Chapter 15.3 - Approximate method for theoretical points of curtailment

In the previous sections we saw the method of finding the theoretical points. We used the method of solving the equation of Bending moment. In this section we will discuss about another method to determine the theoretical cut-off points. 

For this we will first discuss about a special phenomenon that occur at the sections where curtailments are done: We have seen a beam in which there is a total of 8 bars at midspan, and some of these bars were curtailed at various sections. Figs.15.7, 15.8 and 15.9 showed the theoretical sections of curtailment.  

Now let us look at the depth of neutral axis xu which were noted down earlier at each of the sections CL, LL, MM and NN.  • xu,CL = 250.67, • xu,LL = 205.55, • xu,MM = 160.43, • xu,NN = 80.21. We have also noted down the effective depth d at these sections. From these, we can make a table which gives the lever arm z at each of these sections. The table so formed is given below:

Name of Section	d (mm)	xu (mm)	z = d- 0.416xu
CL	533.3	250.67	429.02
LL	545.439	205.55	459.93
MM	554	160.43	487.26
NN	554	80.21	520.63

The next step is to calculate a particular ratio at each of the section where curtailment is made: The ratio of 'z at midspan' to 'z at the section'. These are calculated below:
zCL / zLL = 429.02 / 459.93 = 0.9328
zCL / zMM = 429.02 / 487.26 = 0.8805
zCL / zNN = 429.02 / 520.63 = 0.8240
All the values are close to '1'. So they can be taken approximately to be equal to ‘1’.

When the beam is loaded to the ultimate state,
• the midspan section (section CL in our example ) will reach the ultimate state. 
• The sections at which curtailments are made ( sections LL, MM, and NN in our example ) will also reach the ultimate state. 

In such a situation, the following phenomenon will occur:

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The ratio: [ z at midspan / z at a section where curtailment is done ] can be approximately taken to be equal to '1'. In other words, the lever arm at all the sections where curtailment is done will be approximately equal. They will all be approximately equal to the lever arm at midspan.

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Now, we know that the ultimate moment of resistance MuR at any section is given by MuR = 0.87 fy Ast  z . So we can write:
MuR,CL =0.87 fy Ast,CL  zCL   and   MuR,LL =0.87 fy Ast,LL  zLL 

Taking ratios, we will get
MuR,CL    / MuR,LL   = [ Ast,CL  / Ast,LL] x [ zCL / zLL  ]

But by the special phenomenon that we saw above, [ zCL / zLL  ] =1. 
So we can write:
MuR,CL    / MuR,LL   = [ Ast,CL  / Ast,LL]

We can rearrange this equation as :
• MuR,LL   = MuR,CL  x [ Ast,LL  / Ast,CL]
Similarly we can write:
• MuR,MM   = MuR,CL  x [ Ast,MM  / Ast,CL]  and
• MuR,NN   = MuR,CL  x [ Ast,NN  / Ast,CL]

We can get a general form from the above three equations. For this, let us assign the name ‘XX’ to denote any section at which curtailment is made. Then we can write the general form as:
Eq.15.3:
MuR,XX   = MuR,CL  x [ Ast,XX  / Ast,CL]
   
Ast,XX  is the area of steel remaining in the section after curtailment is made. So, if we intend to curtail a particular number of bars of a particular diameter, we can calculate the area  Ast,XX  which remains after curtailment. So MuR,XX is the only unknown in Eq.15.3.

This gives us an easy method to calculate the MuR at a section where curtailment is made. When we do a curtailment, the remaining bars which continue the journey will be less. As a result, the MuR which remains will also be less. We can calculate this remaining MuR using Eq.15.3. There is no need to do a 'detailed analysis of the section' with the remaining bars and the new value of d.

When MuR at a section is determined, the only step remaining is to find the position at which this much factored bending moment is applied on the beam. We have seen earlier that, this step is done by solving for x in the equation for BM.

So this is an approximate method to calculate the theoretical cut-off points in the beam. Let us now use this method to find the theoretical cut-off points in our beam. We can show the results in a table as given below:

Name of Section	Ast,XX (mm2)	Ast,XX  / Ast,CL	MuR,XX	Distance from support A (mm)
CL	1256.637	_	194.65	_
L1L1	1030.442	0.8200	159.61	1.5358
M1M1	804.248	0.6400	124.57	1.0517
N1N1	402.124	0.3200	62.29	0.4570

Note that the sections are renamed as L1L1, M1M1 and N1N1 , because their positions are slightly different from those of LL, MM and NN that we obtained using the exact method.

Sample calculation:
Let us take section M1M1
• Ast remaining after curtailment at section M1M1 = 804.248 mm²  
• Ratio  [Ast,XX  / Ast,CL] at this section = 804.248 / 1256.637 = 0.6400
• This ratio multiplied by the MuR at midspan will give the MuR at M1M1 (Eq.15.2)
So MuR,M1M1 = 0.6400 x 194.65 = 124.57 kNm

In this ‘approximate method of calculating theoretical cut-off points’, MuR at the midspan (MuR,CL in our example) is the only MuR that we have to calculate using the exact method of analysis that we learned in chapter 3

• Now we can use Eq.15.2 to determine the distance of M1M1 from the support A.
Eq.15.2:
M_u,XX = 150 x – (1.5 x 40 x²)/2 where x is the distance of the section from the support A.
• We can put M_u,XX = 124.57 in the above equation, and obtain x = 1.0517 (the same value in the above table)  and it's symmetric counterpart 3.9483 m. 

The fig. showing all the sections is given below:

Fig.15.18
Final theoretical sections using approximate method

Here is a comparison of the results from the two methods:
Distance from support A:
• Exact Method: LL =1.7609, MM =1.2616, NN =0.5686 m
• Approximate method: L1L1 =1.5358, M1M1 =1.0517, N1N1 =0.4570 m

So we have seen two methods to determine the theoretical curtailment points. In the next section, we will discuss about the modifications that have to be made to the theoretical cut-off points.

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