σ zz Stresses in 3-D σ yy τ yx σ xx z τ zy τ zx τ yz τ xz τ xy

Transcript

1 zz Stresses in -D z z z z z z zz ij z z z z z

2 Matrices Review Matri Multiplication : When the number of columns of the first matri is the same as the number of rows in the second matri then matri multiplication can be performed. Here is an eample of matri multiplication for two matrices Here is an eample of matrices multiplication for a matri When A has dimensions mn, B has dimensions np. Then the product of A and B is the matri C, which has dimensions mp.

3 Transpose of Matrices : The transpose of a matri is found b echanging rows for columns i.e. Matri A (aij) and the transpose of A is: A T (aij) Where i is the row number and j is the column number. For eample, The transpose of a matri would be: In the case of a square matri (mn), the transpose can be used to check if a matri is smmetric. For a smmetric matri A A T

4 The Determinant of a Matri : Determinants pla an important role in finding the inverse of a matri and also in solving sstems of linear equations. Determinant of a matri Assuming A is an arbitrar matri A, where the elements are given b: Determinant of a matri The determinant of a matri is more difficult

5 Inverse Matri For a matri the matri inverse is Eample: Cos A Sin Sin Cos A A Cos Sin Cos Sin Sin Cos Sin Cos A T A Cos Sin For a matri

6 XY Y X X ( ) [T ] Coordinate Transformations in -D T ] [ ] [ T XY Y X Y

7 Theor of Matri Method for Stress Calculations in -D From equations of rotational transformation of ais, we obtain the following: ' ' ' ' T inversel ' ' Hence T Y' T Y T AC Area A AB A BC A A X' X B C '

8 Ug and force equilibrium equation, we obtain epressions for stress transformations as follows: ' {} Σ{ F} F F F {} F F F AB BC AC {} A A A ' A ' {} A A ' ' ' ' Canceling area A out and pre-multipling b transformation T (where T T T I the identit matri. The order of the matri multiplication does matter in the final outcome., we have

9 ' ' ' I {} ' ' ' A A

10 For the forces in the X ais we will use the same procedure. Y Y' X X' B C D BD Area A BC A CD A ' ' ' { } { } {} {} {} Σ ' ' ' ' ' ' BD BC CD A A A A A A F F F F F F F ' ' '

11 ' ' ' ' ' ' Combining the above epressions T T T ' ' ' ' ' ' or ' ' ' ' ' '

12 State of Stresses in Three Dimensions K z The general three dimensional state of stress consists of three unequal principal stresses acting at a point (triaial state of stresses). z z L The plane JKL is assumed to be a principal plane and is the principal stress acting normal to the plane. z z - J - z Letα, β and γ are the angles between the vector and the, and z ais respectivel and k α l β m γ

13 Under equilibrium conditions ( ) z k k z z k ( ) z l z z l l z m m ( ) m z z z z z zz z zz k l m k l m z k k k z l l z z l zz m m m As k, l and m are different than zero (non-trivial solution) k l m The determinant must be equal to zero Solution of the determinant results in a cubic equation in

14 The eigenvalues of the stress matri are the principal stresses. The eigenvectors of the stress matri are the principal directions. m l k > > z z z z z z z z z z z I I I I I I m l k zz z z z z The three roots are the three principal stresses,,. I, I, and I are known as stress invariants as the do not change in value when the aes are rotated to new positions.

15 I z I z z z z z z z I z z z z z I has been seen before for the two dimensional state of stress. It states the useful relationship that the sum of the normal stresses for an orientation in the coordinate sstem is equal to the sum of the normal stresses for an other orientation z z

16 I I I ( ) 6, I A I A ± α α ( ) ( ) A I I I I Cos I I A α Stress Invariants for Principal Stresses I I I,, ma ma The solution are the eigenvalues of the stress tensor

17 Eample: determine the principal stresses for the state of stress (in MPa). Solution: The solution are the eigenvalues of the stress tensor; Substituting: ( 4) ( 4) ( 8) 4 z 4 ( ( 8) ) ( ( ( ) ) (( 4) ( 4)) ) z 8 z z One solution -8MPa is a principal stress because z and z are zero, then the other two principal stresses are eas to find b solving the quadratic equation inside the square brackets for (4) ± 6 ( ) ± ( ) 4(4) 6MPa -6MPa z

18 4 4 8 I 8 I,6 I 6,8, A Cos 96.4Cos 96.4Cos Cos(α ) (.5).86 8 (.5 6) α.5 6. (.5 6)

