Buckling of a Beam The force-engineering strain curve

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Buckling of a Beam

The force-engineering strain curve

BACHELOR THESIS

J.J. Lugthart

Supervisors:

Prof.dr. M. van Hecke and Dr. V. Rottsch¨ afer

November 7, 2013

Leiden Institute of Physics and Mathematisch Instituut,

Leiden University

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Abstract

In this thesis we study the buckling of rubber beams. Buckling is the event where a beam spontaneously bends from straight to curved under a compressive load. The buckling of configurations of one or more rubber beams are analysed theoretically.

We deduce a model which describes this buckling for a beam if we apply a force on it. The model describes the deflection of the beam with respect to the straight line between the ends of the beam. Also it describes the relation between the force and the distance between the two ends of the beam, the force-strain curve.

The model uses Hooke’s law and the balance of the force and bending moment of the beam. It consists of a ordinary differential equation for the deflection and a relation that can be used to calculate the force-strain curve before and shortly after buckling.

After we deduce this model, we use it to calculate the deflection and the force- strain curves for some specific configurations made of rubber beams with different boundary conditions. We first consider a single, hinged beam with or without a rotational spring attached to its endpoint. Next, configurations with more beams are treated. In particular, a configuration of two orthogonal beams and a configuration of many beams is analysed.

The force-strain curve for a single, hinged beam consists of two parts, one steep part that corresponds to when the beam is still straight and one part with a smaller slope that corresponds to a buckled beam. If a spring or a second beam, perpendicular to the first beam, is added, the force-strain curve is similar to the case of a hinged beam. Only the buckling load and the slope after buckling increase. For a large network of beams there appears a peak in the curve at the moment of buckling.

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Preface

Rubber is a widely used material because of its extensibility. It can be compressed by applying force to it. For more compression a larger force is needed. In this thesis we will treat the relationship between the compression and the applied force theoretically. Actual experiments are done by other researchers of the research group

“Complex Media”, a part of the Leiden Institute of Physics. The experimental data and all photos are used with the permission of two researchers of this group, Dr. C.

Coulais and E.J. Vegter, BSc.

Section 1.1 treats Hooke’s law. In section 1.2 we take a piece that is long and slender, called a beam, see figure 1a. The beam is placed vertical. If an increasing

(a) A photo of a beam of 5 centimetres long. The pink part is the beam and the blue part is made of stiffer rubber and is used to fasten the beam.

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35

−0.2 0 0.2 0.4 0.6 0.8 1 1.2

∆l L₀

P(N)

(b) A force-compression curve of a single beam. On the horizontal axis the compression and on the vertical axis the force in Newton. In the steep part where

∆l

L₀ / 0.03, the beam is only compressed, the second part where ^∆l_L

0 ' 0.03, is after buckling, where the beam is compressing and bending.

Figure 1: A single beam and the force-compression curve of the beam.

force is applied to the top of the beam, the beam is not only compressed, but the beam will buckle at some moment. Buckling is the event where a beam spontaneously

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bends from straight to curved. In our case, the ends do not move horizontally.

Questions that arise are: “At which force will the beam buckle?”, “What is the relation between the compression and the force?” and “How important is the way the ends of the beam are fastened, i.e. the boundary conditions of the beam?” These questions are formulated in mathematical way in section 1.2, and the main goal of this thesis is to answer these questions.

Experiments of E.J. Vegter give curves where the force is plotted against the amount of compression, the force-compression curves, for beams such as shown in figure 1a, see figure 1b. For a definition of the compression see equation (1.2.1) and for more information about how these measurements are done, we refer the reader to section 1.3. Note that figure 1b contains of two parts, one steep part for approximately ^∆l_L

0 / 0.03 and one part with a smaller slope for ^∆l_L₀ ' 0.03. These correspond to two regimes, the straight and the buckled regime, respectively.

In chapter 2 we develop a model that describes how we can calculate the deflection and the force-compression curve of a beam. The model consists of a ordinary differential equation for the deflection. The differential equation for the deflection is time-independent, because we study only the steady states. The model consists also of an extra relation that comes from the boundary conditions that is used to calculate the force-strain curve before and shortly after buckling.

After we have developed this model it is used in chapter 3 to get the force- compression curve for a single beam with different boundary conditions.

Chapter 4 will consider configurations with more than one beam. The beams are attached to each other in such a way that the angle between the beams is always 90^◦, so if one beam rotates, all other beams have to rotate too. In section 4.1 we consider a configuration that contains two beams, see figure 2a. With this configuration we find the force-compression curve shown in figure 2b.

(a) A sketch of the configurations of two beams. The blue lines are the beams and the red arrow is the force.

The angle between the two beams is 90^◦.

0 0.05 0.10 0.15

∆l2

L0

P

0 1 2

(b) The force-compression curve.

Figure 2: A sketch and force-compression curve of a configuration made of two beams.

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CHAPTER 0. PREFACE

In section 4.2 we consider rubber networks with holes, as in figure 3a. But a piece

(a) A network of rubber of around 11.5 by 10 centimetres.

0 5 10 15 20

0 10 20 30

∆l (mm)

P (N)

(b) A force-compression curve of a rubber network. For a detailed explanation see figure 4.5.

Figure 3: A photo of a rubber network and the corresponding force-compression curve.

of rubber with holes is not obviously a configuration of many beams. However, if we do not look at the holes but at the rubber between the holes we can approximate these thin pieces by beams. The larger blocks are the connections between the beams. The experimental data of the force-compression curve is shown in figure 3b.

For an explanation of the two different curves and the arrows, we refer the reader to section 4.2. In particular, we will compare the force-compression curve to that of a single beam.

Chapter 5 and 6 contain the discussion and the conclusion.

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Chapter 1

Introduction

1.1 Compression of a piece of rubber

In this section, we will investigate a piece of rubber with length L0, and with constant cross-section over the entire length with a cross-section area A0, as shown in figure 1.1a. When a normal force N is applied at the ends, perpendicular to the cross-section, the piece of rubber shrinks. Let L be the new length after applying a force and ∆L = L0− L the deformation of the length between the uncompressed and the compressed piece, see figure 1.1b.

