{"id":190,"date":"2021-03-01T09:01:30","date_gmt":"2021-03-01T14:01:30","guid":{"rendered":"http:\/\/pressbooks.library.upei.ca\/statics\/?post_type=chapter&#038;p=190"},"modified":"2025-08-01T17:02:53","modified_gmt":"2025-08-01T21:02:53","slug":"3-types-of-internal-forces","status":"publish","type":"chapter","link":"https:\/\/pressbooks.library.upei.ca\/statics\/chapter\/3-types-of-internal-forces\/","title":{"raw":"6.1 Types of Internal Forces","rendered":"6.1 Types of Internal Forces"},"content":{"raw":"When you make a cut in an object, similar to a fixed reaction, we describe what is happening at that point using one horizontal force (called normal force), one vertical force (called shear force), and a bending moment.\r\n\r\n[caption id=\"attachment_1435\" align=\"aligncenter\" width=\"500\"]<a href=\"http:\/\/mechanicsmap.psu.edu\/websites\/6_internal_forces\/6-2_internal_forces_equilibrium\/internal_forces_equilibrium.html\"><img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/internalforces-1.png\" alt=\"Beam with point loads showing bending moment, shear force, and axial force at a section.\" class=\"wp-image-1435 size-full\" width=\"500\" height=\"372\" \/><\/a> Adapted from source: Engineering Mechanics, Jacob Moore, et al. http:\/\/mechanicsmap.psu.edu\/websites\/6_internal_forces\/6-2_internal_forces_equilibrium\/internal_forces_equilibrium.html[\/caption]\r\n<h1>6.1.1 Types of Internal Forces<\/h1>\r\nThere are 3 types of internal forces (&amp; moments):\r\n<ul>\r\n \t<li>Normal force (N) - the horizontal force we calculated in trusses in the last chapter<\/li>\r\n \t<li>Shear force (V) - the vertical force that changes based on the applied loads<\/li>\r\n \t<li>bending moment (M) - changes based on the applied loads and applied moments<\/li>\r\n<\/ul>\r\nNormal force is represented by 'N'. Shear force, the vertical force, is represented by 'V'. Bending moment is 'M'.\u00a0 Normal and shear have units of N or lb, and bending moment has units of Nm or ft-lb. The following table summarizes information on internal forces (and moments).\r\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 70.5882%;height: 92px\" border=\"0\">\r\n<tbody>\r\n<tr style=\"height: 47px\">\r\n<td style=\"width: 20.6656%;height: 47px\"><strong><em>Force\/Moment<\/em><\/strong><\/td>\r\n<td style=\"width: 18.8081%;height: 47px\"><strong><em>Abbreviation<\/em><\/strong><\/td>\r\n<td style=\"width: 16.4861%;height: 47px\"><strong><em>Unit<\/em><\/strong><\/td>\r\n<td style=\"width: 14.6285%;height: 47px\"><strong><em>Directions for a horizontal beam<\/em><\/strong><\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"width: 20.6656%;text-align: center;height: 15px\">Normal Force<\/td>\r\n<td style=\"width: 18.8081%;text-align: center;height: 15px\">N<\/td>\r\n<td style=\"width: 16.4861%;height: 15px\">N\u00a0 or lb<\/td>\r\n<td style=\"width: 14.6285%;height: 15px\">horizontal<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"width: 20.6656%;text-align: center;height: 15px\">Shear Force<\/td>\r\n<td style=\"width: 18.8081%;text-align: center;height: 15px\">V<\/td>\r\n<td style=\"width: 16.4861%;height: 15px\">N\u00a0 or lb<\/td>\r\n<td style=\"width: 14.6285%;height: 15px\">vertical<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"width: 20.6656%;text-align: center;height: 15px\">Moment<\/td>\r\n<td style=\"width: 18.8081%;text-align: center;height: 15px\">M<\/td>\r\n<td style=\"width: 16.4861%;height: 15px\">Nm or ft-lb<\/td>\r\n<td style=\"width: 14.6285%;height: 15px\">rotation<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNote that for a vertical column, the normal force would be vertical. For this reason, the normal force is often called 'axial' as in: along the axis. The shear force for a column would be horizontal and is sometimes called 'transverse'.\r\n\r\nThis is for a 2d analysis of the beam, assuming there is negligible loading in the third dimension.\r\n\r\n&nbsp;\r\n<div class=\"textbox\">\r\n\r\n<span>When a beam or frame is subjected to transverse loadings, the three possible internal forces that are developed are the normal or axial force, the shearing force, and the bending moment, as shown in section\u00a0<\/span><i>k<\/i><span>\u00a0of the cantilever of the figure below<\/span><span>. To predict the behaviour of structures, the magnitudes of these forces must be known. In this chapter, the student will learn how to determine the magnitude of the shearing force and bending moment at any section of a beam or frame and how to present the computed values in a graphical form, which is referred to as the \u201cshearing force\u201d and the \u201cbending moment diagrams.\u201d Bending moment and shearing force diagrams aid immeasurably during design, as they show the maximum bending moments and shearing forces needed for sizing structural members.