19 I I I Eample : Determine the maimum principal stresses and the maimum shear stress for the following triaial stress state. Solution Stress _ Tensor [ ] z I I I z z z z zz MPa z z z z z z z z z z MPa 5 MPa -895 MPa

20 Solution to Eample MPa MPa Sigm a (MPa) MPa Stress (MPa) ma 65.MPa / ( MPa 5.8) 58.5MPa 5.8MPa

21 z, abs ma Normal Stress, (MPa) Shear (MPa)

22 Safet Factor? If the stress state was determined on a steel crankshaft, made of forged SAE45 steel with a ield strength of MPa, what is the factor of safet against ield? S Tresca Criterion: ma 58.5 MPa FS. 6 ma (58.5) Ma Principal Stress Criterion: ma 6.5 MPa Von Mises Criterion: 65.MPa 6.5MPa 5.8MPa e e.mpa e FS S ma [( ) ( ) ( ) ] [( ) ( ) (65. ( 5.8)) ] / / FS S e..9

23 Mohr s Circles for -D Analsis Mohr s circles can make visualization of the stress condition clearer to the designer. Note that the principal stress values are alwas ordered b convention so the is the largest value in the tensile direction and is the largest value in the compressive direction. Note also that there is one dominant peak shear stress in this diagram. Be forewarned the principal stresses and this peak shear stress are going to pla a strong role in determining the factor of safet in mechanical design.

24 A Mohr s circle can be generated for triaial stress states, but it is often unnecessar, as it is sufficient to know the values of the principal stresses. The principal stresses must be ordered from larger to smaller.

25 Compare -D and -D Mohr s Circle. If z is zero, does it have an effect in D?

26 Q P α α Consider then the plane will be an angle α from, in the direction of (anticlockwise). Point P Consider then the plane will be an angle α from, in the direction of (anticlockwise). Point Q The required sstem of stresses, fall within P and Q. Loci determined b the center in

27 β β Consider then the plane will be an angle β from, in the direction of (anticlockwise). Point R Consider then the plane will be an angle β from, in the direction of (clockwise). Point S The required sstem of stresses, fall within R and S. Loci determined b the center in

28 Eample: Use Mohr s Circle to obtain the principal stresses and maimum shear of a component subjected to the following stresses: z 9tension tension 5compression 4ccw ccw counterclockwise

29 Stress on ANY Inclined Plane (-D) S The stress on a plane (S) can be decomposed into its normal component (S n ) and its shear component (S s ). S S n s S s s n s z If α, β and γ are the angles between the vector S n and the, and z ais respectivel and k α l β m γ then

30 [ ] [ ] T zz z z z z z z z z z z T T zz z z z z It can be proved that

31 Mohr s Circles for -D Analsis There is no eas Mohr s circle graphical solution for problems of triaial stress state. Solution for maimum principal stresses and maimum shear stress is analtical. z Consider the, and z ais to coincide with the ais of the principal stresses, and. If α, β and γ are the angles between the normal to the plane and the, and z ais respectivel and k α l β m k l m γ We would like to find graphicall the normal stress and shear stress on the plane.

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33 Octahedral Plane and Stresses An octahedral plane is a plane that makes three identical angles with the principal planes. ( ) ( ) ( ) ( ) 6 9 I I n I n n n n n n n op op op op m l k m m l k l m l k k zz z z z z

34 Mean and Deviatoric Stresses When describing the materials behavior of metals, one concludes that in certain cases some stress components pla a more important role than other components. Plastic behavior of metals, is reported to be independent of the average (mean) normal stress. M M M op z M M I Mean stress matri Deviatoric stress M z z M M The shear components do not change

35 zz z z z zz z zz D Deviatoric Stress Matri Deviatoric stresses pla an important role in the theor of plasticit. The influence the ielding of ductile materials. The principal stresses obtained onl from the deviatoric matri is M M M D P,

36 Failure Theories Wh do mechanical components fail? Mechanical components fail because the applied stresses eceeds the material s strength (Too simple). What kind of stresses cause failure? Under an load combination, there is alwas a combination of normal and shearing stresses in the material.

37 What is the definition of Failure? Obviousl fracture but in some components ielding can also be considered as failure, if ielding distorts the material in such a wa that it no longer functions properl Which stress causes the material to fail? Usuall ductile materials are limited b their shear strengths. While brittle materials (ductilit < 5%) are limited b their tensile strengths. Stress at which point?

