Physics
Young's Modulus
Young's Modulus, also known as the elastic modulus, is a measure of the stiffness of a material. It quantifies how much a material will deform under a given amount of force. It is used to predict how a material will behave under stress and is an important property in engineering and materials science.
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8 Key excerpts on "Young's Modulus"
eBook - PDFEngineering Science N4 Student's Book
TVET FIRST
- MJJ van Rensburg(Author)
- 2022(Publication Date)
- Macmillan(Publisher)
159 Stress, strain and Young’s modulus TVET FIRST 6 Module Stress, strain and Young’s modulus Overview of Module 5 When you have completed this module, you will be able to: Unit 6.1: Stress types ● Name the three main types of stresses (direct, shear and bending). ● Calculate direct stresses including determination of cross-sectional areas or maximum load on a member if the stress limit is known: o Include compound bars. o Mention the nature of stresses. ● Calculate shear stress including determination of shear areas as well as maximum load on a member if the stress limit is known. ● Include calculations on single and double shear. Unit 6.2: Hooke’s law and Young’s modulus ● State Hooke’s law and define Young’s modulus of elasticity. ● Calculate stress and strain and Young’s modulus for the material. ● Draw stress and strain graphs or force-distance graphs (limited to elastic limit) with clear reference to the direct proportionality between stress and strain. ● Calculate change in length or total change in length for compound rods. When a force is applied to a material, it produces a stress, which causes the material to deform and eventually to fail. Strain is the deformation of a material from stress – it is the ratio of change in length compared to the original length. Elasticity is the ability of a material to resist permanent change when a stress is applied to it. The material regains its original shape and size when the stress is removed. Engineers use their understanding of forces, stress, strain and elasticity to create safe designs for buildings, bridges, cars, planes and all kinds of machine parts. Starter activity Discuss the following questions in class: 1. Why do you think it is necessary to know the breaking force of a wire or rope such as the one in Figure 6.1? 2. If you have rods made of different materials, how will you know which one is the strongest? 3. Discuss what you think happens to materials under bending stress.
eBook - PDF- Frank D. Stacey, Paul M. Davis(Authors)
- 2008(Publication Date)
- Cambridge University Press(Publisher)
This is infinitesimal strain theory. It describes elastic deformations in response to stresses that are very small compared with values of the elas-tic moduli. Pressure in the deep interior of the Earth is much too large for this assumption to be even approximately valid. In this situation an elastic modulus must be defined as the ratio of a small change in stress to the consequent small strain increment, superimposed on a much larger ambient compression. The moduli all increase systematically with pressure and a more general, finite strain theory is required. This is a subject of Chapter 18 . Elasticity theory is simplest for materials that are isotropic, that is having properties that are the same in all directions. They include polycrys-talline materials with randomly oriented crys-tals, such as undeformed metals or igneous rocks, as well as glassy or amorphous substances. Elastically isotropic materials have two inde-pendent elastic moduli. There are several ways of choosing these moduli, depending on the form of the strain to be represented, but since only two can be independent it follows that any one of them can be represented in terms of any other two, as summarized in Appendix D . In seismology and much other geophysics we gen-erally prefer to use bulk modulus, K , and rigidity modulus, , as the two independent moduli. The particular reason for geophysical interest in bulk modulus is that it provides a link between seis-mologically observed elastic wave speeds and the density profile of the Earth. With a minor proviso concerning the elasticity of a composite (Section 10.4 ), the combination of P-wave speed, controlled by the modulus 1 (Eq. ( 10.5 )), and shear wave speed, controlled by rigidity, , pro-vides a direct observation of the ratio K / & ¼ d P /d & , where & is density and P is pressure. A pure shear strain with no change in volume, described by , involves no change in temper-ature and is unambiguous.
