Lecture 5 Plane Stress Transformation Equations Stress elements and plane stress. Stresses on inclined sections. Transformation equations. Principal stresses, angles, and planes. Maimum shear stress.
Normal and shear stresses on inclined sections To obtain a complete picture of the stresses in a bar, we must consider the stresses acting on an inclined (as opposed to a normal ) section through the bar. Inclined section Normal section P P Because the stresses are the same throughout the entire bar, the stresses on the sections are uniforml distributed. P Inclined section P Normal section
N P V P The force P can be resolved into components: Normal force N perpendicular to the inclined plane, N P cos Shear force V tangential to the inclined plane V P sin If we know the areas on which the forces act, we can calculate the associated stresses. area A area A area ( A / cos ) area ( A / cos ) 3
Force N Area Area cos P cos A/ cos ( cos ) ma occurs when 0 P A cos Force V P sin P sin cos Area Area A/ cos A sin cos ( sin ) ma ± / occurs when -/ 45 4
Introduction to stress elements Stress elements are a useful wa to represent stresses acting at some point on a bod. Isolate a small element and show stresses acting on all faces. Dimensions are infinitesimal, but are drawn to a large scale. P P / A Area A z PA / PA / 5
Maimum stresses on a bar in tension P a P ma P / A a Maimum normal stress, Zero shear stress 6
Maimum stresses on a bar in tension P ab P / 45 ma / / b Maimum shear stress, Non-zero normal stress 7
Stress Elements and Plane Stress When working with stress elements, keep in mind that onl one intrinsic state of stress eists at a point in a stressed bod, regardless of the orientation of the element used to portra the state of stress. We are reall just rotating aes to represent stresses in a new coordinate sstem. 8
Normal stresses,, z (tension is positive) z z z z z z Shear stresses, z z, z z Sign convention for ab Subscript a indicates the face on which the stress acts (positive face is perpendicular to the positive direction) Subscript b indicates the direction in which the stress acts Strictl,, z zz 9
When an element is in plane stress in the plane, onl the and faces are subjected to stresses ( z 0 and z z z z 0). Such an element could be located on the free surface of a bod (no stresses acting on the free surface). Plane stress element in D, z 0 0
Stresses on Inclined Sections The stress sstem is known in terms of coordinate sstem. We want to find the stresses in terms of the rotated coordinate sstem. Wh? A material ma ield or fail at the maimum value of or. This value ma occur at some angle other than 0. (Remember that for uniaial tension the maimum shear stress occurred when 45 degrees. )
Transformation Equations Stresses Forces A A A sec A sec A tan A tan Forces can be found from stresses if the area on which the stresses act is known. Force components can then be summed. Left face has area A. Bottom face has area A tan. Inclined face has area A sec.
A A A sec A sec A tan Sum forces in the Asec Sum forces in the Using direction : A tan ( A) cos ( A) sin ( A tan ) sin ( A tan ) cos 0 direction : ( A) sin ( A) cos ( A tan ) cos ( Atan ) sin 0 Asec cos sin and simplifing gives : ( ) sin cos ( cos sin ) sin cos 3
Using the following trigonometric identities cos cos cos sin sin cos sin gives the transformation equations for plane stress : ( ) sin cos cos sin For stresses on the face, substitute 90 cos for : sin Summing the epressions for and gives : Can be used to find, instead of eqn above. 4
Eample: The state of plane stress at a point is represented b the stress element below. Determine the stresses acting on an element oriented 30 clockwise with respect to the original element. 50 MPa 80 MPa 80 MPa 5 MPa 50 MPa Define the stresses in terms of the established sign convention: -80 MPa 50 MPa -5 MPa We need to find,, and when -30. Substitute numerical values into the transformation equations: cos 80 50 80 50 cos sin ( 30 ) ( 5) sin ( 30 ) 5.9 MPa 5
cos sin 80 50 80 50 cos 30 5 ( ) sin cos ( 80 50) sin ( ) ( ) sin ( 30 ) 4.5 MPa ( 30 ) ( 5) cos ( 30 ) 68.8 MPa 5.8 MPa 4.5 MPa 60-30 o 5.8 MPa Note that could also be obtained (a) b substituting 60 into the equation for or (b) b using the equation 4.5 MPa 68.8 MPa (from Hibbeler, E. 5.) 6
Principal Stresses The maimum and minimum normal stresses ( and ) are known as the principal stresses. To find the principal stresses, we must differentiate the transformation equations. d d d d tan p ( ) ( sin ) ( cos ) sin cos sin cos 0 0 p are principal angles associated with the principal stresses There are two values of p in the range 0-360, with values differing b 80. There are two values of p in the range 0-80, with values differing b 90. So, the planes on which the principal stresses act are mutuall perpendicular. 7
8 sin cos tan p We can now solve for the principal stresses b substituting for p in the stress transformation equation for. This tells us which principal stress is associated with which principal angle. p ( ) / R R R R p p sin cos R R
9 Substituting for R and re-arranging gives the larger of the two principal stresses: To find the smaller principal stress, use. These equations can be combined to give:, ± Principal stresses To find out which principal stress goes with which principal angle, we could use the equations for sin p and cos p or for.
