General Finite Element Description for Non-uniform

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Arch Appl Mech (2012) 82:43–67DOI 10.1007/s00419-011-0538-8

O RIG INAL

Giuseppe Failla · Nicola Impollonia

General finite element description for non-uniformand discontinuous beam elements

Received: 1 July 2010 / Accepted: 22 March 2011 / Published online: 19 April 2011© Springer-Verlag 2011

Abstract The theory of generalized functions is used to address the static equilibrium problem of Euler–Bernoulli non-uniform and discontinuous 2-D beams. It is shown that if simple integration rules areapplied, the full set of response variables due to end nodal displacements and to in-span loads can be derived,in a closed form, for most common beam profiles and arbitrary discontinuity parameters. On this basis, forfinite element analysis purposes, a non-uniform and discontinuous beam element is implemented, for whichthe exact stiffness matrix and the fixed-end load vector are derived. Upon computing the nodal response, nonumerical integration is required to build the response variables along the beam element.

Keywords Euler–Bernoulli theory · Non-uniform and discontinuous beams · Shape functions · Static Green’sfunctions · Stiffness matrix

1 Introduction

There are frequent circumstances in engineering analysis where beams are non-uniform, in the sense thatgeometry and/or material properties vary along the length. For instance, beam elements of variable crosssection, either continuous or stepwise, are commonly resorted to in engineering design to reduce weight andto optimize strength and stability or to meet specific architectural and functional requirements. Specifically,a continuous variation in the cross section is typical of tapered beams or haunched beams used in bridges,industrial and high-rise buildings, space and aircraft structures, while a stepped variation may result from mod-eling of local damage phenomena [1]. On the other hand, there are also circumstances in engineering analysiswhere beams are discontinuous, in the sense that discontinuities in the response variables shall be accountedfor as a result of structural modeling. Specifically, discontinuities in the displacement or in the rotation resultfrom internal springs modeling elastic connections between adjacent beam segments [2,3] or, quite frequently,concentrated damage due to cracks [4,5].

Analytical tools capable of handling in a comprehensive setting sources of non-uniformity and discontinu-ity may be then certainly attractive for engineering analysis. This will be the target of this paper, with specificattention to static analysis.

G. Failla (B)Dipartimento di Meccanica e Materiali (MECMAT), Università “Mediterranea” di Reggio Calabria,Via Graziella, Località Feo di Vito, 89122 Reggio Calabria, ItalyE-mail: [email protected].: +39-0965875215Fax: +39-0965875201

N. ImpolloniaDipartimento di Architettura (DARC), Università di Catania, P.zza Federico di Svevia, 96100 Siracusa, ItalyE-mail: [email protected]



44 G. Failla, N. Impollonia

Static finite element (FE) analysis of non-uniform beams has been the subject of intensive research effortstartingfrom the early 80’s. In this context, there have been two mainobjectives: (i) building the stiffness matrix(SM) and the fixed-end load vector (LV) and (ii) upon computing the frame nodal response, building the fullset of response variables along the beam length without resorting to numerical integration. In general, both

tasks may be achieved by exact integration of the governing differential equations. This has been performed:based on suitable Bessel functions for specific polynomial tapering [6]; based on the generalized integrals of the unit-step function, for stepped beam elements [7]; and based on an infinite power series representationof the solution for arbitrary polynomial tapering [8], in conjunction with the boundary integral method, forspecific polynomial tapering [9].

Due to the difficulties inherent in the exact integration of the differential equations governing arbitrarynon-uniform beams, however, alternative strategies have been also pursued aiming, in general, to derive theSM and the LV only. Most of these strategies rely on energy principles of elasticity. For instance, enforcingthe stationarity of the complementary energy in conjunction with homogeneous equilibrium states has led tothe SM and LV for beams with arbitrary variable cross section and internal springs [10]; a flexibility methodhas been used to build the LV and the SM (the latter by explicit inversion of the flexibility matrix) for arbitraryvariable cross section [11]; also, the Castigliano theorem and equilibrium equations have been used to derivethe SM for arbitrary variable cross section [12]. As alternative to energy-based methods, the conjugate beammethod has been adopted to derive the SM and the LV for arbitrary variable cross section [13]. There are alsorecent applications of the transfer matrix method to derive the SM for beams with tapered ends [14,15] or theSM and the LV for beams of arbitrary shapes [ 16].

It is now worth remarking that depending on the beam profile, the methods cited above for non-uniformbeams provide either numerical or closed-form solutions. Specifically, closed-form solutions are available forparticular beam profiles where the cross section properties vary according to a polynomial law, of variousorder [6,8,9,11–15]. For most general beam profiles, however, numerical solutions shall be built by numericalintegration [10,11,13,16].

Static analysis of discontinuous beam elements has also attracted the interest of several researchers [2,3,17–21]. For FE analysis purposes, the complementary energy-based method in ref. [10] provides the SMand the LV. As an alternative, both the SM and the LV have been recently proposed by exact integration of thegoverning differential equations [22]. In this case, therefore, the full set of response variables along the beamlength is also available, in a closed form, upon computing the frame nodal response.

Clearly, static FE analysis of beam elements that are both non-uniform and discontinuous poses somedifficulties from a mathematical point of view, since in the governing differential equations simultaneouslyappear variable coefficients (reflecting the variations of the beam profile) and generalized functions (modelingthe discontinuities). To the authors’ knowledge, to date no alternatives exist to the energy-based method inRef. [10]. Such method provides, however, the SM and the LV in a numerical form and requires an a prioriknowledge of the beam flexibility coefficients. Also, upon obtaining the frame nodal response, the full set of response variables can be built only at the expense of a numerical integration of the governing differentialequations.

It is then evident that despite the research effort of the last two decades, there is still the need of an efficientapproach to finite element analysis involving non-uniform and discontinuous beams. This problem will betackled, in this paper, by resorting to the theory of generalized functions. This theory has been traditionallyapplied to discontinuous beam only [2,3,17–21]. Herein it will be used to build, for non-uniform and discontin-uous beams, the full set of response variables due to end nodal displacements as well as to in-span loads, eitherconcentrated or distributed. Based on the latter, the SM and the LV will be also derived. Exact solutions will

be provided for arbitrary discontinuity parameters and for a variety of beam profiles commonly encounteredin engineering applications.

2 Problem statement

Consider the non-uniform and discontinuous 2-D beam in Fig. 1. Be L the length, E A ( x ) and E I ( x ) the axialand flexural stiffness, respectively. Denote by x i the discontinuity locations, 0 < · · · < x i−1 < x i < · · · < Lfor i = 1, 2, . . . , N ; by ζ ui and ζ wi the flexibilities of the internal axial and transverse translational springs,respectively; by ζ ϕi the flexibilities of the internal rotational springs. For generality, each discontinuity locationis represented in Fig. 1 as a multi-discontinuity location where internal springs simultaneously occur. However,any discontinuity pattern may be built at a location x i by proper selection of the discontinuity parameters: forinstance, at a location x i where only a rotational spring occurs, ζ ui = ζ wi = 0 shall be obviously set.



General finite element description for non-uniform and discontinuous beam elements 45

i x

N x

uiζ

wiζ

iϕ ζ

( )S x

( ) M x

( )u x

2 2,S w

1 1,S w

2 2, N u

1 1, N u

2 2, M ϕ

1 1, M ϕ

( ) p x

x

z

( ) ( ), EI x EA x

( ) N x

( )w x

( ) x ϕ

1 x γ = 2

x γ =

Fig. 1 Non-uniform and discontinuous 2-D beam

Be u ( x ) the axial displacement, w ( x ) the transverse deflection, ϕ ( x ) the rotation, N ( x ) the axialforce, M ( x ) the bending moment and S ( x ) the shear force. Positive sign conventions are set in Fig. 1.Further, denote by K the SM, by vT =

u1 w1 ϕ1 u 2 w2 ϕ2

the vector of nodal displacements and by

f T = N 1 S 1 M 1 N 2 S 2 M 2

the LV, for which positive sign conventions are also set in Fig. 1.

