1. Finite Element Analysis Project 1: Design of a Plane Truss
Ryland Ballingham
Sorting #5
Summary
For this project, a plane truss is
analyzed to determine the element
stress and nodal deflections given
the loading conditions.The
following figure represents the
truss with its assigned elements
and nodes. The circled numbers
represent the node numbers and
the othernumbers represent the
element numbers. The horizontal
and vertical truss members have a
length of a=0.3 m and have a cross-sectionalarea of 1𝑐𝑚2
. The diagonal members have a length of √2 ∗ 𝑎 and have
a cross-sectionalarea of 0.25𝑐𝑚2
. All members of the truss are constructed with material with a Young’s modulus
of 100 GPa. After this is complete, two more bays are added and re-analyzed to see how the trusses properties
change.
Introduction
In order to execute this task, a model was created with the exact specifications as the above figure. Once the
model was created with the correct material, section properties, dimensions, boundary-conditions and loads,a job
was created in order to analyze and see the nodal displacements and element forces. Before we can analyze the truss,
the input file has to be modified in order to have the correct node and element numbers. Once the input model is
fixed, we re-run the job with the new input file. In order to see the correct nodal displacements and element forces,
we have to make a field output request that request this information. Once this is completed for case A, the process
is repeated for loading cases B and C.
Phase 1: Loading Cases
The table below shows the different loading values for each loading case. The x and y subscripts represent the
loading direction and the subscript number represents the node at which the load is being applied.
Table 1: Loading cases
Load Case A Load Case B Load Case C
𝐹𝑥13 = 10,000 𝑁 𝐹𝑦13 = 10,000 𝑁 𝐹𝑥13 = 10,000 𝑁
𝐹𝑥14 = 10,000 𝑁 𝐹𝑦14 = 10,000 𝑁 𝐹𝑥14 = −10,000 𝑁
Loading Cases Results
2. Table 2: Elements Stresses/Forces for each loading case
The force values were calculated knowing that the cross-sectionalarea of elements 1-19 is 1 𝑐𝑚2
and elements 20-
31 is 0.25 𝑐𝑚2
. These values were multiplied using excel by the stress values obtained in Abaqus to get the force
values.
Table 3: Load Case A displacements
Node number Magnitude U (m) U1 displacement (m) U2 displacement (m)
1 2.17472E-05 1E-32 2.17472E-05
2 0 1E-32 0
3 0.000280111 0.000278253 3.22118E-05
4 0.00027845 0.000278253 -1.04646E-05
Element
number
Element
Stress
(Case A)
(Pa)
Element
Stress
(Case B)
(Pa)
Element
Stress
(Case C)
(Pa)
Element
Force
(Case C)
(N)
Element
Force
(Case B)
(N)
Element
Force
(Case C)
(N)
