CET 1: Stress Analysis & Pressure Vessels. Lent Term Dr. Clemens Kaminski. Telephone:

Transcription

1 CET : Stress Anlysis & Pressure Vessels Lent Term 5 Dr. Clemens Kminski Telephone: URL: E-mil: clemens_kminski@cheng.cm.c.uk

2 Synopsis Introduction to Pressure Vessels nd Filure Modes. Stresses in Cylinders nd Spheres. Compressive filure. Euler buckling. Vcuum vessels. Tensile filure. Stress Stress Concentrtion & Crcking -D stress nd strin. Elsticity nd Strins-Young's Modulus nd Poisson's Rtio. Bulk nd Sher Moduli. Hoop, Longitudinl nd Volumetric Strins.4 Strin Energy. Overfilling of Pressure Vessels Therml Effects. Coefficient of Therml Expnsion. Therml Effect in cylindricl Pressure Vessels. Two-Mteril Structures 4 Torsion. 4. Sher Stresses in Shfts - τ/r T/J G/L 4. Thin Wlled Shfts 4. Thin Wlled Pressure Vessel subject to Torque 5 Two Dimensionl Stress Anlysis 5. Nomenclture nd Sign Convention for Stresses 5. Mohr's Circle for Stresses 5. Worked Exmples 5.4 Appliction of Mohr's Circle to Three Dimensionl Systems 6 Bulk Filure Criteri 6. Tresc's Criterion. The Stress Hexgon 6. Von Mises' Filure Criterion. The Stress Ellipse 7 Two Dimensionl Strin Anlysis 7. Direct nd Sher Strins 7. Mohr's Circle for Strins 7. Mesurement of Strin - Strin Guges 7.4 Hooke s Lw for Sher Stresses

3 Supporting Mterils There is one Exmples pper supporting these lectures. Two good textbooks for further explntion, worked exmples nd exercises re Mechnics of Mterils (997) Gere & Timoshenko, publ. ITP [ISBN ] Mechnics of Solids (989) Fenner, publ. Blckwell [ISBN -6-8-] This mteril ws tught in the CET I (Old Regultions) Structures lecture unit nd ws exmined in CET I (OR) Pper IV Section. There re consequently lrge number of old Tripos questions in existence, which re of the pproprite stndrd. From 999 onwrds the course ws tught in CET, pper 5. Chpters 7 nd 8 in Gere nd Timoshenko contin lrge number of exmple problems nd questions. Nomenclture The following symbols will be used s consistently s possible in the lectures. E G I J R t T α γ η ν τ Young s modulus Sher modulus second moment of re polr moment of re rdius thickness τορυ therml expnsivity liner strin sher strin ngle Poisson s rtio Norml stress Sher stress

4 A pressure vessel ner you!

5 Ongoing Exmple We shll refer bck to this exmple of typicl pressure vessel on severl occsions. Distilltion column m P 7 br 8 m crbon steel t 5 mm

6 . Introduction to Pressure Vessels nd Filure Modes Pressure vessels re very often sphericl (e.g. LPG storge tnks) cylindricl (e.g. liquid storge tnks) cylindricl shells with hemisphericl ends (e.g. distilltion columns) Such vessels fil when the stress stte somewhere in the wll mteril exceeds some filure criterion. It is thus importnt to be ble to be ble to understnd nd quntify (resolve) stresses in solids. This unit will concentrte on the ppliction of stress nlysis to bulk filure in thin wlled vessels only, where (i) the vessel self weight cn be neglected nd (ii) the thickness of the mteril is much smller thn the dimensions of the vessel (D» t)... Stresses in Cylinders nd Spheres Consider cylindricl pressure vessel L Externl dimeter D internl guge pressure P r L h wll thickness, t The hydrosttic pressure cuses stresses in three dimensions.. Longitudinl stress (xil) L. Rdil stress r. Hoop stress h ll re norml stresses. SAPV LT 5 CFK, MRM

