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Mecˆanica dos S´olidos 3 Flex˜ao inel´astica Flex˜ao inel´astica b h l/2 l/2 P I = bh 3 12 κ = M EI = PL 4EI σt y = σc y = σy εt y = εc y = εy PL 4 σ ε εt y σt y σc y εt y = σt y E εc y εc y = σc y E Fase el´astica: M < My, κ < κy z y b h 2 h 2 h 2 εc max< εc y εt max< εt y σc max< σc y σt max< σt y σ = My I ε = κy κ = M EI ⊳ 1 ⊲ Mecˆanica dos S´olidos 3 Flex˜ao inel´astica In´ıcio do escoamento: M = My, κ = κy z y b h 2 h 2 h 2 εc max= εc y εt max= εt y σc max= σc y σt max= σt y My = bh2 6 σy κy = 2εy h Fase inel´astica: My < M < Mp, κy < κ < ∞ ye ye z y b h 2 h 2 h 2 εc max> εc y εt max> εt y σc max= σc y σt max= σt y εc y εt y εc max = h 2yeεc y > εc y, εt max = h 2yeεt y > εt y κ = εy ye ye −→ 0 ∴ εmax −→ ∞ e κ −→ ∞ y ye σy σ σ = y yeσy ⊳ 2 ⊲ MecAanica dos Sélidos 3 Flexéo inelastica t> Momento inelastico h Ye 2 M= foydA= 2(f oybdy + J oybdy) h Ye 2 M= 2(J 0 ybdy + J'o,ybdy) Ye 2 2 _ bh 1 (Ye M = oe |1— $(Hs) | 2 __ bh h _ & _ Ee M, = -G—-%y: 2 = hy Ye = 2 _ 3 1( Ve M= 3M, |1 — (#5) 2 3 1(* M= ba, |1 — 4(%) |, KK, M M,= 3M,+------------------ 22-52-22 g eee a ! K, AK, " K—0oo 1, M—>3M, <J 3 > Mecanica dos Sdélidos 3 Flexao inelastica Colapso plastico [> Momento fletor limite Y © — 00 Onan Oy ! 2 Zz — Ye - ee ! 2 i 3 2 M = dA| = bdy| = bh p=2 oy =2 a, yody =e A 0 ‘Tensoes e deformacoes residuais carga descarga elastica ‘| ‘| 1/2 (/20 1/2 //2 IK >| I> }—_————| u M} -------------2sss—7 oP ! s | Ry ! s > | I kK K, K,, K <] 4 > Mecˆanica dos S´olidos 3 Flex˜ao inel´astica ye ye z y b h 2 h 2 h 2 εc max= h 2yeεc y εt max= h 2yeεt y εc y εt y deforma¸c˜oes inel´asticas deforma¸c˜oes el´asticas deforma¸c˜oes residuais εc emax= 6M Ebh2 εt emax= 6M Ebh2 εc pmax εt pmax +++ === ye ye z y b h 2 h 2 h 2 σc max= σc y σt max= σt y tens˜oes inel´asticas tens˜oes el´asticas tens˜oes residuais σt emax= 6M bh2 σc emax= 6M bh2 +++ === ⊳ 5 ⊲ Mecˆanica dos S´olidos 3 Flex˜ao inel´astica Comportamento diferente `a tra¸c˜ao e a compress˜ao 0.05 m 0.1 m σ (MPa) ε 70 140 1, 33×10 −3 6, 67×10 −4 E c=210GPa E t=52, 5GPa ⊲ Momento fletor que inicia o escoamento Hip´otese: O escoamento inicia-se na zona de compress˜ao z y b y=? h − y εc max= εc y εt max≤ εt y σc max=140 σt max≤70 2 3y 2 3(h − y) R t R c 