Open Educational Resources

Stress: Examples and Problems

Example 1

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

    \[\sigma = \begin{pmatrix} 1 & -1 & 0 \\ -1 & -5 & 0 \\ 0 & 0 & 4 \\ \end{pmatrix}\]

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

    \[u = \begin{pmatrix} 1 \\ 1 \\ 1 \\ \end{pmatrix}\]

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 u is to normalize it:

    \[n = \frac{u}{||u||} = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ 1\\ 1\\ \end{pmatrix}\]

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

    \[t_n = \sigma n = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 & -1 & 0 \\ -1 & -5 & 0 \\ 0 & 0 & 4 \\ \end{pmatrix} \begin{pmatrix} 1 \\ 1\\ 1\\ \end{pmatrix} = \frac{1}{\sqrt{3}} \begin{pmatrix} 0 \\ -6 \\ 4 \\ \end{pmatrix}\]

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

    \[t_n \cdot n = \frac{-2}{3}\]

The shear stress vector on n:

    \[t_n - \sigma_n n = \frac{1}{\sqrt{3}} \begin{pmatrix} 0 \\ -6 \\ 4 \\ \end{pmatrix} + \frac{2}{3\sqrt{3}} \begin{pmatrix} 1 \\ 1 \\ 1 \\ \end{pmatrix} = \frac{1}{3\sqrt{3}} \begin{pmatrix} 2 \\ -16 \\ 14 \\ \end{pmatrix}\]

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

    \[\tau_n = ||t_n - \sigma_n n || = \frac{2\sqrt{38}}{3}\]

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

    \[\sigma_1 = -5.16 \hspace{5mm} \sigma_2 = 1.16 \hspace{5mm} \sigma_3 = 4\]

The directions of the principal stresses are the associated eigenvectors:

    \[ev_1 = \begin{pmatrix} 0.160 \\ 0.987 \\ 0 \\ \end{pmatrix} \hspace{5mm} ev_2 = \begin{pmatrix} -0.987 \\ 0.160 \\ 0 \\ \end{pmatrix} \hspace{5mm} ev_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \\ \end{pmatrix}\]

Note that the order of the eigenvalues was chosen so that they form a properly oriented coordinate system ev_1\cdot (ev_2\times ev_3)=1. The stress matrix will be diagonal in a coordinate system defined by ev_1, ev_2, and ev_3. The coordinate transformation matrix is:

    \[Q = \begin{pmatrix} 0.160 & 0.987 & 0 \\ -0.987 & 0.160 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}\]

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

    \[\sigma ' = Q\sigma Q^T = \begin{pmatrix} -5.16 & 0 & 0 \\ 0 & 1.16 & 0 \\ 0 & 0 & 4 \\ \end{pmatrix}\]

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

View Python Code

from sympy import Matrix
import sympy as sp
sp.init_printing(use_latex="mathjax")
s = Matrix([[1,-1,0],[-1,-5,0],[0,0,4]])
display("\u03C3 =",s)
u = Matrix([1,1,1])
display("u = ",u)
n = u/u.norm()
display("n = ",n)
t = s*n
display("traction vector = \u03C3n =",t)
n_stress = t.dot(n)
display("normal stress =", n_stress)
shear_stress = t - n_stress*n
display("shear stress vector on n =",shear_stress)
display("\u03C4_n =", shear_stress.norm())
eigensystem = s.eigenvects()
display("eigensystem", eigensystem)
eigenvalues = [i[0] for i in eigensystem]
eigenvectors = [i[2][0].T for i in eigensystem]
neigenvectors = [(i/i.norm()).evalf() for i in eigenvectors]
display("eigenvalues (principal Stresses) =",eigenvalues)
display("eigenvectors (principal directions) =",eigenvectors)
display("Normalized eigenvectors (principal directions) =",neigenvectors)
Q = Matrix(neigenvectors)
# rearrange 
Q = Matrix([Q[2,:],Q[1,:],Q[0,:]])
display("Q =",Q.evalf(3))
sp = Q*s*Q.T
display("\u03C3' = Q*\u03C3*Q^T =",sp.evalf(3))

