European Journal of Scientific Research

ISSN 1450-216X Vol.47 No.4 (2010), pp.562-573

© EuroJournals Publishing, Inc. 2010

http://www.eurojournals.com/ejsr.htm

Design of Sandwich Beam Element with

Partial Composite Action

Pouria Shahani

Faculty of Mechanical, University Technology Malaysia (UTM), Skudai, Johor

E-mail: pouria90@Gmail.com

Abstract

Sandwich beams composed of three layers, the top and button layer as face sheet

with low thickness and high stiffness and the middle layer as core with low modulus of

elasticity are commonly used to span large openings such as in skyscrapers, because it

prepare suitable properties for flexural stiffness with low density and absorbing energy

without weight penalty. This paper deals with the effect of partial composite action on

flexural rigidity of sandwich beams with considering of flexible shear connectors between

interfaces of layers. Then, the relationships that express the equation of elastic stiffness

matrix will be presented. The latter, are used in analysis of beams with finite element

approach based on Timoshenko theory for solving the inconsistency of degree of freedom

that is caused by relative shear slip. The flexural rigidity and shearing rigidity of composite

layers must be taken into account when the finite element approach is used for calculating

the deflection of sandwich beams.

Keywords: Sandwich Beams, Elastic Stiffness Equation, Composite Action, Timoshenko

Theory, Newmark interaction theory

1. Introduction Sandwich beams have better strength to weight ratios and rigidity in the comparison to individual

beams, because they are composed of hard and stiff faces and cores with light- weight materials

(Gdoutos and Daniel, 2008). Such beams can resist more loading than the element of sandwich beams

separately. These structures are used in high-rise buildings, or skyscrapers.

In the state that vertical loads affected on layers of sandwich beams independently, the slabs

hinder as individual element and the relative shear slip take place between the layers. Such structures

are considered as a composite beam without composite action.

Shear connection that happened in interface of layers, determine the behavior of sandwich

structures. Beam manner with full composite action or partially composite action as a result of rigid or

flexible shear connection respectively. If the beams withstand vertical loads unity that means shear

connector can be propagate between layer interfaces for maintaining shear slip depends on that shear

connector (Viest et al., 1997).

The flexural rigidity of the beam with partial composite action depends on two slip strain

furthermore sectional and materials parameters; therefore the relationship between moment and

curvature of the composite beam is nonlinear.

The finite element method is useful for gaining deflection of beams involving complex

geometries, combined loading and material properties, in which the analytical solutions are not

available (Budynas1999), so for composite sandwich structures this approach are used for calculating

Design of Sandwich Beam Element with Partial Composite Action 563

deflection. Account must be taken of the effect of relative shear slip, in finite element analysis of

sandwich beams with partial composite action. For reaching this assumption, at the two ends of such

beams, consider two independent axial degree of freedom, but inconsistency of degree of freedom take

place in finite element analysis. (Faella and et al., 2001). To keep away from this problem, the elastic

stiffness equation of composite beam element, with respect to relative slip, have to be derived (Gue-

Qiang and Jin-Jun, 2009). Because of the nonlinearity of relationship between moment and curvature

of sandwich beam with partial composite action, which happened as a result of existence of sleep strain

in flexural rigidity of such beams, the elastic stiffness equation cannot drive by using the equilibrium

of internal and external moments of beams directly. For solving this problem, according to Newmark

and at el. (1951) the elastic stiffness matrix is derived found on elastic interaction theory through the

solution of the governing differential equilibrium equation of the composite beams.

The elastic stiffness matrix for composite beams that composed of two layers has gained in

previous researches (Gue-Qiang and Jin-Jun, 2009). The majority of those researches especially

consisted on the composite that consist of concrete slabs laid on the steel beams. In this research, this

method will extended for sandwich beams with three layers with different materials and sectional for

top and bottom layers.

The deflection of sandwich beams can gain by finite element analysis considering the

Timoshenko theory (Reddy, 2005), or in the simple state of loading and supporting, the analytical

solution based on Timoshenko Theory can be used (Wang, 1995). But in both of the above approach,

the flexural rigidity and shearing rigidity for composite beams have to be considered. At the end of this

paper, the equations related to flexural and shearing rigidity are derived and explained.

2. The Composite Action Effect on Elastic Stiffness of Composite Sandwich Beams 2.1. Beams with Partial Composite Action

The strain diagram is given in Figure 1 Denote Cc and Cs as the distances from the neutral axes of any

layer components to their top surfaces, respectively. Note that , , , and

.

Figure 1: Composite beam with partial composite action: partial composite action section, strain distribution

along sectional height, and internal forces.