Under the assumption of small displacements, the axial and the bending problems are uncoupled and willbe treated separately in the following. For both problems, the response to end nodal displacements will begiven in Sect. 3, along with the general expression of the SM. The response to in-span loads and the LV will begiven in Sect. 4, based on the Green’s functions (GFs) due to a unit-point load. The overall response along thebeam element will be given in Sect. 5, and a numerical application will be considered in Sect. 6. The paper is

completed by the Appendices, where mathematical details are reported for the ease of reading of the main text.

3 Response to end nodal displacements and stiffness matrix

3.1 Response to end nodal displacements

The response variables due to end nodal displacements are now sought. They are denoted by rµn ( x ) =u ( x ) N ( x )

T and rβn ( x ) =

w ( x ) ϕ ( x ) M ( x ) S ( x )T

for the axial and the bending problems, respec-tively. The key ideas of the method are elucidated for the beam in Fig. 2a, where no releases are encounteredat the beam ends, and the beam profile is continuous and varies with the same law along the length.

In context with the theory of generalized functions, the differential equations governing the responsevariables of the axial problem rµn ( x ) are written as [19]

d N ( x )

d x = 0 (1a)

du ( x )

d x = N ( x )

E A ( x )+

Ni=1

N ( x i ) ζ ui δ ( x − x i ) (1b)

where N ( x i ) ζ ui = u x +i − u

x −i

. Further, the differential equations governing the response variables of the bending problem rβn ( x ) are

dS ( x )

d x = 0 (2a)

d M ( x )

d x = S ( x ) (2b)




dϕ ( x )

d x = − M ( x )

E I ( x )−

Ni=1

M ( x i ) ζ ϕi δ ( x − x i ) (2c)

dw ( x )

d x = ϕ ( x ) +N

i=1

S ( x i ) ζ wi δ ( x − x i ) (2d)

where M ( x i ) ζ ϕi = ϕ x −i− ϕ

x +i

and S ( x i ) ζ wi = w x +i− w

x −i

. In Eqs. (1) and (2), the bar over thedifferentiation symbol denotes generalized derivative and δ (·) is the well-known Dirac’s delta, defined by thegeneralized integral

∞ −∞

δ ( x − x i ) d x = 1 (3)

and related to the Heaviside function or unit-step function

H ( x − x i ) =0 x < x i

1 x > x i(4)

by the generalized integral and derivative

x −∞

δ ( y − x i ) d y = H ( x − x i ) (5a)

d

d x H ( x − x i ) = δ ( x − x i ) (5b)

To the authors’ knowledge, no general method exists to date to integrate Eqs. (1) and (2) in a closed form.To this purpose, it is noted, however, that the continuously varying profile of the beam in Fig. 2a may bethought of as the limit y → 0 of the m -stepped profile shown in Fig. 2b, where each segment of length y = L /m is a uniform segment of constant axial stiffness E Ak = E A ( yk −1) and constant flexural stiffness E I k = E I ( yk −1), for y0 = 0, yk = k y, and k = 1, 2, . . . , m. Recognize in fact that, given the axial andbending flexibility field of the m-stepped beam in Fig. 2b in the form

∆ =∆ =

( ) ( )a

b

Fig. 2 Beam with continuously varying profile (a), thought of as the limit for y→ 0 of the m−stepped profile where eachsegment has length y = L /m (b)




1 E A ( x )= 1

E A

1 +

m−1k =1

µ ( yk )

y H ( x − yk ) y

(6a)

1 E I ( x ) = 1

E I 1 +m−1

k =1

β ( yk )

y H ( x − yk ) y (6b)

where E A = E A (0) , E I = E I (0) and

µ ( yk )

y= E A

y

1

E A ( yk )− 1

E A ( yk − y)

(7a)

β ( yk )

y= E I

y

1

E I ( yk )− 1

E I ( yk − y)

(7b)

the limits y → 0 of Eq. (6) yield

lim y→0

1 E A ( x ) = 1

E A 1 +

L

0

dµ

d y H ( x − y) d y = 1

E A 1 +

x

0

dµ

d y d y = 1

E A ( x ) (8a)

lim y→0

1 E I ( x )= 1

E I

1 + L

0

dβ ( y)

d y H ( x − y) d y

= 1

E I

1 + x

0

dβ ( y)

d yd y

= 1

E I ( x )(8b)

being

dµ ( y)

d y= lim

y→0

µ ( yk )

y= d

d y

E A

E A ( y)

for µ ( y) = E A

E A ( y)(9a)

and

dβ ( y)

d y= lim

y→0

β ( yk )

y= d

d y

E I

E I ( y)

for β ( y) = E I

E I ( y)(9b)

The responses rµn ( x ) and rβn ( x ) of the beam in Fig. 2a can be then built, in general, by taking the limit y → 0 of the corresponding responses, rµn ( x ) and rβn ( x ), of the m-stepped beam in Fig. 2b. That is,

rµn ( x ) = lim y→0

rµn ( x ) (10a)

rβn ( x ) = lim y→0

rβn ( x ) (10b)

This is a crucial statement for the purpose of integrating Eqs. ( 1) and (2) in a closed form. The response of

the m-stepped beam, in fact, can be readily obtained by simple rules of integration for generalized functions;details in this regard may be found in Appendix A, where the responses rµn ( x ) and rβn ( x ) of the m-steppedbeam are given by Eqs (A.4) and (A.6). Such equations depend on integration constants to be set dependingon the boundary conditions. It can be seen that if u (0) = u1 and u ( L) = u2 are set in Eqs. (A.4), whilew (0) = w1, w ( L) = w2, ϕ (0) = ϕ1 and ϕ ( L) = ϕ2 are set in Eq. (A.6), upon taking the limit y → 0Eq. (10) take the general form

rµn ( x ) =2

j=1

(u j )µ ( x ) u j (11a)

rβn ( x ) =2

j=1

(w j )β ( x ) w j +

2i=1

(ϕ j )β ( x ) ϕ j (11b)




where (u j )µ ( x ) ,

(w j )β ( x ) and

(ϕ j )β ( x ) are given by

(u j )µ ( x ) = (−1) j

Gµ ( x ) a(u) + ρ

(u j )µ

(12a)

(w j )β ( x ) = (−1) j Gβ ( x ) b(w) + ρ(w j )β (12b)

(ϕ j )β ( x ) = (−1) j−1

Gβ ( x ) b(ϕ j ) + ρ

(ϕ j )β

(12c)

for j = 1, 2. For clarity, vectors and matrices in the r.h.s. of Eq. ( 12) are reported separately below.

Vectors ρ(u j )µ , ρ

(w j )β and ρ

(ϕ j )β

They are vectors of constants given by

ρ(u j )µ = ( j − 2)

1 0T

(13a)

ρ(w j )β = ( j − 2)

1 0 0 0

T

(13b)

ρ

(ϕ j )β = (2 − j ) x 1 0 0 T

(13c)

Matrices Gµ ( x ) , Gβ ( x )They are matrices of functions, depending on the beam profile and on the discontinuity parameters, given by

Gµ ( x ) =

Gµ11 ( x ) 11 0

(14a)

Gβ ( x ) =

Gβ11 ( x ) Gβ12 ( x ) x 1Gβ21 ( x ) Gβ22 ( x ) 1 0

x 1 0 01 0 0 0

(14b)

where in Eq. (14a),

Gµ11 ( x ) = x + I 10µ ( x ) − I 01

µ ( x )

E A+

Ni=1

ζ ui H ( x − x i ) (15)

and in Eq. (14b),

Gβ11 ( x ) = − x 3 + I 30

β ( x ) + 2 I 03β ( x ) − 3 I 12

β ( x )

6 E I +

Ni=1

ζ wi H ( x − x i ) −N

i=1

ζ ϕi x i R1 ( x − x i ) (16)

Gβ12 ( x ) = − x 2 + I 20

β ( x ) + I 02β ( x ) − 2 I 11

β ( x )

2 E I −

N

i=1

ζ ϕi R1 ( x − x i ) (17)

Gβ21 ( x ) = − x 2 + I 20

β ( x ) − I 02β ( x )

2 E I −

Ni=1

ζ ϕi x i H ( x − x i ) (18)

Gβ22 ( x ) = − x + I 10

β ( x ) − I 01β ( x )

E I −

Ni=1

ζ ϕi H ( x − x i ) (19)

The symbols I rsµ ( x ) and I rs

β ( x ) in Eq. (15) through (19) denote the following integrals

I rsh ( x ) = x r

L 0

dh ( y)

d y ys H ( x − y) d y = x r

x 0

dh ( y)

d y ys d y (20)




where h ( y) is either h ( y) = µ ( y) or h ( y) = β ( y), respectively for the axial and the bending problem.Further, recognize that Eq. (15) through (19) are discontinuous at any location x = x i where an internal springoccurs.

