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..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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Chapter 13 (cont..17) - Shear design solved example

In the previous section we saw that the spacing of stirrups should not be greater than 226.85 mm at any part of the beam that we are now designing. In this section, we will calculate the spacing for the region from A to E.

For this we use the basic Eq.13.58

V_us =V_u -τ_cbd

V_u is the shear force at the critical section =310.5 kN

τ_c = 0.786 N/mm²

So we get V_us = 137668 N

But from Eq.13.39

Putting A_sv = 100.53 mm², (∵ 2-legged stirrups of 8 mm dia.) we get S_v = 145.008 mm. - - - (3)

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Let us check this by using table 62 of SP 16

• Vus/d = 250.305 N/mm = 2.503 kN/cm

2.593  → 14

2.503  → 14.5199

2.420  → 15.  So we get S_v =14.5199 cm =145.199 mm.

This is comparable with the value in (3).

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Thus at the point of greatest shear, the spacing required is 145.008 mm

Upto this point we have the following two informations:

1. At no part of the beam, the spacing should exceed 226.854 mm

2. The spacing at critical section should not exceed 145.008 mm - - - (4)

• So spacing starts from 145.008 mm

• Increases to 226.854 mm

• And there after remains constant at 226.854 mm

We will find the point at which the constant spacing of 226.854 mm begins. For this, we have to work in a sort of 'reverse order'. That is., we usually calculate the spacing required for a given value of V_u that occurs at a particular section. But now we are having the 'spacing', and we are going to calculate the section at which this spacing occurs:

We will put S_v =226.854 mm in the equation for V_us:

From this we get V_us =87,999.155 N

But V_us =V_u -τ_cbd  . So we get V_u =260,831.155 N =260.83 kN

Now we find the point at which this much V_u is being applied. For this, we put the above value for 'y' , and solve for 'x' in the equation (1) 

y =405 -135x. We get x = 1.068 m. To the left of this point, spacing required will be less than 226.854 mm. To the right of the point, the spacing required will be greater than 226.854 mm. But a spacing greater than 226.854 mm cannot be allowed for our beam. So we can say that, towards the right of x= 1.068, the spacing remains constant at 226.854 mm. This is the third information that can be added to the above two in (4)

1. At no part of the beam, the spacing should exceed 226.854 mm

2. The spacing at critical section should not exceed 145.008 mm

3. The spacing remains constant at 226.854 mm after 1.068 m from the support - - - (5)

With the above three information, we can design a satisfactory distribution of spacing along the length of the beam. For this, first we calculate the spacing at regular intervals of 0.2 m along the length of the beam, up to the point of constant spacing. So we have to calculate S_v at 0.2, 0.4, 0.6, 0.8..... m from the left support. 

• The critical section at 0.7 m is an important point. 
• 1.068 m is also an important point because, the S_v remains constant after this point. So we will add these two points also.  

The tabulation is shown below:

We do not need the value of S_v beyond a distance of 1.2 m, as the constant S_v begins at 1.068 m.

All the values in the S_v column are obtained by the same method that we used for calculatiing the S_v at the critical section, as we did in (3)

Now we will make a plot with distance on the X axis and S_v on the Y axis.

This is the plot of the spacing required. From the graph, we can see that when the distance increases, the spacing also increases. This means that we can provide stirrups at 'greater distances apart' at the regions away from supports. The portion beyond 226.883 cannot be considered as part of this plot because spacing greater than it are not acceptable. So it is shown in dotted line.

We will plot another curve on the same graph which will give the spacing that we actually provide. 

• Unlike the above plot, the new plot will have 'horizontal straight line' segments, which will indicate that, spacing provided is constant for various distances along the length of the beam. 

• The new plot should not rise above the above plot at any point. This is because, if it does, it  will mean that the spacing provided is more than the spacing required at that point.

• Even though the new plot will have straight line segments, it should follow the above curve as closely as possible. 

• If it is a simple horizontal line, it would be far too below the above curve at distances away from the support. This would mean that we will be placing the stirrups at smaller spacing unnecessarily at the portions away from the support.

• So the new plot will have some 'steps', so that it will follow the above curve as closely as possible. 

The arrangement should also have ease of placement at the site. A possible plot is shown below:

In the above graph, the plot of 'spacing provided' is given in red colour. The first stirrup is usually placed at a distance of 5 cm from the face of the support. So point 'L' will have an X coordinate of 0.20 m. This is shown in the fig. below:

• In the above graph, the x coordinate of 'L' is 0.20, and the y coordinate is 140. This means that the spacing is 140 mm at the point of the first stirrup. 

• The graph is horizontal from L to M. This means that the spacing of 140 mm, continues upto point 'M'. 

• So the stirrups have the same spacing of 140 mm from L to M. 

• M has an x coordinate of 0.900. So LM = 0.900-0.200 = 0.700 m. 140 x 5 = 700. So there will be five spaces of 140 mm, giving 6 stirrups from L to M.

• At M, there is a step. The plot jumps to N 

Note that some portion of LM is above the 'plot of required spacing'. This means that the spacing provided here is more than the required spacing. This will appear to be 'unsafe'. But we know that the critical section is taken at a distance of d from the face of the support, and the region between the critical section and the face of the support need to be designed for the shear at critical section only. 

Now we look at the segment NO. 'N' and 'O' have y coordinates 160. So the spacing in the region NO is 160 mm. 'O' have x coordinate 1.220. The distance NO = 1.220 -0.900 =0.320 m. 160 x 2 = 320. So there will be two spaces of 160 mm, giving 3 stirrups from N to O. (But the first stirrup among the three is already placed at M, which we discussed in the case of LM )

We have reached 'O' and so, we have crossed the 'section beyond which spacing is constant'. From 'P' onwards, the spacing is given a constant value of 220 mm, upto the center of the beam. 'Q' has an x coordinate of 2.980. So PQ =2.980 -1.220 =1.760 m. 220 x 8 =1760. So there will be 8 spaces of 220 mm, giving 8 stirrups from P to Q.

Q' has an x coordinate of 2.980. So we have covered a distance of 2.980 m from the center of the  left support. This point is just 20 mm (∵ 3.000 -2.980 =0.020) from the center of the beam. We have to provide a mirror image of this arrangement from the right side support also. It will cover a distance of 2.980m on the right side of the beam. So the distance between the middle two stirrups will be 20 x 2 =40 mm.

40 mm is a very small distance which may obstruct the placing of concrete. So when we give the mirror arrangement from the right side support towards the interior, the last stirrup can be avoided. 

A table can be formed showing the spacing in each segment on the left side as follows:

A diagram showing the complete arrangement is given below:

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In the next section we will see another solved example.

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