Chapter 11 (cont..3) Solved example on doubly reinforced beam

In the previous section, we saw the steps in the design of a doubly reinforced beam. We also saw the checks that have to be performed. Now we will see a solved example demonstrating the design process.

Solved example 11.1
A rectangular reinforced concrete beam, located inside a building, is simply supported on two masonry walls 300 mm thick. The clear span between the walls is 4000 mm. The depth and width of the beam are fixed at 400 mm and 230 mm respectively. The beam has to carry a service load of 40.35 kN /m, including it's own weight. Design the beam section, for maximum bending moment at midspan. Assume Fe415 steel and 'moderate' exposure conditions.

Solution:

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Concrete cover and Concrete grade:
Concrete cover for bars of beam
The nominal cover C_c required for the main reinforcing bars can be obtained from cl 26.4.1 and 26.4.2 of the code. The exposure condition is given as 'Moderate'. Based on this, the nominal cover to be provided can be taken as 30 mm. This will also satisfy the requirements for normal fire resistance (cl 26.4.3 and Table 16 of the code) 

The grade of concrete to be used for 'moderate' exposure condition is M25

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Effective depth and Effective span:
Assume 20 mm main bars and 10 mm stirrups. So d = D -(30 +10 +10) = 400 -50 =350 mm.

Effective span = lesser of the following:
Clear span + effective depth = 4000 +350 =4350 mm
c/c distance between supports = 4000 +300 =4300 mm

So effective span l =4300 mm

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Cover for compression steel :
Assume 16 mm bars for compression steel. So d' = 30 +10 +8 =48 mm

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Factored Bending moment:
Service load w per meter length of beam =40.35 kN/m
So factored BM M_u = (1.5wl²)/8 = 139.89 kNm

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Now we can begin the design process. We have the following data:

f_ck =25 N/mm², f_y =415 N/mm², Total depth D =400 mm, Assumed effective depth d =349 mm, d' =46 mm. width =230 mm, M_u = 139.89 kNm

Step 1: Calculate M_ulim. (Eq.3.26)

For Fe415 steel, xumax/d = 0.4791
So we get M_ulim =97041767.93 Nmm =97.815 kNm

Step 2: Calculate A_st,lim. (Eq.3.29)

So we get A_st,lim =966.726 mm²

Step 3: Calculate ΔM_u from the relation: ΔM_u = M_u - M_ulim       
So we get ΔM_u =139.89 -97.815 =42.07 kNm

Step 4: Calculate  ΔA_st using Eq.11.7

So we get ΔA_st = 385.86 mm²
Then A_st = A_st,lim +  ΔA_st = 1352.59 mm²

Step 5: Provide 3 - #25, giving an area of 1472.622 mm²

Step 6: Actual effective depth with 10 mm dia. stirrups = 400 -(30 +10 +12.5) =400 -52.5 =347.5 mm So we can provide d =347 mm. Thus the effective depth has changed by 3 mm. We have to do the steps again:
• M_ulim =95.385 kNm
• A_st,lim =958.440 mm²
• M_u - M_ulim =44.505 kNm
• ΔA_st =412.24 mm²
• A_st = A_st,lim +  ΔA_st =1370.68 mm² - - - (1)
This is the area of tension steel required. So 3-#25 is adequate.

Step 7: Area of steel actually provided = 1472.622 mm²

Step 8: ΔA_st(pd) = 1472.622 - A_st,lim =514.182 mm²

Step 9: d'/d = 48/347 =0.138
This falls between 0.1 and 0.15. So from Table 11.1 we get:
0.100  → 351.88
0.138  → 344.76
0.150  → 342.59
By linear interpolation, we get f_sc =344.76 N/mm²
Step 10: Calculate A_sc using Eq.11.9

So we get A_sc =556.52 mm²
Provide 3-#16 giving an area of 603.186 mm²

The designed beam section is shown in the fig. below:

Fig.11.5
Final beam section

Now we can do the checks:

1. D/b ratio (4.2) = 400/230 =1.74 
Falls between 1.5 and 2. Hence OK

2. Check for deep beam: (4.3)
l/D = 4300/400 = 18.7. Not less than 2. Hence OK

3. Check for slenderness: (4.4)
60b = 60 x 230 = 13800
250b²/d = (250 x 230²)/347 = 38112.4
Lesser of the above = 13800
Clear distance between lateral restraints = 4000 mm
4000 < 13800. Hence OK

4. Spacing of bars:
Minimum clear space required:

For this, we refer cl.26.3 of the code. Also see Eq.4.13

Sh = [b-(2Cc +2Φl +3Φ)] /2

Where b =230 mm, C_c =30 mm, Φl =10 mm, and Φ =25 mm.
So we get Sh = 37.5 mm

Larger bar diameter =25 mm
Max. size of aggregate + 5 mm = 25 mm
37.5 > 25 mm. Hence OK 


Maximum clear space allowable:

● For this, we refer cl.26.3.3 of the code
● From table 15 of the code, (row for Fe 415 steel and column for zero percent redistribution) we get maximum allowable clear distance as 180 mm
37.5 < 180 mm. Hence OK

4. Check for the Minimum area of flexural reinforcement

For this we refer cl.26.5.1.1 of the code. Also see here

So we get A_s = 163.47 mm².
Steel provided = 1472.62 mm² (3-#25) > 163.47 mm². Hence OK

5 (a). Check for maximum area of tension steel
For this we refer cl.26.5.1.b of the code.  Also see here
0.04bd = 0.04 x230 x347 =3192.4 mm² > 1472.62 mm². Hence OK
(b) Check for maximum area of compression steel 
For this we refer cl.26.5.1.2 of the code. Also see here
0.04bd = 0.04 x230 x347 =3192.4 mm² > 603.186 mm². Hence OK

6. Check for deflection control:
For this we refer cl.23.2.1 of the code.  Also see here
For members with span upto 10m, having both tension and compression reinforcements:

(l/d)_actual  ≤  [(l/d)_basic] k_t k_c

For simply supported members, l/d basic = 20. We have to calculate k_t and k_c. First we will calculate k_t:

To find k_t:
• A_st required = 1,370.68  mm² from (1)  • A_st,p = 1472.622 mm²
• f_st = 218.04 N/mm²  • p_t = 1.8452  • From Fig.4 of the code, k_t = 0.9

To find k_c:

We can use the expression given in SP24

 

 Where p_c = 100Asc/bd = (100 x603.186)/(230 x347) =0.7558
So we get k_c = 1.17
• So l/d = 20 x k_t x k_c = 20 x0.9 x1.17 =21.06  
• (l/d) actual = 4300/347 = 12.39
12.39 < 21.06 Hence OK

This completes the design procedure. We must analyse the designed section to find it's M_uR. We must ensure that M_uR < M_u. Also we must prove that it is an under reinforced section. So in the next section, we will discuss about the 'Analysis of doubly reinforced sections'.

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Chapter 11 (cont..2) - Design steps for Doubly reinforced sections

In the previous section, we derived the two equations which relate the extra bending moment ΔM_u, with the force [C_us in the compression steel A_sc] and the force [ΔT_u in the extra tension steel ΔA_st]:

Eq.11.4
ΔM_u = (f_sc - 0.447f_ck)A_sc (d - d'). and
Eq.11.5
ΔM_u = 0.87f_y ΔA_st (d - d')

ΔM_u in the above equations is a known quantity. It is the extra moment capacity that the beam must acquire when it become 'doubly reinforced'. So we can obtain ΔM_u using the following equation:
Eq.11.6
ΔM_u = M_u - M_ulim.

Thus in Eqs.11.4 & 11.5, A_sc and ΔA_st are the unknowns. We can use 11.5 to find ΔA_st. But A_sc should not be calculated using 11.4. We will see the reason for this in a short while.

