{"id":575,"date":"2021-07-21T14:23:51","date_gmt":"2021-07-21T18:23:51","guid":{"rendered":"http:\/\/pressbooks.library.upei.ca\/statics\/?post_type=chapter&#038;p=575"},"modified":"2025-08-01T17:29:45","modified_gmt":"2025-08-01T21:29:45","slug":"6-3-examples","status":"publish","type":"chapter","link":"https:\/\/pressbooks.library.upei.ca\/statics\/chapter\/6-3-examples\/","title":{"raw":"6.3 Examples","rendered":"6.3 Examples"},"content":{"raw":"Here are examples from Chapter 6 to help you understand these concepts better. These were taken from the real world and supplied by FSDE students in Summer 2021. If you'd like to submit your own examples, please send them to the author <a href=\"mailto:eosgood@upei.ca\">eosgood@upei.ca<\/a>.\r\n<h1>Example 6.3.1: Internal Forces, Submitted by Emma Christensen<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\nThe setup that holds the solar panels at the UPEI FSDE is modelled below. Considering beam S (1.9 m length), find the internal forces at point C. Assume the intensity of the solar panel on the beam is 220 N\/m.\r\n\r\nSketch:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-draw-1-1-300x265.jpg\" alt=\"A sketch of the problem.\" class=\"aligncenter wp-image-1124\" width=\"350\" height=\"309\" \/>\r\n\r\nModel:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-draw-2-300x198.jpg\" alt=\"A more detailed sketch of the problem.\" class=\"aligncenter wp-image-1112 size-medium\" width=\"300\" height=\"198\" \/>\r\n\r\n<\/div>\r\n<strong>2. Draw<\/strong>\r\n\r\nFree-body diagram:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-1-300x197.jpg\" alt=\"A FBD of the problem\" class=\"alignnone wp-image-1114\" width=\"394\" height=\"259\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>w = 220 N\/m<\/li>\r\n \t<li>OA = 0.5 m<\/li>\r\n \t<li>AC = 0.2 m<\/li>\r\n \t<li>AB = 0.4 m<\/li>\r\n \t<li>L = 1.9 m<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>N<sub>c<\/sub><\/li>\r\n \t<li>V<sub>c<\/sub><\/li>\r\n \t<li>M<sub>c<\/sub><\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nUse equilibrium equations. First, solve for reaction forces, then make a cut at C and solve for the internal forces.\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\n[latex]w=\\frac{F}{L}\\\\F=wL\\\\F_R=220N\/m\\cdot 1.9m\\\\F_R=418N\\\\\\sum F_X=0=B_X[\/latex]\r\n\r\nFind reaction forces:\r\n\r\n[latex]\\sum M_A=0=B_y(0.4m)-F_R(0.45m)\\\\(0.4m)B_y=418N(0.45m)\\\\B_y=\\frac{188.1 N\\cdot m}{0.4m}\\\\B_y=470.25N[\/latex]\r\n\r\n[latex]\\sum F_y=0=-F_R+A_y+B_y\\\\A_y=F_R-B_y\\\\A_y=418N-470.25N\\\\A_y=-52.25N[\/latex]\r\n\r\nThe answer we got for A<sub>y<\/sub> is negative, which means that the arrow should be drawn in the other direction. We will change it for our next sketch.\r\n\r\nMake a cut at C:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-6-1-300x264.jpg\" alt=\"A sketch showing the internal forces at cut C.\" class=\"alignnone wp-image-1121 size-medium\" width=\"300\" height=\"264\" \/>\r\n\r\nNow solve for the internal forces:\r\n\r\n[latex]\\sum F_x=0\\:\\:;\\:\\:N_c=0\\\\\\sum F_y=0=-A_y-V_c-(w\\cdot L_A)\\\\V_c=-52.25N-(220N\/m\\cdot 0.7m)\\\\V_c=-206.25 N\\\\\\sum M_c=A_y(0.2m)+M_c+(F_{Rc}\\cdot 0.35m)\\\\M_c=52.25 N (0.2m)+(220N\/m\\cdot 0.7m\\cdot 0.35m)\\\\M_c=64.35N\\cdot m[\/latex]\r\n\r\nFinal FBD, showing the arrows in the correct directions:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-7-1-300x216.jpg\" alt=\"Final FBD showing arrows in the correct directions.\" class=\"alignnone wp-image-1122\" width=\"388\" height=\"279\" \/>\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nIt makes sense that A<sub>y<\/sub> and B<sub>y<\/sub> are in different directions because the resultant force Fr of the solar panel on the beam is not between A and B. It also makes sense that the moment at C is in the counterclockwise direction rather than the clockwise direction when you think about the direction of the forces applied to the beam.\r\n\r\n<\/div>\r\n<h1>Example 6.3.2: Shear\/Moment Diagrams, Submitted by Deanna Malone<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\nA beam that is simply supported has two point loads acting on it. One acts 2 m from point A, and the other acts at 2.5 m from C. Point B is in the middle of the beam. The first point load is 500 N and the second is 300 N. What are the internal forces at point B? Solve for reaction forces and include a shear\/moment diagram.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-1-300x257.jpg\" alt=\"A sketch of the problem.\" class=\"aligncenter wp-image-1003\" width=\"307\" height=\"263\" \/>\r\n\r\n<\/div>\r\n<strong>2. Draw<\/strong>\r\n\r\nSketch:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-1-300x257.jpg\" alt=\"A sketch of the problem.\" class=\"alignnone wp-image-1003\" width=\"307\" height=\"263\" \/>\r\n\r\nFree-body diagram:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-2-300x165.jpg\" alt=\"A FBD of the problem.\" class=\"alignnone wp-image-1004\" width=\"407\" height=\"224\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>F1 = 500 N<\/li>\r\n \t<li>F2 = 300 N<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>A<sub>y<\/sub><\/li>\r\n \t<li>A<sub>x<\/sub><\/li>\r\n \t<li>C<sub>y<\/sub><\/li>\r\n \t<li>V<sub>B<\/sub><\/li>\r\n \t<li>M<sub>B<\/sub><\/li>\r\n \t<li>N<sub>B<\/sub><\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nShear\/moment equations, EOM equations\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\nSolve for reaction forces:\r\n\r\n(Ax, Cy)\r\n\r\n[latex]\\begin{aligned}\r\n\\sum F_{x}=0=A_{x}=0 \\\\\r\n\\sum M_{A}=0 &amp;=-F_{1} \\cdot 2 m-F_{2} \\cdot 5.5 m+C_{y} \\cdot 8 m \\\\\r\nC_{y}=&amp;\\frac{F_{1} \\cdot 2 m+F_{2} \\cdot 5.5 m}{8m} \\\\\r\nC_{y} &amp;=\\frac{500 N \\cdot 2 m+300 N \\cdot 5.5 m}{8 m} \\\\\r\nC_{y} &amp;=331.25 \\mathrm{~N}\r\n\\end{aligned}[\/latex]\r\n\r\n(Ay)\r\n\r\n[latex]\\begin{aligned}\r\n\\sum F_{y}=0 &amp;=A_{y}+C_{y}-F_{1}-F_{2} \\\\\r\nA_{y} &amp;=F_{1}+F_{2}-C_{y} \\\\\r\nA_{y} &amp;=500 \\mathrm{~N}+300 \\mathrm{~N} - 331.25 \\mathrm{~N} \\\\\r\nA_{y} &amp;=468.75 \\mathrm{~N}\r\n\\end{aligned}[\/latex]\r\n\r\nCut 1: at B\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-1-300x133.jpg\" alt=\"A sketch showing the cut at B\" class=\"alignnone wp-image-1005\" width=\"399\" height=\"177\" \/>\r\n\r\n[latex]\\begin{aligned}\r\n\\sum F_{X}=0=A_{X} &amp;+N_{B}=0 \\\\\r\n&amp; N_{B}=0 \\\\\r\n\\sum F_{y}=0 &amp;=A_{y}-V_{B}-F_{1} \\\\\r\nV_{B} &amp;=A_{y}-F_{1} \\\\\r\nV_{B} &amp;=468.75 N - 500 N \\\\\r\nV_{B}=-31.25 N\r\n\\end{aligned}[\/latex]\r\n\r\n[latex]\\begin{aligned}\r\n\\sum M_{B}=&amp; 0=-A_{y}(4 m)+F_{1}(2 m)+M_{B} \\\\\r\n&amp; M_{B}=A_{y}(4 m)-F_{1}(2 m) \\\\\r\n&amp; M_{B}=468.75 N(4 m)-500 N(2 m) \\\\\r\nM_{B} &amp;=875 \\mathrm{~N} \\cdot \\mathrm{m}\r\n\\end{aligned}[\/latex]\r\n\r\nCut 2: At the point where F<sub>1<\/sub> is applied\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-2-300x139.jpg\" alt=\"A sketch showing the cut at F1\" class=\"alignnone wp-image-1006\" width=\"376\" height=\"174\" \/>\r\n\r\n[latex]\\begin{aligned}\r\n\\sum M_{1}=0 &amp;=-A_{y}(2 m)+M_{1}=0 \\\\\r\nM_{1} &amp;=A_{y}(2 m) \\\\\r\nM_{1} &amp;=468.75 N(2 m) \\\\\r\nM_{1} &amp;=937.5 \\mathrm{~N} \\cdot m\r\n\\end{aligned}[\/latex]\r\n\r\nCut 3: At the point where F<sub>2<\/sub> is applied\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-3-300x109.jpg\" alt=\"A sketch showing the cut at F2.\" class=\"alignnone wp-image-1007\" width=\"377\" height=\"137\" \/>\r\n\r\n[latex]\\begin{aligned}\r\n\\sum M_{2}=0 =-A_{y}(5.5 \\mathrm{~m})+F_{1}(3.5 \\mathrm{~m})+M_{2} \\\\\r\nM_{2} =A_{y}(5.5 \\mathrm{~m})-F_{1}(3.5 \\mathrm{~m}) \\\\\r\nM_{2} =468.75 \\mathrm{~N}(5.5 \\mathrm{~m})-500 \\mathrm{~N}(3.5 \\mathrm{~m}) \\\\\r\nM_{2} =828.125 \\mathrm{~N} \\cdot \\mathrm{m}\r\n\\end{aligned}[\/latex]\r\n\r\nAnswer: N<sub>B<\/sub> = 0, V<sub>B<\/sub> = -31.25 N, M<sub>B<\/sub> = 875 Nm\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-solve-final-252x300.jpg\" alt=\"A final diagram of the beam, shear and moment diagrams.\" class=\"alignnone wp-image-1008\" width=\"404\" height=\"481\" \/>\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nThe reaction forces make sense as they offset the applied forces. The shear\/moment diagrams returned to zero, so they are correct too. The moment found at B is in the moment diagram; it is smaller than the maximum.\r\n\r\n<\/div>\r\n<h1>Example 6.3.3: V\/M Diagrams, Submitted by Luciana Davila<\/h1>\r\n<div class=\"textbox\">\r\n<ol>\r\n \t<li><strong>Problem\u00a0<\/strong><\/li>\r\n<\/ol>\r\n<div class=\"textbox shaded\">\r\n\r\nFor the 8cm simply supported beam given below, if Fg, which is applied at the center of the beam, is given as 30 N, find the internal forces between the reaction forces and draw the V-M diagram for the beam.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/simply-supported.jpg\" alt=\"An image of a simply supported bridge.\" width=\"551\" height=\"640\" class=\"alignnone wp-image-2205 size-full\" \/>\r\n\r\nSimply supported girder bridge: https:\/\/www.geograph.org.uk\/photo\/2780207\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/chapter-6-unknown-V-m-diagram-question.jpg\" width=\"491\" height=\"192\" class=\"alignnone\" alt=\"A digital sketch of the problem.\" \/>\r\n\r\n<\/div>\r\n<strong>2. Draw\u00a0<\/strong>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/chapter-6-unknown-V-m-diagram-question-2.jpg\" alt=\"A FBD of the problem.\" class=\"alignnone wp-image-1823 size-full\" width=\"608\" height=\"264\" \/>\r\n\r\n&nbsp;\r\n\r\n<strong>3. Knowns and Unknowns\u00a0<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>F<sub>g<\/sub> = 30N<\/li>\r\n \t<li>r <sub style=\"text-align: initial;background-color: initial\">AB<\/sub><span style=\"text-align: initial;background-color: initial;font-size: 1em\"> =\u00a0 r <\/span><sub style=\"text-align: initial;background-color: initial\">BC <\/sub><span style=\"text-align: initial;background-color: initial;font-size: 1em\">=\u00a0 0.04 m<\/span><\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>Reaction forces = R<sub>Ax<\/sub>, R<sub>Ay<\/sub>, R<sub>Cy<\/sub><\/li>\r\n \t<li>For the segment of beam AB nd BC,<\/li>\r\n \t<li>Shear = V, V<sub style=\"text-align: initial;background-color: initial\">2<\/sub><\/li>\r\n \t<li>Bending Moment = M, M<sub style=\"text-align: initial;background-color: initial\">2<\/sub><\/li>\r\n \t<li>Normal force = N, N<sub style=\"text-align: initial;background-color: initial\">2<\/sub><\/li>\r\n<\/ul>\r\n<strong>4. Approach\u00a0<\/strong>\r\n\r\nFind reaction forces from equilibrium equations. Use them to find the shear and bending moment of segment AB and BC. Draw the V-M diagram from the answers.\r\n\r\n<strong>5. Analysis\u00a0<\/strong>\r\n\r\nFinding reaction forces:\r\n\r\n[latex]\\sum F_x=0=R_{Ax}\\\\R_{Ax}=0N[\/latex]\r\n\r\nSolving for RC<sub>y<\/sub>:\r\n\r\n[latex]\\sum M_A=0= -(r_{AB}\\cdot F_g)+(r_{AC}\\cdot R_{Cy})\\\\r_{AB}\\cdot F_{g}=r_{AC}\\cdot R_{Cy}[\/latex]\r\n\r\n[latex]R_{Cy}=\\frac{r_{AB}\\cdot F_g}{r_{AC}}\\\\R_{By}=\\frac{0.04m\\cdot 30 N}{0.08 m}[\/latex]\r\n\r\n<span style=\"text-align: initial;background-color: initial;font-size: 1em\">[latex]R_{Cy}=15 N[\/latex]<\/span>\r\n\r\nSolving for RA<sub>y<\/sub>:\r\n\r\n[latex]\\sum F_y=0=R_{Cy}+R_{Ay} -F_g\\\\R_{Ay}=F_g-R_{Cy}\\\\R_{Ay}= 15 N[\/latex]\r\n\r\nFinding shear and Moment from A to B :\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luciana-beam-AB-3.png\" alt=\"A sketch to find the internal forces between a and b.\" class=\"alignnone wp-image-1713\" width=\"340\" height=\"202\" \/>\r\n\r\n[latex]\\sum F_x=0\\\\N=0N[\/latex]\r\n\r\nSolving for V\r\n\r\n[latex]\\sum F_y=0=R_{Ay}+-V\\\\ V= 15 N[\/latex]\r\n\r\nSolving for M :\r\n\r\n[latex]\\sum M_A=0= -( V\\cdot 0.04) +M\\\\M = 15\\cdot 0.04\\\\M=0.6N[\/latex]\r\n\r\nFinding shear and Moment from B to C:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luciana-BC-4-300x166.png\" alt=\"A sketch to find the internal forces between b and c.\" class=\"alignnone wp-image-1714\" width=\"363\" height=\"201\" \/>\r\n\r\n[latex]\\sum F_x=0\\\\N=0N[\/latex]\r\n\r\nSolving for V\r\n\r\n[latex]\\sum F_y=0=R_{Cy}+V_2\\\\ V_2= -15 N[\/latex]\r\n\r\nSolving for M :\r\n\r\n[latex]\\sum M_C=0= -( V_2\\cdot(0.04) -M_2\\\\M_2 = 15\\cdot 0.04\\\\M=0.6N[\/latex]\r\n\r\nV-M diagram:\r\n\r\nFor the V-x graph, from A to B, there is a constant shear of 15 N, then from B to C, the shear has the same magnitude but a different direction.\u00a0 To plot the M-x graph, substitute x in the equation for M and M<sub>2<\/sub> to find the peak bending moment. Also, because M is the integral of shear force, horizontal plots in shear correspond to a linear plot in the M-x graph.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/V-m-chapter-6.jpg\" alt=\"Shear and moment diagrams\" class=\"alignnone wp-image-1824\" width=\"431\" height=\"823\" \/>\r\n\r\n<strong>6. Review\u00a0<\/strong>\r\n\r\nThe answers make sense. The moment, shear, and reaction forces are equal in both segments of the beam. This is as expected because of the symmetry of the beam.\r\n\r\n<\/div>\r\n<h1>Example 6.3.4: V\/M Diagrams, Submitted by Michael Oppong-Ampomah<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">Roughly sketch the shear and moment diagram for the following structure. Measure from point A.\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Michael1-300x133.png\" alt=\" A cantilever beam with distributed load at the free end.