Vectors a(u), b(w)and b(ϕ j )

They are vectors of constants. Under the assumption of no releases at the beam ends, they are given by

a(u)1 =

L + I 10

µ ( L) − I 01µ ( L)

E A+

Ni=1

ζ ui

−1

(21a)

a(u)2 = 0 (21b)

b(w)1 = − α4

α2α3 − α1α4(22a)

b(w)2 = α3

α2α3 − α1α4(22b)

b

(w)

3 = 0 (22c)b

(w)4 = 0 (22d)

b(ϕ j )1 = −α2v

(ϕ)2 + α4 ( j − 2) L

α2α3 − α1α4(23a)

b(ϕ j )2 = α3 ( j − 2) L + α1

α2α3 − α1α4(23b)

b(ϕ j )3 = 0 (23c)

b(ϕ j )4 = 0 (23d)

where

α1 = Gβ11 ( L) (24a)

α2 = Gβ12 ( L) (24b)

α3 = Gβ21 ( L) (24c)

α4 = Gβ22 ( L) (24d)

Note that, for brevity, the index j is omitted in the superscript of a(u)k in Eq. (21) and in the superscript of b

(w)k

in Eq. (22), since based on Eq. (24), it is seen that a(u)

k , for k = 1, 2, and b

(w)

k , for k = 1, . . .4, do not vary

depending on the nodes 1, 2.To gain insight into Eq. (11a,b) obtained for the response variables, now a few simple cases of interest can

be examined.For instance, consider a uniform beam with internal axial translational springs; if u1 = 0 and u2 = 0 are

set in Eq. (11a), the axial displacement response u ( x ) takes the form

u ( x ) = (−1)Gµ11 ( x ) a

(u)1 − 1

u1 =

1 −

x

E A+

N i=1

ζ ui H ( x − x i )

L

E A+

N i=1

ζ ui

−1 u1 (25)

where as expected u (0) = u1 and u ( L) = 0. In Eq. (25), recognize that a(u)1 =

L E A

+ N i=1 ζ ui

−1is the

constant axial force that developsacrossthe beam when a unit displacementu1 = 1.0issetatnode1and u2 = 0

are set at node 2; correspondingly, Gµ11 ( x ) a(u)1 =

x E A

+ N i=1 ζ ui H ( x − x i )

L E A

+ N i=1 ζ ui

−1is the

contribution to the axial displacement function u ( x ) given by the constant axial force




a(u)1 =

L E A

+ N i=1 ζ ui

−1. This contribution is discontinuous where an internal axial translational spring

occurs, i.e. at any x = x i , and the magnitude of each discontinuity is ζ ui

L E A

+

N i=1 ζ ui

−1. If the beam is

not only discontinuous but also non-uniform, both the axial force that develops across the beam given by a(u)

1

,Eq. (21a), and the axial flexibility given by the first term in Gµ11 ( x ), Eq. (15), do vary according to the beamprofile.

As a further example, consider a uniform beam with internal transverse translational springs. If w1 = 0and w2 = ϕ1 = ϕ2 = 0 are set in Eq. (11b), the displacement response w ( x ) takes the form

w ( x ) = (−1)Gβ11 ( x ) b

(w)1 + Gβ12 ( x ) b

(w)2 − 1

w1

=1 −

Lx 2

4 E I − x 3

6 E I +

N i=1

ζ wi H ( x − x i )

L3

12 E I +

N i=1

ζ wi

−1w1 (26)

where as expected w (0) = w1 and w ( L) = w2. In Eq. (26), recognize that L3

12 E I + N

i=1 ζ wi−1 is the

constant shear force that develops across the beam when w1 = 1.0 and ϕ1 = 0 is set at node 1, w2 = ϕ2 = 0 is

set at node 2; correspondingly,

Lx 2

4 E I − x 3

6 E I + N

i=1 ζ wi H ( x − x i )

L3

12 E I + N

i=1 ζ wi

−1is the contribution

to the displacement function w ( x ) given by the constant shear force

L3

12 E I + N

i=1 ζ wi

−1. This contribution

is discontinuous at any location where a transverse translational spring occurs, that is at any x = x i , and the

magnitude of each discontinuity is ζ wi

L3

12 E I + N

i=1 ζ wi

−1. As in the previous case, if the beam is not only

discontinuous but also non-uniform, both the constant shear force (see Eq. 22a, b) for b(w)1 and b

(w)2 ) and the

transverse flexibility (see Eqs. 16, 17) for Gβ11 ( x ) and Gβ12 ( x )) do vary according to the beam profile.

3.1.1 Closed-form solution for the response to end nodal displacements

It is now evident that the exact solution for(u j )µ ( x ) ,

(w j )β ( x ) and

(ϕ j )β ( x ) given by Eq. (12) can be found

if a solution is found for the integrals I rsh ( x ), Eq. (20). To this purpose, note that the integrals I rs

h ( x ) can begiven the general solution

I rsh ( x ) = x r

x 0

dh ( y)

d y ys d y = x r

Z sh ( x ) − Z sh (0)

(27)

where Z sh (·) is the primitive function

Z sh ( y) = dh ( y)

d y ys d y = s

=0

(−1) h[] ( y)d ys

d y (28)

obtained via a standard integration by parts where dh ( y)/d y is repetitively integrated and ys is repetitively dif-ferentiated until the last derivative vanishes. In Eq. (28) h[] ( y) is the l−th order primitive of the function h ( y).Since s = 1 at most for the axial problem (see Eq. 15) and s = 3 at most for the bending problem (see Eq. 16),then Eq. (12) are available in a closed form if: (i) a 1st order primitiveof the axial flexibility ratio µ ( y) , µ[1] ( y),is known; (ii) primitives up to the 3rd order of the flexural flexibility ratio β ( y) , β[1] ( y) , β[2] ( y), and β[3] ( y),are known. This is the case for most common beam profiles in engineering applications and a few examplescan be found in Appendix B. Obviously, if the flexibility ratios µ ( y) and β ( y) cannot be integrated in a

closed form, (u j )µ ( x ) ,

(w j )β ( x ), and

(ϕ j )β ( x ) in Eq. (12) may be still computed numerically, by building

a numerical solution to the integrals (20).




Table 1 End-released beam elements for different boundary conditions (B.C.): vectors b(w) for the bendingresponse to end nodaldisplacements

B.C. b(w)1 b

(w)2 b

(w)3 b

(w)4

MR1-MR2 0 0 1

L 0

MR2 − 1α2 L−α1

Lα2 L−α1

0 0

SR2 0 0 0 0MRl-SR2 0 0 0 0

MR moment released, SR shear released

Table 2 End-released beam elements for different boundary conditions (B.C.): vectors b(ϕ j ) for the bending response to endnodal displacements

B.C. b(ϕ j )

1 b(ϕ j )

2 b(ϕ j )

3 b(ϕ)4

MR1-MR2 0 0 ( j − 2) 0

MR2 (2− j) L

α2 L−α1

( j−2) L2

α2 L−α10 0

SR2 0 − 1α4

0 0

MR1-SR2 0 0 −

1 0

MR moment released, SR shear released

Finally, two remarks are worth doing.

Remark 1 Equation (12) have been written under the assumption that no releases occur at the beam ends.However, it may be proved that for any release at the beam ends, the response to end nodal displacements is

still given by Eq. (12) where, however, the constants a(u), b(w), and b(ϕ j ) are changed accordingly.

