Graduate Theses, Dissertations, and Problem Reports 2016 Large deflection analysis of composite beams Large deflection analysis of composite beams Sai Manohari Kancharla Follow this and additional works at: https://researchrepository.wvu.edu/etd Recommended Citation Recommended Citation Kancharla, Sai Manohari, "Large deflection analysis of composite beams" (2016). Graduate Theses, Dissertations, and Problem Reports. 5929. https://researchrepository.wvu.edu/etd/5929 This Thesis is protected by copyright and/or related rights. It has been brought to you by the The Research Repository @ WVU with permission from the rights-holder(s). You are free to use this Thesis in any way that is permitted by the copyright and related rights legislation that applies to your use. For other uses you must obtain permission from the rights-holder(s) directly, unless additional rights are indicated by a Creative Commons license in the record and/ or on the work itself. This Thesis has been accepted for inclusion in WVU Graduate Theses, Dissertations, and Problem Reports collection by an authorized administrator of The Research Repository @ WVU. For more information, please contact [email protected].
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Graduate Theses, Dissertations, and Problem Reports
2016
Large deflection analysis of composite beams Large deflection analysis of composite beams
Sai Manohari Kancharla
Follow this and additional works at: https://researchrepository.wvu.edu/etd
Recommended Citation Recommended Citation Kancharla, Sai Manohari, "Large deflection analysis of composite beams" (2016). Graduate Theses, Dissertations, and Problem Reports. 5929. https://researchrepository.wvu.edu/etd/5929
This Thesis is protected by copyright and/or related rights. It has been brought to you by the The Research Repository @ WVU with permission from the rights-holder(s). You are free to use this Thesis in any way that is permitted by the copyright and related rights legislation that applies to your use. For other uses you must obtain permission from the rights-holder(s) directly, unless additional rights are indicated by a Creative Commons license in the record and/ or on the work itself. This Thesis has been accepted for inclusion in WVU Graduate Theses, Dissertations, and Problem Reports collection by an authorized administrator of The Research Repository @ WVU. For more information, please contact [email protected].
The square matrices in Eq. (2.18) are of order 3 x 3 and the stiffness coefficients are defined for
i, j= 1, 2, 6. The �̅�𝑖𝑗(𝑘)
denotes the off-axis material stiffness coefficients of the kth layer. Matrices
in Eq. (2.19) are obtained from,
(𝐴𝑖𝑗 , 𝐷𝑖𝑗 , 𝐹𝑖𝑗) = ∑ ∫ �̅�𝑖𝑗(𝑘)(1, 𝑧2, 𝑧6)𝑑𝑧
𝑧𝑘𝑧𝑘−1
𝑛𝑘=1 (2.21)
where, [A], [D] and [F] are 2 x 2 matrices with i,j = 4, 5.
2.5.4 Reduction of Plate Equations to Beams [Nagappan (2004)]
For isotropic materials beam theories were developed first since they are much simpler than
the corresponding plate theories whereas plate theories were developed first in case of
14
composite materials. Chandrasekaran and Sivaneri (2000) studied an efficient way to reduce
these plate theories to corresponding beam theories. This process for HSDT is outlined in this
section.
For beams, the lateral resultants forces are negligible. Hence, Ny, My,Py, are set to zero in Eq.
(2.18).Similarly, Qy and Ry are set to zero in Eq. (2.19). By rearranging Eq. (2.18) we get,
{
𝑁𝑥𝑁𝑥𝑦𝑀𝑥
𝑀𝑥𝑦
𝑃𝑥𝑃𝑥𝑦000 }
=
[ 𝐴11 𝐴16 𝐵11𝐴16 𝐴66 𝐵16𝐵11 𝐵16 𝐷11
𝐵16 𝐸11 𝐸16𝐵66 𝐸16 𝐸66𝐷16 𝐹11 𝐹16
𝐴12 𝐵12 𝐸12𝐴26 𝐵26 𝐸26𝐵12 𝐷12 𝐹12
𝐵16 𝐵66 𝐷16𝐸11 𝐸16 𝐹11𝐸16 𝐸66 𝐹16
𝐷66 𝐹16 𝐹66𝐹16 𝐻11 𝐻16𝐹66 𝐻16 𝐻66
𝐵26 𝐷26 𝐹26𝐸12 𝐹12 𝐻12𝐸26 𝐹26 𝐻26
𝐴12 𝐴26 𝐵12𝐵12 𝐵26 𝐷12𝐸12 𝐸26 𝐹12
𝐵26 𝐸12 𝐸26𝐷26 𝐹12 𝐹26𝐹26 𝐻12 𝐻26
𝐴22 𝐵22 𝐸22𝐵22 𝐷22 𝐹22𝐸22 𝐹22 𝐻22]
{
휀𝑥
(0)
𝛾𝑥𝑦(0)
휀𝑥(1)
𝛾𝑥𝑦(1)
휀𝑥(3)
𝛾𝑥𝑦(3)
휀𝑦(0)
휀𝑦(1)
휀𝑦(3)}
(2.22)
Representing the above matrix as follows
{{�̅�}{0}
} = [[𝑇11] [𝑇12]
[𝑇21] [𝑇22]] {{휀}̅
{휀�̅�}} (2.23)
It can be observed that
[𝑇21] = [𝑇12]𝑇 (2.24)
Expanding Eq. (2.23)
{�̅�} = [𝑇11]{휀}̅ + [𝑇12]{휀�̅�} (2.25)
{0} = [𝑇21]{휀}̅ + [𝑇22]{휀�̅�}
Omitting {휀�̅�}in Eq. (2.25) we get,
{�̅�} = [𝑇]{휀}̅ (2.26)
where[𝑇] = [[𝑇11] − [𝑇12][𝑇22]−1[𝑇21]] (2.27)
Similarly, Eq. (2.19) can be rearranged and partitioned by setting 𝑄𝑦 = 𝑅𝑦 = 0,
15
{
𝑄𝑥𝑅𝑥⋯00 }
=
[ 𝐴55 𝐷55 ⋮ 𝐴45 𝐷45𝐷55 𝐹55 ⋮ 𝐷45 𝐹45⋯𝐴45𝐷45
⋯𝐷45𝐹45
⋮ ⋯ ⋯⋮⋮
𝐴44𝐷44
𝐷44𝐹44 ]
{
𝛾𝑥𝑧
(0)
𝛾𝑥𝑧(2)
⋯
𝛾𝑦𝑧(0)
𝛾𝑦𝑧(2)}
(2.28)
Say the notations, 𝑤𝑠′ =
𝜕𝑤𝑠
𝜕𝑥 and 𝑤𝑠
𝑦=
𝜕𝑤𝑠
𝜕𝑦 in Eq. (2.13), the strain vector in Eq. (2.28)
becomes,
⌊𝛾𝑥𝑧(0)𝛾𝑥𝑧
(2)𝛾𝑦𝑧(2)𝛾𝑦𝑧
(2)⌋ = ⌊𝑤𝑠′ − 3𝑐1𝑤𝑠
′𝑤𝑠𝑦 − 3𝑐1𝑤𝑠
𝑦⌋ (2.29)
Define
𝐷𝑖𝑗∗ = 𝐴𝑖𝑗 − 6𝑐1𝐷𝑖𝑗 + 9𝑐1
2𝐹𝑖𝑗
𝑄𝑥∗ = 𝑄𝑥 − 3𝑐1𝑅𝑥
𝑄𝑦∗ = 𝑄𝑦 − 3𝑐1𝑅𝑦 (2.30)
Substituting Eq. (2.30) into Eq. (2.28) we get,
{𝑄𝑦∗
𝑄𝑥∗} = 𝐾 [
𝐷44∗ 𝐷45
∗
𝐷45∗ 𝐷55
∗ ] {{𝑤𝑠
𝑦}
{𝑤𝑠′}} (2.31)
A factor K known as shear correction factor in the Eq. (2.31) is introduced even though
HSDT does not require. This is done to obtain FSDT results from the HSDT formulation by
setting 𝑐1=0 and K=5/6. For HSDT case K will be one.
For a beam, 𝑄𝑦∗ is set to 0.Thus, solving Eq. (2.31) we get,
𝑄𝑥∗ =K𝐷55
∗∗𝑤𝑠′ (2.32)
Where 𝐷55∗∗ = 𝐷55
∗ −𝐷45∗2
𝐷44∗ (2.33)
2.5.5 Virtual Strain Energy of a Composite Beam
To obtain strain energy for beam, employ the systematic way discussed above i.e𝑁𝑦 = 𝑀𝑦 =
𝑃𝑦 = 𝑄𝑦 = 𝑅𝑦 = 0. For a rectangular beam of width b and length L, the area integral in Eq.
(2.17) is changed to a line integral along x and further substituting kinematic Eqs. (2.13) and
(2.14) in Eq. (2.17) and following notations are used
16
( )′ = 𝜕( )
𝜕𝑥
( )′ = 𝜕( )
𝜕𝑦
( )" = 𝜕2( )
𝜕𝑥2
( )′𝑦 = 𝜕2( )
𝜕𝑥𝜕𝑦
𝛾0 =𝜕𝑢0
𝜕𝑦+
𝜕𝑣0
𝜕𝑥 (2.34)
Then 𝛿𝑈 becomes,
𝛿𝑈 = 𝑏 ∫ [𝑁𝑥(𝛿𝑢0′ + 𝑤𝑏
′𝛿𝑤𝑏′ + 𝑤𝑠
′𝛿𝑤𝑠′) − 𝑀𝑥𝛿𝑤𝑏
′′ − 𝑃𝑥𝑐1𝛿𝑤𝑠′′ + 𝑁𝑥𝑦 (𝛿𝛾0 + (𝛿𝑤𝑏
′ +𝑙
0
𝛿𝑤𝑠′)(𝑤𝑏
𝑦+ 𝑤𝑠
𝑦) + (𝑤𝑏′ +𝑤𝑠
′)(𝛿𝑤𝑏𝑦+ 𝛿𝑤𝑠
𝑦)) − 2𝑀𝑥𝑦𝛿𝑤𝑏′𝑦− 2𝑃𝑥𝑦𝑐1𝛿𝑤𝑠
′𝑦+ 𝑄𝑥
∗𝛿𝑤𝑠′] 𝑑𝑥 (2.35)
Where,
𝑁𝑥(𝛿𝑢0′ + 𝑤𝑏
′𝛿𝑤𝑏′ + 𝑤𝑠
′𝛿𝑤𝑠′) = 𝑁𝑥𝛿𝑢0
′ +𝑁𝑥𝑤𝑏′𝛿𝑤𝑏
′ + 𝑁𝑥𝑤𝑠′𝛿𝑤𝑠
′ (2.36)
From constitutive equation Eq. (2.22), substituting Nx, Mx, Px, Nxy, Mxy and Pxy in the above Eq.