7 r L r L h h, The longitudinl stress L P L Force equilibrium π D 4 P π D t L if P >, then L P D 4 t L is tensile b, The hoop stress h P h P h P SAPV LT 5 CFK, MRM Force blnce, h P D t D L P h L t

8 c, Rdil stress r r vries from P on inner surfce to on the outer fce r o ( P ) h, L P ( D t ). thin wlled, so D >> t so h, L >> r so neglect r Compre terms d, The sphericl pressure vessel h P P π D P 4 P D h 4 t h π D t SAPV LT 5 CFK, MRM

9 .. Compressive Filure: Bulk Yielding & Buckling Vcuum Vessels Consider n unpressurised cylindricl column subjected to single lod W. Bulk filure will occur when the norml compressive stress exceeds yield criterion, e.g. W bulk W πdt Y Compressive stresses cn cuse filure due to buckling (bending instbility). The criticl lod for the onset of buckling is given by Euler's nlysis. A full explntion is given in the texts, nd the bsic results re summrised in the Structures Tbles. A column or strut of length L supported t one end will buckle if W π EI L Consider cylindricl column. I πrt so the compressive stress required to cuse buckling is or buckle W πdt π EπD t 8L πdt π ED 8L buckle π E ( ) 8 LD SAPV LT 5 CFK, MRM 4

10 where L/D is slenderness rtio. The mode of filure thus depends on the geometry: stress Euler buckling locus y Bulk yield Short L /D rtio Long SAPV LT 5 CFK, MRM 5

11 Vcuum vessels. Cylindricl pressure vessels subject to externl pressure re subject to compressive hoop stresses P D h t Consider length L of vessel, the compressive hoop force is given by, P D L h L t If this force is lrge enough it will cuse buckling. length Tret the vessel s n encstered bem of length πd nd bredth L SAPV LT 5 CFK, MRM 6

12 Buckling occurs when Force W given by. 4π EI W ( π D) P D L 4π EI ( π D) I b t L t p buckle E t D SAPV LT 5 CFK, MRM 7

13 .. Tensile Filure: Stress Concentrtion & Crcking Consider the rod in the Figure below subject to tensile lod. The stress distribution cross the rod long distnce wy from the chnge in cross section (XX) will be uniform, but ner XX the stress distribution is complex. D X X d W There is concentrtion of stress t the rod surfce below XX nd this vlue should thus be considered when we consider filure mechnisms. The rtio of the mximum locl stress to the men (or pprent) stress is described by stress concentrtion fctor K K mx men The vlues of K for mny geometries re vilble in the literture, including tht of crcks. The mechnism of fst frcture involves the concentrtion of tensile stresses t crck root, nd gives the filure criterion for crck of length π K c where Kc is the mteril frcture toughness. Tensile stresses cn thus cuse filure due to bulk yielding or due to crcking. SAPV LT 5 CFK, MRM 8

14 crck K c π stress filure locus length of crck. SAPV LT 5 CFK, MRM 9

15 . -D stress nd strin.. Elsticity nd Yield Mny mterils obey Hooke's lw E Yield Stress pplied stress (P) E Young's modulus (P) strin (-) filure Elstic Limit up to limit, known s the yield stress (stress xis) or the elstic limit (strin xis). Below these limits, deformtion is reversible nd the mteril eventully returns to its originl shpe. Above these limits, the mteril behviour depends on its nture. Consider smple of mteril subjected to tensile force F. F F An increse in length (xis ) will be ccompnied by decrese in dimensions nd. Hooke's Lw ( F / A)/ E

16 The strin in the perpendiculr directions, re given by ν E ; ν E where ν is the Poisson rtio for tht mteril. These effects re dditive, so for three mutully perpendiculr stresses,, ; Giving E ν E ν E ν E + E ν E ν E ν E + E Vlues of the mteril constnts in the Dt Book give orders of mgnitudes of these prmeters for different mterils; Mteril E ν (x 9 N/m ) Steel. Aluminum lloy 7. Brss 5.5

