In the previous section, we discussed the basic principles of analysis. Now we will see a solved example. We will analyse the beam section which we designed in the previous chapter.
Solved example 12.1
Determine the Ultimate moment of resistance MuR of the beam section shown below. Given b =230 mm, d =347 mm, d' =48 mm, Ast =1472.62 mm2, Asc =603.19 mm2, fy =415 N/mm2, fck =25 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 1847 =531689.45 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 25) x 603.19 =211041.101 N
Force in concrete = Cuc = 0.362fckbxu = 2081.5xu
Equating the tensile and compressive forces we get: 211041.101 + 2081.5xu = 531689.45 - - - (1)
So we get xu = 154.047 mm - - - (2)
Let us draw the beam section and the strain diagram based on this value of xu.
Fig.12.2
Calculation of strain in compression steel
We have learned the method for calculating 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 =0.00241 - - - (3)
Similarly, we know how to calculate the strain εst from the above 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 = 0.00438.
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 εst > ε*st.
[There is another method to prove that εst is greater 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 347 =166.248 mm. In the above calculations, we have used xu =154.047 which is less than xu,max. So for this value of xu, the steel would have yielded and so, εst will indeed be greater than ε*st.]
But εsc < ε*st. This means that if at the ultimate state, the position of NA is indeed at the position shown in fig. above, the tension steel will yield as we have assumed. But the compression steel will never yield for that position of NA. This means that:
The stress in Asc is not 0.87fy
So the calculated value of 211041.101 N for Cus is not correct, and the equality in (1) is not valid. Thus the value of xu = 154.047 mm is not correct.
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.
The value that we are having for xu is 154.047 mm from (2) . εsc corresponding to this xu is 0.00241. from (3). Now we want the fsc corresponding to this εsc. This we will get from the table for Fe 415 steel:
0.00241 → 342.80
0.00276 → 351.80
So fsc = 342.80 N/mm2.
Thus, if xu = 154.047 mm, fsc = 342.80 N/mm2.
We will not get equilibrium with these values because if there is to be equilibrium with xu =154.047, fsc must be equal to 0.87fy =361.08 N/mm2. But we have only 342.80 N/mm2 for fsc. So what should be the value of xu if fsc is 342.80 N/mm2 ? For finding that, we do the 1st cycle:
Cycle 1:
we can write a new equality:
(342.80 – 0.447 x 25) x 603.19 + 2081.5xu = 531689.451 N - - - (4).
From this we get xu = 159.335 mm. This is the value of xu which is needed for equilibrium if the stress fsc = 342.8 N/mm2.
If in the fig.12.2 above, xu is equal to 159.335, we will get εsc = 0.00245. Now use the table for Fe415:
0.00241 → 342.80
0.00245 → 343.83
0.00276 → 351.80
So fsc = 343.83 N/mm2
If xu is 159.335 and fsc =342.8 N/mm2, we would have obtained equilibrium. But unfortunately, for our beam section, when xu = 159.335, fsc is never equal to 342.8 N/mm2. It is 343.83 N/mm2. So we don't have an equilibrium. We need an improved value for xu.
Cycle 2:
So what should be the value of xu if fsc is 343.83 N/mm2 ?
For this, we will write a new equality:
(343.83 – 0.447 x 25) x 603.19 + 2081.5 xu = 531689.451 N
From this we get xu = 159.037 mm. This is the value of xu which is needed for equilibrium if the stress in fsc = 343.83 N/mm2.
If in the fig.12.2 above, xu is equal to 159.037, we will get εsc = 0.00244. Now use the table for Fe415:
0.00241 → 342.80
0.00244 → 343.67
0.00276 → 351.80
So fsc = 343.67 N/mm2.
If xu is 159.037 and fsc is 343.83 N/mm2, we would have obtained equilibrium. But unfortunately, for our beam section, when xu = 159.037, fsc is never equal to 343.83 N/mm2. It is 343.67 N/mm2. So we don't have an equilibrium. We need an improved value for xu.
Cycle 3:
So what should be the value of xu if fsc is 343.67 ?
For this, we will write a new equality:
(343.67 – 0.447 x 25) x 603.19 + 2081.5 xu = 531689.451 N - - - (4)
From this we get xu = 159.085 mm. This is the value of xu which is needed for equilibrium if the stress in fsc = 343.67 N/mm2.
If in the fig.12.2 above, xu is equal to 159.085, we will get εsc = 0.00244. Now use the table for Fe415:
0.00241 → 342.80
0.00244 → 343.67
0.00276 → 351.80
[In fact, it is not needed to do the above interpolation because the strain value εsc = 0.00244 is the same as in the previous cycle 2.]
