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)

---

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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Chapter 16.13 - Stairs cantilevering on the sides of Stringer beam

In the previous section we saw the details of transverse stairs which cantilever from the side of a stringer beam.  The following solved example illustrates the design of  such a stair.

Solved example 16.4

The reinforcement details according to the above solved example is shown in the figs.16.81, 82 and 83 given below. The cantilever waist slab has to be properly anchored into the beam. For this, the stirrups of the beam are extended into the slab as shown below in fig 16.81:

Fig.16.81
Extending the stirrups into the slab

When such an arrangement is given, we must take care of the diameter of bars and the spacing requirements. The spacing of the main bars of the slab will be the same as the spacing of the stirrups. So the stirrups of the beam should be having a uniform spacing throughout it's length. And for our case, this spacing should not be greater than 250mm c/c, and the diameter should not be less than 10mm.

If the spacing of 10mm dia. stirrups required for the beam is less than 250mm c/c, then the spacing of bars in the slab will also become less than 250mm c/c. In that case check should be done to ensure that p_t is less than p_t,lim. Other wise, the slab will become over reinforced.

The elevation and sectional views are given below:

Fig.16.82
Sectional elevation

Fig.16.83
Section XX

Now we will discuss some more features of the above figs: In cantilever members, the main tensile steel is given at the top. So in our case, the main bars are given as the top most layer. The distributor bars are given as the second layer from the top. This arrangement will give maximum possible effective depth ‘d’ for the section. So while bending the bars for the stirrups of the beam at the site, care should be taken to see that the extension given to the stirrup will be at the exact required level in the slab. A bar bending schedule should be prepared to give the exact measurements.

As the slab has a thickness of 110mm, a bottom layer consisting of #8 bars at 250mm c/c should be given. This is indicated by the green bars in section XX. In this layer, alternate bars should be given two 90 degree bends at the end. This is for tying them with the top layer. This bottom layer is not shown in the sectional elevation in fig.16.82.

Transverse stairs cantilevering from both sides of a stringer beam:

The following fig. 16.84 shows the part view of a stair

Fig.16.84
Part view of a stair
 

The waist slab is cantilevering on both sides of the stringer beam. The load calculation procedure is same as that of the other types of transverse stairs that we saw before. So we can use the same Eq.16.25 for the calculation of the loads. From this eq., we will get w₁ , the load per 1m length, on a 1m wide strip. If the cantilevering length on both the sides are equal, we need to do the design of only one side as shown in the line diagram below:

Fig.16.85
Line diagram for stairs cantilevering on both sides of a stringer beam

The effective span l of the cantilever (cl.22.2.c) is the length of the cantilever up to the face of the support plus half the effective depth. For initial proportioning, we can assume the thickness of waist slab to be clear length of cantilever/5 . However, this should be finalized only after the complete design and various checks.

In the next section, we will see a solved example of the above type of stairs.

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