\" class=\"alignnone wp-image-1768 size-medium\" width=\"300\" height=\"133\" \/><\/div>\r\n<strong>2. Sketch<\/strong>\r\n\r\nN\/A, Sketch Provided.\r\n\r\n<strong>3. Knows and Unknowns:<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>Distributed Load: w<\/li>\r\n \t<li>Dimensions and shown<\/li>\r\n<\/ul>\r\nUnknown:\r\n<ul>\r\n \t<li>Shear\/Moment Diagram<\/li>\r\n<\/ul>\r\n<strong>4. Approach:<\/strong>\r\n\r\nBreak the beam into sections as forces change.\r\n\r\nDetermine the plotted shape of the shear and moment of each section.\r\n\r\nDraw the V\/M diagrams using these shape diagrams.\r\n\r\n<strong>5. Analysis:<\/strong>\r\n\r\nA distributed load creates a linear section across it, with the magnitude of the shear force increasing across the load. The maximum shear will be at the end of the distributed load. With no other forces being applied except for reactions, section B-C should be constant. Since the beam is static, the reactions must bring the shear back to zero.\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Michael2-300x133.png\" alt=\"Shear diagram of the problem.\" class=\"alignnone wp-image-1769 size-medium\" width=\"300\" height=\"133\" \/>\r\n\r\nThe moment can be drawn using the knowledge that dM = V. Section A-B in the shear diagram is linear; integrating a linear function gives a quadratic function, thus, section A-B in the moment diagram must be curved. Section B-C is constant in shear. Integrating a constant function gives a linear function; therefore, the moment across this section must be linear, and since the shear is negative, the slope of the moment must be negative. Finally, because the beam is static, the moment must return to zero due to the reaction forces.\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Michael3-300x173.png\" alt=\"Moment diagram of the problem.\" class=\"alignnone wp-image-1770 size-medium\" width=\"300\" height=\"173\" \/>\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nUsing principles of shear force (V) and moments and their integral\/derivative relationship, the shear and moment diagrams make sense for their general shape.\r\n\r\n<\/div>\r\n<h1>Example 6.3.5: V\/M Diagrams, Submitted by William Craine<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">Find the reaction forces and create the shear and moment diagrams for the beam.\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William1-4-1024x629.png\" alt=\"A cantilever beam with triangular distributed load at its free end.\" class=\"alignnone wp-image-1780\" width=\"544\" height=\"334\" \/><\/div>\r\n<strong>2. Draw<\/strong>\r\n\r\n[latex]4N\/m \\cdot \\frac{5.5m}{2} = 11N[\/latex]\r\n\r\n[latex]\\bar{x} = 1\/3 \\cdot 5.5m = 1.83m[\/latex]\r\n\r\n[latex]\\bar{x} \\ from\u00a0 A = 4m + 4.5m + 1.83m = 10.3m[\/latex]\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William2-1-1024x545.png\" alt=\"A FBD of the problem.\" class=\"alignnone wp-image-1781\" width=\"597\" height=\"318\" \/>\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>Applied Forces:<\/li>\r\n \t<li>F<sub>B<\/sub> = -20N<\/li>\r\n \t<li>F<sub>C<\/sub> = -400N<\/li>\r\n \t<li>F<sub>D<\/sub> = 500N<\/li>\r\n \t<li>F<sub>Load<\/sub> = -4N\/m<\/li>\r\n \t<li>Dimensions as shown<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>Reaction Forces: R<sub>AX, <\/sub>R<sub>AY, <\/sub>M<sub>A<\/sub><\/li>\r\n \t<li>Shear\/Moment Diagram<\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n<ol>\r\n \t<li>Use the equilibrium equation to find the reaction forces<\/li>\r\n \t<li>Use beam analysis\/cutting between forces to find equations for the shear\/moment diagram.<\/li>\r\n<\/ol>\r\n<strong>5. Analysis<\/strong>\r\n\r\nFirst, use the sum of forces and the sum of moments to determine R<sub>Ay<\/sub> and M<sub>A<\/sub>.\r\n\r\n[latex]\\sum F_x = R_{Ax} = 0[\/latex]\r\n\r\n[latex]\\sum F_y = -R_{Ay} - 20N - 400N - 11N +500N = 0[\/latex]\r\n\r\n[latex]R_{Ay} = 69N[\/latex]\r\n\r\n[latex]\\sum M_A = - M_A - (20N \\cdot 4m) - (400N \\cdot 8.5m) - (11N \\cdot 10.3m) + (500 N \\cdot 14m) = 0[\/latex]\r\n\r\n[latex]M_A = 3406.7Nm[\/latex]\r\n\r\nNext, cut the beam between A and B and analyze the internal forces.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William3-1024x422.png\" alt=\"A sketch of the internal forces from A to B.\" class=\"alignnone wp-image-1788\" width=\"469\" height=\"193\" \/>\r\n\r\n[latex]\\sum F_x = N + 0 = 0 \\\\ N = 0[\/latex]\r\n\r\n[latex]\\sum F_y = -69N -V = 0 \\\\ V= -69N[\/latex]\r\n\r\n[latex]\\sum M_{cut} =\u00a0 M + (69N \\cdot x) - (3406.7Nm) = 0 \\\\ M = -69x + 3406.7[\/latex]\r\n\r\nRepeat the analysis of internal forces from A to C.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William4-1024x367.png\" alt=\"A sketch of the internal forces from A to C.\" class=\"alignnone wp-image-1792 size-large\" width=\"1024\" height=\"367\" \/>\r\n\r\n[latex]\\sum F_x = N + 0 = 0 \\\\ N = 0[\/latex]\r\n\r\n[latex]\\sum F_y = -69N - 20N -V = 0 \\\\ V = -89N[\/latex]\r\n\r\n[latex]\\sum M_{cut} = M + (69N \\cdot x) + (20N \\cdot{(x-4m)}) - 3406.7Nm = 0 \\\\ M+89x-3326.7Nm = 0 \\\\ M = -89x + 3326.7[\/latex]\r\n\r\nFinally,\u00a0 analyze the beam in the area from C to D. Ensure that the value of X used is now taken from point D.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William5-1024x792.png\" alt=\"A sketch of the internal forces from C to D.\" class=\"alignnone wp-image-1796\" width=\"442\" height=\"342\" \/>\r\n\r\n[latex]\\sum F_x = -N = 0 \\\\ N = 0[\/latex]\r\n\r\nSince the section does not reach the end of the distributed load, the slope of the load must be calculated:\r\n\r\n[latex]\\frac{4}{5.5} = \\frac{m}{14-x} \\\\ m = \\frac{56-4x}{5.5}[\/latex]\r\n\r\nThis can be used to find the equivalent point load in terms of x.\r\n\r\n[latex]F_{eq} = \\frac{\\frac{56-4x}{5.5} \\cdot (14-x)}{2} \\\\ F_{eq} = \\frac{4x^2}{11} - \\frac{112x}{11} + \\frac{784}{11}[\/latex]\r\n\r\nThis point load can be used when summing forces to find v.\r\n\r\n[latex]\\sum F_y = 0 = V - F_{eq} + 500 \\\\ V = F_{eq} - 500 \\\\ V = \\frac{4x^2}{11} - \\frac{112x}{11} + \\frac{784}{11} - 500 \\\\ V = \\frac{4x^2}{11} - \\frac{112x}{11} - 428.7[\/latex]\r\n\r\nTo save time, rather than use the sum of moments to determine the equation for moments in this section, the moment can be found by the integral of V.\r\n\r\n[latex]M = \\int \\frac{4x^2}{11} - \\frac{112x}{11} - 428.7 dx\u00a0 \\\\ M = \\frac{4x^3}{33} - \\frac{112x^2}{22} - 428.7x +C[\/latex]\r\n\r\nWe know that for this beam to be static, the moment at x=14m must be 0. This can be used to determine C.\r\n\r\n[latex]M(14) = 0 =\u00a0 \\frac{4(14)^3}{33} - \\frac{112(14)^2}{22} - 428.7(14) +C \\\\ C = 6667 \\\\ Therefore: M_{cd}(x) = \\frac{4x^3}{33} - \\frac{112x^2}{22} - 428.7x + 6667[\/latex]\r\n\r\nWith the values for internal forces found between each change in force, the internal forces can be plotted.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William6.PNG.png\" alt=\"A sketch of the shear and moment diagrams.\" class=\"alignnone wp-image-1991\" width=\"614\" height=\"915\" \/>\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nUsing principles of shear force (V) and moments and their integral\/derivative relationship, the shear and moment diagrams make sense for their general shape. The numerical values calculated align with the variables provided.\r\n\r\n<\/div>\r\n<h1>Example 6.3.6: Internal Forces, Submitted by Riley Fitzpatrick<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div>\r\n<div class=\"textbox shaded\">\r\n<div>A person who weighs 140lbs is standing on the edge of a 10ft long diving board. The board is secured by a pin at 0ft and supported by a roller at 4ft.<\/div>\r\n<div><\/div>\r\n<div>a. Calculate the reaction forces on the board.<\/div>\r\n<div>b. Calculate the internal forces at the midpoint between A and B.<\/div>\r\n<div>c. Calculate the internal forces at the midpoint between B and C.<\/div>\r\n<div>d. Draw the shear and moment diagram.<\/div>\r\n<div><\/div>\r\n<div><img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Riley-chap-6-e1654109214997.jpg\" alt=\"A person at the edge of a diving board.\" class=\"alignnone wp-image-1958 size-full\" width=\"391\" height=\"686\" \/><\/div>\r\n<div>source: https:\/\/www.flickr.com\/photos\/lac-bac\/40229806264<\/div>\r\n<div><\/div>\r\n<div><img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-095515-1-300x131.png\" alt=\"A sketch of the problem.\" width=\"470\" height=\"205\" class=\"alignnone wp-image-2500\" \/><\/div>\r\n<div><\/div>\r\n<\/div>\r\n<strong>2. Draw\u00a0<\/strong>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-100645-300x148.png\" alt=\"A FBD of the problem.\" class=\"alignnone wp-image-2502\" width=\"475\" height=\"234\" \/>\r\n\r\n<strong>3. Knowns and Unknowns\u00a0<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>F<sub>g<\/sub> = 140 lbs<\/li>\r\n \t<li>X<sub>A<\/sub>= 0 ft<\/li>\r\n \t<li>X<sub>B<\/sub>= 4 ft<\/li>\r\n \t<li>X<sub>C<\/sub>= 10 ft<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>R<sub>Ax<\/sub><\/li>\r\n \t<li>R<sub>Bx<\/sub><\/li>\r\n \t<li>R<sub>Ay<\/sub><\/li>\r\n \t<li>Internal forces N, V, M between B and C, &amp; A and B<\/li>\r\n<\/ul>\r\n<strong>4. Approach\u00a0<\/strong>\r\n\r\nCalculate the reaction forces using equilibrium equations. Cut the diving board between A &amp; B, and use equilibrium equations to find internal forces. Do the same to find internal forces between B and C. Use the values to draw the shear and moment diagram.\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\na.\r\n\r\n<\/div>\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-101136-300x143.png\" alt=\"A FBD of the problem.\" class=\"alignnone wp-image-2505\" width=\"445\" height=\"212\" \/>\r\n\r\n[latex]\\sum{F_{x}} = 0=R_{Ax}[\/latex]\r\n\r\n[latex]\\sum{F_{y}}=0=R_{Ay}+R_{By}-140 lb \\\\R_{Ay}+R_{By}=140 lb[\/latex]\r\n\r\n[latex]\\sum{M_{A}}=\\left(X_{B} \\cdot R_{By}\\right)-\\left(X_{c} \\cdot F_{g}\\right)=0[\/latex]\r\n\r\n[latex]R_{By}= \\frac{X_{c}\\cdot F_{g}}{X_{B}}\\\\R_{By}= \\frac{10 ft\\cdot 140 lb}{4 ft}[\/latex]\r\n\r\n[latex]R_{By}=350 lb [\/latex]\r\n\r\nnow, [latex]R_{Ay}=140 lb - R_{By}\\\\ \\kern 1pc=140 lb - 350 lb \\\\ \\kern 1pc =-210 lb [\/latex]\r\n\r\nThe negative sign for [latex]R_{Ay}[\/latex] means that the direction of [latex]R_{Ay}[\/latex]\u00a0 is not +y but in the -y direction.\r\n\r\nThus, the reation forces are [latex]R_{Ay}=-210 lb, R_{By} = 350 lb, R_{Ax}=0[\/latex]\r\n\r\nb. <span style=\"text-decoration: underline\">Finding internal forces between A and B\u00a0<\/span>\r\n\r\nTo find the internal forces between A and B, cut in the middle of A and B, giving [latex]x = \\frac{X_{B}}{2} =2 ft [\/latex]\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-102906-300x135.png\" alt=\"A sketch of the internal forces between A and B\" class=\"alignnone wp-image-2507 size-medium\" width=\"300\" height=\"135\" \/>\r\n\r\n[latex]\\sum{F_{x}} = 0=R_{Ax} + N\\\\ N = 0[\/latex]\r\n\r\n[latex]\\sum{F_{y}}=0=-V - R_{Ay}\\\\ V = -R_{Ay}\\\\V=-210 lb [\/latex]\r\n\r\n[latex]\\sum{M_{A}}=0\\\\ \\kern 2pc =M-xV\\\\ \\kern 2pc = M- \\frac{4 ft}{2}\\left(-210 lb\\right)\\\\\u00a0 M =-420 lb ft[\/latex]\r\n\r\nInternal forces between A and B are [latex]N = 0, V= -210 lb, M=-420 lb ft [\/latex]\r\n\r\nc.\u00a0<span style=\"text-decoration: underline\"> Finding internal forces between B and C<\/span>\r\n\r\nCut at the midpoint between B and C to find the internal forces.\r\n\r\n<span style=\"text-decoration: underline\">Finding x<\/span>\r\n\r\nTo apply the equilibrium equation, you need to find x in this scenario.\r\n\r\n[latex]x= X_{B}+\\frac{X_{c}-X_{B}}{2}\\\\ = 4 ft +\\frac{10 ft -4 ft}{2}\\\\ x =7 ft [\/latex]\r\n\r\n&nbsp;\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-103051-300x142.png\" alt=\"A sketch of the internal forces between B and C.\" class=\"alignnone wp-image-2508 size-medium\" width=\"300\" height=\"142\" \/>\r\n\r\n&nbsp;\r\n\r\n[latex]\\sum{F_{x}} = 0=R_{Ax} + N\\\\ N = 0[\/latex]\r\n\r\n[latex]\\sum{F_{y}}=0=R_{By}-R_{Ay}-V\\\\ \\kern 1pc V=R_{By} -R_{Ay} \\\\ \\kern 1pc V = 350 lb -210 lb\\\\ V= 140 lb[\/latex]\r\n\r\nnow,\r\n\r\n[latex]\\sum{M_{A}}=0[\/latex]\r\n\r\n[latex]M =- \\left( X_{B} \\cdot R_{By} \\right) + xV\\\\ \\kern 1 pc = -\\left(4 ft \\cdot 350 lb\\right) +\\left( 7 ft \\cdot140 lb\\right) \\\\ \\kern 1 pc = 980 lb ft- 1400 lb ft[\/latex]\r\n\r\n[latex]M =-420 lb ft[\/latex]\r\n\r\nInternal forces between C and B are [latex]N = 0, V= 140 lb, M=-420 lb ft[\/latex]\r\n\r\nd. <span style=\"text-decoration: underline\">Shear and Moment diagram\u00a0<\/span>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/riley-chap-6-shear-and-moment-1.jpg\" alt=\"A sketch of the shear and moment diagrams.\" class=\"alignnone wp-image-1957\" width=\"572\" height=\"721\" \/>\r\n\r\n<strong>6. Review\u00a0<\/strong>\r\n\r\nThere is no force in the x direction; therefore, it makes sense that the reaction force, R<sub>Ax, <\/sub>is zero. Because the moment at A is generated by the person's weight, R<sub>By\u00a0<\/sub>should be considerably large. As R<sub>By<\/sub> is larger than the person's weight, R<sub>Ay\u00a0<\/sub>should act downwards and would be of the same magnitude as the difference between R<sub>By<\/sub> and the person's weight\u00a0to make the system stable.\r\n\r\nFor section b, the shear force between A and B is equal and opposite to the reaction force R<sub>Ay.\u00a0<\/sub>Similarly, with the concept of equilibrium, shear is equal and opposite to R<sub>Ax,<\/sub> and the moment in this section is the negative of the moment generated at A.