1. Elastic Compliance and Stiffness constant

2. Toughness and ductility: What are they and how do we measure them?
• Stress and strain: What are they and why are they used instead of load and deformation?
Elastic behavior: When loads are small, how much deformation occurs? What materials deform least?
Plastic behavior: At what point do dislocations cause permanent deformation? What materials are most resistant to permanent deformation?
Toughness and ductility: What are they and how do we measure them?
Ceramic Materials: What special provisions/tests are made for ceramic materials? 2

3. Normal & Shear components of stress
Normal stress is perpendicular to the cross section,  (sigma).
Shear stress is parallel to the cross section,  (tau).
xy
First subscript indicates the axis that is perpendicular to the face y
3D case x y z
yx
Second subscript indicates the positive direction of the shear stress
yz
xy z
zx
zy x
xz
Due to equilibrium condition;
xy =
yx
zy =
yz
zx =
xz

4. Normal & Shear components of stress Two Dimensional Casey
yx
xy x x
xy
yx y

5. Normal Stress Due to Axial Load
A positive sign is used to indicate a tensile stress (tension), a negative sign to indicate a compressive stress (compression)

6. Normal Stress Due to Bending Load
Stress distribution
Maximum stress at the surface
M change in position
Where I is area moment of inertia

7. Shear Stress Due to Torque (twisting)
Torsional stress is caused by twisting a member
Stress distribution
Maximum shear stress at the surface
Where J is polar area moment of inertia

8. Torsional Stress- examples
Power transmission
Mixer
Structural member

9. Combined Stresses - examples
Power transmission, bending and torsion stresses
Bicycle pedal arm and lug wrench, bending and torsion stresses
Trailer hitch, bending and axial stresses
Billboards and traffic signs, bending, axial and torsion stresses

10. Understanding stress in terms of its components
Stress is a Second Order Tensor
It is easier to understand stress in terms of its components and the effect of the components in causing deformations to a unit body within the material
These components can be treated as vectors
Components of a stress: 2D  4 components [2  (tensile) and 2  (shear)] 3D  9 components [3  (tensile) and 6  (shear)]
 written with subscripts not equal implies  (shear stress) e.g. xy  xy
First index refers to the plane and the second to the direction
Direction
Plane

11. 2D

12. Let us consider a body in the presence of external agents (constraints and forces) bringing about stresses in the body. A unit region in the body (assumed having constant stresses) is analyzed. (body forces are ignored)
Note: the directions of the stresses shown are arbitrary (the stresses in general could be compression/tension and shear could be opposite in sign)

14. Elastic Compliance and Stiffness Constants
The generalized Hooke’s law is an assumption, which is reasonably accurate for many material subjected to small strain, for a given temperature, time and location.
A mathematical expression of the stress-strain relation for the elastic deformation of materials was first suggested by Robert Hooke
F is the applied force
 u is the deformation of the elastic body
k is the material dependent spring constant

15. Anisotropy and Isotropy
In a single crystal, the physical and mechanical properties often differ with orientation.
An anisotropic material is a material which does not behave the same way in all directions. Take wood for example. Wood is very strong along the grain. Against the grain, however, it will easily break.  The opposite of an anisotropic material is an isotropic material. Most metals (steel, aluminum) are isotropic materials. They respond the same way in all directions.

16. Figure 7-1 Anisotropic materials: (a) rolled material, (b) wood, (c) glass-fiber cloth in an epoxy matrix, and (d) a crystal with cubic unit cell.

17. Real materials are never perfectly isotropic. In some cases (e. g
Real materials are never perfectly isotropic. In some cases (e.g. composite materials) the differences in properties for different directions are so large that one can not assume isotropic behavior - Anisotropic.
There is need to discuss Hooke’s Law for anisotropic cases in general. This can then be reduced to isotropic cases - material property (e.g., elastic constant) is the same in all directions.

18. In the general 3-D case, there are six components of stress and a corresponding six components of strain.
In highly anisotropic materials, any one component of stress can cause strain in all six components.
For the generalized case, Hooke’s law may be expressed as:
where,
Both Sijkl and Cijkl are fourth-rank tensor quantities.

19. Expansion of either Eqs
Expansion of either Eqs. will produce nine (9) equations, each with nine (9) terms, leading to 81 constants in all.
It is important to note that both ij and  ij are symmetric tensors.
Symmetric tensor Means that the off-diagonal components are equal. For example, in case of stress:

20. We can therefore write:

21. Similarly, the strain tensor can be written as:

22. Elastic Deformation 1. Initial 2. Small load 3. Unload
Elastic means reversible. 22

23. Plastic Deformation (Metals)
1. Initial
2. Small load
3. Unload
Plastic means permanent. 23

24. Typical stress-strain behavior for a metal showing elastic and plastic deformations, the proportional limit P and the yield strength σy, as determined using the strain offset method (where there is noticeable plastic deformation). P is the gradual elastic to plastic transition.
c07f10ab 24

25. Plastic Deformation (permanent)
From an atomic perspective, plastic deformation corresponds to the breaking of bonds with original atom neighbors and then reforming bonds with new neighbors.
After removal of the stress, the large number of atoms that have relocated, do not return to original position.
Yield strength is a measure of resistance to plastic deformation. 25 25

26. c07f25 26

27. The image shows necked region in a fractured sample
(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.
Localized deformation of a ductile material during a tensile test produces a necked region.
The image shows necked region in a fractured sample

28. Permanent Deformation
Permanent deformation for metals is accomplished by means of a process called slip, which involves the motion of dislocations.
Most structures are designed to ensure that only elastic deformation results when stress is applied.
A structure that has plastically deformed, or experienced a permanent change in shape, may not be capable of functioning as intended. 28

29. Yield Strength, sy 29

30. Stress-Strain Diagram
ultimate
tensile strength 3
necking
Slope=E
Strain
Hardening
yield
strength
Fracture 5 2
Elastic region
slope =Young’s (elastic) modulus
yield strength
Plastic region
ultimate tensile strength
strain hardening
fracture
Plastic
Region
Stress (F/A)
Elastic
Region 4 1
Strain ( ) (DL/Lo)

31. Stress-Strain Diagram (cont)
Elastic Region (Point 1 –2)
- The material will return to its original shape
after the material is unloaded( like a rubber band).
- The stress is linearly proportional to the strain in
this region. or
: Stress(psi)
E : Elastic modulus (Young’s Modulus) (psi)
: Strain (in/in)
Point 2 : Yield Strength : a point where permanent
deformation occurs. ( If it is passed, the material will
no longer return to its original length.)

32. Stress-Strain Diagram (cont)
Strain Hardening
- If the material is loaded again from Point 4, the
curve will follow back to Point 3 with the same
Elastic Modulus (slope).
- The material now has a higher yield strength of
Point 4.
- Raising the yield strength by permanently straining
the material is called Strain Hardening.

33. Stress-Strain Diagram (cont)
Tensile Strength (Point 3)
- The largest value of stress on the diagram is called
Tensile Strength(TS) or Ultimate Tensile Strength
(UTS)
- It is the maximum stress which the material can
support without breaking.
Fracture (Point 5)
- If the material is stretched beyond Point 3, the stress
decreases as necking and non-uniform deformation
occur.
- Fracture will finally occur at Point 5.

34. The stress-strain curve for an aluminum alloy.
(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

35. Stress-strain behavior found for some steels with yield point phenomenon.35

36. Toughness
c07f13
Toughness is the ability to absorb energy up to fracture (energy per unit volume of material).
A “tough” material has strength and ductility.
Approximated by the area under the stress-strain
curve.
Lower toughness: ceramics
Higher toughness: metals 36

37. Toughness • Energy to break a unit volume of material
• Approximate by the area under the stress-strain
curve. 21

38. Linear Elastic Properties
c07f05
s = E e
• Hooke's Law:
n = ex/ey
• Poisson's ratio:
metals: n ~ 0.33
ceramics: n ~0.25
polymers: n ~0.40
Modulus of Elasticity, E:
(Young's modulus)
Units:
E: [GPa] or [psi]
n: dimensionless 38

39. Elastic Strain Energy
If the sample obeys Hooke's Law, and is below the elastic limit, the Elastic Strain Energy can be calculated by the formula:
E = ½Fx

40. Example 2: Young’s Modulus - Aluminum Alloy
From the data in Example 1, calculate the modulus of elasticity of the aluminum alloy.

41. Assignment
Determine the deformation of the steel rod shown under the given loads.