Chapter 15.5 - Conditions for doing a curtailment

In the previous section we saw the internal forces at a section in a beam. We will now see the internal forces at a section in a cracked beam.

The fig.15.25 below shows the free body diagram of the 'portion to the right of the crack' shown in fig.15.21 of the previous section.

Fig.15.25
Free body diagram of the portion to the right of the crack

To determine the moment at section XX, all the load on the right side of section XX should be considered. the procedure is same as what we did for the un-cracked beam based on fig.15.23. The expression for bending moment will also be same as before. However, we will write the steps again:
• The moment at XX, due to the reaction R_B is equal to R_B x (anti-clock wise). 
• The moment at XX due to the applied load w_u is equal to (w_u x² )/2 (clock wise) 
• So the net factored moment at XX is given by:

M_u,x = R_B x  – (w_u x² )/2 

Having determined the applied factored bending moment at the section, we can look into the internal forces as shown in the fig.15.26 below:
Fig.15.26
Equilibrium of internal forces

Here also the procedure is same as before. But we will write the steps again:
• The compressive force C_u acts at the centroid of the 'concrete area above the neutral axis'.
• The tensile force T_u acts at the centroid of the steel area.
• These two forces are equal in magnitude but opposite in direction. So they form a couple. The magnitude of this couple is equal to T_u z . It is also equal to C_u z . Here, z is the lever arm.
• This couple resists the applied factored bending moment M_u,x.

The expressions obtained above are same as that of an un-cracked beam. But there is an important difference here. The tensile force T_u that is resisting the bending moment is the T_u at section YY. So, when a crack is formed in this way, the tensile force that we have to consider is the one which is at the beginning of the crack, which is marked as section YY in the fig.15.26. This means that we must not stop the bar at the theoretical point XX. Instead, we must extend it upto YY. Because, if we stop the bar at XX, there will not be any T_u available to form the couple at YY.

So we will now determine the distance between sections XX and YY. This distance is the 'extension' that has to be provided for the bar. To calculate the distance, we will idealise the shape of the crack as a straight line as shown in the fig.15.27 below:

Fig.15.27
Idealised shape of the crack

The crack is idealised into a straight line pq. In the ∆ pqr, the altitude pr is equal to the effective depth d. We want the base qr. For this we can use the relation tanθ = pr⁄qr = d⁄qr  . From this we get qr = d⁄tanθ = d cotθ. We have seen that the diagonal cracks mostly occur at an angle of 45^o. So qr = d cot45 = d x 1 = d (∵ cot45 =1). Thus we can see that the bar should be extended a distance of d from the theoretical cut-off point.

In the code, specifications about this 'extension of bar' is given in the cl.26.2.3.1. According to this clause, the bar should be extended for a distance of d or 12Φ, which ever is greater. Here Φ is the diameter of the bar. The clause also says that this extension is not required at simple supports and at free ends of cantilevers with concentrated loads.
If we denote the extension required as L_a, then the code requirement can be written as:
15.4
L_a ≥ d or 12Φ whichever is larger

Thus we now know how much extension should be given to the bars after the theoretical point of curtailment. Still, we have to learn a lot more details before we do curtailments of bars. In the following discussions we will explore those details: 

Prevention of diagonal cracks
We have seen that (fig.15.21 of the previous section) diagonal cracks can form at the points of curtailment. We have to take steps in order to prevent it. We have to do some 'tests' at the section where we intend to do a curtailment. If the results of any one of these 'tests' are satisfactory, then we can do the curtailment at that section. The details are given in cl.26.2.3.2 of the code.

Test 1:
In the topic on shear design, we have seen the details about the ultimate shear resisting capacity V_uR of a section. At a section where we intend to do a curtailment: 
• First calculate this V_uR. 
• Then calculate two third of this V_uR. - - - (1) 
• Note down the applied external shear force V_u at that section. - - - (2)
• If (2) is less than or equal to (1), we can say that the section has passed this test 1.  
If Test 1 is passed at a section, a curtailment can be made there. So we can write:
15.5:
A curtailment can be done if 2/3 V_uR ≥ V_u

Test 2:
If we intend to do a curtailment at a section, some 'extra stirrups' should be present at that section. We need to know the exact position where these stirrups must be present. We also need to know the quantity required for this stirrups. First we will look at the 'position':

From the end of the curtailed bar, mark of a distance of 0.75d along the length of the curtailed bar. The extra stirrups should be distributed at uniform spacing within this distance. The direction of marking is from the end of the curtailed bar in the direction in which the total area of bars increases. This is shown in the fig.15.28 below:

Fig.15.28
Position of extra stirrups

Now we want the quantity. By 'quantity' we mean the 'area of all the extra stirrups' provided. [This area is equal to N x No. of legs of one stirrup x Area of cross section of the bar of the stirrup. Where N is the total No. of extra stirrups]. According to the code, this area should not be less than:
15.6:

Where b is the width of the beam, s is the spacing at which the extra stirrups are placed, and f_y is the characteristic strength of steel. Out of these, s is an unknown. The code gives us the guidelines to calculate this s. According to the code, the uniform spacing (at which the extra stirrups are distributed with in the space of 0.75d) should not be greater than:
15.7:

Where d is the effective depth and β_b is the ratio:
Area of bars cut-off / Total area of bars at the section

So we can evaluate 15.7, round it down to the nearest multiple of 10 mm, and then substitute it in 14.6. Thus we will get the 'quantity'. These stirrups are to be provided in addition to those provided for shear and torsion.

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Let us look at a simple example to demonstrate the requirements of this condition:
Assume Width b = 230 mm, effective depth d = 404 mm, f_y = 415 N/mm²

• The extra stirrups should be uniformly distributed within a distance of 0.75 x404 =303 mm 
• Let there be a total of 3-16# at the midspan, and let one of them be curtailed at a section away from the midspan.
• Then β_b =  area of one 16 mm bar ⁄ area of three 16 mm bars  = 201⁄603  = 0.33 - - - (1)
• From 15.7, we can say that s should not be greater than  404⁄8 x 0.33   = 153.03 mm - - - (2) Let us provide s = 150 mm.
• Substituting this in 15.6, we can say that the area should not be less than 0.4 x 230 x 150⁄415  = 33.25 mm² - - - (3) 
• Assuming 2-legged stirrups of 8 mm dia., area of one stirrup = 100.5 mm² > 33.25 mm² Hence OK.
• Thus we must provide 2-legged 8 mm dia. stirrups at a spacing of 150 mm over a distance 303 mm.
• If we provide 3 stirrups at a spacing of 152 mm, all the requirements will be satisfied. (Note that with the spacing changed to 152 from 150 mm, (3) will change. But it still will be less than 100.5mm²)
This arrangement is shown in the fig.15.29 below:

Fig.15.29

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If we do this test at a section, and find that there is sufficient additional stirrups, we can do a curtailment at that section. Or conversely, if we want to do a curtailment at a section, we can provide these additional stirrups at that section if they are not already present.

Test 3: This test has two parts.
part 1: At the section that we intend to do the curtailment: 
• First calculate the external applied factored moment M_u. 
• Then calculate the steel required to resist that M_u. - - - (1) 
• Next find how much continuing steel will be available when the intended curtailment is done. - - - (2) 
• If (2) is greater than or equal to twice (1), then we can say that the section has passed the part 1 of this test.

Part 2: At the section that we intend to do the curtailment: 
• First calculate the V_uR. 
• Next calculate three fourth of this V_uR. - - - (1)
• Note down the external factored shear force at the section. - - - (2) 
• If (2) is less than or equal to (1), we can say that, the section has passed part 2 of this test.

The section will pass the Test 3 if and only if it passes both part 1 and part 2. It should be noted that this Test 3 is applicable only for bars whose diameter is less than or equal to 36 mm.

According to cl.26.2.3.2 of the code, if the section passes any one of the above three tests, a curtailment can be done at that section. So this clause gives us the method to safeguard against the formation of diagonal cracks when curtailment is done at a section.

Extension of bars at simply supported ends

Now we will see the details about the extension that has to be given to the bars at simply supported ends. We know that the bottom bars of a simply supported beam will be in tension, and will be in a 'stretched' condition. And so, they will try contract back to their original length. When they try to contract, the ends of bars will try to pull out from the support. This cannot be allowed. We have to ensure that, the concrete can exert 'sufficient grip' on the bars so that, the pulling out is prevented. For that, 'sufficient length' of bars should be available. Our aim is to calculate this 'sufficient length'.

Let us consider a infinitesimal segment pq of a simply supported beam as shown in the fig.15.30.

Fig.15.30
Small segment of a simply supported beam

It can be seen from the bending moment diagram that, the applied bending moment at q will be greater than that at p. We can denote the difference by dM. The free body diagram of the segment pq is shown in the fig.15.31 below:

Fig.15.31
Free body diagram of the beam segment pq

From the free body diagram, the following relations can be obtained:
• M= Tz = Cz - - - (1)    (∵ Moment M is resisted by the couple formed by T and C)
• M+dM = (T+dT)z = (C+dC)z - - - (2)  (∵ Moment M+dM is resisted by the couple formed by T+dT and C+dC)
• Subtracting (1) from (2) we get :
dM = dT z - - - (3)

So we need more details about dT. This we will see in the next section.

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