STRESS TRANSFORMATION AND MOHR S CIRCLE

Chapter 5 STRESS TRANSFORMATION AND MOHR S CIRCLE 5.1 Stress Transformations and Mohr s Circle We have now shown that, in the absence of bod moments, there are si components of the stress tensor at a material point. In Cartesian coordinates, these are as shown below. For clarit, onl the stress components on the positive faces are shown. point 0 z zz z z z z Figure 5.1: Stress Components in Cartesian Coordinates Note that the continuum of interest ma ver often be irregular in shape, so that the choice of a coordinate sstem is quite arbitrar. Nevertheless, this choice of coordinate sstem will determine the planes on which the components of stress will be evaluated. These planes will not alwas be the planes that are phsicall interesting or informative, such as the planes on which cracks will grow in a solid, and there is need to be able to find the components of stress in other coordinate sstems. Consider a column loaded b a compressive load as shown below in the left figure below. Draw two different free-bod diagrams. In free-bod #1, a cut is made normal to the -ais. For free-bod #, a cut is made at some angle θ to the -ais and an - coordinate sstem is defined with at the same angle θ to the -ais. In free-bod 1, onl a normal stress will eist on the cut plane (the shear stress is zero). However, in free-bod #, both a normal stress and shear stress will eist on the inclined 105

106 CHAPTER 5. STRESS TRANSFORMATION AND MOHR S CIRCLE p p p column with compressive load free-bod #1 free-bod # θ Figure 5.: Column Loaded in Compression plane (the traction vector on the inclined surface must have two components so that the resultant force equilibrates the applied load). Consequentl, it is important to be able to define the stress on an plane with an orientation relative to the -ais. In this case, no shear stress eists in the - coordinate sstem, but does eist in the - coordinate sstem. In free-bod #, and must satisf conservation of linear momentum such that the sum of the horizontal components of the forces due to the stresses are zero, while the sum of the vertical components must equal the vertical force due to the applied traction on the column. Thus, it is desirable to develop a method for determining the stresses on arbitrar planes at an point once,, zz, z, z, and have been determined. The process of finding these stresses in a coordinate sstem like -, which is rotated b some angle θ relative to the -ais, is called stress transformation. In this tet, we will consider onl states of stress in which shear stresses are non-zero in at most one plane. This state of stress is termed generalized plane stress. An eample of generalized plane stress in the - plane is shown below. Note that no shear stresses eist in the -z or -z planes for this eample. zz z zz Figure 5.3: Generalized Plane Stress in - Plane Principal stress is defined as the normal stress that eists on a plane (at some angle θ) where

5.1. STRESS TRANSFORMATIONS AND MOHR S CIRCLE 107 all shear stresses are equal to zero. In Figure 5.3 above, zz is a principal stress since no shear stresses are shown on the z face (the face with unit outward normal k). Since the coordinate sstem is arbitrar, note that if a state of generalized plane stress eists at a point, the coordinate sstem can alwas be rotated such that the - stress state is as shown below in Figure 5.4 Consider a solid bod such as that shown below. Suppose that we start with the state of stress defined in - coordinates. F F 1 p ' ' ' ' ' ' point 0 ' ' ' ' ' ' ' ' z Figure 5.4: Two-Dimensional Bod with Applied Loads Now pass a cutting plane through at point 0 which has a unit normal n that is rotated an angle θ counter clockwise from the -ais as shown below. Let the - coordinate sstem be rotated about the z-ais b the same angle θ so that is in the direction n. zz ' t ( n ) n' n ' t ( n ) ' ' = s τ ' ' = n zz ' ' z Figure 5.5: Coordinate Sstem for Stress Transformation

108 CHAPTER 5. STRESS TRANSFORMATION AND MOHR S CIRCLE For the above case the unit outer normal vector n, isgiven b n = cos θi + sin θj (5.1) Therefore, application of Cauch s formula (t (n) = n ) results in the following components for t (n) (the stress tensor is assumed to be smmetric and therefore = ): t (n) = cos θ + sin θ t (n) = cos θ + sin θ (5.) t (n)z = 0 However, these components are of no particular phsical interest since the are neither normal nor parallel to the plane defined b n. The components of t (n) that are normal and parallel to the plane defined b n can be easil obtained through vector calculus. Note that the unit normal n and the -ais have the same direction. The component of t (n) normal to the plane is thus and is given b the dot product of t (n) with n (note: first substitute = in t (n) ): = t (n) n = (( cos θ + sin θ)i +( cos θ + sin θ)j) (cos θi + sin θj) (5.3) = cos θ + sin θ cos θ + sin θ where the - aes are rotated counterclockwise from the -aes b the angle θ. Since the ais is in the direction of n, (stress normal to plane with unit normal n) is often denoted as n. Equations 5. and 5.3 can also be combined b writing t (n) = n and n = = t (n) n so that we have the following vector and matri result: n = = n n = (1 ) [n] ( )( 1) [] [n] (5.4) In order to obtain the component of t (n) parallel to the plane defined b n, itisfirst convenient to define the unit normal in the direction, denoted n,asfollows: n = k n = sin θi + cos θj (5.5) The projection of t (n) on the plane n is found b taking the dot product of t (n) with n,thus ielding the shear stress component, = t (n) n = (5.6) = ( ) sin θ cos θ + (cos θ sin θ) In the same manner, and can be determined (note: this step is not necessar): = t (n ) n = = [ ( sin θ + cos θ) sin θ +( sin θ + cos θ) cos θ] = t (n ) n = = [( sin θ + cos θ) cos θ +( sin θ + cos θ) sin θ] Equations 5.3 and 5.6 can now be used to obtain the normal stress,, and shear stress,,on a plane defined b the angle θ (measured CCW from -ais), given, and : Stress Transformation from - to - Coordinates = cos θ + sin θ cos θ + sin θ (5.7) = ( ) sin θ cos θ + (cos θ sin θ)

