EMT 902/2: ADVANCED ENGINEERING 2

1. EMT 902/2: ADVANCED ENGINEERING 2 by Noraini Othman noraini_othman@unimap.edu.my 04-979 8146 019-5769 184 Lecture 3: Mechanical Properties of Metal

2. Introduction To understand and describe how materials deform (elongate, compress, twist) or break as a function of applied load, time, temperature, and other conditions we need first to discuss the mechanical properties of materials.

3. Types of Loading

4. Concepts of Stress and Strain Engineering stress= σ = F Δl A0 A0 F is load applied perpendicular to specimen cross-section; A0 is cross-sectional area (perpendicular to the force) beforeapplication of the load. A Units of Stress are psi or N/m2 (Pascals) 1 psi = 6.89 x 103 Pa

5. Concepts of Stress and Strain Engineering strain = ε Δl A0 = Change in length Original length A Units of strain are in/in or m/m. Stress and strain are positive for tensile loads, negative for compressive loads

6. Concepts of Stress and Strain (Shear and Torsion) F is load applied parallel to the upper and lower faces each of which has an area A0. Torsion is variation of pure shear. A shear stress in this case is a function of applied torque T, shear strain is related to the angle of twist, φ. Amount of shear displacement Shear strain γ = Distance over which shear acts.

7. Stress-Strain Behavior

8. Stress-Strain Behavior – Elastic Deformation In tensile tests, if the deformation is elastic, the stress strain relationship is called Hooke's law: E is Young's modulusor modulus of elasticity, has the same units as σ, N/m2 or Pa

9. Elastic Deformation – Non-linear elastic behavior In some materials (many polymers, concrete...), elastic deformation is not linear, but it is still reversible.

10. Elastic Deformation – Poisson’s Ratio Materials subject to tension shrink laterally. Those subject to compression, bulge. The ratio of lateral and axial strains is called the Poisson's ratio .  is dimensionless, sign shows that lateral strain is in opposite sense to longitudinal strain Theoretical value for isotropic material: 0.25 Maximum value: 0.50, Typical value: 0.24 - 0.30

11. Poisson’s Ratio Poisson’s ratio =

12. Elastic limit The elastic limit is the highest stress at which all deformation strains are fully recoverable. The material will return to its original shape after The material is unloaded( like a rubber band). For most materials and applications this can be Considered the practical limit to the maximum stress a component can withstand and still function as designed. Beyond the elastic limit, permanent strains are likely to deform the material to the point where its function is impaired. 12

13. Stress-Strain Behavior – Plastic deformation Plastic deformation: • stress and strain are not proportional • the deformation is not reversible • deformation occurs by breaking and re- arrangement of atomic bonds

14. Tensile Properties - Yielding

15. Yield Strength • Yield strength is strength at which • metal or alloy show significant • amount of plastic deformation. • 0.2% offset yield strengthis that • strength at which 0.2% plastic • deformation takes place. • Construction line, starting at 0.2% • strain and parallel to elastic region • is drawn to fiend 0.2% offset yield • strength.

16. Tensile Strength 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. Load cannot be sustained above this point. “Necking” starts to occur in ductile materials. 16

17. Tensile Strength For structural applications, the yield stress is usually a more important property than the tensile strength, since once the yield stress has passed, the structure has deformed beyond acceptable limit 17

18. Question 1: Tensile Strength

19. Tensile Strength Tensile strength is defined as the highest stress that a material can withstand before failure occurs

20. Tensile Strength Material A is stronger than B, because it has a higher tensile strength compared to B

21. Tensile Properties - Ductility 21

22. Percent Elongation • Percent elongation is a measure of ductilityof a material. • It is the elongation of the metal before fracture expressed as percentage of original length. • % Elongation = • Measured using a caliper fitting the fractured metal together. • Example:- Percent elongation of pure aluminum is 35% • For 7076-T6 aluminum alloy it is 11% Final length – Initial Length Initial Length

23. Percent Reduction in Area • Percent reduction area is • also a measure of ductility. • The diameter of fractured • end of specimen is • measured using caliper. Initial area – Final area % Reduction Area Initial area Fig: Stress-strain curves of different metals

24. 24

25. Question 2: Ductility

26. Ductility Ductility is measured by the amount of elongation that can be applied to the material before the failure occurs. Strain is directly proportional to elongation

27. Ductility A is ductile, because the strain under curve A is greater B. This indicates that A shows more elongation than B before failure.

28. Toughness

29. Question 3: Toughness

30. Toughness Toughness is defined as the total area under the stress-strain curve. It indicates the amount of energy absorbed before failure

31. Toughness A is tougher, because the area under curve A is greater than B. this indicates that A absorbs more energy than B before failure

32. (c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license. True Stress - True Strain • True stress The load divided by the actual cross-sectional area of the specimen at that load. • True strain The strain calculated using actual and not original dimensions, given by εt = ln (li / lo). Figure : The relation between the true stress-true strain diagram and engineering stress-engineering strain diagram. The curves are identical to the yield point

33. True Stress and Strain

34. True stress and true strain are based upon instantaneous cross-sectional area and length. True Stress = σt = True Strain = εt = True stress is always greater than engineering stress. True Stress - True Strain F Ai (instantaneous area)

35. Tensile Test • Tensile test is used to evaluate the strength of metals and alloys Load Cell Specimen Extensometer Force data is obtained from Load cell Strain data is obtained from Extensometer.

36. Tensile Test (Cont.) Commonly used Test specimen

37. Tensile specimen in testing (5) (6) (4) (3) (2) (1) s= E e Elastic Fracture Plastic Deformation • beginning of test, no load; • uniform elongation and reduction of cross-sectional area; • continued elongation, maximum load reached; • necking begins, load begins to decrease; and • fracture. • If pieces are put back together as in (6), final length can be measured Properties from Tensile Test- Banjuraizah 37

38. (c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license. When a ductile material is pulled in a tensile test, necking begins and voids form – starting near the center of the bar – by nucleation at grain boundaries or inclusions. As deformation continues a 45° shear lip may form, producing a final cup and cone fracture 38

39. (c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license. Summary

40. Question 4: Tensile Testing of Aluminum Alloy Convert the change in length data in Table 6-1 to engineering stress and strain and plot a stress-strain curve.

41. SOLUTION 4:

42. Question 5: Young’s Modulus of Aluminum Alloy From the calculated data in Question 4, (i) calculate the modulus of elasticity of the aluminum alloy. (ii) Use the modulus to determine the length after deformation of a bar of initial length of 50 in. Assume that a level of stress of 30,000 psi is applied

43. SOLUTION 5:

44. Question 6: Ductility of an Aluminum Alloy The aluminum alloy in Question 4 has a final length after failure of 2.195 in. and a final diameter of 0.398 in. at the fractured surface. Calculate the ductility of this alloy.

45. SOLUTION 6:

46. Question 7: True Stress and True Strain Calculation • Compare engineering stress and strain with true stress and strain for the aluminum alloy in Example 6.1 at • the maximum load; and • fracture. • The diameter at maximum load is 0.497 in. and at fracture • is 0.398 in.

47. SOLUTION 7: