{"id":1821,"date":"2022-12-01T17:00:00","date_gmt":"2022-12-01T16:00:00","guid":{"rendered":"https:\/\/technischemechanik.com\/?p=1821"},"modified":"2023-01-17T14:24:26","modified_gmt":"2023-01-17T13:24:26","slug":"schnittgroessen-mittels-integration","status":"publish","type":"post","link":"https:\/\/technischemechanik.com\/de\/2022\/12\/01\/schnittgroessen-mittels-integration\/","title":{"rendered":"Schnittgr\u00f6\u00dfen mittels Integration"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Herzlich Willkommen!<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Nachdem wir bereits theoretisch \u00fcber <a href=\"https:\/\/technischemechanik.com\/2022\/10\/06\/theorie-schnittgroessen-schnittufer\/\" data-type=\"post\" data-id=\"1672\">Schnittgr\u00f6\u00dfen<\/a> diskutiert haben, uns <a href=\"https:\/\/technischemechanik.com\/2022\/10\/13\/schnittgroessen-an-spezieller-stelle\/\" data-type=\"post\" data-id=\"1683\">Schnittgr\u00f6\u00dfen an speziellen Punkten<\/a> eines Tr\u00e4gers und auch die Berechnung eines <a href=\"https:\/\/technischemechanik.com\/2022\/10\/27\/schnittgroessenverlauf-berechnen\/\" data-type=\"post\" data-id=\"1712\">Schnittgr\u00f6\u00dfenverlaufs<\/a> angesehen haben, m\u00f6chten wir uns nun der Berechnung von Querkraft und Biegemoment mittels Integration widmen. Dazu folgendes Beispiel.<\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p class=\"wp-block-paragraph\">Berechne f\u00fcr den skizzierten Biegetr\u00e4ger die Auflagerreaktionen, sowie die Schnittgr\u00f6\u00dfen Q(x) und M(x). <br>Geg.: q<sub>0<\/sub>, l<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img data-recalc-dims=\"1\" loading=\"lazy\" decoding=\"async\" width=\"712\" height=\"174\" data-attachment-id=\"1822\" data-permalink=\"https:\/\/technischemechanik.com\/de\/2022\/12\/01\/schnittgroessen-mittels-integration\/abb_sg_03\/\" data-orig-file=\"https:\/\/i0.wp.com\/technischemechanik.com\/wp-content\/uploads\/2022\/12\/Abb_SG_03.png?fit=724%2C177&amp;ssl=1\" data-orig-size=\"724,177\" data-comments-opened=\"1\" data-image-meta=\"{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;0&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;,&quot;orientation&quot;:&quot;0&quot;}\" data-image-title=\"Abb_SG_03\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/i0.wp.com\/technischemechanik.com\/wp-content\/uploads\/2022\/12\/Abb_SG_03.png?fit=712%2C174&amp;ssl=1\" src=\"https:\/\/i0.wp.com\/technischemechanik.com\/wp-content\/uploads\/2022\/12\/Abb_SG_03.png?resize=712%2C174&#038;ssl=1\" alt=\"\" class=\"wp-image-1822\" srcset=\"https:\/\/i0.wp.com\/technischemechanik.com\/wp-content\/uploads\/2022\/12\/Abb_SG_03.png?w=724&amp;ssl=1 724w, https:\/\/i0.wp.com\/technischemechanik.com\/wp-content\/uploads\/2022\/12\/Abb_SG_03.png?resize=300%2C73&amp;ssl=1 300w\" sizes=\"auto, (max-width: 712px) 100vw, 712px\" \/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Hinweis: Das Koordinatensystem ist so zu w\u00e4hlen, dass die x-Achse nach rechts, die&nbsp;y-Achse&nbsp;aus der Blattebene heraus und die&nbsp;z-Achse&nbsp;nach unten positiv festgelegt sind.