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In this paper, an analytical and numerical study of strain fields, stress fields and displacements in a rotating hollow cylinder, whose walls were completely made in Functionally Graded Materials (FGM), was conducted. We have considered the rotating hollow cylinder submitted to an asymmetric radial loading. It is assumed that, because of the functional graduation of the material, the mechanical properties such as Young elastic modulus and the density varies in the radial direction, in accordance with a the power law function. The inhomogeneity parameter was selected between -1 and 1. On the basis of the second law of Newton, Hooke’s law and the strain-stress relationship, we established the differential equation which governs the equilibrium for a rotating hollow cylinder. We found the analytical solution and compared to the numerical solution obtained by using the shooting method and the fourth order Runge-Kutta algorithm. The analytical and numerical results lead to the conclusion that the magnitude of the tangential stresses is greater than that of the radial stresses. The changes due to the graduation of FGM does not produce consistent variations in the distribution of radial stresses, but strongly affects the distribution of tangential stresses. The tangential stresses, tangential strains and displacements are much higher at the inner surface of the cylinder wall. The internal radial pressure intensely affects the radial stresses and the radial strain.

Rotary devices such as disks, cylinders and even spheres have been widely used in mechanical applications and engineering including steam turbines and gas rotors, turbine generators and jet engines, internal combustion engines, boat propellers or even reciprocating and centrifugal compressors [

Indeed FGM is a new class of materials that has really emerged in the second half of the 90’s years. They are composite materials in which physical, chemical and mechanical properties vary continuously and which exhibit no discontinuity [

The main objective of this work is to find analytical and numerical solutions for the case of a rotating hollow cylinder whose walls are entirely made on Functionally Graded Materials and submitted to internal and external radial loading.

Analytical and numerical solutions obtained show clearly the effects of different profiles describing the graduation of the material properties. This is the case for the strain fields, stress fields and displacement.

The paper is organized as follows. In the first part we present the model studied. In the second part the equations describing the model are calculated and resolved. The third part presents the analytical and numerical results before concluding the work in the fourth part.

In this part, we analyze the stress and strain states of a rotating hollow cylinder submitted to non-axisymmetric pressure. Let us consider a long hollow FGM circular cylinder with inner radius a and outer radius b subjected to the action of a uniform internal and external pressure loading respectively P i and P 0 , as show in

The most common functionally graded materials in such applications are made of ceramic/metal non homogeneous structure, in which ceramic provides good thermal resistance and metal roles as a superior toughness and hardness [

E ( r ) = E 0 ( r b ) n ; ρ ( r ) = ρ 0 ( r b ) n , (1)

where E_{0} is the reference value of E, r is the radial coordinate and the exponent n is a positive or negative real number, reflecting the degree of non-uniformity of the material.

Due to the axisymmetric load condition of hollow cylinder, the circumferential component of displacement is zero, and the radial component u depends on the radial distance r only [

For the rotating hollow cylinder, the equation of equilibrium is [

d d r ( r σ r ) − σ θ + ρ ( r ) ω 2 r 2 = 0 (2)

where σ r and σ θ are the radial and circumferential components of stress, ω is the angular velocity of the cylinder, ρ the density of the material is assumed to be constant.The strain-stress’s relation for infinitesimally elastic deformation (Hooke’s law) is given by:

σ ¯ ¯ = λ T r ( ε ¯ ¯ ) I ¯ ¯ + 2 μ ε ¯ ¯ (3)

where the Lame’s constant is given by:

λ = E ν ( 1 + ν ) ( 1 − 2 ν ) ; μ = E 2 ( 1 + ν ) (4)

In the case of small deformations and due to the rotational symmetry, the strain-displacement relations are given by:

ε ¯ ¯ = ( ε r = d u r d r 0 0 0 ε θ = u r r 0 0 0 ε z ) (5)

By using Equations (1), (3), (4) and (5), the stress-strain relation can become:

