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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