Kinematic Behaviour of a Bernoulli Curved Beam

Dario Genovese1

1Department of Architecture, Constructions and Structures, Universita Politecnica delle Marche,ItalyE-mail: dario79@email.it

Keywords: curved beam, Bernoulli, virtual work.

SUMMARY. In this article we describe the differences between a straight Bernoulli beam and agenerally curved beam.

Altough the resulting elastic line equations are very different, and obviously more complex in acurved beam for the presence of terms involving initial curvature and torsion of the beam, we cangive, using the right variables, a very simple description of the kinematic behaviour of a curvedbeam.

In particular we can define boundary displacements in the same way of a straight beam, andthe expression of curvatures and twist (kinematic duals of bending moments and torsion) with verysimple corrections, which can be easily geometrically interpreted.

1 INTRODUCTIONThe static and kinematic behaviour of a curved beam has been widely analyzed in the last years.

In this article we focus on the simplest available model for a generally three-dimensional curvedbeam, the linear Bernoulli beam. A reference can be found, for example, in [1].

The approach used here is variational. After defining the equilibrium using the Frenet formulaswe obtain the compatibility and the expression of boundary terms using the virtual work principle,and we give a qualitative interpretation of differences between a straight and curved beam.

The only assumption in the model presented is that displacements are small (linear theory) andthat shear effects are neglectable (Bernoulli beam).

2 MATHEMATICAL VARIABLESThe reference configuration of a generally curved beam is defined by a curve in three-dimensional

space. A curve is a sufficiently smooth function x = x(s), where x ∈ R3 and s ∈ [0;L]. We supposethat the variable s is a curvilinear abscissa, so that the total length of the beam is L.

Along this curve we define the unit-vectors t = t(s) (tangent), n = n(s) (normal) and b = b(s)(binormal). A relation between these vectors is the Frenet formula: t

nb

′ =

0 k 0−k 0 τ0 −τ 0

· tnb

(1)

where k and τ are the curvature and torsion of the curve, and an apex is the differential operatord/ds.

We consider the following functions, defined in [0;L]:

• The load force vector f = f(s). This vector has three components: ft = f · t, fn and fb.

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• The displacement vector d = d(s), defined as the kinematic dual of f , whose components aredt, dn and db

• The load couple vector c = c(s). This vector has three components, but we suppose that theonly not null component is the torque q = c · t

• The rotation φ, dual of c.

• Internal strains N (axial stress), Mn (normal bending moment), Mb (binormal bending mo-ment) and T (torsion). These functions are scalar.

• Kinematic duals of internal strains are respectively e (elongation), curvatures χn and χb, andtwist θ.

• We define also two addictional internal strains, shears Vn and Vb. We neglect shear’s effects,so we will not define their kinematic duals.

3 EQUILIBRIUMWe consider an element of beam of length ds, with applied loads f and c (see Fig.1)

Figure 1: Infinitesimal beam element

Strains are defined as two three-dimensional vectors: forces F and couplesC, whose componentsare written in the Frenet frame along t,n and b:

F =

NVn

Vb

; C =

TMn

Mb

(2)

Equilibrium to translation leads to F ′ = −f , hence

N ′ = (F · t)′ = F ′ · t+ F · t′ = −ft + kVn (3)

V ′n = (F · n)′ = −fn − kN + τVb (4)

V ′b = (F · b)′ = −fb − τVn (5)

Equilibrium to rotation leads to C ′ + t× F + c = 0, and to

C ′ = F × t− qt (6)

We can also write

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M ′n = (C · n)′ = C ′ · n+ C · n′ = F × t · n− kT + τMb = Vb − kT + τMb (7)

M ′b = (C · b)′ = F × t · b− τMn = −Vn − τMn (8)

T ′ = (C · t)′ = F × t · t− q + kMn = −q + kMn (9)

From (7) and (8) we obtain the expressions of shears and their derivatives:

Vn = −M ′b − τMn

Vb = M ′n + kT − τMb

V ′n = −M ′′b − (τMn)′

V ′b = M ′′n + (kT )′ − (τMb)′(10)

Substituting (10) in (3), (4) and (5) we have:

N ′ = −ft − kM ′b − kτMn

−M ′′b − (τMn)′ = −fn − kN + τM ′n + kτT − τ2Mb

M ′′n + (kT )′ − (τMb)′ = −fb + τM ′b + τ2Mn

(11)

Including also (9) and rearranging terms we obtain the equilibriumft = −N ′ − (kMb)′ + k′Mb − kτMn

fn = M ′′b + (2τMn)′ − τ ′Mn − kN + kτT − τ2Mb

fb = −M ′′n − (kT )′ + (2τMb)′ − τ ′Mb + τ2Mn

q = −T ′ + kMn

(12)

The equilibrium may be written in the compact form

f = ∂∗(σC ) (13)

This equation should be read, using Einstein convention on summations, as

fk = ∂∗j (σiCijk) (14)

Where

∂∗ =

1−∂/∂s∂2/∂s2

(15)

σ =

NMn

Mb

T

(16)

f =

ft

fn

fb

q

(17)

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Tensor C is a third order tensor of size 4× 3× 4, and it is written as:

Ci1k =

−k

−kτ −τ ′ τ2 kk′ −τ2 −τ ′

kτ

(18)

Ci2k =

1−2τ

k −2τk 1

(19)

Ci3k =

−11

(20)

where blank spaces indicate null components.

4 COMPATIBILITYOnce defined the equilibrium, we can obtain linear compatibility using the virtual work princi-

ple. Using the same approach described in [2], which essentially consists in standard variationaltechniques, we define

d =

dt

dn

db

φ

(21)

ε =

eχn

χb

θ

(22)

∂ =

1∂/∂s∂2/∂s2

(23)

and we obtain the compatibility

ε = C∂d (i.e. εi = Cijk∂jdk) (24)

The compatibility may be explicitally written ase = d′t − kdn

χn = −d′′b − 2τd′n + kφ+ τ2db − τ ′dn − kτdt

χb = d′′n − 2τd′b + kd′t − τ ′db − τ2dn + k′dt

θ = φ′ + kd′b + kτdn

(25)

From (1) we have

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t′ = knn′ = −kt+ τbn′′ = −k′t+ τ ′b− k2n− τ2nb′ = −τnb′′ = −τ ′n+ kτt− τ2b

(26)

Furthermore

d′t = (d · t)′ = d′ · t+ d · t′d′n = (d · n)′ = d′ · n+ d · n′d′b = (d · b)′ = d′ · b+ d · b′d′′n = (d · n)′′ = d′′ · n+ 2d′ · n′ + d · n′′d′′b = (d · b)′′ = d′′ · b+ 2d′ · b′ + d · b′′

(27)

Through (26) and (27), by proper substitution and after some basic algebra, we can simplify (25)with

e = d′ · tχn = −d′′ · b+ kφχb = d′′ · n− kd′ · tθ = φ′ + kd′ · b

(28)

5 BOUNDARY TERMSFrom terms in (19) and (20), integrating by parts the expression of virtual work principle, we

obtain the expression of boundary forces, which we write here together with the displacement forwhich they do work (see [2] for further details):

(N + kMb) · dt (29)

Mb · d′n (Bending moment) (30)

−Mn · d′b (Bending moment) (31)

(−M ′b − 2τMn) · dn (32)

(M ′n − 2τMb + kT ) · db (33)

T · φ (Torsion) (34)

From the first two equations in (10) we obtain the expressions of M ′n and M ′b, hence (32) and(33) become

(Vn − τMn) · dn (35)

(Vb − τMb) · db (36)

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Finally we substitute in (30) and (31) the expressions d′n = d′ ·n−kdt+τdb and d′b = d′ ·b−τdn.By doing this we have also to rearrange new terms in (29), (35) and (36).