Specifically, for an axial release at node 1, a(u)1 = 0 and a

(u)2 = 1 are found. In fact recognize that Eq. (12a)

yields, as expected, rµn ( x ) = 0 0

if u1 = 1 and u2 = 0 (since neither axial displacements nor axial stress

can be encountered inside the beam element) and rµn ( x ) =

1 0

if u1 = 0 and u2 = 1. Similarly, for an

axial release at node 2, a(u)1

= a

(u)2

= 0 are found.

Expressions of vectors b(w) and b(ϕ j ) for different releases at the beam ends have been summarized inTables 1, 2.

Remark 2 Equation (12) have been written under the assumption that the beam profile is continuous and varieswith the same law along the length. However, if n different variation laws are encountered along the length,as in Fig. 3a, Eq. (12) still hold true and trivial changes apply to the integrals I rs

h ( x ), which read

I rsh ( x ) = x r

λ1

0

dh1 ( y)

d y ys H ( x − y) d y + · · · +

L λn−1

dhn ( y)

d y ys H ( x − y) d y

=

x r n

k =1

Z s

h ( y)min{λn, x }

min{λn−1, x } (29)

where λ0 = 0 and λn = L . Equation (12) still hold true also if stepped variations of the beam profile occur,provided that additional terms are included in the integrals I rs

h ( x ) given by Eq. (20). For instance, for thebeam in Fig. 3b where a step is encountered at x = λ1, it can be readily seen that Eq. (20) for I rs

h ( x ) reads

I rsh ( x ) = x r

λ1 0

dh1 ( y)

d y ys H ( x − y) d y + ρ1λs

1 H ( x − λ1) + L

λ1

dh2 ( y)


= x r

Z sh ( y)

min{λ1, x }0

+ ρ1λs1 H ( x − λ1) + Z sh ( y)

x

min{λ1, x } (30a)




1 1 x λ =

0 0λ =

2 x

2λ

1λ

1 x

1λ

2λ

1 x

0 0λ =

0 0λ =

x

z

x

z

x

z

( ) ( ), EI x EA x

( ) ( ), EI x EA x

( ) ( ), EI x EA x

, E A E I

a

b

c

Fig. 3 Beam with different variation law (a); with one step (b); with two steps enclosing a uniform segment ( c)

where ρ1 = E A/ E A

λ+

1

− E A/ E A

λ−1

for the axial problem and ρ1 =

E I / E I

λ+

1

− E I / E I

λ−1

for the bending problem. Similarly, for the beam in Fig. 3c with a uniform segment, between x = λ1 and

x = λ2, of axial stiffness E A = (1 − µ) E A (0) and bending stiffness E I = 1 − β E I (0), Eq. (20) for

I rsh ( x ) reads

I rsh ( x ) = x r

λ1

0

dh1 ( y)

d y ys H ( x − y) d y +

2k =1

ρk λsk H ( x − λk ) +

L λ2

dh2 ( y)


= x r

Z sh ( y)

min{λ1, x }0

+2

k =1

ρk λsk H ( x − λk ) + Z sh ( y)

x

min{λ2, x }

(30b)

where ρ1 = 1/(1 − µ) − E A/ E A

λ−

1

, ρ2 =

E A/ E A

λ+2

− 1/(1 − µ)

for the axial problem;

ρ1 = 1/

1 − β− E I / E I

λ−1

, ρ2 =

E I / E I

λ+2

− 1/

1 − β for the bending problem. Note that

Eq. (30) do not involve any restriction on the location of the steps, which can also occur where rotation ordeflection discontinuities are encountered due to internal springs.

3.2 Stiffness matrix

The SM of the non-uniform and discontinuous beams may be now built based on Eq. (12), computed for x = 0and x = L . Specifically, taking into account the sign conventions for the nodal forces and the internal stressset in Fig. 1, it yields

K =

a(u)1

0 b(w)1 sym

0 −b(w)2 b

(ϕ1)2

−a(u)1 0 0 a

(u)1

0 −b(w)1 b

(ϕ1)1 0 b

(w)1

0

Lb

(w)1 + b

(w)2

− Lb

(ϕ1)1 + b

(ϕ1)2

0 − Lb

(w)1 + b

(w)2

Lb(ϕ2)1 + b

(ϕ2)2

(31)




The SM holds the form (31) for any release at the beam element ends. In fact, as already remarked at the end

of Sect. 3.1, the vectors a(u) and b(w) and b(ϕ j ) in Eq. (12) shall be set accordingly.

Bearing in mind the sign convention set for the nodal forces in Fig. 1, it can be readily verified that elementsbelonging to each column of the SM (31) do satisfy the beam equilibrium equations.

4 Response to in-span loads and load vector

Next, the elements of the LV f are computed, by deriving first the GFs due to a unit-point force. As in Sect. 3,the key ideas will be presented for the beam in Fig. 2a, where no releases occur at the beam ends and the beamprofile varies with the same law along the length.

4.1 Green’s functions due to a unit axial and transverse point force

Consider the beam in Fig. 2a, acted upon at x = x P , 0 < x P < L , by a point force with a unit axial component

P x = 1 and a unit transverse component P z = 1. The generalized differential equations for the axial problemare written in the form

d N ( x )

d x = −P x δ ( x − x P ) (32a)

du ( x )

d x = N ( x )

E A ( x )+

Ni=1

N ( x i ) ζ ui δ ( x − x i ) (32b)

and the generalized differential equations for the bending problem are

dS ( x )

d x = −P zδ ( x − x P ) (33a)

d M ( x )

d x = S ( x ) (33b)

dϕ ( x )

d x = − M ( x )

E I ( x )−

Ni=1

M ( x i ) ζ ϕi δ ( x − x i ) (33c)

dw ( x )

d x = ϕ ( x ) +

Ni=1

S ( x i ) ζ wi δ ( x − x i ) (33d)

Theresponsevariables in Eqs. (32)and (33) are the GFs due to a point force,and hereinafter, theywillbe denoted

as ga ( x , x P ) = u ( x , x P ) N ( x , x P )T

and gb ( x , x P ) = u ( x , x P ) ϕ ( x , x P ) M ( x , x P ) S ( x , x P )T

. Theymay be computed based on the reasoning followed in Sect. 3, that is by taking the limit y → 0 of the response

variables of the equivalent m-stepped shown in Fig. 2b. In fact, if the m-stepped flexibility fields E A ( x )−1

and E I ( x )−1 given by Eq. (6) are replaced for the continuously varying flexibility fields E A ( x )−1 and E I ( x )−1,Eqs. (32) and (33) can be integrated in a closed form based on the same simple integration rules for generalizedfunctions discussed in Appendix A. Details are omitted for brevity. As a result of the limit y → 0, the soughtGFs take the form

gµ ( x , x P ) = Gµ ( x ) a(P) ( x P ) + pµ ( x , x P ) (34a)

gβ ( x , x P ) = Gβ ( x ) b(P) ( x P ) + pβ ( x , x P ) (34b)

In Eq. (34), the matrices Gµ ( x ) and Gβ ( x ) are given by Eq. (14). The other vectors in the r.h.s of Eq. (34)are reported below.

Vectors pµ ( x , x P ) , pβ ( x , x P )




They are vectors of functions depending on the beam profile, the discontinuity parameters, and the pointforce location, given by

pµ ( x , x P ) =

pu ( x , x P ) − H ( x − x P )

T

(35a)

pβ ( x , x 0) = pw ( x , x P ) pϕ ( x , x P ) − R1 ( x − x P ) − H ( x − x P ) T (35b)

where

pu ( x , x P ) = − R1 ( x − x P ) µ ( x )

E A+

H ( x − x P )˜ I 01µ ( x ) − x P ˜ I 00

µ ( x )

E A

−N

i=1

ζ ui H ( x − x i ) H ( x i − x P ) (36)

pw ( x , x P ) = R3 ( x − x P ) β ( x )

E I

+ H ( x − x P ) 2 I 03β ( x ) − 3 I 12β ( x ) − 3 x P ˜ I 02β ( x ) + 6 x P ˜ I 11β ( x ) − 3 x 2P ˜ I 10β ( x ) + x 3P ˜ I 00β ( x )6 E I

+N

i=1

ζ ϕi R1 ( x − x i ) R1 ( x i − x P ) −N

i=1

ζ wi H ( x − x i ) H ( x i − x P ) (37)

pϕ ( x , x P ) = R2 ( x − x P ) β ( x )

E I −

H ( x − x P )

˜ I 02β ( x ) + x 2P

˜ I 00β ( x ) − 2 x P ˜ I 01

β ( x )

2 E I

+i∈N

ζ ϕi H ( x − x i ) R1 ( x i − x P ) (38)

The symbol ˜ I rsh ( x ) in Eq. (36) through (38) denotes the integrals

˜ I rsh ( x ) = x r

x x P

dh ( y)

d y ys d y = I rs

h ( x ) − I rsh ( x P ) (39)

where I rsh ( x ) is given by Eq. (20).

