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Timestamp: 2019-09-22 02:29:18+00:00

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THIRD EDmON
1ifW'AfIl16,
TEXTB@K
McGraw-Hili Book Company
I ST. LOUIS I SAN FRANCISCO I AUCKLAND I
COLORADO SPRINGS I HAMBURG I USBON I LONDON
MILAN , MONTREAL I NEW DELHI I OKLAHOMA CITY
I sAo PAULO I SINGAPORE I SYDNEY
Almost all engineering materials possess to a certain extent the property
'of elasticity. If the external forces producing deformation do not exceed
a certain limit, the deformation disappears with the removal of the forces.
Throughout this book it will be assumed that the bodies undergoing the
action of external forces are perfectly elastic, i.e., that they resume their
initial form completely after removal of the forces.
Atomic structure will not be considered here. It will be assumed that
the matter of an elastic body is homogeneous and continuously distributed
over its volume so that the sms.llest element cut from the body possesses
the same specific physical properties as the body. To simplify the discussion it will also be assumed that for the most part the body is i8otropic,
Le., that the elastic properties are the same in all directions.
Structural materials do not satisfy the above assumptions completely.
Such an important material as steel, for instance, when studied wit~ a
microscope, is seen to consist of crystals of various kinds and various
orientations. The material is very far from being homogeneous, but
experienge shows that solutions of the theory of elasticity based on the
assumptions of homogeneity and isotropy can be applied to steel structures with very great accuracy. The explanation of this is that the crystals are very small; usually there are millions of them in one cubic inch of
steel. While the elastic properties of a single crystal may be very different in different directions, the crystals are ordinarily distributed at random and the elastic properties of larger pieces of metal represent averages of properties of the crystals. So long as the geometrical dimensions
defining the form of a body are very large in comparison with the dimen1
sions of a single crystal the assumption of homogeneity can be used with
great accuracy, and if the crystals are orientated at random the material
can be treated as isotropic.
When, due to certain technological processes such as rolling, a certain
orientation of the 'crystals in a metal prevails, the elastic properties of the
metal become different in different directions and the condition of anisotropy must be considered. We have such a condition, for instance, in
cold-rolled copper.
Figure 1 indicates a body in equilibrium. Under the action of external
forces PI, . . . , P 7 , internal forces will be produced between the parts
of the body. To study the magnitude of these forces at any point 0,
let us imagine the body divided into two parts A and B by a cross section mm through this point. Considering one of these parts, for instance,
A, it can be stated that it is in equilibrium under the action of external
forces PI, ... , P 7 and the internal forces distributed over the cross
section mm and representing the actions of the material of the part B
on the material of the part A. It will be assumed tha~ these forces are
continuously distributed over the area mm in the same way that hydrostatic pressure or wind pressure is continuously distributed over the surface on which it acts. The magnitudes of such forces are usually defined
by their intensity, i.e., by the amount of force per unit area of the surface
on which they act. In discussing internal forces this .intensity is called
In the simplest case of a prismatical bar submitted to tension by forces
uniformly distributed over the ends (Fig. 2), the internal forces are also
.02-1---1-
uniformly distributed over any cross section mm. Hence the intensity
~f this distribution, i.e., the stress, can be obtained by dividing the total
tensile force P by the cross-sectional area A.
In the case just considered. the stress was uniformly distributed over
- the cross section. In the general case of Fig. 1 the stress is not uniformly distributed over mm. To obtain the magnitude of stress acting
on a small area &A, cut out from the cross section mm at any point 0,
we observe that the forces acting across this elemental area, due to the
action of material of the part B- on the material of the part A, can be
reduced to a resultant &P. If we now continuously contract the elemental area &A, the lilnitipg value of the ratio &P/ &A gives us the magnitude of the stress acting on the cross section mm at the point O. The
limiting direction of the resultant &P is the direction of the stress. In
the general case the direction of stress is inclined to the area &A on which
it acts 'and we can resolve it into two ~omponents: a normal stress perpendicular to the area and a shearing stress acting in the plane of the area &A.
Notation for Forces and Stresses
There are two kinds of external forces which may act on bodies. Forces
distributed over the surface of the body, such as the pressure of one body
on another or hydrostatic pressure, are called surface forces. Forces distributed over the volume of a body, such as gravitational forces, magnetic forces, or in the c,ase of a body in motion, inertia forces, are called
body forces. The surface force per unit area we shall resolve into three
components parallel to carteSian coordinate axes x, y, z, and use for these,
components the notation X, Y, Z. We shall also resolve the body force
per unit volume into three components and denote these components by
X, Y,Z.
We shall use the letter (J' for normal stress and the letter., for shearing
stress. To indicate the direction of the plane on which the stress is act'ing, subscripts to these letters are used. We take a very small cubic ele-
4 I THEORY OF ELASTICITY
ment at a point P (Fig. 1), with sides parallel to the coordinate axes.
The notations for the co~ponents of stress acting on the sides of this
element and the directions taken as positive are as indicated in Fig. 3.
For the sides of the element perpendicular to the y axis, for instance, the
normal components of stress acting on these sides are denoted by (T".
The subscript y indicates that the stress is acting on a plane normal to
the y axis. The norm~ stress is taken positive when it produces tension
and negativ:e when it produces compression.
The shearing stress is resolved into two components parallel to the
coordinate axes. Two subscript ~etters are used in this case, the first
indicating the direction of the normal to the plane under consideration
and the second indicating the direction of the component of the stress.
For instance, if we again consider. the sides perpendicular ~ the y axis,
the component in the x direction is denoted by 1',,: and that in the z direction by 1'".. The positive directions of the components of shearing stress
on any side of the cubic element are taken as the positive directions of
the coordinate axes if a tensile stress on the same side would have the
positive direction of the corresponding axis. If the tensile stress has a
direction opposite to the positive axis, the positive directions of the
shearing-stress components should be reversed. Following this rule, the
positive directions of all the components of stress acting on the right side
of the cubic element (Fig. 3) coincide with the positive directions of the
coordinate axes. The positive directions are all reversed if we are considering the left side of this element.
From the discussion of the previous article, we see that for each pair of
parallel sides of a cubic element, such as in Fig. 3, one symbol is needed
o~-------------y
to denote the normal component of s~ress and two more symbols to
denote the two components of shearing stress. To describe the stresses
acting on the six sides of the element three symbols U:, U,n u. are necessary for normal stresses; and six symbols "Z1/, ",,:, "u, "u, ""., "'''' for
shearing stresses. By a simple consideration of the equilibrium of the
element the number of symbols for shearing stresses can be reduced to
If we take the moments of the forces acting on the element about a line
through the midpoint 0 and parallel to the x axis, for instance, only the
surface stresses shown in Fig. 4 need be considered. Body forces, such
as the weight of the element, can be neglected in this instance because in
reducing the dimensions of the element the body forces acting on it
djmjnish as the cube of the linear dimensions, whereas the surface forces
diminish as the square of the linear dimensions. Hence, for a very small
element; body forces are small quantities of higher order than surface
forces and can be omitted in calculating the moments. Similarly,
moments due to nonuniformity of distribution of normal forces are of
higher order than those due to the shearing forces and vanish in the limit.
