On the Curvature of an Euler–Bernoulli Beam

International Journal of Mechanical Engineering Education 31/2

On the curvature of an EulerBernoulli beamOsman Kopmaza (corresponding author) and mer Gndogduba Department of Mechanical Engineering, Uludag University, 16059 Bursa, Turkeyb Department of Mechanical Engineering, Ataturk University, 25240 Erzurum, TurkeyE-mail: [email protected]; [email protected]

Abstract This paper deals with different approaches to describing the relationship between thebending moment and curvature of a EulerBernoulli beam undergoing a large deformation, from atutorial point of view. First, the concepts of the mathematical and physical curvature are presented indetail. Then, in the case of a cantilevered beam subjected to a single moment at its free end, thedifference between the linear theory and the nonlinear theory based on both the mathematicalcurvature and the physical curvature is shown. It is emphasized that a careless use of the nonlinearmathematical curvature and moment relationship given in most standard textbooks may lead toerroneous results. Furthermore, a numerical example is given for the reader to make a quantitativeassessment.

Keywords beam deflection; beam curvature; bending

Introduction

The EulerBernoulli theory of beams provides a reasonable explanation of thebending behavior of long isotropic beams. It is based on the assumption that a rela-tionship between bending moment and beam curvature exists. This relationship ismathematically stated as

(1)

where k, M and EI denote the beam curvature, the bending moment at any cross-section of the beam, and the bending rigidity of the beam, respectively. Eqn (1) is themost significant result of the EulerBernoulli theory, and it is experimentally verifiedfor isotropic materials in situations where deflection due to shear may be neglected.As will be seen later, eqn (1) is a nonlinear differential equation associated with thedeflection curve of a beam. However, in almost all standard textbooks on the strengthof materials, the linearized form of eqn (1) is used, with the assumption that in mostcommon structural and mechanical engineering applications large deformations arenot allowed, for safety reasons [1, 2]. Indeed, this assumption is valid for a wide rangeof applications. However, the deflections of a leaf spring used in ground vehicles, forinstance, must be studied by using a nonlinear bending curvature equation [3, 4].Moreover, increasing demands for higher operational speeds and lighter, more compliant device constructions are leading to ever more situations in which a linear theory of beams no longer gives satisfactory results. In the past 50 years, especially, developments in aerospace engineering, robotics and manufac-turing have led engineers to excessively use nonlinear models that must be solved

k =MEI

Curvature of an EulerBernoulli beam 133


Fig. 1 A schematic for derivation of eqn (6).

numerically. Consequently, the scientists and engineers who wish to deeply under-stand some physical phenomena that occur in mechanical systems which operate athigh speeds and undergo large deflections must use the nonlinear strain formulas and nonlinear curvature. A considerable amount of literature on nonlinear mechanicsappeared in the late 1970s and early 80s. During this period, researchers used,improved and discussed nonlinear straindeflection formulas needed to study thesystems with large deformations and/or geometric stiffening [314]. Since the aim of this paper is to treat the relationship between curvature and bending moment with a tutorial approach, not all the literature associated with this subject will be citedhere.

Curvature of a planar curve

Consider an in-plane loaded uniform beam with coinciding shear centerline and centroidal line. The curvature of a planar curve is defined as follows:

(2)

where q is the slope angle of the tangent at any point on the elastic curve, and s isthe arc length of the curve measured from an arbitrary starting point (see Fig. 1).

