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GriffithsEnergybalanceapproach

First docu

(1920)Considere

Mechanics

This project assignment is given to Group No. II and the values of (2a/W) selected for

the group are 0.15, 0.25 and 0.35 same for all the problems. The theoretical analyses

and the ANSYS results are compared for better justification. The ultimate goal of both

procedures focuses on finding the stress intensity factor at different values of .

Because the value 2a/W is provided, its possible to fix the value ofW by assigning (a) a

convenient value for crack length. For example, if the value of a is assigned 10mm its

obvious that the value of W will be 133.3mm in the case of the ratio 0.15.

1. Calculate the values of the correction factor for the mode I stress intensity factor

for the center-cracked tension plate having normalized crack lengths of

=2a/W=0.1 to 0.4. Discuss the result in comparison with the mathematical

correction factor Y-model. Use table given in lectures discussion).

3 Calculate the values of the correction factor for the mode I stress intensity factor

for a double edge cracked tension plate having normalized crack lengths of

=2a/h=0.1 to 0.4. Discuss the result in comparison with the mathematical

correction factor Y-model. (Use table given in lectures discussion).

4 Calculate the values of the correction factor for the mode I stress intensity factorfor a single edge cracked tension plate having normalized crack lengths of

=2a/h=0.1 to 0.4. Compare the results with the values of the correction factor Y. (

Use table given in lectures discussion)


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Let us start with number one

1. Calculate the values of the correction factor for the mode I stress intensity factor

for the center-cracked tension plate having normalized crack lengths of

=2a/W=0.1 to 0.4. Discuss the result in comparison with the mathematical

correction factor Y-model. Use table given in lectures discussion).

First let us calculate the theoretical value

a) For 2a/w = 0.15

Given center crack

a = 10 mm

2a/w = 0.15

W =2a/0.15 = 133.33

= 100 mpa

a/w = 10/133.33

from this we can calculate y that is theoretical from lecture note above we can

have a formula for given crack that is

= (1 + 0.5(/)2 + 20.45(/)4 + 81.72(/)6)1/2

= 1.0017

From this we can calculate the theoretical stress intensity factor from the known

formula


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= = 17.7

Let us see the ansys analysis result and compare the result

Step 1 solid

Step 2 Mesh


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Magnifying mesh


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Step 3 Load and Ux

Step4solution


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Pressure and reaction

Pressure and Ux


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Step 5 Stress on the x (Nodal solution)

Magnifying stress


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Key point

Step 6 Nodal calculation


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From this we have found KA ( k ansys) value = 20.077 and from previous calculation

K theoretical Kt = 17.7

Percentage error = (20.07-17.7)/20.07= 11%

From this we can conclude that yt similar y ansys.

b) For 2a/w = 0.25

Given center crack

a = 10 mm

2a/w = 0.25

W =2a/0.25 = 80

= 100 mpa

a/w = 10/80

from this we can calculate y theoretical from lecture note see above we can have a

formula for given crack that is

= (1 + 0.5(/)2 + 20.45(/)4 + 81.72(/)6)1/2

= 1.0065


formula

=

= 17.84


Step 1 material


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Step2mesh

Magnifying mesh

Step 3 apply load and dof


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Ux and pressure


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Step 4 Nodal solution

Magnifying stress


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From this we can conclude

c) For 2a/w = 0.35

Given center crack

a = 10 mm

2a/w = 0.35

W =2a/0.35 = 57.14

= 100 mpa

a/w = 10/57.14


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from this we can calculate y theoretical from lecture note we can have a formula for

given crack that is

= (1 + 0.5(/)2 + 20.45(/)4 + 81.72(/)6)1/2

= 1.018


formula

=

= 18.05


Step 1 Meshing

Magnifying mesh


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Step 2 dof, Ux and pressure


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Step 3 stress in the x (Nodal solution)



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Percentage error= (18.08-16.7)/18.08 = 7.6%

From this we can conclude yt similar to ya.

3 Calculate the values of the correction factor for the mode I stress intensity

factor for a double edge cracked tension plate having normalized crack

lengths of =2a/h=0.1 to 0.4. Discuss the result in comparison with themathematical correction factor Y-model. (Use table given in lectures

discussion).


a) For 2a/w = 0.15

Given double edge crack

W = 100 mm

L = 400 mm , L/2 = 400/2 = 200 mm

= 60 mpa, /2 = 15 mpa considering quarter plate

2a/w = 0.15 , a = wx.15/2 = 7.5 mm, a/w = 0.075


given crack that is


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= (1.12 + 0.41(/)1 4.78(/)2 + 15.44(/)3)

= 1.13


formula

= , = 7.99


Step 1 meshing


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Step 2 Ux and pressure


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Step 3 stress on the x-axis (Nodal solution)

Magnifying stress


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Step 4 Nodal calculation (Kcal)

