Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Introduction
The principles for the direct stiffness method are now in place. In this section of notes we will derive the stiffness matrix, both local and global, for a truss element using the direct stiffness method. Here a local coordinate system will be utilized initially anddirect stiffness method. Here a local coordinate system will be utilized initially and the element stiffness matrix will be transformed into a global coordinate system that is convenient for the overall structure.
Inclined or skewed supports will be discussed.pp
The principle of minimum potential energy will be utilized to re-derive the stiffness matrices.
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
The truss element shown below is assumed to have a constant cross sectional area (A), a modulus of elasticity (E), and an initial length (L). The nodal degrees of freedom are the axial displacements directed along the length of the truss element. In addition the followingaxial displacements directed along the length of the truss element. In addition the following assumptions are made:
1. The truss element cannot sustain shear forces
2 A ff t f t di l t i d
xd2ˆ
d
2. Any effect of transverse displacements are ignored
3. Hooke’s law applies
Deformed shape
x2xd1
ffNode(a truss hinge)
x
Element
xf2xf1
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Assume
This is an approximation function. We are assuming through the use of this function h di l di ib d i i l li f hi l h l
xaau ˆˆ 21 +=
that displacements are distributed in an approximately linear fashion along the element.
( )xu ˆˆ
xAs noted earlier the total number of unknown coefficients must equal the degrees of freedom associated with the element.
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
To establish the unknown coefficients consider the boundary conditions of the truss element
D f d hDeformed shape
Thus
( ) 110ˆ adu x == ( ) LaddLu xx 212ˆˆˆ +==
Thus
−=
Ldda xx 12
2
ˆˆ
and in a matrix format
− dxxdd 1ˆˆˆˆˆˆˆ
−=+=
+=
x
xxx
xxx
d
dLx
LxdNdNx
Ldddu
2
12211
121 ˆ
1ˆˆˆˆˆ
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Here
xNˆ
1−=xNˆ
=
are shape functions. Consider the following guidelines as they relate to one dimensional structural elements when selecting a displacement function.
LN 11 =
LN2 =
g p
1. Common approximation functions are usually polynomials where the function can be expressed in terms of shape functions.
2 The approximation functions must be continuous through the element2. The approximation functions must be continuous through the element.
3. The approximation functions must provide inter-element continuity for all degrees of freedom at each node of every element.
4. The approximation functions must allow for rigid body motion, because rigid body motion can occur in a structure.
With
xaau ˆˆ 21 +=
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
The strain displacement relationship can now be cast as
d ˆ( )
dNNd
xdudx
x1
ˆˆ
=ε
d
dNN
xd
x
x
1
2
21
ˆ11
ˆˆ
=
=
dB
dLL
x
x
1
2
ˆ
ˆ
=
−=
Ldd
dB
xx
x
12
2
ˆˆ
ˆ
−=
=
which is constant along the length of the truss element.
L
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
( ) ( )ˆˆ
The stress-strain relationship becomes
( ) ( )
=
=
d
dBE
xEx
x1
ˆ
ˆˆˆ εσ
−=
LddE
d
xx
x
12
2
ˆˆ
From elementary mechanics
−
=
=
ddAE
AT
xx 12ˆˆ
σ
L
AE
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
The nodal forces become
TfTf xx =−= 21ˆˆ
thus
( )xxx ddL
AEfT 121ˆˆˆ −=−=
ff xx 21
or
( )xxx ddL
AEfT
L
122ˆˆˆ −==
or
( )( )
xxx
AE
ddL
AEf 211ˆˆˆ −=
In a matrix format
df ˆˆ
( )xxx ddL
AEf 122ˆˆˆ −=
−
−=
x
x
x
x
d
dL
AEf
f
2
1
2
1
ˆ1111
ˆ
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Consider the situation where
xx dd 21ˆˆ =
The case where both nodes move the same amount in the same direction. This is definition of rigid body motion. Here
xx 21
121
21
ˆˆˆˆ
ˆˆ
xL
ddd
xaau
xxx
−+=
+=
221 ˆ
ˆˆˆ xL
ddd
L
xxx
−+=
1
ˆ
ˆ0ˆ
d
xL
d x
+=
1
1
ad x
==
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
With
2211ˆˆˆ dNdNu +=
( ) 121
1211
2211
aNNaNaNdNdNu xx
+=+=+
this infers that
1ˆ au =
Thus for rigid body motion
( ) 121 aNN +=
g y
211 NN +=
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
In addition
( )L
ddx xxˆˆ
ˆ 12 −=ε ( )
−=
LddEx xxˆˆ
ˆ 12σ
−=
LddAET xxˆˆ12
Ldd xx
0
ˆˆ22 −
=
−=
LddE xxˆˆ
22
−=
LddAE xx
0
ˆˆ22
S i d h i l f i h l ill ll b f i id b d i
L=
=
LE 0
=
LAE 0
Stress, strain and the axial force in the truss element will all be zero for rigid body motion.
