calculate the effect of forces on objects, including the law of inertia, the relationship between force and acceleration, and the nature of force pairs between objects. Thus, for the net force, we obtain. Solve M B = 0. A fixed support offers a constraint against rotation in any direction, and it prevents movement in both horizontal and vertical directions. Shear force and bending moment in column ED. Cy = Dy = 25 kN, due to symmetry of loading. Did the drapes in old theatres actually say "ASBESTOS" on them? If you remove the eraser, in which direction will the rubber band move? . Because friction acts in the opposite direction, we assign it a negative value. Learn more about Stack Overflow the company, and our products. What is the equation for the normal force for a body with mass m that is at rest on a horizontal surface? A fixed support offers a constraint against rotation in any direction, and it prevents movement in both horizontal and vertical directions. . Because all motion is horizontal, we can assume that no net force acts in the vertical direction, and the problem becomes one dimensional. Identification of the primary and complimentary structure. Support reactions. because these are exerted by the system, not on the system. consent of Rice University. Determine the horizontal reaction at the supports of the cable, the expression of the shape of the cable, and the length of the cable. How are engines numbered on Starship and Super Heavy? Can you still use Commanders Strike if the only attack available to forego is an attack against an ally? You can see evidence of the wheels pushing backward when tires spin on a gravel road and throw rocks backward. In this case, both forces act on the same system and therefore cancel. The force (F) required to move an object of mass (m) with an acceleration (a) is given by the formula F = m x a. Support reactions. Friction f: sin(20) = f/981 N. f = sin(20 . Since the exit mass flow rate is nearly equal to the free stream mass flow rate, and the free . The reactions at the support of the beam can be computed as follows when considering the free-body diagram and using the equations of equilibrium: Shearing force and bending moment functions of beam BC. Using subscript 1 for the left hand side and 2 for the right hand side, we then get two equations: We can then solve all of these simultaneous equations (I'll leave that step to you), and we'll find: NB The plea formula works equally well in tension and compression (assuming no buckling). . Does my answer reflect this? Notice that at the location of concentrated loads and at the supports, the numerical values of the change in the shearing force are equal to the concentrated load or reaction. Use the sum of moments to calculate one of . 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 shearing force and the bending moment diagrams. 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. https://www.texasgateway.org/book/tea-physics are not subject to the Creative Commons license and may not be reproduced without the prior and express written 3.2.5 Fixed Support. View this video to watch examples of Newtons laws and internal and external forces. Support reactions. In addition to the two principal values of bending moment at x = 0 m and at x = 5 m, the moments at other intermediate points should be determined to correctly draw the bending moment diagram. The table applies a 110 N normal reaction force on the box upwards. The normal force is the outward force that a surface applies to an object perpendicular to the surface, and it prevents the object from penetrating it. Since the function for the bending moment is parabolic, the bending moment diagram is a curve. Birds fly by exerting force on air in the direction opposite that in which they wish to fly. Shearing force and bending moment functions of column AB. You can see evidence of the wheels pushing backward when tires spin on a gravel road and throw the rocks backward. The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Second, these forces are acting on different bodies or systems: As force acts on B and Bs force acts on A. The choice of a system is an important analytical step both in solving problems and in thoroughly understanding the physics of the situation (which are not necessarily the same things). { "1.01:_Introduction_to_Structural_Analysis" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.02:_Structural_Loads_and_Loading_System" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.03:_Equilibrium_Structures_Support_Reactions_Determinacy_and_Stability_of_Beams_and_Frames" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.04:_Internal_Forces_in_Beams_and_Frames" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.05:_Internal_Forces_in_Plane_Trusses" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.06:_Arches_and_Cables" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.07:_Deflection_of_Beams-_Geometric_Methods" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.08:_Deflections_of_Structures-_Work-Energy_Methods" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.09:_Influence_Lines_for_Statically_Determinate_Structures" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.10:_Force_Method_of_Analysis_of_Indeterminate_Structures" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.11:_Slope-Deflection_Method_of_Analysis_of_Indeterminate_Structures" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.12:_Moment_Distribution_Method_of_Analysis_of_Structures" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.13:_Influence_Lines_for_Statically_Indeterminate_Structures" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, { "00:_Front_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "01:_Chapters" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "zz:_Back_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, [ "article:topic", "license:ccbyncnd", "licenseversion:40", "authorname:fudoeyo", "source@https://temple.manifoldapp.org/projects/structural-analysis" ], https://eng.