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The material lists reactions and relationships between shear, moment, deflection, and slope at pinned and roller supports, as well as at free ends. A worked example demonstrates finding support reactions, drawing the shear force diagram step by step, then constructing the bending moment diagram using the area under the shear curve, verifying end conditions.",{"@graph":69,"@context":122},[70,84,105],{"@type":71,"itemListElement":72},"BreadcrumbList",[73,77,79,82],{"item":74,"name":75,"@type":76,"position":8},"https://docshare.wps.com","Home","ListItem",{"item":78,"name":9,"@type":76,"position":14},"https://docshare.wps.com/document/",{"item":80,"name":40,"@type":76,"position":81},"https://docshare.wps.com/document/research-report/",3,{"item":83,"name":65,"@type":76,"position":19},"https://docshare.wps.com/document/some-guidelines-for-constructing-shear-force-and-bending-moment-diagrams-sign-conventions-and-boundary-conditions/665449/",{"url":83,"name":65,"@type":85,"image":86,"author":91,"headline":65,"publisher":94,"fileFormat":97,"inLanguage":63,"description":67,"dateModified":98,"datePublished":99,"encodingFormat":97,"isAccessibleForFree":100,"interactionStatistic":101},"DigitalDocument",{"url":87,"@type":88,"width":89,"height":90},"https://docshare.wps.com/thumbnails/some-guidelines-for-constructing-shear-force-and-bending-moment-diagrams-sign-conventions-and-boundary-conditions/665449.png","ImageObject",300,407,{"name":92,"@type":93},"Aurelia","Person",{"url":74,"name":95,"@type":96},"DocShare","Organization","application/pdf","2026-10-30","2026-10-29",true,{"@type":102,"interactionType":103,"userInteractionCount":81},"InteractionCounter",{"@type":104},"ViewAction",{"@type":106,"mainEntity":107},"FAQPage",[108,114,118],{"name":109,"@type":110,"acceptedAnswer":111},"What shear and moment values apply at a pinned left end?","Question",{"text":112,"@type":113},"At the pinned left end, reaction forces Ax and Ay are treated as unknown, with V equal to Ay and M equal to 0 unless an applied moment exists at that point. Deflection y is set to 0 and the slope is unknown.","Answer",{"name":115,"@type":110,"acceptedAnswer":116},"How are boundary conditions handled for a roller support at the right end?",{"text":117,"@type":113},"For a roller support at the right end, Bx equals 0 while By is unknown. The relationship uses V equal to -By, and M equals 0 unless an applied moment exists. Deflection y is set to 0 and the slope is unknown.",{"name":119,"@type":110,"acceptedAnswer":120},"How do you construct the bending moment diagram from the shear force diagram?",{"text":121,"@type":113},"Compute bending moment changes by adding the area under the shear force diagram over each segment. The worked example uses MB = Ma + area(A to B) and continues similarly through C, D, and E, then checks that M matches the expected end conditions (e.g., M = 0 at E).","https://schema.org",{"og:url":83,"og:type":124,"og:title":65,"og:site_name":95,"og:description":67},"article",{"robots":126,"canonical":83},"index,follow",{"doc_id":128,"site_id":62},665449,1793304253,{"code":4,"msg":5,"data":131},{"doc_id":128,"user_id":132,"nickname":92,"user_avatar":133,"doc_module":4,"category_id":39,"category_name":40,"doc_title":65,"doc_description":67,"doc_content":134,"file_id":135,"file_url":136,"file_type":137,"file_size":138,"view_count":81,"is_deleted":4,"is_public":8,"is_downloadable":8,"audit_status":8,"page_count":81,"language":139,"language_code":63,"site_id":62,"html_lang":63,"table_of_contents":140,"faqs":141,"seo_title":142,"seo_description":67,"update_tm":129,"read_time":39},1099514068365,"https://ap-avatar.wpscdn.com/avatar/10000253d8d9f28188e?_k=1776742907772140068","Some Guidelines for Constructing Shear Force and Bending Moment Diagrams  \nSIGN CONVENTIONS  \n+ M  \n+ M -M  \n-V  \n-M  \nBOUNDARY CONDITIONS  \nA B  \nAx   Bx  \nA B  \nAx  \nBx  \nAy   \nBy  \n\n| Pinned LEFT End:\u003Cbr>Reaction force, Ax is unknown.\u003Cbr>Reaction force, Ay is unknown. V (shear force) = Ay (reaction force) M = 0 (unless there is an applied moment at this point) Deflection, y = 0\u003Cbr>Slope, 􀁔 is unknown | Roller Support at Right End:\u003Cbr>Reaction force, Bx =0 .