Chapter 13 (cont..2) - Stresses at Neutral axis

In the previous section we saw how to calculate the stresses and their directions on the particles situated at various parts in the body of the beam. But when we consider the particles situated in the NA, there is a difference. We will discuss about it in this section.

At the NA, there is no stress due to bending, and so, f_x is equal to zero, but q is not equal to zero. In fact, q is maximum at the NA as we saw earlier in the stress distribution (fig.13.5). So for the particles in the NA, the diagram representing the stresses acting on them will be as in fig. 13.7 instead of fig.13.8.

We know that this situation (involving the particle being acted upon only by the shear stress q) is also solved in the text books on 'Strength of Materials'. From those lessons, we know that the plane PQ makes 45^o with the vertical for such a particle. And the stress acting on PQ will have a value numerically equal to q. This is shown in fig.13.13 below:

Fig.13.13 :
Plane PQ of a particle in the NA

The direction of PQ in such a particle is shown in fig.13.14 below:

Fig.13.14 

Direction of plane PQ

Based on the above figs. 13.13 and 13.14, we can easily draw the details of the other principal plane RS. The force on RS will be compressive in nature, and the direction of RS will be perpendicular to PQ. These are shown in figs.13.15 and 13.16 below:

Fig.13.15

Plane RS of a particle in the NA

Fig.13.16

Direction of plane RS, on a particle in the NA

So now we are in a position to calculate the stresses and their directions on any particle in the beam.

Let us consider an actual simply supported beam AB made of a homogeneous material like steel or timber. It's dimensions are shown below in fig.13.17:

Fig.13.17

Simply supported beam

Let this beam carry a udl of 22 kN/m. including self weight.

The reactions at the supports will be: R_A = R_B  = 46.2 kN

The Bending moment at any section along the length of the beam in kNm is given by Eq.13.9: M = 46.2 x - 11 x².

The first differential of this equation will give the equation for SF. Thus:

Eq.13.10: V = 46.2 - 22 x

Where x is the distance of the section measured from support A.

We can consider a number of particles in the body of beam AB, and calculate the stresses and their directions in each of these particles. From Eqs.13.1, 13.5, 13.7 etc.,  we can see that the resultant stress on any particle will depend up on the shear force V and the bending moment M at the section in which the particle is situated. It will also depend on the distance y of the particle from the NA. So we need to know the exact position of the particle in the body of the beam. Let us form a grid as shown in fig.13.18 below:

Fig.13.18

In the above fig., the beam is divided horizontally into 12 equal parts, each of 35 cm width. (35 x 12 = 420 cm). It is also divided vertically into 8 equal parts, each of 5 cm height. (5 x 8 = 40 cm). We are going to take the particles at each 'point of intersection' of the horizontal and vertical grid lines.

We know that, when we consider the vertical grid lines, all particles in any one vertical grid line will be experiencing the same BM, regardless of their vertical distance from the NA. Similarly, all particles in any one vertical grid line will be experiencing the same Shear force V, regardless of their vertical distance from the NA. So we need not consider the horizontal grid lines for the BM and SF. There will be only one unique value for the BM, and another unique value for the SF at a vertical grid line. This is shown in table 13.1 below: (clicking on the figs. will give an enlarged view)

Table 13.1: Values of BM and SF

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Now we are ready to calculate the stresses. We are analysing the particles at the intersection of the grid lines. As the first step, the stress f_x on each of these particles is calculated using Eq.13.4:

Eq.13.4:

The values are given in table 13.2 below:

Table 13.2 : Values of f_x

Sample calculation:

Let us take the particle at the intersection of Grid 315 and Grid 30.

From table 13.1, M =36.383 KNm.

Distance of this particle from the NA = y = 30 -20 = 10 cm = 0.1 m

Moment of inertia of the whole section =I =bD³/12  = 0.0008 m⁴ 

Substituting these values in Eq.13.4 we get,  f_x = 454.78 kN/m² 

This is equal to 0.455 N/mm².

In the above table, the following points can be noted:

• The values are symmetric about the Grid 210, as this grid line marks the midpoint of the beam

• The values are symmetric about the Grid 20, as this grid line marks the NA. But the values have opposite signs. The values below the NA are positive, indicating Tension, and those above NA are negative, indicating compression.

Next we calculate the stress q on each of the particles using Eq.13.1. We will do it in the next section.

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