\r\n\r\nIt also makes sense that the shear between B and C is equal to the difference between R<sub>By\u00a0<\/sub>and R<sub>Ay, <\/sub>and as expected, acting downward, thus it equals the weight of the person.\r\n\r\n<\/div>\r\n<h1>Example 6.3.7: Internal Forces and Shear\/Moment Diagrams, Submitted by Odegua Obehi-Arhebun<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\nA bolt attached to a stair rail is to be removed by a technician. A solid wrench of length 30cm, arched horizontally to the bolt and assumed to be rigidly connected. The technician applies a downward vertical force of 200N at the free end of the wrench. Determine the internal normal force, shear force and bending moment at a point 0.15m along the wrench. Draw the full shear and moment diagram.\r\n\r\n[caption id=\"attachment_2539\" align=\"alignnone\" width=\"300\"]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/bolt-and-wrench-isolated-on-white-3d-rendering-P89T2R-300x215.jpg\" alt=\"A wrench unscrewing a bolt.\" class=\"wp-image-2539 size-medium\" width=\"300\" height=\"215\" \/> Source: https:\/\/www.alamy.com\/bolt-and-wrench-isolated-on-white-3d-rendering-image211658127.html[\/caption]\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-122411-300x216.png\" alt=\"A sketch of the problem.\" width=\"300\" height=\"216\" class=\"alignnone wp-image-2544 size-medium\" \/>\r\n\r\n<\/div>\r\n<strong>2. Draw<\/strong>\r\n\r\nFree-body diagram:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-123303-300x193.png\" alt=\"A FBD of the problem.\" class=\"alignnone wp-image-2545\" width=\"354\" height=\"228\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>F=200N<\/li>\r\n \t<li>L= 30cm =&gt; 0.3m<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>Internal normal force, shear force and bending moment at point B<\/li>\r\n \t<li>Shear + moment diagram<\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nSolve for reaction forces at A, then take arbitrary sections between A and B and between B and C to derive V and M equations. Plot results.\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\nSolve for reaction forces:\r\n\r\n[latex]\\begin{aligned}\r\n\\sum M_{A} &amp;= 0 = -200\\,\\mathrm{N} \\cdot 0.3\\,\\mathrm{m} - M_A \\\\\r\nM_A &amp;= -60\\,\\mathrm{Nm}\r\n\\end{aligned}[\/latex]\r\n\r\n[latex]\\begin{aligned}\r\n\\sum F_{y} &amp;= 0 = A_y - 200\\,\\mathrm{N} \\\\\r\nA_y &amp;= 200\\,\\mathrm{N}\r\n\\end{aligned}[\/latex]\r\n\r\n[latex]\\begin{aligned}\r\n\\sum F_{x} &amp;= 0 = N \\\\\r\nN &amp;= 0\r\n\\end{aligned}[\/latex]\r\n\r\nNo force in the x-direction\r\n\r\nCut 1: With reaction forces known, cut a section between A and B\r\n\r\n0m&lt;x&lt;0.15m\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-133112-300x195.png\" alt=\"A sketch showing the internal forces between 0m to 0.15m.\" width=\"300\" height=\"195\" class=\"alignnone wp-image-2554 size-medium\" \/>\r\n\r\n[latex]\\begin{aligned}\r\n\\sum F_{y} &amp;= 0 = A_y - V \\\\\r\nV &amp;= A_y = 200\\,\\mathrm{N}\r\n\\end{aligned}[\/latex]\r\n\r\n[latex]\\begin{aligned}\r\n\\sum M_x &amp;= 0 = -M_A - A_y \\cdot x +M(x) \\\\\r\nM(x) &amp;= 200x-60\\,\\mathrm{Nm}\r\n\\end{aligned}[\/latex]\r\n\r\nYou can also cut the section between B and C\r\n\r\n0.15m&lt;x&lt;0.3m\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-134813-300x191.png\" alt=\"A sketch showing the internal forces between 0.15m to 0.3m.\" class=\"alignnone wp-image-2557\" width=\"352\" height=\"224\" \/>\r\n\r\n[latex]\\begin{aligned}\r\n\\sum F_{y} &amp;= 0 = A_y - V \\\\\r\nV &amp;= A_y = 200\\,\\mathrm{N}\r\n\\end{aligned}[\/latex]\r\n\r\n[latex]\\begin{aligned}\r\n\\sum M_x &amp;= 0 = M_x - M_A- A_y \\cdot x \\\\\r\nM(x) &amp;= 200x - 60\\,\\mathrm{Nm}\r\n\\end{aligned}[\/latex]\r\n\r\n[latex]\\begin{aligned}\r\n\\text{At } x = 0, \\quad &amp;M = 0 \\\\\r\n\\text{At } x = 0.15\\,\\mathrm{m}, \\quad &amp;M = -30\\,\\mathrm{Nm} \\\\\r\n\\text{At } x = 0.3\\,\\mathrm{m}, \\quad &amp;M = 0\r\n\\end{aligned}[\/latex]\r\n\r\nAnswer: N = 0, V = 200 N, M<sub>x<\/sub> = 200x-60 Nm\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-140735-208x300.png\" alt=\"A sketch of the shear and moment diagrams\" class=\"alignnone wp-image-2561\" width=\"306\" height=\"441\" \/>\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nNegative sign for M<sub>A <\/sub>indicates that it should be in the other direction, as shown in the final diagram above. The shear force remains constant at 200N from A to C, then drops to zero at the end where the load is applied. The shear force is constant, hence it makes sense that the moment is linear.\r\n\r\n<\/div>\r\n<h1>Example 6.3.8: Internal Forces, Submitted by Celina Areoye<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\nA Uniform rectangular light frame of mass 3kg and length 0.8m is suspended from the ceiling by two vertical rods. The rods are 0.2m apart and connected symmetrically to the ceiling plate. Also, there are 5 evenly spaced identical bulbs, each 0.5kg each mounted on the frame. Determine the tension in each vertical rod and the internal forces at the points where the vertical rod is connected to the frame.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture2-300x275.png\" alt=\"A rectangular frame hanging from the ceiling.\" class=\"alignnone wp-image-2564 size-medium\" width=\"300\" height=\"275\" \/>\r\n\r\n<\/div>\r\n<strong>2. Draw<\/strong>\r\n\r\nFree-body diagram:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture3-300x186.png\" alt=\"A FBD of the problem.\" class=\"alignnone wp-image-2565\" width=\"405\" height=\"251\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>Mass of frame (\ud835\udc40\ud835\udc53) = 3kg<\/li>\r\n \t<li>Length of frame (\ud835\udc3f\ud835\udc53) = 0.8m<\/li>\r\n \t<li>Mass of each lightbulb (\ud835\udc40\ud835\udc4f) = 0.5kg<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>Tension in each rod (\ud835\udc47<sub>1<\/sub> \ud835\udc4e\ud835\udc5b\ud835\udc51 \ud835\udc47<sub>2<\/sub>)<\/li>\r\n \t<li>Internal forces at point D (\ud835\udc41<sub>D<\/sub>, \ud835\udc49<sub>D<\/sub>, \ud835\udc4e\ud835\udc5b\ud835\udc51 \ud835\udc40<sub>D<\/sub>)<\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nAssuming the vertical rods are positioned centrally between the bulbs, and the internal forces at points C and D are the same (use any point to get the internal forces). First, find the tension and then use it in calculating internal forces.\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\nWeight of frame (\ud835\udc4a\ud835\udc53) = \ud835\udc40\ud835\udc53\ud835\udc54 = 3kg \u00d7 9.81m\/s<sup>2<\/sup> = 29.43N\r\n\r\nWeight of one bulb (\ud835\udc4a\ud835\udc4f) = \ud835\udc40\ud835\udc4f\ud835\udc54 = 0.5kg \u00d7 9.81m\/s<sup>2<\/sup> = 4.905N\r\n\r\nThe bulbs are evenly spaced, so the distance between 2 bulbs (from centre to centre):\r\n\r\n[latex]\\frac{0.8\\,\\mathrm{m}}{4} = 0.2\\,\\mathrm{m}[\/latex]\r\n\r\nTo get the tension in each rod:\r\n\r\n[latex]+\\uparrow \\sum F_y = 0 \\quad ; \\quad T_1 + T_2 - 5W_b - W_f = 0[\/latex]\r\n\r\nSince the rods are connected symmetrically, they are equal (\ud835\udc47<sub>1<\/sub> = \ud835\udc47<sub>2<\/sub> ), therefore:\r\n<div>[latex]2T - 5(4.905\\,\\mathrm{N}) - 29.43\\,\\mathrm{N} = 0[\/latex]<\/div>\r\n<div><\/div>\r\n<div>[latex]2T = 53.955\\,\\mathrm{N}, \\quad T = 26.98\\,\\mathrm{N}[\/latex]<\/div>\r\n<div>The tension in each vertical rod is <strong>26.98N.<\/strong><\/div>\r\nTo get the internal forces at point D, taking the right side of the diagram:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture4-300x208.png\" alt=\"A sketch showing the internal forces at point D (right side).\" width=\"338\" height=\"234\" class=\"alignnone wp-image-2567\" \/>\r\n\r\nSince the rods are positioned centrally between the bulbs, the distance from point D to the bulb is\r\n\r\n[latex]\\frac{0.2\\,\\mathrm{m}}{2} = 0.1\\,\\mathrm{m}[\/latex]\r\n<div>[latex]+ \\rightarrow \\sum F_x = 0, \\quad N_D = 0\\,\\mathrm{N}[\/latex]<\/div>\r\n<div><\/div>\r\n<div>[latex]+ \\uparrow \\sum F_y = 0; \\quad V_D + T_2 - 2W_b = 0[\/latex]<\/div>\r\n<div><\/div>\r\n<div>[latex]V_D = 2(4.905\\,\\mathrm{N}) - 26.98\\,\\mathrm{N} = -17.17\\,\\mathrm{N}[\/latex]<\/div>\r\n<div><\/div>\r\n<div>[latex]\\text{Therefore, } V_D = 17.17\\,\\mathrm{N} \\text{ (opposite direction)} \\downarrow[\/latex]<\/div>\r\n<div><\/div>\r\n<div>[latex]+ \\text{ anticlockwise } \\sum M_D = 0; \\quad - (4.905\\,\\mathrm{N} \\times 0.3\\,\\mathrm{m}) - (4.905\\,\\mathrm{N} \\times 0.1\\,\\mathrm{m}) - M_D = 0[\/latex]<\/div>\r\n<div><\/div>\r\n<div>[latex]M_D = -1.962\\,\\mathrm{Nm}[\/latex]<\/div>\r\n<div><\/div>\r\n<div>[latex]\\text{Therefore, } M_D = 1.962\\,\\mathrm{Nm} \\text{ (clockwise direction)}[\/latex]<\/div>\r\n<div><\/div>\r\nNo force in the x-direction\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nTo check, taking the left side of the diagram\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture5-300x229.png\" alt=\"A sketch showing the internal forces at point D (left side).\" width=\"300\" height=\"229\" class=\"alignnone wp-image-2568 size-medium\" \/>\r\n<div>[latex]+ \\uparrow \\sum F_y = 0; \\quad V_D + T_1 - 3W_b - W_f = 0[\/latex]<\/div>\r\n<div><\/div>\r\n<div>[latex]17.17\\,\\mathrm{N} + 26.98\\,\\mathrm{N} - 3(4.905\\,\\mathrm{N}) - 29.43\\,\\mathrm{N} = 0[\/latex]<\/div>\r\n<\/div>\r\n<div>\r\n<h1>Example 6.3.9: Internal Forces, Submitted by Okino Itopa Farooq<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\nA uniform horizontal shelf of total length 100 cm and mass 2 kg is supported at both ends by pin connections. The shelf supports a group of 5 books placed together, each with a thickness of 4 cm and a mass of 0.5 kg. The collection of books is positioned such that it is located at a space 50 cm from the left end and 30 cm from the right end of the shelf. Assuming static equilibrium and neglecting the thickness of the shelf, determine the following at a point located 60 cm from the left end of the shelf:\r\n<ol>\r\n \t<li>The shear force<\/li>\r\n \t<li>The normal (axial) force<\/li>\r\n \t<li>The bending moment<\/li>\r\n<\/ol>\r\n[caption id=\"attachment_2577\" align=\"aligncenter\" width=\"388\"]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/StockSnap_LGFGZZC671-300x201.jpg\" alt=\"Books on a shelf.\" class=\"wp-image-2577 \" width=\"388\" height=\"260\" \/> Source: https:\/\/stocksnap.io\/photo\/books-shelf-LGFGZZC671[\/caption]\r\n\r\n<\/div>\r\n<strong>2. Draw<\/strong>\r\n\r\nFree-body diagram:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-191320-300x233.png\" alt=\"A FBD of the problem.\" width=\"411\" height=\"319\" class=\"alignnone wp-image-2609\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>Total length of shelf = 100cm = 1m<\/li>\r\n \t<li>Mass of shelf = 2kg<\/li>\r\n \t<li>Thickness of each book = 4cm = 0.04m<\/li>\r\n \t<li>Mass of each book = 0.5kg<\/li>\r\n \t<li>g = 9.81 m\/s<sup>2<\/sup><\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>The shear force, normal force, and bending moment at 60cm (0.6m) from the left end of the shelf<\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nFirst, resolve the distributed loads, then find reaction forces, and then calculate the internal forces\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\nFinding the force acting per m covered by the books\r\n\r\n[latex]\\left( \\frac{0.5}{0.04} \\right) \\, \\text{kg\/m} = \\left( \\frac{0.5}{0.04} \\right) \\, \\text{kg\/m} \\times 9.81 = 122.625 \\, \\text{N\/m}[\/latex]\r\n\r\nWeight of the shelf\r\n\r\n[latex]2 \\, \\text{kg} \\times 9.81 \\, \\text{m\/s}^2 = 19.62 \\, \\text{N}[\/latex]\r\n\r\nWe first resolve the distributed load\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-193706-300x201.png\" alt=\"A sketch shown resolving the distributed load.\" width=\"300\" height=\"201\" class=\"alignnone wp-image-2610 size-medium\" \/>\r\n\r\n[latex]C = 122.625 \\, \\text{N\/m} \\times 0.2 \\, \\text{m} = 24.525 \\, \\text{N}[\/latex]\r\n[latex]X = \\frac{0.2}{2} = 0.1 \\, \\text{m} \\, \\text{(0.6 m from A)}[\/latex]\r\n\r\nThe new FBD should look like this:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-194707-300x234.png\" alt=\"Updated FBD, including found values.\" width=\"379\" height=\"296\" class=\"alignnone wp-image-2611\" \/>\r\n\r\nUsing equilibrium equations to find the reaction forces\r\n<div>[latex]\\sum F_x = 0 \\Rightarrow A_x = B_x = 0[\/latex]<\/div>\r\n<div>[latex]\\sum F_y = 0 \\Rightarrow A_y + B_y = 19.62 + 24.525[\/latex]<\/div>\r\n<div>[latex]A_y + B_y = 44.145 \\, \\text{N} \\quad \\cdots (i)[\/latex]<\/div>\r\n<div><\/div>\r\n<div>[latex]\\sum M_A^+ = 0 \\Rightarrow B_y(0.1) - 24.525(0.6) - 19.62(0.5) = 0[\/latex]<\/div>\r\n<div>[latex]B_y = \\frac{24.525}{0.1} = 245.25 \\, \\text{N}[\/latex]<\/div>\r\n<div><\/div>\r\n<div>[latex]\\text{Substitute in (i):}[\/latex]<\/div>\r\n<div>[latex]A_y = 44.145 \\, \\text{N} - 245.25 \\, \\text{N} = -201.105 \\, \\text{N}[\/latex]<\/div>\r\n<div><\/div>\r\n<div>A<sub>y <\/sub>= -201.105N, B<sub>y<\/sub> = 245.25N, A<sub>x\u00a0<\/sub>= B<sub>x\u00a0<\/sub>= 0<\/div>\r\n<div><\/div>\r\nNow that we know the Components A<sub>y<\/sub> and B<sub>y<\/sub>, we can start calculating the Shear, Normal and Bending moment at the point 0.6m from A, which will cut into the distributed load\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-201715-300x221.png\" alt=\"A sketch showing the internal forces in the distributed load.