Open Journal of Civil Engineering, 2015, 5, 17-28 Published Online March 2015 in SciRes. http://www.scirp.org/journal/ojce http://dx.doi.org/10.4236/ojce.2015.51003

How to cite this paper: Qissab, M.A. (2015) A New Stiffness Matrix for a 2D-Beam Element with a Transverse Opening. Open Journal of Civil Engineering, 5, 17-28. http://dx.doi.org/10.4236/ojce.2015.51003

A New Stiffness Matrix for a 2D-Beam Element with a Transverse Opening Musab Aied Qissab Department of Civil Engineering, Al-Nahrain University, Baghdad, Iraq Email: musstruct@nahrain-eng.org Received 16 January 2015; accepted 6 February 2015; published 10 February 2015

Copyright © 2015 by author and Scientific Research Publishing Inc. This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/

Abstract Transverse opening in a beam has a reducing effect of the beam stiffness which will cause a sig-nificant increase in beam deflection in the region on the opening. In this paper, a new stiffness matrix for a beam element with transverse opening including the effect of shear deformation has been derived. The strain energy principle is used in the derivation process of the stiffness matrix and the fixed-end force vector for the case of a concentrated or a uniformly distributed load is also derived. The accuracy of the obtained results based on the derived stiffness matrix is examined through comparison with that of the finite element method using Abaqus package and a previous study which show a good agreement with high accuracy.

Keywords Stiffness Matrix, Transverse Opening, Shear Deformation, Inflection Point

1. Introduction In general, transverse openings are made in beams to allow the passage of the service lines through the structure. Depending on the size and location, opening may reduce flexural stiffness of the beam significantly which will cause an increase in beam deflection as well as a decrease in strength of the beam. When a large opening is lo-cated near the support of a beam, the shear stiffness is reduced significantly and the deflection is highly increas- ed in the opening region which will make the beam serviceability unacceptable if it is not taken into account.

There are a limited number of studies that deal with the formulation of stiffness matrix of a beam with a transverse opening or deflection calculation of such beams. Benitez et al. [1] presented procedures for calculat-ing deflection of composite beams with web openings that utilized matrix analysis method. Web opening was simplified and modeled using beam elements connected by rigid links. The analysis considered only the region around the opening assuming the small moment end of the opening to be fixed while the other was assumed to be free for rotation and deflection. The matrix analysis was used to calculate the deflection across the opening

M. A. Qissab

18

only and the results was utilized to derive equations for the total deflection calculation. Donghua et al. [2] derived formulas for determining deflection of simply supported steel I-beams with a web

opening. The formulas were derived by direct solving of differential equations combined by using displacement method. The solution was limited to two loading cases only: a concentrated load at mid span and a uniformly distributed load along the span. The shear deformation was not considered.

The main objective of this paper is to derive an accurate stiffness matrix and a fixed-end force vector for a beam with transverse opening that are useful and simple for matrix analysis and software applications.

2. Statement of the Problem Consider a two dimensional (2D) beam element with a transverse opening as shown in Figure 1. The imperfo-rated segments of the beam are assumed to be connected by upper and lower beams in the opening region which have a flexural rigidity of EIt and EIb respectively.

To simplify the derivation of stiffness matrix, it is assumed that the bending moment in the opening region consists of two parts: primary and secondary moments. The primary moment is caused by the normal ten-sion-compression forces couple acting on the section which has a little effect on the total deflection while the secondary moment is caused by the shear force acting on the upper and lower beams. The secondary moment cause a significant increase in deflection in the opening region as shown in Figure 1. Also, it is assumed that an inflection points are located at the midpoint of the upper and lower beams in the opening region which is utilized to find the secondary moment in the opening region. The previous assumptions have been verified through the finite element analysis carried out in this paper. The assumptions were adopted by Donghua et al. [2].

The stiffness coefficients have been found by using Castigliano’s second theorem Boresi and Schmidt [3], which states that the deflection caused by an external force is equal to the partial derivative of the total strain energy (UT) with respect to that force. The total strain energy caused by bending, axial, and shear forces can be written respectively as follows:

( ) ( ) ( ) ( )2 2 2 2

10

1 1 1 1d d d d2 2 2 2

o o

o

a L a La L

b sa a

po a L

U M x x M x x M x x M x xEI EI EI EI

+ +

+

= + + +∫ ∫ ∫ ∫ (1)

( ) ( ) ( )2 2 2

10

1 1 1d d d2 2 2

o

o

a La

aLa

L

a

U P x x P x x P x xEA EA EA

+

+

+ += ∫ ∫ ∫ (2)

