Shear Force and Bending Moment
23573/0001.jpg cannot be found in /users/23573/0001.jpg. Please contact the submission author.
23287/SF-and-BM-0001.png cannot be found in /users/23287/SF-and-BM-0001.png. Please contact the submission author.
23287/SF-and-BM-0002.png cannot be found in /users/23287/SF-and-BM-0002.png. Please contact the submission author.
Types Of LoadA beam is normally horizontal and the loads vertical. Other cases which occur are considered to be exceptions. A Concentrated load is one which can be considered to act at a point, although in practice it must be distributed over a small area. A Distributed load is one which is spread in some manner over the length, or a significant length, of the beam. It is usually quoted at a weight per unit length of beam. It may either be uniform or vary from point to point.
Types Of SupportA Simple or free support is one on which the beam is rested and which exerts a reaction on the beam. It is normal to assume that the reaction acts at a point, although it may in fact act act over a short length of beam.
23287/SF-and-BM-0005.png cannot be found in /users/23287/SF-and-BM-0005.png. Please contact the submission author.
The Relationship Between W, F, M.In the following diagram is the length of a small slice of a loaded beam at a distance x from the origin O
23287/SF-and-BM-0004.png cannot be found in /users/23287/SF-and-BM-0004.png. Please contact the submission author.
Example - Example 1
- The shearing force suffers sudden changes when passing through a load point. The change is equal to the load.
- The bending Moment diagram is a series of straight lines between loads. The slope of the lines is equal to the shearing force between the loading points.
Uniformly Distributed Loads
Example - Example 3
Example - Example 4
Varying Distributed Loads.
Example - Example 7
The complete diagrams are shown. It can e seen that for a uniformly varying distributed load, the Shearing Force diagram consists of a series of parabolic curves and the Bending Moment diagram is made up of "cubic" discontinuities occurring at concentrated loads or reactions. It has been shown that Shearing Forces can be obtained by integrating the loading function and Bending Moment by integrating the Shearing Force, from which it follows that the curves produced will be of a successively "higher order" in x ( See equations (6) and(7))
Graphical SolutionsNote This method may appear complicated but whilst the proof and explanation is fairly detailed, the application is simple and straight forward. Earlier it was shown that the change of Bending Moment is given by the double Integral of the rate of loading. This integration can be carried out by means of a funicular polygon. See diagram. Suppose that the loads carried on a simply supported beam are are the reactions at the supports. Letter the spaces between the loads and reactions A, B, C, D, E, and F. Draw to scale a vertical line such that. Now take any point "O" to the left of the line and join O to a, b, c, d, and e. This is called The Polar Diagram Commencing at any point p on the line of action of draw pq parallel to Oa in the space "A" , qr parallel to Ob in the space "B" and similarly rs, st, and tu. Draw Of parallel to pu. It will now be shown that fa represents . Also, pqrstu is the Bending Moment diagram drawn on a base pu, M being proportional to the vertical ordinates. is represented by ab and acts through the point q; it can be replaced by forces aO along qp and Ob along qr. Similarly, can be replaced by forces represented by bO along rq and Oc along rs, by cO along sr and Od along st etc. All of these forces cancel each other out except aO along qp and Od along te, and these two forces must be in equilibrium with . This can only be so if is equivalent to a force Oa along pq and fO along up, being equivalent to eO along ut and Of along pu. Hence, is represented by fa and by ef.
23287/SF-and-BM-0031.png cannot be found in /users/23287/SF-and-BM-0031.png. Please contact the submission author.