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margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><br> <b>Оглавление:<= /b><br> <br> Глава I. Основные задачи и определениn= 3;<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 1.1 Введение<o:p></o:p></span><= /p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 1.= 2. Трещина в конструкциl= 0;<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 1.= 3. Напряжения при вершине трещины<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 1.= 4. Критерий <span class=3DSpel= lE>Гриффитса</span><o= :p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 1.= 5. Критерий предельногl= 6; раскрытия трещины<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 1.= 6. Распростраl= 5;ение трещины<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 1.= 7. Заключение<o:p>= </o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'> </span>Глава= II. Механизм роста трещины и ра= 079;рушения<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 2.= 1. Введение<o:p></o:p></span><= /p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 2.= 2. Разрушение сколом<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 2.= 3. Вязкое разрушение<o:p>= </o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 2.= 4. Усталостныk= 7; трещины<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 2.= 5. Образованиk= 7; трещин в материалах под действием окружающей среды<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 2.= 6. Анализ разрушений = 74; условиях эк = 89;плуатации<o:p></o:= p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'> </span>Глава= III. Упругое пол = 77; напряжений при вершине трещины<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 3.= 1. Функция напряжений Эри<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 3.= 2. Комплексныk= 7; функции напряжений<o:p>= </o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 3.= 3. Решение задач о трещине<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 3.= 4. Влияние конечных размеров<o:p></o:p></span><= /p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 3.= 5. Специальныk= 7; случаи<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 3.= 6. Эллиптичесl= 2;ие трещины<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 3.= 7. Некоторые полезные выражения<o:p></o:p><= /span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'> </span>Глава= IV. Пластическk= 2;я зона при вер= 096;ине трещины<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 4.= 1. Поправка <span class=3DSpel= lE>Ирвина</span> на пластичносm= 0;ь<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 4.= 2. Подход <span class=3DSpellE>Д = 72;гдейла</span><o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 4.= 3. Форма зоны пластичносm= 0;и<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 4.= 4. Плоское напряженноk= 7; состояние и плоская деф = 86;рмация<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 4.= 5. Коэффициенm= 0; ограничениn= 3; на пластичн = 86;сть<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 4.= 6. Влияние толщины<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'> </span>Глава= V. Энергетичеl= 9;кий принцип<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 5.= 1. Интенсивноl= 9;ть выделения энергии<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 5.= 2. Критерий роста трещины<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 5.= 3. Сопротивлеl= 5;ие росту трещины (R - кривая)<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 5.= 4. Податливосm= 0;ь<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 5.= 5. J - интеграл<o:p></o:p></span><= /p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'> </span>Глава= VI. Динамика роста трещины и ег= 086; торможение<o:p>= </o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 6.= 1. Скорость распростраl= 5;ения трещины и кинетическk= 2;я энергия<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 6.= 2. Динамическk= 2;я интенсивноl= 9;ть напряжений = 80; интенсивноl= 9;ть выделения энергии<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 6.= 3. Ветвление трещин<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 6.= 4. Основные принципы торможения = 88;оста трещины<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 6.= 5. Торможение трещин на практике<o:p></o:p></span><= /p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 6.= 6. Динамическk= 2;я вязкость разрушения<o:p>= </o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'> </span>Глава= VII. Вязкость разрушения при плоской деформации<o:p>= </o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 7.= 1. Стандартноk= 7; испытание<o:p></o:p><= /span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 7.= 2. Требования = 82; размерам образцов<o:p></o:p></span><= /p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 7.= 3. Нелинейносm= 0;ь<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 7.= 4. Применимосm= 0;ь критериев<o:p></o:p><= /span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'> </span>Глава= VIII. Разрушение при плоском напряженноl= 4; состоянии и = 074; переходной области<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 8.= 1. Введение<o:p></o:p></span><= /p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 8.= 2. Плоское напряженноk= 7; состояние с инженерной точки зрени = 03;<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 8.= 3. Концепция R - кривой<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 8.= 4. Влияние толщины<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 8.= 5. Испытание при плоском напряженноl= 4; состоянии<o:p></o:p><= /span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 8.= 6. Заключение<o:p>= </o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'> </span>Глава= IX. Критерий критическоk= 5;о раскрытия трещины<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 9.= 1. Разрушение после образованиn= 3; общей текучести<o:p></o:p><= /span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 9.= 2. Раскрытие т = 88;ещины при ее вершине<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 9.= 3. Возможные пути использоваl= 5;ия критерия КР = 58;<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 9.= 4. Эксперименm= 0;альное определениk= 7; КРТ<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 9.= 5. Параметры, влияющие на критическоk= 7; значение КР = 58;<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 9.= 6. Ограничениn= 3;. Разрушение при общей текучести<o:p></o:p><= /span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'> </span>Глава= X. Распростраl= 5;ение усталостноl= 1; трещины<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 10= .1. Введение<o:p></o:p></span><= /p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 10= .2. Рост трещин = 99; и коэффициенm= 0; интенсивноl= 9;ти напряжений<o:p>= </o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 10= .3. Факторы, влияющие на процесс рас = 87;ространени&#= 1103; трещины<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 10= .4. Эксплуатацl= 0;онные нагрузки с п= 077;ременной амплитудой<o:p>= </o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 10= .5. Расчет процесса распростраl= 5;ения трещины<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 10= .6. Заключение<o:p>= </o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'> </span>Глава= XI. Сопротивлеl= 5;ие металлов ра = 79;рушению<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 11= .1. Критерии разрушения<o:p>= </o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 11= .2. Критерий разрушения сколом<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 11= .3. Влияние примесей и частиц втор = 86;го рода<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 11= .4. Влияние обработки, анизотропиl= 0;<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 11= .5. Влияние тем = 87;ературы<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 11= .6. Заключение<o:p>= </o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'> </span>Глава= XII. Надежность конструкциl= 1; и допустимо = 89;ть повреждениl= 1;<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 12= .1. Введение<o:p></o:p></span><= /p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 12= .2. Средства обеспечениn= 3; надежности<o:p>= </o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 12= .3. Информация, необходимаn= 3; для примене = 85;ия механики разрушения<o:p>= </o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 12= .4. Заключение<o:p>= </o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'> </span>Глава= XIII. Определениk= 7; коэффициенm= 0;ов интенсивноl= 9;ти напряжений<o:p>= </o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 13= .1. Введение<o:p></o:p></span><= /p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 13= .2. Аналитичесl= 2;ие и численные методы<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 13= .3. Метод конечных элементов<o:p></o:p><= /span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 13= .4. Эксперименm= 0;альные методы<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'> </span>Глава= XIV. Практическl= 0;е вопросы<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 14= .1. Введение<o:p></o:p></span><= /p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 14= .2. Образованиk= 7; сквозных трещин на отверстиях<o:p>= </o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 14= .3. Угловые трещины на отверстиях<o:p>= </o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 14= .4. Трещины, приближающl= 0;еся к отверстию<o:p= ></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 14= .5. <span class=3DSpellE>Нагружен = 80;е</span> смешанного типа<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 14= .6. Вязкость разрушения сварных шво = 74;<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 14= .7. Распростраl= 5;ение трещины при циклическиm= 3; эксплуатацl= 0;онных нагрузках<o:p></o:p><= /span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 14= .8. Анализ разрушений = 74; условиях эк = 89;плуатации<o:p></o:= p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'> </span>Глава= XV. Разрушение конструкциl= 1;<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 15= .1. Введение<o:p></o:p></span><= /p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 15= .2. Емкости высокого давления и трубопровоk= 6;ы<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 15= .3. Критерий «утечки до разрушения&raqu= o;<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 15= .4. Выбор материалов<o:p>= </o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'> </span>Глава= XVI. Оболочечныk= 7; конструкциl= 0;, усиленные ребрами жесткости<o:p></o:p><= /span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 16= .1. Введение<o:p></o:p></span><= /p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 16= .2. Анализ<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 16= .3. Распростраl= 5;ение усталостноl= 1; трещины<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 16= .4. Остаточная прочность<o:p></o:p><= /span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 16= .5. R-кривая и остаточная прочность п = 72;нелей, усиленных ребрами жесткости<o:p></o:p><= /span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 16= .6. Другие методы анализа<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 16= .7. Торможение трещины<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>§ 16= .8. Заключение<o:p>= </o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'> </span>Справ= очные данные<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'>  </span>Поправочнm= 9;е функции для расчета КИН для различных случаев <span class=3DSpellE> = 85;агружения</span> и геометрии конструкциl= 1;<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><span style=3D'mso-spacerun:yes'> </span>Литер= атура<span style=3D'mso-tab-count:1'>         = </span><o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><o:p> </o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'>Книга Д. <span class=3DSpellE>Броека</span> "Основы механики разрушения" является од = 85;ой из первых переводных изданий, которое был = 86; опубликоваl= 5;о на русском языке. Несмо= 090;ря на свой уже достаточно почтенный возраст, это издание и по сей день не потеряло св = 86;ей актуальносm= 0;и.<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt;text-indent:20.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'>Книга Д. <span class=3DSpellE>Броека</span> построена в виде учебного курса, отличается доступным языком изложения и содержит базовые све = 76;ения, необходимыk= 7; для начального изучения Механики Разрушения.<o:p= ></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt'><span style=3D'font-size:10.0pt;font-family:Arial'><= o:p> </o:p></span></p> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt'><span style=3D'font-size:10.0pt;font-family:Arial'>&= #1069;лектронная версия подготовлеl= 5;а <span class=3DSpellE>к.т.н</span>, доцентом Васильевым Дмитрием Владимировl= 0;чем<o:p></o:p></span></p> <span style=3D'font-size:12.0pt;font-family:"Times New Roman";mso-fareast-f= ont-family: "Times New Roman";mso-ansi-language:RU;mso-fareast-language:RU;mso-bidi-lan= guage: AR-SA'><br clear=3Dall style=3D'mso-special-character:line-break;page-break= -before: always'> </span> <p style=3D'margin-top:0cm;margin-right:6.0pt;margin-bottom:0cm;margin-left= :6.0pt; margin-bottom:.0001pt'><span style=3D'font-size:3.0pt;mso-bidi-font-size:10= .0pt; font-family:Arial'><o:p> </o:p></span></p> <h2 style=3D'margin:0cm;margin-bottom:.0001pt'><span style=3D'font-family:A= rial; color:windowtext'>§1.1. Введение<o:p></o:p></span><= /h2> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>В течение многих лет применение материалов = 74; инженерном проектировk= 2;нии ставило перед челов = 77;чеством сложные задачи. В каменном веке задача состояла главным образом в том, чтобы придать материалу нужную форм = 91;. В начале бронзового = 80; железного веков трудность з = 72;ключалась также в производстk= 4;е металлов. В течение многих веко = 74; обработка металлов была трудое = 84;кой и чрезвычайнl= 6; дорогой процедурой. Например, снаряжение рыцаря и его коня стоило столько же, сколько стоил танк «Центурион&raqu= o; во время второй мировой войны.<o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>С развитием искусства обработки металлов он = 80; все чаще стали применятьсn= 3; в конструкциn= 3;х. Опыт показа = 83;, что конструкциl= 0;, построенныk= 7; из этих материалов, не всегда ведут себя удовлетворl= 0;тельно и часто самы= 084; неожиданныl= 4; образом разрушаютсn= 3;. Существуют детальные описания процессов литья и ковки, произ= 074;одимых в средние века. Если судить с современныm= 3; позиций, то за существеннm= 9;е техническиk= 7; недостатки конструкциl= 1; были бы ответственl= 5;ы именно эти методы производстk= 4;а. Должно быть, по этой причине стрелки, поджигая заряд, молились, чтобы доза была отмерена точно, и ство&#= 1083; не взорвалс = 03;.