Recitation 2: Stress/Strain Transformations and Mohr’s Circle

Andy Gish 14 Sept 2012 Structural Mechanics 2.080 Recitation 2 Semester Yr Recitation 2: Stress/Strain Transformations and Mohr’s Circle General Transformation Rules 2D Vector 2.1 General Transformation Rules 2.1.1 2D Vector er a vector v in the (x1 , x2 ) coordinate system. A new coordinate system (x 0 ) is component ined by rotating system by Find Consider athe vectorold v in coordinate the (x1 , x2 ) coordinate system. A newangle coordinate .system (x01 , xthe 2 obtained by rotating the old coordinate system by angle ✓. Find the components of v in new coordinate system. the new coordinate system. geometry, we have: geometry, weFrom have: x01 = x1 cos ✓ + x2 sin ✓ ⇡ = x1 cos ✓ + x2 cos( - ✓) 2 L + x L = x 1 1 11 2 12 1 2 x s = x cos + x sin 0 = x1xcos + x2xcos( ) 2 cos ✓ 2 2 = -x1 sin ✓ + ⇡ cos( ==xx11 L +✓)x+2xL 2 cos 112 + 12✓ (2.1) (2.2) = x1 L21 + x2 L22 x s = x sin + x2 cos 0 x121=cos( ) + x 0 cos L11 = cos(e01 , e= cos(e10 , e+ 1 ), L 2 2 ), L21 = cos(e22 , e1 ), L22 = cos(e2 , e2 ) = x1 L21 + x2 L22 where Lij is the direction cosine of the basis 2 1 vectors: (2.3) Lij is the direction cosine of the basis vectors: 2-1 L = cos(e s, e ), L = cos(e s, e ), L = cos(e s, e ), L = cos(e s, e ) Structural Mechanics 2.1.2 2.080 Recitation 2 Semester Yr General Vector Transformation A vector is a physical quantity, independent of the coordinate system. But its scalar com­ ponents DO depend on the coordinate system. v = vi ei = vi0 e0i (2.4) The scalar components vi0 in the new coordinate system can be expressed in terms of the scalar components vi in the original coordinate system. Multiplying the above equation by e0j gives: vi ei · ej0 = vi0 ei0 · e0j = vi0 = i0 j 0 (2.5) vj0 The indices i, j are arbitrary and may be reversed to give: vi0 = vj ei0 · ej (2.6) Introduce the direction cosine tensor Lij ⌘ e0i · ej = cos(e0i , ej ) then we can write vi0 in index form (i, j = 1, 2, 3) (2.7) vi0 = Lij vj (2.8) v 0 = [L]v (2.9) 3 2 3 v10 v1 7 6 6 7 v 0 = 4 v20 5 , and v = 4 v2 5 v30 v3 (2.10) and in matrix form where 2 Similarly, the inverse transform vi = Lji vj0 (2.11) 0 (2.12) T v = [L] v 2.1.3 Tensor Transformations Using the same definition for Lij defined above, the components of a tensor can be trans­ formed to a new coordinate system by: A0ij = Lik Ljl Akl 2-2 (2.13) Structural Mechanics 2.080 Recitation 2 Semester Yr [A0 ] = [L][A][L]T (2.14) or in matrix notation: The inverse transformation can be found by: Akl = Lik Ljl A0ij (2.15) [A] = [L]T [A0 ][L] (2.16) Note that the transformation matrix [L] is orthogonal, meaning its transpose is equal to its inverse. [L][L]T = I (2.17) 2.2 Eigenvalues and Eigenvecors If there exists a vector ni and a scalar Ai for an arbitrary tensor [A] such that [A]ni = Ai ni (2.18) then Ai and ni are eigenvalues and eigenvectors of tensor [A], respectively. The three eigenvectors are mutually perpendicular, and form a new coordinate system. The ”eigenvalue problem” for tensor [A] can be written as: (A - Ai I)ni = 0 (2.19) This equation has solutions ni 6= 0 only if det(A - Ai I) = 0 (2.20) This equation can be solved for Ai , then Ai substituted into the eigenvalue problem to solve for ni . For the stress (strain) tensor, the eigenvalues represent principal stresses (strains), and eigenvectors represent principal axes (i.e., faces with zero shear stress (strain)). Example Find the principal stresses and principal axes for [i] = " 80 30 30 40 # . Solution: The given stress tensor can be represented graphically by 2-3 Example Find the principal stresses and principal axes for [-] = Structural Mechanics 2.080 Recitation 2 = 80 30 . 30 40 8 Semester Yr Solution: The given stress tensor can be represented graphically by The principal stresses are the eigenvalues ( i ) of the stress tensor, and are found by solvThe principal stresses are the eigenvalues (Ai ) of the stress tensor, and are found by solving: ing: det(i - AI) = 0 # 80 - A 30 det det(I) = 0 = 0 30 40 - A 8 " = (2.21a) (2.21b) (80 80 - A)(40 - A)30 - 302 = 0 det (2.21c) =0 2 A30 - 120A 40+ 2300 = 0 (2.21d) A1 = 96.05 and A2 =23.95 (80 )(40 (2.21e) 2 ) 30 = 0 The principal directions are the eigenvectors, found by substituting the eigenvalues into the 2 original eigenvalue problem: 120 + 2300 = 0 (i - Ai I)ni = 0 (2.22) For A1 = 96.05 MPa, 1 = 96.05 and " 2 = 23.95 #( ) ( ) - 96.05 30 0 The principal directions are80the eigenvectors, foundx1by = substituting the eigenvalues into (2.23) 30 40 96.05 0 y 1 the original eigenvalue problem: (80 - 96.05)x1 + 30y1 = 0 For (2.24) eigenvector. Therefore, we can choose y1 = 1 Any (x1 , y1 ) that satisfies this equation (- is an i I)n i = 0 and solve for x1 , then normalize the vector as follows: 1 = 96.05 MPa: 8 (80 - 96.05)x1 + 30 = 0 80 30= x30 96.05 1 = 16.05 ( 30 40 96.05 30 n1 = q 1 30 2 ( 16.05 ) (80 16.05 12 1 (2.25) x1 = )y1 ( = + 96.05)x 1 + 30y1 = 0 0.88 0.47 0 )0 (2.26) (2.27) eigenvector. Therefore, we can choose y1 = 1 Any (x1 , y1 ) that satisfies this equation is an2-4 and solve for x1 , then normalize the vector as follows: Structural Mechanics 2.080 Recitation 2 Semester Yr Similarly, the second eigenvector is found to be: ( ) -0.47 n2 = 0.88 (2.28) As a check, it is clear that the two eigenvectors are perpendicular (n1 · n2 = 0). 