{ σ r = E 0 ( 1 + ν ) ( 1 − 2 ν ) ( r b ) n [ ν ε z + ν u r r + ( 1 − ν ) d u r d r ] σ θ = E 0 ( 1 + ν ) ( 1 − 2 ν ) ( r b ) n [ ν ε z + ( 1 − ν ) u r r + ν d u r d r ] σ z = E 0 ( r b ) n ε z + ν ( σ r + σ θ ) (6)

Substituting Equation (6) into Equation (2), the equation of displacement is given by:

r d 2 u r d r 2 + ( n + 1 ) d u r d r + [ ν ( n + 1 ) − 1 ] r ( 1 − ν ) u r + n ν ε z 1 − ν + ( 1 + ν ) ( 1 − 2 ν ) ρ 0 ω 2 E 0 ( 1 − ν ) r 2 = 0 (7)

This equation can be written as:

r 2 d 2 u r d r 2 + ( n + 1 ) r d u r d r − [ 1 − ν ( n + 1 ) ] ( 1 − ν ) u r = − n r ν ε z 1 − ν − ( 1 + ν ) ( 1 − 2 ν ) ρ 0 ω 2 E 0 ( 1 − ν ) r 3 (8)

We can note that for a homogenous material ( n = 0 ) , we obtain the governing differential equation found in the literature for a hollow rotating cylinder [

{ r 2 d 2 u r d r 2 + r d u r d r − u r = − ( 1 + ν ) ( 1 − 2 ν ) ρ 0 ω 2 E 0 ( 1 − ν ) r 3 ; n = 0 ; ω ≠ 0 r 2 d 2 u r d r 2 + r d u r d r − u r = 0 ; n = 0 ; ω = 0 (9)

The next part of this work is devoted to the analytical and numerical resolution of Equation (8)

To find exacts solutions to displacements, strain and stress, we rewrite the Equation (8) in a series of three differential equations. One of these equations is homogenous and the others are inhomogeneous.

The Equation (8) becomes:

{ r 2 d 2 u r d r 2 + ( n + 1 ) r d u r d r − [ 1 − ν ( n + 1 ) ] ( 1 − ν ) u r = 0 r 2 d 2 u r d r 2 + ( n + 1 ) r d u r d r − [ 1 − ν ( n + 1 ) ] ( 1 − ν ) u r = − n r ν ε z 1 − ν r 2 d 2 u r d r 2 + ( n + 1 ) r d u r d r − [ 1 − ν ( n + 1 ) ] ( 1 − ν ) u r = − ( 1 + ν ) ( 1 − 2 ν ) ρ 0 ω 2 E 0 ( 1 − ν ) r 3 (10)

1) Resolution of homogenous equation

The homogenous equation is given by:

r 2 d 2 u r d r 2 + ( n + 1 ) r d u r d r − [ 1 − ν ( n + 1 ) ] ( 1 − ν ) u r = 0 (11)

To solve the Equation (11), we make the following change of variable:

r = e λ (12)

By using Equation (12), Equation (11) becomes:

d 2 u d λ 2 + n d u d λ − [ 1 − ν ( n + 1 ) ] 1 − ν u = 0 (13)

The solution of this equation is given by:

u H = C 1 r − ( n + K ) / 2 + C 2 r − ( n − K ) / 2 , C 1 , C 2 ∈ ℜ (14)

where K = n 2 + 4 − 4 n ν 1 − ν and the integration constants are C 1 and C 2

2) Resolution of inhomogeneous equations

According to Equation (10) the first inhomogeneous equation is given by:

r 2 d 2 u r d r 2 + ( n + 1 ) r d u r d r − [ 1 − ν ( n + 1 ) ] ( 1 − ν ) u r = − n ν ε z 1 − ν r (15)

To solve this equation, we search the particular solution u P 1 which has the same form as the second member of the equation:

u P 1 = A r + B (16)

where A and B are the constants.

