Mechanical Metallurgy

Transcript

1 Mechanical Metallurg INME 606 Pabl G. Caceres Valencia Prfessr in Materials Science, B.Sc., Ph.D. (UK)

2 GENERL INFORMTION Curse Number INME 606 Curse Title Mechanical Metallurg Credit Hurs 3 (Lecture: 3hurs) Instructr Dr. Pabl G. Caceres Valencia Office Luchetti Building L Phne Etensin 358 Office Hurs M and Wed :00pm t 5:00pm e mail pcaceres@uprm.edu pcaceres@me.uprm.edu Web site

3 ssessment The curse will be assessed in the fllwing manner: Partial Eam 30% Final Eam 30% Quizzes (3)* 33% ttendance and Class Participatin 7% (*) ttal f three quizzes will be perfrmed. (**) Class ttendance (after the secnd absence pint will be deducted fr each nnauthrized absence). The participatin in class will be taken int accunt. ttendance ttendance and participatin in the lectures are mandatr and will be cnsidered in the grading. Students shuld bring calculatrs, rulers, pen and pencils t be used during the lectures. Students are epected t keep up with the assigned reading and be prepared fr the pp quizzes r t answer questins n these readings during lecture.

4 Tebks G.E. Dieter; Mechanical Metallurg; Mc Graw Hill M.. Meers and K.K. Chawla; Mechanical Metallurg: Principles and pplicatins; Prentice Hall I will als pst m lecture ntes in the web: TENTTIVES DTES Jan/9 Basic Principles Jan/8 Feb/0 Basic Elasticit Feb/8 Dislcatin Ther st Eam March/0 4 Strengthening Mechanisms pril/ 4 Fracture pril/ 5 Mechanical Prperties Jan/4 8 Stress Strain Feb/04 08 Single Crstals Feb/5 9 Strengthening Mechanisms Mar/7 N Class Hl Week pril/7 Mechanical Prperties Eam pr/8 Ma0 Mechanical Prperties Jan/ 5 Basic Elasticit Feb/ 5 Dislcatin Ther March/3 7 Strengthening Mechanisms Mar/4 8 Fracture pril /4 8 Mechanical Prperties

5 Cntent Stress and Strain Relatinships fr Elastic Behavir Elements f the Ther f Elasticit Plastic Defrmatin f Single Crstals Dislcatin Ther Strengthening Mechanisms Fracture Mechanical Prperties

6 The Cncept f Stress Uniaial tensile stress: frce F is applied perpendicular t the area (). Befre the applicatin f the frce, the crss sectin area was O Engineering stress r nminal stress: Frce divided b the riginal area. True stress: Frce divided b the instantaneus area T Engineering Strain r Nminal Strain: Change f length divided b the riginal length l l0 Δl ε l0 l0 True Strain: The rate f instantaneus increase in the instantaneus gauge length. d i ε T ln F 0 F

7 Δ Δ T i T d ln ln ln ε ε Relatinship between engineering and true stress and strain i i i T F F F * * i i ) ( ε Δ Δ i i l l ) ( ε T F We will assume that the vlume remains cnstant. ) ( ε T ) ln( ε ε T

8 Primar Tpes f Lading (a) Tensin (b) Cmpressin (c) Shear (d) Trsin (e) Flein

9 Hke s Law When strains are small, mst f materials are linear elastic. Yung s mdulus Nrmal: Δl F l E Εε Springs: the spring rate k F Δl E l E ε

10 Trsin Lading resulting frm the twist f a shaft. l r z r Z δ δ γ, l r G z r G G z z δ δ γ,, G Shear Mdulus f Elasticit Shear strain Twist Mment r Trque z z r l G r T δ δ,, r J δ rea Plar Mment f Inertia J G l T l J G T r J r T z, Thus: J r T Ma ngular spring rate: l G J T k a

11 Stress Cmpnents Nrmal Stresses,, z Shear Stresses,, z, z, z, z Frm equilibrium principles:, z, z z z Nrmal stress () : the subscript identifies the face n which the stress acts. Tensin is psitive and cmpressin is negative. Shear stress () : it has tw subscripts. The first subscript dentes the face n which the stress acts. The secnd subscript dentes the directin n that face. shear stress is psitive if it acts n a psitive face and psitive directin r if it acts in a negative face and negative directin.

12 Sign Cnventins fr Shear Stress and Strain The Shear Stress will be cnsidered psitive when a pair f shear stress acting n ppsite sides f the element prduce a cunterclckwise (ccw) trque (cuple). shear strain in an element is psitive when the angle between tw psitive faces (r tw negative faces) is reduced, and is negative if the angle is increased.

13 Fr static equilibrium, z z, z z resulting in Si independent scalar quantities. These si scalars can be arranged in a 33 matri, giving us a stress tensr. ij z z z z z The sign cnventin fr the stress elements is that a psitive frce n a psitive face r a negative frce n a negative face is psitive. ll thers are negative. The stress state is a secnd rder tensr since it is a quantit assciated with tw directins (tw subscripts directin f the surface nrmal and directin f the stress).

14 F F 3 F Cube with a Face area ij F F F F prpert f a smmetric tensr is that there eists an rthgnal set f aes, and 3 (called principal aes) with respect t which the tensr elements are all zer ecept fr thse in the diagnal. z z z z z ij 3 ' ' ij

15 Plane Stress r Biaial Stress : When the material is in plane stress in the plane, nl the and faces f the element are subject t stresses, and all the stresses act parallel t the and aes Stresses n Inclined Sectins Knwing the nrmal and shear stresses acting in the element dented b the ais, we will calculate the nrmal and shear stresses acting in the element dented b the ais.

16 O /Cs Equilibrium f frces: cting in O Sin/Cs X O X cs cs O XY sin O Sin Y O Sin Cs YX cs O Sin Cs

17 Eliminating, sec /cs and X X cs Y sin XY sin cs Y X Y XY sin cs sin cs cting in sec sin cs tancs tansin Eliminating, sec /cs and sin cs sin cs cs sin ( )

18 Transfrmatin Equatins fr Plane Stress Using the fllwing trignmetric identities: Cs ½ ( cs ) Sin ½ (- cs ) Sin cs ½ sin cs sin cs sin These equatins are knwn as the transfrmatin equatins fr plane stress.

19 Case : Uniaial stress Special Cases 0 0, Sin Cs Case : Pure Shear 0, Cs Sin cs sin sin cs

20 Case 3: Biaial stress, 0 Cs Sin

21 Eample: n element in plane stress is subjected t stresses 6000psi, 6000psi, and 4000psi (as shwn in figure belw). Determine the stresses acting n an element inclined at an angle 45 (cunterclckwise - ccw). Slutin: We will use the fllwing transfrmatin equatins: sin cs cs sin

22 Numerical substitutin ½( ) ½ ( ) 000psi ½( - ) ½ ( ) 5000psi 4000psi sin sin 90 cs cs 90 0 Then 000psi 5000psi (0) 4000psi () 5000psi - (5000psi) () (4000psi) (0) psi then psi 7,403psi 4,597 psi Ma 6, 403 psi

23 Eample: plane stress cnditin eists at a pint n the surface f a laded structure such as shwn belw. Determine the stresses acting n an element that is riented at a clckwise (cw) angle f 5 with respect t the riginal element. Slutin: We will use the fllwing transfrmatin equatins: sin cs cs sin

24 Numerical substitutin ½( ) ½ (- 46 ) - 7MPa ½( - ) ½ (- 46 ) - 9MPa - 9MPa sin sin (- 30 ) cs cs (- 30 ) then - 7MPa ( - 9MPa)(0.8660) (-9MPa)(- 0.5) - 3.6MPa - (- 9MPa) (- 0.5) (- 9MPa) (0.8660) - 3.0MPa then - 46MPa MPa (- 3.6MPa) -.4MPa

25 Eample : rectangular plate f dimensins 3.0 in 5.0 in is frmed b welding tw triangular plates (see figure). The plate is subjected t a tensile stress f 600psi in the lng directin and a cmpressive stress f 50psi in the shrt directin. Determine the nrmal stress w acting perpendicular t the line r the weld and the shear stress w acting parallel t the weld. (ssume w is psitive when it acts in tensin and w is psitive when it acts cunterclckwise against the weld).

