3D Stress Components. From equilibrium principles: τ xy = τ yx, τ xz = τ zx, τ zy = τ yz. Normal Stresses. Shear Stresses

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1 3D Stress Components From equilibrium principles:, z z, z z The most general state of stress at a point ma be represented b 6 components Normal Stresses Shear Stresses Normal stress () : the subscript identifies the face on which the stress acts. Tension is positive and compression is negative. Shear stress () : it has two subscripts. The first subscript denotes the face on which the stress acts. The second subscript denotes the direction on that face. A shear stress is positive if it acts on a positive face and positive direction or if it acts in a negative face and negative direction. z z z

2 For static equilibrium, z z, z z resulting in s i independent scalar quantities. These si scalars can be arranged in a 33 matri, giving us a stress tensor. ij z z z z z The sign convention for the stress elements is that a positive force on a positive face or a negative force on a negative face is positive. All others are negative. The stress state is a second order tensor since it is a quantit associated with two directions (two subscripts direction of the surface normal and direction of the stress). Same state of stress is represented b a different set of components if aes are rotated. There is a special set of components (when aes are rotated) where all the shear components are zero (principal stresses).

3 A propert of a smmetric tensor is that there eists an orthogonal set of aes, and 3 (called principal aes) with respect to which the tensor elements are all zero ecept for those in the diagonal. ij z z z z z ' igen values ' ij 3 In matri notation the transformation is known as the igen-values. The principal stresses are the new-aes coordinate sstem. The angles between the old-aes and the new-aes are known as the igen-vectors. principal stress Cosine of angle between X and the principal stress Cosine of angle between Y and the principal stress Cosine of angle between Z and the principal stress k l m k l m 3 k3 l3 m3

4 Plane Stress State of stress in which two faces of the cubic element are free of stress. For the illustrated eample, the state of stress is defined b,, and z z z. Sign Conventions for Shear Stress and Strain The Shear Stress will be considered positive when a pair of shear stress acting on opposite sides of the element produce a counterclockwise (ccw) torque (couple).

5 A shear strain in an element is positive when the angle between two positive faces (or two negative faces) is reduced, and is negative if the angle is increased.

6 Stresses on Inclined Sections Knowing the normal and shear stresses acting in the element denoted b the aes, we will calculate the normal and shear stresses acting in the element denoted b the ais. quilibrium of forces: Acting in X AO X A O XY sin A O Sin Y A O Sin Cos YX A O Sin Cos

7 liminating A o, sec / and X X Y sin XY sin Y X Y XY sin sin Acting in A o sec A o sin A o A o tan A o tansin liminating A o, sec / and ( ) sin sin sin

8 Transformation quations for Plane Stress Using the following trigonometric identities: Cos ½ ( ) Sin ½ (- ) Sin ½ sin sin sin sin These equations are known as the transformation equations for plane stress.

9 Case : Uniaial stress Special Cases, Sin Cos Case : Pure Shear, Cos Sin sin sin Case 3: Biaial stress, Sin Cos

10 An element in plane stress is subjected to stresses 6psi, 6psi, and 4psi (as shown in figure below). Determine the stresses acting on an element inclined at an angle 45 o (counterclockwise - ccw). Solution: We will use the following transformation equations: sin sin sin

11 Numerical substitution For -ais Sin Cos 45 (ccw) ( ) Sin( 9 ) ( ) Cos( 9 ) For -ais Sin Cos 45 9 ( ) Sin( 7 ) ( ) Cos( 7 ) (ccw) 4 psi 5 ( 6 6) ( 6 6) psi 5 psi 5( ) 4( ) 5( ) 4( ) () 4( ) 5 psi 5 psi 7 psi Note:

12 A plane stress condition eists at a point on the surface of a loaded structure such as shown below. Determine the stresses acting on an element that is oriented at a clockwise (cw) angle of 5 o with respect to the original element, the principal stresses, the maimum shear stress and the angle of inclination for the principal stresses Solution: We will use the following transformation equations: sin sin 9MPa ( 46 ) ( 46 ) Sin Cos 7MPa 9MPa For -ais ( ) Sin( 3 ).5 ( ) Cos( 3 ). 866 ( 7) ( 9)(.866) ( 9)(.5) ( 9)(.5) ( 9)(.866) 3MPa 5 (cw).4mpa 3.6MPa