Vectors a(P) and b(P)

They are vectors of constants that, for no releases at the beam ends, are given by

a(P)1 = −

L + I 10

µ ( L) − I 01µ ( L)

E A+

N

i=1

ζ ui

−1

pu ( L , x P ), (40a)

a(P)2 = 0 (40b)

and by

b(P)1 = −α2 pϕ ( L , x P ) + α4 pw ( L , x P )

α2α3 − α1α4(41a)

b(P)2 = −α3 pw ( L , x P ) + α1 pϕ ( L , x P )

α2α3 − α1α4(41b)

b(P)3 = 0 (41c)

b(P)4 = 0 (41d)

where αk are given by Eq. (24).




Table 3 End-released beam elements for different boundary conditions (B.C.): vectors b(P) for the GFs (32b) due to a unit-pointtransverse force

B.C. b(P)1 b

(P)2 b

(P)3 b

(P)4

MR1-MR2

L

− x P

L 0 − pw ( L, x P ) L

+α1 ( L

− x P )

L2 0

MR2 α2 ( L− x P )+ pw ( L, x P )

α2 L−α1− pw ( L, x P ) L+α1 ( L− x P )

α2 L−α10 0

SR2 1 − pϕ ( L, x P )+α3

α40 0

MR1-SR2 1 0 − pϕ ( L, x P ) − α3 0 MR moment released, SR shear released

4.1.1 Closed-form solutions for the GFs due to a unit-point force

Equation (39) shows that the integrals ˜ I rsh ( x ) revert to the integrals I rs

h ( x ). Therefore, since s = 1 at mostfor the GFs due to an axial force (see Eq. 36) and s = 3 at most for the GFs due to a transverse force (see Eq.37), it can be concluded that closed-form solutions for the GFs (34) can be built under the same conditions

required in Sect. 3.1 for the response due to end nodal displacements, Eq. (12), to hold in a closed form. That is,closed-form solutions for the GFs (34) hold if: (i) a 1st order primitive µ[1] ( y) is known for the axial flexibilityratio µ ( y) and (ii) if primitives up to the 3rd order, β[k ] ( y) for k = 1, 2, 3, are known for the flexural flexibilityratio β ( y). Obviously, if such primitives are not available, the GFs (32) may be still computed numerically,

by building a numerical solution for the integrals ˜ I rsh ( x ).

Eventually there are a few remarks worth doing, analogous to those at the end of Sect. 3.1. If releases occurat the beam element ends, Eq. (34) hold true as long as the constants a(P) and b(P) are changed accordingly.

For instance, for an axial release at node 1, a(P)1 = 0 and a

(P)2 = − pu ( L , x P ) are found. Expressions for

b(P)are reported in Table 3. Also, Eq. (34) hold true for different variation laws of the beam profile and/or

stepped variations, as long as pertinent changes are made to the integrals ˜ I rsh ( x ), as suggested by Eqs. (29)

and (30) for I rsh ( x ).

4.2 General expression of the response to in-span loads

For generality, consider a piecewise-continuous loading function

p ( x ) = ¯ p ( x ) H ( x − γ 1) − H ( x − γ 2)

for 0 ≤ γ 1 < γ 2 ≤ L (42)

where the axial and the transverse component, p x ( x ) and p z ( x ), are positive if rightward and downward,respectively.

If p x ( x ) and p z ( x ) are applied to the beam in Fig. 2a, based on the GFs (34) the response variables arewritten in the form

rµ f

( x ) =

L

0

gµ

( x , y) p x

( y) d y

= Gµ ( x )

γ 2 γ 1

a(P) ( y) ¯ p x ( y) d y +γ 2

γ 1

pµ ( x , y) ¯ p x ( y) d y (43a)

rβ f ( x ) = L

0

gβ ( x , y) ¯ p z ( y) d y

= Gβ ( x )

γ 2 γ 1

b(P) ( y) ¯ p z ( y) d y +γ 2

γ 1

pβ ( x , y) ¯ p z ( y) d y (43b)




Without lack of generality, it can be assumed that both ¯ p x ( x ) and ¯ p z ( x ) are arbitrary polynomial functions.In this case, a closed-form solution to the integrals in Eq. (43) that involve the vectors a(P) and b(P), that is

γ 2

γ 1

a(P) ( y)

¯ p x ( y) d y, (44a)

γ 2 γ 1

b(P) ( y) ¯ p z ( y) d y (44b)

can be sought based on Eqs. (40) and (41) in Sect. 4.1. Specifically, if the integrals ˜ I rsh ( y) are expressed in

terms of the primitive functions Z sh ( y), (see Eqs. 27, 39), it can be readily seen that computing the integrals(44) involves computing the sum of individual integrals that all revert to the general form

γ 2 γ 1

yk Z sh ( L) − Z sh ( y)

¯ p ( y) d y = Z sh ( L)

γ 2 γ 1

yk ¯ p ( y) d y −s

=0

(−1)

γ 2 γ 1

h[] ( y) yk ¯ p ( y)d ys

d y d y (45)

Note that while the first integral in the r.h.s. of Eq. ( 45) may be readily solved when ¯ p (·) is a polynomial func-tion, the second integral in the r.h.s. of Eq. (45) may be solved by a standard integration by parts where h[] ( y)is repetitively integrated and yk ¯ p ( y) d ys/d y is repetitively differentiated until its derivative vanishes. Forinstance, if ¯ p z ( y) is a linear function in the bending problem, yk ¯ p ( y) d ys/d y will involve powers of y of order 4 at most and, for the second integral in the r.h.s. of Eq. (45) to hold in a closed form, knowledge of the primitives of β ( y) up to the fifth order will be required. Such primitives are generally available for beamprofiles involved in engineering applications, as shown in Appendix B. Clearly, if the primitives of the requiredorder are not available for the flexibility ratios µ ( y) and β ( y), numerical integration may be still performedin Eq. (44).

Next consider the integrals in the r.h.s. of Eq. (43) that involve vectors pµ ( x , y) and pβ ( x , y), that is

γ 2

γ 1

pµ ( x , y)

¯ p x ( y) d y, (46a)

γ 2 γ 1

pβ ( x , y) ¯ p z ( y) d y (46b)

Based on Eq. (35), three general forms may be singled out in the integrals (46):

(i) : ( x )

γ 2 γ 1

yk H ( x − y) ¯ p ( y) d y = ( x )

min(γ 2, x ) min(γ 1, x )

yk ¯ p ( y) d y (47)

where k = 3 at most and ( x ) = h ( x ) , ( x ) = h (0) or ( x ) = x r for r = 3 at most;

(ii) : Rk ( x − x i )γ 2

γ 1

Rs ( x i − y) ¯ p ( y) d y = Rk ( x − x i )min(γ 2, x i )

min(γ 1, x i )

( x i − y)s

s! ¯ p ( y) d y (48)

for k = 1 and s = 1 at most, and

(iii) : x r

γ 2 γ 1

yk H ( x − y) Z sh ( x ) − Z sh ( y)

¯ p ( y) d y

= x r

Z sh ( x )


yk ¯ p ( y) d y −s

=0

(−1)


h[] ( y) yk ¯ p ( y)d ys

d y d y

(49)




which derives from Eq. (39). Note that the integrals (47) and (48) can be readily solved when ¯ p (·) is apolynomial function, while the integral (49) can be solved as already explained for the integral (45). As anexample, the closed-form expression of the deflection response (43b) for a specific beam profile under auniform load will be given in Sect. 6.