26 Slutin Biaial stress weld jint 600psi -50psi 0 Frm the figure tan 3 / 5 arctan(3/5) arctan(0.6) We will use the fllwing cs transfrmatin equatins: sin cs Numerical substitutin ½( ) 75psi ½ ( - ) 45psi 0psi sin sin 6.9 cs cs 6.9 Then 375psi - 375psi then 600 (- 50) 375-5psi 5psi -375psi 375psi sin Θ 30.96

27 Stresses acting n the weld w 5psi w 375psi w -5psi and w 375psi

28 Principal Stresses and Maimum Shear Stresses The sum f the nrmal stresses acting n perpendicular faces f plane stress elements is cnstant and independent f the angle. cs sin X Y X Y s we change the angle there will be maimum and minimum nrmal and shear stresses that are needed fr design purpses. The maimum and minimum nrmal stresses are knwn as the principal stresses. These stresses are fund b taking the derivative f with respect t and setting equal t zer. δ ( )sin cs 0 δ tan P

29 The subscript p indicates that the angle p defines the rientatin f the principal planes. The angle p has tw values that differ b 90. The are knwn as the principal angles. Fr ne f these angles is a maimum principal stress and fr the ther a minimum. The principal stresses ccur in mutuall perpendicular planes. tan P sin P R cs P ( ) R cs fr the maimum stress P sin

30 ( ) ( ) ( ) ( ) But R R R R R R Principal stresses: The plus sign gives the algebraicall larger principal stress and the minus sign the algebraicall smaller principal stress. ( ) ( )

31 Maimum Shear Stress The lcatin f the angle fr the maimum shear stress is btained b taking the derivative f with respect t and setting it equal t zer. sin cs δ δ tan tan S S sin S cs ( )cs S ( ) cs P sin cs P R sin cs tan sin 0 P ct ( 90 P ) ( 90 ) P Therefre, s - p -90 r s p /- 45 The planes fr maimum shear stress ccurs at 45 t the principal planes. The plane f the maimum psitive shear stress ma is defined b the angle S fr which the fllwing equatins appl: P sin cs ( ) R ( P 90 ) ( 90 ) 0 s sin s and s P 45 P

32 The crrespnding maimum shear is given b the equatin nther epressin fr the maimum shear stress The nrmal stresses assciated with the maimum shear stress are equal t Equatins f a Circle General equatin Cnsider Equatin () ( ) VER VER MX MX VER ( ) ( ) ( ) cs sin cs sin cs sin

33 Equatin () ( ) cs sin cs sin Equatin () Equatin () ( ) ( ) cs sin sin cs VER ( ) ( ) ( ) ( ) ( ) cs sin cs sin cs sin cs sin sin cs sin cs R SUM ( ) ( ) R VER

34 P C ( ) R The radius f the Mhr circle is the magnitude R. Mhr Circle ( ) R The center f the Mhr circle is the magnitude ( ) VER ( ) ( ) ( ) VER

35 lternative sign cnversin fr shear stresses: (a) clckwise shear stress, (b) cunterclckwise shear stress, and (c) aes fr Mhr s circle. Frms f Mhr s Circle Nte that clckwise shear stresses are pltted upward and cunterclckwise shear stresses are pltted dwnward. a) We can plt the nrmal stress psitive t the right and the shear stress psitive dwnwards, i.e. the angle will be psitive when cunterclckwise r b) We can plt the nrmal stress psitive t the right and the shear stress psitive upwards, i.e. the angle will be psitive when clckwise. Bth frms are mathematicall crrect. We use (a)

36 Tw frms f Mhr s circle: (a) is psitive dwnward and the angle is psitive cunterclckwise, and (b) is psitive upward and the angle is psitive clckwise. (Nte: The first frm is used here)

37 Cnstructin f Mhr s circle fr plane stress.

38 Eample: t a pint n the surface f a pressurized clinder, the material is subjected t biaial stresses 90MPa and 0MPa as shwn in the element belw. Using the Mhr circle, determine the stresses acting n an element inclined at an angle 30 (Sketch a prperl riented element). Slutin ( 90MPa, 0MPa and 0MPa) Because the shear stress is zer, these are the principal stresses. Cnstructin f the Mhr s circle The center f the circle is aver ½ ( ) ½ (90 0) 55MPa The radius f the circle is R SQR[(( )/) ( ) ] R (90 0)/ 35MPa.