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Feng.libretexts.org%2FBookshelves%2FCivil_Engineering%2FStructural_Analysis_(Udoeyo)%2F01%253A_Chapters%2F1.04%253A_Internal_Forces_in_Beams_and_Frames, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\). Equation 4.1 suggests the following expression: Equation 4.2 states that the change in moment equals the area under the shear diagram. Ask students which forces are internal and which are external in each scenario. Thus. This reaction force is called thrust. The part AC is the primary structure, while part CD is the complimentary structure. The swimmer pushes in the direction opposite that in which she wishes to move. The point of application of the ground reaction force, the position of the ankle, knee and hip joints are known. An octopus propels itself in the water by ejecting water through a funnel from its body, similar to a jet ski. Solution. In other words, the reaction force of a link is in the direction of the link, along its longitudinal axis. Pinned constraint and then its free body diagram shown: Two reaction forces acting perpendicularly in the x and y directions, Moment rotating about fixed constraint (usually a wall), use right hand rule to find its direction, Single reaction force acting in the y direction, This can be the ground that the object rests on as well. When you push on a wall, the wall pushes back on you. Another way to look at this is to note that the forces between components of a system cancel because they are equal in magnitude and opposite in direction. A link has two hinges, one at each end. Imagine a beam extending from the wall. This is exactly what happens whenever one object exerts a force on anothereach object experiences a force that is the same strength as the force acting on the other object but that acts in the opposite direction. The force of friction, which opposes the motion, is 24.0 N. Because they accelerate together, we define the system to be the teacher, the cart, and the equipment. By the end of this section, you will be able to do the following: The learning objectives in this section will help your students master the following standards: [BL][OL] Review Newtons first and second laws. Legal. It restrains the structure from movement in a vertical direction. The 'normal' force is a type of 'contact' force. Want to cite, share, or modify this book? Such force is regarded as compressive, while the member is said to be in axial compression (see Figure 4.2a and Figure 4.2b). When a perfectly flexible connector (one requiring no force to bend it) such as a rope transmits a force, Math: Problem-Solving Strategy for Newtons Laws of Motion. Why? Joint D. Joint A. The best answers are voted up and rise to the top, Not the answer you're looking for? Everyday experiences, such as stubbing a toe or throwing a ball, are all perfect examples of Newtons third law in action. Thus, \[F_{net} = ma = (19.0\; kg)(1.5\; m/s^{2}) = 29\; N \ldotp\], \[F_{prof} = F_{net} + f = 29\; N + 24.0\; N = 53\; N \ldotp\]. Shearing force diagram. feetonwall Write an equation for the horizontal forces: F y = 0 = R A + R B - wL = R A + R B - 5*10 R A + R B = 50 kN. On the other hand, an axial force is considered negative if it tends to crush the member at the section being considered. We solve for Fprof, the desired quantity: The value of f is given, so we must calculate net Fnet. The answer is the normal force. feetonwall The direction is always orthogonal to the motion. Maximum bending moment occurs where the shearing force equals zero. The International System of Units (SI) unit of mass is the kilogram, and the SI unit of acceleration is m/s 2 (meters per second squared). Accessibility StatementFor more information contact us atinfo@libretexts.org. The wall has exerted an equal and opposite force on the swimmer. Two blocks are at rest and in contact on a frictionless surface as shown below, with m1 = 2.0 kg, m2 = 6.0 kg, and applied force 24 N. (a) Find the acceleration of the system of blocks. The vertical reactions of the supports at points A and E are computed by considering the equilibrium of the entire frame, as follows: The negative sign indicates that Ay acts downward instead of upward as originally assumed. However, because we havent yet covered vectors in depth, well only consider one-dimensional situations in this chapter. Add details and clarify the problem by editing this post. Thus, the scale reading gives the magnitude of the packages weight. The net external force on the system is the sum of the external forces: the force