## Stress: Examples and Problems

### Examples and Problems:

#### Example 1

The stress at a point inside a continuum is given by the stress matrix (units of MPa):

Find the normal and shear stress components on a plane whose normal vector is in the direction of the vector

Also, find the principal stresses and their directions. Find the coordinate transformation in which the stress matrix is diagonal and expression the stress matrix in the new coordinate system.

##### Solution:

The first step in finding the stress on the plane whose normal vector is is to normalize it:

The traction vector (units of MPa) on the plane can simply be obtained using the formula:

The normal stress (in MPa) can be obtained using the dot product between and :

The shear stress vector on :

The shear stress (units of MPa) is the norm of the shear stress vector:

The principal stresses and their directions are simply the eigenvalues and eigenvectors of the stress matrix. The principal stresses (units of MPa) are:

The directions of the principal stresses are the associated eigenvectors:

Note that the order of the eigenvalues was chosen so that they form a properly oriented coordinate system . The stress matrix will be diagonal in a coordinate system defined by , , and . The coordinate transformation matrix is:

The stress matrix (units of MPa) in the new coordinate system has the form:

View Mathematica Code:
s = {{1, -1, 0}, {-1, -5, 0}, {0, 0, 4}}
u={1,1,1};
n=u/Norm[u];
tn=Transpose[s].n
sn=tn.n
FullSimplify[tn-sn*n]
tawn=FullSimplify[Norm[tn-sn*n]]
a = N[Eigensystem[s]];
sigma1 = a[[1, 1]]
sigma2 = a[[1, 2]]
sigma3 = a[[1, 3]]
ev1 = a[[2, 1]]/Norm[a[[2, 1]]]
ev2 = a[[2, 2]]/Norm[a[[2, 2]]]
ev3 = a[[2, 3]]/Norm[a[[2, 3]]]
ev1.Cross[ev3, ev2]
Q = {ev1, ev3, ev2};
Q // MatrixForm
sigmaprime = Chop[Q.s.Transpose[Q]];
sigmaprime // MatrixForm


#### Example 2

Show that the following two stress states (units of MPa) have the same hydrostatic stress and .

##### Solution

A quick glance at the stress matrix on the left shows a shear stress of 2 MPa on one plane and three normal stresses of 2, 5 and –5 MPa. The stress matrix on the right shows no shear stresses with three normal stresses of 1, –5 and 6 MPa. A careless observer could assert that the right stress matrix is “more severe” since it shows a plane with a normal stress component of 6 MPa. However, using the stress invariants, we can easily show that both stress matrices are equivalent in terms of severity, assuming that the material failure is independent of the direction of the applied stress. Considering the left stress matrix, the hydrostatic stresses (MPa) for both matrices are:

The von Mises stresses (MPa) for both matrices are:

In fact, we can use Mathematica to show that both matrices represent the same state of stress, but at two different coordinate systems. The following code utilizes the diagonalization procedure for symmetric matrices. The eigenvalues and eigenvectors of the left matrix are obtained. Then, a transformation matrix is formed using the three normalized eigenvectors. The final matrix is shown to be exactly the right stress matrix above.
View Mathematica Code:

S1={{2,2,0},{2,5,0},{0,0,-5}};
ss=Eigensystem[S1];
eigenvector1=ss[[2,1]]/Norm[ss[[2,1]]];
eigenvector2=ss[[2,2]]/Norm[ss[[2,2]]];
eigenvector3=ss[[2,3]]/Norm[ss[[2,3]]];
Q={eigenvector3,eigenvector2,eigenvector1};
S2=FullSimplify[Q.S1.Transpose[Q]];
S2//MatrixForm


#### Example 3

Create a function in Mathematica that returns the value of the von Mises stress of a general stress matrix.

