Lecture 5. Torsion. Module 1. Deformation Pattern in Pure Torsion In Circular Cylinder. IDeALab. Prof. Y.Y.KIM. Solid Mechanics

Transcription

1 Lectue 5. Tosion Module 1. Defomation Patten in Pue Tosion In Cicula Cylinde

2 Defomation Patten Shafts unde tosion ae eveywhee. Candall, An Intoduction to the Mechanics of solid, Mc Gaw-Hill,

3 Defomation Patten Defomation patten analysis # 1 in Pue Tosion of a Cicula Shaft M t M t M t Candall, An Intoduction to the Mechanics of solid, Mc Gaw-Hill, 1999 M t

4 Defomation Patten Defomation patten analysis # in Pue Tosion of a Cicula Shaft

5 Defomation Patten Findings 1 Coss sections otate as igid boundaies They emain plane (no waping) Radii emain staight Candall, An Intoduction to the Mechanics of solid, Mc Gaw-Hill,

6 Defomation Patten Pue Tosion of a Cicula Shaft fixed 5

7 Module. Stain/Stess Analysis

8 Stain/Stess Analysis Stain by Pue Tosion in a Solid Cylinde Fo infinitesimal ba element a Twist ate (ad/length) Angle of twist pe unit length *** Key defomation measue in tosion analysis 1

9 Stain/Stess Analysis Stess y d z d da x dd Aea =A - Calculate stess fom stain G Ga z z -Calculate Resultants M t ( M z ) z da Ga da A A Define J = da: Pola Moment of Inetia, then M ( GJ ) a t

10 Stain/Stess Analysis Stess -Stess in tems of M t z distibution z G ( a) z M G GJ t Mt 1,, J M t J z distibution J = da depending on Coss-sectional shape

11 Stain/Stess Analysis Remak: (One-Dimensional) Constitutive Relation fo Tosion 1) M Da GJa: Moment-Twist Rate Relation t * This elation can be viewed as a maco constitutive Relation fo one-dimensional Tosional Analysis of a Shaft ) D GJ : Tosional Rigidity ) J da ( x y ) da I zz : Pola Moment of Inetia A A 4

12 Stain/Stess Analysis Impotant equations fo Tosion Analysis a Mt ( GJ ) a J da R R 4 0 L L M t adz 0 0 M t Ga J dz GJ -Total twist angle ϕ M t L d L L dz a dz dz 0 dz 0 0 GJ Thus, can be also used fo vaying (GJ) cases shown below 5

13 Stain/Stess Analysis Shaft Design Example: Powe Tansmission Mt o T adius 0 Mt o T Tansmitted powe P : P T [ Watt] T T f t (f: Fequency in Hz) Shaft Design fo Powe Tansmission Given P, f compute T if Allow is known, we can find adius of the shaft, max Mt J T max 0 max 1 J solid cicula 4 0, max 0 ( ) 6

14 Module. Examples

15 Examples Example 1 : Compaison of Solid and Hollow Cylindes against Tosion max Compae GJ and in a solid ba and a hollow cylinde having the same aea ( and same G) cm 5cm 4cm Solid Aea 9 Aea 9 Hollow 1

16 Stain/Stess Analysis Evaluation of J : Pola Moment of Inetia 1 R J I zz da R d R 4 t b a m a b b 4 4 J d b a a b a b a b a b a b a m fo t t m

17 Examples Example 1 (Continued) Sol 1 ( GJ ) hollow ( GJ ) max hollow max solid solid 4 4 (5 4 ) max M t J max M t J 4 hollow solid cm Solid Aea 5cm 4cm Aea 9 9 Hollow ( GJ ) hollow hollow ( GJ ) solid solid max

18 Examples Example : Calculation of Angle of Twist Angle of twist at O fo given M? B M ( ) B Given GJ A B O C L L L L 4

19 Examples Example (Continued) Solution i) Equilibium (FBD) M B M A M C M M M A C B 1 ii) Compatibility A C B B 5

20 Examples Example (Continued) Sol iii) Constitutive Relation (Twisting moment twist ate) a a BA BC M A GJ M C GJ M M M A C B BA BC MAL BA abalba GJ MCL BC abclbc GJ Solving 1,,, 4 yields A B O C 1 M A M B, M C M B L OC M C L MBL OC a 0 BCdz GJ 6GJ 6

21 Examples Example : Angle of Twist in a Conical Shaft d B ( d / ) B B M t A B M t d A ( d / ) A A z L Angle of twist? BA 7

22 Examples Example (Continued) Solution 1) M t = L L Mt ) BA adz dz 0 0 GJ () z ( db d ) A d B z d( z) d A z d A1 1 L d A L 4 J d 4 ( z) M L t dz MtL 1 1 BA G da d z G( d ) B B da d A db 1 1 d L A 8

23 Module 4. Tosion in Stepped o Non-Cicula Stepped Shaft

24 Stepped o Non-cicula Shaft Stess Concentation in a Stepped Ba with a Fillet of Radius Hibble, Mechanics of Mateials, Peason, 010 1

25 Stepped o Non-cicula Shaft Tosion of Non-Cicula Solid Shafts downwads Two impotant obsevations in non-cicula shafts Zeo waping 1 Waping exists. Due to waping, the tosion igidity is educed compaed with the cicula section (fo the same aea of coss-sections) upwads

