Chapter 13 (cont..8) - Direction of increase in Moment and that of Depth of beam

In the previous section we derived the equation for the 'Net shear force' V_u,net for a beam of varying depth. Now we will see a beam in which the depth varies in the opposite direction. Fig.13.38 below shows such a beam PQ of varying depth. Here also the bottom surface makes an angle β with the horizontal.

Fig.13.38

Beam of varying depth

Just as in the previous example, we can get M_u and V_u at any section xx along the length of the beam. Let us see the direction of T_u which is shown in fig.13.39 below:

Fig.13.39

Forces at section XX

As in our previous beam, here also, T_u will make the angle β ,with the horizontal (∵ the bottom surface of the beam is inclined at the angle β with the horizontal). So T_u will have a horizontal component and a vertical component as shown in the next fig.13.40

Fig.13.40

Resolving T_u into vertical and horizontal components

We are concerned about the vertical component because it is acting in the same direction as the vertical shear force V_u. So we have to determine the resultant of the 'vertical component of T_u' and V_u.

Steps for deriving the following equations from 13.22 to 13.28 are same as those of the previous beam AB. But we will do them again for clarity.

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From the triangle oqr in the above fig., we get

Eq.13.22: tan β = qr/or. So we get:

Eq.13.23: qr = or tan β. 

But we have:

• qr = op = the vertical component of T_u , and

• or is the horizontal component of T_u

or, which is the horizontal component of T_u, can be obtained from the basic equation:

Eq.13.24: Bending moment =M_u =C_u x z =(horizontal component of T_u) x z

[∵ C_u = Horizontal component of T_u. (see fig.13.41 below)]

From this we get: 

Eq.13.25: C_u =Horizontal component of T_u =M_u /z

Here C_u is the compressive force, and z is the lever arm as shown in fig.13.41 [Note that we are not using T_u. We are using the 'Horizontal component of T_u'. Because the lever arm z is always measured as the perpendicular distance]

Fig.13.41

Bending Moment at the section

In the above fig., z is measured as the distance between C_u and the horizontal component of T_u. It can be approximately taken to be equal to d, the effective depth at the section.

So we can modify Eq.13.25 as follows:

Eq.13.26: C_u =Horizontal component of T_u =M_u /d

But the horizontal component of T_u is the or in fig.13.36 above. So, substituting 13.26 in place of or in Eq.13.23 we will get qr. This qr is equal to op, the vertical component. So we get:

Eq.13.27: The vertical component of T_u =(M_u /d)tan β

• In a previous fig.13.39, we saw V_u is acting upwards, and

• In the above fig.13.41, we find that the 'vertical component of T_u' is also acting upwards, and it’s magnitude is given by Eq.13.27

• So we get the net shear force V_u,net at the section as:

Eq.13.28:

So, in the case of beams with varying depth, this V_u,net has to be used instead of V_u while calculating the design shear stress τ_v (using Eq.13.14)

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Comparison between the two beams

Let us compare figs.13.36 and 13.40, which shows the resolution of force T_u. Both are shown below for easy comparison:

Fig.13.36:

Fig.13.40:

From the above two figures, we can easily see that for the first beam AB, The force T_u lies in the fourth quadrant, while for the second beam PQ, it lies in the first quadrant.

Now we will compare the bending moment diagrams of the two beams. The following fig.13.42 shows both the beams together:

Fig.13.42

Increase in depth and BM comparison

From the above fig., we can see that for beam AB, direction of increase of depth is same as the direction of increase of BM. But for PQ, the depth decreases in the direction of increase of BM. In the former case we derived Eq.13.21 and for the latter case we derived Eq. 13.28.

Eq.13.21:

Eq.13.28:

The difference between the two equations is in the sign '+' OR '-' between the two terms. Thus we conclude that:

• When the direction of increase of depth is same as the direction of increase of BM, we use eq 13.21 with the '-' sign and

• When the direction of increase of depth is opposite to the direction of increase of BM, we use eq 13.28 with the '+' sign

The above equations can be written together in a general form as:

Eq.13.30:

Thus we get a general equation for the net shear force at the section. So just as we wrote Eq.13.14 for the factored shear stress for a beam of uniform depth, we can write the factored shear stress equation for a beam of varying depth as:

Eq.13.31:

This is the same equation given in cl.40.1.1 of the code. This completes our discussion on the shear stress in beams of varying depth. Before we go to the next section, there are two things that should be mentioned about the above discussion.

• Beams AB and PQ were shown as simply supported. The usual notation of hinged supports were shown in the diagrams. The BM diagrams shown were also of a simply supported beam. In an actual problem, we must carefully examine the type of supports and the type of loading, and draw the BM and SF diagrams accordingly.

• For a beam with uniformly varying depth, the loading diagram of the self weight will be of a trapezoidal shape. In addition, there may be uniform loads coming from the slab, concentrated loads etc., Because of the presence of the trapezoidal loading, the BM equation will be of the third degree (ie., the equation will have a term with x³). And the shear force equation will be of the second degree. So both the BM and SF diagrams will be curves. We used the more familiar shapes of a parabola and straight line for BM and SF for our discussion. This was for simplicity. In the discussion, we just wanted to show that the value of M_u and V_u at the section XX should be taken.  In an actual problem, exact analysis should be done, and correct BM and shear force diagram should be drawn to find M_u and V_u at section XX

Eqs.13.14 and 13.31 gives us the design shear stress at a section. At the first glance we may think that we must provide steel to resist this much stress. But there is another 'stress' which is playing a major role here. This 'stress' acts in the direction opposite to the 'external applied stress'. Because of this, we don’t have to provide steel for the entire quantity given by eq. 13.14 and 13.31. We will discuss about it in the next section.

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