( ) ( ) ( )2 2 223

01

720.6 0.6d d d

1

o

o

a La Lv

sa a L

CU Q x x Q x x Q x x

GA GAhGAh

+

+

= + + −

∫ ∫ ∫ (3)

T b a sU U U U= + + (4)

in which

( ) i iM x Q x M= − (5)

secondary momen2

t os i

LM Q x a

= = − +

(6)

oEI = flexural rigidity of the upper and lower beams at the opening considering the bending of each beam individually due to the secondary moment;

( )o t bEI E I I= + (7)

31

1 flexural rigidity of the beam section at the opening regi 1onhEI EIh

− = = (8)

11 cross-sectional area at the opening regi 1on

hA Ah

−

=

= (9)

M. A. Qissab

19

Figure 1. Beam element with a transverse opening: (a) Nodal forces and de-grees of freedom; (b) Deflected shape.

( ) ( ),i iP x P Q x Q= =

3 51 11 1 11

64 96 320v

h hhh hh

C

− − − = − +

(10)

A = cross-sectional area of the imperforated beam, h = total depth of the beam, b = with of the beam, h1 = depth of the opening, G = shear modulus, E = elastic modulus. After simplifications, carrying out the necessary integrals, and substituting the above expressions into Equation (4), the following expression for the total strain energy can be given:

( )( )

( )( )

3 332

2 2 322 2

2 22

2

1

3

2

2 2 3

1

1 12 3 2 3 2 2

12 2 2 2 3

1 12 3

o oT i i

o

i ii

i i

i i

o oo i i o

o

o o

Q a Q L LU M Q a M a a a

EI EI

Q L L Q aa a a a L M Q a L a

EI EI

M L Q a L aEI

= − + + − + + + − − + + + − + − + + + − +

( )( ) ( )( )( )

( )( ) ( )

22

33

2

2 22

231

1

12

721 1 1 0.62 3 2

1

o i i o

v oo o o

i

i ii

P

M L a L M Q L a LEI

Q C LAQ L a L L L L LEI EA A AG h

h

− + − − +

+ − + + + − + + − −

(11)

The partial derivative of the total strain energy with respect to Pi, Qi, and Mi is equal to the corresponding nodal displacements which can be given as follows:

1

1iTi o

i

PU Au L LP EA A

∂= = + − ∂

(12)

M. A. Qissab

20

( ) ( )i iTi

i

Q MUvQ EI EI

α β∂

= = −∂

(13)

( ) ( )i iTi

i

Q MUM EI EI

θ β γ−∂

= = +∂

(14)

where

( )( ) ( )

3 3 2 232

333 3

231

1

13 3 2 2 2 2

721 2 0.63 2

1

o o o oo

o

o v oo o

L L L La I a a a a a a LI

L C LI EIa a L a L L LI AG h

h

α = + − + + + − − + +

+ + − + − + + + − −

(15)

( )( ) ( )( )2

2 22 2

1

12 2 2o oa I a L a L a L

Iβ = + + − + − + (16)

( )( )1

o oIa L L a LI

γ = + + − + (17)

2.1. Axial Stiffness The axial stiffness coefficients can be found by setting the axial nodal displacement equal to (1) with all other displacements equal to zero (i.e. 1.0iu = , 0iv = , and 0iθ = ) as follows:

1

1.0 1iTi o

i

PU Au L LP EA A

∂= = = + − ∂

(18)

11 21 31

1

, 0, 01

i i i

o

EAk P k Q k MAL LA

= = = = = =

+ −

(19)

The axial force at node (j), (Pj), can be found from equilibrium condition. Accordingly,

41 51 61

1

, 0, 01

j i j j

o

EAk P P k Q k MAL LA

−= = − = = = = =

+ −

(20)

2.2. Translational Stiffness The translational stiffness coefficients corresponding to a unit translational displacement at node (i), (vi), can be found by equating the translational nodal displacement equal to (1) with all other displacements equal to zero (i.e. 0iu = , 1.0iv = , and 0iθ = ) as follows:

1

0 1iTi o

i

PU Au L LP EA A

∂= = = + − ∂

(21)

( ) ( )1.0 i iTi

i

Q MUvQ EI EI

α β∂

= = = −∂

(22)

( ) ( )0 i iTi

i

Q MUM EI EI

θ β γ−∂

= = = +∂

(23)

M. A. Qissab

21

Solving the above equation for Pi, Qi, and Mi, the following stiffness coefficients can be obtained:

