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## CHAPTER OUTLINE

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1. Tension and Compression Test Stress-Strain Diagram Stress-Strain Behavior of Ductile and Brittle Materials Hooke’s Law Poisson’s Ratio CHAPTER OUTLINE

2. The strength of a material can only be determined by experiment One test used by engineers is the tension or compression test This test is used primarily to determine the relationship between the average normal stress and average normal strain in common engineering materials, such as metals, ceramics, polymers and composites 3.1 TENSION & COMPRESSION TEST

3. Before testing, 2 small punch marks identified along specimen’s length Measurements are taken of both specimen’s initial cross-sectional area A0 and gauge-length distance L0; between the two marks 3.1 TENSION & COMPRESSION TEST Performing the tension or compression test do=13 mm Specimen of material is made into “standard” shape and size Lo=50 mm

4. Performing the tension or compression test 3.1 TENSION & COMPRESSION TEST Seat the specimen into a testing machine shown in the figure The machine will stretch specimen at slow constant rate until breaking point At frequent intervals during test, data is recorded of the applied load P.

5. Elongation δ= L−L0 is measured using either a caliper or an extensometer δis used to calculate the normal strain in the specimen Sometimes, strain can also be read directly using an electrical-resistance strain gauge 3.1 TENSION & COMPRESSION TEST

6. A stress-strain diagram is obtained by plotting the various values of the stress and corresponding strain in the specimen Conventional stress-strain diagram Using recorded data, we can determine nominal or engineering stress by P A0 σ = P P 3.2 STRESS-STRAIN DIAGRAM do=13 mm Lo=50 mm Assumption: Stress is constant over the cross-section and throughout region between gauge points

7. Conventional Stress-Strain Diagram Likewise, nominal or engineering strain is found directly from strain gauge reading, or by δ L0  = 3.2 STRESS-STRAIN DIAGRAM Assumption: Strain is constant throughout region between gauge points By plottingσ(ordinate) against(abscissa), we get a conventional stress-strain diagram

8. Conventional stress-strain diagram 3.2 STRESS-STRAIN DIAGRAM Figure shows the characte- ristic stress-strain diagram for steel, a commonly used material for structural members and mechanical elements

9. 3.2 STRESS-STRAIN DIAGRAM Linear Elastic Behavior A straight line Stress is proportional to strain, i.e., linearly elastic Upper stress limit, or proportional limit;σpl If load is removed upon reaching proportional limit, specimen will return to its original shape

10. 3.2 STRESS-STRAIN DIAGRAM Non-linear Elastic Behavior A small curve line Stress is not proportional to strain, i.e., non-linearly elastic Upper stress limit is almost reaching yield stress;σy If load is removed upon reaching elastic limit, specimen will return to its original shape

11. 3.2 STRESS-STRAIN DIAGRAM Yielding. Material deforms permanently; yielding; plastic deformation Yield stress, σY Once yield point reached, specimen continues to elongate (strain) without any increase in load Note figure not drawn to scale, otherwise induced strains is 10-40 times larger than in elastic limit Material is referred to as being perfectly plastic

12. 3.2 STRESS-STRAIN DIAGRAM Strain Hardening. While specimen is elongating, its cross-sectional area will decrease Ultimate stress,σu Decrease in area is fairly uniform over entire gaugelength

13. 3.2 STRESS-STRAIN DIAGRAM Necking. At ultimate stress, cross-sectional area begins to decrease in a localizedregion As a result, a constriction or “neck” tends to form in this region as specimen elongates further Specimen finally breaks at fracture stress,σf

14. 3.2 STRESS-STRAIN DIAGRAM Necking. Specimen finally breaks at fracture stress, σf

15. True stress-strain diagram 3.2 STRESS-STRAIN DIAGRAM Instead of using original cross-sectional area and length, we can use the actual cross-sectional area and length at the instant the load is measured Values of stress and strain thus calculated are called true stress and true strain, and a plot of their values is the true stress-strain diagram

16. Conventionalstress-strain diagram 3.2 STRESS-STRAIN DIAGRAM In strain-hardening range, conventionalσ-diagramshows specimen supporting decreasing load True stress-strain diagram While trueσ-diagram shows material to be sustaining increasing stress

17. Ductile materials Defined as any material that can be subjected to large strains before it ruptures, e.g., mild steel Such materials are used because it is capable of absorbing shock or energy. When overloaded, it will exhibit large deformation before failing Ductility of material is to report its percent elongation or percent reduction in area at time of fracture 3.3 STRESS-STRAIN BEHAVIOR OF DUCTILE & BRITTLE MATERIALS

18. Ductile materials Most metals do not exhibit constant yielding behavior beyond the elastic range, e.g. aluminum 3.3 STRESS-STRAIN BEHAVIOR OF DUCTILE & BRITTLE MATERIALS It does not have well-defined yield point, thus it is standard practice to define its yield strength using a graphical procedure called the offset method

19. Offset method to determine yield strength 3.3 STRESS-STRAIN BEHAVIOR OF DUCTILE & BRITTLE MATERIALS • Normally, a 0.2 % strain is chosen. 2. From this point on the  axis, a line parallel to initial straight-line portion of stress-strain diagram is drawn. 3. The point where this line intersects the curve defines the yield strength.

