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ME362: Stress Analysis

INTRODUCTION TO THE MECHANICS OF DEFORMABLE BODIES

Sridhar Krishnaswamy

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ME362: Stress Analysis

Introduction

1. INTRODUCTION When forces are applied to a body (solid, liquid or gas), internal forces are set up in the body and it deforms and/or moves. The object of our study here is to understand how the applied load relates to these internal forces and deformation/motion of the body. Why do we need to know these things? Let us take an example. Let us say you are asked to design an antenna for a space application. You are required to design the antenna to a specific shape so that a satellite can communicate with earth from the outer edges of the solar system. So you take special care and produce (after spending millions of dollars) a beautiful antenna which has on earth the exact shape that it should have in space. Then you send it up into space and it does not work! The satellite's messages are being beamed to the Oort cloud instead of to Houston. What have you done wrong? You have forgotten a very simple thing --- gravity! A body on earth is always subject to the earth's gravitational attraction. The antenna was perfectly shaped under earth's gravitational force, but in space (or on any other planet) when it is under either zero-g or someother-g, it has quite a different shape! So what happens next? You are fired, of course... Anyway, there are other less esoteric reasons for us to understand the mechanics of deformable bodies and I am sure you can think of hundreds of them. Figure 1 lists a few examples. So, granting that we are embarked on an important mission of discovery and all that, how exactly are we going to characterize the internal forces and deformation of a body? Continuum Assumption: When external forces are applied to a solid body, the atoms or molecules in the body may move apart a little bit from each other. The reason the atoms hopefully do not completely come apart (if the load is sufficiently small) is because they resist the applied external forces by developing internal forces until equilibrium is achieved. If the internal forces cannot resist the external forces, the body breaks or fractures. Though internal forces are due to the atoms or molecules inside a body, it is too complex to study mechanical deformation from the atomistic point of view (even though some people make a living doing so). Therefore, we will now adopt the first of our many approximations, namely the so-called continuum assumption. Under this assumption we can forget about the details of the atomic structure of the solid, and instead treat the solid (equivalently, a fluid) as if it were one continuous thing, whatever that means. In essence, our theory will hold only for length scales which are much larger than atomic distances.

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ME362: Stress Analysis

Introduction

·

stress analysis of dental implants

FIGURE 1: EXAMPLES OF DEFORMATION OF STRUCTURES Sridhar Krishnaswamy 3

ME362: Stress Analysis 2. HOOKE'S EXPERIMENT

Hooke's Experiment

Consider a straight rod of length L and uniform cross-sectional area A made of some material. Let one end of the rod be pinned to a wall, and the other end be subjected to an applied load F. Because of the applied load F, the rod is found to stretch by an amount . If we were to increase the load F, the stretch is found to increase as well. In fact, if we plot the stretch of the rod against the force applied, we

L

F

fracture

proportional limit

F

find a plot that looks something like this:

This experimental fact was discovered by Robert Hooke in 1678. Not being quite sure that he was onto something good, but still wanting to establish priority, he stated his discovery -- "Hooke's law" -- in the form of an anagram: ceiiinosssttuv. This was not of much help to others who did not know what he was talking about. When someone unscrambled the anagram it turned out to be: ut tensio sic vis. I am told that it loosely translates to: force is in proportion to the extension. That is, an axial rod behaves pretty much like a linear spring! Turns out that what Hooke found is quite true but we need to fix it a little bit for it to be of use to us. Problems with Hooke's law as stated by Hooke: · True only for a certain class of materials: for instance, it is true for most engineering materials (steel, aluminum etc), but is not really accurate for rods made out of animal tissue for instance. Materials for which Hooke's law holds are called linear elastic materials.

