Universidad de San Buenaventura
Engineering faculty
Structural Analysis
Workshop 2
Reference:
Finite Element Analysis (FEA)
Final work resume:
Example Exercise and homework exercise
Objective:
To determinate the load and of each member for a truss and its
deformation by using analytical methods based on Finite Element
Analysis Theory.
N-a
1
Tools:
Group Quantities allowed
Example Exercise:
The truss shown below has members with a Young Modulus of 200E9_Pa and a cross section area
of 5E-4_m2, if the structure is subjected to the loads as shown; determine the load on each
member and the truss deformation (motion due to forces of each joint).
Assumptions:
•
•
•
The body force is neglected
The friction on the joints are neglected
Cross section area is not defined
Exercise developed by Jorge Eliécer Gaitán Aroca
Universidad de San Buenaventura
Engineering faculty
Structural Analysis
Workshop 2
Step 1
Give a number for each node (joint) of the truss and determine the location (x,y) for each node by
using a table with respect to a coordinate axis you choose, the coordinate axis may be located on
whatever node.
Position of the nodes
Node
X
Y
1
0
0
2
10
0
3
10
15
4
0
15
Nodes 1 and 4 has no displacement allowed, these are fixed supports, the node 2 has only free
motion on the “y” axis.
Step 2
Give a number or a letter for each bar and give the connectivity on a table, to do this, use the
nodes previously named, you have to take into account the order on which you give the
connections of the bars with respect to the nodes.
Exercise developed by Jorge Eliécer Gaitán Aroca
Universidad de San Buenaventura
Engineering faculty
Structural Analysis
Workshop 2
Bar No
1
2
3
4
Connectivity
Node a
Node b
1
2
2
3
1
3
3
4
Step 3
Find the director cosines for each element (bar) using the following formulas:
௫ ି௫
௟
௬ ି௬
௟
Use the table on step 2 and take into account that nodes on column 1 has coordinates (x1,y1), and
nodes places on column 2 has coordinates (x2,y2) for each bar, then the directors cosines for each
bar are as follows:
Element (bar)
1
2
3
4
c
1
0
0,55
-1
s
0
1
0,83
0
The element 3 has connectivity with nodes 1 and 3, for this case nodea=1 and nodeb=3 the
directors’ cosines are:
ଶ ଵ 10 0
0,55
௘௟௘௠௘௡௧
18
Exercise developed by Jorge Eliécer Gaitán Aroca
Universidad de San Buenaventura
Engineering faculty
Structural Analysis
Workshop 2
‫=ݏ‬
‫ݕ‬ଶ − ‫ݕ‬ଵ 15 − 0
=
= 0,83
݈௘௟௘௠௘௡௧
18
Step 4
Give for each node 2 possible motion on x and y axis, and give a name for each motion, for this
case give odds number on horizontal motions and even numbers on vertical motions. By doing this
we accept that all nodes has at least two freedom degree on the plane.
Thus:
In this case there are some motions and some nodes that are equal to zero, that is on this node
there is a fixed support, there are:
Node
U1
U2
U4
U7
U8
Value
0
0
0
0
0
Exercise developed by Jorge Eliécer Gaitán Aroca
Universidad de San Buenaventura
Engineering faculty
Structural Analysis
Workshop 2
For each element it is necessary to find the truss element stiffness matrix, in general this matrix is
as follows:
∗
c2
cs
-c2
-cs
cs
s2
-cs
-s2
-c2
-cs
c2
cs
-cs
-s2
cs
s2
Where A is the cross section of the bar (element), E is the young modulus for the element, L is the
length of the element and the subscript “e” is the element number.
Then for element 1 there is:
2009_
∗ 5 4_
10_
Node a
U1 U2
1
0
0
0
-1
0
0
0
Node b
U3 U4
-1
0
0
0
1
0
0
0
U1
U2
U3
U4
Here there is an association (connectivity) of the element 1 with the nodes 1 and 2 and then with
its motions U1, U2, U3 and U4, they are placed on the matrix to give a name for each row and
column.
On each matrix it is necessary to mark the rows and columns on which the motions are equal to
zero, in this case for element 1 this is marked with a gray color.
