Chapter 13 (cont..15) - Critical sections for Shear design

In the previous section we saw the critical sections for shear design in simply supported beams and the beams in framed structures. In this section we will see some more cases.

Case 4

In this case we consider the situation in which there is no compressive force acting on the support portion of the beam. A suspended beam is an example of this situation. It is shown in fig 13.70 below:

Fig.13.70

Suspended beam

• Here the shear strength of the beam at the support region is not enhanced. 

• This is because of the absence of any reaction force from a bottom support. 

So the critical section should be taken at the face of the support. This is shown in fig.13.71 below:

Fig.13.71

Critical section for suspended beam

Case 5

In this case also, we consider a situation in which there is no compressive force acting on the support portion of the beam. This time we consider an 'inverted T-beam'. It is shown in fig.13.72 below:

Fig.13.72

Inverted T-beam

• The load will be acting at the top surface of the slab

• But this top surface of the slab is at a lower level than the top surface of the beam. 

• Because of this, the compressive force cannot be achieved at the support portion of the beam. 

So the critical section should be taken at the face of the support as shown in the fig.13.73 below:

Fig.13.73

Critical section for inverted T-beam

Case 6

In this case we consider a beam which is supported on another beam. The supporting beam is called Primary beam and the supported beam is called the secondary beam. This is shown in the fig.13.74 below:

Fig.13.74

Secondary beam supported on a primary beam

In this case, the critical section for the secondary beam is at the face of the support as shown in the next fig.13.75

Fig.13.75

Critical section for the secondary beam

In this case, special care should be taken in the shear design of the primary beam also. It is recommended that extra full depth stirrups, also known as ‘hanger stirrups’ should be provided in the primary beam as shown in the fig.13.76 below:

Fig.13.76

Details of hanger stirrups

When we analyse the secondary beam, we will get the Shear force at the support. This is force acts as a concentrated load on the primary beam. It is multiplied by the suitable load factors to get the factored load. If we denote this factored load as R, then the area A_sv of the extra hanger stirrups can be obtained by the following equation:

Eq.13.65:

The above equation follows from the basic equation: Force (R) = Stress (0.87f_y) x Area (A_sv)

If the force R is large, bent-up bars should also be used in addition to the above extra stirrups. This is shown in the fig.13.77 below:

Fig.13.77

Additional bent up bars when R is large

This completes our discussion on ‘Critical sections’ for shear design. From this discussion it is clear that we must examine each structural element carefully to determine the position of the critical section. We can now proceed to do the complete shear design of a beam.

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Chapter 13 (cont..14) - Critical sections for Shear design

In the previous section we saw the code provisions regarding spacing of stirrups, and the maximum allowable ultimate shear resistance. In this section we will discuss about the 'critical sections for shear design'.

Critical sections for shear design

We know that the shear force applied on the structure is not uniform. It is usually greatest near the supports, and decreases as we move away from the supports. If a beam carries a number of concentrated loads, the shear force is high at the region in between the support and the first concentrated load. We must have a thorough knowledge about selecting the section at which the investigation about the shear force will be made. We must select those sections at which the shear force is the greatest and/or cross sectional area is minimum. Cl.22.6.2 of the code gives us some guidelines for selecting the critical sections in various types of structures. We will now discuss each one of them in detail.

Case 1
Consider a slab supported on two masonry walls and a cross beam as shown in the fig. 13.62 below:


Fig.13.62
Slab supported on masonry walls and a cross beam

We can note the following points:
• The slab will be carrying loads on the entire area of it’s top surface. That is, loads will be present not only on the clear span area, but also on the area directly above the supporting masonry walls.
• Because of this, the cross beam will also be carrying loads not only on it’s the clear span length, but also on the portion directly above the supporting masonry walls. This is shown in the next fig.13.63:


Fig.13.63
Loads present directly above the support

• So at the portion above the supporting masonry walls, the beam will experience a force from the top due to the applied loads, and a force from the bottom due to the reaction from the supports. 
• Thus these regions of the beam at the supports will be experiencing some compression. 
• Because of this compressive force, the shear strength of the region will be enhanced. 

