Overview of Elasticity of Materials/Example Problems

Example 1Edit

Problem Statement:Edit

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:Edit

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

Analytic SolutionEdit

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 SolutionEdit

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

Example 2Edit

Problem Statement:Edit

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:Edit

Example 3Edit

Problem Statement:Edit

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:Edit

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.

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

Example 4Edit

Problem Statement:Edit

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:Edit

Example 5Edit

Problem Statement:Edit

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:Edit

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 6Edit

Problem Statement:Edit

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

 

Solution:Edit

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 7Edit

Problem Statement:Edit

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

 

Solution:Edit

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 8Edit

Problem Statement:Edit

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:Edit

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 9Edit

Problem Statement:Edit

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:Edit

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

 
where   is the Lam  parameter  . Since
 
and
 
we know
 
The stress expression becomes
 
Substituting and solving yields
 
in units of 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 10Edit

Problem Statement:Edit

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:Edit

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 11Edit

Problem Statement:Edit

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:Edit

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 12Edit

Problem Statement:Edit

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:Edit

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