{"id":1449,"date":"2022-07-25T12:40:32","date_gmt":"2022-07-25T10:40:32","guid":{"rendered":"https:\/\/technischemechanik.com\/?p=1449"},"modified":"2022-07-25T12:40:32","modified_gmt":"2022-07-25T10:40:32","slug":"statik-gleichgewicht-am-rahmen-tragwerk","status":"publish","type":"post","link":"https:\/\/technischemechanik.com\/de\/2022\/07\/25\/statik-gleichgewicht-am-rahmen-tragwerk\/","title":{"rendered":"Statik: Gleichgewicht am Rahmen \/ Tragwerk"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Herzlich Willkommen!<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Ein erstes kleines Auslegungsbeispiel nehmen wir uns diesmal vor. Es handelt sich um ein Tragwerk, welches im Lager A maximal mit einer vorgegebenen Kraft belastet werden darf. Die Frage dabei ist, welche Kraft P darf dann am freien Ende des Tragwerks maximal angreifen. <\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\"><p>Bestimme die maximale Kraft P, die auf das Tragwerk aufgebracht werden darf, sodass die Resultierende in A maximal Fmax betr\u00e4gt. <\/p><p>Geg.: Fmax=2kN, l=0.75m, h=0.5m, r=0.1m<\/p><cite>Quelle: Aufgabe 6.74 (S. 351) aus Russell C. Hibbeler, Technische Mechanik 1 Statik, 12. Auflage, 2012 Pearson GmbH, M\u00fcnchen<\/cite><\/blockquote>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full\"><img data-recalc-dims=\"1\" loading=\"lazy\" decoding=\"async\" width=\"712\" height=\"488\" data-attachment-id=\"1450\" data-permalink=\"https:\/\/technischemechanik.com\/de\/2022\/07\/25\/statik-gleichgewicht-am-rahmen-tragwerk\/abb09\/\" data-orig-file=\"https:\/\/i0.wp.com\/technischemechanik.com\/wp-content\/uploads\/2022\/07\/Abb09.jpg?fit=802%2C550&amp;ssl=1\" data-orig-size=\"802,550\" 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=\"Abb09\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/i0.wp.com\/technischemechanik.com\/wp-content\/uploads\/2022\/07\/Abb09.jpg?fit=712%2C488&amp;ssl=1\" src=\"https:\/\/i0.wp.com\/technischemechanik.com\/wp-content\/uploads\/2022\/07\/Abb09.jpg?resize=712%2C488&#038;ssl=1\" alt=\"\" class=\"wp-image-1450\" srcset=\"https:\/\/i0.wp.com\/technischemechanik.com\/wp-content\/uploads\/2022\/07\/Abb09.jpg?w=802&amp;ssl=1 802w, https:\/\/i0.wp.com\/technischemechanik.com\/wp-content\/uploads\/2022\/07\/Abb09.jpg?resize=300%2C206&amp;ssl=1 300w, https:\/\/i0.wp.com\/technischemechanik.com\/wp-content\/uploads\/2022\/07\/Abb09.jpg?resize=768%2C527&amp;ssl=1 768w, https:\/\/i0.wp.com\/technischemechanik.com\/wp-content\/uploads\/2022\/07\/Abb09.jpg?resize=800%2C550&amp;ssl=1 800w\" sizes=\"auto, (max-width: 712px) 100vw, 712px\" \/><\/figure>\n<\/div>\n\n\n<p class=\"wp-block-paragraph\">Wie immer beginnen wir mit einem Freik\u00f6rperbild, wobei wir hier das Seil nur am horizontalen St\u00fcck (links der Rolle) schneiden d\u00fcrfen. Das macht die Rechnung etwas einfacher. Danach stellen wir Kr\u00e4fte- und Momentengleichgewicht auf und l\u00f6sen das entstehende Gleichungssystem. Nachdem in A ein Betrag als Maximalwert vorgegeben ist, wir aber je eine Kraft vertikal und horizontal erhalten, m\u00fcssen wir noch mittels Pythagoras einen Betrag ermitteln. Schlie\u00dflich l\u00e4sst sich eine Gleichung f\u00fcr den maximalen Wert von P finden. Die genaue Rechnung, wie immer gespickt mit einigen Hinweisen und zus\u00e4tzlichen Erkl\u00e4rungen findet ihr im verlinkten Video. Viel Spa\u00df! <\/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\/Q7rMI8jGPZU?