Chapter 2 STRESS, STRAIN AND STRESS-STRAIN RELATIONSHIPS 1. INTRODUCTION 1 There are two main types of stress analyses. The first is conceptual, where the structure does not yet exist and the analyst is given reasonable leeway to define geometry, material, loads, and so on. The preeminent way of doing this nowadays is with the finite element method (FEM). The second analysis is where the structure (or a prototype) exists, and it is this particular structure that must be analyzed. Situations involving real structures and components are, by their very nature, only partially specified. Notation for stress components 1 From Doyle , Modern Experimental Stress Analysis, John Wiley & Sons, 2004. xx yz yx
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Chapter 2
STRESS, STRAIN AND STRESS-STRAIN RELATIONSHIPS 1. INTRODUCTION1 There are two main types of stress analyses. The first is conceptual, where the structure does not yet exist and the analyst is given reasonable leeway to define geometry, material, loads, and so on. The preeminent way of doing this nowadays is with the finite element method (FEM). The second analysis is where the structure (or a prototype) exists, and it is this particular structure that must be analyzed. Situations involving real structures and components are, by their very nature, only partially specified.
Notation for stress components
1 From Doyle , Modern Experimental Stress Analysis, John Wiley & Sons, 2004.
xxyz
yx
2. STRESS AT A POINT At a given point of interest within a body, the magnitude and direction of the resultant stress Tn depends upon the orientation of the plane passing through the point.
cos , cos , cos , 03
nx xx yx zx x
hT A A n x A n y A n z F A
where h = altitude of tetrahedron A = area of base of tetrahedron
xF = average body-force intensity in x-direction
nxT = component of resultant stress in x-direction
where A cos(n,x), A cos(n,y) and A (n,z) are the projections of the area A on the yz-, xz- and xy-planes, respectively.
Letting h 0
cos , cos , cos ,nx xx yx zxT n x n y n z 1.2a
Similarly, one can get
cos , cos , cos ,ny xy yy zyT n x n y n z 1.2b
cos , cos , cos ,nz xz yz zzT n x n y n z 1.2c
The magnitude of the resultant stress Tn is
2 2 2
n nx ny nzT T T T 1.2d
The three direction cosines of the resultant stress Tn are
n
z
Tn
x
y
cos , nxn
n
TT x
T
cos ,ny
n
n
TT y
T
cos , nzn
n
TT z
T
The normal stress n and the shearing stress n which act on the plane considered are
cos ,n n nT n T
and
sin ,n n nT n T
The angle between the resultant stress vector Tn and the vector normal to the plane n is given by
cos , cos , cos , cos , cos , cos , cos ,n n n nT n T x x n T y y n T z z n
Also
cos , cos , cos ,n nx ny nzT x n T y n T z n
and
2 2
n n nT
3. STRESS EQUATION OF EQUILIBRIUM Summation of forces in the x-direction using Fig.1a:2
0
yxxxxx xx yx yx
zxzx zx x
dx dydz dy dxdzx y
dz dxdz F dxdydzz
Dividing through by dxdydz gives
0yxxx zx
xFx y z
1.3a
Similarly, by considering the force and the stress components in the y and z directions, it can be established in a similar fashion that:
2 From Dally, J.W., Riley, W.F., Experimental Stress Analysis, (4th Edition)., College House Enterprise, 2005.
0xy yy zy
yFx y z
1.3b
0yzxz zz
zFx y z
1.3c
where Fx, Fy and Fz are body forces intensities (in N/m3) in the x, y and z directions, respectively. Equations 1.3 are the well known stress equations of equilibrium which any theoretically or experimentally obtained stress distribution must satisfy. A summation of moments about the y-axis gives
2 2 2 2
02 2 2 2
zx zxzx zx
xz xzxz xz
dz dz dz dzdxdy dxdy
z z
dx dx dx dxdydz dydz
x x
or
0zx xzdxdydz dxdydz
Hence zx xz 1.4a
Similarly, the remaining two moment equilibrium conditions with respect to the z an x axis yield:
xy yx 1.4b
and
yz zy 1.4c
The equalities given in Eqs.1.4 reduce the nine Cartesian components of stress to six independent components of the stress tensor:
xx xy zx
xy yy yz
zx yz zz
Assignment 1 (20 July 2010): Follow the above derivation to obtain the stress component for two dimensional case. Provide appropriate sketches.