<o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Ч&= #1088;езвычайно интенсивноk= 7; использоваl= 5;ие металлов в XIX в= ;. привело к тому, что число авари = 81; и несчастны = 93; случаев достигло необычайныm= 3; размеров. В течение десятилетиn= 3;, с 1860 по <st1:metricconverter ProductID=3D"1870 г" w:st=3D"on">1870 г</st1:metricconverter>., число людей, погибших в Великобритk= 2;нии во время железнодорl= 6;жных катастроф, было порядк = 72; двух сотен в год. Большинствl= 6; несчастных случаев про = 80;сходило из-за трещин в колесах, осях или рел= 100;сах. Из отчетов о несчастных случаях за последние 200 лет <span class=3DSpellE>Анде = 88;сон</span> [1] составил интересную сводку. Приводим не = 89;колько выдержек:<o:p></o:p></span>= </p> <p style=3D'margin-top:0cm;margin-right:0cm;margin-bottom:0cm;margin-left:3= 6.0pt; margin-bottom:.0001pt;text-indent:-18.0pt;mso-list:l1 level1 lfo1;tab-stops: list 36.0pt'><![if !supportLists]><span style=3D'font-size:10.0pt;font-fami= ly: Symbol;mso-fareast-font-family:Symbol;mso-bidi-font-family:Symbol'><span style=3D'mso-list:Ignore'>·<span style=3D'font:7.0pt "Times New Roma= n"'>         </span></span></span><![endif]><span style=3D'font-size:10.0pt;font-family:= Arial'>«19 марта <st1:metricconverter ProductID=3D"1830 = г" w:st=3D"on">1830 г</st1:metricconverter>. около 700 человек собралось н = 72; <span class=3DSpellE>монтросс = 82;ом</span> висячем мосту, чтобы наблюдать з = 72; лодочными гонками. В это время одна из основных це = 87;ей разошлась. что привело = 082; определеннm= 9;м людским жертвам»;<o:p></o:p></span>= </p> <p style=3D'margin-top:0cm;margin-right:0cm;margin-bottom:0cm;margin-left:3= 6.0pt; margin-bottom:.0001pt;text-indent:-18.0pt;mso-list:l1 level1 lfo1;tab-stops: list 36.0pt'><![if !supportLists]><span style=3D'font-size:10.0pt;font-fami= ly: Symbol;mso-fareast-font-family:Symbol;mso-bidi-font-family:Symbol'><span style=3D'mso-list:Ignore'>·<span style=3D'font:7.0pt "Times New Roma= n"'>         </span></span></span><![endif]><span style=3D'font-size:10.0pt;font-family:= Arial'>«22 января <st1:metricconverter ProductID=3D"1866 г" w:st=3D"on">1866 г</st1:metricconverter>. провалиласn= 0; часть крыши на железнод = 86;рожной станции в Манчестере, что повлекл = 86; за собой смерть двух людей. Это происшествl= 0;е было вызван = 86; разрушениеl= 4; отлитых из железа подпорок.»;<o:p></o:p= ></span></p> <p style=3D'margin-top:0cm;margin-right:0cm;margin-bottom:0cm;margin-left:3= 6.0pt; margin-bottom:.0001pt;text-indent:-18.0pt;mso-list:l1 level1 lfo1;tab-stops: list 36.0pt'><![if !supportLists]><span style=3D'font-size:10.0pt;font-fami= ly: Symbol;mso-fareast-font-family:Symbol;mso-bidi-font-family:Symbol'><span style=3D'mso-list:Ignore'>·<span style=3D'font:7.0pt "Times New Roma= n"'>         </span></span></span><![endif]><span style=3D'font-size:10.0pt;font-family:= Arial'>«13 декабря <st1:metricconverter ProductID=3D"1898 г" w:st=3D"on">1898 г</st1:metricconverter>. произошло разрушение газового танка в Нью-Й&#= 1086;рке; при этом был= 086; убито и ранено несколько человек и нанесен значительнm= 9;й материальнm= 9;й ущерб»;<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:0cm;margin-bottom:0cm;margin-left:3= 6.0pt; margin-bottom:.0001pt;text-indent:-18.0pt;mso-list:l1 level1 lfo1;tab-stops: list 36.0pt'><![if !supportLists]><span style=3D'font-size:10.0pt;font-fami= ly: Symbol;mso-fareast-font-family:Symbol;mso-bidi-font-family:Symbol'><span style=3D'mso-list:Ignore'>·<span style=3D'font:7.0pt "Times New Roma= n"'>         </span></span></span><![endif]><span style=3D'font-size:10.0pt;font-family:= Arial'>«Водопровl= 6;д высокого давления взорвался в Бостоне 3 января <st1:metricconverter ProductID=3D"1913 г" w:st=3D"on">1913 г</st1:metricconverter>. = и затопил все вокруг.»;<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:0cm;margin-bottom:0cm;margin-left:3= 6.0pt; margin-bottom:.0001pt;text-indent:-18.0pt;mso-list:l1 level1 lfo1;tab-stops: list 36.0pt'><![if !supportLists]><span style=3D'font-size:10.0pt;font-fami= ly: Symbol;mso-fareast-font-family:Symbol;mso-bidi-font-family:Symbol'><span style=3D'mso-list:Ignore'>·<span style=3D'font:7.0pt "Times New Roma= n"'>         </span></span></span><![endif]><span style=3D'font-size:10.0pt;font-family:= Arial'>«Инжинириl= 5;г, февраль <st1:metricconverter ProductID=3D"1866 г" w:st=3D"on">1866 г</st1:metricconverter>. От пятидесяти до шестидесятl= 0; взрывов пар = 86;вых котлов происходит ежегодно в Объединеннl= 6;м Королевствk= 7;, что связано = 089; потерей мно = 75;их жизней и значительнm= 9;м материальнm= 9;м ущербом. Не пришло ли время правительсm= 0;ву создать комиссию дл = 03; расследоваl= 5;ия?»;<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:0cm;margin-bottom:0cm;margin-left:3= 6.0pt; margin-bottom:.0001pt;text-indent:-18.0pt;mso-list:l1 level1 lfo1;tab-stops: list 36.0pt'><![if !supportLists]><span style=3D'font-size:10.0pt;font-fami= ly: Symbol;mso-fareast-font-family:Symbol;mso-bidi-font-family:Symbol'><span style=3D'mso-list:Ignore'>·<span style=3D'font:7.0pt "Times New Roma= n"'>         </span></span></span><![endif]><span style=3D'font-size:10.0pt;font-family:= Arial'>«Наиболее значительнl= 6;е железнодорl= 6;жное происшествl= 0;е педели произошло 20 апреля (1887); оно было вызван = 86; поломкой сцепного бруса. Трое человек был = 86; убито и двое смертельно ранены»;<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:0cm;margin-bottom:0cm;margin-left:3= 6.0pt; margin-bottom:.0001pt;text-indent:-18.0pt;mso-list:l1 level1 lfo1;tab-stops: list 36.0pt'><![if !supportLists]><span style=3D'font-size:10.0pt;font-fami= ly: Symbol;mso-fareast-font-family:Symbol;mso-bidi-font-family:Symbol'><span style=3D'mso-list:Ignore'>·<span style=3D'font:7.0pt "Times New Roma= n"'>         </span></span></span><![endif]><span style=3D'font-size:10.0pt;font-family:= Arial'>«Наиболее значительнl= 6;е железнодорl= 6;жное происшествl= 0;е недели произошло 27 мая (1887). Разруш= 077;ние колеса привело к гибели шест = 80; человек»;<o:p></o:p></span>= </p> <p style=3D'margin-top:0cm;margin-right:0cm;margin-bottom:0cm;margin-left:3= 6.0pt; margin-bottom:.0001pt;text-indent:-18.0pt;mso-list:l1 level1 lfo1;tab-stops: list 36.0pt'><![if !supportLists]><span style=3D'font-size:10.0pt;font-fami= ly: Symbol;mso-fareast-font-family:Symbol;mso-bidi-font-family:Symbol'><span style=3D'mso-list:Ignore'>·<span style=3D'font:7.0pt "Times New Roma= n"'>         </span></span></span><![endif]><span style=3D'font-size:10.0pt;font-family:= Arial'>«Наиболее значительнl= 6;е железнодорl= 6;жное происшествl= 0;е недели произошло 2 июля (1887); оно было вызван = 86; поломкой оси».<o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Н&= #1077;которые из приведенныm= 3; несчастных случаев происходилl= 0;, несомненно, из-за плохог= 086; проектировk= 2;ния. С течением времени ста = 83;о ясно, что трещины и разрушения могут порож = 76;аться пустотами в материале, так называемымl= 0; раковинами. Предотвращk= 7;ние появления таких раков = 80;н повысило бы надежность конструкциl= 1;. Усовершенсm= 0;вование методов производстk= 4;а, накопление знаний и более полно = 77; понимание свойств материалов привели к тому, что количество разрушений значительнl= 6; снизилось.<o:p></o:p>= </span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>С появлением сварных конструкциl= 1; несчастные случаи опят = 00; участились. Из 2500 кораблей типа «Либерти», построенныm= 3; во время вто= 088;ой мировой войны, 145 разломилосn= 0; пополам и по= 095;ти 700 претерпело серьезные разрушения. = 055;одобная участь постигла множество мостов и дру= 075;их конструкциl= 1;. Данные об этих разрушенияm= 3; собраны <span class=3DSpellE> = 40;ндерсоном</span> [1] и более детально <span class=3DSpel= lE>Биггсом</span> [2].<o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Р&= #1072;зрушения часто происходилl= 0; при непреры = 74;ном действии малых напряжений (несколько кораблей разрушилосn= 0; неожиданно, когда находились = 74; гавани), что делало эти разрушения, казалось бы, необъяснимm= 9;ми. В результат = 77; во многих странах, особенно в США, были проведены развернутыk= 7; исследованl= 0;я, которые позволили установить, что в этих случаях ответственl= 5;ыми за разрушение явились рак = 86;вины и концентрацl= 0;и напряжений (= 080; до некоторо = 81; степени внутренние напряжения).<o:= p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Р&= #1072;зрушение происходилl= 6; так, будто материал конструкциl= 0; был хрупким; оно сопровождаl= 3;ось очень малым = 80; пластическl= 0;ми деформацияl= 4;и. Как оказалось, хрупкое разрушение = 89;тали вызывалось низкими температурk= 2;ми и условиями, в которых возникают пространстk= 4;енные напряжения, имеющие место в острой выемке или р= 072;ковине. При этих условиях строительнk= 2;я сталь может растрескивk= 2;ться без заметны = 93; пластическ&= #1080;х деформаций.<o:p= ></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>П&= #1088;и температурk= 7;, превышающеl= 1; некоторую определеннm= 1;ю величину, которая называется температурl= 6;й перехода, ст= 072;ль проявляет свои пластическl= 0;е свойства. Те= 084;пература перехода может возрастать = 86;т теплового воздействиn= 3; при сварке.<o:p></o:p= ></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>В настоящее время хрупкое разрушение сварных конструкциl= 1;, выполненныm= 3; из <span class=3DSpellE>низко = 87;рочных</span> строительнm= 9;х сталей, може= 090; быть предотвращk= 7;но. Необходимо сделать так, чтобы произ = 74;одимый материал имел низкую температурm= 1; перехода, а процесс сварки не вызывал пер = 77;хода из пластичногl= 6; состояния в хрупкое. Сле= 076;ует избегать больших концентрацl= 0;й напряжений = 80; следить за тем, чтобы сварные швы практическl= 0; не имели деф= 077;ктов.<o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>П&= #1086;сле второй мировой войны использоваl= 5;ие высокопрочl= 5;ых материалов увеличилосn= 0;. Эти материалы часто применяют там, где треб&#= 1091;ется уменьшить вес конструкциl= 0;. Развитие ме = 90;одов исследованl= 0;я напряжений дало возмож = 85;ость более надежно определять локальные напряжения. Это в свою очередь позволило уменьшить коэффициенm= 0;ы запаса, что п&#= 1088;ивело к еще больше= 081; экономии веса. Следов= 072;тельно, конструкциl= 0;, выполненныk= 7; из высокопр = 86;чных материалов, имеют лишь небольшой запас прочн = 86;сти. Это означае = 90;, что рабочие напряжения могут быть достаточныl= 4;и для образованиn= 3; трещины (чем= 091; может содействовk= 2;ть агрессивнаn= 3; среда), особенно если в материале с = 089;амого начала имеются большие концентрацl= 0;и напряжений = 80; раковины. Высокопрочl= 5;ые материалы о = 73;ладают малой <span class=3DSpellE>тр = 77;щиностойко&#= 1089;тью</span> (вязкостью разрушения); остаточная прочность при наличии трещин низк = 72;. Даже если им= 077;ются только маленькие трещины, конструкциn= 3;, выполненнаn= 3; из высокопрочl= 5;ых материалов, может разрушитьсn= 3; при напряженияm= 3;, меньших максимальнl= 6;го рабочего напряжения, на которое они были рассчитаны.<o:p= ></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Р&= #1072;зрушения при малых напряженияm= 3;, вызванные маленькими трещинами, в= 086; многих отно = 96;ениях очень похож = 80; на хрупкие разрушения = 89;варных конструкциl= 1; из <span class=3DSpellE>низко = 87;рочных</span> сталей. Тако= 077; разрушение влечет за собой лишь очень маленькие пластическl= 0;е деформации, носит хрупкий характер с техническоl= 1; точки зрения, хотя <s= pan class=3DSpellE>микромех = 72;низм</span> разделения = 74; этом случае такой же, как и в случае пластическl= 6;го разрушения. Случаи разр = 91;шения при низких напряженияm= 3; в высокопрочl= 5;ых материалах стимулировk= 2;ли развитие механики ра = 79;рушения.<o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Т&= #1077;хническая механика разрушения позволяет выработать методику компенсациl= 0; несоответсm= 0;вий в общепринятm= 9;х концепциях = 87;роектирова&#= 1085;ия. Общепринятm= 9;е критерии пр = 86;ектировани&#= 1103; основаны на таких понятиях, ка= 082; напряжение при растяжении, предел текучести и напряжение при изгибе. Эти критери = 80; являются уд = 86;влетворите&#= 1083;ьными для расчета многих инже = 85;ерных сооружений, но они недостаточl= 5;ы, когда имеется вероятностn= 0; возникновеl= 5;ия трещин. Тепе= 088;ь, после примерно двух десятков ле = 90; развития, механика разрушения стала полезным инструментl= 6;м при проектировk= 2;нии изделий из высокопрочl= 5;ых материалов.<o:p= ></o:p></span></p> <p class=3DMsoNormal style=3D'tab-stops:137.25pt 455.0pt'><span style=3D'fo= nt-size: 10.0pt;font-family:Arial'>Настоящ= ая глава является введением в механику разрушения. = 042; <a href=3D"http://www.mysopromat.ru/cgi-bin/index.cgi?n=3D568"><b><span style=3D'color:windowtext'>§ 1.2</span></b></a> приведен обзор задач, которые можно решит = 00; с помощью положений механики разрушения, = 080; очерчены границы области применения = 84;еханики разрушения, которые значительнl= 6; шире, чем принято думать. Остальная часть главы представляk= 7;т собой краткую сводку положений механики разрушения. Более подробно вс = 77; эти вопросы рассмотренm= 9; в последующиm= 3; главах.<!--[if gte vml 1]><v:shapetype id=3D"_x0000_t75" coordsize=3D"21600,21600" o:spt=3D"75" o:preferrelative= =3D"t" path=3D"m@4@5l@4@11@9@11@9@5xe" filled=3D"f" stroked=3D"f"> <v:stroke joinstyle=3D"miter"/> <v:formulas> <v:f eqn=3D"if lineDrawn pixelLineWidth 0"/> <v:f eqn=3D"sum @0 1 0"/> <v:f eqn=3D"sum 0 0 @1"/> <v:f eqn=3D"prod @2 1 2"/> <v:f eqn=3D"prod @3 21600 pixelWidth"/> <v:f eqn=3D"prod @3 21600 pixelHeight"/> <v:f eqn=3D"sum @0 0 1"/> <v:f eqn=3D"prod @6 1 2"/> <v:f eqn=3D"prod @7 21600 pixelWidth"/> <v:f eqn=3D"sum @8 21600 0"/> <v:f eqn=3D"prod @7 21600 pixelHeight"/> <v:f eqn=3D"sum @10 21600 0"/> </v:formulas> <v:path o:extrusionok=3D"f" gradientshapeok=3D"t" o:connecttype=3D"rect"/> <o:lock v:ext=3D"edit" aspectratio=3D"t"/> </v:shapetype><v:shape id=3D"_x0000_i1025" type=3D"#_x0000_t75" alt=3D"" st= yle=3D'width:12.75pt; height:19.5pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image001.png" o:hr= ef=3D"http://www.mysopromat.ru/images/0.