FEA project report

1. Background

In the traditional realm of Finite Elements, there are very few practical applications where thefinite code can be compared to direct measurements. At best, Engineers can usually onlycompare their computer analysis with the results predicted by established mechanical formulae. At worst, Engineers must rely on their experience and intuition to guide them towards a workable “right” answer.

The FEA modeling process requires three types of input data: geometry, material properties, and loading. For the bicycle frame, "geometry" means the overall frame dimensions (such as tube lengths, intersection points, and angles) as well as the tubing specifications (diameters, wall thickness, etc.). The purpose of this analysis is to formulate a finite element model of a bicycle frame and compare theoretical results with those revealed by ANSYS. Eventually, redesigning the structure for safety reasons.

2. Geometry and discretization

The bicycle frame shown in figure is constructed from aluminum tubes. The main part of the frame is made of tubes of a minimum outer diameter of 12mm and 2mm wall thickness.

Figure 1: 3D bicycle frame

2.1 2D model

Keypoint x(m) y(m) z(m)1 0 0.325 02 2.00E-02 0.4 03 0.52 0.4 04 0.4 0 05 0.825 0 0

Table 1 : Key points’ coordinates

Connectivity tableElement i j Θ°

1 1 2 75.062 2 3 03 3 4 764 4 1 140.9065 5 4 06 3 5 129

Table 2 : Connectivity table

L1(m) L2(m) L3(m) L4(m) L5(m) L6(m)0.078 0.500 0.418 0.515 0.425 0.503

Table 3 : Length of elements

2.2 3D model

Keypoint x(m) y(m) z(m)1 0 0.325 02 2.00E-02 0.4 03 0.52 0.4 04 0.4 0 05 0.4 0 0.026 0.4 0 -0.027 0.825 0 -0.058 0.825 0 0.05

Table 4 : Keypoints’ coordinates

Connectivity table Element i j Θ°

1 1 2 75.062 2 3 03 3 4 76

4 4 1 140.9065 4 6 06 6 7 07 8 5 08 5 4 09 3 7 129

10 3 8 129Table 5 : Connectivity table

3. Hand calculations

To validate the ANSYS simulations, we will need to calculate the nodal displacements and stresses and compare them to the ANSYS results. For the first loading case, the loads are assumed to be applied on the saddle and on the pedal. These are applied by the rider who sits down on the bike. Hence, the problem becomes dynamic. To overcome this issue, we are going to multiply the loads by a dynamic load factor G=2. We can then consider the problem as static and we can use static point forces on the nodes. For the second loading case, we will consider a force of 1000 N applied on the front dropout of the bike frame.

3.1 Boundary Conditions and applied forces

Given the above assumptions on loading, we can further assume that the boundary conditions will not change over time. We know from the problem description that the left hand support is fixed by a ball joint and that the rear dropouts have sliding conditions as in Figure 2 .

Figure 2: Boundary conditions for vertical loading

3.2 Calculations

3

1

2

3

45

1

2

45

6

1800 N

600 N

U3

U4

U5

U6

Θ

Θ1

x

y

u2v2

u1v1

Figure 3 : Local displacements on element 1 Figure 4 : Global displacements on element 1

Angle Θ1 Θ3 Θ4 Θ6

Degrees 75.068 76.000 140.906 129.000Radians 1.310 1.326 2.459 2.251

Table 6 : Angles of orientation of the elements

Din(m) Dout(m) A(m2) E(Pa) σ(Pa) ρ(kg/m3) I

0.008 0.012 6.28319E-057000000000

0 210000000 2580 8.16814E-10Table 7: Material and geometry parameters of the pipes

5.67E+07 0 0 -5.67E+07 0 0

0 1.47E+06 5.69E+04 0 -1.47E+06 5.69E+040 5.69E+04 2.95E+03 0 -5.69E+04 1.47E+03

-5.67E+07 0 0 5.67E+07 0 00 -1.47E+06 -5.69E+04 0 1.47E+06 -5.69E+040 5.69E+04 1.47E+03 0 -5.69E+04 2.95E+03

0.257672479 0.966232319 0 0 0 0

-0.96623232 0.257672479 0 0 0 00 0 1 0 0 00 0 0 0.257672479 0.966232319 00 0 0 -0.966232319 0.257672479 00 0 0 0 0 1