So the diagonal crack in such cases will be formed at some distance away from the face of the support. These cracks should have developed near the face of the support, as the shear force is maximum there. But the presence of the compressive forces, alters the situation. The code tells us to take a section located at a distance of d (effective depth of the beam) from the face of the support as the critical section. This is shown in the fig.13.64 below:

Fig.13.64
Critical section

This case is given in the cl.22.6.2.1 of the code. We know that the portion between the face of the support and this critical section will be subjected to a greater shear force than the shear force at the critical section. But the code allows us to use the lesser shear value at the critical section in this portion also. This is shown in the next fig.13.65 below:

Fig.13.65
Portion between the face of the support and critical section

Case 2
Here we consider a framed structure. We saw the 3D view of a framed structure in Chapter 9 – Analysis of singly reinforced Flanged sections. The 3D view is given in the slide 2 of the presentation given at the beginning of that chapter. The framed structure consists of columns, beams and slabs. For analysis of a framed structure, we consider 2D frames consisting of columns and beams. Such a 2D frame is shown in the slides 7 and 8 of the presentation. A portion of such a frame is shown in the fig.13.66 below:

Fig.13.66
View of a column and beam

Just as in the previous case 1, 
• The portion of the beam inside the column, will be experiencing a compressive force due to the downward load from the top column, and the reaction from the bottom column. 
• So the shear strength of the beam in the support portion will be enhanced. 

Thus the code allows us to take the critical section at a distance d from the face of the support. This is shown in the fig.13.67 below:

Fig.13.67
Critical section in the case of framed structure

Also as in the previous case, the portion within the distance d need to be designed only for the shear force at the critical section. This is shown in the fig.13.68 below:

Fig.13.68
Portion between the face of the support and critical section

Case 3
This is a special case within case 2. In this, a heavy concentrated load is applied within a distance of two times d from the face of the support. In the presence of such a heavy concentrated load, the face of the support itself should be taken as the critical section as shown in fig.13.69 below:

Fig.13.69
Critical section when a heavy concentrated load is present within a distance of 2d

In the next section we will continue our discussion on critical sections.

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Chapter 13 (cont..13) - Code provisions regarding Shear design

Recommendations given in the code regarding shear reinforcements:

In the previous section we saw the method of calculations of the spacing S_v of stirrups. In this section we will discuss the code provisions and checks.

According to the code, we must not use a grade of steel higher than Fe415 for making the stirrups or bent-up bars, or any other types of reinforcements intended to resist shear. This is because, the high strength steels may become brittle at the sharp bends. Another reason for giving such a restriction is that, When high strength steel is used, the shear resisting capacity of the beam may increase up to such a high level that the beam will be able to take very high loads with out any risk of developing tension cracks. But at these higher load levels, the other principal stress, namely 'compressive stress' on the particles may reach a level which is beyond the compressive strength of concrete. At this level, the concrete will crush, leading to a brittle failure of the beam.

Another recommendation given by the code is regarding the spacing of stirrups. According to cl.26.5.1.5,

For vertical stirrups, the spacing S_v should not be more than the smallest of the following:

a) 0.75d

b) 300 mm

For inclined stirrups with inclination α = 45^o, the spacing S_v should not be more than the smallest of the following:

a) d 

b) 300 mm

Such a minimum spacing is specified to ensure that every diagonal tension crack that may occur will be intercepted by at least one stirrup.

