TWO CYLINDRICAL SHELL TYPE CRACKS OPENED BY BODY FORCES IN A THICK PLATE

TWO CYLINDRICAL SHELL TYPE CRACKS OPENED BY BODY FORCES IN A THICK PLATE

Mohammad Salim Ahamad

Department of Mathematics, Hindustan College of Science & Technology, Mathura, India P.S.Kushwaha

Department of Mathematics, St. John’s College, Agra(U.P.),India

ABSTRACT

In this paper we have analysed two shell discontinuities in a plate of thickness 2h and radius infinite. The closed form and stress intensity factor are obtained by using Hankel and Fourier transforms. Point body forces are also discussed as a special case. Normal stress, graphical crack shape and solution of Fredholm integral equation are obtained.

Keywords: Stress intensity factor(SIF),body forces(rivets or stiffeners),point force.

1. INTRODUCTION

In the paper we considered a crack (cylindrical shell type crack) opened by body forces. It is very obvious to know the effects caused over physical quantities due to single discontinuity due to the presence of another similar discontinuity. The physical problem is translated to mathematical language in the following way. Discontinuities occupy the region, b< z < c, with r=aand 0≤ ≤θ 2π see figure 1. Then we take the cross-section by θ =0,πsee figure 2 and the following boundary conditions in two dimensions.

( , ) ( , ) 0,

rz r h w r h

σ ^{± = ± = 0 r}≤ < ∞ (1.1)

( , ) 0 0

rz a z z h

σ = = ≤ ≤ (1.2)

with the mixed-boundary condition

( , ) 0, 0 ,

U a z = ≤ z ≤b c≤ z ≤ h (1.3)

( , ) 0, ,

rr a z b z c

σ = < < (1.4)

There are two cracks at distance r=a of lengths (c – b) each. The symmetry of geometry and the presence of symmetric body forces make the boundary conditions (1) – (4) simple. These are now,

( , ) ( , ) 0, 0 ,

rz r h w r h r

σ = = ≤ < ∞ (1.5)

( , ) 0, 0 ,

rz a z z h

σ = ≤ ≤ (1.6)

( , ) 0, 0 , ,

U a z = ≤ ≤z b c≤ ≤ z h (1.7)

( , ) 0, ,

rr a z b z c

σ = < < (1.8)

we checked throughout, see [1],

89

( , ) 0,

U a z > b< z < c (1.9)

which means that cracks really open out and crack faces do not meet each other than at crack tips. The assumptions of previous chapter are also used here. The plan of chapter is as follows :

In next section we shall formulate elasticity problem and reduce to triple series equation whose solution will also be given. Section 3 will give physical quantities. The solution of Fredholm integral equation for special point force will be gives in section 4.

2. FORMULATION,REDUCTION AND SOLUTION OF TRIPLE SERIES

EQUATIONS. FORMULATION

Here also we divide the problem into two namely. Body Force Problem. (It is given in chapter VI) and the Elasticity Problem.

Elasticity Problem— We solve the equations of equilibrium, in the absence of body forces, by using the solution, U^{( )e} ( , ),r z and W^{( )e} ( , ),r z given as,

( ) ( )

1

1 2

( , ) ( , 0) cos( ) ( , ),

2

e e e

c m c m

m

U r z U r z U r

h ^∞ β β

=

= +

∑

while U_c^{( )e} ( ,r β_m) is given by (7.2.6).

( )

5 6

1

( , ) 2 sin( )[ ( ) ( ) ( ) ( )]

e

m m

W r z z A m F r B m F r

Ph ^∞ β

=

=

∑

5 3 1

6 2 4

1 1

2 0

( ) ( 2 ) ( ) ( ),

( ) ( ) ( )

( ) ( ) ,

( ) ( )

m

m

F r F r F r

F r F r F r

F r I r dr F r rI r dr

λ µ

µ β

β

= + + 

= + 

= 

= 

∫

∫

(2.3)

while F r₃( ) and F r4( ) are given by (7.2.12)

[

( )

