Buckling of a Beam the Force-Engineering Strain Curve

Home , Bending moment, Buckling, Deflection (engineering), Second moment of area

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

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.

i Contents

Preface v

1 Introduction 1 1.1 Compression of a piece of rubber ...... 1 1.1.1 Hooke’s law ...... 1 1.2 Buckling of a beam ...... 3 1.2.1 Buckling load ...... 4 1.2.2 Engineering strain ...... 4 1.2.3 Schematic representation ...... 5 1.3 Experiments ...... 6

2 Theory of the buckling of a beam 7 2.1 Moment of the beam ...... 7 2.1.1 Bending moment ...... 7 2.1.2 Calculating the moment ...... 8 2.2 Second moment of area ...... 12 2.3 Euler–Bernoulli, force and moment balance ...... 13 2.3.1 Force and moment balance ...... 13 2.3.2 General solution of the diﬀerential equation ...... 14 2.4 The force-strain curve ...... 15 2.4.1 Straight regime ...... 15 2.4.2 Buckled regime ...... 15 2.4.3 Boundary between the two regimes ...... 16

3 Configurations with a single beam 19 3.1 Hinged beam ...... 19 3.1.1 Solving the differential equation ...... 19 3.1.2 Force-engineering strain curve ...... 20 3.1.3 Relative force ...... 21 3.2 Hinged beam with rotational spring ...... 23 3.2.1 Boundary conditions ...... 23 3.2.2 Solving the differential equation ...... 24

ii CONTENTS

3.2.3 Relative force-engineering strain curve ...... 25 3.2.4 The buckling load and the slope of the force-engineering strain curve after buckling ...... 27

4 Configurations with more beams 31 4.1 Two beams configuration ...... 31 4.1.1 Formulas for the deflections of the beams for a given angle. . 31 4.1.2 The length of the beams ...... 33 4.1.3 Momentum balance ...... 35 4.1.4 Force-engineering strain curve ...... 35 4.2 Network with many beams ...... 37

5 Discussion 41

6 Conclusions 43 6.1 General results from the model ...... 43 6.2 Force-engineering strain curve for speciﬁc conﬁgurations ...... 43

A The slope for a single beam with a spring after buckling 45 A.1 Limits for the spring constant to zero and inﬁnity ...... 46

B List of notation used 49

Bibliography 51

iii CONTENTS

iv Chapter 0 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

1.2

0.8

0.6 (N)

P 0.4

0.2

−0.2 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 ∆l L0 (a) A photo of a beam of 5 (b) A force-compression curve of a single beam. On centimetres long. The pink the horizontal axis the compression and on the verti- part is the beam and the cal axis the force in Newton. In the steep part where blue part is made of stiﬀer ∆l 0.03, the beam is only compressed, the second L0 / rubber and is used to fasten part where ∆l 0.03, is after buckling, where the beam L0 ' 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

v 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 0.03 and one part with a smaller slope for ∆l 0.03. These L0 / L0 ' 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.

P 1

0 0 0.05 0.10 0.15 ∆l2 L0 (a) A sketch of the conﬁgurations of (b) The force-compression curve. two beams. The blue lines are the beams and the red arrow is the force. The angle between the two beams is 90◦.

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

vi CHAPTER 0. PREFACE

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

P (N)

0 0 5 10 15 20 ∆l (mm) (a) A network of rubber of around (b) A force-compression curve of a rubber net- 11.5 by 10 centimetres. work. 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.

vii viii 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.

A0 A0 N

L0 L ∆L ∆L

(a) A piece of rubber with length L0 (b) The compressed piece, N is the and cross-section A0, over the entire normal force perpendicular to the length. 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

1 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 : L0

∆L N ∝ . (1.1.2) L0

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 N , instead of the A0 force gives a proportionality constant independent of the area:

N ∆L ∝ . (1.1.3) A0 L0

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.

N ∆L E ≡ . (1.1.4) A0 L0

Using the Young’s modulus we get:

N ∆L = E . (1.1.5) A0 L0 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: ∆L N = EA0 . (1.1.6) L0 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

2 CHAPTER 1. INTRODUCTION expressed by the local strain and the stress by integrating over the length or the cross-section of the beam: Z L ∆L = dx, (1.1.7) 0 ZZ N = σdA0. (1.1.8) A0 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.

z x z x y y P h L b l End point h L0 b Begin point

(a) A rubber beam with length L0, (b) The compressed, buckled beam, P height h and width b, where h is rel- is the force parallel to the x-axis. L ative small in comparison with L0 is the compressed length and l the dis- and b. tance 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.

3 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 Epo

Deflection Deflection (a) The energy landscape for a (b) The energy landscape for a force smaller than the buckling force larger than the buckling load. load. The only steady state is a There are three steady states. The beam without deflection and this is unstable one is the straight beam, stable one. 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

4 CHAPTER 1. INTRODUCTION uncompressed beam:

∆l L0 − l Engineering strain = = . (1.2.1) L0 L0 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 by inverting relation (1.2.1): L0 ∆l ∆l l = L0 − L0 = L0 1 − . (1.2.2) L0 L0

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-coördinate is not interesting for investigation, and so we can restrict ourselves to the xz-plane. Secondly we will look at the z-coördinate 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

z x y z w l End point x Begin point 0 l (a) A three dimensional beam, the (b) A schematic representation of blue line is the middle of the beam. the beam. The blue line is the middle of the beam. w(x) is the deﬂec- tion 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-coördinate of the beam as function of x, where x ∈ [0, l]. We call this z-coördinate the deflection of the beam, w(x). Note that if the beam is not buckled we have w(x) = 0 for all x.

