Fracture Mechanocs

7/22/2019 Fracture Mechanocs

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Final Report on

Effects of cut-outs, holes and cracks on the

stress induced in the material(BITS - G540 RESEARCH PRACTICE)

In partial fulfilment of the course

Master in Engineering

Design Engineering)Submitted byYogesh Joshi

2012H141031G

Under the supervision ofDr. D.M Kulkarni


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ontentsAbstract ...................................................................................................................................................3

Introducon:...........................................................................................................................................4

Eect of cut-outs (holes) in an innite plate. ..........................................................................................5

Stress concentraon around a circular hole in a plate under biaxial loading.........................................7

Case 1: plate subjected to bi-axial tensile stress:...................................................................................8

Case 2: plate subjected to tensile stress on one edge and compressive stress on adjacent edge.......12

Case 3: Plate subjected to biaxial compressive stress..........................................................................17

Stress concentraons around ellipcal hole in a plate under tensile loading......................................21

Stress Intensity Factor: ..........................................................................................................................26

Finding stress concentraon factor of a plate with a central crack......................................................27

Conclusion:............................................................................................................................................

30

References: ............................................................................................................................................31


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Effects of cut-outs, holes and cracks on the

stress induced in the material

AbstractIn many engineering designs, cut-outs or holes are required to meet some service

requirements, and often on application of load, local stresses around the cut outs or holes

increases to a value which is sometimes even greater than the expected ultimate tensile strength

of the material. These stresses are sometimes unavoidable and in certain cases may even lead

to failure of the component. In this paper we show by using ANSYS software, how a slot

produced is a cantilever beam will affect the stress induced in it and also a plate with a centralhole under bi axial loading is considered in the analysis and the effect on the induced stress in

observed. Further an elliptical hole is analysed with load applied perpendicular to its major

axis and the relationship between maximum stress induced and minor axis is observed. Finally

a term called Stress Intensity Factor is defined at the crack tip and the stress intensity factor

for a plate with a central crack is determined using ANSYS.


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Introducon:The introduction of a cut-out or a hole in a stressed elastic sheet causes a

redistribution of the stress field. This effect is greatest on the boundary of the hole and

diminishes with distance until, in regions remote from the hole, it is negligible. Around

the boundary of the hole the variation of edge or hoop stress may be considerable and

the largest values of the stress are associated with regions of large curvature (i.e.

Corners). The value of the maximum hoop stress is an important design parameter and

its estimation is, in general, very difficult.

Because holes are necessary for fasteners such as bolts, rivets, etc., the need to know

stresses and deformations near them occurs very often and has received a great deal of study.

Plates with cut-outs (or holes) are widely used in structural members. These cut outs provide

stress con-centration in plates. Extensive studies have been carried out on stress concentration

in perforated plates, which consider cut out shapes, boundary conditions, bluntness of cut outs,

and more. This study presents stress concentration analyses of plates with 2 types of cut outs

i.e. for circular hole and elliptical hole, and also plate with a central crack is analysed to find

stress intensity factor.

In general, plates are easily manufactured and are widely used for fabrication of

structural members and eventually for the construction of marine and civil structures.

Especially, a bolt-connection is a way of fastening plates by inserting bolts into the cut outs (or

holes), which are made through designated areas of target plates that are to be connected.

Making cut outs is not merely for connecting but also for reducing the weight of structural

members. This type of connection causes stress concentration near the cut outs, causing high

stresses, and some-times results in the failure of structural members. To quantify the degree of

stress concentration, the concept of stress concentration factor (SCF) is used, which is defined

by the ratio of the maximum stress to nominal stress.

Most research on stress concentration focuses on the structural members that are mostly

subject to (or weakened by) stress concentration. Among such research, the work on stress

concentration on cut outs include: analytical works for flat plates with circular holes and

notches, optimum design of holes and notches by considering fatigue life, stress concentration

analysis of differently materialized circular holes, etc. These works mainly focus on stress

concentration on circularly-perforated plates.


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Eect of cut-outs (holes) in an innite plate.Analysis of plate with a hole with thick cylinder analogy

In the above figure, a is the radius of the hole and at a radius r=b, where b>>a, the stressesare not affected by the presence of the hole.