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D17 height=3D26 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image002.gif" v:shapes=3D"_x0000= _i1025"><![endif]><o:p></o:p></span></p> <p class=3DMsoNormal><span style=3D'font-size:10.0pt;font-family:Arial'><o:= p> </o:p></span></p> <h2 style=3D'margin:0cm;margin-bottom:.0001pt'><span style=3D'font-family:A= rial; color:windowtext'>§ 1.2. Трещин= 072; в конструкциl= 0;<o:p></o:p></span></h2> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Р&= #1072;ссмотрим конструкциn= 2;, в которой развиваетсn= 3; трещина. При действии ци = 82;лических нагрузок ил = 80; при совместном воздействиl= 0; нагрузок и окружающей среды с тече= 085;ием времени эта трещина будет расти. Чем длиннее трещина, тем большую концентрацl= 0;ю напряжений она вызывае = 90;. Это означае = 90;, что скорост = 00; развития трещины с течением времени буд = 77;т увеличиватn= 0;ся. Развитие трещины как = 092;ункцию времени можно представитn= 0; возрастающk= 7;й кривой, как показано на рис. 1.1,<i>а.</i> Из-за наличия трещины прочность конструкциl= 0; уменьшаетсn= 3;; она меньше, чем исходна = 03; прочность, н= 072; которую был = 72; рассчитана. Прочность конструкциl= 0; уменьшаетсn= 3; с ростом размера тре = 97;ины, как показан = 86; схематичесl= 2;и на рис. 1.1,<i>б.</i> Через некоторое время прочность настолько уменьшится, что конструкциn= 3; уже не будет способна выдержать случайные высокие наг = 88;узки, которые могут возникнуть при эксплуа = 90;ации. С этого момента конструкциn= 3; легко разру = 96;ается. Если такие случайные высокие наг = 88;узки не возникаю = 90;, то трещина продолжает расти до тех пор, пока прочность н = 77; становится столь низко = 81;, что разрушение происходит = 87;ри нормальных эксплуатацl= 0;онных нагрузках. Многие конструкциl= 0; рассчитываn= 2;т на такие эксплуатацl= 0;онные нагрузки, которые дос = 90;аточно велики, чтоб= 099; породить трещины, особенно когда имеются раковины ил = 80; концентратl= 6;ры напряжений. Проектировm= 7;ик должен предвидеть = 74;озможность растрескивk= 2;ния и, следовате= 083;ьно, допускать возможностn= 0; разрушения = 82;онструкции. Это означае = 90;, что конструкциn= 3; может иметь лишь ограниченнm= 1;ю долговечноl= 9;ть. Конечно, вероятностn= 0; разрушения должна быть достаточно низкой в течение все = 75;о времени экс = 87;луатации. Для обеспечениn= 3; надежности конструкциl= 0; необходимо предсказатn= 0;, как быстро будут расти трещины и ка= 082; быстро буде = 90; уменьшатьс&= #1103; остаточная прочность. Осуществлеl= 5;ие таких предсказанl= 0;й и развитие м= 077;тодов их получени = 03; являются предметом м = 77;ханики разрушения.<o:p= ></o:p></span></p> <p class=3DMsoNormal><span style=3D'font-size:10.0pt;font-family:Arial'><o:= p> </o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><span style=3D'font-size:10.0pt;font-family:Arial'><!--[if gte vml 1]><v:shape id= =3D"_x0000_i1026" type=3D"#_x0000_t75" alt=3D" " style=3D'width:377.25pt;height:175.5pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image003.png" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00001.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D503 height=3D234 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image004.gif" alt=3D" " v:shapes= =3D"_x0000_i1026"><![endif]><o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><b><span style=3D'font-size:10.0pt;font-family:Arial'>Рис. 1.1. Инженерная задача: <i>а</i></span></b><span style=3D'font-size:10.0pt;font-family:Arial'> — <b>кр = 80;вая роста трещины; <i>б</i></b> ̵= 2; <b>кривая остаточной прочности</b><o:p></o= :p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'><o:p>&nb= sp;</o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>П&= #1086; отношению к рис. 1.1 механика разрушения = 76;олжна ответить на следующие вопросы:<o:p></o:p></span></p> <p class=3Dbase style=3D'margin-left:36.0pt;mso-list:l0 level1 lfo2;tab-sto= ps:list 36.0pt'><![if !supportLists]><span style=3D'font-size:10.0pt;font-family:Arial;mso-fareast-font-family:Arial'>= <span style=3D'mso-list:Ignore'>1.<span style=3D'font:7.0pt "Times New Roman"'>&n= bsp;   </span></span></span><![endif]><span style=3D'font-size:10.0pt;font-family:= Arial'>Какова зависимостn= 0; прочности о = 90; размера тре = 97;ины?<o:p></o:p></span></p> <p class=3Dbase style=3D'margin-left:36.0pt;mso-list:l0 level1 lfo2;tab-sto= ps:list 36.0pt'><![if !supportLists]><span style=3D'font-size:10.0pt;font-family:Arial;mso-fareast-font-family:Arial'>= <span style=3D'mso-list:Ignore'>2.<span style=3D'font:7.0pt "Times New Roman"'>&n= bsp;   </span></span></span><![endif]><span style=3D'font-size:10.0pt;font-family:= Arial'>Какой размер трещины может быть допустим пр = 80; ожидаемых эксплуатацl= 0;онных нагрузках, т. е. каков критическиl= 1; размер трещины?<o:p></o:p></span></p> <p class=3Dbase style=3D'margin-left:36.0pt;mso-list:l0 level1 lfo2;tab-sto= ps:list 36.0pt'><![if !supportLists]><span style=3D'font-size:10.0pt;font-family:Arial;mso-fareast-font-family:Arial'>= <span style=3D'mso-list:Ignore'>3.<span style=3D'font:7.0pt "Times New Roman"'>&n= bsp;   </span></span></span><![endif]><span style=3D'font-size:10.0pt;font-family:= Arial'>Как долго будет продолжатьl= 9;я рост трещин = 99; от определе = 85;ного начального размера до критическоk= 5;о размера?<o:p></o:p></span></p> <p class=3Dbase style=3D'margin-left:36.0pt;mso-list:l0 level1 lfo2;tab-sto= ps:list 36.0pt'><![if !supportLists]><span style=3D'font-size:10.0pt;font-family:Arial;mso-fareast-font-family:Arial'>= <span style=3D'mso-list:Ignore'>4.<span style=3D'font:7.0pt "Times New Roman"'>&n= bsp;   </span></span></span><![endif]><span style=3D'font-size:10.0pt;font-family:= Arial'>Какой размер раковин допустим в начальный момент эксплуатацl= 0;и конструкциl= 0;?<o:p></o:p></span></p> <p class=3Dbase style=3D'margin-left:36.0pt;mso-list:l0 level1 lfo2;tab-sto= ps:list 36.0pt'><![if !supportLists]><span style=3D'font-size:10.0pt;font-family:Arial;mso-fareast-font-family:Arial'>= <span style=3D'mso-list:Ignore'>5.<span style=3D'font:7.0pt "Times New Roman"'>&n= bsp;   </span></span></span><![endif]><span style=3D'font-size:10.0pt;font-family:= Arial'>Как часто следует проверять наличие тре = 97;ин в конструкциl= 0;?<o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>М&= #1077;ханика разрушения дает удовлетворl= 0;тельные ответы на некоторые и = 79; этих вопрос = 86;в и полезные ответы на другие. Как показано на рис. 1.2, для развития методов проектировk= 2;ния с привлечениk= 7;м концепций механики разрушения необходимо использоваm= 0;ь различные отрасли знания. На правом конц = 77; шкалы находится инженерный анализ нагр = 91;зок и напряжени = 81;. Прикладная механика определяет поля напряжений при вершине трещины, а та&#= 1082;же упругие и (до некоторой степени) пластическl= 0;е деформации материала в окрестностl= 0; трещины.<o:p></o:p></span></p> <p class=3DMsoNormal><span style=3D'font-size:10.0pt;font-family:Arial'><o:= p> </o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><span style=3D'font-size:10.0pt;font-family:Arial'><!--[if gte vml 1]><v:shape id= =3D"_x0000_i1027" type=3D"#_x0000_t75" alt=3D" " style=3D'width:375.75pt;height:166.5pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image005.png" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00002.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D501 height=3D222 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image006.gif" alt=3D" " v:shapes= =3D"_x0000_i1027"><![endif]><o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><b><span style=3D'font-size:10.0pt;font-family:Arial'>Рис. 1.2. Отрасли знания, охватываемm= 9;е механикой разрушения</spa= n></b><span style=3D'font-size:10.0pt;font-family:Arial'><o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'><o:p>&nb= sp;</o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>П&= #1088;едсказанну= ;ю прочность можно проверить эксперименm= 0;ально. Структурнаn= 3; механика занимается вопросами разрушения на уровне размеров атомов и дислокаций вплоть до размеров примесей и зерен. Понимание этих процес = 89;ов дало бы возможностn= 0; получить критерии, определяющl= 0;е рост трещин = 080; разрушение. = 069;ти критерии предназначk= 7;ны для предска = 79;ания поведения трещины в заданном поле напряж = 77;ний — деформаци = 81;. Понимание процессов разрушения дает также возможностn= 0; выявить параметры материала, определяющl= 0;е его <span class=3DSpellE>трещ = 80;ностойкост&#= 1100;</span>; эти параметры необходимо знать, если нужно получить материалы с повышенной <span class=3DSpellE>трещинос = 90;ойкостью</span>.<o:p></o= :p></span></p> <p class=3DMsoNormal style=3D'tab-stops:137.25pt 455.0pt'><span style=3D'fo= nt-size: 10.0pt;font-family:Arial'>Для успешного использоваl= 5;ия механики ра = 79;рушения в техническиm= 3; приложенияm= 3; необходимо иметь некоторое понятие о ди= 089;циплинах, приведенныm= 3; на рис. 1.2. В книге сдела = 85;а попытка дат = 00; основы для понимания м = 77;ханики разрушения.<span style=3D'mso-tab-count:1'>        &= nbsp;           &nbs= p;            &= nbsp;            </s= pan><!--[if gte vml 1]><v:shape id=3D"_x0000_i1028" type=3D"#_x0000_t75" alt=3D"" style=3D'width:12.75pt;h= eight:19.5pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image001.png" o:hr= ef=3D"http://www.mysopromat.ru/images/0.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D17 height=3D26 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image002.gif" v:shapes=3D"_x0000= _i1028"><![endif]><o:p></o:p></span></p> <p class=3DMsoNormal style=3D'tab-stops:137.25pt 455.0pt'><span style=3D'fo= nt-size: 10.0pt;font-family:Arial'><span style=3D'mso-tab-count:1'>   = ;            &n= bsp;            = ;            &n= bsp;     </span></span><span style=3D'font-size:10.0pt'><span style=3D'mso-tab-count:1'>  &nbs= p;            &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;             = </span></span><span style=3D'font-size:10.0pt;font-family:Arial'><o:p></o:p></span></p> <p class=3DMsoNormal align=3Dcenter style=3D'text-align:center'><span style=3D'display:none;mso-hide:all'><o:p> </o:p></span></p> <p class=3DMsoNormal><o:p> </o:p></p> <p class=3DMsoNormal align=3Dcenter style=3D'text-align:center'><span style=3D'display:none;mso-hide:all'><o:p> </o:p></span></p> <p class=3DMsoNormal><span style=3D'font-size:10.0pt;font-family:Arial'><sp= an style=3D'mso-tab-count:1'>        &= nbsp;   </span><o:p></o:p></span></p> <h2 style=3D'margin:0cm;margin-bottom:.0001pt'><span style=3D'font-family:A= rial; color:windowtext'>§ 1.3. Напряжения при вершине трещины<o:p></o:p></span></h2> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Р&= #1072;скрытие трещины в твердом тел = 77; может быть осуществлеl= 5;о тремя различными путями, как показано на рис. 1.3. При нормальных напряженияm= 3; возникает трещина тип = 72; «разрыв» (ти= 087; I): перемещениn= 3; берегов трещины перпендикуl= 3;ярны плоскости трещины. При плоском сдвиге образуется трещина тип = 72; II, или трещина типа «сдвиг&raq= uo;: перемещениn= 3; берегов трещины происходят = 74; плоскости трещины и перпендикуl= 3;ярно ее фронталь = 85;ой линии. Трещина тип = 72; «срез», или типа III, образу= ;ется при анти-плоско = 84; сдвиге: перемещениn= 3; берегов трещины совпадают с плоскостью трещины и параллельнm= 9; ее направляющk= 7;й кромке. В общем случа = 77; трещину можно описа = 90;ь этими тремя типами. Наиболее важное знач = 77;ние в технике имеет трещина тип = 72; I, обсуждение&#= 1084; которой мы ограничимсn= 3;.<o:p></o:p></span></p> <div align=3Dcenter> <table class=3DMsoNormalTable border=3D0 cellspacing=3D0 cellpadding=3D0 wi= dth=3D474 style=3D'width:355.5pt;mso-cellspacing:0cm;mso-padding-alt:1.5pt 1.5pt 1.5= pt 1.5pt'> <tr style=3D'mso-yfti-irow:0;mso-yfti-firstrow:yes'> <td width=3D149 valign=3Dtop style=3D'width:111.75pt;padding:1.5pt 1.5pt = 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>Тип I<o:p>= </o:p></span></p> </td> <td width=3D156 valign=3Dtop style=3D'width:117.0pt;padding:1.5pt 1.5pt 1= .5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>Тип II<o:p= ></o:p></span></p> </td> <td width=3D169 valign=3Dtop style=3D'width:126.75pt;padding:1.5pt 1.5pt = 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>Тип III<o:= p></o:p></span></p> </td> </tr> <tr style=3D'mso-yfti-irow:1;mso-yfti-lastrow:yes'> <td width=3D149 valign=3Dtop style=3D'width:111.75pt;padding:1.5pt 1.5pt = 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'><!--[if gte vml 1]><v:shape = id=3D"_x0000_i1029" type=3D"#_x0000_t75" alt=3D" " style=3D'width:43.5pt;height:94.5pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image007.gif" o:= href=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00003.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D58 height=3D126 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image007.gif" alt=3D" " v:shap= es=3D"_x0000_i1029"><![endif]><o:p></o:p></span></p> </td> <td width=3D156 valign=3Dtop style=3D'width:117.0pt;padding:1.5pt 1.5pt 1= .5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'><!--[if gte vml 1]><v:shape = id=3D"_x0000_i1030" type=3D"#_x0000_t75" alt=3D" " style=3D'width:58.5pt;height:95.25pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image008.gif" o:= href=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00004.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D78 height=3D127 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image008.gif" alt=3D" " v:shap= es=3D"_x0000_i1030"><![endif]><o:p></o:p></span></p> </td> <td width=3D169 valign=3Dtop style=3D'width:126.75pt;padding:1.5pt 1.5pt = 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'><!--[if gte vml 1]><v:shape = id=3D"_x0000_i1031" type=3D"#_x0000_t75" alt=3D" " style=3D'width:52.5pt;height:99pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image009.gif" o:= href=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00005.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D70 height=3D132 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image009.gif" alt=3D" " v:shap= es=3D"_x0000_i1031"><![endif]><o:p></o:p></span></p> </td> </tr> </table> </div> <p style=3D'margin:0cm;margin-bottom:.0001pt'><span style=3D'font-size:10.0= pt; font-family:Arial'><o:p> </o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><b><span style=3D'font-size:9.0pt;font-family:Arial'>Рис. 1.3. Типы растрескивk= 2;ния</span></b><span style=3D'font-size:10.0pt;font-family:Arial'><o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><span style=3D'font-size:10.0pt;font-family:Arial'><o:p> </o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Р&= #1072;ссмотрим сквозную трещину тип = 72; I длиной 2<i>a</i> в бесконечноl= 1; пластине, ка= 082; показано на рис. 1.4. Пластина находится под действием растягиваюm= 7;его напряжения <span class=3DSpellE>σ</span><i>, </i>котор= 086;е вызывается приложенныl= 4;и в бесконечн = 86;сти силами. В гл. III = 080; XIII рассмотренl= 6; несколько путей для вычисления поля упруги = 93; напряжений при вершине = 090;рещины. Элемент <span class=3DSpellE><i>d= xdy</i></span> пластины, расположенl= 5;ый на расстоянии <span class=3DSpellE><i>r</i></span> от вершины трещины и составляющl= 0;й с плоскость = 02; трещины уго = 83; <span class=3DSpellE>θ</span>, находится под действием нормальных напряжений <span class=3DSpellE>σ<i><sub>х</sub></i></span> и <span class= =3DSpellE>σ<i><sub>y</sub></i></span>, действующиm= 3; в направлениn= 3;х <span class=3DSpellE><i>x</i></span> и <i>у, </i>и касательноk= 5;о напряжения <span class=3DSpellE>τ<i><sub>ху</sub></i></span><i>. </i>М= ;ожно показать (см. [3—6]), что эти напряжения равны (см. гл. III):<= o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><span style=3D'font-size:10.0pt;font-family:Arial'><!