4.3 General expression of the fixed-end nodal vectors

The elements of the LV f may be now built based on Eq. (43). Taking into account the sign conventions forthe nodal forces and the internal stress set in Fig. 1, they are written in the form

N 1 = −γ 2

γ 1

a(P)1 ( y) ¯ p x ( y) d y (50a)

N 2 =γ 2

γ 1 a

(P)1 ( y) − 1

¯ p x ( y) d y (50b)

S 1 = −γ 2

γ 1

b(P)1 ( y) ¯ p z ( y) d y (50c)

S 2 =γ 2

γ 1

b

(P)1 ( y) − 1

¯ p z ( y) d y (50d)

M 1 =γ 2

γ 1

b

(P)2 ( y)

¯ p z ( y) d y (50e)

M 2 = −γ 2

γ 1

Lb(P)1 ( y) + b(P)

2 ( y) − ( L − y) ¯ p z ( y) d y (50f)

A closed-form solution to Eq. (50) can be built as already explained for Eq. (44). As an example, the LV for aspecific beam profile under a uniform load will be given in Sect. 6.

5 General response of the beam element

Based on the SM K and the LV f given in Sects. 3, 4, the beam element may be obviously assembled withinan arbitrary non-uniform and discontinuous frame structure. Upon solving for the nodal displacements, theresponse variables within each element may be then obtained in the form

rµ ( x ) = rµn ( x ) + rµ f ( x ) (51a)

rβ ( x ) = rβn ( x ) + rβ f ( x ) (51b)

6 Numerical example

Consider a homogeneous beam of rectangular cross section with length L and parabolically varying height,shown in Fig. 4. The flexibility ratios µ ( x ) and β ( x ) are

µ ( x ) = E A/ E A ( x ) = 1 + η x 2

−1(52a)

β ( x ) = E I / E I ( x ) = 1 + η x 2−3

(52b)




( ) ( ), EI x EA x

1 x

1uζ

1wζ

1ϕ ζ

x

z

( ) p x

45°

2 Lγ =

1 0γ =

Fig. 4 Beam with rectangular cross section, length Land parabolically varying height

where E A = Ebh and E I = Ebh3/12, being b and h the cross section dimensions at y = 0; η is a realparameter greater than−1/ L . Assume no releases at the beam ends and the presence of internal springs with

flexibilities ζ u1, ζ w1 and ζ ϕ1located at x = L /4.The elements of the SM (31) can be readily computed via Eqs. (21) through (23). Such equations involve(see Eq. 24) the primitives µ[1] ( y) , β[1] ( y) , β[2] ( y), and β [3] ( y), all given in Appendix B (see Eq. B.7).

As a second step, the LV f is sought for a uniform load ¯ p ( x 0) = ¯ p. The end axial forces are given byEq. (50a,b) assuming no releases at the beam ends:

N 1 = − p x

L + I 10

µ ( L) − I 01µ ( L)

E A+i∈N

ζ ui

−1 γ 2 γ 1

pu ( L , y) d y (53a)

N 2 = N 1 − ¯ p x (γ 2 − γ 1) (53b)

where

γ 2 γ 1

pu ( L , y) d y =

µ ( L) ( L − y)2

2 E A

γ 2

γ 1

+ 1

E A

γ 2 γ 1

˜ I 01µ ( L) − y I 00

µ ( L)

d y +

i∈Nζ ui R1 ( x i − x 0)

γ 2

γ 1

(54)

γ 2 γ 1

˜ I 01µ ( L) − y I 00

µ ( L)

d y =

−µ ( L) ( L − y)2

2− µ[1] ( L) y + µ[2] ( y)

γ 2

γ 1

(55)

Equation (55) involves the primitive µ[2] ( y), also given in Appendix B. Further, the end shear forces and the

end bending moments are given by Eq. (50c–f):

S 1 = ¯ p z

α2α3 − α1α4

−α2

γ 2 γ 1

pϕ ( L , y) d y + α4

γ 2 γ 1

pw ( L , y) d y

(56a)

S 2 = S 1 − ¯ p z (γ 2 − γ 1) (56b)

M 1 = ¯ p z

α2α3 − α1α4

−α3

γ 2 γ 1

pw ( L , y) dy + α1

γ 2 γ 1

pϕ ( L , y) d y

(56c)

M 2 = L S 1 − M 1 + ¯ p z ( L − y)2

2(56d)




The integrals in Eq. (56) are:

γ 2

γ 1

pw ( L , y) d y =

−β ( L) ( L − y)4

24 E I

γ 2

γ 1

+ 1

6 E I

γ 2 γ 1

2 I 03

β ( L) − 3 I 12β ( L) − 3 y I 02

β ( L) + 6 y I 11β ( L) − 3 y2 ˜ I 10

β ( L)+ y3 ˜ I 00β ( L)

d y

×

−i∈N

ζ ϕi ( L − x i ) R2 ( x i − y) +i∈N

ζ wi R1 ( x i − y)

γ 2

γ 1

(57)

where

γ 2 γ 1

2 I 03

β ( L) − 3 I 12β ( L) − 3 y I 02

β ( L) + 6 y I 11β ( L) − 3 y2 ˜ I 10

β ( L) + y3 ˜ I 00β ( L)

d y

=β ( L) ( L − y)4

4− 3β[2] ( L) ( L − y)2 − 12β[3] ( L) y + 6β[3] ( y) ( L − y) + 18β[4] ( y)

γ 2

γ 1

(58)

and

γ 2 γ 1

pϕ ( L , y) d y =

−β ( L) ( L − y)3

6 E I

γ 2

γ 1

− 1

2 E I

γ 1 γ 1

˜ I 02β ( L) + y2 ˜ I 00

β ( L) − 2 y I 01β ( L)

d y

−

i∈Nζ ϕi R2 ( x i − y)

γ 2

γ 1

(59)

where

γ 2 γ 1

˜ I 02β ( L) + y2 ˜ I 00

β ( L) − 2 y I 01β ( L)

d y

=

−β ( L) ( L − y)3

3+ β [1] ( L) ( L − y)2 + 2β[2] ( L) − 2β[3] ( y)

γ 2

γ 1

(60)

Equation (58) requires the 4th order primitive of the flexural flexibility ratio β [4] ( x ), given in Appendix B.The SM and the LV are now computed for the following parameters: E = 3 × 107 kNm−2; L = 4 m;

b = 0.3 m; h = 0.4 m; η = 1/32; ζ u1 = ζ w1 = 10−4 L3/ E I , and ζ ϕ1 = 10−3 L/ E I ; also, ¯ p x = ¯ p z =20 kNm−1, γ 1 = 0, and γ 2 = L. The proposed exact solutions are assessed compared to the correspondingsolutions for a beam with m uniform segments, with increasing m. For the SM and the LV, Tables 4, 5 shows

that at least m = 500 uniform segments are required to achieve a percent difference of 10−2 in the approximateSM and in the LV.