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
The general relationship from the previous page holds for an truss element oriented along the x axis. Thus
[ ]
−
−=
1111ˆ
LAEk
is the local stiffness matrix. We note that the local stiffness matrix is symmetric, i.e.,
and square. The global stiffness matrix and global force matrix are assembled using nodal
jiij kk ˆˆ =
force equilibrium equations, force deformation equations and compatibility equations. Examples of assembling these equations will be given.
[ ] [ ] ( )∑N
ekK ( )∑N
efF[ ] [ ] ( )∑=
=e
ekK1
( )∑=
=e
efF1
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
One can quickly populate the global stiffness matrix for a truss structure using the methodology developed for the spring element. Consider the ith truss element:
bith Element
ith element stiffness propertiestransfers into the global stiffness
ba baa
ii Element
matrix as follows
abaa
kkbkkaba
00LLL
MM
kka
ba
bbba kkb0
LLL
MM
LLL abaa
kkb
kka
00 MM
LLL bbba kkbith element’s stiffness matrix with node numbers a and b
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
I l lIn class example
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Transformation of Vectors
In nearly all finite element analyses it is a necessity to introduce both local (to the element) and global (to the component) coordinate axes.
Y
Gl b l
XZ
Global axes
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
In general for forces, the transformation is as follows:
FGlobal Local
Fy
−=
v
u
Y
X
ff
FF
θθθθ
cossinsincos
which appears as follows when appied to a truss element:
Fx
θ
which appears as follows when appied to a truss element:
ff
ff
y
x
y
x
ˆˆ
00cossin00sincos
1
1
1
1
−
θθθθf2y
fff
ff
y
x
y
y
x
y
ˆˆ
cossin00sincos00
2
2
1
2
2
−=
θθθθ
f1y f2x
ˆ
yf 2xf 2
[ ] ff ˆ*T=f1x
θxf1
yf1
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
For a true truss element, an element that will not sustain shear forces at the end, thenf2y
f1y f2x
02 =yfxf 2
f1xθ
xf101 =yf
and
ˆ00i ff θθ
Global Local
−
−
=
ˆ0
i00sincos0000cossin00sincos
2
1
2
1
1
x
x
x
y
x
f
f
ffff
θθθθ
θθθθ
0cossin002 yf θθ
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
This leads to the following system of equations
000ˆcos ++−= ff θ
0ˆcos00
000ˆsin
000cos
22
11
11
−++=
+++=
++=
xx
xy
xx
ff
ff
ff
θ
θ
θ
solution leads to (show for homework)
0ˆsin00 22 +++= xy ff θ
( )
=
x
y
x
x
x
fff
f
f
2
1
1
2
1
sincos0000sincos
ˆ
ˆ
θθθθ
y
xf
f2
2
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
A similar relationship holds for nodal displacements between local and global coordinate systems.
d2y
dd1y d2x
xd2ˆyd2
ˆd1 Global
d1x
dd
dd
xx
ˆˆ
00cossin00sincos 11
−
θθθθ
θxd1
yd1 Global Local
ddd
ddd
y
x
y
y
x
y
ˆˆ
cossin00sincos0000cossin
2
2
1
2
2
1
−=
θθθθ
θθ
[ ] dd ˆ*T= (9)
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Inverting this last matrix expression yields (show for homework):
−
=
x
y
x
x
y
x
ddd
ddd
2
1
1
2
1
1
sincos0000cossin00sincos
ˆˆˆ
θθθθθθ
or
−
yydd 22 cossin00ˆ θθ
where it is obvious that:
[ ] dd Tˆ =
Similarly
[ ] [ ] 1*TT −=
S a y
(9) [ ] ff Tˆ =
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
The element stiffness matrix can now be formulated in terms of the global coordinate system as follows.
