Introduction to Elasticity

INTRODUCTION TO ELASTICITY David Roylance Department of Materials Science and Engineering Massachusetts Institute of Technology Cambridge, MA 02139 January 21, 2000 Introduction This module outlines the basic mechanics of elastic response — a physical phenomenon that materials often (but do not always) exhibit. An elastic material is one that deforms immediately upon loading, maintains a constant deformation as long as the load is held constant, and returns immediately to its original undeformed shape when the load is removed. This module will also introduce two essential concepts in Mechanics of Materials: stress and strain. Tensile strength and tensile stress Perhaps the most natural test of a material’s mechanical properties is the tension test, in which a strip or cylinder of the material, having length L and cross-sectional area A, is anchored at one end and subjected to an axial load P – a load acting along the specimen’s long axis – at the other. (See Fig. 1). As the load is increased gradually, the axial deflection δ of the loaded end will increase also. Eventually the test specimen breaks or does something else catastrophic, often fracturing suddenly into two or more pieces. (Materials can fail mechanically in many different ways; for instance, recall how blackboard chalk, a piece of fresh wood, and Silly Putty break.) As engineers, we naturally want to understand such matters as how δ is related to P , and what ultimate fracture load we might expect in a specimen of different size than the original one. As materials technologists, we wish to understand how these relationships are influenced by the constitution and microstructure of the material. Figure 1: The tension test. One of the pivotal historical developments in our understanding of material mechanical properties was the realization that the strength of a uniaxially loaded specimen is related to the 1 magnitude of its cross-sectional area. This notion is reasonable when one considers the strength to arise from the number of chemical bonds connecting one cross section with the one adjacent to it as depicted in Fig. 2, where each bond is visualized as a spring with a certain stiffness and strength. Obviously, the number of such bonds will increase proportionally with the section’s area 1 . The axial strength of a piece of blackboard chalk will therefore increase as the square of its diameter. In contrast, increasing the length of the chalk will not make it stronger (in fact it will likely become weaker, since the longer specimen will be statistically more likely to contain a strength-reducing flaw.) Figure 2: Interplanar bonds (surface density approximately 10 19 m −2 ). Galileo (1564–1642) 2 is said to have used this observation to note that giants, should they exist, would be very fragile creatures. Their strength would be greater than ours, since the cross-sectional areas of their skeletal and muscular systems would be larger by a factor related to the square of their height (denoted L in the famous DaVinci sketch shown in Fig. 3). But their weight, and thus the loads they must sustain, would increase as their volume, that is by the cube of their height. A simple fall would probably do them great damage. Conversely, the “proportionate” strength of the famous arachnid mentioned weekly in the SpiderMan comic strip is mostly just this same size effect. There’s nothing magical about the muscular strength of insects, but the ratio of L 2 to L 3 works in their favor when strength per body weight is reckoned. This cautions us that simple scaling of a previously proven design is not a safe design procedure. A jumbo jet is not just a small plane scaled up; if this were done the load-bearing components would be too small in cross-sectional area to support the much greater loads they would be called upon to resist. When reporting the strength of Introduction to Elasticity Introduction to Elasticity By: OpenStaxCollege Netflix On-Demand Media Netflix, Inc is an American provider of on-demand Internet streaming media to many countries around the world, including the United States, and of flat rate DVD-by-mail in the United States (Credit: modification of work by Traci Lawson/Flickr Creative Commons) That Will Be How Much? Imagine going to your favorite coffee shop and having the waiter inform you the pricing has changed Instead of $3 for a cup of coffee, you will now be charged $2 for coffee, $1 for creamer, and $1 for your choice of sweetener If you pay your usual $3 for a cup 1/3 Introduction to Elasticity of coffee, you must choose between creamer and sweetener If you want both, you now face an extra charge of $1 Sound absurd? Well, that is the situation Netflix customers found themselves in—a 60% price hike to retain the same service In early 2011, Netflix consumers paid about $10 a month for a package consisting of streaming video and DVD rentals In July 2011, the