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39 Stress at which point?

40 Failure Theories Load tpe Uniaial Biaial Pure Shear Material Propert Ductile Brittle Application of Stress Static Dnamic Static Loading Maimum Normal Stress Modified Mohr Yield strength Maimum shear stress Distortion energ Dnamic Loading Goodman Gerber Soderberg

41 Static Failure Theories The idea behind the various classical failure theories is that whatever is responsible for failure in the standard tensile test will also be responsible for failure under all other conditions of static loading.

42 Characteristic Failure Stress Important Theories Ductile Material Yield Stress. Maimum Shear Stress. Maimum Octahedral Shear Stress Brittle Material Ultimate Stress. Maimum Normal Stress. Modified Mohr. Maimum Shear Stress Theor Ductile Materials Failure occurs when the maimum shear stress in the part eceeds the shear stress in a tensile test specimen (of the same material) at ield. Hence in a tensile test, ma S

43 For a general state of stresses S ma This leads to an heagonal failure envelop. A stress sstem in the interior of the envelop is considered SAFE The Maimum Shear Stress Theor for Ductile Materials is also known as the Tresca Theor. for design purposes, the failure relation can be modified to include a factor of safet (n): n S

44 Several cases can be analzed in plane stress problems: Case : In this case S S ma Yielding condition Case : S S ma

45 Distortion Energ Theor Based on the consideration of angular distortion of stressed elements. The theor states that failure occurs when the distortion strain energ in the material eceeds the distortion strain energ in a tensile test specimen (of the same material) at ield. Resilience Resilience is the capacit of a material to absorb energ when it is deformed elasticall and then, upon unloading, to have this energ recovered. Modulus of resilience U r If it is in a linear elastic region, U r U r dε ε ε E E

46 For general -D stresses: ( ) ( ) ( ) ν ε ε ε E u u Appling Hooke s Law There are two components in this energ a mean component and deviatoric component. z M M D M D M D,,, The energ due to the mean stress (it gives a volumetric change but not a distortion: ( ) ( ) ( ) [ ] ( ) 6 ν ν ν E E u E u M Mean M M M M M M M M M Mean

47 ( ) ν E u u u Mean D Compare the distortion energ of a tensile test with the distortion energ of the material. ( ) ν ν E u S E u D Tensile S S Plane Stress Von Mises effective stress : Defined as the uniaial tensile stress that creates the same distortion energ as an actual combination of applied stresses.

48 ( ) ( ) ( ) ( ) ( ) 6 z z z z z VM z z z z VM D This simplifies the approach ce we can use the following failure criterion VM VM S n S VM

49 Case of Pure Shear VM S Ma S.577S Brittle Materials Several theories have been developed to describe the failure of brittle materials, such as: Maimum Normal Stress Theor Coulomb-Mohr Theor Modified-Mohr Theor

50 -Sc Maimum Normal Stress Theor Failure occurs when one of the three principal > stresses reaches a permissible strength (TS). Failure is predicted to occur when S t and <-S c Where S t and S c are the tensile and compressive strength For a biaial state of stresses St -Sc St

51 Coulomb-Mohr Theor or Internal Friction Theor (IFT) This theor is a modification of the maimum normal stress theor in the which the failure envelope is constructed b connecting the opposite corners of quadrants I and III. The result is an heagonal failure envelop. Similar to the maimum shear stress theor but also accounts for the uneven material properties of brittle material

52 Mohr s Theor The theor predicts that a material will fail if a stress state is on the envelope that is tangent to the three Mohr s circles corresponding to: a. uni-aial ultimate stress in tension, b. uni-aial ultimate stress in compression, and c. pure shear.

53 Modified Mohr s Theor C T This theor is a modification of the Coulomb-Mohr theor and is the preferred theor for brittle materials.