∆L A0

L₀

(a) A piece of rubber with length L0

and cross-section A0, over the entire length.

A₀

L

∆L N

(b) The compressed piece, N is the normal force perpendicular to the cross-section. L is the compressed length and ∆L is the deformation.

Figure 1.1: A piece of rubber. After applying a normal force N , the piece is compressed.

The rest of section 1.1 describes this relation between the normal force and the deformation, and parametrize the spatial scales of the piece rubber.

1.1.1 Hooke’s law

The physicist Hooke (1635-1703) described that for a spring the force is linear to the compression. This relation is known as Hooke’s law. This theory is described in more detail in the book [TG70]. This first order linear approximation of the relation between the compression and the force is very accurate for a spring, and also for a

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1.1. COMPRESSION OF A PIECE OF RUBBER

lot of other materials. Every material where this first order linear approximation of the response to an applied force is accurate are called linear-elastic or Hookean.

This approximation works only for forces that are not too large, because a material can not be stretched infinitely long, or compress to zero length. Rubber appears to be a Hookean material, so the relation between N and ∆L is linear:

N ∝ ∆L. (1.1.1)

Two equal pieces placed behind each other become both ∆L shorter when the same force as above is applied, the total deformation is two times larger. In general we see that the normal force is linear in the strain, ^∆L_L

0: N ∝ ∆L

L₀ . (1.1.2)

The proportionality constant is now independent of the length of the piece. However, two pieces placed beside each other get both half of the force. In this case both pieces deform half as much as first. Or more general using the stress _A^N

0, instead of the force gives a proportionality constant independent of the area:

N

A₀ ∝ ∆L

L₀ . (1.1.3)

Here the proportionality constant is independent of all the spatial scales of the piece of rubber. We define the hardness of any material, the Young’s modulus, E, as the stress per strain.

E ≡ N A₀

∆L

L₀ . (1.1.4)

Using the Young’s modulus we get:

N A0

= E∆L L0

. (1.1.5)

This relation between the stress and the strain is known as the generalized Hooke’s law.

In this thesis the force is used instead of the stress, so we rewrite (1.1.5) to:

N = EA₀∆L

L₀ . (1.1.6)

Equation (1.1.6) is an important result that tells us how some pieces of rubber responded to a force. The condition that the force is not large is not a problem because in this thesis the forces will be small.

Rather than considering the whole piece of rubber we can look to an infinitesimal piece. Looking to the rate of change of the length at a single point, defined as the local strain, , and to stress, σ, at a point. The compression and the force can be

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CHAPTER 1. INTRODUCTION

expressed by the local strain and the stress by integrating over the length or the cross-section of the beam:

∆L = Z L

0

dx, (1.1.7)

N = Z Z

A0

σdA0. (1.1.8)

For such an infinitesimal piece equation (1.1.5) becomes

σ = E, (1.1.9)

and relation (1.1.9) is the local Hooke’s law.

1.2 Buckling of a beam

In the previous section the response of a thick piece of rubber to a force is described.

It becomes much more interesting when we look to a thin piece. In this thesis a rectangular cuboid piece is used. The height, h, is small in comparison with the length, L0, and the width, b. This piece is called a beam, see figure 1.2a.

L0

h b z x

y

(a) A rubber beam with length L0, height h and width b, where h is relative small in comparison with L0

and b.

Begin point

End point L

l h

b z x

y P

(b) The compressed, buckled beam, P is the force parallel to the x-axis. L is the compressed length and l the distance between the ends of the beam.

Figure 1.2: Sketch of a beam. After applying a force P , the piece is compressed and buckled.

For a force P parallel to the x-axis we expect that by increasing the force, the beam is not only compressing, but that the beam will buckle at some moment.

As in the previous section, L is the length of the beam. This is the real length taking into account the curvature. The distance between the two ends of the beam we call l. See the sketch in figure 1.2b. Note that if the beam is not buckled, L and l are equal, and after buckling it holds that L > l.

The buckling of the beam is an interesting phenomenon. Engineers want to know at which force a steel beam will buckle. This property indicates how strong a steel building is, for when a supporting beam buckles the building collapses.

A second thing is the changing of the response of force in a beam. Because of the difference on a straight and a buckled beam, maybe (rubber) beams can be used to make car bumpers, which are normally hard, but soft on impact.

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1.2. BUCKLING OF A BEAM

1.2.1 Buckling load

We are interested in the question at what force the beam buckles. We will call this force the buckling load, Pc. How to define this force? If you press very careful it is possible to compress the beam more and more without buckling of the beam. This is similar to balancing a ball on the top of a hill. Most of the time the ball will roll down, because the potential energy decreases if the ball rolls to the foot of the hill.

For a force that is larger than the buckling load, a straight beam has a potential energy that is larger than for a buckled beam, so the beam buckles. Figure 1.3 gives a sketch of the potential energy for a force that is smaller than the buckling load, 1.3a, and for a force that is larger than the buckling load, 1.3b.

Epo

Deflection (a) The energy landscape for a force smaller than the buckling load. The only steady state is a beam without deflection and this is stable one.

Epo

Deflection (b) The energy landscape for a force larger than the buckling load.

There are three steady states. The unstable one is the straight beam, and for the stable state the beam is buckled.

Figure 1.3: The energy landscape before and after buckling. The dots are the steady states, the stable states in green and the unstable in red.

For a beam it holds that if the force is lower than the buckling load there is only one possible situation, the straight one. If the force is larger than the buckling load there are three possible situations, a straight and two buckled situations. The beam can buckle to the positive and to negative z-direction. Because of symmetry there are no differences in the potential energy between this two buckled situations.

This observation about the bifurcation allows us to define the buckling load as the force at the bifurcation point. At a lower force there is one steady situation, by a larger force there are three steady situations. The beam buckles because the potential energy of the buckled situations is lower than that of the straight one.

