Chapter 14 (cont..2) - Bends and Hooks to improve anchorage

In the previous section we saw some situations that require proper provision of development lengths. Now we will see some methods to improve the anchorage.

Bends and Hooks to improve anchorage

We have seen the importance of providing sufficient development length for the bars. The provision of this extra length will require space. In many situations, there will not be any space restrictions. Some examples are:
• Bottom bars at the midspan region:
There will be enough space for providing this extra length for the bottom bars at the midspan regions of simply supported and continuous beams. These bars are at the mid span, and there indeed will be enough space on either sides.
• Top bars at interior supports:
These bars are at the interior supports, and so there indeed will be enough space on either sides.

But this may not be so for the top and bottom bars near supports. The length of the beam is coming to an end at the support. So we may not be able to extend the bar to a sufficient distance. When the end of the beam is supported on a column, the bars of the beam will be, and should be, extended into the column. But then also the dimensions of the column may be such that it is not possible to provide the required development length. In such cases Bends or Hooks are provided at the end of bars to improve the anchorage. Fig.14.10 below shows the method of providing a bend in the bar.

Fig.14.10
90^o bend at the end of a bar

AB is the straight portion of the bar. The bend starts at B and extends upto C. OB =OC =r. We can see that the bend BC is a part of a perfect circle. But it has an inner radius and an outer radius. So we can say that it is a part of a 'perfect ring'. It has an inner radius of r and an outer radius of (r + φ). Inner radius r is the one which is important to us. The center of the ring is obtained by drawing the perpendicular BO to AB such that BO = r.

The value of r should not be less than
•  4φ for deformed bars and
•  2φ for plain mild steel bars.

Angle COB is exactly 90^o. So this is a 90^o bend, and the bend is 1/4 of the ring.
Thus we have obtained the bend. But this is not a 'Standard 90^o bend' specified by the code. For that, we have to extend the bar even further from C. But this time, in a straight line. This is shown in fig.14.11 below. The amount of extension required for this extension CD is 4φ.

Fig.14.11
Standard 90^o bend

So we have obtained the standard 90^o bend. In this we have extended the original bar from B to D. How much development length does this bend give? The code (cl.26.2.2.1) says that the anchorage value (that is the development length) contributed by a bend is equal to 4 times the diameter of the bar for each 45^o bend, subject to a maximum of 16 times the diameter of the bar. That is., each 45^o in the bend will give a length of 4φ. But we cannot obtain an indefinite length just by giving more angle. The maximum development length that can be obtained in this way is 16φ. In the 90^o bend that we saw above, there are two 45 degrees. So the anchorage value is equal to 8φ. It should be noted that this contribution of 8φ  is inclusive of the extension CD. That is., we will get this 8φ, only if we provide an extension CD. So 4φ from the portion CD cannot be added to 8φ as if to get a total anchorage of 12φ. The total contribution of the portion from B to D of the ‘standard 90^o bend’ shown in fig.14.11 is equal to 8φ. However, if we provide some length beyond D, that length will be considered as an additional contribution to anchorage.

Now we will see the details about hooks. Fig.14.12 below shows the method of providing a hook in the bar.

Fig.14.12
Hook at the end of a bar

Just as in the case of a 90^o bend, here, the portion BCD is a part of a ring having inner radius r. Angle BOD is exactly 180^o. So it is a 180^o bend, and it is 1/2 of the ring. In fact, DC is an exact mirror image of BC. Thus we have the shape of a 'hook' from B to D.

But this is not a 'Standard U-type hook’ specified by the code. For that, we have to extend the bar even further from D, in a straight line. This is shown in fig.14.13 below. The amount of extension required for this extension DE is 4φ.

Fig.14.13
Standard U-type hook

So we have obtained the standard U-type hook. In this we have extended the original bar from B to E. How much development length does this hook give? As we saw earlier in the case of 90^o bend, the code (cl.26.2.2.1) says that the anchorage value (that is the development length) contributed by a bend is equal to 4 times the diameter of the bar for each 45^o bend, subject to a maximum of 16 times the diameter of the bar. In the 180^o hook that we saw above, there are four 45 degrees. So the anchorage value is equal to 16φ. It should be noted that, as in the case of a bend, this contribution of 16φ is inclusive of the extension DE. That is., we will get this 16φ, only if we provide an extension DE. So 4φ from the portion DE cannot be added to 16φ as if to get a total anchorage of 20φ. The total contribution of the portion from B to E of the ‘standard U-type hook’ shown in fig.14.13 is equal to 16φ.


In the next section we will see how a standard 90 degree bend can be used in a practical situation.

PREVIOUS

NEXT                                          

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