ELASTICITY AND PLASTICITY I

Transcription

1 ELASTICITY AD PLASTICITY I Litertur in Czech Ing. Lenk Lusová LPH 407/1 te enk.usov@vs.cz Literture: Higgeer Sttics nd mechnics of mteri, USA 199 Beer, Johnston, DeWof, zurek echnics of mteris, USA 009, ifth Edition Bend: Stvení sttik I., VŠBTU Ostrv 005 Šmířák: Pružnost psticit I., VUT Brno 1999 Šmířák, Hvinková: Pružnost psticit I, Příkdy, VUT Brno 000 Etern forces, stress Litertur in czech (pdf version) r V r A... orm component of the vector r... Tngenti component of r... Eement of cross section re A norm r σ = im r A 0 A stress (intensity of the intern forces distriuted over given section) V r r A sher r V τ = im r A 0 A r 1

2 Stress Stress: vector, chrcterised y its components Unit: Psc... [P] ormt of the unit: 1. Tension, compression (i od) orm force 0. Bending 3. Torsion 4. Sher P = m The Psc is sm untity, in prctise we use mutipies of this unit m 6 P = 10 P = = 3 k kp = 10 P = m mm R R tensie compression 1. Tension, compression. Bending 3. Torsion 4. Sher Bending moment y, z 0 R z R z compression tension tension compression R z R z 1. Tension, compression. Bending 3. Torsion 4. Sher y 1 z n v = Torsion moment 0 Intern forces:, T (wys), V y, V z, y, z (the memer // with ) V, V z,, z (the memer // with y) V, V y,, y (the memer // with z)

3 1. Tension, compression. Bending 3. Torsion 4. Sher Type of the stress Intern force Stress Sher force V y, V z 0 Ai Loding (tension, compression) σ norm V V V V Bending Sher y, z V y, V z σ norm τ y, τ z sher R z R z Torsion τ y, τ z sher Theorems of superposition nd proportionity Repet!!! Bsic Theorems of Sttics 1) Principe of ction nd rection ) Principe of superposition 3) Principe of proportionity Issc ewton ( ) Types of stresses ) simpe (i oding, ending, torsion, sher) ) comined Comined stresses: gener ending (unsymmetric ending) eccentric i oding torsion comined with tension or compression nd with ending Due to Principe of superposition, which is vid in iner estic rnge, we cn sove the comined stresses. irst y spiting up to sic stresses nd then we cn dd these resuts together. 3

4 Sint Vennt principe of oc effect prt without ffect re of fiure er surroundings Jen Cude SintVennt ( ) kes possie to repce given od y simper one for the purpose of esier ccution the stresses in memer. the stte of stress is infuenced just in ner surroundings of the od frther from this od we hve nery uniform distriution of stress Used for: repcement the surfce od y the od stticy euivent ut simper for soution Sint Vennt principe of oc is not vid in these cses: ) concentrted ods on the end of r: prt without ffect re of fiure re of fiure prt without ffect σ = const. const. ) rs with vrie crosssection re: deduced conditions re vid for rs with grdu chnges of crosssection re. Arupt chnges (nnounce y hoes, nicks or nrrowing) ed to no vidity of condition. d σ Princip terms Stress Strin teri Estic ehvior of mteri Hooke s w Esticity Pstic ehvior of mteri Psticity Ai oding tension, compression The ony one inner force in ech crosssection is n i force. = 0 > 0 tension V V = 0 y = 0 y = z = z R Tension Lod Etern force od (,, ) Temperture od < 0 compression R compression Bsic principes nd condition of soution 4

5 Conditions of soution ) deformted crosssections sty on pne figure nd it is verticy to the is (Bernoui hypothesis) d d Chrcter of condition is deformtiongeometric. Cross sections sty mutuy pre without tpering. efore nd fter deformtion Outcome: y = γ z y z = 0 τ = τ = 0 ) ongitudin fires re not mutuy compressed together γ σ = const. for = const. σ y = σ z Dnie Bernoui ( ) = 0 1. Etern od R ) orm stress under i oding Ais = is of memer i force norm forces norm stress σ (intensity of intern forces distriuted over given section) [P] σ Constnt (Tensie stress positive sign compressive stress negtive sign) ) Strin under i oding deformtion is eongtion or contrction ) Temperture chnges Ai strin (chnges in ength of memer) Lter strin ν 0,5 Poisson s rtio Dimension chnges: eongtion ε = ε = ε = υε y z = y (dimensioness untity) ε y = = h = h h z h h h ε z = h Circe dimeter d? ) stress If there is not defended the deformtion of memer doesn t come up norm force nd stress, ter on indeterminte memers ) Strin (therm strin) ε = ε = ε = α T yt zt T T αt Coefficient of therm epnsion [ C 1 ] = α T. T. T ε T = / = / = h/h = d/d = = h = h h 5