1.2.2 Engineering strain

In section 1.1 we have seen that the strain is a useful quantity. But for a buckled beam we want to look to the distance between the ends of the beam instead of the length of the beam. In analogy with the strain we will define the ‘engineering strain’

as the decrease of the distance between the two ends divided by the length of the

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CHAPTER 1. INTRODUCTION

uncompressed beam:

Engineering strain = ∆l

L₀ = L₀− l

L₀ . (1.2.1)

We have already seen in the preface that the graph of the force versus the compression, or after scaling the engineering strain, contains two parts. One of the reasons why the slope after buckling is much smaller than before buckling is because after buckling the engineering strain increase more than the strain. We will abbre- viate the force-engineering strain curve to ‘f-s curve’, note that the ‘s’ mains the engineering strain and not the strain.

Note that we can express l in ^∆l_L

0 by inverting relation (1.2.1):

l = L₀− L₀∆l L0

= L₀

1 −∆l

L0

. (1.2.2)

1.2.3 Schematic representation

The rest of the thesis we will represent the beam schematically. For this, it is necessary to know in which directions the beam buckles. The beam will buckle in the direction where the minimum force is required.

The second moment of area tells how hard it is to bend the beam. This second moment of area depends on the direction in which the beam is buckled. In section 2.2 we will show that for buckling in the z-direction the second moment of area is the smallest. This because the width is larger than the height. So the beam will buckle in the z-direction.

We can create a schematic representation of the beam. The beam buckles in the z-direction so the y-co¨ordinate is not interesting for investigation, and so we can restrict ourselves to the xz-plane. Secondly we will look at the z-co¨ordinate of the middle between the bottom and top of the beam for all x-values. We can now model the beam as an one dimensional line in the xz-plane, see figure 1.4. We will

Begin point

End point l

z x y

(a) A three dimensional beam, the blue line is the middle of the beam.

w z

x

0 l

(b) A schematic representation of the beam. The blue line is the middle of the beam. w(x) is the deflection of the beam at the point x.

Figure 1.4: From a three dimensional beam to the schematic representation in the xz- plane, where for all x-values we have w(x) as the z-value of the middle of the beam.

describe the z-co¨ordinate of the beam as function of x, where x ∈ [0, l]. We call this z-co¨ordinate the deflection of the beam, w(x). Note that if the beam is not buckled we have w(x) = 0 for all x.

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1.3. EXPERIMENTS

1.3 Experiments

The experimental data used in this thesis, are produced by other researchers of the research group “Complex Media”. The experiments are done using an accurate compression machine from the Instron 5965-series, shown in figure 1.5. It can slowly

Figure 1.5: The Instron 5960-series. The sample is compressed by the Instron controller.

The Instron is used to measure the f-s curves.

compress the beam and measure the force. Because the compression is slow the beam is always in equilibrium (quasi-steady state), so there is no time dependence.

Photos of a beam and a network that we use are shown in figure 1.6.

(a) A front view of a single beam.

(b) A photo of a network what we use. The beams are the parts of rubber between the holes.

Figure 1.6: Two pictures of rubber configurations.

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Chapter 2

Theory of the buckling of a beam

In this chapter we will deduce a model that describes the deflection and the f-s curve of a beam. The model uses the bending moment of the beam, the force and moment balance of a beam in equilibrium and the fact that the stable situation is that situation where the potential energy is minimal.

A important approximation that we make is that the deflection and its derivative are small. This limits the use of the model in such a way that we can only use it before and short after buckling. A second condition will be that the beam is thin.

2.1 Moment of the beam

A bended beam has a tendency to bend back to a straight beam. The quantity that tells how strong this tendency is, is called the bending moment of the beam, or the moment.

First we will describe how we can calculate the moment in general and why a bended beam has moment, and after that we will calculate the moment of a beam as function of the deflection using the curvature of the deflection w.

2.1.1 Bending moment

Suppose we have a lever with a rotation point R and an arm r. Applying a force P at the endpoint of the arm Q, perpendicular to this arm, gives the tendency to rotate, see figure 2.1a. The moment M is equal to the product of the force and the arm:

M = P r. (2.1.1)

In a more general case the force is distributed over an area. In this case the rotation is around a rotation axis, called R. This is shown in figure 2.1b. σ is the stress, the force density, and r the distance between an infinitesimal area dA and the line. We can calculate the moment by integrating the stress times the distance over the whole area:

M = Z Z

A

σrdA. (2.1.2)

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2.1. MOMENT OF THE BEAM

r

R

P Q

M

(a) A lever with a rotation point R and an arm r. There is a force P at the endpoint Q, perpendicular to this arm r. In this case we have M = P r.

A

R M

dA r

σ

(b) An area A that rotate around the line R. The stress σ is orthogonal to the area A. For the moment we have:

M =RR

AσrdA.

Figure 2.1: The moment for a point and for an area by applying a force.

In our case the beam is in equilibrium, so the moment that comes from the applied force have to be compensated by the beam itself. We therefore find for the moment of the beam:

M_beam= − Z Z

A

σrdA. (2.1.3)

To see why a bended beam has a moment we will take a look at figure 2.2. The beam has a height h, so if it is bent the inside is shorter than the outside, s₁ < s₂. A consequence of this is that the inside is under more compression than the outside.

L s2

s₁

h Inside

Outside

σ

M z

x

Figure 2.2: A front view of a bended beam. Since that the inside bend is shorter than the outside bend we have s1 < L < s2. This causes that the stress on the inside is larger than on the outside. If the beam is in equilibrium there is a moment of the beam to compensate this.

From the local Hooke’s low (1.1.9) it follows that the stress at the inside is larger than at the outside. So the stress is not distributed equally over the cross-section, and this difference gives a moment, which the beam has to compensate.

2.1.2 Calculating the moment

The hard part of this subsection is to calculate the difference in the strain between the in- and outside bend. Ones we know this,we can use Hooke’s law, (1.1.9), to get the stress.

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CHAPTER 2. THEORY OF THE BUCKLING OF A BEAM

For an arc the length is known if the radius ρ and the angle in radians θ of the arc:

s = ρθ. (2.1.4)

Let us assume that the beam is curved as an arc, as shown in figure 2.3, with ρ the radius at the middle of the beam and r the distance from a point to the middle of the beam.

s(r) r

θ

ρ

Figure 2.3: A circular beam. ρ is the radius and s(r) is the arc length, r from the middle of the beam.