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

Figure 1.6: Two pictures of rubber conﬁgurations.

6 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: ZZ M = σrdA. (2.1.2) A

7 2.1. MOMENT OF THE BEAM

Q P σ A r dA r

M R R M (a) A lever with a rotation point R and (b) An area A that rotate around the an arm r. There is a force P at the line R. The stress σ is orthogonal to endpoint Q, perpendicular to this arm the area A. For the moment we have: RR r. In this case we have M = P r. M = A σrdA. 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: ZZ Mbeam = − σrdA. (2.1.3) A 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.

Outside z s2 x L σ s1 M Inside h

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.

8 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

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: s(0) − s(r) ρθ − (ρ + r)θ r (r) = = = − . (2.1.5) b s(0) ρθ ρ 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 1 k = 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: dθ 1 k ≡ = . (2.1.6) ds ρ

9 2.1. MOMENT OF THE BEAM

Y k = 0

ρ X 1 k = ρ

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¨ordinatex ˜ as a function of w(x): −1 dθ dθ dx dθ ds k(˜x) = = = . (2.1.7) ds x=˜x dx x=˜x ds x=˜x dx x=˜x dx x=˜x The angle can be calculated using the tangent: dy θ = arctan = arctan(wx), (2.1.8) dx

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

dθ d arctan(wx) 1 = = 2 wxx. (2.1.9) dx dx 1 + wx ds For dx we can use the theorem of Pythagoras: ds2 = dx2 + dy2. (2.1.10)

And this implies:

ds 2 = 1 + w2. (2.1.11) dx x

And so ds p = 1 + w2 (2.1.12) dx x Substituting (2.1.9) and (2.1.12) in (2.1.7) gives: 1 1 k(x) = wxx(x) (2.1.13) 2 p 2 1 + wx(x) 1 + wx(x) wxx(x) = (2.1.14) 2 3/2 (1 + wx(x) )

10 CHAPTER 2. THEORY OF THE BUCKLING OF A BEAM

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: wxx k = (2.1.15) 2 3/2 (1 + wx) 3 2 15 4 = wxx 1 − w + w + ... (2.1.16) 2 x 8 x

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

b(r) = −rwxx. (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 N . The bending gives different stress for different r, and (2.1.18) A0 gives with hooke’s law (1.1.9): N σ = − Erwxx, (2.1.19) A0 Now we can calculate the moment by substituting (2.1.19) in (2.1.3): ZZ N M(x) = − − Erwxx(x) rdA0, (2.1.20) A0 A0 ZZ ZZ N 2 = − rdA0 + Ewxx(x) r dA0. (2.1.21) A0 A0 A0 The integral over the area from r2 is called the second moment of area, I: ZZ 2 I ≡ r dA0. (2.1.22) A0 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:

h b ZZ N Z 2 Z 2 N rdA0 = rdydr. (2.1.23) A0 h b A0 A0 − 2 − 2

The normal force is equal distributed over the area, so N is a constant. A0

h N 1 2 = b r2 = 0 (2.1.24) A0 2 h − 2

11 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:

1 2 h h R

Figure 2.5: A rectangle with height h and width b. The rotation line R is in the middle 1 at height 2 h.

The second moment of area corresponding to the case of ﬁgure 2.5 is called Ih because the bending is in the h direction. Ib corresponds to the case where the area is turned a quarter. Calculating Ih gives:

h b h ZZ Z Z 2 2 2 2 2 1 3 1 3 Ih = r dA0 = r dydz = b r = bh . (2.2.1) A − h − b 3 h 12 0 2 2 − 2

Ib works at the same way and gives:

1 I = hb3. (2.2.2) b 12

12 CHAPTER 2. THEORY OF THE BUCKLING OF A BEAM

And so we get:

2 2 Ih h h = 2 = . (2.2.3) Ib b b For the beam we assume that h is smaller than b, so we get:

Ih < Ib. (2.2.4)

Altogether we know that Ih 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 M0 are 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.

V P

V0 w P0 M x

Figure 2.6: A beam where at x = 0 we apply the horizontal and vertical forces P0 and 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)

13 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:

[EIwxx(x)]xx + P (x)wxx(x) = 0. (2.3.7)

E and I are constants, so we get:

EIwxxxx(x) + P wxx(x) = 0. (2.3.8)

This is the fourth order differential equation that describes the deflection of the 2 P beam. To reduce the writing we introduce the constant k, defined through k ≡ EI : 2 wxxxx + k wxx = 0, (2.3.9) P k2 ≡ . (2.3.10) EI 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 ≡ wxx. (2.3.11)

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

2 vxx + k v = 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)

14 CHAPTER 2. THEORY OF THE BUCKLING OF A BEAM

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 , and because the fact L0 L0 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:

straight ∆l P = EA0 . (2.4.1) L0 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

15 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.