company announced a packaging change Customers wishing to retain both streaming video and DVD rental would be charged $15.98 per month, a price increase of about 60% How would customers of the 14-year-old firm react? Would they abandon Netflix? Would the ease of access to other venues make a difference in how consumers responded to the Netflix price change? The answers to those questions will be explored in this chapter: the change in quantity with respect to a change in price, a concept economists call elasticity Introduction to Elasticity In this chapter, you will learn about: • • • • Price Elasticity of Demand and Price Elasticity of Supply Polar Cases of Elasticity and Constant Elasticity Elasticity and Pricing Elasticity in Areas Other Than Price Anyone who has studied economics knows the law of demand: a higher price will lead to a lower quantity demanded What you may not know is how much lower the quantity demanded will be Similarly, the law of supply shows that a higher price will lead to a higher quantity supplied The question is: How much higher? This chapter will explain how to answer these questions and why they are critically important in the real world To find answers to these questions, we need to understand the concept of elasticity Elasticity is an economics concept that measures responsiveness of one variable to changes in another variable Suppose you drop two items from a second-floor balcony The first item is a tennis ball The second item is a brick Which will bounce higher? Obviously, the tennis ball We would say that the tennis ball has greater elasticity Consider an economic example Cigarette taxes are an example of a “sin tax,” a tax on something that is bad for you, like alcohol Cigarettes are taxed at the state and national levels State taxes range from a low of 17 cents per pack in Missouri to $4.35 per pack in New York The average state cigarette tax is $1.51 per pack The current federal tax rate on cigarettes is $1.01 per pack, but in April 2013 the Obama Administration proposed raising the federal tax nearly a dollar to $1.95 per pack The key question is: How much would cigarette purchases decline? 2/3 Introduction to Elasticity Taxes on cigarettes serve two purposes: to raise tax revenue for government and to discourage consumption of cigarettes However, if a higher cigarette tax discourages consumption by quite a lot, meaning a greatly reduced quantity of cigarettes is sold, then the cigarette tax on each pack will not raise much revenue for the government Alternatively, a higher cigarette tax that does not discourage consumption by much will actually raise more tax revenue for the government Thus, when a government agency tries to calculate the effects of altering its cigarette tax, it must analyze how much the tax affects the quantity of cigarettes consumed This issue reaches beyond governments and taxes; every firm faces a similar issue Every time a firm considers raising the price that it charges, it must consider how much a price increase will reduce the quantity demanded of what it sells Conversely, when a firm puts its products on sale, it must expect (or hope) that the lower price will lead to a significantly higher quantity demanded 3/3 INTRODUCTION TO ELASTICITY David Roylance Department of Materials Science and Engineering Massachusetts Institute of Technology Cambridge, MA 02139 January 21, 2000 Introduction This module outlines the basic mechanics of elastic response — a physical phenomenon that materials often (but do not always) exhibit. An elastic material is one that deforms immediately upon loading, maintains a constant deformation as long as the load is held constant, and returns immediately to its original undeformed shape when the load is removed. This module will also introduce two essential concepts in Mechanics of Materials: stress and strain. Tensile strength and tensile stress Perhaps the most natural test of a material’s mechanical properties is the tension test, in which a strip or cylinder of the material, having length L and cross-sectional area A, is anchored at one end and subjected to an axial load P – a load acting along the specimen’s long axis – at the other. (See Fig. 1). As the load is increased gradually, the axial deflection δ of the loaded end will increase also. Eventually the test specimen breaks or does something else catastrophic, often fracturing suddenly into two or more pieces. (Materials can fail mechanically in many different ways; for instance, recall how blackboard chalk, a piece of fresh wood, and Silly Putty break.) As engineers, we naturally want to understand such matters as how δ is related to P , and what ultimate fracture load we might expect in a specimen of different size than the original one. As materials technologists, we wish to understand how these relationships are influenced by the constitution and microstructure of the material. Figure 1: The tension