--[if gte vml 1]><v:shape id= =3D"_x0000_i1032" type=3D"#_x0000_t75" alt=3D" " style=3D'width:182.25pt;height:156.75pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image010.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00006.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D243 height=3D209 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image010.gif" alt=3D" " v:shapes= =3D"_x0000_i1032"><![endif]><o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><b><span style=3D'font-size:10.0pt;font-family:Arial'>Рис. 1.4. Трещина в бесконечноl= 1; пластине</span></b><span style=3D'font-size:10.0pt;font-family:Arial'><o:p></o:p></span></p> <p style=3D'margin:0cm;margin-bottom:.0001pt'><span style=3D'font-size:10.0= pt; font-family:Arial'><o:p> </o:p></span></p> <p style=3D'margin-top:0cm;margin-right:0cm;margin-bottom:0cm;margin-left:1= .5pt; margin-bottom:.0001pt;tab-stops:228.35pt'><span style=3D'font-size:10.0pt; font-family:Arial'><!--[if gte vml 1]><v:shape id=3D"_x0000_i1033" type=3D"= #_x0000_t75" alt=3D" " style=3D'width:219.75pt;height:170.25pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image011.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00007.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D293 height=3D227 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image011.gif" alt=3D" " v:shapes= =3D"_x0000_i1033"><![endif]><span style=3D'mso-tab-count:1'>  </span>(1.1)<o:p></o:p></span></p> <p style=3D'margin:0cm;margin-bottom:.0001pt'><span style=3D'font-size:10.0= pt; font-family:Arial'><o:p> </o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>К&= #1072;к и следовало ожидать, в упругом случае напр = 03;жения, указанные в (1.= 1), пропорционk= 2;льны внешнему напряжению <span class=3DSpellE>σ</span>. Их величины пропорционk= 2;льны корню квадр = 72;тному из размера трещины и стремятся к бесконечноl= 9;ти в вершине трещины при обращении <span class=3DSpellE><i>r</i></span> в нуль. Зависимостn= 0; <span class=3DSpellE>σ<i><sub>у</sub></i></span> от <span class=3DSpellE><i>r</i></span> при <span class=3DSpellE>&= #952;</span>=3D0 показана на рис. 1.5.<o:p></o:p></span></p> <p style=3D'margin:0cm;margin-bottom:.0001pt;text-indent:15.3pt'><span style=3D'font-size:10.0pt;font-family:Arial'><o:p> </o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><span style=3D'font-size:10.0pt;font-family:Arial'><!--[if gte vml 1]><v:shape id= =3D"_x0000_i1034" type=3D"#_x0000_t75" alt=3D"../images/fracturemechanics/IMG00008.gif" styl= e=3D'width:176.25pt; height:132.75pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image012.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00008.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D235 height=3D177 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image012.gif" alt=3D"../images/fracturemechanics/IMG00008.gif" v:shapes=3D"_x0000_i1034">= <![endif]><o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><b><span style=3D'font-size:10.0pt;font-family:Arial'>Рис. 1.5. Упругое напряжение <span class=3DSpellE>σ<i><sub>y</sub></i></span></span></b><span style=3D'fo= nt-size: 10.0pt;font-family:Arial'> <b>при вершине трещины</b><o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><span style=3D'font-size:10.0pt;font-family:Arial'><o:p> </o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Д&= #1083;я больших значений <span class=3DSpel= lE><i>r</i></span> величина <span class=3DSpel= lE>σ<i><sub>у</sub></i></span> стремится к нулю, в то время как он= 072; должна стре = 84;иться к <span class=3DSpellE>σ</span>. Очевидно, уравнения (1.1) справедливm= 9; только в ограниченнl= 6;й области — вблизи вершины трещины. Каждое из ур= 072;внений представляk= 7;т собой первы = 81; член ряда. В окрестностl= 0;, около вершины трещины, эти первые член = 99; достаточно точно описы = 74;ают поля напряжений, поскольку остальные члены малы п= 086; сравнению с ними. На больших рас = 89;тояниях от вершины трещины следует вводить большее количество членов в уравнения (с= 084;. гл. III).<o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>В уравнениях (1.1) функции координат <span class=3DSpellE><i>r</i></span> и </span><span class=3DSpellE><span style=3D'font-size:10.0pt'>и</span></span><span style=3D'font-size:10= .0pt; font-family:Arial'> имеют простой вид. В обобщенно = 84; виде эти ура= 074;нения можно записать та = 82;:<o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'><o:p>&nb= sp;</o:p></span></p> <p style=3D'margin-top:0cm;margin-right:0cm;margin-bottom:0cm;margin-left:1= .5pt; margin-bottom:.0001pt;tab-stops:228.35pt'><span style=3D'font-size:10.0pt; font-family:Arial'><!--[if gte vml 1]><v:shape id=3D"_x0000_i1035" type=3D"= #_x0000_t75" alt=3D" " style=3D'width:177pt;height:33pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image013.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00009.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D236 height=3D44 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image013.gif" alt=3D" " v:shapes= =3D"_x0000_i1035"><![endif]><span style=3D'mso-tab-count:1'>        &= nbsp;        </span>(1.2)<o:p></o:p></sp= an></p> <p style=3D'margin:0cm;margin-bottom:.0001pt'><span style=3D'font-size:10.0= pt; font-family:Arial'><o:p> </o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>К&= #1086;эффициент <i>K</i><sub>I</sub> называется коэффициенm= 0;ом интенсивноl= 9;ти напряжений, где индекс I обозначает тип разруше = 85;ия I. Когда известен коэффициенm= 0; <i>К</i><sub>I</sub><i>, </i>поле напряжений при вершине трещины пол = 85;остью определено. Две трещины, одна размером <i>4a</i>, а другая размером <i>а</i>, имеют одинаковые поля напряжений при их верши= 085;ах, если первая трещина нагружена н = 72;пряжением <span class=3DSpellE>σ</span><i>, </i>а вторая — напряжениеl= 4; <i>2σ</i>. В этом случа= 077; <i>К</i><sub>I</sub> имеет одинаковые значения дл = 03; обеих трещи = 85;.<o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>У&= #1088;авнение (1.2) есть решение упругой задачи; оно не запрещае = 90; обращения напряжения при вершине трещины в бесконечноl= 9;ть.<o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><span style=3D'font-size:10.0pt;font-family:Arial'><!--[if gte vml 1]><v:shape id= =3D"_x0000_i1036" type=3D"#_x0000_t75" alt=3D" " style=3D'width:221.25pt;height:118.5pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image014.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00010.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D295 height=3D158 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image014.gif" alt=3D" " v:shapes= =3D"_x0000_i1036"><![endif]><o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><b><span style=3D'font-size:10.0pt;font-family:Arial'>Рис. 1.6. Зона пластичносm= 0;и при вершине трещины.</span></b><span style=3D'font-size:10.0pt;font-family:Arial'><o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><b><span style=3D'font-size:10.0pt;font-family:Arial'>Расп&#= 1088;еделение напряжений: <i>= а </i>— принятое; k= 3; — приближеннl= 6;е</span></b><span style=3D'font-size:10.0pt;font-family:Arial'><o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><span style=3D'font-size:10.0pt;font-family:Arial'><o:p> </o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>В действителn= 0;ности этого не может произ = 86;йти: пластическl= 0;е деформации, возникающиk= 7; при вершине трещины, ограничиваn= 2;т напряжения. Точное решение упругой задачи для п= 086;ля напряжении еще не получ= 077;но. Размер зоны пластичносm= 0;и при вершине трещины можно оценить, есл= 080; определить расстояние от вершины трещины <span class=3DSpellE><i>r= </i>*<i><sub>p</sub></i></span>, на котором упругое напряжение <span class=3DSpellE>σ<i><sub>у</sub></i></span> превышает предел текучести <span class=3DSpellE>σ<i><sub>ys</sub></i></span><sub> </sub>(ри= с. 1.6, а) (см. [7,8]). Подставляя <span class=3DSpellE>σ<i><sub>у</sub></i></span><i>=3D</i><span class= =3DSpellE>σ<i><sub>уs</sub></i></span> в уравнение (1.= 1) для <span class=3DSpellE>σ<i><sub>y</sub></i></span> и полагая <span class=3DSpe= llE>θ</span>=3D0, получим<o:p></o:p></span></p> <p style=3D'margin:0cm;margin-bottom:.0001pt'><span style=3D'font-size:10.0= pt; font-family:Arial'><o:p> </o:p></span></p> <p style=3D'margin-top:0cm;margin-right:0cm;margin-bottom:0cm;margin-left:1= .5pt; margin-bottom:.0001pt;tab-stops:228.35pt'><span style=3D'font-size:10.0pt; font-family:Arial'><!--[if gte vml 1]><v:shape id=3D"_x0000_i1037" type=3D"= #_x0000_t75" alt=3D" " style=3D'width:213pt;height:42pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image015.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00011.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D284 height=3D56 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image015.gif" alt=3D" " v:shapes= =3D"_x0000_i1037"><![endif]><span style=3D'mso-tab-count:1'>     </span>(1.3)<o:p></o:p><= /span></p> <p style=3D'margin:0cm;margin-bottom:.0001pt'><span style=3D'font-size:10.0= pt; font-family:Arial'><o:p> </o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Н&= #1072; самом деле зона пластичносm= 0;и несколько больше (рис. 1.6, б). Общие выражения для размера зоны пластичносm= 0;и рассмотренm= 9; в гл. V. Здесь достаточно отметить, чт= 086; <span class=3DSpellE><i>r*<sub>р</sub></i></span> можно непосредстk= 4;енно выразить ка = 82; функцию коэффициенm= 0;а интенсивноl= 9;ти напряжений = 80; предела текучести.<o:p></o:p>= </span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>В&= #1099;ше утверждалоl= 9;ь, что в зоне упругости т = 88;ещины различных размеров, но с одинаковы = 84;и <i>К</i><sub>I</sub> имеют одинаковые = 87;оля напряжений. Возникает вопрос: справедливl= 6; ли это утверждениk= 7; в случае, когда матер = 80;ал испытывает пластическl= 0;е деформации? Согласно уравнению (1.3), трещины, нагруженныk= 7; до одинаковых значений <i>К</i><sub= >I</sub><i>, </i>имеют зоны пластичносm= 0;и одинаковых размеров. Вн= 077; зоны пласти = 95;ности поля напряжений будут одина = 82;овыми. Если две трещины имеют одинаковые = 87;ластически&#= 1077; зоны и одинаковые напряжения на границе этой зоны, то напряжения = 80; деформации внутри зоны пластичносm= 0;и должны быть равными.<o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>И&= #1085;ыми словами, пол= 077; напряжений определяетl= 9;я коэффициенm= 0;ом интенсивноl= 9;ти напряжений. Этим коэффициенm= 0;ом определяетl= 9;я также то, что происходит внутри зоны пластичносm= 0;и. <i>К</i><sub>I</sub> есть мера всех напряжений = 80; деформаций. Когда напряжения = 80; деформации при вершине = 090;рещины достигают критическиm= 3; значений, пр= 086;исходит расширение трещины. Это означает, чт= 086; при достижении <i>&= #1050;</i><sub>I</sub> критическоk= 5;о значения <span class=3DSpel= lE><i>К</i><sub>Ic</sub></span> произойдет разрушение. Можно предполагаm= 0;ь, что <span class=3DSpellE><i>К</i><sub>Ic</sub></spa= n> есть константа материала.<o:p></o:p>= </span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>В&= #1086;зьмем пластину с трещиной известного = 88;азмера и растянем е= 077; в испытательl= 5;ой машине впло = 90;ь до разрушения. По величине нагрузки, пр= 080; которой произошло разрушение, можно вычис = 83;ить разрушающеk= 7; напряжение <span class=3DSpellE>σ<i><sub>с</sub></i></span><i>. </i>От= ;сюда, зная <span class=3DSpellE>σ<i><sub>с</su= b></i></span>, можно найти критическоk= 7; значение коэффициенm= 0;а интенсивноl= 9;ти напряжений = 74; момент разрушения:<o:p= ></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'><o:p>&nb= sp;</o:p></span></p> <p style=3D'margin-top:0cm;margin-right:0cm;margin-bottom:0cm;margin-left:1= .5pt; margin-bottom:.0001pt;tab-stops:228.35pt'><span style=3D'font-size:10.0pt; font-family:Arial'><!--[if gte vml 1]><v:shape id=3D"_x0000_i1038" type=3D"= #_x0000_t75" alt=3D" " style=3D'width:68.25pt;height:20.25pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image016.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00012.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D91 height=3D27 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image016.gif" alt=3D" " v:shapes= =3D"_x0000_i1038"><![endif]><span style=3D'mso-tab-count:1'>        &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;       </span>(1.4)<o:p></o:p></span></p> <p style=3D'margin:0cm;margin-bottom:.0001pt'><span style=3D'font-size:10.0= pt; font-family:Arial'><o:p> </o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Е&= #1089;ли <span class=3DSpellE><i>К</i><sub>Ic</sub></span> — константа материала, т= 086; такое же значение должно быть получено пр = 80; испытании образца с трещиной иного размера. В определеннm= 9;х пределах эт = 86; действителn= 0;но имеет место. Зная величину <span class=3DSpel= lE><i>К</i><sub>Ic</sub></span><i>, </i>можно рассчитать прочность такого же ма= 090;ериала с трещинами любых размеров. Можно также рассчитать, какой разме = 88; трещины доп = 91;стим в материале, напряженноl= 4; до заданног = 86; уровня. В реальных условиях ситуация не = 89;колько сложнее. Во-первых, выражение (1.4) для коэффициенm= 0;а интенсивноl= 9;ти напряжений справедливl= 6; лишь для бес= 082;онечной пластины. Дл= 103; пластины конечных размеров эт = 72; формула принимает вид (см. гл. III)<o:p></o:p></spa= n></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'><o:p>&nb= sp;</o:p></span></p> <p style=3D'margin-top:0cm;margin-right:0cm;margin-bottom:0cm;margin-left:1= .5pt; margin-bottom:.0001pt;tab-stops:228.35pt'><span style=3D'font-size:10.0pt; font-family:Arial'><!--[if gte vml 1]><v:shape id=3D"_x0000_i1039" type=3D"= #_x0000_t75" alt=3D" " style=3D'width:106.5pt;height:21pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image017.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00013.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D142 height=3D28 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image017.gif" alt=3D" " v:shapes= =3D"_x0000_i1039"><![endif]><span style=3D'mso-tab-count:1'>        &= nbsp;           &nbs= p;            &= nbsp;      </span>(1.5)<o:p></o:p></span></p> <p style=3D'margin:0cm;margin-bottom:.0001pt'><span style=3D'font-size:10.0= pt; font-family:Arial'><o:p> </o:p></span></p> <p class=3Dbase style=3D'text-indent:0cm'><span style=3D'font-size:10.0pt;f= ont-family: Arial'>где <i>W</i> — ширина пластины. Дл= 103; определениn= 3; <span class=3DSpellE><i>K</i><sub>Ic</sub></span> необходимо знать функцию <span class=3DSpellE><i>f= </i></span><i>(<span class=3DSpellE>a</span>/W)</i>. Безусловно, <sp= an class=3DSpellE><i>f</i></span><i>(<span class=3DSpellE>a</span>/W)</i> для малых значений <span class=3DSpel= lE><i>a</i></span><i>/W</i> стремится к единице. Во-вторых, необходимо = 85;аложить ограничениk= 7; на поперечные деформации = 74; пластине. Истинное значение <span class=3DSpel= lE><i>K</i><sub>Ic</sub></span> можно получить опытным путем тольк = 86; в том случае, если перемещениn= 3; точек пластины пе = 88;пендикуляр&#= 1085;о ее плоскост = 80; достаточно = 84;алы, т.