6 At home: Stressstrin digrms concrete stee Stressstrin digrm of n ide esticpstic mteri Derivtion for i od σ norm stress (i) ε norm strin (i) σ σ ε est. ε pst. Yied imit f y σ B rctg E = Y α unoding A=C B Tension f e Esticity imit f y... Yied imit f u... Utimte imit Compression Y ε pst. ε B,cek. f y ε e. = σ B /E ε = / Psticity: the iity of mteri to get permnent deformtions without frcture Ductiity: pstic eongtion of roken r (rnge /OT/), stee 15%). Esticpstic rnge Liner estic rnge σ = ε E. Esticpstic rnge Hooke s w in tension nd compression Hooke s w physic retions etween stress nd strin Hooke s w in sher σ =Ν/Α σ f y Derivtion for i tension ε est. Liner estic rnge Y α = rctg E ε Tension Yied imit σ tnα = = E ε ε = / σ = ε.e By sustituting: σ = A ε... Ai strin [] σ... orm stress [P] ε = = EA Hooke s w Other version of Hooke s w E... Young s moduus of esticity in tension nd compression [P] τ z α τ tnα = = G γ = rctg G γ z E G τ z = γ z G = ( 1 υ) γ z... Ange deformtion τ z... Sher stress [P] G... moduus of esticity in sher [P] 6

7 Empes (it is necessery to keep this rues) Construct the digrm of intern forces () Write gener formu, epress numericy (mke sure to hve the right units of untities) Answer wi e highighted Empe 1 The stee rod (see the picture) hs circe crosssection re of dimeter d = 0,05 m. E=, P. ν =0,3 Determineσ, eongtion of the rod, the ter chnges ( in dimensions) nd determine new dimensions of the rod). (Ignore the ded weight). Resuts: A = 490, m σ = 03,718P = 0,0097m =9, m = 9,7mm ε = 9, = 10,0097m = 10 m R ε y = ε z =, d = 7, m = 7, mm d = 0,0499m P = 100 k R Empe Oceová tyč kruhového průřezu d = 0,01 m déky = m je Determine nmáhán the tot thovou deformtion siou = of15 thek. rod in its ength (see the picture). Určete normáové npětí σ, cekové prodoužení příčné zkácení prutu. E = P, ν = 0,3 R Empe 3 The rod is under the temperture od of 0 C. Determine the force so tht the tot deformtion of the rod is 0. C 1 Intermedite dt: A 1 = 314, m A = 78, m 1 = 1, m = 1,364mm = 1, m = 1,1mm σ 1 = 95,49P σ = 17,33P = αt. T.. E.A.A 1 ( ) derive A. 1 A 1. 7

8 Empe 4 Determine the norm force nd the norm stress σ which cuses eongtion 3mm on the stee rod of the circe sross section (dimeter d = 0,05 m, enght = 3 m, E = P). Empe 5 The cue of concret (dimensions = 0,5 m, = 0,40 m, hight h = 1,0 m) is under the compression y the force = 500 k. Determine the chnges in dimensions of the cue. (The sme prctice s the Empe 1). E cm = P, ν = 0, ( h= 0,8mm, = 4,1.10 mm, = 0,1.mm) ( = 41,334k, σ = 10,00P) (h = 0,999m, = 0,50041m, = 0,400656m) Empe 6 The concret coumn of sure crosssection 0,6 0,6 m nd hight h = 3,6 m is uniformy wrmed y T = 75 C. Determine the chnges in dimensions of the coumn cue. α T = C 1 (h = 3,607m, = 0,6005m, = 0,6005m) Empe 7 Determine the chnges in dimensions nd the stress in the stee nd concrete prt of the coumn (see the picture). P 1 =150k d 1 = 0,03m E 1 = P oce 1 ν 1 = 0,3 P =165k eton P h 1 =0,5m h =0,m = 0,15m (d = 30,009mm, = 150,09mm, = 100,019mm, h 1 =499,495mm, h =199,809mm, h = 699,304mm σ 1 = 1,1P, σ = 3,0P) = 0,1m E =33 500P ν = 0, 8