of the floor acting on the teacher, cart, and equipment (in the horizontal direction) and the force of friction. Relationship among distributed load, shear force, and bending moment: The following relationship exists among distributed loads, shear forces, and bending moments. To the right of where force F is applied the opposite is true and the beam is in compression and "wants" to shrink. Other examples of Newtons third law are easy to find: There are two important features of Newtons third law. Newtons third law of motion states that whenever a first object exerts a force on a second object, the first object experiences a force equal in magnitude but opposite in direction to the force that it exerts. That is how you find the direction of any reaction force. Note that this equation is only true for a horizontal surface. Looking Ahead: Every time we model an scenario, we will use reaction forces to show what type of motion is being restrained. An octopus propels itself forward in the water by ejecting water backward through a funnel in its body, which is similar to how a jet ski is propelled. This remarkable fact is a consequence of Newton's third law. At. F = (m dot * V)e - (m dot * V)0. , . He should throw the object upward because according to Newtons third law, the object will then exert a force on him in the opposite direction (i.e., downward). To work this out you need the plea formula: where d is extension, P is axial force, L is the original length, E is Young's modulus and A is cross-sectional area. As an Amazon Associate we earn from qualifying purchases. For instance, at point C where the concentrated load of 10 kips is located in the beam, the change in shearing force in the shear force diagram is 16 k - 6k = 10 kips. We find the net external force by adding together the external forces acting on the system (see the free-body diagram in the figure) and then use Newtons second law to find the acceleration. None of the forces between components of the system, such as between the teachers hands and the cart, contribute to the net external force because they are internal to the system. Similarly, a car accelerates because the ground pushes forward on the car's wheels in reaction to the car's wheels pushing backward on the ground. . The characteristics of a rocker support are like those of the roller support. There are 3 different kinds of constraints we will focus on in this course and they each have different reaction forces and moments: Notice that the Fixed restraint is the most restrictive and the roller is the least restrictive. Draw the shearing force and bending moment diagrams for the compound beam subjected to the loads shown in Figure 4.9a. Draw the shearing force and bending moment diagrams for the frame subjected to the loads shown in Figure 4.11a. By definition, the bending moment at a section is the summation of the moments of all the forces acting on either side of the section. F A common misconception is that rockets propel themselves by pushing on the ground or on the air behind them. Which language's style guidelines should be used when writing code that is supposed to be called from another language? If the problem involves forces, then Newtons laws of motion are involved, and it is important to draw a careful sketch of the situation. Let x be the distance of an arbitrary section from the free end of the cantilever beam (Figure 4.4b). how to determine the direction of support reactions in a truss? Basically: Reaction forces and moments (or constraints) show how motion is restricted, here that is in 2 dimensions. Applying the conditions of equilibrium suggests the following: Shearing force function. Newtons second law can be used to find Fprof. So what you need to work out is the axial force each side of where F is applied. feetonwall https://eng.libretexts.org/Bookshelves/Civil_Engineering/Book%3A_Structural_Analysis_(Udoeyo)/01%3A_Chapters/1.03%3A_Equilibrium_Structures_Support_Reactions_Determinacy_and_Stability_of_Beams_and_Frames. A bending moment is considered positive if it tends to cause concavity upward (sagging). However, the scale does not measure the weight of the package; it measures the force \( \vec{S}\) on its surface. OpenStax is part of Rice University, which is a 501(c)(3) nonprofit. Only external forces are shown on free-body diagrams, not acceleration or velocity. The phrase on either side 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. Reaction forces and moments are how we model constraints on structures. Thus, the expression for the bending moment of the 5 k force on the section at a distance x from the free end of the cantilever beam is as follows: Bending moment diagram. If we select the swimmer to be the system of interest, as in the figure, then Fwall on feet is an external force on this system and affects its motion. A car accelerates forward because the ground pushes forward on the drive wheels, in reaction to the drive wheels pushing backward on the ground. {cos}60^o}{2.0\text{ kg}} \quad \text{(plug in the horizontal . A roller support allows rotation about any axis and translation (horizontal movement) in any direction parallel to the surface on which it rests. We can readily see Newtons third law at work by taking a look at how people move about. If the structure is stable