1 0 1

1

2 ( )

( , ) sin( ) ( ) ( ) ( ) ( )(2 ) (2 ) ( ) ,

e

rz m m m m m m m m

m m

r z z P K A m I r B m I r P r I r

Ph

σ µ β β β β β β β β

β

∞

=

= −

∑

(2.4) The solutions (2.2) and (2.4) satisfy (1.5) identically and (2.4) along with (1.6) gives

( )

1 ( ) 2 ( ) ( , )

2

b

rzs m

m

a A m a B m Ph σ a β

+ = µβ (2.4)

where a1 and a2 are known functions given by

1 1

2 0 1

( ) ( )

( ) 2 (2 ) ( ) /

m

m m m m m

a P Q I a

a I a P a I a

β

β β β β β

= +

= + + +

90

The mixed boundary conditions (1.7) - (1.8) give the following triple series relations

0 ( )

3 ( , )

1

cos( 2) ( ) ( ) ,0 , ,

2 2

b

m a z

m

U h

a m B m U z b c z h

∞ β

=

+

∑

( ) 6

1

( ) ( ) cos( ) ( , ), ,

2

b

m m rr

m

m a m z h a z b z c

β β β σ

∞

=

= − < <

∑

Where U a a₀, ₃, ₆ etc. are given by

1

3 0 2

1

( )

( ) ( ) ( , )

2

m b

m rzs m

m

I a Ph

a m aI a a a

a

β β σ β

µβ

 

= + − + − 

 

0 . (0)

2 U = ah B

1

4 0 0

( )

( ) ( _m) ^m

m

a m I a I a

a

α β β

= + β

5( ) 1 0( _m) 1( _m) 1 _m(1 )

a m =α I aβ +I aβ +aβ +Q

4

6 2 5

1

( ) ( , )

2

b

rzs m

m

a Ph

a m a a a

a σ β

= − + µβ − +

The relations (2.5) - (2.6) are called triple series equation.

SOLUTION OF TRIPLE SERIES After taking,

3( ) ( ) ( )

a m B m =φ m (2.7)

and the trial solution as,

0 1

( ) 2 ^c ( ) sin( ) 1 ^{b h} ( ) sin

m m m

b c

m g t t dt U t dt

β φ = ^

∫

∫ ∫

0 0 1 1 1

2 ^c ( ) 1 ^b ( ) ^h ( ) ( )

bt g t dt U t dt ct U t dt U h

φ = ^

∫

∫

∫

The substitution of (2.7) - (2.9) into (2.5) satisfy it identically if

[

]

( ) ( ) ( ) / ,

c

b g t dt = U b −U c h

∫

where,

( )

1( ) ( , )

2 h b

U z = − U a z (2.11)

And the substitution of (2.7) - (2.8) into (2.6) and then inverting 91

2 0

( ) 1 ( ) ( ) ( , ) , ,

( )

c

g t t t b g t K t dx b t c

h t α

δ ^{ }

= ∆

∫

( )t G b t G t c( , ) ( , ) 1/ 2

δ =   (2.12)b

{

}

0

sin( ) ( , ) ( ) ( )

( ) ( , )

b

c rr

b

qz a z U z z

t dz D

G t z

σ − δ

∆ = −

∫

sin( ) ( ) ( , )

( , ) ,

( , )

c b

qz z M z

K t dz

G t z

δ α

α =

∫

Where M( , )α z is defined as

_{7 7} ^{5 6}

1 5

( ) ( )

( , ) cos( ) sin( ) ( ) ( )

m m ( )

m

a m a m

M z z a m where a m

α ^∞ β αβ a m

=

=

∑

( )

( )

0 0 1 0 1

sin( )

( ) ( ) ( , ) ( )

( , )

b h b h

c c

q d

U z U M z d U

G z

α α α α α α

=

∫ ∫

∫ ∫

The equation (2.12) is called Fredholm Integral equation of second kind. D in (2.13) is an airbitrary constant which will be determined through (2.10).

