Only consider the forces to the left of the section.The shear force is the sum of all the force acting to the left of the section.Since the beam is in equilibrium, it must also be the sum of all the forces acting to the rightIf the beam is cut at this section as shown, a force F must be placed on the end to replace the shear force that was exerted by the material when joined. List all the forces to the left.Remember up is plus.There is a reaction force Ra up.There is a uniform load over the length x metres and this is equivalent to a downwards load of wx Newton.There is a point load F1 acting down.The total load to the left is F = Ra wx F1If the result for F is positive (up) then it produces positive shear.

In the case of uniformly distributed loads, the shear force increases proportional to length. If the u.d.l. is w N/m then the shear force increases by w Newton for every metre length.

Determining the position of greatest bending moment.

It can be shown that where ever the shear force changes from positive to negative, the bending moment is a maximum and where it changes from negative to positive the bending moment is a minimum. This means that anywhere on the SF diagram where the value goes through zero, the bending moment reaches a peak.On more complex beams, the SF diagram might pass through the zero value at several points resulting in several maximum and/or minimum points on the BM diagram. The bending moment that produces maximum stress could be the greatest positive value or the greatest negative value.

The complete diagram reveals that the bending moment is always negative (hogging) sotensile stress would occur on the top layer of the beam. This stress will be a maximum atx = 1 when the bending moment has a value of -500 Nm. The minus sin has no bearingon the stress value and serves to tell us that it is a hogging moment.

What is a Beam ?In a beam transverse load is acted, which in fact comes from the slabs to the column or walls.Length of the beam is much higher than its lateral dimensions. So axial strain developed in a beam will be very small compared to shear strain, or strain induced due to bending.So for design purpose of beams, analysis of shear force and bending moment induced are of the at most importance.

Shear ForceShear 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. So just by applying force balance in vertical direction on the free body diagram, we can determine value of shear force at a particular cross section.Bending MomentBut 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

Resultant of those forces will be zero, but it will produce a moment, to counter balance the external moment. So we can calculate moment induced at any cross section by balancing the external moment acting on the free body diagram.

This sign convention approach is valid for simply supported beam. For cantilever case sign convention is exactly opposite to this.Example of CantileverConsider this case, a cantilever carrying 3 loads.

Here we can start analysis from the free end.Section A-BSo 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.Section B-CBetween B and C effect of force F2also comes. So shear force becomes, F1plus F2. And in bending moment effect of F2also gets added. Similar analysis is done between section C and D alsoNow 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.Distributed load can be assumed as a point load passing through centroid of distributed load.Bending moment (point load) = Force x Distance ( This is actually for a horizontal beam with load acting is a point load) (Kg-mm or Kg-m)

Well a UDL is like a "rectangularly" distributed load, so the load effectively acts at the center of the beam i.e. at a distance of L/2. Since the entire length is L, the load is WL (since UDL = W N/m) so BM = Force*distance = (WL)(L/2)

For a horizontal simply supported beam of length L subject to udl w, the maximum bending moment is at the centre and equal to wL^2/8 distributed along the span parabolically. For a horizontal simply supported beam of length L, and subject to a point load P at mid-span, the maximum bending moment is PL/4. If the point load is applied at aL (0