So fsc = 343.67 N/mm2. This time we are lucky. We have xu = 159.085 mm and fsc = 343.67 N/mm2. So we have an equilibrium, and the correct value of xu is 159.085 mm.
[It may be noted that in the above cycles, the stress fst in tension steel is always taken as 0.87fy. This is because in all the cycles, xu is less than xu,max. But the stress fsc in compression steel varies with the position of xu in each cycle.
It may also be noted that fig.12.2 is drawn only for getting a good understanding about the dependence of strain εsc on the 'position xxof NA'. It is not necessary to draw such a fig. in an analysis problem. We just have to remember the Eq.12.1 for calculating εsc.]
The cycles that we did above can be summarized as follows:
• Assume fsc =fst =0.87fy =361.08 N/mm2
• If fst = fsc = 0.87fy, xu = 154.047 mm
• So, if fsc =fst =0.87fy =361.08, AND xu = 154.047, there would be equilibrium
• But when xu = 154.047, fst is indeed equal to 0.87fy =361.08. But fsc = 342.80
• Assumption that fst = 0.87fy is correct
• Assumption that fsc = 0.87fy is wrong
So use strain compatibility for calculate actual fsc
Cycle 1:
• If fsc = 342.80, AND xu = 159.335, there would be equilibrium
• But when xu = 159.335, fsc = 343.83
Cycle 2:
• If fsc = 343.83, AND xu = 159.037, there would be equilibrium
• But when xu = 159.037, fsc = 343.67
Cycle 3:
• If fsc = 343.67, AND xu = 159.085, there would be equilibrium
• When xu = 159.085, fsc is indeed equal to 343.67
• So final value of xu =159.085 mm.
Now we can obtain MuR
MuR = 0.362fckbxu(d - 0.416xu) + ( fsc - 0.447fck) Asc(d-d')
= 92989662.677 + 59966740.056 =152956402.733 Nmm =152.95 kNm
In the design problem, the external factored bending moment Mu = 139.89 kNm. Here we find that MuR = 152.95 kNm. Thus Mu < MuR. Hence OK.
Also xu = 159.085 mm. xu,max = 0.4791 x 347 =166.25 mm. So we get xu < xu,max. Hence OK
In the next section we will see another solved example.
PREVIOUS
Solved example 12.1
Determine the Ultimate moment of resistance MuR of the beam section shown below. Given b =230 mm, d =347 mm, d' =48 mm, Ast =1472.62 mm2, Asc =603.19 mm2, fy =415 N/mm2, fck =25 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 1847 =531689.45 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 25) x 603.19 =211041.101 N
Force in concrete = Cuc = 0.362fckbxu = 2081.5xu
Equating the tensile and compressive forces we get: 211041.101 + 2081.5xu = 531689.45 - - - (1)
So we get xu = 154.047 mm - - - (2)
Let us draw the beam section and the strain diagram based on this value of xu.
Fig.12.2
Calculation of strain in compression steel
We have learned the method for calculating 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 above 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 = 0.00438.
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 εst > ε*st.
[There is another method to prove that εst is greater 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 347 =166.248 mm. In the above calculations, we have used xu =154.047 which is less than xu,max. So for this value of xu, the steel would have yielded and so, εst will indeed be greater than ε*st.]
But εsc < ε*st. This means that if at the ultimate state, the position of NA is indeed at the position shown in fig. above, the tension steel will yield as we have assumed. But the compression steel will never yield for that position of NA. This means that:
The stress in Asc is not 0.87fy
So the calculated value of 211041.101 N for Cus is not correct, and the equality in (1) is not valid. Thus the value of xu = 154.047 mm is not correct.
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.
The value that we are having for xu is 154.047 mm from (2) . εsc corresponding to this xu is 0.00241. from (3). Now we want the fsc corresponding to this εsc. This we will get from the table for Fe 415 steel:
0.00241 → 342.80
0.00276 → 351.80
So fsc = 342.80 N/mm2.
Thus, if xu = 154.047 mm, fsc = 342.80 N/mm2.
We will not get equilibrium with these values because if there is to be equilibrium with xu =154.047, fsc must be equal to 0.87fy =361.08 N/mm2. But we have only 342.80 N/mm2 for fsc. So what should be the value of xu if fsc is 342.80 N/mm2 ? For finding that, we do the 1st cycle:
Cycle 1:
we can write a new equality:
(342.80 – 0.447 x 25) x 603.19 + 2081.5xu = 531689.451 N - - - (4).
