Shear and moment diagram - Wikipedia, the free encyclopedia Page 1 of 10 Shear and moment diagram From Wikipedia, the free encyclopedia Shear and bending moment diagrams are analytical tools used in conjunction with structural analysis to help perform structural design by determining the value of shear force and bending moment at a given point of a structural element such as a beam. These diagrams can be used to easily determine the type, size, and material of a member in a structure so that a given set of loads can be supported without structural failure. Another application of shear and moment diagrams is that the deflection of a beam can be easily determined using either the moment area method or the conjugate beam method. Shear and moment diagram for a simply supported beam with a concentrated load at mid-span.(right) Contents ◾ 1 Convention ◾ 1.1 Normal convention ◾ 1.2 Alternative drawing convention ◾ 2 Calculating shear force and bending moment ◾ 2.1 Step 1: Compute the reaction forces and moments ◾ 2.2 Step 2: Break beam into segments ◾ 2.3 Step 3: Compute shear forces and moments - first piece ◾ 2.4 Step 4: Compute shear forces and moments - second piece ◾ 2.5 Step 5: Compute shear forces and moments - third piece ◾ 2.6 Step 6: Compute shear forces and moments - fourth piece ◾ 2.7 Step 7: Compute deflections of the four segments ◾ 2.8 Step 8: Apply boundary conditions ◾ 2.9 Step 9: Solve for Mc and Ra ◾ 2.10 Step 10: Plot bending moment and shear force diagrams ◾ 2.11 Relationship between shear force and bending moment http://en.wikipedia.org/wiki/Shear_and_moment_diagram 2/17/2015 Shear and moment diagram - Wikipedia, the free encyclopedia Page 2 of 10 ◾ 3 Relationships between load, shear, and moment diagrams ◾ 4 Practical considerations ◾ 5 See also ◾ 6 References ◾ 7 Further reading ◾ 8 External links Convention Although these conventions are relative and any convention can be used if stated explicitly, practicing engineers have adopted a standard convention used in design practices. Normal convention The normal convention used in most engineering applications is to label a positive shear force one that spins an element clockwise (up on the left, and down on the right). Likewise the normal convention for a positive bending moment is to warp the element in a "u" shape manner (Clockwise on the left, and counterclockwise on the right). Another way to remember this is if the moment is bending the beam into a "smile" then the moment is positive, with compression at the top of the beam and tension on the bottom. [1] This convention was selected to simplify the analysis of beams. Since a horizontal member is usually analyzed from left to right and positive in the vertical direction is normally taken to be up, the positive shear convention was chosen to be up from the left, and to make all drawings consistent down from the Normal positive shear force convention (left) and normal bending moment right. The positive bending convention was convention (right). chosen such that a positive shear force would tend to create a positive moment. Alternative drawing convention In structural engineering and in particular concrete design the positive moment is drawn on the tension side of the member. This convention puts the positive moment below the beam described above. A convention of placing moment diagram on the tension side allows for frames to be dealt with more easily and clearly. Additionally placing the moment on the tension side of the member shows the general shape of the deformation and indicates on which side of a concrete member rebar should be placed, as concrete is weak in tension. [2] Calculating shear force and bending moment With the loading diagram drawn the next step is to find the value of the shear force and moment at any given point along the element. For a horizontal beam one way to perform this is at any point to "chop off" the right end of the beam. The example below includes a point load, a distributed load, and an applied moment. The supports include both hinged supports and a fixed end support. The first drawing shows the beam with the applied forces and displacement constraints. The second drawing is the loading diagram with the reaction values given without the calculations shown or what most people call a free body diagram. The third drawing is the shear force diagram and the fourth drawing is the http://en.wikipedia.org/wiki/Shear_and_moment_diagram 2/17/2015 Shear and moment diagram - Wikipedia, the free encyclopedia Page 3 of 10 bending moment diagram. For the bending moment diagram the normal sign convention was used. Below the moment diagram are the stepwise functions for the shear force and bending moment with the functions expanded to show the effects of each load on the shear and bending functions. The example is illustrated using United States customary units. Point loads are expressed in kips (1 kip = 1000 lbf = 4.45 kN), distributed loads are Loaded beam expressed in k/ft (1 k/ft = 1 kip/ft = 14.6 kN/m), moments are expressed in ft-k (1 ft-k = 1 ft-kip = 1.356 kNm), and lengths are in ft (1 ft = 0.3048 m). Step 1: Compute the reaction forces and moments The first step obtaining the bending moment and shear force equations is to determine the reaction forces. This is done using a free body diagram of the entire beam. The beam has three reaction forces, Ra, Rb at the two supports and Rc at the clamped end. The clamped end also has a reaction couple Mc. These four quantities have to be determined using two equations, the balance of forces in the beam and the balance of moments in the beam. Four unknowns cannot be found given two independent equations in these unknown variables and hence the beam is Free-body diagram of whole beam statically indeterminate. One way of solving this problem is to use the principle of linear superposition and break the problem up into the superposition of a number of statically determinate problems. The extra boundary conditions at the supports have to be incorporated into the superposed solution so that the deformation of the entire beam is compatible. From the free-body diagram of the entire beam we have the two balance equations Summing the forces, we have and summing the moments around the free end (A) we have We can solve these equations for Rb and Rc in terms of Ra and Mc : and If we sum moments about the first support from the left of the beam we have http://en.wikipedia.org/wiki/Shear_and_moment_diagram 2/17/2015 Shear and moment diagram - Wikipedia, the free encyclopedia Page 4 of 10 If we plug in the expressions for Rb and Rc we get the trivial identity 0 = 0 which indicates that this equation is not independent of the previous two. Similarly, if we take moments around the second support, we have Once again we find that this equation is not independent of the first two equations. We could also try to compute moments around the clamped end of the beam to get This equation also turns out not to be linearly independent from the other two equations. Therefore, the beam is statically indeterminate and we will have to find the bending moments in segments of the beam as functions of Ra and Mc. Step 2: Break beam into segments After the reaction forces are found, you then break the beam into pieces. The location and number of external forces on the member determine the number and location of these pieces. The first piece always starts from one end and ends anywhere before the first external force. Step 3: Compute shear forces and moments - first piece Let V1 and M1 be the shear force and bending moment in a cross- section of the first beam segment, respectively. As the section of the beam moves towards the point of application of the external force the magnitudes of the shear force and moment may change. This makes the shear force and bending moment a function of the position of cross-section (in this example x). By summing the forces along this segment and summing the moments, the equations for the shear force and bending moment are obtained. These equations are: Free-body diagram of segment 1 and Therefore, Step 4: Compute shear forces and moments - second piece Taking the second segment, ending anywhere before the second internal force, we have and http://en.wikipedia.org/wiki/Shear_and_moment_diagram 2/17/2015 Shear and moment diagram - Wikipedia, the free encyclopedia Page 5 of 10 Therefore, Free-body diagram of segment 2 Notice that because the shear force is in terms of x, the moment equation is squared. This is due to the fact that the moment is the integral of the shear force. The tricky part of this moment is the distributed force. Since the force changes with the length of the segment, the force will be multiplied by the distance after 10 ft. i.e. (x-10) the moment location is defined in the middle of the distributed force, which is also changing. This is where (x+10)/2 is derived from. Alternatively, we can take moments about the cross-section to get Again, in this case, Step 5: Compute shear forces and moments - third piece Taking the third segment, and summing forces, we have and summing moments about the cross-section, we get Free-body diagram of segment 3 Therefore, and Notice that the distributed force can now be considered one force of 15 kips acting in the middle of where it is positioned.