(2.36) gives,
𝑁𝑥(𝛿𝑢0′ + 𝑤𝑏
′𝛿𝑤𝑏′ + 𝑤𝑠
′𝛿𝑤𝑠′)
= (𝑇11𝑢0′ + 𝑇12𝛾0 − 𝑇13𝑤𝑏
′′ − 2𝑇14𝑤𝑏′𝑦− 𝑐1𝑇15𝑤𝑠
′′ − 2𝑐1𝑇16𝑤𝑠′𝑦)𝛿𝑢0
′
+𝑁𝑥𝑤𝑏′𝛿𝑤𝑏
′ + 𝑁𝑥𝑤𝑠′𝛿𝑤𝑠
′
+ (𝑇112𝑤𝑏′𝛿𝑢0
′𝑤𝑏′ +
𝑇112𝑤𝑠′𝛿𝑢0
′𝑤𝑠′
+ 𝑇12 (𝑤𝑏′𝑤𝑏
𝑦𝛿𝑢0
′ + 𝑤𝑏′𝑤𝑠
𝑦𝛿𝑢0
′ + 𝑤𝑠′𝑤𝑏
𝑦𝛿𝑢0
′ + 𝑤𝑠′𝑤𝑠
𝑦𝛿𝑢0
′ ))
− 𝑀𝑥𝛿𝑤𝑏′′ = −(𝑇31𝑢0
′ + 𝑇32𝛾0 − 𝑇33𝑤𝑏′′ − 2𝑇34𝑤𝑏
′𝑦− 𝑐1𝑇35𝑤𝑠
′′ − 2𝑐1𝑇36𝑤𝑠′𝑦)𝛿𝑤𝑏
′′
− (𝑇312𝑤𝑏′𝛿𝑤𝑏
′′𝑤𝑏′ +
𝑇312𝑤𝑠′𝛿𝑤𝑏
′′𝑤𝑠′
+ 𝑇32 (𝑤𝑏′𝑤𝑏
𝑦𝛿𝑤𝑏
′′ + 𝑤𝑏′𝑤𝑠
𝑦𝛿𝑤𝑏
′′ + 𝑤𝑠′𝑤𝑏
𝑦𝛿𝑤𝑏
′′ + 𝑤𝑠′𝑤𝑠
𝑦𝛿𝑤𝑏
′′))
17
− 𝑃𝑥𝑐1𝛿𝑤𝑠′′ = −𝑐1(𝑇51𝑢0
′ + 𝑇52𝛾0 − 𝑇53𝑤𝑏′′ − 2𝑇54𝑤𝑏
′𝑦− 𝑐1𝑇55𝑤𝑠
′′ − 2𝑐1𝑇56𝑤𝑠′𝑦)𝛿𝑤𝑠
′′
− 𝑐1 (𝑇512𝑤𝑏′𝛿𝑤𝑠
′′𝑤𝑏′ +
𝑇512𝑤𝑠′𝛿𝑤𝑠
′′𝑤𝑠′
+ 𝑇52 (𝑤𝑏′𝑤𝑏
𝑦𝛿𝑤𝑠
′′ + 𝑤𝑏′𝑤𝑠
𝑦𝛿𝑤𝑠
′′ + 𝑤𝑠′𝑤𝑏
𝑦𝛿𝑤𝑠
′′ + 𝑤𝑠′𝑤𝑠
𝑦𝛿𝑤𝑠
′′))
𝑁𝑥𝑦 (𝛿𝛾0 + (𝛿𝑤𝑏′ + 𝛿𝑤𝑠
′)(𝑤𝑏𝑦+ 𝑤𝑠
𝑦) + (𝑤𝑏′ +𝑤𝑠
′)(𝛿𝑤𝑏𝑦+ 𝛿𝑤𝑠
𝑦))
= 𝑁𝑥𝑦𝛿𝛾0 + 𝑁𝑥𝑦 ((𝑤𝑏′ +𝑤𝑠
′)𝛿𝑤𝑏𝑦) + 𝑁𝑥𝑦 ((𝑤𝑏
′ +𝑤𝑠′) 𝛿𝑤𝑠
𝑦)
+ 𝑁𝑥𝑦 ((𝑤𝑏𝑦+ 𝑤𝑠
𝑦)𝛿𝑤𝑏′ ) + 𝑁𝑥𝑦 ((𝑤𝑏
𝑦+𝑤𝑠
𝑦)𝛿𝑤𝑠′)
= (𝑇21𝑢0′ + 𝑇22𝛾0 − 𝑇23𝑤𝑏
′′ − 2𝑇24𝑤𝑏′𝑦− 𝑐1𝑇25𝑤𝑠
′′ − 2𝑐1𝑇26𝑤𝑠′𝑦)𝛿𝛾0
+ (𝑇212𝑤𝑏′𝛿𝛾0𝑤𝑏
′ + 𝑇212𝑤𝑠′𝛿𝛾0𝑤𝑠
′
+ 𝑇22 (𝑤𝑏′𝑤𝑏
𝑦𝛿𝛾0 + 𝑤𝑏
′𝑤𝑠𝑦𝛿𝛾0 + 𝑤𝑠
′𝑤𝑏𝑦𝛿𝛾0 + 𝑤𝑠
′𝑤𝑠𝑦𝛿𝛾0))
+ 𝑁𝑥𝑦 ((𝑤𝑏′ + 𝑤𝑠
′)𝛿𝑤𝑏𝑦) + 𝑁𝑥𝑦 ((𝑤𝑏
′ +𝑤𝑠′) 𝛿𝑤𝑠
𝑦) + 𝑁𝑥𝑦 ((𝑤𝑏
𝑦+𝑤𝑠
𝑦)𝛿𝑤𝑏′)
+ 𝑁𝑥𝑦 ((𝑤𝑏𝑦+ 𝑤𝑠
𝑦)𝛿𝑤𝑠′)
− 2𝑀𝑥𝑦𝛿𝑤𝑏′𝑦= −2(𝑇41𝑢0
′ + 𝑇42𝛾0 − 𝑇43𝑤𝑏′′ − 2𝑇44𝑤𝑏
′𝑦− 𝑐1𝑇45𝑤𝑠
′′ − 2𝑐1𝑇46𝑤𝑠′𝑦)𝛿𝑤𝑏
′𝑦
− 2(𝑇412𝑤𝑏′𝛿𝑤𝑏
′𝑦𝑤𝑏′ +
𝑇412𝑤𝑠′𝛿𝑤𝑏
′𝑦𝑤𝑠′
+ 𝑇42 (𝑤𝑏′𝑤𝑏
𝑦𝛿𝑤𝑏
′𝑦+𝑤𝑏
′𝑤𝑠𝑦𝛿𝑤𝑏
′𝑦+ 𝑤𝑠
′𝑤𝑏𝑦𝛿𝑤𝑏
′𝑦+ 𝑤𝑠
′𝑤𝑠𝑦𝛿𝑤𝑏
′𝑦))
− 2𝑃𝑥𝑦𝑐1𝛿𝑤𝑠′𝑦= −2𝑐1(𝑇61𝑢0
′ + 𝑇62𝛾0 − 𝑇63𝑤𝑏′′ − 2𝑇64𝑤𝑏
′𝑦− 𝑐1𝑇65𝑤𝑠
′′ − 2𝑐1𝑇66𝑤𝑠′𝑦)𝛿𝑤𝑠
′𝑦−
2𝑐1 (𝑇61
2𝑤𝑏′𝛿𝑤𝑠
′𝑦𝑤𝑏′ +
𝑇61
2𝑤𝑠′𝛿𝑤𝑠
′𝑦𝑤𝑠′ + 𝑇62 (𝑤𝑏
′𝑤𝑏𝑦𝛿𝑤𝑠
′𝑦+𝑤𝑏
′𝑤𝑠𝑦𝛿𝑤𝑠
′𝑦+
𝑤𝑠′𝑤𝑏
𝑦𝛿𝑤𝑠
′𝑦+ 𝑤𝑠
′𝑤𝑠𝑦𝛿𝑤𝑠
′𝑦)) (2.37)
2.5.6 Large Deflection Cylindrical Bending of Composite Plates
Consider a laminated plate as show in figure 2.5. Suppose the plate is long in the 𝑥 – direction
and has finite dimensions along the𝑦 – direction and subjected to transverse load 𝑃𝑧(𝑥, 𝑦)
which is uniform at any section parallel to the 𝑦 axis. In such a case plate bends into a
cylindrical surface and is similar to beam bending.
