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Moh'r circle2
- 1. 1 P4 Stress and Strain Dr. A.B. Zavatsky HT08 Lecture 6 Mohr’s Circle for Plane Stress Transformation equations for plane stress. Procedure for constructing Mohr’s circle. Stresses on an inclined element. Principal stresses and maximum shear stresses. Introduction to the stress tensor.
- 2. 2 Stress Transformation Equations σx σx σy τxy τyx τxy τyx x y x y σx1 θ y1 x1 σx1 σy1 σy1 τx y1 1 τy x1 1 τx y1 1 τy x1 1 ( ) θτθ σσ τ θτθ σσσσ σ 2cos2sin 2 2sin2cos 22 11 1 xy yx yx xy yxyx x + − −= + − + + = If we vary θ from 0° to 360°, we will get all possible values of σx1 and τx1y1 for a given stress state. It would be useful to represent σx1 and τx1y1 as functions of θ in graphical form.
- 3. 3 To do this, we must re-write the transformation equations. ( ) θτθ σσ τ θτθ σσσσ σ 2cos2sin 2 2sin2cos 22 11 1 xy yx yx xy yxyx x + − −= + − = + − Eliminate θ by squaring both sides of each equation and adding the two equations together. 2 2 2 11 2 1 22 xy yx yx yx x τ σσ τ σσ σ +⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − =+⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − Define σavg and R 2 2 22 xy yxyx avg R τ σσσσ σ +⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = + =
- 4. 4 Substitue for σavg and R to get ( ) 22 11 2 1 Ryxavgx =+− τσσ which is the equation for a circle with centre (σavg,0) and radius R. This circle is usually referred to as Mohr’s circle, after the German civil engineer Otto Mohr (1835-1918). He developed the graphical technique for drawing the circle in 1882. The construction of Mohr’s circle is one of the few graphical techniques still used in engineering. It provides a simple and clear picture of an otherwise complicated analysis.
- 5. 5 Sign Convention for Mohr’s Circle x y σx1 θ y1 x1 σx1 σy1 σy1 τx y1 1 τy x1 1 τx y1 1 τy x1 1 ( ) 22 11 2 1 Ryxavgx =+− τσσ σx1 τx1y1 R 2θ σavg Notice that shear stress is plotted as positive downward. The reason for doing this is that 2θ is then positive counterclockwise, which agrees with the direction of 2θ used in the derivation of the tranformation equations and the direction of θ on the stress element. Notice that although 2θ appears in Mohr’s circle, θ appears on the stress element.
- 6. 6 Procedure for Constructing Mohr’s Circle 1. Draw a set of coordinate axes with σx1 as abscissa (positive to the right) and τx1y1 as ordinate (positive downward). 2. Locate the centre of the circle c at the point having coordinates σx1 = σavg and τx1y1 = 0. 3. Locate point A, representing the stress conditions on the x face of the element by plotting its coordinates σx1 = σx and τx1y1 = τxy. Note that point A on the circle corresponds to θ = 0°. 4. Locate point B, representing the stress conditions on the y face of the element by plotting its coordinates σx1 = σy and τx1y1 = −τxy. Note that point B on the circle corresponds to θ = 90°. 5. Draw a line from point A to point B, a diameter of the circle passing through point c. Points A and B (representing stresses on planes at 90° to each other) are at opposite ends of the diameter (and therefore 180° apart on the circle). 6. Using point c as the centre, draw Mohr’s circle through points A and B. This circle has radius R. (based on Gere)
- 7. 7 σx σx σy τxy τyx τxy τyx x y σx1 τx1y1 σavg c R A (θ=0) σx τxy A B (θ=90) σy -τxy B
- 8. 8 Stresses on an Inclined Element 1. On Mohr’s circle, measure an angle 2θ counterclockwise from radius cA, because point A corresponds to θ = 0 and hence is the reference point from which angles are measured. 2. The angle 2θ locates the point D on the circle, which has coordinates σx1 and τx1y1. Point D represents the stresses on the x1 face of the inclined element. 3. Point E, which is diametrically opposite point D on the circle, is located at an angle 2θ + 180° from cA (and 180° from cD). Thus point E gives the stress on the y1 face of the inclined element. 4. So, as we rotate the x1y1 axes counterclockwise by an angle θ, the point on Mohr’s circle corresponding to the x1 face moves counterclockwise through an angle 2θ. (based on Gere)
- 9. 9 σx σx σy τxy τyx τxy τyx x y A (θ=0) σx1 τx1y1 c R B (θ=90) A B x y σx1 θ y1 x1 σx1 σy1 σy1 τx y1 1 τy x1 1 τx y1 1 τy x1 1 2θ D D (θ) σx1 τx1y1 σy1 -τx1y1 2θ+180E (θ+90) E
- 10. 10 Principal Stresses σx σx σy τxy τyx τxy τyx x y A (θ=0) σx1 τx1y1 c R B (θ=90) A B x y σ1 θp1 σ1 σ2 σ2 θp2 P1 2θp1 P2 2θp2 σ1σ2
- 11. 11 Maximum Shear Stress σx σx σy τxy τyx τxy τyx x y A (θ=0) σx1 τx1y1 c R B (θ=90) A B 2θs σs τmax τmin x y σs θs σs σs σs τmaxτmax τmax τmax Note carefully the directions of the shear forces.