<\/p>\n<\/blockquote>\n\n\n\n<p class=\"wp-block-paragraph\">Wir beginnen wie gewohnt mit einem Freik\u00f6rperbild, n\u00e4mlich um die Lagerreaktionen berechnen zu k\u00f6nnen. Dann stellen wir die Gleichgewichtsbedingungen auf und berechnen alle Auflagerkr\u00e4fte in A und B. Dabei k\u00f6nnen wir f\u00fcr die Streckenlast eine Kombination aus Rechtecks- und Dreiecksform und deren entsprechende resultierende Einzellasten verwenden. Wenn das erledigt ist widmen wir uns schlie\u00dflich der Berechnung der Querkraft \u00fcber das Integral, genau wie im <a href=\"https:\/\/technischemechanik.com\/2022\/10\/06\/theorie-schnittgroessen-schnittufer\/\" data-type=\"post\" data-id=\"1672\">Theoriebeitrag zu Schnittgr\u00f6\u00dfen<\/a> besprochen. Nat\u00fcrlich m\u00fcssen wir uns dazu noch \u00fcberlegen welche Funktion unsere trapezf\u00f6rmige Streckenlast korrekt beschreibt. Auch hier werden wir wieder bei der Geradengleichung f\u00fcndig. Im Anschluss an die Querkraft k\u00f6nnen wir dann das Schnittmoment bestimmen, indem wir einfach die Querkraft noch einmal integrieren. Zum Schluss zeige ich euch auch noch einen alternativen Weg zur Bestimmung des Schnittmoments und wir diskutieren die Wichtigkeit einer Dimensionskontrolle. Alles im Detail findest ihr wie immer im verlinkten Video sowie auch in der angeh\u00e4ngten pdf-Datei. Viel Spa\u00df damit!<\/p>\n\n\n\n<figure class=\"wp-block-embed is-type-video is-provider-youtube wp-block-embed-youtube wp-embed-aspect-16-9 wp-has-aspect-ratio\"><div class=\"wp-block-embed__wrapper\">\n<span class=\"embed-youtube\" style=\"text-align:center; display: block;\"><iframe loading=\"lazy\" class=\"youtube-player\" width=\"712\" height=\"401\" src=\"https:\/\/www.youtube.com\/embed\/Czull0QiqC8?version=3&#038;rel=1&#038;showsearch=0&#038;showinfo=1&#038;iv_load_policy=1&#038;fs=1&#038;hl=de-DE&#038;autohide=2&#038;wmode=transparent\" allowfullscreen=\"true\" style=\"border:0;\" sandbox=\"allow-scripts allow-same-origin allow-popups allow-presentation allow-popups-to-escape-sandbox\"><\/iframe><\/span>\n<\/div><figcaption class=\"wp-element-caption\"><br><\/figcaption><\/figure>\n\n\n\n<div data-wp-interactive=\"core\/file\" class=\"wp-block-file\"><object data-wp-bind--hidden=\"!state.hasPdfPreview\" hidden class=\"wp-block-file__embed\" data=\"https:\/\/technischemechanik.com\/wp-content\/uploads\/2022\/12\/Statik-A33.pdf\" type=\"application\/pdf\" style=\"width:100%;height:600px\" aria-label=\"Einbettung von Statik-A33.\"><\/object><a id=\"wp-block-file--media-ae65e728-8fa4-4573-b4a9-4516c6dbddfb\" href=\"https:\/\/technischemechanik.com\/wp-content\/uploads\/2022\/12\/Statik-A33.pdf\">Statik-A33<\/a><a href=\"https:\/\/technischemechanik.com\/wp-content\/uploads\/2022\/12\/Statik-A33.pdf\" class=\"wp-block-file__button wp-element-button\" download aria-describedby=\"wp-block-file--media-ae65e728-8fa4-4573-b4a9-4516c6dbddfb\">Herunterladen<\/a><\/div>\n\n\n\n<p class=\"wp-block-paragraph\">Auch hier gilt &#8211; wie schon bei den vorhergehenden Beispielen zu den Schnittgr\u00f6\u00dfen &#8211; dass es sich um ein \u00fcberaus essentielles Kapitel der Technischen Mechanik handelt. Bei Unklarheiten bitte also unbedingt gleich melden.