In the previous section we were discussing the free body diagram of small segment pq in a simply supported beam. We will now continue the discussion and learn more details about dT.
dT is the difference in the tensions at the two ends of the small segment of the bar in pq. [If there are N bars of the same diameter Φ, then dT is the total of the difference in tension in all the N bars.] This small bar segment is shown in the fig.15.32 below:
dT is the difference in the tensions at the two ends of the small segment of the bar in pq. [If there are N bars of the same diameter Φ, then dT is the total of the difference in tension in all the N bars.] This small bar segment is shown in the fig.15.32 below:
Fig.15.32
Free body diagram of the bar segment PQ
From the fig. we can see that the tension at q is greater. So pq must move towards the right. But another force is coming into play here. This force is the one due to the bond stress τb between steel and concrete, and it helps to keep pq in equilibrium. That is., the difference in tensions is neutralised by the bond force. We will try to calculate the magnitude of this force:
•The surface area of the bar (having diameter Φ) within the segment pq = лΦdx.
• This area multiplied by the bond stress τb will give the bond force.
• So the bond force is equal to лΦdxτb
• If there are N bars of the same diameter, then the total surface area = NлΦdx
• and bond force is equal to NлΦdxτb
• This force is the one which neutralizes dT. So we can write:
dT= NлΦdxτb .
• In the previous section we have derived dM =dT z. So substituting for dT we get we get dM = NлΦdxτb z
• This can be rearranged as dM⁄dx = NлΦτb z
But from the basic lessons on strength of materials, we know that dM⁄dx is the shear force V at a section XX through the small segment pq of length dx. This is shown in the fig.15.33 below. [pq has a length dx. But this dx is infinitesimal. So the load effects at section XX is same as the load effects at the position of pq]
Fig.15.33
Section XX passing through pq
So we can write:
V = NлΦτb z which is same as
τb = V ⁄NлΦ z - - - (4)
So we have derived the expression for the bond stress acting on the segment pq.
We will take a small deviation in our discussion here. We are going to focus our discussion on the development length requirements of segment pq. We have earlier analysed the development length requirements of a top bar in a cantilever beam. We did this in chapter 14 on bond stress, and we derived Eq.14.6. The very same procedure can be adopted here also. But we will write the required steps again:
We will take a small deviation in our discussion here. We are going to focus our discussion on the development length requirements of segment pq. We have earlier analysed the development length requirements of a top bar in a cantilever beam. We did this in chapter 14 on bond stress, and we derived Eq.14.6. The very same procedure can be adopted here also. But we will write the required steps again:
Consider fig.15.29 .
• Let the tensile stress in the bar at section XX be σs.
• Then, the total tensile force in all the 'N' number of bars at section XX will be equal to (NлΦ2⁄4 ) σs - - - (5)
• This force tries to pull out the bars from the support. The pulling out force is resisted by the bond force developed along the length lx. Where lx is the length of bar to the left of the section.
• The surface area on which the bond stress acts = NлΦlx - - -(6)
• So, if τb is the bond stress between steel and concrete, then the force resisting the pull out = ( NлΦlx) τb - - - (7)
• So, if τb is the bond stress between steel and concrete, then the force resisting the pull out = ( NлΦlx) τb - - - (7)
• Equating (5) and (7) we get
- - - (8)
• Now we go back to the point where we took the deviation and get the expression for τb. From (4) we get the value of τb. Substituting this in (8), we get
- - - (9)
So we have obtained the expression for the length which will prevent the bars from being pulled out. Let us now see the changes that have to be made to the above expression at the state of impending failure, that is., at the ultimate state. At the ultimate state, the stress σs in steel will be equal to 0.87fy. So we can write the above expression as:
But Ab multiplied by 0.87fy is equal to the ultimate force in the tensile steel (∵ Force =Area x Stress). And this force multiplied by the lever arm z is the 'ultimate moment of resistance of the section' having area of tensile steel Ab. In other words, it is the MuR of the beam at section XX. So we can write :
At ultimate state,
- - - (11)
So, when we calculate MuR ⁄ V , we will get lx, which is the length that prevents the bars from pulling out at the ultimate state. But this length should be sufficient to develop enough grip so that the pulling out can be prevented. As we are considering the pulling out at ultimate state, the length that prevents the pull out should be the unique value that we discussed in chapter 14. So the length obtained from (11) should be greater than or equal to the unique value. If it is less than that, enough grip cannot be developed, and a bond failure will occur causing a pull out.
The next point that we must note is that, while evaluating (11) to get the available length, we must get the ‘least amount of length’ that can possibly occur. [This is because, even the least possible value of lx should be greater than or equal to the unique value of Ld]. If lx is to be as small as possible, V in the denominator in (11) should be as large as possible. We know that the largest possible value of V is the factored shear at support. Let us denote it as Vu.
So it is MuR ⁄ Vu that has to be determined. And, even this least possible value should be greater than or equal to Ld. This can be written as
Ld (unique value) ≤ MuR ⁄ Vu - - - (15)
If we do a dimensional analysis of (15), the two sides will tally as shown below:
If we do a dimensional analysis of (15), the two sides will tally as shown below:
We must note that the expression (for calculating the least available length) has MuR in the numerator. This become significant when curtailment is done near support. If curtailment is done near support, MuR at that section will be low. As MuR is in the numerator, the available length will also become low. So we must take this aspect into account while doing curtailment.
Thus we obtained the least length that is available. We have obtained this length by considering the BM diagram which extends from the point of zero moment in one support to the point of zero moment in the other support. So this length will be a part of the bars which falls within these points of zero moments. But if some extension of the bars is available beyond these points of zero moments, then that length of extension will also contribute towards preventing the pull out. In the next section, we will discuss about this contribution.
Copyright©2016 limitstatelessons.blogspot.com - All Rights Reserved
No comments:
Post a Comment