The next recommendation is regarding the minimum quantity of stirrups to be provided. We have seen that the concrete will give some resistance force against applied shear. Situations can arise wherein, the applied shear force is less than even the 'shear resistance offered by the concrete'. In such a situation, we will naturally think that no shear reinforcement is required. But the code does not allow us to avoid the shear reinforcements altogether in this way. It can be avoided only if a condition is met. This condition is given in cl.26.5.1.6 of the code. In the code, the condition is written in terms of stresses. We will write it here in terms of forces. So, according to this clause, we can avoid the shear reinforcement if the shear force V_u applied at the section is less than half of the resistance force offered to us by the concrete. That is, if V_u < 0.5V_uc, shear reinforcement can be avoided. We can write the converse of this. That is:

13.62 If V_u ≥ 0.5V_uc , it is compulsory to provide stirrups.

This condition can be represented graphically as shown below in the following fig.13.60:

Fig.13.60

Graphical representation of condition 13.62

V_u is less than V_uc but greater than half of V_uc 

In the above graph, the applied shear force V_u is less than V_uc. At the same time, it is greater than half of V_uc. In such a situation, it is compulsory to provide stirrups.

In this condition, a shear force V_u is applied at the section, and this force is of such a low magnitude that, the concrete is able to take up all of this force. So what is the quantity of force, that we want to resist with the stirrups? The answer is that the stirrups in this situation are provided not to take up any load, but to serve the following purposes:

• It prevents the sudden formation of an inclined crack. If at all an inclined crack do form, the minimum shear reinforcement provided will prevent the propagation of that crack.

• If a beam with no shear reinforcement is over loaded to such an extent that shear failure occurs, then, that failure will be an abrupt one. If in such a beam, the minimum shear reinforcement specified by the code is provided, then the stirrups will take up the load, and if load is increased even further, the steel of the stirrups will undergo elongation, thus giving warning about the impending failure.

• The stirrups also help to confine the main bars of the beam together, keeping them in their proper positions. This will enable the main bars to perform their duties in a better way. This also improves the dowel action, thus keeping the portions of concrete on either side of the crack together.

When V_u < V_uc AND V_u ≥ 0.5V_uc, we do not have any force for which the stirrups have to be designed. [∵ V_us is obtained by subtracting V_uc from V_u.] In such a situation, we use cl.26.5.1.6 of the code. The design mainly involves the calculation of spacing S_v of the stirrups. We can calculate S_v using the expression given in this clause:

13.63

From 13.62, it follows that if, V_u < 0.5V_uc, We don't have to consider even the above 13.63. That is., there is no need for stirrups at all. Such regions may be present in a beam at regions away from the supports, where shear force is low. But it is a good design practice to provide stirrups in these regions also, for the same reasons mentioned above.

Limiting value of V_uR:

We know that when a beam section fails by shear at the ultimate state, the steel of the stirrups should have yielded. Such an yielding will give enough warning that the beam is going to fail. In other words, the failure should be ductile in nature. At the beginning of this section, we have seen the reason for not giving high strength bars for making the shear reinforcements. That is, when high strength bars are used, the beam may become strong in diagonal tension and weak in diagonal compression. This very same undesirable condition can arise in another situation also: If we provide an excessive amount of shear reinforcements either in the form of stirrups or bent-up bars, the beam section will become strong in diagonal tension, and weak in diagonal compression. So the strength V_us given to the beam by us, should be kept within a certain limit.

The code gives the method to be implemented so that V_us will be within the safe limit. But it is an indirect method. In this method, the code specifies the upper limit for V_uR, the ultimate shear resistance of the section. We know that V_uR (Eq.13.53) is the total ultimate shear resistance which is the sum of the 'contribution of concrete' and 'contribution of steel reinforcements'. Let us see how this indirect method work, in achieving our objective.

We have

Eq.13.53: V_uR = V_uc + V_us. Where

Eq.13.34: V_uc = τ_cbd

From Eq.13.34, we can see that V_uc is a constant for a given beam section. It can not be increased or decreased. So from 13.53, we can say that, if V_uR has an upper limit, V_us will also have a corresponding upper limit. Thus, when we follow the code method, we keep V_uR within a certain limit. When we do this, we are indirectly keeping V_us also within a desired limit. The only thing is that we are not calculating this upper limit of V_us.

