Technology & Engineering
Shear Modulus
Shear modulus, also known as the modulus of rigidity, is a measure of a material's resistance to deformation when subjected to shear stress. It quantifies the material's ability to withstand shear forces and maintain its shape. It is an important property in engineering and materials science, particularly in the design and analysis of structures and mechanical components.
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4 Key excerpts on "Shear Modulus"
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.
- Barry Goodno, James Gere(Authors)
- 2018(Publication Date)
- Cengage Learning EMEA(Publisher)
Hooke’s Law in Shear The properties of a material in shear can be determined experimentally from direct-shear tests or from torsion tests. The latter tests are performed by twist- ing hollow, circular tubes, thereby producing a state of pure shear. The results of these tests are used to plot shear stress-strain diagrams (that is, diagrams of shear stress t versus shear strain g). These diagrams are similar in shape to tension-test diagrams ( s versus «) for the same materials, although they differ in magnitudes. From shear stress-strain diagrams, you can obtain material properties such as the proportional limit, modulus of elasticity, yield stress, and ultimate stress. Numerical values of these properties in shear are usually about half as large as those in tension. For instance, the yield stress for structural steel in shear is 0.5 to 0.6 times the yield stress in tension. For many materials, the initial part of the shear stress-strain diagram is a straight line through the origin, just as it is in tension. For this linearly elastic region, the shear stress and shear strain are proportional, resulting in the fol- lowing equation for Hooke’s law in shear: G t g 5 (7-13) where G is the Shear Modulus of elasticity (also called the modulus of rigidity). The Shear Modulus G has the same units as the tension modulus E, namely, pascals (or multiples thereof) in SI units. For mild steel, typical values of G are 75 GPa; for aluminum alloys, typical values are 28 GPa. Additional values are listed in Table F-2, Appendix F. The moduli of elasticity in tension and shear are related by 2(1 ) G E 5 1 n (7-14) in which n is Poisson’s ratio. This relationship shows that E, G, and n are not independent elastic properties of the material. Because the value of Poisson’s ratio for ordinary materials is between zero and one-half, from Eq. (7-14) G must be from one-third to one-half of E. The following examples illustrate some typical analyses involving the effects of shear.
eBook - PDFParts Fabrication
Principles and Process
- Richard Crowson(Author)
- 2006(Publication Date)
- CRC Press(Publisher)
Also, if the unit stress in the bar does not exceed the elastic limit of a material, the deformations arc in direct proportion to the stresses. The key to computing the magnitude of the deformation lies in the stiffness of a material. The number that represents the degree of stiffness of a material is known as the modulus of elasticity . We represent this quantity by the letter E and define it as the unit stress divided by the unit deformation (strain), or E f s Where E modulus of elasticity f unit stress s unit deformation FIGURE 1.11 Typical stress–strain diagram for mild steel. ELASTIC LIMIT ULTIMATE STRENGTH YIELD POINT STRESS (PSI) STRAIN (IN./IN.) 10 Parts Fabrication: Principles and Process Example Suppose we place a steel bar with a 1 in. 2 in. cross section in a testing machine and apply a tensile force of 1000 lb. Its length becomes greater (although we cannot see it with the naked eye). If we apply the same force to a piece of wood having the same dimensions as the steel bar, we find that the deformation is greater—probably 20 times greater. We say that the steel has a greater degree of stiffness than the wood (see Figure 1.12). From our discussion on unit stress earlier, we saw that f P A And, from Figure 1.12, where L represents the length of the member and e the total deformation, s the deformation per unit of length would equal the total deformation divided by the original length, or s e L Now, since the modulus of elasticity, E f s , this becomes E P A e L E P L Ae or FIGURE 1.12 Example of stress and strain for calculation of modulus of elasticity E. e L 1 SQ. IN. CROSS SECTION 1,000 LB Principles of Structural Mechanics 11 which can also be written as e PL AE where e total deformation in inches P force in pounds (axial load) L length in inches A cross-sectional area in inches E modulus of elasticity in pounds per square inch. Remember that all of the above is valid only when the unit stress does not exceed the elastic limit of the material.
eBook - PDF- M J Iremonger(Author)
- 2013(Publication Date)
- Butterworth-Heinemann(Publisher)
The Shear Modulus (G) is related to Young's modulus (E) and Poisson's ratio (y) by the equation E = 2G(l + 1/) (4.4) If any two of the constants are known the third can be found from this equation. A fourth, less commonly used, elastic constant is bulk modulus. This represents a volumetric stress divided by volumetric strain. Thus K = -*— (4.5) (ÔV/V) where P is a compressive hydrostatic stress (equal in all directions) and V is the decrease in an initial volume V. Bulk modulus is related to the other elastic constants by the equation E = 3K{ -v) (4.6) 4.5 Torsion of circular shafts Consider the solid circular shaft shown in Figure 4.3 which is held at the left-hand end and subjected to a torque T about the longitudinal axis at the right-hand end. Torsion of circular shafts 35 It is assumed that (1) twisting is uniform along the shaft, (2) cross-sections remain plane and radii straight, (3) neither the length L or diameter D change, (4) the material of the shaft is homogeneous, isotropic and obeys Hooke's law. The analysis requires consideration of equilibrium, compatibility and the stress—strain relationship, i.e. all the 'tools' of stress analysis. Compatibility relates the shear strain at a radius r to the angle of twist Θ over the length L of the shaft. Equilibrium of an elemental tube radius r relates the shear stress r at that radius to the torque T and the polar second moment of area /. When the stress-strain relation, Equation (4.3), is included the following equation can be derived T/r=T/J = Ge/L (4.7) The shear stress r varies linearly with r being zero at the centre of the shaft and a maximum at the outside. The polar second moment of area / is given by the equation J= I r 2 M (4.8) where cL4 is the cross-section area of the elemental tube radius r (cL4 = 2nr dr). Figure 4.3
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