Introduction | ||
Click to view movie (101k) | Susan Placer, a new engineer at AAA Materials Testing Lab, is asked to find the Poisson's ratio for a new epoxy adhesive. The test method used at the lab compresses a 1 in cube specimen under compression. The test frame consists of a 1 in cube hole to place the sample. Then a load is place on the top surface (see figure). The total deflection of the top surface is monitored. The test sample is a 1 in cube which is placed into a perfectly fitting frame. The frame is substantially more stiff than the sample and will not deflect during the test. What is known:
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Question | ||
What is the Poisson's ratio of the new epoxy adhesive? Also, what stresses are generated on the sides of the test frame? | ||
Approach | ||
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One Dimensional Hooke's Law | ||||||||||||||
1-D Hooke's Law | Recall, Hooke's Law in one dimension (uniaxial loading), relates the normal stress and normal strain as
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Poisson's Ratio | ||||||||||||||
Click to view movie (21k) Contraction in y- and z-direction when Stressed in x-direction | When analyzing more than one dimension, interaction between all directions needs to be considered. This is done through Poisson's ratio. Basically, Poisson's ratio is the amount of transverse contraction, or negative strain, when strained in a given direction. For a basic object pulled or strained in the x-direction, the Poisson's Ratio is defined as
For a three dimensional object, Poisson's ratio will occur in equally in both perpendicular directions. If the load is in the x-direction, then strain in the y- and z-direction will be εy = εz = -νεx | |||||||||||||
Two Dimensional Stress-Strain | ||||||||||||||
| If a material is isotropic (homogenous in all directions, such as a solid metal) and is pulled in two directions, then due to Poisson's ratio, the overall normal strain will be the total of the two strains. For example, if there are normal stresses in both the x- and y-directions, then the total normal strain in the x-direction is εx total = εx due to σx + εx due to σy = σx /E - νσy/E εx = (σx - νσy) / E Similarly, the normal strain in the y-direction would be εy = (σy - νσx) / E | |||||||||||||
Click to view movie (29k) | However, Hooke's Law also relates shear strain and shear stress. If the shear stress and strain occurs in a plane then the stress and strain are related as
γ = τ/G = [2(1 + ν)/E] τ where G is the shear modulus (a material property) and γ is the shear strain. The shear strain is defined as the angle (radians) caused by the shear stress as shown in the diagram at the left. The shear modulus is related to Young modulus and Poisson's ratio,
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Three Dimensional Stress-Strain | ||||||||||||||
| Just like 1D or 2D, Hooke's Law can also be applied to material undergoing three dimensional stress (triaxial loading). The development of 3D equations is similar to 2D, sum the total normal strain in one direction due to loads in all three directions. For the x-direction, this gives, εx total = εx due to σx + εx due to σy + εx due to σz = σx /E - νσy/E - νσz/E εx = (σx - νσy - νσy) / E Similarly, the other directions can also be determined. The final equations are summarized in the table below. | |||||||||||||
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| A simple cube of material is inserted into a snug fitting test frame and then a 3,000 lb load is applied on the top surface. The sides of the test frame are stiff and will not deflect at the test load level. Also, the sides are smooth so there is no friction forces as the specimen is loaded. The objective in testing the specimen is to determine the Poisson's ratio of the material. The Young's modulus, E, is known, 2,000 ksi. Also, the total deflection after loading is 0.001 inches. | |
Determining Poisson's ratio | ||
| A good place to start solving this problem is the full 3D Hooke's law equations. While these look complex, many terms are easily determined. The vertical stress can be calculated as σz = F/A = 3000 lb/(1 in)2 = 3,000 psi = 3.0 ksi The other two horizontal stresses are unknown, but must be the same due to symmetric geometry and loading. Thus, σx = σy = p | |
Next, the three strains can be determined. The vertical strain can be found since the deflection is known, giving εz = δ/L = (0.001 in) / (1.0 in) = 0.001 The other two strains are also known since the outside walls do not deflect (due to their high stiffness). Thus, εx = εy = 0.0 Also, Young's modulus is known, E = 2,000 ksi. | ||
Simplifying the 3D equations to two equations give, Next, substituting specific values reduce these equations to The negative signs represent compression for both strain and stress. There are now two unknowns, p and ν. The second equation can be rearranged to give, This can be manipulated to give 3ν2 + 0.5ν - 0.5 = 0 Using the quadratic equation, the positive root gives ν = 0.3333 This equation can also be solved numerically by graphical methods or by using numerical programs like MathCad, Mathematica, MathLab, or even Excel. The equation is graphed at the left which also gives ν = 0.3333 | ||
Stresses | ||
| The stresses in the x and y directions can now be determined, p = σx = σy = -1.5 ksi = 1,500 psi |
Example | ||
A 0.5 cm thick rectangular plate is pulled in tension by two loads in the x and y directions as shown. The total deflection in the x and y direction is 0.021 cm and 0.009 cm, respectively. What is the Poisson's ratio of the material? The z direction has no load and the deflection is not known. | ||
Solution | ||
| This problem involves loading from two directions, and thus requires at least the 2-D Hooke's Law. The 3-D Hooke's Law could be used, but since σz is zero, those equations will reduce to the 2-D equations. The equations are, εx = (σx - ν σy)/E εy = (σy - ν σx)/E The strains and stresses in the x and y direction need to be calculated. σx = Px/Ax = 5/[(0.05)(0.005)] = 20 MPa σy = Py/Ay = 9/[(0.10)(0.005)] = 18 MPa εx = 0.021/10 = 0.0021 cm/cm εy = 0.009/5 = 0.0018 cm/cm Substituting the stresses and strains into the 2-D Hooke's Law equations, gives 0.0021 E = 20 - ν 18 0.0018 E = 18 - ν 20 Solving for ν gives, ν = 0.1876 |