Showing posts with label moment envelope. Show all posts
Showing posts with label moment envelope. Show all posts

Saturday, January 16, 2016

Chapter 15.2 - Curtailment at points of inflection

In the previous section we saw the method of finding the theoretical point of curtailment at more than one section along the length of the beam. In this section we will see the method for the top bars which resist the hogging moment at an intermediate support of a continuous beam.

In fact the method is the same: Find the MuR obtained from the remaining bars, and put it in the equation to obtain 'x'. So we will discuss only the general scheme adopted for the top bars. This is shown in the fig.15.9 below:

Fig.15.11
Curtailment of top bars


• The first curtailment is given to the bars in the bottom most layer (when counted from the top) among the top bars. When they are curtailed, the steel that remains is Ast2
• The second curtailment is given to the second bottom most layer. After this curtailment, the steel that remains is Ast3

When we follow this pattern, greater quantities of steel are available at the regions nearer to the support. This indeed must be the case because, stresses are greater near the supports as far as the 'top bars for hogging moments' are concerned.

Another aspect to note is the position of 'points of inflection'. These are the points where the bending moment changes sign. The hogging nature of the bending moment at the support continues upto the points of inflection. So we have to provide top steel upto the point of inflection. Similarly, the sagging nature of the bending moment at the midspan region also continues upto the points of inflection. So we have to provide the bottom steel upto these points. But we can follow the principles of curtailment, and progressively reduce the quantity of steel at the sections near the points of inflection.

We will now discuss another aspect of the point of inflection. In continuous beams, the maximum bending moment at any section will depend upon the position of the loads along the spans. The DL will be present at all times, and it's position will not change. But the LL will change position. In order to find the maximum bending moment at a section, we have to place the LL in those positions that give maximum bending moment at that section, and then analyse the whole continuous beam. This topic is dealt with in 'influence lines' in Structural analysis classes. 

We had a basic discussion about it in the second section of chapter 7, 'Analysis of a continuous beam'. Let us consider the solved example of a continuous beam ABCDE given in that chapter. We do not require the bending moments or final results obtained in that example. We are interested only in the pattern of loading that we adopted for that beam. This pattern was shown in fig.7.19 and 7.20

Let us consider the support C, and the span CD. We want 
• the maximum hogging moment at support C and 
• the maximum sagging moment in the span CD. 
For obtaining the maximum possible hogging moment at support C, we must place the LL on the adjacent spans BC and CD, and also on alternate spans. But after filling up BC and CD, there are no alternate spans. So we will filled up BC and CD with LL, and then analysed the whole beam. The resulting BM diagram is shown below in fig.15.12:

Fig.15.12
Maximum hogging moment at support C

But this BM diagram is prepared xxx exclusively for the hogging moment at support C. That is., only the hogging moment values at the support C should be obtained from this diagram. So we will remove the unwanted parts to obtain the fig.15.13 shown below:

Fig.15.13
Diagram with unwanted values removed

Now we will look at the span CD. For maximum sagging moment in this span, the LL should be placed on CD and on alternate spans. So we placed it on CD and on the first span AB. The resulting BM diagram is shown in the fig.15.14 below:

Fig.15.14
Maximum sagging moment in span CD

But this BM diagram is prepared exclusively for the sagging moment in span CD. That is., only the sagging moment values in span CD should be obtained from this diagram. [For this particular beam ABCDE, the above loading arrangement for the maximum sagging moment in CD is same as the loading arrangement for the maximum sagging moment in AB also. But we are not considering AB in our present discussion]. So we will remove the unwanted parts to obtain the fig.15.13 shown below:

Fig.15.15
Diagram with unwanted values removed

The figs.15.15 and 15.13 gives the required values. So we can combine the two figs. as shown below:

Fig.15.16
Required values shown in a single diagram.

The above fig.15.16 gives the maximum values of the hogging moment at support C and the sagging moment in span CD. By using the same procedure, we can fill up the remaining supports and spans in the fig. Then, from that single fig., we can obtain the maximum values at any section along the length of ABCDE. Such a fig. is called the 'moment envelope'. Now let us take a closer look at the fig.15.14. This is shown in the fig.15.17 below:

Fig.15.17
Details of the moment envelope at support C and span CD and the points of inflection

We can see that the point of zero moment in the hogging moment envelope which is marked as p0h is different from p0s the point of zero moment in the sagging moment envelope. This is a situation that we often see while analysing continuous beams and continuous one-way slabs.

• The top steel bars should be provided for the hogging moment in the region shown in orange colour in the above fig.15.17. This steel should be available upto the points where the orange graph meets the horizontal axis, which are the points of inflection. In fact, as we will soon see in later sections, the bars should be continued even beyond the points of inflection.

• The bottom steel bars should be provided for the sagging moment in the region shown in yellow colour in the above fig.15.17. This steel should be available upto the points where the yellow graph meets the horizontal axis, which are the points of inflection. In this case also, we will soon see in later sections that, the bars should be continued even beyond the points of inflection.

• So the points of inflection play a major role in the final layout of bars. But these points of inflection for the sagging and hogging moments may not coincide. That is., we cannot expect a 'continuity' at the points of inflection as we saw in the fig.15.11 earlier in this section. The meeting point with the horizontal axis may be different for sagging and hogging moments, just as p0h and p0s in the fig.15.17. This happens when the different positions of LL are considered. This non-coincidence should be considered, and exact positions of various points of inflection should be determined (from the moment envelope) so as to provide a satisfactory lay out of bars.

In the next section, we will discuss another method to determine the theoretical cut-off points.

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Sunday, February 9, 2014

Chapter 7 (cont..2)

In the previous section we completed the analysis of a continuous slab. In this section we will do the analysis of a continuous beam. A view of a four span continuous beam ABCDE is given in fig 7.17 below:

Fig.7.17
View of a continuous beam ABCDE


Total depth of the beam = 400 mm
Width of the beam = 230 mm
Width of all outer supports A,B, D & E = 230 mm
Width of inner support C = 300 mm

In the previous example of a continuous slab, we were required to calculate the loads acting per unit length of the slab (unit length of a 1000mm wide strip). But here, the loads acting per meter length of the beam is given with the problem: 
Span DL LL Factored DL Factored LL
AB 16.63 6.82 24.94 10.23
BC 16.48 6.75 24.72 10.13
CD 16.25 6.64 24.37 9.96
DE 16.63 6.82 24.94 10.23
Effective spans: But unlike in the previous example, here, only the clear spans and the widths of supports are given. We have to calculate the effective span of the various spans.
We will calculate the effective spans by the two different methods: The one based on Eurocode-2, and the other based on IS 456. For this problem, it is convenient to mention before hand that, each of the spans have their support widths less than it's ln/12. So while using the cl.22.2(b) of IS 456, we will not have to look to the portion below the magenta colored dashed line of the chart. (see fig.7a.4)

Also assume dia. of bottom bars = 16 mm, dia. of links =8mm and Cc = 30 mm
So effective depth d = 400 -30 -8 -8 = 354 mm

First we will consider span AB. The calculations based on Eurocode-2 is shown below: (fig.7a.1)

Span AB, ln =4000
Support ASupport B
Type of supportNon-continuous supportContinuous support
Fig. to useFig.(a)Fig.(b)
h400400
t230230
ai = lesser of {h/2; t/2}115115
leff = ln + a1 +a2 =4000 +115+115=4230
The calculations based on IS 456 is given below:
Clear span ln =2850mm.
ln/12 = 4000/12 =333.33. So t1 < ln/12 & t2 < ln/12

As mentioned above, we only need the portion above the magenta colored dashed line for all spans of the beam. This is shown in the fig.7.18 below. This fig. is applicable to all the spans.

