Singly & Doubly ReinforcedDDD Beams.pptx

REINFORCED BEAM

1. RCC DESIGN SINGLY REINFORCED BEAMS VAKA/RCC DESIGN/KR/2023-24

2. VAKA/RCC DESIGN/KR/2023-24 Singly reinforced beam has tension reinforcement provided at the bottom.( Below NA)

3. VAKA/RCC DESIGN/KR/2023-24 Important Points Selection of breadth of the beam b Normally, the breadth of the beam b is governed by: ( i ) proper housing of reinforcing bars and (ii) architectural considerations. It is desirable that the width of the beam should be less than or equal to the width of its supporting structure like column width, or width of the wall etc. Practical aspects should also be kept in mind. Most commonly used beam width dimensions are 150, 200, 230, 250 and 300 mm. Depth Of Beam “D” : Overall depth ‘D’ may be assumed to be effective span/10 to effective span/15. The effective depth has the major role to play in satisfying ( i ) the strength requirements of bending moment and shear force, and (ii) deflection of the beam. 3. Minimum Reinforcement in Beams The minimum reinforcement to be provided in the beam depends on grade of steel ