Solid Mechanics

Introduction
Contents
•
1 Preface
•
2 Introduction
o
2.1 Stress
o
2.2 Strain
o
2.3 Hooke's Law
o
2.4 Poisson's Ratio
o
2.5 Effect of Temperature
o
2.6 Energy Stored due to Deformation
o
2.7 Compatibility Conditions
o
2.8 Finite E
lement Method
o
2.9 Saint

Venant's Principle
o
2.10 Stress Concentration Factors
o
2.11 Elastic and Plastic Deformation
o
2.12 Shear Strain and Modulus
o
2.13 Generalized Hooke's Law
o
2.14 Relationships Between Moduli
o
2.15 Pressure Vessels
•
3 T
orsion
o
3.1 Torsion Formula
o
3.2 Angle of Twist
•
4 Loading of Beams
o
4.1 Examples
o
4.2 Exercises
o
4.3 Calculus for Solving Beam Problems
4.3.1 Examples
o
4.4 Calculus for Point Loads
4.4.1 Exercises
•
5 Stresses and Strains in Beams
o
5.1 Beams with Arbitrary Moments
o
5.2 Shear in Beams
•
6 The General State of Stress
o
6.1 Principal Stresses
o
6.2 Stress Invariants
o
6.3 Hydrostatic Stress
o
6.4 Deviatoric Stresses
o
6.5 Mohr's Circle
6.5.1 Mohr's Circle for Common Cases
o
6.6 Failure Criteria
6.6.1 Maximum Shear Stress Criterion
6.6.2 Maximum Distortion Energy Criterion
6.6.3 Failure of Materials
Preface
This book is a first course in the analysis of structures. Although most of the material should be accessible
t
o all students who have had a mechanics course, a previous exposure to
Engineering Mechanics
would
be useful. There are no mathematical prerequisites, though some
elementary calculus would be useful in
certain sections which can be skipped without affecting the flow of the book.
Introduction
Solid Mechanics is the study of load carrying members in terms of forces, deformations, and stability. In
this book, we take
the continuum mechanics approach, where we take the material properties to be the
same even when we consider infinitesimal areas and volumes. The alternative approach is to build up
material properties from basic equations relating atomic forces and intera
ctions, and extending it to larger
sets of such entities (
e.g.
, molecular dynamics). Other approaches include meso

scale theories like
dislocations to explain behavior of materials, especially for plasticity. However, the approach in this book
is applicabl
e to a large variety of practical problems, where we deal with structural members, usually
under elastic deformation.
The above images show a body which is acted upon by external forces
F
1
,
F
2
, etc. The supports provide
reaction forces (
R
1
,
R
2
) so that the body is stationary.
In addition to external f
orces, which act at points on the surface, we can also have forces which are body
forces,
i.e.
, they act on each point in the body. Gravity is an example of a body force.
Consider a body which has several external forces acting on it. If we take an imagina
ry section of the
body, each of the parts should also be in equilibrium. Now, consider an elemental area
dA
in that section.
Let the force acting on the elemental area so that the whole body is in equilibrium be
dF
. We can resolve
this force in three mutua
lly perpendicular directions. Let the component of the force normal to the surface
be
dF
z
. Let the components in the plane of
dA
be
dF
x
and
dF
y
respectively.
Stress
The stress is defined
on the average
as the force divided by the area of the body over whic
h the force acts.
More precisely, we can talk about a
stress at a point
, or simply a
stress
, in the limiting case where both
the area (an
infinitesimal
area
dA
) and the force (an
infinitesimal
force
dF
) go to zero. The normal stress
occurs due to the infin
itesimal force normal to the infinitesimal area, while shear stresses occur due to the
infinitesimal force in the plane of the infinitesimal area.
Thus,
τ
xx
= dF
x
/dA
is the normal stress, also denoted as
σ
x
. The shear stresses are given by
τ
xy
= dF
y
/dA
and
τ
xz
= dF
z
/dA
respectively. As can be seen, these shear stresses are in the plane
y

z
.
Now, instead of an infinitesimal
area
, consider an infinitesimal
v
olume
at the point in question. Let this
volume be a parallelepiped with the sides
dx
,
dy
, and
dz
. In this case there are in general nine non