4. LAWS OF STRESS TRANSFORMATION
Let n the normal vector at the inclined face, and n’ and n” are two mutually perpendicular directions with respect to n, lying on the inclined plane. The resultant stress Tn acting on the inclined surface can be resolved into the
directions n,n’ and n”, (see sketch above) to yield stresses nn, nn’ and nn” , which can easily be derived to give
cos , cos , cos ,nn nx ny nzT n x T n y T n z
' cos ', cos ', cos ',nn nx ny nzT n x T n y T n z
" cos ", cos ", cos ",nn nx ny nzT n x T n y T n z
Substitution of 1.2 and 1.4 in these equations gives
2 2 2cos , cos , cos ,
2 cos , cos , 2 cos , cos , 2 cos , cos ,
nn xx yy zz
xy yz zx
n x n y n z
n x n y n y n z n z n x
1.5a
' cos , cos ', cos , cos ', cos , cos ',
cos , cos ', cos , cos ',
cos , cos ', cos , cos ',
cos , cos ', cos , cos ',
nn xx yy zz
xy
yz
zx
n x n x n y n y n z n z
n x n y n y n x
n y n z n z n y
n z n x n x n z
1.5b
n
Tn
x
y
z
n”
n'
n
" cos , cos ", cos , cos ", cos , cos ",
cos , cos ", cos , cos ",
cos , cos ", cos , cos ",
cos , cos ", cos , cos ",
nn xx yy zz
xy
yz
zx
n x n x n y n y n z n z
n x n y n y n x
n y n z n z n y
n z n x n x n z
1.5c
Equations 1.5 can be used to determine the normal- and shear-stress components at a point expressed in any set of chosen Cartesian frame of reference if the stresses associated with one frame of reference are known.
Expressions for the stress components x’x’ , y’y’ , z’z’ , x’y’ , y’z’ and z’x’ can be obtained directly from equ. 1.5a or 1.5b by employing the following procedure.
To determine x’x’ , select a plane having an outer normal n coincident with x’. The
resultant stress Tn = Tx’ is associated with this plane. The normal stress x’x’ can be obtained directly from 1.5a by substituting x’ for n :
2 2 2
' ' cos ', cos ', cos ',
2 cos ', cos ', 2 cos ', cos ', 2 cos ', cos ',
x x xx yy zz
xy yz zx
x x x y x z
x x x y x y x z x z x x
1.6a
Following similar procedure, substitution of n with y’ or z’ will yield expressions for y’y’ ,
and z’z’ , respectively
2 2 2
' ' cos ', cos ', cos ',
2 cos ', cos ', 2 cos ', cos ', 2 cos ', cos ',
y y yy zz xx
yz zx xy
y y y z y x
y y y z y z y x y x y y
1.6b
2 2 2
' ' cos ', cos ', cos ',
2 cos ', cos ', 2 cos ', cos ', 2 cos ', cos ',
z z zz xx yy
zx xy yz
z z z x z y
z z z x z x z y z y z z
1.6c
The shear components x’y’ is obtained by selecting a plane having outer normal n coincident with x’ and the in-plane direction n’ coincident with y’.
The shear-stress x’y’ is obtained from 1.5b by substituting x’ for n and y’ for n’ :
' ' cos ', cos ', cos ', cos ', cos ', cos ',
cos ', cos ', cos ', cos ',
cos ', cos ', cos ', cos ',
cos ', cos ', cos ', cos ',
x y xx yy zz
xy
yz
zx
x x y x x y y y x z y z
x x y y x y y x
x y y z x z y y
x z y x x x y z
1.6d
' ' cos ', cos ', cos ', cos ', cos ', cos ',
cos ', cos ', cos ', cos ',
cos ', cos ', cos ', cos ',
cos ', cos ', cos ', cos ',
y z yy zz xx
yz
zx
xy
y y z y y z z z y x z x
y y z z y z z y
y z z x y x z z
y x z y y y z x
1.6e
' ' cos ', cos ', cos ', cos ', cos ', cos ',
cos ', cos ', cos ', cos ',
cos ', cos ', cos ', cos ',
cos ', cos ', cos ', cos ',
z x zz xx yy
zx
xy
yz
z z x z z x x x z y x y
z z x x z x x z
z x x y z y x x
z y x z z z x y
1.6f
These six equations permit the six Cartesian components of stress relative to the Oxyz coordinate system to be transformed into a different set of six Cartesian components of stress relative to an O’x’y’z’ coordinate system. 5. PRINCIPAL STRESSES
If n is selected to coincide with Tn then the plane defined by n is known as the principal plane. The direction given by n is the principal direction, and the normal stress acting on this plane is a principal stress.