13 Determine the normal stress sw acting perpendicular to the line or the weld and the shear stress tw acting parallel to the weld. (Assume sw is positive when it acts in tension and tw is positive when it acts counterclockwise against the weld). Solution: 75psi 45psi psi 375psi 375 psi For -ais 3 tan A rectangular plate of dimensions 3. in 5. in is formed b welding two triangular plates (see figure). The plate is subjected to a tensile stress of 6psi in the long direction and a compressive stress of 5psi in the short direction. 5 psi 5 psi 5psi -375psi 375psi Θ 3.96 o Stresses acting on the weld w 5psi w 375psi 3.96 o w -5psi and w 375psi

14 Principal Stresses and Maimum Shear Stresses The sum of the normal stresses acting on perpendicular faces of plane stress elements is constant and independent of the angle. X X Y sin X Y As we change the angle there will be maimum and minimum normal and shear stresses that are needed for design purposes. X XY sin Y X Y XY Y sin sin The maimum and minimum normal stresses are known as the principal stresses. These stresses are found b taking the derivative of with respect to and setting equal to zero. sin δ ( )sin δ tan P

15 The subscript p indicates that the angle p defines the orientation of the principal planes. The angle p has two values that differ b 9 o. The are known as the principal angles. For one of these angles is a maimum principal stress and for the other a minimum. The principal stresses occur in mutuall perpendicular planes. ( ) tan P ) ( sin R R P P P for the maimum stress sin ( ) ( ) ( ) ( ) But R R R R R R

16 Principal Stresses ( ) R ( ) R Average Average The plus sign gives the algebraicall larger principal stress and the minus sign the algebraicall smaller principal stress. This are the in-plane principal stresses. The third stress is zero in plane stress conditions Maimum Shear Stresses The location of the angle for the maimum shear stress is obtained b taking the derivative of with respect to and setting it equal to zero. sin δ δ tan ( ) S ( ) sin

17 R Therefore, s - p -9 o The planes for maimum shear stress occurs at 45 o to the principal planes. The plane of the maimum positive shear stress ma is defined b the angle S for which the following equations appl: ( ) s sin s and s P 45 R or s p /- 45 o tan sin S S The corresponding maimum shear is given b the equation S ( ) P sin P sin tan P cot o ( 9 P ) o ( 9 ) P P sin MAX o ( P 9 ) o ( 9 ) P ( ) R Another epression for the maimum shear stress MAX ( ) The normal stresses associated with the maimum shear stress are equal to AVR ( )

18 General equation Consider quation () quation () quations of a Circle ( ) sin AVR sin AVR sin sin ( ) sin quation () quation ()

19 ( ) ( ) sin sin AVR ( ) ( ) ( ) ( ) ( ) sin sin sin sin sin sin R SUM ( ) ( ) R AVR

20 P C ( ) R The radius of the Mohr circle is the magnitude R. Mohr Circle ( ) R The center of the Mohr circle is the magnitude ( ) AVR ( ) ( ) ( ) AVR State of Stresses

21 Alternative sign conversion for shear stresses: (a)clockwise shear stress, (b)counterclockwise shear stress, and (c) aes for Mohr s circle. Note that clockwise shear stresses are plotted upward and counterclockwise shear stresses are plotted downward. Forms of Mohr s Circle a) We can plot the normal stress positive to the right and the shear stress positive downwards, i.e. the angle will be positive when counterclockwise or b) We can plot the normal stress positive to the right and the shear stress positive upwards, i.e. the angle will be positive when clockwise. Both forms are mathematicall correct. We use (a)

22 Two forms of Mohr s circle: is positive downward and the angle is positive counterclockwise, and is positive upward and the angle is positive clockwise. (Note: The first form is used here)