2 3h posi¸c˜ao da linha neutra: R c = R t 1 2byσc y = 1 2b(h − y)σt max ∴ σt max σcy = y h−y εc y y = εt max h−y , εc y = σc y Ec, εt max= σt max Et ∴ σt max σcy = Et Ec h−y y , Et Ec = 1 4 1 4 h−y y = y h−y =⇒ 3y2 + 2hy − h2 = 0 ∴ y1 = 1 3h, y2 = −h ⊳ 6 ⊲ Mecˆanica dos S´olidos 3 Flex˜ao inel´astica ⊲ Momento fletor de escoamento My = R c ×2 3h, R c = 1 2σc y 1 3hb = 1 6σc y bh My = 1 6σc y bh×2 3h = 1 9σc y bh 2 = 1 9×140×10 6 ×0,05×0,1 2 = 7778 N.m ⊲ Momento fletor limite z y b y=? h − y 70 140 1 2y 1 2(h − y) 1 2h R t R c posi¸c˜ao da linha neutra: R c = R t 140yb = 70(h − y)b y = 1 2(h − y) ∴ y = 1 3h Mp = R c ×1 2h, R c = σc y ×1 3hb = 1 3σc y bh Mp = 1 3σc y bh×1 2h = 1 6σc y bh 2 = 1 6×140×10 6 ×0,05×0,1 2 = 11667N.m ⊳ 7 ⊲ Mecˆanica dos S´olidos 3 Flex˜ao inel´astica Comportamento diferente `a tra¸c˜ao e a compress˜ao b h σ (MPa) ε 200 150 2×10 −3 1×10 −3 E c=150GPa E t=100GPa ⊲ Momento fletor que inicia o escoamento Hip´otese: O escoamento inicia-se na zona de compress˜ao z y b y=? h − y εc max= εc y εt max≤ εt y σc max=150 σt max≤200 2 3y 2 3(h − y) R t R c 2 3h posi¸c˜ao da linha neutra: R c = R t 1 2byσc y = 1 2b(h − y)σt max ∴ σt max σcy = y h−y εc y y = εt max h−y , εc y = σc y Ec, εt max= σt max Et ∴ σt max σcy = Et Ec h−y y , Et Ec = 2 3 2 3 h−y y = y h−y =⇒ y2 + 4hy − 2h2 = 0 ∴ y1 = ( √ 6 − 2)h, y2 = −( √ 6 + 2)h ⊳ 8 ⊲ Mecˆanica dos S´olidos 3 Flex˜ao inel´astica ⊲ Momento fletor de escoamento My = R c ×2 3h, R c = 1 2σc y ( √ 6 − 2)hb = ( √ 6−2) 2 σc y bh My = ( √ 6−2) 2 σc y bh×2 3h = ( √ 6−2) 3 σc y bh 2 My = ( √ 6−2) 3 ×150×10 6bh 2 = 22474,4871bh 2 kN.m ⊲ Momento fletor que inicia o escoamento na zona tracionada z y b y=? h − y εc max> εc y εt max= εt y σc max=150 σt max=200 2 3 ¯y 1 2(y + ¯y) 2 3(h − y) R t R c 1 R c 2 ¯y=? εc y posi¸c˜ao da linha neutra: R c 1 + R c 2 = R t σc y (y − ¯y)b + 1 2σc y ¯yb = 1 2σt y (h − y)b ∴ σt y σcy = 2y−¯y h−y ∴ 4 3 = 2y−¯y h−y εc y ¯y = εt y h−y ∴ εc y εty = ¯y h−y =⇒ 1 2 = ¯y h−y ∴ ¯y = 1 2(h − y) y = 11 23h, ¯y = 6 23h y1 = 1 2(y + ¯y) = 1 2(11 23 + 6 23)h = 17 46h y2 = 2 3 ¯y = 2 3× 6 23h = 4 23h y3 = 2 3(h − y) = 2 3(1 − 11 23)h = 2 3×12 23h = 8 23h ⊳ 9 ⊲ Mecˆanica dos S´olidos 3 Flex˜ao inel´astica M = R c 1×y1 + R c 2×y2 + R t ×y3 > My R c 1 = σc y (y − ¯y)b = σc y (11 23 − 6 23)bh = 5 23σc y bh R c 2 = 1 2σc y ¯yb = 1 2× 6 23σc y bh = 3 23σc y bh R t = 1 2σt y (h − y)b = 1 2(1 − 11 23)σt y bh = 1 2×12 23σt y bh = 6 23σt y bh M = 5 23×17 46σc y bh 2 + 3 23× 4 23σc y bh 2 + 6 23× 8 23σt y bh 2 M = 109 1058σc y bh 2 + 48 529σt y bh 2 = 33601,1342 bh 2 kN.m ⊲ Momento fletor para um n´ucleo el´astico de altura total de 2 5h z y b y=? h − y εc max> εc y εt max> εt y σc max=150 σt max=200 2 3 ¯y 1 2(y + ¯y) 7 10h − 1 2(y + ¯y) R t 1 R t 2 R c 1 R c 2 ¯y=? εc y y4 y4 = 2 3(2 5h − ¯y) 2 5h − ¯y εt y Posi¸c˜ao do n´ucleo el´astico em rela¸c˜ao a linha neutra εc y ¯y = εt y 2 5h−¯y, εc y εty = ¯y 2 5h−¯y, 1 2 = ¯y 2 5h−¯y ∴ ¯y = 2 15h y1 = 1 2(y + ¯y) = 1 2(y + 2 15h) y2 = 2 3 ¯y = 4 45h y3 = 7 10h − 1 2(y + ¯y) = 7 10h − 1 2(y + 2 15h) = 19 30h − 1 2y y4 = 2 3(2 5h − ¯y) = 2 3(2 5h − 2 15h) = 8 45h ⊳ 10 ⊲ Mecanica dos Soélidos 3 Flexao ineldstica Posigao da linha neutra Cc Cc t t R + R, = R, + Rf, o'(y — g)b+ doy = 4o(2h — Wb +a'((h—y) — 2h—G))b o°(2y —Y) = ai(th — 2y + 9) (2y —y) = 4(8h—2y +9) Sy= Bht+ty yah ¥ = S(y+ eh) =5(Ght+ Sh) = sh _ 19 17, _ 19 11, _ 13 Y3 = 39 — gh = 39h — Bh = zh OF e150 pe 41/105)h <— . ‘hoy alas (23/70)h (4/15)h — (8/45) 13/35)h (22/105)h —> R, o' = 200 Cc Cc t t M= Rxy, + Ryxy, + BR, xy, + Roxy, ec Al R = 1057, bh | R, = 157, bh é 2 4 R, = 52, bh tf 22 + R, = ips?, bh _— Al 23 2,1 4 2 2 8 t272 , 22 13 4 27? M = T05*707, bh + 15*a57, bh + 15% 45° y bh + 105 ~ 350 y bh M = 40439,1534bh’ kN.m <J 11 > Mecˆanica dos S´olidos 3 Flex˜ao inel´astica ⊲ Momento fletor limite z y b y=? h − y σc max=150 σt max=200 R t R c 1 2y 1 2(h−y) h 2 posi¸c˜ao da linha neutra: R c = R t σc yyb = σt y(h − y)b, σt y σcy = y h−y, 4 3 = y h−y ∴ y = 4 7h Mp = R c ×h 2 = σc y 4 7hb×h 2 = 2 7 σc y bh 2 Mp = 42857,1429 bh 2 kN.m ⊳ 12 ⊲ Mecˆanica dos S´olidos 3 Flex˜ao inel´astica O material de uma viga apresenta propriedades lineares el´asticas na tra¸c˜ao e r´ıgido pl´asticas na compress˜ao. Se o n´ıvel m´aximo de tens˜ao em tra¸c˜ao e compress˜ao ´e de σ0, determine o momento fletor Mmax suportado pela viga. Calcule as deforma¸c˜oes m´aximas na tra¸c˜ao e compress˜ao e a curvatura. b h σ σ0 σ0 ε ε0= σ0 E Z.C. Z.T. E 1 z y b y=? h − y εc max εt max σ0 σ0 R t R c y 2 2 3(h − y) posi¸c˜ao da linha neutra: R c = R t σ0yb = 1 2σ0(h − y)b ∴ y = 1 3h R c = σ0yb = 1 3σ0bh, R t = 1 2σ0(h − y)b = 1 3σ0bh Mp = R c ×y 2 + R t ×2 3(h − y) = 1 3σ0bh×1 6h + 1 3σ0bh×4 9h = 11 54 σ0bh 2 curvatura: εt max= κ×(h − y), κ = εt max h−y , εt max= σ0 E ∴ κ = 3σ0 2Eh εc max y = εt max h−y , εc max 1 3h = εt max 2 3h ∴ εc max= 1 2 σ0 E ⊳ 13 ⊲ Mecˆanica dos S´olidos 3 Flex˜ao inel´astica Repita o exerc´ıcio anterior adotando uma se¸c˜ao transversal T como mostra a figura abaixo 10e e 5e e σ σ0 σ0 ε ε0= σ0 E Z.C. Z.T. E 1 10e e 5e e y 11e−y y−e εc max ε0 σ0 σ0 R t R c 2 R c 1 1 2(y−e) 2 3(11e−y) y−e 2 posi¸c˜ao da linha neutra: R c = R t σ0e5e + σ0(y − e)e = 1 2σ0(11e − y)e ∴ y = e R c = 5σ0e2, R t = 1 2σ0(11e − y)b = 5σ0e2 Mp = R c ×e 2 + R t ×2 3(11e − y) = 5σ0e2×1 2e + 5σ0e2×20 3 e = 215 6 σ0e3 curvatura: εt max= κ×(11e − y), κ = εt max 11e−y, εt max= σ0 E ∴ κ = σ0 10Ee εc max y = εt max 11e−y, εc max e = εt max 10e ∴ εc max= σ0 10E ⊳ 14 ⊲ Mecˆanica dos S´olidos 3 Flex˜ao inel´astica Seja o modelo constitutivo dado na figura abaixo, com 2σ0 sendo a resistˆencia `a ruptura a tra¸c˜ao. Determine o momento fletor de ruptura para uma se¸c˜ao retangular b h σ 2σ0 ε ε0= 2σ0 E Z.C. Z.T. E 1 3 2E 1 z y b y=? h − y εc max εt max σc max 2σ0 R t R c 2 3y 2 3(h − y) posi¸c˜ao da linha neutra: R c = R t 1 2σc maxyb = 1 22σ0(h − y)b ∴ σc max σ0 = 2(h−y) y εc max y = εt max h−y , εc max= σc max 3 2E = 2σc max 3E , εt max= 2σ0 E ∴ σc max σ0 = 3y h−y 2(h−y) y = 3y h−y =⇒ y2 + 4hy − 2h2 = 0 y1 = (−2+ √ 6), y2 = −(2+ √ 6)h =⇒ y1 = 0,45h σc max = −6+3 √ 6 3− √ 6 σ0 = 2,45σ0 ⊳ 15 ⊲ Mecˆanica dos S´olidos 3 Flex˜ao inel´astica ⊲ Momento fletor de ruptura Mr = R t ×2 3h R t = 1 2×2σ0(h − y)b = σ0(h − y)b = (3 − √ 6)σ0bh Mr = (3 − √ 6)σ0bh×2 3h = 2(3− √ 6) 3 σ0bh 2 Mr = 0,367σ0bh 2 Aplica-se uma carga P =20 KN na viga mostrada na figura. Cal- cule a curvatura residual, as tens˜oes e deforma¸c˜oes residuais na se¸c˜ao mais solicitada ap´os a retirada desta carga. 