Chapter 12 (cont..4) - Comparison of sections

In the previous section, we completed the 1^st cycle in the analysis of the beam section. We obtained the x_u to be used in the next cycle as x_u =212.335 mm. Now we will continue the iteration process:

Cycle 2:
When the NA is in the position x_u =212.335, ε_sc (Using Eq12.1) =0.00271 The corresponding stress f_sc 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 f_sc =350.483 N/mm².

Similarly, when the NA is in this position, ε_st (using Eq.3.15) =0.00303. The corresponding stress f_st 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 f_st =354.140 N/mm². With these stress values, we can calculate the forces:

C_u =C_us +C_uc =(f_sc -0.447f_ck)A_sc + 0.362f_ckbx_u = (350.483 – 0.447 x 20) x 402.12 + 2172.00 x212.335  = 598533.964 N

T_u =f_stA_st =354.140 x1722.38 = 609963.349 N

These forces are not equal. So we don't have an equilibrium. What is the value of x_u if these were indeed the stresses? We can calculate this by writing the equality: C_u = T_u. Thus we write:

(350.483 – 0.447 x 20) x 402.12 + 2172.00 x_u = 609963.349
From this we get = x_u =217.598 mm. We will take the average of this x_u and the value of x_u 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 x_u =214.966, ε_sc (Using Eq12.1) =0.00272
. The corresponding stress f_sc 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 f_sc =350.483 N/mm².

Similarly, when the NA is in this position, ε_st (using Eq.3.15) =0.00295. The corresponding stress f_st 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 f_st =353.441 N/mm². With these stress values, we can calculate the forces:

C_u =C_us +C_uc =(f_sc -0.447f_ck)A_sc + 0.362f_ckbx_u = (350.732 – 0.447 x 20) x 402.12 + 2172.00 x214.966  = 604348.790 N

T_u =f_stA_st =353.441 x1722.38 = 608759.311 N

These forces are not equal. So we don't have an equilibrium. What is the value of x_u if these were indeed the stresses? We can calculate this by writing the equality: C_u = T_u. Thus we write:

(350.732 – 0.447 x 20) x 402.12 + 2172.00 x_u = 608759.311
From this we get = x_u =216.997 mm. We will take the average of this x_u and the value of x_u 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 x_u =215.982, ε_sc (Using Eq12.1) =0.00272
. The corresponding stress f_sc 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 f_sc =350.827 N/mm².

Similarly, when the NA is in this position, ε_st (using Eq.3.15) =0.00292. The corresponding stress f_st 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 f_st =353.176 N/mm². With these stress values, we can calculate the forces:

C_u =C_us +C_uc =(f_sc -0.447f_ck)A_sc + 0.362f_ckbx_u = (350.827 – 0.447 x 20) x 402.12 + 2172.00 x215.982  = 606592.039 N

T_u =f_stA_st =353.441 x1722.38 = 608302.526 N

These forces are approximately equal. So we have an equilibrium. The final values can be taken as:
• f_sc =350.827 N/mm²  • f_st =353.441 N/mm² • x_u =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 M_uR 
M_uR = 0.362f_ckbx_u(d - 0.416x_u) + ( f_sc - 0.447f_ck) A_sc(d-d')
= 143618673.976 + 47853116.410 =191471790.385 Nmm  =191.5 kNm

Also x_u = 215.982 mm. x_u,max = 0.4791 x 396 =189.724 mm. So we get x_u > x_u,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, A_st =1722.38 mm², A_sc =628.32 mm², f_y =415 N/mm², f_ck =20 N/mm², x_u =188.527 mm, x_u,max =189.72 mm, M_uR =203.53 kNm, x_u < x_u,max, Under reinforced.

• Solved example 3:
b =300 mm, d =396 mm, d' =48 mm, A_st =1722.38 mm², A_sc =402.12 mm², f_y =415 N/mm², f_ck =20 N/mm², x_u =215.982 mm, x_u,max =189.72 mm, M_uR =191.5 kNm. x_u > x_u,max, Over reinforced.

We can see that all the properties except A_sc are the same. The second section have a lesser A_sc than the first. The second section is Over reinforced, while the first section is under reinforced. Let us now discuss the reason for this.

M_ulim of both the sections will be the same. It is given by:

Thus M_ulim =130.66 kNm 

A_st,lim of both the sections will be the same. It is given by 

Thus A_st,lim = 1141.67 mm².
As A_st is same for both sections, ΔA_st will also be the same (Eq.11.8):
A_st = A_st,lim + ΔA_st
Thus we get ΔA_st = 1722.38 -1141.67 =580.71 mm².

If the section under consideration is an under reinforced one, then all the A_st would have yielded at the ultimate state. So the stress in it = 0.87f_y, and the force ΔT_u = 0.87f_yA_st =209666.068 N. The quantity of compression steel A_sc must be such that, the force developed in it should balance ΔT_u. 
So, next we find the stress f_sc in A_sc: 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 f_sc =347.87 N/mm².