1 9.28E+07 1.10E+09 1.00E+08 9275.09 110376 10000
2 9.30E+07 9.00E+08 1.00E+08 9302.36 89985.9 10000
3 9.30E+07 7.00E+08 1.00E+08 9301.33 70000.5 10000
4 9.30E+07 5.00E+08 1.00E+08 9301.33 50000 10000
5 9.30E+07 3.00E+08 1.00E+08 9302.36 30000 10000
6 9.28E+07 1.00E+08 1.00E+08 9275.09 10000 10000
7 9.28E+07 -1.10E+09 -1.00E+08 9275.09 -109624 -10000
8 9.30E+07 -9.00E+08 -1.00E+08 9302.36 -90014.1 -10000
9 9.30E+07 -7.00E+08 -1.00E+08 9301.33 -69999.5 -10000
10 9.30E+07 -5.00E+08 -1.00E+08 9301.33 -50000 -10000
11 9.30E+07 -3.00E+08 -1.00E+08 9302.36 -30000 -10000
12 9.28E+07 -1.00E+08 -1.00E+08 9275.09 -10000 -10000
13 -7.25E+06 -9.62E+07 2.26E-08 -724.905 -9623.92 2.25875E-12
14 -1.42E+07 3.62E+06 0.00E+00 -1422.55 361.94 0
15 -1.40E+07 -1.36E+05 0.00E+00 -1396.31 -13.612 0
16 -1.40E+07 5.12E+03 -1.73E-07 -1397.33 0.511925 -1.73472E-11
17 -1.40E+07 -1.93E+02 3.47E-07 -1396.31 -0.0192527 3.46945E-11
18 -1.42E+07 7.24E+00 -3.47E-07 -1422.55 0.000724101 -3.46945E-11
19 -7.25E+06 -2.83E-01 0.00E+00 -724.905 -2.82552E-05 0
20 4.10E+07 5.44E+08 -1.41E-07 1025.17 13610.275 -3.52365E-12
21 3.95E+07 5.66E+08 -4.34E-08 986.615 14162.15 -1.0842E-12
22 3.95E+07 5.66E+08 8.67E-08 988.065 14141.375 2.16841E-12
23 3.95E+07 5.66E+08 0.00E+00 988.065 14142.175 0
24 3.95E+07 5.66E+08 -5.20E-07 986.615 14142.125 -1.30104E-11
25 4.10E+07 5.66E+08 -2.08E-06 1025.17 14142.125 -5.20418E-11
26 4.10E+07 -5.87E+08 9.76E-08 1025.17 -14674 2.43946E-12
27 3.95E+07 -5.65E+08 -2.17E-08 986.615 -14122.125 -5.421E-13
28 3.95E+07 -5.66E+08 -8.67E-08 988.065 -14142.9 -2.16841E-12
29 3.95E+07 -5.66E+08 8.67E-08 988.065 -14142.1 2.16841E-12
30 3.95E+07 -5.66E+08 6.94E-07 986.615 -14142.125 1.73472E-11
31 4.10E+07 -5.66E+08 2.08E-06 1025.17 -14142.125 5.20418E-11
4. Fig. 1. Deformed displacement plot of load case A.
Fig. 2. Deformed Stress plot of load case A.
5. Fig. 3. Deformed displacement plot of load case B.
Fig. 4. Deformed stress plot of load case B.
6. Fig. 5. Deformed displacement plot of load case C.
Fig. 6. Deformed stress plot of load case C (note bottomof truss is in tension and the top portion is in compression)
Phase 2: Equivalent Properties for 6-bay
If we assume that the truss behaves likes a cantilever beam, cross-sectionalproperties of the truss can be found.
The three properties that we are concerned with include: Axial rigidity (𝐸𝐴 𝑒𝑞 ), flexural rigidity (𝐸𝐼 𝑒𝑞) and shear
rigidity (𝐺𝐴 𝑒𝑞 ). The meaning of these properties can be found in the following table.
Table 5: Equivalent Properties
Axial rigidity (𝐸𝐴 𝑒𝑞 ) A measure of a materials ability to oppose and resist
deformation due to axial forces.
Flexural rigidity (𝐸𝐼 𝑒𝑞) A measure of a materials ability to oppose and resist
deformation due to bending.
Shear rigidity (𝐺𝐴 𝑒𝑞 ) A measure of a materials ability to oppose and resist
deformation due to shearforces.
7. Equivalent Property Results
Load Case A
Using the first loading case, the axial rigidity can be solved due to the axial loading conditions using the following
formula:
𝐸𝐴 𝑒𝑞 =
𝐹𝐿
𝑢 𝑡𝑖𝑝
(1)
Since we know F = 20,000 N (sum of the two forces), L = 1.8 m and the average tip deflection (𝑢 𝑡𝑖𝑝 ) from abaqus =
0.001673 m. Plugging these numbers (1) yields a value of 𝐸𝐴 𝑒𝑞 = 21,518 kN.
Load Case C
Using loading case C, we can solve for the flexural rigidity due to the applied couple at the end of the beam using
the following formula:
𝐸𝐼 𝑒𝑞 =
𝐶𝐿2
2𝑣𝑡𝑖𝑝
(2)
To solve for the applied couple (C), we use the following formula:
𝐶 = 𝐹 × 𝑑
Where F = 10,000 N and d (distance between the applied forces (F)) = 0.3 m. Plugging these numbers in yields a
couple of C=3,000 N-m. The length (L) is the same as loading case A (1.8 m) and the average tip deflection (𝑣𝑡𝑖𝑝 )
from abaqus = 0.010949 m. Plugging these values into (2) yields a value of 𝐸𝐼 𝑒𝑞 = 443,879 N or about 444 kN.
Load Case B
Since we solved for flexural rigidity in load case C, we can solve for the shearrigidity (𝐺𝐴 𝑒𝑞 ) due to the transverse
F using the following formula:
𝐺𝐴 𝑒𝑞 =
𝐹𝐿
𝑣𝑡𝑖𝑝 −
𝐹𝐿3
3𝐸𝐼 𝑒𝑞
(3)
Since F = 20,000 N, L = 1.8 m, 𝐸𝐼 𝑒𝑞 = 443,879 N and 𝑣𝑡𝑖𝑝 from abaqus = 0.106856 m we can solve for the shear
rigidity. Plugging these values into (3) yields a value of 𝐺𝐴 𝑒𝑞 = 1,869,300 Pa. or 1.9 MPa
Phase 3: Extended Bay
In this portion of the project, the beam model was verified by adding two more bays. This was accomplished by
modifying the input file by adding more nodes and more elements in the same manner as the 6-bay design. For this
new model, load cases A-C were applied just as before (except the forces were applied at the new, extended tip).
The following table shows the results that were obtained for the displacements. Note: The new nodes follow the
same pattern as the original models nodes.
