FEA Project 1- Akash Marakani

EML 5526 Finite Element Analysis (Spring, 2015) March 7, 2015
1
Finite Element Analysis and Design of a Plane Truss
Akash Marakani (University of Florida)
Abstract
Finite element analysis for a 6 bay truss model is performed using a FEA software package called ABAQUS. This
software is further used to compute the deflections and the elements forces for the three loading conditions i.e. Tension,
Shear and Bending. The results obtained are used to compute the equivalent section properties and hence verified to
that of the cantilever beam. Finally, a Fully Stressed Design (FSD) analysis is performed under three loading conditions
(tension, shear and bending) to minimize the total weight of the structure. Each element is constrained to have a
minimum gage area of 0.1𝑐𝑚2
, and the maximum allowable stress in each element after considering safety factor is
100 MPa (Mega Pascal).
1. Introduction
Part 1: A Finite element program was used to determine
the deflections and element forces for multiple loading
conditions for a 6 bay type truss frame structure.
Assuming that the truss behaves like a cantilever beam,
the equivalent cross sectional properties of the beam was
computed by substituting the average tip deflections
obtained after FE analysis. Further to verify the FE model,
two additional truss bays are added to the previously
analyzed truss structure. The same load cases were
applied to this 8 bay truss structure and the average tip
deflection were calculated in a similar manner. These
deflections are then compared to the expected tip
displacements calculated from the beam theory using the
section properties which were initially calculated as the
material properties remain the same for previously
analyzed truss as well as the new truss structure with two
more additional bays.
Part 2: For a structure under tension, shear and bending
loading conditions, the method of fully stressed design
proportions the members of the structure such that the
stress in each member is equal to the allowable stress of
100 MPa in at least one loading condition. If analysis
shows that a certain member is overstressed in a critical
loading condition, the fully stressed design analysis
method is performed. For best design, each member of
the structure that is not at its minimum gage area is fully
stressed under at least one of the design load conditions.
This basic concept implies that we should remove
material from members that are not fully stressed unless
prevented by minimum gage constraints. The FSD
technique is usually complemented by a resizing
algorithm based on the assumption that the load
distribution in the structure is independent of the member
sizes. That is, the stress in each member is calculated,
and then the member is resized to bring the stresses to
their allowable stress level assuming that the load carried
by members remained constant. Stress ratio resizing
technique:
Where,
𝐴 𝑛𝑒𝑤, is the resized area of the element member,
𝐴 𝑜𝑙𝑑, is the initial area of the element member,
σ = is the stress in that member,
𝜎 𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 = 100 MPa.
The advantage of using the FSD analysis is that it helps
in reducing the overall weight of the truss structure and
also reduces the final cost and make the structure more
economical.
2. Approach
2.1 Part 1: 6 Bay Truss Analysis
Fig 1. 6 Bay truss model
In abaqus, the FE model analysis is done for the part as
shown in Fig 1. The 6 bay truss model consists of 31
elements and 14 nodes. The finite element analysis is
done for the structure under three loading conditions