zero
stresses. They form a
stress tensor
represented by a 3x3 matrix
Accordingly, in the
plane stress
case (taking, for instance, the plane
x

y
) we will have a four component
stress tensor. Further, in view of the shear stress symmetry (
τ
xy
= τ
yx
=
τ
), the above 3x3 matrix reduces to
the 2x2 symmetric matrix representing a symmetric plane stress tensor
The shear stress sym
metry holds in the three

dimensional case. This makes the above 3x3 matrix

and
hence the corresponding stress tensor

symmetric: only three of its six shear stress components are
independent ones. Consequently, the stress tensor has in general six differe
nt components: three normal
stresses and three shear stresses. As already shown above, in the two

dimensional case the stress tensor
has only three different components: two normal stresses and one shear stress.
Suppose the body is acted upon by the forces
f
x
and
f
y
(per unit volume) in the
x
and
y
directions
respectively. It then can be shown that the stress equilibrium equations in the Cartesian coordinates has
the form
Thus, in the plane stress case, we have two equations involving three unknowns.
Now turn to an example of the simplest one

dimensional lo
ading of a body. Consider a rod being pulled
at its ends along its axis.
If we take a section perpendicular to the axis, it is easy to see that the stress
σ
x
= P/A
where
P
is the load
and
A
is the cross sectional area of the section.
Now consider the case where t
he section makes an angle
θ
with the axis. In this case, we can resolve the
force
P
along and perpendicular to the section, and the components are
P cosθ
and
P sinθ
respectively.
The area of the section is now
A / cosθ
. Thus, the normal stress is
σ
θ
= P cosθ/(A/ cosθ) = (P/A) cos
2
θ
Similarly, the shear stress is
τ
θ
= (P/A) sinθ cosθ
.
Strain
The strain in a material is defined as the fractional change in length, and is a dimensionless quantity. It is
denoted by the symbol
ε
, where
ε = (L − L
0
)/L
0
. Here
L
0
is called the
gage length
.
T
he strain defined above is called the
engineering strain
, and is different from
natural strain
, which is
defined as
Note that the engineer
ing strain and natural strain are the same for small values of
ε
, and for the most part
for the kinds of loads and displacements in this book, the differences are not relevant.
In the infinitesimal case, we have, strain
ε
x
= du/dx
, where
du
is the change in length for the segment
dx
.
Thus, the total change in length i
s given by
Δ = ∫
0
L
du = ∫
0
L
ε
x
dx
.
Hooke's Law
Hooke's law
states that the stress is linearly related to strain for some materials. This is an empirical law
by Robert Hooke, who observed this behavior in springs. Thus, we have,
σ =
E
ε
where
E
is the consta
nt of proportionality called
Young's modulus
. Note that we have considered this
value of
E
to be the same in all directions. Materials with whose properties don't have a directional
variation are called
isotropic
materials. Materials which have different p
roperties in different directions
are called
anisotropic.
The most common cause of anisotropy is the crystalline nature of materials.
However, most common structural materials do not have the same crystal orientation over large ranges.
Another cause for an
isotropy is the kind of processing done on a material. Some processes like drawing
tend to create stresses in a particular direction.
Applying Hooke's law to the definition of length change, we have,
Or,
From mechanics we know that a spring has linear variation of extension w
ith force, the constant of
proportionality usually denoted by
k
. Thus, the spring constant for a beam under axial load is
AE/L
. This
can be extended to components of different shapes, so that a structure is an assembly of springs arranged
in a complicated
manner.
Poisson's Ratio
The longitudinal strain is linearly related to lateral strain in many materials. The ratio of the lateral to the
longitudinal strain is called
Poisson's ratio
and is denoted by
ν
.
The value of Poisson's ratio varies from material to material, and is about 0.3 for most common metals.
The maximum possible value for
Poisson's ratio is 0.5 and occurs for constant volume cases like rubber.
Also, note that constant volume occurs in plastic deformation.
Effect of Temperature
Materials change in length due to temperature change. The amount of change is quantified by
α
, the
coefficient of linear thermal expansion. By definition of strain, we have,
ε = α ΔT
, where
ΔT
is the change
in temperature. Thus, if we have a structural member between rigid supports, a strain of
ΔT
develops in it.
This is accompanied by a stress, which
is called
thermal stress
...
Energy Stored due to Deformation
When a rod is compressed, the work done on it is stored as energy. In the above figure, the energy stored
is shown as the shaded area. Now, we know that the energy stored in a spring is
1/2 k x
2
, for a spring
constant of
k
and extension
x
. For a rod, whose constant is
AE/L
and extension is
PL/AE
, the energy
stored is
1/2 P
2
L / AE
. From the above equation, we see that it is better to use long bolts rather than short
ones as they will have lower peak load
P
for the same dia
meter and material, without doing any
dynamical analysis.
Compatibility Conditions
In an Engineering Mechanics course, we use force equilibrium to solve for forces in individual members.
However, there were cases where it was not possible and such problems
were considered
statically
indeterminate
problems. They require some constitutive equations to solve completely, and Hooke's law
is such an equation.
For instance, consider a column fixed at both ends, and a load
P
applied at the center. Now, we know that
reaction forces
R
1
and
R
2
act
at either ends. Further, using force equilibrium, we have
R
1
+ R
2
= P
. Here
we have two unknowns,
R
1
and
R
2
, but only one equation. We can apply Hooke's law, with displacements
Δ
1
and
Δ
2
, where
Δ
1
= R
1
L/2AE
and
Δ
2
= R
2
L/2AE
. The second equation is
Δ
1
= Δ
2
,
since the supports are
fixed. Thus, we can solve the problem.
Finite Element Method
Consider a rod with different cross sections
A
i
and lengths
L
i
, so that each section has an equivalent spring
constant of
A
i
E/L
i
. If the forces at each node (where the are
a changes) are
F
i
, then we can write equations
in terms of displacements of each node and the spring constant of each section for the force
F
i
. If the
beam is fixed at both ends, then the forces can only act on the other nodes.
In the general case, if you
have loads being applied to connected thin rods, you can use this method to
solve for forces and deformations. The whole structure can be
discretized
, and the nodes numbered so that
we get the matrix
[k
ij
]
.
This procedure of breaking down a complicated pro
blem into a large number of simple ones is called the
Finite Element Method
.
The individual pieces are called elements, and in the above case we hav
e used the line element. For a
general
2