5.1. STRESS TRANSFORMATIONS AND MOHR S CIRCLE 109 Although the above equations are sufficient to perform stress transformations, the are not ver convenient. This is due to the fact that we are often interested in finding the planes defined b θ on which and attain their maima because failure is often initiated on these planes due to the stresses on these planes. Mathematicall, an equation for the plane of maimum (or minimum) stress can be obtained from calculus b appling the following: d dθ d dθ = ( ) sin θ + cos θ =0 (5.8) = sin θ ( ) cos θ =0 Equations (5.8) can be solved for θ that maimizes/minimizes and. Note that one obtains one value from each equation, i.e., θ defining the plane of maimum normal stress (call it θ P ), and a second θ for the maimum shear stress (call it θ S ). It can be shown that θ S = θ P ± π 4. Thus, the plane of maimum shear stress is alwas ±45 from the plane of maimum normal stress. These corresponding two values of θ ma then be substituted into (5.7) to obtain the maimum and minimum values of and. Because equations (5.8) are transcendental, closed form solutions for θ do not eist; however the can be solved numericall or graphicall. It is important to note that numerical evaluations using equations (5.7) and (5.8) must be done with the angle θ in radians, not degrees. Furthermore, θ is positive counterclockwise from the -ais (normal right-hand rule for an - coordinate sstem). In order to deal with the numerical difficult associated with the nonlinear equations, Otto Mohr introduced a graphical technique that is helpful in performing stress transformations. We shall now derive this graphical technique using equations. To accomplish this, first recall the trigonometric half angle formulas: cos θ = 1+cos θ sin θ = 1 cos θ sin θ cos θ = sin θ (5.9) Substituting (5.9) into (5.7) results in = + = ( ( + ) cos θ + sin θ (5.10) ) sin θ + cos θ (5.11) Otto Mohr (1870) discovered that equations (5.10) and (5.11) can be combined b squaring each side of the above two equations and adding the results together to produce the principle of Mohr s Circle represented b the equation: [ ( + Recall that the equation of a circle in - space is given b )] ( ) + = + (5.1) [ a] +[ b] = r (5.13) where a and b are the and intercepts of the center of the circle, respectivel, and r is the radius, as shown below:

110 CHAPTER 5. STRESS TRANSFORMATION AND MOHR S CIRCLE r a b ( a) +( b) = r Figure 5.6: Circle Used in Mohr s Analog Thus b analog of (5.1) to (5.13), we ma effect the following transformation of variables: ( (normal stress) = (shear stress) a + = (center of circle) (5.14) b 0 = r ) + = (radius of circle) thereb producing the following diagram, denoted as Mohr s Circle: (τ s ) r = + (, ) face stresses } C p r A θ s θ p B p1 ( n ) { (, ) face stresses a = + p1, p = principal stresses θ is positive counter clockwise Figure 5.7: Mohr s Circle Parameters

