Chapter 13 (cont..7) - Shear in reinforced concrete beams of Varying Depth

In the previous section we derived the equation for the shear stress that has to be used for the design. This equation was derived for a rectangular beam of 'uniform depth'. Now we will derive the 'modified form' of this equation, which can be used in the case of beams with 'varying depth'.

Fig.13.33 below shows a simply supported beam AB of varying depth. The bottom surface of the beam makes an angle β with the horizontal.

Fig.13.33

Beam of varying depth

In an ordinary beam, the design shear stress is calculated using Eq.13.14 which we saw in the previous section. In that equation we need V_u, b and d only. But for a beam of varying depth, M_u will also come into the picture. Let us see how this is so:

We know that, whatever be the type of loading (udl, concentrated, uniformly varying etc.,) acting on the beam, we can calculate the factored BM  M_u and SF V_u  at any section XX on the beam. This is shown in fig.13.34 below.

Fig.13.34

M_u and V_u  at section XX

Now let us see the direction of the tensile force T_u in the steel reinforcement. For an ordinary beam, T_u would be perfectly horizontal. But our beam, being of varying depth, the steel bar is not horizontal, and so the axial tensile force in the bar is also not horizontal. This is shown in fig.13.35 below.

Fig.13.35

Forces at section XX

We can easily see that T_u will make the same angle β with the horizontal. As T_u is not horizontal, but acting at an angle, it will have a horizontal component and a vertical component as shown in the fig.13.36 below :

Fig.13.36

Resolving T_u into vertical and horizontal components

Note the vertical component. It is acting in the same direction as the vertical shear force V_u. So we have to determine the resultant of: (i) The vertical component of T_u and (ii) V_u .

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

Eq.13.15: tan β = qr/or. So we get:

Eq.13.16: 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.13.17: Bending moment =M_u =C_u x z =(horizontal component of T_u) x z

[∵ C_u = Horizontal component of T_u. (see fig.13.37 below)]

From this we get: 

Eq.13.18: 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.13.37 [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.13.37

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.13.18 as follows:

Eq.13.19: 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 13.19 in place of or in Eq.13.16 we will get qr. This qr is equal to op, the vertical component. So we get:

Eq.13.20: The vertical component of T_u =(M_u /d)tan β

• In a previous fig.13.35, we saw V_u is acting upwards, and

• In the above fig.13.37, we find that the 'vertical component of T_u' is acting downwards, and it’s magnitude is given by Eq.13.20

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

Eq.13.21:

So, in the case of beams with varying depth, this V_u,net has to be used instead of V_u to calculate the design shear stress τ_v . In the next section we will see a type of beam in which the depth varies in the other direction.

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