17 . Bulk nd Sher Moduli These mteril properties describe how mteril responds to n pplied stress (bulk modulus, K) or sher (sher modulus, G). The bulk modulus is defined s P uniform K v i.e. the volumetric strin resulting from the ppliction of uniform pressure. In the cse of pressure cusing expnsion so P [ E ν ν ] P ( ν) E v + + P ( ν) E E K ( ν) For steel, E kn/mm, ν., giving K 75 kn/mm For wter, K. kn/mm For perfect gs, K P ( br, -4 kn/mm) Sher Modulus definition τ Gγ γ - sher strin

18 .. Hoop, longitudinl nd volumetric strins (micro or millistrin) Frctionl increse in dimension: L length h circumference rr wll thickness () Cylindricl vessel: Longitudinl strin L L E Hoop strin: h n E rdil strin - υ h E - υ L E - υ r E PD 4tE - υ ( ) δl L PD 4tE - υ ( ) δd D δr R r [ E r - υ h - υ L ] - PDυ 4ET δt t [frctionl increse in wll thickness is negtive!]

19 [ONGOING EXAMPLE]: L ( E L - υ n ) 6 [ x ] 9 x 6 6 x - (.).4 x millistrin h.486 millistrin r -.57 millistrin Thus: pressurise the vessel to 6 br: L nd D increse: t decreses Volume expnsion Cylindricl volume: V o πd o L o 4 (originl) New volume V π ( 4 D o + δd) ( L o + δl) L o π D o 4 [ + h ] [ + L ] Define volumetric strin v δv V v V - V o V o ( + h ) ( + L ) - ( + h + h ) + L ( ) - v h + L + h + h L + L n Mgnitude inspection: 4

20 mx ( steel) y E Ignoring second order terms, 9 x 6 x 9.95 x smll (b) (c) v h + L Sphericl volume: so Generl result E PD 4Et [ -υ -υ ] ( -υ) h h L r π 6 {( D + δd) - D } o v πd o 6 ( + h ) h + ( ) o v + + ii re the strins in ny three mutully perpendiculr directions. {Continued exmple} cylinder L.4 mstrin n.486 mstrin rr -.57 mstrin v n + L new volume V o ( + v ) Increse in volume π D L 4 Volume of steel o πdlt.77 m v x.86 x - 6 Litres v for steel L + h + rr.4 mstrins increse in volume of steel Strin energy mesure of work done.9 L Consider n elstic mteril for which F k x 5

21 Work done in expnding δx dw Fδx F work done Are x Work done in extending to x w o x Fdx k x dx kx o x F x Smple subject to stress incresed from to : Extension Force: x L o A F W AL o L (no direction here) Strin energy, U work done per unit volume of mteril, U AL o AL ( o ) U Al o Al U o In -D system, U [ + + ] Now E - υ E - υ E So U E ν + + etc [ ( )] Consider uniform pressure pplied: P E U P E ( - υ) P K energy stored in system (per unit vol. 6

22 For given P, U stored is proportionl to /K so pressure test using liquids rther thn gses. {Ongoing Exmple} P 6 brg. δv 6 x - m increse in volume of pressure vessel Incresing the pressure compresses the contents normlly test with wter. V wter? v ( wter) P K 6 x mstrins. x decrese in volume of wter -V o (.7) -5.4 x m Thus we cn dd more wter: Extr spce (L) 76.4 L wter extr spce p p6 7

23 . Therml Effects.. Coefficient of Therml Expnsion Definition: coefficient of therml expnsion α L T Liner Coefficient of therml volume expnsion v αt Volume Steel: α L x -6 K - T o C L -5 rector T 5 o C L 5.5 millistrins (!) Consider steel br mounted between rigid supports which exert stress Het steel: α T - E If rigid: so Eα T (i.e., non buckling) x 9 x x -6 T. x 6 T y 9 MP: filure if T > 8.6 K 8