It is well known from differential calculus [15] that

(3)

and

(4)d d d dd

dx x y yx

x= +( ) = +

2 2 1 22 1 2

1

tan q =dd

yx

kq

=dds

Differentiating eqn (3) and solving for dq provides

(5)

The purpose of rewriting these relationships, which can easily be found in any standard calculus book, is to attract attention to the fact that the x variable here denotes the abscissa of any point of the curve in its final position. Substitutingeqns (4 and 5) into eqn (2) yields the well known curvature formula, as follows:

(6)

Eqn (6) is referred to as an explicit expression of the beam curvature in some text-books, and is used in the literature to obtain large deflections of beams [11, 12]. Eqn (6) is a correct formula for the curvature of planar curves, but there is a little confusion in the application of this expression to problems. In the mechanics ofdeformable bodies, deflections can be referred to either the points of physical space(Eulerian description) or the material points of the bodies (Lagrangian description)[16]. The Lagrangian description is preferably used in texts on strength of materials,where a material point is represented with its coordinates before deformation. Such a coordinate is a kind of label for that point, whereas the x variable in the curvature expression demonstrates the final or instantaneous position of any point.Therefore, when eqn (6) is used, one must take into consideration that the x coor-dinate is defined in a Eulerian sense, because in this case a certain x coordinate rep-resents a different material point. Otherwise, as will be shown later, incorrect resultswill be obtained. The physical curvature of a beam differs from the mathematical curvature in that a certain x coordinate on a physical curve corresponds to the samematerial point, as opposed to the mathematical curvature case [17].

The physical curvature can also be obtained from eqn (2). A different notation willbe used for elastic displacements. Elastic displacements of any point on the medianline of a beam in the x and y directions are denoted as u and v, respectively. ConsiderFig. 2, which shows the final (or instantaneous, in a dynamic case) positions of twopoints which were infinitesimally close to each other before deformation.

It is obvious that the following relationships exist:

(7)

and

tan q =+

dd

dd

v

xu

x1

k =+

dddd

2 yx

yx

2

2 3 2

1

dd

dd

dd

2

qt

yx

yx

=+

2

2

1

134 O. Kopmaz and . Gndogdu




(8)

Note that dr approaches ds at the limit. From eqn (7), it follows that:

(9)

Substituting eqns (8 and 9) into eqn (2) yields the physical curvature as follows:

(10)

Eqn (10) was derived in a different way, based on SerretFrenet unit vectors in [17].If one defines a new variable representing the final coordinate on the x axis of apoint of beam as

(11)it is easy to prove that

(12)k =+

dddd

2

2v

Xv

X1

2 3 2

X x u x= + ( )

k =+

-

+ +

dd

dd

dd

dd

dd

dd

2

2v

x

u

x

v

x

u

x

u

x

v

x

1

12 2 3 2

d =

dd

dd

dd

dd

dd

dd

2 2

q

v

x

u

x

v

x

u

x

u

x

v

x

2 2

2 2

1

1

+ -

+ +

d = 1+ dd

dd

su

x

v

x

+

2 2 1 2

Fig. 2 A schematic for derivation of eqn (10).



is still a valid formula [17]. However, the integration of eqn (12) is difficult becausethe boundaries of the integral are variable. Instead, many authors prefer to use eqn (6) by making successive corrections in the moment expression [10, 11].

A clampedfree beam

In this section, based on the EulerBernoulli theory, the analytical expressions ofthe elastic deflections of a beam will be derived using the mathematical and physi-cal curvatures, and linear theory for comparison purposes. Assuming that the beamis clamped at one end and free at the other, then the beam is subjected to a singlemoment at its free end. The reason for choosing such a basic example is to have asimple deflection curve with a constant radius of curvature as

(13)

Since a planar curve with a constant radius of curvature is known to be a circle, thedeflection curve of the beam will be a circular arc. Then, with the help of Fig. 3, thefollowing relationships can be written:

(14)(15)

(16)u X x R xR

x= - = -sin

X R Rx

R= =

sin sinj

OA OA x R = = = j

RMEI

= =1k

Fig. 3 A beam in deflection.



and

(17)

Eqns (16 and 17) give the horizontal and vertical deflections of a point of thebeam. Furthermore, note that the u deflections are due to transversal deflectionsbecause the median line is inextensional for this kind of loading.

