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Continuum Aspects of Plastic Flow
CHAPTER 12
Continuum Aspects of Plastic Flow
CONTENTS 12.1 Introduction ......................................................................................179 12.2 Onset of yielding and shear yield strength, k.....................................180 12.3 Analyzing the hardness test ............................................................182 12.4 Plastic instability: necking in tensile loading .................................183 Worked Example..........................................................................................187 Examples ..................................................................................................188 Answers ....................................................................................................190 Plastic bending of beams, torsion of shafts, buckling of struts ............191
12.1
INTRODUCTION
Plastic flow occurs by shear. Dislocations move when the shear stress on the slip plane exceeds the dislocation yield strength τy of a single crystal. If this is averaged over all grain orientations and slip planes, it can be related to the tensile yield strength σ y of a polycrystal by σ y ¼ 3τy (Chapter 11). In solving problems of plasticity, however, it is more useful to define the shear yield strength k of a polycrystal. It is equal to σ y/2, and differs from τy because it is an average shear resistance over all orientations of slip plane. When a structure is loaded, the planes on which shear will occur can often be identified or guessed, and the collapse load calculated approximately by requiring that the stress exceed k on these planes. In this chapter, we show that k ¼ σ y/2, and use k to relate the hardness to the yield strength of a solid. We then examine tensile instabilities which appear in the drawing of metals and polymers.
179 Engineering Materials 1: An Introduction to Properties, Applications and Design, Fifth Edition. https://doi.org/10.1016/B978-0-08-102051-7.00011-7 © 2019, David R. H. Jones and Michael F. Ashby. Published by Elsevier Ltd. All rights reserved.
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12.2 ONSET OF YIELDING AND SHEAR YIELD STRENGTH, k A tensile stress applied to a piece of material will create a shear stress at an angle to the tensile stress. Let us examine the stresses in more detail. Resolving forces in Figure 12.1 gives the shearing force as F sin θ. The area over which this force acts in shear is A=cos θ: Thus the shear stress, τ is τ¼
F sin θ F ¼ sin θ cos θ ¼ σ sin θ cos θ A=cosθ A
(12.1)
If we plot this against θ as in Figure 12.2 we find a maximum τ at θ ¼ 45° to the tensile axis. This means that the highest value of the shear stress is found at 45° to the tensile axis, and has a value of σ/2. Now, from what we have said in Chapters 10 and 11, if we are dealing with a single crystal, the crystal will not in fact slip on the 45° plane—it will slip on the nearest lattice plane to the 45° plane on which dislocations can glide (Figure 12.3). In a polycrystal, neighboring grains each yield on their
F
F sin
Area A
F cos
Area A cos
F cos
F sin F
FIGURE 12.1 A tensile stress, F/A, produces a shear stress, τ, on an inclined plane in the stressed material.
12.2
F
= sin cos
Onset of Yielding and Shear Yield Strength, k
Maximum
45° F
1 2
F
F =0 0
45°
90°
=0
F
F
FIGURE 12.2 Shear stresses in a material have their maximum value on planes at 45° to the tensile axis. F
45°
F
F
F = A y
45° 45°
F
F
FIGURE 12.3 In a polycrystalline material the average slip path is at 45° to the tensile axis.
nearest-to-45° slip planes. On a microscopic scale, slip occurs on a zigzag path; but the average slip path is at 45° to the tensile axis. The shear stress on this plane when yielding occurs is therefore τ ¼ σ y/2, and we define this as the shear yield strength k: k ¼ σ y =2
(12.2)
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12.3
ANALYZING THE HARDNESS TEST
The concept of shear yielding—where we ignore the details of the grains in the polycrystal and treat the material as a continuum—is useful in many respects. For example, we can use it to calculate the loads that would make the material yield for all sorts of quite complicated geometries. A good example is the problem of the hardness indenter that we referred to in the hardness test in Chapter 9. Then, we stated that the hardness F H ¼ ¼ 3σ y A
We assume that the material does not work-harden so as the indenter is pushed into the material, the yield strength does not change. For simplicity, we consider a two-dimensional model. (A real indenter is three-dimensional, but the result is, for practical purposes, the same.) As we press a flat indenter into the material, shear takes place on the 45° planes of maximum shear stress shown in Figure 12.4, at a value of shear stress equal to k. By equating the work done by the force F as the indenter sinks a distance u to the work done against k on the shear planes, we get: pffiffiffi Ak Ak u Fu ¼ 2 pffiffiffi u 2 + 2 Ak u + 4 pffiffiffi pffiffiffi 2 2 2
This simplifies to F ¼ 6Ak
from which F ¼ 6k ¼ 3σ y A
F Area A u
u/2 u 2
u/2
u 2 u
This Δ moves sideways by u
u
This Δ moves up by u/2, but also moves u in a direction 2
u A 2 Area 2
Area A This Δ moves down by u
FIGURE 12.4 The plastic flow of material under a hardness indenter modeled by a mechanism of sliding triangles.
