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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