Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter 2 Fundamentals of the Mechanical Behavior of Materials Fetweb.ju.edu.jo/staff/ie/mbarghash.

Similar presentations


Presentation on theme: "Chapter 2 Fundamentals of the Mechanical Behavior of Materials Fetweb.ju.edu.jo/staff/ie/mbarghash."— Presentation transcript:

1 Chapter 2 Fundamentals of the Mechanical Behavior of Materials Fetweb.ju.edu.jo/staff/ie/mbarghash

2 Types of Strain FIGURE 2.1 Types of strain. (a) Tensile, (b) compressive, and (c) shear. All deformation processes in manufacturing involve strains of these types. Tensile strains are involved in stretching sheet metal to make car bodies, compressive strains in forging metals to make turbine disks, and shear strains in making holes by punching.

3 Universal testing machine

4 Tension Test Figure 2.2 (a) Original and final shape of a standard tensile-test specimen. (b) Outline of a tensile-test sequence showing stages in the elongation of the specimen.

5

6

7

8 Understanding shear Not required Definition

9 Imperfections in engineering strain

10

11 First case ln(2Lo/Lo)=ln(2) Second case ln(Lo/2Lo)=ln(1/2)=-ln(2)

12 ln(2)= ln(1.5)+ln(2/1.5)=0.6931

13

14

15

16 Stess in three dimensions n is a vector, so tn is the stress on a plan with n as orthogonal to it

17 Special case n= x, There are other cases where n=y, n=z Thus we have 9 stress vectors

18

19

20 Hydraustaic and deviatoric stress

21

22

23

24 Homework: prove the above relation

25

26 The plastic deformation, the plastic deformation Causes zero volume change Prove the above relation for A unit cube

27

28

29 Mechanical Properties of Materials Table 2.1 Typical mechanical properties of various materials at room temperature.

30 Loading and Unloading FIGURE 2.3 Schematic illustration of loading and unloading of a tensile-test specimen. Note that during unloading, the curve follows a path parallel to the original elastic slope.

31 True Stress- True-Strain Curves in Tension FIGURE 2.5 (a) True stress-true-strain curve in tension. Note that, unlike in an engineering stress-strain curve, the slope is always positive, and the slope decreases with increasing strain. Although stress and strain are proportional in the elastic range, the total curve can be approximated by the power expression shown. On this curve, Y is the yield stress and Y f is the flow stress. (b) True-stress true-strain curve plotted on a log-log scale. (c) True stress-true-strain curve in tension for 1100-O aluminum plotted on a log-log scale. Note the large difference in the slopes in the elastic and plastic ranges. Source: After R. M. Caddell and R. Sowerby.

32 Power Law Material Behavior Table 2.3 Typical values of K and n in Eq. (2.11) at room temperature. K: strength coefficient n: strain hardening coefficient

33 True Stress - True Strain Curves for Various Metals FIGURE 2.6 True-stress-true-strain curves in tension at room temperature for various metals. The point of intersection of each curve at the ordinate is the yield stress Y; thus, the elastic portions of the curves are not indicated. When the K and n values are determined from these curves, they may not agree with those given in Table 2.3, because of the different sources from which they were collected. Source: S. Kalpakjian.

34 Strain Rate Effects Table 2.5 Approximate range of values for C and m in Eq. (2.16) for various annealed materials at true strains ranging from 0.2 to 1.0. C: strength coefficient M: strain rate sensitivity exponent

35 Effect of temperature Power Law Creep One of the most common forms of plastic flow is Power-Law Creep, given by the formula: Strain Rate = C (Stress)n exp(-Q/RT) Let's take each part of the formula in turn: C is a scaling constant. n means that the strain rate increases much faster than stress. Typically n is about 3 but can range from a bit less than 2 to 8. Recall that with viscous deformation stress is proportional to strain rate (n=1). With power-law creep it's faster: the effective viscosity drops with stress. Q is the activation energy required to get crystal dislocations moving. It's typically kilojoules per mole, sometimes up to 500. R is the Universal Gas Constant that turns up everywhere in physical chemistry. In SI units it equals joules/mole-degree Kelvin. T is the temperature in degrees Kelvin. As T increases, Q/RT decreases and thus exp(-Q/RT) increases, though much more slowly than exp(T). At very large T, Q/RT approaches zero and the exponential term approaches 1. This does not happen, though, at geologically realistic temperatures.

