Beams are structural elements constructed of wood, steel and concrete that are designed for buildings, bridges and many types of structures to resist forces that cause bending and shear. Both mechanisms of bending and shear are caused by loads and forces applied to a beam and create a stress (force applied over a unit of area) in that section. The way in which shear forces affect a beam are quite different than that of bending and in some ways can be much more intense.
Shear forces are applied to beams through normal loading conditions and typically occur as a summation of forces that are transferred through a beam at a location of bearing where it is supported which results in a high concentration of forces that meet abruptly that sort of "rip" through the beam. The cross sectional area of and the material strength of the beam is what resists the forces of shear. The cross sectional area is of the beam that transfers the forces of shear is the shear plane. Beams that have less resistance to shear stress can usually be strengthened by increasing its cross sectional area. Beams in general will usually have more than adequate resistance to shear unless the beams are supporting very high loads.
Along the length of the beam, the bending and shear forces and stresses do not look the same. If one considers a cross sectional cut of the beam at one foot increments along the length of the beam, one would find that at each location the level of bending stress and shear stress is different. Of course the profile of these stresses is determined by the loading of the beam, but in general for a simply supported beam loaded with a uniformly distributed load, the location of the maximum bending moment is at the center of the length of beam and the maximum forces of shear are located at the reactions of the beam where it is supported.
The shear stress in a beam is calculated as follows.
Ï = V / A
The shear stress is Ï, the force of shear acting on the beam is V and A is the cross sectional area or shear plane that transfers the force.
An example will help to make this concept more easily understood.
There is a 12" x 18" simply supported concrete beam that is 30 ft long and carries a uniformly distributed load of 300 lb/ft. Determine the shear stress acting on the beam.
If we assume the unit weight of the concrete to be, Î³c = 150 lb/ft3, we have an additional distributed load (self weight) of (1 ft)(1.5 ft)(150 lb/ft3) = 225 lb/ft3. So the total uniform load is 525 lb/ft. The total load acting on the beam is (525 lb/ft)(30 ft) = 15,750 lb, and half of this load is transferred as shear through the concrete beam at each reaction or support location. Therefore the shear force, V = 15,750 lb / 2 = 7,875 lb. The cross sectional area of the beam is (12 in.)(18 in.) = 216 in2. The shear stress can be calculated as follows.
Ï = V / A
Ï = (7,875 lb) / (216 in2)
Ï = 36.5 lb/ in2 = 36.5 psi
Gere, James M. (2004). Mechanics of Materials (6th ed.). USA: Brooks/Cole â Thomson Learning.