:

450 inclined from principal planes

These planes correspond to those on which there is no normal stresses

The shear stresses on these planes (principal planes) are called principal shear stresses

Occur 450 inclination from principal planes

Principal stresses Summary:

Graphical Representation of the stresses at a point in a deformable body

Recall Stresses on an incline plane:

The stress equation might be rewritten as (using trig relations):

Squaring:

Adding:

Equation of Mohr’s Circle:

Circle of radius:

Example:

Direct stress of 160 N/mm2 (tension, x direction)

120 N/mm2 (compression, y direction)

Applied to elastic material on two mutually perpendicular planes

Principal stresses are limited to 200 N/mm2

CALCULATE THE ALLOWABLE VALUE OF SHEAR STRESS ON THE GIVEN

PLANES

DETERMINE ALSO THE VALUE OF THE OTHER PRINCIPAL STRESSES

AND THE MAXIMUM SHEAR STRESS AT THAT POINT

Let’s Look at εx:

Higher order powers of are ignored

Similarly

- Since the six strains are defined in terms of three displacement functions then they must bear some relationship to each other
- Derivation of these equations is described in Megson 1.10.

- We have a 3D compatibility equation and expressions for strain
- We shall concern ourselves with 2D problems (displacement on only one plane, xy)
- Then εz, γxz, γyz become zero

Compatibility equation plain strain

So far, we have three equations of equilibrium for 3D deformable body

& Six strain-displacement relationships

TOTAL: 9 independent equations towards the solution of the 3D stress problem

The total of unknowns is 15:

- 6 stresses

- 6 trains

- 3 displacements

So far we have not made any assumptions regarding the force-displacement relations (i.e. stress-strain relations)

For isotropic materials (homogenous) experiments show that:

Modulus of Elasticity, or Young’s Modulus

For isotropic materials

Sample problem:

E1, A1, L, α1

E2, A2, L, α2

Rigid

Rigid

What are the stresses on the bars when

exposed to a change in temperature

equal toT0?

E1, A1, L, α1

FBD

F1

F1

F2

F2

F3

F3

Sum of the Forces:

2F1 + F2 = 0

Solve for and

Given the information below, determine the limit and ultimate loads of the

structure as well as the corresponding margins of safety.