On the Curvature of an Euler–Bernoulli Beam

On the curvature of an Euler–Bernoulli beam Osman Kopmaza (corresponding author) and Ömer Gündog˘dub Department of Mechanical Engineering, Uludag University, 16059 Bursa, Turkey b Department of Mechanical Engineering, Ataturk University, 25240 Erzurum, Turkey E-mail: [email protected]; [email protected]

a

Abstract This paper deals with different approaches to describing the relationship between the bending moment and curvature of a Euler–Bernoulli beam undergoing a large deformation, from a tutorial point of view. First, the concepts of the mathematical and physical curvature are presented in detail. Then, in the case of a cantilevered beam subjected to a single moment at its free end, the difference between the linear theory and the nonlinear theory based on both the mathematical curvature and the physical curvature is shown. It is emphasized that a careless use of the nonlinear mathematical curvature and moment relationship given in most standard textbooks may lead to erroneous results. Furthermore, a numerical example is given for the reader to make a quantitative assessment. Keywords beam deflection; beam curvature; bending

Introduction The Euler–Bernoulli theory of beams provides a reasonable explanation of the bending behavior of long isotropic beams. It is based on the assumption that a relationship between bending moment and beam curvature exists. This relationship is mathematically stated as k=

M EI

(1)

where k, M and EI denote the beam curvature, the bending moment at any crosssection of the beam, and the bending rigidity of the beam, respectively. Eqn (1) is the most significant result of the Euler–Bernoulli theory, and it is experimentally verified for 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 the deflection curve of a beam. However, in almost all standard textbooks on the strength of materials, the linearized form of eqn (1) is used, with the assumption that in most common structural and mechanical engineering applications large deformations are not allowed, for safety reasons [1, 2]. Indeed, this assumption is valid for a wide range of applications. However, the deflections of a leaf spring used in ground vehicles, for instance, 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 manufacturing have led engineers to excessively use nonlinear models that must be solved International Journal of Mechanical Engineering Education 31/2

Curvature of an Euler–Bernoulli beam

Fig. 1

133

A schematic for derivation of eqn (6).

numerically. Consequently, the scientists and engineers who wish to deeply understand some physical phenomena that occur in mechanical systems which operate at high speeds and undergo large deflections must use the nonlinear strain formulas and nonlinear curvature. A considerable amount of literature on nonlinear mechanics appeared in the late 1970s and early ’80s. During this period, researchers used, improved and discussed nonlinear strain–deflection formulas needed to study the systems with large deformations and/or geometric stiffening [3–14]. 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 cited here. 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: k=

dq ds

(2)

where q is the slope angle of the tangent at any point on the elastic curve, and s is the arc length of the curve measured from an arbitrary starting point (see Fig. 1). It is well known from differential calculus [15] that tan q =

dy dx

(3)

and 12

12

dx = (dx 2 + dy 2 )

dy 2 ˘ È = Í1 + Ê ˆ ˙ dx Î Ë dx ¯ ˚

(4)

International Journal of Mechanical Engineering Education 31/2

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O. Kopmaz and Ö. Gündog˘du

Differentiating eqn (3) and solving for dq provides d2 y dq dx 2 = dy 2 dt 1+ Ê ˆ Ë dx ¯

(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. Substituting eqns (4 and 5) into eqn (2) yields the well known curvature formula, as follows: d2 y dx 2 k= 32 È Ê dy ˆ 2 ˘ 1 + Í Ë dx ¯ ˙ Î ˚

(6)

Eqn (6) is referred to as an explicit expression of the beam curvature in some textbooks, 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 of deformable 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 coordinate is defined in a Eulerian sense, because in this case a certain x coordinate represents a different material point. Otherwise, as will be shown later, incorrect results will 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 same material point, as opposed to the mathematical curvature case [17]. The physical curvature can also be obtained from eqn (2). A different notation will be used for elastic displacements. Elastic displacements of any point on the median line of a beam in the x and y directions are denoted as u and v, respectively. Consider Fig. 2, which shows the final (or instantaneous, in a dynamic case) positions of two points which were infinitesimally close to each other before deformation. It is obvious that the following relationships exist: dv tan q = dx du 1+ dx and International Journal of Mechanical Engineering Education 31/2

(7)


Fig. 2

du 2 dv 2 ˘ È ds = ÍÊ 1+ ˆ + Ê ˆ ˙ ÎË dx ¯ Ë dx ¯ ˚

135

A schematic for derivation of eqn (10).