If we call the strain which is caused by the bending _b(r) we get:

_b(r) = s(0) − s(r)

s(0) = ρθ − (ρ + r)θ

ρθ = −r

ρ. (2.1.5)

There is only one problem, a whole buckled beam does not have the shape of an arc. If we look at a short piece of the beam, we can approximate the shape of that part of the beam with a circle, just like that we can approximate it with the theorem of Taylor. This is done for point X in figure 2.4. For the point Y it will appears that a straight line is the best approximation, i.e. a circle with infinite radius.

The radius of the circle that fits the curve the best is called the radius of curvature. Instead of the radius of curvature we will use the reciprocal of it, the curvature k = ¹_ρ. k can be interpreted as the sharpness of the bend in the curve. For a straight line the curvature is zero.

We can calculate k using (2.1.4). If we take the limit where the arc length s goes to zero and look to how much the angle changes with respect to a change of the arc length we get:

k ≡ dθ ds = 1

ρ. (2.1.6)

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2.1. MOMENT OF THE BEAM

ρ Xk = ¹_ρ

Yk = 0

Figure 2.4: The curvature of a curve at two points. For one point there fits a circle with radius ρ, and for the other point a straight line.

Now we will calculate k at a point with x-co¨ordinate ˜x as a function of w(x):

k(˜x) = dθ ds x=˜x

= dθ dx x=˜x

dx ds x=˜x

= dθ dx x=˜x

ds dx x=˜x

−1

. (2.1.7)

The angle can be calculated using the tangent:

θ = arctan dy dx

= arctan(w_x), (2.1.8)

Where we have used the subscript notation of the derivative: ^dw_dx = wx. So this gives:

dθ

dx = d arctan(w_x)

dx = 1

1 + w_x²w_xx. (2.1.9) For _dx^ds we can use the theorem of Pythagoras:

ds² = dx²+ dy². (2.1.10)

And this implies:

ds dx

2

= 1 + w_x². (2.1.11)

And so

ds

dx =p1 + w²_x (2.1.12)

Substituting (2.1.9) and (2.1.12) in (2.1.7) gives:

k(x) = 1

1 + w_x(x)²w_xx(x) 1

p1 + wx(x)² (2.1.13)

= wxx(x)

(1 + wx(x)²)^3/2 (2.1.14)

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In this thesis we consider straight beams and beams shortly after buckling. This implies that the derivative of w is small for all x. This allows us to approximate k with:

k = wxx

(1 + w_x²)^3/2 (2.1.15)

= wxx

1 −3

2w_x²+15

8 w⁴_x+ ...

(2.1.16)

≈ w_xx (2.1.17)

Substituting (2.1.17) in (2.1.5) we get the bending strain as operator of the deflection:

_b(r) = −rw_xx. (2.1.18)

Now we have to write the stress as function of the strain. The stress can be written as a combination of two parts, one from the normal force applied on the beam and the other part caused by the bending. The normal force gives a stress that is equally distributed over the whole cross-section area, so the stress that comes from the normal force is _A^N

0. The bending gives different stress for different r, and (2.1.18) gives with hooke’s law (1.1.9):

σ = N A0

− Erwxx, (2.1.19)

Now we can calculate the moment by substituting (2.1.19) in (2.1.3):

M (x) = − Z Z

A0

N

A₀ − Erw_xx(x)

rdA₀, (2.1.20)

= − Z Z

A0

N

A₀rdA0+ Ewxx(x) Z Z

A0

r²dA0. (2.1.21) The integral over the area from r² is called the second moment of area, I:

I ≡ Z Z

A0

r²dA₀. (2.1.22)

The beam bends around the rotation line with the lowest second moment of area.

In section 2.2 we will see that the beam will buckle around the centre line parallel to the width.

Calculating the first term of equation (2.1.21) gives:

Z Z

A0

N A0

rdA0 = Z ^h

2

−^h₂

Z ^b

2

−^b₂

N A0

rdydr. (2.1.23)

The normal force is equal distributed over the area, so _A^N

0 is a constant.

= N A₀b 1

2r²

^h

2

−^h

2

= 0 (2.1.24)

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2.2. SECOND MOMENT OF AREA

Substituting the definition of I, (2.1.22), gives:

M (x) = EIwxx(x), (2.1.25)

We will use equation (2.1.25) to get the differential equation in the model of Euler- Bernoulli in section 2.3 and for the boundary conditions to solve the differential equation.

2.2 Second moment of area

The second moment of area depends on the rotation line. This section tells us for which rotation line the second moment of area is minimal.

The beam will bend in the way that the second moment of area is as small as possible. The second moment is the integral of r², and for r² it holds that:

1. it is always non-negative,

2. decreasing for r smaller than zero, 3. increasing for r larger than zero.

And so for a minimal second moment of area the rotation line is the line where all the point of the area are as close as possible by that line. For a rectangle piece this gives that the rotation line is the in the middle of two parallel sides of the rectangle, for R parallel to the width it gives figure 2.5:

h

1 2h b

R

Figure 2.5: A rectangle with height h and width b. The rotation line R is in the middle at height ¹₂h.

The second moment of area corresponding to the case of figure 2.5 is called I_h because the bending is in the h direction. I_b corresponds to the case where the area is turned a quarter.

Calculating I_h gives:

I_h = Z Z

A0

r²dA₀= Z ^h

2

−^h₂

Z ^b

2

−^b₂

r²dydz = b 1 3r³

^h₂

−^h₂

= 1

12bh³. (2.2.1) I_b works at the same way and gives:

Ib = 1

12hb³. (2.2.2)

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And so we get:

I_h I_b = h²

b² = h b

2

. (2.2.3)

For the beam we assume that h is smaller than b, so we get:

I_h< I_b. (2.2.4)

Altogether we know that I_h is the smallest second moment of area, so the beam will bend around the h direction. And so the bending is in the xz-plane.

2.3 Euler–Bernoulli, force and moment balance

Euler and Bernoulli have developed a theory for buckled beams using the balance of the force and the moment. The beam is in equilibrium so the net force and moment have to be zero. For the moment balance we use the result of section 2.1. This gives a differential equation for the deflection w, as we will show in subsection 2.3.1.