The planes on which the principal stresses act are called the principal planes. What shear stresses act on the principal planes? Compare the equations for d d ( ) ( ) sin sin 0 and d cos 0 cos 0 ( ) sin cos 0 d 0 Solving either equation gives the same epression for tan p Hence, the shear stresses are zero on the principal planes. 0
p p Principal Stresses, ± tan p Principal Angles defining the Principal Planes
Eample: The state of plane stress at a point is represented b the stress element below. Determine the principal stresses and draw the corresponding stress element. 50 MPa 80 MPa 80 MPa 5 MPa 50 MPa Define the stresses in terms of the established sign convention: -80 MPa 50 MPa -5 MPa, ±, 80 50 54.6 MPa ± 80 50 84.6 MPa ( 5) 5 ± 69.6
tan tan p p ( 5) 80 50 0.3846 84.6 MPa 54.6 MPa 00.5 o 0.5 o 84.6 MPa p p.0 and.0 80 0.5, 00.5 54.6 MPa But we must check which angle goes with which principal stress. cos 80 50 80 50 cos 0.5 sin ( ) ( 5) sin ( 0.5 ) 84.6 MPa 54.6 MPa with p 00.5-84.6 MPa with p 0.5 3
The two principal stresses determined so far are the principal stresses in the plane. But remember that the stress element is 3D, so there are alwas three principal stresses. p p 3 0 z p,,,, 3 0 Usuall we take > > 3. Since principal stresses can be compressive as well as tensile, 3 could be a negative (compressive) stress, rather than the zero stress. 4
d Maimum Shear Stress To find the maimum shear stress, we must differentiate the transformation equation for shear. d ( ) ( ) tan s sin cos cos sin 0 There are two values of s in the range 0-360, with values differing b 80. There are two values of s in the range 0-80, with values differing b 90. So, the planes on which the maimum shear stresses act are mutuall perpendicular. Because shear stresses on perpendicular planes have equal magnitudes, the maimum positive and negative shear stresses differ onl in sign. 5
6 s ( ) / R ( ) cos sin tan s We can now solve for the maimum shear stress b substituting for s in the stress transformation equation for. R R R s s sin cos ma min ma
Use equations for sin s and cos s or to find out which face has the positive shear stress and which the negative. What normal stresses act on the planes with maimum shear stress? Substitute for s in the equations for and to get s s ma ma s s s ma ma s 7
Eample: The state of plane stress at a point is represented b the stress element below. Determine the maimum shear stresses and draw the corresponding stress element. 50 MPa 80 MPa 80 MPa 5 MPa 50 MPa Define the stresses in terms of the established sign convention: -80 MPa 50 MPa -5 MPa ma ma 80 50 ( 5) 69.6 MPa s s 80 50 5 MPa 8
80 50 tan s ( ) 5 69.0 and 69.0 80 s s 34.5, 55.5.6 5 MPa 5 MPa But we must check which angle goes with which shear stress. ( ) ( 80 50) sin sin cos ma 69.6 MPa with sma 55.5 min -69.6 MPa with smin -34.5 5 MPa 55.5 o -34.5 o 69.6 MPa ( 34.5) ( 5) cos ( 34.5) 69.6 MPa 5 MPa 9
30 Finall, we can ask how the principal stresses and maimum shear stresses are related and how the principal angles and maimum shear angles are related. ma ma, ± p p s p s cot tan tan tan tan
tan s cot p 0 sin cos s p 0 cos sin sin sin cos ( ) s s s s p s p p p p ± 90 s p ± 45 ± 45 cos cos 0 s p 0 So, the planes of maimum shear stress ( s ) occur at 45 to the principal planes ( p ). 3
Original Problem 50 MPa Principal Stresses 54.6 MPa 80 MPa 80 MPa 84.6 MPa 00.5 o 0.5 o 84.6 MPa 5 MPa 50 MPa 54.6 MPa -80, 50, 5 54.6, 0, 3-84.6 Maimum Shear 5 MPa 5 MPa 5 MPa 55.5 o -34.5 o 5 MPa ma ma ma 54.6 69.6 MPa ( 84.6) s s s p ± 45 0.5 ± 45 34.5, 55.5 69.6 MPa ma 69.6, s -5 3
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.
Stress Transformation Equations ( ) sin cos cos sin If we var 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 in graphical form.
