In the previous section, we completed the design procedure for case 2(1). Now we will consider case 2(2).
Step A and Step B are the same. It is at the end of step B that we get to know the case. So the difference will be from Step C.
Step C:
If it is case 2(2):
The BM offered by the section is given by Eq.9.17
Eq.9.17
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. The last term here is not a constant. This is because yf contains xu. (yf = 0.15xu + 0.65Df). So we have to simplify each term. Let us take the first term:
We get a term in (xu/d)2 and another term in (xu/d). Now let us take the second term. It can also be written in terms of (xu/d)2 and (xu/d) (see derivation here):
Where K1 = 0.447fck (bf - bw)
So we can write:
Rearranging, we get:
Eq.10.9
This is a quadratic equation of the form A X2 + B X + C = 0.
A, B and C contain only the known values like Df, bw, d, fck , Mu and K1. 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 changing Df to yf in Eq.9.12
Eq.9.12:
Cuw + Cuf = Cu = 0.362fck bw xu + 0.447fck Df (bf -bw)
Changing Df to yf, we get:
Cuw + Cuf = Cu = 0.362fck bw xu + 0.447fck yf (bf -bw)
The tensile force in the steel 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.10.10
This completes the design procedure if the beam section belongs to case 2(2).
Next we will see some solved examples.
Solved example 10.1
The fig.10.12 shows the cross section of a Continuous T-beam. The cross sectional dimensions are fixed by considering hogging moment at supports. The effective flange width bf is calculated as 1400 mm. Determine the area of steel required to resist a sagging moment of 750 kNm at midspan. Assume moderate exposure conditions. Grade of steel is Fe415.
Fig.10.12
Section of T-beam
Solution:
The data given are:
bf =1400 mm, Df =100 mm, D =700 mm, bw =300 mm, fy = 415 N/mm2, Mu = 750 kNm.
In this problem, all the cross sectional dimensions of the beam section are given. So we do not have to work out any 'preliminary dimensions'. But we do have to calculate the 'most probable' d.
Eq.10.5
z is the lever arm which is taken as the larger of the following:
10.6
♦ 0.9d
♦ d - Df/2
For an initial value of d, we can deduct 50 mm from D. So we get d = 700 -50 =650 mm. So z is the larger of:
♦ 0.9d = 0.9 x 650 = 585 mm
♦ d - Df/2 = 650 - 100/2 =600 mm
Thus z = 600 mm. Substituting the values in 10.5 we get Ast,reqd = 3462.12 mm2. Assuming 3 - 32# and 2 - 28#, we get an area of 3644.25 mm2. These bars have to be arranged in two layers as shown in the fig. below:
Fig.10.13
Arrangement of bars in beam
Check for minimum horizontal distance between bars (details here):
The exposure condition is given as 'moderate'. So we can give a concrete cover of 30 mm. But the maximum size of bar in the beam is 32 mm. So we will give Cc = 32 mm. The the total width of the beam at the bottom most layer of bars can be written as:
300 = 2 x Cc + 2 x Φl + 3 x Φ + 2 x Sh.
⇒ 300 = 2 x 32 + 2 x 8 + 3 x 32 + 2 x Sh.
From this we get Sh = 62 mm
• Bar diameter = 32 mm
• Max. size of aggregate + 5 = 20 + 5 = 25 mm
Larger of the above = 32 mm < 62 mm. Hence OK
Minimum clear vertical distance required between bars:
Ref. fig.4.12 of the section here
• Larger bar diameter = 32 mm
• 15 mm
• 2/3 of Max. size of aggregate = 2/3 x 20 = 13.33 mm
Larger of the above = 32 mm. So spacer bars of 32 mm dia. have to be placed between the two layers.
The above layout of bars will satisfy the requirements of the code. We can now calculate d. It can be calculated based on the fig.10.14 below:
Fig.10.14
Calculation of effective depth
• y is the vertical distance of the c.g of the bar group from the bottom most edge of the bar group.
• y is the vertical distance of each of the individual bars from the bottom most edge of the bar group.
The above derived value of ŷ can be rounded off to 37 mm. So d = 700 -32 -8 -37 =623 mm
Now we can begin the design process. We will do it in the next section.