Also the forces on each side can be considered to be the area of the side
times the stress at the middle. Denoting the dimensions of the small
element in Fig. 4 by dx, dy, dz, the equation of equilibrium of this element, taking moments of forces about 0, is then
"." dx dy dz = "". dx dy dz
The two other equations can be obtained in the same manner.
these equations we find
Hence for two perpendicular sides of a cubic element the components of
6 I THEORY OF ELASTICITY
shearing stress perpendicular to the line of intersection of these sides are
equal. 1
The six quantities u z , u,,' u., 1'Z1/ = 1'p, 1':. = 1'", 1'"a = 1'a1/ are therefore
sufficient to describe the stresses acting on the coordinate planes through
a point; these will be called the component8 of 8tre88 at the point.
It will be shown later (Art. 74) that with these six components the
~tress on any inclined plane through the same point can be determined.
Components of Strain-
In discussing the deformation of an elastic body it will be assumed that
there are enough constraints to prevent the body from moving as a rigid
body so that no displacements of particles of the body are possible without a deformation of it.
In this book, only small deformations such as commonly occur in engineering structures will be considered. The small displacements of particles of a deformed body will first be resolved into components u, v, w
parallel to the coordinate axes x, y, z, respectively. It will be assumed
that these components are very small quantities varying continuously
over the volume of the body. Consider a small element dx dy dz of an
elastic body (Fig. 5). If the body undergoes a deformation and u, v, w
are the components of the displacement of the point P, the displacement
in the x direction of an adjacent point A on the x axis is, to the first
order in dx,
due to the increase (au/ax)-d,x of the function u with increase of the
coordinate x. The increase in length of the element P A due to deformation is therefore (au/ax) dx. Hence the unit elongation at point P in
the x direction is au/ax. In the same manner it can be shown that the
1 There are exceptions, especially when stress is induced by electric and magnetic
fields (see Prob. 2, p. 14).
a,~---------------
v+ dUb
u--L~-.L----~q;c
-r-----J.A'
-.l..JB'
\.U+~d:Y
unit elongations in the y and z directions are given by the derivatives
av/ ay and aw/ az.
Let us consider now the distortion of the angle between the elements
P A and P B, Fig. 6. If u and v are the displacements of the point P in
the x and y directions, the displacement of the point A in the y direction and of the point B in the x direction are v + (av/ax) dx and u +
(au/ay) dy, respectively. Owing to these displacements the new direction pI A I of the element P A is inclined to the initial direction by the
small angle indicated in the figure, equal to av/ax. In the same manner
the direction P'B' is inclined to PB by the small angle au/aYe From
. this it will be seen that the initially right angle APB between the two
elements PA and PB is diminished by the angle av/ax + au/ay. This is
the 8hearing 8train between the .planes xz and yz. The shearing strains
between the planes xy and xz and the planes yx and yz can be obtained
We shall use the letter E for' unit elongation and the letter 'Y for unit
shearing strain. To indicate the directions of strain we shall use the
same subscripts to these letters as for the stress components. Then from
'Ya:J/
'Yu = az
Eu=-
'YU&
= -az
It will be shown later that, having the three unit elongations in three
perpendicular directions and three unit shear strains related to the same
directions, the elongation in any direction and the distortion of the angle
I THEORY OF ELASTICITY
between any two directions can be calculated (see Art. 81). The si
quantities Ez , , 'Yfl. are called the components oj strain.
Linear relations between the components of stress and the componen'
of strain are known generally as Hooke's law. Imagine an element:
rectangular parallelepiped with the sides parallel to the coordinate ax.
and submitted to the action .of normal stress (l'z uniformly distributE
over two opposite sides, as in the tensile test. The unit elongation
the element up to the proportional limit is'given by
in which E is the modulus oj elasticity in tension. Materials used i
engineering structures have moduli which are very large in comparisc
with allowable stresses, and the unit elongation (a) is a yery small qua]
tity. In the case of structural steel, for instance, it is usually sman.
than 0.00l.
This extension of the element in the x direction is accompanied t
lateral strain components (contractions)
in which v is a constant called Poisson's ratio. For many materia
Poisson's ratio can be taken equal to 0.25. For structural steel it
usually taken equal to 0.3.
Equations (a) and (b) can be also used for simple compression. TI
modulus of elasticity and Poisson's ratio in compression are the san
as in tension.
If the above element is submitted simultaneously to the action of no
mal stresses (l'z, CT", CT., uniformly distributed over the sides, the resultaJ
components of strain can be obtained from Eqs. (a) and (b). If v
superpose the strain components produced by each of the three stresSE
Ez = }Ji [CT: v(I'" + .(1'.)]
E1 [(I'll -
E1 [(I'.
v(I':
+ CT.)]
- v(I'z
+ (1',,)]
which have been found consistent with very numerous test measuremen1
In our further discussion we shall often use this method of superpoBitu
in calculating total deformations and stresses produced by several forces.
I t is legitimate as long as the deformations are small and the corresponding small displacements do not affect substantially the action of the
external forces. In such cases we neglect small changes in dimensions
of deformed bodies and also small displacements of points of application
of external forces and base our calculations on initial dimensions and
initial shape of the body. The resultant displacements will then be
obtained by superposition in the form of linear functions of external
forces, as in deriving Eqs. (3). .
There are, however, exceptional cases in which small deformations
cannot be neglected but must be taken into consideration. As an example of this kind, the case of the simultaneous action on a thin bar of axial
and lateral forces may be mentioned. Axial forces alone produce siinple
tension or compression, but they may have a substantial effect on the
bending of the bar if they are acting simultaneously with lateral forces.
In calculating the deformation of bars under such conditions, the effect
of the deflection on the moment of the external forces must be considered, even though the deflections are very small. 1 Then the total
deflection is no longer a linear function of the forces and cannot be
obtained by simple superposition.
In Eqs. (3), the relations between elongations and stresses are completely defined by two physical cOD,stants E and JI. The same constants
can also be used to define the relation between shearing strain and
Let us consider the particular case of deformation of the rectangular
parallelepiped in which (1. = (1, (11/ = - ( 1 , and (1:c = O. Cutting out an
element abed by planes parallel to the x axis anti at 45 to the y and z axes
(Fig. 7), it may be seen from Fig. 7b, by summing up the forces along and
perpendicular to be, that the normal stress on the sides of this element is
1 Several examples of this kind can be found in S. Timoshenko, "Strength of Materials," 3d ed., vol. 2, chap. 2, D. Van Nostrand Company, Inc., Princeton, N.J., 1956.
10 'I THEORY OF ELASTICITY
zero and the shearing stress on the sides is
.,. = ~'2(0': - 0',,) = 0'
Such a condition of stress is called pure shear. The elongation of thE
vertical element Ob is equal to the shortening of the horizontal element:
Oa and Oe, and neglecting a small quantity of the second order we con
clude that the lengths ab and be of the element do not change durinl
deformation. The angle between the sides ab and be changes, and th(
corresponding magnitude of shearing strain l' may be found from th(
triangle Obe. Mter deformation, we have
= tan (! _
1) = 1 + E"
1 + E:
Substituting, from Eqs. (3),
E (0'~ -
E,,= -
= (1 +E v)O'
vO',,)
+E v)O'
and noting that for small 'Y
tan (! _
tani - tan
= _1_-_~
1 + tan! tan 1
2(1 + v)O' . 2(1 + v).,.