The kinematic equations are as follows
18
휀𝑥 = 𝑢
′ + 1
2(𝑤′)2 − 𝑧𝑤′′
휀𝑦 = 0
𝛾𝑥𝑦 = 0
(2.38)
Fig. 2.5 Transverse load 𝑷𝒛(𝒙, 𝒚) on a laminated plate
The constitutive equations are
{
𝜎𝑥𝜎𝑦𝜎𝑧}
𝐾
= [�̅�]𝐾 {
휀𝑥휀𝑦휀𝑧} (2.39)
Substituting Eq. (2.38) in the above gives
(𝜎𝑥)𝐾 = (𝑄11̅̅ ̅̅ ̅)𝐾휀𝑥 = (𝑄11̅̅ ̅̅ ̅)𝐾 [𝑢′ +
1
2(𝑤′)2 − 𝑧𝑤′′] (2.40)
2.5.7 Virtual Strain Energy for Cylindrical Bending
The total virtual strain energy for cylindrical bending is given as,
𝛿U =∭ [𝜎𝑥 𝛿휀𝑥]𝑑𝑉𝑉 (2.41)
where𝛿휀𝑥is the virtual strains and V is the volume of the plate. The rectangular beam has width
bin y direction, length Lin x axis and thickness h in z direction. Rewriting volume integral in
terms of area and thickness integrals we get,
𝛿U = b ∫ ∫ 𝜎𝑥ℎ/2
−ℎ/2
𝐿
0𝛿휀𝑥𝑑𝑥𝑑𝑧 (2.42)
19
Now the𝛿𝑈can be written in terms of the deformation quantities as
𝛿𝑈 = 𝑏 ∫ {[𝐴11 (𝑢′ +
1
2(𝑤′)2) − 𝐵11𝑤
′′] (𝛿𝑢′ + 𝑤′𝛿𝑤′) [−𝐵11 (𝑢′ +
1
2(𝑤′)2) +
𝐿
0
𝐷11𝑤11] 𝛿𝑤′′} 𝑑𝑥 (2.43)
20
3 FINITE ELEMENT METHOD
3.1 Introduction
The finite element method can be employed to solve partial differential equations where the
domain is first discretized into small elements called finite elements. Figure 3.1 shows the
classification of methods to solve partial differential equations. With the increase in number of
elements or nodes or both, the finite element solution approaches analytical solution. Based on
this h version, p version and h-p version finite element formulations are available. In the h-
version, accuracy generally increases with the increase in number of elements, in the p-version
number of internal nodes is increased to improve the accuracy and in the h-p version both
elements and nodes are increased to attain accuracy. In the present research, the h-p version is
used.
Fig. 3.1 Classification of methods to solve partial differential equations
3.2 Finite element shape functions
Sreeram and Sivaneri (1997) conducted a convergence study on an isotropic beam problem and
concluded that four elements with three internal nodes each produce an accurate solution. In
the present research, the composite beam is divided into four elements, each having three
internal nodes and two end nodes. The C0 continuity is satisfied by certain variables and their
shape functions are derived using Lagrangian interpolating polynomials while Hermitian
polynomials are used in deriving other shape functions which obey C1 continuity. For the
Methods
Analytical
Exact Approximate
Numerical
Numerical Solution
Numerical integration
Finite differences
FEM
21
variables that employ C1 continuity, only the end nodes have slope degrees of freedom as the
slope continuity at the internal nodes is automatically satisfied.
Fig. 3.2 Finite element beam representation and representation of an element with natural
coordinates and three internal nodes
Figure 3.2 shows finite element beam representation. The beam considered is divided into four
elements. Each element has three internal nodes and two end nodes. The length of the element
is le, 𝜉is the natural coordinate and xe is the independent variable measured from left end of the
beam. The dependent variables are:
u = axial deformation at mid-plane
γ = mid-plane shear strain
𝑤𝑏= transverse bending deformation
𝑤𝑠= transverse shear deformation
1
𝑧𝑒
𝑥𝑒
𝑙𝑒
1 2 3 4 5
𝜉 𝜉 = −1 𝜉 = 1
22
𝑤𝑏𝑦 = twist angle associated with the bending deformation
𝑤𝑠𝑦 = twist angle associated with the shear deformation
The variables u,𝛾, 𝑤𝑏𝑦, 𝑤𝑠
𝑦 obey C
0 continuity while 𝑤𝑏and𝑤𝑠 have slope degrees of freedom at
the end nodes and thus C1 continuity is assured.
For each element, an intrinsic coordinate 𝜉 ranging from -1 to 1 is described with its origin at the
center as shown in Fig. 3.2. The elemental coordinates can be transformed to the intrinsic
coordinates with the following relation
𝑥𝑒 =
𝑙𝑒2(𝜉 + 1)
𝑑𝑥𝑒 = 𝑙𝑒2𝑑𝜉
(3.1)
where𝑙𝑒 being element length. The shape functions for the variables that follow C0 continuity
and C1 continuity are derived respectively. The axial displacement (u) and lateral displacement
component due to bending (𝑤𝑏)are represented as
𝑢(𝜉) = ∑𝑎𝑖𝜉
𝑖
4
𝑖=0
𝑤(𝜉) = ∑𝑏𝑖𝜉𝑗
6
𝑗=0
(3.2)
The above equations can be written in matrix form as follows
𝑢(𝜉) = ⌊𝜉𝑖⌋{𝑎𝑖}
𝑤(𝜉) = ⌊𝜉𝑗⌋{𝑏𝑗} (3.3)
Where the 𝑎𝑖are determined by the following conditions
𝑢(−1) = 𝑢1
𝑢(−0.5) = 𝑢2
𝑢(0) = 𝑢3 (3.4)
𝑢(0.5) = 𝑢4
𝑢(1) = 𝑢5
23
By solving the above equations and substituting in Eq. (3.3) we get,
𝑢(𝜉) = ⌊𝐻𝐿1(𝜉)…𝐻𝐿5(𝜉)⌋
{
𝑢1...𝑢5}
(3.5)
where 𝐻𝐿1(𝜉), 𝐻𝐿2(𝜉).. 𝐻𝐿5(𝜉)are called as Lagrange shape functions. They are
𝐻𝐿1 = 1
6𝜉 −
1
6𝜉2 −
2
3𝜉3 +
2
3𝜉4
𝐻𝐿2 = − 4
3𝜉 +
8
3𝜉2 +
4
3𝜉3 +
8
3𝜉4
𝐻𝐿3 = 1 − 5𝜉2 + 4𝜉4 (3.6)
𝐻𝐿4 = 4
3𝜉 +
8
3𝜉2 −
4
3𝜉3 −
8
3𝜉4
𝐻𝐿5 = 1
6𝜉 −
1
6𝜉2 −
2
3𝜉3 +
2
3𝜉4
And 𝑏𝑗 can be determined by the following conditions
𝑤𝑏(−1) = 𝑤𝑏1
𝑙𝑒2
𝑑𝑤𝑏𝑑𝜉
(−1) = 𝑤𝑏1′
𝑤𝑏(−0.5) = 𝑤𝑏2
𝑤𝑏(0) = 𝑤𝑏3 (3.7)
𝑤𝑏(0.5) = 𝑤𝑏4
𝑤𝑏(1) = 𝑤𝑏5
𝑙𝑒2
𝑑𝑤𝑏𝑑𝜉
(1) = 𝑤𝑏5′
Solving the above equations for 𝑏𝑗 we get,
24
𝑤𝑏(𝜉) = ⌊𝐻1(𝜉)…𝐻7(𝜉)⌋
{
𝑤𝑏1𝑤𝑏1
′
.
.
.𝑤𝑏5
′}
(3.8)
where 𝐻1(𝜉), 𝐻2(𝜉).. 𝐻5(𝜉)are called as Hermite shape functions. They are
𝐻1 = 1
9(17
4𝜉 − 5𝜉2 −
79
4𝜉3 +
47
2𝜉4 + 11𝜉5 − 14𝜉6)
𝐻2 = 𝑙𝑒6(1
4𝜉 −
1
4𝜉2 −
5
4𝜉3 +
5
4𝜉4 + 𝜉5 − 𝜉6)
𝐻3 = 16
9(−𝜉 + 2𝜉2 + 2𝜉3 − 4𝜉4 − 𝜉5 + 2𝜉6)
𝐻4 = 1 − 6𝜉2 + 9𝜉4 − 4𝜉6 (3.9)
𝐻5 = 16
9(𝜉 + 2𝜉2 − 2𝜉3 − 4𝜉4 + 𝜉5 + 2𝜉6)
𝐻6 = 1
9(−17
4𝜉 − 5𝜉2 +
79
4𝜉3 +
47
2𝜉4 − 11𝜉5 − 14𝜉6)
𝐻7 = 𝑙𝑒6(1
4𝜉 +
1
4𝜉2 −
5
4𝜉3 −
5
4𝜉4 + 𝜉5 + 𝜉6)
3.3 Element stiffness Matrix formulation
The elemental stiffness matrix is obtained from the virtual strain energy equation reduced for
the beam.
𝛿𝑈𝑒 = ⌊𝛿𝑞𝑒⌋[𝐾𝑒]{𝑞𝑒} (3.10)
Where 𝑈𝑒 represents the elemental strain energy, {𝑞𝑒} denotes the vector of element degrees
of freedom and [𝐾𝑒] represents the elemental stiffness matrix. The real and virtual
displacement fields can be represented in terms of shape functions and the corresponding
nodal degrees of freedom.
For example, expressions for mid-plane shear and bending displacements are given as,
𝛾(𝑥) = ⌊𝐻𝐿⌋{𝑞𝛾}
(3.11)
𝛿𝛾(𝑥) = ⌊𝛿𝑞𝛾⌋{𝐻𝐿}
25
𝑤𝑏(𝑥) = ⌊𝐻⌋{𝑞𝑤𝑏}
𝛿𝑤𝑏(𝑥) = ⌊𝛿𝑞𝑤𝑏⌋{𝐻}
where{𝑞𝛾}and {𝑞𝑤𝑏} are the vectors of elemental nodal degrees of freedom for the variables 𝛾
and 𝑤𝑏respectively. Using these above expressions and similar ones for other variables in the
virtual strain energy expression, Eq. (2.35), and then comparing with Eq. (3.10) we get the
stiffness matrix.
The element stiffness matrix has linear and nonlinear parts as the governing equations are
nonlinear. Each stiffness matrix is partitioned into 36 submatrices. The linear stiffness matrix is
a symmetric matrix while the nonlinear stiffness matrix is not symmetrical.
3.3.1 Linear Element Stiffness Matrix for HSDT
The element has a total of 34 degrees of freedom. The stiffness matrix is partitioned into 36
submatrices. The following is the submatrix representation for the linear part of the element
stiffness matrix.
[𝐾𝐿𝑒] =
[ [𝐾𝑢𝑢] [𝐾𝑢𝛾] [𝐾𝑢𝑤𝑏] [𝑘𝑢𝑤𝑠] [𝐾𝑢𝑤𝑏
𝑦] [𝐾𝑢𝑤𝑏𝑦]
[𝐾𝛾𝛾] [𝐾𝛾𝑤𝑏] [𝐾𝛾𝑤𝑠] [𝐾𝛾𝑤𝑏𝑦] [𝐾𝛾𝑤𝑠
𝑦]
[𝐾𝑤𝑏𝑤𝑏] [𝐾𝑤𝑏𝑤𝑠] [𝐾𝑤𝑏𝑤𝑏𝑦] [𝐾𝑤𝑏𝑤𝑠
𝑦]
𝑠𝑦𝑚𝑚 [𝐾𝑤𝑠𝑤𝑠] [𝐾𝑤𝑠𝑤𝑏𝑦] [𝐾𝑤𝑠𝑤𝑠
𝑦]
[𝐾𝑤𝑏𝑦𝑤𝑏𝑦] [𝐾𝑤𝑏
𝑦𝑤𝑠𝑦]
[𝐾𝑤𝑠𝑦𝑤𝑠𝑦]]
(3.14)
Where,
[𝐾𝑢𝑢] = 𝑏∫ 𝑇11
𝑙𝑒
0
{𝐻𝐿′ }⌊𝐻𝐿
′ ⌋ 𝑑𝑥𝑒
[𝐾𝑢𝛾] = 𝑏∫ 𝑇12
𝑙𝑒
0
{𝐻𝐿′ }⌊𝐻𝐿⌋ 𝑑𝑥𝑒
[𝐾𝑢𝑤𝑏] = −𝑏∫ 𝑇13
𝑙𝑒
0
{𝐻𝐿′ }⌊𝐻′′⌋ 𝑑𝑥𝑒
26
[𝐾𝑢𝑤𝑠] = −𝑏𝑐1∫ 𝑇13
𝑙𝑒
0
{𝐻𝐿′ }⌊𝐻′′⌋ 𝑑𝑥𝑒
[𝐾𝑢𝑤𝑏𝑦] = −2𝑏∫ 𝑇14
𝑙𝑒
0
{𝐻𝐿′ }⌊𝐻𝐿
′ ⌋ 𝑑𝑥𝑒
[𝐾𝑢𝑤𝑠𝑦] = −2𝑏𝑐1∫ 𝑇16
𝑙𝑒
0
{𝐻𝐿′ }⌊𝐻𝐿
′ ⌋ 𝑑𝑥𝑒
[𝐾𝛾𝛾] = 𝑏∫ 𝑇22
𝑙𝑒
0
{𝐻𝐿}⌊𝐻𝐿⌋ 𝑑𝑥𝑒
[𝐾𝛾𝑤𝑏] = −𝑏∫ 𝑇23
𝑙𝑒
0
{𝐻𝐿}⌊𝐻′′⌋ 𝑑𝑥𝑒
[𝐾𝛾𝑤𝑠] = −𝑏𝑐1∫ 𝑇25
𝑙𝑒
0
{𝐻𝐿}⌊𝐻′′⌋ 𝑑𝑥𝑒
[𝐾𝛾𝑤𝑏𝑦] = −2𝑏∫ 𝑇24
𝑙𝑒
0
{𝐻𝐿}⌊𝐻𝐿′ ⌋ 𝑑𝑥𝑒
[𝐾𝛾𝑤𝑠𝑦] = −2𝑏𝑐1∫ 𝑇26
𝑙𝑒
0
{𝐻𝐿}⌊𝐻𝐿′ ⌋ 𝑑𝑥𝑒
[𝐾𝑤𝑏𝑤𝑏] = 𝑏∫ 𝑇33
𝑙𝑒
0
{𝐻′′}⌊𝐻′′⌋ 𝑑𝑥𝑒
[𝐾𝑤𝑏𝑤𝑠] = 𝑏𝑐1∫ 𝑇35
𝑙𝑒
0
{𝐻′′}⌊𝐻′′⌋ 𝑑𝑥𝑒
[𝐾𝑤𝑏𝑤𝑏𝑦] = 2𝑏∫ 𝑇34
𝑙𝑒
0
{𝐻′′}⌊𝐻𝐿′ ⌋ 𝑑𝑥𝑒
[𝐾𝑤𝑏𝑤𝑠𝑦] = 2𝑏𝑐1∫ 𝑇36
𝑙𝑒
0
{𝐻′′}⌊𝐻𝐿′ ⌋ 𝑑𝑥𝑒
27
[𝐾𝑤𝑠𝑤𝑠] = 𝑏∫ 𝐷55∗∗
𝑙𝑒
0
{𝐻′}⌊𝐻′⌋ 𝑑𝑥𝑒 + 𝑏𝑐12∫ 𝑇55
𝑙𝑒
0
{𝐻′′}⌊𝐻′′⌋ 𝑑𝑥𝑒
[𝐾𝑤𝑠𝑤𝑏𝑦] = 2𝑏𝑐1∫ 𝑇54
𝑙𝑒
0
{𝐻′′}⌊𝐻𝐿′ ⌋ 𝑑𝑥𝑒
[𝐾𝑤𝑠𝑤𝑠𝑦] = 2𝑏𝑐1
2∫ 𝑇56
𝑙𝑒
0
{𝐻′′}⌊𝐻𝐿′ ⌋ 𝑑𝑥𝑒
[𝐾𝑤𝑏𝑦𝑤𝑏𝑦] = 4𝑏∫ 𝑇44
𝑙𝑒
0
{𝐻𝐿′ }⌊𝐻𝐿
′ ⌋ 𝑑𝑥𝑒
[𝐾𝑤𝑏𝑦𝑤𝑠𝑦] = 4𝑏𝑐1∫ 𝑇46
𝑙𝑒
0
{𝐻𝐿′}⌊𝐻𝐿
′ ⌋ 𝑑𝑥𝑒
[𝐾𝑤𝑠𝑦𝑤𝑠𝑦] = 4𝑏𝑐1
2 ∫ 𝑇66𝑙𝑒0
{𝐻𝐿′ }⌊𝐻𝐿
′ ⌋ 𝑑𝑥𝑒 (3.15)
3.3.2 Nonlinear Element Stiffness Matrix for HSDT
The following is the submatrix representation for the nonlinear part of the stiffness matrix.
[𝐾𝑁𝐿𝑒] =
[ [0] [0] [𝐾𝑢𝑤𝑏] [𝐾𝑢𝑤𝑠] [𝐾𝑢𝑤𝑏
𝑦] [𝐾𝑢𝑤𝑠𝑦]
[0] [0] [𝐾𝛾𝑤𝑏] [𝐾𝛾𝑤𝑠] [𝐾𝛾𝑤𝑏𝑦] [𝐾𝛾𝑤𝑠
𝑦]
[𝐾𝑤𝑏𝑢] [𝐾𝑤𝑏𝛾] [𝐾𝑤𝑏𝑤𝑏] [𝐾𝑤𝑏𝑤𝑠] [𝐾𝑤𝑏𝑤𝑏𝑦] [𝐾𝑤𝑏𝑤𝑠
𝑦]
[𝐾𝑤𝑠𝑢] [𝐾𝑤𝑠𝛾] [𝐾𝑤𝑠𝑤𝑏] [𝐾𝑤𝑠𝑤𝑠] [𝐾𝑤𝑠𝑤𝑏𝑦] [𝐾𝑤𝑠𝑤𝑠
𝑦]
[𝐾𝑤𝑏𝑦𝑢] [𝐾𝑤𝑏
𝑦𝛾] [𝐾𝑤𝑏
𝑦𝑤𝑏] [𝐾𝑤𝑏
𝑦𝑤𝑠] [𝐾𝑤𝑏
𝑦𝑤𝑏𝑦] [𝐾𝑤𝑏
𝑦𝑤𝑠𝑦]
[𝐾𝑤𝑠𝑦𝑢] [𝐾𝑤𝑠
𝑦𝛾] [𝐾𝑤𝑠
𝑦𝑤𝑏] [𝐾𝑤𝑠
𝑦𝑤𝑠] [𝐾𝑤𝑠
𝑦𝑤𝑏𝑦] [𝐾𝑤𝑠
𝑦𝑤𝑠𝑦]]
(3.16)
Where,
[𝐾𝑢𝑤𝑏] = 𝑏∫𝑇112
𝑙𝑒
0
𝑤𝑏′ {𝐻𝐿
′ }⌊𝐻′⌋ 𝑑𝑥𝑒
[𝐾𝑢𝑤𝑠] = 𝑏 ∫𝑇112
𝑙𝑒
0
𝑤𝑠′{𝐻𝐿
′ }⌊𝐻′⌋ 𝑑𝑥𝑒
28
[𝐾𝑢𝑤𝑏𝑦] = 𝑏 ∫ 𝑇12
𝑙𝑒
0
(𝑤𝑏′ + 𝑤𝑠
′){𝐻𝐿′ }⌊𝐻′⌋ 𝑑𝑥𝑒 = [𝐾𝑢𝑤𝑠
𝑦]
[𝐾𝛾𝑤𝑏] = 𝑏 ∫𝑇212
𝑙𝑒
0
𝑤𝑏′ {𝐻𝐿}⌊𝐻
′⌋ 𝑑𝑥𝑒
[𝐾𝛾𝑤𝑠] = 𝑏∫𝑇212
𝑙𝑒
0
𝑤𝑠′{𝐻𝐿}⌊𝐻
′⌋ 𝑑𝑥𝑒
[𝐾𝛾𝑤𝑏𝑦] = 𝑏∫ 𝑇22
𝑙𝑒
0
(𝑤𝑏′ +𝑤𝑠
′){𝐻𝐿}⌊𝐻𝐿⌋ 𝑑𝑥𝑒 = [𝐾𝛾𝑤𝑠𝑦]
[𝐾𝑤𝑏𝑢] = 𝑏∫ 𝑇11
𝑙𝑒
0
𝑤𝑏′ {𝐻′}⌊𝐻𝐿
′ ⌋𝑑𝑥𝑒 + 𝑏∫ 𝑇12
𝑙𝑒
0
(𝑤𝑏𝑦 + 𝑤𝑠
𝑦){𝐻′}⌊𝐻𝐿′ ⌋ 𝑑𝑥𝑒
[𝐾𝑤𝑏𝛾] = 𝑏∫ 𝑇12
𝑙𝑒
0
𝑤𝑏′ {𝐻′}⌊𝐻𝐿
′ ⌋𝑑𝑥𝑒 + 𝑏∫ 𝑇22
𝑙𝑒
0
(𝑤𝑏𝑦 +𝑤𝑠
𝑦){𝐻′}⌊𝐻𝐿′ ⌋ 𝑑𝑥𝑒
[𝐾𝑤𝑏𝑤𝑏] = − 𝑏 ∫ (𝑇312 𝑤𝑏
′ {𝐻′′}⌊𝐻′⌋ − 𝑇13 𝑤𝑏′ {𝐻′}⌊𝐻′′⌋ −
𝑇112(𝑤𝑏
′ )2{𝐻′}⌊𝐻′⌋
𝑙𝑒
0
+ 𝑇23(𝑤𝑏𝑦 + 𝑤𝑠
𝑦){𝐻′}⌊𝐻′′⌋ +𝑇212(𝑤𝑏
𝑦 + 𝑤𝑠𝑦) 𝑤𝑏
′ {𝐻′}⌊𝐻′⌋) 𝑑𝑥𝑒
[𝐾𝑤𝑏𝑤𝑠] = − 𝑏 ∫ (𝑇312 𝑤𝑠
′{𝐻′′}⌊𝐻′⌋ + 𝑐1𝑇15 𝑤𝑏′ {𝐻′}⌊𝐻′′⌋ −
𝑇112(𝑤𝑏
′𝑤𝑠′){𝐻′}⌊𝐻′⌋
𝑙𝑒
0
+ 𝑐1𝑇25(𝑤𝑏𝑦 + 𝑤𝑠
𝑦){𝐻′}⌊𝐻′′⌋ −𝑇212(𝑤𝑏
𝑦 + 𝑤𝑠𝑦) 𝑤𝑠
′{𝐻′}⌊𝐻′⌋) 𝑑𝑥𝑒
[𝐾𝑤𝑏𝑤𝑏𝑦] = 𝑏∫(−𝑇32(𝑤𝑏
′ +𝑤𝑠′){𝐻′′}⌊𝐻𝐿⌋ − 2𝑇14 𝑤𝑏
′ {𝐻′}⌊𝐻𝐿′ ⌋ + 𝑇12𝑤𝑏
′ (𝑤𝑏′ +𝑤𝑠
′){𝐻′}⌊𝐻𝐿⌋
𝑙𝑒
0
− 2𝑇24(𝑤𝑏𝑦 + 𝑤𝑠
𝑦){𝐻′}⌊𝐻𝐿′ ⌋ + 𝑇22(𝑤𝑏
′ + 𝑤𝑠′){𝐻′}⌊𝐻𝐿⌋) 𝑑𝑥𝑒
[𝐾𝑤𝑏𝑤𝑠𝑦] = 𝑏∫(−𝑇32(𝑤𝑏
′ +𝑤𝑠′){𝐻′′}⌊𝐻𝐿⌋ − 2𝑇16 𝑤𝑏
′ {𝐻′}⌊𝐻𝐿′ ⌋ + 𝑇12𝑤𝑏
′ (𝑤𝑏′ +𝑤𝑠
′){𝐻′}⌊𝐻𝐿⌋
𝑙𝑒
0
− 2𝑐1𝑇26(𝑤𝑏𝑦 + 𝑤𝑠
𝑦){𝐻′}⌊𝐻𝐿′ ⌋ + 𝑇22(𝑤𝑏
′ + 𝑤𝑠′){𝐻′}⌊𝐻𝐿⌋) 𝑑𝑥𝑒
29
[𝐾𝑤𝑠𝑢] = 𝑏∫ 𝑇11
𝑙𝑒
0
𝑤𝑠′{𝐻′}⌊𝐻𝐿
′ ⌋𝑑𝑥𝑒 + 𝑏∫ 𝑇12
𝑙𝑒
0
(𝑤𝑏𝑦+𝑤𝑠
𝑦){𝐻′}⌊𝐻𝐿′ ⌋ 𝑑𝑥𝑒
[𝐾𝑤𝑠𝛾] = 𝑏∫ 𝑇12
𝑙𝑒
0
𝑤𝑠′{𝐻′}⌊𝐻𝐿
′ ⌋𝑑𝑥𝑒 + 𝑏 ∫ 𝑇22
𝑙𝑒
0
(𝑤𝑏𝑦 + 𝑤𝑠
𝑦){𝐻′}⌊𝐻𝐿′ ⌋ 𝑑𝑥𝑒
[𝐾𝑤𝑠𝑤𝑏] = 𝑏 ∫ ( −𝑐1𝑇512𝑤𝑏′ {𝐻′′}⌊𝐻′⌋ − 𝑇13 𝑤𝑠
′{𝐻′}⌊𝐻′′⌋ − 𝑇112( 𝑤𝑠
′𝑤𝑏′ ){𝐻′}⌊𝐻′⌋
𝑙𝑒
0
+ 𝑇23(𝑤𝑏𝑦+ 𝑤𝑠
𝑦){𝐻′}⌊𝐻′′⌋ +𝑇212(𝑤𝑏
𝑦+ 𝑤𝑠
𝑦) 𝑤𝑏′ {𝐻′}⌊𝐻′⌋) 𝑑𝑥𝑒
[𝐾𝑤𝑠𝑤𝑠] = 𝑏∫ ( −𝑐1𝑇512𝑤𝑠′{𝐻′′}⌊𝐻′⌋−𝑐1𝑇15 𝑤𝑠
′{𝐻′}⌊𝐻′′⌋ + 𝑇112(𝑤𝑠
′)2{𝐻′}⌊𝐻′⌋
𝑙𝑒
0
− 𝑐1𝑇25(𝑤𝑏𝑦 + 𝑤𝑠
𝑦){𝐻′}⌊𝐻′′⌋ +𝑇212(𝑤𝑏
𝑦 + 𝑤𝑠𝑦) 𝑤𝑠
′{𝐻′}⌊𝐻′⌋) 𝑑𝑥𝑒
[𝐾𝑤𝑠𝑤𝑏𝑦] = 𝑏 ∫(−𝑐1𝑇52(𝑤𝑏
′ +𝑤𝑠′){𝐻′′}⌊𝐻𝐿⌋ − 2𝑇14 𝑤𝑠
′{𝐻′}⌊𝐻𝐿′ ⌋ + 𝑇12𝑤𝑠
′(𝑤𝑏′ +𝑤𝑠
′){𝐻′}⌊𝐻𝐿⌋
𝑙𝑒
0
− 2𝑇24(𝑤𝑏𝑦 + 𝑤𝑠
𝑦){𝐻′}⌊𝐻𝐿′ ⌋ + 𝑇22(𝑤𝑏
′ + 𝑤𝑠′){𝐻′}⌊𝐻𝐿⌋) 𝑑𝑥𝑒
[𝐾𝑤𝑠𝑤𝑠𝑦] = 𝑏∫(−𝑐1𝑇52(𝑤𝑏
′ +𝑤𝑠′){𝐻′′}⌊𝐻𝐿⌋ − 2𝑇16 𝑤𝑠
′{𝐻′}⌊𝐻𝐿′ ⌋ + 𝑇12𝑤𝑠
′(𝑤𝑏′ +𝑤𝑠
′){𝐻′}⌊𝐻𝐿⌋
𝑙𝑒
0
− 2𝑐1𝑇26(𝑤𝑏𝑦+ 𝑤𝑠
𝑦){𝐻′}⌊𝐻𝐿′ ⌋ + 𝑇22(𝑤𝑏
′ + 𝑤𝑠′){𝐻′}⌊𝐻𝐿⌋) 𝑑𝑥𝑒
[𝐾𝑤𝑏𝑦𝑢] = 𝑏∫(𝑇21(𝑤𝑏
′ +𝑤𝑠′){𝐻𝐿}⌊𝐻𝐿
′ ⌋)
𝑙𝑒
0
𝑑𝑥𝑒
[𝐾𝑤𝑏𝑦𝛾] = 𝑏∫(𝑇22(𝑤𝑏
′ + 𝑤𝑠′){𝐻𝐿}⌊𝐻𝐿⌋)
𝑙𝑒
0
𝑑𝑥𝑒
[𝐾𝑤𝑏𝑦𝑤𝑏] = 𝑏∫ (−2
𝑇412𝑤𝑏′ {𝐻𝐿
′ }⌊𝐻′⌋ − 𝑇23(𝑤𝑏′ +𝑤𝑠
′){𝐻𝐿}⌊𝐻′′⌋ +
𝑇212(𝑤𝑏
′ +𝑤𝑠′)𝑤𝑏
′ {𝐻𝐿}⌊𝐻′′⌋)
𝑙𝑒
0
𝑑𝑥𝑒
[𝐾𝑤𝑏𝑦𝑤𝑠] = 𝑏 ∫ (−2
𝑇412𝑤𝑠′{𝐻𝐿
′ }⌊𝐻′⌋ − 𝑐1𝑇25(𝑤𝑏′ +𝑤𝑠
′){𝐻𝐿}⌊𝐻′′⌋ +
𝑇212(𝑤𝑏
′ +𝑤𝑠′)𝑤𝑠
′{𝐻𝐿}⌊𝐻′′⌋)
𝑙𝑒
0
𝑑𝑥𝑒
30
[𝐾𝑤𝑏𝑦𝑤𝑏𝑦] = 𝑏∫(− 2𝑇42(𝑤𝑏
′ +𝑤𝑠′){𝐻𝐿
′}⌊𝐻𝐿⌋ − 2𝑇24(𝑤𝑏′ +𝑤𝑠
′){𝐻𝐿}⌊𝐻𝐿′ ⌋ + 𝑇22(𝑤𝑏
′ +𝑤𝑠′)2{𝐻𝐿}⌊𝐻𝐿⌋)
𝑙𝑒
0
𝑑𝑥𝑒
[𝐾𝑤𝑏𝑦𝑤𝑠𝑦] = 𝑏∫(− 2𝑇42(𝑤𝑏
′ +𝑤𝑠′){𝐻𝐿
′}⌊𝐻𝐿⌋ − 2𝑐1𝑇26(𝑤𝑏′ +𝑤𝑠
′){𝐻𝐿}⌊𝐻𝐿′ ⌋
𝑙𝑒
0
+ 𝑇22(𝑤𝑏′ +𝑤𝑠
′)2{𝐻𝐿}⌊𝐻𝐿⌋) 𝑑𝑥𝑒
[𝐾𝑤𝑠𝑦𝑢] = 𝑏∫(𝑇21(𝑤𝑏
′ +𝑤𝑠′){𝐻𝐿}⌊𝐻𝐿
′ ⌋)
𝑙𝑒
0
𝑑𝑥𝑒
[𝐾𝑤𝑠𝑦𝛾] = 𝑏∫(𝑇22(𝑤𝑏
′ + 𝑤𝑠′){𝐻𝐿}⌊𝐻𝐿⌋)
𝑙𝑒
0
𝑑𝑥𝑒
[𝐾𝑤𝑠𝑦𝑤𝑏] = 𝑏 ∫ (−2𝑐1
𝑇612 𝑤𝑏
′ {𝐻𝐿′ }⌊𝐻′⌋ − 𝑇23(𝑤𝑏
′ +𝑤𝑠′){𝐻𝐿}⌊𝐻
′′⌋ +𝑇212(𝑤𝑏
′ + 𝑤𝑠′)𝑤𝑏
′ {𝐻𝐿}⌊𝐻′′⌋)
𝑙𝑒
0
𝑑𝑥𝑒
[𝐾𝑤𝑠𝑦𝑤𝑠] = 𝑏 ∫ (−2𝑐1
𝑇612 𝑤𝑠
′{𝐻𝐿′ }⌊𝐻′⌋ − 𝑐1𝑇25(𝑤𝑏
′ + 𝑤𝑠′){𝐻𝐿}⌊𝐻
′′⌋
𝑙𝑒
0
+𝑇212(𝑤𝑏
′ + 𝑤𝑠′)𝑤𝑠
′{𝐻𝐿}⌊𝐻′′⌋) 𝑑𝑥𝑒
[𝐾𝑤𝑠𝑦𝑤𝑏𝑦] = 𝑏∫(−2𝑐1𝑇62(𝑤𝑏
′ + 𝑤𝑠′){𝐻𝐿
′ }⌊𝐻𝐿⌋ − 2𝑇24(𝑤𝑏′ + 𝑤𝑠
′){𝐻𝐿}⌊𝐻𝐿′ ⌋
𝑙𝑒
0
+ 𝑇22(𝑤𝑏′ +𝑤𝑠
′)2{𝐻𝐿}⌊𝐻𝐿⌋) 𝑑𝑥𝑒
[𝐾𝑤𝑠𝑦𝑤𝑠𝑦] =
𝑏 ∫ (−2𝑐1𝑇62(𝑤𝑏′ +𝑤𝑠
′){𝐻𝐿′}⌊𝐻𝐿⌋ − 2𝑐1𝑇26(𝑤𝑏
′ +𝑤𝑠′){𝐻𝐿}⌊𝐻𝐿
′ ⌋ + 𝑇22(𝑤𝑏′ +𝑤𝑠
′)2{𝐻𝐿}⌊𝐻𝐿⌋)𝑙𝑒0
𝑑𝑥𝑒(3.16)
where𝑐1 =4
3ℎ2
3.4 Elemental Stiffness matrix for Cylindrical Bending
The expression for elemental stiffness matrix for cylindrical bending is derived following a
procedure similar to the one used for the element stiffness matrix for the HSDT. The number of
dependent variables is two and they are u and w. The C0 continuity is used for u variable
whereas variable w follows C1 continuity. The element has 12 degrees of freedom. The stiffness
31
matrix is subdivided into four submatrices. The stiffness submatrix [𝐾𝑢𝑢] is the only linear part
whereas the other three submatrices contain nonlinear terms.
[𝐾𝑒] = [[𝐾𝑢𝑢] [𝐾𝑢𝑤]
[𝐾𝑤𝑢] [𝐾𝑤𝑤]] (3.17)
Where
[𝐾𝑢𝑢] = 𝑏∫ 𝐴11{𝐻𝐿′ }⌊𝐻𝐿
′ ⌋ 𝑑𝑥𝑒
𝑙𝑒
0
[𝐾𝑢𝑤] = 𝑏∫1
2𝐴11𝑤
′{𝐻𝐿′ }⌊𝐻′⌋ 𝑑𝑥𝑒 − 𝑏∫ 𝐵11{𝐻𝐿
′ }⌊𝐻′′⌋ 𝑑𝑥𝑒
𝑙𝑒
0
𝑙𝑒
0
[𝐾𝑤𝑢] = 𝑏 ∫ 𝐴11𝑤′{𝐻′}⌊𝐻𝐿
′ ⌋𝑑𝑥𝑒 − 𝑏∫ 𝐵11{𝐻′′}⌊𝐻𝐿
′ ⌋ 𝑑𝑥𝑒𝑙𝑒
0
𝑙𝑒
0 (3.18)
[𝐾𝑤𝑤] = 𝑏∫1
2𝐴11(𝑤
′)2{𝐻′}⌊𝐻′⌋ 𝑑𝑥𝑒 − 𝑏∫ 𝐵11𝑤′{𝐻′}⌊𝐻′′⌋ 𝑑𝑥𝑒
𝑙𝑒
0
−
𝑙𝑒
0
𝑏 ∫1
2𝐵11𝑤
′{𝐻′′}⌊𝐻′⌋ 𝑑𝑥𝑒
𝑙𝑒
0
+ 𝑏 ∫ 𝐷11{𝐻′′}⌊𝐻′′⌋ 𝑑𝑥𝑒
𝑙𝑒
0
3.5 Finite element Equation
The elemental stiffness matrix [𝐾𝑒] derived in the previous section is for the individual
elements. The elemental stiffness for all the elements are assembled to form the global
stiffness matrix [𝐾(𝑞)]. The global finite element equations are:
[𝐾(𝑞)]{𝑞} = {𝑄} (3.19) where[𝐾(𝑞)] is the global stiffness matrix. It has both linear and nonlinear stiffness matrices.
Thus, global stiffness matrix is a function of q, {𝑞} is the degrees of freedom vector and {𝑄}is
the load vector. The expression for elemental load vector is as follows
𝑄𝑒 = ∫ 𝑝𝑧{𝐻(𝑥𝑒)}
𝑙𝑒
0
𝑑𝑥𝑒 (3.20)
In natural coordinates,
32
𝑄𝑒 = ∫ 𝑝𝑧
+1
−1
{𝐻(𝜉)}𝑙𝑒2𝑑𝜉 (3.21)
Where the𝐻(𝜉) represent shape functions and pz is the distributed load. Depending on the
variable, the shape functions can be either Hermite or Lagrange polynomials.
The nonlinear finite element equations are solved using the Newton-Raphson iterative process.
The size of global stiffness matrix is 112 x 112, the global degrees of freedom vector is 112 x 1
and global load vector is112 x 1.
3.6 NUMERICAL METHODS
3.6.1 Gaussian Quadrature
The numerical method employed in this research for spatial integration is the Gaussian
Quadrature. This technique is used for computing both the linear and nonlinear element
stiffness matrices as they involve spatial integration. The integration scheme is represented
in the following way
∫𝑓(𝜉)𝑑𝜉 =∑𝑤𝑖𝑓(𝑎𝑖)
𝑚
𝑖=1
1
−1
(3.22)
where m is the number of sampling points,𝑎𝑖 represent the natural coordinate 𝜉 at the
sampling points and 𝑤𝑖 are the corresponding weights.
For a polynomial of order (2m– 1),the Gauss quadrature requires m sampling points to perform
exact integration. In the present work, highest order of polynomial is fourteen. Thus, seven
sampling points are required. They are listed below, along with their respective weights.
Table 3.1Gauss Quadrature sample points and weights
Sampling points Weights
+0.9491079123 0.1294849661
+0.7415311855 0.0797053914
+0.4058451513 0.3813005050
0.0000000000 0.4179591836
33
3.6.2 Newton-Raphson Iterative Method
In general, nonlinear finite element equations are solved using an incremental formulation in
which the variables are updated incrementally with corresponding successive load steps to
trace the path of the solution [Reddy (1997)],[Bathe and Cimento (1980), Felippa (1976), Bathe,
Ramm and Wilson (1973), Bathe (1976,1977,1979)]. In the iterative method, the nonlinear
equations are linearized by evaluating the nonlinear terms with a known solution. In this
approach it is important that the governing finite element equations are satisfied in each load
step. The accuracy of the solution depends on the size of the load step. To attain accurate
result, a small load step need to be taken, thus making the analysis expensive and time
consuming. To overcome this problem, an iterative method should be chosen in such a way
that larger load steps assure the accuracy of the solutions. The Newton-Raphson method,
based on Taylor’s series expansion serves the need. This method is based on the tangent
stiffness matrix, which is symmetric for all structural problems though direct stiffness matrix,
[K], is not symmetric. Also, this method has faster convergence for most applications than the
direct iteration method. Suppose that the solution at rth iteration, {𝑞}𝑟, is known.