Eventually, the response variables along the beam length are computed. Specifically, based on Eq. ( 43b),the deflection and rotation responses due to the in-span transverse load ¯ p z are written as

w ( x ) = ¯ p z

Gβ11 ( x )

γ 2 γ 1

b(P)1 ( y) d y + Gβ12 ( x )

γ 2 γ 1

b(P)2 ( y) d y

+ ¯ p z

γ 2 γ 1

pw ( x , y) d y (62)

ϕ ( x ) = ¯ p z

Gβ21 ( x )

γ 2 γ 1

b(P)1 ( y) d y + Gβ22 ( x )

γ 2 γ 1

b(P)2 ( y) d y

+ ¯ p z

γ 2 γ 1

pϕ ( x , y) d y (63)




Table 4 Elements of the SM for the beam in Fig. 4: exact solution and percent difference error (p.d.) of the approximate solutionfor a m-stepped beam

Exact m = 5 m = 10 m = 50 m = 100 m = 500 m = 103 m = 5 × 103

p.d. p.d. p.d. p.d. (

×10−1) p.d. (

×10−1) p.d. (

×10−2) p.d. (

×10−2)

K 11 =−K 41 = K 44 908705.625 3.412 1.682 0.339 1.712 0.329 1.702 0.308K 22 =−K 52 = K 55 15017.224 11.723 5.771 1.162 5.641 1.232 5.311 1.462K 32 = −K 53 24256.170 7.771 3.769 0.777 3.479 0.782 3.476 0.792K 62 = −K 65 35812.728 14.232 7.127 1.423 7.217 1.415 7.105 1.405K 33 56607.855 5.876 2.930 0.586 2.420 0.456 2.960 0.256K 63 40416.826 10.414 5.224 1.044 5.314 1.134 5.125 1.034K 66 102834.087 15.723 7.858 1.571 7.156 1.491 7.958 1.381

Table 5 Elements of the LV for the beam in Fig. 4: exact solution and percent difference error (p.d.) of the approximate solutionfor a m-stepped beam

Exact m = 5 m = 10 m = 50 m = 100 m = 500 m = 103 m = 5 × 103

p.d. p.d. p.d. (×10−1) p.d. (×10−1) p.d. (×10−2) p.d. (×10−2) p.d. (×10−3)

N 1 −

35.175 1.547 0.720 1.340 0.688 1.311 0.680 1.305

S 1 −36.614 2.109 0.914 1.532 0.764 1.501 0.752 1.516 M 1 −21.468 5.789 2.641 4.734 2.322 4.488 2.182 3.436 N 2 −44.825 1.214 0.565 1.051 0.515 1.029 0.510 1.011S 2 −43.386 1.779 0.771 1.325 0.644 1.266 0.634 1.280 M 2 35.012 5.272 2.205 3.612 1.773 3.527 1.798 4.237

The first two integrals in the r.h.s. of Eqs. (62) and (63) havebeen already given a solution, see for this Eqs. (50c)and (56a), Eqs. (50e), and (56c). The latter integrals in the r.h.s. of Eqs. (62) and (63) are computed in theform

γ 2 γ 1

pw ( x , y) d y

=

−β ( x ) ( x − y)4

24 E I

min{ x ,γ 2}

min{ x ,γ 1}

+ 1

6 E I

β ( x ) ( x − y)4

4− 3β[2] ( x ) ( x − y)2 − 12β[3] ( x ) y + 6β[3] ( y) ( x − y) + 18β[4] ( y)

min{ x ,γ 2}

min{ x ,γ 1}

×

−i∈N

ζ ϕi R1 ( x − x i ) R2 ( x i − y) +i∈N

ζ wi H ( x − x i ) R1 ( x i − y)

γ 2

γ 1

(64)

γ 2 γ 1

pϕ ( x , y) d y

=

−β ( x ) ( x − y)3

6 E I

min{ x ,γ 2}

min{ x ,γ 1}

− 1

2 E I

−β ( x ) ( x − y)3

3+ β [1] ( x ) ( x − y)2 + 2β[2] ( x ) − 2β[3] ( y)

min{ x ,γ 2}

min{ x ,γ 1}

−

i∈Nζ ϕi H ( x − x i ) R2 ( x i − y)

γ 2

γ 1

(65)




0 1 2 3 4

x [m]

1.2E-005

8E-006

4E-006

0

-4E-006

d e f l e c t i o n r e s p o n s e [ m ]

d e f l e c t i o n r e

s p o n s e [ m ]

0.8 0.9 1 1.1 1.2

x [m]

8E-006

7E-006

6E-006

5E-006

4E-006

a

b

Fig. 5 Deflection of beam in Fig. 4 (a) and zoom around discontinuity at x = L /4 (b)

Note that Eqs. (64) and (65) can be readily derived taking into account that γ 2

γ 1 H ( x − y) f ( y) d y =

f [1] ( y)min{ x ,γ 2}

min{ x ,γ 1}, where f [1] ( y) is the first-order primitive of a function f ( y).

The exact deflection and rotation responses (62)–(63) are plotted in Figs. 5, 6. As expected, at x = x 1where the internal springs occur both a deflection and a rotation discontinuity are encountered, correspondingto the generalized functions in Eqs. (62) and (63). Recognize that Eqs. (62) and (63) are closed-form functionsof the discontinuity parameters (location and amplitude) as well as of the beam profile through the primitivesβ[k ] ( x ) given in Appendix B; therefore, they can be readily implemented in a symbolic package.




0 1 2 3 4

x [m]

-8E-006

-4E-006

0

4E-006

8E-006

1.2E-005

r o t a t i o n r e s p o n s e

r o t a t i o n

r e s p o n s e

0.8 0.9 1 1.1 1.2

x [m]

5E-006

5.5E-006

6E-006

6.5E-006

7E-006

7.5E-006

a

b

Fig. 6 Rotation of beam in Fig. 4 (a) and zoom around discontinuity at x = L /4 (b)

7 Conclusions

The theory of generalized functions has been used to solve the static equilibrium problem of Euler–Bernoullinon-uniform and discontinuous beams. Simple integration rules for generalized differential equations have ledto the full set of response variables due to end nodal displacements and to in-span loads, from which the generalexpressions of the stiffness matrix and the load vector have been readily derived. Closed-form expressionshave been given for most common beam profiles.

The proposed formulation proves to be advantageous for FE analysis since, upon computing the nodaldisplacement response via global stiffness matrix inversion, no numerical integration is required to build theresponse variables inside the beam element. Numerical integration would be necessary, instead, if the existingmethods in the literature were resorted to. It is also worth remarking that the proposed formulation appearsparticularly suitable for those applications, such as damage sensitivity, damage identification, or optimization,




where a large number of solutions shall be built for different sets of discontinuity parameters [8,23,24]. In fact,even if the primitives µ[k ] ( y) and β [k ] ( y) are not available for the beam profile, the response variables herederived, Eq. (51), always depend on the discontinuity parameters in a closed form.

Shear effects [13,25] have not been considered in this paper, but they also may be included in the proposed

formulation following the strategy reported in Appendix C.

Appendix A

In Sects. 3, 4, the response of an arbitrary non-uniform and discontinuous beam, as shown in Fig. 2a, has beenbuilt by taking the limit for y → 0 of the response of the corresponding m-stepped shown in Fig. 2b. Theresponse of such a beam is now built in this Appendix, based on simple rules of integration for generalizedfunctions, which apply to derive both the response to end nodal displacements and to a unit-point load.

Consider then the m-stepped in Fig. 2b. The differential equations governing the response variables coin-cide with Eqs. (1) and (2) in the main text, where Eq. (6) are replaced for the axial flexibility E A ( x )−1

and the flexural flexibility E I ( x )−1. To integrate such differential equations in a closed form, the following

generalized integrals are of interest:

x −∞

f ( y) δ ( y − x 0) d y = f ( x 0) H ( x − x 0) (A.1)

which derives directly from Eq. (5). Further, the so-called nth-order ramp functions

Rn ( x − x 0) = x

−∞ Rn−1 ( y − x 0) d y =

x −∞

· · · x

−∞ H ( y − x 0) d y

n−times

= ( x − x 0)n

n! H ( x − x 0) (A.2)

for n = 1, 2. . ., where R0 ( x − x 0) = H ( x − x 0). Finally,

x −∞

f ( y) H ( y − x 0) d y = H ( x − x 0) f [1] ( x ) − f [1] ( x 0)

(A.3)

where f ( x ) is a function for which a first-order primitive f [1] ( x ) exists.

