Chapter 15.7 - Development length requirements at simple supports

In the previous section we saw the least possible length (on which bond stress will be exerted to prevent pulling out) that is available at a simple support. This length is a portion of that 'length of the bar which is coming between the points of zero moments'. If any length is available beyond these points of zero moments, then concrete will exert grip on that length also. So that length will also assist in preventing the pull out.

Let us denote the extension beyond the point of zero moment as L₀. This L₀ can be given in a number of different ways. We will now see each of these methods in detail.

Fig.15.34

Simple extension

In the above fig.15.34, the bar is given a simple extension (that is., with out any bends or hooks) beyond the point of zero moment. But the bar cannot continue indefinitely. It can be continued upto a maximum point where the concrete cover C_c at the end becomes just sufficient. C_c is specified by the code. Details can be seen here. 

In many of the worked out examples that we did until now, 'moderate' exposure conditions were assumed, and so C_c was taken as 30 mm for the bottom bars. This much cover is to be given from the outer surface of the stirrups. So the main bars get even more cover. But in our present case, the cover is taken from the end surface of the beam. Diameter of stirrups does not come into the picture. So we can assume 30 mm from the end of the bar to the end surface of the beam. However, it is important to consider the exposure conditions and other code requirements while fixing up C_c for each design problem. 

From the fig.15.34, we can easily calculate L₀ if the width of support and C_c are known. 

In some cases, the CL of the support may not coincide with the Point of zero moment. This happens when the effective span is different from the c/c distance between the supports. In such cases necessary adjustments should be made in the calculation of L₀.

Now we consider the case when the bar is given a standard 90^o bend at the support as shown in the fig.15.35 below:

Fig.15.35

Extension with a standard 90^o bend

We can see that a small portion of the 'original straight portion' of the bar extends beyond the CL of the support. This is denoted as BA' in the fig.15.36 below. In an actual beam, depending upon the width of the support and C_c, there may or may not be such a small portion. In the above fig.15.35, we have to calculate the length of this portion. 

Before that, let us see the other features of this arrangement: We can see that the bend portion begins at the end of the straight portion. We can also see that after the bend portion, a straight length of 4Φ is given to make it a standard 90^o bend. We know that the radius of the bend is 4Φ. So the calculations can be done based on the fig.15.36 given below:

Fig.15.36

Calculation of L₀

From the fig., we can see that

• C_c + Φ + r + BA’ = half of the width of support = ws/2

• But r = 4Φ. So we can write

• C_c + 5Φ + BA’ = ws/2. From this we will get BA’.

• Now, the anchorage value of the bend from B to D is 8Φ.

• So the value of L₀  = BA’ + 8Φ

Finally we consider the case when an extension is given to the bar even beyond point D in the above fig. This is shown in the fig.15.37 below:

Fig.15.37

Extension beyond point D

In this case, the calculation of BA’ is same as the previous one. Also, the value from B to D is 8Φ.

• So we get  L₀  = BA’ + 8Φ + DE

So now we know how to calculate L₀. This length L₀ will also help to prevent the pull out. So the total length available to us is equal to M_uR ⁄ V_u , + L₀, and this much available length should be greater than or equal to L_d. So we can write:

15.6

L_d (unique value) ≤ M_uR ⁄ V_u + L₀

This is the same expression given in the cl.26.2.3.3 (c) of the code. In the code, the ultimate moment of resistance M_uR is denoted as M1, and the factored shear force V_u is denoted as V.

Next, we will see a modification that has to be made to the above expression, when the simply supported end of the beam satisfies a 'special condition'. In the topic on design for shear, we discussed about the critical sections for shear design. There we saw that, when a 'transverse compression' is present at a support, the shear strength of the beam is increased near the support region. Figs. 13.62 to 13.65 of that section were used to illustrate the support conditions were such compression will occur. The same is applicable for our present discussion also. That is., when such a compression is available, the ends of the bars of the beam which are extended into the support will experience a confining effect. So a greater force will be necessary to pull the bars out. This is in effect, having a longer length of bars to resist the pull out. So the code allows us to increase MuR ⁄ Vu by 30 per cent. Thus, the expression that we can use when the ends of bars are under transverse compression is:

15.7:

In the next section, we will discuss about the development length requirements of the bottom bars of continuous beams.