Let
{𝑅} ≡ [𝐾]{𝑞} − {𝑄} = 0 (3.23)
Where {𝑅}is called the residual, which is a nonlinear function of the unknown solution {𝑞} and
the increment of the solution vector is defined by,
{𝛿𝑞} = −([𝐾({𝑞}𝑟)]𝑡𝑎𝑛)−1{𝑅}𝑟 (3.24) And the total solution at (r+1)th iteration is given by,
{𝑞}𝑟+1 = {𝑞}𝑟 + {𝛿𝑞}
(3.25)
The iteration process is continued by solving Eq. (3.24) until a convergence criterion is satisfied.
The error criterion is of the form
√∑ |𝑞𝐼
𝑟+1 − 𝑞𝐼𝑟|2𝑁
𝐼=1
∑ |𝑞𝐼𝑟+1|2𝑁
𝐼=1
< 𝜖 ( 𝑠𝑎𝑦 10−3) (3.26)
The coefficients of submatrices of element tangent stiffness matrix is defined as follows
34
(𝐾𝑖𝑗
𝛼𝛽)𝑡𝑎𝑛
≡ 𝜕𝑅𝑖
𝛼
𝜕𝑞𝑗𝛽
≡ 𝐾𝑖𝑗𝛼𝛽+ 𝐾𝑖𝑗
𝛼𝛽
(3.27)
Where
𝐾𝑖𝑗𝛼𝛽=∑∑
𝜕𝐾𝑖𝑙𝛼𝛾
𝜕𝑞𝑗𝛽
𝑛
𝑙=1
5
𝛾=1
𝑞𝑙𝛾
(3.28)
The elemental tangential stiffness matrix obtained for HSDT is a symmetrical about its diagonal
and is as show below
[𝐾𝑒]𝑡𝑎𝑛 =
[ [𝐾𝑢𝑢] [𝐾𝑢𝛾] [𝐾𝑢𝑤𝑏] [𝑘𝑢𝑤𝑠] [𝐾𝑢𝑤𝑏
𝑦] [𝐾𝑢𝑤𝑏𝑦]
[𝐾𝛾𝛾] [𝐾𝛾𝑤𝑏] [𝐾𝛾𝑤𝑠] [𝐾𝛾𝑤𝑏𝑦] [𝐾𝛾𝑤𝑠
𝑦]
[𝐾𝑤𝑏𝑤𝑏] [𝐾𝑤𝑏𝑤𝑠] [𝐾𝑤𝑏𝑤𝑏𝑦] [𝐾𝑤𝑏𝑤𝑠
𝑦]
𝑠𝑦𝑚𝑚 [𝐾𝑤𝑠𝑤𝑠] [𝐾𝑤𝑠𝑤𝑏𝑦] [𝐾𝑤𝑠𝑤𝑠
𝑦]
[𝐾𝑤𝑏𝑦𝑤𝑏𝑦] [𝐾𝑤𝑏
𝑦𝑤𝑠𝑦]
[𝐾𝑤𝑠𝑦𝑤𝑠𝑦]]
(3.29)
The coefficients of submatrices of element tangential stiffness matrix for HSDT are derived and
listed below
(𝐾𝑖𝑗11)
𝑡𝑎𝑛= 𝐾𝑖𝑗
11
(𝐾𝑖𝑗12)
𝑡𝑎𝑛= 𝐾𝑖𝑗
12
(𝐾𝑖𝑗13)
𝑡𝑎𝑛= [𝐾𝑢𝑤𝑏] + 𝑏∫ {
𝑇112𝑤𝑏′ {𝐻𝐿
′ }⌊𝐻′⌋ + 𝑇12(𝑤𝑏𝑦+𝑤𝑠
𝑦){𝐻𝐿′ }⌊𝐻′⌋}
𝑙𝑒
0
𝑑𝑥𝑒
(𝐾𝑖𝑗14)
𝑡𝑎𝑛= [𝐾𝑢𝑤𝑠] + 𝑏∫ {
𝑇112𝑤𝑠′{𝐻𝐿
′ }⌊𝐻′⌋ + 𝑇12(𝑤𝑏𝑦+𝑤𝑠
𝑦){𝐻𝐿′}⌊𝐻′⌋}
𝑙𝑒
0
𝑑𝑥𝑒
(𝐾𝑖𝑗15)
𝑡𝑎𝑛= 𝐾𝑖𝑗
15
(𝐾𝑖𝑗16)
𝑡𝑎𝑛= 𝐾𝑖𝑗
16
35
(𝐾𝑖𝑗22)
𝑡𝑎𝑛= 𝐾𝑖𝑗
22
(𝐾𝑖𝑗23)
𝑡𝑎𝑛= [𝐾𝛾𝑤𝑏] + 𝑏∫ {
𝑇212𝑤𝑏′ {𝐻𝐿
′ }⌊𝐻′⌋ + 𝑇22(𝑤𝑏𝑦+𝑤𝑠
𝑦){𝐻𝐿′ }⌊𝐻′⌋}
𝑙𝑒
0
𝑑𝑥𝑒
(𝐾𝑖𝑗2)
𝑡𝑎𝑛= [𝐾𝛾𝑤𝑠] + 𝑏∫ {
𝑇212𝑤𝑠′{𝐻𝐿
′ }⌊𝐻′⌋ + 𝑇22(𝑤𝑏𝑦+𝑤𝑠
𝑦){𝐻𝐿′}⌊𝐻′⌋}
𝑙𝑒
0
𝑑𝑥𝑒
(𝐾𝑖𝑗25)
𝑡𝑎𝑛= 𝐾𝑖𝑗
25
(𝐾𝑖𝑗26)
𝑡𝑎𝑛= 𝐾𝑖𝑗
26
(𝐾𝑖𝑗33)
𝑡𝑎𝑛=
[𝐾𝑤𝑏𝑤𝑏] +
𝑏 ∫ {𝑇11𝑢′{𝐻′}⌊𝐻′⌋ + 𝑇12𝛾{𝐻
′}⌊𝐻′⌋ − 𝑇13𝑤𝑏′′{𝐻′}⌊𝐻′⌋ − 𝑇11(𝑤𝑏
′ )2{𝐻′}⌊𝐻′⌋ −𝑙
0
𝑇31
2𝑤𝑏′ {𝐻′}⌊𝐻′⌋ +
𝑇21
2(𝑤𝑏
𝑦+ 𝑤𝑠
𝑦)𝑤𝑏′ {𝐻′}⌊𝐻′⌋ − 𝐶1𝑇15𝑤𝑏
′′{𝐻′}⌊𝐻′⌋ +
𝑇11
2(𝑤𝑠
′)2{𝐻′}⌊𝐻′⌋ − 2𝑇14𝑤𝑏𝑦′{𝐻′}⌊𝐻′⌋ + 𝑇12𝑤𝑏
𝑦(2𝑤𝑏′ + 𝑤𝑠
′){𝐻′}⌊𝐻′⌋ −
𝑇32𝑤𝑏𝑦{𝐻′}⌊𝐻′⌋ + 𝑇22(𝑤𝑏
𝑦+𝑤𝑠
𝑦)𝑤𝑏
𝑦{𝐻′}⌊𝐻′⌋ − 2𝐶1𝑇16𝑤𝑠𝑦′{𝐻′}⌊𝐻′⌋ +
2𝑇12𝑤𝑏′𝑤𝑠
𝑦{𝐻′}⌊𝐻′⌋ + 𝑇12𝑤𝑠′𝑤𝑠
𝑦{𝐻′}⌊𝐻′⌋ − 𝑇32𝑤𝑠𝑦{𝐻′}⌊𝐻′⌋ + 𝑇22(𝑤𝑏
𝑦+
𝑤𝑠𝑦)𝑤𝑠
𝑦{𝐻′}⌊𝐻′⌋} 𝑑𝑥𝑒
(𝐾𝑖𝑗34)
𝑡𝑎𝑛= [𝐾𝑤𝑏𝑤𝑠]
+ 𝑏∫ {𝑇112𝑤𝑏′𝑤𝑠
′{𝐻𝐿′ }⌊𝐻′⌋ +
𝑇212(𝑤𝑏
𝑦+𝑤𝑠
𝑦)𝑤𝑠′{𝐻′}⌊𝐻′⌋ −
𝑇312𝑤𝑠′{𝐻′}⌊𝐻′⌋
𝑙𝑒
0
+ 𝑇12𝑤𝑏′𝑤𝑏
𝑦{𝐻𝐿′ }⌊𝐻′⌋ + 𝑇22(𝑤𝑏
𝑦+𝑤𝑠
𝑦)𝑤𝑏𝑦{𝐻′}⌊𝐻′⌋ − 𝑇32𝑤𝑏
𝑦{𝐻′}⌊𝐻′⌋
+ 𝑇12𝑤𝑏′𝑤𝑠
𝑦{𝐻′}⌊𝐻′⌋ + 𝑇22(𝑤𝑏𝑦+𝑤𝑠
𝑦)𝑤𝑠𝑦{𝐻′}⌊𝐻′⌋ − 𝑇32𝑤𝑠
𝑦{𝐻′}⌊𝐻′⌋} 𝑑𝑥𝑒
(𝐾𝑖𝑗35)
𝑡𝑎𝑛= [𝐾𝑤𝑏𝑤𝑏
𝑦]
+ 𝑏∫ {𝑇21𝑢′{𝐻′}⌊𝐻𝐿⌋ + 𝑇22𝛾{𝐻
′}⌊𝐻𝐿⌋ − 𝑇23𝑤𝑏′′{𝐻′}⌊𝐻𝐿⌋ +
𝑇212(𝑤𝑏
′ )2{𝐻′}⌊𝐻𝐿⌋𝑙𝑒
0
− 𝐶1𝑇25𝑤𝑠′′{𝐻′}⌊𝐻𝐿⌋ +
𝑇212(𝑤𝑠
′)2{𝐻′}⌊𝐻𝐿⌋ + 𝑇22(𝑤𝑏′ + 𝑤𝑠
′)𝑤𝑏𝑦{𝐻′}⌊𝐻𝐿⌋
+ 𝑇22(𝑤𝑏′ + 𝑤𝑠
′)𝑤𝑠𝑦{𝐻′}⌊𝐻𝐿⌋} 𝑑𝑥𝑒
36
(𝐾𝑖𝑗36)
𝑡𝑎𝑛= [𝐾𝑤𝑏𝑤𝑠
𝑦]
+ 𝑏∫ {𝑇21𝑢′{𝐻′}⌊𝐻𝐿⌋ + 𝑇22𝛾{𝐻
′}⌊𝐻𝐿⌋ − 𝑇23𝑤𝑏′′{𝐻′}⌊𝐻𝐿⌋ +
𝑇212(𝑤𝑏
′ )2{𝐻′}⌊𝐻𝐿⌋𝑙𝑒
0
− 𝐶1𝑇25𝑤𝑠′′{𝐻′}⌊𝐻𝐿⌋ +
𝑇212(𝑤𝑠
′)2{𝐻′}⌊𝐻𝐿⌋ + 𝑇22(𝑤𝑏′ + 𝑤𝑠
′)𝑤𝑏𝑦{𝐻′}⌊𝐻𝐿⌋
+ 𝑇22(𝑤𝑏′ + 𝑤𝑠
′)𝑤𝑠𝑦{𝐻′}⌊𝐻𝐿⌋} 𝑑𝑥𝑒
(𝐾𝑖𝑗44)
𝑡𝑎𝑛=
[𝐾𝑤𝑠𝑤𝑠] +
𝑏 ∫ {𝑇11𝑢′{𝐻′}⌊𝐻′⌋ + 𝑇12𝛾{𝐻
′}⌊𝐻′⌋ − 𝑇13𝑤𝑏′′{𝐻′}⌊𝐻′⌋ −
𝑇11
2(𝑤𝑏
′ )2{𝐻′}⌊𝐻′⌋ −𝑙𝑒
0
𝐶1𝑇15𝑤𝑠′′{𝐻′}⌊𝐻′⌋ + 𝑇11(𝑤𝑠
′)2{𝐻′}⌊𝐻′⌋ +𝑇11
2(𝑤𝑏
𝑦+𝑤𝑠
𝑦)𝑤𝑠′{𝐻′}⌊𝐻′⌋ −
𝐶1𝑇51
2𝑤𝑠′{𝐻′′}⌊𝐻′⌋ − 2𝑇14𝑤𝑏
𝑦′{𝐻′}⌊𝐻′⌋ + 𝑇12𝑤𝑏′𝑤𝑏
𝑦{𝐻′}⌊𝐻′⌋ +
2𝑇12𝑤𝑠′𝑤𝑏
𝑦{𝐻′}⌊𝐻′⌋ + 𝑇22(𝑤𝑏𝑦+𝑤𝑠
𝑦)𝑤𝑏𝑦{𝐻′}⌊𝐻′⌋ − 𝐶1𝑇52𝑤𝑏
𝑦{𝐻′′}⌊𝐻′⌋ −
2𝐶1𝑇16𝑤𝑠𝑦′{𝐻′}⌊𝐻′⌋+ 𝑇12𝑤𝑏
′𝑤𝑠𝑦{𝐻′}⌊𝐻′⌋ + 2𝑇12𝑤𝑠
′𝑤𝑠𝑦{𝐻′}⌊𝐻′⌋ + 𝑇22(𝑤𝑏
𝑦+
𝑤𝑠𝑦)𝑤𝑠
𝑦{𝐻′}⌊𝐻′⌋ − 𝐶1𝑇52𝑤𝑠𝑦{𝐻′′}⌊𝐻′⌋} 𝑑𝑥𝑒
(𝐾𝑖𝑗45)
𝑡𝑎𝑛= [𝐾𝑤𝑠𝑤𝑏
𝑦]
+ 𝑏∫ {𝑇21𝑢′{𝐻′}⌊𝐻𝐿⌋ + 𝑇22𝛾{𝐻
′}⌊𝐻𝐿⌋ − 𝑇23𝑤𝑏′′{𝐻′}⌊𝐻𝐿⌋ +
𝑇212(𝑤𝑏
′ )2{𝐻′}⌊𝐻𝐿⌋𝑙𝑒
0
− 𝐶1𝑇25𝑤𝑠′′{𝐻′}⌊𝐻𝐿⌋ +
𝑇212(𝑤𝑠
′)2{𝐻′}⌊𝐻𝐿⌋ + 𝑇22(𝑤𝑏′ + 𝑤𝑠
′)𝑤𝑏𝑦{𝐻′}⌊𝐻𝐿⌋
+ 𝑇22(𝑤𝑏′ + 𝑤𝑠
′)𝑤𝑠𝑦{𝐻′}⌊𝐻𝐿⌋} 𝑑𝑥𝑒
(𝐾𝑖𝑗46)
𝑡𝑎𝑛= [𝐾𝑤𝑠𝑤𝑠
𝑦]
+ 𝑏∫ {𝑇21𝑢′{𝐻′}⌊𝐻𝐿⌋ + 𝑇22𝛾{𝐻
′}⌊𝐻𝐿⌋ − 𝑇23𝑤𝑏′′{𝐻′}⌊𝐻𝐿⌋ +
𝑇212(𝑤𝑏
′ )2{𝐻′}⌊𝐻𝐿⌋𝑙𝑒
0
− 𝐶1𝑇25𝑤𝑠′′{𝐻′}⌊𝐻𝐿⌋ +
𝑇212(𝑤𝑠
′)2{𝐻′}⌊𝐻𝐿⌋ + 𝑇22(𝑤𝑏′ + 𝑤𝑠
′)𝑤𝑏𝑦{𝐻′}⌊𝐻𝐿⌋
+ 𝑇22(𝑤𝑏′ + 𝑤𝑠
′)𝑤𝑠𝑦{𝐻′}⌊𝐻𝐿⌋} 𝑑𝑥𝑒
(𝐾𝑖𝑗55)
𝑡𝑎𝑛= [𝐾𝑤𝑏
𝑦𝑤𝑏𝑦]
(𝐾𝑖𝑗56)
𝑡𝑎𝑛= [𝐾𝑤𝑏
𝑦𝑤𝑠𝑦]