2
Where,
𝐹𝑥13= Force applied at node 13 in x-direction
𝐹𝑥14= Force applied at node 14 in x-direction
𝐹𝑦13= Force applied at node 13 in y-direction
𝐹𝑦14= Force applied at node 14 in y-direction
The horizontal and the vertical members have length𝑙,
while the inclined members have length√2𝑙. Assume the
young’s modulus (𝐸) =100 GPa, Density ρ =7,830 kg/𝑚3
,
cross sectional area A= 1.0𝑐𝑚2
, and 𝑙=0.3m.
For Load Case A, the boundary conditions at node 2 is
fixed in all directions and at node 1 only the x-direction is
constrained but it is free to move in the y direction due to
roller support. Loads are applied at the 13th and the 14th
node in x-direction with a magnitude of 10,000N
(Tension).
For Load Case B, the boundary conditions are the same
as that of the Case A and the loads are applied at the 13th
and 14th node in y-direction with a magnitude of 1,000N
which results in the shearing effect of the truss structure.
For Load Case C, Similar BC’s are applied to the previous
case but the loads are acting in opposite direction at the
13th and the 14th node which results in a couple and
therefore bending of the truss structure.
Further a FE program is used to determine the deflections
and element force in each truss element. Deflections and
element force at each node are tabulated.
Assuming that the truss behaves like a cantilever beam,
the obtained average tip deflections are used to compute
the equivalent cross sectional properties of the beam i.e.
axial rigidity ( 𝐸𝐴 )eq, flexural rigidity ( 𝐸𝐼 )eq and shear
rigidity (𝐺𝐴)eq using the given formulae.
The deflection of a beam due to an axial force F is given
by:
The transverse deflection due to a transverse force F at
the tip is:
The transverse deflection due to an end couple C is
given by:
Where,
𝑙= Length of the beam (6x0.3 = 1.8 m)
To verify the FE model, two additional bays are added to
the 6 bay truss structure. The same load cases A, B, C
are applied and the average tip deflections were
calculated using an FE program as done before for a 6
bay structure. These deflections are then compared to
the expected tip displacements calculated from the beam
theory formulation using the axial rigidity, flexural rigidity
and shear rigidity calculated in the previous step (where
𝑙=2.4m). Finally, the FE results are compared with the
theoretical results calculated.
2.2 Part 2: FSD Optimization
Fully stressed design is performed on the original 6 bay
structure under three loading conditions (tension, shear
and bending) to find the best design by reducing the
weight and the overall cost of the truss structure. This
basic concept implies that we should remove the material
from the members that are not fully stressed unless
prevented by a minimum gage area of 0.1𝑐𝑚2
.
After performing the analysis for all the three load cases
separately we will have 3 stresses for each member. I
designed the cross sectional area based upon the highest
stress using the resizing algorithm. If the new area Anew
obtained from the resizing formula is smaller than the
minimum gage area (0.1𝑐𝑚2
), the minimum gage area
should be selected rather than the obtained value.
To ensure that the member elements are fully stressed, I
have used Microsoft Excel to perform several iterations
to reduce the area of the member element and increase
the stress value up to the allowable stress of 100MPa.
Therefore, after 10 iterations I could increase the stress
in each member element by reducing the area and
converging the area to the minimum gage area of 0.1𝑐𝑚2
.
After I did this a few times, I found that all the element
stresses are ≤100MPa in any one of the load conditions
and found the weight to be lesser than the initial structure.
Since the truss structure is fully stressed and the final
weight is less we can conclude that the structure is
optimized.

3
3. Results
3.1 Part 1: 6 Bay Truss Analysis
The deformed structure in the three load conditions are
obtained from Abaqus as follows
Load Case A (Tension)
Fig 2. 6 bay truss under tensile loading
Load Case B (Shear)
Fig 3. 6 bay truss under shear loading
Load Case C (Bending)
Fig 4. 6 bay truss under bending
Table 1. Stress (in Pascals) and Force (in Newtons) at
each element for Tension, Shear and Bending Cases.
Element Stress (Pa) Force (N) Stress (Pa) Force (N) Stress (Pa) Force (N)
1 8.10E+07 8.10E+03 1.11E+08 1.11E+04 1.00E+08 1.00E+04
2 8.30E+07 8.30E+03 8.99E+07 8.99E+03 1.00E+08 1.00E+04
3 8.28E+07 8.28E+03 7.00E+07 7.00E+03 1.00E+08 1.00E+04
4 8.28E+07 8.28E+03 5.00E+07 5.00E+03 1.00E+08 1.00E+04
5 8.30E+07 8.30E+03 3.00E+07 3.00E+03 1.00E+08 1.00E+04
6 8.10E+07 8.10E+03 1.00E+07 1.00E+03 1.00E+08 1.00E+04
7 8.10E+07 8.10E+03 -1.09E+08 -1.09E+04 -1.00E+08 -1.00E+04
8 8.30E+07 8.30E+03 -9.01E+07 -9.01E+03 -1.00E+08 -1.00E+04
9 8.28E+07 8.28E+03 -7.00E+07 -7.00E+03 -1.00E+08 -1.00E+04
10 8.28E+07 8.28E+03 -5.00E+07 -5.00E+03 -1.00E+08 -1.00E+04
11 8.30E+07 8.30E+03 -3.00E+07 -3.00E+03 -1.00E+08 -1.00E+04
12 8.10E+07 8.10E+03 -1.00E+07 -1.00E+03 -1.00E+08 -1.00E+04
13 -1.90E+07 -1.90E+03 -8.95E+06 -8.95E+02 -2.80E-07 -2.80E-11
14 -3.59E+07 -3.59E+03 9.37E+05 9.37E+01 -4.34E-08 -4.34E-12
15 -3.41E+07 -3.41E+03 -9.81E+04 -9.81E+00 0.00E+00 0.00E+00
16 -3.44E+07 -3.44E+03 1.03E+04 1.03E+00 -1.73E-07 -1.73E-11
17 -3.41E+07 -3.41E+03 -1075 -1.08E-01 0.00E+00 0.00E+00
18 -3.59E+07 -3.59E+03 112.7 1.13E-02 3.47E-07 3.47E-11
19 -1.90E+07 -1.90E+03 -13.02 -1.30E-03 -6.94E-07 -6.94E-11
20 2.68E+07 2.68E+03 1.27E+07 1.27E+03 3.90E-07 3.90E-11
21 2.40E+07 2.40E+03 1.43E+07 1.43E+03 5.20E-07 5.20E-11
22 2.43E+07 2.43E+03 1.41E+07 1.41E+03 5.20E-07 5.20E-11
23 2.43E+07 2.43E+03 1.41E+07 1.41E+03 6.94E-07 6.94E-11
24 2.40E+07 2.40E+03 1.41E+07 1.41E+03 3.47E-07 3.47E-11
25 2.68E+07 2.68E+03 1.41E+07 1.41E+03 -3.47E-07 -3.47E-11
26 2.68E+07 2.68E+03 -1.56E+07 -1.56E+03 -3.69E-07 -3.69E-11
27 2.40E+07 2.40E+03 -1.40E+07 -1.40E+03 -4.34E-07 -4.34E-11
28 2.43E+07 2.43E+03 -1.42E+07 -1.42E+03 -4.34E-07 -4.34E-11
29 2.43E+07 2.43E+03 -1.41E+07 -1.41E+03 -6.94E-07 -6.94E-11
30 2.40E+07 2.40E+03 -1.41E+07 -1.41E+03 -1.73E-07 -1.73E-11
31 2.68E+07 2.68E+03 -1.41E+07 -1.41E+03 0.00E+00 0.00E+00
LoadCase A LoadCase B LoadCase C

