In
the previous section we completed the discussion on 'Case 1: Df ≥ xu'. In this section we will discuss about the next case. We will call it simply as 'case 2' for the time being. We will soon see that this case has some 'sub cases' within itself. So we will subsequently change the name to something which is more suitable.
Case 2:
Here we take up the condition in which the NA lies in the web. But when we draw the concrete stress block for this condition, we become aware that this case has two sub cases within itself. This can be explained as follows: We know that the concrete stress block has two portions. A rectangular portion (shown in red color in the fig.9.9 below) and a parabolic portion (shown in blue color in the fig.9.9 below). The depth of the rectangular portion is equal to (3/7)xu which when written in decimal form is 0.43xu. If this 0.43xu is greater than Df, then the whole flange will be under the compression from the rectangular stress block only. This will mean that the whole flange will be experiencing a uniform compressive stress. This is shown in the fig.9.9 below:
Fig.9.9
Only the rectangular portion of the stress block is acting on the flange
In the above fig., we can see that 0.43xu (which is written in the fig. as (3/7)xu ) is greater than Df. A 3D view of the stress block of this case is given below:
Fig.9.10
3D view of stress block
Now let us consider the next 'sub case': If 0.43xu is less than Df, then, the rectangular portion of the stress block will get only a small portion (at the top of the flange) to exert it's compressive stress. The remaining portion at the bottom of the flange will be acted upon by the parabolic portion. This is shown in the fig. below:
Fig.9.11
Rectangular portion acts only on a small portion at the top of the flange
In the above fig., we can see that 0.43xu (which is written in the fig. as (3/7)xu ) is lesser than Df. This will mean that a portion at the top of the flange will be experiencing uniform compressive stress, while the remaining portion at the bottom of the flange will be experiencing non-uniform compressive stress. A 3D view of the stress block of this case is given below:
Fig.9.12
3D view of stress block
We will first analyse the former case where Df ≤ 0.43xu. We will call it as 'case 2(1): Df < xu AND Df ≤ 0.43xu'
case 2(1): Df < xu AND Df ≤ 0.43xu:
So this case is related to the situation in which two conditions are satisfied:
• The NA lies in the web (indicated by the relation: Df < xu) and
• The depth of the rectangular stress block is greater than the depth of flange (indicated by the relation: Df ≤ 0.43xu)
We now need to analyse this case. We have already seen the stress distribution for this case in fig.9.9 and the 3D view in fig.9.10. From the 3D view, we can see that, in the middle web region, the full stress block is acting, whereas in the over hanging flange portions, only the rectangular block is acting. So it is convenient to consider the web and the over hangs separately. This is shown in the fig. below:
Fig.9.13
Separation of flange and web portions in a T-beam
Note that while making the separation, the middle web region is given the full depth D. The 'flange' portion will consist of only the overhangs. So the middle web region will have an area of bwD and each of the overhangs will have an area of 0.5(bf -bw)Df.
The method of separating an L-beam in this way is shown in the fig. below:
Fig.9.14
Separation of flange and web portions in a L-beam
So now we have separated the section into web portion and flange portion. We will consider the forces (acting on these portions) also separately. First we take the web portion. We want the compressive force acting on this portion. We know that the full stress block is acting here. So it is evident that this web portion will act exactly like a rectangular beam section. We have already analysed rectangular beam sections in chapter 3. There we derived Eq.3.7 for the compressive force acting on the beam section. Based on that, we can now readily write the following Eq.9.9 for the compressive force Cuw on our web portion as:
Eq.9.9
Cuw = 0.362fck bw xu
Note that the only change made, is to use bw in the place of b. Next we want the compressive force in the flange portion. This is given by the volume of the rectangular stress block acting on the flanges.
• The depth of this block is 0.447fck.
• The thickness is Df.
• And the length is 0.5(bf -bw) + 0.5(bf -bw) = bf -bw.
So we get the volume as: 0.447fck Df (bf -bw). Thus we can write the compressive force acting on the flange as:
Eq.9.10
Cuf = 0.447fck Df (bf -bw)
So we obtained the compressive forces in the two portions. Now we want the moment of resistance offered by the section. We know that Moment = Force x Lever arm. Lever arm is the perpendicular distance between the 'Lines of action' of compressive force and tensile force. The line of action of the tensile force is through the centroid of reinforcing steel bars. But the line of action of the compressive forces are at different levels as shown in the fig. below:
Fig.9.15
Lines of action of Cuf and Cuw
• Line of action of Cuf is at the centroid of the flange which is at a distance of Df/2 from the top most compression fibre.
• Line of action of Cuw the compressive force in the web is at a distance of 0.416xu from the top most compression fibre. This we have derived as Eq.3.8 in chapter 3.
Thus it is clear that the two compressive forces are acting at different levels. So we will take them separately.
• Lever arm for Cuw is d- 0.416xu
• Lever arm for Cuf is d- Df/2
Now we will get the moment contributions.
• Contribution from web = Cuw (d- 0.416xu)
• Contribution from flange = Cuf (d- Df/2)
The sum of the above two will give the total moment of resistance offered by the section. So putting the values of Cuw and Cuf from Eqs. 9.9 and 9.10, and adding, we get
MuR = 0.362fck bw xu(d- 0.416xu) + 0.447fck Df (bf -bw) (d- Df/2)
The first term on the right side of the above equation is the contribution from the web. We can rearrange this part so that we get a more 'familiar form' which is given in the code. This is done by bringing 'd' outside the brackets. Thus we get:
Eq.9.11
To use Eq.9.9 and 9.11, we need the value of xu. In the next section we will learn the method to determine xu.
