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MAE 456 Finite Element Analysis
Beam Elements
Slides from this set are adapted from B.S. Altan, Michigan Technological U.
MAE 456 Finite Element Analysis
Bending of Beams – Definition of Problem
Beam with a straight axis (x-axis), with varying cross-section
(A(x)), loaded transversely (q(x) in y direction) with transverse
deflections (v(x) in y direction).
x
MV
q(x)v(x)
x
x=0 x=L
A(x)
y
z
2
MAE 456 Finite Element Analysis
Bending of Beams – Definition of Problem
ρε
yx −=)(
x
ρ
2
2
dx
vdEyE −== εσ
y
2
21
dx
vd≅
ρ
dAdx
vdEyydA
AA
∫∫ −=2
22σ
3
MAE 456 Finite Element Analysis
Bending of Beams – Definition of Problem
dAdx
vdEyydA
AA
∫∫ −=2
22σ
2
2
dx
vdEIM −=
2
2
dx
vdEI
dx
d
dx
dMV =−=
2
2
2
2
)(dx
vdEI
dx
d
dx
dVxq == Resulting Field Equation
Continued from previous slide.
4
MAE 456 Finite Element Analysis
Bending of Beams – Definition of Problem
Summary
)(4
4
xqdx
vdEI =
Slope:dx
dvx =)(θ
Bending Moment:2
2
)(dx
vdEIxM =
Shear Force:3
3
)(dx
vdEIxV −=
Field equation:
5
MAE 456 Finite Element Analysis
Bending of Beams – Definition of Problem
Boundary Conditions: Four boundary conditions are
necessary to solve a bending problem. Boundary
conditions can be:
deflection:
bending Moment:
shear force:
slope:
)(),0( Lvv
)(),0( Lθθ
)(),0( LMM
)(),0( LVV
6
MAE 456 Finite Element Analysis
Beam Element – Shape Functions
• Recall that shape functions are used to interpolate displacements.7
MAE 456 Finite Element Analysis
Beam Element – Shape Functions
• There are two degrees of freedom (displacements)
at each node: v and θz.
• Each shape function corresponds to one of the
displacements being equal to ‘one’ and all the
other displacements equal to ‘zero’.
• Note that everything we do in this course
assumes that the displacements are small.8
MAE 456 Finite Element Analysis
Beam Element – Shape Functions
Bd
dNNd
Ey
dx
dEy
dx
dEy
dx
vdEyE
−=
−=
−=
−==
2
2
2
2
2
2
εσ
9• Recall that the B row vector is used to interpolate stresses & strains.
MAE 456 Finite Element Analysis
Beam Element – Formal Derivation
• The formal beam element stiffness matrix
derivation is much the same as the bar element
stiffness matrix derivation. From the minim-
ization of potential energy, we get the formula:
• As with the bar element, the strain energy of the
element is given by .kddT
21
10
MAE 456 Finite Element Analysis
Beam Element – Formal Derivation
The result is:
which operates on d = [v1, θz1, v2, θz2]T.
−
−−−
−
−
=
22
22
3
4626
612612
2646
612612
LLLL
LL
LLLL
LL
L
EIk
11
MAE 456 Finite Element Analysis
Beam Element – Formal Derivation
• The moment along the element is given by:
• The stress is given by:
BdyEI
Myx −=−=σ
12
MAE 456 Finite Element Analysis
Beam Element w/Axial Stiffness
• If we combine the bar and beam stiffness
matrices, we get a general beam stiffness
matrix with axial stiffness.
13
MAE 456 Finite Element Analysis
Uniformly Distributed Loads
Laterally:
14
MAE 456 Finite Element Analysis
Equivalent Loadings
15
MAE 456 Finite Element Analysis
Orientating Element in 3-D Space
• Transformation matrices are used to transform
the equations in the element coordinate system
to the global coordinate system, as was shown
for the bar element.
16