24 {Exmple}: stem min, instlled t o C, to contin 6 br stem (4 o C) if ends re rigid, MP filure. must instll expnsion bends... Temperture effects in cylindricl pressure vessels D m. steel construction L m. full of wter t mm Initilly un pressurised full of wter: increse temp. by T: pressure rises to Vessel P. The Vessel Wll stresses (tensile) PD L 8. P 4 t n L 66.7 P 9

25 Strin (volume) L E ν - E L h l + α T υ. E x 9 α x 6 L.585 x - P + x -6 T Similrly h 6.75 x - P + 6 T v L + n 5.8 x - P + x -6 T vessel vol. Strin vessel expnds due to temp nd pressure chnge. The Contents, (wter) Expnds due to T increse Contrcts due to P increse: v, H O α v T P/K H O α v 6 x -6 K - v 6 x -6 T 4.55 x - P Since vessel remins full on incresing T: v (H ) v (vessel) Equting P 75 T Now pressure, rise of.7 br per C increse in temp. n 66.7 P.9 x 6 T n.9 Mp per C rise in Temperture

26 Filure does not need lrge temperture increse. Very lrge stress chnges due to temperture fluctutions. MORAL: Alwys leve spce in liquid vessel. ( v, gs α v T P/K)

27 .. Two mteril structures Bewre, different mterils with different therml expnsivities cn cuse difficulties. {Exmple} Where there is benefit. The Bimetllic strip - controllers 4 mm ( + mm) b mm temperture Cu Fe d L mm b Het by T: Cu expnds more thn Fe so the strip will bend: it will bend in n rc s ll sections re identicl.

28 Cu Fe The different therml expnsions, set up shering forces in the strip, which crete bending moment. If we pply sgging bending moment of equl [: opposite] mgnitude, we will obtin stright bem nd Cu Fe F F cn then clculte the shering forces [nd hence the BM]. Shering force F compressive in Cu Tensile in Fe Equting strins: α cu T - F bde cu α Fe T + F bde Fe F bd E + E So ( α - α ) T cu Fe cu Fe

29 bd x -5 m E cu 9 GP α cu 7 x -6 k - T C F 87 N E Fe GP α Fe x -6 K - (significnt force) F cts through the centroid of ech section so BM F./(d/).774 Nm Use dt book to work out deflection. ML δ EI This is the principle of the bimetllic strip. 4

30 Consider steel rod mounted in upper tube spcer Anlysis relevnt to Het Exchngers Cu Fe. ssembled t room temperture increse T Dt: α cu > α Fe : copper expnds more thn steel, so will generte TENSILE stress in the steel nd compressive stress in the copper. Blnce forces: Tensile force in steel F Fe F cu F Stress in steel F/A Fe Fe copper F/A cu cu Steel strin: FE α Fe T + Fe /E Fe (no trnsvere forces) α Fe T + F/E Fe A Fe copper strin cu α cu T F/E cu A cu 5

31 AFeEFe AcuE cu Strins EQUAL: F + ( ) T sum of strengths α cu -α d Fe So you cn work out stresses nd strins in system. 6

32 4. Torsion Twisting Sher stresses 4.. Sher stresses in shfts τ/r T/J G /L Consider rod subject to twisting: Definition : sher strin γ chnge in ngle tht ws originlly Π/ Consider three points tht define right ngle nd more then: Sher strin γ γ + γ [RADIANS] A γ A B C B γ C Hooke s Lw τ G γ G sher modulus E ( + υ) 7

33 Now consider rod subject to n pplied torque, T. T r. Hold one end nd rotte other by ngle B B B B γ L Plne ABO ws originlly to the X-X xis Plne ABO is now inclined t ngle γ to the xis: tn Sher stress involved Gr τ Gγ L r γ γ L 8

34 Torque required to cuse twisting: τ τ r dr.δt τ Π r.δr r T cf T A τ Π r L dr G r da or G {} J so L τ A r.da T J M I G τ L r E R y τ Gr L DEFN: J polr second moment of re 9

