CRNOGORSKA AKADEMIJA NAUKA I UMJETNOSTI GLASNIK ODJELJENJA PRIRODNIH NAUKA, 16, 2005. QERNOGORSKAYA AKADEMIYA NAUK I ISSKUSTV GLASNIK OTDELENIYA ESTESTVENNYH NAUK, 16, 2005. THE MONTENEGRIN ACADEMY OF SCIENCES AND ARTS PROCEEDINGS OF THE SECTION OF NATURAL SCIENCES, 16, 2005 UDK 539.319 Vlado A. Lubarda MOHR S CIRCLES FOR NON-SYMMETRIC STRESSES AND COUPLE STRESSES A b s t r a c t Determination of the principal values of a non-symmetric stress tensor in plane strain problems of couple stress theory based on Mohr s circle construction is presented. It is shown that the antisymmetric component of the stress tensor affects the maximum shear stress, but not the maximum normal stress. The analysis is then extended to non-symmetric couple stresses under conditions of anti-plane strain. MOHROVI KRUGOVI ZA NESIMETRIČNE NAPONE I NAPONSKE SPREGOVE I z v o d U radu je prezentovana procedura za odredjivanje glavnih napona nesimetricnog tenzora napona u ravnom problemu deformacije na bazi Prof. dr V.A. Lubarda, The Montenegrin Academy of Sciences and Arts, 81000 Podgorica, SCG, and University of California, San Diego, CA 92093-0411, USA.
2 V.A. Lubarda konstrukcije Mohrovog kruga. Pokazano je da antisimetrična komponenta tenzora napona utiče na maksimalni smičući napon, ali ne i na maksimalni normalni napon. Analiza je proširena na slučaj nesimetricnih naponskih spregova u uslovima antiravne deformacije. 1. INTRODUCTION In a micropolar continuum the deformation is described by the displacement vector and an independent rotation vector (Eringen, 1968,1999; Nowacki, 1986). In the couple stress theory, the rotation vector ϕ i is not independent of the displacement vector u i but subject to the constraint ϕ i = 1 2 e ijk ω jk = 1 2 e ijk u k,j, ω ij = e ijk ϕ k, (1.1) as in classical continuum mechanics (Mindlin and Tiersten, 1962; Koiter, 1964). The skew-symmetric alternating tensor is e ijk, and ω ij are the rectangular components of the infinitesimal rotation tensor. The comma designates the partial differentiation with respect to Cartesian coordinates x i. The gradient of the rotation is a non-symmetric curvature tensor κ ij = ϕ j,i = e jkl ɛ ik,l. (1.2) Since ɛ ij is symmetric and e ijk is skew-symmetric, the curvature tensor in couple stress theory is a deviatoric tensor (κ kk = 0). A surface element ds transmits a force vector T i ds and a couple vector M i ds. The surface forces are in equilibrium with a non-symmetric Cauchy stress t ij, and the surface couples are in equilibrium with a non-symmetric couple stress m ij, such that T i = n j t ji, M i = n j m ji, (1.3) where n j are the components of the unit vector orthogonal to the surface element under consideration. In the absence of body forces and body couples, the differential equations of equilibrium are t ji,j = 0, m ji,j + e ijk t jk = 0. (1.4)
