Chapter 14.8 - Conditions for providing splices in bars of beams

In the previous section we saw the safe regions for providing splices. In this section we will discuss about staggering of splices.

Staggering of splices

We have seen how to determine the safe sections where splices can be given according to cl.26.2.5 of the code. This same clause mentions about one more requirement. That is., the splices should be staggered. Staggering of splices is essential because of the following reason: We have seen that at a splice, the force in the stopping bar is transferred to the continuing bar by means of the bond stress in concrete. If more splices are provided at a single section, there will be a cumulative effect from the bond stress in all the spliced bars at that section. This may lead to cracks. 

The method for achieving proper staggering is also mentioned by the clause: 'not more than half the bars shall be spliced at a section'. This is a simple rule to follow. If there are two bars at a section, only one of them can be spliced. If there are 3 bars at a section, then also only one of them can be spliced at that section because if we take more than one, it will become greater than half the number of bars at the section. In this way we can limit the number of splices at a section. Consider the fig.14.42 in the previous section. Suppose in that beam, there are 3 bottom bars of the same diameter, continuing uninterrupted from support to support, and two of them fall short of length. In that case, one can be spliced at the safe zone near support A, and the other can be spliced at the safe zone near support B.

Precautions to be taken when conditions are not satisfied
So we have seen two requirements from this clause. One is about avoiding splices at sections of maximum stress, and the other is about staggering of the splices. In addition to these, the clause also tells us about the precautions to be taken when the above requirements cannot be satisfied. That is., these precautions are to be taken when
• Splice is given at a section of maximum stress. OR
• More than half the bars is spliced at a section.

The precautions to be taken are:
• Increasing the length of the lap AND
• Using spirals or closely spaced stirrups around the length of the splice.
The code does not specify the amount of increase to be given to the lap. We can use the details given in 25.2.5 of SP 24, the explanatory hand book to the code. There, the increase in length to be provided is given in the form of a graph. This is shown in fig.14.44 below:

Fig.14.44
Graph for determining the increase in length

Let us now see the details of this graph. '% of bars spliced at the section' is plotted along the X axis. So if 3 out of 4 bars having the same diameter is spliced at the section that we have under consideration, we will mark a point at 75% on the X axis, and draw a vertical line through it.

Along the Y axis, '% of design stress in steel' is plotted. So we have to first find the stress in the steel at our section. Let us denote it as f_s. It can be calculated using the basic formula: 
Applied factored bending moment = Stress x Area of steel x Lever arm
That is., M_u = f_s x A_st x z . From this, f_s can be calculated.

Now, the maximum stress possible is 0.87f_y. So the percentage of design stress is: (f_s/0.87f_y)x 100.

---

(It may be noted that calculating the percentage of design stress in this way is equivalent to calculating the percentage of applied moment. Which is equal to (M_u/M_uR)x 100.  That is., (f_s/0.87f_y)x 100 = (M_u/M_uR)x 100. The proof for this can be seen here).

---

If we get this as say 60% , we mark this value of 60% on the Y axis and draw a horizontal line through it. The point of intersection of the horizontal and vertical lines is our point of interest. It will fall within one of the four rectangular areas. The lap length to be given will depend upon the rectangle into which the point of intersection falls. For example, the intersection of horizontal line through 60% and vertical line through 75% will fall within the red rectangle as shown in the fig.14.45 below. So the lap length to be provided will be 1.7L_d.

Fig.14.45
Sample calculation

In the above two graphs, we can see that the blue rectangle is specially marked as 'Code recommended area'. This is so because, it falls within the 50% limit of 'both the requirements' of the cl.26.2.5 of the code. We have already seen these two requirements: 
(1) Splice should not be provided at sections where applied factored bending moment is greater than 50% of the moment of resistance, (Y axis) and 
(2) More than 50% of bars should not be spliced at a section, (X axis) 

These two requirements are satisfied in that area. Because it falls within '50' on X axis and also '50' on Y axis. 

The other precautionary measure is to use spirals and/or extra stirrups. We have seen the details of spirals to be provided in the case of bars larger than 36 mm in diameter. It was explained with the help of figs.14.39 and 14.40. The same procedure can be adopted here also.

If spirals cannot be provided, we must use closely spaced stirrups around the length of the splice. The procedure for this is given in SP34, the handbook on concrete reinforcement and detailing. This can be explained based on the fig.14.46 given below.

Fig.14.46
Extra stirrups around the splice

First step of the procedure is to calculate the number of these extra stirrups. This can be done as follows: 
• The total area of cross section of all the stirrups (A_sv) multiplied by the ultimate stress in the steel of the stirrups (0.87f_y) will give the ultimate force which these extra stirrups will be able to resist. 
• This must be equal to the total tension in all the bars that are spliced at the section. 
• So we can write: T = 0.87f_y A_sv. (Where T is the total tension in all the bars that are spliced at the section) From this we will get A_sv. 

[It may be noted that A_sv here is the total area of all the stirrups. So A_sv = No. of stirrups (N) x No. of legs of a single stirrup x Area of cross section of the bar of the stirrup.]

• If we assume the number of legs and the diameter of the bar of a single stirrup, we can calculate the area of one single stirrup. So, A_sv divided by this area of a single stirrup will give us 'N', the number of extra stirrups to be provided.
• Next, one third of the lap length is marked off at both the ends of the lap.
• In the final step, 'N' number of stirrups is provided at each of these marked regions at uniform spacing, starting from the inner side.

It should be noted that
• a minimum of three extra stirrups should be provided at each of the marked off region, and 
• the spacing should not be greater than 150 mm.

If the diameter of the spliced bars in the above fig. is greater than 28 mm, then it is compulsory to provide spirals or compact stirrups even if the extra stirrups are provided. A model showing the shape of compact stirrups is given in the fig.14.47 below. In the fig., the compact stirrups are shown in yellow color.

Fig.14.47
Shape of compact stirrups

So we have had a lengthy discussion about the precautionary measures alone. It will be convenient to have a flowchart like representation of the above discussion, as shown below:



In the next section we will see the provisions in the code, which are related specifically to lap splices.

PREVIOUS       CONTENTS       NEXT                                          

Copyright©2016 limitstatelessons.blogspot.com - All Rights Reserved