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Chapter 13 (cont..19) - Shear Check for RCC slabs

In the previous section we saw that the spacing of stirrups should not be greater than 300 mm at any part of our beam. Now we will calculate the spacing for the region from A to E.

(After completing the design of this beam, we will discuss the 'shear check for slabs', and also the shear in 'beams subjected to axial forces'.)

For calculating the spacing, we use the basic Eq.13.58

V_us =V_u -τ_cbd

V_u is the shear force at the critical section =52.29 kN

τ_c = 0.5745 N/mm²

So we get V_us = -960.405 N

We are getting a negative value. This is so because, as seen from the graph, the critical section falls to the right of E. It can be further explained as follows: We have seen that the portion between A and F need to be designed for 52.29 kN only. But this is less than V_uc which is equal to 53.25 kN. This means the applied factored shear at the critical section is less than even the shear capacity contributed by concrete. So we can give a constant spacing of 300 mm through out the length of the beam. Because, for our beam, we have confirmed that the spacing at any point should not be less than 300 mm.

Some designers consider 300 mm to be excessive, and provide 250 mm. We shall adopt this, and so the final sectional detail can be shown as in the fig below:

---

Design shear strength of concrete in slabs:

The behaviour of slabs and shallow beams under the action of shear forces have been studied through experiments. The results of those experiments show that the slabs and shallow beams can take up a higher shear stress than applicable to beams of usual proportions. This means that the contribution from concrete is more in these cases. So the code allows us to increase τ_c, given in table 19. This increase is effected by multiplying it with a factor k, given in cl.40.2.1.1 of the code.

K depends on only one factor, which is D, the overall depth of the section. From the table, we can see that when D = 300 mm, K becomes equal to 1, which means that at D= 300 mm  and above, the concrete of the section will not give the extra shear strength attributed to slabs and shallow beams.

If we closely examine the values given in the table in the above clause, we can see that the value of k can be written in the following method also:

13.66: 

• k= 1.3 when D is less than or equal to 150 mm

• k= [1.6 -(0.002D)] when D falls between 150 and 300mm (should not be equal to 150 or 300)

• k=1 when D is greater than or equal to 300 mm

Where D is the overall depth of the slab in mm

It should be noted that the above values are applicable only to solid slabs. They cannot be used for ribbed slabs, waffle slabs, flat slabs etc., (Details about these special types of slabs can be seen by scrolling down to items 3, 4 and 5 of the web page given here)

Now we can do a problem demonstrating the shear design of a slab. We will do this for a slab for which we have already done the flexure design. So we will revisit Solved example 6.1 which is given towards the end of chapter 6 – Design of One-way slabs.

Solved example 13.3

Design a One-way slab with a clear span of 3.5 m, simply supported on 230 mm thick masonry walls. The loads other than self weight acting on the slab are the following:

• Live load 2.5 kN/m²

• Surface finish 1 kN/m² 

Assume that the slab is subjected to moderate exposure conditions. Assume Fe 415 steel.

Solution:

We have already done the design for flexure of this problem. From that we will get the following details:

• Total slab thickness  D = 180 mm

• Effective depth d = 145 mm

• f_ck = 25 N/mm²

• p_t at mid span = 0.301

• p_t at the support = 0.301/2 = 0.1505 (Alternate bars are bent up. So only half of the bars at mid span will be present at the support)

• Effective span l= 3645 mm

• Total factored load w_u(DL+LL) = 12 kN/m

Here we are going to check whether the slab is safe in shear. We are going to do this check at the support, which is the section having the maximum value of shear force. If it is safe at the support, it will be safe at all other sections also. So there is no need to consider the different positions of the LL, to obtain the maximum shear force at other sections.

Another point to note is that, the load applied on the slab will be in kN/m². We have converted it into kN/m while doing the flexure design. This same load is to be used for shear check also because, we are checking the shear at the support of a '1000 mm wide strip of the slab', which is considered as a 'beam of width b = 1000 mm and overall depth D'

With the above data, we can do the shear check as shown below:

V_u at the support =w_u l / 2 = (12 x 3.645) /2 = 21.87 kN

From table 19, value of τ_c for a p_t of 0.1505 is equal to 0.2904 N/mm²

This can be increased by multiplying by k. So we have to find the value of k. 

