Chapter 15.4 - Internal forces in an RCC beam

In the previous sections we completed the discussion on the methods to determine the theoretical cut-off points. We saw the exact method as well as the approximate method. Both the methods give us 'theoretical points'. As mentioned earlier, the bars should not be stopped at these theoretical points. They must be continued further. This means that the 'actual' cut-off points will be different from the 'theoretical' ones. 

One of the obvious reasons for not accepting the theoretical points can be stated as follows: We have used the BM diagrams for calculating the points. But these diagrams can undergo unexpected changes. These unexpected changes may occur due to changes in loads, changes in positions of Live loads, sinking of supports etc., So the points obtained from BM diagrams cannot be relied upon. In this and the following sections, we will explore more reasons for continuing the bars beyond the theoretical points, and also 'how much' further we have to take the bars.

Consider a simply supported beam. When the factored load is applied on the beam, flexural cracks (‘flexural cracks’ means the cracks formed due to the bending of the beam) will be first appearing in the midspan region as shown in the fig.15.19 below:

Fig.15.19
Flexural cracks appearing at midspan region

Also, the midspan section will reach the ultimate state. That is., the steel at the midspan section will yield, and the concrete strain at that section will become 0.002. While the midspan section is undergoing these changes, other sections which are at some distance away from the midspan will still be in a safe condition. This is because, these other sections have a 'capacity' (we measure this ‘capacity’ in terms of the ultimate moment of resistance, MuR) which is greater than the applied factored bending moment at these sections.

But when we do 'curtailment of bars', the section or sections at which the curtailment is done, will also become critical sections. This is because, the applied factored bending moment at a section where curtailment is done, will be comparable to the capacity of the section. At such sections flexural cracks will begin to form as shown in the fig.15.20 below:

Fig.15.20
Flexural cracks appearing at a section XX at which curtailment is done

These flexural cracks may develop into diagonal tension cracks as shown in the fig.15.21 below. The possibility of formation of such diagonal cracks is high if heavy shear force is present near the section at which curtailment is done.

Fig.15.21
Flexural cracks developed into a diagonal tension crack

We will now analyse the forces in this cracked beam. For this, we will first look at an ordinary un-cracked beam shown in fig.15.22 below.

Fig.15.22
Simply supported beam with a factored udl of wu kN/m

From the basic lessons of strength of materials, we know how to calculate the applied bending moment at any section XX, at a distance 'x' from either support. We draw a free body diagram of the portion on the right of section XX. This is shown in the fig.15.23 below:

Fig.15.23
Free body diagram of the portion to the right of XX

• The moment at XX, due to the reaction R_B is equal to R_B x (anti-clock wise). 
• The moment at XX due to the applied load w_u is equal to (w_u x² )/2 (clock wise) 
• So the net factored moment at XX is given by:

M_u,x = R_B x  – (w_u x² )/2 

This equation will give us the applied factored bending moment at any section. Now we can look into the internal forces as shown in the fig.15.24 below:

Fig.15.24
Equilibrium of internal forces

• The compressive force C_u acts at the centroid of the 'concrete area above the neutral axis'.
• The tensile force T_u acts at the centroid of the steel area.
• These two forces are equal in magnitude but opposite in direction. So they form a couple. The magnitude of this couple is equal to T_u z . It is also equal to C_u z . Here, z is the lever arm.
• This couple resists the applied factored bending moment M_u,x.

So now we know the details about the internal forces. We can use the same procedure for analysing the internal forces in the cracked beam shown in fig.15.19 above. We will discuss about it in the next section.

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Chapter 13 (cont..5) - Vertical stirrups for Shear reinforcement

In the previous section we completed the discussion on the effects of shear on a homogeneous beam. The discussion that we had about a homogeneous beam, applies to a plain concrete beam (a concrete beam without reinforcement) also. When load is increased in such a beam, tension cracks will develop at the points where tensile stresses are the greatest. In the case of a simply supported beam, it is at the bottom most tension fibre. Once these cracks are formed, we can say that the beam has failed, and it will not take any more load.

Instead of a homogeneous beam, if we are considering a reinforced concrete beam, having steel at the bottom to take up the tension, the situation is quite different. When the first tension cracks form at the bottom most fibres, the loading need not be stopped. The beam can take higher loads because the tension will now be taken by the steel. But when the applied load increases, the shear stress also increases. Because of this, the tension on the principal plane in the particles also increases. We have seen that:
• For a simply supported beam, this tensile stress is maximum for the particles situated at the NA
• At the NA, the tensile stress is equal to q
• At the NA, these planes are inclined at 45^o to the vertical
• For simply supported beams, these stresses are maximum near the supports.

So when the applied load on the simply supported beam is increased, the tension is increased not only in the bottom fibres, but also at the NA. When this stress at the NA exceeds the allowable tensile stress of concrete, cracks will develop. These cracks at the NA will be inclined at 45^o to the vertical. Because of the 45^o inclination, these cracks are also known by the name ‘diagonal tension cracks’. A diagrammatic representation of these cracks is shown in fig.13.24 below. Photographs of such cracks on an actual beam can be seen here.


Fig.13.24
Diagonal tension crack

So we must provide some kind of steel reinforcement to prevent the separation of the concrete along the crack. In other words, we must provide steel reinforcement to keep the two parts together. This is called 'shear reinforcement', and can be provided in three different ways:
• By vertical reinforcement as shown in fig.13.25
• By inclined reinforcement as shown in fig.13.29
• By bent up bars as shown in fig.13.30

Fig.13.25
Vertical shear reinforcement

When the shear reinforcements are provided vertically, they are called vertical stirrups (or simply 'stirrups'). They are formed by bending steel bars around the tensile reinforcement bars as shown in the fig.13.26 below.

Fig.13.26
Closed stirrups

It is necessary to provide additional small diameter bars (usually 12 mm dia.) in the compression zone of the beam, in order to properly anchor the stirrups. These bars are known as 'stirrup-holders' or 'stirrup suspenders'. For a doubly reinforced beam, these bars in the compression zone will be already present, and so there is no need to provide additional bars. In the above fig.13.25, stirrups are shown only near the crack at the support, for clarity. In fact they must be provided through out the length of the beam at suitable spacing. This spacing is determined while we design the stirrups. A 3D view of the stirrups is shown in fig.13.27 below

Fig.13.27
3D view of stirrups

In the fig, the stirrups have two legs. Such stirrups are called two-legged stirrups. In some cases, in order to resist greater shear stresses, it becomes necessary to provide several legged stirrups such as four-legged, six-legged etc.,

In the above figures, closed stirrups are shown. In fact, they need not be of closed type. Simple U-shaped open stirrups can also be used. Their free ends should be anchored properly around the stirrup-holders in the compression zone. This type is shown in fig.13.28 below. But in the case of doubly reinforced beams, we know that the stirrup-holders will be replaced by compression steel, and so, to confine the compression steel properly, we must use closed stirrups. Also it is desirable to locate the hooks of the closed stirrups in the compression zone. In this way, the ends of the stirrup bars will be under confinement. This is necessary for improved anchorage and to avoid crack initiation.

The ends of the bar forming the stirrups should be properly anchored in the concrete, to prevent them from opening out. The code has specified the methods by which this anchorage can be given. We will discuss about it in the topic called 'Anchorage for stirrups and ties'.


Yet another point to note is that the hooks of alternate stirrups should be on opposite sides as shown in the following animation.

Fig.13.28
Open stirrups

In the next section, we will see the details about 'Inclined shear reinforcements'.

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