{"id":45162,"date":"2025-10-12T20:38:41","date_gmt":"2025-10-12T20:38:41","guid":{"rendered":"https:\/\/ati.novanta.com\/?page_id=45162"},"modified":"2025-10-12T20:38:42","modified_gmt":"2025-10-12T20:38:42","slug":"selecao-de-um-sensor-de-colisao","status":"publish","type":"page","link":"https:\/\/ati.novanta.com\/pt\/products\/collision-sensors\/selecao-de-um-sensor-de-colisao\/","title":{"rendered":"Sele\u00e7\u00e3o de um sensor de colis\u00e3o"},"content":{"rendered":"\n\n
\n\t\t\t\t\t
\n\t\t\t\t\"\"\t\t\t<\/figure>\n\t\t\t\n\t\n\t\n
\n\t
\n\t\t\t\t\t

\n\t\t\t\tSENSORES DE COLIS\u00c3O\t\t\t<\/p>\n\t\t\n\t\t

\n\t\t\tSele\u00e7\u00e3o de um sensor de colis\u00e3o\t\t<\/h1>\n\n\t\t

<\/p>\t<\/div>\n<\/div>\n<\/div>\n\n\n\n

\n
\n

Como selecionar um modelo de protetor<\/h2>\n<\/div>\n\n\n\n
\n

Para uma aplica\u00e7\u00e3o bem-sucedida, o Protetor escolhido deve ser dimensionado adequadamente. Para escolher um modelo, considere as cargas produzidas devido ao peso est\u00e1tico da ferramenta, as cargas inerciais impostas pelo movimento do rob\u00f4 e as cargas produzidas pelo efetor final ao executar as tarefas pretendidas. Depois que essas cargas forem calculadas e um modelo espec\u00edfico for escolhido, a configura\u00e7\u00e3o de press\u00e3o nominal para o ponto de ruptura poder\u00e1 ser determinada. A configura\u00e7\u00e3o de press\u00e3o de ar necess\u00e1ria deve estar prontamente dispon\u00edvel com ampla capacidade de ajuste. Por exemplo, uma configura\u00e7\u00e3o de press\u00e3o calculada de 50 psi deve ter uma faixa ajust\u00e1vel de 25 a 75 psi.<\/p>\n\n\n\n

O processo de sele\u00e7\u00e3o \u00e9 o seguinte:<\/p>\n\n\n\n

1. Calcular as cargas aplicadas:<\/h2>\n\n\n\n

A Figura 1 pode ser usada para converter as for\u00e7as que atuam no ferramental do efetor final em momento, torque e cargas axiais resultantes aplicadas ao Protetor. Use o diagrama mostrado na Figura 1 e as f\u00f3rmulas abaixo para calcular as cargas aplicadas no pior caso para sua aplica\u00e7\u00e3o. Todos os tr\u00eas casos de carga – axial, torque e momento – devem ser avaliados quanto aos seus componentes de for\u00e7a est\u00e1tica, din\u00e2mica e de trabalho.<\/p>\n\n\n\n

\"\"
Observa\u00e7\u00e3o: <\/strong>Nem todas as for\u00e7as componentes (Est\u00e1tica, Din\u00e2mica e de Trabalho) est\u00e3o presentes durante todas as fases do programa do rob\u00f4. Como resultado, as piores condi\u00e7\u00f5es de caso para cargas axiais, de torque e de momento podem ocorrer em momentos diferentes do programa.<\/figcaption><\/figure>\n\n\n\n

F\u00f3rmulas:<\/strong><\/p>\n\n\n\n

Carga axial (F) = F2<\/p>\n\n\n\n

Torque (T) = F3*D3<\/em><\/p>\n\n\n\n

Momento (M) = \u221a( (F1D1)2 + (F2*D2)2 )<\/p>\n\n\n\n

(F1, F2 e F3 consistem na soma de seus respectivos componentes de for\u00e7a est\u00e1tica, din\u00e2mica e de trabalho e devem ser sempre positivos para fins de c\u00e1lculo das configura\u00e7\u00f5es de press\u00e3o de ruptura)<\/p>\n\n\n\n

    \n
  • For\u00e7a est\u00e1tica: <\/strong>A carga aplicada pelo peso da ferramenta enquanto o bra\u00e7o do rob\u00f4 est\u00e1 ocioso. Isso inclui o peso de todas as pe\u00e7as anexadas ao Protetor, atuando no centro de gravidade do conjunto ao longo da dire\u00e7\u00e3o da gravidade.<\/li>\n\n\n\n
  • For\u00e7a din\u00e2mica: <\/strong>A for\u00e7a inercial imposta no centro de gravidade do ferramental devido \u00e0 acelera\u00e7\u00e3o do bra\u00e7o do rob\u00f4. Essa for\u00e7a atua na dire\u00e7\u00e3o oposta ao movimento. As for\u00e7as din\u00e2micas s\u00e3o aditivas \u00e0s for\u00e7as est\u00e1ticas e devem ser cuidadosamente consideradas para garantir o dimensionamento adequado do protetor.<\/li>\n\n\n\n
  • For\u00e7a de trabalho:<\/strong> As for\u00e7as s\u00e3o geradas na ponta da ferramenta em condi\u00e7\u00f5es normais de trabalho. Se essas for\u00e7as e sua localiza\u00e7\u00e3o forem conhecidas, elas poder\u00e3o ser convertidas em cargas no Protetor usando a mesma t\u00e9cnica.<\/li>\n<\/ul>\n\n\n\n