• D =180 mm (falls between 150 and 300) 

• So, from 13.66 above, k = [1.6 -(0.002 x180)]= 1.24

• Thus, the contribution from concrete = V_uc = k times τ_c x bd=1.24 x 0.2904 x 1000 x 145 = 52.2 kN.

• This is greater than V_u = 21.87 kN.

So the slab is safe in shear.

---

Members subjected to axial force in addition to shear force

Consider the beam shown in fig.13.77 below:

Fig.13.67
Beam subjected to axial compressive force

It is subjected to an axial compression, in addition to shear and flexure. This type of loading in which axial tensile or compressive loads are present may occur due to a variety of reasons.

• Presence of actual external axial loads

• Longitudinal prestressing of the member

• Restraining axial forces developed due to shrinkage of concrete or temperature changes

The shear strength of a beam will be modified due to the presence of axial tension or compression. This modification in pre-stressed beams is treated by different principles. We will discuss here only about ordinary beams.

When we discussed the basics about shear on a beam, we saw the forces acting on the particles of the beam. We have seen fig.13.8 which showed the various forces on the particles. If an axial force is also acting on the beam, then the forces shown in the fig.13.8 will be modified. The angle α of the principal plane, and the magnitude of the principal tensile stress which causes the diagonal tension crack will also change.

Axial compression will increase the shear capacity of concrete in a beam. And axial tension will cause a decrease in the shear capacity of concrete in a beam.

So when a beam is subjected to an axial compressive force in addition to flexure and shear, the code allows us to increase the value of τ_c by multiplying it with a factor denoted as δ. It’s value is given in cl.40.2.2 of the code as

Eq.13.67:

The value of δ calculated from the above equation should be compared with '1.5'. If it is found to be greater than 1.5, then δ should be taken as equal to 1.5. If the calculated value is less than 1.5, then we can use it. In other words, δ should be taken as the lesser of:

• value calculated using Eq.13.67 and
• 1.5

In the above Eq.13.67,
• p_u is the factored axial compressive force in N
• A_g is the gross area of the section in mm²
• f_ck is the characteristic strength of concrete in N/mm²

When an axial tension is present, we must decrease τ_c using a reduction factor. But the code does not mention about the case of axial tension. We can use the following expression based on ACI code in such a situation:
Eq.13.68:

Here p_u is the factored axial tension (in N), and should be given a negative sign.

---

This completes our discussion on Shear design of beams and slabs. In the next section we will discuss about 'Bond and Development length'.

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Chapter 13 (cont..18) - Second solved example on the shear design of a beam

In the previous section we completed the shear design of a beam. Now we will see another example.

Solved example 13.2

Here we will do the shear design of a beam, the ‘design for flexure’ of which was already done by us. So we revisit the Solved example 4.1 which is given towards the end of chapter 4 (cont..5) . For convenience, the problem, and it’s final designed section diagram are given below: 

A rectangular reinforced concrete beam, located inside a building in a coastal town, is simply supported on two walls. The thickness of both the walls is 230 mm and the centre to centre distance between the walls is 5 m. The beam has to carry a uniformly distributed live load of 10 kN/m and dead load of 5 kN/m. Design the beam section for maximum moment at mid span. Assume Fe 415 steel

The final designed section is:

Shear design:

We have the cross sectional dimensions of the beam, and the details of tension steel provided. In the above fig., the stirrups are indicated as 'links'. Our aim now is to do their actual design.

The effective span of the beam is calculated as 5000 mm, while designing the beam for flexure.

We will write down the available data:

b =230 mm, d =403 mm, A_st =710.00 mm², f_y =415 N/mm², f_ck =25 N/mm², Effective span l =5000 mm, w_u =26.38 kN/m

First of all, we have to check if the condition V_u ≤ V_uR is satisfied

For our beam, maximum shear force V_u will be at the support, and it is equal to the support reaction =65.95 kN

Now we calculate V_uR,lim: (details here)

V_uR,lim = τ_c,max bd

From table 20 of the code, for M25 concrete, τ_c,max =3.1 N/mm²

Thus we get V_uR,lim =287339 N =287.339 kN. This is greater than V_u. Hence OK

We have used the maximum possible value (the support reaction) of V_u for the above check. For the design of stirrups, we will be using a lesser value (the shear force at the critical section, which is at a distance d from the face of the support) of V_u. The check is satisfied for the maximum possible value. So it will be satisfied for the V_u at critical section also. So we can proceed with the design.

