Beam Element under Axial Load (Tension)
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Transcript of Beam Element under Axial Load (Tension)
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ADVANCED STRUCTURAL ANALYSIS
MAE1013
Assignment No.1
22/2/2011
Lecturer: Assoc. Prof. Dr. Jamaludin Mohamad Yatim
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Derive Analytical Procedure of Stiffness Method for Beam Element
Introduction:
Beam element is one of the basic element types in finite element analysis of frame systems, the elastic
stiffness of which in commonplace can be found in many textbooks on structural analysis (McGuire,
Gallagher and Ziemian, 1999; Aslam Kassimali, 1999; Hibbler, 1983). However, the effects of shear
deformation and axial force on the stiffness of beam elements were seldom considered simultaneously
in previous investigations (Tranberg, Meek and Swannell, 1976)
For steel-framed systems, simultaneous effects of shear deformation and axial force on the behavior of
beam elements cannot be ignored in certain cases (Li and Shen, 1995). This section describes the
derivation of elastic stiffness equations from the differential equilibrium equation, for the beam
elements including the above two effects.
The criteria mainly are discussed as a part of advanced structural design in general without any detailing
and further explanation. Hereby I have showed complete steps of derivation of the stiffness matrix of
the beam member under simultaneous effects of shear deformation and axial force.
To distinguish the main body of the reference book [Advanced Analysis and Design of Steel Frame-Guo-
Qiang Li and Jin-Jun Li# 2007 John Wiley & Sons, Ltd. ISBN: 978-0-470-03061-5] from the added parts,
the amended texts are put inside border with the sign of.
Beam Element under Axial Load (Tension)
For the beam elements including the above two effects: = + (2.1)The bending curvature formula of beam element is:
"
=
(2
.2
)
Where E is the elastic modulus, I is the moment of inertia of the cross section and M is the cross-
sectional moment given by: = 1 1 (2.3)The work done by shear in the differential element is (see Figure 2.2; Timoshenko and Gere, 1961) = 12 (2.4)
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and the shear strain energy is = 22 (2.4)where G is the elastic shear modulus, A is the area of the cross section, Q is the shear force of the cross
section and is the shear shape factor of the cross section, considering effects of uneven distribution ofshear
deformation over the cross section, as shown in Figure 2.3.
By energy theory, dWQ = dUQ, one can derive from Equation (2.4) that
= = (2.5)
Substituting Equation (2.3) into Equation (2.5) yields = (1 ) (2.6)and differentiating Equation (2.6) once gives" = " (2.7)
= =
( ) =
= = =
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Combining Equations (2.2) and (2.7), one has" = 11 " (2.8)Let = 1+ (2.9)
" = " = " = " "
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2 = (2.10)and then rewrite Equation (2.8) as
" 2 = 11 (2.11)The solution of Equation (2.11) is = cosh + sinh + 11 (2.12)where a and b are the unknown coefficients depending on boundary conditions. Boundary conditions of
the beam element are
= 0; = 0 (2.13) = + = 1 + (1 ) = 1 1 1 (2.13)
= ; = 2 1,
(2
.13
)
= + = 2 + (1 ) = 1 2 1 (2.13) = 0; = 0Substituting Equation (2.12) into Equations (2.13a)(2.13d) yields four simultaneous linear equations in
terms of a, b, _1 and _2, i.e.
