{"id":399,"date":"2019-11-29T16:33:28","date_gmt":"2019-11-29T16:33:28","guid":{"rendered":"http:\/\/www.aircraftengineer.info\/notes\/?p=399"},"modified":"2019-12-02T02:16:37","modified_gmt":"2019-12-02T02:16:37","slug":"mechanics","status":"publish","type":"post","link":"https:\/\/www.aircraftengineer.info\/notes\/mechanics\/","title":{"rendered":"Mechanics"},"content":{"rendered":"\n<figure class=\"wp-block-embed-youtube wp-block-embed is-type-video is-provider-youtube wp-embed-aspect-16-9 wp-has-aspect-ratio\"><div class=\"wp-block-embed__wrapper\">\n<iframe loading=\"lazy\" title=\"Mechanics\" width=\"604\" height=\"340\" src=\"https:\/\/www.youtube.com\/embed\/Qd_4xiOtwss?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n<\/div><\/figure>\n\n\n\n<!--nextpage-->\n\n\n\n<figure class=\"wp-block-image is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_01-791x1024.jpg\" alt=\"\" class=\"wp-image-400\" width=\"788\" height=\"1019\" srcset=\"https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_01-791x1024.jpg 791w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_01-232x300.jpg 232w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_01-768x994.jpg 768w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_01.jpg 1700w\" sizes=\"auto, (max-width: 788px) 100vw, 788px\" \/><\/figure>\n\n\n\n<!--nextpage-->\n\n\n\n<figure class=\"wp-block-image is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_02-845x1024.jpg\" alt=\"\" class=\"wp-image-401\" width=\"719\" height=\"870\" srcset=\"https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_02-845x1024.jpg 845w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_02-248x300.jpg 248w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_02.jpg 944w\" sizes=\"auto, (max-width: 719px) 100vw, 719px\" \/><\/figure>\n\n\n\n<!--nextpage-->\n\n\n\n<figure class=\"wp-block-image is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_03-792x1024.jpg\" alt=\"\" class=\"wp-image-402\" width=\"604\" height=\"780\" srcset=\"https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_03-792x1024.jpg 792w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_03-232x300.jpg 232w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_03-768x993.jpg 768w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_03.jpg 995w\" sizes=\"auto, (max-width: 604px) 100vw, 604px\" \/><\/figure>\n\n\n\n<!--nextpage-->\n\n\n\n<figure class=\"wp-block-image\"><img loading=\"lazy\" decoding=\"async\" width=\"792\" height=\"1024\" src=\"http:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_03-792x1024.jpg\" alt=\"\" class=\"wp-image-402\" srcset=\"https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_03-792x1024.jpg 792w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_03-232x300.jpg 232w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_03-768x993.jpg 768w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_03.jpg 995w\" sizes=\"auto, (max-width: 792px) 100vw, 792px\" \/><\/figure>\n\n\n\n<!--nextpage-->\n\n\n\n<figure class=\"wp-block-image\"><img loading=\"lazy\" decoding=\"async\" width=\"791\" height=\"1024\" src=\"http:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_04-791x1024.jpg\" alt=\"\" class=\"wp-image-403\" srcset=\"https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_04-791x1024.jpg 791w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_04-232x300.jpg 232w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_04-768x994.jpg 768w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_04.jpg 952w\" sizes=\"auto, (max-width: 791px) 100vw, 791px\" \/><\/figure>\n\n\n\n<!--nextpage-->\n\n\n\n<figure class=\"wp-block-image\"><img loading=\"lazy\" decoding=\"async\" width=\"663\" height=\"869\" src=\"http:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_05.jpg\" alt=\"\" class=\"wp-image-404\" srcset=\"https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_05.jpg 663w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_05-229x300.jpg 229w\" sizes=\"auto, (max-width: 663px) 100vw, 663px\" \/><\/figure>\n\n\n\n<!--nextpage-->\n\n\n\n<figure class=\"wp-block-image\"><img loading=\"lazy\" decoding=\"async\" width=\"792\" height=\"1024\" src=\"http:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_06-792x1024.jpg\" alt=\"\" class=\"wp-image-405\" srcset=\"https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_06-792x1024.jpg 792w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_06-232x300.jpg 232w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_06-768x993.jpg 768w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_06.jpg 893w\" sizes=\"auto, (max-width: 792px) 100vw, 792px\" \/><\/figure>\n\n\n\n<!--nextpage-->\n\n\n\n<figure class=\"wp-block-image\"><img loading=\"lazy\" decoding=\"async\" width=\"784\" height=\"1024\" src=\"http:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_07-784x1024.jpg\" alt=\"\" class=\"wp-image-406\" srcset=\"https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_07-784x1024.jpg 784w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_07-230x300.jpg 230w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_07-768x1003.jpg 768w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_07.jpg 834w\" sizes=\"auto, (max-width: 784px) 100vw, 784px\" \/><\/figure>\n\n\n\n<!--nextpage-->\n\n\n\n<figure class=\"wp-block-image\"><img loading=\"lazy\" decoding=\"async\" width=\"791\" height=\"1024\" src=\"http:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_08-791x1024.jpg\" alt=\"\" class=\"wp-image-407\" srcset=\"https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_08-791x1024.jpg 791w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_08-232x300.jpg 232w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_08-768x994.jpg 768w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_08.jpg 1700w\" sizes=\"auto, (max-width: 791px) 100vw, 791px\" \/><\/figure>\n\n\n\n<!