P
cos ,nx nT n x
cos ,ny nT n y a
cos ,nz nT n z
Substitute eq.1.2 into a, and get
cos , cos , cos , cos ,xx yx zx nn x n y n z n x
cos , cos , cos , cos ,xy yy zy nn x n y n z n y b
cos , cos , cos , cos ,xz yx zz nn x n y n z n z
Rearrangement gives
cos , cos , cos , 0xx n yx zxn x n y n z
cos , cos , cos , 0xy yy n zxn x n y n z e
cos , cos , cos , 0xz yz zz nn x n y n z
Solving for the direction cosines, say cos(n,x):
0
0
0cos ,
yx zx
yy n zy
yz zz n
xx n yx zx
yx yy n zy
xz yz zz n
n x
e
Nontrivial solutions for direction cosines of the principal plane exist only if the determinant of the denominator above is zero:
0
xx n yx zx
yx yy n zy
xz yz zz n
Expanding the determinant gives
3 2
2 2 2
2 2 2 2 0
n xx yy zz n
xx yy yy zz zz xx xy yz zx n
xx yy zz xx yz yy zx zz xy xy yz zx
1.7
The roots of this cubic equation are the three principal stresses. Substitution of the six Cartesian components of stress into the equation will yield three
real roots of n .
Three possible solutions:
1. If 1 . 2 . and 3 are distinct, then n1,n2 and n3 are unique and mutually perpendicular.
2. If 1 = 2 ≠ 3 , then n3 is unique, and every direction perpendicular to n3 is a
principal direction associated with 1 = 2 .
3. If 1 = 2 = 3 , then a hydrostatic state on stress exists and every direction is a principal direction.
Once the three principal stresses have been obtained from equ. 1.7, they can be substituted individually into equs. C to give three sets of simultaneous equations. Note also that the following relationship for the direction cosines prevails3
2 2 2cos , cos , cos , 1n x n y n z f
Equs. (c) along with Equ. (f) can be solved to give the three sets of direction cosines defining the principal planes.
It is very useful to order the principal stresses so that 1 2 3 . The tensile stresses
are considered positive and the compressive stresses are considered negative. Stress Invariants
State of stress can be defined by its six Cartesian stress components with respect to any arbitrary Cartesian coordinate system Oxyz. The relationships between these stresses expressed in one Cartesian coordinate system to another is given by Equ. (1.6). Three other relations exist which are called the three invariants of stress.
3 Prove equ. f (Assignement 2- 20 July 2010)
n
Tn
To appreciate such notion, consider equ. 1.7 which is the cubic equation for the
principal stresses 1, 2, and 3 as the three roots of n in equ. (1.7)
3 2 2 2 2
2 2 2 2 0
n xx yy zz n xx yy yy zz zz xx xy yz zx n
xx yy zz xx yz yy zx zz xy xy yz zx
1.7
The principal stresses 1, 2, and 3 for any state of stress at a point in a structure should be independent of the Cartesian coordinate system chosen!. Therefore, the coefficients of Equ. (1.7) which contains the Cartesian components of the stresses must also be independent or invariant to the coordinate system. Hence, from Equ. (1.7) it is clear that:
1 ' ' ' ' ' 'xx yy zz x x y y z zI
2 2 2
2
2 2 2
' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ''
xx yy yy zz zz xx xy yz zx
x x y y y y z z z z x x x y y z z x
I
1.8
2 2 2
3
2 2 2
' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' '
2
2
xx yy zz xx yz yy zx zz xy xy yz zx
x x y y z z x x y z y y z x z z x y x y y z z x
I
I1, I2 and I3 are known as the first, second and third invariants of stress, respectively. If
the Oxyz coordinate system is chosen to be coincident with the principal stresses 1, 2,
and 3 , Equs. 1.8 reduce to
1 1 2 3I
2 1 2 2 3 3 1I 1.9
3 1 2 3I
Assignment 3 (20 July 2010)
Simplify (Rederive) the above derivation for the three stress invariants for two-dimensional case
6. MOHR’S CIRCLE
For two dimensional stress fields where 0zz zx yz , z’ is coincident with z, and
the angle between x and x’ is , Equs. (1.6a) to (1.6f) reduce to
2 2
' ' cos sin 2 sin cos
cos 2 sin 22 2
x x xx yy xy
xx yy xx yy
xy
1.11a
2 2
' ' cos sin 2 sin cos
cos 2 sin 22 2
y y yy xx xy
xx yy xx yy
xy
1.11b
2 2
' ' cos sin cos sin cos sin
sin 2 cos 22
x y xx yy xy
yy xx
xy
1.11c
' ' ' ' ' ' 0z z z x y z 1.11d
This relationship can graphically be represented by Mohr’s circle.