<\/span>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/1.04-_Internal_Forces_in_Beams_and_Frames-3-1024x615.jpg\" alt=\"Free-body diagram of a beam under two point loads, showing internal axial force N, shear force V, and bending moment M at section k.\" class=\"aligncenter wp-image-1499\" width=\"572\" height=\"343\" \/>\r\n<h2 class=\"h40 lt-eng-17610\"><span class=\"blue1\">Normal Force<\/span><\/h2>\r\n<p class=\"noindent lt-eng-17610\">The normal force at any section of a structure is defined as the algebraic sum of the axial forces acting on either side of the section.<\/p>\r\n\r\n<h3 class=\"h4 lt-eng-17610\"><span class=\"blue1\">Shearing Force<\/span><\/h3>\r\n<p class=\"noindent lt-eng-17610\"><span class=\"right_1\" title=\"90\"><\/span>The shearing force (SF) is defined as the algebraic sum of all the transverse forces acting on either side of the section of a beam or a frame. The phrase \u201con either side\u201d is important, as it implies that at any particular instance, the shearing force can be obtained by summing up the transverse forces on the left side of the section or on the right side of the section.<\/p>\r\n\r\n<h3 class=\"noindent lt-eng-17610\"><span class=\"blue1\">Bending Moment<\/span><\/h3>\r\n<p class=\"noindent lt-eng-17610\">The bending moment (BM) is defined as the algebraic sum of all the forces\u2019 moments acting on either side of the section of a beam or a frame.<\/p>\r\nSource: Internal Forces in Beams and Frames, Libretexts. <a href=\"https:\/\/eng.libretexts.org\/Bookshelves\/Civil_Engineering\/Book%3A_Structural_Analysis_(Udoeyo)\/01%3A_Chapters\/1.04%3A_Internal_Forces_in_Beams_and_Frames\">https:\/\/eng.libretexts.org\/Bookshelves\/Civil_Engineering\/Book%3A_Structural_Analysis_(Udoeyo)\/01%3A_Chapters\/1.04%3A_Internal_Forces_in_Beams_and_Frames<\/a>\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nIn 3 dimensions, there are:\r\n<ul>\r\n \t<li>1 normal force (N)<\/li>\r\n \t<li>2 shear forces (V<sub>1<\/sub> &amp; V<sub>2<\/sub>), and<\/li>\r\n \t<li>3 bending moments (M<sub>1<\/sub>, M<sub>2<\/sub>, &amp; T - torsion).<\/li>\r\n<\/ul>\r\n[caption id=\"attachment_1436\" align=\"aligncenter\" width=\"321\"]<a href=\"http:\/\/mechanicsmap.psu.edu\/websites\/6_internal_forces\/6-2_internal_forces_equilibrium\/internal_forces_equilibrium.html\"><img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/freebodydiagram3D.png\" alt=\"3D internal force diagram showing axial force (N), shear forces (V\u2081, V\u2082), bending moments (M\u2081, M\u2082), and torsion (T).\" class=\"wp-image-1436 size-full\" width=\"321\" height=\"355\" \/><\/a> Source: Engineering Mechanics, Jacob Moore, et al. http:\/\/mechanicsmap.psu.edu\/websites\/6_internal_forces\/6-2_internal_forces_equilibrium\/internal_forces_equilibrium.html[\/caption]\r\n<h1 class=\"noindent lt-eng-17610\">6.1.2 Sign Convention<\/h1>\r\nSo that there is a standard within the industry, a sign convention is necessary so we agree on what is positive and what is negative. On the right, shear-up is positive. Notice that both of the following figures show the identical sign convention.\r\n\r\n[caption id=\"attachment_1497\" align=\"aligncenter\" width=\"515\"]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-7.55.21-PM-1024x460.png\" alt=\"Internal force diagram showing normal force (N), shear force (V), and bending moment (M) at a section cut in a beam.\" class=\"wp-image-1497\" width=\"515\" height=\"231\" \/> <br \/>Positive sign convention adapted from source: https:\/\/eng.libretexts.org\/Bookshelves\/Civil_Engineering\/Book%3A_Structural_Analysis_(Udoeyo)\/01%3A_Chapters\/1.04%3A_Internal_Forces_in_Beams_and_Frames[\/caption]\r\n\r\nWhen you look at the beam as a whole (in the figure below), positive shear is right side down. When you cut into the beam, for it to be in static equilibrium, the positive shear must then be up on the right to be equal and opposite to the overall motion.\r\n\r\n&nbsp;\r\n<div class=\"textbox\">\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/IMG-0633.jpg\" alt=\"Illustration of internal force sign conventions\" class=\"aligncenter wp-image-486 size-full\" width=\"2078\" height=\"928\" \/>\r\n<h2><span class=\"blue1\">Axial (Normal) Force<\/span><\/h2>\r\n<p class=\"noindent lt-eng-17610\">An axial force is regarded as positive if it tends to tier the member at the section under consideration. Such a force is regarded as tensile, while the member is said to be subjected to axial tension. On the other hand, an axial force is considered negative if it tends to crush the member at the section being considered. Such force is regarded as compressive, while the member is said to be in axial compression.