In the previous section, we said that it is important to get a fairly good approximation for the preliminary value of d. Now we will see how this is achieved:
We know the basic equation:
Bending moment = Force x Lever arm
⇒ Bending moment = Area of steel x stress in steel x Lever arm.
⇒ Area of steel = Bending moment / (stress in steel x Lever arm)
This can be written as:
Eq.10.5
• Mu is the factored bending moment which has to be resisted,
• 0.87fy is the stress in steel. (Stress is equal to 0.87fy because, the quantity of steel that we provide must be such that the beam will be under reinforced)
• z is the lever arm
For using in the above Eq.10.5, z can be taken as the larger of the following:
10.6
♦ 0.9d
♦ d - Df/2
So here another problem arises:
• We are using Eq.10.5 above to calculate the approximate area of steel that will be required.
• This is to determine whether the steel will have to be provided in more than one layer.
• From the number of layers, and the dia. of bars, we will get the effective depth d.
• Our ultimate aim here is to determine d.
• But from 10.6 above, we see that d should be known prior to using Eq.10.5.
This forms a loop. The solution to this problem is to use an initial d by deducting 50 mm from the total depth D. This d is to be used in 10.6 to obtain z. Thus we will be able to use Eq.10.5.
From the Eq.10.5, we will get the 'most probable' area of steel that will be required. This area must be converted into 'number and diameters' of bars. (details here.) Then we will know whether the steel will be in single layer or multiple layers. Based on the number of layers and the diameter of bars, we can calculate d. This will be the 'most probable' d.
With the calculation of d, we complete all the calculations on 'preliminary dimensions'. That is., we now have a complete beam section to begin our work.
It may be noted that bf also needs to be calculated at this stage. But it does not come under the category of 'preliminary dimensions'. This can be explained as follows: We have the required equations for calculating bf of any type of T-beams or L-beams. Eqs. 9.5, 9.6 etc., are some of those. We can see that in those equations, the only variable is bw. All others like l0, Df etc., are constants. Now, this bw is not much of a 'variable' because it is a parameter that receives an early 'fix' depending on the width of supports. So bf itself is not a variable.
There are some checks that we have to perform immediately after fixing up the preliminary dimensions. These are same as those for the rectangular sections. We saw them in 4.2, 4.3 and 4.4. For our present flanged beam sections, b should be changed to bw.
Let us now see the different steps involved in the design process.
Step A:
We begin the work by analysing the section that we have. We analyse it first, to find the Limiting moment of resistance, Mulim. If Mulim is less than Mu, we know that the beam section will not serve it's purpose even if we give high quantities of steel. In that case, the section will have to be improved by increasing the depth or making it 'doubly reinforced'.
Step B:
If Mulim is greater than Mu, then we can proceed to the next step. This step also involves an analysis of the section. This time to find xu. From the value of xu, we will know whether
• the NA falls in the flange (case 1) or
• in the web (case 2). If in the web, we will know (from the value of 0.43xu) whether the section belongs to
♦ Case 2(1) or
♦ Case 2(2).
So we get the position of NA. That is., we get the case into which our beam section falls. This will be the 'most probable' case because, we are using the 'most probable' area of steel obtained from Eq.10.5 above.
If it is Case 1, we can design it in just the way we would design a rectangular beam of width bf. We have learned the design procedure here. But the steps 1 & 2 in that procedure can be avoided. This can be explained as follows: In the case of rectangular sections, step 1 involves the calculation of required d. Step 2 involves finalizing D. These two steps are done to ensure that the final section will be capable to resist the load. But here we already do this in step A. In step A, we check whether Mu < Mulim. If Mu is indeed less than Mulim, then what ever be the position of NA, the section will be able to resist Mu if we give the appropriate amount of steel. So Steps 1 and 2 for rectangular sections is equivalent to step A for flanged sections. The remaining steps 3 & 4 are same for both the types. The only change to be made is to change b to bf.