Eqns (1417) should satisfy both eqn (10) and eqn (12), because we obtainedthese relationships by considering the displacement of a certain material point. Infact, when the expressions

(18)

(19)

(20)

and

(21)

are substituted into eqn (10), one finds that k = 1/R, as expected. (Note that u andv are the solutions to eqn (10)). Also, if we express v in terms of X, and substituteinto eqn (12) after taking the necessary derivatives with respect to X, we also obtaink = 1/R.

Replacing the variable y with v, and substituting the derivatives of v with respectto x into eqn (6), will not satisfy k = 1/R:

The reason is that the x variables in eqn (6) and eqn (10) have different meanings,as explained above. Specifically, x represents the final position of a certain materialpoint.

At this point, the question may arise whether the differential equation that isobtained by equating eqn (6) to a constant (radius of curvature) would give an equa-tion for a circle. Of course, the answer will be affirmative. This circle will also beidentical to the one that is used to obtain the u and v deflections in eqns (16 and 17).However, if we want to find the tip deflection, and, for this purpose, if we substi-tute x = L in the expression obtained by integrating eqn (6), the value that we find

k =+

1

1

12

3 2R

x

Rx

RR

cos

sin

dd

2

2v

x Rx

R=

1cos

dd

v

x

x

R=

sin

dd

2

2u

x Rx

R= -

1sin

ddu

x

x

R=

-cos 1

v R Rx

R= -( ) = -

1 1cos cosj



will be incorrect. To clarify this point, consider eqn (6) and equate it to 1/R. The following is obtained:

(22)

where ( ) denotes the derivation with respect to x, for convenience. Integrating eqn (22) yields:

(23)

where we used the boundary condition v(0) = 0 for the cantilevered end of the beam.Arranging and integrating eqn (23) once more gives

(24)

where the boundary condition v(0) = 0 was used, as well. The equation of the circlegiven by eqn (24) can be expressed in the implicit form as follows:

(25)Provided that x in eqn (25) is replaced with X, the true (deformed) material pointlocation, it is obvious that this circle is identical to the one that is used to obtain thetrue u and v deflections. Thus, the x variable in eqn (6) must be considered as X ineqn (12). Ignoring this significant point, for example, if we choose x = L in eqn (24)to find the tip deflection, will give a value of

which might be completely meaningless in the case of L > R. Fig. 4 will be helpfulto understand the mistake caused by using x instead of X. From Fig. 4, we see thatthe deflection from eqn (24) is greater than the actual one, or no real solution mayexist.

Finally, we want to find the deflection curve based on the linear theory; then, wemay write the differential equation of the elastic curve as follows:

(26)

Its solution under the boundary conditions v(0) = v(0) = 0 is:

(27)

In the linear case, of course, the axial displacements, i.e. the u deflections, areassumed to vanish implicitly.

vx M

EIx

R xR

@ = = k

22

2

212 2

@ =vMEI

k

v R R L= - -2 2

v R x R-( ) + =2 2 2

v x R R x Rx

R( ) = - - = - -

2 2

21 1

+ ( )

= =v

v Rx

Rx

x

11 1

2 1 20

d

+ ( ) =

v

v R11

2 3 2



Now, we have three different expressions for the v deflections of the beam, insummary:

(17 repeated)

(24 repeated)

(27 repeated)

where the superscripts PC, MC! and Lin indicate that the deflection formula is basedon the physical curvature, the misused mathematical curvature, and linear theory,respectively. Assuming R > L, and expanding cos(x/R) and [1 - (x/R)]1/2 into powerseries around x = 0 yield

(28)

(29)

After substituting the series given above into eqns (17 and 24), the resulting formulas, along with eqn (27), take the following forms:

(30)

(31)MC!v Rx

Rx

Rx

R=

-

+

12

18

116

2 4 6

...

PC v Rx

Rx

Rx

R=

-

+

-

12

124

1720

2 4 6...

1 1 12

18

116

12 2 4 6

- = -

+

-

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On the Curvature of an Euler–Bernoulli Beam