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:
Concrete cover and Concrete grade:
Concrete cover for bars of beam
The nominal cover Cc 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
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
Cover for compression steel :
Assume 16 mm bars for compression steel. So d' = 30 +10 +8 =48 mm
Factored Bending moment:
Service load w per meter length of beam =40.35 kN/m
So factored BM Mu = (1.5wl2)/8 = 139.89 kNm
Now we can begin the design process. We have the following data:
fck =25 N/mm2, fy =415 N/mm2, Total depth D =400 mm, Assumed effective depth d =349 mm, d' =46 mm. width =230 mm, Mu = 139.89 kNm
Step 1: Calculate Mulim. (Eq.3.26)
For Fe415 steel, xumax/d = 0.4791
So we get Mulim =97041767.93 Nmm =97.815 kNm
Step 2: Calculate Ast,lim. (Eq.3.29)
So we get Ast,lim =966.726 mm2
Step 3: Calculate ΔMu from the relation: ΔMu = Mu - Mulim
So we get ΔMu =139.89 -97.815 =42.07 kNm
Step 4: Calculate ΔAst using Eq.11.7
So we get ΔAst = 385.86 mm2
Then Ast = Ast,lim + ΔAst = 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:
• Mulim =95.385 kNm
• Ast,lim =958.440 mm2
• Mu - Mulim =44.505 kNm
• ΔAst =412.24 mm2
• Ast = Ast,lim + ΔAst =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: ΔAst(pd) = 1472.622 - Ast,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 fsc =344.76 N/mm2
Step 10: Calculate Asc using Eq.11.9
So we get Asc =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, Cc =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 As = 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] kt kc
For simply supported members, l/d basic = 20. We have to calculate kt and kc. First we will calculate kt:
To find kt:
• Ast required = 1,370.68 mm2 from (1) • Ast,p = 1472.622 mm2
• fst = 218.04 N/mm2 • pt = 1.8452 • From Fig.4 of the code, kt = 0.9
To find kc:
We can use the expression given in SP24
Where pc = 100Asc/bd = (100 x603.186)/(230 x347) =0.7558
So we get kc = 1.17
• So l/d = 20 x kt x kc = 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 MuR. We must ensure that MuR < Mu. 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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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:
Concrete cover and Concrete grade:
Concrete cover for bars of beam
The nominal cover Cc 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
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
Cover for compression steel :
Assume 16 mm bars for compression steel. So d' = 30 +10 +8 =48 mm
Factored Bending moment:
Service load w per meter length of beam =40.35 kN/m
So factored BM Mu = (1.5wl2)/8 = 139.89 kNm
Now we can begin the design process. We have the following data:
fck =25 N/mm2, fy =415 N/mm2, Total depth D =400 mm, Assumed effective depth d =349 mm, d' =46 mm. width =230 mm, Mu = 139.89 kNm
Step 1: Calculate Mulim. (Eq.3.26)
So we get Mulim =97041767.93 Nmm =97.815 kNm
Step 2: Calculate Ast,lim. (Eq.3.29)
Step 3: Calculate ΔMu from the relation: ΔMu = Mu - Mulim
So we get ΔMu =139.89 -97.815 =42.07 kNm
Step 4: Calculate ΔAst using Eq.11.7
So we get ΔAst = 385.86 mm2
Then Ast = Ast,lim + ΔAst = 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:
• Mulim =95.385 kNm
• Ast,lim =958.440 mm2
• Mu - Mulim =44.505 kNm
• ΔAst =412.24 mm2
• Ast = Ast,lim + ΔAst =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: ΔAst(pd) = 1472.622 - Ast,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 fsc =344.76 N/mm2
Step 10: Calculate Asc using Eq.11.9
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, Cc =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
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] kt kc
For simply supported members, l/d basic = 20. We have to calculate kt and kc. First we will calculate kt:
To find kt:
• Ast required = 1,370.68 mm2 from (1) • Ast,p = 1472.622 mm2
• fst = 218.04 N/mm2 • pt = 1.8452 • From Fig.4 of the code, kt = 0.9
To find kc:
We can use the expression given in SP24
Where pc = 100Asc/bd = (100 x603.186)/(230 x347) =0.7558
So we get kc = 1.17
• So l/d = 20 x kt x kc = 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 MuR. We must ensure that MuR < Mu. 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'.
PREVIOUS
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