A strain difference can be seen along the sandwich layers interface, which is defined as slip

strain or . In partial composite action restrained slip occurs. The strain diagram is given in figure

1. The slip strain at the two top and bottom layers (faces) with the layers located in the middle (core)

can then be expressed as

564 Pouria Shahani

ϵ����� � ϵ��-ϵ� � D�-C���k � C�k � D�-C�� � C��k (1) ϵ����� � ϵ��-ϵ� � D�-C���k � D-C��k � D� � d-C��-C��k (2)

The compression in the top and bottom slabs and the tension in the core of the beam are given

by N� � k�C��- ��� � E��A�� (3) T � k ��� -C�� E�A� (4) N� � -k �C��- ��� � E��A�� (5)

The equilibrium of N and T, i.e. N=T, results in E�A�C� � E��A��C�� � E��A�� ��� � E�A� �� (6) E�A�C� � E��A��C�� � E�A� �� -E��A�� ��� (7)

and can be expressed with ϵ���� and k as the following :

Combining equations 1 and 6 yields ϵ����� � D�-C�� � C��k ������� ! C� � "#$%&�' � C��-D� (8)

Substituting equation 8 back into equation 6: C�� � �(#)#*(+�)+� ,E�A� �D� � ��� � E��A�� ��� - "#$%&�' E�A�- (9)

Substituting equation 9 back into 8 yields: C� � �(#)#*(+�)+� ,E�A� ���� � E��A�� "#$%&�' - (10)

Substituting 9 back into 3, yields N� � k (+�)+�(#)#(#)#*(+�)+� ���� � ��� - (+�)+�(#)#(#)#*(+�)+� ϵ����� (11)

As shown in the figure 1 r� � ��� � �� (12)

By considering that: �/()00001� � �(#)# � �(+�)+� (13)

From equations 12 and 13 is: N� � k/EA00001�r�-/EA00001�ϵ����� (14)

From the equation 5, can be expressed as: N� � -k �C��- ��� � E��A�� (15)

Combining equations 2 and 7 yields: C� � D� � d-C��- "#$%&�' (16)

Substituting 16 back into 7, denote C�� � �(#)#*(+�)+� ,E�A� �� � E��A�� ��� � E�A�D�-E�A� "#$%&�' - (17)

Substituting equation 17 back into 16 yields: C� � �(#)#*(+�)+� ,E��A�� ��� � E�A� �� � E��A��d-E��A�� "#$%&�' - (18)

Substituting 17 back into 5 yields: N� � k 2 (+�)+�(#)#(+�)+�*(#)#3��� -k 2 (+�)+�(#)#(+�)+�*(#)#3 �� -k 2 (+�)+�(#)#(+�)+�*(#)#3 D� � 2 (+�)+�(#)#(+�)+�*(#)#3 ϵ����� (19)

From the figure 1, can be expressed: r� � ��� � �� (20)

By considering that: �/()00001� � �(#)# � �(+�)+� (21)

Design of Sandwich Beam Element with Partial Composite Action 565

From equations 20 and 21 can be expressed: N� � -k/EA00001�r� � /EA00001�ϵ����� (22)

The equilibrium of internal and external moments with considering the figure 1 gives: M � M�� �M� �M�� � N�r�-N�r� (23.a) M � kE��I�� � kE�I� � kE��I�� � k/EA00001�r��-/EA00001�ϵ�����r� � k/EA00001�r��-/EA00001�ϵ�����r� (23.b)

With respect to the static, can be expressed I66 � 7y�dA I�� � 7 x�dA I � 7 r�dA : ∑Ar�

By approximating and with respect to Radius Zhyrasyvn, can be expressed A�r�� � I� ������� ! /EA00001�r�� � E�I� A�r�� � I� ������� ! /EA00001�r�� � E�I� < � /=>1�?@�A B C D/=E00001�F�����G� � /=E00001�F�����G�H � /=>1�?@�B (23.c)

Where /EI1�?@�A bending stiffness of composite is beam with full composite action and

is the bending stiffness of composite beam with partial composite action. /EI1�?@� � /EI1�?@�A - ,/EA00001� "#$%&�I�' � /EA00001� "#$%&�I�' - (24)

Because of the bending stiffness of the partially composite beam depends on the slip strain in

addition to sectional and material parameters, then the relationship between moment and curvature of

the composite beam is no longer linear. In next Section, the elastic stiffness equation of the partially

composite beam, based on Newmark partial interaction theory, will be derived.

2.2. Elastic Stiffness Equation of Composite Beam Element

2.2.1. Basic Assumptions

The following assumptions are employed in this section:

I. Both faces and core layers are in elastic state.

II. The shear stud is also in elastic state, and the shear–slip relationship for single shear stud is Q � Ks ������� ! s � MN (25)

where K is the shear stiffness of a stud.

III. The composite action is smeared uniformly on the face-core interface, although the actual shear

studs providing composite action are discretely distributed.

IV. The plane section of the faces slabs and the core remain plane independently, which indicates

that the strains are linearly distributed along layers section heights, respectively.

V. Lift-up of shear studs, namely pull-out of shear studs form the face slab, is prevented. The

deflection of the core layer of the beam and the faces slab at the same position along the length

is identical, or the layer components of the composite beam are subjected to the same curvature

in deformation.