- 12. 12 Example: The state of plane stress at a point is represented by the stress element below. Draw the Mohr’s circle, determine the principal stresses and the maximum shear stresses, and draw the corresponding stress elements. 80 MPa 80 MPa 50 MPa x y 50 MPa 25 MPa σ τ 15 2 5080 2 −= +− = + == yx avgc σσ σ c A (θ=0) A B (θ=90)B ( )( ) ( ) 6.692565 251550 22 22 =+= +−−= R R R σ1 σ2 MPa6.84 MPa6.54 6.6915 2 1 2,1 2,1 −= = ±−= ±= σ σ σ σ Rc τmax MPa15 MPa6.69 s max −== == c R σ τ
- 13. 13 84.6 MPa 84.6 MPa 54.6 MPa x y 54.6 MPa 10.5o 100.5 o σ τ c A (θ=0) B (θ=90) R 80 MPa 80 MPa 50 MPa x y 50 MPa 25 MPa σ1 σ2 2θ2 2θ1 °=°= °=°+= °= = − = 5.105.100 2011800.212 0.212 3846.0 1580 25 2tan 21 1 2 2 θθ θ θ θ 2θ
- 14. 14 σ τ c A (θ=0) B (θ=90) R 80 MPa 80 MPa 50 MPa x y 50 MPa 25 MPa 2θ2 2θ 15 MPa 15 MPa 15 MPa x y 15 MPa 69.6 MPa -34.5o 55.5 o 2θsmax °= °=°+= °= 5.55 0.111900.212 0.212 max max 2 s s θ θ θ 2θsmin taking sign convention into account °−= °−=−−= °= 5.34 0.69)0.2190(2 0.212 min min 2 s s θ θ θ τmax τmin
- 15. 15 80 MPa 80 MPa 50 MPa x y 50 MPa 25 MPa A (θ=0) σ τ B (θ=90)25.8 MPa 25.8 MPa 4.15 MPa x y 4.15 MPa 68.8 MPa x1 y1 -30 o Example: The state of plane stress at a point is represented by the stress element below. Find the stresses on an element inclined at 30° clockwise and draw the corresponding stress elements. -60° -60+180° C (θ = -30°) C D (θ = -30+90°) D 2θ2 σx1 = c – R cos(2θ2+60) σy1 = c + R cos(2θ2+60) τx1y1= -R sin (2θ2+60) σx1 = -26 σy1 = -4 τx1y1= -69 2θ θ = -30° 2θ = -60°
- 16. 16 σ τ A (θ=0) B (θ=90) Principal Stresses σ1 = 54.6 MPa, σ2 = -84.6 MPa But we have forgotten about the third principal stress! Since the element is in plane stress (σz = 0), the third principal stress is zero. σ1 = 54.6 MPa σ2 = 0 MPa σ3 = -84.6 MPa σ1σ2σ3 This means three Mohr’s circles can be drawn, each based on two principal stresses: σ1 and σ3 σ1 and σ2 σ2 and σ3
- 18. 18 σx σx σy τxy τyx τxy τyx x y The stress element shown is in plane stress. What is the maximum shear stress? A B σx1 τx1y1 A B σ3 σ1σ2 22 131 )3,1max( σσσ τ = − = 2 21 )2,1max( σσ τ − = 22 232 )3,2max( σσσ τ = − = overall maximum
- 19. 19 ⎟⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎝ ⎛ zzzyzx yzyyyx xzxyxx σττ τστ ττσ y z xσxx σxx σyy σyy σzz τxy τyx τxzτzx τzy τyz Introduction to the Stress Tensor Normal stresses on the diagonal Shear stresses off diagaonal τxy = τyx, τxz = τzx, τyz = τzy The normal and shear stresses on a stress element in 3D can be assembled into a 3x3 matrix known as the stress tensor.