0.257672479 -0.96623232 0 0 0 0

1

21

1

21

U1U2

X

Y

Θ

[k1]=

[R1]=

0.966232319 0.257672479 0 0 0 00 0 1 0 0 00 0 0 0.25767248 -0.96623232 00 0 0 0.96623232 0.257672479 00 0 0 0 0 1

[K1e ]=[R1]T∗[k1]∗[R1]

Horizontal loading

Boundary Conditions and applied forces

Figure 5 : Boundary conditions and loading

4. ANSYS Simulation 2D

4.1 Vertical loading

1

2

3

45

1

32

4 56

2000 N

[R1]T=

14600653.89 -1417630.836 -55018.25273 -14600653.9 1417630.836 -55018.2527

54750215.09 378050.3354 14672.13351 -54750215.1 -378050.3354 14672.133510 56941.01892 2946.507926 0 -56941.01892 1473.253963

-14600653.89 1417630.836 55018.25273 14600653.89 -1417630.836 55018.25273-54750215.09 -378050.3354 -14672.13351 54750215.09 378050.3354 -14672.1335

0 56941.01892 1473.253963 0 -56941.01892 2946.507926

4.1.1 Units

SI system is the default system of units in ANSYS and is the system used to model the problem. The d

4.1.2 Element Type

Given the 3D nature of the structure it should be clear that a 3D analysis is required. The bicycle frame is made of a number of long beams that would be expected to bend due to the applied loads so a truss analysis would not be appropriate. A 3D pipe16 is the most suitable geometric model here. A 3D solid model would overkill and would not estimate bending stress correctly.

4.1.3 Material Properties

As known, the bicycle frame is constructed from aluminum. We will assume a linear material model for aluminum. This is valid considering the problem, as a bicycle would not be expected to deform significantly or yield under operating conditions. We will, however, take a note of the assumed yield stress and use this to ensure that our linear elastic model is valid and to determine the load that causes yielding of the structure.

4.1.4 Meshing

As a first attempt to get preliminary results, we are going to mesh the elements with 3 subdivisions.

4.1.5 Boundary conditions

Ball-joint boundary conditions are used for the front dropout (location1) and sliding conditions for the rear dropouts (location 5). All the translation movements along z are constrained. Node number 1 is constrained from any translational movement and node number 5 is constrained from translation along the y axis.

4.1.6 Results

Figure 6 shows the deformed shape of the structure due to the applied loading. It is clear from the figure that both the saddle and pedal points are displacing downwards and forwards, while the rear wheel attachment point is moving forwards. This deformation seems reasonable considering the loads and boundary conditions applied. In order to quantify the deformation Figure 7 shows a nodal distribution of displacement vector sum. It can be seen here that the maximum displacement is and occurs at the saddle point and at the front o f the top bar. Again, this level of displacement and its locations seem reasonable based on the applied loads. One would expect the displacement of a bicycle frame when the rider sits on it not to be noticeable to the rider and a deformation of 0.0079 mm would certainly not be noticed by the rider.

Figure 6 : Deformed and un-deformed edge

Figure 7: Nodal displacement sum

Error: Reference source not found shows an elements plot of the distribution of Von-Mises stress in the pipe model. This figure is rather difficult to read, as is normally the case for such pipe models, so in order to aid visualization the stress distribution has been converted into a bar graph and mapped onto the pipe model, as shown in Error: Reference source not found. It can be seen from the figure that the maximum stress is and it is located around the front of the bicycle frame.

Von mises contour plot

Von mises magnitude

Axial stresses

The results around the front of the frame and indeed at all of the joints give some cause for concern as stress is changing too rapidly from high values to low values across the joints. This is an indication that a finer mesh is required in these locations and this will be discussed in the next section.

4.2 Horizontal loading

4.2.1 Units

SI system is the default system of units in ANSYS and is the system used to model the problem. The distances are entered in mm and the stresses will be given in MPa.

4.2.2 Element Type

Given the 3D nature of the structure it should be clear that a 3D analysis is required. The bicycle frame is made of a number of long beams that would be expected to bend due to the applied loads so a truss analysis would not be appropriate. A 3D pipe16 is the most suitable geometric model here. A 3D solid model would overkill and would not estimate bending stress correctly.

4.2.3 Material Properties

As known, the bicycle frame is constructed from aluminum. We will assume a linear material model for aluminum. This is valid considering the problem, as a bicycle would not be expected to deform significantly or yield under operating conditions. We will, however, take a note of the assumed yield stress and use this to ensure that our linear elastic model is valid and to determine the load that causes yielding of the structure.