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Vielen Dank und bis bald,<br>Markus <\/p>\n\n\n<div class=\"crowdsignal-feedback-wrapper\" data-crowdsignal-feedback=\"{&quot;feedbackPlaceholder&quot;:&quot;Sag mir bitte was dir besonders gefallen hat und was ich verbessern soll! Vielen Dank!&quot;,&quot;submitText&quot;:&quot;Danke f\\u00fcr dein Feedback!&quot;,&quot;surveyId&quot;:2705235,&quot;title&quot;:&quot;Feedback&quot;,&quot;emailResponses&quot;:false,&quot;emailPlaceholder&quot;:&quot;Deine E-Mail-Adresse&quot;,&quot;emailRequired&quot;:false,&quot;header&quot;:&quot;\\ud83d\\udc4b Hallo!&quot;,&quot;hideBranding&quot;:false,&quot;hideTriggerShadow&quot;:false,&quot;submitButtonLabel&quot;:&quot;Senden&quot;,&quot;toggleOn&quot;:&quot;click&quot;,&quot;triggerLabel&quot;:&quot;Feedback&quot;,&quot;x&quot;:&quot;left&quot;,&quot;y&quot;:&quot;bottom&quot;,&quot;status&quot;:&quot;open&quot;,&quot;TrpContentRestriction&quot;:{&quot;restriction_type&quot;:&quot;exclude&quot;,&quot;selected_languages&quot;:[],&quot;panel_open&quot;:true},&quot;nonce&quot;:&quot;8c46fdffa2&quot;}\"><\/div>\n\n\n<div class=\"wp-block-columns is-layout-flex wp-container-core-columns-is-layout-8f761849 wp-block-columns-is-layout-flex\">\n<div class=\"wp-block-column is-layout-flow wp-block-column-is-layout-flow\" style=\"flex-basis:50%\"><div class=\"crowdsignal-vote-wrapper\" data-crowdsignal-vote=\"{&quot;pollId&quot;:&quot;29311620-3302-4a28-8c62-34c614948b60&quot;,&quot;title&quot;:&quot;Unbenannte Abstimmung 20&quot;,&quot;hideBranding&quot;:false,&quot;pollStatus&quot;:&quot;open&quot;,&quot;size&quot;:&quot;medium&quot;,&quot;borderWidth&quot;:1,&quot;borderRadius&quot;:5,&quot;hideResults&quot;:false,&quot;TrpContentRestriction&quot;:{&quot;restriction_type&quot;:&quot;exclude&quot;,&quot;selected_languages&quot;:[],&quot;panel_open&quot;:true},&quot;apiPollData&quot;:{&quot;id&quot;:11253326,&quot;question&quot;:&quot;&quot;,&quot;note&quot;:&quot;&quot;,&quot;settings&quot;:{&quot;title&quot;:&quot;Unbenannte Abstimmung 20&quot;,&quot;after_vote&quot;:&quot;results&quot;,&quot;after_message&quot;:&quot;&quot;,&quot;randomize_answers&quot;:false,&quot;restrict_vote_repeat&quot;:false,&quot;captcha&quot;:false,&quot;multiple_choice&quot;:false,&quot;redirect_url&quot;:&quot;&quot;,&quot;close_status&quot;:&quot;open&quot;,&quot;close_after&quot;:false},&quot;answers&quot;:[{&quot;answer_text&quot;:&quot;up&quot;,&quot;id&quot;:51404275,&quot;client_id&quot;:&quot;4ae8e890-5309-41a3-bae2-719a23a5da36&quot;},{&quot;answer_text&quot;:&quot;down&quot;,&quot;id&quot;:51404276,&quot;client_id&quot;:&quot;b583d8b0-1c19-492c-a781-9f4dfba236ce&quot;}],&quot;source_link&quot;:&quot;https:\\\/\\\/technischemechanik.com&quot;,&quot;client_id&quot;:&quot;29311620-3302-4a28-8c62-34c614948b60&quot;}}\">\n<div