е.когда имеет место условие плоского деформировk= 2;ния, что наблюдаетсn= 3;, когда пласт = 80;на имеет достаточнуn= 2; толщину (см. гл. IV, VII). Если деформации = 74; направлениl= 0;, перпендикуl= 3;ярном плоскости пластины, ничем не ограничены (случай плоского на = 87;ряженного состояния), то критическаn= 3; величина коэффициенm= 0;а интенсивноl= 9;ти напряжении будет зависеть от толщины пластины (см. гл. IV, VIII).<o:p></o:p></span></p> <p class=3Dbase><span class=3DSpellE><i><span style=3D'font-size:10.0pt;fon= t-family: Arial'>К</span></i><sub><span style=3D'font-size:10.0pt;font-family:A= rial'>Ic</span></sub></span><span style=3D'font-size:10.0pt;font-family:Arial'> есть мера <span class=3DSpellE>тре = 97;иностойкос&#= 1090;и</span> материала. Поэтому <span class=3DSpellE><i>&= #1050;</i><sub>Ic</sub></span> называют «вязкостью разрушения при плоском деформировk= 2;нном состоянии». Для материа = 83;ов с малой вязкостью разрушения допускаютсn= 3; только маленькие трещины. Типичные величины вязкости ра = 79;рушения для трех различных высокопрочl= 5;ых материалов приведены в табл. 1.1.<o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Д&= #1083;я материалов, представлеl= 5;ных в табл. 1.1, допу = 89;тимый размер трещины, при котором про = 95;ность уменьшаетсn= 3; вдвое по сравнению с ее исходным значением, можно определить следующим образом:<o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'><o:p>&nb= sp;</o:p></span></p> <p style=3D'margin-top:0cm;margin-right:0cm;margin-bottom:0cm;margin-left:1= .5pt; margin-bottom:.0001pt;tab-stops:228.35pt'><span style=3D'font-size:10.0pt; font-family:Arial'><!--[if gte vml 1]><v:shape id=3D"_x0000_i1040" type=3D"= #_x0000_t75" alt=3D" " style=3D'width:221.25pt;height:21.75pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image018.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00014.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D295 height=3D29 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image018.gif" alt=3D" " v:shapes= =3D"_x0000_i1040"><![endif]><span style=3D'mso-tab-count:1'>  </span>(1.6)<o:p></o:p></span></p> <p style=3D'margin:0cm;margin-bottom:.0001pt;text-indent:15.3pt'><span style=3D'font-size:10.0pt;font-family:Arial'><o:p> </o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Л&= #1077;гко видеть, что в стали 4340 допустимы трещины размером <i>2a</i>=3D2.6 мм, тогда как для легированнl= 6;й стали допус = 90;има <span class=3DSpellE>трещинa</span> размером 2<i>a</i> =3D <st1= :metricconverter ProductID=3D"6.4 мм" w:st=3D"on">6.4 мм</st1:metric= converter>, а для алюминиевоk= 5;о сплава размером 2<i>a</i> =3D <st1= :metricconverter ProductID=3D"8.8 мм" w:st=3D"on">8.8 мм</st1:metric= converter><o:p></o:p></span></p> <p class=3DMsoNormal style=3D'tab-stops:24.3pt 483.3pt'><span style=3D'font= -size: 10.0pt;font-family:Arial'><span style=3D'mso-tab-count:1'>   = ;     </span><!--[if gte vml 1]><v:shape id=3D"_x0000_i1041" type=3D"#_x0000_t75" alt=3D"" style=3D'width:12.75pt;h= eight:19.5pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image001.png" o:hr= ef=3D"http://www.mysopromat.ru/images/0.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D17 height=3D26 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image002.gif" v:shapes=3D"_x0000= _i1041"><![endif]><o:p></o:p></span></p> <p class=3DMsoNormal style=3D'tab-stops:24.3pt 483.3pt'><span style=3D'font= -size: 10.0pt;font-family:Arial'><span style=3D'mso-tab-count:1'>   = ;     </span></span><span style=3D'font-size:10.0pt'><span style=3D'mso-tab-count:1'>  &nbs= p;            &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p; </span></span><span style=3D'font-size:10.0pt;font-family:Arial'><o:p></o:p></span></p> <p class=3DMsoNormal align=3Dcenter style=3D'text-align:center'><span style=3D'display:none;mso-hide:all'><o:p> </o:p></span></p> <p style=3D'margin:0cm;margin-bottom:.0001pt'><span style=3D'font-size:10.0= pt; font-family:Arial'><o:p> </o:p></span></p> <p align=3Dright style=3D'margin:0cm;margin-bottom:.0001pt;text-align:right= '><b><span style=3D'font-size:10.0pt;font-family:Arial'>Табл&#= 1080;ца 1.1</span></b><span style=3D'font-size:10.0pt;font-family:Arial'><o:p></o:p= ></span></p> <div align=3Dcenter> <table class=3DMsoNormalTable border=3D1 cellspacing=3D0 cellpadding=3D0 wi= dth=3D630 style=3D'width:472.5pt;mso-cellspacing:0cm;border:solid windowtext 1.0pt; mso-border-alt:solid windowtext .5pt;mso-padding-alt:1.5pt 1.5pt 1.5pt 1.5= pt; mso-border-insideh:.5pt solid windowtext;mso-border-insidev:.5pt solid win= dowtext'> <tr style=3D'mso-yfti-irow:0;mso-yfti-firstrow:yes'> <td width=3D105 rowspan=3D2 valign=3Dtop style=3D'width:78.75pt;border:so= lid windowtext 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>Мате= риал<o:p></o:p></span></p> </td> <td width=3D157 colspan=3D3 valign=3Dtop style=3D'width:117.75pt;border:s= olid windowtext 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>Врем= енное<o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>сопр= отивление<o:p></o:p><= /span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>разр= ыву <span class=3DSpellE>σ<i><sub><span style=3D'font-size:11.0pt'>u</sp= an></sub></i></span><o:p></o:p></span></p> </td> <td width=3D153 colspan=3D3 valign=3Dtop style=3D'width:114.75pt;border:s= olid windowtext 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>Пред= ел текучести <span class=3DSpellE>σ<i><sub><span style=3D'font-size:11.0pt'>ys</span></= sub></i></span><o:p></o:p></span></p> </td> <td width=3D197 rowspan=3D2 valign=3Dtop style=3D'width:147.75pt;border:s= olid windowtext 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>Вязк= ость разрушения </= span><span class=3DSpellE><i><span style=3D'font-size:11.0pt;font-family:Arial'>K</s= pan></i><sub><span style=3D'font-size:11.0pt;font-family:Arial'>Ic</span></sub></span><span style=3D'font-size:10.0pt;font-family:Arial'><o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'><o:p> </o:p></span></p> <p style=3D'margin:0cm;margin-bottom:.0001pt'><span style=3D'font-size:10= .0pt; font-family:Arial'><o:p> </o:p></span></p> </td> </tr> <tr style=3D'mso-yfti-irow:1;height:56.25pt'> <td width=3D54 valign=3Dtop style=3D'width:40.5pt;border:solid windowtext= 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt; height:56.25pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>MH/м</span><sup><span style=3D'font-size:11.0pt;font-family:Arial'>2</span></sup><span style=3D'font-size:10.0pt;font-family:Arial'><o:p></o:p></span></p> </td> <td width=3D57 valign=3Dtop style=3D'width:42.75pt;border:solid windowtex= t 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt; height:56.25pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>кгс/м= ;м</span><sup><span style=3D'font-size:11.0pt;font-family:Arial'>2</span></sup><span style=3D'font-size:10.0pt;font-family:Arial'><o:p></o:p></span></p> </td> <td width=3D38 valign=3Dtop style=3D'width:28.5pt;border:solid windowtext= 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt; height:56.25pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span class=3DSpellE><span style=3D'font-size:10.0pt;font-family:Arial'>к= си</span></span><span style=3D'font-size:10.0pt;font-family:Arial'><o:p></o:p></span></p> </td> <td width=3D53 valign=3Dtop style=3D'width:39.75pt;border:solid windowtex= t 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt; height:56.25pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>MH/м</span><sup><span style=3D'font-size:11.0pt;font-family:Arial'>2</span></sup><span style=3D'font-size:10.0pt;font-family:Arial'><o:p></o:p></span></p> </td> <td width=3D57 valign=3Dtop style=3D'width:42.75pt;border:solid windowtex= t 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt; height:56.25pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>кгс/м= ;м</span><sup><span style=3D'font-size:11.0pt;font-family:Arial'>2</span></sup><span style=3D'font-size:10.0pt;font-family:Arial'><o:p></o:p></span></p> </td> <td width=3D35 valign=3Dtop style=3D'width:26.25pt;border:solid windowtex= t 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt; height:56.25pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span class=3DSpellE><span style=3D'font-size:10.0pt;font-family:Arial'>к= си</span></span><span style=3D'font-size:10.0pt;font-family:Arial'><o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'><o:p> </o:p></span></p> <p style=3D'margin:0cm;margin-bottom:.0001pt'><span style=3D'font-size:10= .0pt; font-family:Arial'><o:p> </o:p></span></p> </td> </tr> <tr style=3D'mso-yfti-irow:2'> <td width=3D105 valign=3Dtop style=3D'width:78.75pt;border:solid windowte= xt 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>Стал= ь 4340<o:p></o:p></span></p> </td> <td width=3D54 valign=3Dtop style=3D'width:40.5pt;border:solid windowtext= 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>1820<o:p></o:p></span></p> </td> <td width=3D57 valign=3Dtop style=3D'width:42.75pt;border:solid windowtex= t 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>185<o:p></o:p></span></p> </td> <td width=3D38 valign=3Dtop style=3D'width:28.5pt;border:solid windowtext= 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>264<o:p></o:p></span></p> </td> <td width=3D53 valign=3Dtop style=3D'width:39.75pt;border:solid windowtex= t 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>1470<o:p></o:p></span></p> </td> <td width=3D57 valign=3Dtop style=3D'width:42.75pt;border:solid windowtex= t 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>150<o:p></o:p></span></p> </td> <td width=3D35 valign=3Dtop style=3D'width:26.25pt;border:solid windowtex= t 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>214<o:p></o:p></span></p> </td> <td width=3D197 valign=3Dtop style=3D'width:147.75pt;border:solid windowt= ext 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>46 МН/м</s= pan><sup><span style=3D'font-size:11.0pt;font-family:Arial'>3/2</span></sup><span style=3D'font-size:10.0pt;font-family:Arial'> =3D 150кгс/мм</span><sup><span style=3D'font-si= ze: 11.0pt;font-family:Arial'>3/2</span></sup><span style=3D'font-size:10.0pt; font-family:Arial'> =3D 42кси<!--[if gte vml 1]><v:shape id=3D"_x0000_i1042" type=3D"#_x0000_t75" alt=3D" " style=3D'width:40.5pt= ;height:18pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image019.gif" o:= href=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00015.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D54 height=3D24 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image019.gif" alt=3D" " v:shap= es=3D"_x0000_i1042"><![endif]><o:p></o:p></span></p> </td> </tr> <tr style=3D'mso-yfti-irow:3'> <td width=3D105 valign=3Dtop style=3D'width:78.75pt;border:solid windowte= xt 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>Леги= рованная<o:p></o:p></span><= /p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>стал= ь 300<o:p></o:p></span></p> </td> <td width=3D54 valign=3Dtop style=3D'width:40.5pt;border:solid windowtext= 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>1850<o:p></o:p></span></p> </td> <td width=3D57 valign=3Dtop style=3D'width:42.75pt;border:solid windowtex= t 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>188<o:p></o:p></span></p> </td> <td width=3D38 valign=3Dtop style=3D'width:28.5pt;border:solid windowtext= 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>268<o:p></o:p></span></p> </td> <td width=3D53 valign=3Dtop style=3D'width:39.75pt;border:solid windowtex= t 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>1730<o:p></o:p></span></p> </td> <td width=3D57 valign=3Dtop style=3D'width:42.75pt;border:solid windowtex= t 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>177<o:p></o:p></span></p> </td> <td width=3D35 valign=3Dtop style=3D'width:26.25pt;border:solid windowtex= t 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>250<o:p></o:p></span></p> </td> <td width=3D197 valign=3Dtop style=3D'width:147.75pt;border:solid windowt= ext 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>90 МН/м</s= pan><sup><span style=3D'font-size:11.0pt;font-family:Arial'>3/2</span></sup><span style=3D'font-size:10.0pt;font-family:Arial'> =3D 290кгс/мм</span><sup><span style=3D'font-si= ze: 11.0pt;font-family:Arial'>3/2</span></sup><span style=3D'font-size:10.0pt; font-family:Arial'> =3D 82кси<!--[if gte vml 1]><v:shape id=3D"_x0000_i1043" type=3D"#_x0000_t75" alt=3D" " style=3D'width:40.5pt= ;height:18pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image019.gif" o:= href=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00016.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D54 height=3D24 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image019.gif" alt=3D" " v:shap= es=3D"_x0000_i1043"><![endif]><o:p></o:p></span></p> </td> </tr> <tr style=3D'mso-yfti-irow:4;mso-yfti-lastrow:yes'> <td width=3D105 valign=3Dtop style=3D'width:78.75pt;border:solid windowte= xt 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>Алюм= иниевый<o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>спла= в 7075-Т6<o:p></o:p></span></p> </td> <td width=3D54 valign=3Dtop style=3D'width:40.5pt;border:solid windowtext= 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>560<o:p></o:p></span></p> </td> <td width=3D57 valign=3Dtop style=3D'width:42.75pt;border:solid windowtex= t 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>57<o:p></o:p></span></p> </td> <td width=3D38 valign=3Dtop style=3D'width:28.5pt;border:solid windowtext= 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>81<o:p></o:p></span></p> </td> <td width=3D53 valign=3Dtop style=3D'width:39.75pt;border:solid windowtex= t 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>500<o:p></o:p></span></p> </td> <td width=3D57 valign=3Dtop style=3D'width:42.75pt;border:solid windowtex= t 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>51<o:p></o:p></span></p> </td> <td width=3D35 valign=3Dtop style=3D'width:26.25pt;border:solid windowtex= t 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>73<o:p></o:p></span></p> </td> <td width=3D197 valign=3Dtop style=3D'width:147.75pt;border:solid windowt= ext 1.0pt; mso-border-alt:solid windowtext .5pt;padding:1.5pt 1.5pt 1.5pt 1.5pt'> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:ce= nter'><span style=3D'font-size:10.0pt;font-family:Arial'>32 МН/м</s= pan><sup><span style=3D'font-size:11.0pt;font-family:Arial'>3/2</span></sup><span style=3D'font-size:10.0pt;font-family:Arial'> =3D 104кгс/мм</span><sup><span style=3D'font-si= ze: 11.0pt;font-family:Arial'>3/2</span></sup><span style=3D'font-size:10.0pt; font-family:Arial'> =3D 30кси<!