Chapter 14.7 - Safe regions for providing splices in bars of beams

In the previous section we saw the basic details about 'lap splice'. In this section we will discuss some more details about splices.

Location of splices

We have seen that the splices can be used to increase the length of bars. We may be compelled to give splices wherever there is a shortage of length. But this is not allowed. Their are certain specific locations where splices should not be given. Here we will see some recommendations given by the code regarding the location of splices. These 'recommendations about location' given in cl.26.2.5, are applicable to splices in common (lapped splice, welded splice, splices by mechanical connection).

According to this clause, splices should as far as possible be away from the sections of maximum stress. This is so because, at the sections of maximum stress, the bars will be subjected to a greater amount of tension or compression. So if splices are present at these sections, the concrete will have to transfer a greater force from the stopping bar to the continuing bar. This may give rise to cracks in the concrete at these sections.

So how do we implement this recommendation? The answer is given by the code in the same clause itself:

14.12
'It is recommended that splices in flexural members shall not be at sections where the bending moment is more than 50% of the moment of resistance'. 

This means that, if the 'External factored bending moment' M_u, applied at a section is more than half of the 'Ultimate moment of resistance' M_uR offered by the section, then we cannot provide a splice at that section. So we can write:

14.13
We cannot give splices at a section XX if
M_u,xx   >  0.5M_uR,xx

The converse of this can be written as:

14.14
We can give splices at a section if
M_u,xx   ≤  0.5M_uR,xx

By considering both sides of the inequality in 14.14, we can see that when it is satisfied, even half of the ultimate moment of resistance provided at section XX is sufficient to resist the applied factored moment at that section. So there will not be much stress in the steel at that section, and we can provide splices there. The above discussion can be graphically represented as given in the fig. below:

In the above graph, M_uR,xx  will always be greater than M_u,xx  (∵ this is a primary requirement for any beam. This requirement should be satisfied at all sections of a beam when we do 'Curtailment of bars' also). In case (a), the applied factored bending moment at the section is greater than half of the 'capacity of the beam at ultimate state', M_uR at the section. So splices cannot be given there. But in (b) and in (c), it is less than half of the capacity. So splices can be given for those cases.

This can be further explained by taking a simply supported beam as an example. Fig.14.41 below shows a simply supported beam and it's factored bending moment diagram.

Fig.14.41
Simply supported beam

Suppose in the beam shown in the above fig., 'curtailment of bars' is not done. Then all the bottom bars at the midspan will continue uninterrupted to either supports. So the M_uR will be the same wherever we take a section XX along the length of the beam. But the applied M_u will vary.

• At any section, M_u will be less than M_uR. (∵ this is a primary requirement for any beam.) 
• We want the regions where M_u is less than even the 'half of M_uR'. This can be obtained as follows:

We know that the factored bending moment M_u,xx  at any section of the above beam can be obtained using the equation:

Eq.14.15
M_u,xx  =  R_A x – 0.5w_u x² (Where  R_A  is the reaction at support A, and x is the distance of the section from the support A)

We also know that, as we are not applying any curtailment, the ultimate moment of resistance M_uR  will be constant at all sections. We can find the position of a section XX at which the applied bending moment is equal to half of M_uR, simply by equating them. Thus:

Eq.14.16
M_u,xx  =  R_A x – 0.5w_u x²  = 0.5M_uR

Solving this we will get the value of x. (As the beam is symmetrical, and is symmetrically loaded, there will also be a corresponding section on the other side of the center of the span, that is near support B). In the space between this section and the support, the applied factored bending moment will be less than half of M_uR, and so we can give splices there.

This section can also be determined by superposition of the graph of 0.5M_uR over that of M_u. This is shown in the fig.14.42 below:

Fig.14.42
Method of superposition

In the above fig., the graph of 0.5M_uR is drawn above that of M_u. The points of intersection will give the positions of section XX. We do not need to draw the graph of M_uR. Here it is drawn just to show that when curtailment is not done, M_uR (which is always higher than M_u), will be having a constant value.