Mohr s Circles for Couple Stresses 3 By decomposing the stress tensor into its symmetric and antisymmetric part t ij = σ ij + τ ij, it readily follows from the moment equilibrium equation that τ ij = 1 2 e ijkm lk,l. (1.5) If the gradient of the couple stress vanishes at some point, the stress tensor is symmetric at that point. The normal stress in the plane orthogonal to the direction n is t n = σ n = σ ij n i n j, (1.6) because τ ij n i n j = 0 in view of the symmetry of n i n j. Thus, the principal stresses of the symmetric tensor σ ij are also the principal stresses of the nonsymmetric tensor t ij, although there are shear stresses in the principal planes of t ij due to antisymmetric shear stress components τ ij (Lubarda, 2003). The magnitude of this shear stress is τ 2 n = n i n j τ ik τ jk = τ 2 12 + τ 2 23 + τ 2 31 (n 1 τ 23 + n 2 τ 31 + n 3 τ 12 ) 2. (1.7) Similarly, the couple stress component m n = m ij n i n j is independent of the antisymmetric part of m ij. In the case if isotropic linear elasticity, σ ij = 2µ ɛ ij + λ ɛ kk δ ij, (1.8) m ij = 4α κ ij + 4β κ ji. (1.9) where µ, λ, α, and β are the Lamé-type constants of couple stress elasticity. In this case, since κ kk = 0, the couple stress is a deviatoric tensor (m kk = 0). 2. MOHR S CIRCLE FOR NON-SYMMETRIC STRESSES IN PLANE STRAIN For the plane strain problems, the displacement components are u 1 = u 1 (x 1, x 2 ), u 2 = u 2 (x 1, x 2 ), and u 3 = 0. The stress and couple
4 V.A. Lubarda t 22 m 23 t 21 t rr t 11 m 13 m 13 t 11 m 13 j t rj m r3 t 11 t 21 m 23 t 21 m 23 (a) t 22 t 22 (b) Figure 1: (a) A rectangular material element under conditions of plane strain; (b) an inclined plane at an angle ϕ supports the stresses t ρρ and t ρϕ, and the couple stress m ρ3. stress tensors are accordingly t 11 0 t = t 21 t 22 0 0 0 t 33, m = 0 0 m 13 0 0 m 23 m 31 m 32 0. (2.1) Consider a material element with sides parallel to coordinates directions x 1 and x 2 (Fig. 1a). In an inclined plane whose normal ρ makes an angle ϕ with the direction x 1, the normal and shear stresses are t ρρ and t ρϕ, and the couple stress is m ρ3 (Fig. 1b). From the equilibrium conditions of the triangular element it readily follows that m ρ3 = m 13 cos ϕ + m 23 sin ϕ, (2.2) t ρρ = 1 2 (t 11 + t 22 ) + 1 2 (t 11 t 22 ) cos 2ϕ + 1 2 ( + t 21 ) sin 2ϕ, (2.3) t ρϕ = 1 2 ( t 21 ) + 1 2 ( + t 21 ) cos 2ϕ 1 2 (t 11 t 22 ) sin 2ϕ. (2.4) The planes with the maximum magnitude of t ρρ and t ρϕ are defined
Mohr s Circles for Couple Stresses 5 by tan 2ϕ 1 = + t 21, tan 2ϕ 2 = 2t 11, (2.5) 2t 11 + t 21 with the obvious connection ϕ 2 = ϕ 1 ± π/4. The corresponding extreme values of the stresses are t max ρρ = 1 2 (t 11 + t 22 ) ± 1 (t11 t 22 ) 2 + ( + t 21 ) 2, (2.6) 2 t max ρϕ = 1 2 ( t 21 ) ± 1 (t11 t 22 ) 2 + ( + t 21 ) 2. (2.7) 2 It is noted that t max ρρ depends only on the symmetric part of the stress tensor t ij. Indeed, if we decompose t into its symmetric and antisymmetric parts, σ = σ 11 σ 12 0 σ 12 σ 22 0 0 0 σ 33, τ = where σ 11 = t 11, σ 22 = t 22, σ 33 = t 33, and we can write t max ρρ t max ρϕ 0 τ 12 0 τ 12 0 0 0 0 0, (2.8) σ 12 = 1 2 ( + t 21 ), τ 12 = 1 2 ( t 21 ), (2.9) = σ max ρρ = 1 2 (σ 11 + σ 22 ) ± 1 2 (σ11 σ 22 ) 2 + 4σ12 2, (2.10) = τ 12 ± σρϕ max, σρϕ max = 1 (σ11 σ 22 ) 2 + 4σ 2 12. (2.11) 2 The physical interpretation of Eq. (2.11) is facilitated by observing that the shear stress on any inclined plane due to antisymmetric stress component τ 12 is also equal to τ 12. This follows from the force equilibrium condition for the triangular element shown in Fig. 2b. The moment equilibrium is ensured by the non-uniform field of couple stresses (not shown in Fig. 2). Equations (2.3) and (2.4) can be combined to give t ρρ 1 2 (t 11 + t 22 )] 2 + = 1 4 t ρϕ 1 2 ( t 21 ) ] 2 (t11 t 22 ) 2 + ( + t 21 ) 2], (2.12)