test. One of the pivotal historical developments in our understanding of material mechanical properties was the realization that the strength of a uniaxially loaded specimen is related to the 1 magnitude of its cross-sectional area. This notion is reasonable when one considers the strength to arise from the number of chemical bonds connecting one cross section with the one adjacent to it as depicted in Fig. 2, where each bond is visualized as a spring with a certain stiffness and strength. Obviously, the number of such bonds will increase proportionally with the section’s area 1 . The axial strength of a piece of blackboard chalk will therefore increase as the square of its diameter. In contrast, increasing the length of the chalk will not make it stronger (in fact it will likely become weaker, since the longer specimen will be statistically more likely to contain a strength-reducing flaw.) Figure 2: Interplanar bonds (surface density approximately 10 19 m −2 ). Galileo (1564–1642) 2 is said to have used this observation to note that giants, should they exist, would be very fragile creatures. Their strength would be greater than ours, since the cross-sectional areas of their skeletal and muscular systems would be larger by a factor related to the square of their height (denoted L in the famous DaVinci sketch shown in Fig. 3). But their weight, and thus the loads they must sustain, would increase as their volume, that is by the cube of their height. A simple fall would probably do them great damage. Conversely, the “proportionate” strength of the famous arachnid mentioned weekly in the SpiderMan comic strip is mostly just this same size effect. There’s nothing magical about the muscular strength of insects, but the ratio of L 2 to L 3 works in their favor when strength per body weight is reckoned. This cautions us that simple scaling of a previously proven design is not a safe design procedure. A jumbo jet is not just a small plane scaled up; if this were done the load-bearing components would be too small in cross-sectional area to support the much greater loads they would be called upon to resist. When reporting the strength of materials loaded in tension, it is customary to account for this effect of area by dividing the breaking load by the cross-sectional U = − (n − 1)ACe 2 nr 0 where A is the Madelung constant, C is the appropriate units conversion factor, and e is the ionic charge. 3. Measurements of bulk compressibility are valuable for probing the bond energy function, because unlike simple tension, hydrostatic pressure causes the interionic distance to de- crease uniformly. The modulus of compressibility K of a solid is the ratio of the pressure p needed to induce a relative change in volume dV/V : K = − dp (dV )/V The minus sign is needed because positive pressures induce reduced volumes (volume change negative). (a) Use the relation dU = pdV for the energy associated with pressure acting through a small volume change to show K V 0 =  d 2 U dV 2  V =V 0 where V 0 is the crystal volume at the equilibrium interionic spacing r = a 0 . (b) The volume of an ionic crystal containing N negative and N positive ions can be written as V = cNr 3 where c is a constant dependent on the type of lattice (2 for NaCl). Use this to obtain the relation K V 0 =  d 2 U dV 2  V =V 0 = 1 9c 2 N 2 r 2 · d dr  1 r 2 dU dr  (c) Carry out the indicated differentiation of the expression for binding energy to obtain the expression K V 0 = K cNr 3 0 = N 9c 2 Nr 2 0  −4ACe r 5 0 + n(n +3)B r n+4 0  Then using the expression B = ACe 2 r n−1 0 /n, obtain the formula for n in terms of com- pressibility: n =1+ 9cr 4 0 K ACe 2 4. Complete the spreadsheet below, filling in the values for repulsion exponent n and lattice energy U. 16 type r 0 (pm) K (GPa) A n U(kJ/mol) U expt LiF 201.4 6.710e+01 1.750 -1014 NaCl 282.0 2.400e+01 1.750 -764 KBr 329.8 1.480e+01 1.750 -663 The column labeled U expt lists experimentally obtained values of the lattice energy. 5. Given the definition of Helmholtz free energy: A = U − TS along with the first and second laws of thermodynamics: dU = dQ + dW dQ = TdS where U is the internal energy, T is the temperature, S is the entropy, Q is the heat and W is the mechanical work, show that the force F required to hold the ends of a tensile specimen a length L apart is related to the Helmholtz energy as F =  ∂A ∂L  T,V 6. Show that the temperature dependence of the force needed to hold a tensile specimen at fixed length as the temperature is changed (neglecting thermal expansion effects) is related to the dependence of the entropy on extension as  ∂F ∂T  L = −  ∂S ∂L  T 7. (a) Show that if an ideal rubber (dU =0)ofmassMand specific heat c is extended adiabatically, its temperature will change according to the relation ∂T ∂L = −T Mc  ∂S ∂L  i.e. if the entropy is reduced upon extension, the temperature will rise. This is known as the thermoelastic effect. (b) Use this expression to obtain the temperature change dT in terms of an increase dλ in the extension ratio as dT = σ ρc dλ where σ is the engineering stress (load divided by original area) and ρ is the mass density. 