Now the elementary equations for the plate is as given below;

, r = xcos2 + ysin2 + 2 xysin cos

= xsin2 + ycos2 + 2 xysin cos

r= (x y) sin 2 + xycos 2

Now for this case, we have x= , y= 0, and xy= 0

Thus the above equations reduces to

, r= /2 + /2 cos 2

= /2 - /2 cos 2

r= - /2 sin 2

These are the equation for the stress distribution within r=a and r=b, two components are

constant and depend on .


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If we use thick cylinder analogy, (P0= /2, b = infinity, pi= 0)

, r= [ a2b2 (P0 pi) / (b

2 a2) * 1/r2] [ (a2pi b2P0) / ( b

2 a2) ]

= [ a2

b2

(P0 pi) / (b2

a2

) * 1/r2

] + [ (a2

pi b2

P0) / ( b2

a2

) ]

Which gives

,r' = / 2 [ 1 a2/ r2]

= / 2 [ 1 + a2/ r2]

2ndcomponent is expressed by a stress function { = f cos 2 }

[

+

+

] [

+

+

]= 0

(Compatibility equation)

{ = (A r2+ B r4+ C 1/r2+ D) cos 2}

Thus we get:

r= -(2A + 6C/r4+ 4D/r2) cos 2

= (2A + 6C/r4+ 12Br2) cos 2

r= (2A +6Br2- 6C/r4- 2D/r2) sin 2

Boundary conditions:

1: At r=a, r= 0, r= 0

2: At r=b, r= /2 cos 2, r= - /2 sin 2

Thus we get:

r= /2 (1 - a2/r2) + /2 (1+ 3 a4/r4- 4 a2/r2) cos 2

= /2 (1 + a2/r2) - /2 (1+ 3 a4/r4) cos 2

r= - /2 (1- 3 a

4

/r

4

+ 2 a

2

/r

2

) sin 2


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Stress concentraon around a circular hole in a plate under biaxial

loading.A plate subjected to bi-axial loading is analysed in this section and its effect on the

stress concentration is observed.

The analytical solution is available for plate uniaxial loafing which is given below, for

a plate with a stress 0applied on the side faces, the maximum and minimum stress observed

at the points A and B respectively on the circumference of the hole is given in figure 1:

Figure 1

Maximum stress i.e. stress at A = 3*0 and minimum stress i.e. at point B = - 0

By using the above concept, we can determine the analytical value of stress for any

kind of biaxial loading.

Here we consider 3 types of loading. And for all 3 types we find the stress by using

ANSYS 13 software and also by analytical method using the above concept, and then compare

the 2 results.


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Case 1: plate subjected to bi-axial tensile stress:Here the plate is subjected biaxial tensile stress as shown in the figure 2;

Figure 2

For finding the analytical solution for this problem, the state of stress is split into 2

problems taking one set of stress at a time, finding the magnetite of maximum and minimum

stress for each state and then combining the two results to get the solution for the above

problem. The stress can be split as shown in the figure 3:

Figure 3


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Now using the analogy as mentioned, we can solve this problem by considering it as two

problems of plate subjected to uniaxial stress, it is explained in detail below:

At point A,

r=a, =/2, implies: r=0, r=0

= /2 (1 + a2/r2) - /2 (1+ 3 a4/r4) cos 2

Thus =3

At point B

r=a, =0, implies: r=0, r=0

= /2 (1 + a2/r2) - /2 (1+ 3 a4/r4) cos 2

Thus = -

At point B,


= /2 (1 + a2/r2) - /2 (1+ 3 a4/r4) cos 2

Thus =3

At point A


= /2 (1 + a2/r2) - /2 (1+ 3 a4/r4) cos 2

Thus = -

Thus the state of stress around the hole can be better understood by the figure below


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The model is analysed is ANSYS with applied stress of 50 MPa on the edges. The

uniform, homogeneous plate above is symmetric about horizontal axes in both geometry and

loading. This means that the state of stress and deformation below a horizontal centre line is a

mirror image of that above the centre line, and likewise for a vertical centre line. We can take

advantage of the symmetry and, by applying the correct boundary conditions, use only aquarter of the plate for the finite element model. For small problems using symmetry may not

be too important; for large problems it can save modelling and solution efforts by eliminating

one-half or a quarter or more of the work.

The quarter model of the plate as modelled in ANSYS is as shown in the fig 4:

Figure 4

The model generated in ANSYS is den solved for the given boundary conditions and

applied stress of 50 MPa.

Now the analytical solution for the plate is given below.

Stress at point A = 2* 0which gives 2*50 = 100MPa


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This value is compared with the stress value obtained by ANSYS. The stress plot as

obtained from ANSYS is as shown in the figure 5.