Based on the generalized integrals (A.1)–(A.3), it is readilyseenthatthe response variablesof the m-steppedbeam for the axial problem take the form

ˆ N ( x ) = c1 (A.4a)

u ( x ) = n[1] ( x )

E A+

Ni=1

N ( x i ) ζ ui H ( x − x i ) + c2 (A.4b)

where ci ‘s are integration constants and n[1] ( x ) is the first-order primitive

n[1] ( x ) = ˆ N [1] ( x ) +m+1

k

=1

µ ( yk )

y H ( x − yk )

ˆ N [1] ( x ) − ˆ N [1] ( yk )

y (A.5)




being N [1] ( x ) a first-order primitive of the axial force function (A.4a). Similarly, the response variables of thebending problem are given by

S ( x ) = d1 (A.6a)

ˆ M ( x ) = d1 x + d2 (A.6b)

ϕ ( x ) = − m[1] ( x )

E I −

Ni=1

M ( x i ) ζ ϕi H ( x − x i ) + d3 (A.6c)

w ( x ) = − m[2] ( x )

E I −i∈N

M ( x i ) ζ ϕi R1 ( x − x i ) +N

i=1

S ( x i ) ζ wi H ( x − x i ) + d3 x + d4 (A.6d)

where di ‘s are integration constants; m[1] ( x ) and m[2] ( x ) are the first- and second-order primitives

m[1] ( x ) = ˆ M [1] ( x ) +m+1k =1

β ( yk )

y H ( x − yk )

ˆ M [1] ( x ) − ˆ M [1] ( yk )

y (A.7)

m[2] ( x ) = ˆ M [2] ( x ) +m+1k =1

β ( yk )

y H ( x − yk )

ˆ M [2] ( x ) − ˆ M [2] ( yk ) − ˆ M [1] ( yk ) R1 ( x − yk )

y (A.8)

where ˆ M [i] ( x ) is a i th-order primitive of the bending moment function (A.6b) and R1 (·) is the first-orderramp function.

The integration constants ck and dk can be then computed by setting the boundary conditions. For instance,if no internal releases exist at the beam ends, u (0) = u1, u ( L) = u2, w (0) = w1, w ( L) = w2, ϕ (0) =ϕ1, ϕ ( L) = ϕ2 shall be set, and by taking the limit for y → 0, Eq. (12) are readily derived.

The integration rules (A.1)–(A.3), used to derive the response to end nodal displacements, are involvedalso if the response to a unit-point load is sought. For instance, recognize that the response variables of them-stepped beam subjected to a unit-point load P x

= 1 at x

= x 0 are still given by Eq. (A.4), where

ˆ N ( x ) = c1 − P x H ( x − x 0) (A.9)

and

ˆ N [1] ( x ) = c1 x − P x R1 ( x − x 0) (A.10)

Similar changes shall be applied to Eq. (A.6) to derive the response variables to a unit transverse loadP z = 1. They are omitted for brevity.

Appendix B

In this Appendix, the primitive functions µ[1] and β [k ] (k

= 1, 2, 3) are given for some beam profiles. Such

primitives allow the evaluation of the stiffness matrix and the complete solution of a beam element under pointforces or nodal displacements. Primitives µ[2] and β [4] are also reported in order to solve the problems relatedto constant in-span loads.

Linear taper

If the height of the cross section varies linearly and the other cross section dimension either remains constantor is a constant multiple of the increasing dimension, then

µ ( x ) = E A/ E A ( x ) = [1 + ε x ]−m β ( x ) = E I / E I ( x ) = [1 + ε x ]−(n+2) (B.1)

where m and n are integer number, whereas ε > −1/ L. For most common cross sections, m = n = 1 orm = n = 2. Cross section variation satisfying Eq. (B.1) is summarized in ref. [6]. For the case m = n = 1




(which includes rectangular cross sections with constant base and thin annular cross section with constantthickness), the primitives read:

µ[1] = Log(1+ε x )

η µ[2] = −ε x +(1+ε x )Log(1+ε x )

ε2

β[1] = − 12ε(1+ε x )2 β[2] = 12ε2+2ε3 x

β[3] = Log

2ε2(1+ε x )

2ε3 β[4] = Log(1+ε x )+ε x −1+Log

2ε2(1+ε x )

2ε4

(B.2)

A linear taper with m = n = 2 (e.g. rectangular and triangular cross section with fixed ratio height/baseand circular cross section) is characterized by

µ[1] = − 1ε+ε2 x

µ[2] = −Log[ε(1+ε x )]ε2

β[1] = − 13ε(1+ε x )3 β[2] = 1

6ε2(1+ε x )2

β[3] = − 16ε3(1+ε x )

β[4] = −Log(1+ε x )

6ε4

(B.3)

Parabolically varying height

In the case, the height of the cross section varies with a quadratic law; axial and bending flexibilities are

µ ( x ) = E A/ E A ( x ) = 1 + ε x + η x 2−m

β ( x ) = E I / E I ( x ) = 1 + ε x + η x 2−(n+2)

(B.4)

For m = n = 1, the following primitives are obtained

µ[1] = 2 x 4η − ε2

µ[2] = 2(2η x + ) x −

4η − 2Log[1 + x (η x + )]2η

4η − 2

β[1] = (2η x + )

2η

5 + 3η x 2+ 6η x − 2

2−4η + 2

2(1 + x (η x + ))2

+ 12η2 x 4η − 2

5/2

β[2]

=−

4η − 2

3/2 + 12η(2η x + )(1 + x (η x + )) x

2 4η − 25/2 (1 + x (η x + ))

(B.5)

β[3] = −3η x

4η − 2 + 6η2 x 2 + 2 + η(2 + 6 x ) x

4η − 25/2

β[4] = −12 x (4η x + )

4η − 2 + (2η x + )(1 + x (η x + )) x

4η − 25/2

where

x = ArcTan

2η x +

4η − 2

(B.6)

Clearly, the real parameters η and ε must be such that ε x +

η x 2 > −

1 for 0 < x < L .The numerical application in Sect. 6 refers to the case ε = 0, for which the primitives reduce to

µ[1] = x √ η

µ[2] = xx √ η − Log

1+η x 2

2η

β[1] = 18

x

5+3η x 2

(1+η x 2)2 + 3 x √

η

β [2] = 1

8

− 1

η+η2 x 2 + 3 xx √

η

β[3] = −3

√ η x +1+3η x 2

x

16η3/2 β[4] = x −2

√ η x +1+η x 2

x

16η3/2

(B.7)

Being

x = ArcTan x

√ η

(B.8)

Also, the setting with m = n = 2 is amenable of primitives in closed form, which can be derived bysymbolic software. Here, they are not reported for brevity.




Appendix C

The proposed formulation can be generalized to account for shear effects. For instance, to build the responseto end nodal displacements (with the related SM), the system of governing differential equations can be built

according to the following criteria: the shear equilibrium equation, the bending moment equation, and theflexural curvature equation hold the same form as Eq. (2a), (2b), and (2c), respectively; conversely, for theshear deformation, the following equation replaces Eq. (2d):

dw ( x )

d x = ϕ ( x ) + S ( x )

G A∗ ( x )+

Ni=1

S ( x i ) ζ wi δ ( x − x i ) (C.1)

where G A∗ ( x ) is the shear stiffness varying according to the beam profile. Such a system of differential equa-tions can be then integrated based on the same scheme followed in this paper, that is: (i) by first replacing theflexural flexibility E I ( x )−1 and the shear flexibility G A∗ ( x )−1 by the corresponding stepped expressions forthe m-stepped equivalent beam; (ii) by integrating the shear equilibrium equation and then the moment equi-

librium equation to derive an analytical expression for S ( x ) and ˆ M ( x ) in terms of the integration constants;

(iii) by replacing the latter for ˆ M ( x ) in the flexural curvature equation to derive an analytical expressionfor the rotation ϕ ( x ) in terms of the integration constants, to be subsequently used in the shear deforma-

tion equation along with the analytical expressions for S ( x ) and ˆ M ( x ) to derive an analytical expressionfor the deflection w ( x ) in terms of the integration constants; and (iv) thus, setting the boundary conditionsu (0) = u1, u ( L) = u2 w (0) = w1, w ( L) = w2, ϕ (0) = ϕ1 and ϕ ( L) = ϕ2 and taking the limit y → 0would lead to the sought analytical expressions for the response variables, corresponding to Eq. (11a, b). Then,it can be readily verified that closed-form solutions for the response to end nodal displacements can be builtfor beam profiles for which primitives of the shear flexibility ratio G A∗/G A∗ ( x ) exist up to the first order.Similar criteria could be followed to build the Green’s functions due to unit-point forces and the related LV.

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