Journal of Mathematical Sciences, Vol. 157, No. 1, 2009
SOLUTION OF SOME TWO-DIMENSIONAL PROBLEMS OF ELASTICITY N. Zirakashvili
UDC 517.958:539.3
Abstract. Using the method of boundary elements, we obtain numerical solutions of two-dimensional (plane deformation) boundary-value problems on the elastic equilibrium of infinite and finite homogeneous isotropic bodies having elliptic holes with cracks and cuts of finite length.
In the first part of the work, using the method of boundary elements (see [2]), we solve the two-dimensional (plane deformation) problem on the elastic equilibrium of an infinite, homogeneous, isotropic body with an elliptic hole having cracks of the same length L. The interior surface of the body is free from stresses, and all-round tension is given at infinity. Numerical results are obtained and those of one and two cracks are presented. The second part of the work is devoted to finding in the elliptic coordinates θ, α an elastic equilibrium of a finite body occupying the domain Ω = {θ1 < θ < θ2 , 0 < α < π}. Note that nonzero stresses are given for θ = θ1 and zero stresses are given for θ = θ2 ; the conditions of symmetry and antisymmetry (see [3]) are assigned for α = 0 and α = π, respectively. It is not difficult to understand that this problem coincides with the boundary-value problem on the elastic equilibrium of an elliptic ring with a cut on whose contours the symmetry conditions are fulfilled. I. In an infinite domain Ω = {θ1 < θ < ∞, 0 < α < 2π} with an elliptic hole θ = θ1 containing cracks of length L, find a solution of a system of equations of equilibrium (see [4]) with respect to the unknowns D, K, u, and v: ∂D ∂K − = 0, ∂θ ∂α ∂D ∂K + = 0, ∂α ∂θ
∂u ∂v κ−1 2 + = h D, ∂θ ∂α κμ 0 ∂v ∂u 1 − = h20 K, ∂θ ∂α μ
(1)
with the following boundary conditions: θ = θ1 :
σθθ = 0,
σθα = 0,
(2)
θ→∞:
σθθ = p,
σθα = 0,
(3)
α = 0, 2π :
σαα = 0,
σθα = 0,
θ1 < θ < θ1 + L,
(4)
where u=
2hu , c2
v=
2hv , c2
h0 =
cosh(2θ) − cos(2α),
κ = 4(1 − ν),
μ=
E ; 2(1 − ν)
ν and E are well known constants, u and v are the components of the displacement vector, and σαα , σθθ , and σθα are the components of the stress tensor in the system of elliptic coordinates θ, α (0 ≤ θ < ∞, 0 ≤ α < 2π). If x and y are Cartesian coordinates, then x = c cosh θ cos α and y = c sinh θ sin α. Let c cosh(2θ) − cos(2α) hθ = hα = h = √ 2 Translated from Sovremennaya Matematika i Ee Prilozheniya (Contemporary Mathematics and Its Applications), Vol. 51, Differential Equations and Their Applications, 2008. c 2009 Springer Science+Business Media, Inc. 1072–3374/09/1571–0079
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be the Lam`e coefficients, c = 1 be the scale coefficient;
κ−1 2 h D be the divergence of the displacement κμ 0
1 2 h K be the curl of the displacement vector. μ 0 To obtain a numerical solution of this problem, the combined method of fiction load [2] and displacement discontinuity methods [1, 2] are used. This boundary-value problem generalizes boundary-value problems for infinite bodies containing circular holes and radial cracks (see [7]). To solve the exterior problem with given nonzero stresses at infinity, we formulate boundary conditions in additional stresses. The boundary conditions (2)–(4) in additional stresses can be written as follows:
vector, and
σθθ = −p,
σθα = 0,
α = 0, 2π : σαα = −p,
σθα = 0,
θ = θ1 :
(5) θ1 < θ < θ1 + L.