39 Stresses n an element inclined at 30 B inspectin f the circle, the crdinates f pint D are aver R cs 60 55MPa 35MPa (Cs 60 ) 7.5MPa - R sin 60-35MPa (Sin 60 ) MPa In a similar manner we can find the stresses represented b pint D, which crrespnd t an angle 0 ( 40 ) aver -R cs60 55MPa - 35MPa (Cs 60 ) 37.5MPa R sin 60 35MPa (Sin 60 ) 30.3MPa

40 Eample: n element in plane stress at the surface f a large machine is subjected t stresses 5000psi, 5000psi and 4000psi, as shwn in the figure. Using the Mhr s circle determine the fllwing: a) The stresses acting n an element inclined at an angle 40 b) The principal stresses and c) The maimum shear stresses. Slutin Cnstructin f Mhr s circle: Center f the circle (Pint C): aver ½ ( ) ½ ( ) 0000psi Radius f the circle: R SQR[(( )/) ( ) ] R SQR[(( )/) (4000) ] 6403psi. Pint, representing the stresses n the face f the element ( 0 ) has the crdinates 5000psi and 4000psi Pint B, representing the stresses n the face f the element ( 90 ) has the crdinates 5000psi and psi The circle is nw drawn thrugh pints and B with center C and radius R

41 Stresses n an element inclined at 40 These are given b the crdinates f pint D which is at an angle 80 frm pint. B inspectin the angle CP fr the principal stresses (pint P ) is tan CP 4000/ r Then, the angle P CD is

42 Knwing this angle, we can calculate the crdinates f pint D (b inspectin) aver R cs psi 6403psi (Cs 4.34 ) 480psi - R sin psi (Sin 4.34 ) - 430psi In an analgus manner, we can find the stresses represented b pint D, which crrespnd t a plane inclined at an angle aver - R cs psi psi (Cs 4.34 ) 590psi R sin psi (Sin 4.34 ) 430psi nd f curse, the sum f the nrmal stresses is 480psi 590psi 5000psi 5000psi

43 Principal Stresses The principal stresses are represented b pints P and P n Mhr s circle. 0000psi 6400psi 6400psi 0000psi 6400psi 3600psi The angle it was fund t be r 9.3 Maimum Shear Stresses These are represented b pint S and S in Mhr s circle. The angle CS frm pint t pint S is S This angle is negative because is measured clckwise n the circle. Then the crrespnding S value is 5.7.

44 Eample:t a pint n the surface f a generatr shaft the stresses are -50MPa, 0MPa and - 40MPa as shwn in the figure. Using Mhr s circle determine the fllwing: (a) Stresses acting n an element inclined at an angle 45, (b) The principal stresses and (c) The maimum shear stresses Slutin Cnstructin f Mhr s circle: Center f the circle (Pint C): aver ½ ( ) ½ ((-50) 0) - 0MPa Radius f the circle: R SQR[(( )/) ( ) ] R SQR[((- 50 0)/) (- 40) ] 50MPa. Pint, representing the stresses n the face f the element ( 0 ) has the crdinates -50MPa and - 40MPa Pint B, representing the stresses n the face f the element ( 90 ) has the crdinates 0MPa and 40MPa The circle is nw drawn thrugh pints and B with center C and radius R.

45 Stresses n an element inclined at 45 These stresses are given b the crdinates f pint D ( 90 f pint ). T calculate its magnitude we need t determine the angles CP and PCD. tan CP 40/304/3 CP 53.3 P CD 90 CP Then, the crdinates f pint D are aver R cs MPa 50MPa (Cs ) - 60MPa R sin MPa (Sin ) 30MPa In an analgus manner, the stresses represented b pint D, which crrespnd t a plane inclined at an angle 35 r 70-0MPa 50MPa (Cs ) 0MPa -30MPa nd f curse, the sum f the nrmal stresses is -50MPa0MPa -60MPa 0MPa

46 Principal Stresses The are represented b pints P and P n Mhr s circle. - 0MPa 50MPa 30MPa -0MPa 50MPa - 70MPa The angle CP is P r P 6.6 The angle CP is P 53.3 r P 6.6 Maimum Shear Stresses These are represented b pint S and S in Mhr s circle. The angle CS is S r 7.6. The magnitude f the maimum shear stress is 50MPa and the nrmal stresses crrespnding t pint S is -0MPa.