\" width=\"424\" height=\"312\" class=\"alignnone wp-image-2613\" \/>\r\n\r\nWith V<sub>f<\/sub> = Shear Force, N<sub>f<\/sub> = Normal Force and M<sub>e<\/sub> = Bending Moment. The remaining distributed load is also resolved\r\n\r\n[latex]122.625 \\, \\text{N\/m} \\times 0.1 \\, \\text{m} = 12.2625 \\, \\text{N}[\/latex]\r\n[latex]\\text{(Located } 0.05 \\, \\text{m from the cut E and } 0.55 \\, \\text{m from A)}[\/latex]\r\n\r\nWith all the info gathered, we can calculate the three forces.\r\n\r\nNormal force is zero, as there are no horizontal forces to consider\r\n\r\nFor shear force:\r\n\r\n[latex]\r\n\\sum F_y = 0; \\quad -201.105 - 19.62 - 12.2625 - V_F\u00a0 = 0\\\\\r\nV_F = -233\\, \\text{N} = 233\\, \\text{N} \\uparrow\r\n[\/latex]\r\n\r\nFor Bending Moment:\r\n\r\n[latex]\r\n\\sum M_E = 0; \\quad M_E + 122.625(0.05) + 19.62(0.1) + 201.105(0.6) = 0 \\\\\r\nM_E = -129\\, \\text{N}\\cdot\\text{m}\r\n[\/latex]\r\n\r\nTherefore, V<sub>f\u00a0<\/sub>=-233N and M<sub>E\u00a0<\/sub>=-129Nm\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nNegative signs for V<sub>f <\/sub>and M<sub>E <\/sub>indicate that it should go in the other direction. It makes sense that A<sub>y\u00a0<\/sub>and B<sub>y\u00a0<\/sub>go in opposite directions.\r\n\r\n<\/div>\r\n<\/div>","rendered":"<p>Here are examples from Chapter 6 to help you understand these concepts better. These were taken from the real world and supplied by FSDE students in Summer 2021. If you&#8217;d like to submit your own examples, please send them to the author <a href=\"mailto:eosgood@upei.ca\">eosgood@upei.ca<\/a>.<\/p>\n<h1>Example 6.3.1: Internal Forces, Submitted by Emma Christensen<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>The setup that holds the solar panels at the UPEI FSDE is modelled below. Considering beam S (1.9 m length), find the internal forces at point C. Assume the intensity of the solar panel on the beam is 220 N\/m.<\/p>\n<p>Sketch:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-draw-1-1-300x265.jpg\" alt=\"A sketch of the problem.\" class=\"aligncenter wp-image-1124\" width=\"350\" height=\"309\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-draw-1-1-300x265.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-draw-1-1-1024x903.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-draw-1-1-768x678.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-draw-1-1-65x57.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-draw-1-1-225x198.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-draw-1-1-350x309.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-draw-1-1.jpg 1443w\" sizes=\"auto, (max-width: 350px) 100vw, 350px\" \/><\/p>\n<p>Model:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-draw-2-300x198.jpg\" alt=\"A more detailed sketch of the problem.\" class=\"aligncenter wp-image-1112 size-medium\" width=\"300\" height=\"198\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-draw-2-300x198.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-draw-2-1024x675.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-draw-2-768x506.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-draw-2-65x43.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-draw-2-225x148.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-draw-2-350x231.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-draw-2.jpg 1243w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<\/div>\n<p><strong>2. Draw<\/strong><\/p>\n<p>Free-body diagram:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-1-300x197.jpg\" alt=\"A FBD of the problem\" class=\"alignnone wp-image-1114\" width=\"394\" height=\"259\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-1-300x197.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-1-1024x671.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-1-768x503.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-1-1536x1007.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-1-65x43.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-1-225x147.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-1-350x229.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-1.jpg 1584w\" sizes=\"auto, (max-width: 394px) 100vw, 394px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>w = 220 N\/m<\/li>\n<li>OA = 0.5 m<\/li>\n<li>AC = 0.2 m<\/li>\n<li>AB = 0.4 m<\/li>\n<li>L = 1.9 m<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>N<sub>c<\/sub><\/li>\n<li>V<sub>c<\/sub><\/li>\n<li>M<sub>c<\/sub><\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>Use equilibrium equations. First, solve for reaction forces, then make a cut at C and solve for the internal forces.<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>[latex]w=\\frac{F}{L}\\\\F=wL\\\\F_R=220N\/m\\cdot 1.9m\\\\F_R=418N\\\\\\sum F_X=0=B_X[\/latex]<\/p>\n<p>Find reaction forces:<\/p>\n<p>[latex]\\sum M_A=0=B_y(0.4m)-F_R(0.45m)\\\\(0.4m)B_y=418N(0.45m)\\\\B_y=\\frac{188.1 N\\cdot m}{0.4m}\\\\B_y=470.25N[\/latex]<\/p>\n<p>[latex]\\sum F_y=0=-F_R+A_y+B_y\\\\A_y=F_R-B_y\\\\A_y=418N-470.25N\\\\A_y=-52.25N[\/latex]<\/p>\n<p>The answer we got for A<sub>y<\/sub> is negative, which means that the arrow should be drawn in the other direction. We will change it for our next sketch.<\/p>\n<p>Make a cut at C:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-6-1-300x264.jpg\" alt=\"A sketch showing the internal forces at cut C.\" class=\"alignnone wp-image-1121 size-medium\" width=\"300\" height=\"264\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-6-1-300x264.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-6-1-1024x901.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-6-1-768x676.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-6-1-65x57.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-6-1-225x198.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-6-1-350x308.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-6-1.jpg 1352w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>Now solve for the internal forces:<\/p>\n<p>[latex]\\sum F_x=0\\:\\:;\\:\\:N_c=0\\\\\\sum F_y=0=-A_y-V_c-(w\\cdot L_A)\\\\V_c=-52.25N-(220N\/m\\cdot 0.7m)\\\\V_c=-206.25 N\\\\\\sum M_c=A_y(0.2m)+M_c+(F_{Rc}\\cdot 0.35m)\\\\M_c=52.25 N (0.2m)+(220N\/m\\cdot 0.7m\\cdot 0.35m)\\\\M_c=64.35N\\cdot m[\/latex]<\/p>\n<p>Final FBD, showing the arrows in the correct directions:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-7-1-300x216.jpg\" alt=\"Final FBD showing arrows in the correct directions.\" class=\"alignnone wp-image-1122\" width=\"388\" height=\"279\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-7-1-300x216.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-7-1-1024x738.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-7-1-768x553.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-7-1-65x47.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-7-1-225x162.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-7-1-350x252.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-1-solve-7-1.jpg 1503w\" sizes=\"auto, (max-width: 388px) 100vw, 388px\" \/><\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>It makes sense that A<sub>y<\/sub> and B<sub>y<\/sub> are in different directions because the resultant force Fr of the solar panel on the beam is not between A and B. It also makes sense that the moment at C is in the counterclockwise direction rather than the clockwise direction when you think about the direction of the forces applied to the beam.<\/p>\n<\/div>\n<h1>Example 6.3.2: Shear\/Moment Diagrams, Submitted by Deanna Malone<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>A beam that is simply supported has two point loads acting on it. One acts 2 m from point A, and the other acts at 2.5 m from C. Point B is in the middle of the beam. The first point load is 500 N and the second is 300 N. What are the internal forces at point B? Solve for reaction forces and include a shear\/moment diagram.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-1-300x257.jpg\" alt=\"A sketch of the problem.\" class=\"aligncenter wp-image-1003\" width=\"307\" height=\"263\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-1-300x257.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-1-1024x876.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-1-768x657.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-1-1536x1314.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-1-65x56.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-1-225x192.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-1-350x299.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-1.jpg 1799w\" sizes=\"auto, (max-width: 307px) 100vw, 307px\" \/><\/p>\n<\/div>\n<p><strong>2. Draw<\/strong><\/p>\n<p>Sketch:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-1-300x257.jpg\" alt=\"A sketch of the problem.\" class=\"alignnone wp-image-1003\" width=\"307\" height=\"263\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-1-300x257.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-1-1024x876.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-1-768x657.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-1-1536x1314.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-1-65x56.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-1-225x192.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-1-350x299.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-1.jpg 1799w\" sizes=\"auto, (max-width: 307px) 100vw, 307px\" \/><\/p>\n<p>Free-body diagram:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-2-300x165.jpg\" alt=\"A FBD of the problem.\" class=\"alignnone wp-image-1004\" width=\"407\" height=\"224\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-2-300x165.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-2-1024x565.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-2-768x424.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-2-1536x847.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-2-2048x1129.jpg 2048w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-2-65x36.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-2-225x124.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-draw-2-350x193.jpg 350w\" sizes=\"auto, (max-width: 407px) 100vw, 407px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>F1 = 500 N<\/li>\n<li>F2 = 300 N<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>A<sub>y<\/sub><\/li>\n<li>A<sub>x<\/sub><\/li>\n<li>C<sub>y<\/sub><\/li>\n<li>V<sub>B<\/sub><\/li>\n<li>M<sub>B<\/sub><\/li>\n<li>N<sub>B<\/sub><\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>Shear\/moment equations, EOM equations<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>Solve for reaction forces:<\/p>\n<p>(Ax, Cy)<\/p>\n<p>[latex]\\begin{aligned}  \\sum F_{x}=0=A_{x}=0 \\\\  \\sum M_{A}=0 &=-F_{1} \\cdot 2 m-F_{2} \\cdot 5.5 m+C_{y} \\cdot 8 m \\\\  C_{y}=&\\frac{F_{1} \\cdot 2 m+F_{2} \\cdot 5.5 m}{8m} \\\\  C_{y} &=\\frac{500 N \\cdot 2 m+300 N \\cdot 5.5 m}{8 m} \\\\  C_{y} &=331.25 \\mathrm{~N}  \\end{aligned}[\/latex]<\/p>\n<p>(Ay)<\/p>\n<p>[latex]\\begin{aligned}  \\sum F_{y}=0 &=A_{y}+C_{y}-F_{1}-F_{2} \\\\  A_{y} &=F_{1}+F_{2}-C_{y} \\\\  A_{y} &=500 \\mathrm{~N}+300 \\mathrm{~N} - 331.25 \\mathrm{~N} \\\\  A_{y} &=468.75 \\mathrm{~N}  \\end{aligned}[\/latex]<\/p>\n<p>Cut 1: at B<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-1-300x133.jpg\" alt=\"A sketch showing the cut at B\" class=\"alignnone wp-image-1005\" width=\"399\" height=\"177\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-1-300x133.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-1-1024x453.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-1-768x339.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-1-1536x679.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-1-65x29.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-1-225x99.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-1-350x155.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-1.jpg 1855w\" sizes=\"auto, (max-width: 399px) 100vw, 399px\" \/><\/p>\n<p>[latex]\\begin{aligned}  \\sum F_{X}=0=A_{X} &+N_{B}=0 \\\\  & N_{B}=0 \\\\  \\sum F_{y}=0 &=A_{y}-V_{B}-F_{1} \\\\  V_{B} &=A_{y}-F_{1} \\\\  V_{B} &=468.75 N - 500 N \\\\  V_{B}=-31.25 N  \\end{aligned}[\/latex]<\/p>\n<p>[latex]\\begin{aligned}  \\sum M_{B}=& 0=-A_{y}(4 m)+F_{1}(2 m)+M_{B} \\\\  & M_{B}=A_{y}(4 m)-F_{1}(2 m) \\\\  & M_{B}=468.75 N(4 m)-500 N(2 m) \\\\  M_{B} &=875 \\mathrm{~N} \\cdot \\mathrm{m}  \\end{aligned}[\/latex]<\/p>\n<p>Cut 2: At the point where F<sub>1<\/sub> is applied<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-2-300x139.jpg\" alt=\"A sketch showing the cut at F1\" class=\"alignnone wp-image-1006\" width=\"376\" height=\"174\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-2-300x139.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-2-1024x475.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-2-768x356.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-2-1536x713.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-2-65x30.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-2-225x104.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-2-350x162.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-2.jpg 1884w\" sizes=\"auto, (max-width: 376px) 100vw, 376px\" \/><\/p>\n<p>[latex]\\begin{aligned}  \\sum M_{1}=0 &=-A_{y}(2 m)+M_{1}=0 \\\\  M_{1} &=A_{y}(2 m) \\\\  M_{1} &=468.75 N(2 m) \\\\  M_{1} &=937.5 \\mathrm{~N} \\cdot m  \\end{aligned}[\/latex]<\/p>\n<p>Cut 3: At the point where F<sub>2<\/sub> is applied<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-3-300x109.jpg\" alt=\"A sketch showing the cut at F2.