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:

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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.

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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.

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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..9) - Shear resistance offered by concrete

In the previous section we completed the discussion on 'the calculation of the design shear stress at a beam section'. We do not have to provide reinforcements to resist this exact 'design shear stress'. This is because, there is another 'force' that will act in a direction opposite to the applied external shear force. In this section, we will see the details about this new 'force'.

Design shear strength of concrete in beams 

The new 'force' is none other than the one obtained from the 'interlocking between the aggregates' of the concrete. It is denoted as V_a, and is shown in fig.13.43 below:

Fig.13.43

Force V_a due to interlocking of aggregates

Let us assume that a beam is separated into two parts. Due to the applied external forces, the two parts will slide past each other. But such a sliding will be resisted by the interlocking of the aggregates at the interface of the two parts. So there is a 'resisting force' which opposes the sliding. This resisting force is V_a. The vertical component of V_a, shown in the fig. above as V_ay will contribute towards resisting the external shear force at the section. V_a depends upon the grade of concrete (M20, M25 etc.,), because, stronger the concrete, greater will be the interlocking force.

V_a also depends up on the 'extent of cracking'. Because the aggregates interlocking will not work well along cracks. Now, we have seen the cracks due to diagonal tension, and also the cracks due to bending. The former forms mainly near the supports, and the latter forms near the midspan at the bottom most tension fibres, and propagates upwards. When the loading is increased, this latter type will be seen not only at the midspan region, but also towards the sides, as shown in fig.13.44 below:

Fig.13.44

Flexural cracks

But we know that the 'tensile steel' provided at the bottom of the beam is very effective in controlling this flexural cracks. Thus we can see that the quantity of tensile steel plays a role in controlling cracks, and the cracks play a role in the magnitude of V_a. That is:
• Aggregate interlocking depends on cracks. Greater the cracks, lesser the interlocking.
• Cracks in turn, depends on 'quantity of tensile steel'. Greater the steel, lesser the cracks.
• So the inter locking depends on 'quantity of tensile steel'. We can say that quantity of tensile steel plays a role in determining the magnitude of V_a.

There is yet another force that will resist the sliding of two parts of a beam. It is the dowel force V_d in the tensile steel. This is shown in fig.13.45 below:

Fig.13.45

Dowel force in tensile steel

The tensile steel acts as a 'dowel' or 'pin' connecting the concrete blocks on either sides of the crack. Due to this action, the blocks are prevented from sliding past each other. While performing this task , the steel will experience a vertical force V_d as shown in the fig., and this force is known as the 'dowel force'. Naturally, this dowel force will depend upon the quantity of tensile steel provided.

Thus the new 'force' that we mentioned above, depends upon the grade of concrete, and quantity of tensile steel. This force is commonly referred to as the 'Design shear strength of concrete in beams', and denoted by τ_c. The code uses the following empirical formula to calculate τ_c:

Eq.13.32:

In the above formula, β should be given the largest value among (i) and (ii) given below:

p_t is the ‘percentage of tensile steel’ given by

Eq.13.33:

Eq.13.32 is given in 39.2.1 of SP 24. But we do not have to use this formula directly. The values are given in table 19 of the code, and intermediate values can be obtained by linear interpolation.

An important point is to be noted while using the values in table 19. We know that the quantity of tensile steel is being taken into account, in determining the values given in the table. So if at a region, 'bar cutoff' has been applied, the above values can be used only if detailing requirements are satisfied. We will learn more details about it when we discuss about 'curtailment of bars' in a later chapter.

So, knowing the values of the grade of concrete, and the percentage of tensile steel, we can calculate the stress τ_c (from table 19) which the concrete will be able to take. Multiplying this stress by the area, we will get the force that concrete will offer in resisting the shear. Thus, this force, denoted by V_uc , is obtained using the following equation:

Eq.13.34: V_uc  =τ_cbd

We must note that τ_c given by the code is the 'design' shear stress. That means the required 'factor of safety' has been already applied while using Eq.13.32, and it is the shear stress offered by the concrete at the point of impending failure. The code (clause 40) refers to this ultimate state as ‘ LIMIT STATE OF COLLAPSE : SHEAR’. (ie., the beam is at the point of impending failure because of it’s shear resistance being exceeded by the applied external loads).

[In V_uc, 'u' denotes the ultimate state, and 'c' denotes concrete.]

So we can calculate the force which we will receive from the beam section. We can subtract it from the external design shear. We need to provide stirrups or other reinforcement for that force only, which remains after subtraction.

Design shear resistance offered by stirrups

Now we have to calculate the resistance force which the stirrups will offer. We denote it as V_us. In the following fig.13.46, we can see that some of the stirrups intercept the crack.

Fig.13.46

Stirrups intercepting the crack

So these stirrups will be under tension, as they have to keep the concrete on either sides of the crack together. At the point of impending failure, ie., at the ultimate state, the stress in each of these bars will be equal to 0.87f_y . So if the area of cross section of one stirrup is equal to A_sv, the force in each stirrup is given by:

Eq.13.35: Force in one stirrup =  0.87f_y A_sv (∵ force = stress x Area)

Note that here A_sv is the area of all the legs of a single stirrup. For example, A_sv of a 2-legged stirrup of 8mm dia. Will be given by:

Eq.13.35 gives the force in one stirrup. So if the number of stirrups intercepting the crack is n, then the total force is given by:

Eq.13.36: Total force = n x 0.87f_y A_sv

So we have to find out the value of n. This can be done based on the following fig.13.47:

Fig.13.47

No. of stirrups intercepting the crack

In the fig., p is the horizontal projection of the length of the crack, and S_v is the spacing of stirrups. Then the number of bars within a length of p is given by:

Eq.13.37: n = p /S_v   

The code uses an idealization of the above fig.13.47, in which

• The crack is assumed to be straight

• The inclination of the crack with the horizontal is 45^o

• The crack extends over the full depth of the beam

This is shown in the next fig.13.48 below:

Fig.13.48

Idealized conditions

In the fig., as the inclination of the crack is equal to 45^o, horizontal projection will be same as the vertical projection which is equal to d (as in a 45^o triangle). So Eq.13.37 becomes

Eq.13.38: n = p /S_v. So we can replace n in the Eq.13.36 and write it in the following form:

Eq.13.39:

This is the same equation given in cl.40.4 of the code.

Vertical stirrups are the ones which are most commonly used in practice. When we use a smaller diameter bar for making stirrups, the spacing required will be small. Similarly, when a larger diameter is used, the spacing required will also be higher. It is better to go for the first option. This is because closer spaced stirrups will give better crack control. The diameter of 8, 10 or 12 mm is generally used.

In the next section, we will discuss about 'inclined stirrups'.

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