Poisson's Ratio

3 Key excerpts on "Poisson's Ratio"

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  eBook - ePub

  Woven Textile Structure

  Theory and Applications

  • B K Behera, P K Hari(Authors)
  • 2010(Publication Date)
  • Woodhead Publishing

    (Publisher)

  τ ) within elastic limit is directly proportional to shear strain. Mathematically.

  where G is the constant of proportionality known as shear modulus.

  The SI unit of modulus of elasticity (E ) is the pascal, N/m2 . Modulus can be calculated from the slope of the straight-line portion of the stress–strain curve.

  7.2.5 Poisson ratio

  Tensile or compressive axial strain is accompanied by lateral strain. The lateral strain is a fraction of the linear strain and within the elastic limit bears a constant ratio to the linear strain and is called the Poisson ratio. σ n represents the Poisson ratio for the lateral direction as it gives the contraction due to longitudinal extension. The suffixn is used for the direction in which contraction takes place due to extension in a longitudinal direction.

  7.2.6 Stress–strain diagram

  The stress and corresponding strain when plotted on a graph constitute a stress–strain diagram. The diagram differs for different materials. A brittle material is one which gives very small deformation before fracture as shown in Fig. 7.3 .

  7.3 Stress–strain diagram for a brittle material.

  The yield point is defined as the stress beyond which a material deforms by a relatively large amount for a small increase in the stretching force. Beyond this stress, the material no longer obeys Hooke’s law.

  7.3 Tensile properties of woven fabrics

  The tensile properties of woven fabrics produce several problems and complexities, mainly because the fabric is anisotropic and has a modulus which varies considerably with strain. The variation in the initial modulus is very large and the modulus in the warp and weft directions differs because the cloth is not symmetrical. In fact, the extension which takes place at an angle to the warp or weft is usually much higher and also involves a different mechanism of deformation [1 ]. For example, the modulus at an angle of 45° is mainly determined by the shear behavior. But if the extension is in the warp or weft direction, shear does not play any role. A woven fabric structure consists of fibers and yarns and its deformation results in a series of complex movement of fibers and yarns. The deformation behavior becomes more complex as both fiber and yarn behave in a non-Hookean manner. However, many researchers [2 –4

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  eBook - PDF

  Mechanical Science for Technicians

  Volume 2

  • Ian McDonagh(Author)
  • 2014(Publication Date)
  • Arnold

    (Publisher)

  For metals, Poisson's Ratio can vary between 0.25 and 0.33. Let o x be the stress producing the longitudinal strain e x in fig. 1.1 ; then where E is the modulus of elasticity or Young's modulus for the material. The lateral strain e y induced by o x is The minus sign indicates that e y is compressive. € x = l(o X 'va y ) °v e y = -(a y -υσ χ ) (a) € χ =ϊ~(σ χ +υσ ν ) ■-Ε (σ ν + υο χ (b) Fig. 1.2 For the stressed element shown in fig. 1.2(a), the total strain in the direc-f o x is e x = strain due to o x — strain due to o y tion of o x is i.e. e,, = — °y -1 / -°y .. °χ = 1 also e y = -j - v — = -(σ χ - vo y ) 2 In fig. 1.2(b), σ χ is tensile and o y is compressive: £χ = — + v -f = -(σ χ + w y ) °v °x 1 , and €3, = *- - v — = (o y + vo x ) Example 1 Two mutually perpendicular stresses of magnitude 50 N/mm 2 (tensile) and 35 N/mm 2 (compressive) act at a point in an element. If E = 200 kN/mm 2 and v = 0.3, calculate the strains in the direction of the stresses. where o x = 50 N/mm 2 Oy = —35 N/mm 2 (i.e. compressive) 1; = 0.3 and E = 200 x 10 3 N/mm 2 e* = l -[50 N/mm 2 - 0.3 x (-35 N/mm 2 )] X 200 x 10 3 N/mm 2 = 3.025 x 10~ 4 (i.e. tensile) Also, € y = -(o y - > ) 1 - (-35 N/mm 2 - 0.3 x 50 N/mm 2 ) 200 x 10 3 N/mm 2 = -2.5 x 10~ 4 (i.e. compressive) i.e. the strains are 3.025 x 10~ 4 (tensile) and 2.5 x 10~ 4 (compressive). Example 2 The maximum and minimum mutually perpendicular strains in a stressed component were found to be 8.5 x 10 4 and 1.5 x 10 4 , both tensile. If E = 200 GN/m 2 and v = 0.25, calculate the maximum and mini-mum stresses. 3 and Multiplying equation (i) by E and equation (ii) by Ev and adding the results, (0 GO Also (e x + ve y )E o x l -* 2 _ (ß y +ve x )E 1 -j r where e x =8.5x 10~ 4 e v = 1.5 x 10 4 E = 200 x 10 9 N/mm 2 and i; = 0.25 i.e. the maximum and minimum stresses are respectively 189.3 N/mm 2 and 77.3 N/mm 2 , both tensile.

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  Biomaterials from Nature for Advanced Devices and Therapies

  • Nuno M. Neves, Rui L. Reis, Nuno M. Neves, Rui L. Reis(Authors)
  • 2016(Publication Date)
  • Wiley

    (Publisher)

  As depicted in panel C, the tangent modulus is taken as the slope of the stress–strain curve at some specified level of stress, while the secant modulus represents the slope of a secant drawn from the origin to some given point of curve. Other properties worth characterizing include: the Poisson's Ratio (υ), percent elongation (%EL), and moduli of resilience (Ur) and toughness (U T). The Poisson's Ratio is defined as the negative ratio between the lateral and axial strains that result from an applied axial stress, within the limits of an elastic deformation (for isotropic materials υ = –ϵ x /ϵ z = –ϵ y /ϵ z ; see Fig. 26.3, panel A for axes definition). Theoretically, the Poisson's Ratio for isotropic materials should be 0.25; meanwhile, the maximum value for υ (or that value for which there is no net volume change) is 0.50. The percent elongation is a measure of the degree of plastic deformation that has been sustained at fracture according to: %EL = 100(L f – L 0)/L 0, where L f and L 0 stand for the fracture length and the original gauge length of the sample, respectively. A material that experiences very little or no plastic deformation upon fracture (less than 5%) is termed brittle. Meanwhile, a material showing a large degree of plastic deformation is regarded as ductile. Finally, the moduli of resilience and toughness represent the strain energy per unit volume required to stress a material from an unloaded state up to the points of yielding and fracture, respectively. Numerically speaking, they correspond to the area under the curve from the origin to ϵ YP and ϵ B, respectively (see panel B). 26.2.5.2 Dynamic Tests Polymeric materials in general display a viscoelastic behavior. This is visible in quasi-static tensile tests because there is no linear region in the stress–strain curve (see Fig. 26.3, panel C)

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