∆l ,Pc 11 L0 2c 2

Pe P

0 0.05 0.1 0.15 ∆l L0

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: Z ∆l Epo = P dx. (2.4.2) 0 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

16 CHAPTER 2. THEORY OF THE BUCKLING OF A BEAM 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 Pc. And for the general f-s curve we get: n o P = min P straight,P buckled . (2.4.3)

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

17 2.4. THE FORCE-STRAIN CURVE

18 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.

P 0 x-axis l

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)

19 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 = A sin(kx) + Cx. (3.1.4) w(0) = wxx(0) = 0

And at x = l we get:

w = A sin(kx) + Cx sin(kl) ⇒ w = A sin(kx) − x . (3.1.5) w(l) = 0 l

The last boundary condition gives: sin(kl) 2 0 = A sin(kx) − x = −k A sin(kl). (3.1.6) l xx x=l 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) nπ k = with n ∈ N. (3.1.9) l 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

20 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:  2 2 !2 buckled nπ π n P = EI   = EI . (3.1.10) ∆l ∆l L 1 − L0 1 − L 0 L0 0 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). π 2 1 P buckled = EI . (3.1.11) L 2 0 1 − ∆l L0 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:    ∆l π 2 1  P = min EA0 ,EI . (3.1.12) L L 2  0 0 1 − ∆l   L0 

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: buckled Pc ≈ P (0). (3.1.13)

21 3.1. HINGED BEAM

∆l ,Pc 11 L0 2c 2

Pe P

0 0.05 0.1 0.15 ∆l L0

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 Pe is the Euler load of this beam.

This approximation is called Euler load,

2 buckled π Pe ≡ P (0) = EI . (3.1.14) L0

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

P Pr ≡ . (3.1.15) Pe Using the deﬁnition of the relative force we get for the straight regime:

∆l 2 EA0 A0 L0 ∆l P straight = L0 = . (3.1.16) r 2 I π L EI π 0 L0

For the buckled regime we ﬁnd:

2 π 1 EI L 2 0 1− ∆l buckled L0 1 Pr = 2 = 2 . (3.1.17) EI π 1 − ∆l L0 L0

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.

22 CHAPTER 3. CONFIGURATIONS WITH A SINGLE BEAM

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

buckled ∆l Pr ≈ 1 + 2 . (3.1.18) L0

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:

2 π PrEI 2 2 P PrPe L0 π k = = = = Pr , (3.1.19) EI EI EI L0 or π p k = Pr. (3.1.20) L0

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.

θ P 0 x-axis l

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)

23 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 constantµ ˜:

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.

Mspring ≈ −µ˜ · −wx(l) =µw ˜ x(l). (3.2.4)

The moment of the beam has to balance this moment of the spring, so:

Mbeam = −µw˜ x(l). (3.2.5)

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

EIwxx(l) = −µw˜ x(l), (3.2.6) or

µ˜ wxx(l) = − wx(l). (3.2.7) EI

µL˜ 0 Deﬁning the dimensionless quantity µ ≡ EI gives

µ wxx(l) = − wx(l). (3.2.8) L0 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 ofµ ˜. And (3.2.8) is the fourth boundary condition.

3.2.2 Solving the diﬀerential equation

2 We have to solve the diﬀerential equation wxxxx + k wxx = 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):

sin(kl) w = A sin(kx) − x . (3.2.9) l

The last boundary condition, (3.2.8), gives: sin(kl) µ sin(kl) A sin(kx) − x = − A sin(kx) − x . (3.2.10) l xx x=l L0 l x x=l

24 CHAPTER 3. CONFIGURATIONS WITH A SINGLE BEAM

After some calculations this reduces to: kl A = 0 or tan(kl) = . (3.2.11) 2 L0 1 + k l µ And for a buckled beam it holds that A =6 0 and k > 0, so the solution for the deﬂection is: sin(kl) w = A sin(kx) − x . (3.2.12) l Where k and l such that the following relation holds: kl tan(kl) = . (3.2.13) 2 L0 1 + k l µ

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 (0) hinged (0) of the hinged beam in section 3.1. So P = P . This gives that P (0) = Pe (µ) ∆l (0) and for the relative force, Pr , we have Pr (0) = 1. L0 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 for µ to zero Looking at the right hand side (r.h.s.) of (3.2.13) if µ goes to zero we see that: kl lim = 0, (3.2.14) µ↓0 2 L0 1 + k l µ and we get

tan(kl) = 0. (3.2.15)

sin(α) And because tan(α) = cos(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): ∆l ∆l 1 P (0) = P hinged = . (3.2.17) r L r L 2 0 0 1 − ∆l L0

25 3.2. HINGED BEAM WITH ROTATIONAL SPRING

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

kl lim = kl, (3.2.18) µ→∞ 2 L0 1 + k l µ and so

tan(kl) = kl. (3.2.19)

Calculating wx(l), using (3.2.19) gives: sin(kl) wx(l) = A sin(kx) − x , (3.2.20) l x x=l sin(kl) = A k cos(kl) − , (3.2.21) l cos(kl) = A (kl − tan(kl)) , (3.2.22) l = 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 α rather than a relation of k and l. So solving tan(α) = α gives k = 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 C k = . (3.2.24) l The equations (1.2.2) and (3.1.20) give for the rf-s curve:

∆l C 2 1 P (∞) = . (3.2.25) r L π 2 0 1 − ∆l L0

C 2 This diﬀers only by a factor π from Pr for a hinged beam:

2 C hinged ∆l = Pr (3.2.26) π L0 hinged ∆l = 2.0457... · Pr . (3.2.27) L0

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.