23 Construction of Mohr s circle for plane stress.

24 At a point on the surface of a pressurized clinder, the material is subjected to biaial stresses 9MPa and MPa as shown in the element below. Using the Mohr circle, determine the stresses acting on an element inclined at an angle 3 o (Sketch a properl oriented element). ( 9MPa, MPa and MPa) Because the shear stress is zero, these are the principal stresses. Construction of the Mohr s circle The center of the circle is ( ) ( 9 ) Average 55 The radius of the circle is R MPa 9 ( ) ( ) 35MPa

25 Stresses on an element inclined at 3 o B inspection of the circle, the coordinates of point D are Average R Sin6 Average R Cos6 o 35 R Cos6 o ( Sin6 ) Cos6 3.3MPa Cos6 7.5MPa 37.5MPa

26 An element in plane stress at the surface of a large machine is subjected to stresses 5psi, 5psi and 4psi. Using the Mohr s circle determine the following: a) The stresses acting on an element inclined at an angle 4 o b) The principal stresses and c) The maimum shear stresses. Construction of Mohr s circle: Center of the circle (Point C): ( ) ( 5 5) Average psi Radius of the circle: R 5 5 ( ) ( 4) 643psi Point A, representing the stresses on the face of the element ( o ) has the coordinates 5psi and 4psi Point B, representing the stresses on the face of the element ( 9 o ) has the coordinates 5psi and - 4psi The circle is now drawn through points A and B with center C and radius R

27 Stresses on an element inclined at 4 o These are given b the coordinates of point D which is at an angle 8 o from point A B inspection the angle ACP for the principal stresses (point P ) is : 4 o tan( ACP ) ACP Then, the angle P CD is 8 o o 4.34 o Knowing this angle, we can calculate the coordinates of point D (b inspection) Average R Sin4.34 Average R Cos4.34 o o 643 R Cos4.34 o 643 Cos4.34 o ( Sin4.34 ) 43 psi 643 Cos4.34 o o 48 psi 59 psi

28 And of course, the sum of the normal stresses is 48psi 59psi 5psi 5psi Principal Stresses The principal stresses are represented b points P and P on Mohr s circle. The angle it was found to be o or 9.3 o Average Average R psi R psi

29 Maimum Shear Stresses These are represented b point S and S in Mohr s circle. Algebraicall the maimum shear stress is given b the radius of the circle. The angle ACS from point A to point S is S 5.34 o. This angle is negative because is measured clockwise on the circle. Then the corresponding S value is 5.7 o.

30 3-D stress state Psi Transform to Psi In matri notation the transformation is known as the igen-values. The principal stresses are the new-aes coordinate sstem. The angles between the old-aes and the new-aes are known as the igen-vectors. principal stress Cosine of angle between X and the principal stress Cosine of angle between Y and the principal stress Cosine of angle between Z and the principal stress

31 At a point on the surface of a generator shaft the stresses are -5MPa, MPa and - 4MPa as shown in the figure. Using Mohr s circle determine the following: (a)stresses acting on an element inclined at an angle 45 o, (b)the principal stresses and (c)the maimum shear stresses Construction of Mohr s circle Center of the circle (Point C): Radius of the circle:. ( ) ( 5) ( ) Average R MPa Point A, representing the stresses on the face of the element ( o ) has the coordinates -5MPa and - 4MPa Point B, representing the stresses on the face of the element ( 9 o ) has the coordinates MPa and 4MPa The circle is now drawn through points A and B with center C and radius R. ( ) ( 5) ( ) ( 4) 5MPa

32 Stresses on an element inclined at 45 o These stresses are given b the coordinates of point D ( 9 o of point A). To calculate its magnitude we need to determine the angles ACP and PCD. tan ACP 4/34/3 ACP 53.3 o P CD 9 o 53.3 o o Then, the coordinates of point D are Average R Sin36.87 Average R Cos36.87 o 5 R Cos36.87 ( ) o ( Sin36.87 ) 5 Cos MPa 6MPa o ( ) 5 Cos36.87 MPa And of course, the sum of the normal stresses is -5MPaMPa -6MPa MPa o o o

33 Principal Stresses The are represented b points P and P on Mohr s circle. R 5 3MPa Average Average R 5 7MPa The angle ACP is P 8 o 53.3 o 33.3 o or P 6.6 o The angle ACP is P 53.3 o or P 6.6 o Maimum Shear Stresses These are represented b point S and S in Mohr s circle. The angle ACS is S 9 o 53.3 o 43.3 o or 7.6 o. The magnitude of the maimum shear stress is 5MPa and the normal stresses corresponding to point S is the average stress -MPa.