2.2.2. Differential equilibrium equation of partially composite beam

The strains of the layers components at the interface can be expressed with internal forces as ϵ�� � -O�(+�)+� � P+�(+�Q+� ��� (26.a) ϵ� � O�(#)# � O�(#)# - P#(#Q# �� (26.b) ϵ�� � -O�(+�)+� � P+�(+�Q+� ��� (26.c)

Consider a differential unit of the top and bottom flange (see figure 2.), and the force

equilibrium of the unit in horizontal is

566 Pouria Shahani

Figure 2: Horizontal balances of the top and bottom layers.

For top layer ∑F6 � 0 ������� ! N�-/N� � dN�1-qdx � 0 ������� ! �O��6 � -q (27.a)

For bottom layer ∑F6 � 0 ������� ! N�-/N� � dN�1-qdx � 0 ������� ! �O��6 � -q (27.b)

The shear density transferred by single shear stud on the interface is: q � MU (28)

Combining equations 27, 25 and 28 lead to �O�6 � -q � -MU � -�NU ������� ! s � -UN �O�6 (29)

The slip strain at the interface of layers of the sandwich beams can then be expressed as ϵ���� � ���6 � -UN ��O�6� (30)

Equaling equations 26 to equation 30 result in -UN ��O��6� � -N� V �(+�)+� � �(#)#W � P+�(+�Q+� ��� � P#(#Q# �� - O�(#)# (31)

-UN ��O��6� � -N� V �(+�)+� � �(#)#W � P+�(+�Q+� ��� � P#(#Q# �� - O�(#)# (32)

By considering the assumptions, the moment curvature relationships are P+�(+�Q+� � P#(#Q# � P+�(+�Q+� � K � -yX (33)

And it leads to P+�*P#*P+�(+�Q+�*(#Q#*(+�Q+� � K � -yX (34)

By equation 23.a, one has M � M�� �M� �M�� � N�r�-N�r� (35) N� � �I� DM-M��-M�-M�� � N�r�H (36) N� � -�I� DM-M��-M�-M��-N�r�H (37)

Substituting equations 33, 34, and 36 back into equation 31 leads to the following fourth order

differential equilibrium equation of the partially composite beam between the interface of top and

middle layer �Y��6Y - NU 2 I��/(Q1+Z[&\ � �/()00001�3 ����6� � �/(Q1+Z[&\ ��P�6� - NU/()00001� P/(Q1+Z[&\ � I�/(Q1+Z[&\ ��O��6� - NU/(Q1+Z[&\ � I�/()00001� � I�(#)#�N� � 0 (38)

Substituting equations 33, 34, and 37 back into equation 32 leads to the following fourth order

differential equilibrium equation of the partially composite beam between the interface of bottom and

middle layer: �Y��6Y - NU 2 -I��/(Q1+Z[&\ � �/()00001�3 ����6� � �/(Q1+Z[&\ ��P�6� - NU/()00001� P/(Q1+Z[&\ - I�/(Q1+Z[&\ ��O��6� � NU/(Q1+Z[&\ � I�/()00001� � I�(#)#�N� � 0 (39)

With adding equations 38 and 39, yields 2 �Y��6Y - NU 2 �/()00001� � �/()00001� � I��-I��/(Q1+Z[&\ 3 ����6� � �/(Q1+Z[&\ ��P�6� - NU � �/()00001� � �/()00001�� P/(Q1+Z[&\ � �/(Q1+Z[&\ �r� ��O��6� -r� ��O��6� � � NU/(Q1+Z[&\ , I�(#)# � I�/()00001�-N�- NU/(Q1+Z[&\ , I�(#)# � I�/()00001�-N� � 0 (40)

Design of Sandwich Beam Element with Partial Composite Action 567

From equation 35, can be expressed N�r�-N�r� � M�� �M� �M���-M (41.a)

Two differentiate from the above equation, yields r� ��O��6� -r� ��O��6� � - �Y��6Y /EI1�?@�^ - ��P�6� (41.b) �()0000 � �/()00001� � �/()00001� (42)

�/()00001� � �/()00001� � I��-I��/(Q1+Z[&\ � �()0000 � I��-I��/(Q1+Z[&\ � /(Q1+Z[&\ *()0000I��-I���()0000/(Q1+Z[&\ (43) /EI1�?@�_ � /EI1�?@�^ � EA0000r��-r��� (44) a� � 'U/(Q1+Z[&\ � I�(#)# � I�/()00001�� (45) a� � 'U/(Q1+Z[&\ � I�(#)# � I�/()00001�� (46)

Substituting equations 41.b-46 into equation 40, leads to �Y��6Y - NU 2 /(Q1+Z[&a()0000/(Q1+Z[&\ 3 ����6� � �/(Q1+Z[&\ ��P�6� - NU/()00001 P/(Q1+Z[&\ � a�N�-a�N� � 0 (47)

wherek � NU is the shear modulus of the interface of composite beam. α andβ are parameters that relevant to the material properties and section dimensions, and are

defined as α� � '/(Q1+Z[&a()0000/(Q1+Z[&\ (48) β � '()0000/(Q1+Z[&\ � d�/(Q1+Z[&a (49) a�N�-a�N� � g (50)

2.2.3. Stiffness Equation of Composite Beam Element

The typical forces and deformations of the beam element are as in figure 3.

