Lecture 7 Further Development of Theory and Applications

Transcription

1 P4 Stress and Strain Dr. A.B. Zavatsk HT08 Lecture 7 Further Develpment f Ther and Applicatins Hke s law fr plane stress. Relatinship between the elastic cnstants. lume change and bulk mdulus. Spherical and clindrical pressure vessels.

2 Generalied Hke s Law Appl, get, -ν, -ν Appl, get, -ν, -ν Appl, get, -ν, -ν Fr an istrpic linearl elastic material, / in the,, and directins. Use superpsitin t get : ν ( ν ν ν ν ν etc. The resulting equatins are: ( ν ν ( ν ν ( ν ν

3 Hke s Law fr Plane Stress Fr plane stress, substitute 0 int the generalied Hke s Law equatins t get: ν ( ν ( ν ( HLT, page 08 Remember als the shear strain: γ τ G 3

4 These equatins can be re-written in terms f stresses: ν ν 0 τ Gγ ( ν ( ν HLT, page 08 These equatins cntain three material cnstants:, G, and ν. We can shw that these cnstants are related b the equatin: G HLT, page 0 ( ν 4

5 Cnsider a plane stress element in pure shear and relate the shear strains and stresses t the strains and stresses alng the θ 45 directin. Start with strains. τ Gγ a b h π γ a b π γ d h c lement befre shear is applied. L h c d lement after shear applied. lengthens, ac shrtens Angle changes related t γ 5

6 L L L h L? ΔL L L ΔL h ΔL ( L L ( π γ h Using gemetr, the nrmal strain can be related t the shear strain γ. d a h L 4 b π γ Using the csine rule: ( L ( ( [ ] h h [ cs( π / γ ] ( h h h( h cs( π / cs( π / γ sinγ γ Since ( 0 sinγ γ γ / and γ are small, 6

7 Nw use Mhr s circle and Hke s law t relate strains t stresses. Find the stress alng the θ 45 directin : B τ A B A τ τ - τ θ θ 90 θ 45 θ 45 τ - τ The strain in the directin is: ν τ ( τ ν τ ( ν 7

8 γ G γ / τ ( ν ( ν ( ν Gγ τ ( ν Data frm HLT, page 4 Mild Steel: 0 GPa,ν , G 8 GPa Aluminium 04: 7 GPa,ν 0.33, G 8 GPa Using the equatin just derived: Mild Steel: G 0 / [(0.7] 83 GPa G 0 / [(0.30] 8 GPa Aluminium 04: G 7 / [(0.33] 7 GPa 8

9 9 This equatin is valid fr bth large and small strains. If strains are assumed t be small, the prduct terms all tend t 0. lume Change and Bulk Mdulus Nrmal stresses prduce changes in vlume, whereas shear stresses prduce changes in shape. Nrmal stress causes a change in length dl f each face. Since dl / L, each dl L. a c b a b c ( ( ( ( ( ( ( ( ( ( abc c c b b a a abc Original vlume New vlume

10 0 The unit vlume change ( dilatatin, vlumetric strain is defined as: e e Δ ( Cnsider net tw situatins: spherical stress and uniaial stress. ( Assuming that the strains are small, ( ( Δ Δ Δ The change in vlume Δ is then:

11 Spherical stress is defined as. e e ( ν ν ( ν ν ( ν Define the bulk mdulus f elasticit: e K K e spherical stress vlumetric strain 3 K (Hke s Law (unit vlume change 3 3 ( ν 3( ν HLT, page 0 These equatins als hld fr hdrstatic stress ( -

12 Data frm HLT, page 4 Mild Steel: 0 GPa,ν , K GPa Aluminium 04: 7 GPa,ν 0.33, K 75 GPa Using the equatin fr K just derived: Mild Steel: K 0 / { 3 [-(0.7] } 5 GPa K 0 / { 3 [-(0.30] } 75 GPa Aluminium: K 7 / { 3 [-(0.33] } 7 GPa K 3( ν If ν 0, K / 3. If ν /3, K. If ν /, K. This crrespnds t a rigid material having n change in vlume (that is, the material is incmpressible.

13 Net, cnsider the unit vlume change fr uniaial stress, 0. e ν Substituting 0 int Hke s Law gives: The vlumetric strain is then: ν ν ν e e ( ν Nte here that the maimum pssible value f Pissn s rati fr cmmn materials is 0.5, because a larger value means that the vlume wuld decrease when the material is in tensin, which is cntrar t rdinar phsical behaviur. 3

14 ample (based n Gere, 6th ed, p 537, A slid steel sphere ( 0 GPa, ν 0.3 is subjected t hdrstatic pressure p such that its vlume is reduced b 0.4%. Calculate: (a the bulk mdulus f elasticit K fr the steel (b the pressure p (c the strain energ stred in the sphere if its diameter d 50 mm. Slutin: (a The bulk mdulus f elasticit is fund using the equatin fr K derived earlier. K 3( ν ( ( Pa 75 GPa 4

15 (b e Δ 0.4% K e ( ( Pa 700 MPa ο is the same as the pressure p. (c The strain energ densit u (strain energ per unit vlume is given b the area under the ο K e curve (which is linear. u e K 6 ( (75 0 The vlume f the sphere is.40 0 The strain energ densit U is then simpl U π r π ( u ( ( J m J 3 J m 3 5