df ˆ0101ˆIn local coordinates
Element nodal forces and displacements in local coordinates
−
−
=
x
y
x
x
y
x
ddd
LEA
fff
2
1
1
2
1
1
ˆˆ
010100000101
ˆˆ [ ] df ˆkˆ =→
y
x
y
x
df 2
2
2
2ˆ0000ˆ Element stiffness matrix
in local coordinates (extended)
WithWith
[ ] dd Tˆ = [ ] ff Tˆ =
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
then
[ ] df = ˆkˆ [ ] [ ] [ ][ ]
[ ] [ ][ ] [ ]
df
df
f
=
=− TkT
TkT1
where the element stiffness matrix in terms of global coordinates is
[ ] dkf =
in a matrix format:
[ ] [ ] [ ][ ]TˆT 1 kk −=
[ ] [ ] [ ][ ]
−−−−
−−
== −
θθθθθθθθθθθθθθθθθθ
22
22
22
1
sincoscossincoscossinsincossinsincos
sincoscossincoscos
TˆTL
EAkk
−− θθθθθθθθθθθθ
22 sinsincossinsincossincoscossincoscosL
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
− 0101
In order to load up the component stiffness matrix consider the following simple truss
Node number
2
[ ]
−=
000001010000
1 LEAk
23
2 0000−45
[ ]
−−−−
42
42
42
42
42
42
42
42
EAkP [kN]
11 3
90
L m
[ ]
−−−−
=
42
42
42
42
42
42
42
42
44442 L
k
Elementnumber [ ]
−=
00001010
0000
3 LEAk
L m
[ ]
− 101000003 L
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Create component stiffness matrix
from element stiffness matrices
−
xx
dd
EAff 11
00000101
1,31,1
−=
x
y
x
y
ddd
LEA
fff
3
3
1
3
3
1
000001010000
3,1 3,3 yyf 33 0000
−
y
x
y
x
dd
FF
1
1
1
1
00000101
=
y
x
y
x
ddd
LEA
FFF
2
2
2
2
0101
−
y
x
y
x
dd
FF
3
3
3
3
00000101
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
df 2222
−−−−
−−
=
x
y
x
x
y
x
ddd
LEA
fff
3
2
2
42
42
42
42
42
42
42
42
42
42
42
42
3
2
2 2,32,2
−−
y
x
y
x
dff
3
3
42
42
42
42
4444
3
33,2 3,3
−
y
x
y
x
dd
FF
1
1
1
1
00000101
−−−−
=
y
x
y
y
x
y
ddd
LEA
FFF
2
2
1
222242
42
42
42
42
42
42
42
2
2
1
101
−−−+−−
y
x
y
x
dd
FF
3
3
42
42
42
42
42
42
42
42
3
3
00101
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
df 0000
−=
x
y
x
x
y
x
ddd
LEA
fff
2
1
1
2
1
1
00001010
00002,11,1
−
y
x
y
x
dff
2
2
2
2
1010 2,12,2
−−
y
x
y
x
dd
FF
1
1
1
1
001010010001
−+−−−+−−
−−=
y
x
y
x
ddd
LEA
FFF
2
2
222242
42
42
42
42
42
42
42
2
2
101110
00
−−−+−−
y
x
y
x
dd
FF
3
3
42
42
42
42
4444
3
3
00101
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
−−
y
x
y
x
dd
FF
1
1
1
1
001010010001
+−+−−
−−=
y
x
y
x
ddd
LEA
FFF
2
2
222242
42
42
42
42
42
42
42
2
2
101110
00
−−−+−−
y
x
y
x
dd
FF
3
3
42
42
42
42
42
42
42
42
3
3
00101
vector of nodal displacementsglobal coordinate system
vector of nodal forcesglobal coordinate system [ ] dKF =
Component stiffness matrix
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
• How to deal with the problem of supports (restraints) ?h d h h di l k i h f l i id• These are nodes where the displacements are known, zero in the perfectly rigid
support case.