5.1. STRESS TRANSFORMATIONS AND MOHR S CIRCLE 111 To plot Mohr s circle given stresses,, in the - coordinate sstem: ( ) + 1. Find the center of Mohr s circle; point A:, 0 Note: A = +. Plot stresses on -face as point B (, ) and label as -face. Note that point B corresponds to θ =0 in - plane. 3. Plot stresses on -face as point C (, ) and label as -face. Note that point C corresponds to θ =90 in - plane. 4. Draw the circle. We note that while the -face (θ =0 ) and the -face (θ =90 ) are separated b 90 in the real world - coordinates, the appear as point B and C, respectivel, in Mohr s Circle and are 180 apart on Mohr s Circle. Consequentl, angles on Mohr s Circle are twice that of the real - world. Itisalso important to note that θ =0 in the real world corresponds to the -ais; however, in Mohr s Circle the point θ =0 starts from the line AB since point B represents the stresses on the -face θ =0 ). θ is positive CCW in both cases. 5. Label principal stresses and maimum shear stress. Note: P = A ±r and τ Sma = ±r. There are two principal stresses P 1 and P (planes where shear stress is zero) and two maimum shear stresses τ Sma = ±r (top and bottom of circle). Note that P 1 and P are 180 apart on Mohr s circle, but 90 in the real world. P 1 is the normal stress in the direction where is rotated b the angle θ (CCW) from the -ais. P is in the direction. 6. Identif the relative orientation between -face (B) and the principal planes. Note that point B corresponds to the stresses ( on) the -face, i.e. θ =0 in the real world. For eample, in the above θ P = tan 1 A CCW from the -face (from line A-B on Mohr s Circle). Note: the writing of a specific formula is discouraged. It is far simpler (and usuall less mistakes) to look at Mohr s circle and appl trigonometr to calculate the angle. 7. Identif the relative orientation between -face (B) and the maimum shear planes. From Figure 5.7, θ P +θ S =90.Thusθ S =45 θ P.From Figure 5.7, it easier to identif the principal shear stress planes as being 45 (real world) from the principal normal stress plane (point B, bottom of circle) and +45 from the principal normal stress plane (top of circle). Note that the plane of ma shear stress will generall have non-zero normal stress! 8. Draw Free Bod Diagrams with the -face, principal plane, and the two maimum shear planes. Eample 5-1 Consider the plane stress state given b the stress tensor: [] = [ 50 10 10 0 ] ksi 1. Determine center of circle, point A. A = + = 50 + 0 =35ksi. Determine point B, -face (, )=(50, 10 ) 3. Determine point C, -face (, )=(0, +10 ) 4. Draw Mohr s circle:

11 CHAPTER 5. STRESS TRANSFORMATION AND MOHR S CIRCLE 0 ksi 10 ksi 50 ksi 50 ksi 10 ksi 0 ksi Figure 5.8: ' ' ( τ s ) (0,10) -face (35,0) ' ' ( τ n ) (50,-10) -face Figure 5.9: 5. Compute principal stresses and ma shear stress. r = ( ) + = (50 0 ) +10 =18.03 ksi P = A ± r =35± 18.03 P 1 =35+18.03 =53.03 ksi

5.1. STRESS TRANSFORMATIONS AND MOHR S CIRCLE 113 P =35 18.03 = 16.97 ksi τ Sma = ±r = ±18.03 ksi Note that we have two principal stresses: P 1 and P. These are located 180 on Mohr s circle, but 90 in the real world. In relation to the stress transformation equations (5.7), P 1 in the normal stress in -ais direction (which is oriented at an angle θ P to the -ais) and P is the normal stress in the -ais direction. The planes of maimum shear occur 45 (real world) from the planes of principal stress. Plot the principal stresses and ma shear stresses on Mohr s Circle: ' ' ( τ s ) (0,10) -face τ S ma = 18.03ksi (35,18.03) P = 16.97ksi (35, 0) θ P P1 = 53.03ksi ' ' ( n ) θ S (50,-10) -face τ (35, 18.03) S ma = 18.03ksi Figure 5.10: 6. Identif orientation of planes for principal stresses relative to -face plane. ( ) ( ) θ P = tan 1 10 = tan 1 =33.7 deg A 50 35 θ P =16.85 deg CCW from -face The plane for the second principal stress P is (θ P + 90) deg CCW from -face (real world). Note: In the above calculation, a formula was written down and used for θ P. This is discouraged. It is far simpler (and usuall less mistakes) to look at Mohr s circle and appl trigonometr to calculate the angles. 7. Identif orientation of planes for maimum shear stress (bottom of circle) relative to -face plane. θ S =90deg θ P =90 33.7 =56.3 deg

114 CHAPTER 5. STRESS TRANSFORMATION AND MOHR S CIRCLE θ S =8.15 deg CW from the -face OR, plane of ma shear stress (bottom of circle) is 45 CW in the real world from the principal stress plane with P 1. Note: P + S =45deg (alwas!) Note that the plane of ma shear stress will generall have non-zero normal stress! In this case, the plane of ma shear stress has a normal stress n =35ksi. 8. Draw three free bodies: with the - stresses, with the principal stresses, and with the ma shear stresses. 0 ksi 10 ksi The dark face is the original -face as it is rotated. 50 ksi 50 ksi 10 ksi 0 ksi 16.85 o 16.97 ksi 53.03 ksi θ = 16.85 o P 53.03 ksi principalstress planes 35 ksi 18.03 ksi 35 ksi 35 ksi 8.15 o o 45 18.03 ksi 16.97 ksi θ S = 8.15 o 35 ksi maimum shear stress planes The shear stress on the rotated -face is negative on Mohr s circle and thus positive on thestress cube. OR, negative shear stress on Mohr s circle causes a CCW moment on the stress cube. Figure 5.11: Note the following:

5.1. STRESS TRANSFORMATIONS AND MOHR S CIRCLE 115 To obtain the orientation for principal stresses, the -face has been rotated 16.85 deg CCW in the real world (33.7 deg on Mohr s circle). To obtain the orientation for maimum shear stresses, the -face has been rotated 8.15 deg CW in the real world (56.3 deg on Mohr s circle), OR, equivalentl a rotation of 45 deg CW in the real world from the principal stress plane. Eample 5- Consider the same plane stress state given in Eample 5-1: 0 ksi 10 ksi 50 ksi 50 ksi 10 ksi 0 ksi Figure 5.1: [] = [ 50 10 10 0 ] ksi Values of principal stress and maimum shear stress obtain in Eample 5-1 b Mohr s circle must be identical to what would be obtained b using the stress transformation equations (5.7). This is true because Mohr s circle is simpl a graphical representation of equations (5.7). Use the stress transformations to determine the normal stress and shear stress for values of θ =16.85, 106.85, 8.15 and verif that the correspond to the principal stress and maimum shear stress values obtained in Eample 5-1. For θ =16.85 (same as 33.7 on Mohr s circle from -face, should equal P 1 ) = cos θ + sin θ cos θ + sin θ = (50 ksi) cos 16.85 + (10 ksi) sin 16.85 cos 16.85 + (0 ksi) sin 16.85 =53.03 ksi = ( ) sin θ cos θ + (cos θ sin θ) = (50 ksi 0 ksi) sin 16.85 cos 16.85 +10ksi(cos 16.85 sin 16.85 )=0ksi For θ = 106.85 (same as 13.7 on Mohr s circle from -face, should equal P ) = cos θ + sin θ cos θ + sin θ = (50 ksi) cos 106.85 + (10 ksi) sin 106.85 cos 106.85 + (0 ksi) sin 106.85 =16.97 ksi = ( ) sin θ cos θ + (cos θ sin θ) = (50 ksi 0 ksi) sin 106.85 cos 106.85 +10ksi(cos 106.85 sin 106.85 )=0ksi For θ = 8.15 (same as 56.3 on Mohr s circle from -face, should equal τ ma bottom of circle) = (50 ksi) cos ( 8.15 )+(10 ksi) sin( 8.15 ) cos( 8.15 )+(0 ksi) sin ( 8.15 )=35ksi = (50 ksi 0 ksi) sin( 8.15 ) cos( 8.15 )+10ksi(cos ( 8.15 ) sin ( 8.15 )) = 18.03 ksi

116 CHAPTER 5. STRESS TRANSFORMATION AND MOHR S CIRCLE For θ =40 (same as 80 on Mohr s circle from -face) =47.45 ksi, = 13.04 ksi (real world values). From, Mohr s circle in Eample 5-1, one obtains the same result. Student, ou should verif this! Note that this corresponds to a point 80 degrees CCW from the -face on Mohr s circle (or 46.3 degrees above the n ais on Mohr s circle ( 80, 33.7 )). Wh is the sign negative on the value of calculated from the stress transformation equation? Remember that when ou plot the -face stress, a positive (real world) is plotted as (point B on Mohr s circle). Thus, a positive value of shear stress on Mohr s circle as we have for this case (θ =80 CCW on Mohr s circle) translates to a negative shear stress in the real world. ' ' ( τ s ) τ S ma = 18.03ksi (35,18.03) (0,10) -face 47.45,13.04 P = 16.97ksi (35, 0) 80 ' ' ( n ) P1 = 53.03ksi (50,-10) -face τ 35, 18.03 S ma = 18.03ksi 0 ksi 10 ksi 50 ksi 50 ksi 10 ksi 0 ksi.55 ksi 13.04 ksi 40 o 47.45 ksi θ = 40 o 13.04 ksi 47.45 ksi.55 ksi Figure 5.13:

5.1. STRESS TRANSFORMATIONS AND MOHR S CIRCLE 117 Eample 5-3 Consider the plane stress state given b the stress tensor: 0 ksi 10 ksi 50 ksi 50 ksi 0 ksi 10 ksi Figure 5.14: [ 50 10 [] = 10 0 1. Determine center of circle, point A. ] ksi A = + = 50+( 0) =15ksi. Determine point B, -face (, )=(50, ( 10) ) = ( 50, 10 ) 3. Determine point C, -face (, )=( 0, 10 ) 4. Draw Mohr s circle: 5. Compute principal stresses and ma shear stress. r = ( ) + = (50 ( 0) ) +( 10) =36.4 ksi P = A ± r =15± 36.4 P 1 =15+36.4 =51.4 ksi P =15 36.4 = 1.4 ksi τ Sma = ±r = ±36.4 ksi Note that we have two principal stresses: P 1 and P. These are located 180 on Mohr s circle, but 90 in the real world. In relation to the stress transformation equations (5.7), P 1 in the normal stress in -ais direction (which is oriented at an angle θ P to the -ais) and P is the normal stress in the -ais direction. The planes of maimum shear occur 45 (real world) from the planes of principal stress. Plot the principal stresses and ma shear stresses on Mohr s Circle:

118 CHAPTER 5. STRESS TRANSFORMATION AND MOHR S CIRCLE ' ' ( τ s ) (15, 0) (50,10) -face ' ' ( n ) (-0,-10) -face Figure 5.15: ' ' ( τ s ) (15,36.4) τ S ma = 36.4ksi P = 1.4ksi (-0,-10) -face (15, 0) 74.05 15.95 (50,10) -face P1 = 51.4ksi ' ' ( n ) τ (15, 36.4) S ma = 36.4ksi Figure 5.16: 6. Identif orientation of planes for principal stresses relative to -face plane. ( ) 10 θ P = tan 1 = 15.95 deg 35

5.. GENERALIZED PLANE STRESS AND PRINCIPAL STRESSES 119 θ P =7.97 deg CW from -face The plane for the second principal stress P is (θ P + 90)deg CCW from -face (real world). 7. Identif orientation of planes for maimum shear stress (bottom of circle) relative to -face plane. θ S =90deg θ P =90 15.95 = 74.05 deg θ S =37.03 deg CCW from the -face or, plane of ma shear stress (top of circle) is 45 CCW from plane with P 1. Note that the plane of ma shear stress will generall have non-zero normal stress! In this case, the plane of ma shear stress has a normal stress n =15ksi. 8. Draw three free bodies: with the - stresses, with the principal stresses, and with the ma shear stresses. Note the following: To obtain the orientation for principal stresses, the -face has been rotated 7.97 deg CW in the real world (15.95 deg on Mohr s circle). To obtain the orientation for maimum shear stresses, the -face has been rotated 37.03 deg CCW in the real world (74.05 deg on Mohr s circle), or, equivalentl a rotation of 45 deg CCW in the real world from the principal stress plane. 5. Generalized Plane Stress and Principal Stresses In the - plane, an set of stresses (,, ) can be represented b a Mohr s Circle as was shown previousl and as shown in Figure 5.18 below (the dark circle). Corresponding to these - stresses, there are two principal stresses ( P 1, P ) which also define the same Mohr s Circle with coordinates of ( P 1, 0) and ( P, 0). Thus, the circle is (or can be) defined b an two principal stresses as the form the diameter of the circle. It follows that the Mohr s circle for the -z plane would also be formed b the principal stresses in the -z plane, in this case P 1 and P 3. Similarl, Mohr s circle in the -z plane is defined b P and P 3.Wecan draw all three of these circles on one diagram as shown below. If the value of the third principal stress is less than the second, or greater than the first, we obtain a larger circle than the circle obtain for the - plane and hence the maimum shear stress is also larger. For the case in the figure above, even if the out-of-plane normal stress zz =0,one obtains a larger maimum shear stress than was obtained from the - plane. For a general stress tensor (3-D), it can be shown that the principal stresses are defined b the following eigenvalue problem: [] P [I] =0 ( P ) z ( P ) z z z ( zz P ) =0 After the 9 components of stress (,,..., zz ) are substituted into the above, the determinant can be epanded to obtain a cubic polnomial in P which will ield the three principal stress values P 1, P, P 3.For-D (- plane), the determinant reduces to ( P ) ( P ) = 0 which results in two principal stress values P 1, P.

10 CHAPTER 5. STRESS TRANSFORMATION AND MOHR S CIRCLE 10 ksi 0 ksi 50 ksi 50 ksi 10 ksi 0 ksi 51.4 ksi 1.4 ksi 7.97 principalstress planes 51.4 ksi 1.4 ksi 45 36.4 ksi 15 ksi 15 ksi 36.4 ksi maimumshear stress planes 15 ksi 15 ksi 37.03 Figure 5.17: Eample 5-4 Same as Eample 5-1 ecept a z component of stress is added to make the problem a generalized plane stress problem. The generalized plane stress state is given b the stress tensor:

5.. GENERALIZED PLANE STRESS AND PRINCIPAL STRESSES 11 ( ) ' ' τ s τ ( generalized plane stress) S ma r (, ) P3 ( zz,0) P θ P ( ) ' ' n P1 θ S (, ) τ S ma Figure 5.18: Mohr s Circle for Generalized Plane Stress (plane stress in - plane) [] = 50 10 0 1 0 0 10 ksi Steps 1-7 will be identical to Eample 5-1 for the - plane (since both eamples have eactl the same stress components in the - plane). Hence, for the - plane we obtain Mohr s circle: To complete the solution we note that there are no shear stresses in the z plane and hence zz is automaticall a principal stress. Thus, P 3 = zz = 10. Adding the third principal stress to Mohr s circle, we obtain: Hence, the principal stresses are P = 10, 16.97, 53.03 ksi. The maimum shear stress is the radius of the largest Mohr s circle. Thus, τ Sma = 53.03+10 = 31.5 ksi. In order to change circles, we need to be in the principal orientation, because this is the same on all three circles. The sketches below illustrate this changing process. Note that when changing views, there is NOT a change in the stresses, onl in the wa the are viewed. If we had rotated 45 off the principal plane P 1 P, we would have still been on the smaller circle and therefore could never attain the maimum shear stress.