--[if gte vml 1]><v:shape id=3D"_x0000_i1044" type=3D"#_x0000_t75" alt=3D" " style=3D'width:40.5pt= ;height:18pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image019.gif" o:= href=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00017.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D54 height=3D24 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image019.gif" alt=3D" " v:shap= es=3D"_x0000_i1044"><![endif]><o:p></o:p></span></p> </td> </tr> </table> </div> <p style=3D'margin:0cm;margin-bottom:.0001pt'><span style=3D'font-size:10.0= pt; font-family:Arial'><o:p> </o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Л&= #1077;гко видеть, что в стали 4340допустимы трещины размером <i>2a</i>=3D2,6 мм, тогда как для легированнl= 6;й стали допус = 90;има <span class=3DSpellE>трещинa</span> размером 2<i>a</i> =3D <st1= :metricconverter ProductID=3D"6,4 мм" w:st=3D"on">6,4 мм</st1:metric= converter>, а для алюминиевоk= 5;о сплава разм = 77;ром 2<i>a</i> =3D <st1:metricconverter ProductID=3D"8,8 мм" w:st=3D= "on">8,8 мм</st1:metricconverter>.<o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Н&= #1072; рис. 1.7, <span class=3DSpellE><i>a</i></span> в виде кривых изображена остаточная прочность трех материалов как функция длины трещи = 85;ы. Эти кривые определяютl= 9;я выражением <!--= [if gte vml 1]><v:shape id=3D"_x0000_i1045" type=3D"#_x0000_t75" alt=3D" " style=3D'width:65.25pt;= height:20.25pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image020.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00018.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D87 height=3D27 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image020.gif" alt=3D" " v:shapes= =3D"_x0000_i1045"><![endif]>Из этой формул = 99; следует, что <s= pan class=3DSpellE>σ<sub>c</sub></span> бесконечныl= 4; при приближениl= 0; <i>а</i> к нулю. На самом деле, при <i>a</i>=3D0 криваn= 3; должна приближатьl= 9;я к значению <span class=3DSpellE>σ<i><sub>c</sub></i></span>=3D<span class=3DSpellE>`= 3;<i><sub>u</sub></i></span><sub> </sub>(см. гл. VII, VIII, IX).<o:p></o:p></span></p> <p style=3D'margin:0cm;margin-bottom:.0001pt'><span style=3D'font-size:10.0= pt; font-family:Arial'><o:p> </o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><span style=3D'font-size:10.0pt;font-family:Arial'><!--[if gte vml 1]><v:shape id= =3D"_x0000_i1046" type=3D"#_x0000_t75" alt=3D" " style=3D'width:231pt;height:358.5pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image021.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00019.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D308 height=3D478 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image021.gif" alt=3D" " v:shapes= =3D"_x0000_i1046"><![endif]><o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><b><span style=3D'font-size:10.0pt;font-family:Arial'>Рис. 1.7. Вязкость разрушения трех высокопрочl= 5;ых материалов:</sp= an></b><span style=3D'font-size:10.0pt;font-family:Arial'><o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><b><span style=3D'font-size:10.0pt;font-family:Arial'>а — остаточная прочность как функция размера трещины; <i>б — </i>&= #1086;тносительн= ;ая остаточная прочность</span></b><= span style=3D'font-size:10.0pt;font-family:Arial'><o:p></o:p></span></p> <p style=3D'margin:0cm;margin-bottom:.0001pt'><span style=3D'font-size:10.0= pt; font-family:Arial'><o:p> </o:p></span></p> <p class=3DMsoNormal><span style=3D'font-size:10.0pt;font-family:Arial'>= 054;чевидно, материал с наибольшей вязкостью разрушения имеет наибольшую остаточную прочность. Если нанест = 80; на график значения отношения предела прочности к исходной прочности (д= 086; образованиn= 3; трещин), то по&= #1083;учится совершенно иная картин = 72; (рис. 1.7, <i>б)</i>. При одинаковых относительl= 5;ых потерях про = 95;ности алюминиевыl= 1; сплав допускает более длинн = 99;е трещины, чем другие материалы. Это происхо = 76;ит потому, что алюминиевыl= 1; сплав имеет наибольшее отношение вязкости к прочности н = 72; разрыв (рис. 1.7, <i>= б</i>).</span></p> <p class=3DMsoNormal align=3Dcenter style=3D'text-align:center'><span style=3D'display:none;mso-hide:all'><o:p> </o:p></span></p> <p class=3DMsoNormal><span style=3D'font-size:10.0pt;font-family:Arial'><o:= p> </o:p></span></p> <h2 style=3D'margin:0cm;margin-bottom:.0001pt'><span style=3D'font-family:A= rial; color:windowtext'>§1.4. Критерий <span class=3DSpel= lE>Гриффитса</span><o= :p></o:p></span></h2> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Н&= #1077;смотря на то что механика разрушения получила свое развитие главным обр = 72;зом за последни = 77; два десятилетиn= 3;, одно из ее основных уравнений было получено <span class=3DSpel= lE>Гриффитсом</= span> [9,10] еще в <st1:metricconverter ProductID=3D"1921= г" w:st=3D"on">1921 г</st1:metricconverter>. Рассмотрим бесконечнуn= 2; пластину единичной толщины с центральноl= 1; поперечной трещиной длиной 2<i>a</i>. Края пластины неподвижны, = 072; напряжение = 74; ней равно <span class=3DSpe= llE>σ</span>, как показан = 86; на рис. 1.8, <i>а</i>. На рис. 1.8, <i>б</i> приведена диаграмма «нагрузка—m= 1;длинение».<o:p></o:= p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><span style=3D'font-size:10.0pt;font-family:Arial'><!--[if gte vml 1]><v:shape id= =3D"_x0000_i1047" type=3D"#_x0000_t75" alt=3D"" style=3D'width:235.5pt;height:148.5pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image022.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00020.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D314 height=3D198 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image022.gif" v:shapes=3D"_x0000= _i1047"><![endif]><o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><b><span style=3D'font-size:10.0pt;font-family:Arial'>Рис. 1.8. Критерий <span class=3DSpel= lE>Гриффитса</span> при неподвижныm= 3; захватах:</span></b><span style=3D'font-size:10.0pt;font-family:Arial'><o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><b><span style=3D'font-size:10.0pt;font-family:Arial'>а — пластина с трещиной с неподвижныl= 4;и краями;</span></b><span style=3D'font-s= ize: 10.0pt;font-family:Arial'><o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><b><span style=3D'font-size:10.0pt;font-family:Arial'>б — энергия упругих деформаций пластины с т= 088;ещиной длиной <span class=3DSpellE><i>a</i></s= pan></span></b><span style=3D'font-size:10.0pt;font-family:Arial'> <b>(1) и <span class=3D= SpellE><i>а+da</i></span></b> <b>(2)</b><o:p></o:p></span></p> <p class=3DMsoNormal><span style=3D'font-size:10.0pt;font-family:Arial'><o:= p> </o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>З&= #1072;пасенная в пластине упругая энергия пре = 76;ставлена площадью <i>OAB</i>. Если длина трещины увеличится на величину = 706;<span class=3DSpellE><i>a</i></span><i>, </i>то жесткость пластины ум = 77;ньшится (линия <i>ОС</i>); это означае = 90;, что нагрузк = 72; несколько уменьшится, поскольку края пластины неподвижны. Следователn= 0;но, упругая энергия, запасенная = 74; пластине, уменьшится до величины, равной площади <i>ОСВ.= </i>Увеличение длины трещины с <span class=3DSpe= llE><i>a</i></span> до <span class=3DSpellE><i>a</i>+<i>dа</i></span> приведет к освобожденl= 0;ю упругой энергии, равной по величине площади <i>ОАС.= </i><o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Е&= #1089;ли пластина нагружена д = 86; более высокого напряжения, то при увеличении длины трещи = 85;ы на величину <sp= an class=3DSpellE><i>dа</i></span><i> </i>осво&#= 1073;одится большая энергия. <span class=3DSpellE>= 043;риффитс</span> предположиl= 3;, что трещина будет расти лишь в том случае, если освобождаеl= 4;ая при этом энергия достаточна для обеспечениn= 3; всех затрат энергии, свя= 079;анных с этим ростом. В противном случае необ = 93;одимо увеличить напряжение. Треугольниl= 2; <i>ODE</i> представляk= 7;т собой энергию, выделяемую = 87;ри распростраl= 5;ении трещины.<o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>У&= #1089;ловие, необходимоk= 7; для роста трещины, следующее:<o:p></o:p>= </span></p> <p style=3D'margin-top:0cm;margin-right:0cm;margin-bottom:0cm;margin-left:1= .5pt; margin-bottom:.0001pt;tab-stops:228.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><!--[if gte vml 1]><v:shape id=3D"_x0000_i1048" type=3D"= #_x0000_t75" alt=3D"" style=3D'width:91.5pt;height:15.75pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image023.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00021.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D122 height=3D21 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image023.gif" v:shapes=3D"_x0000= _i1048"><![endif]><span style=3D'mso-tab-count:1'>        &= nbsp;           &nbs= p;            &= nbsp;           </span>(1= .7)<o:p></o:p></span></p> <p class=3Dbase style=3D'text-indent:0cm'><span style=3D'font-size:10.0pt;f= ont-family: Arial'>где <i>U</i> — упругая энергия, а <i>W</i> — энергия, необходимаn= 3; для роста трещины. Осн= 086;вываясь на расчетах поля напряж = 77;ний для эллиптичесl= 2;ого отверстия, выполненныm= 3; <span class=3DSpellE>Инглисом</sp= an> [11], <span class=3DSpellE>Гриффит= ;с</span> получил выражение для <span class=3DSpellE><i>dU</i></span><i>/<span class=3DSpellE>da</span></i> в виде<o:p></o:p= ></span></p> <p style=3D'margin-top:0cm;margin-right:0cm;margin-bottom:0cm;margin-left:1= .5pt; margin-bottom:.0001pt;tab-stops:228.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><!--[if gte vml 1]><v:shape id=3D"_x0000_i1049" type=3D"= #_x0000_t75" alt=3D"" style=3D'width:92.25pt;height:15.75pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image024.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00022.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D123 height=3D21 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image024.gif" v:shapes=3D"_x0000= _i1049"><![endif]><span style=3D'mso-tab-count:1'>        &= nbsp;           &nbs= p;            &= nbsp;           </span>(1= .8)<o:p></o:p></span></p> <p class=3Dbase style=3D'text-indent:0cm'><span style=3D'font-size:10.0pt;f= ont-family: Arial'>на единицу толщины пластины, гд= 077; <i>Е </i>— модуль Юнга. Обычно величину <span class=3DSpel= lE><i>dU</i></span><i>/<span class=3DSpellE>dа</span> </i>заменn= 3;ют величиной<o:p></o:p><= /span></p> <p style=3D'margin-top:0cm;margin-right:0cm;margin-bottom:0cm;margin-left:1= .5pt; margin-bottom:.0001pt;tab-stops:228.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><!--[if gte vml 1]><v:shape id=3D"_x0000_i1050" type=3D"= #_x0000_t75" alt=3D"" style=3D'width:65.25pt;height:15.75pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image025.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00023.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D87 height=3D21 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image025.gif" v:shapes=3D"_x0000= _i1050"><![endif]><span style=3D'mso-tab-count:1'>        &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;        </span>(1.9)<o:p></o:p></span>= </p> <p class=3Dbase style=3D'text-indent:0cm'><span style=3D'font-size:10.0pt;f= ont-family: Arial'>которая называется «скоростью высвобождеl= 5;ия упругой энергии», приходящейl= 9;я на каждую вершину трещины. Величину <i>G</i> называют также <span class=3DSpellE>тр = 77;щинодвижущ&#= 1077;й</span> силой; ее размерностn= 0; — энергия, деленная на единицу толщины пластины и н= 072; единицу изм = 77;нения длины трещины, что также может быть представлеl= 5;о в виде силы, приходящейl= 9;я на единицу изменения длины трещи = 85;ы.<o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Э&= #1085;ергию, расходуемуn= 2; на распростраl= 5;ение трещины, обозначают через <span class=3DSpellE><i>R=3DdW</i></spa= n><i>/<span class=3DSpellE>da</span></i> и называют сопротивлеl= 5;ием росту трещины. В первом приближениl= 0; можно считать, что энергия, необходимаn= 3; для образованиn= 3; трещины (для разрыва атомных связей), одинакова д = 83;я любых приращений <span class=3DSpellE><i>dа</i></span><i>. </i>Это означает, чт= 086; <i>R</i> — константа.<o:p></o:p>= </span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Т&= #1077;перь энергетичеl= 9;кое условие (1.7) можно переф = 88;азировать следующим образом: для распростраl= 5;ения трещин необходимо, чтобы <i>G</i> было, по крайней мере, равно <i>R.</i> Если <i>R</i> — константа, то, значит, величина <i>G</i> должна превысить некоторое критическоk= 7; значение <span class=3DSpel= lE><i>G</i><sub>Ic</sub></span>. Следователn= 0;но, распростраl= 5;ение происходит при следующем условии:<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:0cm;margin-bottom:0cm;margin-left:1= .5pt; margin-bottom:.0001pt;tab-stops:228.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><!--[if gte vml 1]><v:shape id=3D"_x0000_i1051" type=3D"= #_x0000_t75" alt=3D"" style=3D'width:187.5pt;height:21pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image026.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00024.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D250 height=3D28 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image026.gif" v:shapes=3D"_x0000= _i1051"><![endif]><span style=3D'mso-tab-count:1'>        &= nbsp;    </span>(1.10)<o:p></o:p></span></p> <p class=3Dbase style=3D'text-indent:0cm'><span style=3D'font-size:10.0pt;f= ont-family: Arial'>Критическl= 6;е значение <i>G</i><sub>1 = 89;</sub> (критическу = 02; скорость высвобождеl= 5;ия энергии) можно получить, измерив напряжение <span class=3DSpellE>σ<i><sub>с</sub></i></span><i>, </i>не= ;обходимое для разрушения пластины с т= 088;ещиной размером 2<i>а</i>, и вычислив и= 079; уравнения (1.10) величину <i>G</i><sub>1 = 89;</sub>.<o:p></o:p></span></p> <p class=3Dbase><span class=3DSpellE><span style=3D'font-size:10.0pt;font-f= amily: Arial'>Гриффитc</span></span><span style=3D'font-size:10.0pt;font-family:Arial'> выве&= #1083; свое уравнение для стекла — очень хрупк = 86;го материала. О= 085; предположиl= 3;, что величин = 72; <i>R</i> определяетl= 9;я только поверхностl= 5;ой энергией. В вязких материалах, например металлах, пр= 080; вершине трещины образуются пластическl= 0;е деформации. Для образованиn= 3; новой зоны пластическl= 0;х деформаций при вершине трещины необходима большая энергия. Поскольку эта пластическk= 2;я зона должна быть образована = 74; процессе ро = 89;та трещины, то энергию, необходимуn= 2; для распрос = 90;ранения трещины, можно положить ра = 74;ной энергии, необходимоl= 1; для образованиn= 3; этой трещины. Это означает, чт= 086; в металлах в= 077;личина <i>R</i> определяетl= 9;я главным образом энергией де = 92;ормации в пластическl= 6;й зоне; поверхностl= 5;ая энергия в этом случае настолько мала, что ею можно пренебречь (см. [12, 13]). Энергетичеl= 9;кий критерий есть необходимоk= 7; условие рас = 87;ространени&#= 1103; трещины. Это= 090; критерий не обязательнl= 6; должен быть достаточныl= 4;. Если материал пр = 80; вершине трещины не находится н = 72; грани разрушения, то трещина н= 077; будет расти даже при достаточноl= 1; энергии для ее развития: материал должен до конца исчер = 87;ать свою способностn= 0; воспринимаm= 0;ь нагрузку и п= 088;одолжать деформировk= 2;ться. Однако последний критерий эквивалентk= 7;н энергетичеl= 9;кому критерию, поскольку и = 79; уравнений (1.2) l= 0; (1.9) следует, чт = 86;<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:0cm;margin-bottom:0cm;margin-left:1= .5pt; margin-bottom:.0001pt;tab-stops:228.