26 CHAPTER 3. CONFIGURATIONS WITH A SINGLE BEAM

General µ

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): q (µ) ∆l q π Pr 1 − (µ) ∆l L0 tan π Pr 1 − = . (3.2.28) L π2 (µ) ∆l 0 1 + Pr 1 − µ L0

In ﬁgure 3.5 relation (3.2.28) is plotted for ∆l = 0.05 and µ = 20. The blue line is L0 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).

4 ∆l tan π√Pr 1 − 1 1 L0 22 3 ∆l π√Pr 1− L0 π2 ! ∆"l 1+ Pr 1− µ L0 2 Solutions! "

−1

−2

−3 0 2 4 6 8 10 12 14 16 18 20 22 Pr

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 gives the rf-s curves for that specific spring constant. L0 For four µ values this curve is plotted in figure 3.6. (µ) ∆l Now the relation between Pr and , and so between k and l, is known, it L0 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 wx(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 (µ) with the Pr-axis, with other words Pr (0). This approximation as function of mu is plotted as the blue line in ﬁgure 3.8.

27 3.2. HINGED BEAM WITH ROTATIONAL SPRING

2.5

r 1.5 P

1 µ = ∞ µ = 10 = 2 0.5 µ µ = 0

0 0 0.05 ∆l 0.1 0.15 L0

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 ∞.

3.5 (µ) Pr (0) ′ P (µ) 3 r (0)

2.5

1.5

0.5 0 10 20 30 40 50 60 70 µ

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.

(0) If µ is zero, the buckling load is equal to the Euler load, so Pr (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 (µ)0 rf-s curve at zero strain as function of µ, Pr (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,

28 CHAPTER 3. CONFIGURATIONS WITH A SINGLE BEAM

1 µ = ∞ µ = 10 µ = 2 0.75 µ = 0

w 0.5

0.25

0 0 1 2 x 3 4 5

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 (µ) ∆l ˜(µ) ∆l L0 Pr = (µ) , (3.2.29) L0 P (0) instead of P (µ) ∆l P (µ) ∆l (µ) ∆l L0 L0 Pr = = (0) . (3.2.30) L0 Pe P (0)

If the force is scaled in this way the f-s curve becomes figure 3.9. Because of the (∞) ∆l (0) ∆l fact that Pr is a multiple of Pr , this two lines collapse. The lines for L0 L0 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:

(µ)0 P (0) ≡ r f(µ) (µ) . (3.2.31) Pr (0)

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.

29 3.2. HINGED BEAM WITH ROTATIONAL SPRING

1.4

1.3

1.2 (∞)

) P˜r µ ( r (10)

˜ P˜r P (2) 1.1 P˜r (0) P˜r

0.9 0 0.05 0.1 0.15 ∆l L0

Figure 3.9: The rescaled f-s curves of a hinged beam with a rotational spring. The curves for µ = 0 and for µ = ∞ coincide. The slopes for µ equal to 2 and 10 are lower than the slopes for µ equal to zero or inﬁnity.

2 lim f(µ) = 2 µ→∞ 1.95

1.9 ) µ ( f 1.85

1.8

1.75 0 10 20 30 40 50 60 70 µ

Figure 3.10: The slope of the f-s curve versus µ short after buckling, if the force is scaled with P (µ)(0). If µ = 0 or µ = ∞ the slope is 2, and for µ ∈ (0, ∞) it is smaller than 2.

30 Chapter 4 Conﬁgurations with more beams

In this chapter we will consider configurations with two or more beams. The beams are attached to each other in such a way that the angle between the beams is 90◦, so if the end of a beam rotates the beams that are attached to that beam have to rotate with it. In the experiments the angle between the beams is fixed because the beams are connected to small cubes of rubber. These cubes are hard to bend because they are thick and short. So the beams will bend instead of the cubes and the angle remains 90◦. In section 4.1 we will discuss a configuration that contains two beams and in section 4.2 we will investigate a network containing several beams.

4.1 Two beams conﬁguration

The configuration has a horizontal and a vertical beam, respectively beam 1 and beam 2, see figure 4.1. A force P is pressing on the top of beam 1. The begin point of beam 1 has the freedom to move horizontally. We expect that the buckling load is larger than that of a single hinged beam, because beam 1 works like the spring described in section 3.2. The angle between the x1-axis and beam 1 at x1 = l1 will be denoted with θ. ◦ Since the angle between the two beams is 90 the angle between the x2-axis and beam 2 at x2 = l2 is also θ. Instead of decreasing l and then calculating the corresponding k, as we have done in chapter 3, we will calculate a pair k, l2 for a given value of the angle θ in this section. This gives a point in the rf-s plane. We will do this for different values of θ, and all these points together will give us an rf-s curve.