Example 2

Show that the following two stress states (units of MPa) have the same hydrostatic stress and \sigma_{vM}.

    \[\sigma_a = \begin{pmatrix} 2 & 2 & 0 \\ 2 & 5 & 0 \\ 0 & 0 & -5 \\ \end{pmatrix} \hspace{5mm} \sigma_b = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 6 \\ \end{pmatrix}\]

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:

    \[p_a = \frac{2+5-5}{3} = \frac{2}{3} \hspace{5mm} p_b = \frac{1-5+6}{3} = \frac{2}{3}\]

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

    \[\sigma_{vMa} = \sqrt{91} \hspace{5mm} \sigma_{vMb} = \sqrt{91}\]

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

View Python Code

from sympy import Matrix, N
import sympy as sp
sp.init_printing(use_latex="mathjax")
s = Matrix([[2,2,0],[2,5,0],[0,0,-5]])
display("\u03C3 =",s)
eigensystem = s.eigenvects()
display("eigensystem", eigensystem)
eigenvalues = [i[0] for i in eigensystem]
eigenvectors = [i[2][0].T for i in eigensystem]
neigenvectors = [(i/i.norm()).evalf() for i in eigenvectors]
display("eigenvalues (principal Stresses) =",eigenvalues)
display("eigenvectors (principal directions) =",eigenvectors)
display("Normalized eigenvectors (principal directions) =",neigenvectors)
Q = Matrix(neigenvectors)
# rearrange 
Q = Matrix([Q[1,:],Q[0,:],Q[2,:]])
display("Q =",Q.evalf(3))
sp = N(Q*s*Q.T,chop=True)
display("\u03C3' =",sp)

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]]

View Python Code

from sympy import Matrix
import sympy as sp
sp.init_printing(use_latex="mathjax")
s1 = Matrix([[2,2,0],[2,5,0],[0,0,-5]])
s2 = Matrix([[2,2,1],[2,4,-3],[1,-3,7]])
# vonMises function
def vonMises(M):
    return sp.sqrt(1/2*((M[0,0] - M[1,1])**2+(M[1,1] - M[2,2])**2+
                       (M[2,2] - M[0,0])**2+6*(M[0,1]**2+M[0,2]**2+M[1,2]**2)))
# example of a rotation matrix
t = sp.symbols("\u03B8")
Qx = Matrix([[1, 0, 0], 
            [0, sp.cos(t), -sp.sin(t)], 
            [0, sp.sin(t), sp.cos(t)]])
Qx = Qx.subs({t:30*sp.pi/180})
# outputs
display("s_1 =",s1)
display("Von Mises =",vonMises(s1))
display(vonMises(Qx*s1*Qx.T))
display("s_2 =",s2)
display("Von Mises =",vonMises(s2))
display(vonMises(Qx*s2*Qx.T).evalf())

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):

    \[\sigma = \begin{pmatrix} 25 & 40 & 0 \\ 40 & 20 & 30 \\ 0 & 30 & 20 \\ \end{pmatrix}\]

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

Soltuion:

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

    \[\tau_{max} = \frac{| \sigma_y |}{2} = 100 MPa\]

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:

    \[\frac{\max {\{|71.7+28.5|, |71.7-21.8|, |-28.5-21.8|}\}}{2} = 50.1 MPa < \tau_{max}\]

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 \sigma_{max}=|\sigma_y|=200MPa. In this case, and utilizing the von Mises stress expression as a function of the principal stresses:

    \[\sigma_{vM} = \sqrt{\frac{(71.7+28.5)^2+(71.7-21.8)^2+(-21.8-28.5)^2}{2}} = 86.75 MPa < \sigma_{max}\]