If it is case 2(1) or case 2(2), much more calculations are involved. Here also, the 'calculation of d required' and 'finalizing D' can be avoided because we do this in step A. So we will proceed to the next step.
Step C:
If it is case 2(1):
The BM offered by the section is given by Eq.9.11
Eq.9.11
This should be equal to or greater than Mu. For obtaining a solution, We will first equate it to Mu.
For solving, we must simplify it. Note that the last term is constant. We will put it equal to K. Thus we get:
Let us simplify it further:
Eq.10.7
A, B and C contain only the known values b, d, fck , Mu and K. So if we are using a spread sheet program, it will calculate A, B and C just when the input data are given. The program will then directly give the two solutions of the quadratic equation. That is., we will get two values for xu/d. Out of these, one will be greater than 1, and the other will be less than 1. We must choose the lesser value because xu is always less than d. If we choose the value which is greater than 1, when we multiply it with d, we will get a value for xu which is greater than d.
This step comes to an end with the calculation of xu . So step C can be summarized as:
• Calculate the values of A, B and C and then solve xu
The next step is the calculation of Ast.
Step D: With the calculation of xu, we can split the section into an upper part and a lower part. The compressive force in the upper part is given by Eq.9.12
Eq.9.12:
Cuw + Cuf = Cu = 0.362fck bw xu + 0.447fck Df (bf -bw)
The tensile force in the steel in the lower part is given by :
Tu = 0.87 fy Ast
Where Ast is the area of steel. The stress in steel is equal to 0.87 fy because, it is an under reinforced section, and so the steel would have yielded at the ultimate state.
For equilibrium, Cu = Tu. So, equating the two, we get
Eq.10.8
This completes the design procedure if the beam section belongs to case 2(1). In the next section, we will now discuss the case 2(2).
PREVIOUS
We know the basic equation:
Bending moment = Force x Lever arm
⇒ Bending moment = Area of steel x stress in steel x Lever arm.
⇒ Area of steel = Bending moment / (stress in steel x Lever arm)
This can be written as:
Eq.10.5
• 0.87fy is the stress in steel. (Stress is equal to 0.87fy because, the quantity of steel that we provide must be such that the beam will be under reinforced)
• z is the lever arm
For using in the above Eq.10.5, z can be taken as the larger of the following:
10.6
♦ 0.9d
♦ d - Df/2
So here another problem arises:
• We are using Eq.10.5 above to calculate the approximate area of steel that will be required.
• This is to determine whether the steel will have to be provided in more than one layer.
• From the number of layers, and the dia. of bars, we will get the effective depth d.
• Our ultimate aim here is to determine d.
• But from 10.6 above, we see that d should be known prior to using Eq.10.5.
This forms a loop. The solution to this problem is to use an initial d by deducting 50 mm from the total depth D. This d is to be used in 10.6 to obtain z. Thus we will be able to use Eq.10.5.
From the Eq.10.5, we will get the 'most probable' area of steel that will be required. This area must be converted into 'number and diameters' of bars. (details here.) Then we will know whether the steel will be in single layer or multiple layers. Based on the number of layers and the diameter of bars, we can calculate d. This will be the 'most probable' d.
With the calculation of d, we complete all the calculations on 'preliminary dimensions'. That is., we now have a complete beam section to begin our work.
It may be noted that bf also needs to be calculated at this stage. But it does not come under the category of 'preliminary dimensions'. This can be explained as follows: We have the required equations for calculating bf of any type of T-beams or L-beams. Eqs. 9.5, 9.6 etc., are some of those. We can see that in those equations, the only variable is bw. All others like l0, Df etc., are constants. Now, this bw is not much of a 'variable' because it is a parameter that receives an early 'fix' depending on the width of supports. So bf itself is not a variable.