12

(8)

Note that dr approaches ds at the limit. From eqn (7), it follows that: d 2 v Ê du ˆ dv d 2 u 1+ dx 2 Ë dx ¯ dx dx 2 dq = 2 2 Ê 1 + du ˆ + Ê dv ˆ Ë dx ¯ Ë dx ¯

(9)

Substituting eqns (8 and 9) into eqn (2) yields the physical curvature as follows: d 2 v Ê du ˆ dv du 1+ dx 2 Ë dx ¯ dx dx k= 32 ÈÊ du ˆ 2 Ê dv ˆ 2 ˘ 1 + + ÍË dx ¯ Ë dx ¯ ˙ Î ˚

(10)

Eqn (10) was derived in a different way, based on Serret–Frenet unit vectors in [17]. If one defines a new variable representing the final coordinate on the x axis of a point of beam as X = x + u( x )

(11)

it is easy to prove that d2v dX 2 k= 32 È Ê dv ˆ 2 ˘ 1 + Í Ë dX ¯ ˙ Î ˚

(12)


136


Fig. 3

A beam in deflection.

is still a valid formula [17]. However, the integration of eqn (12) is difficult because the 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 clamped–free beam In this section, based on the Euler–Bernoulli theory, the analytical expressions of the elastic deflections of a beam will be derived using the mathematical and physical curvatures, and linear theory for comparison purposes. Assuming that the beam is clamped at one end and free at the other, then the beam is subjected to a single moment at its free end. The reason for choosing such a basic example is to have a simple deflection curve with a constant radius of curvature as R=

1 M = k EI

(13)

Since a planar curve with a constant radius of curvature is known to be a circle, the deflection curve of the beam will be a circular arc. Then, with the help of Fig. 3, the following relationships can be written: OA¢ = OA = x = Rj x X = R sin j = R sinÊ ˆ Ë R¯ x u = X - x = R sinÊ ˆ - x Ë R¯ International Journal of Mechanical Engineering Education 31/2

(14) (15) (16)


137

and x ˆ Ê v = R(1 - cos j ) = R 1 - cosÊ ˆ Ë R¯¯ Ë

(17)

Eqns (16 and 17) give the horizontal and vertical deflections of a point of the beam. Furthermore, note that the u deflections are due to transversal deflections because the median line is inextensional for this kind of loading. Eqns (14–17) should satisfy both eqn (10) and eqn (12), because we obtained these relationships by considering the displacement of a certain material point. In fact, when the expressions du x = cosÊ ˆ - 1 Ë R¯ dx

(18)

d 2u 1 x = - sinÊ ˆ dx 2 R Ë R¯

(19)

dv x = sinÊ ˆ Ë R¯ dx

(20)

and d2v 1 x = cosÊ ˆ 2 Ë dx R R¯

(21)

are substituted into eqn (10), one finds that k = 1/R, as expected. (Note that u and v are the solutions to eqn (10)). Also, if we express v in terms of X, and substitute into eqn (12) after taking the necessary derivatives with respect to X, we also obtain k = 1/R. Replacing the variable y with v, and substituting the derivatives of v with respect to x into eqn (6), will not satisfy k = 1/R: 1 x cos 1 R R k = 32 π x R Ê 1 + sin 2 ˆ Ë 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 material point. At this point, the question may arise whether the differential equation that is obtained by equating eqn (6) to a constant (radius of curvature) would give an equation for a circle. Of course, the answer will be affirmative. This circle will also be identical 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 substitute x = L in the expression obtained by integrating eqn (6), the value that we find International Journal of Mechanical Engineering Education 31/2

138


will be incorrect. To clarify this point, consider eqn (6) and equate it to 1/R. The following is obtained: v ¢¢ 1 = 2 32 R (1 + v ¢ )

(22)

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

1 1 v¢ = Ú dx = x 2 12 R0 R (1 + v ¢ )

(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 È x 2˘ v(x ) = R - R 2 - x 2 = R Í1 - 1 - Ê ˆ ˙ Ë R¯ ˚ Î

(24)

where the boundary condition v(0) = 0 was used, as well. The equation of the circle given by eqn (24) can be expressed in the implicit form as follows: (v - R)2 + x 2 = R 2

(25)

Provided that x in eqn (25) is replaced with X, the true (deformed) material point location, it is obvious that this circle is identical to the one that is used to obtain the true u and v deflections. Thus, the x variable in eqn (6) must be considered as X in eqn (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 v = R - R 2 - L2 which might be completely meaningless in the case of L > R. Fig. 4 will be helpful to understand the mistake caused by using x instead of X. From Fig. 4, we see that the deflection from eqn (24) is greater than the actual one, or no real solution may exist. Finally, we want to find the deflection curve based on the linear theory; then, we may write the differential equation of the elastic curve as follows: v ¢¢ @ k =

M EI

(26)

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

x2 1 M 2 R Ê x ˆ 2 = x = 2 2 EI 2 Ë R¯

(27)

In the linear case, of course, the axial displacements, i.e. the u deflections, are assumed to vanish implicitly. International Journal of Mechanical Engineering Education 31/2


Fig. 4

139

Comparison of different cases.