D
problem, we would need to use triangular elements at least. The matrix for a typical problem
will contain hundreds of such elements and the size of the matrix be same as the number of elements.
However, as most of
the members are zero, the matrix is called a
sparse matrix
, and it can be solved
much faster than the typical matrix you would encounter, say, in a mathematics course.
Saint

Venant's Principle
So far we have been dealing with point loads acting at the ends of beams. We have a
ssumed that the stress
is the same as
σ
av
= P/A
. However, this is not true close to the point of application of the load.
Saint

Venant's principle
(by the French elastician Saint

Venant) states that the stress is the same as average
stress at a distance
a
along the axis of the beam, where
a
is
the width of the beam.
In the figure shown, for instance, we would be concerned about stress variations in the section
A

A
, but
not in the section
B

B
. Note that this is applicable for elastic deformations, and near the point load, either
the idealization
as a point load fails, or the material flows.
Stress Concentration Factors
So far we have assumed that the stress at all points in a cross section is the same, and by and large that is
correct. However, there is an increase in stress near the point loads
and discontinuities. The peak loads are
a function of the average stress calculated by methods stated previously. The correct peak loads can only
be obtained by methods like finite element analysis. However, charts have been published for several
typical s
cenarios. As mentioned earlier, the peak stresses
σ
p
is a function of the average stress
σ
av
. Now,
we have
σ
p
= K×σ
av
, where
K
is called the
stress concentration factor
and can be looked up from tables
for common scenarios. The study of stress concentration factors is also a design guideline, and indicate
s
the types of features that you should avoid in a real world situation. For instance, the stress concentration
factor for a fillet radius increases rapidly beyond
r < 0.1 d
, so you should specify large fillet radii for your
shafts or eliminate the change
in diameter.
The above schematic shows the possible values of the stress concentration ratios
K
for a plate with a hole
in it. The stress at the narrowest section is calculated based on
d = D − 2r
, where
r
is the radius of the
hole and
D
is the width of the slab.
As can be seen, the stress concentration is not significant till about
r/d
= 0.3
.
Note that local stresses might not be as high as that implied by the stress concentration factors due to
plastic flow of the material, since the stress concentration analysi
s is done for elastic flow.
Elastic and Plastic Deformation
So far, we have considered members which deform elastically when a load is applied. Some of them obey
Hooke's law, so that the relationship between stress and strain is linear. Elastic deformation
refers to the
ability of the material to regain its original shape after the external load is removed. However, we know
that for large loads, the material deformation is permanent, and this is called plastic deformation. Metals
can undergo last plastic de
formation, and metals with such a property are called
ductile
. Materials which
cannot undergo plastic deformation are called
brittle
.
Materials for which the elastic region is very small are called
plastic
materials. The above image shows
the idealization for such a materia
l. The material
flows
when a certain stress state is reached.
Materials which are elastic for large deformations are called
elastic
materials. The above image shows an
ideal elastic material

