[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"doc-seo-665456-105":3,"detail-sidebar-cat-0-en-105":81,"doc-detail-665456-en":130},{"code":4,"msg":5,"data":6},0,"ok",{"site_id":7,"language":8,"slug":9,"title":10,"keywords":11,"description":12,"schema_data":13,"social_meta":74,"head_meta":76,"extra_data":78,"updated_unix":80},105,"en","hw6-solutions-shear-and-moment-diagrams","HW6-solutions - Shear and Moment Diagrams","","Homework solutions for structural analysis problems focused on constructing shear-force and bending-moment diagrams for beams and shafts. Multiple cases are covered, including reactions at supports, piecewise equilibrium leading to shear and moment functions V(x) and M(x), and plotted diagram values. The content shows governing equations, intermediate results, and final expressions for shear and moment as functions of position, presented in different units for each problem.",{"@graph":14,"@context":73},[15,34,56],{"@type":16,"itemListElement":17},"BreadcrumbList",[18,23,27,31],{"item":19,"name":20,"@type":21,"position":22},"https://docshare.wps.com","Home","ListItem",1,{"item":24,"name":25,"@type":21,"position":26},"https://docshare.wps.com/document/","Document",2,{"item":28,"name":29,"@type":21,"position":30},"https://docshare.wps.com/document/research-report/","Research & Report",3,{"item":32,"name":10,"@type":21,"position":33},"https://docshare.wps.com/document/hw6-solutions-shear-and-moment-diagrams/665456/",4,{"url":32,"name":10,"@type":35,"image":36,"author":41,"headline":10,"publisher":44,"fileFormat":47,"inLanguage":8,"description":12,"dateModified":48,"datePublished":49,"encodingFormat":47,"isAccessibleForFree":50,"interactionStatistic":51},"DigitalDocument",{"url":37,"@type":38,"width":39,"height":40},"https://docshare.wps.com/thumbnails/hw6-solutions-shear-and-moment-diagrams/665456.png","ImageObject",300,407,{"name":42,"@type":43},"Riley West","Person",{"url":19,"name":45,"@type":46},"DocShare","Organization","application/pdf","2026-10-30","2026-10-29",true,{"@type":52,"interactionType":53,"userInteractionCount":55},"InteractionCounter",{"@type":54},"ViewAction",5,{"@type":57,"mainEntity":58},"FAQPage",[59,65,69],{"name":60,"@type":61,"acceptedAnswer":62},"What is the main objective of these solutions?","Question",{"text":63,"@type":64},"To draw shear-force (V) and bending-moment (M) diagrams for beams and shafts, and to determine V(x) and M(x) as functions of position.","Answer",{"name":66,"@type":61,"acceptedAnswer":67},"How are reactions at supports used in the process?",{"text":68,"@type":64},"Support reactions are determined from equilibrium (free-body diagrams), and those forces are then used to compute the shear and moment equations.",{"name":70,"@type":61,"acceptedAnswer":71},"Why do some problems use piecewise expressions for V and M?",{"text":72,"@type":64},"Because loads or boundary conditions change at specific positions x, requiring different equilibrium relations over different x intervals.","https://schema.org",{"og:url":32,"og:type":75,"og:title":10,"og:site_name":45,"og:description":12},"article",{"robots":77,"canonical":32},"index,follow",{"doc_id":79,"site_id":7},665456,1793304273,{"code":4,"msg":82,"data":83},"success",[84,88,92,96,100,105,110,114,119,122,126],{"id":22,"doc_module":4,"doc_module_name":25,"category_name":85,"show_sort_weight":86,"slug":87},"Story & Novel",90,"story-novel",{"id":26,"doc_module":4,"doc_module_name":25,"category_name":89,"show_sort_weight":90,"slug":91},"Literature",80,"literature",{"id":33,"doc_module":4,"doc_module_name":25,"category_name":93,"show_sort_weight":94,"slug":95},"Exam",70,"exam",{"id":55,"doc_module":4,"doc_module_name":25,"category_name":97,"show_sort_weight":98,"slug":99},"Comic",60,"comic",{"id":101,"doc_module":4,"doc_module_name":25,"category_name":102,"show_sort_weight":103,"slug":104},6,"Technology",50,"technology",{"id":106,"doc_module":4,"doc_module_name":25,"category_name":107,"show_sort_weight":108,"slug":109},7,"Healthcare",40,"healthcare",{"id":111,"doc_module":4,"doc_module_name":25,"category_name":29,"show_sort_weight":112,"slug":113},8,30,"research-report",{"id":115,"doc_module":4,"doc_module_name":25,"category_name":116,"show_sort_weight":117,"slug":118},9,"Religion & Spirituality",20,"religion-spirituality",{"id":117,"doc_module":4,"doc_module_name":25,"category_name":120,"show_sort_weight":117,"slug":121},"World Cup","world-cup",{"id":123,"doc_module":4,"doc_module_name":25,"category_name":124,"show_sort_weight":123,"slug":125},10,"Lifestyle","lifestyle",{"id":127,"doc_module":4,"doc_module_name":25,"category_name":128,"show_sort_weight":55,"slug":129},19,"General","general",{"code":4,"msg":82,"data":131},{"doc_id":79,"user_id":132,"nickname":42,"user_avatar":133,"doc_module":4,"category_id":111,"category_name":29,"doc_title":10,"doc_description":12,"doc_content":134,"file_id":135,"file_url":136,"file_type":137,"file_size":138,"view_count":55,"is_deleted":4,"is_public":22,"is_downloadable":22,"audit_status":22,"page_count":139,"language":140,"language_code":8,"site_id":7,"html_lang":8,"table_of_contents":141,"faqs":142,"seo_title":143,"seo_description":12,"update_tm":80,"read_time":144},1099523885074,"https://ap-avatar.wpscdn.com/davatar_9964176cb1d06d4a9deccf72a44ae3dc","© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.  \n\n| 6–2. Draw the shear and moment diagrams for the shaft. The bearings at A and D exert only vertical reaction on the shaft. The loading is applied to the pulleys at B and C and E.\u003Cbr>| \u003Cbr>110 lb |  |\n| --- | --- | --- |\n| Ans:\u003Cbr>\u003Cbr>\u003Cbr>x\u003Cbr>V (lb) 82.2\u003Cbr>􀀃50\u003Cbr>M (lb􀀂in) |  |  |\n| |  | x |\n\n© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.  \n\n| 6–5. Draw the shear and moment diagrams for the beam.\u003Cbr>| \u003Cbr>\u003Cbr>\u003Cbr>2 m\u003Cbr>3 m\u003Cbr>10 kN\u003Cbr>8 kN\u003Cbr>\u003Cbr>15 kN􀀂m\u003Cbr>\u003Cbr>\u003Cbr>|\n| --- | --- |\n| Ans:\u003Cbr>\u003Cbr>M (kN􀀂m)\u003Cbr>\u003Cbr>􀀃75 |  |\n\n© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.  \n\n| 6–18. Draw the shear and moment diagrams for the beam, and determine the shear and moment throughout the beam as functions of x.\u003Cbr>Support Reactions: As shown on FBD.