In the previous section we saw the method of finding the theoretical point of curtailment. In this section we will see the theoretical curtailment at more than one point in a beam.
Fig.15.6
Simply supported beam
The maximum factored BM at mid span will be equal to (1.5 x 40 x 52) /8 = 187.5 kNm. Assuming M20 concrete, and Fe 415 steel, the beam can be designed for this BM. The result is: Ast,CL = 1227.3 mm2, at an effective depth of 525 mm. Let us give 'sufficiently large number' of bars so that we will get the option to do curtailments at many sections along the length of the beam. Carrying out curtailment at a least of 3 points along the same beam, will give us a better understanding about the theoretical cut-off points.
So we will give 4-16# and 4-12# in three layers. This gives an area of 1256.637 mm2. The effective depth d will be equal to 533.3 mm. The section showing the arrangement of bars, and the details of calculation of d can be seen here.
When we analyse the beam section, by the methods that we discussed in chapter 3, we will get the ultimate moment of resistance MuR,CL equal to 194.65 kNm. This is safe, as it is greater than 187.5kNm.
While we do the analysis, let us note down the value of xu, the depth of neutral axis. This will be used in a discussion on another topic in a later section. Here, the calculated value of xu,CL =250.67 mm
Now, let it be decided to curtail the two 12 mm dia. bars in the third layer at a section LL. We have to calculate the suitable position of the section LL to do this curtailment. This calculation can be done as follows:
When the two 12 mm dia bars are curtailed, the remaining bars are 4-16# and 2-12#. This will give an area of 1030.442 mm2. The c.g of this bar group will be different. So d will also be different. The details of the calculations of the new values can be seen here.
So at LL, we have a beam section with an area of steel equal to 1030.442 mm2 at an effective depth d of 545.439 mm. By the same method that we used to calculate MuR,CL, we can obtain MuR,LL = 171.11 kNm.
Just as before we will note down the depth of NA. Here, the calculated value of xu,LL =205.55 mm.
So now we find the position of the section at which the applied factored bending moment is equal to 171.11 kNm:
The factored reactions RA and RB = 150 kN.
The equation for BM at any section on the beam can be written as
Eq.15.2:
Mu,XX = 150 x – (1.5 x 40 x2)/2 where x is the distance of the section from the support A.
We can put Mu,XX = 171.11 in the above equation, and obtain x = 1.7609 and 3.2391 m.
These two values are symmetric (∵ 1.7609 + 3.2391 = 5.0m) about the section CL passing through the mid span. So we will concentrate on one single value, 1.7609. We must place section LL at this distance from support A. So the two 12 mm dia. bars in the third layer can be stopped at this section. Rest of the bars can continue their journey towards the supports. This is shown in fig.15.7 below:
Fig.15.7
Curtailment at section LL
When this two 12 mm dia bars are also curtailed, the beam will have 4-16#. This will give an area of 804.248 mm2. The c.g of this bar group will be different. So d will also be different. But as there is only one layer, calculations are easier. The c.g is at a distance of 16/2 = 8 mm from the bottom most edge of the bar group. So d = 600-(38+8) = 554 mm.
So at MM, we have a beam section with an area of steel equal to 804.248 mm2 at an effective depth d of 554 mm. By the same method that we used to calculate MuR,CL and MuR,LL , we can obtain MuR,MM = 141.49 kNm.
Just as before we will note down the depth of NA. Here, the calculated value of xu,MM =160.43 mm.
So now we find the position of the section at which the applied factored bending moment is equal to 141.49 kNm:
The factored reactions RA and RB = 150 kN.
We can use the same equation 15.2 for BM at any section on the beam.
Mu,XX = 150 x – (1.5 x 40 x2)/2 where x is the distance of the section from the support A.