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<p class=\"wp-block-paragraph\">Wenn Fragen oder Unklarheiten auftauchen, freue ich mich jederzeit auf eure Kommentare &#8211; entweder hier oder direkt auf YouTube. <\/p>\n\n\n\n<p class=\"wp-block-paragraph\">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;77006c9619&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;:11165226,&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;:51093530,&quot;client_id&quot;:&quot;4ae8e890-5309-41a3-bae2-719a23a5da36&quot;},{&quot;answer_text&quot;:&quot;down&quot;,&quot;id&quot;:51093531,&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;:11165227,&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;:51093532,&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! Ein erstes kleines Auslegungsbeispiel nehmen wir uns diesmal vor. Es handelt sich um ein Tragwerk, welches im Lager A maximal mit einer vorgegebenen Kraft belastet werden darf. Die Frage dabei ist, welche Kraft P darf dann am freien Ende des Tragwerks maximal angreifen. Bestimme die maximale Kraft P, die auf das Tragwerk aufgebracht &hellip; <a href=\"https:\/\/technischemechanik.com\/de\/2022\/07\/25\/statik-gleichgewicht-am-rahmen-tragwerk\/\" class=\"more-link\"><span class=\"screen-reader-text\">Statik: Gleichgewicht am Rahmen \/ Tragwerk<\/span> weiterlesen<\/a><\/p>\n","protected":false},"author":110396568,"featured_media":1419,"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":[2602546,1584145,172484],"tags":[25466,729611524,100678,414019,35890,5167337,750913306,727961753,350894,166867,113216,290680250,163250810,2517859,8207,662707,720191,1213491,143003,10126,73726871,750913302,161115,750913295,2411,1121513,727213064,729611525,2326513,2722,1268153,271082537,727961757,729611523],"class_list":["post-1449","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-ubungsaufgaben","category-gleichgewicht","category-statik","tag-beispiel","tag-ebenes-zentrales-kraftesystem","tag-einfuhrung","tag-erklart","tag-featured","tag-festigkeitslehre","tag-freikorperbild","tag-gleichgewicht","tag-grundlagen","tag-hochschule","tag-kraft","tag-kraftsystem","tag-kraftvektor","tag-krafte","tag-lernen","tag-maschinenbau","tag-mechanik","tag-nachhilfe","tag-newton","tag-physik","tag-physik-nachhilfe","tag-reduktion","tag-screencast","tag-statik","tag-studium","tag-technische-mechanik","tag-technische-mechanik-1","tag-technische-mechanik-grundlagen","tag-tm-1","tag-uni","tag-verstehen","tag-zentrales-kraftsystem","tag-zentrales-kraftesystem","tag-zentralkraftsystem"],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"https:\/\/i0.wp.com\/technischemechanik.com\/wp-content\/uploads\/2022\/07\/YouTube-Thumbnail-Gleichgewicht-512-%C3%97-512-px.png?fit=512%2C512&ssl=1","jetpack_likes_enabled":true,"jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/p7UJuu-nn","jetpack-related-posts":[],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/technischemechanik.com\/de\/wp-json\/wp\/v2\/posts\/1449","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=1449"}],"version-history":[{"count":3,"href":"https:\/\/technischemechanik.com\/de\/wp-json\/wp\/v2\/posts\/1449\/revisions"}],"predecessor-version":[{"id":1453,"href":"https:\/\/technischemechanik.com\/de\/wp-json\/wp\/v2\/posts\/1449\/revisions\/1453"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/technischemechanik.com\/de\/wp-json\/wp\/v2\/media\/1419"}],"wp:attachment":[{"href":"https:\/\/technischemechanik.com\/de\/wp-json\/wp\/v2\/media?parent=1449"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/technischemechanik.com\/de\/wp-json\/wp\/v2\/categories?post=1449"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/technischemechanik.com\/de\/wp-json\/wp\/v2\/tags?post=1449"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}