12.4
Plastic Instability: Necking in Tensile Loading
But F/A ¼ H so H ¼ 3σ y
(12.3)
(Strictly, shear occurs not just on the shear planes we have drawn, but on myriad 45° planes near the indenter. If our assumed geometry for slip is wrong it can be shown rigorously by a theorem called the upper-bound theorem that the value we get for F at yield—the so-called “limit” load—is always on the high side.) Similar treatments can be used for all sorts of two-dimensional problems: for calculating the plastic collapse load of structures of complex shape, and for analyzing metal-working processes like forging, rolling, and sheet drawing.
12.4 PLASTIC INSTABILITY: NECKING IN TENSILE LOADING We now turn to the other end of the stress–strain curve and explain why, in tensile straining, materials eventually start to neck, a name for plastic instability. It means that flow becomes localized across one section of the specimen or component, as shown in Figure 12.5, and (if straining continues) the material fractures there. Plasticine necks readily; chewing gum is very resistant to necking. We analyze the instability by noting that if a force F is applied to the end of the specimen, then any section must carry this load. But is it capable of doing so? Suppose one section deforms a little more than the rest, as the figure shows. Its section is less, and the stress in it is therefore larger than elsewhere. If workhardening has raised the yield strength enough, the reduced section can still carry the force F; but if it has not, plastic flow will become localized at the neck Section area
Section area A
A
Section area A
L
Section area A – dA + d
FIGURE 12.5 The formation of a neck in a bar of material that is being deformed plastically.
L+ dL
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and the specimen will fail there. Any section of the specimen can carry a force Aσ, where A is its area, and σ its current strength. If Aσ increases with strain, the specimen is stable. If it decreases, it is unstable and will neck. The critical condition for the start of necking is that Aσ ¼ F ¼ constant
Then Adσ + σdA ¼ 0
or dσ dA ¼ σ A
But volume is conserved during plastic flow, so
dA dL ¼ ¼ dε A L
(prove this by differentiating AL ¼ constant). So dσ ¼ dε σ
or dσ ¼σ dε
(12.4)
This equation is given in terms of true stress and true strain. As we saw in Chapter 9, tensile data are usually given in terms of nominal stress and strain. From Chapter 9: σ ¼ σ n ð1 + εn Þ ε ¼ lnð1 + εn Þ
If these are differentiated and substituted into the necking equation we get dσ n ¼0 dεn
(12.5)
In other words, on the point of instability, the nominal stress–strain curve is at its maximum as we know experimentally from Chapter 9. To see what is going on physically, it is easier to return to our first condition. At low stress, if we make a little neck, the material in the neck will work-harden and will be able to carry the extra stress it has to stand because of its smaller area; load will therefore be continuous, and the material will be stable. At high stress, the rate of work-hardening is less as the true stress–true strain curve shows: that is,
12.4
Plastic Instability: Necking in Tensile Loading
Onset of necking Slope,
d = de
d = de
e
FIGURE 12.6 The condition for necking.
the slope of the σ/ε curve is less. Eventually, we reach a point at which, when we make a neck, the work-hardening is only just enough to stand the extra stress. This is the point of necking (Figure 12.6) with dσ ¼σ dε
At still higher true stress dσ=dε, the rate of work-hardening decreases further, becoming insufficient to maintain stability—the extra stress in the neck can no longer be accommodated by the work-hardening produced by making the neck, and the neck grows faster and faster, until final fracture takes place.