\" class=\"alignnone wp-image-1007\" width=\"377\" height=\"137\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-3-300x109.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-3-1024x371.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-3-768x278.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-3-1536x556.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-3-2048x741.jpg 2048w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-3-65x24.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-3-225x81.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/deanna-2-solve-3-350x127.jpg 350w\" sizes=\"auto, (max-width: 377px) 100vw, 377px\" \/><\/p>\n<p>[latex]\\begin{aligned}  \\sum M_{2}=0 =-A_{y}(5.5 \\mathrm{~m})+F_{1}(3.5 \\mathrm{~m})+M_{2} \\\\  M_{2} =A_{y}(5.5 \\mathrm{~m})-F_{1}(3.5 \\mathrm{~m}) \\\\  M_{2} =468.75 \\mathrm{~N}(5.5 \\mathrm{~m})-500 \\mathrm{~N}(3.5 \\mathrm{~m}) \\\\  M_{2} =828.125 \\mathrm{~N} \\cdot \\mathrm{m}  \\end{aligned}[\/latex]<\/p>\n<p>Answer: N<sub>B<\/sub> = 0, V<sub>B<\/sub> = -31.25 N, M<sub>B<\/sub> = 875 Nm<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-solve-final-252x300.jpg\" alt=\"A final diagram of the beam, shear and moment diagrams.\" class=\"alignnone wp-image-1008\" width=\"404\" height=\"481\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-solve-final-252x300.jpg 252w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-solve-final-859x1024.jpg 859w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-solve-final-768x915.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-solve-final-1289x1536.jpg 1289w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-solve-final-65x77.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-solve-final-225x268.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-solve-final-350x417.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-2-solve-final.jpg 1365w\" sizes=\"auto, (max-width: 404px) 100vw, 404px\" \/><\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>The reaction forces make sense as they offset the applied forces. The shear\/moment diagrams returned to zero, so they are correct too. The moment found at B is in the moment diagram; it is smaller than the maximum.<\/p>\n<\/div>\n<h1>Example 6.3.3: V\/M Diagrams, Submitted by Luciana Davila<\/h1>\n<div class=\"textbox\">\n<ol>\n<li><strong>Problem\u00a0<\/strong><\/li>\n<\/ol>\n<div class=\"textbox shaded\">\n<p>For the 8cm simply supported beam given below, if Fg, which is applied at the center of the beam, is given as 30 N, find the internal forces between the reaction forces and draw the V-M diagram for the beam.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/simply-supported.jpg\" alt=\"An image of a simply supported bridge.\" width=\"551\" height=\"640\" class=\"alignnone wp-image-2205 size-full\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/simply-supported.jpg 551w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/simply-supported-258x300.jpg 258w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/simply-supported-65x75.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/simply-supported-225x261.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/simply-supported-350x407.jpg 350w\" sizes=\"auto, (max-width: 551px) 100vw, 551px\" \/><\/p>\n<p>Simply supported girder bridge: https:\/\/www.geograph.org.uk\/photo\/2780207<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/chapter-6-unknown-V-m-diagram-question.jpg\" width=\"491\" height=\"192\" class=\"alignnone\" alt=\"A digital sketch of the problem.\" \/><\/p>\n<\/div>\n<p><strong>2. Draw\u00a0<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/chapter-6-unknown-V-m-diagram-question-2.jpg\" alt=\"A FBD of the problem.\" class=\"alignnone wp-image-1823 size-full\" width=\"608\" height=\"264\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/chapter-6-unknown-V-m-diagram-question-2.jpg 608w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/chapter-6-unknown-V-m-diagram-question-2-300x130.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/chapter-6-unknown-V-m-diagram-question-2-65x28.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/chapter-6-unknown-V-m-diagram-question-2-225x98.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/chapter-6-unknown-V-m-diagram-question-2-350x152.jpg 350w\" sizes=\"auto, (max-width: 608px) 100vw, 608px\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><strong>3. Knowns and Unknowns\u00a0<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>F<sub>g<\/sub> = 30N<\/li>\n<li>r <sub style=\"text-align: initial;background-color: initial\">AB<\/sub><span style=\"text-align: initial;background-color: initial;font-size: 1em\"> =\u00a0 r <\/span><sub style=\"text-align: initial;background-color: initial\">BC <\/sub><span style=\"text-align: initial;background-color: initial;font-size: 1em\">=\u00a0 0.04 m<\/span><\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>Reaction forces = R<sub>Ax<\/sub>, R<sub>Ay<\/sub>, R<sub>Cy<\/sub><\/li>\n<li>For the segment of beam AB nd BC,<\/li>\n<li>Shear = V, V<sub style=\"text-align: initial;background-color: initial\">2<\/sub><\/li>\n<li>Bending Moment = M, M<sub style=\"text-align: initial;background-color: initial\">2<\/sub><\/li>\n<li>Normal force = N, N<sub style=\"text-align: initial;background-color: initial\">2<\/sub><\/li>\n<\/ul>\n<p><strong>4. Approach\u00a0<\/strong><\/p>\n<p>Find reaction forces from equilibrium equations. Use them to find the shear and bending moment of segment AB and BC. Draw the V-M diagram from the answers.<\/p>\n<p><strong>5. Analysis\u00a0<\/strong><\/p>\n<p>Finding reaction forces:<\/p>\n<p>[latex]\\sum F_x=0=R_{Ax}\\\\R_{Ax}=0N[\/latex]<\/p>\n<p>Solving for RC<sub>y<\/sub>:<\/p>\n<p>[latex]\\sum M_A=0= -(r_{AB}\\cdot F_g)+(r_{AC}\\cdot R_{Cy})\\\\r_{AB}\\cdot F_{g}=r_{AC}\\cdot R_{Cy}[\/latex]<\/p>\n<p>[latex]R_{Cy}=\\frac{r_{AB}\\cdot F_g}{r_{AC}}\\\\R_{By}=\\frac{0.04m\\cdot 30 N}{0.08 m}[\/latex]<\/p>\n<p><span style=\"text-align: initial;background-color: initial;font-size: 1em\">[latex]R_{Cy}=15 N[\/latex]<\/span><\/p>\n<p>Solving for RA<sub>y<\/sub>:<\/p>\n<p>[latex]\\sum F_y=0=R_{Cy}+R_{Ay} -F_g\\\\R_{Ay}=F_g-R_{Cy}\\\\R_{Ay}= 15 N[\/latex]<\/p>\n<p>Finding shear and Moment from A to B :<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luciana-beam-AB-3.png\" alt=\"A sketch to find the internal forces between a and b.\" class=\"alignnone wp-image-1713\" width=\"340\" height=\"202\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luciana-beam-AB-3.png 248w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luciana-beam-AB-3-65x39.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luciana-beam-AB-3-225x133.png 225w\" sizes=\"auto, (max-width: 340px) 100vw, 340px\" \/><\/p>\n<p>[latex]\\sum F_x=0\\\\N=0N[\/latex]<\/p>\n<p>Solving for V<\/p>\n<p>[latex]\\sum F_y=0=R_{Ay}+-V\\\\ V= 15 N[\/latex]<\/p>\n<p>Solving for M :<\/p>\n<p>[latex]\\sum M_A=0= -( V\\cdot 0.04) +M\\\\M = 15\\cdot 0.04\\\\M=0.6N[\/latex]<\/p>\n<p>Finding shear and Moment from B to C:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luciana-BC-4-300x166.png\" alt=\"A sketch to find the internal forces between b and c.\" class=\"alignnone wp-image-1714\" width=\"363\" height=\"201\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luciana-BC-4-300x166.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luciana-BC-4-65x36.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luciana-BC-4-225x125.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luciana-BC-4-350x194.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luciana-BC-4.png 373w\" sizes=\"auto, (max-width: 363px) 100vw, 363px\" \/><\/p>\n<p>[latex]\\sum F_x=0\\\\N=0N[\/latex]<\/p>\n<p>Solving for V<\/p>\n<p>[latex]\\sum F_y=0=R_{Cy}+V_2\\\\ V_2= -15 N[\/latex]<\/p>\n<p>Solving for M :<\/p>\n<p>[latex]\\sum M_C=0= -( V_2\\cdot(0.04) -M_2\\\\M_2 = 15\\cdot 0.04\\\\M=0.6N[\/latex]<\/p>\n<p>V-M diagram:<\/p>\n<p>For the V-x graph, from A to B, there is a constant shear of 15 N, then from B to C, the shear has the same magnitude but a different direction.\u00a0 To plot the M-x graph, substitute x in the equation for M and M<sub>2<\/sub> to find the peak bending moment. Also, because M is the integral of shear force, horizontal plots in shear correspond to a linear plot in the M-x graph.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/V-m-chapter-6.jpg\" alt=\"Shear and moment diagrams\" class=\"alignnone wp-image-1824\" width=\"431\" height=\"823\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/V-m-chapter-6.jpg 511w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/V-m-chapter-6-157x300.jpg 157w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/V-m-chapter-6-65x124.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/V-m-chapter-6-225x430.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/V-m-chapter-6-350x668.jpg 350w\" sizes=\"auto, (max-width: 431px) 100vw, 431px\" \/><\/p>\n<p><strong>6. Review\u00a0<\/strong><\/p>\n<p>The answers make sense. The moment, shear, and reaction forces are equal in both segments of the beam. This is as expected because of the symmetry of the beam.<\/p>\n<\/div>\n<h1>Example 6.3.4: V\/M Diagrams, Submitted by Michael Oppong-Ampomah<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">Roughly sketch the shear and moment diagram for the following structure. Measure from point A.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Michael1-300x133.png\" alt=\"A cantilever beam with distributed load at the free end.\" class=\"alignnone wp-image-1768 size-medium\" width=\"300\" height=\"133\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Michael1-300x133.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Michael1-65x29.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Michael1-225x100.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Michael1-350x155.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Michael1.png 685w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/div>\n<p><strong>2. Sketch<\/strong><\/p>\n<p>N\/A, Sketch Provided.<\/p>\n<p><strong>3. Knows and Unknowns:<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>Distributed Load: w<\/li>\n<li>Dimensions and shown<\/li>\n<\/ul>\n<p>Unknown:<\/p>\n<ul>\n<li>Shear\/Moment Diagram<\/li>\n<\/ul>\n<p><strong>4. Approach:<\/strong><\/p>\n<p>Break the beam into sections as forces change.<\/p>\n<p>Determine the plotted shape of the shear and moment of each section.<\/p>\n<p>Draw the V\/M diagrams using these shape diagrams.<\/p>\n<p><strong>5. Analysis:<\/strong><\/p>\n<p>A distributed load creates a linear section across it, with the magnitude of the shear force increasing across the load. The maximum shear will be at the end of the distributed load. With no other forces being applied except for reactions, section B-C should be constant. Since the beam is static, the reactions must bring the shear back to zero.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Michael2-300x133.png\" alt=\"Shear diagram of the problem.\" class=\"alignnone wp-image-1769 size-medium\" width=\"300\" height=\"133\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Michael2-300x133.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Michael2-65x29.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Michael2-225x100.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Michael2-350x155.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Michael2.png 652w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>The moment can be drawn using the knowledge that dM = V. Section A-B in the shear diagram is linear; integrating a linear function gives a quadratic function, thus, section A-B in the moment diagram must be curved. Section B-C is constant in shear. Integrating a constant function gives a linear function; therefore, the moment across this section must be linear, and since the shear is negative, the slope of the moment must be negative. Finally, because the beam is static, the moment must return to zero due to the reaction forces.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Michael3-300x173.png\" alt=\"Moment diagram of the problem.\" class=\"alignnone wp-image-1770 size-medium\" width=\"300\" height=\"173\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Michael3-300x173.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Michael3-65x37.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Michael3-225x130.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Michael3-350x201.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Michael3.png 608w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>Using principles of shear force (V) and moments and their integral\/derivative relationship, the shear and moment diagrams make sense for their general shape.<\/p>\n<\/div>\n<h1>Example 6.3.5: V\/M Diagrams, Submitted by William Craine<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">Find the reaction forces and create the shear and moment diagrams for the beam.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William1-4-1024x629.png\" alt=\"A cantilever beam with triangular distributed load at its free end.\" class=\"alignnone wp-image-1780\" width=\"544\" height=\"334\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William1-4-1024x629.png 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William1-4-300x184.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William1-4-768x472.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William1-4-1536x944.png 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William1-4-2048x1258.png 2048w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William1-4-65x40.