Fig.7.18
Application of chart to span AB

Now we calculate the following:
• c/c distance between the supports = 4000 +115 +115 =4230
• clear span + effective depth = 4000 +354 =4354
Effective span = leff = Lesser of the above = 4230mm

Thus we calculated leff of span AB using the two methods.

Now we will consider span BC. The calculations based on Eurocode-2 is shown below:

Span BC, ln =3900
Support BSupport C
Type of supportContinuous supportContinuous support
Fig. to useFig.(b)Fig.(b)
h400400
t230300
ai = lesser of {h/2; t/2}115150
leff = ln + a1 +a2 =3900 +115+150=4165
The calculations based on IS 456 is given below:
Clear span ln =3300mm.
ln/12 = 3900/12 =325.00. So t1 < ln/12 & t2 < ln/12

Fig.7.18 is applicable here also. Now we calculate the following:
• c/c distance between the supports = 3900 +115 +150 =4165
• clear span + effective depth = 3900 +354 =4254
Effective span = leff = Lesser of the above = 4165mm

Thus we calculated leff of span BC using the two methods.

Now we will consider span CD. The calculations based on Eurocode-2 is shown below:

Span CD, ln =3800
Support CSupport D
Type of supportContinuous supportContinuous support
Fig. to useFig.(b)Fig.(b)
h400400
t300230
ai = lesser of {h/2; t/2}150115
leff = ln + a1 +a2 =3800 +150+115=4065
The calculations based on IS 456 is given below:
Clear span ln =3800mm.
ln/12 = 3800/12 =316.67. So t1 < ln/12 & t2 < ln/12

Fig.7.18 is applicable here also. Now we calculate the following:
• c/c distance between the supports = 3800 +150 +115 =4065
• clear span + effective depth = 3800 +354 =4154
Effective span = leff = Lesser of the above = 4065mm

Thus we calculated leff of span CD using the two methods.

Now we will consider span DE. The calculations based on Eurocode-2 is shown below:

Span DE, ln =4000
Support DSupport E
Type of supportContinuous supportNon-Continuous support
Fig. to useFig.(b)Fig.(a)
h400400
t230230
ai = lesser of {h/2; t/2}115115
leff = ln + a1 +a2 =4000 +115+115=4230
The calculations based on IS 456 is given below:
Clear span ln =4000mm.
ln/12 = 4000/12 =333.33. So t1 < ln/12 & t2 < ln/12

Fig.7.18 is applicable here also. Now we calculate the following:
• c/c distance between the supports = 4000 +115 +115 =4230
• clear span + effective depth = 4000 +354 =4354
Effective span = leff = Lesser of the above = 4230mm

Thus we calculated leff of span DE using the two methods. All the results from the two methods are tabulated below:

Effective spans:
Name of spanBased on Euro codeBased on IS456
AB42304230
BC41654165
CD40654065
DE42304230
Now the beam is analysed for the ten possible combinations of the loads, using Kani's method. The first combination (where LL is applied on span AB only) is denoted as case1, and it's line diagram is shown in fig.7.19 below:

Fig.7.19
Load arrangement for case 1


In this way, a total of ten combinations are analysed. The results are shown in the fig.7.20 below:

Fig.7.20
Results of analysis
Results of the structural analysis of a continuous beam







Extreme values are picked from the above fig., and are tabulated below in fig.7.21:

Fig.7.21
Extreme values of BM and SF

In the above figs., '1' denotes the maximum BM at supports; '2' denotes the maximum BM at midspans, and '3' denotes the maximum SF at supports.

*2.a denotes the minimum BM at midspans. We have to take special care about minimum BM in BC and CD because, for some load combinations (cases 8 and 9), these spans are bending upwards. 

So now we have the BM and SF at the various points of the continuous beam. Just as in the previous example, let us calculate them again using the coefficients given in the code. We will do this in the next section

                                                         
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