So the force in A_sc =( f_sc - 0.447f_ck) A_sc = (347.87 -0.447 x20)A_sc

[It may be noted that Table.11.1 can be used to find f_sc 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 f_sc]

This must be equal to the force in ΔT_u. So we can write 209666.068 = (347.87 -0.447 x20)A_sc. From this we get A_sc = 618.61 mm². This is the 'ideal' quantity of A_sc. If we provide this exact quantity, x_u will be equal to x_u,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 x_u will decrease from x_u,max, and so, it will be an under reinforced section. If A_sc 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 A_sc =628.32 >618.61 mm². So it is an under reinforced section.
• The second one have an A_sc =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'.

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Chapter 12 - Analysis of Doubly reinforced beams

In the previous section, we completed the discussion on design of a doubly reinforced beam section. In this section we will discuss the analysis. When we are given a doubly reinforced beam section for analysis, we have to find the depth of the neutral axis x_u , and the ultimate moment of resistance M_uR.

We know that when we analyse the section using the Limit State Method, we are calculating the forces in the section at the point of impending failure. The concrete at the extreme compression fibre will be about to crush. And also, if it is an under reinforced section, the steel in the tension zone would have yielded.

So now we are going to find the forces in a doubly reinforced section, at the above mentioned point of impending failure. We can base our discussions on the fig.12.1 given below:

Fig.12.1
Stresses, strains and forces

The above fig. shows all the stresses and forces at the ultimate state. We will now see the details of each component:

The depth of NA is denoted as x_u. It's value can be less than or greater than x_u,max, depending on the quantities of steel. It is one of our aims to calculate this x_u of the given beam section.

We have derived the expression for ε_sc in the previous chapter. We used the triangles above the NA in the strain diagram. We have the same strain diagram here also. The only difference is that x_u,max is now changed to x_u. So from Eq.11.1, we get:

Eq.12.1:

When we know ε_sc, we can calculate the stress f_sc in A_sc. But here we cannot use the table 11.1 because, the values in it correspond to the condition of x_u = x_u,max. Here we are having x_u, which may be having a value higher or lower than x_u,max. So we have to use the original tables for calculating f_sc.

The net compressive force in steel is obtained as:
C_us = (f_sc - 0.447f_ck) A_sc
The compressive force in concrete C_uc = 0.362f_ckbx_u 

So the total compressive force is given by:
C_u = 0.362f_ckbx_u  + (f_sc - 0.447f_ck) A_sc

The total tensile force is given by T_u = f_st A_st
equating the above two, we get 

0.362f_ckbx_u  + (f_sc - 0.447f_ck) A_sc = f_st A_st

From this we get
Eq.12.2

But from Eq.12.1, we can see that f_sc depends on x_u. So the above Eq.12.2 will not give a closed form solution. We have to use strain compatibility method.

Another point to note is that, f_st also depend on x_u. If the tension steel has not yielded at ultimate state, we cannot take f_st as equal to 0.87f_y. So for such sections also, we will have to use strain compatibility method. We will discuss the details of the method when we do a solved example.

It may be noted that the above Eq.12.2 will give a closed form solution if both the tension steel and compression steel have yielded. Because, then f_sc and f_st will be having a constant value of 0.87f_y 

Once we determine x_u, we can calculate M_uR, the ultimate moment of resistance of the doubly reinforced section. For this we will use the compressive forces. We have two compressive forces:

• C_u = 0.362f_ckbx_u. Lever arm of this force = d - 0.416x_u. So the contribution from this force = 0.362f_ckbx_u (d - 0.416x_u).
• C_us = (f_sc - 0.447f_ck) A_sc.   Lever arm of this force = d-d'. So the contribution from this force = (f_sc - 0.447f_ck) A_sc (d-d') 

Sum of the above contributions will be the M_uR. So we get:
Eq.12.3: M_uR = 0.362f_ckbx_u(d - 0.416x_u) + ( f_sc - 0.447f_ck) A_sc(d-d')

Under reinforced or Over reinforced Doubly reinforced sections

In the case of singly reinforced sections, we have seen the details of 'balanced sections'. We discussed it based on the fig.3.26. This fig. is shown again below:

Fig.3.26
Stresses and strains in singly reinforced balanced sections

From the strain diagram in this fig., we saw that x_u,Bal, (which is same as x_u,max) is the particular depth of NA at which the strain in tension steel is equal to ε*_st, the yield strain. If in a beam section, x_u is less than x_u,max, the strain ε_st will be greater than ε*_st. Then it is an under reinforced section. If x_u is greater than x_u,max, ε_st will be less than ε*_st. That means the steel will not yield at ultimate state, and it is an over reinforced section.

We can extend the above discussion to Doubly reinforced sections also. As the plane sections remain plane after bending (Details here), the strain diagram for a doubly reinforced section will also be a straight line similar the strain diagram in the above fig. Thus the ε_st of 'tension steel' in a doubly reinforced section will be greater than ε*_st if x_u is less than x_u,max. It will be less than ε*_st if x_u is greater than x_u,max. 

So we can say that while designing a doubly reinforced section, just as in the case of a singly reinforced section, we must ensure that x_u is less than x_u,max. So that, if over loading occurs, and the beam fails, the failure will be a tension failure.

In the next three sections we will see analysis examples based on the above discussions. Links to these sections are given below.

• Solved example 12.1  • Solved example 12.2  • Solved example 12.3 (over reinforced)  

After the above solved examples, a comparison between 'Under reinforced and Over reinforced' doubly reinforced beam sections is given here. 


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