<\/p>\r\n\r\n<h3 class=\"h4 lt-eng-17610\"><span class=\"blue1\">Shear Force<\/span><\/h3>\r\n<p class=\"noindent lt-eng-17610\">A shear force that tends to move the left of the section upward or the right side of the section downward will be regarded as positive. Similarly, a shear force that has the tendency to move the left side of the section downward or the right side upward will be considered a negative shear force.<\/p>\r\n\r\n<h3 class=\"h4 lt-eng-17610\"><span class=\"blue1\">Bending Moment<\/span><\/h3>\r\n<p class=\"noindent lt-eng-17610\">A bending moment is considered positive if it tends to cause concavity upward (sagging). If the bending moment tends to cause concavity downward (hogging), it will be considered a negative bending moment.<\/p>\r\nSource: Internal Forces in Beams and Frames, Libretexts. <a href=\"https:\/\/eng.libretexts.org\/Bookshelves\/Civil_Engineering\/Book%3A_Structural_Analysis_(Udoeyo)\/01%3A_Chapters\/1.04%3A_Internal_Forces_in_Beams_and_Frames\">https:\/\/eng.libretexts.org\/Bookshelves\/Civil_Engineering\/Book%3A_Structural_Analysis_(Udoeyo)\/01%3A_Chapters\/1.04%3A_Internal_Forces_in_Beams_and_Frames<\/a>\r\n\r\n<\/div>\r\n<h1>6.1.3 Calculating the Internal Forces<\/h1>\r\nTo solve the internal forces at a certain point along the beam,\r\n\r\n[caption id=\"attachment_1498\" align=\"alignright\" width=\"408\"]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-7.55.21-PM-1-1024x460.png\" alt=\"Internal force diagram showing normal force (N), shear force (V), and bending moment (M) at a section cut in a beam.\" class=\"wp-image-1498\" width=\"408\" height=\"183\" \/> Positive sign convention adapted from https:\/\/eng.libretexts.org\/Bookshelves\/Civil_Engineering\/Book%3A_Structural_Analysis_(Udoeyo)\/01%3A_Chapters\/1.04%3A_Internal_Forces_in_Beams_and_Frames[\/caption]\r\n\r\n&nbsp;\r\n<ol>\r\n \t<li>Find the external &amp; reaction forces<\/li>\r\n \t<li>Make a cut.<\/li>\r\n \t<li>In a FBD of one side of the cut, add the internal forces (and moments) using the positive sign convention.<\/li>\r\n \t<li>Use the equilibrium equations to solve for the unknown internal forces and moments.<\/li>\r\n<\/ol>\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\nExample: For the following distributed load, a) what are the reaction forces? b) What are the internal forces at the midpoint B between the reaction forces?\r\n\r\n[caption id=\"attachment_1451\" align=\"aligncenter\" width=\"608\"]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/1.04-_Internal_Forces_in_Beams_and_Frames.jpg\" alt=\"Simply supported beam with a 100 lb\/ft uniform distributed load, supported at points A and C.\" class=\"wp-image-1451 size-full\" width=\"608\" height=\"291\" \/> Adapted from: Source: Engineering Mechanics, Jacob Moore, et al. http:\/\/mechanicsmap.psu.edu\/websites\/6_internal_forces\/6-3_axial_torque_diagrams\/axial_torque_diagrams.html[\/caption]\r\n\r\n1. Solve external forces:\r\n\r\n[caption id=\"attachment_1452\" align=\"aligncenter\" width=\"673\"]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/1.04-_Internal_Forces_in_Beams_and_Frames-2.jpg\" alt=\"FBD of the problem\" class=\"wp-image-1452 size-full\" width=\"673\" height=\"236\" \/> Adapted from: Source: Engineering Mechanics, Jacob Moore, et al. http:\/\/mechanicsmap.psu.edu\/websites\/6_internal_forces\/6-3_axial_torque_diagrams\/axial_torque_diagrams.html[\/caption]\r\n\r\n$latex \\sum F_{X}=A_{x}=0$\r\n$latex \\sum F_{y}=A_{y}+C-\\omega L=0$\r\n$latex \\sum M_{A}=-(\\omega L)\\left(\\frac{L}{2}\\right)+d_{A C} C=0$\r\n[latex]C = \\left(\\frac{\\omega L^2}{2d_{A C}}\\right) = \\frac{(100 \\frac{lb}{ft} )*(7ft)^2}{2 * (4ft)} = 612.5 lb \\text{ (+j direction)} [\/latex]\r\n[latex]A_y = \\omega*L- C = (100 \\frac{lb}{ft})*(7 ft) - 612.5 lb = 87.5 lb \\text{ (+j direction) }[\/latex]\r\n[latex]\\underline{A_x = 0 \\qquad A_y = 87.5 \\text{ (+j )} \\qquad C = 612.5 lb \\text{ (+j )} }[\/latex]\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/admin-ajax.php_.png\" alt=\"Solving for distributed load.\" class=\"aligncenter wp-image-1458\" width=\"165\" height=\"159\" \/>\r\n\r\n2. Make a cut at B.\r\n\r\n3. In a FBD of one side of the cut, add the internal forces (and moments) using the positive sign convention.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-12.30.15-PM-1024x606.png\" alt=\"Solving for internal forces.\" class=\"aligncenter wp-image-1454\" width=\"287\" height=\"170\" \/>\r\n\r\n4. Use the equilibrium equations to solve for the unknown internal forces and moments.