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Read Chapter 1. Basic Elasticity - Equilibrium Equations - Plane Stresses - Principal Stresses - Mohr’s Circle - Stress Strain Relationships. Stress. Equilibrium under external forces Continuous and deformable material Forces are transmitted throughout its volume. Stress.

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Read chapter 1

Read Chapter 1

Basic Elasticity

- Equilibrium Equations

- Plane Stresses

- Principal Stresses

- Mohr’s Circle

- Stress Strain Relationships


Stress

Stress

  • Equilibrium under external forces

  • Continuous and deformable material

  • Forces are transmitted throughout its volume


Stress1

Stress

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


Stress2

Stress

Normal or direct stress

Shear stress


Equilibrium equations

Equilibrium Equations

First two terms Taylor Series Expansion

Plane Stresses

Body Forces per unit mass times density


Equilibrium equations1

Equilibrium Equations

  • 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:


Determination of stresses on an incline plane

Determination of stresses on an incline plane

Body forces ignored (second order term)

δx and δy are small  stress distribution is assumed uniform


Determination of stresses on an incline plane1

Determination of stresses on an incline plane

Sum of the Forces perpendicular to ED:

Sum of the Forces parallel to ED:


Example

Example

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


Example cont d

Example cont’d


Example cont d1

Example cont’d

θ = 300

σy = 0


Example cont d2

Example cont’d


Principal stresses

Principal Stresses

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


Principal stresses1

Principal Stresses

  • Derivation on page 16 Megson.

Maximum principal stress

Minimum principal stress

If negative (compressive) could be numerically larger that σI


Principal stresses2

Principal Stresses

  • 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


Principal stresses3

Principal Stresses

Occur 450 inclination from principal planes

Principal stresses Summary:


Mohr s circle

Mohr’s Circle

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):


Mohr s circle1

Mohr’s Circle

Squaring:

Adding:

Equation of Mohr’s Circle:


Mohr s circle2

Mohr’s Circle

Circle of radius:


Mohr s circle3

Mohr’s Circle

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


Mohr s circle4

Mohr’s Circle


Strain

Strain

Let’s Look at εx:


Strain1

Strain

Higher order powers of are ignored

Similarly


Compatibility equations

Compatibility Equations

  • 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.


Plain strain

Plain Strain

  • 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


Principal strains

Principal Strains


Stress strain relationships

Stress-Strain Relationships

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


Stress strain relationships1

Stress-Strain Relationships

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


Stress strain relationships2

Stress-Strain Relationships

For isotropic materials


Material constitutive equations

Material Constitutive Equations

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


Sample problem plane stresses

Sample Problem (Plane Stresses)

FBD

F1

F1

F2

F2

F3

F3

Sum of the Forces:

2F1 + F2 = 0

Solve for and


Sample problem fos

Sample Problem FOS

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

structure as well as the corresponding margins of safety.


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