26 Stepped o Non-cicula Shaft Tosion of Non-Cicula Solid Shafts Hibble, Mechanics of Mateials, Peason, 010 This esult equies elasticity solutions (beyond the scope of this class) * Obseve the locations of maximum shea stess!! <Table> (No need to emembe)

27 Stepped o Non-cicula Shaft Example 4 : Tosional Rigidity of a Non-Cicula Shaft * Compae Tosional igidities of a cicula section and an equilateal tiangle. same aea 0 a 40mm mm 4

28 Stepped o Non-cicula Shaft Example 4 (Continued) Solution i) igidity atio 4 ag ( GJ ) tiangle 46 Table 4 0 ( GJ ) cicula ( GJ ) tiangle a 46 ( GJ ) cicula 4 0 G ii) stess atio M M 0M t max tiangle a Table t 0 t max cicula Jcylinde max cicula a 1 max tiangle 0 5

29 Module 5. Thin-walled Tubes of Closed Section Thin-walled open sections Thin-walled closed sections

30 Thin-walled Tubes of Closed Section Mt ( T) s z n ts () Assumption : 1 Only shea stess sz is pesent Stess vaiation along the thickness diection is negligible. Thin-walled Tubes of Closed Section M t 1

31 Thin-walled Tubes of Closed Section Thin-walled Tubes of Closed Sections Summay: 1 q M A t m : unifom egadless of thickness values Aea A m sz M t ta m t t() s D 4 m A G ds t (Tosional Rigidity) (we only give the esult without deivation equies futhe analysis) o ML 4A G t m ds t

32 Thin-walled Tubes of Closed Section Example 5 : Stess and Angle of Twist in a Thin-Walled Box Shaft 1), ) Stesses at A and B =? ) Angle of twist at C =? G 8GPa Hibble, Mechanics of Mateials, Peason, 010

33 Thin-walled Tubes of Closed Section Example 5 (Continued) <FBD> 60 N m 5 N m 5 N m C D B E Between DE, applied moment, Between CD, applied moment, M t =5 Nm M t =60 Nm Hibble, Mechanics of Mateials, Peason, 010 M To solve fo 1) and ), t sz, M t DE 5 Nm ta Am A (0.05)(0.057) 0.00 m, MPa, B.9MPa (0.005)(0.00) (0.00)(0.00) m t A t B 4

34 Thin-walled Tubes of Closed Section Example 5 (Continued) To solve fo ) Angle of twist at C?, use ML 4A G t m m 1 4A G ds t ds t ( ) (810 ) Rad M L M L t DE DE t CD CD Hibble, Mechanics of Mateials, Peason, 010 5

35 Module 6. Inelastic Tosion

36 Inelastic Tosion Inelastic Tosion Y stain Y stain 0 O 0 stess O Y stess (a) Entiely elastic stain (b) Onset of yield 0 O stess 0 stess (c) Patially plastic (d) Fully plastic 1

37 Inelastic Tosion Inelastic Tosion To calculate T T Y fo which at, 0 Y da 1 dd

38 Inelastic Tosion Inelastic Tosion To find T Y fo case (c), 0 0 d Y 0 0 y d y d Y Y Y 40 Y 6 1 Y TY 1 4 Y Y 0 T T Y 0 O Y (c) Patially plastic stess

39 Inelastic Tosion Inelastic Tosion To find T T p fo case (d), one can use with 0 Y 0 stess 4 Tp TY Y 0 4 (d) Fully plastic 0 0 o Tp d 0 0 Y d Remak: The angle of twist cannot be uniquely detemined fo the given elasticpefect plastic elation because Y does not coespond to any unique value of shea stain. 4

40 Inelastic Tosion Example 6 : Tosion Involving Inelastic Defomation Example : 75 MPa 0mm Y Given find T =? L 1.5m 0.6 ad, Sol. a d dz max d 0.6 dz max 0.0 Y 5

41 Inelastic Tosion Example 6 (Continued) Recall: max Shea stain Distibution Vs Y Shea stess Distibution Y O A Y max O B Y 1 Fom A: one can compute Y, Once Y is detemined, 6

42 Module 7. Residual Stess

43 Residual Stess Residual Stess A shaft subjected to plastic shea stain toque emoval causes esidual shea stain! 1

44 Residual Stess Residual Stess Case 1 : Unloading fom fully-plastic state Y T p T p Y Y a) plastic toque Tp applied b) unloading elastic shea stain thoughout the shaft c) Residual shea stess

45 Residual Stess Residual Stess Tp Y0 (deived ealie)? T 4 p 0 0 Y 0 Y J 4 0 Case : Unloading patially-plastic state O 0 ( modulus of uptue ) Y Y T ep T ep max max

46 Residual Stess Example 7 : Plastic Toque and Residual Stess in a hollow cylinde? i MPa Y o i o 50mm 5mm O Y ad Y 84 MPa, Y 0.00 Length L=1.5m 1 plastic toque Tp =? esidual shea stess distibution if unloaded fom the fully plastic state? 4

47 Residual Stess Example 7 (Continued) Solution 1 T p Y evey whee T p Y Y i o da d Y o i 19.4 kn m

48 Residual Stess Example 7 (Continued) Solution fo unloading, conside T p linea vaiation : the max. shea stess T p o J MPa 104.5MPa 5.6MPa

49 Residual Stess Example 7 (Continued) Solution Finally supeposing two fields yields MPa MPa 7