Topic 8
The Two- oded
Truss Element­
Updated
Lagrangian
Formulation
Contents:
• Derivation of updated Lagrangian truss element
displacement and strain-displacement matrices from
continuum mechanics equations
• Assumption of large displacements and rotations but
small strains
• Physical explanation of the matrices obtained directly by
application of the principle of virtual work
• Effect of geometric (nonlinear strain) stiffness matrix
• Example analysis: Prestressed cable
Textbook:
Section 6.3.1
Examples:
6.15,6.16
Topic Eight 8-3
TRUSS ELEMENT DERIVATION
A truss element is a structural member
which incorporates the following
assu mptions:
Transparency
8-1
• Stresses are transmitted only in the
direction normal to the cross-section.
• The stress is constant over the cross­
section.
• The cross-sectional area remains
constant during deformations.
We consider the large rotation-small
strain finite element formulation for a
straight truss element with constant
cross-sectional area.
2
Elastic material with
Young's modulus E
Cross-sectional area A
1
Element lies in the Xl - X2
plane and is initially aligned
with the X1 axis.
1
L
'I
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8-2
8-4 The 'l\vo-Noded Truss Element - UL.F.
Transparency
The deformations of the element are
specified by the displacements of its
nodes:
8-3
time 0
t Ul
Xl
Our goal is to determine the element
deformations at time t + ~t.
!!Pdated Lagrangian formulation:
Transparency
8-4
The derivation is simplified if we
consider a coordinate system aligned
with the truss element at time t.
Topic Eight 8-5
Written in the rotated coordinate system,
the equation of the principle of
virtual work is
JVt+Ll~s~ot+Ll~Ei}
tdV
Transparency
8-5
= t+Lltm
As we recall, this may be linearized to
obtain
r tC~ijrS te
Jtv
=
rs
Otei} tdV +
t+ Lltm
_
r tf }odli} tdV
Jtv i
r tTi}Otei} tdV
Jtv
Because the only non-zero stress
component is tf11, the linearized
equation of motion simplifies to
r
Jtv
tC 1111 te11 Ote11 tdV +
= t+Lltm -
r tf OtTl11 tdV
Jtv 11
r tf Ote11 tdV
Jtv 11
Notice that we need only consider one
component of the strain tensor.
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8-6
8-6 The Two-NocIecl Truss Element - V.L.F.
Transparency
We also notice that:
tC 1111
8-7
t-
=E
tp
T11 = A
tv = AL
The stress and strain
states are constant
along the truss.
Hence the equation of motion becomes
(EA)
= HAtm.
'(Jt
Transparency
8-8
L
t811 8 t 811
-
+ tp 8dl11 L
tp
S::.
Ute11
L
To proceed, we must express the strain
increments in terms of the (rotated)
displacement increments:
where
~
0=
This form is analogous
to the form used in
the two-dimensional
element formulation.
Topic Eight 8-7
·
- 11 = tU1,1
• + 21 ((.
SInce
tE
tU1,1 )2 + (.
tU2,1 )2) ,
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8-9
we recognize
t«311
= tU1,1
d111
= ~ ((tU1,1)2 + (tU2,1)2)
and
= 8tu1,1 tU 1,1 + 8tU2,1 tU2,1
8dl11
= [8tU1,1
8 tU2,1]
.
[t~1'1]
t 2,1
matrix form
We can now write the displacement
derivatives in terms of the
displacements (this is simple because
all quantities are constant along the
truss). For example,
•
tU1,1
aU1
= ax1 =
t
aU1
a tX1
=
U~ - U~
L
Hence we obtain
[:~::~] = ~ [ - 6 - ~ 6 ~ ]
• 11
U
• 21
U
•2
U
1
2
•2
U
-s-u-"
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8-10
8-8 The 'IWo-Noded 'lruss Element - D.L.F.
