analysis of stress and strain i - deu.edu.trkisi.deu.edu.tr/emine.cinar/asm16-analysis of stress and...
TRANSCRIPT
ANALYSIS OF STRESS AND STRAIN I
When dealing with a new design or design improvement
problem the design engineer knows that the design must
(1) function according to some prescribed requirements and
(2) have an acceptable level of structural integrity. Stress
analysis is that part of the design process which strives to
ensure that each element of a given system will not fail to
meet the structural requirements of the design throughout
the specified life of the system.
Stress analysis tools range from the countless theoretical
techniques employed in elementary and advanced mechanics
of materials and the mathematical theory of elasticity, to
computer-based numerical procedures such as the finite
element method and to the wide variety of available
experimental techniques.
Practical stress analysis problems range from problems
which closely resemble simple models which have known
closed form solutions to complex problems which are not
easily adaptable to the various classical techniques. For the
complex problem, the analysit must either obtain an
approximation from a theoretical analysis, seek a numerical
solution, perform an experimental analysis or carry out a
combination of these approaches. In order to be competent,
the analysit should have a good working knowledge of the
theoretical, numerical and experimental procedures
employed in the field of stress analysis.
Experimental stress and strain analysis requires thorough
understanding of stresses and strains developed in any
structural member. Therefore, relationships between
deformations, strains, and stresses developed in a body
must be well understood. Further, generalized Hooke’s law,
equilibrium equations and compatibility equations, and Airy’s
stress functions for determining stresses at any point in a
member subjected to known loads/stresses on its boundary
are fundamental concepts. Plane stress state, plane strain
state, and three-dimensional displacement field along with
strain matrix should also be discussed.
DEFINITION AND COMPONENTS OF STRESS
Stress is simply an internally distributed force within a body. Let’s consider a body in
equilibrium subject to a system of external forces, Figure 1. Under the action of
these forces, internal forces are developed within the body. To examine the stresses
at some interior point Q, we use an imaginary plane to cut the body at a section a-a
through Q, dividing the body into two parts. As the forces acting on the entire body
are in equilibrium, the forces acting on one part alone must also be in equilibrium;
this requires the presence of forces on plane a-a. These internal forces, applied to
both parts, are distributed continuously over the cut surface. This process, referred
to as the Method of Sections, is relied on as a first step in solving all problems
involving the investigation of internal forces. The resulting figure is the Free Body
Diagram.
Figure 1 Method of sections: (a) Sectioning of a loaded body; (b) free body with external and internal forces; (c) enlarged area DA with components of the force D F.
An element of area DA, located at point Q on the cut surface, is acted on by force DF.
In general DF will not lie along x, y or z axis. Decomposing into components parallel
to x, y and z, we define the normal stress sx and the shearing stresses txy and txz.
Figure 1 Method of sections: (a) Sectioning of a loaded body; (b) free body with external and internal forces; (c) enlarged area DA with components of the force D F.
dA
dF
A
F
dA
dF
A
F
dA
dF
A
F
zz
Axz
yy
Axy
xx
Ax
D
D
D
D
D
D
DD
D
00
0
limlim
lim
tt
s
Written in integral form
A
xzz
A
xyy
A
xx dAFdAFdAF tts ,,
On a given surface only one normal stress and one shear stress can exist (Figure 2). The net
tangential force acting on the surface will be equal to ; the net shear stress
will then be equal to
22
zy FF DD
22
xzxynetx ttt
Figure 2 Stress components acting on a differential element Net shear stress
Stress components
To describe the state of stress at point Q completely it would be necessary to examine
other surfaces by making different planar slices. Since different planar slices would
necessitate different coordinates and different free body diagrams, the stresses on each
slice would be, in general, quite different. So, to understand the complete state of stress
at point Q, every possible surface intersecting point Q should be examined. However,
this would require an infinite number of slices surrounding point Q. That is, if point Q
were completely isolated from the body, it would be described by an infinitesimal
sphere. Naturally, this would be impossible to do, so a coordinate transformation is
necessary for this.
