Recommendations given in the code regarding shear reinforcements:
In the previous section we saw the method of calculations of the spacing Sv of stirrups. In this section we will discuss the code provisions and checks.
According to the code, we must not use a grade of steel higher than Fe415 for making the stirrups or bent-up bars, or any other types of reinforcements intended to resist shear. This is because, the high strength steels may become brittle at the sharp bends. Another reason for giving such a restriction is that, When high strength steel is used, the shear resisting capacity of the beam may increase up to such a high level that the beam will be able to take very high loads with out any risk of developing tension cracks. But at these higher load levels, the other principal stress, namely 'compressive stress' on the particles may reach a level which is beyond the compressive strength of concrete. At this level, the concrete will crush, leading to a brittle failure of the beam.
Another recommendation given by the code is regarding the spacing of stirrups. According to cl.26.5.1.5,
For vertical stirrups, the spacing Sv should not be more than the smallest of the following:
a) 0.75d
b) 300 mm
For inclined stirrups with inclination α = 45o, the spacing Sv should not be more than the smallest of the following:
a) d
b) 300 mm
Such a minimum spacing is specified to ensure that every diagonal tension crack that may occur will be intercepted by at least one stirrup.
The next recommendation is regarding the minimum quantity of stirrups to be provided. We have seen that the concrete will give some resistance force against applied shear. Situations can arise wherein, the applied shear force is less than even the 'shear resistance offered by the concrete'. In such a situation, we will naturally think that no shear reinforcement is required. But the code does not allow us to avoid the shear reinforcements altogether in this way. It can be avoided only if a condition is met. This condition is given in cl.26.5.1.6 of the code. In the code, the condition is written in terms of stresses. We will write it here in terms of forces. So, according to this clause, we can avoid the shear reinforcement if the shear force Vu applied at the section is less than half of the resistance force offered to us by the concrete. That is, if Vu < 0.5Vuc, shear reinforcement can be avoided. We can write the converse of this. That is:
This condition can be represented graphically as shown below in the following fig.13.60:
Fig.13.60
Graphical representation of condition 13.62
Vu is less than Vuc but greater than half of Vuc
In the above graph, the applied shear force Vu is less than Vuc. At the same time, it is greater than half of Vuc. In such a situation, it is compulsory to provide stirrups.
In this condition, a shear force Vu is applied at the section, and this force is of such a low magnitude that, the concrete is able to take up all of this force. So what is the quantity of force, that we want to resist with the stirrups? The answer is that the stirrups in this situation are provided not to take up any load, but to serve the following purposes:
• It prevents the sudden formation of an inclined crack. If at all an inclined crack do form, the minimum shear reinforcement provided will prevent the propagation of that crack.
• If a beam with no shear reinforcement is over loaded to such an extent that shear failure occurs, then, that failure will be an abrupt one. If in such a beam, the minimum shear reinforcement specified by the code is provided, then the stirrups will take up the load, and if load is increased even further, the steel of the stirrups will undergo elongation, thus giving warning about the impending failure.
• The stirrups also help to confine the main bars of the beam together, keeping them in their proper positions. This will enable the main bars to perform their duties in a better way. This also improves the dowel action, thus keeping the portions of concrete on either side of the crack together.
When Vu < Vuc AND Vu ≥ 0.5Vuc, we do not have any force for which the stirrups have to be designed. [∵ Vus is obtained by subtracting Vuc from Vu.] In such a situation, we use cl.26.5.1.6 of the code. The design mainly involves the calculation of spacing Sv of the stirrups. We can calculate Sv using the expression given in this clause:
13.63
From 13.62, it follows that if, Vu < 0.5Vuc, We don't have to consider even the above 13.63. That is., there is no need for stirrups at all. Such regions may be present in a beam at regions away from the supports, where shear force is low. But it is a good design practice to provide stirrups in these regions also, for the same reasons mentioned above.
Limiting value of VuR:
We know that when a beam section fails by shear at the ultimate state, the steel of the stirrups should have yielded. Such an yielding will give enough warning that the beam is going to fail. In other words, the failure should be ductile in nature. At the beginning of this section, we have seen the reason for not giving high strength bars for making the shear reinforcements. That is, when high strength bars are used, the beam may become strong in diagonal tension and weak in diagonal compression. This very same undesirable condition can arise in another situation also: If we provide an excessive amount of shear reinforcements either in the form of stirrups or bent-up bars, the beam section will become strong in diagonal tension, and weak in diagonal compression. So the strength Vus given to the beam by us, should be kept within a certain limit.
The code gives the method to be implemented so that Vus will be within the safe limit. But it is an indirect method. In this method, the code specifies the upper limit for VuR, the ultimate shear resistance of the section. We know that VuR (Eq.13.53) is the total ultimate shear resistance which is the sum of the 'contribution of concrete' and 'contribution of steel reinforcements'. Let us see how this indirect method work, in achieving our objective.
We have
Eq.13.53: VuR = Vuc + Vus. Where
From Eq.13.34, we can see that Vuc is a constant for a given beam section. It can not be increased or decreased. So from 13.53, we can say that, if VuR has an upper limit, Vus will also have a corresponding upper limit. Thus, when we follow the code method, we keep VuR within a certain limit. When we do this, we are indirectly keeping Vus also within a desired limit. The only thing is that we are not calculating this upper limit of Vus.
But we do have to calculate the upper limit of VuR. This is denoted as VuR,lim and is obtained by the following:
VuR,lim = τc,max bd
The value of τc,max is given in the table 20 of the code.
So we know how to calculate VuR,lim of a given section. We will see how to implement this in a design procedure:
At the time of taking up the shear design of a beam section, b and d will be known. So we calculate VuR,lim using table 20 and Eq.13.64. Then we compare this with the applied shear Vu. What if Vu is of such a high magnitude that, Vu > VuR,lim ?
Then the basic requirement that Vu ≤ VuR will never be satisfied because the maximum value that VuR can take is VuR,lim. This can be represented graphically as shown in the following fig.13.61:
Fig.13.61
Maximum value that VuR can take is VuR,lim
In the above graph, Vu has a higher magnitude. So we must increase VuR so that the condition Vu ≤ VuR will be satisfied. But VuR cannot be increased beyond VuR,lim.
In such a situation, the only solution is to revise the design. This can be done either by using any one or both of the methods given below:
• Increase the dimensions b and d of the beam, thus increasing VuR,lim (Eq.13.64) or
• Use a higher grade of concrete, thus increasing τc,max, which will increase VuR,lim (Eq.13.64)
So we have seen the various code recommendations that we have to consider while doing a shear design. In the next section, we will discuss about the ‘critical sections’ in a structure at which we have to check the shear.
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