Chapter 14.9 - Calculation of lap length at splices

In the previous section we saw the conditions for providing splices. We also saw the precautions to be taken while giving splices. In this sections we will discuss some code provisions which are related specifically to lap splices.

These are given in cl.26.2.5.1 of the code. 
• The sub clause (a) of this clause is related to the lap splices of bars larger than 36 mm in dia. We have already discussed about it with the help of figs.14.39 and 14.40. 
• The next sub clause (b) is about ‘staggering of laps’. We shall discuss about it after seeing the sub clause (c).

Cl.26.2.5.1(c)
When the bar is under flexural tension, (a discussion on the difference between flexural tension and direct tension is given here) the lap length (fig.14.48) should not be less than the largest of the following:

(a) L_d
(b) 30Ф
Here L_d is the unique value of development length that we saw earlier.

Fig.14.48
Lap length in Flexural tension

If for the same bar shown in the above fig.14.48, which is under flexural tension, hooks are provided at the lapped ends, the fig. can be modified as shown below:

Fig.14.49
Lap length in flexural tension when hooks are provided

We have discussed about the anchorage value of hooks and bends in a previous section, and we know that it is equal to 16Ф for hooks. So this can be added to the 'straight length’ of the lap. 

• Thus the 'Total contribution' [A] = 'Straight length' [B] + 16Ф. [C]
• [A] should not be less than the largest of 
    ♦ L_d
    ♦ 30Ф
• [B] should not be less than the largest of
    ♦ 200 mm
    ♦ 15Ф 
So we have to simultaneously check two parameters [A] and [B] and ensure that they do not fall below required values. These requirements are clearly shown in the fig.14.49

This same fig. can be used for bends also, just by changing the hooks to bends. Also, in the calculations, the anchorage value should be taken as 8Ф. All the other details remain the same. It is shown in fig.14.50 below:

Fig.14.50
Lap length in flexural tension when bends are provided

• Thus the 'Total contribution' [A] = 'Straight length' [B] + 8Ф. [C]
• [A] should not be less than the largest of 
    ♦ L_d
    ♦ 30Ф
• [B] should not be less than the largest of
    ♦ 200 mm
    ♦ 15Ф 
So we have to simultaneously check two parameters [A] and [B] and ensure that they do not fall below required values. These requirements are clearly shown in the fig.14.50

Next we will see the lap length requirements when the bar is in 'direct tension'. In fact, with a small modification, the above three figures can be used to show the lap length requirements in direct tension also. The modification is that, 2L_d is to be used in place of L_d when direct tension is considered. Thus, corresponding to the above three figs.14.48, 14.49 and 14.50, we get the three figs.14.51, 14.52 and 14.53 given below:

Fig.14.51
Lap length in Direct tension

Fig.14.52
Lap length in Direct tension when hooks are provided

• 'Total contribution' [A] = 'Straight length' [B] + 16Ф. [C]
• [A] should not be less than the largest of 
    ♦ 2L_d
    ♦ 30Ф
• [B] should not be less than the largest of
    ♦ 200 mm
    ♦ 15Ф 
So we have to simultaneously check two parameters [A] and [B] and ensure that they do not fall below required values. These requirements are clearly shown in the fig.14.52

Fig.14.53
Lap length in Direct tension when bends are provided

• 'Total contribution' [A] = 'Straight length' [B] + 8Ф. [C]
• [A] should not be less than the largest of 
    ♦ 2L_d
    ♦ 30Ф
• [B] should not be less than the largest of
    ♦ 200 mm
    ♦ 15Ф 
So we have to simultaneously check two parameters [A] and [B] and ensure that they do not fall below required values. These requirements are clearly shown in the fig.14.53

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The clause which we are discussing now is the sub clause (c) of 26.2.5.1. It gives us the guidelines for calculating the lap length. We have seen the details above. 

Now, this sub clause in turn, have two sub clauses. Sub clause (1) and sub clause (2). These sub clauses gives us the guidelines for 'increasing the lap length' when the concrete cover provided for the steel bars are below certain specified values.

