In the previous section, we completed the 1st cycle in the analysis of the beam section. We obtained the xu to be used in the next cycle as xu =212.335 mm. Now we will continue the iteration process:
Cycle 2:
When the NA is in the position xu =212.335, εsc (Using Eq12.1) =0.00271 The corresponding stress fsc is obtained from from the same table for Fe 415 steel:
0.00241 → 342.800
0.00271 → 350.483
0.00276 → 351.800. So we get fsc =350.483 N/mm2.
Similarly, when the NA is in this position, εst (using Eq.3.15) =0.00303. The corresponding stress fst is obtained from the same table for Fe 415 steel:
0.00276 → 351.800
0.00303 → 354.140
0.00380 → 360.900. So we get fst =354.140 N/mm2. With these stress values, we can calculate the forces:
Cu =Cus +Cuc =(fsc -0.447fck)Asc + 0.362fckbxu = (350.483 – 0.447 x 20) x 402.12 + 2172.00 x212.335 = 598533.964 N
Tu =fstAst =354.140 x1722.38 = 609963.349 N
These forces are not equal. So we don't have an equilibrium. What is the value of xu if these were indeed the stresses? We can calculate this by writing the equality: Cu = Tu. Thus we write:
(350.483 – 0.447 x 20) x 402.12 + 2172.00 xu = 609963.349
From this we get = xu =217.598 mm. We will take the average of this xu and the value of xu used in this cycle. It is equal to 0.5 x(217.598 +212.335) =214.966 mm. This can be used as the trial value for the next cycle.
Cycle 3:
When the NA is in the position xu =214.966, εsc (Using Eq12.1) =0.00272
. The corresponding stress fsc is obtained from from the same table for Fe 415 steel:
0.00241 → 342.800
0.00272 → 350.732
0.00276 → 351.800. So we get fsc =350.483 N/mm2.
Similarly, when the NA is in this position, εst (using Eq.3.15) =0.00295. The corresponding stress fst is obtained from the same table for Fe 415 steel:
0.00276 → 351.800
0.00295 → 353.441
0.00380 → 360.900. So we get fst =353.441 N/mm2. With these stress values, we can calculate the forces:
Cu =Cus +Cuc =(fsc -0.447fck)Asc + 0.362fckbxu = (350.732 – 0.447 x 20) x 402.12 + 2172.00 x214.966 = 604348.790 N
Tu =fstAst =353.441 x1722.38 = 608759.311 N
These forces are not equal. So we don't have an equilibrium. What is the value of xu if these were indeed the stresses? We can calculate this by writing the equality: Cu = Tu. Thus we write:
(350.732 – 0.447 x 20) x 402.12 + 2172.00 xu = 608759.311
From this we get = xu =216.997 mm. We will take the average of this xu and the value of xu used in this cycle. It is equal to 0.5 x(216.997 +214.966) =215.982 mm. This can be used as the trial value for the next cycle.
Cycle 4:
When the NA is in the position xu =215.982, εsc (Using Eq12.1) =0.00272
. The corresponding stress fsc is obtained from from the same table for Fe 415 steel:
0.00241 → 342.800
0.00272 → 350.827
0.00276 → 351.800. So we get fsc =350.827 N/mm2.
Similarly, when the NA is in this position, εst (using Eq.3.15) =0.00292. The corresponding stress fst is obtained from the same table for Fe 415 steel:
0.00276 → 351.800
0.00292 → 353.176
0.00380 → 360.900. So we get fst =353.176 N/mm2. With these stress values, we can calculate the forces:
Cu =Cus +Cuc =(fsc -0.447fck)Asc + 0.362fckbxu = (350.827 – 0.447 x 20) x 402.12 + 2172.00 x215.982 = 606592.039 N
Tu =fstAst =353.441 x1722.38 = 608302.526 N
These forces are approximately equal. So we have an equilibrium. The final values can be taken as:
• fsc =350.827 N/mm2 • fst =353.441 N/mm2 • xu =215.982 mm
[It may be noted that fig.12.5 is drawn only for getting a good understanding about the extreme positions of the NA. It is not necessary to draw such a fig. in an analysis problem.]
Now we can obtain MuR
MuR = 0.362fckbxu(d - 0.416xu) + ( fsc - 0.447fck) Asc(d-d')
= 143618673.976 + 47853116.410 =191471790.385 Nmm =191.5 kNm
Also xu = 215.982 mm. xu,max = 0.4791 x 396 =189.724 mm. So we get xu > xu,max. So this is an over reinforced section.
Thus we complete the analysis process.
Let us compare the sections in the last two solved examples:
• Solved example 2:
Given b =300 mm, d =396 mm, d' =50 mm, Ast =1722.38 mm2, Asc =628.32 mm2, fy =415 N/mm2, fck =20 N/mm2, xu =188.527 mm, xu,max =189.72 mm, MuR =203.53 kNm, xu < xu,max, Under reinforced.
