In the previous section, we analysed a doubly reinforced beam section. Now we will analyse a doubly reinforced beam section which is 'over reinforced'. We analyse the same section which we saw in the previous example. With a minor difference: The compression steel now consists of 2-#16 instead of 2-#20.
Solved example 12.3
Determine the Ultimate moment of resistance MuR of the beam section shown below. Given 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
Solution:
So we are going to analyse the given beam section to find the stresses, strains, forces and depth of NA at the ultimate state. Our main aim is to determine the MuR. We will see the various steps involved in the analysis in detail:
Initially, we assume that 'both the tension steel and the compression steel will yield at the ultimate state'.
We will now check if this really happens at the ultimate state. For making this check, first we find xu the depth of NA. We have assumed that the tension steel has yielded. So the stress in tension steel =0.87fy. Thus the force in tension steel =Tu =0.87fyAst = 0.87 x 415 x 1722.38 =621865.299 N
We have assumed that the compression steel has also yielded. So the stress in compression steel =fsc =0.87fy. Thus the force in compression steel =Cus = (fsc -0.447fck)Asc = (0.87fy – 0.447 x 20) x 402.12 =141590.473 N
Force in concrete = Cuc = 0.362fckbxu = 2172.00xu
Equating the tensile and compressive forces we get: 141590.473 + 2172.00xu = 621865.299 N - - - (1)
So we get xu = 221.121 mm - - - (2)
We have learned the method for calculating (from the strain diagram) the strain εsc at the level of Asc. We use the two similar triangles above the NA. The equation is Eq.12.1
From this we get εsc =0.00274 - - - (3)
Similarly, we know how to calculate the strain εst from the strain diagram. The equation is Eq.3.14 :
Rearranging this, we get:
Note that we are using the equation for εst,u because we have assumed that the tension steel has yielded. Using this equation we get εst = 0.00277.
Now we compare the values with the yield strain ε*st. We know that, for Fe415 steel, ε*st = 0.00380. Comparing the values, we find that εsc < ε*st and εst < ε*st.
[There is another method to prove that εst is less than ε*st: We know that, for Doubly reinforced sections also, if xu ≤ xu,max, the tension steel would have yielded. In our case xu,max =0.4791 x 396 =189.724 mm. In the above calculations, we have used xu =221.121 which is greater than xu,max. So for this value of xu, the steel would not yield, and so, εst will indeed be less than ε*st.]
So, if at the ultimate state, the NA is at the position xu = 221.12 mm, the tension steel and compression steel would never yield. This means that:
• The stress in Ast is not 0.87fy
• The stress in Asc is not 0.87fy
So the equality in (1) is not valid. Thus the value of xu = 221.121 mm is not correct.
As xu > xu,max, it is an over reinforced section. We have to use iteration method using strain compatibility. We want an improved value of xu. We will try to obtain the new value, based on the value that we are having.
Fig.12.5
Positions of NA
In the fig above, 'Line p' shows the position of the NA when xu = xu,max =189.724 mm. The NA of our beam section will never occupy this position because, it is an 'Over reinforced section'. The other 'Line q' shows the position when the tension steel has yielded. This position will also never occur because an over reinforced section will fail (by compression of concrete in the extreme fibre) before the yielding of it's tension steel occurs. The actual NA lies some where between these two lines. So we can take the average of these two values as the initial 'trial value'. Thus xu for the 1st cycle = 0.5 x(189.724 +221.121) =205.422 mm.
Cycle 1:
When the NA is in the position xu =205.422, εsc (Using Eq12.1) =0.00268. The corresponding stress fsc is obtained from from the table for Fe 415 steel:
0.00241 → 342.800
0.00268 → 349.799
0.00276 → 351.800. So we get fsc =349.799 N/mm2.
Similarly, when the NA is in this position, εst is obtained using Eq.3.15 :
Rearranging this we get:
[The equation involving xu,o and εst,o is used here because the section is over reinforced. But it may be noted that Eq.3.14 and 3.15 are the same with only the subscripts changed from 'u' to 'o'.]
So we will get εst =0.00325. The corresponding stress fst is obtained from the same table for Fe 415 steel:
0.00276 → 351.800
0.00325 → 356.062
0.00380 → 360.900. So we get fst =356.062 N/mm2. With these stress values, we can calculate the forces:
Cu =Cus +Cuc =(fsc -0.447fck)Asc + 0.362fckbxu = (349.799 – 0.447 x 20) x 402.12 + 2172.00 x205.422 = 583243.352 N
Tu =fstAst =356.062 x1722.38 = 613273.929 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:
(349.799 – 0.447 x 20) x 402.12 + 2172.00 xu = 613273.929.