\r\n\r\nFor just this portion, the force from intensity is: F<sub>w<\/sub> = ( 100 lb\/ft ) * ( 2 ft) = 200 lb and acts 1 ft from the left, so the moment due to intensity is: M<sub>w<\/sub> = w * 2 ft * 1 ft = F<sub>w<\/sub> * 1 ft = ( 100 lb\/ft ) * ( 2 ft) * (1 ft)\u00a0 =\u00a0 200 ft-lb\r\n<p style=\"text-align: left\">[latex]\\sum F_y = 87.5 lb - 200 lb - V = 0 \\\\ V = -112.5 lb \\text{ (- indicates going up not down)}[\/latex]<\/p>\r\n&nbsp;\r\n<p style=\"text-align: left\">[latex]\\sum M_A = - (w * 2 ft) * (1 ft) - V * (2 ft)\u00a0 + M = 0 \\\\ M = (100 \\frac{lb}{ft}) * 2 ft + (-112.5 lb) * (2 ft) \\\\ M = 200 ft \\cdot lb - 225 ft \\cdot lb \\\\ M = -25 ft \\cdot lb \\text{ (- indicates going reverse direction)}[\/latex]<\/p>\r\n&nbsp;\r\n\r\n[latex]\\underline{N = 0 \\qquad V = -112.5 lb \\text{ (+j )} \\qquad M = -25 ft \\cdot lb \\text{ (clockwise)} }[\/latex]\r\n\r\n&nbsp;\r\n<div class=\"textbox textbox--key-takeaways\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Key Takeaways<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\n<em>Basically<\/em>: The internal forces (and moments) for a 2d beam are: shear, normal, and bending moment. There is a positive sign convention to use when making a cut along a beam to determine the forces inside: on the left, shear down, normal out, moment up.\r\n\r\n<em>Application<\/em>: A bridge that has different loads applied (from cars, trucks, lampposts, etc). Use this method to calculate the internal loads at a particular point of interest.\r\n\r\n<em>Looking Ahead<\/em>: In the next section, we'll look at how to calculate the internal force across the whole beam and display the results graphically.\r\n\r\n<\/div>\r\n<\/div>","rendered":"<p>When you make a cut in an object, similar to a fixed reaction, we describe what is happening at that point using one horizontal force (called normal force), one vertical force (called shear force), and a bending moment.<\/p>\n<figure id=\"attachment_1435\" aria-describedby=\"caption-attachment-1435\" style=\"width: 500px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/mechanicsmap.psu.edu\/websites\/6_internal_forces\/6-2_internal_forces_equilibrium\/internal_forces_equilibrium.html\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/internalforces-1.png\" alt=\"Beam with point loads showing bending moment, shear force, and axial force at a section.\" class=\"wp-image-1435 size-full\" width=\"500\" height=\"372\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/internalforces-1.png 500w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/internalforces-1-300x223.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/internalforces-1-65x48.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/internalforces-1-225x167.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/internalforces-1-350x260.png 350w\" sizes=\"auto, (max-width: 500px) 100vw, 500px\" \/><\/a><figcaption id=\"caption-attachment-1435\" class=\"wp-caption-text\">Adapted from source: Engineering Mechanics, Jacob Moore, et al. http:\/\/mechanicsmap.psu.edu\/websites\/6_internal_forces\/6-2_internal_forces_equilibrium\/internal_forces_equilibrium.html<\/figcaption><\/figure>\n<h1>6.1.1 Types of Internal Forces<\/h1>\n<p>There are 3 types of internal forces (&amp; moments):<\/p>\n<ul>\n<li>Normal force (N) &#8211; the horizontal force we calculated in trusses in the last chapter<\/li>\n<li>Shear force (V) &#8211; the vertical force that changes based on the applied loads<\/li>\n<li>bending moment (M) &#8211; changes based on the applied loads and applied moments<\/li>\n<\/ul>\n<p>Normal force is represented by &#8216;N&#8217;. Shear force, the vertical force, is represented by &#8216;V&#8217;. Bending moment is &#8216;M&#8217;.\u00a0 Normal and shear have units of N or lb, and bending moment has units of Nm or ft-lb. The following table summarizes information on internal forces (and moments).<\/p>\n<table class=\"grid aligncenter\" style=\"border-collapse: collapse;width: 70.5882%;height: 92px\">\n<tbody>\n<tr style=\"height: 47px\">\n<td style=\"width: 20.6656%;height: 47px\"><strong><em>Force\/Moment<\/em><\/strong><\/td>\n<td style=\"width: 18.8081%;height: 47px\"><strong><em>Abbreviation<\/em><\/strong><\/td>\n<td style=\"width: 16.4861%;height: 47px\"><strong><em>Unit<\/em><\/strong><\/td>\n<td style=\"width: 14.6285%;height: 47px\"><strong><em>Directions for a horizontal beam<\/em><\/strong><\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"width: 20.6656%;text-align: center;height: 15px\">Normal Force<\/td>\n<td style=\"width: 18.8081%;text-align: center;height: 15px\">N<\/td>\n<td style=\"width: 16.4861%;height: 15px\">N\u00a0 or lb<\/td>\n<td style=\"width: 14.6285%;height: 15px\">horizontal<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"width: 20.6656%;text-align: center;height: 15px\">Shear Force<\/td>\n<td style=\"width: 18.8081%;text-align: center;height: 15px\">V<\/td>\n<td style=\"width: 16.4861%;height: 15px\">N\u00a0 or lb<\/td>\n<td style=\"width: 14.6285%;height: 15px\">vertical<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"width: 20.6656%;text-align: center;height: 15px\">Moment<\/td>\n<td style=\"width: 18.8081%;text-align: center;height: 15px\">M<\/td>\n<td style=\"width: 16.4861%;height: 15px\">Nm or ft-lb<\/td>\n<td style=\"width: 14.6285%;height: 15px\">rotation<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Note that for a vertical column, the normal force would be vertical. For this reason, the normal force is often called &#8216;axial&#8217; as in: along the axis. The shear force for a column would be horizontal and is sometimes called &#8216;transverse&#8217;.