Plot


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Percentage error = (ka-kt)/ka = ( 8.2882-7.99)/8.2882 = 3.6%

From this we can conclude yt=1.13 and ya =1.034.


b) For 2a/w = 0.25


W = 100 mm

L = 400 mm , L/2 = 400/2 = 200 mm

= 120 mpa, /2 = 60mpa

2a/w = 0.25

a = wx.25/2 = 12.5 mm

a/w = 0.125


given crack that is

= (1.12 + 0.41(/)1 4.78(/)2 + 15.44(/)3)

= 1.127


formula

=

= 18.94



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Step 1 Meshing

Magnifying mesh


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Step 3 solution


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Step 4 Stress on the x

Step 5 Nodal Calculation (Kcal)


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Node plot



(18.94 -12.08)/18.94 = 36%

From this we can conclude ya =1.59 yt = 1.127 .


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c)For 2a/w = 0.35


W = 100 mm

L = 400 mm , L/2 = 400/2 = 200 mm

= 60 mpa, /2 = 30 mpa

2a/w = 0.35

a = wx.35/2 = 17.5mm

a/w = 0.175


given crack that is

= (1.12 + 0.41(/)1 4.78(/)2 + 15.44(/)3)

= 1.128


formula

=

= 15.86



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Step1 Meshing



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Step3 Solution


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Step4Stressonthex

Magnifying stress


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Nodal Plot


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(Ka-kt)/ka = (8.28-7.93)/8.28 = 4.2%

From this we can conclude ya = 1.177 , yt 1.128

4 Calculate the values of the correction factor for the mode I stress intensity factor

for a single edge cracked tension plate having normalized crack lengths of

=2a/h=0.1 to 0.4. Compare the results with the values of the correction factor Y. (

Use table given in lectures discussion)


a) For 2a/w = 0.15

Given single edge crack

W = 100 mm

L = 400 mm , L/2 = 400/2 = 200 mm

= 160 mpa, /2 = 80 mpa

2a/w = 0.15

a = wx.15/2 = 7.5 mm

a/w = 0.075


given crack that is


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= (1.12 0.23(/)1 + 10.55(/)2 21.71(/)3

+ 30.38(

)4)

= 1.154


formula

=

= 14.81


Step 1 meshing



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Step 3 Solution


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Step 5 Nodal calculation (Kcal)


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Node plot


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(Ka-kt)/ka = (16.234-14.167)/16..234 = 12.7 %

From this we can conclude ya = 1.33, yt = 1.154.


b) For 2a/w = 0.25


W = 100 mm

L = 400 mm , L/2 = 400/2 = 200 mm

= 160 mpa, /2 = 80 mpa

2a/w = 0.25

a = wx.25/2 = 12.5 mm

a/w = 0.125


given crack that is

= (1.12 0.23(/)1 + 10.55(/)2 21.71(/)3

+ 30.38(

)4)

= 1.22


formula


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=

= 19.3


Step1



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Step 4 Solution


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Nodal Plot



(kt-ka)/kt = 15.5%

From this we can conclude ya= 1.0151, yt= 1.22,


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c)For 2a/w = 0.35


W = 100 mm

L = 400 mm , L/2 = 400/2 = 200 mm

= 160 mpa, /2 = 80 mpa

2a/w = 0.35

a = wx.35/2 = 17.5mm

a/w = 0.175


given crack that is

= (1.12 0.23(/)1 + 10.55(/)2 21.71(/)3

+ 30.38(

)4)

= 1.315


formula

=

= 24.6



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Step 1 Meshing



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3 ) Solution

Step 4 stress on the x


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Nodal plot



(kt-ka)/kt = 44%

From this we can conclude ya .748, yt 1.315,


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Part II

Q6) 60, DEGREE CRACK ANGEL .

Given crack angel at 60o

W = 250 mm

L = 1270 mm , L/2 = 635mm

= 670kpa /2 =335kpa

a = 25.4

Determine the shear stress and normal stress at the crack tip.And disscuse on stress

displacement result.


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1 ) meshing


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2) load

3

2) Press ure DOF of UX


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3 solution


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4 Magnify stress on the x axiss


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Stress 0n the x axis

Stress 0n the y axis


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Disscution

From the above figures we have stress in the x and stress in the y axis. Since the plate has a crack

angle there is maximum shear stress and maximum normal stress occurred at the tip of the crack .

x= 14.323 MPa

(1) n=xcos 600= 7.16 MPa

(2) xy=xsin 600= 12.4 MPa

(3 ) To plot the curve vs /y

= /2*y a

Using the above formulas and the value getting from Ansyse evaluating along the y-

axis starting from at the tip of the crack taking an arbitrary distance or indicating the

different colors respectively by /y = 1.14, 1.328,1.59,1.977. and

=0.25,0.29,0.3379,0.4043


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