4.1.1 Formulas for the deﬂections of the beams for a given angle. Beam 1 There is no force acting on beam 1, so with (2.3.15) we get:

3 2 w1(x1) = ax1 + bx1 + cx1 + d (4.1.1)

31 4.1. TWO BEAMS CONFIGURATION

x2-axis w2 = 0

Beam 2

Beam 1 θ l2 θ w1 = 0 x1-axis 0 l1

Figure 4.1: Sketch of the two beams. l1 and l2 are the distances between the two end of respectively beam 1 and beam 2. θ is angle between the beams and the axes at the endpoints of the beams. The begin point of beam 1 can move freely in the horizontal direction, so there is no horizontal force on beam 1.

The boundary conditions at x1 = 0 are w1(0) = 0; w1x1x1 (0) = 0, so: w (0) = 0 1 ⇒ b = d = 0. (4.1.2) w1x1x1 (0) = 0 3 This gives: w1(x1) = ax1 + cx1. The boundary conditions at x1 = l1 are w1(l1) = 0 and w1x1 (l1) = − arctan(θ) ≈ −θ for small θ: 3 w1(l1) = 0 ⇒ al1 + cl1 = 0 θ θ 2 ⇒ a = − 2 , c = . (4.1.3) w1x1 (l1) = −θ ⇒ 3al1 + c = −θ 2l1 2 Altogether we have: 2! θ 3 θ θ x1 w1(x1) = − 2 x1 + x1 = x1 1 − . (4.1.4) 2l1 2 2 l1

Beam 2 There is a force acting on beam 2, so (2.3.14) gives:

w2(x2) = A sin(kx2) + B cos(kx2) + Cx2 + D. (4.1.5) The four boundary conditions are:

w2(0) = w2(l) = w2x2x2 (0) = 0, (4.1.6)

w2x2 (l) = − arctan(θ) ≈ −θ. (4.1.7)

32 CHAPTER 4. CONFIGURATIONS WITH MORE BEAMS

The ﬁrst three boundary conditions are the same as for the hinged beam in section 3.1, so we can use result (3.1.5): sin(kl2) w2(x2) = A sin(kx2) − x2 . (4.1.8) l2 The fourth boundary condition gives: θ w2x2 (l2) = −θ ⇒ A = − . (4.1.9) k cos(kl ) − sin(kl2) 2 l2 If we combine (4.1.8) and (4.1.9) we get the result for the deﬂection of beam 2: sin(kl2) w2(x2) = A sin(kx2) − x2 , (4.1.10) l2 with θ A = − . (4.1.11) k cos(kl ) − sin(kl2) 2 l2

4.1.2 The length of the beams We can get more information about the deflection of the beams if we know the length of them, i.e. L1 and L2. To calculate the length of the beams we will deduce a general formula for the length of a beam as function of its deflecting. After that we will substitute the deflections of beam 1 and 2 in this general formula to get the length of the beams.

General beam length Let w(x) be the deﬂection of a beam between zero and l. For the length L we have:

Z Z l ds L = ds = dx. (4.1.12) 0 dx

ds Using the expression of dx that we get in section 2.1, (2.1.12), gives:

Z l p 2 L = 1 + wxdx. (4.1.13) 0

Using the ﬁrst order Taylor approximation of the square root, assuming that wx is small, gives:

Z l 1 2 L ≈ 1 + wxdx, (4.1.14) 0 2 Z l 1 2 = l + wxdx. (4.1.15) 2 0

33 4.1. TWO BEAMS CONFIGURATION

The length of beam 1

For beam 1 there is no force to compress the beam, so we have L1 = L0. Now using formula (4.1.4), we get:

!2 Z l1 1 θ 3 θ L0 = L1 = l1 + − x + x1 dx1. (4.1.16) 2 2l2 1 2 0 1 x1

After expanding the brackets and computing the integral we ﬁnd:

L0 l1 = θ2 . (4.1.17) 1 + 10

The length of beam 2 Equation (4.1.15) together with equation (4.1.10) gives:

!2 Z l2 1 sin(kl2) L2 = l2 + A sin(kx2) − x2 dx2. (4.1.18) 2 0 l2 x2

We compute this integral and ﬁnd:

2 1 A 2 2 L2 = l2 + −2 + 2k l2 + 2 cos(2kl2) + kl2 sin(2kl2) . (4.1.19) 8 l2

Now we have a relation between L2 and l2, but we don’t know L2. However, we can approximate L2 using the force P and L0. We know L2 if we know the normal force. Formula (1.1.6): N = EA ∆L gives 0 L0

N L2 = L0 1 − . (4.1.20) EA0

For small deﬂections the normal force is by approximation equal to the force P .

P L2 ≈ L0 1 − . (4.1.21) EA0

If we now use (2.3.10), the deﬁnition of k, we get for L2

I 2 L2 = L0 1 − k . (4.1.22) A0

Now combing (4.1.19) and (4.1.22), we get a relation for l2:

2 I 2 1 A 2 2 L0 1 − k = l2 + −2 + 2k l2 + 2 cos(2kl2) + kl2 sin(2kl2) . (4.1.23) A0 8 l2

34 CHAPTER 4. CONFIGURATIONS WITH MORE BEAMS

4.1.3 Momentum balance

The point (l1, l2) is hinged, so there is no net momentum,

M1(l1) + M2(l2) = 0. (4.1.24)

For the momentum we know from (2.1.25) that M(x) = EIwxx(x). This gives for M1 and M2:

M (l ) θ θ θ 1 1 − 3 − = 2 x1 + x1 = 3 (4.1.25) EI 2l1 2 x x l1 1 1 x1=l1

M2(l2) sin(kl2) = A sin(kx2) − x2 (4.1.26) EI l2 x2x2 x2=l2 2 = −Ak sin(kl2) (4.1.27)

Balancing of the moment gives: −3θ A = 2 (4.1.28) l1k sin(kl2)

4.1.4 Force-engineering strain curve Beam 2 can be compressed without affecting beam 1, so all formulas in the straight regime are equal to those for a single beam computed in section 2.4.1. From this we find the green line shown in figure 4.2. For the buckled regime we will find a pair of k and l2 for different values of θ. For this we need a system of two equations as there are two unknowns, k and l2. From equations (4.1.11) and (4.1.28) we get: θ  − 2 A = sin(kl2)  k cos(kl2)− 1 l1k l2 ⇒ k cos(kl ) − sin(kl ) + = 0. (4.1.29) −3θ 2 2 A = 2 l2 3 l1k sin(kl2)  Rewriting equation (4.1.29) gives a relation similar as in (3.2.19):

l2k tan(kl2) = 1 2 . (4.1.30) 1 + 3 l1l2k

And this gives some values for k for given values of l1 and l2. By substituting (4.1.17) into (4.1.30) we can remove the l1 dependence: , 1 L0 2 tan(kl2) = l2k 1 + θ l2k . (4.1.31) 3 1 + 10 And we can use the equations (4.1.11) and (4.1.23) to get another relation between k, θ and l2, 2! I 2 sin(kl2) l2 − L0 1 − k 8l2 k cos(kl2) − + A0 l2 2 2 2 θ −2 + 2k l2 + 2 cos(2kl2) + kl2 sin(2kl2) = 0. (4.1.32)

35 4.1. TWO BEAMS CONFIGURATION

Combining (4.1.31) and (4.1.32) gives the following system:   tan(kl ) =l k 1 + 1 L0 l k2  2 2 3 1+ θ 2  10 2 I 2 sin(kl2) (4.1.33) 0 = l2 − L0 1 − k 8l2 k cos(kl2) −  A0 l2  2 2 2  +θ −2 + 2k l2 + 2 cos(2kl2) + kl2 sin(2kl2) So we have to solve this system for different positive values of θ and k. If θ =6 0 we can solve system (4.1.33) numerically with Mathematica. This system has multiple solutions, but we need only the first solution with positive k. Because the beam will become by approximation half of a standing wave we expect that k is around π . After computing the pairs of k and l we can use (1.2.2) and (3.1.20) to l2 2 get the rf-s curve, as shown with a blue line in figure 4.2.

Pr 1

0 0 0.05 0.10 0.15 ∆l2 L0

∆l2 Figure 4.2: Pr as function of , the green line is the straight regime, the blue one is the L0 bended regime. The red point is the buckling point.

We will now approximate the buckling load and the slope of the rf-s curve shortly after buckling. Beam 1 prevents beam 2 to bend by applying a moment to the endpoint of beam 2 and this is similar to the spring in section 3.2. So if we know the spring constant corresponding by beam 1 we can use ﬁgure 3.8 to get an approximation of the buckling load and slope. The endpoint of beam 1 gives a moment given by equation (4.1.25): 3EI M1(l1) = − θ (4.1.34) l1 For a spring we used equation (3.2.3):

Mspring = −µθ˜ (4.1.35)

This means we can replace beam 1 with a spring with spring constant 3EI µ˜ = . (4.1.36) l1

36 CHAPTER 4. CONFIGURATIONS WITH MORE BEAMS

µL˜ 0 Using the same deﬁnition of µ as in section 3.2: µ ≡ EI gives

3L0 µ = . (4.1.37) l1

Equation (4.1.17) gives us l1 as function of θ and L0. Using this we get:

3 µ = 3 + θ2. (4.1.38) 10

So beam 1 is like a spring where the spring constant increases if the angle increases. At the moment of buckling the angle is zero, so we get

µ = 3. (4.1.39)

And now we can use ﬁgure 3.8 to get the approximation of the buckling load and slope. For µ = 3 we ﬁnd that the buckling load is equal to 1.41 and the slope of the relative force-stain curve equal to 2.47.

4.2 Network with many beams

We will now consider a conﬁguration consisting of one piece of rubber of about 10 by 10 centimetres with 99 holes in it, as shown in ﬁgure 4.3a.

(a) A piece of rubber with 99 holes in it. (b) The network after buckling.

Figure 4.3: The network made from one piece of rubber with holes, before and after buckling.

But a piece 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, see ﬁgure 4.4.

37 4.2. NETWORK WITH MANY BEAMS

Figure 4.4: The network with holes, and the approximation with blue beams between the holes and the blocks of rubber to connect them.

The experimental data of the f-s curve of the compression of the network is shown in ﬁgure 4.5. Both the compression and the release are measured.

P (N)

0 0 5 10 15 20 ∆l (mm)

Figure 4.5: A f-s curve of a large network. Increasing the engineering strain gives the upper line with a peak. Releasing the network gives the bottom line without a peak.