<\/p>\n<p>This is for a 2d analysis of the beam, assuming there is negligible loading in the third dimension.<\/p>\n<p>&nbsp;<\/p>\n<div class=\"textbox\">\n<p><span>When a beam or frame is subjected to transverse loadings, the three possible internal forces that are developed are the normal or axial force, the shearing force, and the bending moment, as shown in section\u00a0<\/span><i>k<\/i><span>\u00a0of the cantilever of the figure below<\/span><span>. To predict the behaviour of structures, the magnitudes of these forces must be known. In this chapter, the student will learn how to determine the magnitude of the shearing force and bending moment at any section of a beam or frame and how to present the computed values in a graphical form, which is referred to as the \u201cshearing force\u201d and the \u201cbending moment diagrams.\u201d Bending moment and shearing force diagrams aid immeasurably during design, as they show the maximum bending moments and shearing forces needed for sizing structural members.<\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/1.04-_Internal_Forces_in_Beams_and_Frames-3-1024x615.jpg\" alt=\"Free-body diagram of a beam under two point loads, showing internal axial force N, shear force V, and bending moment M at section k.\" class=\"aligncenter wp-image-1499\" width=\"572\" height=\"343\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/1.04-_Internal_Forces_in_Beams_and_Frames-3-1024x615.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/1.04-_Internal_Forces_in_Beams_and_Frames-3-300x180.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/1.04-_Internal_Forces_in_Beams_and_Frames-3-768x461.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/1.04-_Internal_Forces_in_Beams_and_Frames-3-65x39.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/1.04-_Internal_Forces_in_Beams_and_Frames-3-225x135.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/1.04-_Internal_Forces_in_Beams_and_Frames-3-350x210.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/1.04-_Internal_Forces_in_Beams_and_Frames-3.jpg 1353w\" sizes=\"auto, (max-width: 572px) 100vw, 572px\" \/><\/p>\n<h2 class=\"h40 lt-eng-17610\"><span class=\"blue1\">Normal Force<\/span><\/h2>\n<p class=\"noindent lt-eng-17610\">The normal force at any section of a structure is defined as the algebraic sum of the axial forces acting on either side of the section.<\/p>\n<h3 class=\"h4 lt-eng-17610\"><span class=\"blue1\">Shearing Force<\/span><\/h3>\n<p class=\"noindent lt-eng-17610\"><span class=\"right_1\" title=\"90\"><\/span>The shearing force (SF) is defined as the algebraic sum of all the transverse forces acting on either side of the section of a beam or a frame. The phrase \u201con either side\u201d is important, as it implies that at any particular instance, the shearing force can be obtained by summing up the transverse forces on the left side of the section or on the right side of the section.<\/p>\n<h3 class=\"noindent lt-eng-17610\"><span class=\"blue1\">Bending Moment<\/span><\/h3>\n<p class=\"noindent lt-eng-17610\">The bending moment (BM) is defined as the algebraic sum of all the forces\u2019 moments acting on either side of the section of a beam or a frame.<\/p>\n<p>Source: Internal Forces in Beams and Frames, Libretexts. <a href=\"https:\/\/eng.libretexts.org\/Bookshelves\/Civil_Engineering\/Book%3A_Structural_Analysis_(Udoeyo)\/01%3A_Chapters\/1.04%3A_Internal_Forces_in_Beams_and_Frames\">https:\/\/eng.libretexts.org\/Bookshelves\/Civil_Engineering\/Book%3A_Structural_Analysis_(Udoeyo)\/01%3A_Chapters\/1.04%3A_Internal_Forces_in_Beams_and_Frames<\/a><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>In 3 dimensions, there are:<\/p>\n<ul>\n<li>1 normal force (N)<\/li>\n<li>2 shear forces (V<sub>1<\/sub> &amp; V<sub>2<\/sub>), and<\/li>\n<li>3 bending moments (M<sub>1<\/sub>, M<sub>2<\/sub>, &amp; T &#8211; torsion).<\/li>\n<\/ul>\n<figure id=\"attachment_1436\" aria-describedby=\"caption-attachment-1436\" style=\"width: 321px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/mechanicsmap.psu.edu\/websites\/6_internal_forces\/6-2_internal_forces_equilibrium\/internal_forces_equilibrium.html\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/freebodydiagram3D.png\" alt=\"3D internal force diagram showing axial force (N), shear forces (V\u2081, V\u2082), bending moments (M\u2081, M\u2082), and torsion (T).