2.3.1 Force and moment balance

This subsection is closely analogous to the first chapter of the book [BC10], where the derivation of the differential equation is calculated in a more general setting.

Let’s take a beam like the one in figure 2.6. Let P₀, V₀ and M₀are the horizontal force, the vertical force and the moment at the endpoint of the beam respectively.

P (x), V (x) and M (x) are the forces and moment as a function of x.

P₀ w

V P V0

M0

x M

Figure 2.6: A beam where at x = 0 we apply the horizontal and vertical forces P⁰and V0

and a moment M0. For every x value there are P , V , and M . These forces and moments have to be balanced.

The beam is in equilibrium, so the net force and moment are zero. The balance of the forces is quite easy:

P (x) = P₀, (2.3.1)

V (x) = V₀. (2.3.2)

So this implies that the horizontal and the vertical force are constant:

Px(x) = 0, (2.3.3)

Vx(x) = 0. (2.3.4)

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2.3. EULER–BERNOULLI, FORCE AND MOMENT BALANCE

For the balance of the moment we have to take both the moment and the forces into account. The forces play a role because the forces have an arm to the rotation point, so they give a moment too. Let’s look at a rotation around the begin point of the beam by x = 0. The forces P0 and V0 have no arm with respect to the rotation point so they do not give a moment. For the vertical force V (x) for general x the length of the arm is x, and for P (x) the length of the arm is w(x). So together with the moment M0 and M (x) it gives:

M (x) + P (x)w(x) + V (x)x − M0= 0. (2.3.5) P (x) is a constant, so we will write P instead. There are two unknown constants in this equation, M0 and V (x) = V0. We will differentiate equation (2.3.5) twice to remove the constants. This increases the order of our differential equation. The advance is that we now can use other boundary conditions than V0 and M0 too.

Differentiating twice gives:

Mxx(x) + P wxx(x) = 0. (2.3.6) Now we can substitute the equation for the moment, (2.1.25), found in section 2.1, to get:

[EIw_xx(x)]_xx+ P (x)w_xx(x) = 0. (2.3.7) E and I are constants, so we get:

EIw_xxxx(x) + P w_xx(x) = 0. (2.3.8) This is the fourth order differential equation that describes the deflection of the beam. To reduce the writing we introduce the constant k, defined through k² ≡ _EI^P :

wxxxx+ k²wxx= 0, (2.3.9)

k² ≡ P

EI. (2.3.10)

2.3.2 General solution of the differential equation

A brief look to (2.3.9) tells us that the fourth order differential equation reduces to a second order equation if we introduce v as the second derivative of w:

v ≡ w_xx. (2.3.11)

Substituting v in equation (2.3.9) gives a very well known second order differential equation:

vxx+ k²v = 0. (2.3.12)

If k is not zero, or equivalently if P > 0, the solution is:

v(x) = ˜A sin(kx) + ˜B cos(kx). (2.3.13)

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Now we can integrate (2.3.13) twice to get w, this adds a linear and a constant term:

w(x) = A sin(kx) + B cos(kx) + Cx + D, if P 6= 0. (2.3.14) Sometimes there is no pressing force P . In that case the solution of (2.3.9) is a third degree polynomial:

w(x) = ax³+ bx²+ cx + d, if P = 0. (2.3.15)

In (2.3.14) we see that k is not only a constant which reduces writing, but that k can be interpreted as the wave number, so a larger force gives more waves in the deflection of the beam.

2.4 The force-strain curve

Now that we know the differential equation that describes the deflection of the beam for a given force it is time to calculate the f-s curve. We will see that a non-zero deflection is only possible if the force and the engineering strain are related to each other in a special way. This relation is the key to the f-s curve.

2.4.1 Straight regime

For a straight beam the deflection is zero, w = 0. The straight beam satisfies the differential equation (2.3.8) for all P because both sides will equal zero, and generally w = 0 satisfies the boundary conditions too.

In the straight regime the slope can be calculated by using (1.1.6). This formula appliers because the beam is straight, L = l, and so ^∆L_L

0 = ^∆l_L

0, and because the fact that the normal force is parallel to the x-axis. The second observation gives that the normal force N is equal to the force P . Substituting this two things in (1.1.6) gives:

P^straight= EA0

∆l L0

. (2.4.1)

So for the straight regime the f-s curve is a line through the origin.

2.4.2 Buckled regime

For a buckled beam, we can solve differential equation (2.3.9), by using the general solution for non-zero force, (2.3.14). We have to calculate the four constants for some given boundary conditions. There are four constants, so we need four boundary conditions. For example a beam whose ends are mounted on hinges and the hinges on the x-axis. This gives a beam where the deflection and the moment are zero at the ends: w(0) = w(l) = M (0) = M (l) = 0. This is elaborated in section 3.1. This boundary conditions with (2.3.14) gives a system of four equations.

If we now solve this system the trivial solution where the beam is a straight line, w = 0, is always allowed and for some specific wave numbers there is also a non-zero

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2.4. THE FORCE-STRAIN CURVE

solution. By these k’s the last boundary conditions is always satisfied and there is no unique solution.

For a buckled beam, k has to be one of these specific numbers. The specific values of k depend on l, so this gives a relation between k and l. If we change l, k has to be changes in such a way that the relation is still true. Rewriting this equation between k and l using the definitions of k (2.3.10), and the engineering strain (1.2.1) gives a relation between the force and the engineering strain. This will give the f-s curve for a buckled beam.

2.4.3 Boundary between the two regimes

Now we have two f-s curves, one for the straight regime and one for the buckled regime. For a hinged beam in section 3.1 the curves are shown in figure 2.7, the green and the blue lines correspond to the straight and the buckled regime respectively.

0 0.05 0.1 0.15

∆l L0

P

11∆l L0

2

c, Pc

2

Pe

Figure 2.7: The f-s curve of a hinged beam. Here the green line corresponds to the straight regime and the blue line corresponds to the buckled regime. The red point is the moment where the beam buckles.