36 Barreling In Compression FIGURE 2.15 Barreling in compression of a round solid cylindrical specimen (7075-O aluminum) between flat dies. Barreling is caused by interfaces, which retards the free flow of the material. See also Figs. 6.1 and 6.2. Source: K. M. Kulkarni and S. Kalpakjian. FIGURE 2.16 Schematic illustration of the plane-strain compression test. The dimensional relationships shown should by satisfied for this test to be useful and reproducible. This test give the yield stress of the material in plane strain, Y’. Source: After A. Nadai and H. Ford. Plane Strain Compression

37

38

39

40

41

42 Toughness It is the Area under the stress strain curve

43

44 Derive the instability point for plain stress case with equal stresses

45 Principle stresses Since x, y, z are optional coordinates, then we can obtain Certain coordinates where the shear stresses disappear Example, the normal directions in simple tension

46

47 Two dimensional-plain stress case

48

49

50

51

52 Homework, (don’t submitted it, just solve it) Find the maximum stress case for the simple tension And Find the principle stresses for the simple shear case

53

54 Not required

55

56

57

58 Plain strain transformation

59 Not required

60

61

62

63

64

65

66 Plane Strain A state of plane strain exits when the strains are confined to a single plane, such as the x-y plane. This generally means that the stresses in the other direction; eg., the z direction, are non-zero. Plane Strain occurs in thick sections that “constrain” out of plane deformations

67 Transformations in plane strain The state of plane strain at a point p is given by  x  y  xy.  Determine the principal strains and the maximum in-plane shear strain and show the orientations of the elements subjected to these strains. Also determine the absolute maximum shear strain.  =

68 Solution tan2 p  xy  x  y  p 1 2 atan  xy  x  y   1 2 atan  ( )

69 Principal Strains  ' x  x  y 2  x  y 2 cos2  p   xy 2 sin2  p   ' y  x  y 2  x  y 2 cos2  p   xy 2 sin2  p   ' x 22 cos  2 sin º º =  ’ y =

70 Maximum Shear Strain  test  ' x  ' y  ' x  ' y =  max = Use diameters of circles!, Mohr’s circle plots , so  is the diameter (=2x radius).

71 Max In-Plane Shear Strain =9.2º =

72 Mohr’s Circle Plot normal strain on the x-axis Plot ½ the shear strain on the y-axis Solve as you would for plane stress problem  

73 Strain energy For 3D case For principle stress case For the elastic region

74 Plane Strain A state of plane strain exits when the strains are confined to a single plane, such as the x-y plane. This generally means that the stresses in the other direction; eg., the z direction, are non-zero. Plane Strain occurs in thick sections that “constrain” out of plane deformations

75 Transformations in plane strain The state of plane strain at a point p is given by  x  y  xy.  Determine the principal strains and the maximum in-plane shear strain and show the orientations of the elements subjected to these strains. Also determine the absolute maximum shear strain.  =

76 Solution tan2 p  xy  x  y  p 1 2 atan  xy  x  y   1 2 atan  ( )

77 Principal Strains  ' x  x  y 2  x  y 2 cos2  p   xy 2 sin2  p   ' y  x  y 2  x  y 2 cos2  p   xy 2 sin2  p   ' x 22 cos  2 sin º º =  ’ y =

78 Maximum Shear Strain  test  ' x  ' y  ' x  ' y =  max = Use diameters of circles!, Mohr’s circle plots , so  is the diameter (=2x radius).

79 Max In-Plane Shear Strain =9.2º =

80 Mohr’s Circle Plot normal strain on the x-axis Plot ½ the shear strain on the y-axis Solve as you would for plane stress problem  

81

82 *Prove the above relations *prove that equation 21 and 22 are the same (for the plain stress case)

83 Principle stresses in three direction

84 We want to determine the normal and shear stresses on the plane. The normal stresses are easiest. Unfortunately, we can't add stresses, only forces, so we have to determine the forces the stresses exert, add them up, then convert back to stress. Consider stress S1. It acts along the X1 axis,.but the stress "sees" only the area of the plane visible along the X1 axis, which is c1. So F1 = S1c1. Similarly, F2 = S2c2 and F3 = S3c3. The force normal to the plane exerted by F1 is F1c1, and the total force normal to the plane is F1c1 + F2c2 + F3c3. Since F1 = S1c1, we find: Fn = S1c12 + S2c22 + S3c32 Furthermore, stress = force/area, but the area of the plane is one, so we have Sn = S1c12 + S2c22 + S3c32 Determining shear stress can be a lot messier, if we do things the brute force way. Or we can do it the easy way.