(6)
If the boundary of the domain is divided into N segments of small lengths, then it is possible to i = −p (or σ i = −p) and assume that on every element i constant normal and tangential stresses σθθ αα i σθα = 0, respectively, act. Thus, the boundary conditions (5) and (6) in additional stresses can be written in the following form: i σθθ = −p,
θ = θ1 :
i σθα = 0,
i = −p, α = 0, 2π : σαα
i σθα = 0,
(7) θ1 < θ < θ1 + L.
(8)
For every boundary element, we choose uniformly distributed concentrated forces. For example, for the jth element, we assume that the tangential Psj and normal Pnj stresses are distributed continuously. Thus, for the jth element we have applied fictitious stresses Psj and Pnj and also real stresses σsj and σnj which are caused by the stresses applied to all boundary elements. Using the solution of the Kelvin problem for the plane deformation (see [6]) and transformation formulas of coordinates (see [5]) (with regard for the segment orientation), we can calculate real stresses σsi and σni at the midpoint of each segment, i = 1, . . . , N1 . Thus, we obtain the following formulas: σsi
≡
i σθα
=
i = σni ≡ σθθ
N1
j ij j Aij ss Ps + Asn Pn ,
j=1 N1
j
(9)
j ij Aij ns Ps + Ann Pn ,
j=1
ij ij ij where Aij ss , Asn , Ans , and Ann are the boundary coefficients of influence of stresses for the problem under consideration. For example, the coefficient Aij ns supplies with the real normal stress at the center of the ith segment, caused by the constant unit tangential load (Psj = 1), applied to the jth segment. Now consider that part of the domain which contains a crack having two contours. The method of fictitious loads does not fit for the solution of crack problems, since the influence of elements on one contour is indistinguishable from that on the other contour. For the solution of problems of such type, we can apply the boundary-element method which is also called the method of discontinuous displacements (see [2]). This method is based on the analytic solution of the problem on an infinite plane whose displacements have discontinuity constant in size in the limits of an infinite segment. Analytic solutions of that problem have been obtained by S. L. Crouch (see [1]). If the crack is divided into N2 segments (elements) of small length, then we can assume that the displacement discontinuity is constant in the limits of every element length. On the basis of the analytic solution obtained by Crouch, one can find the influence of a separate elementary displacement discontinuity on the displacements and stresses at an arbitrary point of an infinite rigid body. For
80
example, the tangential and normal stresses at the ith element can be expressed in terms of the components of the jth element displacement discontinuity. If the elementary displacement discontinuity is placed on each of the segments along the crack, then we obtain ⎧ N ij j ⎪ ⎪ i i ij j ⎪ C , ≡ σ = D + C D σ ⎪ s ss s sn n θα ⎪ ⎨ j=N1 +1 i = N1 + 1, . . . , N = N1 + N2 , (10) N ⎪ ⎪ ⎪ i i ij j ij ⎪ Ds + Cnn Dnj , Cns ⎪ ⎩ σn ≡ σαα = j=N1 +1
ij , Css
ij ij Csn , Cns , and Cnn are the boundary coefficients of influence for stress. For example, the where ij supplies with the normal stress (σni ) at the center of the ith segment, which is caused coefficient Cns by the constant unit displacement discontinuity directed tangentially along the jth element (Dsj = 1).