In physical systems with multiple possible configurations the most stable situation is that situation with the lowest potential energy. The potential energy can by calculated by integrating the force over the distance over which the endpoint of the beam has moved:

E_po= Z _∆l

0

P dx. (2.4.2)

So the lowest potential energy implies that the force was minimal during the whole compression. For small engineering strain the force from the straight regime is the

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lowest, and for large engineering strain the force for the buckled regime. The point where the two forces are equal is the point where the beam becomes buckled, the red point in figure 2.7. The corresponding force is the buckling load P_c.

And for the general f-s curve we get:

P = minn

P^straight, P^buckledo

. (2.4.3)

The slope of P^straightis much larger than the slope of P^buckled. This implies that P^buckled(0) is a good approximation of the buckling load.

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2.4. THE FORCE-STRAIN CURVE

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Chapter 3

Configurations with a single beam

In this chapter we will start to address the main goal of this thesis. The model that is deduced in chapter 2 can be used to calculate the f-s curves for several configurations.

The difference between the situations is the way in which the ends of the beam are fastened to the press, represented by different boundary conditions.

In the first section we will treat a beam fastened with two hinges, such that the ends of the beam can rotate freely. In the second section a rotational spring is added at one end. We will see that this increases the buckling load.

3.1 Hinged beam

A schematic overview of a hinged beam is given in figure 3.1. The hinges are depicted as triangles.

x-axis

0 l

P

Figure 3.1: Sketch of hinged beam. The deflection on the ends is zero, but both ends can rotate freely.

3.1.1 Solving the differential equation

To solve the differential equation for chapter 2, given in (2.3.9), it is needed to write the boundary conditions in terms of w. For a hinged beam, figure 3.1, the hinges are placed on the x-axis and fastened on the ends of the beam, so the deflections at the ends are zero:

w(0) = w(l) = 0. (3.1.1)

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3.1. HINGED BEAM

A hinge can rotate, so the moment on the endpoints is zero. From equation (2.1.25) it follows that

EIwxx(0) = EIwxx(l) = 0. (3.1.2) So:

wxx(0) = wxx(l) = 0. (3.1.3) Now the boundary conditions are known, it is possible to solve (2.3.14) with the boundary conditions (3.1.1) and (3.1.3). For x = 0:

w = A sin(kx) + B cos(kx) + Cx + D w(0) = wxx(0) = 0

⇒ w = A sin(kx) + Cx. (3.1.4) And at x = l we get:

w = A sin(kx) + Cx w(l) = 0

⇒ w = A

sin(kx) −sin(kl)

l x

. (3.1.5) The last boundary condition gives:

0 =

A

sin(kx) −sin(kl)

l x

xx

x=l

= −k²A sin(kl). (3.1.6) Since we apply a force k is not zero and equation (3.1.6) implies

A = 0 or sin(kl) = 0. (3.1.7)

If A = 0 the deflections reduces to w(x) = 0. But then the beam is not buckled. With other words, for a buckled beam is A non-zero. From that we get that sin(kl) = 0, so kl = nπ with n ∈ Z. If n = 0 we get w = A sin(0) = 0, so for the same reason as why A differs from zero, n differs from zero.

The minus sign of n can be absorbed in A, so without lose of generalization we can take n ∈ N = {1, 2, ...}. In conclusion, the deflection is:

w(x) = A sin (kx) , (3.1.8)

k = nπ

l with n ∈ N. (3.1.9)

Note that the A in formula (3.1.8) is undetermined. In some similar problems calculating a higher order of w using perturbation theorem gives more information about an undetermined constant. In this thesis this is not done.

3.1.2 Force-engineering strain curve

For the straight regime there is found that the f-s curve is given by (2.4.1). For the buckled regime the f-s curve can be calculated with the relation between k and

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CHAPTER 3. CONFIGURATIONS WITH A SINGLE BEAM

l given in equation (3.1.9) to derive the f-s curve. We use equations (2.3.10) and (1.2.2) to eliminate k and l to get:

P^buckled = EI





nπ L₀

1 −^∆l_L

0





2

= EI π L0

2

n 1 −^∆l_L

0

!2

. (3.1.10)

Equation (3.1.10) gives the force as a function of the engineering strain, but there is a dependence on the number n. The number n indicates the mode of the beam, if n is equal to one the beam is in the first mode, n = 2 gives the second mode, and so on. The first three modes are plotted in figure 3.2.

0 l

Figure 3.2: The first three modes of a beam, the blue line is the first mode where n = 1, the red line the second mode and the green line is the third mode.

Now the question is: ‘With mode is seen in in an experiment?’ In section 2.4 there is noted that the most stable configuration is those with the lowest potential energy and this implies that the beam is in the configuration corresponding to the lowest force. For the case of the modes of the beam this implies that the beam prefers to be in the lowest mode. Taking n equal to one gives the formula in (3.1.11).

P^buckled = EI π L₀

2

1

1 −^∆l_L

0

2. (3.1.11)

This formula is sketched as the blue line in figure 3.3, the green line is the f-s curve for a straight beam and the red point the buckling point.

Substituting the force for a straight beam (2.4.1), and that for a buckled beam (3.1.11) in equation (2.4.3) gives the f-s curve before and after buckling:

P = min







EA₀∆l

L₀, EI π L₀

2

1

1 −^∆l_L

0

2







. (3.1.12)

3.1.3 Relative force

It is useful to scale the force such that all beams buckle around the same value, even if the beams have different dimensions or are made of rubber with different Young’s modulus. An approximation of the buckling load is used to scale the force.

The slope before buckling is much larger than after buckling, so the force for a buckled beam with engineering strain zero is a good approximation for the buckling load:

Pc≈ P^buckled(0). (3.1.13)

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3.1. HINGED BEAM

0 0.05 0.1 0.15

∆l L0

P

11∆l L0

2

c, Pc

2

Pe

Figure 3.3: The f-s curve of a hinged beam, where the green line is the straight regime, and the blue one the buckled regime. The red point is the buckling point and Peis the Euler load of this beam.