35 Consider rod of circulr section: J o R π π. r r dr R 4 y r x 4 πd J Now r x + y It cn be shown tht J I xx + I yy [perpendiculr xis thn] see Fenner this gives n esy wy to evlute I xx or I yy in symmetricl geometrics: I xx I yy πd 4 /64 (rod)

36 Rectngulr rod: d b I I xx yy bd db J bd [ b + d ]

37 Exmple: steel rod s drive shft D 5 mm L.5 m Filure when τ τ y 95 MP G 8 GP Now τ r mx mx 4 πd J T J 95 x.5 8 x 8 6 m 4 so T 9 Nm G T L J 7.6 x 9 9 8x.5 ( ) From.4 rd 8. Sy shft rottes t 45 rpm: power Tω π 9x x kw

38 4.. Thin wlled shfts (sme Eqns pply) Consider brcket joining two Ex. Shfts: T 9 Nm D min.5m Wht is the minimum vlue of D for connector? r mx D/ J (π/){d } τ r y T 95 x J D 6 x 9 π 4 4 ( D -. ) mx 5 D x -5 D D 4.5 cm

39 4.. Thin wlled pressure vessel subject to torque τ T r J 4 [ ] 4 - π D now cylinder J ( D + t) π π D t 4 [ 8D t + 4 D t +...] so τ 4T D πd t T τ π D t 4

40 CET, SAPV 5. Components of Stress/ Mohr s Circle 5. Definitions Sclrs tensor of rnk Vectors tensors of rnk r r F m hence: F F F or : m m m F i m i Tensors of rnk p i T j ij q j i, j,, or : p p p T T T q q q + T + T + T q q q + T + T + T q q q Axis trnsformtions The choice of xes in the description of n engineering problem is rbitrry (s long s you choose orthogonl sets of xes!). Obviously the physics of the problem must not depend on the choice of xis. For exmple, whether pressure vessel will explode cn not depend on how we set up our co-ordinte xes to describe the stresses cting on the 4

41 CET, SAPV vessel. However it is cler tht the components of the stress tensor will be different going from one set of coordintes x i to nother x i. How do we trnsform one set of co-ordinte xes onto nother, keeping the sme origin? x x x ' ' ' x x x... where ij re the direction cosines Forwrd trnsformtion: x ' x New in terms of Old i j ij j Reverse trnsformtion: x x Old in terms of New i j ji j We lwys hve to do summtions in co-ordinte trnsformtion nd it is conventionl to drop the summtion signs nd therefore these equtions re simply written s: x ' i ij x j x i ji x j 5

42 CET, SAPV Tensor trnsformtion How will the components of tensor chnge when we go from one coordinte system to nother? I.e. if we hve sitution where p i T q T q j ij j ij j (in short form) where T ij is the tensor in the old co-ordinte frme x i, how do we find the corresponding tensor T ij in the new co-ordinte frme x i, such tht: p ' T ' q ' T i j ij j ij ' q j ' (in short form) We cn find this from series of sequentil co-ordinte trnsformtions: p' p q q' Hence: p ' i ik p k p k T kl q l q l jl q j ' Thus we hve: 6

43 CET, SAPV ' ' ' ' ' j ij j kl jl ik j jl kl ik i q T q T q T p For exmple: ' T T T T T T T T T T T T T j i j i j i j i j i j i j i j i j i l jl i l jl i l jl i ij Note tht there is difference between trnsformtion mtrix nd nd rnk tensor: They re both mtrices contining 9 elements (constnts) but: Symmetricl Tensors: T ij T ji 7

44 8 CET, SAPV

45 CET, SAPV We cn lwys trnsform second rnk tensor which is symmetricl: ' : such tht ' T T T T T T ij ij ij Consequence? Consider:,, then q T p q T p q T p q T p j ij i The digonl T, T, T is clled the PRINCIPAL AXIS. If T, T, T re stresses, then these re clled PRINCIPAL STRESSES. 9