Figure 5

The comparison of the results obtained from ANSYS with the analytical solution is as

given in the table below:

Case 1

Analycal result

Result from ANSYS

At point A 100MPa 106.293MPa

At point B 100MPa 106.293MPa


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Case 2: plate subjected to tensile stress on one edge and compressive

stress on adjacent edge.

The state of stress which is applied on the plate in as shown in the figure 6:

Figure 6

Even for this case, for finding the analytical solution for this problem, the state of stress

is split into 2 problems taking one set of stress at a time, finding the magnetite of maximum

and minimum stress for each state and then combining the two results to get the solution for

the above problem. The stress can be split as shown in the figure 7:

Figure 7


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The problem is solved as given below:

At point A,


= /2 (1 + a2/r2) - /2 (1+ 3 a4/r4) cos 2

Thus =3

At point B


= /2 (1 + a2/r2) - /2 (1+ 3 a4/r4) cos 2

Thus = -

At point B,

r=a, =/2, =- implies: r=0, r=0

=-/2 (1 + a2/r2) - /2 (1+ 3 a4/r4) cos 2

Thus =-3

At point A

r=a, =0, =- implies: r=0, r=0

= /2 (1 + a2/r2) - /2 (1+ 3 a4/r4) cos 2

Thus =


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14

The state of stress on the circumference of the hole is gives as follows:

The component is modelled in ANSYS and the stress is applied at the edges. On one

edge tensile stress is applied and on the adjacent face compressive stress is applied. And even

for this case, the symmetry of the plate is taken into account and only one quarter of the plate

is taken for analysis as it will be symmetrical about the horizontal and vertical axes,

The component after modelling in ANSYS is as shown in the figure 8:


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15

Figure 8


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16

The model generated in ANSYS is den solved for the given boundary conditions and applied

stress of 100 MPa.


Stress at point A = 4* 0which gives 4*100 = 400MPa



Figure 9

The results obtained from ANSYS is compared with that of the analytical solution and

the comparison is as given in the table below:

Case 2: Theorecal result Result by ANSYS

At point A 400MPa 366.429MPa

At point B -400MPa -366.429MPa


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17

Case 3: Plate subjected to biaxial compressive stress.In this case the plate is subjected to compressive stress on all of its faces. The state of

stress is as shown in the figure 10

Figure 10

Even for this case, for finding the analytical solution for this problem, the state of stress

is split into 2 problems taking one set of stress at a time, finding the magnetite of maximum

and minimum stress for each state and then combining the two results to get the solution for

the above problem. The stress can be split as shown in the figure 11:

Figure 11


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18

The problem is solved as follows:

At point A,

r=a, =/2, =- implies: r=0, r=0

= /2 (1 + a2/r2) - /2 (1+ 3 a4/r4) cos 2

Thus = -3

At point B


= /2 (1 + a2/r2) - /2 (1+ 3 a4/r4) cos 2

Thus =

At point B,

r=a, =/2, =- implies: r=0, r=0

=-/2 (1 + a2/r2) - /2 (1+ 3 a4/r4) cos 2

Thus = -3

At point A

r=a, =0, =- implies: r=0, r=0

= /2 (1 + a2/r2) - /2 (1+ 3 a4/r4) cos 2

Thus =

The state of stress on the circumference of the hole is given as follows:


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19

The component is modelled in ANSYS and the stress is applied at the edges. On one

edge tensile stress is applied and on the adjacent face compressive stress is applied. And even

for this case, the symmetry of the plate is taken into account and only one quarter of the plate

is taken for analysis as it will be symmetrical about the horizontal and vertical axes,

The component after modelling in ANSYS is as shown in the figure 12:

Figure 12

The model generated in ANSYS is den solved for the given boundary conditions and applied

stress of 50 MPa.


Stress at point A = -2* 0which gives -2*50 = -100MPa


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20

Stress at point B= -2* 0which gives -2*50 = -100MPa



Figure 13

The results obtained from ANSYS are compared with analytical results and the

comparison\ is as shown in the table:

Case 3:

Analycal results

Results by ANSYS

At point A -100MPa -106.293MPa

At point B -100MPa -106.293MPa


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21

Stress concentraons around ellipcal hole in a plate under tensile

loadingIn this section, we model and analyse a plate with elliptical hole with load applied

perpendicular to major axis. The analysis for elliptical hole is important then circular hole

because unlike circular hole, elliptical hole will not have equal spacing from the centre and the

maximum stress induced depends on whether the stress is applied perpendicular to major or

minor axis, it is found that when the stress applied is perpendicular to major axis, the maximum

stress in quite high and so it is important to analyse this type if state of stress so as this is more

seviour case.