The first part of the graph is very flat, this is because the press was too high at the beginning. The steep part is the compressing of the horizontal beams, like in the case of the single beam. At the moment that one beam buckles, its neighbours have to buckle too, and so all the beams will buckle at the same moment, see figure 4.3b. This moment corresponds with the peak in the graph. Directly after the first buckling the forces decreases. After the returning point the force is always lower than before and there is only a small peak, the straight regime is almost the same. The two striking things are the peak and the absence of the peak at the return. The peak appears always, so it is not the unstable branch in the bifurcation. It is

38 CHAPTER 4. CONFIGURATIONS WITH MORE BEAMS local stable and at the return the conﬁguration is on a other stable branch with a lower force. The peak might be caused by the blocks of rubber, by the fact that there are many beams or some other cause.

39 4.2. NETWORK WITH MANY BEAMS

40 Chapter 5 Discussion

The model that we have deduced in this thesis describes three observables, the f-s curve before buckling, the buckling load and the f-s curve after buckling. The linear approximation of the response of the force on the strain, given in equation (2.4.1), is accurate. Because of this, the approximation of the f-s curve is accurate for the straight regime. Equation (2.4.1) is, after all, the only thing that has been used for the straight regime. The estimation of the buckling load is accurate. The extra relation between the force and the engineering strain which comes from the model, is indeed satisﬁed at the buckling moment. The results for the f-s curve after buckling do not agree with experiments. Using the same relation that we used to calculate the buckling load, we get for the rf-s curve a slope of around two. In experiments on the other hand the slope is less than a half. So the relation between the force and the engineering strain is not valid after the moment of buckling itself. Also directly after buckling the slope of the calculated rf-s curve is too large. It is not clear why this relation does not hold for a buckled beam. An option is that the neglected higher orders of w changes this relation in such a way that the slope after buckling decreases.

41 42 Chapter 6 Conclusions

In this thesis we have deduced a model to describe the buckling load, the deﬂection and the force-engineering strain curve for a slender, rubber beam. If we apply a force on a rubber beam, the beam will compress. For a larger force the beam will buckle.

6.1 General results from the model

For a small force the configuration of a straight beam is stable. For a large force the straight beam is an unstable configuration and a bended, or buckled, beam is the stable configuration. This is described by a pitchfork bifurcation with the force as bifurcation parameter. The buckling load of the beam is defined as the bifurcation point. The buckling of beams is described by the following ordinary differential equation for the deflection of the beam: 2 wxxxx+k wxx = 0, (6.1.1) P k2 = , (6.1.2) EI with the general solution: w(x) = A sin(kx) + B cos(kx) + Cx + D. (6.1.3) In addition, we derive a relation that can be used to calculate the force-strain curve before and shortly after buckling. This relation follows from the boundary conditions.

6.2 Force-engineering strain curve for speciﬁc conﬁgurations

The force-engineering strain curve for a hinged beam is shown in ﬁgure 6.1. The buckling load of the hinged beam is the Euler load: π 2 Pe = EI . (6.2.1) L0

43 6.2. FORCE-ENGINEERING STRAIN CURVE FOR SPECIFIC CONFIGURATIONS

∆l ,Pc 11 L0 2c 2

Pe P

0 0.05 0.1 0.15 ∆l L0

Figure 6.1: The force-engineering strain 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 Pe is the Euler load of this beam.

If we scale the force with the Euler load we get the relative force. The slope of the relative force-engineering strain curve after buckling is equal to 2. With a rotational spring added at one end of the beam, the buckling load and the slope of the relative force-engineering strain curve increase as function of the spring constant µ. This is shown in ﬁgure 6.2, the blue line is the relative buckling load and the green line is the slope of the relative force-engineering strain curve after buckling as function of the spring constant µ.

3.5 (µ) Pr (0) ′ P (µ) 3 r (0)

2.5

1.5

0.5 0 10 20 30 40 50 60 70 µ

Figure 6.2: The blue line is the relative buckling load as function of µ. The green line is the slope of the relative force-engineering strain curve after buckling as function of the spring constant µ.

If a second beam perpendicular to the ﬁrst beam is added instead of a spring, this beam behaves as a spring with spring constant of around 2.5, section 4.1.

44 Appendix A

The slope for a single beam with a spring after buckling

(µ) ∆l For the slope of the rf-s curve we will assume that Pr is a function of . Equa- L0 (µ) ∆l tion (3.2.28) gives a relation between Pr and : L0

q (µ) ∆l q π Pr 1 − (µ) ∆l L0 tan π Pr 1 − = . (A.0.1) L π2 (µ) ∆l 0 1 + Pr 1 − µ L0

We will use implicit diﬀerentiation of this equation to get the slope of the rf-s curve. Diﬀerentiating of the l.h.s. of (A.0.1) with respect to ∆l gives: L0

" s !# (µ) ∆l ∆l tan π Pr 1 − L0 L0 ∆l L0    (µ)0 q (µ)0 2 Pr (µ) Pr ∆l = q π  q −  Pr + q  . (µ) ∆l (µ) (µ) L0 cos 2π Pr 1 − + 1 2 Pr 2 Pr L0 (A.0.2)