• Solved example 3:
b =300 mm, d =396 mm, d' =48 mm, Ast =1722.38 mm2, Asc =402.12 mm2, fy =415 N/mm2, fck =20 N/mm2, xu =215.982 mm, xu,max =189.72 mm, MuR =191.5 kNm. xu > xu,max, Over reinforced.
We can see that all the properties except Asc are the same. The second section have a lesser Asc than the first. The second section is Over reinforced, while the first section is under reinforced. Let us now discuss the reason for this.
Mulim of both the sections will be the same. It is given by:
Thus Mulim =130.66 kNm
Ast,lim of both the sections will be the same. It is given by
Thus Ast,lim = 1141.67 mm2.
As Ast is same for both sections, ΔAst will also be the same (Eq.11.8):
Ast = Ast,lim + ΔAst
Thus we get ΔAst = 1722.38 -1141.67 =580.71 mm2.
If the section under consideration is an under reinforced one, then all the Ast would have yielded at the ultimate state. So the stress in it = 0.87fy, and the force ΔTu = 0.87fyAst =209666.068 N. The quantity of compression steel Asc must be such that, the force developed in it should balance ΔTu.
So, next we find the stress fsc in Asc: For this we can use Table 11.1.
d'/d = 48/396 =0.121.
0.100 → 342.90
0.121 → 347.87
0.150 → 342.40. So we get fsc =347.87 N/mm2.
So the force in Asc =( fsc - 0.447fck) Asc = (347.87 -0.447 x20)Asc
[It may be noted that Table.11.1 can be used to find fsc only in the condition when tension steel has yielded. If this steel has not yielded at the ultimate state, we have to use strain compatibility to find fsc]
This must be equal to the force in ΔTu. So we can write 209666.068 = (347.87 -0.447 x20)Asc. From this we get Asc = 618.61 mm2. This is the 'ideal' quantity of Asc. If we provide this exact quantity, xu will be equal to xu,max. But as explained based on fig.11.4 , this exact quantity cannot be provided. The actual quantity that we give should be greater than 618.61 mm2. When we give this higher quantity, there will be 'more material' towards the top side of the beam, and so the NA will move up wards. Thus xu will decrease from xu,max, and so, it will be an under reinforced section. If Asc is less than 618.61, then the NA will move downwards, and it will be an over reinforced section.
In the two solved examples,
• The first one have an Asc =628.32 >618.61 mm2. So it is an under reinforced section.
• The second one have an Asc =402.12 <618.61 mm2. So it is an over reinforced section.
This completes the comparison between the two sections. In the next chapter, we will discuss a new topic: 'Shear in beams'.
PREVIOUS
Cycle 2:
When the NA is in the position xu =212.335, εsc (Using Eq12.1) =0.00271 The corresponding stress fsc is obtained from from the same table for Fe 415 steel:
0.00241 → 342.800
0.00271 → 350.483
0.00276 → 351.800. So we get fsc =350.483 N/mm2.
Similarly, when the NA is in this position, εst (using Eq.3.15) =0.00303. The corresponding stress fst is obtained from the same table for Fe 415 steel:
0.00276 → 351.800
0.00303 → 354.140
0.00380 → 360.900. So we get fst =354.140 N/mm2. With these stress values, we can calculate the forces:
Cu =Cus +Cuc =(fsc -0.447fck)Asc + 0.362fckbxu = (350.483 – 0.447 x 20) x 402.12 + 2172.00 x212.335 = 598533.964 N
Tu =fstAst =354.140 x1722.38 = 609963.349 N
These forces are not equal. So we don't have an equilibrium. What is the value of xu if these were indeed the stresses? We can calculate this by writing the equality: Cu = Tu. Thus we write:
(350.483 – 0.447 x 20) x 402.12 + 2172.00 xu = 609963.349
From this we get = xu =217.598 mm. We will take the average of this xu and the value of xu used in this cycle. It is equal to 0.5 x(217.598 +212.335) =214.966 mm. This can be used as the trial value for the next cycle.
Cycle 3:
When the NA is in the position xu =214.966, εsc (Using Eq12.1) =0.00272
. The corresponding stress fsc is obtained from from the same table for Fe 415 steel:
0.00241 → 342.800
0.00272 → 350.732
0.00276 → 351.800. So we get fsc =350.483 N/mm2.
Similarly, when the NA is in this position, εst (using Eq.3.15) =0.00295. The corresponding stress fst is obtained from the same table for Fe 415 steel:
0.00276 → 351.800
0.00295 → 353.441
0.00380 → 360.900. So we get fst =353.441 N/mm2. With these stress values, we can calculate the forces:
Cu =Cus +Cuc =(fsc -0.447fck)Asc + 0.362fckbxu = (350.732 – 0.447 x 20) x 402.12 + 2172.00 x214.966 = 604348.790 N
Tu =fstAst =353.441 x1722.38 = 608759.311 N
These forces are not equal. So we don't have an equilibrium. What is the value of xu if these were indeed the stresses? We can calculate this by writing the equality: Cu = Tu. Thus we write:
(350.732 – 0.447 x 20) x 402.12 + 2172.00 xu = 608759.311
From this we get = xu =216.997 mm. We will take the average of this xu and the value of xu used in this cycle. It is equal to 0.5 x(216.997 +214.966) =215.982 mm. This can be used as the trial value for the next cycle.