- D R Moore, S Turner(Authors)
- 2001(Publication Date)
- Woodhead Publishing(Publisher)
3 Modulus and stiffness: general principles 3.1 ‘Modulus’ relates to a material and translates into the stiffness of an article made from that material when the dimensions and load configuration are taken into account. For some classes of material, modulus can be regarded as a physical property the value of which should therefore be independent of the dimensions of the test specimen. However, this is not usually so for plastics because the properties vary with the degree of molecular order and other structural factors which, in turn, depend on the processing conditions and the flow geometry, see Chapter 1, Sections 1.5 and 1.7. Consequently, the stiffness of a plastics test specimen should be expressed only as the modulus of the material as manifest in that particular test piece. 3.2 Apart from the limitation embodied in Section 3.1, the geometrical rela-tionships between the modulus of a plastics material and the stiffness of a structure made from it are determined by the laws of mechanics and are identical to those that emerge in the theories of elasticity, plasticity, etc. The relatively large deformations or strains that may develop in some plastics test specimens and service items in some situations may require the use of higher order terms in the elastic equations, the imposition of special constraints on the test configurations and even redefinitions of strain, see Supplement S3.2.1. Also, local yielding may occur at stress concentrations, e.g. at loading points, thereby requiring adjustments to the calculations or further constraints on the test configurations. Some of the linear elastic relationships commonly used in the plastics industry’s measurement of modulus are given in Supplement S3.2. There are many such formulae, but most of them were developed initially as a means of calculating the stiffness of engineering structures when the elastic properties of the material of construction are known. However, when 50
eBook - PDFStrength of Materials & Structures N5 SB
TVET FIRST
- L Maraschin, VH Dookhi A Knott(Authors)
- 2021(Publication Date)
- Macmillan(Publisher)
Strain Stress F 2 F 3 F extension 0 1 x 2 x 3 x Stress Strain E = Figure 1.28: Young’s modulus of elasticity ductile: able to be deformed or drawn out into thin wire without losing toughness 21 Introduction to the strength of materials TVET FIRST The constant has a definitive value for each material. This value does not depend on the size of the material’s cross sectional area. This constant is called Young’s modulus or the modulus of elasticity and is indicated by the symbol E . We calculate the modulus of elasticity as follows: σ ∝ ε , therefore: Remember ∝ means ‘in direct proportion to’. σ = ε × constant σ = ε × E E = σ _ ε where: E is Young’s modulus of elasticity, usually in gigapascals (GPa). σ is the stress in pascals (Pa). ε is the strain (no units used). But remember: ε = x _ L and σ = F _ A Therefore: Formulae for calculating the change in length using Young’s modulus of elasticity ( E ) The following are VERY important formulae: x = σ ⋅ L _ E or x = F ⋅ L _ A ⋅ E Figure 1.29 shows the following: • When you have a small stress with large-strain material, the material is easy to stretch. This gives the material a low modulus of elasticity value. • When you have a large stress with small-strain material, the material gets harder to stretch. This gives the material a high modulus of elasticity value. Stress ( σ ) Strain ( ε ) Strain ( ε ) Stress ( σ ) 0 0 Small stress with large-strain material easy to stretch: E has low numerical value Large stress with small-strain material hard to stretch: E has high numerical value Figure 1.29: Values of Young’s modulus of elasticity modulus: a numerical value that tells us the degree to which a material has a certain physical property Young’s modulus: the numerical value that tells us how easy it is to deform (stretch) a material 22 TVET FIRST Module 1 Example 1.9 A tensile load of 300 kN acts on a steel tube that is 200 mm long. The outside diameter of the tube is 90 mm and its wall thickness is 10 mm.