\u003Cbr>Reaction force, By is unknown. V (shear force) = -By (reaction force) M = 0 (unless there is an applied moment at this point) Deflection, y = 0\u003Cbr>Slope, 􀁔 is unknown |\n| --- | --- |\n|  |  |\n\nMA Ax   \nA B  \nFree End: V (shear force) = 0 (unless there is an applied point force at this end)  \nM = 0 (unless there is an applied moment at this point) Deflection, y is unknown Slope, 􀁔 is unknown  \nExample:  \nDraw the shear force and bending moment diagram for the following beam:  \nStep 1: Find the reaction forces at A and D.  \nDraw F.B.D.:  \n􀂦 Fx = 0, Ax = 0  \n􀂦 MA = 0, (20 kips)(6 ft) + (12 kips)(14 ft) + (1.5 kips/ft)(8 ft)(28ft)–(Dy)(24 ft) = 0  \nDy = 26 kips  \n􀂦 Fy = 0, Ay + Dy – 20 kips – 12 kips –(1.5 kips/ft)(8 ft) = 0, Ay = 18 kips Construction of the shear force diagram  \n20 kips 12 kips 1.5 kips/ft  \nV  \n18 kips  \n\n| |  | 12 kips | | x |\n| --- | --- | --- | --- | --- |\n|  |  |  |  |  |\n| -2 kips |  |  |  |  |\n|  | -14 kips |  |  |  |\n|  |  |  |  |  |\n\na) V = Ay = 18 kips at point A.  \nb) The shear force is constant until there is another load applied at B.  \nc) The shear force decreases by 20 kips to –2 kips at B because of the applied 20 kip force in the negative y direction.  \nd) The shear force is constant until there is another load applied at C.  \ne) The shear force decreases by 12 kips to –14 kips at C because of the applied 12 kip force in the negative y direction.  \nf) The shear force is constant until there is another load applied at D.  \ng) The shear force increases by 26 kips to 12 kips at D because of the 26 kip reaction force in the positive y direction.  \nh) The shear force decreases linearly from D to E because there is a constant applied load in the negative y-direction.  \ni) The change in shear force from D to E is equal to the area under the load curve between D and E, -12 kips,[ADE = (-1.5 kips/ft)(8 ft) = -12 kips]  \nj) The shear force at E = 0 as expected by inspection of the boundary conditions.  \nConstruction of the Bending Moment diagram  \nV  \n18 kips  \n\n|  B\u003Cbr>6 ft\u003Cbr>| | 8 ft | C | 10 ft\u003Cbr>\u003Cbr>12 kips | E 8 ft | x\u003Cbr>|\n| --- | --- | --- | --- | --- | --- | --- |\n|  |  |  |  |  | \u003Cbr>|  |\n|  |  |  |  |  |  |  |\n| -2 kips |  |  |  |  |  |  |\n|  | -14 |  | kips |  |  |  |\n|  |  |  |  |  |  |  |\n\nM  \n\n| \u003Cbr>108 kip-ft |  |  |  | x\u003Cbr>|\n| --- | --- | --- | --- | --- |\n| | \u003Cbr>92 kip | -ft |  |  |\n|  |  |  | |  |\n|  |  | -48 kip-ft |  |  |\n\na) M = 0 at point A because it is a pinned end with no applied bending moment.  \nb) MB = Ma + (the area under the shear force diagram between A and B.)  \nc) MB = 0 + (18 kips)(6 ft) = 108 kip-ft  \nd) MC = MB + (the area under the shear force diagram between B and C.)  \ne) MC = 108 kip-ft- (2 kips)(8 ft) = 92 kip-ft  \nf) MD = MC + (the area under the shear force diagram between C and D.)  \ng) MD = 92 kip-ft- (14 kips)(10 ft) = -48 kip-ft  \nh) ME = MD + (the area under the shear force diagram between D and E.)  \ni) ME = -48 kip-ft + 1/2 (12 kips)(8 ft) = 0 kip-ft (as expected)","cbCaikkDF3mKHr4Y","https://ap.wps.com/l/cbCaikkDF3mKHr4Y","pdf",135540,"English","# Sign Conventions\n# Boundary Conditions\n# Worked Example: Reaction Forces\n# Construction of the Shear Force Diagram\n# Construction of the Bending Moment Diagram","[{\"question\":\"What shear and moment values apply at a pinned left end?\",\"answer\":\"At the pinned left end, reaction forces Ax and Ay are treated as unknown, with V equal to Ay and M equal to 0 unless an applied moment exists at that point. Deflection y is set to 0 and the slope is unknown.\"},{\"question\":\"How are boundary conditions handled for a roller support at the right end?\",\"answer\":\"For a roller support at the right end, Bx equals 0 while By is unknown. The relationship uses V equal to -By, and M equals 0 unless an applied moment exists. Deflection y is set to 0 and the slope is unknown.\"},{\"question\":\"How do you construct the bending moment diagram from the shear force diagram?\",\"answer\":\"Compute bending moment changes by adding the area under the shear force diagram over each segment. The worked example uses MB = Ma + area(A to B) and continues similarly through C, D, and E, then checks that M matches the expected end conditions (e.g., M = 0 at E).\"}]","Some Guidelines for Constructing Shear Force and Bending Moment Diagrams - Sign Conventions and Boundary Conditions | PDF"]