1'= , E
Thus the relation between shearing strain and shearing stress is defined
by the constants E and v. Often the notation
G = 2(1
'Y=a
The constant G, defined by Eq. (5), is called the modulus of-elasticity in
shear, or the modulus of rigidity.
If shearing stresses act on all the sides of an element, as shown in
Fig. 3, the distortion of the angle between any two intersecting sides
depends only on the corresponding shearing-stress component. We have
The elongations (3) and the distortions (6) are independent of each other.
The general case of strain, produced by three normal and three shearing,
components of stress, can be' obtained by superposition: on the three
elongations given by Eqs. (3) are superposed three shearing strains given
by Eqs. (6).
Equations (3) and (6) give the components of strain as functions of the
components of stress. Sometimes the components of stress expressed as
functions of the components of strain are needed. Th'ese can be obtained
as follows. Adding Eqs. (3) together and using the notations
= Ez + Ell + E,l
we obtain the following relation between the volume expansion e and the
sum of normal stresses:
1 -- 2v
e=--e
In the case of a uniform hydrostatic pressure of the amount p we have
and Eq. (8) gives
3(1 -- 2v)p
which represents the relation between unit volume expansion e and
hydrostatic pressure p.
The quantity E13(1 -- 2v) is called the modulu8 of volume expansion.
Using notations (7) and solving Eqs. (3) for (Tz, (Til' (T., we find
+ v)(l
+ vH1 -- 2v) e + 1 +
v)(l __ 2v) e + 1 +
V E,l
__ 2r) e
+ v)(l -- 2J1)
and Eq. (5), these become
+ 2GEz
= ~e + 2GEJI
= ~e + 2GEIl
The notation already introduced for components of force, stress, displacement, and
strain is one that has become well established in many countries, particularly for
engineering purposes. It will be used throughout this book. For the concise representation of general equations and the theorems derived from them, however, the
alternative ~ndex notation is advantageous and is often encountered. The displacement components for instance are written 'Ul, 'Us, 'Ua, or collectively as 'Ui, with the
understanding that the index i can be 1, 2, or 3. The coordinates themselves are
written Xl, XI, X" or simply Xi, instead of X, y, z.
In Fig. 3 nine stress components appear. They can be arranged as in the table or
array on the left below.
"'11:"
'Tli:S
'T;I&;I&
'T;I&"
"',,11:
(1'"
'TI/II
'T"II:
'TJ/"
'TJ/'
(1',
'T,z
'TIIJ/
Writing 1';1&;1& instead of (1';1&, "'11" for (1'", and 'Tn for (I'll, we have the middle array above.
Here the first suffix indicates the direction of the normal to the side of the element on
which the component acts and the second suffix indicates the axis to which the stress
component arrow is parallel. In the array on the right, above, the suffi.xes are ohanged
to the corresponding numerical indices. To write the nine components collectively we
need now two indices i and i, each being 1, 2, 3 independently. Then all nine components are comprised in
with i, i = 1, 2, or 3
The relations (1) which reduced the nine components to six distinct numbers (but we
still have nine entries in the array), can now be expressed as
= Tii
If we permit i = i we have merely three identities such as 'Til = 'TU.
In place of the strain-displacement relations (2) we can take nine strain components
Eii (with Eii = Eii, as the definition of shearing strain requires) according to the
EOO="
= i = 1 this reproduces the first of (2) in the form of the first of the three
= 1, i = 2 we get from
(d) the first of the three relations
- l+aul)
l au.)
-+axl ax.
We observe that 2En,. 2Eu, 2f11 are the same as "},;I&", "}'ZII,."}'"II in (2). Thus EU is half the
reduction of the original right angle between line elements dxl , dxl , at X., XI, Xa.
To express the sum of the three terms appearing in the first of (~) we may write
i= 1.2.3
But in this notation it is customary to suppress the summation sym bol, and write simply
The summation is implied by the repeated index.
This is known as the 8'Umma-
tion convention. Thus, in the stress components,
Use of j (or any other literal index we may introduce) instead of i does not change the
meaning. For this reason such a repeated index is often referred to as a "dummy"
The six stress components are expressed in terms of the six strain components by
(11) together with (6). To put these together under the index notation we require
This is written ~ii' Evidently this symbol means zero when i ~ i, and it means unity
when i = j = 1 or 2 or 3. It is referred to as the "Kronecker delta." The six
relations obtainable from
= 1 or 2 or 3
reproduce the six relations (11) with (6). The symbol EU means, of course,. a sum like
in (h). But the reader will see here the necessity of using a dummy index k distinct
from i and j. For instance, to reproduce the first of (11) we take i = 1, j = 1, and
find from (j)
= ~8I1EU + 2GEll
= ~EU + 2GEll
and EU means the same thing as e, by (7).
Differentiation with respect to coordinates, as for instance in (d), is often expressed
more concisely by the use of commas. Thus (d) may be written as
Writing 3T for the sum in (h), T is the mean of the three normal stress components.
The stress Til can be regarded as a superposition of the two stress states
The first, often called simply the mean 8tre8S,. can be represented by T8ij. The second,
called the deuiatoric stress, or stress deuiator, can be represented by Td where
Similarly we can separate the strain Eij into a mean strain Eid3 or e/3, and a deuiatoric
8train Ei/ where
Ei/ = Eii
3-~dii
The six equations expressing Hooke's law are equivalent to
with 3-r
= (3~ + 2G)e
It is a simple exercise to deduce these from Eqs. (jj, or conversely to begin with (p)
and recover (J).
If T = -p, P
> 0, it is a hydrostatic pressure p.
The form (p) is particularly convenient as an ingredient of the theory of plasticit~
or the theory of viscoelasticity. The constant 3X + 2G is often written 3K. Then l!
is the modulus of fJolume ezpansion, already introduced on p. 11.
1. Show that Eqs. (1) continue to hold if the element of Fig. 4 is in motion and ha
an angular acceleration like a rigid body.
2. Suppose an elastic material contains a large number of evenly distributed sma
magnetized particles, so that a magnetic field exerts on any element clx dy clz
moment p clx dy dz about an axis parallel to the z axis. What modification wi
be needed in Eqs. (I)?
3. Give some reasons why the formulas (2) will be valid for small strains only.
4. An elastic layer is sandwiched between two perfectly rigid plates, to which it j
bonded. The layer is compressed between the plates, the direct stress being (I'
Supposing that the attachment to the plates prevents lateral strain Ea, Ell con:
pletely, fi~d the apparent Young's modulus (that is, (I'./E.) in terms of E and I
Show that it is many times E if the material of the layer has a Poisson's ratio oni
slightly less than 0.5, e.g., rubber.
5. Prove that Eq. (8) follows from Eqs. (11), (10), and (5).
If a thin plate is loaded by forces applied at the boundary, parallel to
the plane of the plate and distributed uniformly over the thickness (Fig.