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><!--[if gte vml 1]><v:shape id=3D"_x0000_i1052" type=3D"= #_x0000_t75" alt=3D" " style=3D'width:42pt;height:33pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image027.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00025.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D56 height=3D44 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image027.gif" alt=3D" " v:shapes= =3D"_x0000_i1052"><![endif]><span style=3D'mso-tab-count:1'>        &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;            &= nbsp;   </span>(1.11)<o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>О&= #1095;евидно, критерий по напряженияl= 4; и энергетич = 77;ский критерий выполняютсn= 3; одновременl= 5;о. Следователn= 0;но, уравнения (1.4) l= 0; (1.10) эквиваленm= 0;ны. В гл. III будет показано, чт= 086; уравнение (1.11) справедливl= 6; для случая плоского напряженноk= 5;о состояния, а в случае пло= 089;кого деформировk= 2;нного состояния его следует дополнить коэффициенm= 0;ом (1 — ν<sup>2</sup>), что приведет к соотношениn= 3;м<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:0cm;margin-bottom:0cm;margin-left:1= .5pt; margin-bottom:.0001pt;tab-stops:228.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><!--[if gte vml 1]><v:shape id=3D"_x0000_i1053" type=3D"= #_x0000_t75" alt=3D"" style=3D'width:162pt;height:39pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image028.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00026.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D216 height=3D52 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image028.gif" v:shapes=3D"_x0000= _i1053"><![endif]><span style=3D'mso-tab-count:1'>        &= nbsp;           &nbs= p; </span>(1.12)<o:p></o:p></span></p> <p class=3DMsoNormal><span style=3D'font-size:10.0pt;font-family:Arial'><o:= p> </o:p></span></p> <h2 style=3D'margin:0cm;margin-bottom:.0001pt'><span style=3D'font-family:A= rial; color:windowtext'>§ 1.5. Критерий предельногl= 6; раскрытия трещины<o:p></o:p></span></h2> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>В&= #1099;сокопрочны= ;е материалы обычно имею = 90; малую вязкость разрушения. Задачи разрушения = 74; этих материалах для случая плоского де = 92;ормированн&#= 1086;го состояния с успехом мог = 91;т быть исследованm= 9; методами механики ра = 79;рушения, приведенныl= 4;и в § 1.6 и 1.7. Эти методы известны ка = 82; концепции линейной уп = 88;угой механики разрушения (ЛУМР), поскол&= #1100;ку они основан = 99; на уравнениях, описывающиm= 3; упругие пол = 03; напряжений, которые можно испол = 00;зовать только в том случае, если размер пластическl= 6;й зоны мал по сравнению с = 088;азмером трещины. Из уравнения (1.3) следует, что размер пластическl= 6;й зоны пропорционk= 2;лен <!--[if gte vml 1]><v:shape id=3D"_x0000_i1054" type=3D"#_x0000_t75" alt=3D= "" style=3D'width:43.5pt;height:20.25pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image029.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00027.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D58 height=3D27 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image029.gif" v:shapes=3D"_x0000= _i1054"><![endif]><span class=3DSpellE>Низкопро = 95;ные</span> материалы с малым пределом текучести обычно обладают большой вязкостью. Это означае = 90;, что при разр= 091;шении (<span class=3DSpellE><i>K</i><sub>I</sub>=3D<i>K</i><sub>Ic</sub></span>) размер пластическl= 6;й зоны может быть настол = 00;ко велик по сравнению с размером трещины, что ЛУМР применять нельзя. Последнее и = 84;еет место, если отношение <span class=3DSpellE>σ<i><sub>с</sub></i></span>/<span class=3DSpellE>= σ<i><sub>ys</sub></i></span> порядка единицы [из второго уравнения (1.3) следует, что размер пластическl= 6;й зоны пропор = 94;ионален отношению (<span class=3DSpellE>σ<i><sub>с</sub></i></span>/<span class=3DSpellE>= σ<i><sub>ys</sub></i></span>)<sup>2</sup>].<o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>В настоящее время не существует общего мето = 76;а исследованl= 0;я проблем, связанных с трещинами в материалах = 89; большой вязкостью. Д= 083;я таких материалов <span class=3DSpellE>Уэлсом</span> [14, 15] было введен = 86; понятие «раскрытие трещины» (РТ). = <span class=3DSpellE>Уэлс</span> сделал предположеl= 5;ие, что распростраl= 5;ение трещины будет иметь место в том случае, если пластическk= 2;я деформация = 74; вершине тре = 97;ины достигнет максимальнl= 6;го допустимогl= 6; значения. Деформацию при вершине трещины можно выраз = 80;ть через ее раскрытие (см. гл. IX), которое явл = 03;ется измеримой величиной.<o:p></o:p>= </span></p> <p class=3DMsoNormal style=3D'tab-stops:317.75pt'><span style=3D'font-size:= 10.0pt; font-family:Arial'>Предпола= гается, что распростраl= 5;ение трещины или разрушение происходит тогда, когда раскрытие трещины превышает критическуn= 2; величину. Легко показать (см. гл. IX), что в слу= 095;ае применения ЛУМР критерий РТ эквивалентk= 7;н критерию, связанному = 89; понятиями <span class=3DSpellE><i>К</i><sub>Ic</sub></span> и <span class=3DSpe= llE><i>G</i><sub>Ic</sub></span>. Это определеннm= 9;м образом обосновываk= 7;т предположеl= 5;ие об общей применимосm= 0;и. На настояще = 84; этапе одним из препятствиl= 1; для развити = 03; критерия РТ является то = 90; факт, что он не позволяе = 90; вычислить напряжение, при котором происходит разрушение. Для <span class=3DSpellE>низк = 86;прочных</span> материалов = 89; большой вязкостью критическоk= 7; значение РТ = 212; просто относительl= 5;ый параметр вязкости.<span style=3D'mso-tab-count:1'>        &= nbsp;          </span><!--[if = gte vml 1]><v:shape id=3D"_x0000_i1055" type=3D"#_x0000_t75" alt=3D"" style=3D'width:12.75pt;h= eight:19.5pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image001.png" o:hr= ef=3D"http://www.mysopromat.ru/images/0.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D17 height=3D26 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image002.gif" v:shapes=3D"_x0000= _i1055"><![endif]><o:p></o:p></span></p> <p class=3DMsoNormal><o:p> </o:p></p> <h2 style=3D'margin:0cm;margin-bottom:.0001pt'><span style=3D'font-family:A= rial; color:windowtext'>§ 1.6. Распростраl= 5;ение трещины<o:p></o:p></span></h2> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>К&= #1072;к было показано в § 1.3, коэффициенm= 0; интенсивноl= 9;ти напряжений есть мера напряжений = 80; деформаций = 74; окрестностl= 0; вершины трещины. Коэффициенm= 0; интенсивноl= 9;ти напряжений сохраняет свое значение лишь тогда, когда пластическk= 2;я зона мала. В этом случае можно также ожидать, что степень распростраl= 5;ения трещины за ц= 080;кл определяетl= 9;я коэффициенm= 0;ом интенсивноl= 9;ти напряжений. Если две различные трещины имеют два одинаковых распределеl= 5;ия напряжений, т. е. равные коэффициенm= 0;ы интенсивноl= 9;ти напряжений, то они должн= 099; распростраl= 5;яться с одной и той же скорость = 02;.<o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Е&= #1089;ли циклическаn= 3; нагрузка меняется от нуля до некоторой положительl= 5;ой величины (по= 089;тоянной амплитуды), то коэффициенm= 0; интенсивноl= 9;ти напряжений меняется в интервале Δ<i>K = =3D</i> <span class=3DSpellE><i>K</i><sub>max</sub></span> ÷ <span class=3DS= pellE><i>K</i><sub>min</sub></span>, где <span class=3DSpellE><i>K</i><sub>min</sub></span> = =3D 0. Следователn= 0;но, распростраl= 5;ение трещины за один цикл пр= 080; циклическоl= 4; процессе <span class=3DSpel= lE>нагружения</= span> (скорость распростраl= 5;ения трещины) ест= 100; величина, зависящая о = 90; амплитуды изменения интенсивноl= 9;ти напряжений [= 6;<i>K</i>:<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:0cm;margin-bottom:0cm;margin-left:1= .5pt; margin-bottom:.0001pt;tab-stops:228.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><!--[if gte vml 1]><v:shape id=3D"_x0000_i1056" type=3D"= #_x0000_t75" alt=3D"" style=3D'width:158.25pt;height:24pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image030.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00028.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D211 height=3D32 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image030.gif" v:shapes=3D"_x0000= _i1056"><![endif]><span style=3D'mso-tab-count:1'>        &= nbsp;           &nbs= p;  </span>(1.13)<o:p></o:p></span></p> <p class=3Dbase style=3D'text-indent:0cm'><span style=3D'font-size:10.0pt;f= ont-family: Arial'>где <span class=3DSpellE><i>S<sub>a</sub></i></spa= n> — амплитуда изменения напряжения (символ <i>S</i> — общепринятl= 6;е в литератур = 77; обозначениk= 7; циклически&#= 1093; напряжений). <s= pan class=3DSpellE>Пэрис</span>, <span class=3DSp= ellE>Гомез</span> и <span class=3DSpellE>Андерс = 86;н</span> [16] первыми пришли к этому вывод = 91; и проверили = 077;го на практике. Если использоваm= 0;ь результаты только одного испытания, т= 086; уравнение (1.13), очевидно, удовлетворl= 0;тся автоматичеl= 9;ки: в этом случа= 077; любая зависимостn= 0; <span class=3DSpellE><i>dа</i></span><i>/<span class=3DSpellE>dn</span> </i= >от Δ<i>K</i> подтвердит уравнение (1.13).<o:p= ></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Р&= #1072;ссмотрим результаты двух испытаний н = 72; распростра&= #1085;ение трещин, изображеннm= 9;х на рис. 1.9, <i>а. </i>Ам&#= 1087;литуды изменения напряжений были одинак = 86;выми и постояннымl= 0; в каждом испытании. Скорость распростраl= 5;ения трещины, оче= 074;идно, увеличивалk= 2;сь с ростом трещины. Ско= 088;ость <span class=3DSpellE><i>dа</i></span><i>/<span class=3DSpellE>dn</spa= n> </i>можно определить из наклона кривых. Вели= 095;ина Δ<i>K</i> получается из соотношениn= 3; <!--[if gte vml 1]><v:shape id=3D"_x0000_i1057" type=3D"#_x0000_t75" alt=3D"" style=3D'width:1in;heigh= t:20.25pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image031.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00029.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D96 height=3D27 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image031.gif" v:shapes=3D"_x0000= _i1057"><![endif]>при подстановкk= 7; соответствm= 1;ющего значения <i>а</i>. На рис. 1.9, <i>б </i>гр= 072;фик зависимостl= 0; <span class=3DSpellE><i>dа</i></span><i>/<span class=3DSpellE>dn</span> </i= >от Δ<i>K</i> изображе = 85; в логарифмичk= 7;ском масштабе по обеим осям. Данные, полученные при больших амплитудах изменений напряжений, указывают н = 72; сравнительl= 5;о большие значения Δ<i>K</i> = 080; <span class=3DSpellE><i>dа</i></span><i>/<span class=3DSpellE>dn</spa= n> </i>в начале процесса. Другие данные получены пр = 80; малых величинах Δ<i>K<= /i> и <span class=3DSpellE><i>dа</i></span><i>/<span class=3DSpellE= >dn</span>, </i>которые, однако, достигают таких же бол= 100;ших значений, ка= 082; и в первом испытании.<o:p></o:p>= </span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er; text-indent:15.3pt'><span style=3D'font-size:10.0pt;font-family:Arial'><!--= [if gte vml 1]><v:shape id=3D"_x0000_i1058" type=3D"#_x0000_t75" alt=3D"" style=3D'width:312pt;hei= ght:234pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image032.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00030.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D416 height=3D312 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image032.gif" v:shapes=3D"_x0000= _i1058"><![endif]><o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><b><span style=3D'font-size:10.0pt;font-family:Arial'>Рис. 1.9. Распростраl= 5;ение усталостноl= 1; трещины</span></b><span style=3D'font-size:10.0pt;font-family:Arial'><o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><b><span style=3D'font-size:10.0pt;font-family:Arial'>а — кривые рост = 72; трещины; б — скорость распростраl= 5;ения трещины</span></b><span style=3D'font-size:10.0pt;font-family:Arial'><o:p></o:p></span></p> <p class=3DMsoNormal><span style=3D'font-size:10.0pt;font-family:Arial'><o:= p> </o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Д&= #1072;нные двух испытаний, выполненныm= 3; при различн = 99;х условиях, располагаюm= 0;ся на одной кри= 074;ой, что подтверждаk= 7;т полезность уравнения (1.13). Очевидно, между двумя испытаниямl= 0;, из которых в одном имеется маленькая трещина и большое напряжение, = 072; в другом — длинная трещина и малое напряжение, нет никакой разницы, есл= 080; величины Δ<i>K</i> = 074; них одинаковы; в обоих испытаниях скорость распростраl= 5;ения трещины одн = 72; и та же.<o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Н&= #1072; графике зависимостl= 0; <span class=3DSpellE><i>dа</i></span><i>/<span class=3DSpellE>dn</span> </i= >от Δ<i>K</i>, построенноl= 4; в логарифмичk= 7;ском масштабе по обеим осям, эксперименm= 0;альные точки часто ложатся на прямую лини = 02;. Поэтому уравнение (1.13) было принят = 86; в виде:<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:0cm;margin-bottom:0cm;margin-left:1= .5pt; margin-bottom:.0001pt;tab-stops:228.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><!--[if gte vml 1]><v:shape id=3D"_x0000_i1059" type=3D"= #_x0000_t75" alt=3D"" style=3D'width:88.5pt;height:18pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image033.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00031.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D118 height=3D24 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image033.gif" v:shapes=3D"_x0000= _i1059"><![endif]><span style=3D'mso-tab-count:1'>        &= nbsp;           &nbs= p;            &= nbsp;            </s= pan>(1.14)<o:p></o:p></span></p> <p class=3Dbase style=3D'text-indent:0cm'><span style=3D'font-size:10.0pt;f= ont-family: Arial'>где <i>C</i> и <span class=3DSpellE><i>n</i>= </span> — константы. Было получено большое кол = 80;чество значений <span class=3DSpel= lE><i>п</i></span><i>, </i>которые обычно лежали в пределах от 2 до 4. Однако уравнение (1.14), как оказалось, плохо согла = 89;уется с данными испытаний. П= 072; практике гр = 72;фик зависимостl= 0; <span class=3DSpellE><i>dа</i></span><i>/<span class=3DSpellE>dn</span> </i= >от Δ<i>K</i> имеет форму буквы <i>= S</i> или, по крайней мер = 77;, состоит из участков ра = 79;ного наклона (см. [17, 18]= ). В испытания = 93;, связанных с ограниченнm= 9;м диапазоном изменения Δ<i>K<= /i>, получена экспоненциk= 2;льная зависимостn= 0; типа (1.14); в этом случае значение <span class=3DSpel= lE><i>п</i></span><i> </i>зависит от величины амплитуды Δ<i>K<= /i> (большие, малые и промежуточl= 5;ые значения Δ<i>K</i>). Когда трещи = 85;а достигает критическоk= 5;о размера, при котором отношение <span class=3DSpellE><i>dа</i></span><i>/<span class=3DSpellE>dn</span></i> обращается = 74; бесконечноl= 9;ть, при определ = 77;нии максимальнl= 6;го значения амплитуды Δ<i>K<= /i> могут появиться погрешностl= 0;. Общее разру = 96;ение происходит за один цикл, в котором ин= 090;енсивность напряжений достигает <span class=3DSpellE><i>K</i><sub>Ic</sub></span>.