Beams are structural members, which are most commonly used in buildings. Beams have numerous other applications in case of bridges, automobiles or in mechanical systems. In this article we will see how we can do strength analysis of a beam.Following article gives detailed description of the video lecture.
What is a Beam ?
In a beam transverse load is acted, which in fact comes from the slabs to the column or walls. It is clear from following figure that, beams are integral part of of building structure. In all of the beams load acted is transverse, as shown.
|Fig.1 In beam transverse load is acted, and it is an integral part of building structure|
|Fig.2 Building structure consists of many beam-column systems, beam is a part of beam-column system|
|Fig.3 Axial strain in beam is negligible compared to shear strain|
Both shear force and bending moment are induced in beam in order to balance external load acting on it. We will go through details of it separately.
Shear force is the internal resistance created in beam cross sections, in order to balance transverse external load acting on beam. Consider following beam, it does not matter from where you take a section, when you add forces acting on it, it should be in equilibrium. Shear force is induced exactly for this purpose, to bring the section to equilibrium in vertical direction. It acts parallel to cross section.
|Fig.4 Shear force is induced in a section to balance the external load|
|Fig.5 Usual sign convention of shear force|
But balance of transverse forces alone does not guarantee equilibrium of a section. There is another possibility of beam rotation, if moment acting on it is not balanced. If this is the case a bending moment will be induced in cross section of beam, to arrest this rotation. It will be induced as normal forces acting on fiber cross section as shown.
|Fig.6 Bending moment is induced in section to balance external moment, section is zoomed in left figure for better viewing|
Sign convention of bending moment is as follows.
|Fig.7 Sign convention of bending moment for simply supported case|
With these concepts developed, we can easily calculate distribution of shear force and bending moment along the length of the beam. We will see few examples.
Example of Cantilever
Consider this case, a cantilever carrying 3 loads.
|Fig.8 Analysis of cantilever carrying 3 loads|
So for between A and B, if you take a section the only external force acting on it is F1. So a shear force should induce in section to balance this force. So value of shear force between A and B is F1. But force balance alone does not guarantee equilibrium of the section. There is an external moment on the section. So a bending moment will be induced in section, in order to balance the external moment. Since value of external moment is F into x, bending moment will vary linearly.
Between B and C effect of force F2 also comes. So shear force becomes, F1 plus F2. And in bending moment effect of F2 also gets added. Similar analysis is done between section C and D also. So SFD and BMD of this problem would look like this.
|Fig.9 SFD and BMD of cantilever beam|
Simply Supported Case
Now consider this problem. A simply supported beam with uniformly distributed load. First step here would be determination of reaction forces. Since the problem is symmetrical reaction forces will be equal, and will be half of total load acting on beam.
|Fig.10 A simply supported beam with uniformly distributed load in it|
Lets start analysis from point A. If you take section between point A and B, it should be in equilibrium. So shear force will have equal magnitude of Reaction force. Bending moment gives a linear variation.
But after point B, effect of point load and distributed load come. Effect of distributed load is something interesting. Take a section in BC. In this section, along with two point loads there is a distributed load also. This distributed load can be assumed as a point load passing through centroid of distributed load. Value of point load is U( x - L/3). And it is at a distance (x - L/3)/ 2 from section line. So shear force will have one more term, which comes from distributed load. From the equation its clear that shear force varies linearly.
You can easily predict, how bending moment varies along length, from the same force diagrams. Since this equation is quadratic it will have a parabolic shape. Same procedure is repeated in remaining section. Since this problem is symmetrical in nature, S.F.D and B.M.D would also be symmetrical. It is shown in figure below.
|Fig.11 SFD and BMD of simply supported beam|