46 CET, SAPV Mohr s circle Consider n elementry cuboid with edges prllel to the coordinte directions x,y,z. y Fxy y fce Fx x z fce Fxz Fxx z x fce The fces on this cuboid re nmed ccording to the directions of their normls. There re thus two x-fces, one fcing greter vlues of x, s shown in Figure nd one fcing lesser vlues of x (not shown in the Figure). On the x-fce there will be some force F x. Since the cuboid is of infinitesiml size, the force on the opposite side will not differ significntly. The force F x cn be divided into its components prllel to the coordinte directions, F xx, F xy, F xz. Dividing by the re of the x-fce gives the stresses on the x-plne: τ τ xx xy xz It is trditionl to write norml stresses s nd sher stresses s τ. Similrly, on the y-fce: τ,, yx yy τ yz 4

47 CET, SAPV nd on the z-fce we hve: τ, τ, zx zy zz There re therefore 9 components of stress; ij τ τ xx yx zx τ τ xy yy zy τ xz τ yz zz Note tht the first subscript refers to the fce on which the stress cts nd the second subscript refers to the direction in which the ssocited force cts. y x xx yy τ yx τ xy xx τ xy τ yx yy But for non ccelerting bodies (or infinitesimlly smll cuboids): nd therefore: ij τ τ xx yx zx τ τ xy yy zy τ xz xx τ yz τ xy zz τ xz τ τ xy yy yz τ xz τ yz zz Hence ij is symmetric! 4

48 CET, SAPV This mens tht there must be some mgic co-ordinte frme in which ll the stresses re norml stresses (principl stresses) nd in which the off digonl stresses (sher stresses) re. So if, in given sitution we find this frme we cn pply ll our stress strin reltions tht we hve set up in the previous lectures (which ssumed there were only norml stresses cting). Consider cylindricl vessel subject to sher, nd norml stresses ( h, l, r ). We re usully interested in shers nd stresses which lie in the plne defined by the vessel wlls. Is there trnsformtion bout zz which will result in sher Would relly like to trnsform into co-ordinte frme such tht ll components in the x i : ij ij ' So stress tensor is symmetric nd rnk tensor. Imgine we re in the coordinte frme x i where we only hve principl stresses: ij Trnsform to new co-ordinte frme x i by rottoin bout the x xis in the originl co-ordinte frme (this would be, in our exmple, z-xis) 4

49 CET, SAPV The trnsformtion mtrix is then: cos sin sin cos ij Then: sin cos sin cos sin cos sin cos sin cos sin cos cos sin sin cos cos sin sin cos ' kl jl ik ij 4

50 CET, SAPV Hence: )sin ( sin cos sin cos ' )cos ( ) ( cos sin ' )cos ( ) ( sin cos '

51 Yield conditions. Tresc nd Von Mises Mohrs circle in three dimensions. y z x Sher stresses τ y,z plne x, y plne x,z plne norml stresses

52 We cn drw Mohrs circles for ech principl plne. 6. BULK FAILURE CRITERIA Mterils fil when the lrgest stress exceeds criticl vlue. Normlly we test mteril in simple tension: P P y P yield A τ This mteril fils under the stress combintion ( y,, ) τ mx.5 y 95 Mp for steel We wish to estblish if mteril will fil if it is subject to stress combintion (,, ) or ( n, L, τ)

53 Filure depends on the nture of the mteril: Two importnt criteri (i) (ii) Tresc s filure criterion: brittle mterils Cst iron: concrete: cermics Von Mises criterion: ductile mterils Mild steel + copper 6.. Tresc s Filure Criterion; The Stress Hexgon (Brittle) A mteril fils when the lrgest sher stress reches criticl vlue, the yield sher stress τ y. Cse (i) Mteril subject to simple compression: Principl stresses (-,, ) τ -

54 M.C: mc psses through (,), (,), (,) τ mx occurs long plne t 45 to Similrly for tensile test. τ mx / Cse (ii) < < τ - M.C. Fils when - τ mx τ y y i.e., when - y mteril will not fil. 4