--nextpage-->\n\n\n\n<figure class=\"wp-block-image\"><img loading=\"lazy\" decoding=\"async\" width=\"551\" height=\"1024\" src=\"http:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_09-551x1024.jpg\" alt=\"\" class=\"wp-image-408\" srcset=\"https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_09-551x1024.jpg 551w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_09-161x300.jpg 161w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_09-768x1427.jpg 768w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_09.jpg 1190w\" sizes=\"auto, (max-width: 551px) 100vw, 551px\" \/><\/figure>\n\n\n\n<!--nextpage-->\n\n\n\n<figure class=\"wp-block-image\"><img loading=\"lazy\" decoding=\"async\" width=\"632\" height=\"1024\" src=\"http:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_10-632x1024.jpg\" alt=\"\" class=\"wp-image-409\" srcset=\"https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_10-632x1024.jpg 632w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_10-185x300.jpg 185w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_10-768x1244.jpg 768w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_10.jpg 978w\" sizes=\"auto, (max-width: 632px) 100vw, 632px\" \/><\/figure>\n\n\n\n<!--nextpage-->\n\n\n\n<figure class=\"wp-block-image\"><img loading=\"lazy\" decoding=\"async\" width=\"670\" height=\"1024\" src=\"http:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_11-670x1024.jpg\" alt=\"\" class=\"wp-image-410\" srcset=\"https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_11-670x1024.jpg 670w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_11-196x300.jpg 196w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_11-768x1173.jpg 768w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_11.jpg 1037w\" sizes=\"auto, (max-width: 670px) 100vw, 670px\" \/><\/figure>\n\n\n\n<!--nextpage-->\n\n\n\n<figure class=\"wp-block-image\"><img loading=\"lazy\" decoding=\"async\" width=\"791\" height=\"1024\" src=\"http:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_12-791x1024.jpg\" alt=\"\" class=\"wp-image-411\" srcset=\"https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_12-791x1024.jpg 791w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_12-232x300.jpg 232w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_12-768x994.jpg 768w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_12.jpg 1700w\" sizes=\"auto, (max-width: 791px) 100vw, 791px\" \/><\/figure>\n\n\n\n<!--nextpage-->\n\n\n\n<figure class=\"wp-block-image\"><img loading=\"lazy\" decoding=\"async\" width=\"727\" height=\"1024\" src=\"http:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_13-727x1024.jpg\" alt=\"\" class=\"wp-image-412\" srcset=\"https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_13-727x1024.jpg 727w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_13-213x300.jpg 213w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_13-768x1082.jpg 768w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_13.jpg 1632w\" sizes=\"auto, (max-width: 727px) 100vw, 727px\" \/><\/figure>\n\n\n\n<!--nextpage-->\n\n\n\n<figure class=\"wp-block-image\"><img loading=\"lazy\" decoding=\"async\" width=\"791\" height=\"1024\" src=\"http:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_14-791x1024.jpg\" alt=\"\" class=\"wp-image-413\" srcset=\"https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_14-791x1024.jpg 791w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_14-232x300.jpg 232w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_14-768x994.jpg 768w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_14.jpg 1700w\" sizes=\"auto, (max-width: 791px) 100vw, 791px\" \/><\/figure>\n\n\n\n<!--nextpage-->\n\n\n\n<figure class=\"wp-block-image\"><img loading=\"lazy\" decoding=\"async\" width=\"656\" height=\"1024\" src=\"http:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_15-656x1024.jpg\" alt=\"\" class=\"wp-image-414\" srcset=\"https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_15-656x1024.jpg 656w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_15-192x300.jpg 192w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_15.jpg 740w\" sizes=\"auto, (max-width: 656px) 100vw, 656px\" \/><\/figure>\n\n\n\n<!--nextpage-->\n\n\n\n<figure class=\"wp-block-image\"><img loading=\"lazy\" decoding=\"async\" width=\"792\" height=\"1024\" src=\"http:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_16-792x1024.jpg\" alt=\"\" class=\"wp-image-415\" srcset=\"https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_16-792x1024.jpg 792w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_16-232x300.jpg 232w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_16-768x993.jpg 768w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_16.jpg 842w\" sizes=\"auto, (max-width: 792px) 100vw, 792px\" \/><\/figure>\n\n\n\n<!--nextpage-->\n\n\n\n<figure class=\"wp-block-image\"><img loading=\"lazy\" decoding=\"async\" width=\"872\" height=\"1024\" src=\"http:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_17-872x1024.jpg\" alt=\"\" class=\"wp-image-416\" srcset=\"https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_17-872x1024.jpg 872w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_17-256x300.jpg 256w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_17-768x902.jpg 768w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_17.jpg 1471w\" sizes=\"auto, (max-width: 872px) 100vw, 872px\" \/><\/figure>\n\n\n\n<!