Transparency
and
8-11
t«311 =
(i [-1
0
~
1 0])
'-------v.:======~.s
t .s--,B
L
t-
,. .-,- ---"'-----"
,BNL
~~11 &~T (tri -~)(H-6 ~~. 6
=
_______________'
[8t O1,1
[,01,1 ]
8,0 2 ,1]
,0 2 ,1
Using these expressions,
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8-12
(EA)
te11
Bte11 L
\
BOT
-
~AL -1~ g0
o
0
-1
o
1
o
0
0
0
0
(setting successively each virtual nodal point
displacement equal to unity)
Topic Eight 8-9
Transparency
8-13
\
,
~T
()!!
and
10-1
0])
0
( ~o
tp 0
L -1
1
0
-1
0 -1
1
0
0
1
-
Transparency
8-14
8-10 The 1\vo-Noded Truss Element - U.L.F.
Transparency
8-15
We have now obtained the required
element matrices, expressed in the
coordinate system aligned with the
truss at time t.
To determine the element matrices in
the stationary global coordinate system,
we must express the rotated
displacement increments
in terms of
the unrotated displacement increments
u
O.
We can show that
[~:] = [_~~~~e cS~~~:] [~:]
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8-16
Hence
-1
U1
-1
U2
-2
U1
-2
U2
coste sinte
-sinte coste
a
a
u~
a
a
coste
sinte
u1
u~
0
0
-sinte coste
u~
~
v
Q
T
-
a
a
'~
0
-
Topic Eight 8-11
Using this transformation in the
equation of motion gives
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8-17
BUT ~KL
U~ BOT TT ~KL
Ta
-
-
-
\,,-
~KL
,- 4
-/-
-/
~F
Transparency
8-18
Performing the indicated matrix
multiplications gives
-
-
(COSt8)2 (cost8)(sint8)
-(COSt8)2 -(cost8)(sint8)
t
-(cost8)(sint8)
(sin 8)2
-(sin t8)2
symmetric
'---
I
(COSt8)2
(cost8)(sint8)
(sin t8)2
-
8-12 The 'I\vo-Noded Truss Element - U.L.F.
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8-19
1 0
--1
1
0
1
symmetric
and
~F =
The vector
o
-1
o
1
-coste
-sinte
tp
coste
sinte
:F makes physical sense:
Transparency
8-20
Hence, at equilibrium,
tR - :F = Q
Topic Eight 8-13
We note that the ~KNL matrix is unchanged by
the coordinate transformation.
• The nonlinear strain increment is
related only to the vector magnitude
of the displacement increment.
Transparency
8-21
~ + (O~f~~(:,~J + (:I~~r)L
)(1
./'
Physically, :~NL gives the required
change in the externally applied nodal
point forces when the truss is rotated.
Consider only O~ nonzero.
For small o~, this gives a rotation
about node 1.
Moment equilibrium:
(dA) (L) = CA) (Ci~)
~R
or AA _ tp -2
L.1
-
-
1-
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8-22
U2
entry (4,4)
of lK
,-.-::l
- NL,-
For small Q,
X~,-//'-
:~NL
internal
force tp
Q=
L\R
t+~tB
- tA
8·14 The 1\vo-Noded Truss Element - U.L.F.
Example: Prestressed cable
Transparency
8-23
r Applied load 2 tR
Initial
ten~ion =
0p
i
Length 2 L
Young's modulus E
Area A
Finite element model (using symmetry):
Transparency
8-24
Of particular interest is the configuration
at time 0, when te = 0:
(O~) u~ = ~tR
The undeformed cable stiffness is given
solely by ~KNL.
'Ibpic Eight 8-15
The cable stiffens as load is applied:
~K =
EA
,L
(sin t O)2
Transparency
8-25
+ tp
. ~
~KL
~KL increases as
~KNL
to
increases (the truss
provides axial stiffness as to increases).
As
to ~ 90°, the stiffness approaches Et,
but constant L and A means here that
only small values of to are permissible.
Using: L = 120 in , A = 1 in 2 ,
E = 30 x 106 psi , 0p = 1000 Ibs
we obtain
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8-26
200.
Applied
force
(Ibs)
o.o........,=--
+_
Deflection (inches)
2.5
8-16 The Two-Noded Tmss Element - U.L.F.
Transparency
8-27
We also show the stiffness matrix
components as functions of the applied
load:
200.0
Stiffness
(Iblin)
o.o~-----------+--
0.0
Applied force (Ibs)
200.0
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Resource: Finite Element Procedures for Solids and Structures
Klaus-Jürgen Bathe
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