<content> Utilidades Fatiga </content>

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&[DATE] &[TIME] - &[FILENAME]

<content> Utilidades Fatiga </content> <content> Utilidades Fatiga </content>

x a n fsn v x a - (n^x a > (* n 0 ≥ (* : v 0 v 0 ≈ v cases : 21 line :

<content> Cálculos Fatiga </content>

Cálculo del límite de resistencia a la fatiga sin corregirpara aceros y hierros.Sut: Valor de límite de tracción en ksi o MPaMaterial: "steel", "iron", "aluminum" o "cooper alloy" Sut material S.e' s 0.5 Sut * Sut 200 ksi * < 100 ksi * cases : material steel ≡ s 0.4 Sut * Sut 60 ksi * < 24 ksi * cases : material iron ≡ s 0.4 Sut * Sut 60 ksi * < 19 ksi * cases : s s at 5*10⁸ cicles 12 mat : 21 line material aluminum ≡ s 0.4 Sut * Sut 60 ksi * < 14 ksi * cases : s s at 5*10⁸ cicles 12 mat : 21 line material cooper alloy ≡ s no definido : cases s 21 line :

tipoCarga C.load 1 tipoCarga bend ≡ (tipoCarga torsion ≡ (| 0.7 cases :

d C.size 1 d 8 mm * ≤ 1.189 d 1 mm * / (0.097 -^* d 8 mm * > (d 250 mm * ≤ (& 0.6 cases :

d type d.equiv A.95 0.0766 d 2^* type rotatory ≡ 0.010462 d 2^* cases : A.95 0.0766 / sqrt 21 line :

b h type d.equiv A.95 0.05 b * h * : A.95 0.0766 / sqrt 21 line :

Sut finish C.surf A 1.58 4.51 57.7 27214 mat : transpose b 0.085 - 0.265 - 0.718 - 0.995 - 14 mat : transpose id 1 finish ground ≡ 2 finish machined ≡ (finish cold-rolled ≡ (| 3 finish hot-rolled ≡ 4 finish as forged ≡ no definido cases : c A id el Sut 1 MPa * / (b id el^* : c 1 > c 1 : c if 51 line :

T C.temp 1 T 450 °C * ≤ 1 0.0058 T °C / 450 - (* - T 450 °C * > (T 550 °C * ≤ (& no definido cases :

Neuber.steels 344.74 MPa * 0.13 379.21 MPa * 0.118 413.69 MPa * 0.108 482.63 MPa * 0.093 551.58 MPa * 0.08 620.53 MPa * 0.07 689.48 MPa * 0.062 758.42 MPa * 0.055 827.37 MPa * 0.049 92 mat :

Neuber.AnnealedAluminum 68.948 MPa * 0.5 103.42 MPa * 0.341 137.9 MPa * 0.264 172.37 MPa * 0.217 206.84 MPa * 0.18 241.32 MPa * 0.152 275.79 MPa * 0.126 310.26 MPa * 0.111 82 mat :

r C.conf conf 50909599 99.9 99.99 99.999 99.9999 81 mat : C 1 0.897 0.868 0.814 0.753 0.702 0.659 0.620 81 mat : conf C r ainterp 31 line :

Neuber.HardenedAluminum 103.422 MPa * 0.475 137.896 MPa * 0.38 206.844 MPa * 0.278 275.792 MPa * 0.219 344.74 MPa * 0.186 413.688 MPa * 0.162 482.636 MPa * 0.144 551.584 MPa * 0.131 620.532 MPa * 0.122 92 mat :

material s_ut r# q dat Neuber.steels material steel ≡ Neuber.AnnealedAluminum material AnnAlum ≡ Neuber.HardenedAluminum material HardAlum ≡ no definido cases : √a dat 1 col MPa / dat 2 col s_ut MPa / ainterp : 11 √a r# in / sqrt / + / 31 line :

<content> Parámetros para gráfica SN </content>

Inserta una Gráfica tipo Zed Graph, luego, sobre ella: click derecho - Format...
en el campo "Name" cambia el valor zGraph por SN

nm SN :

xa {nm}.XAxis :

ya {nm}.YAxis :

tl {nm}.Title :

{xa}.Scale.Max 109^setprop 0

{ya}.Scale.Max 103^setprop 0

{xa}.Scale.Min 1 setprop 0

{ya}.Scale.Min 1 setprop 0

{xa}.Scale.Min 1 setprop 0

{ya}.Scale.Min 1 setprop 0

{xa}.Type Log setprop 0

{ya}.Type Log setprop 0

{xa}.Title.Text N[Ciclos] setprop 0

{ya}.Title.Text S[MPa] setprop 0

{tl}.IsVisible True setprop 0

{tl}.Text Diagrama S-N setprop 0

sc 106 -^:

<content> Parámetros para la línea de Goodman </content>

Inserta una Gráfica tipo X-Y Plot, luego, sobre ella: click derecho - Format...
en el campo "Name" cambia el valor "XYPlot" por "Goodman" (sin comillas)

s_ut s_y s_f ZonaGoodman Parámetros Gráfica nm Goodman : xa {nm}.XAxis : ya {nm}.YAxis : {nm}.Title.Text Línea de Goodman Modificada setprop {xa}.Max s_ut MPa / setprop {xa}.Min 0 setprop {xa}.Tick 100 setprop {nm}.XYLabel.XLabel σm setprop {ya}.Max s_f MPa / setprop {ya}.min 0 setprop {ya}.Tick 100 setprop {nm}.XYLabel.YLabel σa setprop 131 line x2 s_ut / y2 s_f / + 1 ≡ x2 s_y / y2 s_y / + 1 ≡ 21 sys x2 y2 21 sys Solve Assign polygon 000 s_f MPa / x2 MPa / abs y2 MPa / abs s_y MPa / 042 mat blue solid 1 #200000FF 61 mat 11 mat 11 mat 31 line :

<content> Cálculos fs </content>

Se calcula el fs para los 4 casos de carga:1)σa constante y σm variable2)σa variable y σm constante3)σa y σm varían en proporción constante4)σa y σm varían libremente.Para el caso 4 se usa la distancia del punto (σm,σa) hasta la línea de Goodman y la distancia de (0,0) hasta (σm,σa) s_ut s_y σ_a σ_m s_f fs Se evalúa si σa está por encima o por debajo del punto de intersección de la línea Sy-Sy con Sf-Sut σ_m 0 ≡ f1 no definido : s_y 1 σ_m s_y / - (* s_f 1 σ_m s_ut / - (* > f1 s_ut σ_m / 1 σ_a s_f / - (* : f1 s_y σ_m / 1 σ_a s_y / - (* : if 11 line if Aquí se calcula el punto de coordenadas (x1,y1) más cercano desde (σm,σa) hasta la línea de Goodman d σ_m s_ut / σ_a s_f / + 1 - abs 1 s_ut / (2^1 s_f / (2^+ sqrt / : x1 s_ut / y1 s_f / + 1 ≡ x1 σ_m - (2^y1 σ_a - (2^+ d 2^≡ 21 sys x1 y1 21 sys Solve Assign f1 s_f σ_a / 1 σ_m s_ut / - (* s_f s_ut * σ_a s_ut * σ_m s_f * + / σ_m 2^σ_a 2^+ sqrt d + σ_m 2^σ_a 2^+ sqrt / datos d x1 y1 24 mat 81 line :

Geometric Stress-Concentration Factor Kt for a Shaft with a Shoulder Fillet in Bending Dd rd Kt_C2 D_d 6.00 3.00 2.00 1.50 1.20 1.10 1.07 1.05 1.03 1.02 1.01 111 mat transpose : A 0.87868 0.89334 0.90879 0.93836 0.97098 0.95120 0.97527 0.98137 0.98061 0.96048 0.91938 111 mat transpose : b 0.33243 - 0.30860 - 0.28598 - 0.25759 - 0.21796 - 0.23757 - 0.20958 - 0.19653 - 0.18381 - 0.17711 - 0.17032 - 111 mat transpose : A# D_d A Dd ainterp : b# D_d b Dd ainterp : A# rd (b#^* 61 line :

Geometric Stress-Concentration Factor Kt for a Shaft with a Shoulder Fillet in Torsion Dd rd Kt_C3 D_d 2.00 1.33 1.20 1.09 14 mat transpose : A 0.86331 0.84897 0.83425 0.90337 14 mat transpose : b 0.23865 - 0.23161 - 0.21649 - 0.12692 - 14 mat transpose : A# D_d A Dd ainterp : b# D_d b Dd ainterp : A# rd (b#^* 61 line :

Geometric Stress-Concentration Factor Kt for a Filleted Flat Bar in Axial Tension Dd rd Kt_C9 D_d 2.00 1.50 1.30 1.20 1.15 1.10 1.07 1.05 1.02 1.01 110 mat transpose : A 1.09960 1.07690 1.05440 1.03510 1.01420 1.01300 1.01450 1.02610 1.02590 0.97662 110 mat transpose : b 0.32077 - 0.29558 - 0.27021 - 0.25084 - 0.23935 - 0.21535 - 0.19366 - 0.17085 - 0.16978 - 0.10656 - 110 mat transpose : A# D_d A Dd ainterp : b# D_d b Dd ainterp : A# rd (b#^* 61 line :