Extended Bay Results
10. Fig. 7. Extended bay deformed displacements of load case B
Fig. 8. Extended bay deformed displacements of load case C
FEA model and equivalent property comparison
Table 9: 6-bay equivalent properties
Property 6-bay model
Axial rigidity (𝐸𝐴 𝑒𝑞 ) 21,518 kN
Flexural rigidity (𝐸𝐼 𝑒𝑞) 443,879 N
Shear rigidity (𝐺𝐴 𝑒𝑞 ) 1.869 MPa
Table 10: End node displacement comparison
Load Case FEA model method Equivalent Property calculation
Load Case A 0.002229 m 0.002231 m
Load Case B 0.232093 m 0.233302 m
Load Case C 0.0192 m 0.019493 m
The equivalent properties were calculated the same way as in phase 2 of this report. For the 8-bay design,the length
was increased to 2.4 m from 1.8 m and the tip deflections were different as well (these tip deflections were the
average deflections at nodes 17 and 18 instead of at nodes 13 and 14). Table 10 shows that whether doing the
calculations for the tip deflections by hand or by using the FEA model, the results are similar.
Load Case B (6-bay) Fully-Stressed Design Results
Given that the allowable stress is 600 MPa and that we know the old stress/area values for load case B, a new
minimum cross-sectionalarea can be calculated using the following formula (4):
𝐴 𝑛𝑒𝑤
𝑒
=
𝜎 𝑜𝑙𝑑
𝑒
𝜎 𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒
𝑒 𝐴 𝑜𝑙𝑑
𝑒
(4)
11. Table 10: Load Case B fully stressed design
Element number Old Stress (𝜎 𝑜𝑙𝑑)
(MPa)
Old Area (𝐴 𝑜𝑙𝑑 )( 𝑚2
) New Stress
(𝜎 𝑛𝑒𝑤) (MPa)
New Area
(𝐴 𝑛𝑒𝑤 )( 𝑐𝑚2
)
1 1103.76 0.0001 602.974 1.8396
2 899.859 0.0001 596.796 1.499765
3 700.005 0.0001 601.607 1.166675
4 500 0.0001 598.716 0.833333333
5 300 0.0001 601.174 0.5
6 100 0.0001 598.507 0.166666667
7 -1096.24 0.0001 -597.005 1.827066667
8 -900.141 0.0001 -603.203 1.500235
9 -699.995 0.0001 -418.893 1.166658333
10 -500 0.0001 -601.308 0.833333333
11 -300 0.0001 -598.826 0.5
12 -100 0.0001 -601.493 0.166666667
13 -96.2392 0.0001 -565.894 0.160398667
14 3.6194 0.0001 428.446 0.006032333 (0.01)
15 -0.13612 0.0001 -306.708 0.000226867 (0.01)
16 0.00511925 0.0001 79.0079 8.53208E-06 (0.01)
17 -0.000192527 0.0001 -50.2687 3.20878E-07 (0.01)
18 7.24101E-06 0.0001 33.8309 1.20684E-08 (0.01)
19 -2.82552E-07 0.0001 -24.8756 4.7092E-10 (0.01)
20 544.411 0.000025 565.894 0.226837917
21 566.486 0.000025 628.792 0.236035833
22 565.655 0.000025 588.753 0.235689583
23 565.687 0.000025 606.537 0.235702917
24 565.685 0.000025 596.478 0.235702083
25 565.685 0.000025 601.493 0.235702083
26 -586.96 0.000025 -631.633 0.244566667
27 -564.885 0.000025 -571.127 0.23536875
28 -565.716 0.000025 -611.246 0.235715
29 -565.684 0.000025 -593.463 0.235701667
30 -565.685 0.000025 602.974 0.235702083
31 -565.685 0.000025 596.796 0.235702083
Above shows the new stress values (from abaqus)and the new minimum cross-sectionalareas to achieve a fully
stressed design.Element numbers 14, 15, 16,17,18,19 would have to have a minimum cross-sectionalarea of 0.01
𝑐𝑚2
(since any area smaller than 0.01 𝑐𝑚2
cannot be achieved due to manufacturing reasons).
12. Fig. 9. Deformed fully-stressed load case B
Fig. 10. Deformed displacements of fully-stressed load case B
Discussion
It appears that increasing the length of a beam (while keeping everything else constant)increases axial rigidity
and flexural rigidity and decreases shearrigidity. This needs to be taken into consideration when designing
beams/trusses in order to prevent potential failure.
Anotherphenomena that I noticed is that when the bay size is increased, the magnitude of the average nodal
displacements increases as well for all loading cases.
In addition, whether using equivalent properties method or a FEM model, the values for the displacements are
similar. This shows that Abaqus does a great job at conveying correct data when performing analysis.
13. References
[1] EML 4507 Project 1.pdf
[2] Beer, Ferdinand P., and E. Russell Johnston. Mechanics of Materials.New York: McGraw-Hill, 1981. Print.