Consequences of plastic instability Plastic instability is very important in processes like deep drawing sheet metal to form car bodies, cans, and so on. Obviously we must ensure that the materials and press designs are chosen carefully to avoid instability. Mild steel is a good material for deep drawing in the sense that it flows a great deal before necking starts. It can therefore be drawn very deeply without breaking (Figure 12.7). Aluminum alloy is much less good (Figure 12.8)—it can only be drawn a little before instabilities form. Pure aluminum is not nearly as bad, but is much too soft to use for most applications. Polythene shows a kind of necking that does not lead to fracture. Figure 12.9 shows its σ n/εn curve. At quite low stress dσ n !0 dεn
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n
en
FIGURE 12.7 Mild steel can be drawn out a lot before it fails by necking. n
en
FIGURE 12.8 Aluminum alloy quickly necks when it is drawn out.
and necking begins. However, the neck never becomes unstable—it simply grows in length—because at high strain the material work-hardens considerably, and is able to support the increased stress at the reduced cross-section of the neck. This odd behavior is caused by the lining up of the polymer chains in the neck along the direction of the neck—and for this sort of reason drawn (i.e., “fully necked”) polymers can be made to be very strong indeed—much stronger than the undrawn polymers. Finally, mild steel can sometimes show an instability like that of polythene. If the steel is annealed, the stress–strain curve looks like that in Figure 12.10. A stable neck, called a L€ uders Band, forms, and propagates (as it did in polythene) without causing fracture because the strong work-hardening of the later part of the stress–strain curve prevents this. L€ uders Bands are a problem when sheet steel is pressed because they give lower precision and disfigure the pressing.
Worked Example
n
Necking
en
FIGURE 12.9 Polythene forms a stable neck when it is drawn out; drawn polythene is very strong. n
Unstable necking
Lüders Bands– stable necking
y
en
FIGURE 12.10 Mild steel often shows both stable and unstable necks.
WORKED EXAMPLE Referring to Figure 12.9, we can easily demonstrate stable necking in polymers using strips cut from a polythene bag. You need to cut the strips across the bag (the polymer chains are already partially lined up along the length of the bag because of the forming process). You then grip each end of the strip between thumb and index finger, and pull out the strip as shown in the diagram.
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After drawing Specimen 8
55
Extrusion direction
EXAMPLES 12.1 Determine whether the bolt passing through the plate will fail, when loaded in tension, by yielding of the shank or shearing-off of the head. (Assume no work-hardening.) 12.2 A metal bar of width w is compressed between two hard anvils as shown in the diagram. The third dimension of the bar, L, is much greater than w. Plastic deformation takes place as a result of shearing along planes, defined by the dashed lines in the diagram, at a shear stress k. Find an upper bound for the load F when there is no friction between anvils and bar. Note that there are three distinct types of blocks in the model, labeled A, B, and C. As each anvil moves down/up by a small distance u, A does nothing—the points at top and bottom of A are simply blunted. Blocks C move sideways by distance 2u, but apart from this do nothing. The four B blocks move sideways by distance u, but also move down/up by distance u, which means that they shear against blocks A and C by a distance √ 2u. The eight shearing interfaces between blocks B and A/C consume shear work, which is provided by the work input from the two moving anvils. 12.3. Show that the same result can be obtained from simple force equilibrium. Why?