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

View Python Code

from sympy import Matrix,N
import sympy as sp
sp.init_printing(use_latex="mathjax")
s1 = Matrix([[25,40,0],[40,20,30],[0,30,20]])
eVals = Matrix([i[0] for i in s1.eigenvects()]).evalf()
#Added chop=True to remove the small numbers.
display("eigenvalues (principal Stresses) =",N(eVals,chop=True))
e1, e2, e3 = eVals
maxShearStress = max([abs(e1-e2),abs(e2-e3),abs(e2-e3)])/2
display("max shear stress =",maxShearStress)
MisesStress = sp.sqrt((abs(e1-e2)**2+abs(e1-e3)**2+abs(e2-e3)**2)/2)
display("Von Mises stress =",MisesStress.evalf())

Example 5

A metallic cylindrical pressure vessel with capped ends is used to hold a gas with an internal pressure P 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.

n
n
Solution:

As shown in the sketch, the orthonormal basis vectors are chosen in this example such that e_1 is aligned with the longitudinal axis of the pipe and e_2 is vertical. The analyzed part of the pressure wall is taken to be perpendicular to e_3. The normal stress component in the longitudinal direction is:

    \[\sigma_{11} = \frac{P \pi r^2}{2 \pi rt} = \frac{Pr}{2t} = 150 MPa\]

The normal stress component in the direction of the vector e_2 is:

    \[\sigma_{22} = \frac{Pr}{t} = 300 MPa\]

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

    \[\sigma = \begin{pmatrix} 150 & 0 & 0 \\ 0 & 300 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} MPa\]

The von Mises stress is:

    \[\sigma_{vM} = 150 \sqrt{3} = 259.81 MPa\]

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]

View Python Code

from sympy import Matrix
import sympy as sp
sp.init_printing(use_latex="mathjax")
display("Units are in MPa")
P = 3
d = 2000
t = 10
r = d/2
def vonMises(M):
    return sp.sqrt(1/2*((M[0,0] - M[1,1])**2+(M[1,1] - M[2,2])**2+
                       (M[2,2] - M[0,0])**2+6*(M[0,1]**2+M[0,2]**2+M[1,2]**2)))
s_11 = (P*sp.pi*r**2)/(2*sp.pi*r*t)
display("\u03C3_11 =",s_11)
s_22 = P*r/t
display("\u03C3_22 =",s_22)
s = Matrix([[s_11,0,0],[0,s_22,0],[0,0,0]])
display("\u03C3 =",s)
display("Von Mises =",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.

n
SOLUTION

The plane of maximum shear is oriented at 45^\circ 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 \sigma_{11}=\sigma_{33}, the maximum shear stresses will occur on a plane oriented at 45^\circ to the vertical direction. The shear stress on that plane will have the value:

    \[\frac{\max{\{|-500+100|,|-500+100|,|-500+500|}\}}{2} = 200MPa\]

View Python Code

import sympy as sp
sp.init_printing(use_latex="mathjax")
e1, e2, e3 = -500, -100, -500
maxShearStress = max([abs(e1-e2),abs(e2-e3),abs(e2-e3)])/2
display("max shear stress (MPa) =",maxShearStress)

Problems

  1. Write the traction vectors on the boundary surfaces AB, BC, CD, and DE of the structure shown below.

    n

  2. The Cauchy stress at a point inside a continuum is given by (Units of MPa)

    \sigma = \begin{pmatrix} 100 & 200 & 0 \\ 200 & 300 & 0 \\ 0 & 0 & 400 \\ \end{pmatrix}

    Determine the following:

    • The traction vector acting on the plane with the normal e_1+2e_2+2e_3 which passes through the point.

    • The angle between the traction vector and the normal to the plane.

    • The normal and shear stress components acting on the plane and indicate whether the normal stress is tension or compression.

  3. The Cauchy stress at a point inside a continuum is given by (Units of MPa):

    \sigma = \begin{pmatrix} 400 & 300 & 0 \\ 300 & -400 & 0 \\ 0 & 0 & 200 \\ \end{pmatrix}

    Determine the following:

    • The traction vector acting on the plane passing through the point and defined by the equation 3x_1+x_2=c where c is a constant.