Now, we have three different expressions for the v deflections of the beam, in summary: PC

x v = R(1 - cos j ) = RÊ 1 - cos ˆ Ë R¯

MC!

Lin

(17 repeated)

È x 2˘ v(x ) = R - R 2 - x 2 = R Í1 - Ê ˆ ˙ Ë R¯ ˚ Î

v =k

(24 repeated)

x2 1 M 2 R Ê x ˆ 2 = x = 2 2 EI 2 Ë R¯

(27 repeated)

where the superscripts PC, MC! and Lin indicate that the deflection formula is based on 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 power series around x = 0 yield cos

x 1 x 2 1 Ê xˆ4 1 Ê x ˆ 6 ... = 1- Ê ˆ + + R 2 Ë R¯ 24 Ë R ¯ 720 Ë R ¯

x 2 1 x 2 1 x 4 1 x 6 1 - Ê ˆ = 1 - Ê ˆ + Ê ˆ - Ê ˆ ... for Ë R¯ 2 Ë R¯ 8 Ë R¯ 16 Ë R ¯

(28) x <1 R

(29)

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

È1 x 2 1 Ê x ˆ 4 1 Ê x ˆ 6 ...˘ v = RÍ Ê ˆ + - ˙ 24 Ë R ¯ 720 Ë R ¯ Î 2 Ë R¯ ˚

MC!

È1 x 2 1 x 4 1 x 6 ˘ v = R Í Ê ˆ - Ê ˆ + Ê ˆ ...˙ 8 Ë R¯ 16 Ë R ¯ Î 2 Ë R¯ ˚

(30)

(31)


140


Lin

È1 x 2 ˘ v = RÍ Ê ˆ ˙ Î 2 Ë R¯ ˚

(27 repeated)

A comparison of these formulas justifies that for small deflections the linear theory will give sufficiently accurate results, but at large deflections the results deviate markedly. To compare the results of eqns (17, 24 and 27), let us assign some reasonable values to the physical parameters of the beam, which will give the radius of curvature. In order that the stress and strain stay in the elastic region, we must choose a suitable bending moment value. For standard steel, E = 2.1 ¥ 1011 N/m2 and sSafety = 1.4 ¥ 108 N/m2, and it will be assumed that the cross-section is solid circular. Based on the condition smax £ sSafety, we find that M should be equal to pD3sSafety/32. Furthermore, EI = pD4E/64. Consequently, R = EI/M = ED/2sSafety. If we substitute the numerical values of E and sSafety, we obtain R = 750D. If we prefer substituting the value of D in mm, the relationship between R and D becomes as follows: R[ m] = 0.75 D[ mm] Let us take D = 5 mm; then R will be 3.75 m. Assume that L = 3 m. The exact deflections of the tip point then are PC

v x = L = 1.337 m

PC

u x = L = -0.309 m

MC! Lin

v x = L = 1.500 m

v x = L = 1. 200 m

As is seen from these results, the difference between PCv and Linv is about 10% while that between MC!v and PCv is around 12%. The formula for MC!vx=L gives the real values for D £ 4 mm. A discussion on curvature formulas and conclusions It should be remembered that the curvature formula given by eqn (6) is based on the final (or instantaneous) state of the elastic curve of a beam. Otherwise, this formula will lead to incorrect results, as shown in the preceding section. Some authors use this formula in an iterative manner. Specifically, they solve the differential eqn (6) starting from the initial bending moment function. Then, they correct the moment expression according to the new beam configuration by using an arc integral, and continue this process so as to reach final true elastic deflections within a certain error tolerance [10, 11]. When the differential eqn (10) is integrated, it will give the direct solution. However, there is a difficulty in integrating it. In this equation, two different dependent variables, u and v, appear. For the inextensional case, u can be expressed in terms of v and its derivatives. For this special case, there is a relationship between u and v: 1 = (1 + u ¢)2 + v ¢ 2 International Journal of Mechanical Engineering Education 31/2

(32)


141

Fig. 5 Point at which eqns (36 and 37) fail.

or u ¢ 2 + 2u ¢ + v ¢ 2 = 0

(33)

Solving eqn (33) for u yields: u ¢ = -1 ± 1 - v ¢ 2

(34)

If it is assumed that u¢ 2 << u¢, we obtain u from eqn (33) as 1 u¢ = - v¢ 2 2

(35)

which is used as an approximation to u¢ to find the potential of a normal force acting on a beam in many publications. If the expression (eqn 34) with a positive signed square root term is substituted in eqn (10), one obtains the following for an inextensional case: k =

v ¢¢ 1 - v¢

2

(36)

Eqn (36) is nothing but the curvature formula expressed in terms of arc length, in which s is replaced with x because, for an inextensional case, the coordinate of any material point in terms of s, in any configuration, is always equal to its initial coordinate x in the Lagrangian sense [18]. In fact, the curvature formula in which s is the independent variable is given by [18] as d2v ds 2 k = dv 2 1- Ê ˆ Ë ds ¯

(37)

The denominators of both eqn (36) and eqn (37) become zero for dv/dx = 1 and dv/ds = 1, that means q = 90°, as shown in Fig. 5, because sin q = dv/ds. Then, (d2v/ds2)ds International Journal of Mechanical Engineering Education 31/2

142


= cos q dq. For q = 90°, cos q = 0 and hence, d2v/ds2 = 0. It is seen that eqns (36 and 37) lead to an indeterminacy for dv/ds = 1 and dv/dx = 1, respectively [13]. This difficulty can be overcome by controlling the values of dv/ds and d2v/ds2 at every step of the numerical integration, and by changing the integration step so as to skip that critical point and so avoid numerical indeterminacy. Acknowledgement We would like to express our sincere appreciation to Dr K. S. Anderson for his support and valuable comments during the preparation of this paper. References [1] J. M. Gere and S. P. Timoshenko, Mechanics of Materials (PWS-KENT, Boston, 1990). [2] F. P. Beer and E. R. Johnston Jr, Mechanics of Materials (McGraw-Hill, New York, 1981). [3] H. J. Barten, ‘On the deflections of a cantilever beam,’ Quarterly of Applied Mathematics, 2(2) (1945), 168–171 (some part of this work was corrected by the author, ibid., 3(3) (1945), 275–276). [4] K. E. Bisshopp and D. C. Drucker, ‘Large deflection of cantilever beams’, Quarterly of Applied Mathematics, 3(3) (1945), 272–275. [5] A. C. Eringen, ‘On the non-linear vibrations of elastic bars’, Quarterly of Applied Mathematics 9(4) (1952), 361–369. [6] S. Woinowsky-Krieger, ‘Effect of the axial force on the vibrations of hinged bars’, Journal of Applied Mechanics, 17 (1950), 35–36. [7] H. Wagner, ‘Large amplitude free vibrations of a beam’, Journal of Applied Mechanics, 32 (1965), 887–892. [8] M. Epstein and D. W. Murray, ‘Large deformation in-plane analysis of elastic bars’, Computers and Structures, 6(1–A) (1976), 1–9. [9] K. R. V. Kaza and R. Kvaternik, ‘Nonlinear flap-lag-axial equations of a rotating beam’, AIAA Journal, 15(6) (1977), 871–874. [10] G. Lewis and F. Monasa, ‘Large deflections of cantilever beams of nonlinear materials’, Computers and Structures, 14(5–6) (1991), 357–360. [11] F. Monasa, ‘Large deflections of point loaded cantilevers with nonlinear behavior’, Journal of Applied Mathematics and Physics (ZAMP), 34 (January 1983), 123–130. [12] P. S. Theocaris and D. E. Panayotounakos, ‘Exact solution of the nonlinear differential equation concerning the elastic line of a straight rod due to terminal loading’, Int Journal of Non-Linear Mechanics, 17(5–6) (1982), 395–402. [13] D. H. Hodges, ‘Comment on “Linear and nonlinear analysis of a non-conservative frame of divergence instability”,’ AIAA Journal, 20(11) (1982), 1629–1630. [14] C. Venkatesan and V. T. Nagaraj, ‘Non-linear flapping vibrations of rotating blades’, Journal of Sound and Vibration, 84(4) (1982), 549–556. [15] G. B. Thomas Jr, R. L. Finney, Calculus and Analytic Geometry, 7th edn (Addison-Wesley, Reading, MA, 1990). [16] L. E. Malvern, Introduction to the Mechanics of a Continuous Medium (Prentice-Hall, New Jersey, 1969). [17] D. H. Hodges, ‘Proper definition of curvature in nonlinear beam kinematics’, AIAA Journal 22(12) (1984) 1825–1827. [18] V. V. Bolotin, The Dynamic Stability of Elastic System (Holden-Day, San Francisco, 1964).


On the Curvature of an Euler–Bernoulli Beam

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