\u003Cbr>Shear and Moment Function:\u003Cbr>For 0 … x 6 6 ft:\u003Cbr>+ c ©Fy = 0; 30.0-2x-V = 0\u003Cbr>V = {30.0-2x} kipa +©MNA = 0; M + 216 + 2xa ~~ ~~x2b-30.0x = 0\u003Cbr>M = { -x2 + 30.0x-216} kip \\# ft For 6 ft 6 x … 10 ft:\u003Cbr>+ c ©Fy = 0; V-8 = 0 V = 8.00 kip\u003Cbr>a +©MNA = 0; -M-8(10-x) -40 = 0 M = {8.00x-120} kip \\# ft | \u003Cbr>\u003Cbr>6 ft 4 ft\u003Cbr>8 kip\u003Cbr>10 kip\u003Cbr>40 kip􀀂ft\u003Cbr>\u003Cbr>\u003Cbr>\u003Cbr>~~ ~~ x |  |  |\n| --- | --- | --- | --- |\n|  | Ans.\u003Cbr>Ans.\u003Cbr>Ans.\u003Cbr>Ans. | |  |\n| Ans:\u003Cbr>For 0 … x 6 6 ft: V = {30.0-2x} kip, M = { -x2 + 30.0x-216} kip \\# ft, For 6 ft 6 x … 10 ft: V = 8.00 kip, M = {8.00x-120} kip \\# ft |  |  |  |\n| \u003Cbr>18.0\u003Cbr>8.00\u003Cbr>6 10\u003Cbr>V (kip) 30.0\u003Cbr>0\u003Cbr>x (ft) |  |  | \u003Cbr>\u003Cbr>0\u003Cbr>􀀃216\u003Cbr>10 x (ft)\u003Cbr>M\u003Cbr>(kip􀀂ft) |\n\n© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.  \n\n| 6–25. Draw the shear and moment diagrams for the beam and determine the shear and moment in the beam as functions of x, where 4 ft \u003C x \u003C 10 ft. | 200 lb􀀂ft\u003Cbr>\u003Cbr>200\u003Cbr> A B\u003Cbr>~~ ~~ x\u003Cbr>~~ ~~ 4 ft  6 ft  4 ft ~~ ~~\u003Cbr>|  | lb􀀂ft\u003Cbr>|\n| --- | --- | --- | --- |\n| + c ©Fy = 0; -150(x-4) -V + 450 = 0\u003Cbr>V = 1050-150x\u003Cbr>(x-4)\u003Cbr>a +©M = 0; -200-150(x-4) 2 -M + 450(x-4) = 0\u003Cbr>M = -75x2 + 1050x-3200\u003Cbr>|  | Ans.\u003Cbr>Ans. |  |\n| Ans:\u003Cbr>V = 1050-150x\u003Cbr>M = -75x2 + 1050x-3200\u003Cbr>\u003Cbr>V (lb)\u003Cbr>x |  |  |  |\n| \u003Cbr>M (lb􀀂ft) 475\u003Cbr>􀀃200\u003Cbr>~~ ~~ x\u003Cbr>􀀃200 |  |  |  |\n\n© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.  \n\n| 6–33. The shaft is supported by a smooth thrust bearing at A and smooth journal bearing at B. Draw the shear and moment diagrams for the shaft. |  |  | \u003Cbr>\u003Cbr>\u003Cbr>\u003Cbr>A\u003Cbr>B\u003Cbr>\u003Cbr>\u003Cbr>\u003Cbr>\u003Cbr>1 m 1 m 1 m\u003Cbr>400 N􀀂m\u003Cbr>\u003Cbr>900 N |\n| --- | --- | --- | --- |\n| Equations of Equilibrium: Referring to the free-body diagram of the shaft shown in Fig. a, |  |  |  |\n| a +©MA = 0;\u003Cbr>+ c ©Fy = 0; | By(2) + 400-900(1) = 0 By = 250 N\u003Cbr>Ay + 250-900 = 0\u003Cbr>Ay = 650 N |  |  |\n| Shear and Moment Diagram: As shown in Figs. b and c. |  |  |  |\n| |  | |  |\n| |  |  | Ans:\u003Cbr>V (N)\u003Cbr>650\u003Cbr>0\u003Cbr>1 2 3\u003Cbr>x (m)\u003Cbr>􀀃250\u003Cbr>\u003Cbr>M (N􀀂m)\u003Cbr>650 400\u003Cbr>0\u003Cb","cbCairyTZzCoOSgT","https://ap.wps.com/l/cbCairyTZzCoOSgT","pdf",2036426,14,"English","# Shear and moment diagrams\n## Shafts supported by bearings\n## Beams with support reactions\n## Piecewise shear and moment functions","[{\"question\":\"What is the main objective of these solutions?\",\"answer\":\"To draw shear-force (V) and bending-moment (M) diagrams for beams and shafts, and to determine V(x) and M(x) as functions of position.\"},{\"question\":\"How are reactions at supports used in the process?\",\"answer\":\"Support reactions are determined from equilibrium (free-body diagrams), and those forces are then used to compute the shear and moment equations.\"},{\"question\":\"Why do some problems use piecewise expressions for V and M?\",\"answer\":\"Because loads or boundary conditions change at specific positions x, requiring different equilibrium relations over different x intervals.\"}]","HW6-solutions - Shear and Moment Diagrams | PDF",35]