Geometric Stress-Concentration Factor Kt for a Filleted Flat Bar in Bending Dd rd Kt_C10 D_d 6.00 3.00 2.00 1.30 1.20 1.10 1.07 1.05 1.03 1.02 1.01 111 mat transpose : A 0.89579 0.90720 0.93232 0.95880 0.99590 1.01650 1.01990 1.02260 1.01660 0.99528 0.96689 111 mat transpose : b 0.35847 - 0.33333 - 0.30304 - 0.27269 - 0.23829 - 0.21548 - 0.20333 - 0.19156 - 0.17802 - 0.17013 - 0.15417 - 111 mat transpose : A# D_d A Dd ainterp : b# D_d b Dd ainterp : A# rd (b#^* 61 line :

Geometric Stress-Concentration Factor Kt for a Flat Bar with Transverse Hole in Axial Tension d_W Kt_C13 d_W 0.65 ≤ 3.0039 3.753 d_W * - 7.9735 d_W 2^* + 9.2659 d_W 3^* - 1.8145 d_W 4^* + 2.9684 d_W 5^* + no definido if 11 line :

Stress concentration factors Kt for bending of a shaft of circular cross sectionwith a semicircularend keyseat (based on data of Fessler et al. 1969a,b). r_d d Kt_keywayFlex KtB 1.426 0.1643 0.1 r_d / (* + 0.0019 0.1 r_d / (2^* - r_d 0.005 0.04 lele (d 6.5 in * ≤ (& 1.426 0.1643 0.1 0.0208 / (* + 0.0019 0.1 0.0208 / (2^* - Con este valor
se sugiere usar
r/d=0.0208 21 mat d 6.5 in * > no definido cases : 1.6 KtB Max 21 line :

Stress concentration factors Kt , Kts for a torsion shaft with a semicircular end keyseat (Leven1949; Okubo et al. 1968). r_d Kt_keywayTorsion KtB 1.953 0.1434 0.1 r_d / (* + 0.0021 0.1 r_d / (2^* - r_d 0.005 0.07 lele (no definido cases : 3.4 KtB Max 21 line :

EXEMPLO 6-1

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Determinação de diagramas S-N estimados para materiais ferrosos

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Problema

Construa um diagrama S-N estimado para uma barra de aço e defina suas equações. Quantos ciclos de vida podem ser esperados se a tensão alternada é de 100 MPa?

Dados

O S_ut obtido experimentalmente é 600 MPa. A barra quadrada tem 150 mm de lado e tem acabamento superficial de laminado a quente. A temperatura máxima de operação é de 500°C. O carregamento aplicado é flexão pura alternada.

Hipóteses

A vida infinita é requerida e pode ser obtida, pois este aço dúctil apresentará limite de fadiga. Será considerado um fator para confiabilidade de 99,9%.

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S.ut 600 MPa * : σ.max 100 MPa * : σ.min 100 MPa * : 13 mat

Solução:

1) Como nenhuma informação sobre o limite de fadiga ou resistência à fadiga é fornecido, estima-se S_e'com base no limite de ruptura usando a Equação 6.5a.

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(a) S.e' 0.5 S.ut * : MPa 300

3) A peça é maior que o corpo de prova e não tem seção circular, portanto um diâmetro equivalente baseado em 95% da tensão máxima (A₉₅) deve ser determinado e usado para encontrar o fator de tamanho. Para uma seção retangular sob flexão sem rotação, a área A₉₅ é definida na Figura 6.25c e o diâmetro equivalente é obtido por meio da Equação 6.7d:

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b 150 mm * :

h 150 mm * :

A.95 0.05 b * h * : mm 2^1125

(c) d.b A.95 0.0766 / sqrt : mm 121.2

Tabela.secao6 Polegada Métrica 21 mat :

"Polegada"
"Métrica"

Unidade de Medida:

B <content> Tabelas e fórmulas da Seção 6.6 </content>

Equações 6a@d - Páginas 329 e 330 Limite_Fadiga.1 Aços 0.5 S.ut * 700 MPa * 1400 MPa * Ferros 0.4 S.ut * 160 MPa * 400 MPa * Alumínios 0.4 S.ut * 130 MPa * 330 MPa * Ligas de Cobre 0.4 S.ut * 100 MPa * 280 MPa * 44 mat :

Equação(6.7a) - Página 330 C.carreg.matrix Flexão 1 Força Normal 0.7 22 mat :

Equação (6.7b) - Página 331 C.tamanho.matrix para d ≤ 0,3in (8mm) 1 para 0,3in < d ≤ 10in 0.896 d.b in / (0.097 -^* para 8mm < d ≤ 250mm 1.189 d.b mm / (0.097 -^* 32 mat :

Tabela 6-3 - Página 333 B 1 el Métrica ≡ CS 1.58 0.085 - 4.51 0.265 - 57.7 0.718 - 272 0.995 - 42 mat : CS 1.34 0.085 - 2.7 0.265 - 14.4 0.718 - 39.9 0.995 - 42 mat : if