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ME362: Stress Analysis

Hooke's Experiment

·

True only upto a point: beyond a certain amount of stretch the force is no longer proportional to the stretch, but may be non-linearly related, and at some point the rod will break (fracture). · Cannot distinguish between

F geometry 1 geometry 2 same material

F

material 1 material 2 same geometry

material and geometric effects: That is if we did two sets of experiments where in one set we test several rods of the same material but different geometry (cross sectional area A and length L), and in the other we test several rods of the same geometry but made from different materials, we get different proportionality constants for the force-stretch relation. So we recognize that the force is proportional to the stretch, but the proportionality constant depends both on the material and the geometry. In order to separate the effect of geometry and material, we scale out the geometric effect by defining:

Stress: = F/A = (force) / (cross-sectional area) Strain: = /L = (stretch) / (original length)

[N.m-2] [dimensionless]

Then, if we were to plot stress vs strain, we find:

material 1 (independent of A,l) material 2 (independent of A,l)

That is, the stress is proportional to the strain in axial stretching of a rod, and the proportionality constant is just a material property. Amended version of Hooke's law: Stress is proportional to strain for a certain class of materials and for small deformations:

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ME362: Stress Analysis

Hooke's Experiment

=E where E is called the Young's modulus [N.m-2], and it is a property of the material. It will have different values for different materials.

The Deeper Meaning of Stress and Strain: Both stress and strain have a deeper meaning, and have significant character! Stress turns out to be a useful measure of the intensity with which the atoms or molecules of a material resist the applied load. The reason that one part of the rod does not break away (hopefully) from the other is because of these resisting forces that arise from atomic / molecular forces. Consider two rods A and B made of the same material but A is fatter than B. Since, a fatter rod has more atoms/molecules across which to spread out the force that is needed to resist the applied load, for a given force F, rod A is stressed less than B. However, the F F F maximum stress at which a fat rod and a F thin rod made of the same material will break - or go non-linear or "plastic" will be the same and depends only on the material. Strain also has a deeper meaning. Rather than merely measuring the total elongation of the rod, it characterizes the change in length of any "local" line segment. That is, if we were to draw a line of length 'l' axially as shown on the rod before applying the load, and then find that the line segment elongates by l due to the load, we find that for this l L uniform rod we have: = l / l = / L, and is independent of length 'l'.

l

These ideas will crystallize further when we look at the following examples.

F

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ME362: Stress Analysis

Axial Stretching of Rods

3. AXIAL STRETCHING OF RODS

Case 1: Uniform Rod in Tension · · ·

L

uniform cross-sectional area A, original length L Neglect weight of rod itself Young's modulus E

F

Since the stress: = E --> P/A = E /L we have: P = [AE/L] or = PL/AE

where AE/L is called the axial stiffness (and is the equivalent of the spring constant for a linear spring).

Remarks (i)If the force F were to act into the rod, it is called compressive, and the rod shrinks or compresses. Compression can be thought of as opposite (negative) of tension. For most materials, the amount of rod compression is proportional to the applied compressive force, and the proportionality constant is the same as that of tension. Therefore, the above axial-force vs stretch relation can be used in both compression and tension, where we treat negative forces and negative stretches as meaning compression..

(ii) Note that a uniform rod under axial load is like a linear spring which exerts a force that is proportional to its stretch from its unstretched state :F = k where k [N/m] is called the spring constant of the spring.

F

Case 2: 2-Stepped Rods:

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ME362: Stress Analysis

Axial Stretching of Rods

· ·

A1, L1 1 A2, L2 2

Rod of two segments of length L1 and L2 and cross-sectional area A1 and A2 made of material whose Young's modulus is E1 and E2 respectively. Neglect weight of the rod itself. Determine total stretch of the rod due to the applied load F.

