Linear Algebra in Mechanics of
Materials
Chance Stupack, Quinn Carvalho
05/10/2012
1
Abstract
The purpose of this paper is to explain how useful linear algebra is in
Mechanics of Materials. Because Mechanics of Materials is such a broad
topic our main focus will be on the stresses and strains in an element.
2
1
A Brief Overview
Mechanics of materials, also known as strengths of materials, is one of the basic
engineering subjects that is essential to being a successful engineer. As you
might have guessed by the name, mechanics of materials is concerned with the
integrity and design of structures and the materials used to build them. Statics,
another basic engineering class, is essential to accurately analyze the structures
and forces acting upon them. Linear algebra comes into play in the forces on
the structure; you see, all forces have a magnitude and direction and as such
can be expressed as vectors. Also, vectors can be used to calculate stresses and
strains that would have otherwise been quite laborious to find. Although linear
algebra isn’t a required course for most engineering degrees it’s applications can
be found in nearly all.
2
Statics
Before we begin to calculate the strengths of materials and the stresses and
strains acting upon them we must be able to find the forces that are acting on
the structure in question. We do this by using statics. Statics is the study of
objects in static equilibrium and the forces that are acting upon them.
2.1
External Forces
All forces that are acting on a structure can be expressed as vectors. The
basic external forces that we will be calculating are forces in the x(horizontal)
direction, forces in the y(vertical) direction and moments. Forces in the x
and y directions are simply as they sound, forces with magnitudes pointing
horizontally or straight vertically. Moments however are a little different. A
moment is the tendency of a force to twist an object around a point. A single
force vector can in effect have influence on one two or all three of these. Because
the object is in equilibrium we can say that adding all of the forces acting upon
the object will equal to zero, we use this idea to find unknown forces.
3
Linear Algebra
In the following subsections we will give a very brief overview of what linear
algebra is and how it will be applied.
3.1
RREF
RREF is an acronym for Reduced Row Echelon Form. For a matrix to be in
RREF it must satisfy 5 conditions. First, all nonzero values must be located
along the ”diagonal” of the matrix. The ”diagonal” of a matrix is the ith row
and ith column of a matrix. In this way we will end up only having one nonzero
value per column. It is possible to achieve these conditions by performing row
3
operations. Row operations are a series of procedures that we perform on each
row of the matrix. There are four basic procedures: addition, subtraction,
multiplication, and division. Any row of a matrix may be multiplied or divided
by a constant and then any row can be added to or subtracted from by another
row. By performing these operations we can convert our matrix to RREF. It
is possible to make a matrix by using the leading coefficients of the elements
of a system of linear equations. RREF is very valuable to us because if we
convert this matrix into RREF it will give us the solutions to the linear system
of equations.
Here is an example of converting a linear system of equations to RREF. Our
equations will be 3x + 2y + 4z = 0 5x + 2y + 3z = 3 and 5x + 6y + 4z = 7. We
convert this into a 3 × 4 matrix as shown.


3 2 4 0
H = 5 2 3 3 .
5 6 4 7
Now we will perform row operations on this matrix until it meets the requirements to be considered in RREF. After the row operations are performed the
resulting matrix will be as follows.