in this case, it also obey's Hooke's law, and is called a
linearly elastic
material.
Note that a
substance can be elastic and still not obey Hooke's law (
non

linearly elastic
).
Most metals undergo elastic deformation for small loads and deform plastically for larger loads. There is
certain amount of increase in stress when the material deforms plastically for
most metals. However, the
elastic plastic approximation shown above is good enough simple analysis.
Note that material which exhibit such behavior will show what is known as
elastic recovery
, where the
sample will show some tendency to the earlier state. In t
he above image, the amount of recovery is
Δ
after
the load has been removed. Note that there is a residual deformation in the material in the unloaded state.
Some materials show increased stress during plastic flow, with a phenomenon called
strain hardening
. In
this case they can be approxima
ted with a stress strain curve with two slopes, but this kind of analysis is
rarely used in practice as most structural deformation is in the elastic zone, and plastic flow will most
likely be considered a failure mode.
Shear Strain and Modulus
Earlier we introduced stress

strain relationsh
ip for materials under an axial load, where we found that the
stress
σ
was proportional to the strain
ε
. If we plot the shear stress
τ
against the shear strain
γ
, we get a
straight line for material which obey Hooke's law. Thus, Hooke's law can be extended to shear stress
using the following equation:
τ = Gγ
, where
G
is the
shear modulus, and
γ
is the shear strain.
Generalized Hooke's Law
Poisson's ratio shows the amount of change in dimension that occurs in another direction when a certain
change occurs in one direction. The strain in each direction is the superposition of
all the strains

the
direct strains and the ones due to Poisson's ratio. We can therefore state the
generalized
Hooke's law as
follows:
ε
x
=
σ
x
/E −
νσ
y
/E −
νσ
z
/E
ε
y
= −
νσ
x
/E +
σ
y
/E −
νσ
z
/E
ε
z
= −
νσ
x
/E −
νσ
y
/E +
σ
z
/E
γ
xy
=
τ
xy
/G
γ
yz
=
τ
yz
/G
γ
zx
= τ
zx
/G
Note
that these equations apply only to
isotropic
materials (steel, aluminum, copper, etc.) whose elastic
properties are identical in all directions.
Relationships Between Moduli
Consider an element which has stresses
σ
1
and
σ
2
acting on it along the
x
and
y
d
irections respectively.
Now consider further that
σ
1
= σ
and
σ
2
= −σ
.
The shear modulus can be expressed in terms of the Young's modulus:
Consider a body undergoing compression, so that the strains are
ε
x
,
ε
y
, and
ε
z
respectively. The fractional
change in volume is thus
ε
x
+ ε
y
+ ε
z
. Applying the generalized version of Hooke's law, we have, the bulk
modulus
k
is given by
Pressure Vessels
Pressure vessels are thin walled vessels containing (usually) gas or steam at high pressure. The design
decision is to find the value for the
thickness
t
of the pressure vessel for a given maximum pressure
p
.
Since the thickness is much smaller than the radius of the vessel
r
, the inner and outer radius are taken to
be equal. For the longitudinal direction, we have the force due to pressure
πr
2
p
. This is balanced by the
stress developed in the walls, acting on the circumference of the circle
2πrtσ
. Thus the stress in the
longitudinal direction is
pr/2t
. For the tangential direction, the pressure force is
π2rl
and the stress force is
2tlσ
. The
stress is
pr/t
.
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