--nextpage-->\n\n\n\n<figure class=\"wp-block-image\"><img loading=\"lazy\" decoding=\"async\" width=\"791\" height=\"1024\" src=\"http:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_18-791x1024.jpg\" alt=\"\" class=\"wp-image-417\" srcset=\"https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_18-791x1024.jpg 791w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_18-232x300.jpg 232w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_18-768x994.jpg 768w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_18.jpg 1250w\" sizes=\"auto, (max-width: 791px) 100vw, 791px\" \/><\/figure>\n\n\n\n<!--nextpage-->\n\n\n\n<figure class=\"wp-block-image\"><img loading=\"lazy\" decoding=\"async\" width=\"612\" height=\"792\" src=\"http:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_19.jpg\" alt=\"\" class=\"wp-image-418\" srcset=\"https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_19.jpg 612w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_19-232x300.jpg 232w\" sizes=\"auto, (max-width: 612px) 100vw, 612px\" \/><\/figure>\n\n\n\n<!--nextpage-->\n\n\n\n<figure class=\"wp-block-image\"><img loading=\"lazy\" decoding=\"async\" width=\"723\" height=\"935\" src=\"http:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_20.jpg\" alt=\"\" class=\"wp-image-419\" srcset=\"https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_20.jpg 723w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_20-232x300.jpg 232w\" sizes=\"auto, (max-width: 723px) 100vw, 723px\" \/><\/figure>\n\n\n\n<!--nextpage-->\n\n\n\n<figure class=\"wp-block-image\"><img loading=\"lazy\" decoding=\"async\" width=\"741\" height=\"1024\" src=\"http:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_21-741x1024.jpg\" alt=\"\" class=\"wp-image-420\" srcset=\"https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_21-741x1024.jpg 741w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_21-217x300.jpg 217w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_21.jpg 748w\" sizes=\"auto, (max-width: 741px) 100vw, 741px\" \/><\/figure>\n\n\n\n<!--nextpage-->\n\n\n\n<figure class=\"wp-block-image\"><img loading=\"lazy\" decoding=\"async\" width=\"791\" height=\"1024\" src=\"http:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_22-791x1024.jpg\" alt=\"\" class=\"wp-image-421\" srcset=\"https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_22-791x1024.jpg 791w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_22-232x300.jpg 232w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_22-768x994.jpg 768w, https:\/\/www.aircraftengineer.info\/notes\/wp-content\/uploads\/2019\/11\/Matter-module2_Page_8_Page_22.jpg 1700w\" sizes=\"auto, (max-width: 791px) 100vw, 791px\" \/><\/figure>\n\n\n\n<!--nextpage-->\n\n\n\n<h1 class=\"wp-block-heading\"><a>1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;\nMechanics<\/a><\/h1>\n\n\n\n<h2 class=\"wp-block-heading\"><a>1.1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;\nForces,\nMoments and Couples<\/a><\/h2>\n\n\n\n<h3 class=\"wp-block-heading\"><a>1.1.1&nbsp;&nbsp;&nbsp;&nbsp;\nScalar\nand Vector Quantities<\/a><\/h3>\n\n\n\n<p>Before introducing force as a measurable\nquantity, we should discuss how we identify that quantity. <\/p>\n\n\n\n<p>Quantities\nare thought of as being either<strong> scalar<\/strong>\nor <strong>vector<\/strong>. The term scalar means\nthat the quantity possesses magnitude (size) ONLY.&nbsp; Examples of scalar quantities include mass,\ntime, temperature, length etc.&nbsp; These\nquantities, as the name \u201cscalar\u201d indicates, may only be represented graphically\nto some form of scale. <\/p>\n\n\n\n\n\n<p><br>\nTHUS\na temperature of 15<sup>\u00b0<\/sup>C may be\nrepresented as: <\/p>\n\n\n\n<p><strong>Fig 2.1&nbsp;\nScalar representation of 15\u00baC<\/strong><\/p>\n\n\n\n<p>Vector\nquantities are different in that they possess both magnitude AND direction and,\nif either change, the vector quantity changes. Vector quantities include force,\nvelocity and any quantity formed from these. <\/p>\n\n\n\n<p>A force\nis a vector quantity, and as such, possesses magnitude and direction.&nbsp; In specifying a force, therefore, you must\nspecify both the size of the force and the direction in which it is applied.&nbsp; This can be shown on a diagram by a line of a\nspecific length with the direction indicated by an arrow. The most convenient\nmethod is to represent the force by means of a <strong>vector <\/strong>as shown in the diagram. If the point of application of a\nforce is important it may be shown in a space diagram.<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"\"><tbody><tr><td><\/td><\/tr><tr><td><\/td><td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><strong>Vector\nDiagram&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp; Space Diagram<\/strong><\/p>\n\n\n\n<p><strong>Fig 2.2 Vector Representation of a Force<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><a>1.1.2&nbsp; Triangle of Forces<\/a><\/h3>\n\n\n\n<p>The total\neffect, or resultant, of a number of forces acting on a body may be determined\nby vector addition. Conversely, a single force may be resolved into components,\nsuch that these components have the same total effect as the original force. It\nis often convenient to replace a force by its two components at <em>right angles<\/em>.<\/p>\n\n\n\n<p>Two or more forces can be added or\nsubtracted to produce a <strong>Resultant Force<\/strong>.&nbsp; If two forces are equal but act in opposite\ndirections, then obviously they cancel each other out, and so the resultant is\nsaid to be zero.&nbsp; Two forces can be added\nor subtracted mathematically or graphically, and this procedure often produces\na <strong>Triangle of Force<\/strong>.