data-crowdsignal-vote-item=\"{&quot;answerId&quot;:&quot;4ae8e890-5309-41a3-bae2-719a23a5da36&quot;,&quot;type&quot;:&quot;up&quot;,&quot;TrpContentRestriction&quot;:{&quot;restriction_type&quot;:&quot;exclude&quot;,&quot;selected_languages&quot;:[],&quot;panel_open&quot;:true}}\"><\/div>\n\n<div data-crowdsignal-vote-item=\"{&quot;answerId&quot;:&quot;b583d8b0-1c19-492c-a781-9f4dfba236ce&quot;,&quot;type&quot;:&quot;down&quot;,&quot;TrpContentRestriction&quot;:{&quot;restriction_type&quot;:&quot;exclude&quot;,&quot;selected_languages&quot;:[],&quot;panel_open&quot;:true}}\"><\/div>\n<\/div><\/div>\n\n\n\n<div class=\"wp-block-column is-layout-flow wp-block-column-is-layout-flow\" style=\"flex-basis:50%\"><div class=\"crowdsignal-applause-wrapper\" data-crowdsignal-applause=\"{&quot;pollId&quot;:&quot;94052372-60ee-4246-bdb3-b44cfaeeec3b&quot;,&quot;title&quot;:&quot;Unbenannter Applaus&quot;,&quot;answerId&quot;:&quot;4951fcb8-b48f-428e-82e4-90330cd7aed6&quot;,&quot;hideBranding&quot;:false,&quot;size&quot;:&quot;medium&quot;,&quot;pollStatus&quot;:&quot;open&quot;,&quot;TrpContentRestriction&quot;:{&quot;restriction_type&quot;:&quot;exclude&quot;,&quot;selected_languages&quot;:[],&quot;panel_open&quot;:true},&quot;apiPollData&quot;:{&quot;id&quot;:11253327,&quot;question&quot;:&quot;&quot;,&quot;note&quot;:&quot;&quot;,&quot;settings&quot;:{&quot;title&quot;:&quot;Unbenannter Applaus&quot;,&quot;after_vote&quot;:&quot;results&quot;,&quot;after_message&quot;:&quot;&quot;,&quot;randomize_answers&quot;:false,&quot;restrict_vote_repeat&quot;:false,&quot;captcha&quot;:false,&quot;multiple_choice&quot;:false,&quot;redirect_url&quot;:&quot;&quot;,&quot;close_status&quot;:&quot;open&quot;,&quot;close_after&quot;:false},&quot;answers&quot;:[{&quot;answer_text&quot;:&quot;clap&quot;,&quot;id&quot;:51404277,&quot;client_id&quot;:&quot;4951fcb8-b48f-428e-82e4-90330cd7aed6&quot;}],&quot;source_link&quot;:&quot;https:\\\/\\\/technischemechanik.com&quot;,&quot;client_id&quot;:&quot;94052372-60ee-4246-bdb3-b44cfaeeec3b&quot;}}\"><\/div><\/div>\n<\/div>\n\n\n\n<p class=\"wp-block-paragraph\"><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Herzlich Willkommen! Nachdem wir bereits theoretisch \u00fcber Schnittgr\u00f6\u00dfen diskutiert haben, uns Schnittgr\u00f6\u00dfen an speziellen Punkten eines Tr\u00e4gers und auch die Berechnung eines Schnittgr\u00f6\u00dfenverlaufs angesehen haben, m\u00f6chten wir uns nun der Berechnung von Querkraft und Biegemoment mittels Integration widmen. Dazu folgendes Beispiel. Wir beginnen wie gewohnt mit einem Freik\u00f6rperbild, n\u00e4mlich um die Lagerreaktionen berechnen zu k\u00f6nnen. &hellip; <a href=\"https:\/\/technischemechanik.com\/de\/2022\/12\/01\/schnittgroessen-mittels-integration\/\" class=\"more-link\"><span class=\"screen-reader-text\">Schnittgr\u00f6\u00dfen mittels