3 PHYSICAL QUANTITIES

The quantities of interest in fracture mechanics are fracture-design parameters. Two important fracture design paremeters are stress intensity-factor and crack shape.

STRESS-INTENSITY FACTORS The stress-intensity factors at crack tips are defined as,

lim ( )^e ( , ),

b rr

z b

K ₋ b zσ a z

= → − (3.1)

lim ( )^e ( , ),

c rr

z c

K ₊ z cσ a z

= → − (3.2)

lim ( )^e ( , ) 0,

b rz

z b

N ₋ b zσ a z

= → − = (3.3)

lim ( )^e ( , ) 0,

c rz

z c

N ₊ z cσ a z

= → − = (3.4)

Nb and Nc are zero because in our paper we took σ_rz( , )a z =0. But σ_rz^{( )b} ( , )a z does not possess singularity at crack tips therefore Nb, Nc will be zero.

NORMAL STRESS

Normal stress, σ^{( )}_rz^b ( , )a z , in the vicinity of crack tips, ( , ), ( , )a b a c is evaluated through the value of series on lefthand side of (2.6) while, 0≤ <z b c, < ≤z h,

( )

1

( ) sin( )

( , ) ( ),

( , )

e c

rr b

g t qt

a z dz z

G t z

σ =

∫

92

( )

1 1

0

1 sin( )

( ) ( ) ( , ) ( ) ( , ) ,

( , )

c b h

b c

z g M z d U M z q d

h G z

α α α α α α α

α

 

∆ = − + ′  + 

 

∫ ∫ ∫

Now, we substitute the value of g(t) from (2.12) into first part of right hand side of (3.5) and evaluate the integrals,

2 1

( )

2 1

( ) ( ), ( , ) 1

( ) ( )

e rr

z z

a z h z z

σ π

∆ + ∆

= −∆ + ∆ (3.7)

2 0

sin( )

( ) ( ) ( ) ( , ) ,

( )

c b

z qz z g K z d

z α α α

δ ^{ }

∆ = ∆ +

∫

Now using (3.7) into (3.1) - (3.2) and evaluate the limit,

3

3

( ), ( ),

b

c

K b

h K c

h π

π

∆ 

= 

∆ 

= − 

(3.9)

3( ) 0( ) 0( ) ^c ( ) ( , ) ,

b

x ψ x ^ x g α K α x dα^

∆ = ∆ +

∫

[ ]

0( )x qsin(qx G b c) ( , ) x b c q, , h

ψ = ⁻ = = π (3.11)

CRACK SHAPE

Crack shape is the graph of crack opening displacement U^{( )e} ( , ),a z b< <z c.

( )

( , ) ^c ( ) 1( ), ,

e

U a z =h

∫

4. SOLUTION OF FREDHOLM

INTEGRAL EQUATION

The numerical solution of Fredholm integral equation will be given for a special type of point force see figure 3, 4. Before we go for solution of Fredholm integral equation we evaluate the constant D. To evaluate constant D we take,

2 0

( ) 1 ( )

g t ( ) t

h δ t

= ∆ (4.1)

Then we use (2.10). The double singular integral is to be evaluated numerically.

(

)

1 1

( ) ( ) sin( / 2)

( ) ( ) ,

2,

t t

D qc h U b U c

F h π µ π

 ∆ + ∆ 

=     − − 

(4.3)

93

2 ( , ) (0, ) G b c

G c

µ = (4.4)

( )

3

sin( ) ( ) ( , )

( ) ( , )

c b

rr b

qz z a z

t dz

G z t

∆ =

∫

4 0 1 1

5 1 1

1

( ) ( ) sin( ) ( )

2 ( , ) ( ) ( ) sin( ) ( )

2 ( , ) , / sin( / 2)

2

b

h c

t U q F h d

G t

t U qh F h d

G t

F F qc

α α δ α α

α

α δ α α

α π µ

  

∆ = ′  −  

  

  

∆ = ′  +  

  

  