Examples
F
20
9
32
Dimensions in mm
F
B d = w/2
C
u
B A
u C
B
B
F
W
12.4. Repeat the calculation of Example 12.2, but instead assume there is sufficient friction to effectively weld the anvils to the bar. Show that the solution satisfies the general formula w F 2wLk 1 + 4d
which defines upper bounds for all integral values of w/2d. 12.5 A composite material used for rock-drilling bits consists of tungsten carbide cubes (each 20 μm in size) stuck together with a thin layer of cobalt. The
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material is required to withstand compressive stresses of 4000 MN m2 in service. Use the general formula in Example 12.4 to estimate an upper limit for the thickness of the cobalt layer. You may assume that the compressive yield stress of tungsten carbide is well above 4000 MN m2, and that the cobalt yields in shear at k ¼ 175 MN m2. 12.6 Discuss the assumption that when a piece of metal is plastically deformed at constant temperature, its volume is unchanged. 12.7 A ductile metal wire of uniform cross-section is loaded in tension until it just begins to neck. Assuming that volume is conserved, derive a differential expression relating the true stress to the true strain at the onset of necking. 12.8 The curve of true stress against true strain for the metal wire approximates to σ ¼ 350ε0:4 MN m2
12.9 12.10 12.11 12.12
Estimate the tensile strength of the wire. Remember that the tensile strength is the nominal stress at onset of necking. You will first need to find the true strain at onset of necking, then the true stress at onset of necking, then use the equations in Section 12.4 to convert between true and nominal stresses and strains. Referring to Example 12.8, estimate the work required to take 1 m3 of the wire to the point of necking. To keep things simple, work with true stress and strain. If the true stress–true strain curve for a material is defined by σ ¼ Aεn, where A and n are constants, find the tensile strength σ TS. For a nickel alloy, n ¼ 0.2 and A ¼ 800 MN m2. Evaluate the tensile strength of the alloy. Evaluate the true stress in an alloy specimen loaded to σ TS. The indentation hardness, H, is given by H 3σ y where σ y is the true yield stress at a nominal plastic strain of 8%. If the true stress–strain curve of a material is given by σ ¼ Aεn
and n ¼ 0.2, calculate the tensile strength of a material for which the indentation hardness is 600 MN m2. You may assume that σ TS ¼ Ann/εn.
ANSWERS 12.1 Tensile yield force in shank is πr2σ y ¼ 314 σ y. Shear yield force in head is 2πrtk ¼ πrtσ y ¼ 282 σ y. The head will shear off first. 12.2 Equating work input 2Fu to work consumed by the 8 sliding interfaces gives F ¼ 2wLk. 12.3 Because F ¼ wLσ y. See also Equation (12.2). There is no friction between anvils and bar, so the metal behaves as if it is an unconstrained compression specimen. See Figure 9.8. (Interestingly, because force equilibrium and work balance both give the same answer, in this particular case the upper bound analysis is exact.) 12.4 The four B blocks also shear against the anvils by a distance u, and this must be added to the work, giving F ¼ 3wLk.
Plastic Bending of Beams, Torsion of Shafts, Buckling of Struts
12.5 12.6 12.7 12.8 12.9 12.10 12.11 12.12
0.48 μm. See Section 9.4. Also see Figures 12.2, 12.4, etc. See Section 12.4, Equation (12.4). 163 MN m–2. 69.3 MJ. σ TS ¼ Ann/en. Also see Example 12.8. 475 MN m–2. 580 MN m–2. 198 MN m–2. Remember to convert the strain from 8% to 0.08. Note that 198/600 ¼ 0.33, in good agreement with Equation (9.9).
PLASTIC BENDING OF BEAMS, TORSION OF SHAFTS, BUCKLING OF STRUTS Bending of beams Mp
Mp
Neutral axis (no strain) = y (compression)
= y (tension) A 2
dA q
A 2
Mp =
Ú
A
y dA q
Neutral axis of area A
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Plastic moments
a
Mp =
ya3 4
a
d
Mp =
ybd 2 4
b
r
Mp =
r1
Mp =
r2
4yr 3 3
4y (r 13 - r 23) 3
r
Mp = 4ytr 2 t (<< r)
Centroid (center of gravity) of area A A e x
Mpx = yAe x
Neutral axis of complete section
Plastic Bending of Beams, Torsion of Shafts, Buckling of Struts
Shearing torques T r
T=
2kr 3 3
T=
2k(r 13 − r 23) 3
T r1 r2
T r
T = 2kr 2t t (<< r)
Plastic buckling The results given in Chapter 7 for the elastic buckling of beams can be applied to situations where the buckling is plastic, provided Young’s modulus E is replaced in the equations by the tangent modulus Et which is defined in the following diagram.