We can put Mu,XX = 141.49 in the above equation, and obtain x = 1.2616 and 3.7384 m. These two values are symmetric (∵ 1.2616 + 3.7384 = 5.0m) about the section CL passing through the mid span. So we will concentrate on one single value, 1.2616. We must place section MM at this distance from support A. So the two 12mm dia. bars in the middle layer can be stopped at this section. Rest of the bars can continue their journey towards the supports. This is shown in fig.15.8 below:
Fig.15.8
Curtailment at section MM
Now, let us curtail the two middle 16 mm dia. bars in the bottom most layer at a section NN. We have to calculate the suitable position of the section NN to do this curtailment. This calculation can be done as follows:
When this two middle 16 mm dia bars are also curtailed, the beam will have 2-16# left. This will give an area of 402.124 mm2. The c.g of this bar group will be same as that at section MM that we saw just before. So d will also be same, and is equal to 554 mm.
So at NN, we have a beam section with an area of steel equal to 402.124 mm2 at an effective depth d of 554 mm. By the same method that we used to calculate MuR,CL, MuR,LL, and MuR,MM, we can obtain MuR,NN = 75.59 kNm.
Just as before we will note down the depth of NA. Here, the calculated value of xu,NN =80.21 mm.
So now we find the position of the section at which the applied factored bending moment is equal to 75.59 kNm:
The factored reactions RA and RB = 150 kN.
We can use the same equation 15.2 for BM at any section on the beam.
Mu,XX = 150 x – (1.5 x 40 x2)/2 where x is the distance of the section from the support A.
We can put Mu,XX = 75.59 in the above equation, and obtain x = 0.5686 and 4.4314 m. These two values are symmetric (∵ 0.5686 + 4.4314 = 5.0m) about the section CL passing through the mid span. So we will concentrate on one single value, 0.5686. We must place section NN at this distance from support A. So the two middle 16mm dia. bars in the bottom most layer can be stopped at this section. Rest of the bars can continue their journey towards the supports. This is shown in fig.15.9 below:
Fig.15.9
Curtailment at section NN
So we have completed all the possible theoretical curtailments. It may be noted that the positions of the above sections can be determined by graphical methods also. For this we have to plot the length of the beam along the X axis, and the bending moment along the Y axis. The fig.15.10 below shows the position of two sections LL and NN determined graphically. MM can also be determined using the same procedure.
Fig.15.10
Graphical method
From the green coloured grid lines in the above fig., we can see that bending moment is plotted at a scale of 1 unit = 100 kNm, and the length of the beam is plotted at a scale of 1 unit = 1m. The horizontal magenta coloured dotted lines are drawn through -171.11 (MuR,LL) and -75.59 (MuR,NN) in the Y axis. The points of intersection of these lines with the plot of the bending moment will give the positions of the sections.
We have obtained the positions of the sections for curtailment. Consider the sections LL, MM and NN on the left side of CL. Each of them has a definite position. None of them should be moved towards the right. That is., towards the CL. If any of them is moved towards CL, the bars will be curtailed earlier. This may seem to be economical. But the undesirable effect that we discussed earlier based on fig.15.5 in the previous section, will occur. Similarly none of L'L', M'M' and N'N' should not be moved towards the left.
Now let us see the relation between ‘the position of curtailment’ and the ‘percentage of steel left at the section after curtailment’. For this we can prepare a table as shown below:
Name of Section
|
Area of steel after curtailment (mm2)
|
(B) expressed as % of total area at section CL
|
Distance of section from support (m)
|
| (A) | (B) | (C) | (D) |
| LL | 1030.442 | (1030.442/1256.637)x100 = 81.99% | 1.7609 |
| MM | 804.248 | (804.248/1256.637)x100 = 64.00% | 1.2616 |
| NN | 402.124 | (402.124/1256.637)x100 =32.00% | 0.5686 |
• When the ‘percentage of steel left after curtailment at a section’ is higher, the section will be at a region where the applied factored bending moment is higher.
• When the ‘percentage of steel left after curtailment at a section’ is lower, the section will be at a region where the applied factored bending moment is lower.
This completes the discussion on the method to determine theoretical points of curtailment in a simply supported beam acted upon by a udl. In the next section we will see a general scheme for the curtailment of top bars at an intermediate support in a continuous beam.
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