20. Brittle Materials Tension failure of A brittle material Compression causes Material to bulge out 3.3 STRESS-STRAIN BEHAVIOR OF DUCTILE & BRITTLE MATERIALS Material that exhibit little or no yielding before failure are referred to as brittle materials, e.g., gray cast iron Brittle materials do not have a well-defined tensile fracture stress, since appearance of initial cracks in a specimen is quite random

21. Brittle Materials Instead, the average fracture stress from a set of observed tests is generally reported s – e diagram for typical concrete mix 3.3 STRESS-STRAIN BEHAVIOR OF DUCTILE & BRITTLE MATERIALS

22. E represents the constant of proportionality, also called the modulus of elasticity or Young’s modulus σ = E 3.4 HOOKE’S LAW Most engineering materials exhibit a linear relationship between stress and strain in the elastic region Discovered by Robert Hooke in 1676 using springs, known as Hooke’s law E has units of stress, i.e., Pascals, MPa or GPa.

23. As shown above, most grades of steel have same modulus of elasticity,Est = 200 GPa 3.4 HOOKE’S LAW Modulus of elasticity is a mechanical property that indicates the stiffness of a material Materials that are still have large E values, while spongy materials (vulcanized rubber) have low values

24. IMPORTANT Modulus of elasticity E, can be used only if a material has linear-elastic behavior. Also, if stress in material is greater than the proportional limit, the stress-strain diagram ceases to be a straight line and the equation is not valid 3.4 HOOKE’S LAW

25. When body is subjected to axial tensile force (tension), it elongates and contracts laterally 3.5 POISSON’S RATIO d = longitudinal elongation d ’ = lateral contraction Similarly, it will contract and its sides expand laterally when subjected to an axial compressive force

26. Strains of the bar for tension: 3.5 POISSON’S RATIO δ L long = positive δ’ r negative lat= ̶ Strains of the bar for compression: δ L negative long= ̶ δ’ r positive lat=

27. ν is unique for homogenous and isotropic material Why negative sign? 3.6 POISSON’S RATIO Early 1800s, S.D. Poisson realized that within elastic range, ration of the two strains is a constant value, since both are proportional. Longitudinal elongation cause lateral contraction (-ve strain) and vice versa Lateral strain is the same in all lateral (radial) directions Poisson’s ratio is dimensionless, 0 ≤ ν ≤ 0.5 lat long Poisson’s ratio,ν =−

28. Bar is made of A-36 steel and behaves elastically. Determine change in its length and change in dimensions of its cross section after load is applied. EXAMPLE 3-1

29. Normal stress in the bar is P A σz = EXAMPLE 3-1 = 16.0(106) Pa = 16 MPa From tables, Est = 200 GPa, strain in z-direction is σz Est 16.0(106) 200(109) = 80(10−6) m/m z= =

30. Using νst = 0.32, contraction strains in both x and y directions are EXAMPLE 3-1 Axial elongation of the bar is, δz = zLz = [80(10−6)](1.5 m) = 120(10-6) m x = y = −νstz = −0.32[80(10−6)] = −25.6(10-6) m/m

31. EXAMPLE 3-1 Thus changes in dimensions of cross-section are δx = xLx =−[25.6(10−6)](0.1 m) = −2.56(10-6) m δy = yLy =−[25.6(10−6)](0.05 m) = −1.28(10-6) m

32. y tyx txy x y gxy 2 gxy 2  2 gxy x 3.7 SHEAR STRESS-STRAIN DIAGRAM When an element of material is subjected to a pure shear in equilibrium, the equal shear stresses must be developed on four faces of the element. txy = tyx The shear stress will distort the element uniformly.

33. Material will exhibit linear-elastic behavior till its proportional limit, τpl Strain-hardening continues till it reaches ultimate shear stress, τu Material loses shear strength till it fractures, at stress of τf 3.6 SHEAR STRESS-STRAIN DIAGRAM

34. τ = Gγ E 2(1 + ν) G = 3.6 SHEAR STRESS-STRAIN DIAGRAM • Hooke’s law for shear G is shear modulus of elasticity or modulus of rigidity • G can be measured as slope of line on τ-γ diagram, G = τpl/ γpl The three material constants E, ν, and G is related by

35. Tension test is the most important test for determining material strengths. Results of normal stress and normal strain can then be plotted. Many engineering materials behave in a linear-elasticmanner, where stress is proportional to strain, defined by Hooke’s law,σ = E. E is the modulus of elasticity, and is measured from slope of a stress-strain diagram When material stressed beyond yield point, permanent deformation will occur. CHAPTER REVIEW

36. Strain hardening causes further yielding of material with increasing stress At ultimate stress, localized region on specimen begin to constrict, and starts “necking”. Fracture occurs. Ductilematerials exhibit both plastic and elastic behavior. Ductility specified by permanent elongation to failure or by the permanent reduction in cross-sectional area Brittle materials exhibit little or no yielding before failure CHAPTER REVIEW • Yield point for material can be increased by strain hardening, by applying load great enough to cause increase in stress causing yielding, then releasing the load. The larger stress produced becomes the new yield point for the material

37. Poisson’s ratio (ν), a dimensionless property that measures the lateral strain to the longitudinal strain [0 ≤ ν≤ 0.5] For shear stress vs. strain diagram: within elastic region, τ= Gγ, where G is the shearing modulus, found from the slope of the line within elastic region. G can also be obtained from the relationship ofG= E/[2(1+ ν)] CHAPTER REVIEW