In order to enable the determination of the stresses on an infinite number of planes
passing through a point Q, thus defining the stresses at that point, we need only specify
the stress components on three mutually perpendicular planes passing through the point
(Figure 3). These three planes, perpendicular to the coordinate axes, contain three hidden
sides of an infinitesimal cube. It should be emphasized that when moving from point Q to
point Q the values of stress will, in general, change. Also, body forces, such as weight of a
body acting through the whole volume of the body can exist. These stresses lead to the
stress tensor.
Figure 3 Element subjected to three- dimensional stress. All stresses have positive sense.
STRESS TENSOR
All stresses acting on a cube of infinitesimally small dimensions, i.e. ∆x = ∆y = ∆z → 0 are
identified by the diagram of a cube. The first subscript on normal stress σ or shear stress τ
associates the stress with the plane of the stress, i.e. subscript defines the direction of the
normal to the plane, and the second subscript identifies the direction of stress as
txx = normal stress on yz plane in x-direction
txy = shear stress on yz plane in y-direction
txz = shear stress on yz plane in z-direction.
Similarly the stresses on xz plane and xy planes are identified as shown in Figure 1. Stress
tensor τij can be written as
Figure 4 Stresses on a cube element
zzzyzx
yzyxyx
xzxyxx
ij
ttt
ttt
ttt
t
But, generally the normal stress is designated by σ.
So, txx = σxx = σx , tyy = σyy = σy , tzz = σzz = σz
Stress tensor can be written as
Stress tensor is symmetric, i.e. tij = tji
Complementary shear stresses are txy and tyx; tzy and tyz ; tzx and txz
Complementary shear stresses meet at diametrically opposite corners of an element to satisfy the equilibrium conditions.
zzyzx
yzyyx
xzxyx
ij
stt
tst
tts
t
NORMAL STRAIN
Stresses on a rectangular parallelepiped DxDyDz element cause a change of the element both
dimensionally and in shape. The normal stresses cause the element to expand and/or shrink in the
x, y and z directions so that the element remains a rectangular parallelepiped. The shear stresses do
not basically cause dimensional changes, but shear causes the element to change shape from a
rectangular parallelepiped to a rhombohedron.
Initally let’s consider only sx applied to the element as shown in Figure 5. The element increases in
length in x direction and decreases in length in y and z directions. The dimensionless rate of
increase in length is defined as the normal strain where
ex, ey and ez represent the normal strains in x, y
and z directions. Thus the new length in any
direction is equal to its original length plus the
rate of increase (normal strain) times its
original length.
Figure 5 Deformations due to sx
zzz
yyy
xxx
z
y
x
DDD
DDD
DDD
e
e
e
By using Hooke’s law,
As the element elongates in x direction, it contracts in y and z directions. If the material is linear,
these contractions are also directly proportional to the normal stress. It is common practice to
express the contractions in terms of the primary normal strain, in this case ex using the Poisson’s
ratio n. If the material is isotropic,
Ex
x
se
Ex
xzy
snneee
The normal strains caused by sy and sz are
similar to the strains caused by sx:
E
E
E
E
zzyx
zz
y
yzx
y
y
snneee
se
snneee
se
For an element undergoing sx, sy and sz simultaneously, the effect of each stress can be added
using the concept of linear superposition. Thus the general stress-strain relationship for a linear,
homogeneous, isotropic material is
yxzz
zxyy
zyxx
E
E
E
ssnse
ssnse
ssnse
1
1
1
If the strains are known, stresses in terms of strains can be expressed as
yxzz
zxyy
zyxx
E
E
E
eenennn
s
eenennn
s
eenennn
s
1211
1211
1211
SHEAR STRAIN
The change in shape of the element caused by shear stresses can be illustrated first by examining
the effect of txy alone, Figure 6. The shear strain gxy is a measure of the deviation of the stresses
element from a rectangular parallelepiped. It is defined as
DABBADBADxyDg
Figure 6 Pure shear
where gxy is in dimensionless radians. For a linear, homogeneous,
isotropic material the shear strain is directly related to the shear
stress by
G
xy
xy
tg
Similarly, the remaining shear strans are related to the corresponding
shear stresses.
GGzx
zx
yz
yz
tg
tg ,
n
12E
G
It can be shown that for a linear, homogeneous, isotropic material the
shear modulus is related to the modulus of elasticity and Poisson’s
ratio by
Strain tensor in a biaxial case will be (Figure 7)
Figure 7 Shear stresses and shear strains on a differential element
y
xy
xy
x
eg
ge
2
2
STRAIN TENSOR
Three-dimensional strain tensor will be
z
zyzx
yz
y
yx
yzxy
x
egg
ge
g
gge
22
22
22
PLANE STRESS CONDITION
Under plane stress condition, stresses σx, σy, τxy are represented in one plane, i.e. central plane of a
thin element in yx plane, as shown shaded in Figure 8.
Figure 8 Plane stress state
In this case stresses σz = txz = tyz = 0.
If E and v are Young’s modulus and
Poisson’s ratio of the material, respectively,
then strains are
yxz
xyy
yxx
E
E
E
ssne
nsse
nsse
1
1
Shear strain
But shear strain gxy is equally divided about x and y axes both.
Strain tensor for plane stress condition will be as follows:
For a particular set of three orthogonal planes, where shear stresses are zero and normal stresses on
these planes are termed as principal stresses σ1, σ2, and σ3, stress tensor will be
G
xy
xy
tg
z
y
xy
xy
x
e
eg
ge
00
02
02
3
2
1
00
00
00
ss
s
PLANE STRAIN CONDITION
For strain condition shown in Figure 9 the strain tensor is
Figure 9 Plane strain state
In this, strain
Therefore,
To satisfy the condition of plane strain
Moreover,
y
yx
xy
x
eg
ge
2
2
0
0
yxz
z
z
EEssns
e
e
yxz ssns
0 zxyz gg
Example
A 100 mm cube of steel is subjected to a uniform pressure of 100 MPa
acting on all faces. Determine the change in dimensions between parallel
faces of cube if E = 200 GPa, v = 0.3.
DEFORMATIONS IN CARTESIAN COORDINATES
There are two types of strains resulting from deformation of a body, i.e. (a) linear or extensional
strains and (b) shear strains, resulting in change of shape. Let us consider an element of dimensions
∆x, ∆y, with change in x- and y-direction as ∆u and ∆v as shown in Figure 10. A body of dimensions
∆x and ∆y, in x y plane, represented by ABCD, deforms to AB′C′D as shown in Figure 10 (a).
Strain in x-direction,
Again it is deformed to A′B′CD as shown in Figure 10 (b), strain in y-direction,
.
Now the body in x-y plane is deformed to A′B′C′D, as shown in Figure 10 (c).
Figure 10 (a) Small element (b) Normal strain (c) Shear strains
xu
x DD
e
yv
y DD
e
Shear strain,
Moreover, linear strains are on both side of coordinate axes
In order to study the deformation, or change in shape, one has to consider a displacement field(s) for
a point in a body subjected to deformation.
A point P is located at position vector OP = r and displaced to position vector OP′ = r′. The
displacement vector S = r′ - r. The displacement vector (Figure 11) a function of x, y, z, has
components u, v, w in x-, y-, z-directions, such that
where S is a function of initial coordinates of point P.
u = u (x, y, z)
v = v (x, y, z)
w = w (x, y, z)
function of x, y, and z.
Figure 11 Displacement vectors
D C
D' C'
Dx
Dx+(du/dx) Dx
(dv/dx) Dx
(du/dy) Dy
y
yyu
x
xxv
D
D
D
D
/tan,
/tan 21
or,
kwjviuS
These strains at point P are defined by
and shear strains are
In x-y plane, or in a two-dimensional case, strains are
Example
The displacement field for a given point of a body is
at point P (1, −2, 3) determine displacement components in x-, y-, z-directions.
What is the deformed position?
4322 10624332 kzxjyxiyxS
Example 1.1
The displacement field for a body is given by
write down the strain components and strain matrix at point (2,1,2).
422 103242 kyxjziyxS
Figure 12 Stresses and elastic displacements in cylindrical coordinates
DEFORMATIONS IN CYLINDRICAL COORDINATES
In many problems the geometry of the component may not be suitable for the use of a rectangular
(Cartesian) coordinate system, and is more practical to use a different coordinate system. Problems
like thin or thick-walled pressure vessels, circular rings, curved beams, and half plane problems are
more suitable to a cylindrical coordinate system.
In figure 12 (a), an infinitesimal element is
constructed using r coordnates and the
corresponding normal and shear stresses are
shown. The depth of the element in the z
direction is Dz. It is assumed that both Dr and D
are approaching zero.
The stress-strain relations are the same as in
rectangular coordinates:
ssnse
ssnse
ssnse
rzz
zr
zrr
E
E
E
1
1
1
Figure 12 Stresses and elastic displacements in cylindrical coordinates
and
zrzr
zz
r
G
r
E
E
E
tn
g
tn
g
tn
g
12
12
12
The strain-displacement relations in cylindrical
coordinates are determined similar to Cartesian
coordinates but due to the complexity of the
deformed element, the deformation due to radial
and tangential displacements are viewed
separately, Figure 9 (b) and (c).
The initial and deformed elements are indicated
by dotted and solid lines, respectively. The net
radial strain is due to the radial displacement
only and is given by
r
u
r
urruu rrrrr
D
D
/e
The net tangential strain is
where the first term is due to the radial displacement and
the second to tangential displacement. Simplifying gives
e
D
D
D
DD
r
uuu
r
rur r /
e
u
rr
ur 1
The shear strain gr is equal to a + b. Thus
r
rrrurruu
r
uuu rrrr D
DD
D
D
///
g
Simplifying
r
u
r
uu
rr
r
g
1
The strains in z and zr planes are developed in a similar manner.
z
uzz
e
r
u
z
u
z
uu
rzr
zrz
z
g
g
1
A special case is encountered for problems symmetric with respect to the longitudinal z axis, which
are referred to as axisymmetric problems (such as pressurized circular disks), where vartiations
with respect to are zero and due to symmetry u = 0 everywhere. Thus, for axisymmetric
problems we can write
r
urer
urr
e
z
uzz
e
r
u
z
u zrzrzr
ggg 00
GENERALIZED HOOKE’S LAW
For a simple prismatic bar subjected to axial stress σx and axial strain εx (Figure 13), Hooke’s
law states that stress ∝ strain are related as
where E is the proportionality constant and Young’s modulus of elasticity of the material.
Figure 13 Simple bar subjected to axial stress and axial strain
xx
xx
E es
es
xx es ,
But for an elastic body subjected to six stress components σx, σy, σz, tyx, tyz, tzx and six strain
components, i.e. εx, εy, εz, gxy, gyz, gzx, the generalized Hooke’s law can be expressed as
σx = A11 εx + A12 εy + A13 εz + A14 gxy + A15 gyz + A16 gzx
σy = A21 εx + A22 εy + A23 εz + A24 gxy + A25 gyz + A26 gzx
σz = A31 εx + A32 εy + A33 εz + A34 gxy + A35 gyz + A36 gzx
τxy = A41 εx + A42 εy + A43 εz + A44 gxy + A45 gyz + A46 gzx
τyz = A51 εx + A52 εy + A53 εz + A54 gxy + A55 gyz + A56 gzx
τxz = A61 εx + A62 εy + A63 εz + A64 gxy + A65 gyz + A66 gzx
where A11, A12, …, A65, A66 are 36 elastic constants for a given material.
For homogeneous linearly elastic material, above noted six equations are known as
generalized Hooke’s law. Similarly, strains can be expressed in terms of stresses as follows:
For a fully anisotropic material there are 21 constants
In generalized Hooke’s law and for orthotropic materials there are nine constants in Hooke’s law.
Say for an isotropic material having the same elastic properties in all directions and which has no
directionally varying property, there are three principal stresses σ1, σ2, σ3 and three principal
strains ε1, ε2 and ε3, then Hooke’s law can be written as
where A, B, and C are elastic constants.
zxyzxyzyxzx
zxyzxyzyxx
BBBBBB
BBBBBB
tttsssg
tttssse
666564636261
161514131211
3211 eees CA
In the above equation ε1 is the longitudinal strain along σ1 but ε2 and ε2 are lateral strains so,
constants B = C, therefore
σ1 = A ε1 + B ε2 + C ε3
= A ε1 + B ε1 − B ε1 + B ε2 + B ε3
= (A − B)ε1 + B(ε1 + ε2 + ε3)
where ε1 + ε2 + ε3 = ∆, a cubical dilatation = first invariant of strain σ1 = (A − B)ε1 + ΔB
Let us denote (A − B) by 2 μ and B by λ, then
σ1 = λ Δ + 2με1
Similarly σ2 = λ Δ + 2με2
σ3 = λ Δ + 2με3
λ and μ are called Lame’s coefficients.
σ1 + σ2 + σ3 = 3λ Δ + 2μ(ε1 + ε2 + ε3)
= 3λ Δ + 2μ Δ
= (3λ + 2μ) Δ
Adding s1, s2 and s3
Principal stress
Solving for ε1, we get
sss
23321
D
13211
11
223
2
esss
s
es
D
3211 223
ss
s
e
From elementary strength of materials
Therefore, Young’s modulus
Poisson’s ratio,
3211
1 ssnse E
23E
n
2
Example
For steel, Young’s modulus E = 208000 MPa and Poisson’s ratio,
n = 0.3. Find Lame’s coefficients λ and μ.
ELASTIC CONSTANTS K AND G From the previous part we know that
Putting the value into above equation of n, we get for E
From the elementary strength of materials we know that
Therefore Lame’s coefficient μ = G, shear modulus.
If σ1 = σ2 = σ3 = p, hydrostatic stress or volumetric stress
Therefore, Bulk modulus is,
Shear modulus, , Lame’s coefficient.
2
23EorE
n
n
2
2or
n
n
12
22E
n 12GE
ModulusBulkKp
p
,3
23
23
3
D
D
3
23 K
G
Example 1.2
For aluminium, E = 67000 MPa, Poisson’s ratio, v = 0.33. Determine
the Bulk modulus and shear modulus of aluminium.
GENERAL THREE DIMENSIONAL STRESS TRANSFORMATIONS
As stated earlier, a state of stress not only depends on position within a structure, but it also
depends on the orientation of the surface containing the stresses. The analysis of stress
variation with respect to orientation can be carried out by using coordinate transformation. This
relates an arbitrary stress surface to a set of known stresses on three mutually orthogonal
surfaces defined by the stress matrix
zzyzx
yzyyx
xzxyx
stt
tst
tts
s
Let us consider the state of stress at a point, Figure 14. If
an arbitrary oblique plane is passes through the solid so
that the plane intersects the three mutually perpendicular
reference planes, a tetrahedral element about the point will
be isolated as shown in Figure 15 (a).
Figure 14 General state of stress
Let us consider the x' axis to be normal to the oblique plane and y '
and z ' axes to be tangent to the plane. The orientation of the normal
to the oblique surface can be established by the angles x ' x, x ' y and
x ' z which relate the axis to the x, y and z axes, Figure 15 (b).
Given the coordinates of the vertices A (a, 0, 0), B (0, b, 0) and
C (0, 0, c), the equation for the oblique plane is
Figure 15 Stresses on oblique surface
01cz
b
y
ax
The direction cosines of the surface normal are proportional to
the gradients (derivatives ) of the surface equation with respect
to the x, y and z directions, respectively. For the equation of a
plane the derivatives are simply the coefficients of the x, y and
z terms and are called directional numbers. The direction
cosines are thus,
cK
bK
aK zxyxxx
1cos
1cos
1cos
Where K is the proportional constant. The sum of the squares of
the directional cosines is unity.
1222
cK
bK
aK
Solving for K yields 222 )()()( abacbc
abcK
Thus x ' x is 222 )()()(
cosabacbc
bcaK
xx
From vector mechanics, the oblique area A0 is equal to the
cross product ACAB21
222
0 )()()(21
abacbcA
Since bcAx 21
xxxx
xx AAA
A
cos
2
2cos 0
0
Similarly zxzyxy AAAA cos,cos 00
Direction cosines can shortly be denoted as
zxzxyxyxxxxx nnn coscoscos
So zxzyxyxxx nAAnAAnAA 000
To determine the normal stress sx' acting on the oblique surface,
a summation of forces in the x' direction is necessary. The force
due to each stress is obtained by multiplying the stress by the
area over which the stress acts. Next, the component of the
force in the x ' direction is determined. For example, the force
due to sx is . The component of this force in the
x ' direction is . Summing all forces in x' direction
for equilibrium
xxxxx nAA 0ss
xxxxx nnA 0s
0000
000
0000
yxzxyzxxzxzxzxyxyz
xxyxxyzxxxzxyxxxxy
zxzxzyxyxyxxxxxx
nnAnnAnnA
nnAnnAnnA
nnAnnAnnAA
ttt
ttt
ssss
Simplifying results in
For a complete transformation of stresses with respect to the arbitrary
oblique surface, it is necessary to determine the shear stresses t x'y'
and t x‘z' .If we define the direction cosines of the y' axis relative to the
xyz coordinate system as ny‘x , ny‘y and ny‘z, a summation of forces in
the y' direction for equilibrium yields
xxzxzxzxyxyzyxxxxyzxzyxyxxxx nnnnnnnnn tttssss 222222
00000
000000
yyzxyzxyzxzxzyyxyzxyyxxy
zyxxzxyyxxxyzyzxzyyyxyxyxxxyx
nnAnnAnnAnnA
nnAnnAnnAnnAnnAA
tttt
ttssst
Simplifying results in
xyzxzyxxzx
yyzxzyyxyzxyyxyyxxxyzyzxzyyyxyxyxxxyx
nnnn
nnnnnnnnnnnnnn
t
ttssst
The final shear stress t z’x‘ on the oblique surface is found in a similar manner. Defining the direction
cosines of the z' axis relative to the xyz coordinate system as nz‘x , nz‘y and nz‘z, equilibrium of forces in
the z' direction yields
xzzxzzxxzx
yzzxzzyxyzxzyxyzxxxyzzzxzyzyxyxzxxxxz
nnnn
nnnnnnnnnnnnnn
t
ttssst
This set of equations are completely sufficient for the determination of the state of stress on any internal
surface in which an arbtrarily selected tangential set of coordinates is used (in this case y' z' coordinates).
xyzyzxzyyyyzyyxyxyzyzyyyxyxy nnnnnnnnn tttssss 222222
For a complete transformation of the stress element to that of a rectangular element oriented by the x' y' z'
coordinate system, the six stresses on the two surfaces with normals in the y' and z' directions must also
be determined (sy‘ , t y’z‘ , t y’x‘ , sz‘ , t z’y‘ , t z’x‘ ). Since it is assumed that no body force acts on the body the
cross shear will be equal to each other (t y’x‘ = t x’y‘, t y’z‘ = t z’y‘, t z’x‘ = t x’z‘). Thus with cross shears being
equal, it is only necessary to evaluate sy‘ , sz‘ and t y’z‘ .
xzzzzxzzyzyzyzxzxyzzzyzyxzxz nnnnnnnnn tttssss 222222
xzzyzzxyzx
yzzyzzyyyzxzyyyzxyxyzzzyzyzyyyxzxyxzy
nnnn
nnnnnnnnnnnnnn
t
ttssst
Example
The state of stress at a point relative to an xyz coordinate sytem is given by
σx = -8 MPa, σy = 4 MPa, σz = -5 MPa, τxy = 6 MPa, τyz =2 MPa, τzx = -2 MPa.
Determine the state of stress on an element that is oriented by first rotating the
xyz axes 45° about the z axis and then rotate the resulting axes 30 ° about the
new x axis.
Example 1.3
The state of stress at a point relative to an xyz coordinate sytem is given by
Determine the state matrix relative to a coordinate system defined by first rotating
the xyz axes 45° about the x axis and then rotate the resulting axes -45 ° about
the new z axis.
MPa
101525
154030
25300
s
PLANE STRESS TRANSFORMATIONS
For the state of plane stress shown in Figure 16, sz t yz = t zx =0. Plane stress
transformations are normally performed in the xy plane as shown in Figure 16
(b). The angles relating the x'y'z' axes are
09090
9090
9090
zzyzxz
zyyyxy
zxyxxx
Figure 16 Plane stress transformations
Thus,
100
0cossin
0sincos
zzyzxz
zyyyxy
zxyxxx
nnn
nnn
nnn
Stresses are obtained as
tsst
tsss
tsss
22
22
22
sincoscossin
sincos2cossin
sincos2sincos
xyyxyx
xyyxy
xyyxx