( )( )

( )( )22 32 122 2

, , 0i i iEI EI

k Q k M k Pγ β

αγ β αγ β= = = = = =

− − (24)

Similarly, from equilibrium, the following stiffness coefficients can be given:

42 52 22 62 22 320, ,j j i jk P k Q Q k k M k L k= = = = − = − = = − (25)

2.3. Rotational Stiffness In the same previous procedure, the rotational stiffness corresponding to a unit rotation at node ( )i , ( )iθ , with all other degrees of freedom fixed (i.e. 0iu = , 0iv = , and 1.0iθ = ) as follows:

1

0 1iTi o

i

PU Au L LP EA A

∂= = = + − ∂

(26)

( ) ( )0 i iTi

i

Q MUvQ EI EI

α β∂= = = −

∂ (27)

( ) ( )1.0 i iTi

i

Q MUM EI EI

θ β γ−∂= = = +

∂ (28)

The rotational stiffness coefficients can be given as follows:

( )( )

( )( )33 23 132 2

, , 0i i iEI EI

k M k Q k Pα β

αγ β αγ β= = = = = =

− − (29)

From equilibrium,

43 53 23 63 23 330, ,j j i jk P k Q Q k k M k L k= = = = − = − = = − (30)

Making use of the symmetry of the stiffness matrix and from equilibrium requirements, the other stiffness matrix coefficients can be obtained as follows:

14

1

1o

EAkAL LA

−=

+ −

(31)

24 340, 0k k= = (32)

44

1

1o

EAkAL LA

=

+ −

(33)

54 64 150, 0, 0k k k= = = (34)

( )( )25 2

EIk

γαγ β−

=−

(35)

( )( )35 2

EIk

βαγ β−

=−

(36)

45 0k = (37)

( )( )55 2

EIk

γ

αγ β=

− (38)

M. A. Qissab

22

65 25 35k k L k= − (39)

16 0k = (40)

26 22 32k k L k= − (41)

36 23 33k k L k= − (42)

46 0k = (43)

56 32 22k k k L= − (44)

66 26 36k k L k= − (45)

The (6 × 6) stiffness matrix can be written as follows:

[ ]

11 12 13 14 15 16

21 22 23 24 25 26

31 32 33 34 35 36

41 42 43 44 45 46

51 52 53 54 55 56

61 62 63 64 65 66

k k k k k kk k k k k kk k k k k kk k k k k kk k k k k kk k k k k k

=

K (46)

3. Fixed-End Forces for a Beam with a Transverse Opening Due to a Concentrated Load (P)

Consider a fixed-end beam with a transverse opening is acting upon by a concentrated load (P) as shown in Figure 2.

Making use of superposition and the condition that the sum of nodal displacements in each direction must equal to zero at the fixed end and neglecting the effect of axial deformation (i.e. 0iPF = ), the flexibility matrix corresponding to node (i) can be obtained as follows:

11 12

21 22

i i

i i

QF PMF P

δ δθδ δ∆

=

(47)

where iP∆ , iPθ are the vertical displacement and rotation at node (i) due to the applied load assuming node (i) is free and node (j) is fixed respectively.

( )( ) ( )( )

( )( ) ( )( )

( )( )

3 33 3 3 311

1

2 22 2 212 21

1

221

3 4

2

io o o

O

io o

io o

QF I Ia L a L a L a LEI I I

QF Ia a L a L a LEI I

MF Ia L L a LEI I

δ

δ δ

δ

= + + + − + − +

− = = + + − + − +

= + + − +

(48)

Figure 2. Fixed-end beam with a transverse opening under a concentrated load.

M. A. Qissab

23

( )( ) ( )( )( )( )( ) ( )( )( )

33 23 23 2

3 23 2

1

1 2 33 2 6 12

2 3 , 06

oi o o

o

o o

LP a ea IP L a L e L a LEI I

I a L a e a L a e aI

∆ = − + − + − − + +

+ + − − + − ≤ ≤

(49)

11 12 13 14 ,i oP P P P P a e a L∆ = ∆ + ∆ + ∆ + ∆ ≤ ≤ + (50)

in which

( ) ( ) ( ) ( )

( )

2 2

11

2 2 3 3 2 2 2

2 2 3 3

12 4 2 2 2

1 12 4 3 2

12 2 2 6 2 2

o o o

t

o o

o o oto

b

e L a L LPP e aEI

L aLe a e a a e a a e a

L L I L Le a e a e a

I

+ − ∆ = − + − + − − − + − − + −

+ − − + − − − + −

(51a)

