Normal Stress Shear Stress - Cleveland State University · If the stress components on opposite...

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


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.


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


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


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


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


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


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


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


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


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.

yzyzyxzz

xzxzzxyy

xyxyzyxx

EE

EE

EE

11

11

11


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.


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.


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


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

,,**,,

,,**,,

,,**,,


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.


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


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

****


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 *


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


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


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


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.


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*


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


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


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


zzyzxz

yyzyxy

xyxzxx

Bzyx

Bzyx

Byzx

0

0

0

yzyzyxzz

xzxzzxyy

xyxyzyxx

EE

EE

EE

11

11

11

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


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.


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

2

2

2

2

2

22

2

2

2

2

2

12

2

2

2

2

2

)(0

)(0

)(0

Bz

wy

wx

wzw

yv

xu

z

Bz

vy

vx

vzw

yv

xu

y

Bz

uy

ux

uzw

yv

xu

x


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.

yu

xuuyx

yuyxC

yxuyxB

xuyxA

,,,,

),(),(2),( 2

22

2

2


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.

Normal Stress Shear Stress - Cleveland State University · If the stress components on opposite...

Documents

Transcript of Normal Stress Shear Stress - Cleveland State University · If the stress components on opposite...