Figure 3: The typical forces and deformation of the beam element

The moment at an arbitrary location distance x away from end 1 can be expressed with the end

moment and the end shearQ1 ∑M � 0 ∶ M-M� � Q�x � 0 ������� !M � M�-Q�x (51)

The force balance also determines ∑F� � 0 ∶ Q� � -Q� (52) ∑M� � 0 ∶ Q� � P*P�� (53)

Substituting equations 48-53 into equation 47, yield �Y��6Y -α� ����6� -βM�-Q�x� � g � 0 (54)

The solution of the fourth order differential equation 54 is Z � yX � c� cosh/αx1 � c� sinh/αx1- md M�-Q�x� � nd (55)

where and are integration constants.

568 Pouria Shahani

Integrating equation 55 twice results in the deflection of the composite beam element with slip

as y � ��d� cosh/αx1 � ��d� sinh/αx1 - �/(Q1+Z[&a �P�� x�- M�o xp� � nd 6�� � cpx � cq (56)

Where and are also integration constants.

Consider the following boundary conditions with considering the figure 3. rst � 0 uv � w/57r1vz � {�/57|1

rst � } uv � ~� C ~�/58r1vz � {�/58|1 With use above boundary conditions into equation 56 with considering the crammer rules

yields four simultaneous algebra equations as following: C� � d�*d���_�/d�1-��?��/d�1 2αcosh/αl1-1�δ�-δ�� ��?��/d�1-�/(Q1+Z[&a �P�� l�- M�o lp� α-gl�cosh/αl1-1�-lαθ�cosh/αl1-1�-θ�-θ��sinh/αl1-αl�- ��_�/d�1-d�/(Q1+Z[&a �M�l- ����� � �n���_�/d�1-d��d 3 (59.a) C� � d�*d���_�/d�1-��?��/d�1 2-αsinh/αl1�δ�-δ�� � cosh/αl1-1�θ�-θ�� � �?��/d�1-�/(Q1+Z[&a �M�l- M�� l�� �

1-cosh/αl1� n�d - ��_�/d�1/(Q1+Z[&a α �P�� l�- M�o lp� � sinh/αl1 n��� � αl sinh/αl1θ�3 (59.b)

�p � 12 � �}����/�}1 C 2�w��/�}1 ������/�}1�/~� C ~�1 C /�w��/�}1 C 11/{� C {�1 C �w��/�}1 C 1/=>1�?@�� V<�} C ��2 }�W� /cosh/�}1 C 11�}� � sinh/�}1/=>1�?@�� � V<�2 }� C ��6 }pW C sinh/�}1 �}�2 C �} sinh/�}1{�� � {�

(59.c) Cq � �dD�*d���_�/d�1-��?��/d�1H 2α1- cosh/αl1�δ�-δ�� � sinh/αl1-αl�θ�-θ��- �?��/d�1-�/(Q1+Z[&a �P�� l�- M�o lp� α � gl�cosh/αl1-1� �cosh/αl1-1�lαθ� � ��_�/d�1-d�/(Q1+Z[&a �M�l- M�� l�� - ��_�/d�1-d�d gl3 (59.d)

In most cases, the middle layer are connected to columns fixedly, and when the anchor-hold of

negative reinforcement bars in top and bottom slabs has good performance, it is reasonable to assume

that the slip between the middle layer and 2 up and bottom slabs at the ends of composite beams is

negligible, namely s�|6�^ � s�|6�^ � 0 (60.a) s�|6�� � s�|6�� � 0 (60.b)

Substituting equation 60 into equation 29 leads to �O��6 |6�^ � �O��6 |6�^ � 0 (61.a) �O��6 |6�� � �O��6 |6�� � 0 (61.b)

Differentiate from equation 41.a, one has �O��6 r�- �O��6 r� � ��6 M�� �M� �M���- ��6M (62)

Substituting equations 34 and 51 into above equation, yields �O��6 r�- �O��6 r� � - ����6� /EI1�?@�^ � Q� (63)

Three differentiate from equation 56, leads to ����6� � C�α sinh/αx1 � C�α cosh/αx1 - �/(Q1+Z[&a D-Q�H (64)

Substitute equation 64 back into 63, result in - �O��6 r� � �O��6 r� � /EI1�?@�^ �C�α sinh/αx1 � C�α cosh/αx1� � 2/(Q1+Z[&\/(Q1+Z[&a -13 Q� (65)

Substituting equation 65 into equation 61.a, yields

Design of Sandwich Beam Element with Partial Composite Action 569

0 � /EI1�?@�^ C�α � 2/(Q1+Z[&\/(Q1+Z[&a -13 Q� (66.a)

Substituting equation 65 into equation 61.b, yields 0 � /EI1�?@�^ �C�α sinh/αl1 � C�α cosh/αl1� � 2/(Q1+Z[&\/(Q1+Z[&a -13 Q� (66.b)

From equation 59 and 66.a, with considering the following assumptions, M1, M2, Q1, and Q2

calculated:

Assumption: a� � 2 � αlsinh/αl1-2cosh/αl1 (67.a) a� � Dα cosh/αL1 -1�δ�-δ��H-Dsinh/αL1 -αLHDθ�-θ�H-αLcosh/αL1 -1�θ� � ,L ���_�/d�1-d�d � -L�cosh/αL1 -1�- g (67.b) ap � D-α sinh/αL1Hδ�-δ�� � Dcosh/αL1 -1HDθ�-θ�H � �Lα sinh/αL1�θ� � ,L �- �?��/d�1d � �� ��_�/d�1� - g (67.c) g� � L ���_�/d�1-d�d � -L�cosh/αL1 -1� (67.d) gp � L �- �?��/d�1d � �� ��_�/d�1� (67.e)

Substitute equation 59.b into 66.a, yields

0 � /EI1�?@�^ α dU� �ap � ��?��/d�1-���- ��_�/d�1�$�� �P�/(Q1+Z[&a � 2�- �?��/d�1�$��*��_�/d�1�$�� 3M�/(Q1+Z[&a � � 2/(Q1+Z[&\/(Q1+Z[&a -13 Q� (68)

Assume that is equal to coefficient of in equation 68 b�' � �d�/(Q1+Z[&a ,���d�� � 2� - ���d�� � 2� cosh/αl1 � ���d�o � lα� sinh/αl1- -D2 � αl sinh/αl1 -2 cosh/αl1H 2/(Q1+Z[&a/(Q1+Z[&\ 3 (69)

From equation 68 with considering to 69, one has 0 � �U�/(Q1+Z[&a ,apα�/EI1�?@�^ /EI1�?@�_ � α�/EI1�?@�^ �lcosh/αl1-1�- ��d� sinh/αl1�M� � α�/EI1�?@�^ /EI1�?@�_ b�' Q�- (70)

Now, the equation 70 has to summarized as following T � α�/EI1�?@�^ /EI1�?@�_ |� � |�¢£ � ,�¤�¥�� � 2� C �¤�¥�� � 2� cosh/�}1 � �¤�¥�o � }�� sinh/�}1- /=>1�?@�^ C �2 � �} sinh/�}1 C 2 cosh/�}1�/=>1�?@�� (71.a) p� � ,-lα� � lα� cosh/αl1 - ��d�� sinh/αl1- /EI1�?@�^ (71.b) Z� � apα�/EI1�?@�^ /EI1�?@�_ (71.c)

Then rewrite equation 70 with considering the following assumptions p�M� � b�Q� � -apα�/EI1�?@�^ /EI1�?@�_ (72)

So, one has p�M� � b�Q� � -Z� (73)

Substitute assumption 67 in equation 66.b, lead to ��� sinh/�}1 � �§�/¨©1+Z[&ª «,/=>1�?@�� r��� sinh/�}1 � �C}�� sinh�/�}1 C ¤�� �p sinh/�}1 � ¤�� �p sinh/�}1�-<� � ,¤�¥�� sinh�/�}1 C¤�¥�p sinh/�}1 C ¤�¥�o sinh/�}1 cosh/�}1-��¬ (74.a) C�αcosh/αl1 ��U�/(Q1+Z[&a ,/EI1�?@�_ apα� cosh/αL1 � ,-lα� cosh/αL1 � lα� cosh�/αl1 - ��d�� cosh/αL1 sinh/αl1-M� �,��� α� cosh/αL1 - ��� α� cosh�/αl1 � ��d�o cosh/αL1 sinh/αl1- Q�- (74.b)

From equations 74.a and 74.b, yields C�α sinh/αl1 � C�αcosh/αl1 ��U�/(Q1+Z[&a ,/a� sinh/αl1 � ap cosh/αl11α�/EI1�?@�_ � lα� ,1- ��α sinh/αl1 - cosh/αl1-M� ���� α� ,-1-��dp sinh/αl1 � cosh/αl1- θ�- (74.c)

Substitute 74.c into back 66.b, yields 0 � /(Q1+Z[&\U�/(Q1+Z[&a ,/a� sinh/αl1 � ap cosh/αl11α�/EI1�?@�_ � lα� ,1- �d� sinh/αl1 - cosh/αl1-M� � ��� α� �-1- ��dp sinh/αl1 �cosh/αl1� θ�- � 2/(Q1+Z[&a/(Q1+Z[&\ -13 Q� (75)

Extract the coefficient of from equation 75, yields: b�' � �/(Q1+Z[&\ «,�2- ��d�� � � �-��d�p � lα� sinh/αl1 � ���d�� -2� cosh/αl1- /EI1�?@�^ -2 � lα sinh/αl1 -2 cosh/αl1�/EI1�?@�_ ¬ (76)