8. Show that the end-to-end distance r 0 of a chain composed of n freely-jointed links of length a is given by r o = na 2 . 17 9. Evalute the temperature rise in a rubber specimen of ρ = 1100 kg/m 3 , c = 2 kJ/kg·K, NkT = 500 kPa, subjected to an axial extension λ =4. 10. Show that the initial engineering modulus of a rubber whose stress-strain curve is given by Eqn. 14 is E =3NRT. 11. Calculate the Young’s modulus of a rubber of density 1100 gm/mol and whose inter- crosslink segments have a molecular weight of 2500 gm/mol. The temperature is 25 ◦ C. 12. Show that in the case of biaxial extension (λ x and λ y prescribed), the x-direction stress based on the original cross-sectional dimensions is σ x = NkT  λ x − 1 λ 3 x λ 2 y  and based on the deformed dimensions t σ x = NkT  λ 2 x − 1 λ 2 x λ 2 y  where the t subscript indicates a “true” or current stress. 13. Estimate the initial elastic modulus E, at a temperature of 20C, of an elastomer having a molecular weight of 7,500 gm/mol between crosslinks and a density of 1.0 gm/cm 3 .What is the percentage change in the modulus if the temperature is raised to 40C? compelling advantages for engineered materials that can be made stronger in one direction than another (the property of anisotropy). If a pressure vessel constructed of conventional isotropic material is made thick enough to keep the hoop stresses below yield, it will be twice as strong as it needs to be in the axial direction. In applications placing a premium on weight this may well be something to avoid. Example 1 Figure 6: Filament-wound cylindrical pressure vessel. Consider a cylindrical pressure vessel to be constructed by filament winding, in which fibers are laid down at a prescribed helical angle α (see Fig. 6). Taking a free body of unit axial dimension along which n fibers transmitting tension T are present, the circumferential distance cut by these same n fibers is then tan α. To balance the hoop and axial stresses, the fiber tensions must satisfy the relations hoop : nT sin α = pr b (1)(b) axial : nT cos α = pr 2b (tan α)(b) Dividing the first of these expressions by the second and rearranging, we have tan 2 α =2,α=54.7 ◦ This is the “magic angle” for filament wound vessels, at which the fibers are inclined just enough to- ward the circumferential direction to make the vessel twice as strong circumferentially as it is axially. Firefighting hoses are also braided at this same angle, since otherwise the nozzle would jump forward or backward when the valve is opened and the fibers try to align themselves along the correct direction. Deformation: the Poisson effect When a pressure vessel has open ends, such as with a pipe connecting one chamber with another, there will be no axial stress since there are no end caps for the fluid to push against. Then only the hoop stress σ θ = pr/b exists, and the corresponding hoop strain is given by Hooke’s Law as:  θ = σ θ E = pr bE Since this strain is the change in circumference δ C divided by the original circumference C =2πr we can write: δ C = C θ =2πr pr bE 4 The change in circumference and the corresponding change in radius δ r are related by δ r = δ C /2π, so the radial expansion is: δ r = pr 2 bE (4) This is analogous to the expression δ = PL/AE for the elongation of a uniaxial tensile specimen. Example 2 Consider a compound cylinder, one having a cylinder of brass fitted snugly inside another of steel as shown in Fig. 7 and subjected to an internal pressure of p =2MPa. Figure 7: A compound pressure vessel. When the pressure is put inside the inner cylinder, it will naturally try to expand. But the outer cylinder pushes back so as to limit this expansion, and a “contact pressure” p c develops at the interface between the two cylinders. The inner cylinder now expands according to the difference p − p c , while the outer cylinder expands as demanded by p c alone. But since the two cylinders are obviously going to remain in contact, it should be clear that the radial expansions of the inner and outer cylinders must be the same, and we can write δ b = δ s −→ (p − p c )r 2 b E b b b = p c r 2 s E s b s where the a and s subscripts refer to the brass and steel cylinders respectively. Substituting numerical values and solving for the unknown contact pressure p c : p c = 976 KPa Now knowing p c , we can calculate the radial expansions and the stresses if desired. For instance, the hoop stress in the inner brass cylinder is σ θ,b = (p −p c )r b b b =62.5 MPa (= 906 psi) Note that the stress is no longer independent of the material properties (E b and E s ), depending as it does on the contact pressure p c which in turn depends on the material stiffnesses. This loss of statical determinacy