( )( )

( )( ) ( )( ) ( )( )

( ) ( )( ) ( ) ( )( )

2 22 2

12

3 23 2 2

3 3 2 2

3 3

1 12 4 2 2 2 2 4

1 12 2

1

2 2

6

3 4

6

o o o oo

t

oo o o

oo o

o o

b

t

e L a L L LPP e a a L eEI

aLa L e a a L e a a L e

LL a e L a e

I L L e aI

+ − − = − − + + + − − + − + + − − + + −

− − − − + − − −

− − − − +

(51b)

( ) ( )( ) ( ) ( )( )3 23 213

13 2o o

P eP L a L L a LEI

∆ = − + − − +

(51c)

( ) ( )( ) ( ) ( )( )3 23 214

1

13 2o o

P eP a L e a L eEI

= + − − + −

(51d)

( ) ( )( ) ( ) ( )( )( )3 3 2 22 3 ,6i o

PP L e e L e a L e LEI

∆ = − − − + ≤ ≤ (52)

for 0 e a≤ ≤ :

( )( ) ( )( ) ( )( )( )2

2 22 2

1

12 2i o o o o

P a IP ea a L a eL L a L e L a LEI I

θ = − + + − − + − + − − +

(53)

for oa e a L≤ ≤ + :

( )( ) ( ) ( )( ) ( )( )2 22 2

1

1 12 2i o o

P IP e a e e a L a L e L a LEI I

θ = − − − + − + − − +

(54)

for oa L e L+ ≤ ≤ :

( ) ( )( ) ( )2 212i

PP L e e L eEI

θ = − − −

(55)

Solving Equation (47) for QFi, MFi yields the following:

M. A. Qissab

24

( ) ( )

( ) ( )

22 1211 22 12

12 1111 22 12

2

2

1

1

i i i

i i i

QF P P

MF P P

δ δ θδ δ δ

δ δ θδ δ δ

= −

= −

∆ −

+ ∆−

(56)

4. Fixed-End Forces for a Beam with a Transverse Opening Due to a Uniformly Distributed Load (W)

Consider a fixed-end beam with a transverse opening is acting upon by a concentrated load (W) as shown in Figure 3.

Following the same previous procedure, the following expressions for ( )iP∆ and ( )iPθ can be obtained:

( )( )

( )( ) ( )( )

( )( ) ( )( )

2 2 234 3

2 22 42 4

34 44 4

1

2 18 2 2 8 2 6 2

2 1 12 2 8 2 8

2

o o o oi o o

t

o oo o

t

oo o

b

L aL L LW I aP a aL a a L aEI I

aL LI a a L a a L aI

aL I I a L a L a LI I

∆ = + + + + − + + −

− + + + − + + −

+ + + − +

+

−

(57)

( )( ) ( )( )3 33 3 3

16i o oW Ia a L a L a LEI I

θ

∆ = + + − + − +

(58)

Substituting Equations (48), (57) and (58) into Equation (56) the values of QFi and MFi for a uniformly dis-tributed load (W) can be obtain.

5. Numerical Examples To examine the correctness of the derived stiffness matrix and the fixed-end force vectors, the simply supported steel I-beam shown in Figure 4. (which was previously analyzed by Donghua et al. [2]) is analyzed under two loading cases; a concentrated load and a uniformly distributed load. The obtained results are compared with that obtained by Donghua et al. [2] which are shown in Figure 5 and Figure 6 and a good agreement is obtained.

Another verification for the obtained results is made by the finite element analysis of the same beam using Abaqus package [4]. The (C3D8IH) 8-node linear brick, hybrid, linear pressure, incompatible modes element is used in the finite element analysis. The finite element mesh of the beam is shown in Figure 7. The deformed shape of the beam for the two loading cases is shown in Figure 8 and Figure 9 and the results are shown in Figure 10 and Figure 11. The present finite element analysis shows a good agreement with the obtained results using the derived matrix and the fixed-end force vectors.

Figure 12 and Figure 13 show the effect of shear deformation for the two loading cases and it is found that the maximum increase in deflection due to shear deformation is (6.96%) at midspan of the beam under the con-centrated load. The increase in deflection is (4.48%) at midspan of the beam under the uniformly distributed load.

Figure 3. Fixed-end beam with a transverse opening under a uniformly distributed load.

M. A. Qissab

25

Figure 4. Steel I-beam with a transverse opening under two loading cases.

Figure 5. Beam deflection profile of the present analysis versus Donghua et al. [2] analysis (concentrated load).