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56 Eample: Given the material S Y,, v and find the safet factors for all the applicable criteria. a. Pure aluminum SY MPa MPa MPa MPa MPa MPa Ma MPa Is Al ductile or brittle? Ductile - Use either the Maimum Shear Stress Theor (MSST) or the Distortion Theor (DT) MSST Theor n S ( ).5 DT Theor VM 7.MPa n S VM MPa 7.MPa.7

57 b..%c Carbon Steel MSST Theor SY 65 Ksi 5Ksi 5Ksi Ksi n S In the plane XY the principal stresses are -.97Ksi and -8.Ksi with a maimum shear stress in the XY plane of 8.Ksi In an orientation Ksi.97Ksi Ma 9. Ksi 65 ( 8.).7 Ductile 8.Ksi DT Theor VM n S VM 65Ksi.7 8.MPa 8.Ksi

58 C. Gra Cast Iron Sut Ksi Suc Ksi 5Ksi Ksi Ksi Ksi Ksi. Ksi Ma 5 5Ksi -5 MNST Theor (tensile) MNST Theor (compression) Brittle Use Maimum Normal Stress Theor (MNST), Internal Friction Theor (IFT), Modified Mohr Theor (MMT) n Sut. n Suc 5.4

59 MMT n n ) ( ()() 5) ( _ 4 n S S S S n S S S quadrant ut uc ut uc ut uc ut th IFT n S S n uc ut 4 ut uc uc th S S S equation line quadrant C T

60 Eample The cantilever tube shown is to be made of 4 aluminum allo treated to obtain a specified minimum ield strength of 76MPa. We wish to select a stock size tube (according to the table below). Ug a design factor of n4. The bending load is F.75kN, the aial tension is P9.kN and the torsion is T7N.m. What is the realized factor of safet? Consider the critical area ( top surface).

61 S.76 GPa. GPa VM n 4 69 Maimum bending moment F P A Mc I z 9kN A Tr J mm.75kn I d 7 J 6d J d VM ( ) z For the dimensions of that tube

62 For 4 Inside diameter mm Thickness 4mm Outside diameter48mm Area.66cm I,87cm 4 J5.65cm 4 For 45 n S VM

63 Eample : A certain force F is applied at D near the end of the 5-in lever, which is similar to a socket wrench. The bar OABC is made of AISI 5 steel, forged and heat treated so that it has a minimum (ASTM) ield strength of 8kpsi. Find the force (F) required to initiate ielding. Assume that the lever DC will not ield and that there is no stress concentration at A. Solution: ) Find the critical section The critical sections will be either point A or Point O. As the moment of inertia varies with r 4 then point A in the in diameter is the weakest section.

64 ) Determine the stresses at the critical section ) Chose the failure criteria. The AISI 5 is a ductile material. Hence, we need to emplo the distortion-energ theor. F VM S VM M I d M πd 64 F 4in πd Tr J d T πd 6 F 5in π (in) z 4 46lbf z 94.5F F F

65 Appl the MSS theor. For a point undergoing plane stress with onl one non-zero normal stress and one shear stress, the two nonzero principal stresses ( A and B ) will have opposite signs (Case ). ( ) ( ) ( ) lbf F F F S S z z B A z B A ma ±

66 Eample : A round cantilever bar is subjected to torsion plus a transverse load at the free end. The bar is made of a ductile material having a ield strength of 5psi. The transverse force (P) is 5lb and the torque is lb-in applied to the free end. The bar is 5in long (L) and a safet factor of is assumed. Transverse shear can be neglected. Determine the minimum diameter to avoid ielding ug both MSS and DET criteria. Solution ) Determine the critical section The critical section occurs at the wall.

67 4 64 d PL d d PL I Mc π π 4 6 d T d d T J Tc π π ( ) ( ) ( ) ( ) ± ± ± ± ±,,, d T PL PL d d T d PL d PL π π π π π

68 645 d 645 d MAX 98.8 d 98.8 d MSS d. in MAX The stresses are in the wrong order.. Rearranged to 645 S n ( 98.8) 75 d d 5 5,.4 DET VM 695 VM d d.5in S n d 98.8 d 645 d 98.8 d

69 Eample 5: The factor of safet for a machine element depends on the particular point selected for the analsis. Based upon the DET theor, determine the safet factor for points A and B. This bar is made of AISI 6 cold-drawn steel (S 8MPa) and it is loaded b the forces F.55kN, P8.kN and TN.m Solution: d Fl 4 Point A Mc P P Fl P 4 I Area πd πd πd πd 64 4 ( )(.55 )(.) π (.) ( )( 8 ) MPa π (.)

70 n VM S VM Tr J 6T πd 6( ) (.) 9. π [ ] ( ) ( 9.) Point B ( )( 4P 4 8 ) n VM πd 6T πd (.) 6( ) A π (.) [ 5.47 (.4) ] V π MPa ( )(.55 ) 4 π 4 (.) MPa.MPa 45.MPa.4MPa