=   

∫

∫

In∆₄( )t α<t which gives cos(qα)>cosqt, therefore,

0

1 1 cos( )

cos( ) cos( ) cos( ) cos( ) ,

l

l

qt

qα qt qα qα

∞

=

− = +

∑

And, α > ⇒t cos(qα)<cos( ),qt then,

0

1 1 cos( )

cos( ) cos( ) cos( ) cos( ) ,

n

n

q

q qt qt qt

α α

∞

=

− = −

∑

Then,

2 1

4 1 0 1 0 1 1

0

cos( ) sin( ). ( )

( ) ( ) sin( )

( ) 2 ( ) cos( )

l

c b c b

b b l

l

dt h qt q

t F U q d dt u d

t t q

α δ α

α α α α

δ δ α

∞

= +

∆ = ′ −

∑

∫ ∫ ∫ ∫

And,

2 5

1 1 1

0

( ) ( ) sin( ) ( ) ( ) sin( ) cos( )

( ) 2 cos( ) ( )

c h c h l

b c b c

n

t dt h

F U q d dt U q q d

t α α α qt t α δ α α α α

δ δ

∞

=

∆ = ′ +

∑

∫ ∫ ∫ ∫

(4.10) Non of the integral is singular. Now for ∆₃( ) :t z>t, cos(qz)<cos( )qt

( ) ( )

( )

1 2 2

sin( ) ( ) ( , ) sin( ) ( , ) sin( ) ( , )

( , )

( ) ( , ) ( ) ( )

b b

c c rr c b c rr

b b b rr b

qz z a z qz a z

dt qz G b z

dz F a z d E E

t G z t z z

δ σ σ σ

δ ⁼ δ ⁻ δ ⁺

∫ ∫ ∫ ∫

(4.11) Where E=E( / 2, ),π µ complete elliptic integral of second type.

( )

2 1

0

sin( ). ( , ) cos ( ) ( ).

cos( ) ( )

b n

c c

rr

b n b

n

qz a z qz

E t dt d z

qt z

σ δ

δ

∞

= +

=

∑∫ ∫

Ifz<t, then in (4.11) replace E2 by E2 and given as, 94

( )

1 2

2 1

0

sin( ) ( , ) ( ) cos( )

( ) cos( )

l b

c c

rr

b b l

l

qz a z d

E t qt dt

z qz

δ σ

δ

∞

= +

= −

∑∫ ∫

Thus, the evaluation of D does not involve any singular integral. The same logic we shall apply when we will be solving Fredholm integral equation numerically. The quantities

( ) ( )

( , ), ( , )

b b

U a z σrz a z and σ_rr^{( )b} ( , )a z are given by (7.5.1), (7.5.3) and (7.5.5), respectively.

( , )

M t z is given by (7.5.8) which will be used.

( ) ( ) ( ) ( )

2 2 2 2 4 2 2 4 4 2 2 4

( ) 0 3

2 2 4 4 4

2 2 2 2 2 2 2 2 2

(0) 4 16 16 24 256 96

( , )

2 4 16 4 16

b aP d h z h z h h z z h h z z

U a z d

h h z h z h z h z

µ

 + + + + + + 

 

=  − + − − − + − 

1, 2:

d d distance of point force from r=0.

( )

( )

2 2 2 2

1 2

3 2

2 2

2 1 2

4

5 3

( ) 2 ! K+1 ! 3

2 16

2 (0)

k k

k

d d

d k k

d d

d a

d λ µ d

+ − +

=

 − 

=  + 

 

= +

( ) ( ) (

)

6 1 2

3 32 2

d =16 λ µ+ + λ+ µ d +d 

( )

2

7 12 2 3(1)

d = a λ+ µ d

3

8 1 3

(0) (1)

8 16

d d = aµ d − 

3 2 9

(1)

16 16

d d a

h

πµ 

=  − 

2 3 10

(1) 256 a d

d h

=π µ

(

) ^{( )}

11 1 10 3(1)

16

d a a d

h π λ µ

= + +

( ) ( ) ( )

( ) ( )

( ) ( )