There are some checks that we have to perform immediately after fixing up the preliminary dimensions. These are same as those for the rectangular sections. We saw them in 4.2, 4.3 and 4.4. For our present flanged beam sections, b should be changed to bw.
Let us now see the different steps involved in the design process.
Step A:
We begin the work by analysing the section that we have. We analyse it first, to find the Limiting moment of resistance, Mulim. If Mulim is less than Mu, we know that the beam section will not serve it's purpose even if we give high quantities of steel. In that case, the section will have to be improved by increasing the depth or making it 'doubly reinforced'.
Step B:
If Mulim is greater than Mu, then we can proceed to the next step. This step also involves an analysis of the section. This time to find xu. From the value of xu, we will know whether
• the NA falls in the flange (case 1) or
• in the web (case 2). If in the web, we will know (from the value of 0.43xu) whether the section belongs to
♦ Case 2(1) or
♦ Case 2(2).
So we get the position of NA. That is., we get the case into which our beam section falls. This will be the 'most probable' case because, we are using the 'most probable' area of steel obtained from Eq.10.5 above.
If it is Case 1, we can design it in just the way we would design a rectangular beam of width bf. We have learned the design procedure here. But the steps 1 & 2 in that procedure can be avoided. This can be explained as follows: In the case of rectangular sections, step 1 involves the calculation of required d. Step 2 involves finalizing D. These two steps are done to ensure that the final section will be capable to resist the load. But here we already do this in step A. In step A, we check whether Mu < Mulim. If Mu is indeed less than Mulim, then what ever be the position of NA, the section will be able to resist Mu if we give the appropriate amount of steel. So Steps 1 and 2 for rectangular sections is equivalent to step A for flanged sections. The remaining steps 3 & 4 are same for both the types. The only change to be made is to change b to bf.
If it is case 2(1) or case 2(2), much more calculations are involved. Here also, the 'calculation of d required' and 'finalizing D' can be avoided because we do this in step A. So we will proceed to the next step.
Step C:
If it is case 2(1):
The BM offered by the section is given by Eq.9.11
Eq.9.11
This should be equal to or greater than Mu. For obtaining a solution, We will first equate it to Mu.
For solving, we must simplify it. Note that the last term is constant. We will put it equal to K. Thus we get:
Eq.10.7
There is a term in (xu/d)2 and another term in (xu/d). It is a quadratic equation. We can write it as:
A X2 + B X + C = 0
Where A = 0.1506 bw d2fck ; B=-0.362bwd2fck ; C = Mu -K and X = xu/d
A X2 + B X + C = 0
Where A = 0.1506 bw d2fck ; B=-0.362bwd2fck ; C = Mu -K and X = xu/d
A, B and C contain only the known values b, d, fck , Mu and K. So if we are using a spread sheet program, it will calculate A, B and C just when the input data are given. The program will then directly give the two solutions of the quadratic equation. That is., we will get two values for xu/d. Out of these, one will be greater than 1, and the other will be less than 1. We must choose the lesser value because xu is always less than d. If we choose the value which is greater than 1, when we multiply it with d, we will get a value for xu which is greater than d.
This step comes to an end with the calculation of xu . So step C can be summarized as:
• Calculate the values of A, B and C and then solve xu
The next step is the calculation of Ast.
Step D: With the calculation of xu, we can split the section into an upper part and a lower part. The compressive force in the upper part is given by Eq.9.12
Eq.9.12:
Cuw + Cuf = Cu = 0.362fck bw xu + 0.447fck Df (bf -bw)
The tensile force in the steel in the lower part is given by :
Tu = 0.87 fy Ast
Where Ast is the area of steel. The stress in steel is equal to 0.87 fy because, it is an under reinforced section, and so the steel would have yielded at the ultimate state.
For equilibrium, Cu = Tu. So, equating the two, we get
Eq.10.8
This completes the design procedure if the beam section belongs to case 2(1). In the next section, we will now discuss the case 2(2).
PREVIOUS
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