2m 2m 20 KN carga 2m 2m 20 KN descarga el´astica P L 4 = 20KN.m P L 4 = 20KN.m κ M 20 KN.m descarga el´astica κp κe κ = 3.39×10 −2 ⊳ 16 ⊲ MecAanica dos Sélidos 3 Flexao ineldstica . E = 200 GPa e Secao transversal o,= 240 MPa 0.1m M = 20 KN.m T = 3.33.10 ° m! 0.04m 1 2 1 2 6 1 2 1 2 6 M, <M< M, = Flexao inelastica ! e Calculo do nucleo elastico 2 — 8 1 [ Ye m= 3M,(1~3(#)) 2 2M _ 1 _ (4) 3 My 3 h 2 Ye) — 3 9M (4) = 3-24 =f /3— 9M — 001) [3 _ 9,20 — 9.0354 m Ye 9 M, 2 <16 deformacoes deformacoes deformacoes y inelasticas elasticas residuais oe =e -¢ — 6M > maz Qyo~y mat Boz ova | a z ------- - —~---+------ ---=---- h Ye b C= ae oO pone Crna <J 17 > Mecˆanica dos S´olidos 3 Flex˜ao inel´astica • deforma¸c˜oes residuais εp max = εmax − εe max εp max = 0.1 2×0.0354×240×106 200×109 − 6×20×103 200×109×0.04×0.12 εp max = 1.76×10 −3 − 1.5×10 −3 = 2.6×10 −4 • curvatura residual: κ = εy ye e κe = M EI κp = κ − κe κp = (240×106/200×109) 0.0354 − 20000 200×109×3.33×10−6 κp = 0.0339 − 0.03 = 3.9×10 −3 m−1 • tens˜oes residuais ye ye z y b h 2 h 2 h 2 240 MPa 240 MPa tens˜oes inel´asticas tens˜oes el´asticas tens˜oes residuais 300 MPa 300 MPa 212.4MPa 212.4MPa 60 MPa 60 MPa 27.6 MPa 27.6 MPa +++ === σe max = 6M bh2 = 6×20000 0.04×0.12 = 300 MPa σe ye = Mye I = 20000×0.0354 3.33×10−6 = 212.4 MPa ⊳ 18 ⊲ Mecˆanica dos S´olidos 3 Flex˜ao inel´astica Seja uma se¸c˜ao transversal retangular de concreto armado como mostra a figura abaixo. Assume-se a resistˆencia a compress˜ao do concreto de fck= 25MPa, a tens˜ao de escoamento do a¸co de fy = 500MPa e o m´odulo de elasticidade do a¸co de E = 210GPa. H´a duas barras de a¸co na zona tracionada. O diˆametro de cada barra ´e de φ = 2.86cm. Determine o m´aximo momento fletor que a se¸c˜ao suporta, a curvatura e a m´axima deforma¸c˜ao na zona comprimida. b b 55 cm 25 cm 0.025 m h−y−0.025 y εy εc max 0.85fck 0.5y R c R t h−0.25y−0.025 posi¸c˜ao da linha neutra: R c = R t 0.85fck0.5yb = 2fyA y = 2fyA 0.85fck0.5b = 2×500×π φ2 4 0.5×0.85×25×b y = 2×500×π 0.02862 4 0.5×0.85×25×0.25 = 0.24185m R c = 0.5×0.24185×0.85×25×10 6 ×0.25 = 642.42kN R t = 2×500×10 6 ×π 0.0286 2 4 = 642.42kN momento fletor de ruptura M = R c ×(h−0.25y−0.025) M = 642.42×(0.55 − 0.025 − 0.25×0.24185) = 298.43kN.m ⊳ 19 ⊲ Mecˆanica dos S´olidos 3 Flex˜ao inel´astica m´axima deforma¸c˜ao de compress˜ao εc max y = εy h−y−0.025 εc max = y h−y−0.025εy = y h−y−0.025 fy E εc max = 0.24185 0.55−0.24185−0.025×500×10 6 210×109 = 2.34×10 −3 curvatura εy = κ(h − y − 0.025) κ = εy h−y−0.025 = fy E h−y−0.025 κ = 500×106 210×109 0.55−0.24185−0.025 = 8.41×10 −3m−1 ⊳ 20 ⊲