Chapter 13 (cont..11) - Shear resistance offered by Bent up bars

In the previous section we completed the discussion on 'inclined stirrups'. Now we will see 'bent-up bars'.

Design shear resistance offered by bent-up bars

We have seen earlier that the bent-up bars also offer resistance against shear. We will now determine this resistance. Fig.13.53 below shows a bent-up bar. It is intercepting the crack, and the angle of bent-up is α. 

Fig.13.53

Bent-up bar

The fig. shown above is a sectional elevation. There can be more than one bent-up bar in the fig., which may be aligned with the one that we see. Because of the perfect alignment, the other bent-up bars will not be visible in the sectional elevation. Usually we mention about the number of bent-up bars in the sectional elevation itself. However, to get a better understanding, we will see some 3D views below:

Fig.13.54

View of a beam having a single bent-up bar

Fig.13.55

View of a beam having 2 bent-up bars 

Some points should be noted about the above 3D views.

• All the stirrups of the beam are not shown in the views. For clarity, only a few random stirrups are shown.

• In fig.13.55, the beam has an increased width. In such cases it may be necessary to provide 4-legged or 6-legged stirrups.

• Though not related to our present discussion, it may be noted that, in the fig.13.55, the distance between the bottom bars is greater towards the end of the beam. In such a situation, it should be checked that this distance provided between the bars does not exceed the maximum distance allowed by the code. Details about this topic were discussed in a previous chapter.

Now that we have a better understanding about the arrangement of bent-up bars, we will analyse the resistance force in the bent-up bar in fig.13.53. For resolving the forces, we can use the same fig.13.50, that we used for inclined stirrups. Based on this fig., we derived Eq.13.41 for the vertical component of the force in the inclined stirrup.

Eq.13.41:

Vertical component =0.87f_y A_sv sinα

Our bent-up bar will also carry the same vertical force because the method of resolution of force is the same as in fig.13.50. The only difference about which we have to be aware about is the quantity A_sv.

When we consider inclined stirrups, A_sv is the area of cross section of the bar used for making the stirrup multiplied by the 'number of legs' of the stirrup.

When we consider the bent-up bars, A_sv is the area of cross section of the bar used for the bent-up bar. If there is more than one bent-up bar at the same section as shown in the view in fig. 13.55, then:
Eq.13.48:
A_sv = Total area of all the bent-up bars at that section

Here, the quantity n does not come into the picture because, the bent-ups are not distributed along the length of the beam. Instead, they are provided at a particular section.

So for a single bar, or a single group of parallel bars, all bent up at the same cross section, We get the resistance offered in the vertical direction as:

Eq.13.49:

V_us =0.87f_y A_sv sinα

Where A_sv is calculated based on Eq.13.48 given above. This is the same equation given in cl.40.4.c.c of the code.

---

Now we discuss about another 'type of arrangement' of bent-up bars. In this arrangement, the bent-ups are not at the same section. They are given at a regular spacing as shown in the fig.13.56 below:

Fig.13.56

Series of bent-up bars

A 3D view of a beam having 2 bent-up bars in series is shown in fig.13.57 below:

Fig.13.57

3D view of a beam having two bent-up bars in series

From the 3D view we will get a better understanding about the arrangement of bent-up bars in series. It can be seen that in this arrangement, only one bent up can be provided at a particular section. So the area of bar at a section will be equal to the area of one bar. With this point in mind, we can proceed to calculate n, the number of bars intercepting the crack. The calculations are similar to those of inclined stirrups. So we will add some triangles to fig.13.56 above. The modified fig.13.58 is given below:

Fig.13.58

Calculation of horizontal projection

The calculations are identical to those we did earlier based on fig.13.52 for inclined stirrups. So we get n, the number of stirrups intercepting the crack as:

Eq.13.50:

Now we can multiply the force in a single bent-up bar (given by eq.13.41) by n the total number of stirrups intercepting the crack. The product is the 'total resistance force'. This total resistance is denoted as V_us. So we get

Eq.13.51:

Here, A_sv is the area of a single bent-up bar. And, to use the equation, all the bent-up bars in the series should be of the same diameter.

Just as in the case of inclined stirrups, if for convenience, we assume that the crack is inclined at θ =45^o  to the horizontal, the above Eq.13.51 will be simplified to the form:

Eq.13.52:

This is the same equation given in cl.40.4.c.b of the code.

In general, the bent-up bars give a lower shear strength. Major inclined cracks are likely to occur at bent points. The position of bends in the beam should be exactly at the designed positions. Otherwise the bent-up bar will not be effective. So they can be considered effective only if there is a series of closely spaced bent-ups. If there are only a few bent-ups, widely scattered along the span of the beam, their shear strength contribution is ignored, and the design is done using stirrups.

So now we are in a position to calculate the resistance offered by concrete and that by the reinforcements. In the next section we will see their combined effect.

PREVIOUS

NEXT                                          

Copyright ©2015 limitstatelessons.blogspot.com - All Rights Reserved