)
Et = d d critical
Critical buckling stress E
193
Continuum Aspects of Plastic Flow
CONTENTS 12.1 Introduction ......................................................................................179 12.2 Onset of yielding and shear yield strength, k.....................................180 12.3 Analyzing the hardness test ............................................................182 12.4 Plastic instability: necking in tensile loading .................................183 Worked Example..........................................................................................187 Examples ..................................................................................................188 Answers ....................................................................................................190 Plastic bending of beams, torsion of shafts, buckling of struts ............191
12.1
INTRODUCTION
Plastic flow occurs by shear. Dislocations move when the shear stress on the slip plane exceeds the dislocation yield strength τy of a single crystal. If this is averaged over all grain orientations and slip planes, it can be related to the tensile yield strength σ y of a polycrystal by σ y ¼ 3τy (Chapter 11). In solving problems of plasticity, however, it is more useful to define the shear yield strength k of a polycrystal. It is equal to σ y/2, and differs from τy because it is an average shear resistance over all orientations of slip plane. When a structure is loaded, the planes on which shear will occur can often be identified or guessed, and the collapse load calculated approximately by requiring that the stress exceed k on these planes. In this chapter, we show that k ¼ σ y/2, and use k to relate the hardness to the yield strength of a solid. We then examine tensile instabilities which appear in the drawing of metals and polymers.
179 Engineering Materials 1: An Introduction to Properties, Applications and Design, Fifth Edition. https://doi.org/10.1016/B978-0-08-102051-7.00011-7 © 2019, David R. H. Jones and Michael F. Ashby. Published by Elsevier Ltd. All rights reserved.
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12.2 ONSET OF YIELDING AND SHEAR YIELD STRENGTH, k A tensile stress applied to a piece of material will create a shear stress at an angle to the tensile stress. Let us examine the stresses in more detail. Resolving forces in Figure 12.1 gives the shearing force as F sin θ. The area over which this force acts in shear is A=cos θ: Thus the shear stress, τ is τ¼
F sin θ F ¼ sin θ cos θ ¼ σ sin θ cos θ A=cosθ A
(12.1)
If we plot this against θ as in Figure 12.2 we find a maximum τ at θ ¼ 45° to the tensile axis. This means that the highest value of the shear stress is found at 45° to the tensile axis, and has a value of σ/2. Now, from what we have said in Chapters 10 and 11, if we are dealing with a single crystal, the crystal will not in fact slip on the 45° plane—it will slip on the nearest lattice plane to the 45° plane on which dislocations can glide (Figure 12.3). In a polycrystal, neighboring grains each yield on their
F
F sin
Area A
F cos
Area A cos
F cos
F sin F
FIGURE 12.1 A tensile stress, F/A, produces a shear stress, τ, on an inclined plane in the stressed material.
12.2
F
= sin cos
Onset of Yielding and Shear Yield Strength, k
Maximum
45° F
1 2
F
F =0 0
45°
90°
=0
F
F
FIGURE 12.2 Shear stresses in a material have their maximum value on planes at 45° to the tensile axis. F
45°
F
F
F = A y
45° 45°
F
F
FIGURE 12.3 In a polycrystalline material the average slip path is at 45° to the tensile axis.
nearest-to-45° slip planes. On a microscopic scale, slip occurs on a zigzag path; but the average slip path is at 45° to the tensile axis. The shear stress on this plane when yielding occurs is therefore τ ¼ σ y/2, and we define this as the shear yield strength k: k ¼ σ y =2
(12.2)
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12.3
ANALYZING THE HARDNESS TEST
The concept of shear yielding—where we ignore the details of the grains in the polycrystal and treat the material as a continuum—is useful in many respects. For example, we can use it to calculate the loads that would make the material yield for all sorts of quite complicated geometries. A good example is the problem of the hardness indenter that we referred to in the hardness test in Chapter 9. Then, we stated that the hardness F H ¼ ¼ 3σ y A
We assume that the material does not work-harden so as the indenter is pushed into the material, the yield strength does not change. For simplicity, we consider a two-dimensional model. (A real indenter is three-dimensional, but the result is, for practical purposes, the same.) As we press a flat indenter into the material, shear takes place on the 45° planes of maximum shear stress shown in Figure 12.4, at a value of shear stress equal to k. By equating the work done by the force F as the indenter sinks a distance u to the work done against k on the shear planes, we get: pffiffiffi Ak Ak u Fu ¼ 2 pffiffiffi u 2 + 2 Ak u + 4 pffiffiffi pffiffiffi 2 2 2
This simplifies to F ¼ 6Ak
from which F ¼ 6k ¼ 3σ y A
F Area A u
u/2 u 2
u/2
u 2 u
This Δ moves sideways by u
u
This Δ moves up by u/2, but also moves u in a direction 2
u A 2 Area 2
Area A This Δ moves down by u
FIGURE 12.4 The plastic flow of material under a hardness indenter modeled by a mechanism of sliding triangles.