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William1-4-225x138.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William1-4-350x215.png 350w\" sizes=\"auto, (max-width: 544px) 100vw, 544px\" \/><\/div>\n<p><strong>2. Draw<\/strong><\/p>\n<p>[latex]4N\/m \\cdot \\frac{5.5m}{2} = 11N[\/latex]<\/p>\n<p>[latex]\\bar{x} = 1\/3 \\cdot 5.5m = 1.83m[\/latex]<\/p>\n<p>[latex]\\bar{x} \\ from\u00a0 A = 4m + 4.5m + 1.83m = 10.3m[\/latex]<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William2-1-1024x545.png\" alt=\"A FBD of the problem.\" class=\"alignnone wp-image-1781\" width=\"597\" height=\"318\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William2-1-1024x545.png 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William2-1-300x160.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William2-1-768x409.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William2-1-1536x817.png 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William2-1-2048x1090.png 2048w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William2-1-65x35.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William2-1-225x120.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William2-1-350x186.png 350w\" sizes=\"auto, (max-width: 597px) 100vw, 597px\" \/><br \/>\n<strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>Applied Forces:<\/li>\n<li>F<sub>B<\/sub> = -20N<\/li>\n<li>F<sub>C<\/sub> = -400N<\/li>\n<li>F<sub>D<\/sub> = 500N<\/li>\n<li>F<sub>Load<\/sub> = -4N\/m<\/li>\n<li>Dimensions as shown<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>Reaction Forces: R<sub>AX, <\/sub>R<sub>AY, <\/sub>M<sub>A<\/sub><\/li>\n<li>Shear\/Moment Diagram<\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<ol>\n<li>Use the equilibrium equation to find the reaction forces<\/li>\n<li>Use beam analysis\/cutting between forces to find equations for the shear\/moment diagram.<\/li>\n<\/ol>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>First, use the sum of forces and the sum of moments to determine R<sub>Ay<\/sub> and M<sub>A<\/sub>.<\/p>\n<p>[latex]\\sum F_x = R_{Ax} = 0[\/latex]<\/p>\n<p>[latex]\\sum F_y = -R_{Ay} - 20N - 400N - 11N +500N = 0[\/latex]<\/p>\n<p>[latex]R_{Ay} = 69N[\/latex]<\/p>\n<p>[latex]\\sum M_A = - M_A - (20N \\cdot 4m) - (400N \\cdot 8.5m) - (11N \\cdot 10.3m) + (500 N \\cdot 14m) = 0[\/latex]<\/p>\n<p>[latex]M_A = 3406.7Nm[\/latex]<\/p>\n<p>Next, cut the beam between A and B and analyze the internal forces.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William3-1024x422.png\" alt=\"A sketch of the internal forces from A to B.\" class=\"alignnone wp-image-1788\" width=\"469\" height=\"193\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William3-1024x422.png 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William3-300x124.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William3-768x317.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William3-1536x633.png 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William3-2048x844.png 2048w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William3-65x27.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William3-225x93.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William3-350x144.png 350w\" sizes=\"auto, (max-width: 469px) 100vw, 469px\" \/><\/p>\n<p>[latex]\\sum F_x = N + 0 = 0 \\\\ N = 0[\/latex]<\/p>\n<p>[latex]\\sum F_y = -69N -V = 0 \\\\ V= -69N[\/latex]<\/p>\n<p>[latex]\\sum M_{cut} =\u00a0 M + (69N \\cdot x) - (3406.7Nm) = 0 \\\\ M = -69x + 3406.7[\/latex]<\/p>\n<p>Repeat the analysis of internal forces from A to C.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William4-1024x367.png\" alt=\"A sketch of the internal forces from A to C.\" class=\"alignnone wp-image-1792 size-large\" width=\"1024\" height=\"367\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William4-1024x367.png 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William4-300x107.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William4-768x275.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William4-1536x550.png 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William4-2048x734.png 2048w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William4-65x23.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William4-225x81.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William4-350x125.png 350w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/p>\n<p>[latex]\\sum F_x = N + 0 = 0 \\\\ N = 0[\/latex]<\/p>\n<p>[latex]\\sum F_y = -69N - 20N -V = 0 \\\\ V = -89N[\/latex]<\/p>\n<p>[latex]\\sum M_{cut} = M + (69N \\cdot x) + (20N \\cdot{(x-4m)}) - 3406.7Nm = 0 \\\\ M+89x-3326.7Nm = 0 \\\\ M = -89x + 3326.7[\/latex]<\/p>\n<p>Finally,\u00a0 analyze the beam in the area from C to D. Ensure that the value of X used is now taken from point D.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William5-1024x792.png\" alt=\"A sketch of the internal forces from C to D.\" class=\"alignnone wp-image-1796\" width=\"442\" height=\"342\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William5-1024x792.png 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William5-300x232.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William5-768x594.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William5-1536x1188.png 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William5-65x50.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William5-225x174.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William5-350x271.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William5.png 1851w\" sizes=\"auto, (max-width: 442px) 100vw, 442px\" \/><\/p>\n<p>[latex]\\sum F_x = -N = 0 \\\\ N = 0[\/latex]<\/p>\n<p>Since the section does not reach the end of the distributed load, the slope of the load must be calculated:<\/p>\n<p>[latex]\\frac{4}{5.5} = \\frac{m}{14-x} \\\\ m = \\frac{56-4x}{5.5}[\/latex]<\/p>\n<p>This can be used to find the equivalent point load in terms of x.<\/p>\n<p>[latex]F_{eq} = \\frac{\\frac{56-4x}{5.5} \\cdot (14-x)}{2} \\\\ F_{eq} = \\frac{4x^2}{11} - \\frac{112x}{11} + \\frac{784}{11}[\/latex]<\/p>\n<p>This point load can be used when summing forces to find v.<\/p>\n<p>[latex]\\sum F_y = 0 = V - F_{eq} + 500 \\\\ V = F_{eq} - 500 \\\\ V = \\frac{4x^2}{11} - \\frac{112x}{11} + \\frac{784}{11} - 500 \\\\ V = \\frac{4x^2}{11} - \\frac{112x}{11} - 428.7[\/latex]<\/p>\n<p>To save time, rather than use the sum of moments to determine the equation for moments in this section, the moment can be found by the integral of V.<\/p>\n<p>[latex]M = \\int \\frac{4x^2}{11} - \\frac{112x}{11} - 428.7 dx\u00a0 \\\\ M = \\frac{4x^3}{33} - \\frac{112x^2}{22} - 428.7x +C[\/latex]<\/p>\n<p>We know that for this beam to be static, the moment at x=14m must be 0. This can be used to determine C.<\/p>\n<p>[latex]M(14) = 0 =\u00a0 \\frac{4(14)^3}{33} - \\frac{112(14)^2}{22} - 428.7(14) +C \\\\ C = 6667 \\\\ Therefore: M_{cd}(x) = \\frac{4x^3}{33} - \\frac{112x^2}{22} - 428.7x + 6667[\/latex]<\/p>\n<p>With the values for internal forces found between each change in force, the internal forces can be plotted.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William6.PNG.png\" alt=\"A sketch of the shear and moment diagrams.\" class=\"alignnone wp-image-1991\" width=\"614\" height=\"915\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William6.PNG.png 3706w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William6.PNG-201x300.png 201w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William6.PNG-687x1024.png 687w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William6.PNG-768x1145.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William6.PNG-1031x1536.png 1031w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William6.PNG-1374x2048.png 1374w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William6.PNG-65x97.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William6.PNG-225x335.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/William6.PNG-350x522.png 350w\" sizes=\"auto, (max-width: 614px) 100vw, 614px\" \/><\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>Using principles of shear force (V) and moments and their integral\/derivative relationship, the shear and moment diagrams make sense for their general shape. The numerical values calculated align with the variables provided.<\/p>\n<\/div>\n<h1>Example 6.3.6: Internal Forces, Submitted by Riley Fitzpatrick<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div>\n<div class=\"textbox shaded\">\n<div>A person who weighs 140lbs is standing on the edge of a 10ft long diving board. The board is secured by a pin at 0ft and supported by a roller at 4ft.<\/div>\n<div><\/div>\n<div>a. Calculate the reaction forces on the board.<\/div>\n<div>b. Calculate the internal forces at the midpoint between A and B.<\/div>\n<div>c. Calculate the internal forces at the midpoint between B and C.<\/div>\n<div>d. Draw the shear and moment diagram.<\/div>\n<div><\/div>\n<div><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Riley-chap-6-e1654109214997.jpg\" alt=\"A person at the edge of a diving board.\" class=\"alignnone wp-image-1958 size-full\" width=\"391\" height=\"686\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Riley-chap-6-e1654109214997.jpg 391w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Riley-chap-6-e1654109214997-171x300.jpg 171w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Riley-chap-6-e1654109214997-65x114.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Riley-chap-6-e1654109214997-225x395.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Riley-chap-6-e1654109214997-350x614.jpg 350w\" sizes=\"auto, (max-width: 391px) 100vw, 391px\" \/><\/div>\n<div>source: https:\/\/www.flickr.com\/photos\/lac-bac\/40229806264<\/div>\n<div><\/div>\n<div><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-095515-1-300x131.png\" alt=\"A sketch of the problem.\" width=\"470\" height=\"205\" class=\"alignnone wp-image-2500\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-095515-1-300x131.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-095515-1-65x28.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-095515-1-225x98.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-095515-1-350x153.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-095515-1.png 669w\" sizes=\"auto, (max-width: 470px) 100vw, 470px\" \/><\/div>\n<div><\/div>\n<\/div>\n<p><strong>2. Draw\u00a0<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-100645-300x148.png\" alt=\"A FBD of the problem.\" class=\"alignnone wp-image-2502\" width=\"475\" height=\"234\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-100645-300x148.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-100645-65x32.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-100645-225x111.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-100645-350x173.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-100645.png 691w\" sizes=\"auto, (max-width: 475px) 100vw, 475px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns\u00a0<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>F<sub>g<\/sub> = 140 lbs<\/li>\n<li>X<sub>A<\/sub>= 0 ft<\/li>\n<li>X<sub>B<\/sub>= 4 ft<\/li>\n<li>X<sub>C<\/sub>= 10 ft<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>R<sub>Ax<\/sub><\/li>\n<li>R<sub>Bx<\/sub><\/li>\n<li>R<sub>Ay<\/sub><\/li>\n<li>Internal forces N, V, M between B and C, &amp; A and B<\/li>\n<\/ul>\n<p><strong>4. Approach\u00a0<\/strong><\/p>\n<p>Calculate the reaction forces using equilibrium equations. Cut the diving board between A &amp; B, and use equilibrium equations to find internal forces. Do the same to find internal forces between B and C. Use the values to draw the shear and moment diagram.<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>a.<\/p>\n<\/div>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-101136-300x143.png\" alt=\"A FBD of the problem.\" class=\"alignnone wp-image-2505\" width=\"445\" height=\"212\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-101136-300x143.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-101136-65x31.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-101136-225x107.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-101136-350x167.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-101136.png 683w\" sizes=\"auto, (max-width: 445px) 100vw, 445px\" \/><\/p>\n<p>[latex]\\sum{F_{x}} = 0=R_{Ax}[\/latex]<\/p>\n<p>[latex]\\sum{F_{y}}=0=R_{Ay}+R_{By}-140 lb \\\\R_{Ay}+R_{By}=140 lb[\/latex]<\/p>\n<p>[latex]\\sum{M_{A}}=\\left(X_{B} \\cdot R_{By}\\right)-\\left(X_{c} \\cdot F_{g}\\right)=0[\/latex]<\/p>\n<p>[latex]R_{By}= \\frac{X_{c}\\cdot F_{g}}{X_{B}}\\\\R_{By}= \\frac{10 ft\\cdot 140 lb}{4 ft}[\/latex]<\/p>\n<p>[latex]R_{By}=350 lb[\/latex]<\/p>\n<p>now, [latex]R_{Ay}=140 lb - R_{By}\\\\ \\kern 1pc=140 lb - 350 lb \\\\ \\kern 1pc =-210 lb[\/latex]<\/p>\n<p>The negative sign for [latex]R_{Ay}[\/latex] means that the direction of [latex]R_{Ay}[\/latex]\u00a0 is not +y but in the -y direction.<\/p>\n<p>Thus, the reation forces are [latex]R_{Ay}=-210 lb, R_{By} = 350 lb, R_{Ax}=0[\/latex]<\/p>\n<p>b. <span style=\"text-decoration: underline\">Finding internal forces between A and B\u00a0<\/span><\/p>\n<p>To find the internal forces between A and B, cut in the middle of A and B, giving [latex]x = \\frac{X_{B}}{2} =2 ft[\/latex]<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-102906-300x135.png\" alt=\"A sketch of the internal forces between A and B\" class=\"alignnone wp-image-2507 size-medium\" width=\"300\" height=\"135\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-102906-300x135.