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Section 2: STRONG FORMULATIONWashkewicz College of Engineering
Concept of Stress at a Point
Consider a point within an arbitrarily loaded deformable body
Define
AF
StressNormal
n
Alim
0
AF
StressShear
S
Alim
0
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
x
y
z
xy
xz
yx
zx
yz
zy
xy
xzx
y
x
y
zyx
yz
zx
zy
z
With the previous two definitions for stress we are in a position to represent the state of stress at a point graphically using this differential volume:
We take our deformable body and virtually slice around a point creating a differential volume represented by a cube. The cube has faces perpendicular to the three coordinate axes.
Six faces, three components of stress per face (= 18?). The state of stress at a point can be represented by nine stress components, six of which are independent as we will see shortly.
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
Initially let’s assume that body forces are minimal relative to the loads on the component generating the state of stress at the point we are investigating. The previous figure reflects this. Summing moments about an axis through the centroid of the x-face and parallel to the x-axis yields
Similar summations of moments about the an axis through the centroid of the y-face and parallel to the y-axis, as well as an axis through the centroid of the z-face and parallel to the z-axis yields
yzzy
yzzy
x
dydzdxdzdydx
Mc
0
xzzx
xyyx
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
If the stress components on opposite faces are equal and opposite (18 --> 9), then these equalities among the shear stresses allows us to represent the state of stress at a point with a 3 x 3 symmetric matrix, i.e.,
Recall from linear algebra that the eigenvalues of symmetric matrices are all real. From undergraduate strength of materials we know that the eigenvalues of the matrix above are the principle stresses at that point.
zzyzxz
yzyyxy
xzxyxx
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
Previously we assumed that the stress components do not change from one face of the differential cube representing stress at a point to the corresponding face on the other sided of the differential cube.
We now relax this assumption and allow stress gradients to exist. Consider the tapered bar subjected to an axial force depicted in the figure to the left. Axial stresses change along the length. Body forces are still neglected.
In order for the axial stress to change from one cross section to the next along the x-axis, then
Differential Equations of Motion
0
xx
x P
1
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
x
y
z
Bx
xy
xzx dx
xx
x
dxxxy
xy
dxxxz
xz
In general we could allow for body forces as is shown in the figure.
Focusing on the x-faces of the differential volume representing the state of stress at a point we would see the following
In general we design components so that a stress gradient
is small. There are situations where this is not possible, e.g., at sharp corners. Here stress values change rapidly and intensify.
xx
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
x
y
z
By
yxyz
y
dyy
yy
dyyyx
yx
dyyyz
yz
The stresses acting on the y-faces of the differential cube representing the state of stress at a point is portrayed in the figure to the right.
A body force, By, is shown acting in the y-direction
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
x
y
zBz
zyzx
z
dzz
zz
dzzzy
zy
dzzzx
zx
And the z-faces of the differential cube representing the state of stress at a point generally appears as follows
A body force, Bz, is shown acting in the z-direction
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
Summing forces in the x-direction using the last three figures yields
xyxzxx
xyxzxx
xyx
yxyx
zxzxzx
xxx
x
Byzx
dzdydxByzx
dzdydxBdzdxdyy
dzdx
dydxdzz
dydx
dzdydxx
dzdy
F
0
0
0
x
y
z
xF
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
Similar expressions are obtained by summing forces in the y-direction and the z-direction
The body forces Bx, By and Bz are per unit volume quantities. From force equilibrium we now have a system of three coupled partial differential equations in six unknown stress components at a point in a component..
zzyzxz
z
yyzyxy
y
Bzyx
F
Bzyx
F
0
0
zzyzxz
yyzyxy
xyxzxx
Bzyx
Bzyx
Byzx
0
0
0
x
y
z
x
y
z
yF
zF
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
Six unknowns and three equations is an underdetermined (inconsistent) system of equations. So we search for more equations to use in solving for stress at a point a solid mechanics problem. Let’s introduce six equations stipulating the relationship between stress and strain (Hooke’s Law).
However this introduces six more unknowns, i.e., the components of the strain tensor. So now we now have nine equations and twelve unknowns. We do not appear to be gaining ground here. Since we have looked into the concept of stress in detail, it is time to do the same with strain.
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xzxzzxyy
xyxyzyxx
EE
EE
EE
11
11
11
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
x
y
z
dsy
xz
P Q
The Differential Definition of Strain
We define strain by monitoring the effect loads and boundary conditions have on virtual lines segments arbitrarily assigned to positions within a component. That position is considered a “point” because the line segments have only a differential length.