The other matrixes are:
2009_
∗ 5 4_
15_
Node a
U3 U4
0
0
0
1
0
0
0
-1
Node b
U5 U6
0
0
0
-1
0
0
0
1
U3
U4
U5
U6
Exercise developed by Jorge Eliécer Gaitán Aroca
Universidad de San Buenaventura
Engineering faculty
Structural Analysis
Workshop 2
2009_
∗ 5 4_
18_
2009_
Node a
U1
U2
0,302
0,425
0,424
0,68
-0,302 -0,425
-0,425
-0,68
Node a
U5 U6
1
0
0
0
-1
0
0
0
∗ 5 4_
10_
Node b
U5
U6
-0,302 -0,425
-0,425
-0,68
0,302
0,425
0,425
0,68
Node b
U7 U8
-1
0
0
0
1
0
0
0
U1
U2
U5
U6
U5
U6
U7
U8
Step 5
Generate the general truss element stiffness matrix.
As some motions are equal to zero the general matrix is generated by using the remaining motions
U3, U5 and U6.
2009_ ∗ 5 4_
10 ∗ 15 ∗ 18_
U3
270
0
0
U5
U6
0
0
315,3 63,75
63,75 282
U3
U5
U6
Here the common denominator for the 4 matrices is (10*15*18)_m.
To find the value of the U3,U3 there is necessary to find the same location (U3,U3) on all matrices.
In the matrix K1 has U3,U3, the value of the box is “1”, the denominator is 10_m, then the
common denominator (10*15*18) is divided into 10 and this value multiplied with the value of the
box “1”, then,
U3,U3 = 1*15*18 = 270
If there is more than 1 value on all matrices, then the value of each one is added:
Exercise developed by Jorge Eliécer Gaitán Aroca
Universidad de San Buenaventura
Engineering faculty
Structural Analysis
Workshop 2
For example U5,U5 is places on matrices K2, K3 and K4, for matrix K2 the value is zero, then no
value is extracted from matrix K2, for matrices K2, K3 and K4 it is:
For matrix K2 the denominator is 15 and the value of the box is 0, then:
0*10*18 = 0
For K3 matrix the denominator is 18 and the value of the box is 0,302, then:
0,302*10*15 = 45,3
For K4 matrix the denominator is 10 and the value of the box is 1, then:
1*15*18 = 270
Then the total value of U5,U5 is:
U5,U5 = 0 + 45,3 + 270 = 315,3
Step 6
Determine the forces matrix and solve the structural equilibrium equation and find the motions of
each node.
The forces matrix is as follows according to nodes are:
F=
5000
0
-10000
U3
U5
U6
As the structural equilibrium equation is given by:
[K]*[U] = [F]
Where [k] is the truss element stiffness matrix, [F] is the forces matrix and [U] is the motion
matrix, it is:
5000
0
-10000
[F]
2009_ ∗ 5 4_
10 ∗ 15 ∗ 18_
=
[K]
270
0
0
0
0
315,3 63,75
63,75 282
*
*
U3
U5
U6
[U]
Exercise developed by Jorge Eliécer Gaitán Aroca
Universidad de San Buenaventura
Engineering faculty
Structural Analysis
Workshop 2
Then:
U3
U5 =
U6
[U] =
U3
U5 =
U6
_∗ _
∗ ∗ _
*
270
0
0
[K]-1
0
0
315,3 63,75
63,75 282
-1
*
*
5000
0
-10000
[F]
5000
0,0000001
0
0
*
0
0 8,97343E-08 -2,02857E-08
-10000
0 -2,02857E-08 1,00331E-07
U3
U5 =
U6
0,0005
0,00020286
-0,00100331
Step 7
Find the stress on each element
The equation for the stress on each member is related with the displacements on each node that
has connectivity with such element, thus.
, , , ∗ ∗
Where [Ue] is the matrix of the motions (displacements) related with such element, thus for
element 1 the stress is:
1, 0, 1, 0 ∗
0
0
0,0005
0
2009_
10_
Exercise developed by Jorge Eliécer Gaitán Aroca
Universidad de San Buenaventura
Engineering faculty
Structural Analysis
Workshop 2
_
For elements 2, 3 and 4 the stress is:
0, 1, 0,1 ∗
0,0005
0
0,00020286
-0,00100331
2009_
15_
-13,377_Mpa
0.55 , 0.83 , 0.55 , 0.83 ∗
0
0
0,00020286
-0,00100331
2009_
18_
0,00020286
-0,00100331
0
0
2009_
10_
- 8,013_Mpa
1 , 0 , 1 , 0 ∗
4,057_Mpa
Element
1
2
3
4
Stress [Mpa]
10
-13,377
- 8,013
4,057
Exercise developed by Jorge Eliécer Gaitán Aroca
Universidad de San Buenaventura
Engineering faculty
Structural Analysis
Workshop 2
Step 8
Plot results
The following is a scheme of the displacements and stress of each node and element.
Exercise developed by Jorge Eliécer Gaitán Aroca