4
Table 2: Displacements at each node for the 3 load cases
(All the displacements are in metres in Table 2.)
Section Properties Calculations are as follows:
From FEA results in Table 2, the average tip deflection
for load case A is:
=
1.48 ×10−3+1.48 ×10−3
2
= 1.48 × 10−3
𝑚
Computing the axial rigidity (𝐸𝐴) 𝑒𝑞 from the above
obtained deflection:
(𝐸𝐴) 𝑒𝑞=
(20000)(1.8)
1.48 ×10−3 = 2.43 × 107
𝑁
The average tip deflection for load case C is:
=
1.08 ×10−2+ 1.08 ×10−2
2
= 1.08 × 10−2
𝑚
The equivalent flexural rigidity (𝐸𝐼) 𝑒𝑞 using the above
calculated deflection:
(𝐸𝐼) 𝑒𝑞 =
𝐶𝑙2
2𝑣 𝑡𝑖𝑝
=
0.3×10000×1.8
2×1.08×10−2 = 4.50 × 105
𝑁. 𝑚2
Where, 𝐶 = (10,000)(0.3) 𝑁. 𝑚
The average tip deflection for load case B is:
=
9.1025 ×10−3+ 9.1025 ×10−3
2
= 9.1025 × 10−3
𝑚
The equivalent shear rigidity (𝐺𝐴) 𝑒𝑞 is calculated using
the above tip deflection and the previously
calculated (𝐸𝐼) 𝑒𝑞:
Where, 𝐹 = 2000 𝑁
𝑙 = 1.8 𝑚
(𝐺𝐴) 𝑒𝑞 = 7.78 × 106
𝑁
Therefore the section properties are as follows
(𝐸𝐴) 𝑒𝑞= 2.43 × 107
𝑁
(𝐸𝐼) 𝑒𝑞 = 4.50 × 105
𝑁. 𝑚2
(𝐺𝐴) 𝑒𝑞 = 7.78 × 106
𝑁
To verify the FE model, two more additional truss bays
are added to the 6 bay structure, the same load cases A
through C were applied and the average tip deflection
were calculated. These deflections are then compared to
the expected tip displacements from the beam theory
using the axial rigidity, flexural rigidity and shear rigidity
which were evaluated in the previous step for 6 bay truss
Node U1 U2
1 0 5.69E-05
2 0 0
3 2.43E-04 8.23E-05
4 2.43E-04 -2.55E-05
5 4.92E-04 7.96E-05
6 4.92E-04 -2.28E-05
7 7.41E-04 8.00E-05
8 7.41E-04 -2.31E-05
9 9.89E-04 7.96E-05
10 9.89E-04 -2.28E-05
11 1.24E-03 8.23E-05
12 1.24E-03 -2.55E-05
13 1.48E-03 5.69E-05
14 1.48E-03 3.43E-12
Load Case A
Node U1 U2
1 0.000 2.6859E-05
2 0 0
3 3.33E-04 4.27E-04
4 -3.27E-04 4.30E-04
5 6.03E-04 1.44E-03
6 -5.97E-04 1.44E-03
7 8.13E-04 2.94E-03
8 -8.07E-04 2.94E-03
9 9.63E-04 4.79E-03
10 -9.57E-04 4.79E-03
11 1.05E-03 6.89E-03
12 -1.05E-03 6.89E-03
13 1.08E-03 9.10E-03
14 -1.08E-03 9.10E-03
Load Case B
Node U1 U2
1 0 0
2 0 0
3 3.00E-04 3.00E-04
4 -3.00E-04 3.00E-04
5 6.00E-04 1.20E-03
6 -6.00E-04 1.20E-03
7 9.00E-04 2.70E-03
8 -9.00E-04 2.70E-03
9 1.20E-03 4.80E-03
10 -1.20E-03 4.80E-03
11 1.50E-03 7.50E-03
12 -1.50E-03 7.50E-03
13 1.80E-03 1.08E-02
14 -1.80E-03 1.08E-02
Load Case C