By making a cut in each segment, we note that the force in each segment in this case is F. But the stresses and stretches are different. stretch of segment 1 is 1 = FL1/A1E1 stretch of segment 2 is 2 = FL2/A2E2

F

Total stretch of the stepped rod is: = 1 + 2= FL1/A1E1 + FL2/A2E2

Remark: In this case, the forces on each segment were the same. But this need not be the case if an additional force were to act at, say, just under segment 1. In this case, = 1 + 2= P1L1/A1E1 +P2L2/A2E2

P 1 = 1 A1 = F 1 + F 2

A1 , L 1 1

F1

A1 , L 1 1

F1

P 2 = 2 A2 = F 2

A2 , L 2 2

2

A2 , L 2 2

F2

F2

F2

Case 3: N-Stepped Rods: We can easily extend the above to a rod with N-steps.

=

AE

i =1 i

N

Pi Li

i

where Pi is the net axial force (also called the net internal force) on the ith segment whose length is Li, cross-sectional area is Ai, and is made of a material whose Young's modulus is Ei. The idea is that we treat each segment of the rod as a uniform rod over which the net cross-sectional force as well as the area are constant. Case 4: Rods with Continuously Varying Cross-Section and/or Loading:

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ME362: Stress Analysis

Axial Stretching of Rods

Next, let us consider the case of a rod with varying cross-section such as the conical rod shown. By making cuts at several locations along the stalactite, convince yourself that in this case, even though the net cross sectional force is the same everywhere, the stress at each cross-section of the rod is not the same. Or else consider a rod with uniform crosssection but whose weight is not negligible. In this case, the net cross-sectional force is different at different locations.

x model as

x

F

F

It is possible for us to approximate the continuously varying rod as being made up of N segments each of which is uniform and is of length x. The stretch of any segment is just: P = x . EA

Then the total stretch of the rod is just the sum of the stretches of each segment. The approximation becomes exact as we shrink the segment lengths x --> 0. In this limit, total stretch is just L P P = = x dx as x 0 EA all segments all segments EA 0 {Recall the meaning of an integral as a sum}

Example 1: Stretching of a Uniform Rod Under its Own Weight:

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ME362: Stress Analysis ·

L

Axial Stretching of Rods Rod with uniform cross sectional area A and length L Density of material is . Young's modulus of material is E. The only load is due to its own weight.

x

P (x ) =

n et cro s s s e c tio n a l f o r c e

· · ·

x

W

=

{ A

x }

g

Locate the origin of the x axis at the

bottom as shown.

To get the net cross-sectional force at any cross-section, imagine making a "cut" at a distance x from the bottom and look at the part below the cut. The net cross-sectional force P(x) due to the internal forces must balance the weight of the chunk of material below it. So P(x) = gAx. The total stretch of the rod under its own weight is therefore: L L P(x ) gAx gL2 = dx = dx = . EA EA 2E o o

Sridhar Krishnaswamy

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ME362: Stress Analysis Example 2: Consider the two trusses shown in the figures:

B C B

Axial Stretching of Rods

D

C

A

A

P

P

Case a

Case b

We can solve for the forces carried by the truss members in case (a) from conditions for static equilibrium - all you have to do is consider the equilibrium of the pin at A. There are two unknown forces in the rods AB and AC which can be solved for in terms of the applied load P and the angle . You have done things like this in CE-B12 or Physics courses and you should right now do this on your own to see if you remember how! However, static equilibrium considerations that you learnt in B12 are insufficient to obtain the forces carried by truss (b)! This is because we now have unknown forces in the three rods AB, AC and AD, but we still have only two useful equations from static equilibrium (the moment equilibrium equation is trivially satisfied.) Problems such as this are called statically indeterminate. Turns out, the only way we can get the forces in the members of truss (b) is if we look into the deformation suffered by the members of the truss due to the applied loads. That is, we can no longer afford to neglect the deformation of structures - as you did in B12 by assuming that structures were rigid. Now back to the problem. Given that all three rods are linear elastic of Young's modulus E, and cross-sectional area A, and that the rods AB and AC are of length L, and rod AD is vertical, determine the forces carried by the three rods due to the applied load P at A. (Neglect the weight of the rods). From the equilibrium of the pin at A, we get Fx = 0 P1 = P2 : Fy = 0 P1 cos + P2 cos = P which are two equations for three unknowns (insufficient). Problem is statically indeterminate. Need to look for an additional condition. Requiring that the three rods not break apart, we find that the stretches of the rods are not independent but are related.