<\/p>\n\n\n\n<p>Firstly, it is important to realise that a\nforce has three important features; magnitude (size), direction and line of\naction.<\/p>\n\n\n\n<p>Force is therefore a <strong>vector<\/strong> quantity, and as such, it can be represented by an arrow,\ndrawn to a scale representing magnitude and direction.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><a>1.1.3&nbsp;&nbsp;&nbsp;\nGraphical\nMethod<\/a><\/h3>\n\n\n\n\n\n<p>Consider two forces A and B.&nbsp; Choose a starting point O and draw OA to\nrepresent force A, in the direction of A.&nbsp;\nThen draw AB to represent force B.<\/p>\n\n\n\n<p>Fig. 2.3\nTriangle of Forces<\/p>\n\n\n\n\n\n<p>The line OB represents the resultant of two forces.<\/p>\n\n\n\n<p>Note that the line representing force B\ncould have been drawn first, and force A drawn second; the resultant would have\nbeen the same.<\/p>\n\n\n\n<p>The two forces added together have formed 2\nsides of the triangle; the resultant is the third side.<\/p>\n\n\n\n\n\n<p>If a <strong>third<\/strong>\nforce, <strong>equal<\/strong> in <strong>length<\/strong> but <strong>opposite<\/strong> in <strong>direction<\/strong> to the resultant is added to\nthe resultant, it will cancel the effect of the two forces.&nbsp; This third force would be termed the\nEquilabrant. <\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><a>1.1.4&nbsp;&nbsp;&nbsp;\nPolygon\nof Forces<\/a><\/h3>\n\n\n\n<p>This topic just builds on the previous\nTriangle of Forces.<\/p>\n\n\n\n\n\n<p>Consider three forces A, B and C as shown in the\ndiagram.&nbsp; A and B can be added and by\ndrawing a triangle, the resultant is produced.<\/p>\n\n\n\n\n\n<p>If force C is joined to this resultant, a\nfurther or &#8220;new&#8221; resultant is created, which represents the effect of\nall three forces.<\/p>\n\n\n\n\n\n<p><br>\nNow this procedure can be repeated\nmany times; the effect is to <\/p>\n\n\n\n<p>produce a Polygon of <a>Forces<\/a><a href=\"#_msocom_1\">[SJ1]<\/a>&nbsp;.<\/p>\n\n\n\n\n\n<h3 class=\"wp-block-heading\"><a>1.1.5&nbsp;&nbsp;&nbsp;&nbsp; Coplanar Forces<\/a><\/h3>\n\n\n\n<p>Forces\nwhose lines of action all lie in the same plane are called<strong> coplanar forces<\/strong>.&nbsp; The\nfollowing laws relating to coplanar forces are of importance and should be\nnoted carefully. However, it must also be remembered that these laws are\napplicable ONLY to two dimensional problems.<\/p>\n\n\n\n<p>The line of action of the resultant of any two coplanar forces must pass\nthrough the point of intersection of the lines of action of the two forces.<\/p>\n\n\n\n<p>If any number of coplanar forces act on a body and are not in\nequilibrium, then they can always be reduced to a single resultant force and a\ncouple.<\/p>\n\n\n\n<p>If three forces acting on a body are in equilibrium, then their lines of\naction must be concurrent, &#8211; that is, they must all pass through the same\npoint.<\/p>\n\n\n\n<p>Forces acting at the same point are called CONCURRENT forces.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><a>1.1.6&nbsp;&nbsp;&nbsp;&nbsp;\nEffect\nof an Applied Force<\/a><\/h3>\n\n\n\n<p>If a Force is applied to a body, it will\ncause that body to move or rotate.&nbsp; A\nbody that is already moving will change its speed or direction.&nbsp; Note that the term &#8216;change its speed or\ndirection&#8217; implies that an acceleration has taken place.<\/p>\n\n\n\n<p>This is usually summarised in the\nformula;&nbsp;&nbsp; F&nbsp; = ma<\/p>\n\n\n\n<p>Where F is the force,&nbsp; m = mass of body and a = acceleration. <\/p>\n\n\n\n<p>The units of force should be kg.m\/s<sup>2<\/sup>\nbut the SI Unit used is the Newton.<\/p>\n\n\n\n<p>Hence, &#8220;A Newton is the unit of force\nthat when applied to a mass of 1 kg. causes that mass to accelerate at a rate\nof 1 m\/s<sup>2<\/sup>.<\/p>\n\n\n\n<p>Applied forces can also cause changes in\nshape or size of a body, which is important when analysing the behaviour of\nmaterials.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><a>1.1.7&nbsp;&nbsp;&nbsp;&nbsp;\nEquilibriums<\/a><\/h3>\n\n\n\n<p>Earlier it was defined that a force applied\nto a body would cause that body to accelerate or change direction.<\/p>\n\n\n\n<p>If at any stage a <strong>system<\/strong> of forces is applied to a body, such that their resultant is\nzero, then that body will <strong>not<\/strong>\naccelerate or change direction.&nbsp; The\nsystem of forces and the body are said to be in the equilibrium.<\/p>\n\n\n\n<p><strong>Note:<\/strong> This does <strong>not<\/strong> mean that there are no forces acting; it is just that their\ntotal resultant or effect is zero.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><a>1.1.8&nbsp;&nbsp;&nbsp;&nbsp;\nResolution\nof Forces<\/a><\/h3>\n\n\n\n\n\n<p>This topic is important, but is really the opposite to\nAddition of forces. Recalling that two forces can be <strong>added<\/strong> to give a single force known as the Resultant, it is obvious\nthat this single force can be considered as the addition of the two original\nforces.<\/p>\n\n\n\n\n\n<p>Therefore, the single force can be\nseparated or Resolved into two components.<\/p>\n\n\n\n\n\n<p>&nbsp;It\nshould be appreciated that almost always the single force is resolved into two\ncomponents, that are <strong>mutually\nperpendicular<\/strong>.<\/p>\n\n\n\n<p>This technique forms the basis\nof the mathematical methods for adding forces.<\/p>\n\n\n\n\n\n<p>Note that by drawing the right-angled triangle, with\nthe single force F, and by choosing angle q relative to a datum, the two components\nbecome F Sin q and F Cos q.