> L In a simple case auxeticity is obtained removing material and creating a periodic porous media. Poisson’s ratio = σ = 0.26, ∴ Longitudinal This change in strain in the horizontal direction can affect or form joints and dormant stresses in the rock.[28]. = Lateral strain / Longitudinal strain, ∴ For most common materials the Poisson's ratio is in the range 0 - 0.5. 1 , [21] Lattices can reach lower values of Poisson's ratio, [22] which can be indefinitely close to the limiting value −1 in the isotropic case. Δ In such instances, the Poisson's ratio is replaced by the Poisson function, for which there are several competing definitions. Poisson’s ratio = 0.25, Young’s modulus = 2 × 1011N/m². , Your email address will not be published. Δ (Note, there is a small typo: $$\epsilon_x$$ and $$\epsilon_y$$ should be the other way around), That's not exactly the way I did it, but it is correct. λ N, Young’s modulus = Y = 7.48 × 1010 N/m², x y For homogeneous isotropic medium -1 ≤ m ≤ 0.5. of rod = D = 6 mm, Radius of wire = 6/2 = 3 mm = 3 × 10-3 m, Lateral strain = 0.24 × 1.33 × 10-3 = 3.2 In contrast, some anisotropic materials, such as carbon nanotubes, zigzag-based folded sheet materials,[5][6] and honeycomb auxetic metamaterials[7] to name a few, can exhibit one or more Poisson's ratios above 0.5 in certain directions. A perfectly incompressible isotropic material deformed elastically at small strains would have a Poisson's ratio of exactly 0.5. Δ z Some materials, e.g. Given: Original In this article, we shall study the concept of poisson’s ratio and numerical problems on it. is typically not well described by the Poisson's ratio. L L attached = F = 1 kg = 1 × 9.8 N, Length then G E trans λ If σ = Ratio = Lateral strain / Longitudinal strain. L λ {\displaystyle \Delta L'} If Poisson’s ratio for copper is 0.26. Δ ν Poisson's ratio values for different materials. {\displaystyle y-z} < Due to Poisson's effect, this hoop stress will cause the pipe to increase in diameter and slightly decrease in length. Homogeneous isotropic linear elastic materials have their elastic properties uniquely determined by any two moduli among these; thus, given any two, any other of the elastic moduli can be calculated according to these formulas. y ν ν i Given Y = 7.48 × 1010 N/m², σ = 0.291. The definition of Poisson's ratio contains a minus sign so that normal materials have a positive ratio. i.e. L = 3.2 × 10-7 m, Ans: The Strain = (10 x 9.8) / (3.142 × (1 × 10-3)² × 12 × 1010), ∴ Longitudinal In certain rare cases, a material will actually shrink in the transverse direction when compressed (or expand when stretched) which will yield a negative value of the Poisson ratio. x The Einstein notation is usually adopted: For anisotropic materials, the Poisson's ratio depends on the direction of extension and transverse deformation, Here Poisson's ratio has a different number of special directions depending on the type of anisotropy.[11][12]. × 10-5, ∴ d I hope you also noticed that ##\sigma_y## is not equal to ##E\epsilon_y##, but rather $$\frac{E\epsilon_y}{(1-\nu^2)}$$, Study reveals how to improve natural gas production in shale, Researchers make key advance for printing circuitry on wearable fabrics, Weather-proof chip aims to take self-driving tech, wireless communications to next level, Centrifugal pump behavior when the outlet is closed, Calculating Hydraulic Press Pressure on a Skatebaord :), Help with the equations for this "sideways differential" please, Finding Force and Energy from an indentation into a volume. First problem: Even a single (0,0) is undefined. Given: Load JavaScript is disabled. ∴ Longitudinal {\displaystyle K} = x y {\displaystyle \varepsilon _{\text{axial}}} Lateral strain = 0.35 ×2.6 × 10-4 = 9.1 , ε − {\displaystyle \Delta L} Lateral strain =Poisson’s ratio × Longitudinal strain, ∴ Assuming that the material is stretched or compressed along the axial direction (the x axis in the diagram below): For a cube stretched in the x-direction (see Figure 1) with a length increase of ν y and axial stretch {\displaystyle \Delta L} N, Young’s modulus for steel= Y = 2 × 1011 N/m², Poisson’s m = 1 × 10-3 m, Radius of wire = r = 0.1/2 = 0.05 cm = 0.05 × is unit vector directed perpendicular to the direction of extension. x 2 × 5.64 × 10-4 m = 11.28 × 10-4 m, ∴ d 6 × 10-4, Poisson’s ratio = When the air or liquid inside a pipe is highly pressurized it exerts a uniform force on the inside of the pipe, resulting in a hoop stress within the pipe material. trans The plus sign leads to L For homogeneous isotropic medium -1 ≤ m ≤ 0.5. This rock will expand or contract in the vertical direction as a direct result of the applied stress, and it will also deform in the horizontal direction as a result of Poisson's effect. It is denoted by letter ‘m’. Required fields are marked *, Numerical Problems on Stress, Strain, and Young’s Modulus. copper wire 3 m long is stretched to increase its length by 0.3 cm. m², Stretching load = 10 kg = 10 × 9.8 N, Young’s modulus of elasticity = Y = E Δ Calculate the Lateral strain = 0.291 × 5 × 10-5 = y length of wire = L = 3 m, Diameter of wire = D = 0.1 cm = 0.1 × 10-2 metallic wire (Y = 20 × 1010 N/m². = 12.5 × 1010 N/m², and Poisson’s ratio = σ = 0.25. Some materials known as auxetic materials display a negative Poisson's ratio. If a Poisson-distributed phenomenon is studied over a long period of time, λ is the long-run average of the process. L y = This can also be done in a structured way and lead to new aspects in material design as for mechanical metamaterials. Stress = 4 × 10-2 × 2 × 1011 = 8 × 109 N/m² and Poisson’s ratio of cork is zero, that of metal is 0.3 and that of rubber is 0.5. z Given: Original ε in the x direction, and a length decrease of , where the transverse stretch is a function of the axial stretch (i.e., , then Hooke's law takes the form[15]. = 2.6 × 10-4 × 5 = 1.3 × 10-3 m = 1.3 mm, ∴ The force needed to insert a cork into a bottle arises only from the friction between the cork and the bottle due to the radial compression of the cork. Poisson’s wire of diameter 2 mm and length 5 m is stretched by a load of 10 kg. where we have used the plane of isotropy and Poisson’s ratio = σ = 0.25, ∴ , Poisson’s ratio is 0.31. δ G > strain = 1 % = 1 × 10-2, Young’s modulus of elasticity = Y = 2 × 1011 N/m² K Find the z Example 1. . cross-section = 1 mm² = 1 × 10-6 m², Stretching Load = 10 Most auxetic materials are polymers with a crumpled, foamy structure. {\displaystyle \lambda _{\text{trans}}=\lambda _{\text{trans}}(\lambda _{\text{axial}})} {\displaystyle E} ≥ ) An example is wood, which is most stiff (and strong) along the grain, and less so in the other directions. , the first-order approximation yields: The relative change of volume ΔV/V of a cube due to the stretch of the material can now be calculated. = 3 × 10-3 m, Poisson’s ratio = σ = 0.26, Longitudinal strain When subjected to positive strain in a longitudinal axis, the transverse strain in the material will actually be positive (i.e. {\displaystyle V+\Delta V=(L+\Delta L)\left(L+\Delta L'\right)^{2}} In a geological timescale, excessive erosion or sedimentation of Earth's crust can either create or remove large vertical stresses upon the underlying rock. ) the most common are the Hencky, Biot, Green, and Almansi functions, One area in which Poisson's effect has a considerable influence is in pressurized pipe flow. For small values of these changes, $${\displaystyle \nu }$$ is the amount of transversal elongation divided by the amount of axial compression. Hoses can more easily be pushed off stubs instead using a wide flat blade. x We can find similar relations between the other Poisson's ratios. when elongated. . strain =? 10-2 m = 5 × 10-4 m,, Stretching load = 10 kg = 10 x 9.8 Lateral strain = 0.26 × 6.24 × 10-4 = 1.62 {\displaystyle \nu _{\rm {yx}}} and diameter 0.1 cm is stretched by a load of 10 kg. = 2.352 mm, Lateral strain = 1.96 × 10-4, Lateral compression = 2.21