12.4
Plastic Instability: Necking in Tensile Loading
But F/A ¼ H so H ¼ 3σ y
(12.3)
(Strictly, shear occurs not just on the shear planes we have drawn, but on myriad 45° planes near the indenter. If our assumed geometry for slip is wrong it can be shown rigorously by a theorem called the upper-bound theorem that the value we get for F at yield—the so-called “limit” load—is always on the high side.) Similar treatments can be used for all sorts of two-dimensional problems: for calculating the plastic collapse load of structures of complex shape, and for analyzing metal-working processes like forging, rolling, and sheet drawing.
12.4 PLASTIC INSTABILITY: NECKING IN TENSILE LOADING We now turn to the other end of the stress–strain curve and explain why, in tensile straining, materials eventually start to neck, a name for plastic instability. It means that flow becomes localized across one section of the specimen or component, as shown in Figure 12.5, and (if straining continues) the material fractures there. Plasticine necks readily; chewing gum is very resistant to necking. We analyze the instability by noting that if a force F is applied to the end of the specimen, then any section must carry this load. But is it capable of doing so? Suppose one section deforms a little more than the rest, as the figure shows. Its section is less, and the stress in it is therefore larger than elsewhere. If workhardening has raised the yield strength enough, the reduced section can still carry the force F; but if it has not, plastic flow will become localized at the neck Section area
Section area A
A
Section area A
L
Section area A – dA + d
FIGURE 12.5 The formation of a neck in a bar of material that is being deformed plastically.
L+ dL
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and the specimen will fail there. Any section of the specimen can carry a force Aσ, where A is its area, and σ its current strength. If Aσ increases with strain, the specimen is stable. If it decreases, it is unstable and will neck. The critical condition for the start of necking is that Aσ ¼ F ¼ constant
Then Adσ + σdA ¼ 0
or dσ dA ¼ σ A
But volume is conserved during plastic flow, so
dA dL ¼ ¼ dε A L
(prove this by differentiating AL ¼ constant). So dσ ¼ dε σ
or dσ ¼σ dε
(12.4)
This equation is given in terms of true stress and true strain. As we saw in Chapter 9, tensile data are usually given in terms of nominal stress and strain. From Chapter 9: σ ¼ σ n ð1 + εn Þ ε ¼ lnð1 + εn Þ
If these are differentiated and substituted into the necking equation we get dσ n ¼0 dεn
(12.5)
In other words, on the point of instability, the nominal stress–strain curve is at its maximum as we know experimentally from Chapter 9. To see what is going on physically, it is easier to return to our first condition. At low stress, if we make a little neck, the material in the neck will work-harden and will be able to carry the extra stress it has to stand because of its smaller area; load will therefore be continuous, and the material will be stable. At high stress, the rate of work-hardening is less as the true stress–true strain curve shows: that is,
12.4
Plastic Instability: Necking in Tensile Loading
Onset of necking Slope,
d = de
d = de
e
FIGURE 12.6 The condition for necking.
the slope of the σ/ε curve is less. Eventually, we reach a point at which, when we make a neck, the work-hardening is only just enough to stand the extra stress. This is the point of necking (Figure 12.6) with dσ ¼σ dε
At still higher true stress dσ=dε, the rate of work-hardening decreases further, becoming insufficient to maintain stability—the extra stress in the neck can no longer be accommodated by the work-hardening produced by making the neck, and the neck grows faster and faster, until final fracture takes place.
