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ME362: Stress Analysis

INTRODUCTION TO THE

MECHANICS OF DEFORMABLE

BODIES

Sridhar Krishnaswamy 1



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 gravitationalforce, 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 thisassumption 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.




ME362: Stress Analysis Introduction


• stress analysis of dental

implants

FIGURE 1: EXAMPLES OF DEFORMATION OF STRUCTURES



ME362: Stress Analysis Hooke's Experiment

2. 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 δ. If wewere 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

find a plot that looks

something like this:

F

δ

L

F

δ

proportional limit

fracture

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 looselytranslates 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.





• 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

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 theother 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.

F

δ

geometry 2

geometry 1

same material

F

δ

material 2

material 1

same geometry

In order to separate the effect of geometry and material, we scale out the

geometric effect by defining:

Stress: σ = F/A = (force) / (cross-sectional area) [N.m-2]

Strain: ε = δ /L = (stretch) / (original length) [dimensionless]

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

σ

ε

material 2 (independent of A,l)

material 1 (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:





σ = E ε

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 theintensity 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, theum stress at which a fat rod and a

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.

maximFF

F F

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 δl due to the load, we find that for this

uniform rod we have: ε = δl / l = δ / L, and is independent of length

'l'.

F

δ

Ll

δ l

These ideas will crystallize further when we look at the following

examples.




ME362: Stress Analysis Axial Stretching of Rods

3. AXIAL STRETCHING OF RODS

Case 1: Uniform Rod in Tension

• uniform cross-sectional area A, original length L• Neglect weight of rod itself

• Young's modulus E

Since the stress: σ = E ε --> P/A = E δ /Lwe have:

P = [AE/L] δ or δ = PL/AE

where AE/L is called the axial stiffness (and is the equivalent of the

spring constant for a linear spring).

F

δ

L

Remark s(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 δ where k [N/m]is called the spring constant of the spring. F

δ

Case 2: 2-Stepped Rods:





• 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 δ1 = FL1 /A1E1

stretch of segment 2 is δ2 = FL2 /A2E2

Total stretch of the stepped rod is: δ = δ1 + δ2= FL1 /A1E1 + FL2 /A2E2

F

A 1 , L 1

A 2 , L 2

1

2

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,δ = δ1 + δ2= P1L1 /A1E1 +P2L2 /A2E2

F2

A1, L1

A2, L2

1

2

F1

F2

A1, L1

A2, L2

1

2

F1

P1 = σ1 A1 = F 1+ F 2

F2

2

P2 = σ2 A2 = F 2

Case 3: N-Stepped Rods:

We can easily extend the above to a rod with N-steps.

δ =Pi Li

Ai E ii=1

N

∑

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:





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 cross-

section but whose weight is not negligible. In this case, the net cross-sectional force is

different at different locations.

F

x

F

x

model as

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:

∆δ =P

EA ∆ x .

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

δ = ∆δ all segments

∑ =P

EA ∆ x

all segments

∑ as ∆ x →0 ⎯ → ⎯ ⎯ ⎯ ⎯ P

EA dx

0

L

∫

{Recall the meaning of an integral as a sum}

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





• Rod with uniform cross sectional

area A and length L

• Density of material is ρ.

• Young's modulus of material is E.

• The only load is due to its own

weight.

Locate the origin of the x axis at the

bottom as shown.

L

W = { A x } ρ g

P ( x ) = n e t c o s ss e c t i o n a l f o r c e

x

x

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) = ρgAx. The total stretch of the rod under its own weight is therefore:

δ =P x( ) EA

dx

o

L

∫ =ρ gAx

EAdx =

ρ gL2

2 E o

L

∫ .





Example 2: Consider the two trusses shown in the figures:

P

A

B C

α α

D

P

A

B C

α α

Case a Case b

We can solve for the forces carried by the truss members in case (a) from conditions forstatic 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 α. 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

:F

x

= 0 ⇒ P1

= P2

∑ F y = 0 ⇒ P1 cosα + P2∑ cosα = 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.





We recognize that from the symmetry of the problem (rods AB and AC are identical), rod

AD stretches by δAD such that it is still vertical. Rods AB and AC then must also stretch

appropriately.

If we assume that the

deformations are smallcompared to the lengths of

the rods, then, we find

from the figure (shown

grossly exaggerated) that

the stretches are

approximately related

through:

δ AB = δ AC = δ AD cosα ' ≈ δ AD cosα

where again the

assumption of small

deformation allows us to

say that the angle α' is approximately the same as α.

P

B C

α α

D

A'

A

α' α'

A'

A

δAB

δAC

δAD

Recasting this in terms of the forces, we have:

P1 L

AE =

P2 L

AE =

P3 ( Lcosα )

AE cosα

which gives us the additional restriction needed to solve for the forces in the rods.