The expression of boundary terms assume then the familiar appearance

N · dt (Axial stress)Mb · (d′ · n) (Bending moment)Vb · db (Shear)−Mn · (d′ · b) (Bending moment)Vn · dn (Shear)T · φ (Torsion)

(37)

6 ELASTIC LINETo obtain the differential equations of the elastic line, we have to define a relation (the constitutive

law) between stress and strain. In linear elasticity this relation can be written as

σ = Hε (38)

where H is a 4× 4 positive symmetric matrix.In this paper, without going further, we suppose here that, if

• The section is small in respect to 1/k, 1/τ and L

• The material of which the beam is made up is isotropic

• The section has a shape which is compatible with a Bernoulli beam

• The spatial derivative of the constitutive law is small (H′ � 1)

then

H =

EA 0 0 00 Jnn Jnb 00 Jnb Jbb 00 0 0 GJt

(39)

where J∗∗ are the inertia moments, A is the area and Jt is the torsional inertia of the crosssection, and E and G are the Young and shear modulus. The elastic line may be written, using theequilibrium, the constitutive law and the compatibility, as

f = ∂∗(H(C∂d)C ) (i.e. fk = ∂∗j (Hii′(Ci′j′k′∂j′dk′)Cijk) ) (40)

7 DIFFERENCES BETWEEN A STRAIGHT AND A CURVED BEAMWe have seen in (37) that boundary terms can be written in the same way both in a straight and

a curved beam.The compatibility of a curved beam, as written in (28), has some difference. The axial strain

has the same appearance, but twist and curvatures has each a correction addendum, which does notappear in a straight beam.

In this section we will give a qualitative interpretation of these three correction terms.

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7.1 Normal curvatureThe correction +kφ, which appears in the expression of χn, can be explained in Fig.2, where

on the left is plotted the section of the beam, with center O, and the Frenet frame (normal n andbinormal b vectors).

Figure 2: Normal curvature correction

The point A, on the axis b and at a distance z from O, and the point B on the axis n, after aninfinitesimal rotation φ, go respectively in A′ and B′. We suppose that the section is undeformable.

On the right is plotted the beam seen from the axis b. The radius of the fiber SA through A isr0 = 1/k, while the radius of the fiber SA′ through A′ is r1 = 1/k + zφ. The elongation is then+kφz. This is the same elongation of a fiber of a straight beam where an element of length ds has acurvature v′′ = kφ, assuming that v is the transversal displacement (see Fig.3).

Figure 3: Curvature in a straight beam

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This correction terms does not appear for the fiber SB through B, thus it affects only the normalcurvature χn

7.2 Binormal curvatureThe correction −kd′ · t in the binormal curvature χb is explained referring to Fig.4.

Figure 4: Binormal curvature correction

The point B, in the deformed configuration, goes in B′. We have z/r0 = z/r1, where r0 is theinitial curvature radius and r1 is the final one. We have also

• z ' ds

• z = z + (d′ · t)ds

• r0 = 1/k

• r1 = 1/(k + ∆k)

where ∆k is the variation of curvature. Thus we obtain

kds = (ds+ d′ · tds)(k + ∆k) (41)

and, with some basic algebra and supposing ∆k � k

1 + d′ · t =1

1 + ∆k/k' 1−∆k/k (42)

∆k = −kd′ · t (43)

7.3 TorsionFor an explanation of the correction +kd′ · b in the expression of θ, refer to Fig.5An initial circular arch of length ds and radius r = 1/k (dashed bold line) has null displacement

at one side, while at the other side the displacement field is only along the binormal and is d′ · bds(continuos bold line is the deformed rod). The section has no rotation (φ = 0).

Even if φ = 0, there is a twist θ such that

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Figure 5: Twist correction

θds =(d′ · b)ds

r(44)

from which we obtain the correction θ = kd′ · b.

8 CONCLUSIONSWe have presented a simple interpretation of the kinematical behaviour of a generally curved

beam in three-dimensional space. We want to remark the fact that the only assumptions made arethe small displacements and the neglectable effects of shear.

Thus what exposed in section 7 is not a justification of this behaviour, but only a geometricalinterpretation. The justification is the equilibrium and the virtual work principle.

We could have started from the compatibility (28) and obtained the equilibrium as a consequence,but we should have given a clear justification of the compatibility. The equilibrium is far more easyto handle and to justify.

References[1] Jacqueline Sanchez Hubert and Evarisre Sanchez Palencia. Statics of curved rods on account of

torsion and flexion. Eur. J. Mech. A/Solids, 18:365–390, 1999.

[2] D.Genovese. A generalized method for the static analysis of a monodimensional prestressedcontinuum. Proc. of XIX AIMETA Conference, 2009.

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