Figure 6. Beam deflection profile of the present analysis versus Donghua et al. [2] analysis (uniformly distributed load).

0 600 1200 1800 2400 3000 3600 4200300 900 1500 2100 2700 3300 3900

Beam Length (mm)

-14

-12

-10

-8

-6

-4

-2

0

2

-13

-11

-9

-7

-5

-3

-1

1

Def

lect

ion

(mm

)

Present (Stiffness Method)Donghua Zhou et al. [2](method 2)

0 600 1200 1800 2400 3000 3600 4200300 900 1500 2100 2700 3300 3900

Beam Length (mm)

-12

-10

-8

-6

-4

-2

0

2

-11

-9

-7

-5

-3

-1

1

Def

lect

ion

(mm

)

Donghua Zhou et al.[2] (method 2)Present (Stiffness Method)

M. A. Qissab

26

Figure 7. Finite element mesh of the beam under consideration.

Figure 8. Deformed shape of the beam under the action of a concentrated load.

Figure 9. Deformed shape of the beam under the action of a uniformly distributed load.

Figure 10. Beam deflection profile of the present analysis versus finite element analysis (concentrated load).

0 600 1200 1800 2400 3000 3600 4200300 900 1500 2100 2700 3300 3900

Beam Length (mm)

-16

-14

-12

-10

-8

-6

-4

-2

0

2

-15

-13

-11

-9

-7

-5

-3

-1

1

Def

lect

ion

( mm

)

present (Abaqus)Present (Stiffness Method)

M. A. Qissab

27

Figure 11. Beam deflection profile of the present analysis versus finite element analysis (uniformly distributed load).

Figure 12. Beam deflection profile of the present analysis showing the shear deformation effect (concentrated load).

Figure 13. Beam deflection profile of the present analysis showing the shear deformation effect (uniformly distributed load.

0 600 1200 1800 2400 3000 3600 4200300 900 1500 2100 2700 3300 3900

Beam Length (mm)

-14

-12

-10

-8

-6

-4

-2

0

2

-13

-11

-9

-7

-5

-3

-1

1

Def

lect

ion

(mm

)

present (Abaqus)Present (Stiffness Method)

0 600 1200 1800 2400 3000 3600 4200300 900 1500 2100 2700 3300 3900

Beam Length (mm)

-16

-14

-12

-10

-8

-6

-4

-2

0

2

-15

-13

-11

-9

-7

-5

-3

-1

1

Def

lect

ion

(mm

)

Present (Stiffness Method)Present (Stiffness Method with Shear Deformation)

0 600 1200 1800 2400 3000 3600 4200300 900 1500 2100 2700 3300 3900

Beam Length (mm)

-12

-10

-8

-6

-4

-2

0

2

-11

-9

-7

-5

-3

-1

1

Def

lect

ion

(mm

)

Present (Stiffness Method)Present (Stiffness Method with Shear Deformation)

M. A. Qissab

28

6. Conclusions In this article, a new stiffness matrix and fixed-end force vectors for a 2D-beam element have been derived in-cluding the effect of shear deformation. It has been found that the existence of the opening enlarges the maximum deflection of the beam significantly for the studied cases. The ratio of the maximum deflection (at the opening region) to the maximum deflection of the solid beam (no opening) is 3.32 for the case of a concentrated load at midspan and 4.39 for a uniformly distributed load. This is due to the effect of secondary moments acting on the upper and lower beams at the opening region. The maximum ratio for the increase in deflection due to the effect of shear deformation is found in the range of 4.48% - 6.96% for the cases under consideration. Through the verification of results, a good agreement has been observed between the obtained results and that obtained from the finite element analysis in addition to that available in the literature. The derived stiffness matrix and the fixed-end force vectors are useful and simple to use in the matrix structural analysis packages.

References [1] Benitez, M.A., Darwin, D. and Donahey, R.C. (1990) Deflection of Composite Beams with Web Openings. A Report

on Research, University of Kansas, Lawrence. [2] Donghua, Z., Longqi, L., Jürgen, S., Wolfgang, K. and Peng, W. (2012) Elastic Deflection of Simply Supported Steel

I-Beams with a Web Opening. International Conference on Advances in Computational Modeling and Simulation. Procedia Engineering, 31, 315-323.

[3] Boresi, A.P. and Schmidt, R.J. (2003) Advanced Mechanics of Materials. 6th Edition, Wiley, New York. [4] (2009) Abaqus Theory Manual. Dassault Systèmes Simulia Corp. Providence.