First we will see ΔA_st. Substituting 11.6 in 11.5 and rearranging, we get:
Eq.11.7

When we add this ΔA_st to A_st,lim, we will get the total tension steel A_st provided in the beam section. So we can write:
Eq.11.8
A_st = A_st,lim + ΔA_st

Once we calculate A_st we can fix the number and diameter of bars for all the bars on the 'tension side' of the beam. We must choose the number of bars and their diameters in such a way that, the total area obtained is as close as possible, and at the same time, a little greater than what is obtained from Eq.11.8.

When the tension bars are finalized, the d may differ from what we had assumed. In that case, M_ulim, A_st,lim, and A_st should be recalculated. This is one of the reasons why we should not use Eq.11.4 for the calculation of A_sc. The d that we use in it may change after we calculate A_st. The second reason is as follows:

We get a value for A_st from our calculations. Based on that value, we try different combinations of 'numbers and diameters' of bars to make up the A_st. We can provide only discrete number of bars to make up a certain A_st. If we use bundled bars, then each bar in a bundle should be discrete. For example, we can provide '3' sepatate bars of 20 mm bars giving an A_st of 942 mm². We can not provide numbers like 3¹⁄₂ or 3¹⁄₄ or 3¹⁄₈ of 20 mm bars. This is shown in the fig. below:

Fig.11.4
Fractions of bar area cannot be provided

Also only a certain diameters are available in the market. So it is clear that a combination giving the exact A_st as obtained in the calculations cannot be achieved. Out of all the combinations which give values close to A_st, some will be less than A_st, and others will be greater than A_st. We must never use those combinations which are less. Always select a combination which give an area greater than but close to A_st.

Based on the above discussion, we can now get a better understanding about the 'quantity' of ΔA_st. That is., we subtract A_st,lim from A_st (the final A_st calculated from the actual bars provided) to get ΔA_st actually provided, and it will be greater than the calculated ΔA_st.

We can denote this ΔA_st which is actually provided as: ΔA_st(pd). We must ensure that this  ΔA_st(pd) is also self contained in itself. That is., there is equilibrium between the force in  ΔA_st(pd) and the force in A_sc. So it is clear that it is the force in  ΔA_st(pd) that we must equate to the force in A_sc, and not ΔA_st.

So we can write: (f_sc - 0.447f_ck)A_sc = 0.87f_y ΔA_st(pd). Rearranging this we get:
Eq.11.9

Thus we obtained the equation for A_sc. Now we have to discuss an important point about this A_sc. As in the case of A_st, A_sc also cannot be given the exact value obtained in 11.9. We must choose the combination which give an area greater than but close to A_sc.

So in the final beam section we will be giving an A_sc which is a little higher than what is obtained from Eq.11.9. So the force in the A_sc will also be greater than what is obtained in the calculations. That is., we calculated the A_sc in such a way that it will balance the force in  ΔA_st(pd). But now it is greater the calculated A_sc. So the force will also be greater. Then what about the equilibrium? The answer is that, to get equilibrium at the ultimate state, the compressive force in concrete will be reduced by the upward shifting of NA, thus reducing concrete area. So the depth of NA will become less than x_u,max. This will ensure that the steel A_st on the tension side will yield at the ultimate state, and thus, the beam section will indeed be an 'under reinforced' one.

