Lecture 6 Mohr s Circle for Plane Stress

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1 P4 Stress and Strain Dr. A.B. Zavatsk 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 maimum shear stresses. Introduction to the stress tensor.

2 Stress Transformation Equations θ + ( ) + sin θ + cos θ + cos θ sin θ If we var θ from 0 to 360, we will get all possible values of and for a given stress state. It would be useful to represent and as functions of θ in graphical form.

3 3 To do this, we must re-write the transformation equations. ( ) θ θ θ θ cos sin sin cos Eliminate θ b squaring both sides of each equation and adding the two equations together Define avg and avg + +

4 Substitue for avg and to get ( ) + avg which is the equation for a circle with centre ( avg,0) and radius. This circle is usuall referred to as Mohr s circle, after the German civil engineer Otto Mohr (835-98). He developed the graphical technique for drawing the circle in 88. 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 analsis. 4

5 Sign Convention for Mohr s Circle ( ) + avg θ θ avg Notice that shear stress is plotted as positive downward. The reason for doing this is that θ is then positive counterclockwise, which agrees with the direction of θ used in the derivation of the tranformation equations and the direction of θ on the stress element. Notice that although θ appears in Mohr s circle, θ appears on the stress element. 5

6 Procedure for Constructing Mohr s Circle. Draw a set of coordinate aes with as abscissa (positive to the right) and as ordinate (positive downward).. Locate the centre of the circle c at the point having coordinates avg and Locate point A, representing the stress conditions on the face of the element b plotting its coordinates and. Note that point A on the circle corresponds to θ Locate point B, representing the stress conditions on the face of the element b plotting its coordinates and. Note that point B on the circle corresponds to θ 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 80 apart on the circle). 6. Using point c as the centre, draw Mohr s circle through points A and B. This circle has radius. (based on Gere) 6

7 B B (θ90) A - c A (θ0) avg 7

8 Stresses on an Inclined Element. On Mohr s circle, measure an angle θ counterclockwise from radius ca, because point A corresponds to θ 0 and hence is the reference point from which angles are measured.. The angle θ locates the point D on the circle, which has coordinates and. Point D represents the stresses on the face of the inclined element. 3. Point E, which is diametricall opposite point D on the circle, is located at an angle θ + 80 from ca (and 80 from cd). Thus point E gives the stress on the face of the inclined element. 4. So, as we rotate the aes counterclockwise b an angle θ, the point on Mohr s circle corresponding to the face moves counterclockwise through an angle θ. (based on Gere) 8

9 B B (θ90) E (θ+90) θ+80 - A c θ D (θ) A (θ0) E θ D 9

10 Principal Stresses B B (θ90) θ p A c θ p A (θ0) P θ p θ p P 0

11 Maimum Shear Stress B B (θ90) min A θ s ma c A (θ0) s Note carefull the directions of the shear forces. ma s s ma θ s ma s ma s

12 Eample: The state of plane stress at a point is represented b the stress element below. Draw the Mohr s circle, determine the principal stresses and the maimum shear stresses, and draw the corresponding stress elements. c avg 65 + ( 50 ( 5) ) + ( 5) A (θ0),, c ± 5 ± MPa 84.6 MPa B 50 MPa c B (θ90) 80 MPa 80 MPa A 5 MPa ma ma s 69.6 MPa c 5 MPa 50 MPa

13 50 MPa 80 MPa 80 MPa 5 tan θ θ.0 θ MPa 5 MPa θ 00.5 θ 0.5 A (θ0) 54.6 MPa θ θ c B (θ90) 00.5 o 84.6 MPa 84.6 MPa 0.5 o θ 54.6 MPa 3

14 50 MPa 80 MPa 80 MPa 5 MPa 50 MPa A (θ0) θ smin min θ.0 θ s min (90.0) 69.0 θs min 34.5 taking sign convention into account θ 5 MPa 5 MPa o 5 MPa 5 MPa 69.6 MPa θ 55.5 o θ sma c ma B (θ90) 4 θ.0 θ s ma θs ma 55.5

15 Eample: The state of plane stress at a point is represented b the stress element below. Find the stresses on an element inclined at 30 clockwise and draw the corresponding stress elements. 50 MPa 80 MPa 80 MPa 5 MPa 50 MPa A (θ0) C (θ -30 ) θ -60 c cos(θ +60) c + cos(θ +60) - sin (θ +60) MPa 4.5 MPa -30 o 68.8 MPa 4.5 MPa C D 5.8 MPa D (θ ) B (θ90) θ θ -30 θ -60 5

16 Principal Stresses 54.6 MPa, MPa But we have forgotten about the third principal stress! Since the element is in plane stress ( z 0), the third principal stress is zero MPa 0 MPa MPa A (θ0) This means three Mohr s circles can be drawn, each based on two principal stresses: 3 B (θ90) and 3 and and 3 6

17

18 The stress element shown is in plane stress. What is the maimum shear stress? B B A 3 A ma(,) 3 ma(,3) overall maimum 3 ma(,3) 8

19 Introduction to the Stress Tensor z z z z z z z z zz z zz Normal stresses on the diagonal Shear stresses off diagaonal, z z, z z The normal and shear stresses on a stress element in 3D can be assembled into a 33 matri known as the stress tensor. 9

20 From our analses so far, we know that for a given stress sstem, 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, z z z z zz In mathematical terms, this is the process of matri diagonalization in which the eigenvalues of the original matri are just the principal stresses. 0

21 Eample: The state of plane stress at a point is represented b the stress element below. Find the principal stresses. 50 MPa 80 MPa 80 MPa M MPa 5 MPa We must find the eigenvalues of this matri. emember 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 matri. Ar λr 0 or (A λi) r 0 where I is the identit matri. 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 λ and λ.

22 80 λ 5 det λ ( 80 λ)(50 λ) ( 5)( 5) λ + 30λ 465 λ 84.6, 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. 80 λ λ is one eigenvector.

23 80 λ 5 80 ( 84.6) λ ( 84.6) is the other eigenvector. Before finding the angles at which the principal stresses act, we can check to see if the eigenvectors are correct D D C M C C C A T det C C where A matri of M co - factors

24 D To find the angles, we must calculate the unit eigenvectors: And then assemble them into a rotation matri so that det det The rotation matri has the form cosθ sinθ sinθ cosθ D (0.983)(0.983) (0.83)( 0.83) T M So θ 0.5, as we found earlier for one of the principal angles. 4

25 Using the rotation angle of 0.5, the matri M (representing the original stress state of the element) can be transformed to matri D (representing the principal stress state). D D D T M MPa 54.6 MPa So, the transformation equations, 00.5 o 84.6 MPa 0.5 o Mohr s circle, and eigenvectors all give the same result for the principal stress element MPa 5

26 Finall, we can use the rotation matri approach to find the stresses on an inclined element with θ -30. M M M cos( 30 ) sin( 30 ) T M sin( 30 ) cos( 30 ) MPa 4.5 MPa Again, the transformation equations, Mohr s circle, and the stress tensor approach all give the same result. 4.5 MPa -30 o 68.8 MPa 5.8 MPa 6

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