4.2.4 Meshing

As a first attempt to get preliminary results, we are going to mesh the elements with 3 subdivisions.

4.2.5 Boundary conditions

Ball-joint boundary conditions are used for the rear dropout (location 5) and no translation along y-axis for the front dropouts (location 1). All the translation movements along z are constrained.

4.2.6 Results

5. Result Verification

5.1 Comparison with calculations

Table 8 and Table 9 show a comparison of the displacements previously calculated and the results given by ANSYS simulation. The error is generally around 7% which is a reasonable result for a preliminary study using a coarse mesh with 3 subdivisions. Generally, the ANSYS results tend to slightly underestimate the theoretical results in both loading cases.

Table 8 : Comparison of results for vertical loading case

DOF ANSYS Theory ErrU2 -0.0055446 -0.0060 -8%V2 0.0014724 0.0016 -8%U3 -0.0056491 -0.0060 -6%V3 -0.0026222 -0.0029 -11%U4 -0.0026462 -0.0028 -6%V4 -0.0034917 -0.0037 -6%U5 -0.0025435 -0.0027 -6%

Table 9: comparison of results for the horizontal loading case

DOF ANSYS Theory ErrU1 0.002588 0.0027 -4%U2 -0.002236 -0.00230 -3%V2 0.0012812 0.00130 -1%U3 -0.0023258 -0.00240 -3%V3 -0.0019129 -0.00210 -10%U4 0.00013707 0.00013 4%V4 -0.0027396 -0.00280 -2%

5.2 Convergence analysis

0.0010%

0.0100%

0.1000%

1.0000%

10.0000%

100.0000%

1000.0000%

10000.0000%Vertical loads: VNMS min error %

Vertical loads: VNMS min errorNumber of subdivisions

Erro

r per

cent

age

3 7 12 18 25 33 40 50 60 80100

120140

2200.00%

0.20%

0.40%

0.60%

0.80%

1.00%

1.20%

1.40%

1.60%

Horizontal loads: VNMSmax error %

VNMSmax error %

Number of subdivisions

Erro

r per

cent

age

1. ANSYS Simulation 3D

The structure and loading are symmetrical about the mid-plane; however, given the simple nature of the geometry it would probably take more time to build a symmetric model than the full model, so in this case the full geometry will be modeled.

2. Design optimization2.1 Failure criteria calculations

Equivalent Von-Mises stress

σ axial=FxA

σ b=−M∗C c

I

M : applied bending moment

I : moment of inertia of cross-section about neutral axis

Cc : distances from neutral axis

σx=σaxial+σb

Maximum shear stress

τ xy=2∗FyA

In 2-D

σ=√σ x2+σ xσ y+σ y2+3 τ xy2

σ y=0

σ=√σ x2+3 τ xy2

In 3-D

σ=√¿¿¿¿¿ ¿

Yielding

We will assume that the yielding takes place once the von mises stress exceeds the uniaxial yield stress for the material σ y.

Given the axial yield strength σ y=210MPa and a safety factor n=1.5

σ max=σ y1.5

σ max=2101.5

=140MPa

Buckling

B=S y∗l

2

C∗π2∗E

k=√ IASy : yield strength of column

k: Radius of gyration of column

I: Moment of inertia of column

A: Area of column cross-section

C=4 (Welded joints)

Conclusion

Appendices

Element N⁰2

8796459.43 0 0 -8796459.43 0 0

0 5488.990684 1372.24767 0 -5488.99068 1372.247670 1372.247671 538.136342 0 -1372.24767 228.707945

-8796459.43 0 0 8796459.43 0 00 -5488.99068 -1372.24767 0 5488.990684 -1372.247670 1372.247671 228.707945 0 -1372.24767 538.136342

Element N⁰3

1.05E+07 0 0 -1.05E+07 0 0

0 9.42E+03 1.37E+03 0 -9.42E+03 1.37E+030 1.37E+03 5.48E+02 0 -1.37E+03 2.74E+02

-1.05E+07 0 0 1.05E+07 0 00 -9.42E+03 -1.37E+03 0 9.42E+03 -1.37E+030 1.37E+03 2.74E+02 0 -1.37E+03 5.48E+02

0.241921896 0.970295726 0 0 0 0

-0.97029573 0.241921896 0 0 0 00 0 1 0 0 00 0 0 0.241921896 0.970295726 00 0 0 -0.970295726 0.241921896 00 0 0 0 0 1