From this we get xu = 159.335 mm. This is the value of xu which is needed for equilibrium if the stress fsc = 342.8 N/mm2.
If in the fig.12.2 above, xu is equal to 159.335, we will get εsc = 0.00245. Now use the table for Fe415:
0.00241 → 342.80
0.00245 → 343.83
0.00276 → 351.80
So fsc = 343.83 N/mm2
If xu is 159.335 and fsc =342.8 N/mm2, we would have obtained equilibrium. But unfortunately, for our beam section, when xu = 159.335, fsc is never equal to 342.8 N/mm2. It is 343.83 N/mm2. So we don't have an equilibrium. We need an improved value for xu.
Cycle 2:
So what should be the value of xu if fsc is 343.83 N/mm2 ?
For this, we will write a new equality:
(343.83 – 0.447 x 25) x 603.19 + 2081.5 xu = 531689.451 N
From this we get xu = 159.037 mm. This is the value of xu which is needed for equilibrium if the stress in fsc = 343.83 N/mm2.
If in the fig.12.2 above, xu is equal to 159.037, we will get εsc = 0.00244. Now use the table for Fe415:
0.00241 → 342.80
0.00244 → 343.67
0.00276 → 351.80
So fsc = 343.67 N/mm2.
If xu is 159.037 and fsc is 343.83 N/mm2, we would have obtained equilibrium. But unfortunately, for our beam section, when xu = 159.037, fsc is never equal to 343.83 N/mm2. It is 343.67 N/mm2. So we don't have an equilibrium. We need an improved value for xu.
Cycle 3:
So what should be the value of xu if fsc is 343.67 ?
For this, we will write a new equality:
(343.67 – 0.447 x 25) x 603.19 + 2081.5 xu = 531689.451 N - - - (4)
From this we get xu = 159.085 mm. This is the value of xu which is needed for equilibrium if the stress in fsc = 343.67 N/mm2.
If in the fig.12.2 above, xu is equal to 159.085, we will get εsc = 0.00244. Now use the table for Fe415:
0.00241 → 342.80
0.00244 → 343.67
0.00276 → 351.80
[In fact, it is not needed to do the above interpolation because the strain value εsc = 0.00244 is the same as in the previous cycle 2.]
So fsc = 343.67 N/mm2. This time we are lucky. We have xu = 159.085 mm and fsc = 343.67 N/mm2. So we have an equilibrium, and the correct value of xu is 159.085 mm.
[It may be noted that in the above cycles, the stress fst in tension steel is always taken as 0.87fy. This is because in all the cycles, xu is less than xu,max. But the stress fsc in compression steel varies with the position of xu in each cycle.
It may also be noted that fig.12.2 is drawn only for getting a good understanding about the dependence of strain εsc on the 'position xxof NA'. It is not necessary to draw such a fig. in an analysis problem. We just have to remember the Eq.12.1 for calculating εsc.]
The cycles that we did above can be summarized as follows:
• Assume fsc =fst =0.87fy =361.08 N/mm2
• If fst = fsc = 0.87fy, xu = 154.047 mm
• So, if fsc =fst =0.87fy =361.08, AND xu = 154.047, there would be equilibrium
• But when xu = 154.047, fst is indeed equal to 0.87fy =361.08. But fsc = 342.80
• Assumption that fst = 0.87fy is correct
• Assumption that fsc = 0.87fy is wrong
So use strain compatibility for calculate actual fsc
Cycle 1:
• If fsc = 342.80, AND xu = 159.335, there would be equilibrium
• But when xu = 159.335, fsc = 343.83
Cycle 2:
• If fsc = 343.83, AND xu = 159.037, there would be equilibrium
• But when xu = 159.037, fsc = 343.67
Cycle 3:
• If fsc = 343.67, AND xu = 159.085, there would be equilibrium
• When xu = 159.085, fsc is indeed equal to 343.67
• So final value of xu =159.085 mm.
Now we can obtain MuR
MuR = 0.362fckbxu(d - 0.416xu) + ( fsc - 0.447fck) Asc(d-d')
= 92989662.677 + 59966740.056 =152956402.733 Nmm =152.95 kNm
In the design problem, the external factored bending moment Mu = 139.89 kNm. Here we find that MuR = 152.95 kNm. Thus Mu < MuR. Hence OK.
Also xu = 159.085 mm. xu,max = 0.4791 x 347 =166.25 mm. So we get xu < xu,max. Hence OK
In the next section we will see another solved example.
PREVIOUS
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