5
structure (where 𝑙 = 2.4 𝑚). The comparison of the FEA
results and beam theory calculations are tabulated below.
Table 3: Comparisons of tip displacements (in metres)
from FEA results and Beam Theory formulation for 8 bay
truss structure
Therefore, we can conclude that the FE model accurately
predicts the tip deflection for the 8 bay truss element.
3.2 Part 2: 6 bay Truss Optimization
An optimized truss structure was designed after
performing 10 iteration on Microsoft Excel to reduce the
area of each element and to ensure that the stress in
each element is fully stressed and has a maximum value
of 100MPa (≤100 MPa). The final obtained stress values
for the optimized case have been shown in Table 4.
Table 4: Element Stresses (in Pascals) for the final
iteration.
The most critical load case for each element and the
corresponding areas have been tabulated in Table 5.
Element Load Case A Load Case B Load Case C
1 8.61E+07 1.01E+08 8.84E+07
2 9.76E+07 8.85E+07 1.00E+08
3 9.76E+07 7.03E+07 1.00E+08
4 9.75E+07 4.97E+07 1.00E+08
5 9.76E+07 3.00E+07 1.00E+08
6 9.72E+07 9.68E+06 1.00E+08
7 9.14E+07 -9.94E+07 -9.36E+07
8 9.76E+07 -9.15E+07 -1.00E+08
9 9.76E+07 -6.97E+07 -1.00E+08
10 9.76E+07 -5.03E+07 -1.00E+08
11 9.76E+07 -3.00E+07 -1.00E+08
12 9.72E+07 -1.03E+07 -1.00E+08
13 -2.43E+07 -6.07E+07 8.56E+05
14 -4.84E+07 2.40E+07 7.12E+05
15 -4.86E+07 -1.19E+07 -1.19E+05
16 -4.89E+07 7.64E+05 2.01E+04
17 -4.82E+07 -2.44E+06 -2329
18 -5.22E+07 -3.00E+06 1388
19 -2.84E+07 -3.17E+06 -267
20 3.43E+07 8.59E+07 -1.21E+06
21 2.13E+07 1.02E+08 1.27E+05
22 2.46E+07 9.72E+07 -2.42E+04
23 2.37E+07 9.93E+07 3855
24 2.34E+07 9.85E+07 -1633
25 2.70E+07 9.81E+07 253.9
26 1.81E+07 -1.04E+08 -6.37E+05
27 2.78E+07 -9.75E+07 1.65E+05
28 2.42E+07 -1.02E+08 -2.38E+04
29 2.53E+07 -1.00E+08 4107
30 2.40E+07 -1.01E+08 -1672
31 2.98E+07 -1.02E+08 279.9
FE Model Beam Theory
- -
- -
- -
Load Case A
Load Case B
Load Case C
1.97 × 10−3
1.97 × 10−3
2.1 × 10−2
2.1 × 10−2
1.92 × 10−2 1.92 × 10−2