Consider the points P(x, y, z) and Q(x+ds, y, z) that both lie on a line parallel to the x-axis.
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
x
y
z
ds* > dsy*
x*z*
P*Q*
As the component deforms the original points P(x, y, z) and Q(x+ds, y, z) move to new positions defined by P*(x*, y*, z*) and Q*(x*+dx*, y*+dy*, z*+dz*)
x* + dx*
z* + dz*
y* + dy*
The initial and final positions of points P and Q can be used to define the axial strain at a point, x, in terms of the change in length of the differential line segment.
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
The initial squared length of the differential line segment PQ is
The axial strain in the x-direction is defined as the change in length over the original length and we manipulate that relationship as follows
2
2222 00
dx
dxds
1*
1*
*
x
x
dsds
dsds
dsdsds
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
Squaring both sides of the last expression yields
If we designate u, v and w (three more unknowns – the list grows longer) as the displacements at the point we are trying to characterize strain, then the following relationships can be easily developed from the geometry in the previous figures
The relationships above underscores that the three new unknowns, i.e., the displacements, are functions of position.
222 1* dsds x
wzz
vyy
uxx
*
*
*
zyxxxzyxww
zyxxxzyxvv
zyxxxzyxuu
,,**,,
,,**,,
,,**,,
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
Since x* = x*(x, y, z) then by the chain rule of calculus
With x* a function of position, then recall that from the definition of the line segment PQ
Thus
dzz
xdyy
xdxx
xdx
****
00
dzdy
dxxu
dxx
ux
dxx
xdx
1
**
The partial derivative of u with respect to x represents the change in the displacement u with respect to x(y and z are held constant). This is known as a displacement gradient.
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
Similarly
where it is assumed
This implies the coordinate variables defining the initiating point for line segment PQ are independent variables, i.e.,
dxxv
dxx
vy
dxx
y
dzz
ydyy
ydxx
ydy
*
****
0xy
xfy
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
Since the choice of the location of line segment PQ is arbitrary (but the initial direction is not) the assumption of independence is reasonable. Similarly
By Pythagorean theorem’s the squared length
dxxw
dxx
wz
dxx
z
dzz
zdyy
zdxx
zdz
*
****
ds
dzdydxds
x2
2222
1
****
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
Substituting dx*, dy* and dz* from above yields
We also note that
since in the original figure
222
2 1*
dxxwdx
xvdx
xuds
22
222
1
1*
dx
dsds
x
x
dxds *
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
Thus
and from this
22
2222
2222
1
1
1*
dx
dxxw
xv
xu
dxxwdx
xvdx
xuds
x
222
2
2222
112
11
xw
xv
xu
xw
xv
xu
xx
x
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
Strains are quite small in most engineering applications. Taking a quantity that is small and squaring it produces something even smaller. So if terms quadratic in strain (left side of the equal sign) are neglected
If displacement gradients are small, then squaring them will be even smaller. Neglecting quadratic displacement gradient terms yields
Note that if strains and displacement gradients are not small none of what appears on this page can be implemented.
222
222
222
21
1212
112
xw
xv
xu
xu
xw
xv
xu
xu
xw
xv
xu
x
x
x
xu
x
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
In a similar fashion
If displacement gradients are small
These are the normal strains. The next step is defining shear strains. Normal strains correspond to changes in length. Shear strains correspond to changes in angles. For angles we need to differential line segments to define an angle.
222
222
21
21
zw
zv
zu
zw
yw
yv
yu
yv
z
y
zwyv
z
y
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
Now consider two differential line segments. Both originally lie in an arbitrary x-y plane contained within the deformable body and are initially perpendicular to one another.
x
y
z
ds1y
xz
ds2The intersection of the two line segments defines our “point” in the component.
Initially differential line segments ds1 and ds2 are “unit” lengths.