- 20. 20 From our analyses so far, we know that for a given stress system, it is possible to find a set of three principal stresses. We also know that if the principal stresses are acting, the shear stresses must be zero. In terms of the stress tensor, ⎟⎟ ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎜ ⎝ ⎛ zzzyzx yzyyyx xzxyxx σττ τστ ττσ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ 3 2 1 00 00 00 σ σ σ In mathematical terms, this is the process of matrix diagonaliza- tion in which the eigenvalues of the original matrix are just the principal stresses.
- 21. 21 80 MPa 80 MPa 50 MPa x y 50 MPa 25 MPa Example: The state of plane stress at a point is represented by the stress element below. Find the principal stresses. ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − −− =⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 5025 2580 yyx xyx M στ τσ We must find the eigenvalues of this matrix. Remember the general idea of eigenvalues. We are looking for values of λ such that: Ar = λr where r is a vector, and A is a matrix. Ar – λr = 0 or (A – λI) r = 0 where I is the identity matrix. For this equation to be true, either r = 0 or det (A – λI) = 0. Solving the latter equation (the “characteristic equation”) gives us the eigenvalues λ1 and λ2.
- 22. 22 6.54,6.84 0462530 0)25)(25()50)(80( 0 5025 2580 det 2 −= =−+ =−−−−−− =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −− −−− λ λλ λλ λ λ So, the principal stresses are –84.6 MPa and 54.6 MPa, as before. Knowing the eigenvalues, we can find the eigenvectors. These can be used to find the angles at which the principal stresses act. To find the eigenvectors, we substitute the eigenvalues into the equation (A – λI ) r = 0 one at a time and solve for r. ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −− −−− ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −− −−− 0 0 6.545025 256.5480 0 0 5025 2580 y x y x λ λ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛− −= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −− −− 1 186.0 186.0 0 0 64.425 256.134 yx y x is one eigenvector.
- 23. 23 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −−− −−−− ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −− −−− 0 0 )6.84(5025 25)6.84(80 0 0 5025 2580 y x y x λ λ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − 1 388.5 388.5 0 0 6.13425 256.4 yx y x is the other eigenvector. Before finding the angles at which the principal stresses act, we can check to see if the eigenvectors are correct. CMCD MCD 1 5025 2580 11 388.5186.0 6.840 06.54 −= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − −− =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛− =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛− = == − − 033.0179.0 967.0179.0 factors-coofmatrixwhere det 1 1 1 C AA C C T
- 24. 24 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛− ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − −− ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛− = 6.840 06.54 11 388.5186.0 5025 2580 033.0179.0 967.0179.0 D To find the angles, we must calculate the unit eigenvectors: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ →⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛− →⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛− 183.0 938.0 1 388.5 983.0 183.0 1 186.0 And then assemble them into a rotation matrix R so that det R = +1. 1)183.0)(183.0()983.0)(983.0(det 983.0183.0 183.0983.0 =−−=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = RR RMRDR T=′⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = θθ θθ cossin sincos The rotation matrix has the form So θ = 10.5°, as we found earlier for one of the principal angles.
- 25. 25 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛− =′ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − −− ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − =′ =′ 6.540 06.84 983.0183.0 183.0983.0 5025 2580 983.0183.0 183.0983.0 D D RMRD T Using the rotation angle of 10.5°, the matrix M (representing the original stress state of the element) can be transformed to matrix D’ (representing the principal stress state). 84.6 MPa 84.6 MPa 54.6 MPa x y 54.6 MPa 10.5o 100.5 o So, the transformation equations, Mohr’s circle, and eigenvectors all give the same result for the principal stress element.
- 26. 26 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −− −− =′ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − −− ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − =′ =′ 1 1 15.48.68 8.688.25 866.05.0 5.0866.0 5025 2580 866.05.0 5.0866.0 yyx xyx T M M RMRM στ τσ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ °−°− °−−°− = 866.05.0 5.0866.0 )30cos()30sin( )30sin()30cos( R 25.8 MPa 25.8 MPa 4.15 MPa x y 4.15 MPa 68.8 MPa x1 y1 -30 o Finally, we can use the rotation matrix approach to find the stresses on an inclined element with θ = -30°. Again, the transformation equations, Mohr’s circle, and the stress tensor approach all give the same result.