    • The angle between the traction vector and the normal to the plane.

    • The normal and shear stress components acting on the plane and indicate whether the normal stress is tension or compression.

    Can we consider the normal to the plane as an eigenvector for the stress tensor?

    If so, what is the corresponding eigenvalue? (Hint: The unit normal to the plane defined by f(x_1,x_2,x_3)=c is given by n=\frac{\nabla f}{\|f\|} in which \nabla f = \frac{\partial f}{\partial x_1}e_1+\frac{\partial f}{\partial x_2}e_2+\frac{\partial f}{\partial x_3}e_3)

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

    \sigma = \begin{pmatrix} 45 & -12 & 16 \\ -12 & 8 & 23 \\ 16 & 23 & -10 \\ \end{pmatrix}

    Determine the following:

    • The components of the traction vector on a plane with unit normal vector n=\frac{1}{3}\left(\begin{array}{c}-2\\1\\-2\end{array}\right)

    • The norm of t_n.

    • The normal and shear stresses on the plane with normal n.

    • The angle of inclination of the force per unit area acting on the plane with normal n with respect to n.

  5. 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 P of 1.3MPa is maintained inside the vessel.

    n

    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.

  6. Consider a thin closed cylindrical pressure vessel that is manufactured using Filament Winding with filaments wound at an angle \theta=60^\circ. The vessel is 1m long, has an internal diameter of 60cm, and a wall thickness of 10mm. If the cylinder is pressurized to 1.5MPa, determine:

    • The shear force per unit length along the fibre-matrix interface.

    • The normal force per unit length along the fibre-matrix inetrface.

    • The normal force per unit length along the fibre direction.

  7. The shown cuboid is under the state of stress represented by the Cauchy stress matrix below. The coordinates of points C, B, and D are \{3,0,0\}, \{3,0,2\}, and \{0,1,0\}, respectively. The cuboid is composed of two halves glued together on the plane ABCD. 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?

    n

    \sigma = \begin{pmatrix} 100 & 200 & 30 \\ 200 & -100 & -50 \\ 30 & -50 & 100 \\ \end{pmatrix}

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

    Determine:

    • The Cauchy stress matrix \sigma in the orthonormal basis set B=\{e_1,e_2,e_3\}.

    • The Cauchy stress matrix \sigma’ in the orthonormal basis set B=\{e'_1,e'_2,e'_3\}.

    • The coordinate system in which the Cauchy stress matrix is diagonal. Find the Cauchy stress matrix \sigma'' 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 \sigma and \sigma'.

  9. A machine component was analyzed by engineer A in the orthonormal basis set B=\{e_1,e_2,e_3\} 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 B'=\{e'_1,e'_2,e'_3\}. 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 \sigma' 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.)

    n

    \sigma = \begin{pmatrix} 100 & 200 & 30 \\ 200 & -100 & -50 \\ 30 & -50 & 100 \\ \end{pmatrix}

  10. 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.

    n

    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.

  11. A metal plate is subjected to a shear state of stress in a certain coordinate system represented by the matrix below.

    \sigma = \begin{pmatrix} 0 & t & 0 \\ t & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}

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

    • t=\frac{\sigma_y}{\sqrt{3}} if the von Mises yield criterion is used.

      • t=\frac{\sigma_y}{2} if the Tresca yield criterion is used.

  12. Two plates are connected together using a bolt through a hole as shown in the figure.

    n

    If the diameter of the bolt is 20mm, find the force F 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.

  13. The Cauchy stress at a point inside a conitnuum is given by (Units of MPa):

    \sigma = \begin{pmatrix} 500 & 300 & 0 \\ 300 & 300 & 0 \\ 0 & 0 & 400 \\ \end{pmatrix}

    Find:

    • The hydrostatic stress.

    • The principal stresses and the principal directions.

    • The von Mises stress.