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
If the line segments deform such that their lengths are unchanged (ds1 = ds1* and ds2= ds2*) but their orientations in Cartesian three space are altered, then the original line segments could assume the following locations
x
y
z
ds1*y*
x*z*
ds2*12
x* + dx1*
z* + dz1*
y* + dy1*
y* + dy2*
z* + dz2*x* + dx2*
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
xw
xv
xu
dxxwdx
xvdx
xu
dzdydxds
,,1
1,1,11
**,*,* 1111
yw
yv
yu
dyywdy
yvdy
yu
dzdydxds
,1,
1,11,1
**,*,* 2222
Earlier the chain rule from Calculus was employed, and now applying this rule to ds1*yields the following vector components
when we treat the line segment ds1* as a positional vector. Similarly
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
Next a dot product is formed from the two positional vector that yields the following scalar value which equal to the angle formed between the two vectors, i.e.,
After introducing a third line segment ds3* parallel to the z-axis, then derivations similar to the one above yield
xy
yw
xw
yv
xv
yu
xu
xv
yu
yw
xw
yv
yv
yu
xu
dsds
11
**cos 2112
zw
yw
zv
yv
zu
yu
yw
zv
zw
xw
zv
xv
zu
xu
xw
zu
yz
xz
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
Under the assumption of small deformation gradients
to go along with
We have six equations in terms of three additional unknowns, i.e., the displacements u, v, and w.
Viola! We have 15 equations in terms of 15 unknowns when the stress components embedded in the force equilibrium are included with the linear relationships between stress and strain. A solvable system.
yw
zv
xw
zu
xv
yu
yz
xz
xy
zw
yv
yu
zyx
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
zzyzxz
yyzyxy
xyxzxx
Bzyx
Bzyx
Byzx
0
0
0
yzyzyxzz
xzxzzxyy
xyxyzyxx
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11
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yw
zv
zw
xw
zu
yv
xv
yu
yu
yzz
xzy
xyx
Hooke’s Law – 6 equations, 12 unknowns
Force Equilibrium – 3 equations
Strain-Displacement Relationships- 6 equations, 3 unknowns
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
These equations must hold at every point in the component for a valid solution. When boundary conditions are added (force and/or displacement boundary conditions) these 9 linear, somewhat non-homogeneous (no body forces and then all the equations are homogeneous) linear, partial differential equations must be solved along with the 6 algebraic equations stipulating the relationship between stress and strain. This is known as the strong formulationof the problem.
Solving partial differential equations is not as easy as solving ordinary differential equations. You are invited to take a graduate course on the topic. Solving a system of partial differential equations at every point in a component is even more difficult.
A cautionary note: finite element analysis does not solve the strong formulation of the problem using numerical methods.
As we will see, a “weak formulation” for the solid mechanics problem is posed and finite element analysis solves this formulation at select points (nodes) throughout the component. Everywhere other than these select points the solution is an approximation.
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
Navier’s Equations
The previous section of notes demonstrated that the strong formulation of a solid mechanics problem involved solved 15 equations in 15 unknowns. In a course in Elasticity (CVE 604, MME 604) or Continuum Mechanics you will learn how to compress the number of equations to three, and these equations will be in terms of the unknown displacements u, v, and w.
The three equations are known as Navier’s problem in partial differential equations
Solution of this system is still daunting. Boundary conditions for this system of equations?
32
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2
2
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2
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2
2
2
2
2
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2
2
2
2
2
)(0
)(0
)(0
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wy
wx
wzw
yv
xu
z
Bz
vy
vx
vzw
yv
xu
y
Bz
uy
ux
uzw
yv
xu
x
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
Strong Formulation – Closing RemarksCommonly in engineering problems the governing partial differential equations contain derivatives up to and including the second order, i.e., the most general form in two dimensions is of the form
Note the first order and zero order derivatives to the right of the equal sign. This notation is a mathematics artifact. Equation formulation are classified as follows
B2 – AC < 0 EllipticB2 – AC = 0 ParabolicB2 – AC > 0 Hyperbolic
A wide range of problems from elasticity, acoustics, atmospheric science and hydraulics are governed by hyperbolic partial differential field equations.
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xuuyx
yuyxC
yxuyxB
xuyxA
,,,,
),(),(2),( 2
22
2
2
Section 2: STRONG FORMULATIONWashkewicz College of Engineering
Parabolic equations arise in heat conduction problems.
The strong formulation places continuity requirements on the dependent field variables –in our case the displacement field. Whatever function is developed for the solution to the hyperbolic partial differential equation(s) the solution must be continuously differentiable through second order derivatives. In finite element analysis this is identified as C2
continuity.
As noted earlier obtaining exact solutions, or any solutions, for a strong formulation of a solid mechanics problem can be a difficult task. One can employ finite difference methods to solve a system of equations of the strong form and obtain an approximate solution. However, this approach only works well for problems with simple and regular geometry and boundary conditions.
Since most solid mechanics problems do not fit into this scheme most engineering problems are solved using finite element methods applied to a weak formulation of the problem. The weak form can be obtained through energy principles. This approach fits into concept of variational methods.