6 V.A. Lubarda j t t 12 12 (a) (b) Figure 2: (a) A rectangular material element carrying the antisymmetric shear stress τ 12 ; (b) the corresponding shear stress at an arbitrarily inclined plane is also τ 12. which defines Mohr s circle for the non-symmetric stress components under conditions of plane strain (Fig. 3). The normal stress t ρρ = (t 11 + t 22 )/2 acts in the planes where t ρϕ attains its maximum or minimum value, while the shear stress t ρϕ = ( t 21 )/2 acts in the planes where t ρρ has its maximum or minimum value. If = t 21, Eq. (2.12) defines the classical Mohr s circle for a symmetric stress tensor (Timoshenko and Goodier, 1970). 2.1. Eigenvalue Analysis The extreme values of the stresses t ρρ and t ρϕ can also be determined by an eigenvalue analysis. If n = {n 1, n 2 } is the unit vector perpendicular to the plane which supports t ρρ and t ρϕ = ( t 21 )/2, we can write n j t ji = λ n i + 1 2 ( t 21 ) ˆn i, (i, j = 1, 2). (2.13) The unit vector orthogonal to n is ˆn = { n 2, n 1 }. The system of
Mohr s Circles for Couple Stresses 7 t max rj 0 -t ( t 22, 21) 1 -( t11 22) 2 + 2j 2 2j 0 1-2 ( t12 - t ) 21 t max rr t 2j 1 t rr ( t 11, ) t min rj t rj Figure 3: Mohr s circle for non-symmetric stresses in plane strain. The center of the circle is at the point with the coordinates 1 2 (t 11 + t 22 ), ( t 21 )]. The radius of the circle is 1 2 (t 11 t 22 ) 2 +( +t 21 ) 2. The angles ϕ 1, ϕ 2, and ϕ 0 specify the planes of t max ρρ, t max ρϕ, and t ρϕ = 0. equations (2.13) has a nontrivial solution for (n 1, n 2 ) if λ is given be the right-hand side of Eq. (2.6). The corresponding planes are defined by tan ϕ 1 = t 11 t 22 + t 21 ± ( ) ] t11 t 2 1/2 22 1 +, (2.14) + t 21 in agreement with the first expression from Eq. (2.5). Similarly, for the plane which supports the stresses t ρϕ and t ρρ = (t 11 + t 22 )/2, we have n j t ji = λ ˆn i + 1 2 (t 11 + t 22 ) n i, (i, j = 1, 2). (2.15) The system of equations (2.15) has a nontrivial solution for (n 1, n 2 ) if λ is given by the right-hand side of Eq. (2.7). The corresponding
8 V.A. Lubarda planes are defined by tan ϕ 2 = + t 21 t 11 t 22 ± ( ) ] t12 + t 2 1/2 21 1 +, (2.16) t 11 t 22 in agreement with the second expression from Eq. (2.5). The planes with the vanishing t ρϕ (if they exist) are found by solving the eigenvalue problem which specifies λ = t ρρ = 1 2 (t 11 + t 22 ) ± 1 2 The corresponding planes are defined by n j t ji = λ n i, (2.17) (t11 t 22 ) 2 + 4 t 21. (2.18) tan ϕ 0 = 1 2t 21 { (t 11 t 22 ) ± (t 11 t 22 ) 2 + 4 t 21 }. (2.19) There is one such plane if (t 11 t 22 ) 2 = 4 t 21, and two such planes if (t 11 t 22 ) 2 > 4 t 21. The planes with the vanishing t ρρ (if they exist) are found by solving the eigenvalue problem which specifies λ = t ρϕ = 1 2 ( t 21 ) ± 1 2 The corresponding planes are defined by n j t ji = λ ˆn i, (2.20) (t12 + t 21 ) 2 4t 11 t 22. (2.21) tan ϕ 0 = 1 2t 22 { ( + t 21 ) ± ( + t 21 ) 2 4t 11 t 22 }, (2.22) provided that ( + t 21 ) 2 4t 11 t 22.
