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Homework 2: Middle crack in plate under uniaxial

Tension

T.Arriessa Sukhairi

(108013235855)

Known :

h = 1 mm,

b = 1 mm,

a = 0.2 mm,

E = 210 GPa,

V = 0.3 1 = 50 + 2 5

1 = 60

2

=2 1

210000

60

0.2 [ 1

0.2 1

1 0.2

2 1 + 0.326(0.2 1 )

2

]

= 0.000721 [0.8944

1 0.1 + 0.013]

= 0.000721 [0.9796]

= 0.000706

Picture 1.Global Mesh

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Picture 2. Crack Tip Mesh

1. Assuming elastic material behaviour and make a contour plot of the von Mises stress

distribution at the crack tip. Plot the von Mises stress (MISES) and the stress in 2-direction

(S22). Discuss the differences between MISES/S22-stress components in bullet points.

Picture 3. Contour Plot at the crack Tip

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Picture 4. XY-Data plot at path Z: MISES/S22 (u = 0.000706 mm)

Picture 3. Contour Plot at the crack Tip

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Picture 4. XY-Data plot at path Z: MISES/S22 (u = 0.00706 mm)

The value of stress that is calculated by using u = 0.000706 mm still in the

elastic zone of the materials. The maximum stress value is 155 Mpa.

The maximum Mises Stress value is bigger than S22 stress value.

Because I use the student version, I have the limitation of the node that I can

use, then I change the displacement value u = 0.00706 mm, so I can get the

equivalent plastic strain value for question number 2.

However there is no difference in the global Behaviour, only the maximum

stress value now reach 1500 Mpa.

2. Now change the material behaviour from elastic to plastic with the material parameters

shown in the stress-strain curve and make a contour plot of the von Mises stress distribution

at the crack tip. Plot the von Mises stress (MISES), the stress in y-direction (S22) and the

equivalent plastic strain (PEEQ) for plane stress on path Z. Discuss the differences by

comparing the results of the elastic and plastic material model in bullet points.

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Picture 5. Contour plot at the crack tip: MISES

Picture 6. XY-Data plot at path Z: MISES/S22

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Picture 7. XY-Data plot at path Z: PEEQ

With the input of Plasticity in the material, as can be seen from the picture 6, that the

value of S22 > Mises. In the case

The maximum value of the stress now is on 700 Mpa, it is smaller than if we did not

apply the plasticity as in the question 1 with the value 1500 Mpa.

The maximum value of the plasticity region is in the crack tip and then getting smaller

until 0.13 mm from the crack tip.

3. Change the crack length (a = 0.1mm, a = 0.2mm, a = 0.4mm) and plot the von Mises stress,

the stress in 2-direction (S22) and the equivalent plastic strain for different crack lengths on

path Z by using constant displacement, as calculated in question 1. Discuss the influence of

crack length on the stress and the plastic zone at the crack tip in bullet points.

Picture 8. XY-Data plot at path Z: MISES cracks length (a = 0.1)

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Picture 9. XY-Data plot at path Z: MISES cracks length (a = 0.2)

Picture 10. XY-Data plot at path Z: MISES cracks length (a = 0.4)

Picture 11. XY-Data plot at path Z: MISES 3 cracks Combine

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Picture 12. XY-Data plot at path Z: PEEQ cracks length (a = 0.1)

Picture 13. XY-Data plot at path Z: PEEQ cracks length (a = 0.2)

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Picture 14. XY-Data plot at path Z: PEEQ cracks length (a = 0.4)

Picture 15. XY-Data plot at path Z: PEEQ 3 cracks Combine

Bigger crack length can make the stress intensity factor value increase, as a

result the stress is also getting bigger. As can be seen in the picture 11, where

the brown colour is for 0.1 crack length (620 Mpa), the orange colour is for 0.1

crack length (675 Mpa), and the red colour is for 0.1 crack length (685 Mpa).

Higher crack length also can make x distance along the path smaller. As can

be seen in the picture 11, where the brown colour is for 0.1 crack length (90

mm), the orange colour is for 0.1 crack length (80 mm), and the red colour is

for 0.1 crack length (60 mm).

Higher crack length also give some influence for the plasticity region, as can be

seen from the picture 15 that the plasticity region for 0.1 mm, 0.2 mm, and 0.4

mm is 0.02 mm, 0.08, and 0.13 mm respectively.