Consequences of plastic instability Plastic instability is very important in processes like deep drawing sheet metal to form car bodies, cans, and so on. Obviously we must ensure that the materials and press designs are chosen carefully to avoid instability. Mild steel is a good material for deep drawing in the sense that it flows a great deal before necking starts. It can therefore be drawn very deeply without breaking (Figure 12.7). Aluminum alloy is much less good (Figure 12.8)—it can only be drawn a little before instabilities form. Pure aluminum is not nearly as bad, but is much too soft to use for most applications. Polythene shows a kind of necking that does not lead to fracture. Figure 12.9 shows its σ n/εn curve. At quite low stress dσ n !0 dεn
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n
en
FIGURE 12.7 Mild steel can be drawn out a lot before it fails by necking. n
en
FIGURE 12.8 Aluminum alloy quickly necks when it is drawn out.
and necking begins. However, the neck never becomes unstable—it simply grows in length—because at high strain the material work-hardens considerably, and is able to support the increased stress at the reduced cross-section of the neck. This odd behavior is caused by the lining up of the polymer chains in the neck along the direction of the neck—and for this sort of reason drawn (i.e., “fully necked”) polymers can be made to be very strong indeed—much stronger than the undrawn polymers. Finally, mild steel can sometimes show an instability like that of polythene. If the steel is annealed, the stress–strain curve looks like that in Figure 12.10. A stable neck, called a L€ uders Band, forms, and propagates (as it did in polythene) without causing fracture because the strong work-hardening of the later part of the stress–strain curve prevents this. L€ uders Bands are a problem when sheet steel is pressed because they give lower precision and disfigure the pressing.
Worked Example
n
Necking
en
FIGURE 12.9 Polythene forms a stable neck when it is drawn out; drawn polythene is very strong. n
Unstable necking
Lüders Bands– stable necking
y
en
FIGURE 12.10 Mild steel often shows both stable and unstable necks.
WORKED EXAMPLE Referring to Figure 12.9, we can easily demonstrate stable necking in polymers using strips cut from a polythene bag. You need to cut the strips across the bag (the polymer chains are already partially lined up along the length of the bag because of the forming process). You then grip each end of the strip between thumb and index finger, and pull out the strip as shown in the diagram.
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After drawing Specimen 8
55
Extrusion direction
EXAMPLES 12.1 Determine whether the bolt passing through the plate will fail, when loaded in tension, by yielding of the shank or shearing-off of the head. (Assume no work-hardening.) 12.2 A metal bar of width w is compressed between two hard anvils as shown in the diagram. The third dimension of the bar, L, is much greater than w. Plastic deformation takes place as a result of shearing along planes, defined by the dashed lines in the diagram, at a shear stress k. Find an upper bound for the load F when there is no friction between anvils and bar. Note that there are three distinct types of blocks in the model, labeled A, B, and C. As each anvil moves down/up by a small distance u, A does nothing—the points at top and bottom of A are simply blunted. Blocks C move sideways by distance 2u, but apart from this do nothing. The four B blocks move sideways by distance u, but also move down/up by distance u, which means that they shear against blocks A and C by a distance √ 2u. The eight shearing interfaces between blocks B and A/C consume shear work, which is provided by the work input from the two moving anvils. 12.3. Show that the same result can be obtained from simple force equilibrium. Why?