34 5 4 4 MPa 3-D stress state Transform to 3 7 MPa In matri notation the transformation is known as the igen-values. The principal stresses are the new-aes coordinate sstem. The angles between the old-aes and the new-aes are known as the igen-vectors. principal stress Cosine of angle between X and the principal stress Cosine of angle between Y and the principal stress Cosine of angle between Z and the principal stress

35 Hooke s Law for Plane Stress The stress transformations equations were derived solel from equilibrium conditions and the are material independent. Here the material properties will be considered (strain) taking into account the following: a)the material is uniform throughout the bod (homogeneous) b)the material has the same properties in all directions (isotropic) c)the material follows Hooke s law (linearl elastic material) Hooke s law: Linear relationship between stress and strain For uniaial stress: ( modulus of elasticit or Young s modulus) Poisson s ratio: lateral strain υ aial strain transverse longitudinal For pure shear : (G Shear modulus of elasticit) G

36 lement of material in plane stress ( z ). Consider the normal strains,, z in plane stress. All are shown with positive elongation. The strains can be epressed in terms of the stresses b superimposing the effect of the individual stresses. For instance the strain in the direction: a)due to the stress is equal to /. b)due to the stress is equal to ν /. The resulting strain is: ν ν z ν ν

37 The shear stress causes a distortion of the element such that each z face becomes a rhombus. XY G The normal stresses and have no effect on the shear strain ν XY or rearranging the equations: ν ν ν ν XY G XY ( ) ( ) Y ( ) ( ) XY ν ν ν ν z ν These equations are known collectivel as Hooke s Law for plane stress These equations contain three material constants (, G and ν) but onl two are independent because of the relationship: G ( ν ) XY G

38 Special cases of Hooke s law ( z ) z υ Uniaial stress : G z Pure Shear : Biaial stress : - - υ υ υ υ z When a solid object undergoes strains, both its dimensions and its volume will change. Consider an object of dimensions a, b, c. The original volume is V o abc and its final volume is V (a a ) (b b ) (c c z ) V abc ( ) ( ) ( z ) Volume Change

39 Upon epanding the terms: For small strains: V V o ( z z z z ) V V o ( z ) The volume change is ΔV V V o V o ( z ) The unit volume change e, also known as dilatation is defines as: e ΔV / V o z Positive strains are considered for elongations and negative strains for shortening, i.e. positive values of e for an increase in volume. e ΔV V ν ν ( ) ( υ ) z ν ν For uniaial stress ( υ ) e We can notice that the maimum possible value of Poisson s ratio is.5, because a larger value means that the volume decreases when the material is in tension (contrar to phsical behavior).

40 STRAIN NRGY DNSITY IN PLAN STRSS The strain energ densit u is the strain energ stored in a unit volume of the material.because the normal and shear strains occur independentl, we can add the strain energ of these two elements to obtain the total energ. Work done Force distance bc a Work done in the - direction Work done in the - direction ( )( ) ( ) ( )( ac) ( b ) The sum of the energies due to normal stresses: U abc ( ) Then the strain energ densit (strain per unit volume) u ( ) Similarl, the strain energ densit associated with the shear strain: u B combining the strain energ densities for the normal and shear strains: u ( )

41 The strain energ densit in terms of stress alone: G u XY Y X Y X υ The strain energ densit in terms of strain alone: ( ) ( ) XY Y X Y X G u υ υ For the special case of uniaial stress: For the special case of pure shear: or u u υ or G u G u An element of the material subjected to normal stresses, and z acting in three mutuall perpendicular directions is said to be in a state of triaial stress. Since there is no shear in, or z faces then the stresses, and z are the principal stresses in the material. TRIAXIAL STRSS