45 A.1. LIMITS FOR THE SPRING CONSTANT TO ZERO AND INFINITY

And for the r.h.s. we get:  r  (µ) ∆l ∆l π Pr 1 −  L0 L0    π2 (µ) ∆l ∆l 1 + Pr 1 −  µ L0 L0 ∆l L0 0 0 2 (µ) q (µ) π (µ) ∆l Pr (µ) Pr ∆l 1 + Pr 1 − π √ − Pr + √ µ L0 (µ) 2 P L0 2 Pr = 2 π2 (µ) ∆l 1 + Pr 1 − µ L0 q 0 0 (µ) ∆l π2 (µ) (µ) ∆l (µ) π Pr 1 − Pr − Pr − Pr L0 µ L0 − 2 (A.0.3) π2 (µ) ∆l 1 + Pr 1 − µ L0 Since the l.h.s. and the r.h.s. are equal as functions, the derivatives have to be equal too. So we know that the r.h.s.’s of the equations (A.0.2) and (A.0.3) are the same. We will know the slope just after buckling, so the engineering strain is small. 0 ∆l (µ) Approximate the slope by substituting = 0 and isolating Pr gives: L0

2 (µ) (µ) P (µ)(0) π 2Pr (0) − Pr (0) r µ q π2 (µ) + 2 (µ) 2 (µ) 1+ Pr (0) π cos 2π Pr (0) +1 µ 1+ µ Pr (0) (µ)0 Pr (0) = 2 . (A.0.4) P (µ)(0) π 1 1 r µ − 2 + 2 q (µ) π (µ) π2 (µ) 2 1+ Pr (0) cos 2π Pr (0) +1 µ 1+ µ Pr (0)

(µ) (µ) So now we have the slope expressed in µ and Pr (0). And we know Pr (0), this is the blue line in ﬁgure 3.8. So now we have the slope of the rf-s curve short after buckling as function of µ. And this gives the green line in ﬁgure 3.8.

A.1 Limits for the spring constant to zero and inﬁnity

To see that formula (A.0.4) can be true, we will consider the behaviour while taking the limits of µ to zero and inﬁnity. We have already seen in equation (3.2.17) that for µ ↓ 0, equation (A.0.4) has to reduce to the case of a hinged beam. Taking the limit of µ to zero gives:

2 (µ) (µ) P (µ)(0) π 2Pr (0) − Pr (0) r µ q π2 (µ) + 2 (µ) 2 (µ) 1+ Pr (0) π cos 2π Pr (0) +1 µ 1+ µ Pr (0) (0)0 Pr (0) = lim 2 (A.1.1) µ↓0 P (µ)(0) π 1 1 r µ − 2 + 2 q (µ) π (µ) π2 (µ) 2 1+ Pr (0) cos 2π Pr (0) +1 µ 1+ µ Pr (0) (0) = 2Pr (0). (A.1.2) And P (0)(0) is equal to 1 so we get that the slope after buckling is 2. This is the same as what we get for the hinged beam.

46 CHAPTER A. THE SLOPE FOR A SINGLE BEAM WITH A SPRING AFTER BUCKLING

For the limit of µ to infinity we know that the the force only differs a factor of approximately 2.04 with respect to the hinged beam, see (3.2.27). So for the differentiation we get:

" 2 #0 0 C P (∞) (0) = P hinged (0) (A.1.3) r π r 2 2 C 0 C = P hinged (0) = 2 . (A.1.4) π r π

(∞) And for Pr (0) we have:

C 2 C 2 P (∞)(0) = P hinged(0) = . (A.1.5) r π r π

And so the slope is two times as large as the relative buckling load. And indeed, if we take the limit for µ → ∞ we get:

2 (µ) (µ) P (µ)(0) π 2Pr (0) − Pr (0) r µ q π2 (µ) + 2 (µ) 2 (µ) 1+ Pr (0) π cos 2π Pr (0) +1 µ 1+ µ Pr (0) (∞)0 Pr (0) = lim 2 (A.1.6) µ→∞ P (µ)(0) π 1 1 r µ − 2 + 2 q (µ) π (µ) π2 (µ) 2 1+ Pr (0) cos 2π Pr (0) +1 µ 1+ µ Pr (0) (∞) = 2Pr (0). (A.1.7)

47 A.1. LIMITS FOR THE SPRING CONSTANT TO ZERO AND INFINITY

48 Appendix B List of notation used

Symbol Meaning Calculation wwwwwwDimensions of the beam L0 Inertial length L Compressed length l Distance between the two ends h Height b Width w Deﬂection r Distance to the middle line of the beam A0 Cross-section (area) A0=h · b I Second moment of area I=RR r2dA A0 0 wwwwwwStrain ∆L Strain ∆L = L0−L L0 L0 L0 ∆l Engineering strain ∆l = L0−l L0 L0 L0 ∆l Critical engineering strain L0 c Local strain b Local strain caused by the bending b=−rwxx wwwwwwForces N Normal force P Force Pc Buckling load 2 P Euler load P =EI π e e L0 P Relative force P = P r r Pe σ Stress σ= N − Erw A0 xx M Moment M=EIwxx wwwwwwGeneral . E Young’s modulus E= N ∆L A0 L0 µL˜ 0 µ spring constant µ= EI

49 50 Bibliography

[BC10] Zdenek P Bazant and Luigi Cedolin. Stability of structures: elastic, inelastic, fracture and damage theories. World Scientiﬁc Publishing Company, 2010.

[TG70] SP Timoshenko and JN Goodier. Theory of elasticity 3rd edition, 1970.