BME402 Lab 4 - Lab PDF

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National University of Ireland, Galway

Abaqus Assignment 4 Report 2019-2020

Modulus Code/ Name: BME402, Computational Methods in Engineering Analysis Instructors: Dr. William Ronan Class: Biomedical Engineering

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Introduction: This report looks at the comparison between theoretical calculations from Abaqus and hand calculations of linear and non-linear components. Two problems are presented in the assignment. The first problem investigates the uniaxial tension of a single element with and without the NLGEOM function. When the NLGEOM function is turned on, it will include changes in the geometry during the analysis. In linear geometry, all the outputs are of the original geometry. This means that nominal stress, 𝜎𝑥 𝑁 , is the force on the original cross-sectional area of the model. This is the change in length with respect to original length. Whilst in non-linear geometry, all outputs are shown in terms of current deformed geometry. The true stress is the force on the current area, whereas true strain is the change in length with respect to the current length. Problem 2 looks at the bending of the beam with different element types. These element types are c3d8r and c3d8i. Strain will vary linearly across the beam whilst it is bending. It is stated that linear quadrilateral elements will capture the correct strain gradient in the transverse and axial direction of the beam in bending. Incompatible element modes are good in bending due to the stretching of the handles and separate bending.

Problem 1: Young’s Modulus, E Poisson’s Ratio, 𝛾 Original Length, Lo Displacement U

110 0.35 1.4101 3

Equations: Linear: Nominal strain 11, 𝜀11 𝑁

𝑈 𝐿𝑜

Nominal Strain 22, 𝜀22 𝑁

(𝜀11 𝑁 ) ∙ ( 𝛾) 𝐿𝑜 2

Area, 𝐴 Nominal Stress 11, 𝜎11 𝑁

(𝐸) ∙ (𝜀11 𝑁 )

Force, 𝐹

(𝜎11 𝑁 ) ∙ (𝐴)

Non-linear: Nominal strain 11, 𝜀11 𝑁 Tensile strain 11, 𝜀11 𝑇

𝑈 𝐿𝑜 ln (1 + 𝜀11 𝑁 )

Tensile strain 22, 𝜀22 𝑇

(𝜀11 𝑇 ) ∙ (−𝛾) 𝑇

Nominal strain 22, 𝜀22 𝑁

(𝑒 𝜀22 ) − 1

(𝜀11 𝑇 ) ∙ ( 𝑤𝑜 + 𝑤𝑜 )

Current width, 𝑤𝑖

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Tensile stress 11, 𝜎11 𝑇

(𝜀11 𝑁 ) ∙ (𝐸) (𝑤𝑖 )2

Area, 𝐴

1) Strain 11 plot for the linear and non-linear strain component:

Fig.3 & Fig.4: S11 contour plot for the linear and non-linear cube.

The theoretical strain calculated on the linear model in Fig.3 has a value of 2.340 x 102, This is the same as the hand calculated strainin Fig.1, 𝜎11 𝑁 = 234.025. The theoretical stress calculated on the non-linear model in Fig.4 is at 1.252 x 102. This matches the hand calculated stress results in Fig.2, 𝜎11 𝑇 = 125.422. These values are identcial to eachother.

2) Plot for the linear (E11) and non-linear (LE11) strain component:

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Fig.5 & Fig.6: Linear (E11) and non-linear (LE11) strain plot.

The linear strain component in Fig.5, E11, of the model is estimated at 2.128. The hand calculated result in Fig.1 is that same, 𝜀11 𝑁 = 2.2175.

The theoretical non-linear strain component in Fig.6, LE11, is estimated to be 1.140. The hand calculated results in Fig.2 are similar at 𝜀11 𝑇 = 1.1402.

3) Plot for the linear (E22) and non-linear (LE22) strain component:

Fig.7 & Fig.8: Linear (E22) and non-linear (LE22) strain plot.

The theoretical linear strain component in Fig.7, E22, is estimated at -7.446 x 10-1. The hand calculated result is that same in Fig.1, 𝜀22 𝑁 = -0.7446.

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The theoretical non-linear strain component in Fig.8, LE22, is estimated to be -3.986 x 10-1. The hand calculated results are similar in Fig.2, 𝜀22 𝑇 = -0.3991.

4) Plot of the linear and non-linear Reaction Forces:

Fig.9 & Fig.10: Linear and non-linear Reaction Forces plot

XY Plot of Reaction forces: Linear:

Fig.11: Linear plot of Reaction Forces – Force vs. Time

Non-linear:

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Fig.12: Non-linear plot of Reaction Forces – Force vs. Time

XY plot of Displacement: Linear:

Fig.13: Linear plot of Displacement vs. Time

Non-Linear: 6

Fig.14: Non-linear plot of Displacement vs. Time

Linear Force vs. Displacement Plot:

Force (N)

Hand Calculated linear & non-linear 500 450 400 350 300 250 200 150 100 50 0

linear non-linear

0

0.5

1

1.5

2

2.5

3

3.5

Displacement (mm)

Fig.15: Linear & Non-linear Force vs. Displacement Plot

The maximum force applied on both the linear and non-linear models does not match for the theoretical and hand-calculations. For the linear model, the theoretical maximum force applied to the cube is approximately 120 N, however the hand calculations state the maximum force is 465.312 N. Similarly, for the non-linear model, the theoretical maximum force is approximate 27 N however the hand calculated maximum force is 110 N from the graph.

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Problem 2: Problem2_c3d8r: SS11 plot of un-deformed shape:

Fig. 16: S11 Contour plot of un-deformed shape for c3d8r

S11 plot of deformed shape:

Fig. 17: S11 Contour plot of deformed shape for c3d8r

Path along beam of the variable U2:

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Fig.18: Path along U2 beam for c3d8r

Problem2_c3d8i: SS11 plot of un-deformed shape:

Fig.19: S11 contour plot of un-deformed shape for c3d8i

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SS11 plot of deformed shape:

Fig.20: S11 contour plot of deformed shape for c3d8i

Path along beam of the variable U2:

Fig.21: Path along U2 beam for c3d8i

Excel Plot of Displacement vs True Distance along the Path

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Fig.22: Hand calculated displacement vs true distance along the path for c3d8i & c3d8r

From Fig.22 the true distance values match up with the figures in Fig.18 and Fig.21, however the theoretical displacement value does not match up with the hand calculated values from the Excel plot. The displacement value in Fig.18 goes to -40 whilst in Fig.21. the displacement value reaches 30. This is a big difference to the excel plot of -55 for c3d8i and -71 for c3d8r. The error could be due to incorrect loads being applied to beam. In Abaqus the force was multiplied by four nodes and then re done by applying the force on two nodes and multiplying the two nodes. Both ways gave similar results however did not match the hand calculated results on excel.

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