<\/p>\n\n\n\n\n\n\n\n<p>From your mathematics, <\/p>\n\n\n\n\n\n<h3 class=\"wp-block-heading\"><a>1.1.9&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;\nGraphical\nSolutions<\/a><\/h3>\n\n\n\n<p>This topic looks at deriving <strong>graphical<\/strong> solutions to problems\ninvolving the Addition of Vector Quantities.<\/p>\n\n\n\n<p>Firstly, the quantities must be vector\nquantities.&nbsp; Secondly, they must all be\nthe same, i.e. all forces, or all velocities, etc. (they cannot be mixed-up).<\/p>\n\n\n\n<p>Thirdly, a suitable scale representing the <strong>magnitude<\/strong> of the vector quantity should\nbe selected.<\/p>\n\n\n\n<p>Finally, before drawing a Polygon of\nvectors, a reference or datum direction should be defined.<\/p>\n\n\n\n<p>To derive a solution (i.e. a resultant),\nproceed to draw the lines representing the vectors (be careful to draw all\nlines with reference to the direction datum).<\/p>\n\n\n\n<p>The resultant is determined by measuring\nthe magnitude and direction of the line drawn from the <strong>start<\/strong> point to the <strong>finish<\/strong>\npoint.<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"\"><tbody><tr><td><\/td><\/tr><tr><td><\/td><td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>\nNote that the <strong>order<\/strong> in which the individual vectors are drawn is <strong>not<\/strong> important.\n\n<\/p>\n\n\n\n<p><strong>Fig 2.7 Adding Vector Quantities<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><a>1.1.10&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;\nMoments\nand Couples<\/a><\/h3>\n\n\n\n<p>In para 2.1.6, it was stated that if a\nforce was applied to a body, it would move (accelerate) in the direction of the\napplied force.<\/p>\n\n\n\n<p>Consider that the body cannot move from one\nplace to another, but can <strong>rotate<\/strong>.&nbsp; The applied force will then cause a\nrotation.&nbsp; An example is a door.&nbsp; A force applied to the door cause it to open\nor close, rotating about the hinge-line.&nbsp;\nBut what is important to realise is that the force required to move the\ndoor is dependent on how far from the hinge the force is applied.<\/p>\n\n\n\n\n\n<p>So the <strong>turning\neffect<\/strong> of a force is a combination of the magnitude of the force and its\ndistance from the point of rotation.&nbsp; The\nturning effect is termed the Moment of a Force.<\/p>\n\n\n\n\n\n<p>From the diagram it can be seen that the moment is a\nresult of the formula:<\/p>\n\n\n\n<p><strong>Moment of a force (F) about a point (O)&nbsp; =&nbsp; F\nx&nbsp; y<\/strong><\/p>\n\n\n\n<p>[where \u2018y\u2019 is the perpendicular distance between the\nforce and the point &#8216;O&#8217;&nbsp; often referred\nto as the &#8216;moment arm&#8217; ].<\/p>\n\n\n\n<p>Using SI units, the units are Newton&nbsp; x&nbsp;\nmetres = Newton Metres or Nm<\/p>\n\n\n\n<p><strong>Note<\/strong>:&nbsp; It\nis important to realise that the \u201cdistance\u201d is perpendicular to the line of\naction of the force.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><a>1.1.11&nbsp;&nbsp;\nClockwise\nand Anti-Clockwise Moments<\/a><\/h3>\n\n\n\n<p>Fig 2.9\nClockwise and Anti-Clockwise Moments<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"\"><tbody><tr><td><\/td><\/tr><tr><td><\/td><td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>\nThe moment or turning effect of a\nforce about a specific point can be clockwise or anti-clockwise depending on\nthe direction of the force. In the diagram shown, Force B produces a clockwise\nmoment about point O and Force A produces an anti-clockwise moment.\n\n<\/p>\n\n\n\n<p>When several forces are involved, <strong>equilibrium<\/strong> concerns not just the\nforces, but moments as well.&nbsp; If\nequilibrium exists, then clockwise (positive) moments are balanced by\nanticlockwise (negative) moments. It is normal to say:<\/p>\n\n\n\n<p>Clockwise\nMoments = Anti-clockwise Moments<\/p>\n\n\n\n<p><strong>Beam\nExample 1:<\/strong><\/p>\n\n\n\n<p>The diagram shows a light beam pivoted at\npoint B with vertical forces of 50N and 125N acting at the ends. The 50N force\nproduces an anti-clockwise moment of 50 x 3 = 150Nm about point B and the 125N\nforce produces a clockwise moment of 125 x Y = 125Y Nm.<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"\"><tbody><tr><td><\/td><\/tr><tr><td><\/td><td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>\nFig\n2.10&nbsp; Simple Beam\n\n<\/p>\n\n\n\n<p>If the beam is in equilibrium, Clockwise\nmoments = Anti-clockwise moments, so:<\/p>\n\n\n\n<p>125Y = 150, or Y = 1.2m<\/p>\n\n\n\n<p><strong>Note:<\/strong>&nbsp; In the previous beam example, if the beam is\nin equilibrium, we have stated that the CWM = ACWM. As well as this, the total\nforce acting downwards, must equal the total upwards force. There is a vertical\n\u201creaction\u201d acting at point B. The magnitude of this reaction is equal to the\nsum of the other two forces i.e. 175N. We do not need to include this value in\nthe calculation, because it does not produce a turning moment if we assume the\nbeam is pivoted at this point. (175 x 0m = 0Nm)&nbsp;\n<\/p>\n\n\n\n<p><strong>Beam Example 2:<\/strong><\/p>\n\n\n\n\n\n<p><br>\nThe diagram shows a beam with a total\nlength of 8m pivoted at point F. Three forces A, B and C are shown acting on\nthe beam. What additional force must be applied to the beam at D to maintain\nequilibrium. As no further information is given, we assume the beam has\nnegligible mass.<\/p>\n\n\n\n<p>The statement \u201cto maintain\nequilibrium\u201d means that the clockwise moments must be balanced by the\nanti-clockwise moments i.e.&nbsp; CWM = ACWM.\nAt this point we do not know if the force at D will be acting upwards or downwards.