2 2 2 2

( ) 0

5 2 2 2 2 2 2

4 2 2 4 4 2 2 4

6 2 2 4 2 2 4

3 2 3 3

7 2 2 4 2 2 4

4 16

( , )

2 2 4 16

16 24 256 96

4 16

6 48 2

4 16

b rz

aP h z h z

a z d

h h z h z

h h z z h h z z

d

h z h z

h z hz h z hz d

h z h z

σ µ λ µ

 + +

=  + −

+  − −

+ + + + + −

− −

+ + + 

− − 

95

( ) ( )

( ) ( ) ( )

( ) 0

8 8 1

10 2 3 11 4

( , ) 2 ,

2

, , ,

b rr

a z P d d h z

h

d h z h z d h z

σ θ

µ λ µ

θ θ θ

= − +  +

− + − 

The numerical evaluation of singular integral

2

1 1

0

sin( ) ( ) cos ( ) sin( ) ( ) cos ( ) sin( ) ( )

( ) ( , ) ( ) cos( ) ( ).cos( )

n n

c t c

n n

b b t

n

qz F z d qt qz F z dz qz qz F z dz

z G z t z qz z qt

δ δ δ

∞

+ +

=

 

=

∑

∫ ∫ ∫

Now we substitute,

2 2 2

2 sin (qz/ 2)=2 sin (qc/ 2)−G b c( , ) sin θ (4.14) Then

1 1

/ 2

1 1 0

0

sin( ) ( ) ( ) 1

cos( ) ( ) , ( ) ,

( ) ( , ) , ( ) cos( )

c n n

n n

b n

qz F z dz d

qt d

z G z t qt

π θ

θ

φ θ θ φ θ φ θ θ

δ φ θ

∞

+ +

=

 

 

= −

 

 

∫ ∑ ∫ ∫

Where ( )F z changes to ( )φ θ after using (4.14) and

2 1( ) cos(qc) G b c( , ) sin

φ θ = + θ (4.16)

None of the integrals is singular, there fore can easily be evaluated numerically. The Fredholm integral equation (2.12) will reduce to n linear algebraic equations in n variables which are n values of ( ),g t i=1,.... ,n giving, g_i =g t( )_i i.e., n gi's.

96

Fig-1. Geometry of two shell type discontinuities in a plate of thickness 2h and radius Infinite.

Fig.2. : Cross Section of Figure 1 by θ =0,π_.

Fig. 3. Point forces in ring shape which are of opening mode.

97

Fig. 4. Point forces with crack in two dimension.

References:

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2) Kushwaha,P.S.(1974), some two dimensional crack problems in the mathematical theory of elasticity,Ph.D. thesis,I.I.T., Bombay.

3) Love,A.E.H.(1944),a Treatise on the mathematical theory of elasticity, Dover publication, New York.

4) Sih,G.C.(1975), Three dimensional crack problems, Mechanics of fracture, Noordhoof international publishing,Leyden, Netherlands.

5) Tweed,J.(1970). The determination of stress- intensity factor of a partially closed Griffith crack, Int.J. Engg. Scid., 793-803.

6) Varughesh,P.P.(1992), Shear pull out of cracked interface of FRP cylinder, Ph.D.

Thesis, Agra University,Agra.

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8) Kushwaha,P.S.and Ahmad,M.S.(2006)the stress intensity factors of two shell type cracks opened by body forces,J.Rajasthan Acad. Phy. Sci.,Vol. 5, No. 4,pp. 399- 414.

9) Stable,P.,(1983), On the small crack fracture mechanics, Int.J.fracture.22 pp.203- 216.

10) Selvadurai,A.P.S. and Singh,B.M., (1996)The matrix fibrecrack in an elastic solid., trans. ASME Vol.63, No.3

11)Zhigang Suo and Hutchinson, J.W.,(1990) Interface cracks between two elastic layers., Int.J.Frac 43(1),pp 1-18.

12) Dirgantara T and Aliabadi MH (2002), Stress intensity factors for cracks in thin plates,Engineering Fracture Mechanics,69 , 1465-1486.

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98