When curtailments are done:
When we design the beam, we design the section at the midspan, where the bending moment is the largest. But we have seen earlier that all the bars provided at the midspan region need not be given through out the entire length of the beam, because the BM progressively decreases on either side of the midspan. So, after designing the beam at the midspan, we follow the principles of 'Curtailment of bars' and do the necessary bar cut-off at various sections. When this is done, different sections of the beam will be having different values of M_uR. We will learn more details about curtailment in the next chapter. When we follow the principles of 'curtailment of bars' strictly, the M_uR at any section, (denoted as M_uR,xx) will be greater than or equal to the applied factored moment at that section (denoted as M_u,xx ). That is., M_uR,xx ≥ M_u,xx. But the steel at many sections may be just sufficient for making up the required M_uR,xx. That is., M_u,xx at many sections will be 'just below' the M_uR,xx at those sections. It will not be 'too much below' to be less than even 'half of M_uR,xx'. At such sections, the stress in steel will be maximum. So we cannot give splice at those sections. Superposition of M_uR over M_u for this case is shown below:

Fig.14.43
Superposition when curtailment is done

We can see that at any given section M_uR should be greater than M_u. But M_uR is not constant through out the length of the beam. If we want to give a splice at a particular place, we must superimpose the graph of 0.5M_uR and check whether it is safe to give splice in that region.

The graph of M_uR in the above fig. is only a schematic one. The actual graph will depend on the actual curtailments done in the beam. So a lot of calculations are involved.

---

It may be noted that we have done the above discussions based on a simply supported beam. So safe regions for lapping the 'bottom bars' were considered. In the case of an intermediate support of a continuous beam, the 'top bars' will have to be considered. There, in the bending moment diagram, the hogging moment region at that support should be carefully studied. The curtailment if any, provided for the top bars should also be considered.

---

In general:• For Sagging moments (bottom bars)
   ♦ M_u is greater near mid spans and lesser near supports
   ♦ So safe regions for splices may be available near supports.
   ♦ Additional calculations are required if curtailments are done

• For Hogging moments (top bars)
   ♦ M_u is greater near supports and lesser away from supports
   ♦ So  splices may be provided away from supports.
   ♦ Additional calculations are required if curtailments are done

In any case, each structure should be carefully studied to determine whether it is safe or not to give a splice at a section. And relevant clauses of various codes should be satisfied.


In the next section, we will discuss about the Staggering of splices.

PREVIOUS       CONTENTS       NEXT                                          

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

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.

---

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.

---

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.

---

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.

---

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.

PREVIOUS       CONTENTS       NEXT                                          

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

Chapter 14.5 - Bearing stress at bends

In the previous section we saw the various methods that are used to give anchorage at the ends of stirrups. In this section we will discuss about 'Bearing stress'.

Bearing stress at bends

We have seen the 'bond stress' induced in concrete when stresses are developed in reinforcement bars. The bond stress is a 'gripping' stress. Now we will see another type of stress that is induced in concrete. This stress is induced due to the presence of bends provided at the ends of bars. When tension develop in a bar which is provided with bends, the bar is kept in it's exact position because, the bend bears against the concrete. This 'bearing' will give rise to 'bearing stresses' in the concrete. Fig. below shows the bearing stress at a bend.

F_bt is the tensile force in the bar. Due to this force, a bearing stress σ will develop at the bend. Cl.26.2.2.5 of the code gives us the procedure for ensuring that the bearing stress produced in concrete is within the acceptable limit. The procedure is simple involving just two steps. The details are as follows:


First we calculate the bearing stress σ developed in concrete by using the formula:

Eq.14.8

• F_bt is the design tensile force in the bar
• r is the inner radius of the bend
• Ф is the diameter of the bar. If it is a bundle of bar, then Ф is the size of the bar of equivalent area.

---

This equation is derived as follows:
The bend at which σ acts, is part of a circular shape. The 'projected length' of this circular portion is r. This is shown in the fig. below:

So the 'projected area' will be equal to rФ. Now, stress = Force/Area. Thus we get Eq.14.8. As r is in the denominator, σ will decrease when r increases.

