In the previous section we saw the development length required at the inflection points in continuous beams. In this section we will see some code requirements that has to be satisfied at simple supports and at inflection points.
This clause gives requirements about three parameters. They are :
(i) Length of the embedment.
(ii) quantity of the bars which are embedded.
(iii) Position of the bars which are embedded.
We will see each of these parameters in detail:
(i) Length of embedment: Each of the continuing bars that get embedded inside the support should have an embedment length not less than Ld ⁄ 3 . If it is not possible to obtain this much embedment, bends can be provided.
(ii) Quantity of embedment: The continuing bars which get embedded inside the support of a simply supported beam should have a total area of Ast ⁄ 3 . Where Ast is the steel provided at the midspan to resist the maximum sagging moment.
An example demonstrating this requirement about 'quantity' is given below:
• Let there be 3- 16# and 2- 12# at midspan (of a simply supported beam) giving an area of 829.38 mm2.
• One third of this is equal to 276.46 mm2
• Let 1 -16# and 2 -12# be curtailed at various suitable points. So that only 2 -16# goes into the support.
• Area of these continuing bars = 402.12 mm2. This is greater than 276.46 mm2. Hence OK.
(iii) Position of continuing bars: The bars that we intend to embed in the support, 'should extend along the same face into the support'. This can be best explained by the example of bent-up bars. We know that in some cases, one or two bars are 'bent up' near the support region. At the midspan region, these bars are at the bottom face of the beam. After the point of bent, they deviate from the bottom face, and move towards the top face of the beam. So they will finally become top bars at the support, and will get embedded at the top face in the support. Such bars may satisfy (i) and (ii) above. But they do not satisfy (iii). So the bars that are compulsorily required to be embedded must not be bent up. In fact, they must not take any deviation, and 'must continue along the same face into the support'.
The figs. below shows the above requirements for a simply supported beam.
Fig.15.42
Bars having simple extension
Fig.15.43
Bars having a standard 90 degree bend
In the above fig., the anchorage value from B to D is 8Φ. So the total embedded length = A’B + 8Φ. This length should be greater than or equal to Ld ⁄ 3
Fig.15.44
Bars having an extension beyond the bend
In the above fig., the anchorage value from B to D is 8Φ. So the total embedded length = A’B + 8Φ + DE. This length should be greater than or equal to Ld ⁄ 3
Earlier we saw the application of the expression:
Ld (unique value) ≤ MuR ⁄ Vu + L0
The final arrangement of bars should be in such a way that, each bottom bar at the simple support satisfies the above expression.
Once this is checked, we must proceed to check the code requirements given in cl.26.2.3.3(a). That is., we must compare the final layout of bars with the fig.15.42, 43 or 44, which ever is applicable.
• If it is found that the embedded length is less than Ld ⁄ 3 then, more length should be provided in such a way that each bar has an embedded length greater than or equal to Ld ⁄ 3.
♦ If the length provided is found to be greater than that indicated in the figs., no reduction should be made.
• If it is found that the embedded area is less than Ast ⁄ 3, then more bars should be brought into the support to increase the area.
♦ If the area provided is found to be greater than that indicated in the figs., no reduction should be made.
• Finally, the position of bars should also be checked to ensure that 'the bars continue along the same face into the support'.
The clause gives the requirements of the three parameters related to continuous beams also. These can be summarized as follows:
(i) Length of embedment : For continuous beams, this is same as that of simply supported members, which is Ld ⁄ 3
(ii) Quantity of embedment: For continuous beams, this is equal to Ast ⁄ 4
(iii) Position of continuing bars: For continuous beams, this requirement is same as that of simply supported members. That is., 'bars must extend along the same face'.
For continuous beams, the bars will not require hooks or 90 degree bends, as enough space is available to extend the bars. So the above requirements can be shown in a single fig. as given below:
Fig.15.45
Extension required for bottom bars of continuous beams
Earlier we saw the application of the expression:
Ld (unique value) ≤ MuR ⁄ Vu + L0
The final arrangement of bars should be in such a way that, each bottom bar at the continuous support satisfies the above expression.
Once this is checked, we must proceed to check the code requirements given in cl.26.2.3.3(a). That is., we must compare the final layout of bars with the fig.15.45.
• If it is found that the embedded length is less than Ld ⁄ 3 then, more length should be provided in such a way that each bar has an embedded length greater than or equal to Ld ⁄ 3.
♦ If the length provided is found to be greater than that indicated in the figs., no reduction should be made.
• If it is found that the embedded area is less than Ast ⁄ 4, then more bars should be brought into the support to increase the area.
♦ If the area provided is found to be greater than that indicated in the figs., no reduction should be made.
• Finally, the position of bars should also be checked to ensure that 'the bars continue along the same face into the support'.
So we have discussed the sub clause (a) of 26.2.3.3. It is related to the bottom bars at simple supports and continuous supports. In the next section, we will discuss the sub clause (b) which is related to the bottom bars at various supports in a lateral load resisting frame.
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