Please forward this error screen sfd and bmd for simply supported beam pdf 158. Shear and Bending moment diagram for a simply supported beam with a concentrated load at mid-span.

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. 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 right. The positive bending convention was chosen such that a positive shear force would tend to create a positive moment. 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.

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 example is illustrated using United States customary units.

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.

0 which indicates that this equation is not independent of the previous two. Once again we find that this equation is not independent of the first two equations. 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. 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.

Let V1 and M1 be the shear force and bending moment respectively in a cross-section of the first beam segment. 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. By summing the forces along this segment and summing the moments, the equations for the shear force and bending moment are obtained. 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. Notice that the distributed force can now be considered one force of 15 kips acting in the middle of where it is positioned. By plotting each of these equations on their intended intervals, you get the bending moment and shear force diagrams for this beam. We now use the Euler-Bernoulli beam theory to compute the deflections of the four segments. E is the Young’s modulus and I is the area moment of inertia of the beam cross-section. Now we will apply displacement boundary conditions for the four segments to determine the integration constants.

Also, the slopes of the deflection curves at this point are the same, i. We can now calculate the reactions Rb and Rc, the bending moments M1, M2, M3, M4, and the shear forces V1, V2, V3, V4. These expressions can then be plotted as a function of length for each segment. It is important to note the relationship between the two diagrams.

The moment diagram is a visual representation of the area under the shear force diagram. That is, the moment is the integral of the shear force. Another note on the shear force diagrams is that they show where external force and moments are applied. With no external forces, the piecewise functions should attach and show no discontinuity.