1. Chapter 3 Mechanical Properties of Materials
What do they tell us? Relationship between stress and strain.
How do we find them? Uniaxial stress/strain testing.
Stress-strain characteristic is inherent to the material

2. Test Specimens Need to follow some standard procedure
Concept of engineering stress/true stress
σengstress = P/A0
A0  Original undeformed area of cross section
σtruestress = P/A
A = deformed area of cross section

3. Testing set-up

4. Videos Tensile Test MaterialsScience2000

5. Stress-Strain Diagrams
Nominal Stress: σ = P/A0
Where A0 is original cross section
Nominal Strain: ε = δ/L0
Where L0 is the original gauge length and
δ is the change in gauge length
No two stress-strain diagrams for a particular material will be exactly the same since the results depend on:
material’s composition
microscopic imperfections,
way material is manufactured
rate of loading
temperature

6. Stress/strain curve

7. Testing

8. Necking

9. Stress/strain diagram for mild steel

10. Ductile/brittle behaviour
A ductile material is subjected to large strains before it ruptures
Two measures for ductility
Percent elongation =((Lf – L0)/ L0)) x (100%)
For mild steel this value is 38%
Percent reduction in area = ((A0 – Af)/ A0)) x(100%)
For mild steel this value is 60%

11. Concept of offset method for yield strength
When yield point is not well defined (☺ ?)  use 0.2% strain criterion

12. Yield Strength
Yield Strength is not a physical property of the material.
We will use approach that
Yield strength
Yield point
Elastic limit
Proportional limit
All coincide unless otherwise stated

13. What about rubber?
No yield point
Non-linear elastic behavior

14. Exhibit no yielding before failure
Brittle materials
Exhibit no yielding before failure

15. Grey Cast Iron

16. Concrete
Reinforced with steel to give the tensile resistance

17. Effect of temperature

18. 3.4 Hooke’s Law Linear elastic behavior leads to:
σ = E ε (Hooke’s law)
E = Young’s modulus (Thomas Young-1807) (aka: modulus of elasticity)
E = (σpl / εpl)
Typically E = 210 GPa (steel); 70 GPa(aluminium)

20. Modulus of Elasticity
Modulus of Elasticity (E) indicates stiffness of a material.
Stiff material -> large E
(e.g. steel: E = 200 GPa)
Spongy material -> small E
(for vulcanized rubber, E = 0.70 MPa)

21. Strain Hardening
Strain hardening is used to establish a higher yield point for a material
The modulus of elasticity stays the same.
The ductility decreases.

22. Strain Energy Density (u)
u = (1/2) (σ2 / E) (linear elastic)
Modulus of Resilience (ur)
ur = (1/2) σpl εpl = (1/2) (σ2pl / E)
=ability to absorb energy w/o permanent damage
Modulus of Toughness  entire area under the stress-strain curve
=total energy absorbed before fracture

23. Strength, Toughness and Ductility
Strength ~ related to the height of the curve
Ductility ~ related to the width of the curve
Toughness ~ related to area under the curve

24. Chapter 3 Lecture Example 1
A tension test for a steel alloy results in the stress-strain diagram shown. Calculate the modulus of elasticity and the yield strength based on a 0.2% offset. Identify on the graph the ultimate stress and the fracture stress.

25. Chapter 3 Lecture Example 1

27. Chapter 3 Lecture Example 2
The stress-strain diagram for an aluminum alloy that is used for making aircraft parts is shown. If a specimen of this material is stressed to 600 MPa, determine the permanent strain that remains in the specimen when the load is released. Also, find the modulus of resilience both before and after the load application.

29. Don’t cover in detail in class
Don’t cover in detail in class. Mention: same force, different A, different stress. AB elastic, BC yields !

30. Poisson’s Ratio 
A tensile force causes a deformable body to longitudinally elongate and to laterally contract.
εlong = ( δ/L)
εlat = ( δ’/r)

31. Tension

32. Compression

33. Poisson’s Ratio

34. Poisson’s Ratio  Value of  is positive
Value of  same in tension and compression
Typical value of  ~ 0.25,
Range is between 0 and 0.5
 constant only in the elastic range
Only Longitudinal force is acting to cause the lateral strain.

35. Lecture Example 3: A steel pipe (Stainless 304) has a length of 1
Lecture Example 3: A steel pipe (Stainless 304) has a length of 1.5 m and inside and outside diameters of di=12 cm and do=15cm. It is compressed by an axial force P=500 kN.
Before you calculate, make predictions a. the pipe gets longer/shorter. b. The pipe’s outer diameter increases/decreases/stays the same. c. The pipe’s wall thickness increases/decreases/stays the same.
Now find the change in length of the pipe, the lateral strain, the new inner and outer diameters and the change in wall thickness.

36. Shear stress-strain behavior
Can be determined by subjecting circular tubes to torsional loading

37. Shear Stress –Strain Diagram

38. Modulus of Rigidity – linear region
G – how much a material resists twisting

39. Lecture Example 4: Determine the shear modulus, G, the proportional limit, and the ultimate shear stress. Find the maximum d where the material behaves elastically. What is the magnitude of V to cause d?

40. Bending test

41. Bending test

42. Impact Test

43. Failure of turbine blade due to creep
Failure of steam pipe due to creep

44. Creep
Creep is the time-related deformation of a material for which temperature and stress play an important role.
Creep results from sustained loading below the measured yield point.
Members are designed to resist the effects of creep based on their creep strength, which is the highest amount of stress a member can withstand during a specified amount of time without experiencing creep strain.

45. Fatigue
Fatigue occurs in metals when stress or strain is cycled. It causes a brittle fracture to occur.
Engineering Design: stress in the member should not exceed the material’s endurance or fatigue limit.
This the maximum stress that the member can resist when enduring a specified number of cycles.

46. Material Properties
Strength – Capacity to resist loads – yield stress. Higher y, higher strength.
Resilience – Measure of energy absorbed without permanent damage .
Toughness – Measure of energy absorbed before fracturing.
Ductility – Property of material which allows large deformation before fracture.
Brittleness – Property of material which allows little or no yielding before fracture.
Endurance – Ability to sustain cyclic loads.
Stiffness – Mechanical property indicated by E. Higher E means stiff material. Steeper slope in the Stress – Strain diagram.
Rigidity - Mechanical property of material indicated by G.

47. 3.27 A short cylindrical block of 2014-T6 Aluminum having an original diameter of 0.5 in and an original length of 1.5 in is placed in the smooth jaws of a vise and a compressive force of 800 lb is applied. Determine (a) The decrease in length. (b) The new diameter.