Sridhar Krishnaswamy

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ME362: Stress Analysis

Axial Stretching of Rods

We recognize that from the symmetry of the problem (rods AB and AC are identical), rod AD stretches by AD such that it is still vertical. Rods AB and AC then must also stretch appropriately. If we assume that the deformations are small compared to the lengths of the rods, then, we find from the figure (shown grossly exaggerated) that the stretches are approximately related through:

B

D

C

A

A

AD

A'

AB

' ' A'

AC

AB = AC = AD cos ' AD cos

where again the assumption of small deformation allows us to

P

say that the angle ' is approximately the same as .

Recasting this in terms of the forces, we have: P L P2 L P3 (L cos ) 1 = = cos AE AE AE which gives us the additional restriction needed to solve for the forces in the rods. P cos2 ; P = P2 = 1 1 + 2cos 3 P3 = P . 1+ 2cos3

Remark: If you study what we have just done carefully, you will notice that there are three things needed to solving these problems. We impose (a) equilibrium (b) compatibility - or the geometric constraints (c) the material response (Hooke's law) In fact, to determine the forces (stresses) and stretches (displacements) of all structures, we follow exactly the same procedure. An N-stepped rod or a system of N linear springs connected along a line can be analyzed by the process above. Trusses, which are made of rods but now in two or three-dimensions can also be analyzed in exactly the same way, except that now the compatibility (geometry) part can become somewhat complicated.

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ME362: Stress Analysis

Multi-axial Stress and Strain

4. MULTI-AXIAL STATE OF STRESS AND STRAIN Lateral Contraction of Rods Under Axial Stretch:

When a rod is axially under tension, it stretches. The axial strain is given by: axial = F/AE where A is the cross-sectional area and E is the Young's modulus. In addition to the axial stretching, it is found that the rod also laterally contracts. If we were to draw a line segment of length llateral prior to applying the axial load, we find that after deformation, L this line segment contracts by an amount lateral. This leads to a lateral strain lateral = lateral/llateral What is more, for linear elastic rods, it is experimentally found that the lateral strain and the axial strain are linearly related through: lateral= - axial , where is called the Poisson's ratio, a property of the material. F Most materials have a Poisson's ratio between 0 and 0.5. The negative sign says that when the rod is in axial tension it compresses laterally and when the rod is in axial compression it expands laterally.

Biaxial Stress in a Thin-walled Cylindrical Pressure Vessel:

t r

Consider a thin-walled (thickness t << radius r) cylindrical pressure vessel with flat-end caps (unrealistic), containing gas under pressure p. The cylinder does not blow apart (hopefully) because of internal forces. These can be characterized by their intensities, namely the stresses.

L

First, imagine making a lateral cut as shown. The forces due to the pressure of the gas on the cylindrical surfaces balance out by themselves. However, the pressure of the gas on the flat end cap must be balanced by the internal forces on the thin rim on the cut face. That is: 1 p ( r2) = 1 (2 r t) That is, 1 = pr / 2t and this acts axially as shown.

2 L

2

2 1 1 Sridhar Krishnaswamy 2

Next, consider making a longitudinal cut as shown. Internal "hoop" stresses 2 on the cut rim must counter the pressure of the gas that wants to blow the top half away. That is;

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ME362: Stress Analysis

Multi-axial Stress and Strain

2 {t L}(2) = p {l r d }sin = p (2rL ) ==> 2 = pr / t

0

The sides of the pressure vessel are said to be in a state of biaxial stress. Shear Stress: Consider the lap joint shown. This consists of a central plate glued to two side plates as shown.

