1. Read Chapter 1 Basic Elasticity - Equilibrium Equations
- Plane Stresses
- Principal Stresses
- Mohr’s Circle
- Stress Strain Relationships

2. Stress Equilibrium under external forces
Continuous and deformable material
Forces are transmitted throughout its volume

3. 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

4. Stress
Normal or direct stress
Shear stress

5. Equilibrium Equations
First two terms Taylor Series Expansion
Plane Stresses
Body Forces per unit mass times density

6. 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:

7. Determination of stresses on an incline plane
Body forces ignored (second order term)
δx and δy are small  stress distribution is assumed uniform

8. Determination of stresses on an incline plane
Sum of the Forces perpendicular to ED:
Sum of the Forces parallel to ED:

9. 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

10. Example cont’d

11. Example cont’d
θ = 300
σy = 0

12. Example cont’d

13. 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

14. Principal Stresses Derivation on page 16 Megson.
Maximum principal stress
Minimum principal stress
If negative (compressive) could be numerically larger that σI

15. 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

16. Principal Stresses Occur 450 inclination from principal planes
Principal stresses Summary:

17. 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):

18. Mohr’s Circle
Squaring:
Adding:
Equation of Mohr’s Circle:

19. Mohr’s Circle
Circle of radius:

20. 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

21. Mohr’s Circle

22. Strain
Let’s Look at εx:

23. Strain
Higher order powers of are ignored
Similarly

24. 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.

25. 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

26. Principal Strains

27. 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

28. 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

29. Stress-Strain Relationships
For isotropic materials

30. 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

31. Sample Problem (Plane Stresses)
FBD F1 F1 F2 F2 F3 F3
Sum of the Forces:
2F1 + F2 = 0
Solve for and

32. Sample Problem FOS
Given the information below, determine the limit and ultimate loads of the
structure as well as the corresponding margins of safety.