PREVIOUS
Case 2:
Here we take up the condition in which the NA lies in the web. But when we draw the concrete stress block for this condition, we become aware that this case has two sub cases within itself. This can be explained as follows: We know that the concrete stress block has two portions. A rectangular portion (shown in red color in the fig.9.9 below) and a parabolic portion (shown in blue color in the fig.9.9 below). The depth of the rectangular portion is equal to (3/7)xu which when written in decimal form is 0.43xu. If this 0.43xu is greater than Df, then the whole flange will be under the compression from the rectangular stress block only. This will mean that the whole flange will be experiencing a uniform compressive stress. This is shown in the fig.9.9 below:
Fig.9.9
Only the rectangular portion of the stress block is acting on the flange
In the above fig., we can see that 0.43xu (which is written in the fig. as (3/7)xu ) is greater than Df. A 3D view of the stress block of this case is given below:
Fig.9.10
3D view of stress block
Now let us consider the next 'sub case': If 0.43xu is less than Df, then, the rectangular portion of the stress block will get only a small portion (at the top of the flange) to exert it's compressive stress. The remaining portion at the bottom of the flange will be acted upon by the parabolic portion. This is shown in the fig. below:
Fig.9.11
Rectangular portion acts only on a small portion at the top of the flange
In the above fig., we can see that 0.43xu (which is written in the fig. as (3/7)xu ) is lesser than Df. This will mean that a portion at the top of the flange will be experiencing uniform compressive stress, while the remaining portion at the bottom of the flange will be experiencing non-uniform compressive stress. A 3D view of the stress block of this case is given below:
Fig.9.12
3D view of stress block
We will first analyse the former case where Df ≤ 0.43xu. We will call it as 'case 2(1): Df < xu AND Df ≤ 0.43xu'
case 2(1): Df < xu AND Df ≤ 0.43xu:
So this case is related to the situation in which two conditions are satisfied:
• The NA lies in the web (indicated by the relation: Df < xu) and
• The depth of the rectangular stress block is greater than the depth of flange (indicated by the relation: Df ≤ 0.43xu)
We now need to analyse this case. We have already seen the stress distribution for this case in fig.9.9 and the 3D view in fig.9.10. From the 3D view, we can see that, in the middle web region, the full stress block is acting, whereas in the over hanging flange portions, only the rectangular block is acting. So it is convenient to consider the web and the over hangs separately. This is shown in the fig. below:
Fig.9.13
Separation of flange and web portions in a T-beam
Note that while making the separation, the middle web region is given the full depth D. The 'flange' portion will consist of only the overhangs. So the middle web region will have an area of bwD and each of the overhangs will have an area of 0.5(bf -bw)Df.
The method of separating an L-beam in this way is shown in the fig. below:
Fig.9.14
Separation of flange and web portions in a L-beam
So now we have separated the section into web portion and flange portion. We will consider the forces (acting on these portions) also separately. First we take the web portion. We want the compressive force acting on this portion. We know that the full stress block is acting here. So it is evident that this web portion will act exactly like a rectangular beam section. We have already analysed rectangular beam sections in chapter 3. There we derived Eq.3.7 for the compressive force acting on the beam section. Based on that, we can now readily write the following Eq.9.9 for the compressive force Cuw on our web portion as:
Eq.9.9
Cuw = 0.362fck bw xu
Note that the only change made, is to use bw in the place of b. Next we want the compressive force in the flange portion. This is given by the volume of the rectangular stress block acting on the flanges.
• The depth of this block is 0.447fck.
• The thickness is Df.
• And the length is 0.5(bf -bw) + 0.5(bf -bw) = bf -bw.
So we get the volume as: 0.447fck Df (bf -bw). Thus we can write the compressive force acting on the flange as:
Eq.9.10
Cuf = 0.447fck Df (bf -bw)
So we obtained the compressive forces in the two portions. Now we want the moment of resistance offered by the section. We know that Moment = Force x Lever arm. Lever arm is the perpendicular distance between the 'Lines of action' of compressive force and tensile force. The line of action of the tensile force is through the centroid of reinforcing steel bars. But the line of action of the compressive forces are at different levels as shown in the fig. below:
Fig.9.15
Lines of action of Cuf and Cuw
• Line of action of Cuf is at the centroid of the flange which is at a distance of Df/2 from the top most compression fibre.
• Line of action of Cuw the compressive force in the web is at a distance of 0.416xu from the top most compression fibre. This we have derived as Eq.3.8 in chapter 3.
Thus it is clear that the two compressive forces are acting at different levels. So we will take them separately.
• Lever arm for Cuw is d- 0.416xu
• Lever arm for Cuf is d- Df/2
Now we will get the moment contributions.
• Contribution from web = Cuw (d- 0.416xu)
• Contribution from flange = Cuf (d- Df/2)
The sum of the above two will give the total moment of resistance offered by the section. So putting the values of Cuw and Cuf from Eqs. 9.9 and 9.10, and adding, we get
MuR = 0.362fck bw xu(d- 0.416xu) + 0.447fck Df (bf -bw) (d- Df/2)
The first term on the right side of the above equation is the contribution from the web. We can rearrange this part so that we get a more 'familiar form' which is given in the code. This is done by bringing 'd' outside the brackets. Thus we get:
Eq.9.11
To use Eq.9.9 and 9.11, we need the value of xu. In the next section we will learn the method to determine xu.
PREVIOUS
nice
ReplyDeleteIn below fig.9.11 It should 0.43Xu(3/7Xu) is lesser than Df. I think you made a mistake there....
ReplyDeleteThanks for drawing attention to the mistake. It is now corrected.
Delete