Substituting equation 76 into equation 75

570 Pouria Shahani

|� � /=>1�?@�^ |�¢ � ,�2 C ¤�¥�� � � �­¤�¥�p � }�� sinh/�}1 � �¤�¥�� C 2� cosh/�}1- /=>1�?@�^ C /2 � }� sinh/�}1 C2 cosh/�}11/=>1�?@�� (77.a) p� � lα� ,1- �d� sinh/αl1 - cosh/αl1- /EI1�?@�^ (77.b) Z� � /a� sinh/αl1 � ap cosh/αl11α�/EI1�?@�^ /EI1�?@�_ (77.c)

Substituting equation 77 into equation 75, yields p�M� � b�Q� � -Z� (78)

From equation 73 and 78 with considering to crammer rules, one has

M� � ®-¯� °�-¯� °�®±�� °��� °�± (79)

Q� � ®�� -¯��� -¯�®±�� °��� °�± (80)

Denominator of equations 79 and 80 are k� � 2p� b�p� b�3 � p�b�-p�b� (81)

Now, have to solve equation 81 k� �lα�/EI1�?@�^ � ,-4 � ��p lpαp-2lα� sinh/αl1 � 8 cosh/αl1 � �YdYq sinh�/αl1 -4 cosh�/αl1 - �p lpαp sinh/αl1 cosh/αl1- �lα�/EI1�?@�^ /EI1�?@�_ D�4 � 2lα sinh/αl1 -8 cosh/αl1 � 4 cosh�/αl1 -2lα sinh/αl1 cosh/αl1H (82)

Numerator of equation 79 is M� � Z�b�-Z�b� (83)

With considering equations 71.a, 71.c, 77.a, and 77.c yields ³�|� � ��/=>1�?@�^ /=>1�?@�� /r� sinh/�}1 � rp cosh/�}11 ,�¤�¥�� � 2� C �¤�¥�� � 2� cosh/�}1 � �¤�¥�o � }�� sinh/�}1- /=>1�?@�^ C�2 � }� sinh/�}1 C 2 cosh/�}1�/=>1�?@�� (84.a) Z�b� �apα�/EI1�?@�^ /EI1�?@�_ � ,�-��d�� � 2� � �-��d�p � lα� sinh/αl1 � ���d�� -2� cosh/αl1- /EI1�?@�^ -D2 �lα sinh/αl1 -2 cosh/αl1 /EI1�?@�_ H (84.b)

After solving equation 83 by using equation 84 with considering equations 67.b, 67.c, 67.d, and

67.e, the coefficient of equation 83, can be written as Coefficient of δ�-δ��: t� �,-4α sinh/αl1 � 4α sinh/αl1 cosh/αl1 � ���d�o -2lα� α sinh�/αl1- /EI1�?@�^ -D-4α sinh/αl1 -2lα� sinh�/αl1 �4α sinh/αl1 cosh/αl1H/EI1�?@�_ (85.a)

Coefficient of θ�: t� �,�p��d�� � 2- �YdYp � sinh�/αl1 � lpαp-2lα� sinh/αl1 cosh/αl1 � ���d�� � 2� coshp/αl1 � �3lα- ¶��d�o � sinh/αl1 � �-��d�� -2� cosh/αl1 sinh�/αl1 ���d�� sinh�/αl1 cosh�/αl1 � ����d�p -lα� cosh�/αl1 sinh/αl1 � ���d�o � lα� sinhp/αl1 � �-p��d�� -2� cosh�/αl1 � l�α�-2� cosh/αl1 � 2- ��d�� - /EI1�?@�^ �D-2 sinh�/αl1 � lα sinh/αl1 cosh/αl1 -lα sinhp/αl1 � 2 sinh�/αl1 cosh/αl1 � lα cosh�/αl1 sinh/αl1 � 2 cosh�/αl1 �2 cosh/αl1 -2 coshp/αl1-3lα sinh/αl1 -l�α� sinh�/αl1 -2H/EI1�?@�_ (85.b)

Coefficient of θ�: sp � ,¤�¥�� � /}� � }p�p1 sinh/�}1 � �6 C ¤�¥�� � cosh/�}1 C /2 � }p�p1 sinh/�}1 cosh/�}1 � �¤�¥�� � ¤Y¥Yo � sinh�/�}1 � �p¤�¥�� � cosh�/�}1 � �2 � ¤�¥�o �}�� sinh/�}1 cosh�/�}1 � �2 � ¤�¥�� � cosh/�}1 sinh�/�}1 C �}� � ¤�¥�o � sinhp/�}1 C �2 � ¤�¥�� � coshp/�}1- /=>1�?@�^ C �}� sinh/�}1 � 2 cosh/�}1 C2}� sinh/�}1 cosh/�}1 � }� sinh/�}1 cosh�/�}1 � 2 sinh�/�}1 cosh/�}1 C 2 coshp/�}1 C �} sinh�/�}1�/=>1�?@�� (85.c)