2.3 Mohr’s Circle Mohr’s circle (named after Otto Mohr (1835-1918)) is a graphical technique to transform stress (strain) from one coordinate system to another, and to find maximum normal and shear stresses (strains). Constructing 2D Mohr’s Circle: 1. Establish a rectangular coordinate system with x =normal stress, y =shear stress. Scales must be identical. 2. Plot stresses for 2 orthogonal adjacent faces (values from the original stress (strain) tensor). 3. Connect the 2 points to find center of the circle, C. 4. Draw circle through 2 points with center C. 5. Principal stresses (strains) are values where the circle crosses the x-axis. 6. Max shear stress (strain) is max y-value on the circle. Sign convention for Mohrs circle: Positive shear stress on a face causes clockwise rotation of the unit square. The stress state for a face rotated an angle ✓ from an original coordinate axis may be found by rotating an angle 2✓ on Mohrs circle in the same direction from the original coordinate axis. Example1 Solve the above example using Mohr’s circle. Solution: The given stress tensor is represented graphically by 1 Example 1.3 from Ugural and Fenster, Advanced Strength and Applied Elasticity, 2003 2-5 mple1 the above example using Mohr’s circle. Structural Mechanics 2.080 Recitation 2 Semester Yr olution: The given stress tensor is represented graphically by Example1 Solve the above example using Mohr’s circle. Solution: The given stress tensor is represented graphically by Theon stress state ononthe x-face = +80 MPa, and 0 and = MPa M30a(because MPa (because of stress state the positive x-face isisisi+== +80 MPa, = s30 (becau The stress state thepositive positive x-face +80 MPa, and ⌧ = -30 of theMPa the sign convention defined above). sign convention defined above). gn convention defined above). The stateononthethe positive y-face is +40 + = MPa, +40 MPa, and = +30 MPa. The stress stress state positive y-face is i = and ⌧ = +300 MPa. he stress state on the positive y-face isis as shown: =as+40 The Mohr’s Mohrs circle forfor thethe given stress state state The circle given stress is shown:MPa, and a = +30 MPa. he Mohr’s circle for the given stress state is as shown: TheThe center, C, C, is is located 60MPa the + axis. center, locatedatat(40 (40 + + 80)/2 80)/2 ==60 MPa ononthe i axis. The principal principal stresses represented by A B1 . The of thoseofpoints B1 . coordinates The coordinates those are points The stressesareare represented by A1 and 1 and are found from geometry as: +1,2 = 60 ± or p (802-6 M 60)2 + 302 MPa Structural Mechanics 2.080 Recitation 2 Semester Yr found from geometry as: i1,2 = 60 ± or p (80 - 60)2 + 302 MPa i1 = 96.05 MPa and i2 = 23.95 MPa The principal directions are found by: (2.29) (2.30) The principal directions are found by: 2(2✓p ==arctan arctan p 30 o 30 = 56.3 ° = 56.3 (80-660) 60) (80 ° ✓(p = =28.15 28.15o p (2.31a) (2.31b) The principal stress state is as shown below: The principal stress state is as shown below: 3D Mohrs Circle To draw Mohrs circle for a general 3D stress state, the principal stresses and directions 3D Mohr’s must firstCircle be evaluated (by solving the eigenvalue problem). Then the Mohrs circle can be constructed shownfor below: To draw Mohr’sascircle a general 3D stress state, the principal stresses and directions must first be evaluated (by solving the eigenvalue problem). Then the Mohr’s circle can The stress state for any rotation will be represented by a point either on one of the 3 circles, be constructed as shown below: or in the shaded green area between the inner and outer circles. 2-7 p = 28.15 The principal stress state is as shown below: Structural Mechanics 2.080 Recitation 2 Semester Yr 3D Mohr’s Circle To draw Mohr’s circle for a general 3D stress state, the principal stresses and directions must first be evaluated (by solving the eigenvalue problem). Then the Mohr’s circle can be constructed as shown below: The stress state for any rotation will be represented by a point either on one of the 3 circles, or in the shaded green area between the inner and outer circles. 7 2-8