To show how the maximum will increase for an elliptical hole and how it depends on

the geometry of the ellipse and on the orientation of the load direction with respect to hole

geometry I analysed the plate by keeping the axis of the major axis of the ellipse constant andalso load and its direction which is tensile in nature and perpendicular to major axis, and only

varying minor axis from maximum to minimum, noted down the maximum stress value

observed for each case, found the stress concentration factor for each case and then plot a

graph with SCF vs. minor axis to show how the SCF increases with decrease in minor axis and

at one point when the minor axis is very small as compared to major axis, we get the highest

value of SCF.

Analytically for an elliptical hole under tensile stress 0, the maximum stress inducedin the hole is given by the relation:

max = 0[1 +2

]

o= max* (1+ 2b/a)

Where 2b = major axis and 2a = minor axis

Now it is clear from the above equation that maxdepends on a/b ratio and as b i.e. minoraxis decreases, the maximum stress also increases.


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22

The plate with elliptical hole with load applied perpendicular to major axis and the

stress distribution that will occur in the hole is as shown in the figure 14.

Figure 14

The plate is modelled in ANSYS and applied boundary conditions and load. On the

similar lines as we did for plate with circular hole, where we considered only quarter section

of the plate for analysis considering the symmetry of the geometry, for plate with elliptical

hole, we proceed in the similar manner. Considering only quarter section of the plate, the model

along with the boundary conditions as modelled in ANSYS is as shown in the figure 15,


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23

Figure 15

The plate is solved in ANSYS and the maximum stress in recorded, also stress concentration

factor is calculated, and the stress plot for the above plate is as shown in the figure 16:

Figure 16


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Now for this case when we compare the maximum stress obtained from ANSYS and

maximum stress obtained by analytical formula, we find they are very similar, the comparison

for this case is given in the table below:

Maximum stress analycal Maximum stress by ANSYS Stress concentraon factor

250 MPa 234.712 MPa 5

Now in the similar manner, I analysed the plate for different value of minor axis and

solved it in ANSYS to get the value of maximum value of stress for each case and then found

stress concentration factor for each case, and then used those values to plot a graph of SCF vs.

Minor axis and from the graph it is clear that maximum stress and thus stress concentration

factor increases with decrease in minor axis. The graph is as shown in the figure 17,

Figure 17


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25

Now from the graph it can be seen that SCF is maximum for small value of minor axis,

now as minor axis tends to zero, the maximum stress tends to infinity, i.e. at this point ellipse

is more like a crack, and the stress at the crack tip intensifies and this state is worse for the

material. For this purpose, the study of stress around crack tip requires the use of a different

parameter called Stress Intensity Factor. This is discussed in detail in the sext section.


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Stress Intensity Factor:In practical engineering problems, different types and numbers of cracks are often

randomly distributed and encountered in the structure components under various loading and

boundary conditions. So, the strength of a structure could be severely affected by the presence

of crack and the defects are unavoidable in a cost effective manufacturing process. In most ofthe structures, cracks may be located or developed during the service in the zones of stress

concentration and the crack may be large enough so that the crack-tip may be closer to a

boundary. The crack may grow to cause structure failure due to low stress, which acts to a

structure. Stress intensity factor (SIF) is a most important single parameter in fracture

mechanics, which can be used to examine if a crack, would propagate in a cracked structure

under particular loading condition, i.e. it controls the stability of the crack. For relatively

simple components and loadings, SIF can be calculated very easily, but problem might arise

when one is dealing with complicated cracked structures.

Stress Intensity Factor, K, is used in fracture mechanics to more accurately predict the

stress state ("stress intensity") near the tip of a crack caused by a remote load or residual

stresses. When this stress state becomes critical a small crack grows ("extends") and the

material fails. The load at which this failure occurs is referred to as the fracture strength. The

experimental fracture strength of solid materials is 10 to 1000 times below the theoretical

strength values, where tiny internal and external surface cracks create higher stresses near these

cracks, hence lowering the theoretical value of strength.

Unlike "stress concentration", Stress Intensity, K, as the name implies, is a parameter

that amplifies the magnitude of the applied stress that includes the geometrical parameter Y

(load type). These load types are categorized as Mode-I, -II, or -III. The Mode-I stress intensity

factor, KIcis the most often used engineering design parameter in fracture mechanics and hence

must be understood if we are to design fracture tolerant materials used in bridges, buildings,

aircraft, or even bells. Polishing just won't do if we detect a crack. Typically for most materials

if a crack can be seen it is very close to the critical stress state predicted by the "Stress Intensity

Factor".


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27

Finding stress concentraon factor of a plate with a central crack. In this section, I have considered a plate with a central crack under tensile loading and

used ANSYS to solve the problem and get the stress intensity factor for the plate. Further

analytical solution is used to find the stress intensity factor for the plate using an empirical

formula and the results are compared.

The plate with central crack is as shown in the figure 18,

Figure 18

Let the youngs modulus E=200 GPa, poissons ratio v=0.3. Let the dimensions of the

plate be 2W=0.2m 2a=0.02m. The applied stress is 100 MPa.

We can take advantage of the symmetry of the plate and use only quarter portion of the

plate for our analysis.


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28

The quarter portion is modelled in ANSYS, given proper boundary conditions and

applied load, the model is as shown in the figure 19,

Figure 19

The problem is solved in ANSYS to get a value for stress intensity factor as 26.53 MPa

Now this value is compared with the value calculated by the below given empirical formula

Analytical method:The analytical solution for this problem can be found out by using the formula:

K1 = f()**(a)

(As given in the text book on fracture mechanics by Prashant Kumar)

Where f()= 1+0.128 -0.288 2+1.523 3

= a/w


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= applied stress = 100MPa

By solving the above equation, we get the value of K1=25.35 MPa.The two values of stress intensity factors are compared and the comparison is as shown in the

table below:

Analycal result result from ANSYS Percentage error

25.35 MPa 26.53 MPa 4.6%

Thus the two results are very similar with an error of 4.6%.

The figure 20 shows the stress plot as obtained for the plate. Clearly it can be seen that

maximum stress occurs at the crack tip, and its here where the stress is intensified, and it may

even lead to the growth of crack and finally to the failure of the material.

Figure 20


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Conclusion: In this paper, we analysed plates with hole subjected to different orientation of loads

and obtained stress concentration for the plates, and compared it with analyticalsolution.

We then considered an elliptical hole in a plate and analysed it for maximum stress onload applied perpendicular to major axis

We then considered ellipse with varying minor axis and observed how maximum stressincreased with decreasing minor axis.

We defined stress intensity factor and saw how important it is in deciding the failure ofthe component.

We analysed a plate with central crack and found out its stress intensity factor andcompared it with the analytical formula.

Thus we conclude that stress intensity and fracture toughness are critically importantfracture mechanics parameters used by materials engineers and designers. There are a

lot of factors that determine fracture of a material. KIc is a unique material property

that is used by engineers to design and manufacture products for durability and safe

operation.


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References:1. Referred paper Effect of Cutout Orientation on Stress Concentration of Perforated

Plates with Various Cutouts and Bluntness by Jinho Woo and Won-Bae Na

Department of Ocean Engineering, Pukyong National University, Busan 608-737,

Korea2. Referred paper stress concentration around a hole in a radially inhomogeneous plate

by Mohsen Mohammadi,, John R. Dryden, Liying Jiang, Department of Mechanical

and Materials Engineering, The University of Western Ontario, London, Ontario,

Canada N6A 5B

3. Journal of Mechanics of materials and structures4. Paper Introduction to Fracture Mechanicsby C.H. Wang, Airframes and Engines

Division Aeronautical and Maritime Research Laboratory

5. Paper Stress-Intensity Factors for Elliptical Cracks Emanating From CountersunkRivet Holes

6. Paper Stress-Concentration Factors for Rounded Rectangular Holes in Infinite SheetsBy A. J. 8OBEY, M.A.

7. Paper Stress Concentrations for Straight-Shank and Countersunk Holes in PlatesSubjected to Tension, Bending and Pin Loading by K. N. Shivakumar Analytical

Services & Materials, Inc. Hampton, Virginia

8. ANSYS Tutorial, Release 8.09. Paper STRESS CONCENTRATION AROUND A CIRCULAR HOLE IN A FG PLATE

UNDER BIAXIAL LOADINGby Mohsen Mohammadi and John Dryden, Faculty of

Engineering, University of Western Ontario, London, Canada, N6A 5B9

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