PREVIOUS
Step A and Step B are the same. It is at the end of step B that we get to know the case. So the difference will be from Step C.
Step C:
If it is case 2(2):
The BM offered by the section is given by Eq.9.17
Eq.9.17
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. The last term here is not a constant. This is because yf contains xu. (yf = 0.15xu + 0.65Df). So we have to simplify each term. Let us take the first term:
We get a term in (xu/d)2 and another term in (xu/d). Now let us take the second term. It can also be written in terms of (xu/d)2 and (xu/d) (see derivation here):
Where K1 = 0.447fck (bf - bw)
So we can write:
Rearranging, we get:
Eq.10.9
This is a quadratic equation of the form A X2 + B X + C = 0.
A, B and C contain only the known values like Df, bw, d, fck , Mu and K1. 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 changing Df to yf in Eq.9.12
Eq.9.12:
Cuw + Cuf = Cu = 0.362fck bw xu + 0.447fck Df (bf -bw)
Changing Df to yf, we get:
Cuw + Cuf = Cu = 0.362fck bw xu + 0.447fck yf (bf -bw)
The tensile force in the steel 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.10.10
This completes the design procedure if the beam section belongs to case 2(2).
Next we will see some solved examples.
Solved example 10.1
The fig.10.12 shows the cross section of a Continuous T-beam. The cross sectional dimensions are fixed by considering hogging moment at supports. The effective flange width bf is calculated as 1400 mm. Determine the area of steel required to resist a sagging moment of 750 kNm at midspan. Assume moderate exposure conditions. Grade of steel is Fe415.
Fig.10.12
Section of T-beam
Solution:
The data given are:
bf =1400 mm, Df =100 mm, D =700 mm, bw =300 mm, fy = 415 N/mm2, Mu = 750 kNm.
In this problem, all the cross sectional dimensions of the beam section are given. So we do not have to work out any 'preliminary dimensions'. But we do have to calculate the 'most probable' d.
Eq.10.5
z is the lever arm which is taken as the larger of the following:
10.6
♦ 0.9d
♦ d - Df/2
For an initial value of d, we can deduct 50 mm from D. So we get d = 700 -50 =650 mm. So z is the larger of:
♦ 0.9d = 0.9 x 650 = 585 mm
♦ d - Df/2 = 650 - 100/2 =600 mm
Thus z = 600 mm. Substituting the values in 10.5 we get Ast,reqd = 3462.12 mm2. Assuming 3 - 32# and 2 - 28#, we get an area of 3644.25 mm2. These bars have to be arranged in two layers as shown in the fig. below:
Fig.10.13
Arrangement of bars in beam
Check for minimum horizontal distance between bars (details here):
The exposure condition is given as 'moderate'. So we can give a concrete cover of 30 mm. But the maximum size of bar in the beam is 32 mm. So we will give Cc = 32 mm. The the total width of the beam at the bottom most layer of bars can be written as:
300 = 2 x Cc + 2 x Φl + 3 x Φ + 2 x Sh.
⇒ 300 = 2 x 32 + 2 x 8 + 3 x 32 + 2 x Sh.
From this we get Sh = 62 mm
• Bar diameter = 32 mm
• Max. size of aggregate + 5 = 20 + 5 = 25 mm
Larger of the above = 32 mm < 62 mm. Hence OK
Minimum clear vertical distance required between bars:
Ref. fig.4.12 of the section here
• Larger bar diameter = 32 mm
• 15 mm
• 2/3 of Max. size of aggregate = 2/3 x 20 = 13.33 mm
Larger of the above = 32 mm. So spacer bars of 32 mm dia. have to be placed between the two layers.
The above layout of bars will satisfy the requirements of the code. We can now calculate d. It can be calculated based on the fig.10.14 below:
Fig.10.14
Calculation of effective depth
• y is the vertical distance of the c.g of the bar group from the bottom most edge of the bar group.
• y is the vertical distance of each of the individual bars from the bottom most edge of the bar group.
The above derived value of ŷ can be rounded off to 37 mm. So d = 700 -32 -8 -37 =623 mm
Now we can begin the design process. We will do it in the next section.
PREVIOUS
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