This completes the discussion on the design procedure. We can now write the procedure as various steps in sequential order:

Step 1: Calculate M_ulim. (Eq.3.26) 
Step 2: Calculate A_st,lim. (Eq.3.29)
Step 3: Calculate ΔM_u from the relation: M_u = M_ulim + ΔM_u      
Step 4: Calculate  ΔA_st using Eq.11.7. Then A_st = A_st,lim +  ΔA_st
Step 5: Provide suitable combination of bars to obtain the above A_st. The area so obtained must be greater than but close to A_st. 
Step 6: Check whether actual d provided is same as the assumed d. If there is change, the above steps should be repeated.
Step 7: Find the area of this combination of bars actually provided.
Step 8: Subtract A_st,lim from this to obtain  ΔA_st(pd)
Step 9: Calculate d'/d and use it to determine f_sc from Table 11.1
Step 10: Use Eq.11.9 to calculate A_sc
Step 11: Provide a suitable combination of bars which give an area greater than but close to A_sc. 

A doubly reinforced beam designed by the above process will have all the required properties:
• The ultimate moment of resistance M_uR of the section will be greater than the applied factored moment M_u
• The depth of NA x_u at the ultimate state will be less than x_umax.

However we must always do an analysis of the newly designed doubly reinforced section, to verify that these conditions are satisfied. For this we must learn how to analyse a doubly reinforced rectangular beam section. This will be discussed in the next chapter.


At present we will look into the other essential aspects of the design procedure.

The basic concepts of the process of design of singly reinforced rectangular sections were discussed earlier. There, in the introduction part, we discussed about:
1.concrete cover 
2.Minimum distance to be provided between the bars of beams
3.Maximum spacing allowable between bars of beams
4.Beams with overall depth greater than 750 mm
5.Minimum area of flexural reinforcements in beams
6.Maximum allowable area of flexural reinforcements in beams
7.Deflection control of singly reinforced beams and
8.Guide lines for fixing up the dimensions of beams
Links to each of the above items can be seen here.

All the above eight topics are applicable to Doubly reinforced beams also. However, some minor modifications have to be made to some of them. We will now look at the required modifications:

No.6: Maximum allowable area of flexural reinforcements in flanged beams:
We have seen earlier that, according to cl 26.5.1(b), area of the tension steel Ast provided should not be more than 4% of the total cross sectional area of the beam ie., A_st ≯ 0.04b_wD  The code specifies an allowable limit for the compression steel also. According to cl 26.5.1.2, area of compression steel provided should not be more than 4% of the total cross sectional area of the beam ie., A_sc ≯ 0.04b_wD  


From the above, we can see that, if both and are provided in a beam at their maximum allowable limits, the area of steel will take up about 8% of the total cross sectional area of the beam. This is rather excessive. In such cases, the high steel areas should be avoided by using improved grades of steel and concrete. Increasing the size of the beam will also result in a lesser quantity of required steel.


No.7: Deflection control
We have seen the modification factors that have to be applied to the basic l/d ratio. There we mentioned that the modification factor  kc  will be discussed when we take up the design of doubly reinforced sections.

So we have to learn about this new factor kc. For this we look at cl. 23.2.1(d) of the code. According to this clause, we must calculate kc from fig. 5 of the code, and then the basic l/d ratio must be multiplied by this kc.

Another method to determine kc is to use the following formula given by SP 24:

Where pc is the percentage of compression reinforcement given by 

Now the final equation for a doubly reinforced rectangular beam section can be written as follows:

Thus we can write the deflection control equations in the final form:

11.10
For doubly reinforced beams with span less than 10m,

(l/d)_actual  ≤  [(l/d)_basic] k_t k_c

11.11
For doubly reinforced beams with span greater than 10m,

(l/d)_actual  ≤  [(l/d)_basic] α k_t k_c

If the beam is a cantilever with span greater than 10 m, actual deflection calculations should be made.

No.8: Guide lines for fixing up the dimensions of beams
We have seen at the beginning of this chapter that, generally, we make the decision to make a beam 'doubly reinforced', when we find that the cross sectional dimensions of the beam cannot be increased further to provide the required bending moment capacity as a singly reinforced beam. This means that the dimensions are already fixed when we start to design it as a singly reinforced beam. So we do not have to learn any new methods to fix up the preliminary dimensions for doubly reinforced beams.


In the next section we will see a solved example.

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