\nUsing the known forces:<\/p>\n\n\n\n<p>CWM are&nbsp; (1000 x 1) + (250 x 3.5) = 1875Nm<\/p>\n\n\n\n<p>ACWM are (500 x 3) = 1500Nm<\/p>\n\n\n\n\n\n<p>At this point we know that the\nforce at D must produce an ACWM of 375Nm to produce equilibrium. The value of D\nwill be . It must therefore act vertically upwards. It also follows\nthat if vertical equilibrium exists, downward forces must equal upwards forces,\nso:<\/p>\n\n\n\n<p>Downwards forces = 500N + 1000N +\n250N = 1750N<\/p>\n\n\n\n<p>Upwards forces = F + D. If D =\n75N, F must be 1750 \u2013 75 = 1675N.<\/p>\n\n\n\n<p><strong>Beam Example 3:<\/strong><\/p>\n\n\n\n\n\n<p><br>\nAssuming the beam shown is in\nequilibrium, find the value of the two supports R1 and R2.<\/p>\n\n\n\n<p>The beam shown above has loads A-F\nacting vertically downwards. The two forces R1 and R2 are acting vertically\nupwards. Our first thought are that as we have two unknown values, we cannot\nsolve the problem. We can start to solve it by first taking moments about one\nof the points R1 or R2. We assume the beam can rotate about point R1, the\nmoment at point R1 is 0, and say CWM = ACWM:<\/p>\n\n\n\n<p>Total CWM = (2000 x 1) + (10000 x\n2) + (5000 x 3.5) + (5000 x 4.5) + (1000 x 5.5) = 67,500Nm<\/p>\n\n\n\n<p>Total ACWM = (R2 x 6.25) + (1000 x\n0.5)<\/p>\n\n\n\n<p>So if CWM = ACWM<\/p>\n\n\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 67,500\n= (6.25 x R2) + 500&nbsp; so&nbsp; <\/p>\n\n\n\n<p>The value of the vertical force at\nR2 is therefore 10,720N.<\/p>\n\n\n\n<p>As we have stated the beam is in\nequilibrium, not only do the CWM = ACWM, but also the total downwards forces\nare balanced by upwards forces. The total value of R1 + R2 must be 1,000+ 2000\n+10000 + 5000 + 5000 + 1000 = 24,000N.<\/p>\n\n\n\n<p>We have calculated the value of R2\nto be 10,720N, it follows that R1 must be 13,280N.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><a>1.1.12&nbsp;&nbsp;\nCouples<\/a><\/h3>\n\n\n\n\n\n<p>When two equal but opposite forces are present, whose\nlines of action are not coincident, then they cause a rotation.<\/p>\n\n\n\n<p>Together, they are termed a Couple, and the\nmoment of a couple is equal to the magnitude of a force F, multiplied by the\ndistance between them.<\/p>\n\n\n\n<p>The basic principles of moments and couples\nare used extensively in aircraft engineering<\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><br>\ncentre of gravity<\/h2>\n\n\n\n\n\n<p>Consider\na body as an accumulation of many small masses (molecules), all subject to\ngravitational attraction.&nbsp; The total\nweight, which is a <strong>force<\/strong>, is equal\nto the sum of the individual masses, multiplied by the gravitational\nacceleration&nbsp;&nbsp; g&nbsp; =&nbsp; 9.81\nm\/s<sup>2<\/sup>).<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>W&nbsp;\n=&nbsp; mg<\/strong><\/p>\n\n\n\n\n\n<p>The diagram shows that the individual forces all act\nin the same <strong>direction<\/strong>, but have\ndifferent lines of action.<\/p>\n\n\n\n\n\n\n\n<p>There must be datum position, such that the total\nmoment to one side, causing a clockwise rotation, is <strong>balanced <\/strong>by a total moment, on the other side, which causes an\nanticlockwise rotation.&nbsp; In other words,\nthe total weight can be considered to act through that datum position (=&nbsp; line of action).<\/p>\n\n\n\n<p>If the body is considered in two different\nposition, the weight acts through two lines of action, W<sub>1<\/sub> and W<sub>2<\/sub>\nand these interact at point G, which is termed the Centre of Gravity.<\/p>\n\n\n\n\n\n<p>Hence, the Centre of Gravity is the point\nthrough which the Total Mass of the body may be considered to act.<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <br>\nFor a 3-dimensional body, the centre of gravity can be determined practically\nby several methods, such as by measuring and equating moments, and this is done\nwhen calculating Weight and Balance of aircraft.<\/p>\n\n\n\n<p>A 2-dimensional body (one of negligible\nthickness) is termed a lamina, which only has area (not volume).&nbsp; The point G is then termed a Centroid.&nbsp; If a lamina is suspended from point P, the\ncentroid G will hang vertically below \u2018P<sub>1<\/sub>\u2019.&nbsp; If suspended from P<sub>2<\/sub>&nbsp; G will hang below P<sub>2<\/sub>.&nbsp; Position G is at the intersection as shown.<\/p>\n\n\n\n\n\n\n\n<p>A\nregular lamina, such as a rectangle, has its centre of gravity at the\nintersection of the diagonals.<\/p>\n\n\n\n\n\n<p>A triangle has its centre of gravity at the\nintersection of the medians.<\/p>\n\n\n\n\n\n\n\n<p>The<strong> centre of gravity<\/strong> of a solid object is\nthe point about which the total weight <em>appears<\/em>\nto act. Or, put another way, if the object is balanced at that point, it will\nhave no tendency to rotate.&nbsp; In the case\nof hollow or irregular shaped objects, it is possible for the centre of gravity\nto be in free space and not within the objects at all. The most important\napplication of centre of gravity for aircraft mechanics is the weight and\nbalance of an aircraft.<\/p>\n\n\n\n<p>If an\naircraft is correctly loaded, with fuel, crew and passengers, baggage, etc. in\nthe correct places, the aircraft will be in balance and easy to fly.&nbsp; If, for example, the baggage has been loaded\nincorrectly, making the aircraft much too nose or tail heavy, the aircraft\ncould be difficult to fly or might even crash.<\/p>\n\n\n\n<p>It is\nimportant that whenever changes are made to an aircraft, calculations MUST be\nmade each time to ensure that the centre of gravity is within acceptable limits\nset by the manufacturer of the aircraft. These changes could be as simple as a\nnew coat of paint, or as complicated as the conversion from passenger to a\nfreight carrying role.