<o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>Ц&= #1080;клическое напряжение определяетl= 9;я двумя параметрамl= 0;: амплитудой <span class=3DSpellE><i>S<sub>a</sub></i></span> и средним напряжениеl= 4; <span class=3DSpellE><i>S<sub>m</sub></i></span>. Если <s= pan class=3DSpellE><i>S<sub>m</sub></i></span> =3D <span class=3DSpellE><i>S<su= b>a</sub></i></span>, то минимальноk= 7; напряжение за цикл равн= 086; нулю. Это означает, чт= 086; максимальнk= 2;я интенсивноl= 9;ть напряжений за цикл <span class=3DSpellE><i>K</i><s= ub>max</sub></span> =3D Δ<i>K</i>. Если <span class=3DSpellE><i>S<= sub>m</sub></i></span> > <span class=3DSpellE><i>S<sub>a</sub></i></span>, то l= 4;аксимальна= 103; интенсивноl= 9;ть напряжений <!--= [if gte vml 1]><v:shape id=3D"_x0000_i1060" type=3D"#_x0000_t75" alt=3D"" style=3D'width:106.5pt;h= eight:20.25pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image034.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00032.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D142 height=3D27 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image034.gif" v:shapes=3D"_x0000= _i1060"><![endif]>превышае&= #1090; значение Δ<i>K</i>. Не вызывает сомнений, чт= 086; скорость роста трещи = 85;ы зависит от максимальнl= 6;й интенсивноl= 9;ти напряжений. Поэтому более общей формой урав = 85;ений (1.13) является соотношениk= 7;<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:0cm;margin-bottom:0cm;margin-left:1= .5pt; margin-bottom:.0001pt;tab-stops:228.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><!--[if gte vml 1]><v:shape id=3D"_x0000_i1061" type=3D"= #_x0000_t75" alt=3D"" style=3D'width:159.75pt;height:63.75pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image035.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00033.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D213 height=3D85 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image035.gif" v:shapes=3D"_x0000= _i1061"><![endif]><span style=3D'mso-tab-count:1'>        &= nbsp;           &nbs= p; </span>(1.15)<o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>и называется коэффициенm= 0;ом асимметрии = 94;икла (см. гл. X).<o:p></o:p></span></p> <p class=3Dbase><span class=3DSpellE><span style=3D'font-size:10.0pt;font-f= amily: Arial'>Докритичеl= 9;кий</span></span><span style=3D'font-size:10.0pt;font-family:Arial'> медленный рост раковины может происходитn= 0; не только по= 076; действием циклическиm= 3; нагрузок, но и за счет других механизмов, из которых наиболее важным является механизм коррозионнl= 6;го растрескивk= 2;ния под напряже = 85;ием. Как и в случае рост = 72; усталостноl= 1; трещины, скорость роста коррозионнl= 6;й трещины при заданных условиях взаимодейсm= 0;вия материала с = 86; средой (а, следователn= 0;но, и время до разрушения) определяетl= 9;я коэффициенm= 0;ом интенсивноl= 9;ти напряжений. Одинаковые = 86;бразцы с одинаковымl= 0; начальными трещинами, н= 086; нагруженныk= 7; до различны = 93; напряжений (разные нача= 083;ьные значения <i>К</i>)<i>= , </i>разрушаютс= ;я через различное время (см. [19]), ка&= #1082; показано схематичесl= 2;и на рис. 1.10. Образец, наг= 088;уженный до значения <sp= an class=3DSpellE><i>К</i><sub>Ic</sub></span>, разрушаетсn= 3; сразу. Образцы, нагруженныk= 7; до значений <i>= К</i>, меньших определеннl= 6;го порогового уровня, не разрушаютсn= 3; никогда; это пороговое значение обозначают через <span class=3DSpellE><i>K</i><sub>I = 82;рн</sub></span>, где индекс «<sp= an class=3DSpellE>крн</span>» означает коррозионнl= 6;е растрескивk= 2;ние под напряжениеl= 4;.<o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><span style=3D'font-size:10.0pt;font-family:Arial'><!--[if gte vml 1]><v:shape id= =3D"_x0000_i1062" type=3D"#_x0000_t75" alt=3D"" style=3D'width:126pt;height:119.25pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image036.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00034.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D168 height=3D159 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image036.gif" v:shapes=3D"_x0000= _i1062"><![endif]><o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><b><span style=3D'font-size:10.0pt;font-family:Arial'>Рис. 1.10. Зависимостn= 0; времени до разрушения = 74; процессе коррозионнl= 6;го растрескивk= 2;ния под напряжениеl= 4; от начального <span class=3DSpellE>зна</span> <span class=3DSpellE>ч= 077;ния</span> коэффициенm= 0;а интенсивноl= 9;ти напряжений <i>K= </i></span></b><span style=3D'font-size:10.0pt;font-family:Arial'><o:p></o:p></span></p> <p class=3DMsoNormal><span style=3D'font-size:10.0pt;font-family:Arial'><o:= p> </o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><span style=3D'font-size:10.0pt;font-family:Arial'><!--[if gte vml 1]><v:shape id= =3D"_x0000_i1063" type=3D"#_x0000_t75" alt=3D"" style=3D'width:135pt;height:117pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image037.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00035.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D180 height=3D156 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image037.gif" v:shapes=3D"_x0000= _i1063"><![endif]><o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><b><span style=3D'font-size:10.0pt;font-family:Arial'>Рис. 1.11. Коррозионнl= 6;е растрескивk= 2;ние под напряже = 85;ием</span></b><span style=3D'font-size:10.0pt;font-family:Arial'><o:p></o:p></span></p> <p class=3DMsoNormal><span style=3D'font-size:10.0pt;font-family:Arial'><o:= p> </o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>В процессе коррозионнl= 6;го растрескивk= 2;ния под напряжениеl= 4; нагрузка может оставаться постоянной. Поскольку трещина расширяетсn= 3;, интенсивноl= 9;ть напряжений непрерывно = 91;величивает&#= 1089;я. В результат = 77; скорость ро = 89;та трещины за единицу времени <span class=3DSpellE><i>d= a</i></span><i>/<span class=3DSpellE>dt</span></i> увеличиваеm= 0;ся в соответствl= 0;и с уравнение = 84;<o:p></o:p></span></p> <p style=3D'margin-top:0cm;margin-right:0cm;margin-bottom:0cm;margin-left:1= .5pt; margin-bottom:.0001pt;tab-stops:228.0pt'><span style=3D'font-size:10.0pt; font-family:Arial'><!--[if gte vml 1]><v:shape id=3D"_x0000_i1064" type=3D"= #_x0000_t75" alt=3D"" style=3D'width:74.25pt;height:16.5pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image038.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00036.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D99 height=3D22 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image038.gif" v:shapes=3D"_x0000= _i1064"><![endif]><span style=3D'mso-tab-count:1'>        &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;     </span>(1.16)<o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>К&= #1086;гда трещина достигает размера, при котором <i>К </i>с= 090;ановится равным <span class=3DSpellE><i>К<= /i><sub>Ic</sub></span>, происходит окончательl= 5;ое разрушение, как показан = 86; на рис. 1.11.<o:p></o:p></span></p> <p class=3Dbase><span style=3D'font-size:10.0pt;font-family:Arial'>П&= #1086;роговое значение коэффициенm= 0;а <span class=3DSpellE><i>K</i><sub>Iкрн</sub></span> для процесс = 72; коррозионнl= 6;го растрескивk= 2;ния под напряжениеl= 4; и скорость роста трещи = 85;ы зависят от материала и = 091;словий окружающей среды. Из рис. = 1.12 следует, что деталь с трещиной определеннl= 6;го размера, нагруженнаn= 3; до такого напряжения <span class=3DSpellE>σ</span>, что <!--[if gte vml 1]><v:s= hape id=3D"_x0000_i1065" type=3D"#_x0000_t75" alt=3D"" style=3D'width:60.75pt;h= eight:20.25pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image039.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00037.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D81 height=3D27 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image039.gif" v:shapes=3D"_x0000= _i1065"><![endif]>разрушае&= #1090;ся в самом начале процесса <span class=3DSpel= lE>нагружения</= span>. В деталях, нагруженныm= 3; до значений <i>= К</i>, равных или больших <span class=3DSpellE><i>K= </i><sub>Iкрн</sub></span> (заштрихова = 85;ная область), трещина будет расти вплоть до разрушения. Положения механики разрушения применимы к коррозионнl= 6;му растрескивk= 2;нию под напряжениеl= 4;, однако ее возможностl= 0; в этом плане пока еще вес= 100;ма ограниченнm= 9;. Поэтому в настоящей книге задачам коррозионнl= 6;го растрескивk= 2;ния под напряжениеl= 4; уделяется небольшое внимание.<o:p></o:p></span>= </p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er; text-indent:15.3pt'><span style=3D'font-size:10.0pt;font-family:Arial'><!--= [if gte vml 1]><v:shape id=3D"_x0000_i1066" type=3D"#_x0000_t75" alt=3D"" style=3D'width:164.25pt;= height:141.75pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image040.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00038.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D219 height=3D189 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image040.gif" v:shapes=3D"_x0000= _i1066"><![endif]><o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><b><span style=3D'font-size:10.0pt;font-family:Arial'>Рис. 1.12. Зависимостn= 0; длины трещины от напряжения</spa= n></b><span style=3D'font-size:10.0pt;font-family:Arial'><o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><b><span style=3D'font-size:10.0pt;font-family:Arial'>при коррозионнl= 6;м растрескивk= 2;нии под напряже = 85;ием:</span></b><span style=3D'font-size:10.0pt;font-family:Arial'><o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><b><span style=3D'font-size:10.0pt;font-family:Arial'>1 — пороговое напряжение при коррозионнl= 6;м растрескивk= 2;нии под напряжениеl= 4;: </span></b><span style=3D'font-size:10.0pt;font-family:Arial'><br> <!--[if gte vml 1]><v:shape id=3D"_x0000_i1067" type=3D"#_x0000_t75" alt=3D= "" style=3D'width:87.75pt;height:21pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image041.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00039.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D117 height=3D28 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image041.gif" v:shapes=3D"_x0000= _i1067"><![endif]><o:p></o:p></span></p> <p align=3Dcenter style=3D'margin:0cm;margin-bottom:.0001pt;text-align:cent= er'><b><span style=3D'font-size:10.0pt;font-family:Arial'>2</span></b><span style=3D'fon= t-size: 10.0pt;font-family:Arial'> — <b>корро&#= 1079;ионное растрескивk= 2;ние под напряже = 85;ием;</b><o:p></o:p></span></p> <p class=3DMsoNormal style=3D'tab-stops:458.25pt'><b><span style=3D'font-si= ze:10.0pt; font-family:Arial'>3</span></b><span style=3D'font-size:10.0pt;font-family:= Arial'> — <b>окончател&= #1100;ное разрушение: </b= ><!--[if gte vml 1]><v:shape id=3D"_x0000_i1068" type=3D"#_x0000_t75" alt=3D"" style=3D'width:1in;heigh= t:20.25pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image042.gif" o:hr= ef=3D"http://www.mysopromat.ru/images/fracturemechanics/IMG00040.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D96 height=3D27 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image042.gif" v:shapes=3D"_x0000= _i1068"><![endif]><span style=3D'mso-tab-count:1'>        &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;   </span><!--[if gte vml 1]><v:shape id=3D"_x0000_i1069" type=3D"#_x0000_t75" alt=3D"" style=3D'width:12.75pt;h= eight:19.5pt'> <v:imagedata src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image001.png" o:hr= ef=3D"http://www.mysopromat.ru/images/0.gif"/> </v:shape><![endif]--><![if !vml]><img border=3D0 width=3D17 height=3D26 src=3D"D_Boek_Osn_mexan_razrushen_1-.files/image002.gif" v:shapes=3D"_x0000= _i1069"><![endif]><o:p></o:p></span></p> <p class=3DMsoNormal><span style=3D'font-size:10.0pt;font-family:Arial'><o:= p> </o:p></span></p> <h2 style=3D'margin:0cm;margin-bottom:.0001pt'><span style=3D'font-family:A= rial; color:windowtext'>§ 1.7. Заключение<o:p>= </o:p></span></h2> <p class=3DMsoNormal style=3D'tab-stops:317.75pt'><span style=3D'font-size:= 10.0pt; font-family:Arial'>Было показано, чт= 086; процессы распростраl= 5;ения трещины и разрушения определяютl= 9;я коэффициенm= 0;ом интенсивноl= 9;ти напряжении. = 069;тот коэффициенm= 0; играет в механике разрушения определяющm= 1;ю роль. В принципе, зная коэффициенm= 0; интенсивноl= 9;ти напряжении для трещины = 074; данном элементе конструкциl= 0;, можно рассчитан, процесс роста трещины и время до разрушения. Иными словами, на все вопросы, поставленнm= 9;е в § 1.2, могут быть даны ответы. К сожалению, н= 072; практике встречаетсn= 3; так много ос= 083;ожнений, что применить, кажется, простые пол = 86;жения, рассмотренl= 5;ые в данной главе, не все&#= 1075;да представляk= 7;тся возможным. Однако во многих случаях можно получить полезные ре = 79;ультаты. Для правильной оценки области применения механики разрушения = 74; технике про = 77;ктировщик и инженер должны обладать достаточныl= 4;и сведениями = 86; физических принципах и допущениях, лежащих в ее основе. Наук= 072; «Механика разрушения&raqu= o; еще далека о= 090; завершения = 80; не является простым инструментl= 6;м проектировk= 2;ния. В последующиm= 3; главах буду = 90; выявлены до = 89;тоинства и недостатк = 80; этих положений.<o:p></o:p>= </span></p> <p class=3DMsoNormal><o:p> </o:p></p> </div> </body> </html> ------=_NextPart_01C4AB06.63706400 Content-Location: file:///C:/9688824D/D_Boek_Osn_mexan_razrushen_1-.files/image001.png Content-Transfer-Encoding: base64 Content-Type: image/png iVBORw0KGgoAAAANSUhEUgAAAAEAAAABAQMAAAAl21bKAAAABlBMVEUrMjYAAADGrQChAAAAAnRS TlP/AOW3MEoAAAABYktHRACIBR1IAAAABGdJRmcBAAAAJDBjKAAAAAxjbVBQSkNtcDA3MTIAAAAH T223pQAAAApJREFUGNNjaAAAAIIAgacBuhAAAAAaZ0lGeFBJQU5ZR0lGMi4wQ3JvcCBvZiBJbWFn ZQEANsmMygAAAABJRU5ErkJggk== ------=_NextPart_01C4AB06.63706400 Content-Location: file:///C:/9688824D/D_Boek_Osn_mexan_razrushen_1-.files/image002.gif Content-Transfer-Encoding: base64 Content-Type: image/gif R0lGODlhEQAaAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAAAAAAB AAEAgAAAAAECAwICRAEAOw== ------=_NextPart_01C4AB06.63706400 Content-Location: 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