P/2 P P/2

The force P from the central plate is transferred to the side plates by means of shear internal forces which act parallel to D C the contact surface (ABCD on the top face and a similar region on the P B A bottom). We define an average shear stress: = P / (2AABCD) where AABCD is the area of the lap joint surface, and the 2 is because we have the top and the bottom surfaces carrying this shear stress.

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ME362: Stress Analysis

Multi-axial Stress and Strain

Shear and Normal Stress: Consider a short stud embedded in a wall. A cable is attached to the stud and it carries a load P through the pulley arrangement shown. Assuming that the pulley is frictionless, this means that the stud is subjected to a load P making an angle as shown. Imagine making a cut a shown. To balance the applied load P, both normal stresses and shear stresses must develop. On average, they are: P P cos normal stress: = n = A A Pt Psin = shear stress: = A A P where A is the area of the cross-section.

Remarks: (i) Note that both normal stresses (ie stresses perpendicular to a plane) and shear (tangential to P a plane) stresses can occur simultaneously. (ii) In this example, both the normal and shear stress measures given above are only true in an average sense. This is because, moment balance will not be satisfied if these stresses were uniform as shown in the figure. Later we will see the actual distribution of these stresses that must exist so as to satisfy moment equilibrium as well.

Stretching of a Uniform Rod (Revisited): Let us go back to the problem of a nice uniform rod under axial tension as shown in the figure. Let us see if there are any shear stresses in this case.

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ME362: Stress Analysis

Multi-axial Stress and Strain

P

P

P

P

Pn P Pt P

First, consider a transverse cut as shown. Internal forces must exist on this cut face to counter the applied load. Since this internal force is normal (perpendicular) to the cut face, we have: P = A =0 Now consider making an oblique cut. Once again, the net internal force must balance the applied load P as shown. However, this force can be resolved into a component normal and tangential to the cut surface. Thus: Pn P = = sin2 (A / sin ) A Pt P = = sin cos (A / sin ) A Remarks: What we learn is that the intensity of internal forces can not only vary from point to point in a body (as we saw earlier), but it also depends on the plane across which we measure it. So, figuring out only the normal or only the shear stress at a point across just one plane is inadequate. For instance, in the above example, the shear stress takes its largest value along a 45o plane, and failure of the structure could occur in shear along this plane. Alternately, failure can occur in tension along the transverse (=90o) direction. Which mode of failure occurs depends on the maximum values of shear and normal tensile stresses that the material can withstand. In general, therefore, our task is to figure out the internal stresses at every point in a body and across every plane at each point. These ideas will lead us to the concept of the stress tensor.

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ME362: Stress Analysis

SUMMARY

Summary of Introductory Material

It is now time for us to leave this somewhat discursive introduction to the mechanics of deformable continua and venture into the complete theory. Let us summarize what all we have figured out so far. (i) For a certain class of materials, the extension of a cylindrical rod is proportional to the applied axial force as long as the deformation is small (Hooke's experiment). (ii) A better statement of the above is obtained by scaling out geometric effects. This is done by defining: stress, = (force) / (area) [N/m2] strain, = (elongation) / (original length) [dimensionless] Then we have Hooke's' law: For a class of solids called linear elastic materials, the stresses and strains are linearly related as long as the deformations are small: = E, where E is the Young's modulus of the material. (iii) Deeper meaning of stresses: they characterize the internal forces that develop in a body to resist the applied external forces. We have encountered normal stresses that act perpendicular to a surface, and shear stresses that act tangential to a surface. (iv) Deeper meaning of strains: they characterize the change in geometry due to deformation of a body. We have so far seen only normal strains which we defined as change in length (size) to original length of line segments that we imagine drawing on the body. There are things called shear strains as well (we have not seen them yet), and these characterize change in shape (due to twisting or shearing). (v) Stresses and strains can, and in general do, vary from point to point in a body. In addition, they depend on the orientation and direction under consideration. (vi) In general, our goal is to obtain the internal forces (stresses) and geometry of deformation (strains) of a body. To do this, we use: Newton's laws (equilibrium) any constraints on the deformation (compatibility of deformation) Material response (Hooke's law for elastic materials) These three concepts are used in more and more complex settings, but the basic ideas are all in here. To crystallize these thoughts further, we will next revisit the topic of axial stretching of a rod, but now come at it from an angle that is typically used in continuum mechanics.