and determinate, proceed to the next step of the analysis. First, compute the reactions at the support. Cable. This reaction force, which pushes a body forward in response to a backward force, is called. A graphical representation of the bending moment acting on the beam is referred to as the bending moment diagram. Next, as in Figure 4.10, use vectors to represent all forces. floor These techniques also reinforce concepts that are useful in many other areas of physics. The numerical value of the change should be equal to the value of the concentrated load. The reactions at the supports are shown in the free-body diagram of the beam in Figure 4.7b. The bending moment diagram is a curve in portion AB and is straight lines in segments BC and CD. It is important to remember that there will always be a sudden change in the shearing force diagram where there is a concentrated load in the beam. [BL][OL][AL] Demonstrate the concept of tension by using physical objects. F we get 5*10 = 50 kN. This force is significantly less than the 150-N force the professor exerted backward on the floor. 6.9 A cable subjected to a uniform load of 300 N/m is suspended between two supports at the same level 20 m apart, as shown in Figure P6.9. Determining forces in members due to applied external load. This page titled 5.6: Newtons Third Law is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. F Fx = ma. A 45.0 kg box is pulled with a force of 205 N by a rope held at an angle of 46.5 degrees to the horizental. This is a graphical representation of the variation of the shearing force on a portion or the entire length of a beam or frame. 4.1. Shear force: The shear force at any section of a beam is determined as the summation of all the transverse forces acting on either side of the section. Whenever a first body exerts a force on a second body, the first body experiences a force that is twice the magnitude but acts in the direction opposite the direction of the applied force. The free-body diagram of the beam is shown in Figure 4.10a. Draw the shearing force and bending moment diagrams for the cantilever beam subjected to the loads shown in Figure 4.6a. For example, the force exerted by the teacher on the cart is of equal magnitude but in the opposite direction of the force exerted by the cart on the teacher. This means that the rocket exerts a large force backward on the gas in the rocket combustion chamber, and the gas, in turn, exerts a large force forward on the rocket in response. We start with, The magnitude of the net external force on System 2 is. The word tension . Boolean algebra of the lattice of subspaces of a vector space? They actually work better in a vacuum, where they can expel exhaust gases more easily. is an external force on the swimmer and affects her motion. Note that this applies only to 2d restraints. The student is expected to: He should throw the object upward because according to Newtons third law, the object will then exert a force on him in the same direction (i.e., upward). Once the system is identified, its possible to see which forces are external and which are internal (see Figure 4.10). Ask students what the difference is between the two. Horizontal. The reactions at the supports of the frame can be computed by considering the free-body diagram of the entire frame and part of the frame. Because the swimmer is our system (or object of interest) and not the wall, we do not need to consider the force Not all of that 150-N force is transmitted to the cart; some of it accelerates the professor. Engineering Mechanics: Statics by Libby (Elizabeth) Osgood; Gayla Cameron; Emma Christensen; Analiya Benny; and Matthew Hutchison is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted. Because the package is not accelerating, application of the second law yields, \[\vec{S} - \vec{w} = m \vec{a} = \vec{0},\]. 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 (see Figure 4.2c and Figure 4.2d). Note that because the shearing force is a constant, it must be of the same magnitude at any point along the beam. Label the forces carefully, and make sure that their lengths are proportional to the magnitude of the forces and that the arrows point in the direction in which the forces act. The mass of the system is the sum of the mass of the teacher, cart, and equipment. then you must include on every digital page view the following attribution: Use the information below to generate a citation. See the free-body diagram in the figure. In Pfafian form this constraint is y = 0 and y = 0. Shearing force and bending moment functions. Because acceleration is in the same direction as the net external force, the swimmer moves in the direction of feetonwall Canadian of Polish descent travel to Poland with Canadian passport, A boy can regenerate, so demons eat him for years. (b) Suppose that the blocks are later separated. So what you need to work out is the axial force each side of where F is applied. Another chapter will consider forces acting in two dimensions. [AL] Start a discussion about action and reaction by giving examples. Figure 5.6.3: The runner experiences Newton's third law. net F Helicopters create lift by pushing air down, thereby experiencing an upward reaction force. Let the shear force and bending moment at a section located at a distance of x from the left support be V and M, respectively, and at a section x + dx be V + dV and M + dM, respectively. 