This approximation is called Euler load,

Pe≡ P^buckled(0) = EI π L0

2

. (3.1.14)

We now define the relative force as the force divided by the Euler load:

Pr ≡ P

P_e. (3.1.15)

Using the definition of the relative force we get for the straight regime:

P_r^straight= EA0∆l L0

EI

π L0

2 = A0

I

L0

π

2

∆l L0

. (3.1.16)

For the buckled regime we find:

P_r^buckled = EI

π L0

2 1

1−^∆l

L0

2

EI

π L0

2 = 1

1 −^∆l_L

0

2. (3.1.17)

Note that after this scaling the relative force is independent of the Young’s modulus and in the buckled regime the force does not depend on the dimensions of the beam.

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Short after buckling is the engineering strain small, so a Taylor approximation of P_r^buckled gives the slope of the relative force-engineering strain curve (rf-s curve) after buckling:

P_r^buckled ≈ 1 + 2∆l

L₀. (3.1.18)

So the slope of the rf-s curve is 2 after buckling.

Sometimes it is useful to write k as function of the relative force. We see that:

k² = P

EI = PrPe

EI =

PrEI

π L0

2

EI = Pr

π L0

2

, (3.1.19)

or

k = π L0

pPr. (3.1.20)

3.2 Hinged beam with rotational spring

In this section a rotational spring is added to the configuration of section 3.1. A rotational spring gives a moment against the rotation that is proportional to the rotational angle. See the sketch in figure 3.4. The spring is represented by the spiral and θ is the angle of the beam with the x-axis at x = l. The spring is placed at the point x = l because this gives an easier formula for w(x) later, but x = 0 would give the same f-s curve.

x-axis

0 l

θ P

Figure 3.4: Sketch of hinged beam with a rotational spring to one endpoint of the beam.

The deflections on the ends are zero. The spring applies a momentum which causes that the beam buckles at a larger force.

3.2.1 Boundary conditions

At both endpoints the deflection is zero because of the hinges,

w(0) = w(l) = 0. (3.2.1)

At x = 0 we know that the moment of the beam is zero because of the hinge, M (0) = 0. Equation (2.1.25) gives:

wxx(0) = 0. (3.2.2)

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3.2. HINGED BEAM WITH ROTATIONAL SPRING

The spring applies a moment at the point x = l. The moment is proportional to the angle θ with the spring constant ˜µ:

Mspring= −˜µθ. (3.2.3)

It holds that θ = arctan(−wx(l)). The deformations are small, so the linear approximation of the tangent is accurate.

M_spring≈ −˜µ · −w_x(l) = ˜µw_x(l). (3.2.4) The moment of the beam has to balance this moment of the spring, so:

Mbeam= −˜µwx(l). (3.2.5)

Substituting the moment of the beam, (2.1.25), gives:

EIw_xx(l) = −˜µw_x(l), (3.2.6)

or

w_xx(l) = − µ˜

EIw_x(l). (3.2.7)

Defining the dimensionless quantity µ ≡ ^µL^˜_EI⁰ gives wxx(l) = − µ

L0

wx(l). (3.2.8)

This µ can be interpreted as the strength of the spring compared to the strength of the beam. In the rest of this section µ is called the spring constant instead of ˜µ.

And (3.2.8) is the fourth boundary condition.

3.2.2 Solving the differential equation

We have to solve the differential equation w_xxxx+ k²w_xx = 0 for this boundary conditions. There is a force P so we can use the general solution given in equation (2.3.14). Three of the four boundary conditions are the same as in section 3.1, this gives with (3.1.5):

w = A

sin(kx) −sin(kl)

l x

. (3.2.9)

The last boundary condition, (3.2.8), gives:

A

sin(kx) −sin(kl)

l x

xx

x=l

= − µ L0

A

sin(kx) −sin(kl)

l x

x

x=l

. (3.2.10)

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After some calculations this reduces to:

A = 0 or tan(kl) = kl

1 + k²l^L_µ⁰. (3.2.11) And for a buckled beam it holds that A 6= 0 and k > 0, so the solution for the deflection is:

w = A

sin(kx) −sin(kl)

l x

. (3.2.12)

Where k and l such that the following relation holds:

tan(kl) = kl

1 + k²l^L_µ⁰. (3.2.13)

3.2.3 Relative force-engineering strain curve

With a spring, the force is not only depending on the engineering strain, but also on µ. To exhibit the underlying µ-dependence for the force an upper index (µ) is added: P^(µ). If µ is equal to zero the spring has no effect and the situation is that of the hinged beam in section 3.1. So P⁽⁰⁾ = P^hinged. This gives that P⁽⁰⁾(0) = P_e and for the relative force, Pr^(µ)

∆l L0

, we have Pr⁽⁰⁾(0) = 1.

Equation (3.2.13) is a relation between k and l. Which can not be solved ana- lytical. We will consider the limit cases where the spring constant becomes zero or infinity and later we will solve it numerical. In the limit µ → 0 the relation (3.2.13) must simplify to the relation for a hinged beam. For the limit µ → ∞ the endpoint of the beam can not rotate, and the endpoint of the beam has to be tangent to the x-axis.

The limit forµ to zero

Looking at the right hand side (r.h.s.) of (3.2.13) if µ goes to zero we see that:

limµ↓0

kl

1 + k²l^L_µ⁰ = 0, (3.2.14)

and we get

tan(kl) = 0. (3.2.15)

And because tan(α) = _cos(α)^sin(α), this implies

sin(kl) = 0, (3.2.16)

and (3.2.16) is equal to (3.1.7), so relation (3.2.13) reduces to the case for a hinged beam. And with equation (3.1.17):

P_r⁽⁰⁾ ∆l L₀

= P_r^hinged ∆l L₀

= 1

1 −^∆l_L

0

2. (3.2.17)

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The limit for µ to infinity

For the limit µ to infinity, the r.h.s. of equation (3.2.13) reduces to:

µ→∞lim kl

1 + k²l^L_µ⁰ = kl, (3.2.18) and so

tan(kl) = kl. (3.2.19)

Calculating w_x(l), using (3.2.19) gives:

w_x(l) =

A

sin(kx) −sin(kl)

l x

x

x=l

, (3.2.20)

= A

k cos(kl) −sin(kl) l

, (3.2.21)

= Acos(kl)

l (kl − tan(kl)) , (3.2.22)

= 0. (3.2.23)

So in this case the derivative at x = l is zero so the endpoint of the beam will be tangent to the x-axis.

