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Overview of Elasticity of Materials/Example Problems

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Example 1

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Problem Statement:

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Given a point on an elastic body with stress state , determine the principal stresses and the angle between the original orientation and the orientation with maximum shear stress. Solve this problem both using the analytical equations and using Mohr's circle construction. Stresses are given in MPa.

Solution:

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We're going to solve this by two methods. First using the analytic solution then using Mohr's circle.

Analytic Solution

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We have the equations

and

where the prime stresses are in the rotated reference frame. When the system is in its principal orientation and and . Setting and solving yields

and upon substituting the stresses given in this problem we find °. Substituting this value of and the given values for yields MPa MPa and MPa, which verifies that this is indeed the principal axis.

If we are given a stress state that is the principal orientation then in our rotated reference frame above the term is zero. We find the direction with maximum shear by taking the derivative of , setting it to zero, and solving for .

and the maximum occurs when °. This means the maximum or minimum occurs at °. Substituting this into our equations along with yields MPa, MPa and MPa, the minimum. By the symmetry of the system rotating an additional ° yields the maximum, MPa, MPa, and MPa.

Mohr's Circle Solution

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Diagram showing the Mohr's circle representation of this example problem.
Diagram showing the Mohr's circle representation of this example problem.

Following the diagram given here,

and

Given , , and , we concluded that and . Therefore, and °.From the principal orientation MPa and MPa. The minimum and maximum shear stress is MPa and are found by rotating the system ° and .

Example 2

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Problem Statement:

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Given a point on an elastic body with stress state , determine the principal stresses. Stresses are given in MPa. [Hint: There is only one way to draw the Mohr's circle representation. Use this to simplify your work. You'll find that there is almost no math once you draw the picture.]

Solution:

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Example 3

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Problem Statement:

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Given a point on an elastic body with stress state , and that rotating about yields stress state , fully determine both stress states and the unknown parameter . Stresses are given in MPa.

Solution:

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Diagram showing the Mohr's circle representation of this example problem. The shaded regions are similar triangles.
Diagram showing the Mohr's circle representation of this example problem. The shaded regions are similar triangles.

We have the equations

and

where the prime stresses are in the rotated reference frame.

This immediately allows us to substitute for and to find . This give MPa and . The invariant relation

allows

with substitutions determines MPa. At this point all the parameters are determined except and simply substituting into either the or allows us to find MPa. The resulting stress tensors, in units of MPa, are

The Mohr's circle representation of this solution is shown here. Note that the shaded triangles are similar and therefore can be used to simplify the solution if sought graphically.

Example 4

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Problem Statement:

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Given a point on an elastic body with stress state , determine the principal stresses and the angle between the original orientation and the orientation with maximum shear stress. Stresses are given in MPa.

Solution:

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Example 5

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Problem Statement:

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Both mathematically and with words explain the impact of applying the transformation tensors , ,, and on vector . Apply the same transformation to rank two tensor .

Solution:

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Remembering the definition of tensor multiplication and the implicit summations used in Einstein notation, we know that the rank two tensor acting on the vector will necessarily result in terms that look like

which is similar to "normal" matrix multiplication as you've seen before. In contrast applying the rank two transformation tensor on the rank two tensor requires a double sum which results in terms that look similar to

When working with these type of 9-term sums it is usually advisable to use a software package to simply the work. The results are given here.

Transformation tensor 1 results in

which corresponds to the identity transformation, i.e., leave the tensors unmodified.

Transformation tensor 2 results in

This transformation is the inversion transformation. This can be seen in the behavior of the vector. Interestingly enough it leaves unmodified.

Transformation tensor 3 results in

This transformation involves mirroring across the and directions. This is equivalent to rotation around the -axis by °.

Transformation tensor 4 results in This transformation involves rotation around the -axis. Substituting ° for ° yields identical results to transformation tensor 3.

Example 6

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Problem Statement:

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Consider the stress state . Write a transformation tensor that rotates the reference frame into the principal directions. Stresses are given in MPa.

Solution:

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We know from Example 1 that the solution is to rotate by around the axis and we know the form of the rotation transformation matrix from Example 5, therefore the solution is

where .

Example 7

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Problem Statement:

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Given the displacement tensor , identify the rotation tensor and strain tensor.