55 Lets do n esy exmple. 5

56 6. Von Mises Filure (ductile mterils) Criterion; The stress ellipse Tresc s criterion does not work well for ductile mterils. Erly hypothesis mteril fils when its strin energy exceeds criticl vlue (cn t be true s no filure occurs under uniform compression). Von Mises : filure when strin energy due to distortion, U D, exceeds criticl vlue. U D difference in strin energy (U) due of compressive stress C equl to the men of the principl stresses. C [ + + ] U D E υ + + G {( - ) + ( - ) + ( - ) } M.C. ( ) - [ ] E C + 6υC τ 6

57 squre of {τ, τ b, τ c } criticl vlue Tresc filure when mx (τ I ) τ y ) Von Mises filure when root men τ y Compre with ( y,, ). simple tensile test filure Filure if {( - ) + ( + ) + ( - ) } > { y + + } y G {( - ) + ( + ) + ( - ) } > y G 7

58 8

59 Lets do simple exmple. 9

60 Exmple Tresc's Filure Criterion The sme pipe s in the first exmple (D. m, t'.5 m) is subject to n internl pressure of 5 brg. Wht torque cn it support? Clculte stresses Mohr's Circle L PD 4t' h PD t' 5N / mm N / mm nd r τ (,τ) 5 s (5, τ) Circle construction s 75 N/mm t (5+τ) The principl stresses, s ± t Thus my be positive (cse A) or negtive (cse B). Cse A occurs if τ is smll.

61 Cse A τ (,τ) 5 s (5, τ) Cse B τ (,τ) β 5 s (5, τ) We do not know whether the Mohr's circle for this cse follows Cse A or B; determine which cse pplies by tril nd error. Cse A; 'minor' principl stress is positive ( > ) Thus filure when τ mx y 5N / mm

62 For Cse A; ( ) [ ] τ mx τ τ 5 5 τ.7n / mm Giving N / mm ; 6N / mm Cse B; We now hve τmx s the rdius of the originl Mohr's circle linking our stress dt. Thus τ mx 5 + τ 5 τ.98n / mm Principl stresses, 75 ±5 8N / mm ; N / mm Thus Cse B pplies nd the yield stress is.98 N/mm. The torque required to cuse filure is T πd t' τ / knm Filure will occur long plne t ngle β nticlockwise from the y (hoop) direction; tn(λ) λ 76. ; 75 β 9-λ β 6.9

63 Exmple More of Von Mises Filure Criterion From our second Tresc Exmple h N / mm L 5N / mm r N / mm Wht torque will cuse filure if the yield stress for steel is N/mm? Mohr's Circle τ (,τ) 5 s (5, τ) Giving s + t τ s t τ At filure U D G ( ) +( ) +( ) { }

64 Or ( ) +( ) +( ) ( y ) + () + ( y ) 4t + (s t) + (s + t) y s + 6t y s + t y 75 + (5 + τ ) τ N / mm The tube cn thus support torque of T πd t' τ 5kNm which is lrger thn the vlue of knm given by Tresc's criterion - in this cse, Tresc is more conservtive. 4

65 7. Strins 7.. Direct nd Sher Strins Consider vector of length lx lying long the x-xis s shown in Figure. Let it be subjected to smll strin, so tht, if the left hnd end is fixed the right hnd end will undergo smll displcement δx. This need not be in the x-direction nd so will hve components δxx in the x-direction nd δxy in the y-direction. γ δ x δ xy We cn define strins xx nd xy by, l x δ xx Figure xx δ l xx x ; xy δ l xy x xx is the direct strin, i.e. the frctionl increse in length in the direction of the originl vector. xy represents rottion of the vector through the smll ngle γ where, γ tn γ l x δ xy + δ xx δ l xy x xy Thus in the limit s δx, γ xy. Similrly we cn define strins yy nd yx γ by, δ yy δ yy ; yx l l y yx y