eBook - PDF- Thomas Flynn(Author)
- 2004(Publication Date)
- CRC Press(Publisher)
As shown in Fig. 4.23, rod bending also involves Young’s modulus. The stress along a bent rod is s z ¼ E ; e z ¼ Ex = r ð 4 : 7 Þ where x is a coordinate perpendicular to z , and r is the curvature radius of the neutral surface near the origin. 6.4. Shear Modulus, G Sometimes called the torsional modulus, rigidity modulus, or transverse modulus, the shear modulus, G , relates the shear stress, t in simple shear to the shear strain, g . Thus, t ¼ G g ð 4 : 8 Þ Figure 4.23 Mechanical deformations with which Young’s modulus is associated. Mechanical Properties of Solids 275 5367-4 Flynn Ch04 R2 090904 Figure 4.24 illustrates the mechanical deformations with which the shear modulus is associated. A cube sheared on one face has a shear strain related to the length change of the face diagonal by g ¼ 2 D ‘=‘ ð 4 : 9 Þ In torsion (see Fig. 4.24), t is the torsional stress. For isotropic materials, simple-shear and torsional G ’s are identical. Both shear and torsion conserve volume but change shape. The bulk modulus, as described above, describes volume change without shape change. 6.5. Poisson’s Ratio, n Poisson’s ratio, n , is not an elastic modulus, but a dimensionless ratio of two elastic compliances. Usually defined by a uniaxially stressed rod (Young’s-modulus) experi-ment, Poisson’s ratio is the negative ratio of transverse ( x ) and longitudinal ( z ) strains: n ¼ e x = e z ð 4 : 10 Þ Figure 4.25 illustrates the deformation and strains involved. A typical value for metals is 1 = 3, 0.28–0.42 being the observed range for most materials. Such v values Figure 4.24 Mechanical deformations with which shear modulus and torsional modulus are associated. Figure 4.25 Mechanical deformation and strains with which Poisson’s ratio is associated. 276 Chapter 4 5367-4 Flynn Ch04 R2 090904 mean that a material tends to maintain constant volume during uniaxial deforma-tion. Liquids exhibit constant volume during deformation, and some solids (such as rubber) tend to.
eBook - PDF- George C. King(Author)
- 2023(Publication Date)
- Wiley(Publisher)
pascals. Note that we can rearrange Equation (9.6) as F AE l Δl , (9.7) when it has an analogous form to Hooke’s law. In previous discussions, we saw that a solid is composed of atoms that are held together by interatomic forces, which give a solid its rigidity. And we modelled these bonds as Hooke’s law springs. Roughly speak- ing, when we stretch a solid, we are pulling the atoms of the solid further apart. Then when the stress is removed, the atoms return to their equilibrium separations. Typical values of Young’s modulus for some constructional materials are shown in Table 9.1, together with values of their other elastic moduli. For example, Young’s modulus for steel is 20 10 10 Pa. If a car of mass 1.5 10 3 kg is suspended from a factory roof by a steel rod of length 2.0 m and diameter 30 mm, the rod extends by the amount Δl Fl EA 1 5 10 3 9 8 2 0 20 10 10 π 15 10 3 2 2 1 10 4 m 0 21 mm, F F l Δl Area, A Wall Figure 9.2 The figure shows a bar of material that initially has length l and cross-sectional area A. It is fixed to a rigid wall at one end and a longitudinal force F is applied to the other end. There is a reactive force F at the wall that has the same magnitude as the applied force but acts in the opposite direction. The applied force increases the length of the bar to l + Δl. The ratio of the applied force F to the cross-sectional area A is the tensile stress. The fractional change Δl/l is the tensile strain. A stretched bar also suffers a contraction, not shown, in the direction at right angles to the stretch, as described in Section 9.2. The elastic properties of solids 311 or by 0.01% of its original length. In engineering applications, tensile strain rarely exceeds 0.1 % . If the forces acting on the bar in Figure 9.2 push rather than pull the bar, the bar is said to be under compressional stress.
