Resumo - Torção Inelástica - 2023-2

Mecˆanica dos S´olidos 3 Tor¸c˜ao inel´astica Tor¸c˜ao inel´astica r0 T l Se¸c˜ao Circular J = π 2r4 0 +++ T +++ ϕ = T l GJ dϕ dz = ϕ l γ τ τy γy τy = 1 2σy (Tresca) τy = 1 √ 3σy (Von-Mises) Fase el´astica: T < Ty, ϕ < ϕy deforma¸c˜oes angulares γmax= r0 dϕ dz = r0 ϕ l < γy tens˜oes de cisalhamento γmax< γy r0 r0 τmax< τy τmax< τy τmax = Tr0 J < τy ϕ = Tl GJ < ϕy γmax = r0 ϕ l < γy ⊳ 1 ⊲ Mecˆanica dos S´olidos 3 Tor¸c˜ao inel´astica In´ıcio do escoamento: T = Ty, ϕ = ϕy deforma¸c˜oes angulares γmax= r0 dϕ dz = r0 ϕy l = γy tens˜oes de cisalhamento γmax= γy r0 r0 τmax= τy τmax= τy τmax = Tyr0 J = τy =⇒ Ty = π 2τyr3 0 ϕ = Tyl GJ = ϕy =⇒ ϕy = τyl Gr0 γmax = r0 ϕy l = γy Fase inel´astica: Ty < T < Tp, ϕy < ϕ < ∞ deforma¸c˜oes angulares γmax= r0 dϕ dz = r0 ϕ l > γy tens˜oes de cisalhamento γmax> γy r0 r0 re re τmax= τy τmax= τy γy γy γmax = r0 reγy > γy =⇒ ϕ = τyl Gre > ϕy re −→ 0 ∴ γmax −→ ∞ e ϕ −→ ∞ ⊳ 2 ⊲ Mecanica dos Sdélidos 3 Torcao inelastica > Torque inelastico r y yy r y dA = 2rrdr PA T r. r K_| — tt dr T,, T= rely T= {[rrdA A Te "9 T= [rr2nrdr + | rr2nrdr 0 Te re "9 — fz T= J yo%yr2nrdr + J tyr 2ardr e€ _ 2 3(1 — l(tey? P= 307,15 (1 i(*) ) _— 1, »3 _— tyl _ Ty! L, = BT yF o> To = Gey’ Me Gy —_ 4 _ 1¥yy3 l= 32, (1 i() ), P> Py T T= $1, }----------- =o ee I Tb ! | | | l I l I l I | I y ~, 3D, . 4 po .. T3537, <J 3 > Mecanica dos Sdélidos 3 Torcao inelastica Colapso plastico: T’ = T), y = 00 > Torque limite T, "9 dA = 2nrdr T dr\ y "0 _ _ _ 2 3 T,= JrrdA= J 7,r2ardr = 397,r° A 0 ‘Tensoes e deformacoes residuais torque inelastico descarga elastica / | T | T —> <— S S et 4 T 1 _ Tl | P=-G <J 4 > Mecˆanica dos S´olidos 3 Tor¸c˜ao inel´astica ϕ ϕp ϕe ϕ T T descarga el´astica ϕ = ϕp + ϕe tens˜oes inel´asticas tens˜oes el´asticas tens˜oes residuais τy τy T T τ e max= 2T πr3 0 τ e max= 2T πr3 0 +++ === deforma¸c˜oes inel´asticas deforma¸c˜oes el´asticas deforma¸c˜oes residuais γmax= r0 re γy > γy γmax γe max γe max= r0 ϕe l γp max γp max +++ === ϕ = τyl Gre > ϕy ϕe = T l GJ ϕp = ϕ − ϕe ⊳ 5 ⊲ Mecˆanica dos S´olidos 3 Tor¸c˜ao inel´astica Dadas as seguintes rela¸c˜oes constitutivas, mostradas abaixo, deter- mine o m´aximo torque que pode ser aplicado ao eixo cil´ındrico tal que a m´axima deforma¸c˜ao angular n˜ao exceda γmax ≤ 1×10 −3 rad. r0 T l Se¸c˜ao Circular J = π 2r4 0 r0 = 0.01 m G = 82 GPa 0.5×10 −3 1×10 −3 70 MPa 1×10 −3 70 MPa τ γ τ γ 1 G τ = Cγ 1 2 Caso A Caso B Caso A deforma¸c˜oes angulares γmax= 1×10 −3 tens˜oes de cisalhamento γmax> γy r0 r0 re re τmax= 70 MPa τmax= 70 MPa γy= 0,5×10 −3 γy n´ucleo el´astico: γmax r0 = γy re , re = γy γmaxr0, re = 0,5×10 −3 1×10−3 0,01 re = 5×10 −3 ⊳ 6 ⊲ Mecanica dos Sdélidos 3 Torcao inelastica [> Momento torcor maximo 5x10” x10 __ r T= J (7,2 )r2nrdr + |7,r2ardr 0 5x10° 5x10” 3 1x10 ) __ r T= J 2a7,—dr + f2a7,r dr 0 5x10? —3 —2 4 5x10 3 1x10 _—_ r r _—_ T =2nr, Ei +3 | = 142,0262 N.m 0 5x10? Caso B deformagoes angulares tensoes de cisalhamento y= 1x10 T,= 70 MPa Vmax T na 70 MPa 7 1 ff x10” mar = 2, -y = Tat, y= py = 0.1r z 3 3 relagao constitutiva: T= Cy, 7T=C(0.1r)°, 7 = V0.1Cr 3 V0.1 70x 10° T=kr°, k=VJ01C, C= V1x10° -1 6 7 7 4 k= 4/24, 70x10’ = 70x10, 7 = 70x10'r? 1x10 <J 7 > Mecanica dos Sdélidos 3 Torcao inelastica [> Momento torcor maximo —~2 —2 =2 1x10 1x10 4 1x105 T = [ rr2ardr = f[ kr? r2ardr = 2k f[ r7dr 0 0 0 -2 -2 1x10 1x10 5 4d T = 2rk C | = Snkr = 125,6637 N.m 0 0 O eixo cilindrico mostrado na figura esta submetido a um torque de 8000 N.m. Calcule as tensoes de cisalhamento residuais, as deformagoes angulares residuais e a rotagao y,, residual ao retirar este torque. Secao Circular B aa NN {= 1.5m J= ar y(z) = 7Ypa yMH= 0.03 m -— Pra G = 80 GPa T, = 150 MPa dp _ Ppa dz _ « 3 _ 7 6 3 T, = a7" 6 = 5x 15010 «0.03 = 6361, 7251 N.m 2 3 2 6 3 I, = 3171) = 37 x 150x10 «0.03 = 8482, 3002 N.m T,<T <T, = Torgao inelastica dq 8 > Mecanica dos Sdélidos 3 Torcao inelastica e Calculo do nucleo elastico r, _ 4 1/Te\3 P= 47,(1—38)) BT _ y 1h aT, A\r) aT _ 4 (t2)° Ty "0 rey — 4 9T (*) Ly 1 r\3 r= 1, (4 — 34) y 3 _ _3x8000_)\" r= 0.08(4— ee) = 0.0183 m e Deformacao de cisalhamento residual r, _ Pe _ _2T Ine Vy > Vy naz "oT = FG ye YL -? OL Vina Vox Va 0.03. 150x10° -3 Vmax = 0.0183" “sp. 107 3, 072x10 e — __2x8000 9. 358,10 * Vmax ~ 7380x109 «0.088 —3 —3 —4 VP = Vinax ~ Vonn = 3, 072x10 — — 2, 358x10 © = 7, 14x10 dq 9 > Mecˆanica dos S´olidos 3 Tor¸c˜ao inel´astica • Rota¸c˜ao ϕBA residual γmax = r0 reγy, γmax = r0 ϕBA l , γy = τy G ⇒ ϕBA = τyl Gre ϕBA = τyl Gre = 150×106×1.5 80×109×0.0183 = 0.1536 rad ≈ 8.8 o ϕe BA = Tl GJ = 8000×1.5 80×109×π 2 ×0.034 = 0.1179 rad ≈ 6.8 o ϕp BA = ϕBA − ϕe BA = 0.1536 − 0.1175 = 0.0357 rad ≈ 2 o • Tens˜ao de cisalhamento residual tens˜oes inel´asticas tens˜oes el´asticas tens˜oes residuais τy=150MPa τy T T τ e max= 2T πr3 0 = 2×8000 π×0.033 =188.63MPa 38.63MPa 34.94MPa τ e max= 2T πr3 0 τ = T re J = 8000×0.0183 π 2 ×0.034 = 115.06MPa +++ === τ τ ⊳ 10 ⊲