Mohr’s circle is a geometric representation of the 2-D transformation of stresses and is very useful to perform quick and efficient estimations, checks of more extensive work, and other such uses. CONSTRUCTION: Given the following state of stress:
with the definition (by Mohr) of positive and negative shear: “Positive shear would cause a clockwise rotation of the infinitesimal element about the element center.” Thus, from the illustration above, σ12 is plotted negative on Mohr's circle, and σ21 is plotted positive on Mohr's circle. (Note: a similar formulation can be used for tensorial strain)
Then complete the circle by doing Step 4: 4. Draw a circle of diameter of the line AB about the point where the line AB crosses the horizontal axis (denote this as point C)
xx
yy
xy
yx
xx
yy
xy
yx
Begin the construction by doing the following: 1. Plot σ11, - σ12 as point A 2. Plot σ22, σ21 as point B 3. Connect A and B
USE OF THE CONSTRUCTION To read off stresses for a rotated system: 1. Note that the vertical axis is the shear stress axis and the horizontal axis is the extensional stress axis. 2. Positive rotations are measured counterclockwise as referenced to the original system and thus to the line AB.
3. Rotate line AB about point C by the angle 2θ where θ is the angle between the unrotated and rotated systems. 4. The points D and E where the rotated line intersects the circle are used to read off the
stresses in the rotated system. The vertical location of D is - 12 ; the horizontal location of D is
11 . The vertical location of E is σ21 , the horizontal location of E is 22 (Recall Mohr definition
with regard to negative/positive sense of shear stress on Mohr's circle).
We can immediately see the following: 5. The principal stresses, σI and σII , are defined by the points F and G (along the horizontal axis where σ12 = 0). The rotation angle to the principal axis is θp which is 1/2 the angle from the line AB to the horizontal line FG. 6. The maximum shear stress is defined by the points H and H’ which are the endpoints of the vertical line. The line is orthogonal to the principal stress line and thus the maximum shear stress acts along a plane 45° (= 90°/2) from the principal stress system.
Full two-dimensional stress transformation equations (θ as on figure on p.13):
F G F G
H’
H
but this does yield the same set of operating equations! 7. SPECIAL STATES OF STRESS Two states of stress occur very frequently in practice that they have been classified. They are: 1. Pure Shearing Stress 2. Hydrostatic state of stress 1. Pure Shearing Stress
In a state of pure sthear stress in one particular set of axes Oxyz, xx= , yy= zz= 0. It can be shown that this particular set of axes Oxyz exists if and only if the first invariant of stress, I1 = 0. Two of the infinite number of arrays which represent a state of pure shearing stress are:
0
0
0
xy xz
xy yz
xz yz
or
0
xx xy xz
xy yy xx yz
xz yz
It can be shown that the second form can be converted to the first form by a suitable rotation. 2. Hydrostatic state of stress
A state of stress is hydrostatic if xx= yy= zz= -p , and all the shearing stresses are zero.
0 0
0 0
0 0
p
p
p
3. General state of stress in terms of pure shear and hydrostatic state of stress One important property of these two states of strss is that they can be combined to form a general state of stress.
0 0
0 0
0 0
xx xy xz xx xy xz
xy yy yz xy yy yz
xz yz zz xz yz zz
p p
p p
p p
1.17
Or: General state of stress = hydrostatic state of stress + state of pure shearing stress However, the third array represents a state of pure shear stress if and only if its first stress invariant is zero, i.e.