##### Solution

The function shown in the code can be very helpful to quickly calculate the von Mises stress of a stress matrix using Mathematica.
View Mathematica Code:

VonMises[sigma_]:=Sqrt[1/2*((sigma[[1,1]]-sigma[[2,2]])^2+(sigma[[2,2]]-sigma[[3,3]])^2+(sigma[[3,3]]-sigma[[1,1]])^2+6*(sigma[[1,2]]^2+sigma[[1,3]]^2+sigma[[2,3]]^2))]
stress1={{2, 2, 0},{2, 5, 0},{0, 0,-5}};
stress2={{2, 2, 1},{2, 4, -3},{1, -3, 7}};
Q=RotationMatrix[30Degree,{2,1,1}];
Qt=Transpose[Q];
VonMises[stress1]
FullSimplify[VonMises[Q.stress1.Qt]]
VonMises[stress2]
FullSimplify[VonMises[Q.stress2.Qt]]


#### Example 4

A metal yields in a uniaxial state of stress when the uniaxial stress reaches 200MPa. Will the metal yield when the stress is described by the matrix (units of MPa):

Assuming that the metal follows either the Tresca yield criterion or the von Mises yield criterion.

##### Solution

If the material follows the Tresca yield criterion, the first step is to find . Since in a uniaxial state, the metal will fail when the uniaxial stress reaches 200MPa, therefore:

The eigenvalues of the matrix describing the given state of stress are: 71.7MPa, -28.5MPa, and 21.8MPa. The maximum shear stress in this case is:

Therefore, according to the Tresca yield criterion, the state of stress is well below yield. If the metal follows the von Mises yield criterion, then . In this case, and utilizing the von Mises stress expression as a function of the principal stresses:

Therefore, according to the von Mises yield criterion, the state of stress is well below yield.

#### Example 5

A metallic cylindrical pressure vessel with capped ends is used to hold a gas with an internal pressure of 3 MPa. The vessel’s average diameter is 2 m, and its thickness is 10mm. Find the von Mises stress at a point on the cylindrical wall.

##### Solution:

As shown in the sketch, the orthonormal basis vectors are chosen in this example such that is aligned with the longitudinal axis of the pipe and is vertical. The analyzed part of the pressure wall is taken to be perpendicular to . The normal stress component in the longitudinal direction is:

The normal stress component in the direction of the vector is:

All the remaining components of the stress matrix are zero. Thus, the stress matrix has the following form:

The von Mises stress is:

View Mathematica Code:
P=3;r=1000;t=10;Ee=210000;Nu=0.3;
s=Table[0,{i,1,3},{j,1,3}];
s[[1,1]]=P*r/2/t
s[[2,2]]=P*r/t
VonMises[sigma_]:=Sqrt[1/2*((sigma[[1,1]]-sigma[[2,2]])^2+(sigma[[2,2]]-sigma[[3,3]])^2+(sigma[[3,3]]-
sigma[[1,1]])^2+6*(sigma[[1,2]]^2+sigma[[1,3]]^2+sigma[[2,3]]^2))];
a=VonMises[s]

#### Example 6

A soil specimen is in a confined compression test as shown in the figure. If the soil material is weak in shear, find the potential orientation of the failure plane and find the shear stress on that plane.

##### Solution

The plane of maximum shear is oriented at to the principal planes. Since the specimen is in a confined compression test, therefore, in the current coordinate system, there are no shear stresses. Since , the maximum shear stresses will occur on a plane oriented at to the vertical direction. The shear stress on that plane will have the value:

#### Problems:

1. If the components of the Cauchy stress tensor at some point in a body are given by:

Determine the following:

• The components of the traction vector on a plane with unit normal vector .
• The norm of .
• The normal and shear stresses on the plane with normal .
• The angle of inclination of the force per unit area acting on the plane with normal with respect to .
2. A 1m average diameter pressure vessel is constructed by butt-welding a 10mm plate with a spiral seam weld, as shown in the figure. A constant pressure of 1.3MPa is maintained inside the vessel.

Determine:

• The traction vector on the spiral seam weld plane in the coordinate system shown in the figure.
• The normal stress on the spiral seam weld.
• The shearing stress on the spiral seam weld.
3. The shown cuboid is under the state of stress represented by the Cauchy stress matrix below. The coordinates of points , , and are , , and , respectively. The cuboid is composed of two halves glued together on the plane . The maximum shear stress that the glue can take is 100MPa, while the maximum normal tensile stress is 200MPa. Will the glue fail in shear and/or in tension?