We will now draw the Shear force diagram for the beam:

The data required for drawing the SF diagram are the following:

• Total Factored load w_u =26.38 kN/m 

• Effective span l =5000 mm

• Support reaction = w_u l /2 =65.95 kN

In the above diagram, points A and C have a 'y' coordinate of 65.95 kN, which is the support reaction.

It may be noted that the equation of the above shear force plot is: y =65.95 -26.38x - - - (1)

• 65.95 is the support reaction

• 26.38 is the udl

Now we will consider V_uc, the contribution from concrete:

The bottom tensile steel is given uninterrupted from support to support. So A_st will be the same at whichever section that we take.  This means that the shear resistance force contributed by concrete will be the same at whichever section that we take. So the graph of V_uc will be a horizontal line.

We have V_uc =τ_cbd (Details here). We have to calculate τ_c. For this, we look at table 19 of the code.

pt = 100Ast/bd =0.766

0.750  → 0.5700

0.766  → 0.5745

1.000  → 0.6400. So we get τ_c =0.5589 N/mm².

Thus V_uc = 53248 N =53.248 kN

So the plot of V_uc will be as follows:

In the above diagram, points D and D' have a 'y' coordinate of 53.25 which is the value of V_uc.

It may be noted that the equation of the above plot is: 

• y =53.25 upto the mid span and 

• y = -53.25 from the mid span to the end support. 

Now we superimpose V_uc on V_u, as shown in the fig. below:

In the above fig., the point of intersection of the two plots is at 'E'.  It has a y coordinate of 53.25. This we can obtain without any calculations because 'E' lies on the plot of V_uc, on which all points have same y value

Now we will find the x coordinate. For this, we put the above value for 'y' , and solve for 'x' in the equation (1) for V_u that we wrote earlier:

y =65.95 -26.38x

Solving we get x = 0.481 m

After point E, the applied shear force V_u becomes less than V_uc. 

Now we look at the critical section of the above beam. The critical section is at a distance of d from the face of the support. The support details are shown below. [We can see that the beam is symmetrical about the midpoint. So we need to consider only one half for the shear design. This will give us more space to work on.]

The distance of the critical section from the center of support will be equal to 0.115 + 0.403 =0.518 m. The applied factored shear force at this critical section is obtained by putting x = 0.518 in (1). 

So putting x =0.518 in 'y =65.95 -26.38x', we get, y = V_u =52.29 kN

The portion between A and F need to be designed for 52.29 kN only.

Before calculating the spacing of stirrups for the above V_u of 52.29 kN, We will work out the constraints that have to be considered.

The first constraint that we have to calculate is:

Portions of our beam for which stirrups should be compulsorily provided. (details here)

So we have to make a plot of 0.5V_uc and super impose it on the above graph. This is shown in the fig. below:

V_uc =53.25 kN. So 0.5V_uc =26.625 kN

The point of intersection is 'G'. It has a y coordinate of 26.625. This we can obtain without any calculations because 'G' lies on the plot of 0.5V_uc, on which all points have same y value. 

Now we find the x coordinate. For this, we put the above value for 'y' , and solve for 'x' in the equation (1). Solving we get x = 1.491 m.

From the above graph, we can make the following inferences:

• The concrete of the beam section can take up the shear beyond point E. 

• So theoretically there is no need to provide stirrups beyond E. 

• But the code does not allow us to do so. According to the code, we must provide stirrups even beyond E, upto point G.

The amount of stirrups required from E to G can be obtained from:

Assuming 2 legged 8mm dia. stirrups, A_sv = 100.53 mm²

Substituting in the above expression, we get S_v should be less than or equal to 394.53 mm - - - (2)

The code does not require us to provide stirrups in the region from G to B. But it is a good design practice to provide stirrups there also.

Now we look at the second restraint that we must calculate:
The maximum spacing of stirrups allowed by the code.

Spacing should not be more than the smallest of the following:

(a) 0.75d =412.5 mm

(b) 300 mm

Comparing with (2), we can say that, the spacing at any point of the beam should not be less than 300 mm

Now we have to calculate the spacing for the region from A to E. We will do this in the next section.

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