1 CHAPTER 5. STRESS TRANSFORMATION AND MOHR S CIRCLE ' ' ( τ s ) τ S ma = 18.03ksi (35,18.03) (0,10) -face P = 16.97ksi (35,0) 16.85 P1 = 53.03ksi ' ' ( n ) θ S (50,-10) -face τ (35, 18.03) S ma = 18.03ksi Figure 5.19: ' ' ( τ s ) τ S ma = 31.5ksi (35,18.03) (0,10) -face P3 = 10ksi (35,0) P = 16.97ksi 16.85 P1 = 53.03ksi '' ( n ) θ S (50,-10) -face (35, 18.03) τ S ma = 31.5ksi Figure 5.0: Eample 5-5 Given the stress state below, use Mohr s circle to find the principal planes and maimum shear

5.. GENERALIZED PLANE STRESS AND PRINCIPAL STRESSES 13 0 ksi 50 ksi 10 ksi -10 ksi 10 ksi 50 ksi z 0 ksi 16.97 ksi 8.45 o -10 ksi 53.03 ksi principalstress planes P 53.03 ksi P 1 16.97 ksi P ( inzdirection) 3 AA Vie w AA: 53.03 ksi principalstress planes 10 ksi 16.97 ksi 10 ksi P 1 P P 3 53.03 ksi 1.515 ksi 31.5 ksi 45 16.97 ksi 1.515 ksi maimumshear stress planes 1.515 ksi 1.515 ksi Figure 5.1: stresses. Uniaial Stress [] = 0 0

14 CHAPTER 5. STRESS TRANSFORMATION AND MOHR S CIRCLE 5MPa z B -face.5.5 C 45 = n =.5 B = τ s =.5 -face (0,0).5 C Figure 5.: Maimum shear traction at 45 from -ais Maimum normal traction along -ais Eample 5-6 Given: The state of stress at each point in a bar can be represented b the stress tensor []: [] = 0 0 = 10 kn m 0 0 z Figure 5.3: Required : a) Construct Mohr s Circle for the given stress tensor. b) What are the traction vectors eperienced each face of the cube?

5.. GENERALIZED PLANE STRESS AND PRINCIPAL STRESSES 15 c) What are the normal and shear components of these traction vectors? d) What is the maimum shear stress eperienced b the bar? e) What is the orientation of the surface that eperiences this maimum shear stress? f) What is the maimum normal stress eperienced b the bar? Solution a) Use the steps mentioned in the above tet. ' ' MaimumShear (5,5) Maimum Normal -face C(0,0) A(5,0) Maimum Normal -face B(10,0) ' ' (5,-5) MaimumShear Figure 5.4: b) t i =10i t j = 0 t k = 0 c) =10 = z = zz = =0 d) 5 kn m at the points ( 5, 5) and (5, 5) e) The surface is tilted at 45 to the -ais (90 in Mohr space is 45 in real space). f) 10 kn m at the points ( 10, 0) and (0, 0) Eample 5-7 Given: -D Stress State Ignore the z components of stress. Required : [] = 0 0 = a) Construct Mohr s Circle for the given stress tensor. 6 0 6 0 MPa b) What is the orientation of the surface when the maimum shear stress is obtained? c) What is the maimum shear stress eperienced b the bar?

16 CHAPTER 5. STRESS TRANSFORMATION AND MOHR S CIRCLE =6 = = =6 z Figure 5.5: ' ' Maimum Shear -face B(6,) Maimum Normal A(6,0) Maimum normal ' ' C(6,-) -face Maimum Shear Figure 5.6: Solution a) b) 0 c) MPa at points ( 6, ) and (6, )

5.. GENERALIZED PLANE STRESS AND PRINCIPAL STRESSES 17 Deep Thought

18 CHAPTER 5. STRESS TRANSFORMATION AND MOHR S CIRCLE 5.3 Questions 5.1 Describe in our own words the meaning of principal stress. 5. What is the angle between the planes of principal stress and maimum shear stress in a bod? Wh is it that wa? 5.3 What is the main point of Mohr s circle? 5.4 What does the term t (n) in Cauc s formula phsicall represent? 5.5 When considering Mohr s circle, eplain phsicall the stress state defined b the point on Mohr s Circle that lies on the horizontal ais. 5.4 Problems 5.6 GIVEN :Amaterial point is known to be in a state of generalized plane stress as given b: 4 0 [] = 5 0 MPa 0 0 3 REQUIRED: Draw all important sketches, showing planes of maimum shear as well as principal planes. 5.7 GIVEN : The stress tensor as follows: REQUIRED: [] = 3 0 0 0 19 0 ksi (a) compute the three principal stresses and maimum shear stress; (b) indicate the maimum shear direction. 5.8 The state of stress at a certain point of a bod is given b 1 0 [] = 7 0 MPa Determine on which planes: i) principal stresses occur ii) shearing stress is greatest. 5.9 [] = 0 0 psi Find maimum shear and normal stresses. 5.10 (a) What is a principal plane?