Mohr s Circles for Couple Stresses 9 3. MOHR S CIRCLE FOR COUPLE STRESSES IN ANTI-PLANE STRAIN For the anti-plane strain problems, the displacement components are u 1 = u 2 = 0, and u 3 = u 3 (x 1, x 2 ). The stress and couple stress tensors have the components 0 0 t 13 t = 0 0 t 23 t 31 t 32 0, m = m 11 m 12 0 m 21 m 22 0 0 0 0. (3.1) A material element under conditions of anti-plane strain is shown in Fig. 4a. In an inclined plane whose normal ρ makes an angle ϕ with the longitudinal direction x 1, the shear stress is t ρ3 and the couple stresses are m ρρ and m ρϕ (Fig. 4b). From the equilibrium conditions of the triangular element it readily follows that t ρ3 = t 13 cos ϕ + t 23 sin ϕ, (3.2) m ρρ = 1 2 (m 11 + m 22 ) + 1 2 (m 11 m 22 ) cos 2ϕ + 1 2 (m 12 + m 21 ) sin 2ϕ, (3.3) m ρϕ = 1 2 (m 12 m 21 ) + 1 2 (m 12 + m 21 ) cos 2ϕ 1 2 (m 11 m 22 ) sin 2ϕ. (3.4) The planes with the maximum magnitude of m ρρ and m ρϕ are defined by tan 2ϕ 1 = m 12 + m 21 2m 11, tan 2ϕ 2 = 2m 11 m 12 + m 21. (3.5) The two angles are related by ϕ 2 = ϕ 1 ± π/4. extreme values of the couple stress components are The corresponding m max ρρ = 1 2 (m 11 + m 22 ) ± 1 (m11 m 22 ) 2 + (m 12 + m 21 ) 2, (3.6) 2
10 V.A. Lubarda m 22 m 21 m rr m 11 m 12 t 13 t 23 t 23 t 13 m 12 m 11 m 11 m 12 m rj 13 tt r3 23 j t m 21 m 21 m 22 m 22 (a) (b) Figure 4: (a) A rectangular material element under conditions of antiplane strain; (b) an inclined plane at an angle ϕ supports the shear stress t ρ3 and couple stresses m ρρ and m ρϕ. m max ρϕ = 1 2 (m 12 m 21 ) ± 1 (m11 m 22 ) 2 + (m 12 + m 21 ) 2. (3.7) 2 Again, it may be noted that m max ρρ does not depend on the antisymmetric part of the couple stress tensor, i.e., on the component (m 12 m 21 )/2. Equations (3.3) and (3.4) can be combined to give m ρρ 1 2 2 (m 11 + m 22 )] + m ρϕ 1 ] 2 2 (m 12 m 21 ) = 1 4 (m11 m 22 ) 2 + (m 12 + m 21 ) 2]. (3.8) This defines Mohr s circle for the couple stresses in anti-plane strain (Fig. 5). The couple stress component m ρρ = (m 11 + m 22 )/2 acts in the planes where m ρϕ attains its maximum or minimum value, while m ρϕ = (m 12 m 21 )/2 acts in the planes where m ρρ has its maximum
Mohr s Circles for Couple Stresses 11 0,-m ( m22 21) m max rj 1-2 ( m11+ m ) 22 2j 0 1-2 ( m12-m ) 21 m max rr 2j 1 2j 2 m rr ( m 11, m 12 ) m min rj m rj Figure 5: Mohr s circle for couple stresses in anti-plane strain. The center of the circle is at the point with the coordinates 1 2 (m 11 + m 22 ), (m 12 m 21 )]. The radius of the circle is 1 2 (m 11 m 22 ) 2 + (m 12 + m 21 ) 2. The angles ϕ 1, ϕ 2, and ϕ 0 specify the