8), the stress components u., 'Tz., 'T". are zero on both faces of the plate,
and it may be assumed, tentatively, that they are zero also within the
plate. The state of stress is then specified by U z , UrI, 'Tzu only, and is called
plane stress. It may also be assumed tentatively! that these three components are independent of z, i.e., they do not vary through the thickness. They are then functions of x and yonly.
A similar simplification is possible at the other extreme when the dimension of the body in the z direction is very large. If a long cylindrical or
prismatical body is loaded by forces that are perpendicular to the longitudinal elements and do not vary along the length, it may be assumed
that all cross sections are in the same condition. It is simplest to suppose at first that the end sections. are confined between fixed smooth
rigid planes, so that displacement in the axial direction is prevented.
The effect of removing these will be examined later. Since there is no
axial displacement at the ends and, by symmetry, at the midsection, it
may be assumed that the same holds at every cross section.
There are many important problems of this kind, for instance, a retain1 The assumptions made here are examined critically in Art. 98.
Variation of stress
does occur, but in a sufficiently thin plate it can be ignored, like the meniscus on the
column of fluid in the capillary tube of a thermometer.
~---+-.%
ing wall with lateral pressure (Fig. 9), a culvert or tunnel (Fig. 10), a
cylindrical tube with internal pressure, a cylindrical roller compressed by
forces in a diametral plane as in a roller bearing (Fig. 11). In each case,
of course, the loading must not vary along the length. Since conditions
are the same at all cross sections, it is sufficient to consider only a slice
between two sections unit distance apart. The components u and v of
the displacement are functions of x and y but are independent of the
longitudinal coordinate z. Since the longitudinal displacement w is zero,
Eqs. (2) give
"17/'
= -az + -ay = 0
az + ax =
E,=-=O
"1= =
7P"YJIJYJ"C;"CCWXCY:,J\VJA;";XCO"C'""W",P;.,,
The longitudinal normal stress 0". can be found in terms of
by means of Hooke's law, Eqs. (3). Since E. = 0 we find
+ 0"1/) = 0
v(O"s + 0",,)
0"1/
v(O"s
These normal stresses act over the cross sections, including the ends,
where they represent forces required to maintain the plane strain and
provided by the fixed smooth rigid planes.
By Eqs. (a) and (6), the stress components Tu and ",,8 are zero, and, by
Eq. (b), 0". can be found from O"~ and 0"1/. Thus the plane strain problem,
like the plane stress problem, reduces to the determination of O"s, 0"1/' and
"SrI as functions of x and y only.
Knowing the stress components O"s, 0"1/' T z" at any point of a plate in a
condition of plane stress or plane strain, the stress acting on any plane
through this point perpendicular to the plate and inclined to the x and
y axes can be calculated from the equations of statics. Let P be a point
of the stressed plate and suppose the stress components o"z, 0"1/' "SrI are
~------------~Z
~----------------~X
known (Fig. 12). We take a plane Be parallel to the z axis, at a smaI
distance from P, so that this plane together with the coordinate planef
cu~ out from the plate a very small triangular prism PBO. Since thE
stresses vary continuously over the volume of the body, the stress actin@
on the plane BO will approach the stress on the parallel plane through F
as the element is made smaller.
In discussing the conditions of equilibrium of the small triangulal
prism, the body force can be neglected as a small quantity of a highel
order. Likewise, if the element is very small, we can neglect the variation of the stresses over the sides and assume that the stresses are uniformly distributed. The forces acting on the triangular prism can therefore be determined by multiplying the stress components by the areas of
the sides. Let N be the direction of the normal to the plane BC. and
denote the cosines of the angles between the normal N and the axes x
and y by
cos Nx = I
cos Ny
Then, if A denotes the are$ of the side BO of the element, the areas of the
other two sides are Al and Am.
If we denote by X 81ld Y the components of stress acting on the side
BO, the equations of equilibrium of the prismatical element give
X = 1Hz + mTztI
Y = ?nulf + lTztI
Thus the components of stress on any plane defined by the direction
cosines I and m can easily be calculated from Eqs. (12), provided the
three components of stress (TZ1 (T1f' TZIf at the point P are known.
Letting a be the angle between the normal N and the x axis, so that
1 = cos a and m = sin a, the normal and shearing components of stress
on the plane Be are (from Eqs. 12)
= X cos a + Y sin a = Uz cos2 a + u" sin2 a + 2Tz" sin a
= Y cos a - X sin a = Tzu(COS 2 a - sin2 a}
+ (u" -
u z)
I t may be see~ that the angle a can be chosen in such a manner that the
. shearing stress T becomes equal to zero. For this case we have
Tz,,(COS 2
a - sin! a} + (u" - uz ) sin a cos a
- - - - - - - = - an a
cos 2 a - sin 2 a
From this equation, two perpendicular directions can be found for which
the shearing stress is zero. These directions are called principal directions and the correspdnding normal stresses principal stresses.
If the principal directions are taken as the x and y axes, T ZII is zero and
Eqs. (13) are simplified to
+ u" sin! a
.,. = ~~(u" - uz ) sin 2a
The variation of the stress components u and T, as we vary the angle a,
can be easily represented graphically by making a diagram in which u and
T are taken as coordinates. 1
For each plane there will correspond a point
on this diagram, the coordinates of which represent the values of u and .,.
for this plane. Figure 13 represents such a diagram. For the planes
perpendicular to the principal directions we obtain points A and B with
This graphical method is due to O. Mohr, Zilliling';"ieur, 1882, p. 113. See also
his "Technische Mechanik," 2d ed., 1914.
~------~~------~
-.'C
I I H I:.ORY OF ELASTICITY
abscissas U:z: and UlI, respectively. Now it can be proved that the stre:
components for any plane BC with an angle a (Fig. 12) will be reprl
sented by coordinates of a point on the circle having AB as a diamete
To find this point it is only necessary to measure from the point A in tl
same direction as a is measured in Fig. 12 an arc sub tending an ang
equal to 2a. If D is the point obtained in this manner, then, from tl
= OC + CF = U z +2 U 11 + U:z; -2 UlI cos .2a = Uz cos 2 a + UlI sin
DF = CD sin 2a
= ~(u:z: -
Comparing with Eqs. (13'), it is seen that the coordinates of point
give the numerical values of stress components on the plane BC at til
angle a. To bring into coincidence the sign of the shearing componen'
we take T positive in the upward direction (Fig. 13) and consider shearin
stresses as positive when they give a couple in the clockwise direction, s
on the sides be and ad of the element abed (Fig. 13b). Shearing stresSE
of opposite direction, as on the sides ab and de of the element, are cor
sidered as negative. 1
As the plane BC rotates about an axis perpendicular to the xy plan
(Fig. 12) in the clockwise direction, and a vaties from 0 to r/2, the poin
D in Fig. 13 moves from A to B, so that the lower half of the circle detel
mines the stress variation for all values of a within these limits. Th
upper half of the circle gives stresses for r/2 ~ a ~ r.
Prolonging the radius CD to the point Dl (Fig. 13), i.e., taking th
angle 1r + 2a, instead of 2a, the stresses on the' plane perpendicular t
BC (Fig. 12) are obtained. This shows that the shearing stresses on tw
perpendicular planes are numerically equal, as previously proved. A
for normal stresses, we see from the figure that OFl + OF = 20C, that iE
the sum of the normal stresses over two perpendicular cross section
remains constant when the angle a changes.
The maxiinum shearing stress is given in the diagram (Fig. 13) by th
maximum ordinate of the circle, Le., is equal to the radius of the circlE
It acts on the plane for which a = r/4, that is, on the plane bisecting th.
angle between the two principal stresses.
The diagram can also be used in the case when one or both principa
stresses are negative (compression). I t is only necessary to change th.
1 This rule is used only in the construction of Mohr's circle.
Otherwise the rul
given on p. 5 holds.
sign of the abscissa for compressive stress. In this manner Fig. 14a
represents the case when both principal stresses are negative and Fig.
14b the case of pure shear.
From Figs. 13 and 14 it is seen that the stress at a point can be resolved into two
parts: one, biaxial tension or compression, the two components being equal and of
magnitude given by the abscissa of the center of the circle; and the other, pure shear,
the magnitude of which is given by the radius of the circle. When several plane stress
distributions are superposed, the uniform tensions or compressions can be added
together algebraically. The pure shears must be added together by taking into
account the directions of the planes on which they are acting. It can be shown that if
we superpose two systems of pure shear whose planes of maximum shear make an
angle of fJ with each other, the resulting system will be another case of pure shear.
For example, Fig. 15 represents the determination of stress on any plane defined by
I---+--~r----t--,
.----0;
Fig.l6
, produced by two pure shears of magnitude 'Tl and T'2 acting one on the planes zz and
yz (Fig. 15a) and the other on the planes inclined to xz and yz by the angle fJ (Fig. 15b).
In Fig. 15a the coordinates of point D represent the shear and normal stress on plane
CB produced by the first system, whereas the coordinate of Dl (Fig. 15b) gives the
stresses on this plane for the second system. Adding OD and OD l geometrically we
obtain OG, the resultant stress on the plane due to both systems, the coordinates of G
giving us the shear and normal stress. Note that the magnitude of OG does not
depend upon . Hence, as the result of the superposition of two shears, we obtain a
Mohr circle for pure shear, the magnitude of which is given by OG, the planes of
maximum shear being inclined to the zz and yz planes by an angle equal to half the
angle GOD.
A diagram, such as is shown in Fig. 13, can also be used for determining principal stresses if the stress components tT:, tT111 T:fI for any two
perpendicular planes (Fig. 12) are known. We begin in such a case with
the plotting of the two points D and D l , representing stress conditions
on the two coordinate planes (Fig. 16). In this manner the diameter
DDI of the circle is obtained. Constructing the circle, the principal
stresses tTl and tT2 are obtained from the intersection of the circle with the
abscissa axis. From the figure we find
v, = OC + CD = v. ~ v. + ~(v. ; vr +
v.. OC - CD = v. ~ v. - ~(v. ; vr +
Tzu'
The maximum shea'ring stress is given by the radius of the circle, Le.,
In this manner, all necessary features of the stress distribution at a point
can be obtained if only the three stress components U:, tT111 T:1/ are known.
When the strain components E:, Ern 'YsrI at a point are known, the unit
elongation for any direction and the decrease of a right angle-the shearing strain-of any orientation at the point can be ~ound. A line element
'PQ (Fig. 17a) between the points (x,y), (x + dx, y + dy) is translated,
stretched (or contracted), and rotated into the line elementP'Q' when
the deformation occurs. The displacement components of Pare u, v,
. and those of Q are
u+-dx+-dy
v+-dx+-dy
If P'Q' in Fig.17a is now translated so that P' is brought back to P,
it is in the position PQ" of Fig. 17b, and QR, RQ" represent the components of the displacement of Q relative to P. Thus
QR = ax dx
+ 81J
= ax dx + ay dy
The components of this relative displacement QS, SQ", normal to PQ"
and along PQ", can be found from these as
QS = -QR sin 8 + RQ" cos8
QR cos 8 + RQ" sin 8 (b)
ignoring the small angle QPS in comparison with 8. Since the short line
QS may be identified with an arc of a circle with center P, SQ" gives the
stretch of PQ. The unit elongation of P'Q', denoted by ES, is SQ"/PQ.
Using (b) and (a) we have
cos 8 (au dx
= -av, cos 2 8
iJ,y) + sin 8 (ov dx + ov dY)
ay ds
oy ds
. 8 cos 8 + -av SIn
- + -av) SIn
cos 8 + 'YZrI sin 8 cos 8 +
which giv~ the unit elongation for any direction 8.
The angle 1/Is through which PQ is rotated is QS/PQ. Thus from (b)
and (a),
1/19 = - sin 8 (a ddx
= ax cos
~u ddSY) + cos 8 (ov
dx + av dY)
+ (av
ax SIn 8 cos 8 - au
The line element PT at right angles to PQ makes an angle 8 + (1(/2)
with the x direction, and its rotation ,pS+r/2 is therefore given by (d)
when 8 + (w,/2) is substituted for 8. Since cos [8 + (11"/2)1 = - sin 8,
(~~'l
~%~~dY)
sin [8
+ (1r/2)]
';6+fI'/2
= cos 8, we find
. 2 8 - (av
ay - au).
ax sln 8 cos 8 - au
ay cos2 8
The shear strain 'Y6 for the directions PQ, PT is 1/16 - 1/16+T/2, so
'Y6 =
ax + au)
ay (cos2 8 _
sin2 8)
~'Y6 = ~'YztI(COS2 8 - sin 2 8)
ay - au)
ax 2 sin 8 cos 8
+ (Etl -
E~) sin 8 cos 8
Comparing (c) and (f) with (13), we observe that they may be obtained
from (13) by replacing tT by E6, 'T by 'Y6/2, tT~ by E~, tTtI by Er" 'T~ by 'Y~/2,
and a by 8. Consequently, for each deduction made from (13) as to tT
and'T, there is a corresponding deduction from (c) and (f) as to E6 and
'Y6/2. There are thus two values of 8, differing by 90, for which 'Y6 is
zero. They are given by
~=tan28
The corresponding strains E6 are principal strains. A Mohr circle diagram analogous to Fig. 13 or 16 may be drawn, the ordinates representing
. 'Y6/2 and the abscissas EI. The principal strains El, E2 will be the algebraically greatest and least values of ES as a function of 8. The greatest
value of 'Y6/2 will be represented by the radius of the circle.. Thus the
greatest shearing strain 'Y6 max is given by
'Y6max = El - E2
Measurement of Surface Strains
The strains, or unit elongations, on a surface are usually most conveniently measured by means of electric-resistance strain gauges. 1 The
1 A detailed account of this method is given in M. Hetenyi (ed.), "Handbook of
Experimental Stress Ana1ysis," Chaps. 5 and 9, John Wiley & Sons, Inc., New York,
OL.____
-L-=~
simplest form of such a gauge is a short length of wire insulated from
and glued to the surface. When stretching occurs the resistance of the
wire is increased, and the strain can thus be measured electrically: The
effect 'is usually magnified by looping the wires backward and forward
several times, to form several gauge lengths connected in series. The
wire is glued between two tabs of paper, and the assembly is glued to the
The use of th~e gauges is simple when the principal directions are
known. One gauge is placed along each principal direction and direct
measurements of EI, E2 are obtained. The principal stresses O'b 0'2 may
then be calculated from Hooke's law, Eqs. (3), with O'~ = 0'1, 0'" = 0'2,
0'. = 0, the last holding on the assumption that there is no stress acting
on the surface to whl:ch the gauges are attached. Then
When the principal directions are not known in advance, three measurements are needed. Thus the state of strain is completely determined
if E~, E", 'Yzu can be measured. But since the strain gauges measure extensions, and not shearing strain directly, it is convenient to measure the
unit elongations in three directions at the point. Such a set of gauges is
called a "strain rosette." The Mohr circle can be drawn by the simple
construction l given in Art. 13, and the principal strains can then be read
off. The three gauges are represented by the three full lines in Fig. 18a.
The broken line represents the (unknown) direction of the larger principal strain EI, from which the direction of the first gauge is obtained by a
clockwise rotation f/J.
If the x and y directions for Eqs. (c) and (f) of Art. 11 had been taken
as the principal directions, E~ would be El) E" would be E2, and 'Y~" would be
Glenn Murphy, J. Appl. Mech., vol. 12, p. A-209, 1945; N. J. Hoff, ibid.
26 I THEORY OF ELASTICITY
zero. The equations would then be
cos 2 8 + E2 sin 2 8
3-2'Y6 = -
where 8 is the angle measured from the direction of
and these values are represented by the point P on the circle in Fig. 18c.
If 8 takes the value f/J, p. corresponds to the point A on the circle in Fig.
18b, the angular displacement from the E6 axis being 2f/J. The abscissa of
this point is ~, which is known. If 8 takes the value f/J + a, P moves to
B, through a further angle AFB = 2a, and the abscissa is the known
value Ecr+4l' If 8 takes the value f/J + a + fj, P moves on to C, through
a further angle BFC = ~, and the abscissa is EQ+JI+/I'
The problem is to draw the circle when these three abscissas and the
two angles a, {3 are known.
Construction of Mohr Strain Circle for Strain Rosette
A temporary horizontal E axis is drawn horizontally from any origin 0',
Fig. 18b, and the three measured strains ~, Ecr+4lJ Ecr+JI+4I are laid oft' along it.
Verticals are drawn through these points. Selecting any point D on the
vertical through Ecr+/l, lines DA, DC are drawn at angles a and fj to the
vertical at D as shown, to meet the other two verticals at A and C. The
circle drawn through D, A, and C is the required circle. Its center F is
determined by the intersection of the perpendicular bisectors of CD, DA.
The points representing the three gauge directions are A, B, and C. The
angle AFB, being twice the angle ADB at the circumference, is 2a, and
BFC is 2fj. Thus A, B, C are at the required angular intervals round the
circle and have the required abscissas. The E6 axis can now be drawn as
OF, and the distances from 0 to the intersections with the circle give EI, E2.
The angle 2f/J is the angle of F A below this axis.
We now consider the equilibrium of a small rectangular block of edges h,
The stresses acting on the faces 1, 2, 3, 4, and
their positive directions are indicated in the figure. On account of the
variation of stress throughout the material, the value of, for instance,
u~ is not quite the same for face 1 as for face 3. The symbols U~, UlI, T~
refer to the point x, y, the ~dpoint of the rectangle in Fig. 19. The
values at the midpoints of the faces are denoted by (uzh, (uzh, etc. Since
k, and unity (Fig. 19).
(O"Y)4
(r:ac:y4
r:cy)3
(0:.~}3
(:;e,Y)
(O"x
fz-.xy).l
(r:cy)Z
(0",,)2
the faces are very small, the corresponding forces are obtained by multiplying these values by the areas of the faces on which they act.!
The body force on the block, which was neglected as a small quantity
of higher order in considering the equilibrium of the triangular. prism of
Fig. 12, must be taken into consideration, because it is of the same order
of magnitude as the terms owing to the variations of the stress components that are now under consideration. If X, Y denote the components
of body force per unit volume, the equation of equilibrium for forces in
the x direction is
(uzhk - (uzhk
+ (T'zu)2h -
(T'zu)4h
+ Xhk =
or, dividing by hk,
(uz )1 ~ (uzh
(T'zuh ~ (T'ZU)4
If now the block is taken smaller and smaller, that is, h -+ 0, k -+. 0, the
limit of l(O'zh - (uz)a]/h is au:,Jax by the definition of such a derivative.
Similarly [(Tzuh - (Tzu)4)/k becomes aTzu/ay. The equation of equilibrium for forces in the y direction is obtained in the same manner. Thus
+ a.,zu
+ aT'zu + Y
These are the differential equations of equilibrium for two-dimensional
1 More precise considerations would introduce terms of higher order that vanish in
the final limiting process.
28 I THEORY OF ELASTICITY
. In many practical applications the weight of the body is usually the
only body force. Then, taking the y axis downward and denoting by p
the mass per unit volume of the body, Eqs. (18) become
au: + Ur:w =
+ aT:"
ax- pg
Equations (18) or (19) must be satisfied at all points throughout the volume of the body. The stress components vary over the volume of the
plate; and when we arrive at the boundary they must be such as to be
in equilibrium with the external forces on the boundary of the plate, so
that external forces may be regarded as a continuation of the internal
~tress distribution. These conditions of equilibrium at the boundary
ban be obtained from Eqs. (12). Taking the small triangular prism P Be
(Fig. 12), so that the side Be coincides with the boundary of the plate,
as shown in Fig. 20, and denoting by X and Y the components of the
surface forces per unit area at this point of the boundary, we have
14: + mT:w
= mu"
+ IT:w
in which land m are the direction cosines of the normal N to the boundary.
In the particular case of a rectangular plate, the coordinate axes are
usually taken parallel to the sides of the plate and the boundary condi""'
tions (20) can be simplified. Taking, for instance, a side of the plate
parallel to the x axis we have for this part of the boundary the normal
N parallel to the y axis; hence l = 0 and m = 1. Equations (20) then
X = T:w
Y = u"
Here the positive sign should be taken if the normal N has the positive
directIon of the y axis and the negative sign for the opposite direction
of N. It is seen from this that at the boundary the stress components
become equal to the components of the surface forces per unit area of the
Compatibility Equatio~s
It is a fUndamental problem of the theory of elasticity to determine the
state of stress in a body submitted to the action of given forces. In a .
two-dimensional problem it is necessary to solve the differential equations of equilibrium (18), and the solution must be such as to satisfy the
boundary conditions (20). These equations, derived by application of
the equations of statics and containing three stress components U z ,
'rZI"
are not sufficient for .the determination of these components. The problem is a statically indeterminate one, and in order to obtain the solution
the elastic deformation of the body must also be considered.