\" class=\"wp-image-1436 size-full\" width=\"321\" height=\"355\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/freebodydiagram3D.png 321w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/freebodydiagram3D-271x300.png 271w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/freebodydiagram3D-65x72.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/freebodydiagram3D-225x249.png 225w\" sizes=\"auto, (max-width: 321px) 100vw, 321px\" \/><\/a><figcaption id=\"caption-attachment-1436\" class=\"wp-caption-text\">Source: Engineering Mechanics, Jacob Moore, et al. http:\/\/mechanicsmap.psu.edu\/websites\/6_internal_forces\/6-2_internal_forces_equilibrium\/internal_forces_equilibrium.html<\/figcaption><\/figure>\n<h1 class=\"noindent lt-eng-17610\">6.1.2 Sign Convention<\/h1>\n<p>So that there is a standard within the industry, a sign convention is necessary so we agree on what is positive and what is negative. On the right, shear-up is positive. Notice that both of the following figures show the identical sign convention.<\/p>\n<figure id=\"attachment_1497\" aria-describedby=\"caption-attachment-1497\" style=\"width: 515px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-7.55.21-PM-1024x460.png\" alt=\"Internal force diagram showing normal force (N), shear force (V), and bending moment (M) at a section cut in a beam.\" class=\"wp-image-1497\" width=\"515\" height=\"231\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-7.55.21-PM-1024x460.png 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-7.55.21-PM-300x135.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-7.55.21-PM-768x345.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-7.55.21-PM-1536x689.png 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-7.55.21-PM-65x29.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-7.55.21-PM-225x101.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-7.55.21-PM-350x157.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-7.55.21-PM.png 1662w\" sizes=\"auto, (max-width: 515px) 100vw, 515px\" \/><figcaption id=\"caption-attachment-1497\" class=\"wp-caption-text\">Positive sign convention adapted from source: https:\/\/eng.libretexts.org\/Bookshelves\/Civil_Engineering\/Book%3A_Structural_Analysis_(Udoeyo)\/01%3A_Chapters\/1.04%3A_Internal_Forces_in_Beams_and_Frames<\/figcaption><\/figure>\n<p>When you look at the beam as a whole (in the figure below), positive shear is right side down. When you cut into the beam, for it to be in static equilibrium, the positive shear must then be up on the right to be equal and opposite to the overall motion.<\/p>\n<p>&nbsp;<\/p>\n<div class=\"textbox\">\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/IMG-0633.jpg\" alt=\"Illustration of internal force sign conventions\" class=\"aligncenter wp-image-486 size-full\" width=\"2078\" height=\"928\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/IMG-0633.jpg 2078w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/IMG-0633-300x134.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/IMG-0633-1024x457.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/IMG-0633-768x343.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/IMG-0633-1536x686.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/IMG-0633-2048x915.jpg 2048w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/IMG-0633-65x29.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/IMG-0633-225x100.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/IMG-0633-350x156.jpg 350w\" sizes=\"auto, (max-width: 2078px) 100vw, 2078px\" \/><\/p>\n<h2><span class=\"blue1\">Axial (Normal) Force<\/span><\/h2>\n<p class=\"noindent lt-eng-17610\">An axial force is regarded as positive if it tends to tier the member at the section under consideration. Such a force is regarded as tensile, while the member is said to be subjected to axial tension. On the other hand, an axial force is considered negative if it tends to crush the member at the section being considered. Such force is regarded as compressive, while the member is said to be in axial compression.<\/p>\n<h3 class=\"h4 lt-eng-17610\"><span class=\"blue1\">Shear Force<\/span><\/h3>\n<p class=\"noindent lt-eng-17610\">A shear force that tends to move the left of the section upward or the right side of the section downward will be regarded as positive. Similarly, a shear force that has the tendency to move the left side of the section downward or the right side upward will be considered a negative shear force.<\/p>\n<h3 class=\"h4 lt-eng-17610\"><span class=\"blue1\">Bending Moment<\/span><\/h3>\n<p class=\"noindent lt-eng-17610\">A bending moment is considered positive if it tends to cause concavity upward (sagging). If the bending moment tends to cause concavity downward (hogging), it will be considered a negative bending moment.