For the boundary conditions that must hold on θ = θ1 , we use formulas (9), which are obtained by the method of fictitious loads, while for the cracks we use formulas (10), obtained by the method of discontinuous displacements. Thus, we obtain the following system of 2N linear equations with 2N unknowns (N = N1 + N2 ): ⎧ N1 N ij j ⎪ ⎪ ij j ij j ij j ⎪ A + Css Ds + Csn P + A P Dn = 0, ⎪ ss s sn n ⎪ ⎨ j=1 j=N1 +1 i = 1, . . . , N. (11) N1 N ⎪ ⎪ ⎪ j ij j ij j ij ⎪ Aij Cns Ds + Cnn Dnj = −p, ⎪ ns Ps + Ann Pn + ⎩ j=1
j=N1 +1
The stresses Psj and Pnj in these equations are fictitious. They were introduced as auxiliary unknowns and do not have physical sense. Along with what was said above, linear combinations of fictitious loads (9) provide us with real tangential and normal stresses which are aimed at satisfying the boundary conditions, while the unknowns Dsj and Dnj represent discontinuous displacements. If system (11) is solved by any numerical method (here by the Gaussian method), we can express displacements and stresses at an arbitrary point of the body. Using the MATLAB computing system, we have obtained numerical results and constructed graphs for the boundary-value problem (1)–(4) (or (1), (7), (8)) for ν = 0.3, E = 7 × 104 N/m2 , N1 = 360, N2 = 20, p = 10 N/m2 , θ1 = 1 m. In particular, we obtained numerical results for three values of crack length L = 0.1, 0.5, and 0.8 m. In Table 1, we can see the values of σαα for θ = θ1 , 0 < α < 2π, and of one crack along α = 0 and of two cracks along α = 0 and α = π of length L = 0.1, L = 0.5, and L = 0.8. Relying on the table, we can draw the following conclusions. The number of cracks and their length increase, the concentration of stresses σαα , strange as it may seem, decreases. Knowing this fact, sometimes engineers make themselves the so-called technical cracks to reinforce tunnel constructions. Now we solve the boundary-value problem on the elastic equilibrium of a elliptic ring with a cut. The problem is formulated as follows.
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Table 1. Stresses σαα on the contour θ = θ1 . 1 crack α 0.0349 0.1396 0.8378 1.1170 1.2566 1.3963 1.5359 1.6057 1.6755 1.8151 2.3736 3.1067 3.1765 3.2114 3.7699 4.6077 4.6775 4.7473 4.8869 5.0265 5.1662 5.3058 6.1436 6.2657
L = 0.1
L = 0.5
2 cracks L = 0.8 L = 0.1
32.224 46.484 84.109 10.767 −13.489 −12.653 18.330 10.219 5.149 16.398 12.780 9.523 15.753 13.422 11.028 15.347 13.933 12.283 15.181 14.440 13.419 15.188 14.721 13.978 15.256 15.032 14.545 15.577 15.768 15.746 19.436 20.898 22.443 26.363 28.984 32.000 26.386 29.011 32.032 26.328 28.944 31.954 21.193 22.986 24.950 15.283 15.115 14.689 15.199 14.796 14.118 15.177 14.508 13.559 15.313 13.994 12.429 15.689 13.490 11.196 16.305 12.873 9.730 17.149 11.958 7.879 12.353 −12.792 −12.195 32.224 46.484 84.109
25.279 8.168 18.942 18.090 16.464 11.669 11.595 25.279 5.971 14.125 18.954 11.595 25.279 6.9954 18.769 6.940 11.595 25.279 10.606 16.117 17.962 18.674 9.425 25.279
L = 0.5
L = 0.8
39.476 −11.266 2.226 −2.350 −6.247 −9.871 −2.228 39.476 −12.052 −8.558 2.313 −2.228 39.476 −10.167 1.018 −11.723 −2.228 39.476 −10.322 −6.729 −2.817 0.427 −10.788 39.476
71.497 −10.337 −5.431 −6.757 −7.753 −9.404 10.225 71.497 −10.236 −8.617 −5.401 10.225 71.497 −4.445 −5.820 −10.487 10.225 71.497 −9.720 −7.898 −6.876 −5.999 −10.046 71.497
II. In the domain Ω1 = {θ1 < θ < θ2 , 0 < α < π}, we seek a solution of the system of equilibrium equations (1) with the following boundary conditions: α (a) θ = θ1 : σθθ = p cos , σθα = 0, (12a) 2 (12b) (b) θ = θ2 : σθθ = 0, σθα = 0, α=0:
v = 0,
σθα = 0,
(12c)
(d) α = π :
u = 0,
σαα = 0.