occurs here because the problem has a mixture of some load boundary values (the internal pressure) and some displacement boundary values (the constraint that both cylinders have the same radial displacement.) If a cylindrical vessel has closed ends, both axial and hoop stresses appear together, as given by Eqns. 2 and 3. Now the deformations are somewhat subtle, since a positive (tensile) strain in one Example 1 To illustrate how volumetric strain is calculated, consider a thin sheet of steel subjected to strains in its plane given by  x =3, y =−4, and γ xy = 6 (all in µin/in). The sheet is not in plane strain, since it can undergo a Poisson strain in the z direction given by  z = −ν( x +  y )=−0.3(3 − 4) = 0.3. The total state of strain can therefore be written as the matrix []=   36 0 6−40 000.3   ×10 −6 where the brackets on the [] symbol emphasize that the matrix rather than pseudovector form of the strain is being used. The volumetric strain is: ∆V V =(3−4+0.3) ×10 −6 = −0.7 ×10 −6 Engineers often refer to “microinches” of strain; they really mean microinches per inch. In the case of volumetric strain, the corresponding (but awkward) unit would be micro-cubic-inches per cubic inch. Finite strain The infinitesimal strain-displacement relations given by Eqns. 3.1–3.3 are used in the vast major- ity of mechanical analyses, but they do not describe stretching accurately when the displacement gradients become large. This often occurs when polymers (especially elastomers) are being con- sidered. Large strains also occur during deformation processing operations, such as stamping of steel automotive body panels. The kinematics of large displacement or strain can be complicated and subtle, but the following section will outline a simple description of Lagrangian finite strain to illustrate some of the concepts involved. Consider two orthogonal lines OB and OA as shown in Fig. 4, originally of length dx and dy,alongthex-yaxes, where for convenience we set dx = dy = 1. After strain, the endpoints of these lines move to new positions A 1 O 1 B 1 as shown. We will describe these new positions using the coordinate scheme of the original x-y axes, although we could also allow the new positions to define a new set of axes. In following the motion of the lines with respect to the original positions, we are using the so-called Lagrangian viewpoint. We could alternately have used the final positions as our reference; this is the Eulerian view often used in fluid mechanics. After straining, the distance dx becomes (dx)  =  1+ ∂u ∂x  dx Using our earlier “small” thinking, the x-direction strain would be just ∂u/∂x. But when the strains become larger, we must also consider that the upward motion of point B 1 relative to O 1 , that is ∂v/∂x, also helps stretch the line OB. Considering both these effects, the Pythagorean theorem gives the new length O 1 B 1 as O 1 B 1 =   1+ ∂u ∂x  2 +  ∂v ∂x  2 We now define our Lagrangian strain as 5 Figure 4: Finite displacements.  x = O 1 B 1 − OB OB = O 1 B 1 − 1 =  1+2 ∂u ∂x +  ∂u ∂x  2 +  ∂v ∂x  2 −1 Using the series expansion √ 1+x =1+x/2+x 2 /8+··· and neglecting terms beyond first order, this becomes  x ≈  1+ 1 2  2 ∂u ∂x +  ∂u ∂x  2 +  ∂v ∂x  2  − 1 = ∂u ∂x + 1 2   ∂u ∂x  2 +  ∂v ∂x  2  (11) Similarly, we can show  y = ∂v ∂y + 1 2   ∂v ∂y  2 +  ∂u ∂y  2  (12) γ xy = ∂u ∂y + ∂v ∂x + ∂u ∂y ∂u ∂x + ∂v ∂y ∂v ∂x (13) When the strains are sufficiently small that the quadratic terms are negligible compared with the linear ones, these reduce to the infinitesimal-strain expressions shown earlier. Example 2 The displacement function u(x) for a tensile specimen of uniform cross section and length L,fixedat one end and subjected to a displacement δ at the other, is just the linear relation u(x)=  x L  δ The Lagrangian strain is then given by Eqn. 11 as 6  x = δ L + 1 2  δ L  2 The first term is the familiar small-strain expression, with the second nonlinear term becoming more important as δ becomes larger. When δ = L, i.e. the conventional strain is 100%, there is a 50% difference between the conventional and Lagrangian strain measures. The Lagrangian strain components can be generalized using index notation as  ij = 1 2 (u i,j + u j,i + u r,i u r,j ). A pseudovector form is also convenient occasionally:       x  y γ xy      =      u ,x v ,y u ,y + v ,x  ... with respect to a change in price, a concept economists call elasticity Introduction to Elasticity In this chapter, you will learn about: • • • • Price Elasticity of Demand and Price Elasticity. .. Cases of Elasticity and Constant Elasticity Elasticity and Pricing Elasticity in Areas Other Than Price Anyone who has studied economics knows the law of demand: a higher price will lead to a lower... world To find answers to these questions, we need to understand the concept of elasticity Elasticity is an economics concept that measures responsiveness of one variable to changes in another