Mohr's Circle for 2-D Stress Analysis

Transcription

1 Mohr's Circle for 2-D Stress Analysis If you want to know the principal stresses and maximum shear stresses, you can simply make it through 2-D or 3-D Mohr's cirlcles! You can know about the theory of Mohr's circles from any text books of Mechanics of Materials. The following two are good references, for examples. 1. Ferdinand P. Beer and E. Russell Johnson, Jr, "Mechanics of Materials", Second Edition, McGraw-Hill, Inc, James M. Gere and Stephen P. Timoshenko, "Mechanics of Materials", Third Edition, PWS-KENT Publishing Company, Boston, The 2-D stresses, so called plane stress problem, are usually given by the three stress components σ x, σ y, and τ xy, which consist in a two-by-two symmetric matrix (stress tensor): What people usually are interested in more are the two prinicipal stresses σ 1 and σ 2, which are the two eigenvalues of the two-by-two symmetric matrix of Eqn (1), and the maximum shear stress τ max, which can be calculated from σ 1 and σ 2. Now, see the Fig. 1 below, which represents that a state of plane stress exists at point O and that it is defined by the stress components σ x, σ y, and τ xy associated with the left element in the Fig. 1. We propose to determine the stress components σ xθ, σ yθ, and τ xyθ associated with the right element after it has been rotated through an angle θ about the z axis. (1)

2 Fig. 1 Plane stresses in different orientations Then, we have the following relationship: and σ xθ = σ x cos 2 θ + σ y sin 2 θ + 2 τ xy sin θ cos θ τ xyθ = -(σ x σ y ) cos 2 θ + τ xy (cos 2 θ sin 2 θ) Equivalently, the above two equations can be rewritten as follows: σ xθ = (σ x + σ y )/2 + (σ x σ y )/2 cos 2θ + τ xy sin 2θ (4) and τ xyθ = -(σ x σ y )/2 sin 2θ + τ xy cos 2θ (5) (2) (3)

3 The expression for the normal stress σ yθ may be obtained by replacing the θ in the relation for σ xθ in Eqn. 3 by θ + 90 ο, it turns out to be σ yθ = (σ x + σ y )/2 (σ x σ y )/2 cos 2θ τ xy sin 2θ (6) From the relations for σ xθ and σ yθ, one obtains the circle equation: (σ xθ σ ave ) 2 + τ 2 xyθ = R 2 m (7) where σ ave = (σ x + σ y )/2 = (σ xθ + σ yθ )/2 ; Rm = [(σ x σ y ) 2 / 4 + τ 2 xy] 1/2 (8) This circle is with radius R 2 m and centered at C = (σ ave, 0) if let σ = σ xθ and τ = τ xyθ as shown in Fig. 2 below - that is right the Mohr's Circle for plane stress problem or 2-D stress problem!

4 Fig. 2 Mohr's circle for plane (2-D) stress In fact, Eqns. 4 and 5 are the parametric equations for the Mohr's circle! In Fig. 2, one reads that the point X = (σ x, -τ xy ) which corresponds to the point at which θ = 0 and the point A = (σ 1, 0 ) which corresponds to the point at which θ = θ p that gives the principal stress σ 1! Note that tan 2θ p = 2τ xy /(σ x σ y ) and the point (9) (10) (11)

5 Y = (σ y, τ xy ) which corresponds to the point at which θ = 90 ο and the point B = (σ 2, 0 ) which corresponds to the point at which θ = θ p + 90 ο that gives the principal stress σ 2! To this end, one can pick the maxium normal stressess as σ max = max(σ 1, σ 2 ), σ min = min(σ 1, σ 2 ) (12) (13) (14) Besides, finally one can also read the maxium shear stress as τ max = Rm = [(σ x σ y ) 2 / 4 + τ 2 xy] 1/2 (15) which corresponds to the apex of the Mohr's circle at which θ = θ p + 45 ο! (The end.) Mohr's Circles for 3-D Stress Analysis The 3-D stresses, so called spatial stress problem, are usually given by the six stress components σ x, σ y, σ z, τ xy, τ yz, and τ zx, (see Fig. 3) which consist in a three-bythree symmetric matrix (stress tensor):

6 (16) What people usually are interested in more are the three prinicipal stresses σ 1, σ 2, and σ 3, which are eigenvalues of the three-by-three symmetric matrix of Eqn (16), and the three maximum shear stresses τ max1, τ max2, and τ max3, which can be calculated from σ 1, σ 2, and σ 3. Fig. 3 3-D stress state represented by axes parallel to X-Y-Z Imagine that there is a plane cut through the cube in Fig. 3, and the unit normal vector ν of the cut plane has the direction cosines v x, v y, and v z, that is ν = (v x, v y, v z )