Solution:

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We can break any tensor into a fully symmetric and fully anti-symmetric tensor resulting in an anti-symmetric rotation tensor

and symmetric strain tensor

resulting ultimately in .

Example 8

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Problem Statement:

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Within linear, homogeneous, isotropic elasticity theory, for a given stress state (units in MPa) determine the strain state given the Poisson ratio is 0.40 and the shear modulus is 50 GPa. Identify the hydrostatic stress, deviatoric stresses, strain dilatation, and strain deviator.

Solution:

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We're given the stress, so returning a solution requires determining the strain, which can be determined by

We're given the Poisson ratio, and shear modulus but to use this equation we need the elastic modulus . This can be found from

which can be inverted to

Upon substitution this yields GPa. The resulting strain tensor is

The hydrostatic stress is

The strain dilatation is

which results in a mean strain of

(Note that when strain is small , which yields . Unfortunately, in this example the strain is relatively large.) The deviatoric stress and strain are then determined by subtracting the mean stress and strain from their respective tensors diagonal.

and

Example 9

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Problem Statement:

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Within linear, homogeneous, isotropic elasticity theory, for a given strain state determine the stress state given the bulk modulus of 100 GPa and Lam parameter of 50 GPa. Identify the hydrostatic stress, deviatoric stresses, strain dilatation, and strain deviator.

Solution:

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We are given the strain, so returning a solution requires determining the stress, which can be determined by

where is the Lamé parameter . Since

and

we know

The stress expression becomes

Substituting and solving yields

GPa

The hydrostatic stress is

The strain dilatation is

which results in a mean strain of

An interesting observation is that the strain dilation and hydrostatic (mean) stress are related by the bulk modulus , which in this case is . When performing calculations it is important to use known checkpoints such as these to validate your work.

The deviatoric stress and strain are then determined by subtracting the mean stress and strain from their respective tensors diagonal.

and

Example 10

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Problem Statement:

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For a single crystal of cubic zirconia, which has a elastic constants approximately , , and GPa, determine the elastic energy required to apply a uniaxial strain of 0.001 in the direction. Determine the elastic energy required to apply a uniaxial strain of 0.001 in the direction. Compute the Zener anisotropy ratio. Using isotropic elasticity theory, the elastic modulus of 200 GPa, and Poisson ratio of 0.3 the elastic modulus of and the Poisson's ratio of to compute the elastic energy to apply a uniaxial strain of 0.001.

Solution:

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Say that the is in the direction. The elastic energy, , is

so

In the case of anisotropic elasticity theory

In this case all except . Therefore all except , , and .

Substituting into the above equationsː

Example 11

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Problem Statement:

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For a single crystal of cubic zirconia, which has a elastic constants approximately , , and GPa, determine the elastic energy required to apply a uniaxial strain of 0.001 in the direction followed by a shear strain of 0.001 on the face in the direction.

Solution:

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Obeying the given order of operation, first we apply a uniaxial load of in the direction. (Also covered in Example 10)

Note that our general equations are , and .

Restating the strain tensorː

From this we get the non-zero stress tensorː

Thus the only non-zero is , and by integrating our equation for we getː

Now let's apply the shear on the face in the direction.

Our new stress tensor isː

Here, the only non-zero terms are , and . As these are equivalent terms, due to symmetry, we can solve for one and multiply the answer by two.


Once again utilizing our basic energy equation we getː

Finally, adding and together to get the total energy of this combined transformation gives us a final answer of .

Example 12

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Problem Statement:

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For a polycrystal specimen of cubic zirconia, which has a elastic constants approximately , , and GPa, use isotropic elasticity theory and the elastic modulus of and the Poisson's ratio of to compute the elastic energy to apply a strain state

If the polycrystal material has a porosity of 2% approximately how much will this change the elastic modulus? Approximately how much will this change the elastic energy for this applied strain?

Solution:

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Note that our general equations are , and .

Here, we can put the stress in terms of the elastic modulus and the Poisson's ratioː

Where the latter constant is equivalent to the Lamé Constant (

Briefly solving for the Lamé Constant yields usː

Keeping in mind that equals thanks to symmetry, the non-zero stresses for this problem areː

Therefore, we can write out the non-zero energy terms asː