<\/p>\n\n\n\n\n\n\n\n<h2 class=\"wp-block-heading\"><a>1.2&nbsp;&nbsp;&nbsp;&nbsp; Stress, Strain and\nElastic Tension<\/a><\/h2>\n\n\n\n<h3 class=\"wp-block-heading\"><a>1.2.1&nbsp;&nbsp;&nbsp;\nStress<\/a><\/h3>\n\n\n\n<p>When an engineer designs a component or\nstructure he needs to know whether it is strong enough to prevent failure due\nto the loads encountered in service.&nbsp; He\nanalyses the <strong>external<\/strong> forces and\nthen deduces the forces or <strong>stresses<\/strong>\nthat are induced <strong>internally<\/strong>.<\/p>\n\n\n\n<p>Notice the introduction of the word\nstress.&nbsp; Obviously a component which is\ntwice the size is stronger and less likely to fail due an applied load.&nbsp; So an important factor to consider is not\njust force, but size as well.&nbsp; Hence <strong>stress<\/strong> is load divided by area (size).<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; s<sub> (sigma)<\/sub>&nbsp; =&nbsp; &nbsp;(=\nNewtons per second metre).<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"\"><tbody><tr><td><\/td><\/tr><tr><td><\/td><td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>\nComponents fail due to being <strong>over-stressed<\/strong>, not over-loaded. \n\n<\/p>\n\n\n\n<p>The external forces induce internal\nstresses which oppose or balance the external forces.<\/p>\n\n\n\n<p>Stresses can occur in differing forms,\ndependent on the manner of application of the external force.<\/p>\n\n\n\n<p>There are\nfive different types of stress in mechanical bodies.&nbsp; They are <strong>tension<\/strong>,\n<strong>compression<\/strong>, <strong>torsion,<\/strong> <strong>bending<\/strong> and <strong>shear.<\/strong><\/p>\n\n\n\n<h4 class=\"wp-block-heading\">1.2.1.1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;\nTension or Tensile Stress<\/h4>\n\n\n\n<p>Tensile\nstress describes the effect of a force that tends to pull an object apart.&nbsp; Flexible steel cable used in aircraft control\nsystems is an example of a component that is in designed to withstand tension\nloads.&nbsp; Steel cable is easily bent and\nhas little opposition to other types of stress, but, when subjected to a purely\ntensile load, it performs exceptionally well.<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"\"><tbody><tr><td><\/td><\/tr><tr><td><\/td><td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n\n\n\n\n\n\n<h4 class=\"wp-block-heading\">1.2.1.2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;\nCompression or Compressive Stress<\/h4>\n\n\n\n\n\n<p><br>\nCompression\nis the resistance to an external force that tries to push an object\ntogether.&nbsp; Aircraft rivets are driven\nwith a compressive force. When compression stress is applied to a rivet, the\nrivet firstly expands until it fills the hole and then the external part of the\nshank spreads to form a second head, which holds the sheets of metal tightly\ntogether.<\/p>\n\n\n\n\n\n<h4 class=\"wp-block-heading\">1.2.1.3&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;\nTorsion<\/h4>\n\n\n\n<p>A\ntorsional stress is applied to a material when it is twisted. Torsion is\nactually a combination of both tension and compression. For example, when a\nobject is subjected to torsional stress, tensional stresses operate diagonally\nacross the object whilst compression stresses act at right angles to the\ntension stress.<\/p>\n\n\n\n\n\n\n\n<p>An engine crankshaft\nis a component whose primary stress is torsion.&nbsp;\nThe pistons pushing down on the connecting rods rotate the crankshaft\nagainst the opposition, or resistance of the propeller. The resulting stresses\nattempt to twist the crankshaft.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">1.2.1.4&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;\nBending\nStress<\/h4>\n\n\n\n\n\n<figure class=\"wp-block-table\"><table class=\"\"><tbody><tr><td><\/td><\/tr><tr><td><\/td><td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>\nIf a beam is anchored at one end and\na load applied at the other end, the beam will bend in the direction of the\napplied load.\n\n<\/p>\n\n\n\n<p>An\naircraft wing acts as a cantilever beam, with the wing supported at the\nfuselage attachment point.<\/p>\n\n\n\n<p>When the\naircraft is on the ground the force of gravity causes the wing to bend in a\nsimilar manner to the beam shown in Fig. 2.21.&nbsp;\nIn this case, the top of the wing is subjected to tension stress whilst\nthe lower skin experiences compression stress. In flight, the force of lift\ntries to bend an aircraft&#8217;s wing upward.&nbsp;\nWhen this happens the skin on the top of the wing is subjected to a\ncompressive force, whilst the skin below the wing is pulled by a tension force.\nThe following diagram illustrates this.&nbsp; <\/p>\n\n\n\n\n\n\n\n<p><strong><br>\n<\/strong><\/p>\n\n\n\n<h4 class=\"wp-block-heading\">1.2.1.5&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;\nShear<\/h4>\n\n\n\n<p>A shear stress attempts to slice, (or shear) a body apart.&nbsp; A clevis bolt in an aircraft control system\nis designed to withstand shear loads. These are made of high-strength steel and\nare fitted with a thin nut that is held in place with a split pin. Whenever a\ncontrol cable moves, shear forces are applied to the bolt. However, when no\nforce is present, the clevis bolt is free to turn in its hole. The other\ndiagram shows two sheets of metal held together with a rivet. If a tensile load\nis applied to the sheets (as would happen to the top skin of an aircraft wing,\nwhen the aircraft is on the ground), the rivet is subjected to a shear load.