[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"detail-sidebar-cat-0-en-105":3,"doc-seo-656584-105":59,"doc-detail-656584-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","shear-forces-and-bending-moments-in-beams-lecture-notes","Shear Forces and Bending Moments in Beams - Lecture Notes","","Lecture notes on shear forces and bending moments in beams, focusing on reactions, shear force diagrams, and moment diagrams. The material presents equilibrium-based methods for constructing V and M diagrams, including equilibrium checks for upward and downward forces. Multiple worked problems guide calculation of support reactions, zero-shear locations, maximum bending moments, and diagram drawing, including cases with point loads, distributed loading, and triangularly distributed loads.",{"@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/shear-forces-and-bending-moments-in-beams-lecture-notes/656584/",{"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/shear-forces-and-bending-moments-in-beams-lecture-notes/656584.png","ImageObject",300,407,{"name":92,"@type":93},"dawugda","Person",{"url":74,"name":95,"@type":96},"DocShare","Organization","application/pdf","2026-11-03","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 reactions computed for beams in these notes?","Question",{"text":112,"@type":113},"Reactions at supports are obtained by applying static equilibrium, using moment equations about selected points to solve for support forces.","Answer",{"name":115,"@type":110,"acceptedAnswer":116},"What method is used to construct shear force (V) and moment (M) diagrams?",{"text":117,"@type":113},"The notes emphasize the equilibrium method to determine V and M, with examples also referencing a semi-graphical approach.",{"name":119,"@type":110,"acceptedAnswer":120},"How do you find the location where shear force V equals zero?",{"text":121,"@type":113},"Set the shear-force expression or diagram segment equal to zero and solve for the distance from the reference support, as demonstrated in the step-by-step problems.","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},656584,1793061153,{"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},3985747870947,"https://ap-avatar.wpscdn.com/davatar_3d24733baf745e90a7e4bdd5f77d97b2","Dr. Mohammed E. Haque, P.E.  \nLecture Notes  \nShear Forces and Bending Moments in Beams   \nREACTIONS  \nSHEAR FORCE DIAGRAMS   \nMOMENT DIAGRAMS   \n[EXAMPLES]   \n• Equilibrium Method for V and M Diagrams  \n• Semi-graphicalMethodfor V and M Diagrams  \nCOSC321Haque 1  \nPDF_C8_b (Shear Forces and Bending Moments in Beams)  \nDr. Mohammed E. Haque, P.E.  \nLecture Notes  \nEquilibrium Method for V and M Diagrams  \nQ1:  \nB  \nRA = 6 (12 –3)/ 12 = 4.5k  \nRB = 6x3 /12=1 .5k  \nA  \n12 (ft)  \n4.5k  \nV  \n4.5x3=13.5k-ft  \n-1.5k  \n3 ft P=6k  \nM ~~   ~~  \nCOSC321Haque 2  \nPDF_C8_b (Shear Forces and Bending Moments in Beams)  \nDr. Mohammed E. Haque, P.E.  \nLecture Notes  \nQ2:  \nV  \nM  \n5 k  \nA B  \n2 ft 10 ft  \nRA = 5 (10+2) /10 RB = 5(2) /10=1k  \n= 6k  \n1k  \n-5k  \nM = -5x2=-10 k-ft  \nCOSC321Haque 3  \nPDF_C8_b (Shear Forces and Bending Moments in Beams)  \nDr. Mohammed E. Haque, P.E.  \nLecture Notes  \nQ3: Find reactions, Shear Force, Location of zero shear forced, Maximum Moment, Mid-span moment.  \nw = 2 k/ft  \nA  \n10 ft  \n2x10=20 k  \nw = 2 k/ft  \nA  \nC  \nRA = 15 k 10 ft  10 ft  \nTake a moment about C and Find Reaction at ARA = (20x15)/20 = 15 kips  \nRC = 5 k  \nB  \nB  \nTakea moment about A and Find Reaction at CRc =(20x5)/20 = 5 kips  \n[CHECK: Sum of all the forces Upward = sum of the all the forces downward 15 + 5 = 20 OK]  \nCOSC321Haque 4  \nPDF_C8_b (Shear Forces and Bending Moments in Beams)  \nDr. Mohammed E. Haque, P.E.  \nLecture Notes  \n2x10=20 k  \nw = 2 k/ft  \nA  \nB   \nC  \nRA = 15 k  \n15  \nRC = 5 k  \nM at B = 5 x 10 = 50 k-ft  \n10 ft 10 ft  \n-5  \nV (kips)  \nM (k-ft)  \n\n|  |  |  |  |\n| --- | --- | --- | --- |\n| V=0\u003Cbr>|  |  |  |\n|  |  |  |  |\n|  |  |  |  |\n| X = 15/2 = 7.5 ft |  |  |  |\n|  | \u003Cbr>Mmax =\u003Cbr>| 0.5(15 x 7 .5) = 56.25 k-ft |  |\n|  |  | |  |\n|  |  |  |  |\n\nCOSC321Haque 5  \nPDF_C8_b (Shear Forces and Bending Moments in Beams)  \nDr. Mohammed E. Haque, P.E.  \nQ4: Find reactions, Support Moment; Draw Sear Force and Moment diagrams.  \nLecture Notes  \n4 k  \nMb = (4x2) +(2X5) + (0 .25x10x5) = 30.5 k-ft (Anti-clockwise)  \nRbh = 4 k (towards left) Rbv= 2+(0 .25x10) =4 .5 k (up)  \n\n| 4.5 –(0.25x5)= 3.25 k |  |  |\n| --- | --- | --- |\n|  |  |  |\n|  | 3.25 – 2\u003Cbr>= 1.25 |  |\n\n4 k  \nA  \nRBH  \nMB  \nA  \nRBV  \n2 k  \nCOSC321Haque  \n 2 ft  \n6  \n0.25 k/ft  \nB C  \n4.5 k  \n1.25 –(0.25x5)=0  \nV (kips)  \n4x2=8 k-ft  \n30.5 k-ft  \nM (k-ft)  \n8 + (0 .5x1 .25x5)=11 .125 k-ft  \nPDF_C8_b (Shear Forces and Bending Moments in Beams)  \nDr. Mohammed E. Haque, P.E.  \nLecture Notes  \nQ5: Calculate Reactions at A and B. Draw Shear Force Diagram. Find location of Zero Shear  \nB  \nCOSC321Haque  \n7  \nforce. Find the  \nMaximum Moment. Draw the Moment Diagram.  \nM (k-ft)  \nA  \n12 (ft)  \nRA = [12x8 + (4x12x6)]/12 = 32k  \nP=12k  \n4 ft  \n| \u003Cbr>| | | | | | | | | | |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- |\n\n4 k/ft  \nRB = [12x4 + (4x12x6)]/12 = 28k  \n\n|  |  |  |\n| --- | --- | --- |\n| | 16 | (12-5)=7 ft\u003Cbr>|\n|  | 4 | V=0 |\n| X=4 | /4 = 1ft  |  |\n|  | | -28 |\n| | | |\n|  |  |  |\n\n32  \n(k)  \nV  \nMmax=28(12-5)– 4(12-5)(12-5)/2 = 98 k-ft  \nPDF_C8_b (Shear Forces and Bending Moments in Beams)  \nDr. Mohammed E. Haque, P.E.  \nLecture Notes  \nQ6: A simply supported beam with a triangularly distributed downward load is shown in Fig. Calculate reaction; draw shear force diagram; find location of V=0; calculate maximum moment,  \nand draw the moment diagram.  \n6k/ft  \nRB = (27k)(6)/9= 18k  \nA  \nRA = (27k)(9-6)/9= 9k  \n9k  \nV  \n-18k  \nM  \n\n|  |  |\n| --- | --- |\n| | V=0 |\n| \u003Cbr>X= 5.2’\u003Cbr>Mmax = 31.2 k-ft\u003Cbr>| |\n| | |\n|  |  |\n\nFind X = ?  \n(Location at V=0)  \n(6/9)(X)(X/2) = 9 X = √(9x18/6) =5 .2 ft  \nMmax = 9(5.2)– [0.5(6/9)(5.2)(5.2)](5.2/3) = 31.2 k-ft  \n9 ft  \nCOSC321Haque 8  \nPDF_C8_b (Shear Forces and Bending Moments in Beams)  \nDr. Mohammed E. Haque, P.E.  \nLecture Notes  \nQ7:  \n4 k/ft  \n4 k/ft  \nA B  \n16 ft  \n4 ft  \n| \u003Cbr>| | \u003Cbr>| | \u003Cbr>| \u003Cbr>| | |\n| --- | --- | --- | --- | --- | --- | --- | --- |\n\nA  \nC (hinge)  \nB  \nD  \n16 ft  \n4 ft 8 ft  \nCalculate Reaction","cbCaio5aIXYzIfTQ","https://ap.wps.com/l/cbCaio5aIXYzIfTQ","pdf",215573,13,"English","# Reactions\n# Shear Force Diagrams\n# Moment Diagrams\n## Equilibrium Method for V and M Diagrams\n## Semi-graphical Method for V and M Diagrams","[{\"question\":\"How are reactions computed for beams in these notes?\",\"answer\":\"Reactions at supports are obtained by applying static equilibrium, using moment equations about selected points to solve for support forces.\"},{\"question\":\"What method is used to construct shear force (V) and moment (M) diagrams?\",\"answer\":\"The notes emphasize the equilibrium method to determine V and M, with examples also referencing a semi-graphical approach.\"},{\"question\":\"How do you find the location where shear force V equals zero?\",\"answer\":\"Set the shear-force expression or diagram segment equal to zero and solve for the distance from the reference support, as demonstrated in the step-by-step problems.\"}]","Shear Forces and Bending Moments in Beams - Lecture Notes | PDF",1792969492,33]