By substituting Equation (16) into Equation (15) we obtain:

A = − ν ε z ; B = 0 (17)

Then Equation (16) becomes:

u P 1 = − ν ε z r (18)

According to Equation (10) the second inhomogeneous equation is given by:

r 2 d 2 u r d r 2 + ( n + 1 ) r d u r d r − [ 1 − ν ( n + 1 ) ] ( 1 − ν ) u r = − ( 1 + ν ) ( 1 − 2 ν ) ρ 0 ω 2 E 0 ( 1 − ν ) r 3 (19)

To solve Equation (19) we search the particular solution u P 2 which has the same form as the second member of the equation:

u P 2 = D r 3 (20)

where D is constant.

By substituting Equation (20) into Equation (19) we obtain the constant:

D = − ( 1 + ν ) ( 1 − 2 ν ) ρ 0 ω 2 [ 8 + 3 n − 2 ( 4 + n ) ν ] E 0 (21)

By substituting Equation (21) into Equation (20) we obtain:

u P 2 = − ( 1 + ν ) ( 1 − 2 ν ) ρ 0 ω 2 [ 8 + 3 n − 2 ( 4 + n ) ν ] E 0 r 3 (22)

The general solution of Equation (10) is given by:

u r = C 1 r − ( n + K ) / 2 + C 2 r − ( n − K ) / 2 − ( 1 + ν ) ( 1 − 2 ν ) ρ 0 ω 2 [ 8 + 3 n − 2 ( 4 + n ) ν ] E 0 r 3 − ν ε z r (23)

In the case of homogenous material, Equation (23) becomes:

{ u r = C 1 r − 1 + C 2 r − ν ε z r ; n = 0 ; ω = 0 u r = C 1 r − 1 + C 2 r − ( 1 + ν ) ( 1 − 2 ν ) ρ 0 ω 2 8 ( 1 − ν ) E 0 r 3 − ν ε z r ; n = 0 ; ω ≠ 0 (24)

By substituting Equation (24) into Equation (5) the components of strain tensor are given by:

{ ε r = − C 1 ( n + K ) 2 r − ( n + K + 2 ) / 2 − C 2 ( n − K ) 2 r − ( n − K + 2 ) / 2 − 3 ( 1 + ν ) ( 1 − 2 ν ) ρ 0 ω 2 [ 8 + 3 n − 2 ( 4 + n ) ν ] E 0 r 2 − ν ε z ε θ = C 1 r − ( n + K + 2 ) / 2 + C 2 r − ( n − K + 2 ) / 2 − ( 1 + ν ) ( 1 − 2 ν ) ρ 0 ω 2 [ 8 + 3 n − 2 ( 4 + n ) ν ] E 0 r 2 − ν ε z (25)

By substituting Equation (22) into Equation (6), the components of stress tensor are given by:

These solutions don’t exist when n = n c = − 8 ( 1 − ν ) 3 − 2 ν .

3) Boundary conditions

The boundary conditions are used to determine the constants C 1 and C 2 . For the hollow cylinder submitted to uniform pressures P 0 and P i on the inner and outer surfaces respectively, the mechanical boundary conditions can be expressed as:

σ r ( b ) = − P 0 ; σ r ( a ) = − P i (27)

By substituting Equation (27) into Equation (26), the constants are given by:

{ C 1 = − 2 ( 1 + ν ) ( 1 − 2 ν ) [ 8 + 3 n − 2 ν ( 4 + n ) ] [ ( n + K ) ( 1 − ν ) − 2 ν ] E 0 ( a K − b K ) { C 11 + C 12 + C 13 } C 2 = − 2 ( 1 + ν ) ( 1 − 2 ν ) [ 8 + 3 n − 2 ν ( 4 + n ) ] [ ( n − K ) ( 1 − ν ) − 2 ν ] E 0 ( a K − b K ) { C 21 + C 22 + C 23 } (28)