6
Table 5: Element stress (in Pascals), Critical load case
and element areas (m2).
Weight of the truss = ∑ (volume of all elements) * ρ
𝑉𝑜𝑙 𝑚𝑒 = 𝐴𝑟𝑒𝑎 × 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡
𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 = 0.3 𝑚
𝐷𝑒𝑛𝑠𝑖𝑡𝑦 (𝜌) = 7,830
𝑘𝑔
𝑚3
Final Weight of the Truss = 3.43 𝒌𝒈
4. Conclusions
After the FEA analysis, the tip displacements for the 6 bay
plane truss structure were computed which were further
used to find out the sectional properties using the beam
theory formulation. Hence, we can conclude that the truss
structure behaves like a cantilever beam. Moreover,
using the Fully Stressed Design (FSD) technique
unwanted material was removed from the frame structure
which helped in reducing the overall weight of the truss
structure which helped us in coming with the most
optimized design for the structure. The final weight of the
6 bay truss is computed to be 3.43 𝒌𝒈𝒔.
5. Appendix
The input file for 6 bay truss element has been elaborated
as follows;
Firstly, the nodes are defined in the similar manner. Node
1 location is specified at (0, 0).
*Node
1, 0, 0.
Next, the element type is specified along with the element
connectivity table which specifies which two nodes are
connected by the element. Example, Element 1 connects
node 1 and node 3.
*Element, type=T2D2
1, 1, 3
*Nset, nset=SET-1, generate
1, 14, 1 (Specifies the total number of nodes)
*Elset, elset=SET-1, generate
1, 31, 1 (Specifies the total number of elements)
Next, after the assembly of the part the cross sectional
area and material properties are defined and assigned to
the respective element.
** Section: Section-1-SET-1
*Solid Section, elset=SET-1, material=MATERIAL-1
0.0001,
** MATERIALS
**
*Material, name=MATERIAL-1
*Density
7830.,
*Elastic
Element Max StressCritical Case Area (m^2)
1 1E+08 B 0.000113924
2 1E+08 C 9.99755E-05
3 1E+08 C 0.000100001
4 1E+08 C 9.99991E-05
5 1E+08 C 9.99994E-05
6 1E+08 C 1E-04
7 9.9E+07 B 0.000106071
8 1E+08 C 0.00010001
9 1E+08 C 9.99936E-05
10 1E+08 C 9.99999E-05
11 1E+08 C 9.99951E-05
12 1E+08 C 0.0001
13 6.1E+07 B 0.00001
14 4.8E+07 A 0.00001
15 4.9E+07 A 0.00001
16 4.9E+07 A 0.00001
17 4.8E+07 A 0.00001
18 5.2E+07 A 0.00001
19 2.8E+07 A 0.00001
20 8.6E+07 B 0.00001
21 1E+08 B 1.63069E-05
22 9.7E+07 B 1.36647E-05
23 9.9E+07 B 1.4511E-05
24 9.8E+07 B 1.41179E-05
25 9.8E+07 B 1.45904E-05
26 1E+08 B 1.96951E-05
27 9.7E+07 B 1.19782E-05
28 1E+08 B 1.46198E-05
29 1E+08 B 1.37731E-05
30 1E+08 B 1.41935E-05
31 1E+08 B 1.36934E-05

7
1e+11, 0.
Next, the boundary conditions and the loads are assigned
** BOUNDARY CONDITIONS
**
** Name: Disp-BC-1
Type: Symmetry/Antisymmetry/Encastre
*Boundary
SET-3, ENCASTRE
** Name: Disp-BC-2 Type: Displacement/Rotation
*Boundary
SET-4, 1, 1
** ----------------------------------------------------------------
**
** STEP: Step-1
**
*Step, name=Step-1, nlgeom=NO, perturbation
*Static
**
** LOADS
**
** Name: CFORCE-1 Type: Concentrated force
*Cload
SET-5, 1, 10000.
** Name: CFORCE-3 Type: Concentrated force
*Cload
SET-6, 1, 10000.
**
Next, the stress value and the nodal displacements which
are needed for the calculation are printed in the output
requests section
** OUTPUT REQUESTS
**
**
** FIELD OUTPUT: F-Output-1
*EL PRINT
S
*NODE PRINT
U
**
*Output, field, variable=PRESELECT
**
** HISTORY OUTPUT: H-Output-1
**
*Output, history, variable=PRESELECT
*End Step
In the similar manner another step is created for the shear
and bending case.
To do the analysis for an 8 bay truss element the input
file is edited according to our requirement to perform the
analysis.
For an FSD analysis for a 6 bay truss element, the steps
are almost the same just that each element needs to be
assigned a different cross sectional property and each
element needs to be defined separately. As the cross
sectional area needs to be changed at each and every
iteration.
For Example;
Firstly, element number 12 has been defined
*Elset, elset=SET-12
12,
Next, a cross sectional area of 0.0001 m2 is assigned to
that respective element.
** Section: Section-12-SET-12
*Solid Section, elset=SET-12, material=MATERIAL-1
0.0001,
This needs to be followed for 31 elements in case of FSD
analysis and each area needs to be changed in every
iteration and analysis needs to be performed.

FEA Project 1- Akash Marakani

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