= +
= + = = ( )
" = "
" + " = " + " = " + = + + " = + +
" = +
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+1 =
0
(2
.14
)
1 = 1 1 1 (2.14)
cosh + sinh + 11 = 2 1 (2.14)
sinh + cosh 1 = 1 2 1 (2.14)Then, a and b are obtained from Equations (2.14a) and (2.14c) as = 1 (2.15)
= 1 1 (cosh 1) + 1 + (2 1) (2.15)Substituting Equations (2.15a) and (2.15b) into Equations (2.14b) and (2.14d) yields
sinh = cosh
+ ( )
= cosh + sinh =
= cosh + sinh + = ; =
(sinh()) = cosh()(cosh()) = sinh() = cosh + sinh + = cosh + sinh = ; =
=
=
= cosh + sinh +
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sinh = cosh + ( )
=
sinh (cosh ) + + ( ) = = sinh (cosh ) + + ( ) + = (cosh )sinh + ( )sinh + sinh +
= + =
= (cosh )sinh + ( )sinh + sinh + ( )
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1 = 1 (cosh 1) + 1 + ( 1 ) + 21
2 = 1 1 + 1 1 + 21 The equilibrium of elemental moments can be written as 1 + 2 1 (2 1) = 0
sinh + cosh = = ; = sinh (cosh ) + + ( ) sinh + sinh (cosh ) + + ( ) cosh =
= sinh +coshsinh (cosh ) + + ( ) = sinh + coshsinh (cosh ) + + ( )
= sinh sinh +
cosh sinh
coshsinh
+
coshsinh
+
coshsinh ( = cosh sinh sinh sinh coshsinh + coshsinh + coshsinh ( cosh sinh =
= + = =
sinh cosh
sinh
+ cosh
sinh
+ cosh
sinh( ) + (
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by which one obtains 1 = 12 (21) From Equation (2.10), axial force can be expressed as
= 2 = ()22 Substituting Equation (2.17) into Equations (2.16a) and (2.16b) yields
= sinh (cosh ) + sinh + sinh = sinh (cosh ) + + sinh + sinh = sinh cosh sinh + sinh + + + sinh+ sinh
= cosh sinhsinh + sinhsinh
= sinh cosh() sinh sinh () sinh
= ; = +
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1 21 = ()2 1 ()2 22 21 = ()2 1 ()2 2
= cosh sinh + cosh sinh + cosh sinh
= +
( ) ; = =
()
= cosh sinh +coshsinh + + coshsinh = coshsinh + + coshsinh + coshsinh +
= sinh coshsinh + coshsinh + coshsinh + + coshsinh + = () sinh + () coshsinh + () +
= () sinh + () coshsinh () ( = () sinh sinh () sinh + cosh () sinh (
= sinh
sinh cosh
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= () sinhsinh cosh + sinh () sinh
= () sinhsinh cosh () sinhsinh cosh + () sinhsinh cosh si( = sinh () sinh () sinhsinh cosh () sinhsinh cosh+ () sinhsinh cosh sinh () sinh sinh cosh () sinh
= sinh () sinh () sinhsinh cosh sinh () sinh () sisinh + sinh () sinh () sinhsinh cosh sinh () sinh sinh co() sinh
= (sinh )() sinh (sinh )() sinh (() sinh)(sinh cosh)
+ sinh () sinh () sinhsinh cosh sinh () sinh sinh () s
= (sinh ) (sinh ) (sinh cosh) + sinh () sinh sinh sinh cosh
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= (sinh ) (sinh ) (sinh cosh) + (sinh ) (sinh ( () sinh)(sinh
(sinh cosh) (sinh cosh) (sinh ) + (sinh (sinh cosh)(sinh ) (sinh cosh)(() sinh)(sinh cosh)() sinh(sinh cosh) (sinh cosh) (sinh ) + (sinh(sinh ) (sinh cosh )() sinh(sinh cosh)(sinh ) (sinh cosh) () sinh(sinh cosh)(sinh ) (sinh cosh)
() sinh(sinh )(sinh ) (sinh cosh) + () sinh(sinh (sinh ) (sinh c() sinh(sinh cosh)(sinh ) (sinh cosh) () sinh(sinh (sinh ) (sinh co+ () sinh + cosh(sinh ) (sinh cosh) =
(sinh ) (sinh cosh) = sinh + () sinh sinh () cosh
(sinh ) (sinh cosh) = () sinh () cosh + (sinh ) (sinh cosh) = sinh cosh + sin
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2 = 241 + 432 62 21 2 = 16 ()2(cosh 1)
(sinh)(sinh cosh) sinh cosh + sinh cosh (sinh)(sinh sinh cosh + si+ () sinh cosh sinh
sinh cosh + sinh cosh
=
= (sinh)(sinh cosh) sinh (cosh )+ sinhcosh (sinh)(sinh sinh (cosh )+ () sinh cosh sinh sinh (cosh )+ sinhcosh cosh sinh =
= (sinh)(sinh cosh) sinh sinh + sinh cosh (sinh)(sinh sinh sinh + sin+ () sinh cosh sinh sinh sinh + sinh cosh
= sinh ( cosh sinh)sinh ( + sinh cosh) + sinh (sinh )sinh ( + sinh co+ sinh ( cosh)()sinh ( + sinh cosh)
= (sinh ) cosh + sinh + ( cosh sinh) cosh + sinh + (cosh )( cosh +
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3 = 14 ( cosh sinh)
4 = 12
(sinh )
= 2 2 cosh + sinhSubstituting Equations (2.20a), (2.20b) and (2.18) into Equation (2.17) yields
Where:
Equations (2.20a)(2.20c) are factually the elastic stiffness equations for beam elements considering
effects of shear deformation and axial force simultaneously, which can be expressed in matrix form as
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