This can be explained as follows: We know that, for the lapped splice to work properly, sufficient bond should develop between steel and concrete. So there must be sufficient 'mass of well compacted concrete' all around the bars. The 'inner bars' of various members will have sufficient concrete all around. But the bars which are near the exposed surfaces will be having only limited concrete (which we denote as the concrete cover) on the exposed side. This is a cause for concern. If the concrete cover is below certain specified values, an increase in lap length is required. The sub clauses (1) and (2) give the guidelines for this increase. We will discuss about them with the help of an example. Consider the continuous beam shown in the fig.14.54 below:

Fig.14.54
Sectional Elevation of a continuous beam

Fig.14.55
Cross sectional views

Consider the sectional view XX. This section shows the bars at the midspan region. We can see that, there are 3 bars at the top. These bars are named as a, a' and b. We know that at the intermediate support of a continuous beam, there will be hogging moment, and so, these three top bars will be in tension. If the cover c_t at the top is less than twice the diameter of the bar, then each of these bars is an example of the bar specified in the sub clause (1). So if lapped splices need to be provided for these types of bars, and c_t is less than 2Ф, we must take the following steps:

• First, find L by following the 'rules for obtaining the lap length for bars in flexural tension' that we discussed above. (Figs.14.48 to 14.50). 
• Then obtain the modified value by multiplying this L by 1.4. 

This modified value should be provided for the lap. The need for such an increase in the lap length can be explained as follows:
When concrete is poured into the member, and compacted at the time of casting, water and air will rise towards the top of the concrete mass, and will get trapped at the under side of the horizontal reinforcement. This will weaken the bond between steel and concrete. So we must increase the lap length.

The sub clause 26.2.5.1.c (1) that we discussed above is specifically intended for top bars of a beam. All the top bars will be having a top cover of c_t. It is the value of this c_t which will determine whether this sub clause (1) is applicable to the top bar under consideration.

But the top bars will be having cover from sides and bottom also. How do these covers affect the lap length calculations? 
• Bottom covers: We know that, the bottom covers of the top bars will be large. So they do not cause much concern.
• Side covers: If slab is present at the top portion of the beam, so as to form a T-beam, then side bars a and a' will also be having large covers. The middle bar b will be having enough side covers whether or not it is a T-beam.
    ♦ But if it is not a T-beam, the side covers for a and a' will also cause concern. Then some other clauses (which we will discuss soon) will also be applicable to a and a'.

So in general, if a bar is such that, only the sub clause (1) is applicable (for computing required increments in lap lengths based on covers), then:
■ It should be a top bar
■ Only c_t should cause concern.
■ c_b and c_s should be sufficiently large so as not to cause any concern.  

Next we consider the section YY. This section shows the bars at the midspan region. We can see that there are 3 bottom bars in the midspan region. These bars (named as c, c' and d) will be in tension. Consider the corner bars c and c'. These two bars have clear cover from two faces. The side vertical face, and the bottom horizontal face of the beam. These covers are denoted as c_s and c_b. If any of these covers is less than 2Ф, then these two bars are examples of the bars specified in the sub clause (2). So if lapped splices need to be provided for any of these two bars, we must take the following steps:

• First, find L by following the 'rules for obtaining the lap length for bars in flexural tension' that we discussed above. (Figs.14.48 to 14.50). 
• Then obtain the modified value by multiplying this L by 1.4.

The sub clause 26.2.5.1.c (2) that we discussed above is specifically intended for corner bars of a beam. All the corner bars will be having two covers: 
• The cover from top horizontal surface c_t (if it is a top corner bar) or the cover from bottom horizontal surface c_b (if it is a bottom corner bar).
• The cover from the exposed vertical side c_s  

Both these values should be checked. If any of the two is less than 2Ф, then this sub clause is applicable to that bar.

It may be noted that, if it is not a T-beam, the top bars a and a' will also be corner bars.

This sub clause (2) gives specifications for some other situations also. We will see them in the next section.

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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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