1 0 0 17/20
H = 0 1 0 53/40  .
0 0 1 −13/10
This implies that x = 17/20 y = 53/40 and z = −13/10
3.2
Transformations
A linear transformation is a function, or mapping, that assigns each vector x
in Rn to a vector T (x) in Rm . Linear Transformations consist of a matrix
being multiplied by a vector to give a new matrix or vector depending on the
size of the initial matrix and vector. This resultant matrix or vector is T (x).
Transformations are useful to us when calculating plane stress and strain.
Here is an example
4
4.1
Calculating Plane Stress and Strain
Definitions
Stress is a measure of the average force per unit area of a surface within a
body on which internal forces are acting. These forces might include tension,
compression, bending, or torsion. There are two types of stresses, direct stress
and shear stress. Direct stress tends to change the volume of the body, one
example of this would be hydrostatic pressure inside of a pressurized water tank.
Shear stress deforms the body without actually changing the overall volume, a
truss pushing against the bolt that secures it to the flange would be exerting a
4
shear stress on the bolt. Strain is a measure of the actual deformation that takes
place in the body. Unlike stress, strain is dimensionless because it is actually a
measure of elongation per unit length. Plane stress and plane strain are a more
general state that we use to analyze the stresses and strains on inclined planes.
Such as a helical weld along a propane tank.
This is an example of the effects shear stress.
4.2
The Applications
If we know the stresses and strains of a planar section, such as a weld going
around the diameter of a propane tank, how would we determine the stresses
and strains of a weld that goes around the tank in an angular fashion? We want
to find the stresses and strains at different angles within the object because the
stresses and strains have different values at different angles. Also, it can be very
valuable if there are two objects attached together, such as two steal beams
welded together or two pieces of wood glued together. Not only can knowing
these values for different orientations show us what forces will be acting on a
weld that is already in place but if we know the stresses and strains that will be
acting on a weld we can choose the plane that will have the least stress acting
upon it for to make our weld.
4.3
4.3.1
Derivation of the Equations
Stress Transformation Equations
First we will solve for the stresses on the inclined plane, σx0 , σy0 , τx0 y0 . By
applying our transformation matrix, which in this case is the rotation matrix R,
5
to the stress tensor matrix T we come up with the equations for the stresses on
a plane oriented at an angle Θ. σx , σy , and τxy are the stresses on the original
plane. σx and σy refer to the direct stresses in the x and y directions associated
with the object. τxy refers to the shear stress associated with the object.
cos(Θ) sin(Θ)
R=
(1)
− sin(Θ) cos(Θ)
−τxy sin(Θ) − σx sin(Θ)
T =
(2)
−σy sin(Θ) − τxy cos(Θ)
σx0
cos(Θ) sin(Θ) −τxy sin(Θ) − σx sin(Θ)
+
=0
(3)
τx0 y0
− sin(Θ) cos(Θ) −σy sin(Θ) − τxy cos(Θ)
By doing the matrix multiplication in equation (3) we obtain the following:
σx0
−2τxy sin(Θ) cos(Θ) − σx cos2 (Θ) − σy sin2 (Θ)
+
=0
τx0 y0
τxy sin2 (Θ) + σx sin(Θ) cos(Θ) − σy sin(Θ) cos(Θ) − τxy cos2 (Θ)
(4)
Now if we do some simplifying of equation (4) we get
σx0
2τxy sin(Θ) cos(Θ) + σx cos2 (Θ) + σy sin2 (Θ)
=
(5)
τx0 y0
(σy − σx ) sin(Θ) cos(Θ) + τxy (cos2 (Θ) − sin2 (Θ))
Given that the y axis is plus 90 degrees to the x axis, and we know the
equation for σx0 . We can use this knowledge to our advantage to solve σy0 with.
σy0 (Θ) = σx0 (Θ + 90)
(6)
This implies that
σy0 = 2τxy sin(Θ + 90) cos(Θ + 90) + σx cos2 (Θ + 90) + σy sin2 (Θ + 90)
(7)
When equation (7) is simplified it becomes
σy0 = −2τxy sin(Θ) cos(Θ) + σx sin2 (Θ) + σy cos2 (Θ)
Finally we have our three transformation equations for σ , σ , and τ
x0
y0
2
2
(8)
x0 y 0
:
σx0 = 2τxy sin(Θ) cos(Θ) + σx cos (Θ) + σy sin (Θ)
(9)
σy0 = −2τxy sin(Θ) cos(Θ) + σx sin2 (Θ) + σy cos2 (Θ)
(10)
2
2
τx0 y0 = (σy − σx ) sin(Θ) cos(Θ) + τxy (cos (Θ) − sin (Θ))
(11)
This can be expressed as a vector and a matrix
 
 
cos2 (Θ)
sin2 (Θ)
2 sin(Θ) cos(Θ)
σx0
σx
 σy 0  = 
sin2 (Θ)
cos2 (Θ)
−2 sin(Θ) cos(Θ)   σy  (12)
τx0 y0
τxy
− sin(Θ) cos(Θ) sin(Θ) cos(Θ) cos2 (Θ) − sin2 (Θ)

Here is an example of plane stresses when rotated at an angle of Θ.
6
4.3.2
Strain Transformation Equations
Now we will solve for the strains on the inclined plane, εx0 , εy0 and γx0 y0 . εx0 and
εy0 refer to the change of length in the x and y directions whereas γx0 y0 refers
to the change in the angle α. We find the strain transformation equations in
the exact same way that we found the stress transformation equations, actually,
because they are found in the exact same way we can just substitute εx for
σx , εy for σx , and finally γxy for τxy with the exception that because of the
when substituting in γxy we must multiply it by 1/2 because of the classical
definition of shear strain. After substituting these in and simplifying as we did
in equations (3) through (11) we attain equations (12) through (14).
εx0 = γxy sin(Θ) cos(Θ) + εx cos2 (Θ) + εy sin2 (Θ)
2
2
(13)
εy0 = −γxy sin(Θ) cos(Θ) + εx sin (Θ) + εy cos (Θ)
(14)
γx0 y0 = 2(εy − εx ) sin(Θ) cos(Θ) + γxy (cos2 (Θ) − sin2 (Θ))
(15)
Another way of obtaining strain equations is by transforming the stress equations into a matrix A and then multiplying it by the strain vector which we will
name ε.