5.4. PROBLEMS 19 75 psi z Problem 5.9 (b) Are the following stress tensors generalized plane stress cases? Answer in es/no. i) iii) 0 0 1 0 3 3 0 5 1 0 0 0 0 1 0 1 6 ii) 0 1 0 4 3 0 1 iv) 0 3 7 7 5 5.11 GIVEN : [] = 0 0 0 8 0 0 0 10 ksi (a) Write out the three principal stresses and compute the maimum shear stress. (b) What is the maimum shear direction and on what plane is it acting in? 5.1 GIVEN : [] = [ 5 5 5 5 ] MPa (a) Draw the corresponding stress cube. (b) Construct the three Mohr s circles for the given stress state. (c) Compute the two principal stresses and maimum shear stress. (d) Using the formula given below, determine the normal stress, which acts upon a surface corresponding to a 30 clockwise rotation of the -face. ( ) ( ) + n = cos θ + sin θ + 5.13 GIVEN :Amaterial in the following state of plane stress: 50 [] = ksi REQUIRED: Draw all important sketches, showing planes of maimum shear stress as well as principal planes.

130 CHAPTER 5. STRESS TRANSFORMATION AND MOHR S CIRCLE 5.14 Using the following -D stress state at a point in a bod, 50 0 0 [] = 0 50 0 MPa Find: ( ) a) t n for n = i, n = j, and n = 1 3i + j b) If the coordinate aes were rotated 45 degrees clockwise, to -, solve for the new values of,,. Use Cauch s Formula. c) Construct the three Mohr s circles for the given stress state. 5.15 The state of stress at a certain point of a bod is given b [] = 1 0 ksi 3 1 4 4 0 i) Find the traction vector on the plane whose normal is in the direction i +j +k ii) Find the traction vector on the plane whose normal is in the direction 4i 3j + k iii) Determine the normal and shearing components of the traction vector on the above planes. 5.16 For a stress state defined b: =10MPa, = =40MPa, =50MPa. Define: a) the value of the Principal Stress b) the maimum Shear Stress and corresponding Normal Stress 5.17 GIVEN : A thin-walled clindrical pressure vessel is subjected to internal pressure, p θ p =500psi r = 10in t =0.5in t =wall thickness r =clinder radius pr t pr t Problem 5.17 It can It can be shown that the stress state is given b a hoop (circumferential) stress θθ = pr t and an aial stress (along the ais of pressure vessel) of = pr t : REQUIRED: Draw all important sketches, showing planes of maimum shear as well as principal planes. 5.18 GIVEN : Suppose that the pressure vessel of 5.17 is also subjected to an end torque, M t. The stress state be shown b: REQURIED: Draw all important sketches, showing planes of maimum shear as well as principal planes.

5.4. PROBLEMS 131 pr t Mr t J M t M t = 1, 000, 000 lb-in 3 J = π r t pr t Problem 5.18 5.19 GIVEN : The following stress tensor in Cartesian coordinates [] = 1 0 MPa 5 0 0 0 3 REQUIRED: a) Sketch the stress cube representation b) Construct the three Mohr s circles c) Determine the traction vectors: t (i), t (j), t (k) d) Make a sketch of t (i), t (j), t (k) acting on the appropriate plane(s) 5.0 GIVEN :Amaterial point is known to be in a state of generalized plane stress as given b: 1 0 [] = 3 0 ksi 0 0 4 REQUIRED: 1) Construct the three Mohr s circles. ) Find the three principal stresses and the maimum shear stress. 3) Draw all important sketches, showing planes of maimum shear as well as principal planes with the associated coordinate rotations labeled properl. 5.1 GIVEN : The state of stress at a certain point on a bod is given b: 4 0 a) [] = 3 1 0 1 0 b) [] = 0 3 7 7 5 REQUIRED: Determine for both a) and b) on which of the three coordinate planes: 1) Normal stress is greatest ) Shear stress is greatest