eBook - PDFElasticity and Fluid Dynamics
Volume 3 of Modern Classical Physics
- Kip S. Thorne, Roger D. Blandford(Authors)
- 2021(Publication Date)
- Princeton University Press(Publisher)
(11.18) Here K is called the bulk modulus and μ the shear modulus, and the factor 2 is included for purely historical reasons. The first minus sign (with K > 0) ensures that the isotropic part of the stress, − K g , resists volume changes; the second minus sign (with μ > 0 ) ensures that the symmetric, trace-free part, − 2 μ Σ , resists shape changes (resists shearing). 11.3 Stress, Elastic Moduli, and Elastostatic Equilibrium 581 Hooke’s law (Figs. 11.1 and 11.4) can be expressed in this same stress-proportional-to-strain form. The stress, when the rod is stretched, is the force F that does the stretching divided by the rod’s cross sectional area A , the strain is the rod’s fractional change of length / , and so Hooke’s law takes the form F/A = − E/ , (11.19) with E an elastic coefficient called Young’s modulus . In Sec. 11.4, we show that E is a combination of the bulk and shear moduli: E = 9 μK/( 3 K + μ) . Lam´ e coefficients In many treatments and applications of elasticity, the shear tensor Σ is paid little attention. The focus instead is on the strain S ij and its trace S kk = , and the elastic stress tensor (11.18) is written as T = − λ g − 2 μ S , where λ ≡ K − 2 3 μ . In these treatments μ and λ are called the first and second Lam´ e coefficients and are used in place of μ and K . We shall not adopt this viewpoint. 11.3.4 11.3.4 Energy of Deformation Take a wire of length and cross sectional area A , and stretch it (e.g., via the “Hooke’s-law experiment” of Figs. 11.1 and 11.4) by an amount ζ that grows gradually from 0 to . When the stretch is ζ , the force that does the stretching is [by Eq. (11.19)] F = EA(ζ /) = (EV/ 2 )ζ ; here V = A is the wire’s volume, and E is its Young’s modulus. As the wire is gradually lengthened, the stretching force F does work W = 0 F dζ = 0 (EV/ 2 )ζ dζ = 1 2 EV (/) 2 . This tells us that the stored elastic energy per unit volume is U = 1 2 E(/) 2 .
eBook - ePub- Shutilov, Vladimir Alexandrovich Shutilov, Yelena Vladimirovna Tcharnaya(Authors)
- 2020(Publication Date)
- CRC Press(Publisher)
Since strains along the y and z axes in a continuous medium must be accompanied by corresponding stresses, the initial condition of one-dimensional stress can be satisfied only in the presence of free lateral surfaces. Therefore, the case examined is realized for extension of a rod oriented along the x -axis. The coefficient in front of the stress in the first of equations (I.24), by definition, represents the coefficient of extension of the rod being stretched and the inverse quantity is the effective modulus of elasticity, which in this case is called Young’s modulus : E = (3 λ + 2 μ) μ λ + μ. (I.25) Substituting (I.25), the first equation of (I.24) assumes the form ϵ 11 = σ / E. Thus Young’s modulus characterizes the rigidity of the rod relative to longitudinal extension (compression) and determines the mechanical stress with which the magnitude of the strain must equal 1, i.e. , the length of the rod changes by a factor of two (of course, with Hooke’s law remaining valid). The values of Young’s modulus E for some isotropic bodies are presented in Table 2. Table 2 Values of Young’s Modulus. Poisson’s Ratio, and the Shear Modulus for Some Isotropic Solids Material E · 10 −10. N/m 2 v 0 G · 10 −10. N/m 2 Tungsten 36. 0 0. 27 13. 3 Steel 3 22–24 0. 30 8. 5–8. 8 Iron 21 0. 28 8. 2 Copper 12. 0 0. 35 4. 6 Brass 9–10 0. 35 3. 0–3. 7 Gold 8. 0 0. 41 2. 9 Aluminum 7. 0 0. 34 2. 6 Tin 5. 4 0. 33 2. 0 Lead 1.6 0. 44 0. 6 Fused quartz 7. 4 0. 18 3. 2 Glass crown 7. 2 0. 25 2. 9 flint 5. 5 0. 23 2. 4 Porcelain 6. 0 0. 23 2. 4 Ice 1. 0 0. 33 0. 4 Plexiglass 0. 5 0. 35 0. 15 The absolute ratio of the transverse deformation of the rod to the longitudinal deformation, i.e. , the relative elongation due to the longitudinal stress, is called Poisson’s ratio (v 0) : ν 0 = | ϵ 2 2 ϵ 1 1 | = | ϵ 3 3 ϵ 11 | = ϵ 2 2 E σ = λ 2 (λ + μ). (I.26) Thus ϵ 22 = ϵ 33 = – ν 0 σ / E, i.e
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