cos2 (Θ)
sin2 (Θ)
2 sin(Θ) cos(Θ)
(16)
A=
sin2 (Θ)
cos2 (Θ)
−2 sin(Θ) cos(Θ) 
− sin(Θ) cos(Θ) sin(Θ) cos(Θ) cos2 (Θ) − sin2 (Θ)
 
εx
ε =  εy 
(17)
γxy
2
Given an initial strain, , listed as equation (17) we want to find the strain
if the system is turned some angle Θ. In order to do this we must create a
7
new equation which eliminates the γxy /2 and creates γx0 y0 . In order to do this
we first multiply the original transformation matrix A by B and B −1 to give
matrix C:




1 0 0
1 0 1
−1
B = 0 1 0
B = 0 1 0 
0 0 2
0 0 12
BAB −1 = C
cos2 (Θ)

C=
sin2 (Θ)
−2 sin(Θ) cos(Θ)

sin2 (Θ)
cos2 (Θ)
2 sin(Θ) cos(Θ)

sin(Θ) cos(Θ)
− sin(Θ) cos(Θ) 
cos2 (Θ) − sin2 (Θ)
(18)
Now if we multiply C by the strain vector ε we should obtain our resultant
strain vector ε0
ε0 = Cε
 
cos2 (Θ)
εx0
 εy 0  = 
sin2 (Θ)
γx0 y 0
−2 sin(Θ) cos(Θ)
sin2 (Θ)
cos2 (Θ)
2 sin(Θ) cos(Θ)
 
sin(Θ) cos(Θ)
εx
− sin(Θ) cos(Θ)   εy 
γxy
cos2 (Θ) − sin2 (Θ)
(19)
 


εx cos2 (Θ)
εy sin2 (Θ)
γxy sin(Θ) cos(Θ)
εx0
 εy 0  = 
εx sin2 (Θ)
εy cos2 (Θ)
−γxy sin(Θ) cos(Θ) 
γx0 y0
−2εx sin(Θ) cos(Θ) 2εy sin(Θ) cos(Θ) γxy cos2 (Θ) − sin2 (Θ)
(20)
Equation (20) takes an initial strain on an object and rotates it Θ degrees. It
then provides the distance that the object has been deformed in the x and y
directions given as x0 and y 0 . Although different variables are used the following
image shows how a plane on a surface can deform from certain stresses acting
upon it.

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4.4
A Worked Out Stress and Strain Example
For our example we will calculate the plane stresses involved where two pieces
of steel are welded together. Our sign convention defines forces causing the
object to be in tension positive and forces causing the object to be in compression
negative. As we can see from the picture our σx = −4.5[M P a] and σy =
10[M P a]. Steel has an average Young’s modulus of elasticity of E = 200[GP a]
and an average Poisson’s ratio of ν = .285 Because it is not defined we will
assume that initially there is no shear stress, τxy = 0.
First we must find the angle of the plane of which we will be calculating the
stresses. For that we just use basic trigonometry.
Θ = tan−1 (
.1
) = 21.8o
.25
Now we can substitute our values of σx , σy , τxy and Θ in to equation (12)
to find our values of σx0 , σy0 and τx0 y0 .
 
cos2 (21.8)
σx0
 σy0  = 
sin2 (21.8)
τx0 y0
− sin(21.8) cos(21.8)

sin2 (21.8)
cos2 (21.8)
sin(21.8) cos(21.8)


2 sin(21.8) cos(21.8)
−4.5
−2 sin(21.8) cos(21.8)   10 
0
cos2 (21.8) − sin2 (21.8)

 

σx0
−2.5
 σy0  =  8.0 
τx0 y0
5.0
This implies that the direct stress in the x direction along the weld is equal
to 2.5 [MPa] in compression, the direct stress in the y direction perpendicular
to the weld is equal to 8.0 [Mpa] in tension, and the shear stress on the weld is
5.0 [MPa]. Now that we have found the stresses acting on the weld it is time to
find what the strains upon the weld will be.
The initial strains are not given to us so we must use what we are given to
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calculate them. By using Hooke’s Law for plane stress we can find the εx and
εy . Hooke’s Law for plane stress uses Young’s of elasticity, Poisson’s ratio, and
the initial stresses to find the initial strains. Hooke’s Law for plane stress is as
follows:
Eεx + Eνεy = σx − ν 2 σx
Eεy + Eνεx = σy − ν 2 σy
Eγxy = τxy
Now our knowledge of Reduced Row-Echelon Form comes into play. As you
can see, what we have here is a system of three equations and three unknowns.
After we substitute in the known values we can make a matrix using the leading
coefficients.