But we do have to calculate the upper limit of V_uR. This is denoted as V_uR,lim and is obtained by the following:

Eq.13.64:

V_uR,lim = τ_c,max bd  

The value of τ_c,max is given in the table 20 of the code.

So we know how to calculate V_uR,lim of a given section. We will see how to implement this in a design procedure:

At the time of taking up the shear design of a beam section, b and d will be known. So we calculate V_uR,lim using table 20 and Eq.13.64. Then we compare this with the applied shear V_u. What if V_u is of such a high magnitude that, V_u > V_uR,lim ?

Then the basic requirement that V_u  ≤ V_uR  will never be satisfied because the maximum value that V_uR can take is V_uR,lim. This can be represented graphically as shown in the following fig.13.61:

Fig.13.61

Maximum value that V_uR can take is V_uR,lim

In the above graph, V_u has a higher magnitude. So we must increase V_uR so that the condition V_u ≤ V_uR will be satisfied. But V_uR cannot be increased beyond V_uR,lim.

In such a situation, the only solution is to revise the design. This can be done either by using any one or both of the methods given below:

• Increase the dimensions b and d of the beam, thus increasing V_uR,lim (Eq.13.64) or

• Use a higher grade of concrete, thus increasing τ_c,max, which will increase V_uR,lim (Eq.13.64)

So we have seen the various code recommendations that we have to consider while doing a shear design. In the next section, we will discuss about the ‘critical sections’ in a structure at which we have to check the shear.

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Chapter 13 (cont..12) - Calculation of spacing of stirrups

In the previous section we completed the discussion on 'bent-up bars'. With that, we know how any of the steel reinforcements (vertical stirrups, inclined stirrups, or bent-up bars) will contribute towards the shear capacity of a beam section. We also know the contribution V_uc from the concrete. So the sum of these two contributions will give the total shear capacity of the beam section.

Both V_uc and V_us are the values at the 'ultimate state'.
• V_uc is the shear resistance offered by the concrete when the beam section is at the 'point of impending failure' due to it's shear capacity being exceeded.
• V_us is the shear resistance offered by the shear reinforcements at this 'point of impending failure'.

This is shown in the fig. below:

If we are given a beam section and it's properties, and also the details of the shear reinforcements, we can calculate both V_uc and V_us. Their sum will be the total 'ultimate shear resistance' that the beam section can offer. We denote this ultimate shear resistance as V_uR. Thus we can write:

Eq.13.53:

V_uR =V_uc +V_us

This Eq.13.53 can be used for analyzing the shear capacity of a beam. That is, if we are given a beam section, and asked to calculate the ultimate shear resistance, we can use this Eq.

If we look at the Eq. from another point of view, we can use it for design purpose also. This can be explained as follows: When we design a new beam section for shear, we have to make sure that the factored shear force V_u applied at the section is less than or equal to the ultimate shear resistance V_uR of the section. That is, the following relation should be satisfied:

13.54:

V_u ≤  V_uR

Substituting for V_uR from Eq.13.53, we get

13.55:

V_u ≤ V_uc +V_us

While designing, we can first assume the left side to be equal to the right side. That is:

Eq.13.56:

V_u = V_uc + V_us

Here

• V_u is a known value given to us.

• V_uc can be calculated by using Eq.13.34.: V_uc = τ_cbd

• V_us is calculated based on the type of shear reinforcement that we intend to use for the beam. Thus we have:

  ♦ Eq.13.39 (for vertical stirrups):

  ♦ Eq.13.47 (for inclined stirrups):

  ♦ Eq.13.49 (for  single or parallel bent-up bars):

         V_us = 0.87f_y A_sv sinα  

  ♦ Eq.13.52 (for bent-up bars in series):

From the above equations for V_us, we can see that, in Eq.13.56, there will be only one unknown, and that is S_v. [A_sv is not an unknown because the diameter of stirrups and no. of legs will be already assumed.]