42 If an inclined plane parallel to the z-ais is cut through the element, the onl stress of the inclined face are the normal stress and the shear stress, both of which act parallel to the plane. Because these stresses are independent of the z, we can use the transformation equations of plane stress, as well as the Mohr s circle for plane stress, when determining the stresses and in triaial stress. Maimum Shear Stress The same general conclusion hold for normal and shear stresses acting on inclined planes cut through the element parallel to the and aes. For a material in triaial stress, the maimum shear stresses occur on elements oriented at angles of 45 o to the, and z aes. for the inclined plane // z-ais for the inclined plane // -ais for the inclined plane // -ais ( ) MAX ( ) MAX ( ) MAX Z X Y ± ± ± X Y X The absolute maimum of the shear stress is the numericall largest of the above. Y Z Z

43 The stresses acting on elements oriented at various angles to the, and z aes can be visualized with the aid of the Mohr s circle. In this case > > z

44 Hooke s Law for Triaial Stress If Hooke s law is obeed, it is possible to obtain the relationship between normal stresses and normal strains using the same procedure as for plane stress. z z z z ν ν ν ν ν ν Or ( )( ) ( ) [ ] ( )( ) ( ) [ ] ( )( ) ( ) [ ] z z z z υ υ υ υ υ υ υ υ υ υ υ υ υ υ υ The are known as the Hooke s law for triaial stress. Unit Volume Change The unit volume change (or dilatation) for an element in triaial stress is obtained in the same manner as for plane stress. ( ) ( ) Z Y X O Z Y X O V V e V V e υ Δ Δ then appl, laws s Hooke' If

45 Strain nerg Densit The strain energ densit for an element in triaial stress is obtained b the same method used for plane stress. ( ) ( ) z z z z z u u υ - : stresses In terms of ( )( ) ( )( ) ( ) [ ] z z z u υ υ υ υ In terms of the strains: Spherical Stress A special case of triaial stress, called spherical stress, occurs whenever all three normal stresses are equal: z The Mohr s circle is reduced to a single point. An plane cut through the element will be free of shear stress and will be subjected to the same normal stress so and it is a principal plane. The normal strains in spherical stress are also the same in all directions, provided the material is isotropic and if Hooke s law applies: ( ) ( ) O O O υ υ 3 3 e change volume The O

46 K bulk or volume modulus of elasticit K K 3 ( υ) e lement in spherical stress. If ν /3 then K If ν then K /3 If ν / then K infinite (rigid material having no change in volume) These formulas also appl to and element in uniform compression, for eample rock deep within the earth or material submerged in water (hdrostatic stress).

47 PLAIN STRAIN Strains are measured b strain gages. A material is said to be in a state of plain strain if the onl deformations are those in the plane, i.e. it has onl three strain components, and. Plain stress is analogous to plane stress, but under ordinar conditions the do not occur simultaneousl ception when - and when ν Strain components,, and in the plane (plane strain).

48 Comparison of plane stress and plane strain.

49 APPLICATION OF TH TRANSFORMATION QUATIONS The transformation equations for plane stress are valid even when z is not zero, because z does not enter the equations of equilibrium.therefore, the transformations equations for plane stress can also be used for stresses in plane strain. Similarl, the strain transformation equations that will be derived for the case of plain strain in the plane are valid even when z is not zero, because the strain z does not affect the geometric relationship used for the derivation.therefore, the transformations equations for plane strain can also be used for strains in plane stress. Transformation quations for Plain Strain We will assume that the strain, and associated with the plane are known. We need to determine the normal and shear strains ( and ) associated with the ais. can be obtained from the equation of b substituting 9 for.