<\/p>\n<p>Source: Internal Forces in Beams and Frames, Libretexts. <a href=\"https:\/\/eng.libretexts.org\/Bookshelves\/Civil_Engineering\/Book%3A_Structural_Analysis_(Udoeyo)\/01%3A_Chapters\/1.04%3A_Internal_Forces_in_Beams_and_Frames\">https:\/\/eng.libretexts.org\/Bookshelves\/Civil_Engineering\/Book%3A_Structural_Analysis_(Udoeyo)\/01%3A_Chapters\/1.04%3A_Internal_Forces_in_Beams_and_Frames<\/a><\/p>\n<\/div>\n<h1>6.1.3 Calculating the Internal Forces<\/h1>\n<p>To solve the internal forces at a certain point along the beam,<\/p>\n<figure id=\"attachment_1498\" aria-describedby=\"caption-attachment-1498\" style=\"width: 408px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-7.55.21-PM-1-1024x460.png\" alt=\"Internal force diagram showing normal force (N), shear force (V), and bending moment (M) at a section cut in a beam.\" class=\"wp-image-1498\" width=\"408\" height=\"183\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-7.55.21-PM-1-1024x460.png 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-7.55.21-PM-1-300x135.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-7.55.21-PM-1-768x345.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-7.55.21-PM-1-1536x689.png 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-7.55.21-PM-1-65x29.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-7.55.21-PM-1-225x101.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-7.55.21-PM-1-350x157.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-7.55.21-PM-1.png 1662w\" sizes=\"auto, (max-width: 408px) 100vw, 408px\" \/><figcaption id=\"caption-attachment-1498\" class=\"wp-caption-text\">Positive sign convention adapted from https:\/\/eng.libretexts.org\/Bookshelves\/Civil_Engineering\/Book%3A_Structural_Analysis_(Udoeyo)\/01%3A_Chapters\/1.04%3A_Internal_Forces_in_Beams_and_Frames<\/figcaption><\/figure>\n<p>&nbsp;<\/p>\n<ol>\n<li>Find the external &amp; reaction forces<\/li>\n<li>Make a cut.<\/li>\n<li>In a FBD of one side of the cut, add the internal forces (and moments) using the positive sign convention.<\/li>\n<li>Use the equilibrium equations to solve for the unknown internal forces and moments.<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>Example: For the following distributed load, a) what are the reaction forces? b) What are the internal forces at the midpoint B between the reaction forces?<\/p>\n<figure id=\"attachment_1451\" aria-describedby=\"caption-attachment-1451\" style=\"width: 608px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/1.04-_Internal_Forces_in_Beams_and_Frames.jpg\" alt=\"Simply supported beam with a 100 lb\/ft uniform distributed load, supported at points A and C.\" class=\"wp-image-1451 size-full\" width=\"608\" height=\"291\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/1.04-_Internal_Forces_in_Beams_and_Frames.jpg 608w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/1.04-_Internal_Forces_in_Beams_and_Frames-300x144.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/1.04-_Internal_Forces_in_Beams_and_Frames-65x31.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/1.04-_Internal_Forces_in_Beams_and_Frames-225x108.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/1.04-_Internal_Forces_in_Beams_and_Frames-350x168.jpg 350w\" sizes=\"auto, (max-width: 608px) 100vw, 608px\" \/><figcaption id=\"caption-attachment-1451\" class=\"wp-caption-text\">Adapted from: Source: Engineering Mechanics, Jacob Moore, et al. http:\/\/mechanicsmap.psu.edu\/websites\/6_internal_forces\/6-3_axial_torque_diagrams\/axial_torque_diagrams.html<\/figcaption><\/figure>\n<p>1. Solve external forces:<\/p>\n<figure id=\"attachment_1452\" aria-describedby=\"caption-attachment-1452\" style=\"width: 673px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/1.04-_Internal_Forces_in_Beams_and_Frames-2.jpg\" alt=\"FBD of the problem\" class=\"wp-image-1452 size-full\" width=\"673\" height=\"236\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/1.04-_Internal_Forces_in_Beams_and_Frames-2.jpg 673w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/1.04-_Internal_Forces_in_Beams_and_Frames-2-300x105.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/1.04-_Internal_Forces_in_Beams_and_Frames-2-65x23.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/1.04-_Internal_Forces_in_Beams_and_Frames-2-225x79.