85 Here we are looking in the plane of the normal and shear forces. It's obvious from the vector diagram that F2 = Fn2 + Fs2. Since the plane has an area of one and stress = force per unit area, we have F2 = Sn2 + Ss2. Note that it's only the magnitudes of the stresses that we are adding. Stresses do not add vectorially! The total force F can be found from the three vectors F1, F2 and F3 above. Since these three components are mutually perpendicular, we have F2 = F12 + F22 + F32 or Sn2 + Ss2 = S12c12 + S22c22 + S32c32 (this will be very useful a bit later)

86 So we have: Ss2 = F2 - Sn2 = F12 + F22 + F32 - (S1c12 + S2c22 + S3c32)2 = S12c12 + S22c22 + S32c32 - (S1c12 + S2c22 + S3c32)2 = S12c12 + S22c22 + S32c32 - S12c14 - S22c24 - S32c34 - 2S1S2c12c22 - S2S3c32c22 - S3S1c12c32 We regroup terms to get: Ss2 = S12c12(1 - c12) + S22c22(1 - c22) + S32c32(1 - c32) - 2S1S2c12c22 - S2S3c32c22 - S3S1c12c32 Now, since 1 - c12 = c22 + c32, we can rewrite the above as: Ss2 = S12c12(c22 + c32) + S22c22(c32 + c12) + S32c32(c22 + c12) - 2S1S2c12c22 - S2S3c32c22 - S3S1c12c32 Gathering terms, we get = (S12 - 2S1S2 + S22)c12c22 + (S22 - 2S2S3 + S32)c32c22 + (S32 - 2S3S1 + S12)c12c32 Ss2 = (S1 - S2)2c12c22 + (S2 - S3)2c32c22 + (S1 - S3)2c12c32

87

88

89   Triaxial Stress State (+ve sense shown)

90 3D Principal – Triaxial Stress

91 3D Stress – Principal Stresses The three principal stresses are obtained as the three real roots of the following equation: where I 1, I 2, and I 3 are known as stress invariants as they do not change in value when the axes are rotated to new positions.

92 Stress Invariants for Principal Stress Zero shear stress on principal planes

93 Derive the two dimensional principle stress case using the Equations for the tri-axial stress case

94 Mohr’s Circle? There is no Mohr’s circle solution for problems of triaxial stress state Solution for maximum principal stresses and maximum shear stress is analytical Either closed form solution or numerical solution (or computer program) are used to solve the eigenvalue problem.

95 Maximum Shear Stresses Absolute max shear stress is the numerically larger of: Normal Stress,  33 11  y’z ’,  abs  max 22  x’y’  y’z’

96 3D Mohr’s Circle – Plane Stress A Case Study – The two principal stresses are of the same sign  33 11 22 

97 3D Mohr’s Circle – Plane Stress A Case Study – The two principal stresses are of opposite sign  33 11 22 

98 For the following state of stress, find the principal and critical values. Tensor shows that:  z = 0 and  xz =  yz = 0 80 MPa 120 MPa 50 MPa y x Example :

99 120 MPa 0 MPa 80 MPa 0 MPa y z x z The other 2 faces:

100 Shear Stress, MPa   max = 77 MPa 3-D Mohr’s Circles

101 Example: triaxial stress state, not plane stress Determine the maximum principal stresses and the maximum shear stress for the following triaxial stress state. (+ve values as defined in slide 1)  MPa

102 Solution   MPa = –10 = 40 MPa = MPa = MPa Solve

103 Results

104 Mohr’s circles Normal Stress,  MPa   3 =63.5  1 =  y’z ’,  abs  max =58.5  2 =26.5 Shear (MPa)

105 Safety Factor? If the stress state was determined on a steel crankshaft, made of forged SAE1045 steel with a yield strength of 300 MPa, what is the factor of safety against yield? 1.Tresca Criterion:  max = 58.5 MPa 2. Max Principal Stress Criterion:  max = 63.5 MPa Not Required

106 3. Von Mises Criterion: = MPa NOT REQUIRED

107 Yield criterion

108 For simple tension

109 Tresca yield locus

110

111

112

113

114

115

116

117

118

119

120

121

122

123

124

125

126

127 Flow rules

128

129

130

131 Prove it Ignoring the Shear strains

132

133

134 Efficiency = ideal work / (total work)

135

136 Next lecture problems and solutions for two lectures And tutorials

137 Using abacus, the program is a finite element software for Modelling and analysis of Manufacturing processes We will use it to understand the operations and its analysis

138


Download ppt "Chapter 2 Fundamentals of the Mechanical Behavior of Materials Fetweb.ju.edu.jo/staff/ie/mbarghash."

Similar presentations


Ads by Google