[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"detail-sidebar-cat-0-en-105":3,"doc-seo-656573-105":59,"doc-detail-656573-en":130},{"code":4,"msg":5,"data":6},0,"success",[7,13,18,23,28,33,38,43,48,51,55],{"id":8,"doc_module":4,"doc_module_name":9,"category_name":10,"show_sort_weight":11,"slug":12},1,"Document","Story & Novel",90,"story-novel",{"id":14,"doc_module":4,"doc_module_name":9,"category_name":15,"show_sort_weight":16,"slug":17},2,"Literature",80,"literature",{"id":19,"doc_module":4,"doc_module_name":9,"category_name":20,"show_sort_weight":21,"slug":22},4,"Exam",70,"exam",{"id":24,"doc_module":4,"doc_module_name":9,"category_name":25,"show_sort_weight":26,"slug":27},5,"Comic",60,"comic",{"id":29,"doc_module":4,"doc_module_name":9,"category_name":30,"show_sort_weight":31,"slug":32},6,"Technology",50,"technology",{"id":34,"doc_module":4,"doc_module_name":9,"category_name":35,"show_sort_weight":36,"slug":37},7,"Healthcare",40,"healthcare",{"id":39,"doc_module":4,"doc_module_name":9,"category_name":40,"show_sort_weight":41,"slug":42},8,"Research & Report",30,"research-report",{"id":44,"doc_module":4,"doc_module_name":9,"category_name":45,"show_sort_weight":46,"slug":47},9,"Religion & Spirituality",20,"religion-spirituality",{"id":46,"doc_module":4,"doc_module_name":9,"category_name":49,"show_sort_weight":46,"slug":50},"World Cup","world-cup",{"id":52,"doc_module":4,"doc_module_name":9,"category_name":53,"show_sort_weight":52,"slug":54},10,"Lifestyle","lifestyle",{"id":56,"doc_module":4,"doc_module_name":9,"category_name":57,"show_sort_weight":24,"slug":58},19,"General","general",{"code":4,"msg":60,"data":61},"ok",{"site_id":62,"language":63,"slug":64,"title":65,"keywords":66,"description":67,"schema_data":68,"social_meta":123,"head_meta":125,"extra_data":127,"updated_unix":129},105,"en","beams-sfd-and-bmd-lecture-11","Beams - SFD and BMD - Lecture 11","","Lecture notes on beam analysis using shear force and bending moment diagrams. Covers shear–moment relationships, including how the slope of the shear diagram relates to applied loading and how integrating shear or loading functions yields expressions for shear force V and bending moment M. Discusses conditions for validity when the load intensity is continuous, treatment of discontinuities at points of applied loading or concentrated loads, and how maximum or minimum bending moment occurs when shear changes sign. Includes step-by-step guidance for drawing SFD and BMD and solving example problems using equilibrium.",{"@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":30,"@type":76,"position":81},"https://docshare.wps.com/document/technology/",3,{"item":83,"name":65,"@type":76,"position":19},"https://docshare.wps.com/document/beams-sfd-and-bmd-lecture-11/656573/",{"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/beams-sfd-and-bmd-lecture-11/656573.png","ImageObject",300,407,{"name":92,"@type":93},"Marglet","Person",{"url":74,"name":95,"@type":96},"DocShare","Organization","application/pdf","2026-10-31","2026-10-25",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},"How are the slope of the shear diagram and applied loading related?","Question",{"text":112,"@type":113},"The slope of the shear diagram equals the negative value of the applied loading. This establishes the link between w(x) and the shear-force function V(x).","Answer",{"name":115,"@type":110,"acceptedAnswer":116},"How can V and M be expressed using integration?",{"text":117,"@type":113},"V can be obtained by integrating the loading function (or from the area under the loading curve, with sign). M can be obtained by integrating shear (or from the area under the shear diagram), using integration constants defined at x0.",{"name":119,"@type":110,"acceptedAnswer":120},"When do bending moment maxima or minima occur?",{"text":121,"@type":113},"Bending moment reaches a maximum or minimum where V passes through zero and V is continuous with nonzero dV/dx (i.e., nonzero loading). Critical values can also occur at discontinuous crossings of the shear axis for cases like concentrated loads.","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},656573,1793030660,{"code":4,"msg":5,"data":131},{"doc_id":128,"user_id":132,"nickname":92,"user_avatar":133,"doc_module":4,"category_id":29,"category_name":30,"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":139,"language":140,"language_code":63,"site_id":62,"html_lang":63,"table_of_contents":141,"faqs":142,"seo_title":143,"seo_description":67,"update_tm":144,"read_time":145},2336477982702,"https://ap-avatar.wpscdn.com/davatar_9964176cb1d06d4a9deccf72a44ae3dc","Beams – SFD and BMD   \nShear and Moment Relationships  \nw 􀀠 􀀐 ~~ ~~ddVx  \nSlope of the shear diagram = -Value of applied loading  \ndM V 􀀠 ~~  ~~dx  \nSlope of the moment curve = Shear Force  \nBoth equations not applicable at the point of loading because of discontinuity produced by the abrupt change in shear.  \nME101-Division III Kaustubh Dasgupta 1  \nBeams – SFD and BMD   \nw 􀀠 􀀐 ~~ ~~ddVx dM  \nV 􀀠 ~~  ~~dx  \nDegree of V in x is one higher than that of w  \nDegree of M in x is one higher than that of V  \n􀃆 Degree of M in x is two higher than that of w  \nCombining the two equations 􀃆  \nd 2 Mdx 2 􀀠 􀀐w  \n􀃆 M :: obtained by integrating this equation twice  \n􀃆 Method is usable only if w is a continuous function of x (other cases not part of this course)  \nME101-Division III Kaustubh Dasgupta 2  \nBeams – SFD and BMD   \nShear and Moment Relationships  \nExpressing V in terms of w by integrating w 􀀠 􀀐 ~~ ~~ddVx  \n􀂳VV 0  \ndV 􀀠 􀀐 􀂳xx 0wdx  \nOR V = V0 + (the negative of the area under  \nthe loading curve from x0 to x)  \nV0 is the shear force at x0 and V is the shear force at x Expressing M in terms of V by integrating V 􀀠 dMdx  \n􀂳MM 0dM 􀀠 􀂳xx 0Vdx OR M = M0 + f( area undrom x 0toerx)the shear diagram  \nM0 is the BM at x0 and M is the BM at x  \nME101-Division III Kaustubh Dasgupta 3  \nV = V0 + (negative of area under the loading curve from x0 to x)  \nM = M0 + (area under the shear diagram from x0 to x)  \nIf there is no externally applied moment M0 at x0 = 0,  \ntotal moment at any section equals the area under the shear diagram up to that section  \nWhen V passes through zero and is a continuous  \nfunction of x with dV/dx ≠ 0 (i.e. , nonzero loading) dM 􀀠 0  \ndx  \n􀃆 BM will be a maximum or minimum at this point  \nCritical values of BM also occur when SF crosses the zero axis discontinuously (e.g. , Beams under concentrated loads)  \nME101-Division III Kaustubh Dasgupta 4  \nMaximum BM occurs where Shear changes the direction  \n• Draw the SFD and BMD.  \n• Determine reactions at supports.  \n• Cut beam at C and consider member AC, V 􀀠 􀀎 P2 M 􀀠 􀀎 Px2  \n• Cut beam at E and consider member EB, V 􀀠 􀀐 P2 M 􀀠 􀀎 P􀀋L􀀐 x􀀌2  \n• For a beam subjected to concentrated loads, shear is constant between loading points and moment varies linearly  \nME101-Division III Kaustubh Dasgupta 5  \nDraw the shear and bending moment diagrams for the beam and loading shown.  \nSolution: Draw FBD and find out the support reactions using equilibrium equations  \nME101-Division III Kaustubh Dasgupta 6  \nSFD and BMD: Example (2)   \nUse equilibrium conditions at all sections to get the unknown SF and BM  \n􀂦 Fy 􀀠 0 : 􀀐 20 kN 􀀐V1 􀀠 0 V1 􀀠 􀀐20 kN  \n􀂦 M1 􀀠 0 : 􀀋20 kN􀀌􀀋0 m􀀌􀀎 M1 􀀠 0 M1 􀀠 0 V2 = -20 kN; M2 = -20x kNm  \nV3 = +26 kN; M3 = -20x+46×0 = -20x 􀃆 MB= -50 kNm V4 = +26 kN; M4 = -20x+46×(x-2.5) 􀃆 MC= +28 kNm V5 = -14 kN; M5 = -20x+46×(x-2.5)-40×0 􀃆 MC= +28 kNm  \nV6 = -14 kN; M6 = -20x+46×(x-2.5)-40×(x-5.5)  \n􀃆 MD= 0 kNm  \nME101-Division III Kaustubh Dasgupta 7  \nDraw the SFD and BMD for the beam acted upon by a clockwise couple at mid point  \nSolution: Draw FBD of the beam and Calculate the support reactions  \nDraw the SFD and the BMD starting From any one end  \nC   \nl  \nV  \nC  \nl  \nC  \nC 􀀐  \nl  \n2  \nME101-Division III Kaustubh Dasgupta 8  \n Beams – SFD and BMD: Example (3)   \nDraw the SFD and BMD for the beam  \nSolution: Draw FBD of the beam and Calculate the support reactions  \n∑MA = 0 􀃆 RA = 60 N  \n∑MB = 0 􀃆 RB = 60 N  \nDraw the SFD and the BMD starting from any one end  \n|  | 120 Nm |  |\n| --- | --- | --- |\n|  | 60 N |  |\n| |  |  |\n|  |  |  |\n| |  |  |\n|  | -120 Nm |  |\n\n60 N  \nV  \n-60 N  \nM  \nME101-Division III Kaustubh Dasgupta 9  \nDraw the SFD and BMD for the beam  \nSolution:  \nDraw FBD of the entire beam and calculate support reactions using equilibrium equations  \nReactions at supports: RA 􀀠 RB 􀀠 wL2  \nDevelop the relations between loading, shear force, and bending moment and plot the SFD and BMD  \nME101-Division III Kaustubh Dasgupta 10","cbCaicaUsNRfvjtS","https://ap.wps.com/l/cbCaicaUsNRfvjtS","pdf",1667686,21,"English","# Shear and Moment Relationships\n## Relations among w, V, and M\n## Validity conditions and discontinuities\n## Drawing SFD and BMD for beams\n## Worked examples","[{\"question\":\"How are the slope of the shear diagram and applied loading related?\",\"answer\":\"The slope of the shear diagram equals the negative value of the applied loading. This establishes the link between w(x) and the shear-force function V(x).\"},{\"question\":\"How can V and M be expressed using integration?\",\"answer\":\"V can be obtained by integrating the loading function (or from the area under the loading curve, with sign). M can be obtained by integrating shear (or from the area under the shear diagram), using integration constants defined at x0.\"},{\"question\":\"When do bending moment maxima or minima occur?\",\"answer\":\"Bending moment reaches a maximum or minimum where V passes through zero and V is continuous with nonzero dV/dx (i.e., nonzero loading). Critical values can also occur at discontinuous crossings of the shear axis for cases like concentrated loads.\"}]","Beams - SFD and BMD - Lecture 11 | PDF",1792969390,53]