13 CHAPTER 5. STRESS TRANSFORMATION AND MOHR S CIRCLE 5. Use Cauch s Formula, t n = n, for the following stress tensor [] = 0 5 0 5 6 0 ksi n = cos θi + sin θj +0k REQUIRED: a) The plane at which the shear stress equals zero, i.e., the angle in the unit normal which describes the plane for the following stress tensor. b) The normal stress on the face described b the plane found in part a). What is this normal stress called? c) At what angle is the plane of maimum shear stress from the principal stress plane. d) What is the value of the maimum shear stress for the stress tensor given above. 5.3 GIVEN : The state of stress at each point in a bar can be represented b the stress tensor []: [] = 0 0 = 8 kn m 0 0 REQUIRED: z =8 [ ] kn m Problem 5.3 a) What are the traction vectors eperienced b faces in other orientations? b) What are the normal and shear components of these traction vectors? c) What is the maimum shear stress eperienced b the bar? d) What is the orientation of the surface that eperiences this maimum shear stress? e) What is the maimum normal stress eperienced b the bar? 5.4 GIVEN : [] = 0 0 = 3 1 0 1 3 0 MPa REQUIRED: a) Construct Mohr s circle for the given stress tensor. b) What is the orientation of the surface when the maimum shear stress is obtained? c) What is the maimum shear stress eperienced b the bar?

5.4. PROBLEMS 133 =3 =1 =1 z =3 Problem 5.4 5.5 The state of stress at a certain point of a bod is given b 1 0 [] = 3 0 MPa 0 0 4 a) Calculate the values of the corresponding principal stresses. b) Indicate the magnitude of the maimum shear stress in the - plane. c) Draw the - plane including 1) principal stresses and ) ma shear stress. In each case, indicate the magnitude of all stresses and the angle the element is rotated relative to the original -face. d) Determine the magnitude of the maimum shear stress considering all planes. e) If the material s failure criteria is given b a6mpanormal stress and 3 MPa shear stress. Indicate if the material will fail under the applied load and if so, whether it is due to shear or normal stress. 5.6 Repeat problem 5.5 for: [] = 5 3 0 3 0 MPa Remember to consider the problem as a case of Generalized Plane Stress. 5.7 GIVEN : stress tensors as follows 5 0 0 (1) 0 0 () REQUIRED: 4 0 0 0 4 0 0 0 4 (3) 3 3 0 3 0 0 0 0 (a) sketch the stress cube aligned with the coordinate aes (Cartesian coordinate sstem) and the values of all shear and normal stresses on its surfaces; (b) construct the three Mohr s circles for the stress state; (c) sketch a stress cube oriented so that it is aligned with the principal aes and indicate its angle relative to the -face and the principal stresses which act upon it; (d) determine the maimum shear and normal stresses eperienced b the material which is subjected to this stress state and for each of these stresses, indicate the orientation of the surface which eperiences this stress;

134 CHAPTER 5. STRESS TRANSFORMATION AND MOHR S CIRCLE (e) determine the shear and normal stresses that act upon a surface which corresponding to a30 counter-clockwise rotation of the -face about z-ais. 5.8 For the given simpl supported beam as pictured below, with length L = 10 m, thickness h =0.1 m,cross sectional area A =0.01 m and a distributed load P o =10 kn m,astress analsis indicates a stress of = MPa, = 1 MPa, =4MPaat =5m, = 0.05 m. Repeat the stress analsis described in problem 5.7. P o h L Problem 5.8 5.9 Derive an equation for b hand. 5.30 Using Scientific Workplace: a) evaluate,,, b) verif that = (assume = ) c) Graph (θ), (θ) for 0 <θ<π for the two cases below and discuss our findings. 5 10 Problem 5.30 5.31 GIVEN : REQUIRED: a) Find the stress tensor, []. b) Find the principal stresses. c) Find the principal direction. 5.3 GIVEN :Athin plate is loaded on its four faces so that it develops the following stress state. [] = 7 ( 3 ) 7 0 7 7 3 3 0 MPa

5.4. PROBLEMS 135 5MPa 3MPa 10 MPa Problem 5.31 h h L Problem 5.3 REQUIRED: 1) Verif that the conservation of angular momentum and linear momentum are satisfied under static conditions, in the absence of bod forces, at ever point (, )onthe plate. ) Find the components of the traction vector applied on the faces: =0, = L, = h, = h 5.33 For the following structure and the applied load F, the stress state in a rectangular Cartesian coordinate sstem at point P is given b the specified stress tensor []. 5 0 [] = 1 0 MPa 0 0 3 a) Sketch the stress cube representation. Label all the components, planes and points of interest. b) Determine the values of the Principal Stresses. c) Specif the required rotation (in terms of the normal to the -face) in order to achieve maimum shear stress in the face. Make a sketch of the face showing the components of stress and the angle rotated.

136 CHAPTER 5. STRESS TRANSFORMATION AND MOHR S CIRCLE F z P Problem 5.33 d) According to the material s properties, the failure criteria is given b a maimum stress of 4 MPa for shear and 6 MPa for normal stress. Predict if the material will fail due to the applied load. Wh?