Coeficiente_Superficie.matrix Retificado CS 11 el CS 12 el Usinado ou Estirado a Frio CS 21 el CS 22 el Laminado a Quente CS 31 el CS 32 el Forjado CS 41 el CS 42 el 43 mat :

Coeficiente_Superficie.matrix Retificado 1.58 0.08 - Usinado ou Estirado a Frio 4.51 0.26 - Laminado a Quente 57.7 0.72 - Forjado 2721 - 43 mat

Equação 6.7f - Página 335 C.temperatura.matrix T ≤ 450°C (840°F) 1 para 450°C < T ≤ 550°C 1 0.0058 T °C / 450 - (* - para 840°F < T ≤ 1020°F 1 0.0032 T °F / 840 - (* - 32 mat :

Tabela 6-4 - Página 335 C.confiabilidade.matrix 50190 0.897 95 0.868 99 0.814 99.9 0.753 99.99 0.702 99.999 0.659 99.9999 0.620 82 mat :

"Aços"
"Ferros"
"Alumínios"
"Ligas\0020\de\0020\Cobre"

Limite de Fadiga não Corrigido:

Limite_Fadiga.2

Limite_Fadiga.2 Aços 3108^* kg m s 2^* / * 7108^* kg m s 2^* / * 1.4 109^* kg m s 2^* / * 14 mat

S.ut Limite_Fadiga.2 4 el < S.e' 0.5 S.ut * : S.e' Limite_Fadiga.2 3 el : if

S.e' MPa 300

"Flexão"
"Força\0020\Normal"

Fator devido à solicitação:

C.carreg

C.carreg 1

"para\0020\d\0020\\2264\\0020\0\002C\3in\0020\\0028\8mm\0029\"
"para\0020\0\002C\3in\0020\\003C\\0020\d\0020\\2264\\0020\10in"
"para\0020\8mm\0020\\003C\\0020\d\0020\\2264\\0020\250mm"

Fator de Tamanho:

C.tamanho

ATENÇÃO! FALTA INSERIR DIÂMETRO EQUIVALENTE PARA PERFIS
DIFERENTE DE REDONDOS (VER PAG.333)

d.b mm 121.19

d.b in 4.77

C.tamanho 0.747

"Retificado"
"Usinado\0020\ou\0020\Estirado\0020\a\0020\Frio"
"Laminado\0020\a\0020\Quente"
"Forjado"

Tabela 6-3 Coeficientes para a equação do fator de superfície:

Coeficiente_Superficie

A Coeficiente_Superficie 2 el : 57.7

b Coeficiente_Superficie 3 el : 0.718 -

B 1 el Métrica ≡ C.superficie A S.ut MPa / (b^* : C.superficie A S.ut ksi / (b^* : if 0.584

C.superficie 0.584

"T\0020\\2264\\0020\450\00B0\C\0020\\0028\840\00B0\F\0029\"
"para\0020\450\00B0\C\0020\\003C\\0020\T\0020\\2264\\0020\550\00B0\C"
"para\0020\840\00B0\F\0020\\003C\\0020\T\0020\\2264\\0020\1020\00B0\F"

Fator de Temperatura:

C.temp

T 500 °C * :

C.temp 0.71

50
90
95
99
99.9
99.99
99.999
99.9999

Fator de Confiabilidade:

C.conf

C.conf 0.753

Limite de Fadiga não Corrigido: S.e' MPa 300

Limite de Resistência Corrigido: S.e S.e' C.carreg * C.tamanho * C.superficie * C.temp * C.conf * :

S.e MPa 70

8) Para criar o diagrama S-N é necessário estimar a tensão Sm em 103 ciclos de acordo com a Equação 6.9 para flexão.

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(i) S.m 0.9 S.ut * : MPa 540

9) O diagrama S-N estimado é mostrado na Figura 6-34 com os valores de S_m e de S_e acima. As equações das duas linhas são obtidas pelas Equações de 6.10a a 6.10c assumindo que S_e começa aos 10⁶ ciclos.

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(j) b 13 / S.m S.e / log10 * - : a S.m 103^(b^/ : 21 line

a MPa 4169.31

b 0.29589 -

σ 100 MPa * :

N σ a N b^* ≡ 1000 FindRoot : 2.987 105^*

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10 0 ^ S.ut sc * 10 3 ^ S.m sc * 10 6 ^ S.e sc * 10 9 ^ S.e sc * 4 2 mat 2.987 10 5 ^ * 100 MPa * sc * 1 2 mat 2 1 sys

10) O número de ciclos de vida para qualquer nível de tensão alternada pode ser determinado com as Equações (k). Para a tensão alternada aplicada de 100MPa tem-se:

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N 2.987 105^*

ciclos

A Figura 6-34 mostra a interseção da linha de estresse alternado com a linha de falha em N = 3E5 ciclos.

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