50 For an element of size d, d In the direction: the strain produces an elongation d. The diagonal increases in length b d. In the direction: the strain produces an elongation d. The diagonal increases in length b d sin. The shear strain in the plane produces a distortion of the element such that the angle at the lower left corner decreases b an amount equal to the shear strain. Consequentl, the upper face moves to the right b an amount d. This deformation results in an increase in the length of the diagonal equal to: d

51 The total increase Δδ of the diagonal is the sum of the preceding three epressions, thus: Δd d ( Cos ) d ( Sin ) d ( Cos ) But ( Cos ) ( Sin ) ( Cos ) d ds Δd ds Cos d ds d ds Sin d ds Cos Sin Shear Strain associated with aes. d ds Sin Cos This strain is equal to the decrease in angle between lines in the material that were initiall along the and aes. Oa and Ob were the lines initiall along the and ais respectivel. The deformation caused b the strains, and caused the Oa and Ob lines to rotate and angle α and β from the and ais respectivel. The shear strain is the decrease in angle between the two lines that originall were at right angles, therefore, αβ.

52 The angle α can be found from the deformations produced b the strains, and. The strains and produce a clockwise rotation, while the strain produces a counterclockwise rotation. Let us denote the angle of rotation produced b, and as α, α and α 3 respectivel. d α Sin ds d α Cos ds d α Sin ds d ds Cos d ds Sin ( ) SinCos α α α α3 Sin

53 The rotation of line Ob which initiall was at 9 o to the line Oa can be found b substituting 9 for in the epression for α. Because β is positive when clockwise. Thus: α β β β ( ) Sin( 9) Cos( 9) Sin ( 9) ( ) SinCos Cos ( ) [ ] Sin Cos Cos Sin Transformation quations for Plain Strain Using the following trigonometric identities: ( Cos ) ( ) ½ ( Cos ) Sin ½ (- Cos ) ( ) Sin ½ Sin sin In variant ( ) Average sin

54 PRINCIPAL STRAINS The angle for the principal strains is : The value for the principal strains are ( ) P tan ( ) ( ) Maimum Shear The maimum shear strains in the plane are associated with aes at 45 o to the directions of the principal strains ( ) or Ma Ma For isotropic materials, at a given point in an stressed bod, the principal strains and principal stresses occur in the same directions.

55 MOHR S CIRCL FOR PLAN STRAIN It is constructed in the same manner as the Mohr s circle for plane stress with the following similarities:

56 Strain Measurements An electrical-resistance strain gage is a device for measuring normal strains () on the surface of a stressed object. The gages are small (less than ½ inch) made of wires that are bonded to the surface of the object. ach gage that is stretched or shortened when the object is strained at the point, changes its electrical resistance. This change in resistance is converted into a measurement of strain. From three measurements it is possible to calculate the strains in an direction. A group of three gages arranged in a particular pattern is called a strain rosette. Because the rosette is mounted in the surface of the bod, where the material is in plane stress, we can use the transformation equations for plane strain to calculate the strains in various directions. 45 strain rosette, and element oriented at an angle to the aes.

57 General quations Other Strain Rosette

58 An element of material in plane strain undergoes the following strains: 34-6 ; -6 ; 8-6. Determine the following quantities: (a) the strains of an element oriented at an angle 3 o ; (b) the principal strains and (c) the maimum shear strains. (a) lement oriented at an angle 3 o ( 6 o ) ( ) ( ) ( ) Cos 6 sin 8 sin 8 Sin 6 Cos 6 Sin μ μ 36μ 9μ Average 5 μ

59 (b) Principal Strains and Angle of Rotation ( ) μ μ μ μ μ μ μ μ , ± tan P P μ μ μ

60 (c) In-Plane Maimum Shear Strain Ma Ma ( ) 37μ 8μ 9μ 34μ -μ 8μ 45μ Average 5 μ (d) Out-of-Plane Maimum Shear Strain ( ) 37μ μ μ 37 Ma 3

61 ( ) sin sin sin XY Y X Y X sin sin XY Y X X sin sin XY Y X Y Transformation quations [ ] [ ] Y X Y X XY Y X XY Y X Y X Y X T T [ ] ( ) sin sin sin sin sin sin sin T ( ) sin sin sin sin sin sin sin XY Y X Y X Y X