[K2e]=

[K3]=

[R3]=

[R3]T=

0.241921896 -0.97029573 0 0 00.970295726 0.241921896 0 0 0

0 0 1 0 00 0 0 0.2419219 -0.970295730 0 0 0.97029573 0.2419218960 0 0 0 0

Element N⁰4

-0.76604444 -0.64278761 0 0 0 0

0.64278761 -0.76604444 0 0 0 00 0 1 0 0 00 0 0 -0.76604444 -0.64278761 00 0 0 0.64278761 -0.76604444 00 0 0 0 0 1

Element N⁰5

[K4]=

[R4]T=

[K3e]=

[K4e]=

[R4]=

625277.4135 2470063.07 -1331.48605 -625277.413 -2470063.07 -1331.486052470063.07 9916303.387 331.976758 -2470063.07 -9916303.39 331.976758

-1331.48605 331.9767578 547.672283 1331.48605 -331.976758 273.836141-625277.413 -2470063.07 1331.48605 625277.413 2470063.07 1331.48605-2470063.07 -9916303.39 -331.976758 2470063.07 9916303.387 -331.976758-1331.48605 331.9767578 273.836141 1331.48605 -331.976758 547.672283

8.53E+06 0 0 -8.53E+06 0 00 5.01E+03 1.29E+03 0 -5.01E+03 1.29E+030 1.29E+03 4.44E+02 0 -1.29E+03 2.22E+02

-8.53E+06 0 0 8.53E+06 0 00 -5.01E+03 -1.29E+03 0 5.01E+03 -1.29E+030 1.29E+03 2.22E+02 0 -1.29E+03 4.44E+02

-0.76604444 0.64278761 0 0 0 0-0.64278761 -0.766044443 0 0 0 0

0 0 1 0 0 00 0 0 -0.766044443 0.64278761 00 0 0 -0.64278761 -0.766044443 00 0 0 0 0 1

5009806.659 -4199521.76 -830.139692 -5009806.66 4199521.757 -830.139692-4199521.76 3528828.675 -989.32196 4199521.76 -3528828.68 -989.32196-830.139692 -989.32196 443.748438 830.139692 989.3219599 221.874219-5009806.66 4199521.757 830.139692 5009806.66 -4199521.76 830.1396924199521.757 -3528828.68 989.32196 -4199521.76 3528828.675 989.32196-830.139692 -989.32196 221.874219 830.139692 989.3219599 443.748438

10348775.8 0 0 -10348775.8 0 00 8937.904636 1899.30474 0 -8937.90464 1899.304740 1899.304735 538.136342 0 -1899.30474 269.068171

-10348775.8 0 0 10348775.8 0 00 -8937.90464 -1899.30474 0 8937.904636 -1899.304740 1899.304735 269.068171 0 -1899.30474 538.136342

[K5e]=

Element N⁰6

-0.62932039 0.777145961 0 0 0

0

-0.77714596 -0.629320391 0 0 0 00 0 1 0 0 00 0 0 -0.629320391 0.777145961 00 0 0 -0.777145961 -0.629320391 00 0 0 0 0 1

3466264.434 -4273822.31 -1053.75376 -3466264.43 4273822.314 -1053.75376

-4273822.31 5283122.386 -853.312966 4273822.31 -5283122.39 -853.312966-1053.75376 -853.312966 443.748438 1053.75376 853.3129661 227.343882

[K6]=

[R6]=

[R6]T=

[R6e]=

8.74E+06 0 0 -8.74E+06 0 00 5.39E+03 1.36E+03 0 -5.39E+03 1.36E+030 1.36E+03 4.44E+02 0 -1.36E+03 2.27E+02

-8.74E+06 0 0 8.74E+06 0 00 -5.39E+03 -1.36E+03 0 5.39E+03 -1.36E+030 1.36E+03 2.27E+02 0 -1.36E+03 4.44E+02

-0.62932039 -0.77714596 0 0 00.777145961 -0.62932039 0 0 0

0 0 1 0 00 0 0 -0.62932039 -0.777145960 0 0 0.77714596 -0.629320390 0 0 0 0

-3466264.43 4273822.314 1053.75376 3466264.43 -4273822.31 1053.753764273822.314 -5283122.39 853.312966 -4273822.31 5283122.386 853.312966-1053.75376 -853.312966 227.343882 1053.75376 853.3129661 443.748438

References