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/1.04-_Internal_Forces_in_Beams_and_Frames-2-350x123.jpg 350w\" sizes=\"auto, (max-width: 673px) 100vw, 673px\" \/><figcaption id=\"caption-attachment-1452\" class=\"wp-caption-text\">Adapted from: Source: Engineering Mechanics, Jacob Moore, et al. http:\/\/mechanicsmap.psu.edu\/websites\/6_internal_forces\/6-3_axial_torque_diagrams\/axial_torque_diagrams.html<\/figcaption><\/figure>\n<p>[latex]\\sum F_{X}=A_{x}=0[\/latex]<br \/>\n[latex]\\sum F_{y}=A_{y}+C-\\omega L=0[\/latex]<br \/>\n[latex]\\sum M_{A}=-(\\omega L)\\left(\\frac{L}{2}\\right)+d_{A C} C=0[\/latex]<br \/>\n[latex]C = \\left(\\frac{\\omega L^2}{2d_{A C}}\\right) = \\frac{(100 \\frac{lb}{ft} )*(7ft)^2}{2 * (4ft)} = 612.5 lb \\text{ (+j direction)}[\/latex]<br \/>\n[latex]A_y = \\omega*L- C = (100 \\frac{lb}{ft})*(7 ft) - 612.5 lb = 87.5 lb \\text{ (+j direction) }[\/latex]<br \/>\n[latex]\\underline{A_x = 0 \\qquad A_y = 87.5 \\text{ (+j )} \\qquad C = 612.5 lb \\text{ (+j )} }[\/latex]<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/admin-ajax.php_.png\" alt=\"Solving for distributed load.\" class=\"aligncenter wp-image-1458\" width=\"165\" height=\"159\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/admin-ajax.php_.png 283w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/admin-ajax.php_-65x62.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/admin-ajax.php_-225x216.png 225w\" sizes=\"auto, (max-width: 165px) 100vw, 165px\" \/><\/p>\n<p>2. Make a cut at B.<\/p>\n<p>3. In a FBD of one side of the cut, add the internal forces (and moments) using the positive sign convention.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-12.30.15-PM-1024x606.png\" alt=\"Solving for internal forces.\" class=\"aligncenter wp-image-1454\" width=\"287\" height=\"170\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-12.30.15-PM-1024x606.png 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-12.30.15-PM-300x178.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-12.30.15-PM-768x455.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-12.30.15-PM-65x38.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-12.30.15-PM-225x133.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-12.30.15-PM-350x207.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/03\/Screen-Shot-2021-08-29-at-12.30.15-PM.png 1142w\" sizes=\"auto, (max-width: 287px) 100vw, 287px\" \/><\/p>\n<p>4. Use the equilibrium equations to solve for the unknown internal forces and moments.<\/p>\n<p>For just this portion, the force from intensity is: F<sub>w<\/sub> = ( 100 lb\/ft ) * ( 2 ft) = 200 lb and acts 1 ft from the left, so the moment due to intensity is: M<sub>w<\/sub> = w * 2 ft * 1 ft = F<sub>w<\/sub> * 1 ft = ( 100 lb\/ft ) * ( 2 ft) * (1 ft)\u00a0 =\u00a0 200 ft-lb<\/p>\n<p style=\"text-align: left\">[latex]\\sum F_y = 87.5 lb - 200 lb - V = 0 \\\\ V = -112.5 lb \\text{ (- indicates going up not down)}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: left\">[latex]\\sum M_A = - (w * 2 ft) * (1 ft) - V * (2 ft)\u00a0 + M = 0 \\\\ M = (100 \\frac{lb}{ft}) * 2 ft + (-112.5 lb) * (2 ft) \\\\ M = 200 ft \\cdot lb - 225 ft \\cdot lb \\\\ M = -25 ft \\cdot lb \\text{ (- indicates going reverse direction)}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>[latex]\\underline{N = 0 \\qquad V = -112.5 lb \\text{ (+j )} \\qquad M = -25 ft \\cdot lb \\text{ (clockwise)} }[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<div class=\"textbox textbox--key-takeaways\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Key Takeaways<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p><em>Basically<\/em>: The internal forces (and moments) for a 2d beam are: shear, normal, and bending moment. There is a positive sign convention to use when making a cut along a beam to determine the forces inside: on the left, shear down, normal out, moment up.<\/p>\n<p><em>Application<\/em>: A bridge that has different loads applied (from cars, trucks, lampposts, etc). Use this method to calculate the internal loads at a particular point of interest.<\/p>\n<p><em>Looking Ahead<\/em>: In the next section, we&#8217;ll look at how to calculate the internal force across the whole beam and display the results graphically.<\/p>\n<\/div>\n<\/div>\n","protected":false},"author":74,"menu_order":1,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-190","chapter","type-chapter","status-publish","hentry"],"part":60,"_links":{"self":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/190","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\/74"}],"version-history":[{"count":25,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/190\/revisions"}],"predecessor-version":[{"id":2863,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/190\/revisions\/2863"}],"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\/190\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/media?parent=190"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapter-type?post=190"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/contributor?post=190"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/license?post=190"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}