Cycle 4:
When the NA is in the position xu =215.982, εsc (Using Eq12.1) =0.00272
. The corresponding stress fsc is obtained from from the same table for Fe 415 steel:
0.00241 → 342.800
0.00272 → 350.827
0.00276 → 351.800. So we get fsc =350.827 N/mm2.
Similarly, when the NA is in this position, εst (using Eq.3.15) =0.00292. The corresponding stress fst is obtained from the same table for Fe 415 steel:
0.00276 → 351.800
0.00292 → 353.176
0.00380 → 360.900. So we get fst =353.176 N/mm2. With these stress values, we can calculate the forces:
Cu =Cus +Cuc =(fsc -0.447fck)Asc + 0.362fckbxu = (350.827 – 0.447 x 20) x 402.12 + 2172.00 x215.982 = 606592.039 N
Tu =fstAst =353.441 x1722.38 = 608302.526 N
These forces are approximately equal. So we have an equilibrium. The final values can be taken as:
• fsc =350.827 N/mm2 • fst =353.441 N/mm2 • xu =215.982 mm
[It may be noted that fig.12.5 is drawn only for getting a good understanding about the extreme positions of the NA. It is not necessary to draw such a fig. in an analysis problem.]
Now we can obtain MuR
MuR = 0.362fckbxu(d - 0.416xu) + ( fsc - 0.447fck) Asc(d-d')
= 143618673.976 + 47853116.410 =191471790.385 Nmm =191.5 kNm
Also xu = 215.982 mm. xu,max = 0.4791 x 396 =189.724 mm. So we get xu > xu,max. So this is an over reinforced section.
Thus we complete the analysis process.
Comparison between 'Under reinforced' and 'Over reinforced' Doubly reinforced sections.
Let us compare the sections in the last two solved examples:
• Solved example 2:
Given b =300 mm, d =396 mm, d' =50 mm, Ast =1722.38 mm2, Asc =628.32 mm2, fy =415 N/mm2, fck =20 N/mm2, xu =188.527 mm, xu,max =189.72 mm, MuR =203.53 kNm, xu < xu,max, Under reinforced.
• Solved example 3:
b =300 mm, d =396 mm, d' =48 mm, Ast =1722.38 mm2, Asc =402.12 mm2, fy =415 N/mm2, fck =20 N/mm2, xu =215.982 mm, xu,max =189.72 mm, MuR =191.5 kNm. xu > xu,max, Over reinforced.
We can see that all the properties except Asc are the same. The second section have a lesser Asc than the first. The second section is Over reinforced, while the first section is under reinforced. Let us now discuss the reason for this.
Mulim of both the sections will be the same. It is given by:
Thus Mulim =130.66 kNm
Ast,lim of both the sections will be the same. It is given by
Thus Ast,lim = 1141.67 mm2.
As Ast is same for both sections, ΔAst will also be the same (Eq.11.8):
Ast = Ast,lim + ΔAst
Thus we get ΔAst = 1722.38 -1141.67 =580.71 mm2.
If the section under consideration is an under reinforced one, then all the Ast would have yielded at the ultimate state. So the stress in it = 0.87fy, and the force ΔTu = 0.87fyAst =209666.068 N. The quantity of compression steel Asc must be such that, the force developed in it should balance ΔTu.
So, next we find the stress fsc in Asc: For this we can use Table 11.1.
d'/d = 48/396 =0.121.
0.100 → 342.90
0.121 → 347.87
0.150 → 342.40. So we get fsc =347.87 N/mm2.
So the force in Asc =( fsc - 0.447fck) Asc = (347.87 -0.447 x20)Asc
[It may be noted that Table.11.1 can be used to find fsc only in the condition when tension steel has yielded. If this steel has not yielded at the ultimate state, we have to use strain compatibility to find fsc]
This must be equal to the force in ΔTu. So we can write 209666.068 = (347.87 -0.447 x20)Asc. From this we get Asc = 618.61 mm2. This is the 'ideal' quantity of Asc. If we provide this exact quantity, xu will be equal to xu,max. But as explained based on fig.11.4 , this exact quantity cannot be provided. The actual quantity that we give should be greater than 618.61 mm2. When we give this higher quantity, there will be 'more material' towards the top side of the beam, and so the NA will move up wards. Thus xu will decrease from xu,max, and so, it will be an under reinforced section. If Asc is less than 618.61, then the NA will move downwards, and it will be an over reinforced section.
In the two solved examples,
• The first one have an Asc =628.32 >618.61 mm2. So it is an under reinforced section.
• The second one have an Asc =402.12 <618.61 mm2. So it is an over reinforced section.
This completes the comparison between the two sections. In the next chapter, we will discuss a new topic: 'Shear in beams'.
PREVIOUS
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