62 ( ) μ μ sin 3 sin 33 sin 33 sin sin sin 33 3 sin 3 Y X Y X XY Y X Y X Y X For Θ3 degrees

63 [] Tensor Strain zz z z z z _ [] μ μ _ Values igen zz z z z z ample

64 A 45 o strain rosette (rectangular rosette) consists of three electrical-resistance strain gages, arranged to measure strains in two perpendicular directions and also at a 45 o angle (as shown below). The rosette is bonded to the surface of the structure before it is loaded. Gages A, B and C measure the normal strains a, b and c in the directions of the lines Oa, Ob and Oc, respectivel. plain how to obtain the strains, and, associated with an element oriented at an angle to the aes. a sin sin Angles with respect to -ais: (a) is zero ; (b) is 45 degrees CCW ; (c) 9 degrees CCW a b b 45 9 sin sin sin 45 9 sin ( sin ) ( sin 9 9 ) a b b a c

65 The following results are obtained from a 6 strain gauge rosette: Strain in direction of strain gauge A 75μ; Strain in direction of SG B, 6 to A 35μ; Strain in direction of SG C, to A μ. Determine the principal strains and their directions. 75 μ b b b b 6 a sin 6 ( sin 6 6 ) (.5 ) (.75 ) (.433 ) sin (.5 ) (.75 ) (.433 ) ( sin ) 5μ 89μ

66 [] _ 779 _ Vectors igen Values igen zz z z z z μ μ ArcCos(angle).98 Angle.degress

67 [ ] (A) Using the transformation equations define the maimum and minimum principal strains, maimum shearing strain and principal angles given X 35μ ; Y 7μ and XY -5μ (B) Repeat using the Mohr s circle igen _ Values.984 igen _ Vectors.78 μ μ (d) Out-of-Plane Maimum Shear Strain ArcCos(angle).984 Angle.8degress (c) In-Plane Maimum Shear Strain 3595.μ 64.8μ Ma ( ) 3595.μ μ μ Ma 3 ( ) μ

68 The state of stress at a point in a structural member is determined to be as shown. Knowing that for this material GPa and ν.3, use the Mohr s circle to determine: () the principal stresses; () the in-plane maimum shear stress; (3) the absolute maimum shear stress; (4) principal angles; (5) the strains and the principal strains; (6) the maimum shear strain; (7) the principal angles. 4MPa.MPa 56MPa 56. [ ] Stress _ Tensor. 4 MPa R. Principal Stresses, ± Average ( 4) 56 ( 4) Average (.) MPa MPa MPa igen _ Values ( ) ± R 3.8MPa MPa. 3.MPa 58.8MPa MPa 58.8

69 . In-Plane Maimum Shear Stress R 3. MPa 3. The Absolute Maimum Shear Stress (Out of plane) Ma 8 ( 58.8) 3 Ma 9. 4MPa igen _Vectors tan ( ) 4. Angle between the -ais and the Principal Stresses ( In _ Plane) tan P. P deg. ( P ) ( ) 56 ( 4) ( ) For ArcCos(angle). Angle 9degress For -.MPa ArcCos(angle).45 Angle 76degress For MPa ArcCos(angle).97 Angle 4degress

70 5. Strains and Principal Strains z z z z ν ν ν ν ν ν G XY XY ( ) ( ) ( ) ( ) ( ) ( ) ( ) μ μ μ μ μ μ z μ G ( ) GPa G ) ( ν [] μ μ _ Values igen zz z z z z ( ) R Average ± ±,, ( ) μ μ R Average μ μ

71 6. Maimum Shear Strain and Absolute Maimum Shear Strain Ma Ma Ma Ma ( in _ plane) 94.8μ R ( out _ of _ plane) 364μ ( 64) 8μ 47.4μ 7. In-plane and Out-of-plane angles tan P 39 ( ) ( ) igen _Vectors For μ ArcCos(angle). Angle 9degress For 3.8μ ArcCos(angle).45 Angle 76degress tan ( ) P P deg For 3-64μ ArcCos(angle).97 Angle 4degress