  14. A metal plate is subjected to a state of stress represented by the following matrix (units of MPa):

    \sigma = \begin{pmatrix} 100 & 200 & 0 \\ 200 & 0 & 0 \\ 0 & 0 & 100 \\ \end{pmatrix}

    Assume that the yield stress of the plate material (when the plate material is subjected to a uniaxial state of stress) is equal to 400MPa.

    • Using the von Mises yield criterion, determine whether the plate has yielded or not.

    • Using the Tresca yield criterion, determine whether the plate has yielded or not.

    • If the plate has not reached yield based on the von Mises yield criterion, then, calculate the value of t>0 in the following additional stress that when added to the existing stress state would bring the plate to yield:

      • \Delta \sigma = \begin{pmatrix} 0 & t & 0 \\ t & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}

    • Is it possible to bring about the yielding of the plate by adding the following state of stress to the existing state of stress? Explain your answer.

      • \Delta \sigma = \begin{pmatrix} t & 0 & 0 \\ 0 & t & 0 \\ 0 & 0 & t \\ \end{pmatrix}

  15. Let a and b\in\mathbb{R}^3 be two unit vectors defining distinct directions at a point P in a body under stress. The vectors t_a and t_b are the traction vectors at P associated with areas having unit normals a and b respectively. Show that t_a\cdot b=t_b\cdot a.

  16. If S is a scalar having the units of stress and n\in\mathbb{R}^3 is a unit vector, show that the stress tensor whose components are given by: \sigma_{ij}=Sn_in_j represents a state of uniaxial tension S in the direction n. (Hint: Use a coordinate transformation into an orthonormal basis set B =\{n,a,b\} where a and b are unit vectors orthogonal to each other and to n.

  17. Let the components of a stress matrix be expressed as:

    \sigma_{ij} = an_in_j = bm_im_j

    Where, a, b\in\mathbb{R} while n,m\in\mathbb{R}^3 is a pair of orthonormal vectors. Find the directions of the principal stresses and show that the principal stresses are a, b, and 0.

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

    \sigma = \begin{pmatrix} 12 & 0 & 43 \\ 0 & 20 & -14 \\ 43 & -14 & 25 \\ \end{pmatrix}

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

    F = \begin{pmatrix} 1.1 & 0 & 0 \\ 0.2 & 1.2 & 0 \\ 0.1 & 0 & 1.3 \\ \end{pmatrix}

    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.

  19. A state of deformation is described by F=\lambda I, where \lambda is constant and I is the unit matrix. The state of stress is hydrostatic, and the Cauchy stress is given by \sigma=pI where p is the hydrostatic pressure. Find the first and second Piola-Kirchhoff stress tensors.

  20. The engineering strain-engineering stress curve for a steel material follows the numbers in the table shown. Assuming that E=210,000MPa and Poisson’s ratio \nu=0.3, plot the true strain – true stress curve.

    Strain (%)00.0850.172.065.029.952027.07
    Stress(MPa)0178.4356.4382428.5463.8472.5456.2

  21. For a special type of solid mechanics problems, there could be a need to have three different configurations, rather than simply the reference and deformed configurations. Assume that a deformed configuration is described by the function x_d(X) and a deformation gradient F_d. Assume the reference configuration to have coordaintes X. Let there be another intermediate configuration described by the relationship x_r(X) and the deformation gradient F_r. Find the relationship between the first Piola Kirchhoff stress tensor in the reference configuration P and the first Piola Kirchhoff stress tensor in the intermediate configuration P_r in terms of F_r and F_d.

    22. Page Comments

    1. Danny Panduro says:
      August 1, 2024 at 2:27 pm

      I think the normal stress answer for example 5.6.1 problem 1 is missing the square root over 3. It says -2/3 but I think it’s -2/sqrt(3).

      Reply
      1. Samer Adeeb says:
        August 1, 2024 at 5:34 pm

        The answer in the text is correct. When you take the dot product between t_n and n, you multipley the square root of 3 by another square root of 3 and you obtain 3 in the denominator.

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