unknown known
−−
y
x
EAFFF
22221
1
000
00001010010001support
Reactions
known supportDisplacements
−+−−−+−−
−−=
x
y
x
dL
EAFF
342
42
42
42
42
42
42
42
4444
2
2
00
101110
00
0
−−
− ydP 34
242
42
4200
k d lknown applied nodal loads unknown nodal Displacements
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
With the stiffness matrix partitioned as follows
f1 0
=
x
y
x
KKEAfff
12112
1
1
000
−x
y
dd
KKL
P
f
3
3
22212 00
where ydP 3
01100001
−
0001
−
−
=42
4200
0110
11K
−=42
42
00
12K
+−
421
4210
−
42
42
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
−−2201
−+ 221
−
=
42
4200
4401
21KJSK =
−
=
42
42
4422
44
d
then determine nodal displacements first by solving the following system of equations:
[ ]
+
−
=
−
−
01
0
122
1
223
3
L
PK
dd
y
x
−
−
−
−+=
1
01
42
42
42
42
PL
PEAL
−−=
221AE
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Next determine the nodal forces:
F 0010001
−−−
−
PLEAFFF
x
y
x
000
00001010010001
42
42
42
42
2
1
1
−=
−+−−−+−−
=
dd
AEPL
LEA
F
x
y
x
2211
0101
1100 3
222242
42
42
42
42
42
42
42
4444
2
2
−−=
−−
−
P
dP y
0
22100 342
42
42
42
−
=PP
0
− P0
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
In general, with [ ] dd Tˆ =
−
=
y
x
y
x
ddd
dd
1
1
1
1
i0000cossin00sincos
ˆˆˆ
θθθθθθ
−
y
x
y
x
dd
dd
2
2
2
2
cossin00sincos00
ˆ θθθθ
000011d x
The for element #1 in the truss
PL
ˆ
−−−
=
2211
0
100001000010
ˆˆˆ
3
3
1
AEPL
ddd
y
x
y ( )
−−
−=
AEPL
AEd
d
y
x
221ˆ3
3
y
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
( ) dBx ˆ=ε
The strain in element #1 is
( )
( )PLAEPL
LL
dBx
22111
−
−=
ε
( )
PAE
PLLL
22
221
−=
−−
EA=
Negative sign indicates compression. Calculate the stress and the force in element #1
( ) ( )P
xEx
22−=
= εσ ( ) ( )P
AxxF
22−=
= σ
A−=
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
I l lIn class example
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Potential Energy Approach in Deriving Bar Element Equations
The principle of minimum potential energy is now used to derive the bar element equations presented earlier. Recall that
Ω+Uπ
For the internal strain energy component consider the following figure:
Ω+= Upπ
P
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
The differential of strain energy is given by the expression:
( ) ( ) ( )( ) ( ) ( )dVd
dxzydU
xx
xx
εσεσ
=∆∆∆=
For the entire bar:
dVdUV xx
x
εσε
∫ ∫
= ~
( ) dVdEV xx
V
x
εεε
∫ ∫
∫ ∫
=
~~0
0
( ) dVEV
xx
εε
∫
=
2
~
0
2
0
( )( ) dVdVExV xV
xx εσεε∫∫ =
=
21
2
0
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
The potential energy term is associated with externally applied loads and is given by the expression
where the first term represents contributions from any body forces, the second term
∑∫∫=
−−−=ΩM
iixixS xV b dfdSuTdVuX
1
ˆˆˆˆˆˆ
p y y ,represents any contribution from surface tractions, and the third term represents the contribution from concentrated nodal forces.
The total potential energy for a bar element of length L and constant cross sectional area p gy gA becomes:
p U Ω+=π
L
M
iixixS xV bxV x
A
dfdSuTdVuXdV1
ˆˆˆˆˆˆ
ˆˆˆˆˆˆ21
−−−=
∫∫∫
∑∫∫∫=
εσ
xxxxS xV bxx dfdfdSuTdVuXxdA2211
0
ˆˆˆˆˆˆˆˆˆ2
−−−−= ∫∫∫ εσ
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
With( )
APxσ −=
( )[ ] ( )xD
xEA
εε
==
and
[ ] [ ]ED =
then formally in matrix notation
[ ] D εσ = [ ] xx D εσ =
dPdPdSTudVXuxdAxxS x
T
V bT
L
xT
xpˆˆˆˆˆˆˆ
2 210
−−−−= ∫∫∫ εσπ
PddSTudVXuxdA T
S xT
V bT
L
xT
xˆˆˆˆˆˆ
2 0
0
−−−= ∫∫∫ εσ
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
which leads to
[ ] [ ] [ ] ( ) ddBDBdA LTTTT ˆˆˆ ∫ [ ] [ ] [ ] ( )
[ ] [ ] dSTNdudVXNdu
xddBDBdA
S xTTT
V bTTT
xTTT
xp
ˆˆˆˆˆˆ
ˆ2 0
=−
=−
=
==
∫∫
∫ εσπ
PdT
SV
ˆ−
∫∫
or
[ ] [ ] [ ] [ ] dVXNdxddBDBdAV b
TTL
TTT
pˆˆˆˆˆ
2−= ∫∫π
[ ] PddSTNdT
S xTT
Vp
ˆˆˆ
2 0
−− ∫
∫∫
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Integration yields
[ ] [ ] [ ] A LTTT ˆˆ ∫ [ ] [ ] [ ]
