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.

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(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).

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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.

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Chapter 14.6 - Splices for bars

In the previous section we saw the details about 'bearing stress'. In this section we will discuss about splices.

Splicing of bars

Sometimes, a bar placed in a structural member will fall short of it's required length. In such a situation, the bar can be extended by splicing. When this is done, the axial force in the bar should be transferred effectively from the bar which fell short of length, to the new continuing bar. There are three methods for splicing.
• Lap splice
• Welded splice
• Mechanical connection.

First we will see the details about Lap splice. In this method, the two bars to be spliced are overlapped over a certain distance. The lapped bars are placed in contact and tied together lightly, so that they stay in place when the concrete is placed and compacted. In this type of splice, the force in the first bar is first transferred to the concrete by the action of bond stress. From there, the force is transferred to the continuing bar, again by the action of bond stress.

Details of a model of a lapped splice which is used in situation other than Reinforced concrete work can be seen here.

It should be noted that, lap splicing is not permitted for bars having diameter greater than 36 mm. The following figs., give some details about lap splice.

Fig.14.32
Overlapping of bars

In the above fig., Bar A is the one which falls short of length. The end portion of this bar from A to D is shown in the fig. Bar B is the continuing bar. We can see that the centre lines of the two bars do not coincide. This will give rise to eccentricity. We will now see the procedure for avoiding this type of eccentricity at the region of overlapping. The next fig.14.33 shows a small deviation given to the bar AD.

Fig.14.33
Deviation to the bar

For making the deviation, the bar is given a small bend at B. The bend should be in such a way that, the slope of the portion BD is not greater than 1 in 6. In the fig, the slope of BD is given exactly equal to 1 in 6 as shown by the blue triangle. (The position of B at which we must give the bend, will be clear after a few more steps). Now the bar is deviating away from the original center line. So we give it one more bend, but this time, in the reverse direction. This is shown in the next fig.14.34

Fig.14.34
Bend in the portion BD

From the fig.14.34, we can see that the second bend is given at point C. This point lies at the intersection of the portion BD (in the previous fig. 14.33) with the center line. The bend at C should be given in such a way that the portion CD becomes parallel to the center line. When this is done, the whole portion from C to D lies just below the center line.

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We can calculate the length BC from the similar triangles (colored in orange) in the fig.14.35 given below:
Fig.14.35
Calculation of length BC

From similar triangles in BCC’ we get

So BC’ = 3Ф
BC’ is the altitude, and C’C is the base of the right angled triangle BCC’ . So we get

From the above, we can see that the 'slope of the bend' and Ф, are the only two parameters that determine the length of BC.

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So we have obtained the final shape of Bar A. Now, Bar B comes from the opposite side. So we will make a mirror image of Bar A, as shown below in fig.14.36. The mirror line is vertical. But the bent portion should be in the upper side. So we will make another mirror image with a horizontal mirror line. This gives the final shape of Bar B.
Fig.14.36
Final shape of Bar B

So this Bar B can be lapped onto Bar A as shown in the fig.14.37 below:

Fig.14.37
Lapped bars

When this is done, centre lines of both the bars will coincide, and thus there will not be any eccentricity. The over lapping length L shown in the fig.13.37 must be equal to the development length L_d of the bar.

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We have seen in fig.14.33 a little while earlier that, the slope of portion BC should not be greater than 1 in 6. The fig.14.38 given below shows the same bar, but with a slope of 2 in 6.

Fig.14.38
Slope of BC greater than 1 in 6

We can see that when a greater slope is given, BC becomes steeper, and it’s  length is reduced. This will not give an effective transfer of force. So any slope greater than 1 in 6 is not permitted.

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Now we will see a special case of bars with larger diameters. Lap splices should not be used for bars of diameter greater than 36 mm. For such bars, welded splices are recommended. But when welding is not practicable, lap splices can be used for such bars as per cl.26.2.5.1a of the code. According to this clause, such lap splices should be given additional spirals as shown in the fig.14.39 below.

Fig.14.39
Additional spirals at lap

The bars used for these spirals should have a minimum diameter of 6 mm, and the pitch should not be greater than 100 mm. This is shown in the elevation view of the above fig.

Fig.14.40
Maximum pitch and minimum diameter for spirals

In the next section we will see more details about lap splice.

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