(12d)
(c)
The above-formulated problem is solved by the boundary-element method. At the characteristic points of the domain, we obtain stresses for ν = 0.3, E = 7 × 104 N/m2 , θ1 = 1 m, θ2 = 10 m, p = 10 N/m2 . The semiellipse θ = θ1 and θ = θ2 are divided into 180 equal arcs, whereas the linear parts of the boundary α = 0 and α = π are divided into 40 equal segments. It should be noted that the problem considered above can be solved analytically. This solution is given below. III. The method of separation of variables yields an analytical (exact) solution of the boundary-value problem (1), (12) for θ2 → ∞. The solution is constructed by means of the general representation of 82
a solution through two harmonic functions ϕ1 and ϕ2 . The components of the stress tensor have the following form:
∂ 2 ϕ2 ∂ 3 ϕ1 ∂ 2 ϕ1 h2 2 σθθ = 2 sinh θ1 coth θ −2 sinh θ cos α − μ ∂α3 ∂θ∂α ∂α2
∂ 3 ϕ1 ∂ 2 ϕ1 ∂ 2 ϕ2 − 2 cosh2 θ1 tanh θ − − 2 cosh θ sin α ∂θ∂α2 ∂α2 ∂θ∂α
∂ 2 ϕ1 2 [cosh(2θ1 ) − cosh(1θ)] ∂ 2 ϕ1 sinh θ cos α , cosh θ sin α − − cosh(2θ) − cos(2α) ∂α2 ∂θ∂α
h2 ∂ 3 ϕ1 ∂ 2 ϕ1 ∂ 2 ϕ2 2 σαα = 2 cosh θ1 tanh θ cosh θ sin α +3 −2 μ ∂θ∂α2 ∂α2 ∂θ∂α
∂ 3 ϕ1 ∂ 2 ϕ1 ∂ 2 ϕ2 2 − 2 sinh θ1 coth θ sinh θ cos α +3 −2 ∂α3 ∂θ∂α ∂α2
2 [cosh(2θ1 ) − cosh(1θ)] ∂ 2 ϕ1 ∂ 2 ϕ1 + sinh θ cos α , cosh θ sin α − cosh(2θ) − cos(2α) ∂α2 ∂θ∂α
h2 ∂ 3 ϕ1 ∂ 2 ϕ1 ∂ 2 ϕ2 2 cosh θ sin α σθα = − 2 cosh θ1 tanh θ −2 − μ ∂α3 ∂θ∂α ∂α2
∂ 3 ϕ1 ∂ 2 ϕ1 ∂ 2 ϕ2 2 − 2 sinh θ1 coth θ cosh θ sin α − −2 ∂θ∂α2 ∂α2 ∂θ∂α
2 [cosh(2θ1 ) − cosh(1θ)] ∂ 2 ϕ1 ∂ 2 ϕ1 + cosh θ sin α . sinh θ cos α + cosh(2θ) − cos(2α) ∂α2 ∂θ∂α The boundary conditions (12c) and (12d) are satisfied if ϕi =
2
Aij e−k(θ−θ2 ) sin(kα),
i = 1, 2,
j=1
k=
2j − 1 . 2
The constants Aij are defined if they satisfy the boundary conditions (12a) and (12b). It is important to indicate that the solution found in the domain Ω1 can be continuously extended through the boundary α = π. As a result, we obtain the domain Ω2 = {θ1 < θ < θ2 , 0 < α < 2π} with a cut. On the contours of the cut α = 0 and α = 2π, the conditions v = 0, σθα = 0 hold. Thus, we have obtained exact and approximate solutions in the annular domain with the cut on the contours of which the symmetry conditions hold. Comparison of results obtained by the boundary-element method and the exact solution shows that the results are in good agreement (See Table 2). This circumstance allows us to conclude that the application of the method of boundary elements proved to be correct for the solution of boundary-value problems considered in the present paper, as well as for the circular ring (see [1]). In Table 2, we can see the values of exact and approximate solutions of the problem for the stresses σαα , σθα , and σθθ for α = π/3, θ1 < θ < θ2 . REFERENCES 1. S. L. Crouch, “Solution of plane elasticity problems by the displacement discontinuity method,” Int. J. Num. Methods Engng., 10, 301–343 (1976). 2. S. L. Crouch and A. M. Starfield, Boundary Element Methods in Solid Mechanics, George Allen, London–Boston–Sydney (1983). 3. N. Khomasuridze, “Thermoelastic equilibrium of bodies in generalized cylindrical coordinates,” Georgian Math. J., 5, No. 6, 521–544 (1998). 83
Table 2. Stresses for α = π/3 (θ1 < θ < θ2 ). σθθ θ 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.0 11.0 12.0 13.0 14.0 15.0 16.0 17.0 18.0 19.0 20.0
σαα
σθα
Approx.