A second thing is that for µ → ∞ relation (3.2.13) reduces to a relation of kl rather than a relation of k and l. So solving tan(α) = α gives k = ^α_l. tan(α) = α has infinity many solutions, but as for the hinged beam in section 3.1 the first non-zero solution is the one that is seen in the experiments, and is called C,numerically we find that C = 4.4934..., and so

k = C

l. (3.2.24)

The equations (1.2.2) and (3.1.20) give for the rf-s curve:

P_r^(∞) ∆l L0

= C π

2

1

1 −^∆l_L

0

2. (3.2.25)

This differs only by a factor ^C_π2

from P_r for a hinged beam:

= C π

2

P_r^hinged ∆l L0

(3.2.26)

= 2.0457... · P_r^hinged ∆l L0

. (3.2.27)

So the buckling load and the slope after buckling for a beam with an infinitely strong spring are around 2 times as large as the buckling load for a hinged beam.

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Generalµ

For general µ a numerical approximation is needed. To get the rf-s curve, first equation (3.2.13) is written in terms of the engineering strain and the relative force by using (1.2.2) and (3.1.20):

tan

π

q Pr^(µ)

1 −∆l

L0

= π

q Pr^(µ)

1 −^∆l_L

0

1 +^π_µ²Pr^(µ)

1 −^∆l_L

0

. (3.2.28)

In figure 3.5 relation (3.2.28) is plotted for ^∆l_L

0 = 0.05 and µ = 20. The blue line is the l.h.s. and the green line is the r.h.s. The red dots are the intersections of this two lines and so the solutions of relation (3.2.28).

0 2 4 6 8 10 12 14 16 18 20 22

−3

−2

−1 0 1 2 3 4

Pr

tan1 π√P_r1

1−^∆lL₀

22

π√Pr

!1−_L0^∆l

"

1+^π² µPr

!1−_L0^∆l

"

Solutions

Figure 3.5: Graph of the l.h.s. of (3.2.28) in blue, and the r.h.s. in green. The intersections are shown in red.

Taking the first non-zero solution for a fixed spring constant µ and several values of the engineering strain ^∆l_L

0 gives the rf-s curves for that specific spring constant.

For four µ values this curve is plotted in figure 3.6.

Now the relation between Pr^(µ) and ^∆l_L

0, and so between k and l, is known, it can be used to calculate the deflection of the beam as function of x given in equation (3.2.12). The result is shone in figure 3.7. The limit cases are the blue and the purple lines. The blue line for µ = 0 gives a perfect sine, and the purple line for µ = ∞ with w_x(l) = 0.

3.2.4 The buckling load and the slope of the force-engineering strain curve after buckling

An approximation of the buckling load is the intersection of the rf-s curve is figure 3.6 with the Pr-axis, with other words Pr^(µ)(0). This approximation as function of mu is plotted as the blue line in figure 3.8.

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0 0.05 0.1 0.15

0 0.5 1 1.5 2 2.5

∆l L0

Pr

µ = ∞ µ = 10 µ = 2 µ = 0

Figure 3.6: The rf-s curves for a hinged beam with rotational springs of different strength.

The blue, green, red and purple lines are with a strength of respectively 0, 2, 10 and ∞.

0 10 20 30 40 50 60 70

0.5 1 1.5 2 2.5 3 3.5 4

µ

Pr^(µ)(0) Pr^(µ)

′(0)

Figure 3.8: The blue line is the relative buckling load as function of µ. The green line is the slope of the rf-s curve after buckling. If the beam is hinged, i.e. µ is zero, the relative buckling load is one and the slope is two. For positive µ the relative buckling load and the slope increase.

If µ is zero, the buckling load is equal to the Euler load, so Pr⁽⁰⁾(0) = 1. For µ larger than zero the relative buckling load increases.

For the slope of the rf-s curve after buckling the theorem of implicit differentiation is used. This is done for equation (3.2.28) in appendix A. A plot of the slope of the rf-s curve at zero strain as function of µ, Pr^(µ)

0(0), is given by the green line in figure 3.8. If µ = 0, the slope equals 2, agreeing which our result from section 3.1,

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0 1 2 3 4 5

0 0.25 0.5 0.75 1

x

w

µ = ∞ µ = 10 µ = 2 µ = 0

Figure 3.7: w(x) for different spring constants. The blue, green, red and purple lines are with a constant of respectively 0, 2, 10 and ∞. For µ = 0 it is a sine and for µ = ∞ the slope is zero at the end.

for larger µ the slope is steeper.

The relative force is the force scaled with the Euler load, an approximation of the buckling load for a hinged beam. Now the buckling load is a function of µ, an other scaling is the scaling with the approximation of the buckling load of a hinged beam with a spring with spring constant µ:

P˜_r^(µ) ∆l L₀

= P^(µ)

∆l L0

P^(µ)(0) , (3.2.29)

instead of

P_r^(µ) ∆l L₀

= P^(µ)

∆l L0

P_e =

P^(µ)

∆l L0

P⁽⁰⁾(0) . (3.2.30)

If the force is scaled in this way the f-s curve becomes figure 3.9. Because of the fact that Pr^(∞)

∆l L0

is a multiple of Pr⁽⁰⁾

∆l L0

, this two lines collapse. The lines for µ ∈ (0, ∞) are lower than the lines for µ equal to zero or infinity.

It appears that this rescaling has the nice property that the slope after buckling is around 2 for all µ. A graph of the slope after buckling with this rescaling is given in figure 3.10. Here f (µ) is the rescaled slope after buckling as function of µ:

f (µ) ≡ Pr^(µ) 0(0)

Pr^(µ)(0). (3.2.31)

The limits of f (µ) for µ to zero or to infinity are 2. For the calculations we refer the reader to appendix A.1. For µ ∈ (0, ∞), f (µ) is smaller than two.

Buckling of a Beam The force-engineering strain curve