200 × 109 57 × 109 0 −4.1345 × 106
 57 × 109 200 × 109 0 9.18775 × 106 
0
0
0
0
Now if we convert this matrix to RREF it will tell us the initial strains on
the element.

200 × 109
RREF  57 × 109
0
57 × 109
200 × 109
0
0
0
0
 
−4.1345 × 106
1 0 0
9.18775 × 106  = 0 1 0
0
0 0 0

−3.675 × 10−5
5.6413 × 10−5 
0
This refers to εx = −3.675 × 10−5 , εy = 5.6413 × 10−5 and γxy = 0. Now
we can substitute these values into equation (19).
 
cos2 (21.8)
εx0
 εy0  = 
sin2 (21.8)
0
0
γx y
−2 sin(21.8) cos(21.8)

sin2 (21.8)
cos2 (21.8)
2 sin(21.8) cos(21.8)


sin(21.8) cos(21.8)
−3.675 × 10−5
− sin(21.8) cos(21.8)   5.6413 × 10−5 
0
cos2 (21.8) − sin2 (21.8)

 
εx0
−2.3902 × 10−5
 εy0  =  4.3565 × 10−5 
6.4247 × 10−5
γx0 y 0

Now we know that with the stresses of 10 [MPa] tension in the y-direction
and 4.5 [MPa] compression in the x-direction we have a deformation of −2.3902×
−5 m
[ m ] perpendicular to the weld.
10−5 [ m
m ] along the weld and 4.3565 × 10
10
5
Conclusion
Linear algebra is not only useful to the specific engineering topic of mechanics
of materials but to all of engineering. One might even go as far as to say that
linear algebra is as important to an engineer as is calculus. We only showed
the importance of linear algebra in our derivation of the stress and strain transformation equations and the usage of RREF but linear algebra can be applied
to a number of other topics in engineering as well. Eigenvectors are extremely
important when studying trajectories of objects and as such can be quite useful
in the design of things from jet engines to satellites. The usage of ordinary
differential equations becomes a necessity in engineering very often and very extensively; equations such as the Bernoulli and Riccati equations, in which linear
algebra can come in very handy when solving, are essential for topics such as
fluid mechanics, thermodynamics and heat transfer. Finally, the Laplace Transform can be very advantageous for electrical engineers when designing digital
control systems. Although there are only a few uses listed here there are a
multitude of uses for linear algebra in engineering.
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This picture contains an example of an improperly designed structure. The wind
blowing across the Tacoma Narrows bridge caused it to oscillate at its natural
frequency and eventually collapse. The study of transformation equations can
both help with the study of natural frequencies and as we showed, the stresses
involved.
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References
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[2] Rohan Abeyaratne. ”Lecture Notes on The Mechanics of Elastic Solids”
Rohan Abeyaratne. 1987. Web. 15 March 2012
[3] Brooks/Cole. Typical Problem. 2001. Thomson Learning Inc. Web. 15
March 2012.
[4] Paul Dawkins. ”Linear Transformations”. Paul’s Online Math Notes. n.p.
n.d. 15 March 2012
[5] ”Strain (materials science)”. Absolute Astronomy. n.p. n.d. 15 March 2012
[6] ”Principal Stress Calculator”. efunda. n.p. n.d. 15 March 2012
[7] Brooks/Cole. ”Chapter 7: Analysis of Stress and Strain” Thomson Learning Inc. 2001. Web. 15 March 2012
[8] Dr. Mysore Narayanan. ”Importance of Linear algebra in Engineering Design Methodology” Miami University. n.d. 06 May 2012
[9] Sergio Barrachina, Peter Benner, Enrique S. Quintana-Orti. ”Efficient algorithms for generalized algebraic Bernoulli equations based on the matrix
sign function” Springer Science + Business Media, LLC . 1 December 2007.
06 May 2012
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