<strong><\/strong><\/p>\n\n\n\n\n\n\n\n<h3 class=\"wp-block-heading\"><a>1.2.2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Strain<\/a><\/h3>\n\n\n\n<p>When the\nmaterial of a body is in a state of stress, deformation takes place so that the\nsize and shape of the body is changed.&nbsp;\nThe manner of deformation will depend on how the body is loaded, but a\nsimple tension member tends to stretch and a simple compression member tends to\ncontract.&nbsp; If the member has a uniform\ncross section, the intensity of stress will be the same throughout its length,\nso that each unit of length will extend or contract by the same amount.&nbsp; The total change in length, corresponding to\na given stress, will thus depend on the original length of the member. <\/p>\n\n\n\n<p>Deformation\ndue to an internal state of stress is called<strong> strain (\u03b5)<\/strong>. Any measurement of strain must be related to the\noriginal dimension involved.<\/p>\n\n\n\n<p>Example:<\/p>\n\n\n\n<p>Intensity\nof strain (<strong>\u03b5<\/strong>)&nbsp; =&nbsp;\nchange in length (x) \/ original length (L) <\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>\u03b5<\/strong> =&nbsp; x \/ L<\/p>\n\n\n\n<p>Where x is the extension <em>or compression<\/em> of the member.<\/p>\n\n\n\n\n\n<p><br>\nNote:\nSince strain is simply the ratio between two lengths, it is dimensionless.&nbsp; It is, however, usually expressed as a percentage..\n<\/p>\n\n\n\n<p><strong>Example\nof Stress and Strain<\/strong><\/p>\n\n\n\n<p>A steel rod 20 mm diameter and 1m carries a\nload of 45 kN. This causes an extension of 1.8mm. Calculate the stress and\nstrain in the rod.<\/p>\n\n\n\n\n\n\n\n<p>Note that there are no units for strain.\nStrain may also be indicated as a percentage. To show strain as a percentage\nyou simply multiply by 100. So in the above example the strain as a percentage\nis 0.0018 x 100 = 0.18%.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><a>1.2.3&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Elasticity<\/a><\/h3>\n\n\n\n<p>Engineering\nmaterials must, of necessity, possess the property of <strong>elasticity<\/strong>.&nbsp; This is the\nproperty that allows a piece of the material to regain its original size and\nshape when the forces producing a state of strain are removed.&nbsp;&nbsp; If a bar of elastic material of uniform\ncross-section, is loaded progressively in tension, it will be found that, up to\na point, the corresponding extensions will be proportional to the applied\nloads. <\/p>\n\n\n\n<p>This\nproportionality is known as <strong>Hooke&#8217;s Law<\/strong>.&nbsp; However, to be meaningful, loads and\nextensions must be related to a particular bar of known cross-sectional area\nand length. A more general statement of this law may be made in terms of the\nstress and strain in the material of the bar.<\/p>\n\n\n\n<p><strong>Within the limit of proportionality, the\nstrain is directly proportional to the stress producing it.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"\"><tbody><tr><td><\/td><\/tr><tr><td><\/td><td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>\nIf\nwe plot the graph of stress against strain, we will produce a straight line\npassing through the origin as shown below.&nbsp;\nThe slope of the graph, stress\/strain, is&nbsp; a constant for a given material.&nbsp; This constant is known as Young&#8217;s Modulus of\nElasticity and is always denoted by the capital letter <strong>E<\/strong>.&nbsp;&nbsp;&nbsp;&nbsp; Once the line plotted\nbegins to curve towards horizontal the material is said to have passed its <strong>elastic limit<\/strong> and will NOT return to\nits original length. It will have a permanent stretch.\n\n<\/p>\n\n\n\n\n\n<p>Young&#8217;s\nModulus of Elasticity (<strong><em>E<\/em>)<\/strong> = &nbsp;=&nbsp; the slope of stress\/strain graph<\/p>\n\n\n\n<p>The value\nof&nbsp; <strong><em>E <\/em><\/strong>&nbsp;for any given material can only be obtained by\ncarrying out tests on specimens of the material.&nbsp; <\/p>\n\n\n\n<p>For\nexample:&nbsp;&nbsp;  For Mild Steel, <strong><em>E<\/em><\/strong> = 200 x 10<sup>9<\/sup> N\/m2&nbsp; &nbsp;=&nbsp; 200\nGN\/m2<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;\n&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; For Aluminium, <strong><em>E<\/em><\/strong>\n=&nbsp; 70 x 10<sup>9<\/sup> N\/m2&nbsp; =&nbsp;&nbsp; 70\nGN\/m2 <\/p>\n\n\n\n<p>Since strain is a ratio and so dimensionless, it follows\nthat <strong><em>E<\/em><\/strong>\nhas the same units as stress<\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p>&nbsp;<a href=\"#_msoanchor_1\">[SJ1]<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-399","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.aircraftengineer.info\/notes\/wp-json\/wp\/v2\/posts\/399","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.aircraftengineer.info\/notes\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.aircraftengineer.info\/notes\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.aircraftengineer.info\/notes\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.aircraftengineer.info\/notes\/wp-json\/wp\/v2\/comments?post=399"}],"version-history":[{"count":7,"href":"https:\/\/www.aircraftengineer.info\/notes\/wp-json\/wp\/v2\/posts\/399\/revisions"}],"predecessor-version":[{"id":434,"href":"https:\/\/www.aircraftengineer.info\/notes\/wp-json\/wp\/v2\/posts\/399\/revisions\/434"}],"wp:attachment":[{"href":"https:\/\/www.aircraftengineer.info\/notes\/wp-json\/wp\/v2\/media?parent=399"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.aircraftengineer.info\/notes\/wp-json\/wp\/v2\/categories?post=399"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.aircraftengineer.info\/notes\/wp-json\/wp\/v2\/tags?post=399"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}