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ME362: Stress Analysis

Continuum Mechanics Approach to Stretching of a Rod

5. CONTINUUM MECHANICS APPROACH TO STRETCHING OF RODS

Consider a linear elastic rod of uniform cross-sectional area A and initial length L. It is pinned at one end and carries an axial load P at the other end. Let be the density and E the Young's modulus of the material of the rod. Imagine scooping out a segment of the rod abcd of length x at a distance x from the pin. Draw the (x) A x free body diagram of this scooped out element. Note that the b a b a L external force acting on this c d segment is only the weight of the W=g A x x segment which I have lumped at its center of mass. Internal forces c d act on the cut faces ab and cd. Indeed the applied load P and the (x+ x) A weight of the rod below cd is P passed on to the element abcd by means of the internal forces along cd. A similar situation holds for the internal forces along ab, where now the additional weight of the segment abcd must also be passed on to the rod above ab. Thus, the internal forces (and therefore the stresses, since in this case we have assumed that the crosssectional area is uniform) must be different at ab and cd. We recognize that in this case the stress varies with position, ie (x).

Force balance of the segment abcd yields:

(x + x)A - (x)A + gA x = 0

Canceling out the area A and dividing by x, we get:

(x + x) - (x) + g = 0 x As we shrink the element abcd, ie as x-->0, the above becomes: d + g = 0 dx (*)

The above is called the equation of equilibrium in one-dimension. It is easy to integrate the above to get: = -gx+c1 , where c1 is a constant of integration. To get this constant, we need a boundary condition, we need to know the stress at some point along the rod, which we do. At the very end of the rod, we only have the load P acting on it: (x=L) = P/A.

Sridhar Krishnaswamy

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ME362: Stress Analysis Thus:

Continuum Mechanics Approach to Stretching of a Rod

P + g(L - x) A Note that the above is consistent with what we got earlier.

(x) =

(**)

To obtain the geometry of deformation, we use Hooke's law to get the strains: (x) P g (x) = = + (L - x) E AE E

(#)

The above tells us that the strain depends on where we draw the line segments. To get a better handle on this, we now define a quantity u(x) called displacement. This measures the amount by which any cross-section has displaced from its original undeformed position. That is, the cross-section cd which was originally at (x+x) from the top, will have displaced by u(x+x) due to the deformation. It is easy to see that the elongation of the element abcd which was originally of length x is just u(x+x)-u(x). Therefore the average strain over this segment is: u(x + x) - u(x) x And the above expression becomes exact as we make the length x -->0, when we get: du (##) dx The above is called the strain-displacement relation. Using (#) in (##), we can integrate for the displacements: g P x2 u(x) = (x) dx = x+ Lx - + c2 2 E AE where the constant of integration c2 is obtained from the boundary condition u(x=0) = 0 since the rod is pinned at this point. Therefore,

(x) =

P gL g 2 u(x) = + x x - AE 2E E In particular, the displacement of the tip u(x=L) is the stretch of the rod , and this is:

= u(L) =

PL g 2 + L AE 2E

Check that this is consistent with what we obtained earlier. You can recover the solution for the case where we neglect the weight by setting g -->0 in the above. In summary, the continuum mechanics approach to stretching of a uniform rod leads to the following one-dimensional continuum mechanics equations: the equation of equilibrium (Newton's law): d + g = 0 dx

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ME362: Stress Analysis

Continuum Mechanics Approach to Stretching of a Rod

the strain-displacement relation: and the material response (Hooke's law):

(x) =

du dx =E

When we consider more complex situations where the deformation is in general threedimensions, you can expect that the above equations will look somewhat more complicated.

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