3.4.2 Roller Support. Determine the position and the magnitude of the maximum bending moment. Draw the shearing force and the bending moment diagrams for the frames shown in Figure P4.12 through Figure P4.19. The original material is available at: In this case, both forces act on the same system, so they cancel. For accurate plotting of the bending moment curve, it is sometimes necessary to determine some values of the bending moment at intermediate points by inserting some distances within the region into the obtained function for that region. The sign convention adopted for shear forces is below. Figure out which variables need to be calculated; these are the unknowns. You put a force to show how the restraint restricts motion. . If the astronaut in the video wanted to move upward, in which direction should he throw the object? Since the beam is constrained we know that the total elongation/deformation is 0. Have you searched on here? Support reactions. and you must attribute Texas Education Agency (TEA). In this case, there are two different systems that we could choose to investigate: the swimmer or the wall. It restrains the structure from movement in a vertical direction. The strategy employed to find the force of tension is the same as the one we use to find the normal force. Shearing force and bending moment diagrams. The computed values of the shearing force and bending moment are plotted in Figure 4.6c and Figure 4.6d. Considering the equilibrium of part CDE of the frame, the horizontal reaction of the support at E is determined as follows: Again, considering the equilibrium of the entire frame, the horizontal reaction at A can be computed as follows: Shear and bending moment of the columns of the frame. Next, make a list of knowns and unknowns and assign variable names to the quantities given in the problem. The spring force is called a restoring force because the force exerted by the spring is always . Calculate the acceleration produced when the professor exerts a backward force of 150 N on the floor. Rockets move forward by expelling gas backward at a high velocity. Explain how forces can be classified as internal or external to the system of interest. \(Fig. The following section provides a second explanation on reactions & supports: A pin support allows rotation about any axis but prevents movement in the horizontal and vertical directions. This decision is important, because Newtons second law involves only external forces. F Equation 4.3 suggests the following expression: Equation 4.4 states that the change in the shear force is equal to the area under the load diagram. This book uses the It only takes a minute to sign up. of 150 N. According to Newtons third law, the floor exerts a forward force Similarly, a car accelerates because the ground pushes forward on the car's wheels in reaction to the car's wheels pushing backward on the ground. In contrast, the force Ffeet on wall acts on the wall, not on our system of interest. We know from Newtons second law that a net force produces an acceleration; so, why is everything not in a constant state of freefall toward the center of Earth? What are the forces acting on the first peg? If you are redistributing all or part of this book in a print format, How to draw force diagrams Statics Reactions. To push the cart forward, the teachers foot applies a force of 150 N in the opposite direction (backward) on the floor. What is this brick with a round back and a stud on the side used for? Note that forces acting in opposite directions have opposite signs. . Use the questions in Check Your Understanding to assess whether students have mastered the learning objectives of this section. The expression also shows that the shearing force varies linearly with the length of the beam. F When a person pulls down on a vertical rope, the rope pulls up on the person (Figure \(\PageIndex{2}\)). citation tool such as, Authors: Paul Peter Urone, Roger Hinrichs. Whenever a first body exerts a force on a second body, the first body experiences a force that is twice the magnitude and acts in the direction of the applied force. For the situation shown in Figure 5.2.5, the third law indicates that because the chair is pushing upward on the boy with force \(\vec{C}\), he is pushing downward on the chair with force \( \vec{C}\). Draw the shearing force and bending moment diagrams for the beam with an overhang subjected to the loads shown in Figure 4.7a. He also rips off an arm to use as a sword. Bending moment: The bending moment at a section of a beam can be determined by summing up the moment of all the forces acting on either side of the section. foot In equation form, we write that. Creative Commons Attribution License F Namely, we use Newton's second law to relate the motion of the object to the forces involved. cart A diagram showing the variation of the shear force along a beam is called the shear force diagram. Ra. The reaction force R is at right angles to the ramp. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Note that the distance x to the section in the expressions is from the right end of the beam.

David Spence Obituary, Presentation About Doctor Career, Articles H