P1 = P2 =P cos

2α

1 + 2cos3α ; P3 =

P

1+ 2cos3α .

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.




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: εaxial = F/AE where A is the cross-sectional area and E isthe 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,

this line segment contracts by an amount δlateral. This leads to a

lateral strain εlateral = δlateral /llateral

What is more, for linear elastic rods, it is experimentally found that

the lateral strain and the axial strain are linearly related through:

εlateral= -ν εaxial ,

where ν is called the Poisson's ratio, a property of the material.

Most materials have a Poisson's ratio between 0 and 0.5. Thenegative sign says that when the rod is in axial tension it compresses

laterally and when the rod is in axial compression it expands laterally.

F

L

Biaxial Stress in a Thin-walled Cylindrical Pressure Vessel:

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.

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:

L

r

t

p (π r2) = σ1 (2 π r t)

That is, σ1 = pr / 2t and this acts axially as shown.

σ1

Next, consider making a

longitudinal cut as shown.

Internal "hoop" stresses σ2 on

the cut rim must counter the

pressure of the gas that wants

to blow the top half away.

That is;

σ2σ2

L

σ2

σ1σ1

σ2




σ 2 t L{ }(2)= p l r dθ { }0

π

∫ sinθ = p (2rL) ==> σ2 = pr / t

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.

PP/2

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

the contact surface (ABCD on the

top face and a similar region on the

bottom). We define an average

shear stress:

A B

CD

P

τ = 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.





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 α 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:

normal stress: σ = Pn

A =

P cosα

A

shear stress: τ = Pt

A =

Psinα

A

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

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.

α

P

α

P

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.





P

P

PP

P

Pn

Pt θθ

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:

σ =P

A

τ = 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 normaland tangential to the cut surface. Thus:

σ θ =Pn

( A /sinθ )=

P

Asin2 θ

τ θ =Pt

( A /sinθ )=

P

Asinθ cosθ

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 itslargest value along a 45

oplane, and failure of the structure could occur in shear along this

plane. Alternately, failure can occur in tension along the transverse (θ=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.




ME362: Stress Analysis Summary of Introductory Material

SUMMARY

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, σ = (force) / (area) [N/m2]

strain, ε = (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: σ = Eε,

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 areall 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.




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 ρ 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

free body diagram of this scooped

out element. Note that the

external force acting on this

segment is only the weight of the

segment which I have lumped at

its center of mass. Internal forces

act on the cut faces ab and cd.

Indeed the applied load P and theweight of the rod below cd is

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 cross-

sectional area is uniform) must be different at ab and cd. We recognize that in this case

the stress varies with position, ie σ(x).

P

Lc d

a b a b

c d

W=ρg A ∆ x

σ(x) A

σ(x +∆ x) A

x

∆ x

Force balance of the segment abcd yields:

σ ( x + ∆ x) A −σ ( x) A + ρ gA ∆ x = 0

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

σ ( x + ∆ x) − σ ( x)

∆ x⎧⎨⎩

⎫⎬⎭

+ ρ g = 0

As we shrink the element abcd, ie as ∆x-->0, the above becomes:

d σ

dx + ρ g = 0 (*)

The above is called the equation of equilibrium in one-dimension. It is easy to integrate

the above to get: σ = -ρgx+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: σ(x=L) =

P/A.





Thus:

σ ( x) =P

A+ ρ g( L − x) (**)

Note that the above is consistent with what we got earlier.

To obtain the geometry of deformation, we use Hooke's law to get the strains:ε ( x) =

σ ( x)

E =

P

AE +ρ g

E ( L − x) (#)

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:

ε ≈ u( x + ∆ x) − u( x)∆ x

And the above expression becomes exact as we make the length ∆x -->0, when we get:

ε ( x) =du

dx(##)

The above is called the strain-displacement relation. Using (#) in (##), we can integrate

for the displacements:

u( x) = ε ( x) d x∫ =P

AE x +

ρ g

E Lx −

x2

2

⎧⎨⎩

⎫⎬⎭

+ c2

where the constant of integration c2 is obtained from the boundary condition u(x=0) = 0since the rod is pinned at this point. Therefore,

u( x) =P

AE +ρ gL

E

⎧⎨⎩

⎫⎬⎭ x −

ρ g

2 E x

2

In particular, the displacement of the tip u(x=L) is the stretch of the rod δ, and this is:

δ = u( L) =PL

AE +ρ g

2 E L

2

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 ρg -->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 σ

dx+ ρ g = 0





the strain-displacement relation: ε ( x) =du

dx

and the material response (Hooke's law): σ = E ε

When we consider more complex situations where the deformation is in general three-

dimensions, you can expect that the above equations will look somewhat morecomplicated.

S idh K i h 20

lec intro me362

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