Exact
Approx.
Exact
Approx.
Exact
4.07e + 03 7.13e + 02 1.31e + 02 2.46e + 01 5.22e + 00 1.13e + 00 2.47e − 01 5.44e − 02 1.20e − 02 2.65e − 03 5.86e − 04 1.29e − 04 2.87e − 05 6.37e − 06 1.42e − 06 3.26e − 07 7.83e − 08 2.09e − 08 6.32e − 09 1.58e − 09
4.07e + 03 7.13e + 02 1.24e + 02 2.46e + 01 5.24e + 00 1.15e + 00 2.54e − 01 5.66e − 02 1.26e − 02 2.82e − 03 6.28e − 04 1.40e − 04 3.13e − 05 6.98e − 06 1.56e − 06 3.47e − 07 7.75e − 08 1.73e − 08 3.86e − 09 8.61e − 10
−2.85e + 03 −6.18e + 02 −1.13e + 02 −2.41e + 01 −5.86e + 00 −1.16e + 00 −2.33e − 01 −5.13e − 02 −1.13e − 02 −2.50e − 03 −5.52e − 04 −1.22e − 04 −2.69e − 05 −5.93e − 06 −1.30e − 06 −2.81e − 07 −5.83e − 08 −1.06e − 08 −9.25e − 10 5.46e − 10
−2.86e + 03 −6.18e + 02 −1.17e + 02 −2.40e + 01 −5.19e + 00 −1.14e + 00 −2.54e − 01 −5.66e − 02 −1.26e − 02 −2.82e − 03 −6.28e − 04 −1.40e − 04 −3.13e − 05 −6.98e − 06 −1.56e − 06 −3.47e − 07 −7.75e − 08 −1.73e − 08 −3.86e − 09 −8.61e − 10
−8.24e − 13 −2.77e + 02 −7.65e + 01 −1.79e + 01 −4.04e + 00 −9.02e − 01 −2.27e − 01 −5.06e − 02 −1.13e − 02 −2.52e − 03 −5.62e − 04 −1.26e − 04 −2.80e − 05 −6.27e − 06 −1.40e − 06 −3.17e − 07 −7.28e − 08 −1.76e − 08 −4.78e − 09 −1.19e − 09
−3.92e − 13 −2.78e + 02 −7.65e + 01 −1.80e + 01 −4.08e + 00 −9.16e − 01 −2.05e − 01 −4.58e − 02 −1.02e − 02 −2.28e − 03 −5.09e − 04 −1.13e − 04 −2.53e − 05 −5.65e − 06 −1.26e − 06 −2.81e − 07 −6.28e − 08 −1.40e − 08 −3.12e − 09 −6.97e − 10
4. N. Khomasuridze and N. Zirakashvili, “Some two -dimensional elastic equilibrium problems of elliptic bodies,” Proc. I. Vekua Inst. Appl. Math., 49, 39–48 (1999). 5. N. I. Muskhelishvili, Some Basic Problems of the Mathematical Theory of Elasticity, Noordhoff, Groningen (1958). 6. I. S. Sokolnikoff, Mathematical Theory of Elasticity, McGraw-Hill, New York (1956). 7. N. Zirakashvili, “The application of the boundary element method to the solution of boundary value problems for elastic body containing circular hole and radial cracks,” Bull. Georgian Acad. Sci., 173, No. 2 (2006). N. Zirakashvili I. Vekua Institute of Applied Mathematics, Tbilisi, Georgia E-mail:
[email protected]
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