Sunday, November 22, 2015

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 Vuc from the concrete. So the sum of these two contributions will give the total shear capacity of the beam section.

Both Vuc and Vus are the values at the 'ultimate state'.
 Vuc 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.
• Vus is the shear resistance offered by the shear reinforcements at this 'point of impending failure'.

This is shown in the fig. below:

The applied shear force on a reinforced concrete beam is resisted by the shear stress in concrete and the force in stirrups

If we are given a beam section and it's properties, and also the details of the shear reinforcements, we can calculate both Vuc and Vus. Their sum will be the total 'ultimate shear resistance' that the beam section can offer. We denote this ultimate shear resistance as VuR. Thus we can write:
Eq.13.53:
VuR =Vuc +Vus

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 Vu applied at the section is less than or equal to the ultimate shear resistance VuR of the section. That is, the following relation should be satisfied:
13.54:
Vu   VuR
Substituting for VuR from Eq.13.53, we get
13.55:
Vu  Vuc +Vus

While designing, we can first assume the left side to be equal to the right side. That is:
Eq.13.56:

Vu = Vuc + Vus

Here
• Vu is a known value given to us.
• Vuc can be calculated by using Eq.13.34.Vuc = τcbd
• Vus 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):
         Vus = 0.87fy Asv sinα  
  ♦ Eq.13.52 (for bent-up bars in series):
From the above equations for Vus, we can see that, in Eq.13.56, there will be only one unknown, and that is Sv[Asv 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:
Vu = Vuc + Vus(stirrups) + Vus(bent-up)

When single or parallel bent-ups are used, Eq.13.49 above can be used for Vus(bent-up). But in Eq.13.49, Sv is not coming into the picture at all. Asv of the bent-up bars will already be fixed while designing the beam for flexure. So in this case also Sv, 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, Sv 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 Vus to the left side because it is the term which contains the unknown. Thus we get:
Vus = Vu - Vuc  Vus = Vu -τcbd

Once we do the above subtraction and obtain VusSv can be calculated. We will have to round off the spacing Sv to multiples of 5 or 10 mm. When this rounding off is done, the final Sv provided should be less than the Sv calculated. This will give more area of steel to resist the shear. Because of providing more area than required, the Vus  part of Eq.13.56 or 13.57 will be more than what is required to resist Vu. Thus 13.56 will become an inequality as shown below:
13.59:
Vu <  Vuc + Vus
But the right side of 13.59 is the ultimate shear resistance VuR provided. So we can write:
13.60: Vu < VuR
This satisfies our basic requirement given in 13.54.

We will now do a problem just to demonstrate the calculation of Sv. [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, Ast =603.186 mm2fy =415 N/mm2fck =20 N/mm2Vu =75 kN

First we calculate Vus using Eq.13.58:
Vus = Vu - τcbd

So we must calculate τc. (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 τc =0.5298 N/mm2.

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

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

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

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

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

• Vuc is obtained using Eq.13.34: Vuc = τcbd. We get Vuc = 48741.945 N.

Adding the above two we get VuR =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 Sv. We will discuss about the code provisions and various checks in the next sections.


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