7 (17) then the normal stress on this plane can be represented by σ ν = σ x v 2 x+ σ y v 2 y+ σ z v 2 z+ 2 τ xy v x v y + 2 τ yz v y v z + 2 τ xz v x v z (18) There exist three sets of direction cosines, ν 1, ν 2, and ν 3 - the three principal axes, which make σ ν achieve extreme values σ 1, σ 2, and σ 3 - the three principal stresses, and on the corresponding cut planes, the shear stresses vanish! The problem of finding the principal stresses and their associated axes is equivalent to finding the eigenvalues and eigenvectors of the following problem: (σι 3 Τ 3 )ν = 0 (19) The three eigenvalues of Eqn (19) are the roots of the following characteristic polynomial equation: det(σι 3 Τ 3 ) = σ 3 Aσ 2 + Bσ C = 0 (20) where Α = σ x + σ y + σ z (21) B = σ x σ y + σ y σ z + σ x σ z τ 2 xy τ 2 yz τ 2 xz (22) C = σ x σ y σ z + 2 τ xy τ yz τ xz σ x τ 2 yz σ y τ 2 xz σ z τ 2 xy (23) In fact, the coefficients A, B, and C in Eqn (20) are invariants as long as the stress state is prescribed(see e.g. Ref. 2). Therefore, if the three roots of Eqn (20) are σ 1, σ 2, and σ 3, one has the following equations: σ 1 + σ 2 + σ 3 = A (24) σ 1 σ 2 + σ 2 σ 3 + σ 1 σ 3 = B (25)

8 σ 1 σ 2 σ 3 = C (26) Numerically, one can always find one of the three roots of Eqn (20), e.g. σ 1, using line search algorithm, e.g. bisection algorithm. Then combining Eqns (24)and (25), one obtains a simple quadratic equations and therefore obtains two other roots of Eqn (20), e.g. σ 2 and σ 3. To this end, one can re-order the three roots and obtains the three principal stresses, e.g. σ 1 = max(σ 1, σ 2, σ 3 ) (27) σ 3 = min(σ 1, σ 2, σ 3 ) (28) σ 2 = (A σ 1 σ 2 ) (29) Now, substituting σ 1, σ 2, or σ 3 into Eqn (19), one can obtains the corresponding principal axes ν 1, ν 2, or ν 3, respectively. Similar to Fig. 3, one can imagine a cube with their faces normal to ν 1, ν 2, or ν 3. For example, one can do so in Fig. 3 by replacing the axes X,Y, and Z with ν 1, ν 2, and ν 3, respectively, replacing the normal stresses σ x, σ y, and σ z with the principal stresses σ 1, σ 2, and σ 3, respectively, and removing the shear stresses τ xy, τ yz, and τ zx. Now, pay attention the new cube with axes ν 1, ν 2, and ν 3. Let the cube be rotated about the axis ν 3, then the corresponding transformation of stress may be analyzed by means of Mohr's circle as if it were a transformation of plane stress. Indeed, the shear stresses excerted on the faces normal to the ν 3 axis remain equal to zero, and the normal stress σ 3 is perpendicular to the plane spanned by ν 1 and ν 2 in which the transformation takes place and thus, does not affect this transformation. One may therefore use the circle of diameter AB to determine the normal and shear stresses exerted on the faces of the cube as it is rotated about the ν 3 axis (see Fig. 4). Similarly, the circles of diameter BC and CA may be used to determine the stresses on the cube as it is rotated about the ν 1 and ν 2 axes, respectively.

9 Fig. 4 Mohr's circles for space (3-D) stress What if the rotations are about the axes rather than principal axes? It can be shown that any other transformation of axes would lead to stresses represented in Fig. 4 by a point located within the area which is bounded by the bigest circle with the other two circles removed! Therefore, one can obtain the maxium/minimum normal and shear stresses from Mohr's circles for 3-D stress as shown in Fig. 4! Note the notations above (which may be different from other references), one obtains that σ max = σ 1 (30) σ min = σ 3 (31)

10 τ max = (σ 1 σ 3 )/2 = τ max2 (32) Note that in Fig. 4, τ max1, τ max2, and τ max3 are the maximum shear stresses obtained while the rotation is about ν 1, ν 2, and ν 3, respectively. (The end.) Mohr's Circles for Strain and for Moments and Products of Inertia Mohr's circle(s) can be used for strain analysis and for moments and products of inertia and other quantities as long as they can be represented by two-by-two or three-by-three symmetric matrices (tensors). (The end.)