3 To do this, we must re-write the transformation equations. ( ) cos sin sin cos Eliminate b squaring both sides of each equation and adding the two equations together. Define avg and R avg R
Substitue for avg and R to get ( ) avg R which is the equation for a circle with centre ( avg,0) and radius R. 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
Sign Convention for Mohr s Circle ( ) avg R avg R Notice that shear stress is plotted as positive downward. The reason for doing this is that is then positive counterclockwise, which agrees with the direction of used in the derivation of the tranformation equations and the direction of on the stress element. Notice that although appears in Mohr s circle, appears on the stress element. 5
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 0. 3. 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 0. 4. 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 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 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 R. (based on Gere) 6
B B (90) A - c R A (0) avg 7
Stresses on an Inclined Element. On Mohr s circle, measure an angle counterclockwise from radius ca, because point A corresponds to 0 and hence is the reference point from which angles are measured.. The angle 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 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, the point on Mohr s circle corresponding to the face moves counterclockwise through an angle. (based on Gere) 8
B B (90) E (90) 80 - A c R D () A (0) E D 9
Principal Stresses B B (90) p A c R p A (0) P p p P 0
Maimum Shear Stress B B (90) min A s ma c R A (0) s Note carefull the directions of the shear forces. ma s s ma s ma s ma s
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 R avg R 65 ( 50 ( 5) ) ( 5) 5 69.6 80 50 5 A (0),, c ± R 5 ± 69.6 54.6 MPa 84.6 MPa B 50 MPa c R B (90) 80 MPa 80 MPa A 5 MPa ma ma s R 69.6 MPa c 5 MPa 50 MPa
50 MPa 80 MPa 80 MPa 5 tan 0.3846 80 5.0.0 80 0 50 MPa 5 MPa 00.5 0.5 A (0) 54.6 MPa c R B (90) 00.5 o 84.6 MPa 84.6 MPa 0.5 o 54.6 MPa 3
50 MPa 80 MPa 80 MPa 5 MPa 50 MPa A (0) smin min.0 s min (90.0) 69.0 s min 34.5 taking sign convention into account 5 MPa 5 MPa -34.5 o 5 MPa 5 MPa 69.6 MPa 55.5 o sma c ma R B (90) 4.0 s ma.0 90.0 s ma 55.5
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 (0) C ( -30 ) -60 c R cos( 60) c R cos( 60) -R sin ( 60) -6-4 -69 5.8 MPa 4.5 MPa -30 o 68.8 MPa 4.5 MPa C D 5.8 MPa -6080 D ( -3090 ) B (90) -30-60 5
Principal Stresses 54.6 MPa, -84.6 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. 54.6 MPa 0 MPa 3-84.6 MPa A (0) This means three Mohr s circles can be drawn, each based on two principal stresses: 3 B (90) and 3 and and 3 6
3 3 3 3 3 3 3 7
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
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
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 0 0 0 0 0 0 3 In mathematical terms, this is the process of matri diagonalization in which the eigenvalues of the original matri are just the principal stresses. 0
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 80 5 5 50 50 MPa 5 MPa We must find the eigenvalues of this matri. 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 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.
80 λ 5 det 0 5 50 λ ( 80 λ)(50 λ) ( 5)( 5) λ 30λ 465 λ 84.6, 54.6 0 0 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 5 80 54.6 5 5 50 λ 0 0 5 50 54.6 0 0 34.6 5 5 4.64 0.86 0.86 0 0 is one eigenvector.
80 λ 5 80 ( 84.6) 5 5 50 λ 0 0 5 50 ( 84.6) 0 0 4.6 5 5 34.6 5.388 5.388 0 0 is the other eigenvector. Before finding the angles at which the principal stresses act, we can check to see if the eigenvectors are correct. 54.6 0 D 0 84.6 D C M C C C A T det C 0.79 0.79 C 0.967 0.033 0.86 5.388 where A matri of M co - factors 80 5 5 50 3
D 0.79 0.79 0.967 80 0.033 5 5 0.86 50 5.388 54.6 0 0 84.6 To find the angles, we must calculate the unit eigenvectors: 0.86 0.83 0.983 5.388 0.938 0.83 And then assemble them into a rotation matri R so that det R. 0.983 0.83 R det R 0.83 0.983 The rotation matri has the form R cos sin sin cos D (0.983)(0.983) (0.83)( 0.83) R T M So 0.5, as we found earlier for one of the principal angles. R 4
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 R T M 0.983 0.83 84.6 0 R 0.83 80 0.983 5 0 54.6 5 0.983 50 0.83 0.83 0.983 84.6 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. 54.6 MPa 5
Finall, we can use the rotation matri approach to find the stresses on an inclined element with -30. R M M M 0.866 0.5 cos( 30 ) sin( 30 ) R T M R 5.8 68.8 sin( 30 ) cos( 30 ) 0.5 80 0.866 5 68.8 4.5 5 0.866 50 0.5 0.866 0.5 0.5 0.866 0.5 0.866 5.8 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