---

Next step is to calculate the maximum permissible bearing stress using the formula:

Eq.14.9
Permissible bearing stress =

If σ calculated using Eq.14.8 is less than or equal to the value calculated using Eq.14.9, then it is safe.
In Eq.14.9, a is a new parameter. It is calculated as follows:

a can take two values based on the position of the bar inside the structure. 
• If the bar under consideration is an end bar which is adjacent to the face of the member, then a is equal to the clear cover plus Ф, the diameter of the bar. 
• If the bar is an internal bar, then a is equal to the c/c distance between the bars.

All other parameters on the right side of Eq.14.9 are constants. So we find that the permissible bearing stress for an 'end bar' is different from that of an 'internal bar'. 

The above points will be clear when we look at the details of an rcc structural member shown below:

Fig.14.26
Part view of a structural member

If the concrete was transparent, we will be able to see the reinforcing bars inside as shown below:

Fig.14.27
View showing the reinforcing bars

The reinforcement consists of U-type bars. Note that secondary reinforcement bars are not shown in the above fig.14.27. Let us assume that both legs of the U-type bars are subjected to a tensile force of F_bt. This is shown in the sectional view XX given below:

Fig.14.28
Sectional view

The sectional view XX shows clearly, the bearing stress σ, the inner radius of the bend r, and the two values of a that are to be taken. The above figs.14.26 to 14.28 give a clear understanding of how Eqs.14.8 and 14.9 can be used for checking the bearing stress.

We will see one more example which shows the 'necessary details required' for computing the bearing stress at the bend, and checking whether it is safe. Fig.14.29 below shows the view of a structural member.

Fig.14.29
Part view of a structural member

As before, if the concrete was transparent, we will be able to see the reinforcements as shown in the fig.14.30 below:

Fig.14.30
View showing the reinforcing bars

The reinforcement consists of L-shaped bars. It has a standard 90^o bend. Note that secondary reinforcement bars are not shown in the above fig.14.30. The horizontal segment of the L-shaped bar is subjected to a tensile force of F_bt. This is shown in the sectional view XX given below:

Fig.14.31
Sectional view

The sectional view shows clearly, the bearing stress σ, the inner radius of the bend r, and the two values of a that are to be taken. Thus The above figs.14.26 to 14.28 give a clear understanding of how Eqs.14.8 and 14.9 can be used for checking the bearing stress.

The L-bars should be formed by standard 90^o bend. So the radius of the bend should not be less than 4Ф. When we are given a problem for analysis, r will be given. We must first check and ensure that this r is not less than 4Ф. Then we must proceed to use Eqs.14.8 and 14.9 

So the above procedure can be used when we analyse a structural member to check whether the bearing stress at the bends falls within the safe limit. But when we are designing a new member, we can ensure this at the design stage itself. This is done as follows:

The maximum tensile 'stress' (at the ultimate state) that the bar will be able to carry is 0.87f_y. So the maximum 'force' can be written as

Eq.14.10

Substituting this in Eq.14.8, we get

Eq.14.11

But σ should be less than or equal to he value in Eq.14.9. So we can write

We can rearrange this, and bring r to one side to obtain a final form as:
14.11

So while designing a bend, we must ensure that r, the radius of the bend satisfies the above relation 14.11. And at the same time r should not be less than the values specified for the standard types of bends.

In the previous section, we saw that, the bends in stirrups need not have large radii. We discussed this based on fig.14.25. We said that the reason for this is the presence of longitudinal main bars of the beams. These main bars will be having considerable diameters. So they will help to reduce the bearing stress. But in the figs. that we saw above, the 'bars with bends' them selves are the main bars. The longitudinal bars (not shown in the figs.) which will be coming at the corners will be secondary or distributor bars, which will generally be having smaller diameters. So they will not help in reducing the bearing stress on concrete. Even if the longitudinal bars at corners have larger diameters, it is compulsory to check the bearing stress at bends of main bars. 

In the next section, we will discuss about the 'splicing' of bars.

PREVIOUS       CONTENTS       NEXT                                          

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