Coefficient of t4: sq � ,�� �¤�¥�� � 2� sinh/�}1 C �� �¤�¥�� � 2� sinh/�}1 cosh/�}1 � �� �¤¥�o � }�� sinh�/�}1 � �p �¤�¥�� � 2� cosh/�}1 C �p �¤�¥�� � 2� cosh�/�}1 � �p �¤¥�o �}�� sinh/�}1 cosh/�}1 � �p �2 C ¤�¥�� � � �p �}� C ¤�¥�p � sinh/�}1 � �p �¤�¥�� C 2� cosh/�}1- /=>1�?@�^ C�2�� sinh/�}1 � }��� sinh�/�}1 C 2�� sinh/�}1 cosh/�}1 � 2�p cosh/�}1 � }��p sinh/�}1 cosh/�}1 C 2�p cosh�/�}1 � 2�p � }��p sinh/�}1 C2�p cosh/�}1�/=>1�?@�� (85.d)

Substituting equation 85 into equation 83, yields M� � Z�b�-Z�b� � α�/EI1�?@�^ /EI1�?@�_ Dt�δ�-δ�� � t�θ� � tpθ� � tqgH (86)

From equation 79, one has

Design of Sandwich Beam Element with Partial Composite Action 571

<� � ��/=>1�?@�^ /=>1�?@�� �·�/~� C ~�1 � ·�{� � ·p{� � ·q�� (87)

Equation 87, is the solution of equation 79, with considering the following formula φ� � ¹�'# (88.a) φ� � ¹�'# (88.b) φ� � ¹�'# (88.c) φq � ¹Y'# (88.d)

That t1-t4 in equation 88 have gained by equation 85, and Ks have gained by equation 82.

If the equation 80 has been considered

At first, p1Z2 and p2Z1 with considering equations 71.b, 71.c, 77.b, and 77.c, will be calculated.

Then Qs has been gained p�Z� � lα� ,-1 � cosh/αl1 - �d� sinh/αl1- /EI1�?@�^ α�/EI1�?@�^ /EI1�?@�_ �a� sinh/αl1 � ap cosh/αl1� (89.a) p�Z� � lα� ,1- �d� sinh/αl1 - cosh/αl1- /EI1�?@�^ apα�/EI1�?@�^ /EI1�?@�_ (89.b)

Substituting equation 89 into equation 80 and considering equations 67 yields Q� � p�Z�-p�Z� � αql/EI1�?@�^ �/EI1�?@�_ ,2α sinh/αl1 cosh/αl1 -2α sinh/αl1�δ�-δ�� � V�-1 � ��d�� �sinh�/αl1 � p�d� sinh/αl1 �sinh�/αl1 cosh/αl1 - �d� sinhp/αl1 � cosh�/αl1 � cosh/αl1 - coshp/αl1 � �d� cosh�/αl1 sinh/αl1 -1W θ� � V�1- ��d�� �sinh�/αl1 �lα sinh/αl1 cosh/αl1 - sinh�/αl1 cosh/αl1 � �d� sinhp/αl1 -3 cosh/αl1 � cosh�/αl1 � coshp/αl1 - p�d� cosh�/αl1 sinh/αl1 � �d� sinh/αl1 �1W θ� � V�g�- �d� gp� sinh/αl1 � �-g� � �d� gp�sinh/αl1 cosh/αl1 � �d� g� sinh�/αl1 -gp cosh�/αl1 � gpW g- (90)

From equations 82 and 90, one has Q� � -Q� � α�/EI1�?@�^ /EI1�?@�_ Dφ¶δ�-δ�� � φoθ� �φºθ� � φ»gH (91) φ¶ � /(Q1+Z[&\ �d�'# D2α sinh/αl1 cosh/αl1 -2α sinh/αl1H (92.a) φo �/(Q1+Z[&\ �d�'# ,�1- ��d�� � sinh�/αl1 � lα sinh/αl1 cosh/αl1 - sinh�/αl1 cosh/αl1 � �d� sinhp/αl1 -3 cosh/αl1 �cosh�/αl1 � coshp/αl1 - p�d� cosh�/αl1 sinh/αl1 � �d� sinh/αl1 � 1- (92.b) φº � /(Q1+Z[&\ �d�'# ,�-1 � ��d�� � sinh�/αl1 � p�d� sinh/αl1 � sinh�/αl1 cosh/αl1 - �d� sinhp/αl1 � cosh�/αl1 �cosh/αl1 - coshp/αl1 � �d� cosh�/αl1 sinh/αl1 -1- (92.c) φ» � /(Q1+Z[&\ �d�'# ,�g�- �d� gp� sinh/αl1 � �-g� � �d� gp� sinh/αl1 cosh/αl1 � �d� g� sinh�/αl1 -gp cosh�/αl1 � gp- (92.d)