Integration<\/span> weiterlesen<\/a><\/p>\n","protected":false},"author":110396568,"featured_media":1826,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_coblocks_attr":"","_coblocks_dimensions":"","_coblocks_responsive_height":"","_coblocks_accordion_ie_support":"","_crdt_document":"","advanced_seo_description":"","jetpack_seo_html_title":"","jetpack_seo_noindex":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":true,"jetpack_social_options":{"image_generator_settings":{"template":"highway","default_image_id":0,"font":"","enabled":false},"version":2},"jetpack_post_was_ever_published":false},"categories":[2602546,1584145,750913509,172484],"tags":[129538,25466,728229938,100678,35890,727961753,750913794,8207,727263877,750913355,662707,574221697,7091101,750913348,132249310,750913789,16082175,750913792,750913790,750913788,750913787,750913786,750913791,750913295,172005,1121513,727213064,727213067,750913347,750913793,2722],"class_list":["post-1821","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-ubungsaufgaben","category-gleichgewicht","category-schnittgroessen","category-statik","tag-aufgabe","tag-beispiel","tag-digitalerunterricht","tag-einfuhrung","tag-featured","tag-gleichgewicht","tag-grundlagen-mechanik","tag-lernen","tag-markus-orthaber","tag-markus-orthaber-statik","tag-maschinenbau","tag-maschinenbau-studieren","tag-maschinenbauingenieur","tag-mechanik-grundlagen","tag-normalkraft","tag-querkraft","tag-schnittgrosen","tag-schnittgroessen-berechnen","tag-schnittgroessen-bestimmen","tag-schnittgroessenverlauf","tag-schnittgroessenverlauf-zeichnen","tag-schnittmoment","tag-schnittufer","tag-statik","tag-studieren","tag-technische-mechanik","tag-technische-mechanik-1","tag-technische-mechanik-2-2","tag-theorie","tag-traeger","tag-uni"],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"https:\/\/i0.wp.com\/technischemechanik.com\/wp-content\/uploads\/2022\/12\/Schnittgroessen-A33.png?fit=1280%2C720&ssl=1","jetpack_likes_enabled":true,"jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/p7UJuu-tn","jetpack-related-posts":[],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/technischemechanik.com\/de\/wp-json\/wp\/v2\/posts\/1821","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/technischemechanik.com\/de\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/technischemechanik.com\/de\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/technischemechanik.com\/de\/wp-json\/wp\/v2\/users\/110396568"}],"replies":[{"embeddable":true,"href":"https:\/\/technischemechanik.com\/de\/wp-json\/wp\/v2\/comments?post=1821"}],"version-history":[{"count":4,"href":"https:\/\/technischemechanik.com\/de\/wp-json\/wp\/v2\/posts\/1821\/revisions"}],"predecessor-version":[{"id":1929,"href":"https:\/\/technischemechanik.com\/de\/wp-json\/wp\/v2\/posts\/1821\/revisions\/1929"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/technischemechanik.com\/de\/wp-json\/wp\/v2\/media\/1826"}],"wp:attachment":[{"href":"https:\/\/technischemechanik.com\/de\/wp-json\/wp\/v2\/media?parent=1821"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/technischemechanik.com\/de\/wp-json\/wp\/v2\/categories?post=1821"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/technischemechanik.com\/de\/wp-json\/wp\/v2\/tags?post=1821"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}