where

{ C 11 = − [ 8 + 3 n − 2 ν ( 4 + n ) ] P i a ( 2 + K − n ) / 2 b ( n + K ) C 12 = [ 8 + 3 n − 2 ν ( 4 + n ) ] P 0 a K b ( n + K + 2 ) / 2 C 13 = { ( 3 − 2 ν ) ( b ( n + K ) a ( n + K + 6 ) / 2 − a K b ( 3 n + K + 6 ) / 2 ) } ρ 0 ω 2 C 21 = [ 8 + 3 n − 2 ν ( 4 + n ) ] P o b ( 2 + K + n ) / 2 C 22 = − [ 8 + 3 n − 2 ν ( 4 + n ) ] P i b n a ( 2 + K − n ) / 2 C 23 = { ( 3 − 2 ν ) ( a ( n + K + 6 ) / 2 b n − b ( 3 n + K + 6 ) / 2 ) } ρ 0 ω 2 (29)

In the case of homogenous hollow cylinder, Equation (29) becomes:

{ C 1 = [ 8 ( 1 − ν ) ( P 1 − P 2 ) + ( b 2 − a 2 ) ( 3 − 2 ν ) ρ ω 2 ] ( 1 + ν ) a 2 b 2 8 ( a 2 − b 2 ) ( ν − 1 ) E 0 C 2 = − [ 8 ( 1 − ν ) ( P 1 − P 2 ) + ( b 2 − a 2 ) ( 3 − 2 ν ) ρ ω 2 ] ( 1 + ν ) ( 2 ν − 1 ) 8 ( a 2 − b 2 ) ( ν − 1 ) E 0 ; n = 0 , ω ≠ 0 (30)

In order to determine the constant ε z , we use the expression of the z-component of the force given by:

F z ( α , β ) = ∫ α β r σ z d r = B 1 [ α ( n − K + 2 ) / 2 − β ( n − K + 2 ) / 2 ] + B 2 [ α ( n + K + 2 ) − β ( n + K + 2 ) ] + B 3 [ α ( n + 4 ) − β ( n + 4 ) ] + B 4 [ α ( n + 2 ) − β ( n + 2 ) ] ε z (31)

where:

B 1 = − ν E 0 ( K + n − 2 ) C 1 2 ( 1 + ν ) ( 1 − 2 ν ) b n ; B 2 = − ν E 0 ( K − n − 2 ) C 2 2 ( 1 + ν ) ( 1 − 2 ν ) b n ; B 3 = − 4 ν ρ 0 ω 2 8 + 3 n − 2 ν ( 4 + n ) b n ; B 4 = E o b n

We suppose that the ends of the cylinders are free. It contracts as is rotates. The net force F_{z} in the axial direction should be equal to be zero [

ε z = − 2 ( n + 2 ) B 1 ( n − K + 2 ) B 4 [ α − ( n + K + 2 ) / 2 − β − ( n + K + 2 ) / 2 ] − 2 ( n + 2 ) B 2 ( n + K + 2 ) B 4 [ α − ( n − K + 2 ) / 2 − β − ( n − K + 2 ) / 2 ] − ( n + 2 ) B 3 ( n + 4 ) B 4 [ α 2 − β 2 ] (32)

In this part we study the influence of the inhomogeneous parameter n on the stress, strain and displacement in the rotation hollow cylinder. We use the constants given by the properties of an aluminum alloy (7075-T6) [

a = 40 cm ; b = 60 cm ; ν = 0.3 ; E 0 = 79.0 × 10 9 Pa ; P o = 10 5 Pa ; ρ 0 = 2800 kg ⋅ m − 3 ; ω = 250 rad ⋅ s − 1 ; a ≤ r ≤ b

To validate the analytical results we make the comparison with the numerical one. Equation (2) can be written as:

d σ r d r + σ r − σ θ r + ρ ( r ) ω 2 r = 0 (33)

The solution of this equation can be efficiently handled by using a special stress-function that automatically satisfies the equilibrium Equation (33). A particular stress-stress function relation is given by:

σ r = F r ; σ θ = d F d r + ρ ( r ) ω 2 r 2 , (34)

where F = F ( r ) is the stress function. The Strain compatibility equation is given by:

ε r = d d r ( r ε θ ) (35)

The strain-stress relation can be written:

{ ε θ = 1 E ( r ) ( σ θ − ν σ r ) ε r = 1 E ( r ) ( σ r − ν σ θ ) (36)

Substitution of Equations (34) and (36) in the compatibility relation (35) gives the following governing equation:

r 2 d 2 F d r 2 + r ( 1 − n ) d F d r − ( 1 − ν n ) F = ρ ω 2 r n + 3 ( 3 + ν ) b n . (37)

with boundary conditions:

F ( a ) = a P i ; F ( b ) = b P o , (38)

the Equation (36) can be transformed as:

Y ′ 2 = 1 − n r Y 2 + 1 − ν n r 2 Y 1 + ρ ω 2 r n + 1 ( 3 + ν ) b n , (39)

where Y 1 = F ; Y 2 = d F d r = F ′ = f ( r , Y 1 , Y 2 ) .

The Equation (36) is transformed into a set of coupled time-dependent discrete differential equations that are solved by using the shooting method and the fourth order Runge-Kutta algorithm. The analytical and numerical solutions are thus presented.

In ^{−4} m and 3.22 × 10^{−4} m for P_{i} = 10^{5} Pa. We conclude that radial displacement decrease when the internal pressure increases. More ever the radial displacement curves save the parallelism when internal pressure P_{i} and the inhomogeneous parameter n changes. In

The variation of radial strain is depicted in

In

For n = 1, the radial strain decreases gradually along the radial direction and the maximum value of strain is 7.6 × 10^{−4} m.

In

For Pi = Po tangential strains increase with increasing inhomogeneous parameter and reaches to maximum value of 4.8 × 10^{−4} m (n = −1). For P_{i} = 100Po, tangential strains decrease with increasing inhomogeneous parameter and the maximum value of tangential strain is 8.3 × 10^{−4} m (n = 1).

In

reaches to its maximum value 3.75 Mpa at r = 0.5 and then decrease to zero. In addition, radial stress decreases when inhomogeneous parameter n. They have lower value for n = 1 when compared with the results for n = 0 and n = −1. It is due to the fact that Young modulus of elasticity decreases when inhomogeneous parameter (n) increases.

The variation of tangential stress obtained from analytical and numerical solutions is depicted in

In

inner surface. For P_{i} = P_{o}, tangential stress decreases with increasing inhomogeneous parameter and reaches to its maximum value of 55 Mpa (n = −1). For P_{i} = 100P_{o}, tangential strains decrease with increasing inhomogeneous parameter and maximum value of tangential stress is 82 Mpa (n = −1). They have lower value for n = 1when compared with the results for n = 0 and n = −1.

In this paper we analyze stress and strains along the radial direction of inhomogeneous hollow rotating cylinder under axisymmetric radial loading. Analytical and numerical results lead us to conclude that:

- The stress, strain and displacement obtained from analytical and numerical solution are in good correlation.

- The magnitudes of tangential stress are higher than those of radial stress.

- Change in the gradient of the FGM tube does not produce a substantial variation of the radial stress, but strongly affects the distribution of the tangential stress.

- Tangential stress, tangential strain and displacements are higher at the inner surface.

- Internal radial pressure strongly affects the radial stresses and radial strain.

The results obtained are helpful in designing FGM cylindrical vessels to prevent failure. This investigation permits us to optimize the elastic response of cylinders under pressure by tailoring the thickness variation of the elastic properties and to reduce manufacturing costs given by the technological limitations that occur to produce entire functionally graded walls.

Atangana Nkene, E.R., Mambou Ngueyep, L.L., Ndop, J., Djiokeng, E.S. and Ndjaka, J.-M.B. (2018) Displacements, Strains, and Stresses Investigations in an Inhomogeneous Rotating Hollow Cylinder Made of Functionally Graded Materials under an Axisymmetric Radial Loading. World Journal of Mechanics, 8, 59-72. https://doi.org/10.4236/wjm.2018.83005