16 Pressure essels Clsed structures cntaining liquids r gases under pressure. amples are tanks, pipes, pressuried cabins in aircraft, etc. Pressure vessels are cnsidered t be thin-walled when the rati f radius r t wall thickness t is greater than 0. Assume that the internal pressure p in eceeds the pressure p ut acting n the utside f the vessel (usuall atmspheric pressure. If p ut > p in, the vessel culd cllapse inward due t buckling. We are interested in the stresses and strains that develp in the walls f pressure vessels. We will derive equatins based n the net r gauge pressure p, where p p in p ut. 6

17 Spherical Pressure essels T determine the stresses in the (thin walls f a spherical pressure vessel with inner radius r and wall thickness t, first cut thrugh the sphere n a vertical diametral plane. Net, islate half f the sphere and its fluid cntents as a single free bd. tensile stress in the walls fluid pressure p p r t Fluid frce in hrintal directin P p (πr Tensile frce in hrintal directin T (πr m t where r m r t/ 7

18 Fr equilibrium, frces in the hrintal directin must balance. ΣF (π r m t p( π r 0 pr r m t Fr thin-walled vessels, r r m and the equatin becmes pr t * Nte that using r instead f r m actuall gives a result clser t the thereticall eact result. Since the same equatin fr tensile stresses wuld result frm an slice thrugh the centre f the sphere, we cnclude that the wall f a spherical pressure vessel is subjected t unifrm tensile stress in all directins. Smetimes called membrane stresses 8

19 Stresses at the Outer Surface pr / t 0 τ 0 When new stress elements n the sphere are btained frm rtating this element abut the ais, the nrmal stresses remain the same and there are n shear stresses. S, ever plane tangent t the sphere is a principal plane, and ever directin a principal directin. The principal stresses are pr / t, 3 0. The maimum (ut-f-plane shear stress is τ ma ( 3 / pr / 4t Draw the three Mhr s circles t cnvince urself f this. 9

20 Cmments n the ther. The wall thickness must be small in cmparisn t the ther dimensins. The rati r/t shuld be 0 r mre.. The internal pressure must eceed the eternal pressure t avid inward buckling. 3. The analsis is based nl n the effects f (net internal pressure. The effects f eternal lads, reactins, the weight f the cntents, and the weight f the structure are nt included. 4. The frmulae derived are valid thrughut the (thin wall f the vessel, ecept near pints f stress cncentratin. 0

21 A Clindrical Pressure essels m n p q b B The thin-walled clindrical tank is subjected t a net internal pressure p. Lngitudinal Stress A m tensile stress L in the walls fluid pressure p Fluid frce in hrintal directin P p (πr n Tensile frce in hrintal directin T L (πr t

22 Fr equilibrium, frces in the hrintal directin must balance. The lngitudinal stress is then: Circumferential (r Hp Stress tensile stress in the walls h fluid pressure p m n b L p r q pr t t Tensile frce T h (bt Fluid frce P p (br Fr equilibrium, frces in the hrintal directin must balance. The circumferential (r hp stress is then: h pr t

23 A m p B b n The nrmal stresses and are principal stresses since n shear stresses are acting. Circumferential (Hp Directin h pr Lngitudinal Directin L t pr t An bvius discntinuit eists at the ends f the clinder, where the ends (usuall plates r hemispheres are attached, because the gemetr f the structure changes abruptl. q 3

24 Stresses at the Outer Surface pr / t pr / t 3 0 τ 0 The maimum (in-plane shear stress is τ ( pr / t ( pr / t ma(, The maimum (ut-f-plane shear stresses are 3 ( pr / t 0 pr τma(,3 verall ma shear stress t 3 ( pr / t 0 pr τma(,3 4t pr 4t 4

25 ample A spherical pressure vessel having 450 mm inside diameter and 6 mm wall thickness is t be cnstructed b welding tgether tw aluminium hemispheres. Frm tests, it is fund that the ultimate and ield stresses in tensin at the weld are 65 MPa and 0 MPa, respectivel. The tank must have a factr f safet f. with respect t the ultimate stress and.5 with respect t the ield stress. What is the maimum permissible pressure in the tank? (Gere and Timshenk, 3 rd ed, p 43 r d/ 0.450/ 0.5 m t m r / t 37.5 (>0, s thin-walled assumptin ka The allwable stress based n the ultimate stress is: allw ult / n 65/ MPa The allwable stress based n the ield stress is: allw / n 0/ MPa The latter is lwer, s it is the mst critical and gverns the design. 5

26 Tensin at the weld fr the spherical vessel is pr / t p t allw / r (0.006( / 0.5 p Pa 3.9 MPa S, the maimum allwable pressure is 3.9 MPa. (Nte that fr safet reasns we have runded dwn here, nt up. 6

27 ample A clindrical pressure vessel is cnstructed with a helical weld that makes an angle f 55 with the lngitudinal ais. The tank has inside radius.8 m and wall thickness 8 mm. The maimum internal pressure is 600 kpa. Find the circumferential and lngitundal stresses, the abslute maimum shear stress, and the nrmal and shear stresses acting perpendicular and parallel t the weld. (Gere and Timshenk, 3 rd ed, p r / t.8/ > 0 (s thin-walled assumptin ka Circumferential and lngitudinal stresses L pr / t ( (.8/( MPa h pr / t 35 MPa 7

28 Abslute maimum shear stress h, L, 3 0 (at the uter surface τ ma ( - 3 / 67.5 MPa Need stresses perpendicular and parallel t the weld. Cnsider the stress element belw and use either the transfrmatin equatins r Mhr s circle with θ 35 (wh nt 55? t find 89.7 MPa,.8 MPa, τ 3.7 MPa