We have seen that when bent-up bars are used to resist shear, stirrups should also be provided, and these stirrups should take up equal to or greater than 50% of the factored shear force. Thus, when bent-up bars are provided, Eq.13.56 can be modified as:

Eq.13.57:

V_u = V_uc + V_us(stirrups) + V_us(bent-up)

When single or parallel bent-ups are used, Eq.13.49 above can be used for V_us(bent-up). But in Eq.13.49, S_v is not coming into the picture at all. A_sv of the bent-up bars will already be fixed while designing the beam for flexure. So in this case also S_v, which is coming in the second term of Eq.13.57 is the only unknown.

When a series of bent-up bars are provided, Eq.13.52 is to be used in Eq.13.57. In this case, S_v can be assumed to be the same for both stirrups and the bent-up bars, and so it becomes the only unknown.

So in Eq.13.56, we can bring V_us to the left side because it is the term which contains the unknown. Thus we get:

Eq.13.58:

V_us = V_u - V_uc ⇒ V_us = V_u -τ_cbd

Once we do the above subtraction and obtain V_us, S_v can be calculated. We will have to round off the spacing S_v to multiples of 5 or 10 mm. When this rounding off is done, the final S_v provided should be less than the S_v calculated. This will give more area of steel to resist the shear. Because of providing more area than required, the V_us  part of Eq.13.56 or 13.57 will be more than what is required to resist V_u. Thus 13.56 will become an inequality as shown below:

13.59:

V_u <  V_uc + V_us

But the right side of 13.59 is the ultimate shear resistance V_uR provided. So we can write:

13.60: V_u < V_uR

This satisfies our basic requirement given in 13.54.

We will now do a problem just to demonstrate the calculation of S_v. [It may be noted that the result obtained in this problem is not a valid result, as it does not do the various checks and code provisions. Here we are interested in the 'steps involved in the calculation' only. we will discuss about the checks and code provisions the next sections.]

Problem:

Find the spacing of stirrups in a beam section, to resist a factored shear force of 75 kN. The beam has a width of 230 mm and an effective depth of 400 mm. The tensile steel consists of 3 - #16. Assume Fe 415 steel and M20 concrete.

Solution:

We will write the given data:

b =230 mm, d =400 mm, A_st =603.186 mm², f_y =415 N/mm², f_ck =20 N/mm², V_u =75 kN

First we calculate V_us using Eq.13.58:

V_us = V_u - τ_cbd

So we must calculate τ_c. (Details here). For this we look at table 19 of the code.

pt = 100Ast/bd =0.6556

0.5000  → 0.4800

0.6556  → 0.5298
0.7500  → 0.5600. So we get τ_c =0.5298 N/mm².

Substituting all the known values in Eq.13.58 we get V_us =26258.055 N

V_us for vertical stirrups is given by Eq.13.39. Assuming 2-legged stirrups of  8 mm dia., we get A_sv =100.53 mm². 

Substituting all the known values in Eq.13.39 we get, S_v = 552.92 mm. Rounding this off to multiples of 5 or 10 mm, we get S_v =550 mm.

This completes the design part. We can now do an analysis:

• The actual V_us obtained at the section can be obtained by using Eq.13.39 again. But this time, V_us is the 'unknown'. We get V_us =26397.350 N

• V_uc is obtained using Eq.13.34: V_uc = τ_cbd. We get V_uc = 48741.945 N.

Adding the above two we get V_uR =75139.295 N =75.13 KN. This is greater than the applied factored shear force. So the section is safe.

But it should be noted that the spacing of 550 mm exceeds the maximum spacing allowed by the code. Also various checks have to be carried out in the design process. The above problem is just a demonstration of the method of calculation of S_v. We will discuss about the code provisions and various checks in the next sections.

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