Examples
F
20
9
32
Dimensions in mm
F
B d = w/2
C
u
B A
u C
B
B
F
W
12.4. Repeat the calculation of Example 12.2, but instead assume there is sufficient friction to effectively weld the anvils to the bar. Show that the solution satisfies the general formula w F 2wLk 1 + 4d
which defines upper bounds for all integral values of w/2d. 12.5 A composite material used for rock-drilling bits consists of tungsten carbide cubes (each 20 μm in size) stuck together with a thin layer of cobalt. The
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Continuum Aspects of Plastic Flow
material is required to withstand compressive stresses of 4000 MN m2 in service. Use the general formula in Example 12.4 to estimate an upper limit for the thickness of the cobalt layer. You may assume that the compressive yield stress of tungsten carbide is well above 4000 MN m2, and that the cobalt yields in shear at k ¼ 175 MN m2. 12.6 Discuss the assumption that when a piece of metal is plastically deformed at constant temperature, its volume is unchanged. 12.7 A ductile metal wire of uniform cross-section is loaded in tension until it just begins to neck. Assuming that volume is conserved, derive a differential expression relating the true stress to the true strain at the onset of necking. 12.8 The curve of true stress against true strain for the metal wire approximates to σ ¼ 350ε0:4 MN m2
12.9 12.10 12.11 12.12
Estimate the tensile strength of the wire. Remember that the tensile strength is the nominal stress at onset of necking. You will first need to find the true strain at onset of necking, then the true stress at onset of necking, then use the equations in Section 12.4 to convert between true and nominal stresses and strains. Referring to Example 12.8, estimate the work required to take 1 m3 of the wire to the point of necking. To keep things simple, work with true stress and strain. If the true stress–true strain curve for a material is defined by σ ¼ Aεn, where A and n are constants, find the tensile strength σ TS. For a nickel alloy, n ¼ 0.2 and A ¼ 800 MN m2. Evaluate the tensile strength of the alloy. Evaluate the true stress in an alloy specimen loaded to σ TS. The indentation hardness, H, is given by H 3σ y where σ y is the true yield stress at a nominal plastic strain of 8%. If the true stress–strain curve of a material is given by σ ¼ Aεn
and n ¼ 0.2, calculate the tensile strength of a material for which the indentation hardness is 600 MN m2. You may assume that σ TS ¼ Ann/εn.
ANSWERS 12.1 Tensile yield force in shank is πr2σ y ¼ 314 σ y. Shear yield force in head is 2πrtk ¼ πrtσ y ¼ 282 σ y. The head will shear off first. 12.2 Equating work input 2Fu to work consumed by the 8 sliding interfaces gives F ¼ 2wLk. 12.3 Because F ¼ wLσ y. See also Equation (12.2). There is no friction between anvils and bar, so the metal behaves as if it is an unconstrained compression specimen. See Figure 9.8. (Interestingly, because force equilibrium and work balance both give the same answer, in this particular case the upper bound analysis is exact.) 12.4 The four B blocks also shear against the anvils by a distance u, and this must be added to the work, giving F ¼ 3wLk.
Plastic Bending of Beams, Torsion of Shafts, Buckling of Struts
12.5 12.6 12.7 12.8 12.9 12.10 12.11 12.12
0.48 μm. See Section 9.4. Also see Figures 12.2, 12.4, etc. See Section 12.4, Equation (12.4). 163 MN m–2. 69.3 MJ. σ TS ¼ Ann/en. Also see Example 12.8. 475 MN m–2. 580 MN m–2. 198 MN m–2. Remember to convert the strain from 8% to 0.08. Note that 198/600 ¼ 0.33, in good agreement with Equation (9.9).
PLASTIC BENDING OF BEAMS, TORSION OF SHAFTS, BUCKLING OF STRUTS Bending of beams Mp
Mp
Neutral axis (no strain) = y (compression)
= y (tension) A 2
dA q
A 2
Mp =
Ú
A
y dA q
Neutral axis of area A
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Plastic moments
a
Mp =
ya3 4
a
d
Mp =
ybd 2 4
b
r
Mp =
r1
Mp =
r2
4yr 3 3
4y (r 13 - r 23) 3
r
Mp = 4ytr 2 t (<< r)
Centroid (center of gravity) of area A A e x
Mpx = yAe x
Neutral axis of complete section
Plastic Bending of Beams, Torsion of Shafts, Buckling of Struts
Shearing torques T r
T=
2kr 3 3
T=
2k(r 13 − r 23) 3
T r1 r2
T r
T = 2kr 2t t (<< r)
Plastic buckling The results given in Chapter 7 for the elastic buckling of beams can be applied to situations where the buckling is plastic, provided Young’s modulus E is replaced in the equations by the tangent modulus Et which is defined in the following diagram.
)
Et = d d critical
Critical buckling stress E
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