From equation 53, one has Q� � P�*P�� ������� !M� � lQ�-M� (93)

Combining equations 91, 92, 87, and 88, lead to M� � α�/EI1�?@�^ /EI1�?@�_ Dlφ¶-φ��δ�-δ�� � lφo-φ��θ� � lφº-φp�θ� �lφ»-φq�gH (94)

The matrix expression of equations 87, 91, and 92 is

α�/EI1�?@�^ /EI1�?@�_¼½½½¾ -φ¶ φo φ¶ φº-φ� φ� φ� φpφ¶ -φo -φ¶ -φº-lφ¶*φ� lφo-φ� lφ¶-φ� lφº-φp¿À

ÀÀÁ Âδ�θ�δ�θ�à �

ÂQ�M�Q�M�à � α�/EI1�?@�^ /EI1�?@�_

ÄÅÆ φ»φq-φ»lφ»-φqÇÈ

Ég (95)

Or �k���Êδ�Ë � Êf�Ë (96)

Where Êδ�Ë � �δ� θ� δ� θ��Í (97)

572 Pouria Shahani

�k��� � α�/EI1�?@�^ /EI1�?@�_¼½½½¾ -φ¶ φo φ¶ φº-φ� φ� φ� φpφ¶ -φo -φ¶ -φº-lφ¶*φ� lφo-φ� lφ¶-φ� lφº-φp¿À

ÀÀÁ (98)

Equation 95 or 96 is the elastic stiffness equation for the composite beam element with partial

composite action for three nonsymmetrical layers and �k��� is the corresponding elastic stiffness matrix

of the element.

3. Equivalent Flexural and Shearing Rigidity For calculating the deflection of composite beams like sandwich beams, all of the formulas that use for

individual beam can be used, but the different only is related to the flexural rigidity, EI, and shearing

rigidity, . The equivalent values have to be used instead of the constant values of them. In the

following, the relationships between them are summarized.

In the table 1 and 2, is assumed that layers are composed of steel (St), aluminum (Al), and brass

(Br). The letters h1, h2, and h3 is the height of each layer. x, t, and m are the width of layers. E, and G

are modulus of elasticity, and shear modulus respectively, and I is moment of inertia.ks is introduced to

account for the difference in the constant state of shear stress in the Timoshenko beam theory and the

parabolic variation of the actual shear stress through the beam depth. The values of ks for various cross-

sectional shapes are given in Gere and Timoshenko (1991).

When the deflections of composite beams like sandwich structures are calculated by finite

element method, these two tables are useful. Because of sandwich beams categorize in wide beams,

then, for calculating the deflection of them, Timoshenko theory has to be used.

Table 1: Equivalent flexural rigidity

Table 2: Equivalent shearing rigidity

Design of Sandwich Beam Element with Partial Composite Action 573

4. Conclusion Equation 90 or 91 is the elastic stiffness equation for the sandwich composite beam element with

partial composite action for three nonsymmetrical layers and is the corresponding elastic

stiffness matrix of the element, for sandwich beams with partial composite action. This equation can

solve the inconsistency problem during the using of finite element analysis.

The deflection resultant relationships for single span sandwich beams can be calculated by

considering tables 1 and 2 and using the familiar formula that presented in any books related to

Timoshenko theory like Reddy 3rd

edition (2005). The above mentioned tables facilitate using

individual beam solutions by engineering designers for the calculation of the deflection of composite

beams.

References [1] Budynas R. G. (1999) advanced strength and applied stress analysis, 3

rd edition, Prentica Hall

[2] Faella, C., Martinelli, E. and Nigro, E. (2001). One-dimensional finite element approach for the

analysis of steel concrete composite frames, Proceedings of the First International Conference

on Steel and Composite Structures, Pusan, Korea, 1245–52.

[3] Gdoutos E.E., Daniel I.M., (2008). Nonlinear and Eformation Behaviour of Composite

Sandwich Beams, Applied Mechanics and Materials Vols 13-14 pp91-98.

[4] Gere,J,and Timoshenko S.P.(1991). Mechanics of materials.3rd

Ed., Chapman and Hall, Ltd,

London, England.

[5] Guo-Qiang Li, Jin-Jun Li, (2009). Advanced Analysis and Design of Steel Frames, John Wiley

& Sons.

[6] Newmark, N. M., Siess, C. P., and Viest, I. M. (1951). Tests and analysis of composite beams

with incomplete interaction, Proceedings of Society of Experimental Stress Analysis, V9 (1),

75–92.

[7] Reddy J.N. (1995), an introduction to finite element method, 3rd

edition, McGraw-Hill, Inc.

Chapter 5, part 5.3 pp 261-274.

[8] Viest, I. M., Colaco, J. P., Furlong, R.W. et al. (1997). Composite Construction Design for

Buildings, ASCE and McGraw Hill, New York.

[9] Wang C.M. (1995) Timoshenko Beam-Bending Solution in Terms of Euler-Bernoulli solution,

journal of engineering mechanics.