    Escolha um modelo de Protetor:<\/h2>\n\n\n\n

    Depois que as cargas aproximadas forem conhecidas na etapa 1, escolha um modelo que tenha um momento nominal e uma classifica\u00e7\u00e3o de torque acima das cargas calculadas, tanto em condi\u00e7\u00f5es din\u00e2micas quanto de trabalho.<\/p>\n\n\n\n

    3. Obtenha a configura\u00e7\u00e3o de press\u00e3o necess\u00e1ria:<\/p>\n\n\n\n

    Para um determinado modelo com cargas conhecidas, o ajuste de press\u00e3o necess\u00e1rio pode ser aproximado a partir da f\u00f3rmula a seguir:<\/p>\n\n\n\n

    p = PM + PT + PF<\/p>\n\n\n\n

    Em que PM, PT e PF s\u00e3o os componentes de press\u00e3o relacionados aos componentes de momento, torque e carga de for\u00e7a esperados na ruptura. PM, PT e PF s\u00e3o calculados usando as f\u00f3rmulas das tabelas a seguir, em que M, T e F s\u00e3o as cargas esperadas no rompimento da press\u00e3o definida.<\/p>\n\n\n\n

    Tabela 1: C\u00e1lculos de ajuste da press\u00e3o do protetor<\/h2>\n\n\n\n
    Modelo<\/th>Momento<\/th>Torque<\/th>Axial<\/th><\/tr><\/thead>
    SR-101<\/strong><\/td>(M x 0,0517)<\/td>(T x 0,0495) – 0,2<\/td>F x 0,00228<\/td><\/tr>
    SR-131<\/strong><\/td>(M x 0,0183)<\/td>(T x 0,0199) – 0,1<\/td>F x 0,00132<\/td><\/tr>
    SR-176<\/strong><\/td>(M x 0,0077)<\/td>(T x 0,0075) – 0,2<\/td>F x 0,00070<\/td><\/tr>
    SR-221<\/strong><\/td>(M x 0,0032)<\/td>(T x 0,0040) + 0,5<\/td>F x 0,00045<\/td><\/tr>
    SR-48<\/strong><\/td>(M x 1,0874) – 0,5<\/td>(T x 0,9267) – 0,2<\/td>F x 0,01435<\/td><\/tr>
    SR-61<\/strong><\/td>(M x 0,2294) – 0,2<\/td>(T x 0,2708) – 0,4<\/td>F x 0,00719<\/td><\/tr>
    SR-81<\/strong><\/td>(M x 0,1052)<\/td>(T x 0,1027) – 0,1<\/td>F x 0,00361<\/td><\/tr>
    SR-82<\/strong><\/td>(M x 0,1052)<\/td>(T x 0,1027) – 0,1<\/td>F x 0,00361<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    Exemplo: <\/strong>Para um SR-81 com um momento de 100 lb-in, torque de 50 lb-in e uma carga axial de 20 lbs, e uma acelera\u00e7\u00e3o de 2 G’s, a configura\u00e7\u00e3o de press\u00e3o \u00e9 calculada da seguinte forma:<\/p>\n\n\n\n

    P = ((100*0.172) – 0.2) + ((50*0.168) – 0.8) + (20*0.233) + ((100*2*0.172) – 0.2)
    = 17 + 7.6 + 4.66 + 34.2
    = 63.46<\/p>\n\n\n\n

    \u00c9 necess\u00e1ria uma configura\u00e7\u00e3o de press\u00e3o de ar nominal de 63 psi.<\/p>\n<\/div>\n<\/div>\n\n\n\n

    <\/p>\n","protected":false},"excerpt":{"rendered":"","protected":false},"author":36,"featured_media":0,"parent":32924,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_acf_changed":false,"footnotes":""},"class_list":["post-45162","page","type-page","status-publish","hentry"],"acf":[],"yoast_head":"\nSele\u00e7\u00e3o de um sensor de colis\u00e3o - ATI Industrial Automation<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ati.novanta.com\/pt\/products\/collision-sensors\/selecao-de-um-sensor-de-colisao\/\" \/>\n<meta property=\"og:locale\" content=\"pt_BR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Sele\u00e7\u00e3o de um sensor de colis\u00e3o\" \/>\n<meta property=\"og:url\" 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