From this we get = xu =219.249 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(205.422 +219.249) =212.335 mm. This can be used as the trial value for the next cycle.
So we have completed one cycle. We will see the rest of the cycles in the next section.
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Solved example 12.3
Determine the Ultimate moment of resistance MuR of the beam section shown below. Given 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
So we are going to analyse the given beam section to find the stresses, strains, forces and depth of NA at the ultimate state. Our main aim is to determine the MuR. We will see the various steps involved in the analysis in detail:
Initially, we assume that 'both the tension steel and the compression steel will yield at the ultimate state'.
We will now check if this really happens at the ultimate state. For making this check, first we find xu the depth of NA. We have assumed that the tension steel has yielded. So the stress in tension steel =0.87fy. Thus the force in tension steel =Tu =0.87fyAst = 0.87 x 415 x 1722.38 =621865.299 N
We have assumed that the compression steel has also yielded. So the stress in compression steel =fsc =0.87fy. Thus the force in compression steel =Cus = (fsc -0.447fck)Asc = (0.87fy – 0.447 x 20) x 402.12 =141590.473 N
Force in concrete = Cuc = 0.362fckbxu = 2172.00xu
Equating the tensile and compressive forces we get: 141590.473 + 2172.00xu = 621865.299 N - - - (1)
So we get xu = 221.121 mm - - - (2)
We have learned the method for calculating (from the strain diagram) the strain εsc at the level of Asc. We use the two similar triangles above the NA. The equation is Eq.12.1
Similarly, we know how to calculate the strain εst from the strain diagram. The equation is Eq.3.14 :
Rearranging this, we get:
Note that we are using the equation for εst,u because we have assumed that the tension steel has yielded. Using this equation we get εst = 0.00277.
Now we compare the values with the yield strain ε*st. We know that, for Fe415 steel, ε*st = 0.00380. Comparing the values, we find that εsc < ε*st and εst < ε*st.
[There is another method to prove that εst is less than ε*st: We know that, for Doubly reinforced sections also, if xu ≤ xu,max, the tension steel would have yielded. In our case xu,max =0.4791 x 396 =189.724 mm. In the above calculations, we have used xu =221.121 which is greater than xu,max. So for this value of xu, the steel would not yield, and so, εst will indeed be less than ε*st.]
So, if at the ultimate state, the NA is at the position xu = 221.12 mm, the tension steel and compression steel would never yield. This means that:
• The stress in Ast is not 0.87fy
• The stress in Asc is not 0.87fy
So the equality in (1) is not valid. Thus the value of xu = 221.121 mm is not correct.
As xu > xu,max, it is an over reinforced section. We have to use iteration method using strain compatibility. We want an improved value of xu. We will try to obtain the new value, based on the value that we are having.
Fig.12.5
Positions of NA
In the fig above, 'Line p' shows the position of the NA when xu = xu,max =189.724 mm. The NA of our beam section will never occupy this position because, it is an 'Over reinforced section'. The other 'Line q' shows the position when the tension steel has yielded. This position will also never occur because an over reinforced section will fail (by compression of concrete in the extreme fibre) before the yielding of it's tension steel occurs. The actual NA lies some where between these two lines. So we can take the average of these two values as the initial 'trial value'. Thus xu for the 1st cycle = 0.5 x(189.724 +221.121) =205.422 mm.
Cycle 1:
When the NA is in the position xu =205.422, εsc (Using Eq12.1) =0.00268. The corresponding stress fsc is obtained from from the table for Fe 415 steel:
0.00241 → 342.800
0.00268 → 349.799
0.00276 → 351.800. So we get fsc =349.799 N/mm2.
Similarly, when the NA is in this position, εst is obtained using Eq.3.15 :
[The equation involving xu,o and εst,o is used here because the section is over reinforced. But it may be noted that Eq.3.14 and 3.15 are the same with only the subscripts changed from 'u' to 'o'.]
So we will get εst =0.00325. The corresponding stress fst is obtained from the same table for Fe 415 steel:
0.00276 → 351.800
0.00325 → 356.062
0.00380 → 360.900. So we get fst =356.062 N/mm2. With these stress values, we can calculate the forces:
Cu =Cus +Cuc =(fsc -0.447fck)Asc + 0.362fckbxu = (349.799 – 0.447 x 20) x 402.12 + 2172.00 x205.422 = 583243.352 N
Tu =fstAst =356.062 x1722.38 = 613273.929 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:
(349.799 – 0.447 x 20) x 402.12 + 2172.00 xu = 613273.929.
From this we get = xu =219.249 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(205.422 +219.249) =212.335 mm. This can be used as the trial value for the next cycle.
So we have completed one cycle. We will see the rest of the cycles in the next section.
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