Th'e mathematical formulation' of the condition for compatibility of
stress distribution with the existence of continuous functions u, v, w
defining the deformation will be obtained from Eqs. (2). For twodimensional problems we consider three strain components, namely,
EII=-
These three strain components are expressed by two functions u and v;
hence, they cannot be taken arbitrarily, and there exists a certain relation between the strain components that can easily be obtained from (a).
Differentiating the first of the Eqs. (a) twice with respect to y, the second
twice with respect to x, and the third once with respect to x and once
with respect to y, we find
a + iJ2EII = a2"(zll
iJx 2
iJx iJy
This differential relation, called the condition of compatibility, must be
satisfied by the strain components to secure the existence of functions
u and v connected with the strain components by Eqs. (a). By using
Hooke's law, [Eqs. (3)], the condition (21)' can be transformed into a
relation between the components of stress.
In the case of plane stress distribution (Art. 8), Eqs. (3) reduce to
Substituting in Eq. (21), we find
ay2 (us - vO'1/) + ax 2 (Uti - vO'z)
= 2(1 + v) ax ~~
This equation can be written in a different form by using the equations
of equilibrium. For the case when the weight of the body is the only
body force, differentiating the first of Eqs. (19) with respec~ to x and the
second with respect to y.and adding them, we find
2 a2.rz~ = _ a20's _ a20'tI .
Substituting in Eq. (b), the compatibility equation in terms of stress
Proceeding in the same manner with the general equations of equilib.
rium (18), we find
( ax! + ay2) (O'z + Uti) = -(1 + v) ax + ay
In the case of plane strain (Art. 9), we have
= v(O'z
+ Uti)
and from Hooke's law (Eqs. 3), we find
'YZ1I
E1 [(1
-- v2)O'z - v(1
= E [(1 - v2)0"1/ - v(l
+ JI)O"tI]
+ v)O"z]
Substituting in Eq. (21) and using, as before, the equations of equilibrium (19), we find that the compatibility equation (24) holds also for
plane strain. For the general case of body forces we obtain from Eqs.
(21) and (18) the compatibility equation in the following form:
( ax2 + ay'l) (0': + Uti) = - 1 - JI ax + ay
The equations of equilibrium (18) or (19) together with the boundary
conditions (20) and one of the above compatibility equations give us a
system of equations that is usually sufficient for the complete determi-
nation of the stress distribution in a two-dimensional problem. 1 The
particular cases in which certain additional considerations are necessary
will be discussed later (page 133). It is interesting to note that in the
case of constant body forces the equations determining stress distribution
do not contain the elastic constants of the material. Hence, the stress
distribution is the same for all isotropic materials, provided the equations
are sufficient for the complete determination of the stresses. The conelusion' is of practical importance: we shall see later that in the case of
transparent materials, such as glass or xylonite,' it is possible to determine stresses by an optical method using polarized' light (page 150).
From the above discussion it is evident that experimental results obtained
with a transparent material in most cases can be applied immediately to
any other material, such as steel.
It should-;also be noted that in the case of constant body forces the
compatibility equation (24) holds both for the case of plane stress and
for the case of plane strain. The stress distribution is hence the same in
these two cases, provided the shape of the boundary and the external
forces are the same. 2
It has been shown that a solution of two-dimensional problems reduces
to the integration of the differential equations of equilibrium together
with the compatibility equation and the boundary conditions. If we
begin with the case when the weight of the body is the only body force,
the equations to be satisfied are [see Eqs. (19) and (24)]
+ a'l":u =
+ a'l":"
ax + pg =
(:;. + :;.) (v_ + v.)
To these equations the boundary conditions (20) should be added. The
usual method of solving these equations is by introducing a new function,
1 In plane stress there are.compatibility conditions other tlm.n (21) that are in fact
violated by our assumptions. It is shown in Art. 131 that in spite of this the method
of the present chapter gives good approximations for thin plates.
t This statement may require modification when the plate or cylinder has holes, for
then the problem can be correctly solved only by considering the displacements as
well as the stresses. See Art. 43.
called the 8tre88 Junction. l As is easily checked, Eqs. (a) are satisfied by
taking any function q, of x and y and putting the following expressions
for the stress components:
a2q,
ay2 - pgy
In this manner we can get a variety of ~olutions of the equations of equilibrium (a). The true solution of the problem is that which satisfies also
the compatibility equation (b). Substituting expressions (29) for the
stress components into Eq. (b), we find that the stress function q, must
Thus, the solution of a two-dimensional problem, when the weight of the
body is the only body force, reduces to finding a solution ~f Eq. (30) that
satisfie~ the boundary conditions (20) of the problem. In the following
chapters, this method of solution will be applied to several examples of
Let us now consider a more general case of body forces and assume that these
forces have a potential. Then the components X and Yin Eqs. (18) are given by the
x = _av
in which V is the potential function. Equations (18) become
.2.... (l1z
-. V)
.2.... (11f1
+ M~II
aTZfI
These equations are of the same form as Eqs. (a) and can be satisfied by taking
= a2q,
11f1 _
~fI -
in which q, is the stress function. Substituting expressions (31) in the compatibility
equation (25) for plane stress distribution, we find
+ 2 az. ay2 + ayC = -
(1 - II)
(a 2 V
a:a v )
An analogous equation can be obtained for the case of plane strain.
1 This function was introduced in the solution of two-dimensional problems by
G. B. Airy, Brit. Assoc. Advan. Sci. Rept., 1862, and is sometimes called the Airy stress
When the body force is simply the weight, the potential V is -pgy. In this case
the right-hand side of Eq. (32) reduces to zero. By taking the solution q, = 0 of (32),
or of (30), we find the stress distribution from (31), or (29),
-pgy
= -pgy
'r." = 0
as a possible state of stress due to gravity. This is a state of hydrostatic pressure pgy
in two dimensions, with zero stress at y = o. It can exiSt in a plate or cylinder of any
shape provided the corresponding boundary forces are applied. Considering a boundary element as in Fig. 12, Eqs. (13) show that there must be a normal pressure pgy on
the boundary, and zero shear stress. If the plate or cylinder is to be supported in some
other manner we have to superpose a boundary normal tension pgy and the new supporting forces. The two together will be in equilibrium, and the determination of their .
effects is a problem of boundary forces only, without body forces.!
1. Show that Eqs. (12) remain valid when the element of Fig. 12 has acceleration.
2. Find graphically the principal strains and their directions from rosette measuremen~
= 2 X 10-'
= 1.35 X 10-1
Ea~#
= 0.95 X 10-1 in. per in.
where a: = fJ = 45.
3. Show that the line elements at the point x, y that have the maximum and minimum
rotation are those in the two perpendicular directions 8 determined by
tan 28 = av/ay - au/ax
av/ax + au/ay
4. The stresses in a rotating disk (of unit thickness) can be regarded as due to centrifugal force as body force in a stationary disk. Show that this body force is
derivable from the potential V = - Hp",'(x' + y'), where p is the density and '"
the angular velocity of rotation (about the origin).
5. A disk with its axis horizontal has the gravity stress represented by Eqs. (d) of
Art. 16. Make a sketch showing the boundary forces that support its weight.
Show by another sketch the auxiliary problem of boundary forces that must be
solved when the weight is entirely supported by the reaction of a horizontal surface
on which the disk stands.