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-102906-65x29.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-102906-225x101.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-102906-350x157.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-102906.png 688w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>[latex]\\sum{F_{x}} = 0=R_{Ax} + N\\\\ N = 0[\/latex]<\/p>\n<p>[latex]\\sum{F_{y}}=0=-V - R_{Ay}\\\\ V = -R_{Ay}\\\\V=-210 lb[\/latex]<\/p>\n<p>[latex]\\sum{M_{A}}=0\\\\ \\kern 2pc =M-xV\\\\ \\kern 2pc = M- \\frac{4 ft}{2}\\left(-210 lb\\right)\\\\\u00a0 M =-420 lb ft[\/latex]<\/p>\n<p>Internal forces between A and B are [latex]N = 0, V= -210 lb, M=-420 lb ft[\/latex]<\/p>\n<p>c.\u00a0<span style=\"text-decoration: underline\"> Finding internal forces between B and C<\/span><\/p>\n<p>Cut at the midpoint between B and C to find the internal forces.<\/p>\n<p><span style=\"text-decoration: underline\">Finding x<\/span><\/p>\n<p>To apply the equilibrium equation, you need to find x in this scenario.<\/p>\n<p>[latex]x= X_{B}+\\frac{X_{c}-X_{B}}{2}\\\\ = 4 ft +\\frac{10 ft -4 ft}{2}\\\\ x =7 ft[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-103051-300x142.png\" alt=\"A sketch of the internal forces between B and C.\" class=\"alignnone wp-image-2508 size-medium\" width=\"300\" height=\"142\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-103051-300x142.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-103051-65x31.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-103051-225x106.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-103051-350x166.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-07-08-103051.png 691w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>[latex]\\sum{F_{x}} = 0=R_{Ax} + N\\\\ N = 0[\/latex]<\/p>\n<p>[latex]\\sum{F_{y}}=0=R_{By}-R_{Ay}-V\\\\ \\kern 1pc V=R_{By} -R_{Ay} \\\\ \\kern 1pc V = 350 lb -210 lb\\\\ V= 140 lb[\/latex]<\/p>\n<p>now,<\/p>\n<p>[latex]\\sum{M_{A}}=0[\/latex]<\/p>\n<p>[latex]M =- \\left( X_{B} \\cdot R_{By} \\right) + xV\\\\ \\kern 1 pc = -\\left(4 ft \\cdot 350 lb\\right) +\\left( 7 ft \\cdot140 lb\\right) \\\\ \\kern 1 pc = 980 lb ft- 1400 lb ft[\/latex]<\/p>\n<p>[latex]M =-420 lb ft[\/latex]<\/p>\n<p>Internal forces between C and B are [latex]N = 0, V= 140 lb, M=-420 lb ft[\/latex]<\/p>\n<p>d. <span style=\"text-decoration: underline\">Shear and Moment diagram\u00a0<\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/riley-chap-6-shear-and-moment-1.jpg\" alt=\"A sketch of the shear and moment diagrams.\" class=\"alignnone wp-image-1957\" width=\"572\" height=\"721\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/riley-chap-6-shear-and-moment-1.jpg 626w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/riley-chap-6-shear-and-moment-1-238x300.jpg 238w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/riley-chap-6-shear-and-moment-1-65x82.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/riley-chap-6-shear-and-moment-1-225x284.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/riley-chap-6-shear-and-moment-1-350x441.jpg 350w\" sizes=\"auto, (max-width: 572px) 100vw, 572px\" \/><\/p>\n<p><strong>6. Review\u00a0<\/strong><\/p>\n<p>There is no force in the x direction; therefore, it makes sense that the reaction force, R<sub>Ax, <\/sub>is zero. Because the moment at A is generated by the person&#8217;s weight, R<sub>By\u00a0<\/sub>should be considerably large. As R<sub>By<\/sub> is larger than the person&#8217;s weight, R<sub>Ay\u00a0<\/sub>should act downwards and would be of the same magnitude as the difference between R<sub>By<\/sub> and the person&#8217;s weight\u00a0to make the system stable.<\/p>\n<p>For section b, the shear force between A and B is equal and opposite to the reaction force R<sub>Ay.\u00a0<\/sub>Similarly, with the concept of equilibrium, shear is equal and opposite to R<sub>Ax,<\/sub> and the moment in this section is the negative of the moment generated at A.<\/p>\n<p>It also makes sense that the shear between B and C is equal to the difference between R<sub>By\u00a0<\/sub>and R<sub>Ay, <\/sub>and as expected, acting downward, thus it equals the weight of the person.<\/p>\n<\/div>\n<h1>Example 6.3.7: Internal Forces and Shear\/Moment Diagrams, Submitted by Odegua Obehi-Arhebun<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>A bolt attached to a stair rail is to be removed by a technician. A solid wrench of length 30cm, arched horizontally to the bolt and assumed to be rigidly connected. The technician applies a downward vertical force of 200N at the free end of the wrench. Determine the internal normal force, shear force and bending moment at a point 0.15m along the wrench. Draw the full shear and moment diagram.<\/p>\n<figure id=\"attachment_2539\" aria-describedby=\"caption-attachment-2539\" style=\"width: 300px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/bolt-and-wrench-isolated-on-white-3d-rendering-P89T2R-300x215.jpg\" alt=\"A wrench unscrewing a bolt.\" class=\"wp-image-2539 size-medium\" width=\"300\" height=\"215\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/bolt-and-wrench-isolated-on-white-3d-rendering-P89T2R-300x215.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/bolt-and-wrench-isolated-on-white-3d-rendering-P89T2R-1024x734.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/bolt-and-wrench-isolated-on-white-3d-rendering-P89T2R-768x551.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/bolt-and-wrench-isolated-on-white-3d-rendering-P89T2R-65x47.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/bolt-and-wrench-isolated-on-white-3d-rendering-P89T2R-225x161.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/bolt-and-wrench-isolated-on-white-3d-rendering-P89T2R-350x251.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/bolt-and-wrench-isolated-on-white-3d-rendering-P89T2R.jpg 1300w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><figcaption id=\"caption-attachment-2539\" class=\"wp-caption-text\">Source: https:\/\/www.alamy.com\/bolt-and-wrench-isolated-on-white-3d-rendering-image211658127.html<\/figcaption><\/figure>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-122411-300x216.png\" alt=\"A sketch of the problem.\" width=\"300\" height=\"216\" class=\"alignnone wp-image-2544 size-medium\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-122411-300x216.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-122411-65x47.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-122411-225x162.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-122411-350x252.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-122411.png 625w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<\/div>\n<p><strong>2. Draw<\/strong><\/p>\n<p>Free-body diagram:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-123303-300x193.png\" alt=\"A FBD of the problem.\" class=\"alignnone wp-image-2545\" width=\"354\" height=\"228\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-123303-300x193.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-123303-65x42.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-123303-225x145.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-123303-350x225.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-123303.png 652w\" sizes=\"auto, (max-width: 354px) 100vw, 354px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>F=200N<\/li>\n<li>L= 30cm =&gt; 0.3m<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>Internal normal force, shear force and bending moment at point B<\/li>\n<li>Shear + moment diagram<\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>Solve for reaction forces at A, then take arbitrary sections between A and B and between B and C to derive V and M equations. Plot results.<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>Solve for reaction forces:<\/p>\n<p>[latex]\\begin{aligned}  \\sum M_{A} &= 0 = -200\\,\\mathrm{N} \\cdot 0.3\\,\\mathrm{m} - M_A \\\\  M_A &= -60\\,\\mathrm{Nm}  \\end{aligned}[\/latex]<\/p>\n<p>[latex]\\begin{aligned}  \\sum F_{y} &= 0 = A_y - 200\\,\\mathrm{N} \\\\  A_y &= 200\\,\\mathrm{N}  \\end{aligned}[\/latex]<\/p>\n<p>[latex]\\begin{aligned}  \\sum F_{x} &= 0 = N \\\\  N &= 0  \\end{aligned}[\/latex]<\/p>\n<p>No force in the x-direction<\/p>\n<p>Cut 1: With reaction forces known, cut a section between A and B<\/p>\n<p>0m&lt;x&lt;0.15m<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-133112-300x195.png\" alt=\"A sketch showing the internal forces between 0m to 0.15m.\" width=\"300\" height=\"195\" class=\"alignnone wp-image-2554 size-medium\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-133112-300x195.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-133112-65x42.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-133112-225x146.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-133112-350x227.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-133112.png 631w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>[latex]\\begin{aligned}  \\sum F_{y} &= 0 = A_y - V \\\\  V &= A_y = 200\\,\\mathrm{N}  \\end{aligned}[\/latex]<\/p>\n<p>[latex]\\begin{aligned}  \\sum M_x &= 0 = -M_A - A_y \\cdot x +M(x) \\\\  M(x) &= 200x-60\\,\\mathrm{Nm}  \\end{aligned}[\/latex]<\/p>\n<p>You can also cut the section between B and C<\/p>\n<p>0.15m&lt;x&lt;0.3m<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-134813-300x191.png\" alt=\"A sketch showing the internal forces between 0.15m to 0.3m.\" class=\"alignnone wp-image-2557\" width=\"352\" height=\"224\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-134813-300x191.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-134813-65x41.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-134813-225x143.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-134813-350x223.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-134813.png 632w\" sizes=\"auto, (max-width: 352px) 100vw, 352px\" \/><\/p>\n<p>[latex]\\begin{aligned}  \\sum F_{y} &= 0 = A_y - V \\\\  V &= A_y = 200\\,\\mathrm{N}  \\end{aligned}[\/latex]<\/p>\n<p>[latex]\\begin{aligned}  \\sum M_x &= 0 = M_x - M_A- A_y \\cdot x \\\\  M(x) &= 200x - 60\\,\\mathrm{Nm}  \\end{aligned}[\/latex]<\/p>\n<p>[latex]\\begin{aligned}  \\text{At } x = 0, \\quad &M = 0 \\\\  \\text{At } x = 0.15\\,\\mathrm{m}, \\quad &M = -30\\,\\mathrm{Nm} \\\\  \\text{At } x = 0.3\\,\\mathrm{m}, \\quad &M = 0  \\end{aligned}[\/latex]<\/p>\n<p>Answer: N = 0, V = 200 N, M<sub>x<\/sub> = 200x-60 Nm<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-140735-208x300.png\" alt=\"A sketch of the shear and moment diagrams\" class=\"alignnone wp-image-2561\" width=\"306\" height=\"441\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-140735-208x300.png 208w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-140735-65x94.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-140735-225x324.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-140735-350x504.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-140735.png 644w\" sizes=\"auto, (max-width: 306px) 100vw, 306px\" \/><\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>Negative sign for M<sub>A <\/sub>indicates that it should be in the other direction, as shown in the final diagram above. The shear force remains constant at 200N from A to C, then drops to zero at the end where the load is applied. The shear force is constant, hence it makes sense that the moment is linear.<\/p>\n<\/div>\n<h1>Example 6.3.8: Internal Forces, Submitted by Celina Areoye<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>A Uniform rectangular light frame of mass 3kg and length 0.8m is suspended from the ceiling by two vertical rods. The rods are 0.2m apart and connected symmetrically to the ceiling plate. Also, there are 5 evenly spaced identical bulbs, each 0.5kg each mounted on the frame. Determine the tension in each vertical rod and the internal forces at the points where the vertical rod is connected to the frame.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture2-300x275.png\" alt=\"A rectangular frame hanging from the ceiling.\" class=\"alignnone wp-image-2564 size-medium\" width=\"300\" height=\"275\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture2-300x275.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture2-1024x940.png 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture2-768x705.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture2-65x60.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture2-225x207.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture2-350x321.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture2.png 1135w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<\/div>\n<p><strong>2. Draw<\/strong><\/p>\n<p>Free-body diagram:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture3-300x186.png\" alt=\"A FBD of the problem.\" class=\"alignnone wp-image-2565\" width=\"405\" height=\"251\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture3-300x186.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture3-1024x634.png 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture3-768x475.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture3-65x40.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture3-225x139.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture3-350x217.