(𝐾𝑖𝑗66)
𝑡𝑎𝑛= [𝐾𝑤𝑠
𝑦𝑤𝑠𝑦] (3.30)
37
3.7 Boundary Conditions
The boundary conditions considered in this research are given below.
3.7.1 Boundary Conditions for HSDT
In the formulation based on HSDT, ,𝛾, 𝑤𝑏 ,𝑤𝑏′ , 𝑤𝑠, 𝑤𝑠
′, 𝑤𝑏𝑦
and 𝑤𝑠𝑦
are the end nodal degrees of
freedom. The boundary conditions for a pinned end is
𝑢 = 𝑤𝑏 = 𝑤𝑠 = 𝑤𝑏𝑦= 𝑤𝑠
𝑦= 0
3.7.2 Boundary Conditions for Cylindrical Bending
In the formulation based on Cylindrical Bending, u, w and w’ are the end nodal degrees of
freedom. The boundary conditions for a pinned end is
u = w = 0
38
4 RESULTS and DISCUSSION
A code is written in MATLAB to solve the nonlinear finite element equations of beam bending
based on HSDT using the Newton-Raphson technique. The formulation and the code are
verified first by considering a simpler large-deflection problem of cylindrical bending of plates
and comparing with existing solutions. The present results for the HSDT case of large deflection
are for a pinned-pinned composite beam subjected to positive and negative uniformly-
distributed loads; a comparison with the linear case is also made.
4.1 Cylindrical Bending Verification
The MATLAB code incorporating the Newton-Raphson iterative technique for nonlinear analysis is verified by solving the large-deflection problem of cylindrical bending considered by Sun and Chin (1988). The problem consists of a pinned-pinned rectangular composite laminated plate, [04/904], subjected to a uniform transverse load. The material properties of the graphite-epoxy composite are presented in Table 4.1 as follows
Table 4.1 Material properties [Sun and Chin (1988)]
Property Value
Composite Graphite-epoxy
E1 (msi) 20
E2(msi) 1.4
ν12 0.3 G12,G13, G23 (msi) 0.7
The geometric properties of the composite beam are tabulated in Table 4.2
Table 4.2 Geometric properties [Sun and Chin (1988)]
Property Value
Length, L(in.) 9
Width, b(in.) 0.05
Total thickness of beam, h(in.) 0.04
Lay-ups considered [04/904]
The composite laminated pinned-pinned beam is discretized into four elements with three
internal nodes each. The beam is subjected to an axial load of 1 lb/in. The Newton-Raphson
iterative method is employed to find out the deflection as the function of position of beam.
Figure 4.1 shows the out-of-plane non-dimensional deflection (w/h) along the non-dimensional
length of beam (x/L) when subjected to the axial load of 1 lb/in.
39
Fig. 4.1 Out-of-plane deflection of [04/904] laminate subjected to uniform in-plane load
Nx=1 lb/in
The comparison indicates that the present result agrees well with that of Sun and Chin (1988).
The maximum normalized transverse deflection for Nx=1 lb/in. is 0.006587.
Fig. 4.2Load-deflection curveof a pinned-pinned composite beam under transverse load
-0.01
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
-1.5 -1 -0.5 0 0.5 1 1.5
w/h
x/L
Present
Sun and Chin (1988)
40
The second verification case pertains to a pinned-pinned beam with the same material and
geometric properties and the same lay up as before but the loading is a uniform transverse
load. Figure 4.2 shows the plot of the normalized maximum deflection (w0/h) as a function of
the load intensity. . The graph is plotted for the load varying from 0 to 0.1lb/in.2. Table 4.2
presents the comparison between the analytical results of Reddy (1997) and the present work.
The results show that the agreement between the two sets of results is excellent.
Table 4.3 Transverse deflections, w0/h, of cylindrical bending of a [04/904]laminate under
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