planes of m max ρρ, m max ρϕ, and m ρϕ = 0. or minimum value. The planes for which m ρϕ = 0 are defined by tan ϕ 0 = 1 2m 21 { (m 11 m 22 ) ± (m 11 m 22 ) 2 + 4m 12 m 21 }. (3.9) There is one such plane if (m 11 m 22 ) 2 = 4m 12 m 21, and two such planes if (m 11 m 22 ) 2 > 4m 12 m 21. The normal component of the couple stress in these planes is m ρρ = 1 2 (m 11 + m 22 ) ± 1 2 (m11 m 22 ) 2 + 4m 12 m 21. (3.10) The planes for which m ρρ = 0 are defined by
12 V.A. Lubarda m max rj,-m ( m22 21) 0 2j 2 2j 0 1 -( m12 m ) 2 - m max 21 rr 2j 1 m rr ( m 11, m 12 ) m min rj m rj Figure 6: Mohr s circle for couple stresses in anti-plane strain in the case when the couple stress is a deviatoric tensor. The center of the circle is along the m ρϕ axis at the distance 1 2 (m 12 m 21 ) from the origin. The radius of the circle is m max ρρ = m 2 11 + 1 4 (m 12 + m 21 ) 2. tan ϕ 0 = 1 2m 22 { (m 12 + m 21 ) ± (m 12 + m 21 ) 2 4m 11 m 22 }, (3.11) provided that (m 12 + m 21 ) 2 4m 11 m 22. The non-vanishing couple stress in these planes is m ρϕ = 1 2 (m 12 m 21 ) ± 1 2 (m12 + m 21 ) 2 4m 11 m 22. (3.12) The extreme values of the couple stresses m ρρ and m ρϕ can also be determined by an eigenvalue analysis. The derivation is analogous to that presented in section 2. The resulting formulas can be obtained from Eqs. (2.13) (2.22) by replacing the stress symbol t with the couple stress symbol m. In the case when the couple stress is a deviatoric tensor (e.g., isotropic linear elasticity), the results simplify due to the
Mohr s Circles for Couple Stresses 13 condition m 11 + m 22 = 0. The corresponding Mohr s circle is shown in Fig. 6. Acknowledgment: Research support from the Montenegrin Academy of Sciences and Arts is kindly acknowledged. REFERENCES Eringen, A.C. (1968) Theory of micropolar elasticity. In Fracture An Advanced Treatise (Ed. H. Liebowitz), pp. 621 729. Academic Press, New York. Eringen, A.C. (1999) Microcontinuum Field Theories, Springer-Verlag, New York. Koiter, W.T. (1964) Couple-stresses in the theory of elasticity. Proc. Ned. Akad. Wet. (B) 67, Part I: 17 29, Part II: 30 44. Lubarda, V.A. (2003) Circular inclusions in anti-plane strain couple stress elasticity. Int. J. Solids Struct. 40, 3827 3851. Mindlin, R.D. and Tiersten, H.F. (1962) Effects of couple-stresses in linear elasticity. Arch. Ration. Mech. Anal. 11, 415 448. Nowacki, W. (1986) Theory of Asymmetric Elasticity, (translated by H. Zorski). Pergamon Press, Oxford, and PWN - Polish Sci. Publ., Warszawa. Timoshenko, S. and Goodier, J.N. (1970) Theory of Elasticity. McGraw- Hill, New York.