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture3.png 1354w\" sizes=\"auto, (max-width: 405px) 100vw, 405px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>Mass of frame (\ud835\udc40\ud835\udc53) = 3kg<\/li>\n<li>Length of frame (\ud835\udc3f\ud835\udc53) = 0.8m<\/li>\n<li>Mass of each lightbulb (\ud835\udc40\ud835\udc4f) = 0.5kg<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>Tension in each rod (\ud835\udc47<sub>1<\/sub> \ud835\udc4e\ud835\udc5b\ud835\udc51 \ud835\udc47<sub>2<\/sub>)<\/li>\n<li>Internal forces at point D (\ud835\udc41<sub>D<\/sub>, \ud835\udc49<sub>D<\/sub>, \ud835\udc4e\ud835\udc5b\ud835\udc51 \ud835\udc40<sub>D<\/sub>)<\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>Assuming the vertical rods are positioned centrally between the bulbs, and the internal forces at points C and D are the same (use any point to get the internal forces). First, find the tension and then use it in calculating internal forces.<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>Weight of frame (\ud835\udc4a\ud835\udc53) = \ud835\udc40\ud835\udc53\ud835\udc54 = 3kg \u00d7 9.81m\/s<sup>2<\/sup> = 29.43N<\/p>\n<p>Weight of one bulb (\ud835\udc4a\ud835\udc4f) = \ud835\udc40\ud835\udc4f\ud835\udc54 = 0.5kg \u00d7 9.81m\/s<sup>2<\/sup> = 4.905N<\/p>\n<p>The bulbs are evenly spaced, so the distance between 2 bulbs (from centre to centre):<\/p>\n<p>[latex]\\frac{0.8\\,\\mathrm{m}}{4} = 0.2\\,\\mathrm{m}[\/latex]<\/p>\n<p>To get the tension in each rod:<\/p>\n<p>[latex]+\\uparrow \\sum F_y = 0 \\quad ; \\quad T_1 + T_2 - 5W_b - W_f = 0[\/latex]<\/p>\n<p>Since the rods are connected symmetrically, they are equal (\ud835\udc47<sub>1<\/sub> = \ud835\udc47<sub>2<\/sub> ), therefore:<\/p>\n<div>[latex]2T - 5(4.905\\,\\mathrm{N}) - 29.43\\,\\mathrm{N} = 0[\/latex]<\/div>\n<div><\/div>\n<div>[latex]2T = 53.955\\,\\mathrm{N}, \\quad T = 26.98\\,\\mathrm{N}[\/latex]<\/div>\n<div>The tension in each vertical rod is <strong>26.98N.<\/strong><\/div>\n<p>To get the internal forces at point D, taking the right side of the diagram:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture4-300x208.png\" alt=\"A sketch showing the internal forces at point D (right side).\" width=\"338\" height=\"234\" class=\"alignnone wp-image-2567\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture4-300x208.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture4-1024x710.png 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture4-768x533.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture4-65x45.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture4-225x156.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture4-350x243.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture4.png 1195w\" sizes=\"auto, (max-width: 338px) 100vw, 338px\" \/><\/p>\n<p>Since the rods are positioned centrally between the bulbs, the distance from point D to the bulb is<\/p>\n<p>[latex]\\frac{0.2\\,\\mathrm{m}}{2} = 0.1\\,\\mathrm{m}[\/latex]<\/p>\n<div>[latex]+ \\rightarrow \\sum F_x = 0, \\quad N_D = 0\\,\\mathrm{N}[\/latex]<\/div>\n<div><\/div>\n<div>[latex]+ \\uparrow \\sum F_y = 0; \\quad V_D + T_2 - 2W_b = 0[\/latex]<\/div>\n<div><\/div>\n<div>[latex]V_D = 2(4.905\\,\\mathrm{N}) - 26.98\\,\\mathrm{N} = -17.17\\,\\mathrm{N}[\/latex]<\/div>\n<div><\/div>\n<div>[latex]\\text{Therefore, } V_D = 17.17\\,\\mathrm{N} \\text{ (opposite direction)} \\downarrow[\/latex]<\/div>\n<div><\/div>\n<div>[latex]+ \\text{ anticlockwise } \\sum M_D = 0; \\quad - (4.905\\,\\mathrm{N} \\times 0.3\\,\\mathrm{m}) - (4.905\\,\\mathrm{N} \\times 0.1\\,\\mathrm{m}) - M_D = 0[\/latex]<\/div>\n<div><\/div>\n<div>[latex]M_D = -1.962\\,\\mathrm{Nm}[\/latex]<\/div>\n<div><\/div>\n<div>[latex]\\text{Therefore, } M_D = 1.962\\,\\mathrm{Nm} \\text{ (clockwise direction)}[\/latex]<\/div>\n<div><\/div>\n<p>No force in the x-direction<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>To check, taking the left side of the diagram<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture5-300x229.png\" alt=\"A sketch showing the internal forces at point D (left side).\" width=\"300\" height=\"229\" class=\"alignnone wp-image-2568 size-medium\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture5-300x229.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture5-1024x783.png 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture5-768x587.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture5-65x50.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture5-225x172.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture5-350x268.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture5.png 1364w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<div>[latex]+ \\uparrow \\sum F_y = 0; \\quad V_D + T_1 - 3W_b - W_f = 0[\/latex]<\/div>\n<div><\/div>\n<div>[latex]17.17\\,\\mathrm{N} + 26.98\\,\\mathrm{N} - 3(4.905\\,\\mathrm{N}) - 29.43\\,\\mathrm{N} = 0[\/latex]<\/div>\n<\/div>\n<div>\n<h1>Example 6.3.9: Internal Forces, Submitted by Okino Itopa Farooq<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>A uniform horizontal shelf of total length 100 cm and mass 2 kg is supported at both ends by pin connections. The shelf supports a group of 5 books placed together, each with a thickness of 4 cm and a mass of 0.5 kg. The collection of books is positioned such that it is located at a space 50 cm from the left end and 30 cm from the right end of the shelf. Assuming static equilibrium and neglecting the thickness of the shelf, determine the following at a point located 60 cm from the left end of the shelf:<\/p>\n<ol>\n<li>The shear force<\/li>\n<li>The normal (axial) force<\/li>\n<li>The bending moment<\/li>\n<\/ol>\n<figure id=\"attachment_2577\" aria-describedby=\"caption-attachment-2577\" style=\"width: 388px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/StockSnap_LGFGZZC671-300x201.jpg\" alt=\"Books on a shelf.\" class=\"wp-image-2577\" width=\"388\" height=\"260\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/StockSnap_LGFGZZC671-300x201.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/StockSnap_LGFGZZC671-1024x687.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/StockSnap_LGFGZZC671-768x516.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/StockSnap_LGFGZZC671-1536x1031.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/StockSnap_LGFGZZC671-2048x1375.jpg 2048w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/StockSnap_LGFGZZC671-65x44.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/StockSnap_LGFGZZC671-225x151.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/StockSnap_LGFGZZC671-350x235.jpg 350w\" sizes=\"auto, (max-width: 388px) 100vw, 388px\" \/><figcaption id=\"caption-attachment-2577\" class=\"wp-caption-text\">Source: https:\/\/stocksnap.io\/photo\/books-shelf-LGFGZZC671<\/figcaption><\/figure>\n<\/div>\n<p><strong>2. Draw<\/strong><\/p>\n<p>Free-body diagram:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-191320-300x233.png\" alt=\"A FBD of the problem.\" width=\"411\" height=\"319\" class=\"alignnone wp-image-2609\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-191320-300x233.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-191320-65x50.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-191320-225x175.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-191320-350x272.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-191320.png 733w\" sizes=\"auto, (max-width: 411px) 100vw, 411px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>Total length of shelf = 100cm = 1m<\/li>\n<li>Mass of shelf = 2kg<\/li>\n<li>Thickness of each book = 4cm = 0.04m<\/li>\n<li>Mass of each book = 0.5kg<\/li>\n<li>g = 9.81 m\/s<sup>2<\/sup><\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>The shear force, normal force, and bending moment at 60cm (0.6m) from the left end of the shelf<\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>First, resolve the distributed loads, then find reaction forces, and then calculate the internal forces<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>Finding the force acting per m covered by the books<\/p>\n<p>[latex]\\left( \\frac{0.5}{0.04} \\right) \\, \\text{kg\/m} = \\left( \\frac{0.5}{0.04} \\right) \\, \\text{kg\/m} \\times 9.81 = 122.625 \\, \\text{N\/m}[\/latex]<\/p>\n<p>Weight of the shelf<\/p>\n<p>[latex]2 \\, \\text{kg} \\times 9.81 \\, \\text{m\/s}^2 = 19.62 \\, \\text{N}[\/latex]<\/p>\n<p>We first resolve the distributed load<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-193706-300x201.png\" alt=\"A sketch shown resolving the distributed load.\" width=\"300\" height=\"201\" class=\"alignnone wp-image-2610 size-medium\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-193706-300x201.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-193706-65x44.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-193706-225x151.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-193706-350x235.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-193706.png 666w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>[latex]C = 122.625 \\, \\text{N\/m} \\times 0.2 \\, \\text{m} = 24.525 \\, \\text{N}[\/latex]<br \/>\n[latex]X = \\frac{0.2}{2} = 0.1 \\, \\text{m} \\, \\text{(0.6 m from A)}[\/latex]<\/p>\n<p>The new FBD should look like this:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-194707-300x234.png\" alt=\"Updated FBD, including found values.\" width=\"379\" height=\"296\" class=\"alignnone wp-image-2611\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-194707-300x234.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-194707-65x51.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-194707-225x175.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-194707-350x272.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-194707.png 749w\" sizes=\"auto, (max-width: 379px) 100vw, 379px\" \/><\/p>\n<p>Using equilibrium equations to find the reaction forces<\/p>\n<div>[latex]\\sum F_x = 0 \\Rightarrow A_x = B_x = 0[\/latex]<\/div>\n<div>[latex]\\sum F_y = 0 \\Rightarrow A_y + B_y = 19.62 + 24.525[\/latex]<\/div>\n<div>[latex]A_y + B_y = 44.145 \\, \\text{N} \\quad \\cdots (i)[\/latex]<\/div>\n<div><\/div>\n<div>[latex]\\sum M_A^+ = 0 \\Rightarrow B_y(0.1) - 24.525(0.6) - 19.62(0.5) = 0[\/latex]<\/div>\n<div>[latex]B_y = \\frac{24.525}{0.1} = 245.25 \\, \\text{N}[\/latex]<\/div>\n<div><\/div>\n<div>[latex]\\text{Substitute in (i):}[\/latex]<\/div>\n<div>[latex]A_y = 44.145 \\, \\text{N} - 245.25 \\, \\text{N} = -201.105 \\, \\text{N}[\/latex]<\/div>\n<div><\/div>\n<div>A<sub>y <\/sub>= -201.105N, B<sub>y<\/sub> = 245.25N, A<sub>x\u00a0<\/sub>= B<sub>x\u00a0<\/sub>= 0<\/div>\n<div><\/div>\n<p>Now that we know the Components A<sub>y<\/sub> and B<sub>y<\/sub>, we can start calculating the Shear, Normal and Bending moment at the point 0.6m from A, which will cut into the distributed load<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-201715-300x221.png\" alt=\"A sketch showing the internal forces in the distributed load.\" width=\"424\" height=\"312\" class=\"alignnone wp-image-2613\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-201715-300x221.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-201715-65x48.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-201715-225x165.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-201715-350x257.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-10-201715.png 726w\" sizes=\"auto, (max-width: 424px) 100vw, 424px\" \/><\/p>\n<p>With V<sub>f<\/sub> = Shear Force, N<sub>f<\/sub> = Normal Force and M<sub>e<\/sub> = Bending Moment. The remaining distributed load is also resolved<\/p>\n<p>[latex]122.625 \\, \\text{N\/m} \\times 0.1 \\, \\text{m} = 12.2625 \\, \\text{N}[\/latex]<br \/>\n[latex]\\text{(Located } 0.05 \\, \\text{m from the cut E and } 0.55 \\, \\text{m from A)}[\/latex]<\/p>\n<p>With all the info gathered, we can calculate the three forces.<\/p>\n<p>Normal force is zero, as there are no horizontal forces to consider<\/p>\n<p>For shear force:<\/p>\n<p>[latex]\\sum F_y = 0; \\quad -201.105 - 19.62 - 12.2625 - V_F\u00a0 = 0\\\\  V_F = -233\\, \\text{N} = 233\\, \\text{N} \\uparrow[\/latex]<\/p>\n<p>For Bending Moment:<\/p>\n<p>[latex]\\sum M_E = 0; \\quad M_E + 122.625(0.05) + 19.62(0.1) + 201.105(0.6) = 0 \\\\  M_E = -129\\, \\text{N}\\cdot\\text{m}[\/latex]<\/p>\n<p>Therefore, V<sub>f\u00a0<\/sub>=-233N and M<sub>E\u00a0<\/sub>=-129Nm<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>Negative signs for V<sub>f <\/sub>and M<sub>E <\/sub>indicate that it should go in the other direction. It makes sense that A<sub>y\u00a0<\/sub>and B<sub>y\u00a0<\/sub>go in opposite directions.<\/p>\n<\/div>\n<\/div>\n","protected":false},"author":60,"menu_order":3,"template":"","meta":{"pb_show_title":"on","pb_short_title":"6.3 Examples","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-575","chapter","type-chapter","status-publish","hentry"],"part":60,"_links":{"self":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/575","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/users\/60"}],"version-history":[{"count":69,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/575\/revisions"}],"predecessor-version":[{"id":2865,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/575\/revisions\/2865"}],"part":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/parts\/60"}],"metadata":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/575\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/media?parent=575"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapter-type?post=575"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/contributor?post=575"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/license?post=575"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}