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.

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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.

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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.

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Chapter 14.4 - Anchorage at the ends of stirrups

In the previous section we saw one method to give the required anchorage at the end of stirrups. Now we will see the other methods.

Method 2: Using 180^o bend 

In this, we use a type of bending similar to the standard U-type hook that we discussed earlier. As before, the hook is given at both the ends of the bar of the stirrup. This is shown in the fig.14.19 below:

Fig.14.19

Anchorage of stirrups: Method 2

Here, the extension DE beyond the bent portion is the same 4φ that we saw earlier in the case of the ‘standard U-type hook’. [Note that, in the Method 1, where we used standard 90^o bend at the ends of stirrups, the length of DE was 8φ.] The resulting final shape of the stirrup is shown in fig.14.20 below.

Fig.14.20

Resulting shape of stirrup

Method 3: Using 135^o bend

In this case we use a type of bending that we have not discussed before. We know that the ‘standard 90^o bend’ has an angle of 90^o, and a ‘standard U-type hook’ has an angle of 180^o. The new type which we are going to see has an angle which is the average of these two values. That is : (90 + 180) / 2 = 135^o. This is shown in the fig.14.21 below:

Fig.14.21

135^o bend in a bar

Here angle DOB = 135^o. So the portion BCD is the exact 3/8 of a ring (∵ 135/360 = 3/8) having inner radius r. As usual, we must extend it beyond D. Here, for the purpose of anchoring stirrup ends, this extension required is equal to 6φ  . Note that 6φ is the average of 8φ (Method 1) and 4φ (Method 2). Fig.14.22 shows the extension.

Fig.14.22

135^o bend for anchoring the stirrup

Note that angle EDO is 90^o. We will apply this bend to the two ends of the stirrup bar as shown below:

Fig.14.23

Anchorage of stirrups: Method 3

The resulting final shape of the stirrup is shown in fig.14.24 below:

Fig.14.24

Resulting shape of stirrup

This completes the discussion about the anchorage to be provided to the stirrups. A few important additional points are given below:

The grip is exerted by the concrete mainly on the straight portion DE coming after the bend, at the ends of the stirrup bar. In the previous section we saw the method of anchorage for stirrups by 90^o bend (Method 1). We discussed it based on fig.14.17. In this method, the straight portion is nearer to the outer surface of a concrete member. (In methods 2 and 3, DE is embedded into the mass of concrete of the beam). So, if the defect known as ‘spalling’ occurs in the concrete member, the outer cover for DE may fall off. (More details about spalling can be seen here) This will reduce the anchorage provided to the ends of the stirrup. In such a situation, if the stirrup bars undergo higher tensions, it may open out due to the non availability of concrete to keep the ends in position. So we must adopt Method 2 or Method 3 in situations where concrete cover around the stirrups is not restrained against spalling.

The next point that we have to note, is the radius of the bends at the four corners of the stirrups. In the figs. 14.17, 14.19, and 14.23, that we saw for the three methods of anchorage, the diameter of the bar of the stirrup is about 10 mm. We have seen that, the radius of the bend should not be less than 4 times the diameter. So a radius of 40 mm is given. Then the diameter of the bend will be 80 mm. This means that the bar shown in blue colour at the corner of the stirrup has a diameter of 80 mm. But bar diameters greater than 36 mm are not generally used in practice. And it is not practical to give such large diameter bends at each corner of a beam. 

The main reason for specifying a large radius for the bend is to reduce the 'bearing stress' on concrete. But in a beam, there will be a longitudinal bar at each corner of the stirrup. So, even if the radius is small, the concrete will not be subjected to much bearing stress at the corners. We will learn more details about bearing stress in a later section.

We used a large value of bending radius in the figs. of stirrups, just for the clarity of showing the bending procedure, and for showing the similarity with the standard bends. Fig.14.25 below shows the correct procedure. A 10 mm dia. bar is given a 135^o bend around a normal diameter bar. The larger radius bend is also shown along side for comparison. In the fig., the dia. of the bar of stirrup is the same in both the cases.

Fig.14.25

Comparison between small and large radius of bend

In the above fig.,

• The stirrup bar AE has the same diameter in both the cases.

• The length of the extension DE is the same in both the cases.

• Angle BOD is equal to 135^o in both the cases.

• BCD in both the cases are exact 3/8 of their corresponding rings.

• The only difference is that, the radius of the ring on the left side is small, and that on the right side is large.

3D views of the stirrups having small diameter of bend is shown in the fig. below:

In the above fig., (a) shows a stirrup with both ends bent through 90^o, (b) shows 180^o and (c) shows 135^o.

In the next section, we will discuss about the 'bearing stress' caused in concrete by the forces in the bends.

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