66 δ yy δ yx δ y Figure l y γ s in Figure. γ l x Or, in generl terms: ij δ ij ; where i, j,, l i The ENGINEERING SHEAR STRAIN is defined s the chnge in n ngle reltive to set of xes originlly t 9. In prticulr γxy is the chnge in the ngle between lines which were originlly in the x- nd y- directions. Thus, in our exmple (Figure bove): γ xy (γ + γ ) xy + yx or γ xy (γ + γ ) depending on sign convention. A τ yx A' τ xy B C Figure B' C' Figure b Positive vlues of the sher stresses τxy nd τyx ct on n element s shown in Figure nd these cuse distortion s in Figure b. Thus it is sensible to tke γxy s +ve when the ngle ABC decreses. Thus

67 γ xy +(γ + γ ) Or, in generl terms: γ + ) ij nd since τ ij τ ji, we hve γ ij γ ji. ( ij ji Note tht the TENSOR SHEAR STRAINS re given by the verged sum of sher strins: γ ij ( ij + ji ) ( γ + γ ) γ ji 7. Mohr s Circle for Strins The strin tensor cn now be written s: ij y y y y y y y y y y y y where the digonl elements re the stretches or tensile strins nd the off digonl elements re the tensor sher strins. Thus our strin tensor is symmetricl, nd:

68 4 ji ij This mens there must be co-ordinte trnsformtion, such tht: : such tht ' ij ij ij we only hve principl (longitudinl) strins! Exctly nlogous to our discussion for the trnsformtion of the stress tensor we find this from: sin cos sin cos sin cos sin cos sin cos sin cos cos sin sin cos cos sin sin cos ' kl jl ik ij And hence:

69 5 )sin ( sin cos sin cos ' ' )cos ( ) ( cos sin ' )cos ( ) ( sin cos ' γ For which we cn drw Mohr s circle in the usul mnner:

70 Note, however, tht on this occsion we plot hlf the sher strin ginst the direct strin. This stems from the fct tht the engineering sher strins differs from the tensoril sher strins by fctor of s s discussed. 7. Mesurement of Stress nd Strin - Strin Guges It is difficult to mesure internl stresses. Strins, t lest those on surfce, re esy to mesure. Glue piece of wire on to surfce Strin in wire strin in mteril As the length of the wire increses, its rdius decreses so its electricl resistnce increses nd cn be redily mesured. In prctice, multiple wire ssemblies re used in strin guges, to mesure direct strins. _ strin guge 45 Strin rosettes re employed to obtin three mesurements: Strin Rosette Three direct strins re mesured C B A principl strin Mohr s circle for strins gives 6

71 γ/ B A B C A rdius t so we cn write C s + t cos( ) A s + t cos( + 9) s t sin( ) B circle, centre s, C s t cos( ) equtions in unknowns Using strin guges we cn find the directions of Principl strins γ/ 7

72 7.4 Hooke s Lw for Sher Stresses St. Vennt s Principle sttes tht the principl xes of stress nd strin re co-incident. Consider -D element subject to pure sher (τxy τyx τo). y τ o The Mohr s circle for stresses is Y τo Q τ o P τ o x X where P nd Q re principl stress xes nd pp τ o qq τ o τ pq τ qp Since the principl stress nd strin xes re coincident, pp E ν E τ o E ( + ν) qq E ν E τ o E ( + ν) nd the Mohr s circle for strin is thus γ/ Q qq X Y (, γ ) xy P pp the Mohr s circle shows tht xx γ xy τ o E ( + ν) 8

73 So pure sher cuses the sher strin γ τ o τ o γ τ o E (+ν ) γ/ γ/ nd But by definition τo Gγ so G τ o γ E (+ν) Use St Vennts principl to work out principl stress vlues from knowledge of principl strins. Two Mohrs circles, strin nd stress. E ν E ν E ν E + E ν E ν E ν E + E 9

74 So using strin guges you cn work out mgnitudes of principl strins. You cn then work out mgnitudes of principl stresses. Using Tresc or Von Mises you cn then work out whether your vessel is sfe to operte. ie below the yield criteri