6. A cylinder with its axis horizontal has the gravity streBB represented by Eqs. (d)
of Art. 16. Its ends are confined between smooth fixed rigid planes that maintain
the condition of plane strain. Sketch the forces acting on its surface, including
7. Using the stress-strain relations, and Eqs. (a) of Art. 15 in the equations of equilibrium (18), show that in the absence of body forces the displacements in problems
of plane stress must satisfy
+ atu + 1 + v .!. (au + av) = 0
and a companion equation.
1 This problem, and the general case of a potential V such that the right-hand side
of Eq. (32) vanishes, have been discUssed by M. Biot, J. Appl. M ech., 1935, p. A-41.
IN ENGINEERING ANALYSIS
Department 0/ Mechanical Engineering
Massachusetts Institute 0/ Technology
P.RENTICE-HALL, INC., Englewood Cliffs, New Jersey 07632
(b) One-quarter of structure
FIGURE 4.13 Analysis of a cyclicly symmetric structure.
4.2.3 Generalized Coordinate Models
In Section 4.2.2 the finite element discretization procedure and derivation
of the equilibrium equations was presented in general, i.e., a general threedimensional body was considered. As shown in Examples 4.2 to 4.7, the general
equations derived must be specialized in specific analyses to the specific stress
and strain conditions considered. The objective in this section is to discuss and
summarize how the finite element matrices that correspond to specific problems
can be obtained from the general finite element equations (4.8) to (4.23).
Although in theory any body may be understood to be three-dimensional,
for practical analysis it is in many cases imperative to reduce the dimensionality
of the problem. The first step in a finite element analysis is therefore to decide
what kind of problem is at hand. This decision is based on the assumptions used
in the theory of elasticity approach to the solution of specific problems. The
classes of problems that are encountered may be summarized as (I) truss, (2)
beam, (3) plane stress, (4) plane strain, (5) axisymmetric, (6) plate bending, (7)
thin shell, (8) thick shell, and (9) general three-dimensional. For each of these
problem cases, the general formulation is applicable; however, only the appropriate displacement, stress and strain variables must be used. These variables
are summarized in Tables 4.2 and 4.3 together with the stress-strain matrices
to be employed when considering an isotropic material. Figure 4.14 shows
various stress and strain conditioris considered' in the formulation of finite
element matrices. 11 .12
THE DISPLACEMENTBASED FINITE ELEMENT METHOD
TABLE 4.2 Corresponding kinematic and static variables iff
Pl ane stress
Plane strai n
~isymmetri c
Noratioll:
[ fxx ]
[K xxl
[f xx f yy YXy]
(f xx fJ'J' YXY]
= ax' fJ'J' = dr'
1M,,]
['t" xx 't"n T xy]
['t" xx Tn
T xy]
[f xx El'y Yxl' Ezz ]
[T xx T yy T xy T:z ]
[f xx f l'y f u Y Xl' Yy: Y1 X)
['t" x ... 't" yy T:: T xl' T"t T: ...
[K u
S Iren Vector
K yy K xy]
au + ax'
av ... ,K x ... = - dx'l'
[M xx M yy
KJ'J'
alw K ... y =
= - qy'i'
In Examples 4.2 to 4.7 we deve lo ped already so me specific finite element
matrices. Referring to Example 4. 5, in which we considered a plane stress condition, we used for th e u and v displacements simple linea r pol ynomial ass umptions, where we identified th e unkn ow n coefficients in th e polynomials as
generalized coordinates. The number o f unkn own coefficients in th e polynomials
was equal to the number of element nodal po int displacements. Expressi ng the
ge nera li zed coordinates in terms of the element nodal point di splacements,
we found that, in ge nera l, each polynomial coefficient is not an actual phys ical
displacement but is eq ual to a linear combination of th e element nodal point
Finite element matrices which areformlliat ed by assuming that the displacements
vary in the form of o/tmcliolJ whose unknown coefficients ore trealed as generalized
coordinates are referred to as generalized coordinale finit e element models. A
rath er natura l class of function s to use for approximating element displacements
are polynomials, beca use pol ynomials a re commonly employed to approxi mate
unkn own function s, and the hi gher the deg ree of the polynomial, th e better is
the approximation that we can ex pect. ln add ition, polynomials are easy to
differentiate; i.e., if th e polynomials approximate the d isplacements of the
structure, we ca n evaluate th e strains with relative ease.
Using polynomial displacement ass umptio ns, a very large number of finite
elements for practica ll y all problems in structural mechanics have been developed.
The objecti ve in this secti o n is to describe the form ulati o n of a variety of
generalized coordinate finite clement models that use polynomials to app roximate the displacement field s. Other function s wo uld, in principle, be used in
th e sam e way and their use ca n be effecti ve in specific app licatio ns (see Example 4. 16). In th e presentati on, emphas is is given to th e ge nera l fo rmu lat io n
rat her tha n to numeri cally effecti ve finite elements. Therefore, th is secti o n serves
primarily to enh ance our genera l und erstandin g o f th e fi ni te clement meth od.
The curren tl y most effective fi nite elements for ge neral appli cati o n arc believed
to be the isoparametri c a nd related elements th at are desc ri bed in Chapter 5.
FORM ULATION OF THE FIN ITE ELEMENT MET HOD
TABLE 4.3 Generalized stress-strain matrices for
isotropic materials and the problems in table 4.2.
Material Matrix C
I - v2
0 1 ; .]
E(1 - v)
- 2v) 1 ~ v
+ v)(1
1 -o 2v
1E(1 - v)
+E(1v)(1-
2(1 - v)
I - v 1- v
1- v 1- v
shown are zeros
12(1 - V2) 0 0 1
Notation: E
= Young's modulus, v =
Poisson's ratio, h = thickness of plate, I = moment
In the following derivations the displacements of the finite elements are
always described in the local coordinate systems shown in Fig. 4.14. Also, since
we consider one specific element, we shall leave out the superscript (m) used in
For one-dimensional bar elements (truss elements) we have
= ~1 + ~2X + ~lX2 + ...
(a) Uniaxial stress condition: frame under concentrated loads
TxX' T yy , Txy are uniform
across the th ickness
(b) Plane stress conditions: membrane and beam under in-plane
u(x, V), v(x, y) are non-zero
'Yyz = 'Yzx = 0
w = 0, EZZ =
(c) Plane strain condition: long dam subjected to water pressure
Various stress and strain conditions with illustrative examples.
are axisymmetric
1<t- -_ .
.I <t
Tyz.= Tzx = 0
(d) Axisymmetric condition: cylinder under internal pressure
(e) Plate and shell structures
FIGURE 4.14 (cont.)
1. Blok cell n x n
2. Di cell pertama ketik: @minverse( matrix[A])
3. tekan shift+control enter
2. Di cell pertama ketik: @mmult( matrix[A],matrix[B])
determinan [D]
1. Blok satu cell
2. Di cell ketik: @mdetermin( matrix[A])
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References: Art. 81
 Art. 98
 Art. 13
 Art. 11
 Art. 131
 Art. 43

Art. 16
 Art. 16
 Art. 15