4. A material point is subjected to a state of plane stress as shown in the figure.

Determine:

• The Cauchy stress matrix in the orthonormal basis set .
• The Cauchy stress matrix in the orthonormal basis set .
• The coordinate system in which the Cauchy stress matrix is diagonal. Find the Cauchy stress matrix in this new coordinate system.
• The coordinate system producing the maximum shear. Evaluate the stress matrix in this new coordinate system.
• The hydrostatic stress, the von Mises stress, the maximum shear stress for both and .
5. A machine component was analyzed by engineer A in the orthonormal basis set using a finite element analysis package. The critical stress state was found to be represented by the Cauchy stress matrix  shown below. Another engineer (engineer B) conducted a separate analysis in which the machine component was oriented as shown in the figure with the orthonormal basis set . Notice that point C is the new origin of the coordinate system chosen by engineer B. Assuming that the loads and the boundary conditions of both parts have the same orientation with respect to the machine component, find the critical Cauchy stress matrix  obtained by engineer B and show that the hydrostatic stress and the von Mises stress are equal in both cases. (Hint: Find the transformation matrix between the coordinate systems.)

6. The shown cylinder is subjected to a confined compression test in which the horizontal stress is kept constant at –200MPa, while the magnitude of the compressive vertical stress is increased above the value of –200MPa.
Find the magnitude of the compressive vertical stress at which the material will fail in the following two separate cases:

• The maximum shear stress that the sample can withstand is 200MPa.
• The maximum von Mises stress that the sample can withstand is 400MPa.
7. A metal plate is subjected to a shear state of stress in a certain coordinate system represented by the matrix below.

If the yield stress of the material under uniaxial stress is , show that the maximum value for at which the material will yield is:

• if the von Mises yield criterion is used.
• if the Tresca yield criterion is used.
8. Two plates are connected together using a bolt through a hole as shown in the figure.
If the diameter of the bolt is 20mm, find the force that will cause yielding in the bolt if the yield stress of the bolt material is 400MPa in the following two cases:

• Using the von Mises yield criterion.
• Using the Tresca yield criterion.
9. Let and be two unit vectors defining distinct directions at a point P in a body under stress. The vectors and are the traction vectors at P associated with areas having unit normals and respectively. Show that .
10. If is a scalar having the units of stress and is a unit vector, show that the stress tensor whose components are given by: represents a state of uniaxial tension in the direction . (Hint: Use a coordinate transformation into an orthonormal basis set where and are unit vectors orthogonal to each other and to .
11. Let the components of a stress matrix be expressed as:

Where, , while is a pair of orthonormal vectors. Find the directions of the principal stresses and show that the principal stresses are , , and 0.

12. The stress at a point is given by the following Cauchy stress tensor (Units of MPa).

If the material point deforms under the action of the following linear mapping:

Determine:

1. The three principal stresses and the unit vectors in the principal directions.
2. The hydrostatic and von Mises stresses.
3. The first and second Piola-Kirchhoff stress tensors.
4. The area vectors after deformation of the three planes perpendicular to the coordinate system axes.
5. The force vectors after deformation on the three planes perpendicular to the coordinate system axes.
6. The action of the first Piola-Kirchoff stress matrix on the three planes perpendicular to the coordinate system axes.
7. The action of the second Piola-Kirchhoff stress matrix on the three planes perpendicular to the coordinate system axes.
13. A state of deformation is described by , where is constant and is the unit matrix. The state of stress is hydrostatic, and the Cauchy stress is given by where is the hydrostatic pressure. Find the first and second Piola-Kirchhoff stress tensors.
14. The engineering strain-engineering stress curve for a steel material follows the numbers in the table shown. Assuming that MPa and Poisson’s ratio , plot the true strain – true stress curve.
 Strain (%) 0 0.085 0.17 2.06 5.02 9.95 20 27.07 Stress(MPa) 0 178.4 356.4 382 428.5 463.8 472.5 456.2

1. hanan says:

how can i find the traction at p on the plane normal to x1-axis

Sorry, I don’t understand the question.

2. muzaffar says:

when is max. radial stress at test pressure is equal to negative test pressure