[ ] [ ] PdSTNdVXNd
xddBDBdA
S xT
V bTT
TTT
p
ˆˆˆ
ˆ2 0
−−−
=
∫∫
∫π
[ ] [ ] [ ] fddBDBdAL TTTT
SV
ˆˆˆˆ2
−=
∫∫
where
[ ] [ ] PdSTNdVXNf TT ++= ∫∫ ˆˆˆ
now one can definitely see that
[ ] [ ] PdSTNdVXNfS xV b ++= ∫∫
( )( ) ( )d
dd
p
xxpp
ˆ, 21
π
ππ
=
=
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
The minimization of potential energy requires
0ˆ0ˆ21
=
∂
∂=
∂
∂
x
p
x
p
ddππ
With
[ ] [ ] [ ] ˆˆ* TTTdBDBdU [ ] [ ] [ ]
[ ] 121 ˆ
ˆ111
ˆˆ
*
xxx
dELdd
dBDBdU
−
−
=
=
[ ]
( )2221
21
2
21
ˆˆˆ2ˆ
ˆ1x
xx
ddddE
dLLL
+−=
( )22112 2 xxxx ddddL
+
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
and
xxxx
Tfdfdfd 2211ˆˆˆˆˆˆ +=
then
xxxx fff 2211
( ) ( )
( )
xxxxxxxxxx
p
EAL
fdfdddddLEAL
dd 22112
2212
1211
ˆˆˆˆˆˆˆ2ˆ2ˆˆ0
+−+−
∂∂
=∂
∂=
π
( ) xxx fddLEAL
1212ˆˆ2ˆ2
2−
−=
( ) ( )xxxxxxxxxx
p fdfdddddLEAL
dd 22112
2212
1222
ˆˆˆˆˆˆˆ2ˆ2ˆˆ0
+−+−
∂∂
=∂
∂=
π
( ) xxx fddLEAL
2212ˆˆ2ˆ2
2−
+−=
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
These last two equations. can be put into a matrix format as follows
=
−
−
−=
∂
∂00
ˆ
ˆ
ˆ
ˆ
1111
ˆ2
1
2
1
x
x
x
xp
f
f
d
dL
AEdπ
or
−
xx dAEf 11ˆ11ˆ
where
−
=
xx dLf 22ˆ11ˆ
where
[ ]
−
−=
1111ˆ
LAEk
as before.
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
I l lIn class example
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Comparison of a Finite Element Solution to an Exact Solution
It is time to make comparisons. In the previous in-class problem we solved the problem using first one and then two elements. Here we will make comparisons of two, four and eight element models to an exact solution. The exact solution for axial di l t i bt i d f th l ti f th f ll i idisplacements is obtained from the solution of the following expression:
( ) ( )∫ dP1
with
( ) ( )∫= dxxPAE
xu
( ) ( )1021 xxxP
=
25x=
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
then
( ) 251 dxxxu = ∫( )
1
3
35 CAEx
AE
+=
∫
Evaluating this expression at the support yields
( ) 0L( )
1
3
35
0
CAEL
Lxu
+=
==
or
LC 5 3
( )3335x
AELC
35
1 −= ( )33
35 LxAExu −=
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
The following figure plots the exact expression for the displacement along the rod along with the finite element calculations:
The finite element analysis match the exact solutions at the node points. Although the nodal displacements match the exact solution, the displacement values at i di l i b i li di l f iintermediate locations are poor because we are using linear displacement functions within each element. As we increase the number of elements we converge on the exact solution.
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
The exact solution for stress is:
( ) dEduEExP =
===
δεσ
22
5.22
5 xxdx
Edx
EEA
==
=
=== εσ
Since stress is derived from the slope of the displacement, and since the displacement is linear within each element, then the stress is constant in each element. This can be seen in the following figure:
2
seen in the following figure:
The best approximation of the stress occurs at the midpoint of the element, not at the nodes The reason for this is that theat the nodes. The reason for this is that the derivative of displacement is better predicted between the nodes then at the nodes.
Section 4: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
As we saw in the example problem the axial stress is not continuous across element boundaries. Recall that equilibrium is not satisfied within each element, only at the
d A h b f l i h di i i i d d hnodes. As the number of elements increase the discontinuity in stress decreases and the approximation of equilibrium improves.
Finally in the figure below the convergence of the axial stress at the fixed end is y g gdepicted. Again, as the number of elements increase the stress at the fixed end converges to the exact solution from below.