Chapter 13 - Shear stress in beams

Upto the previous section, we were discussing the effects of bending on a beam or slab. When a beam or a slab is loaded, it bends. The beam resists the bending by developing an 'internal moment of resistance'. We discussed about it's details in the previous chapters. We analysed beam and slab sections to determine how much resisting moment they will offer at ultimate state. We designed beam and slab sections in such a way as to make them capable to resist the factored bending moments acting on them.

In addition to the bending moment, shear forces also act on the beam. We have learned about it in the 'Strength of Materials' classes. There we learned to draw shear force diagrams for various loading conditions like concentrated loads, uniformly distributed loads, uniformly varying loads etc., We also learned how to draw the shear force diagrams for the above types of loads, for various support conditions like simply supported, continuous, cantilever support etc., In this chapter we will learn how to make beam and slab sections capable to resist the factored shear forces acting on them. We will also learn how to analyse and determine the 'shear resistance' that a given beam is able to offer at ultimate state. Reinforced concrete beam is a non-homogeneous material consisting of concrete and steel. We will first discuss about the shear forces in a homogeneous material like timber or steel.

Consider a load applied on the beam in fig.13.1. It is made up of a number of wooden planks stacked one above the other.

Fig.13.1:

Beam made up of stacked wooden planks.

The planks are simply stacked, with no bonding between them. When a load is applied, the beam bends. The bending is shown in the animation in fig.13.2 below:

Fig.13.2:

Bending of the stacked planks

Fig.13.3:

Final deflected shape

We can see that there is some horizontal movement for each of the individual planks. That is., the planks slide past each other. So there exists a horizontal force which move the planks in the horizontal direction. If these planks were glued together, then the planks would not have moved horizontally. But then, the glue at the interfaces between the planks will experience a shearing force. Note that the force experienced by the glue is of 'shearing' in nature. It is not 'tensile' or 'compressive'. If it was tensile or compressive, then the planks will be pulling away from each other or pushing onto each other. But here, they are sliding past each other.

In the text books on 'Strength of Materials', a formula is derived for calculating this shear force:

Eq.13.1: q = ^VQ⁄_Ib

The details of this formula are shown in the fig. below:

Fig.13.4:

Horizontal shear acts on any horizontal plane EF

• EF is a horizontal plane at a distance of 'y' from the NA

• V is the Shear force at  the section

• Q is the moment of area (of the portion above EF) about the NA. It can be easily  calculated from the known values D, which is the total depth of the section, b, the width of the section, and y.

• I is the 'moment of inertia' of the section about the NA. For the rectangular section shown in the fig.,

Eq.13.2: I =bD³/12

We also know that this shear stress q have a parabolic distribution across the section. The maximum value will be at the peak of the parabola. It is at the NA, and the value there is 3V/2bD. This is shown in fig.13.5 below:

Fig.13.5:

Horizontal shear stress distribution across the section

So if we take any elementary particle from a beam, except from the top and bottom most fibres, there will be a horizontal shear stress on top and bottom of that particle, as shown in fig.13.6 given below:

Fig.13.6 

Horizontal shear stresses on a particle

But a particle acted upon by the forces shown in fig.13.6 will not be in equilibrium. It will spin. So there will be equal and opposite shear stresses on the vertical sides also. We have learned about this as the principle of 'Complementary shears'. So fig.13.6 can be modified as fig.13.7 given below:

Fig.13.7 

All the shear stresses acting on the particle

But these shear stresses are not the only stresses acting on the particle. We know that when the beam bends due to applied loads, the fibres above the NA will be in compression and the fibres below will be in tension. This stress is in a direction along the length of the beam. That is., it acts horizontally. We can denote it by f_x, and determine it from the basic bending equation:

Eq.13.3

From this we get

Eq.13.4

Where

• M = Bending moment acting on the section,

• y = distance of the particle from the NA, and

• I = Moment of inertia of the section about the NA, given by eq.13.2

So the particle that we consider will experience f_x also, in addition to the above shear stresses, and so the fig.13.7 can be modified again to get fig.13.8 given below:

Fig.13.8 

All the stresses acting on the particle

So now we know all the stresses acting on a particle of the beam. We have to learn about the changes that the particle will be subjected to, due to these stresses. The combined effect of these six stresses shown in fig.13.8 may result in any one of the following:
(i) The particle may be pulled from all directions, and thus it may be split up into two particles, or
(ii) It may be pushed from all directions and thus it may be crushed. 

The result will depend on the relative magnitude of the stresses, and the strength of the material. Some materials have greater tensile strength. Particles of such a material will not fail under a pulling force. Some materials have greater compressive strength. Particles of such a material will not fail under a pushing force. In the next section, we will try to make a mathematical formulation of the effect of the stresses.

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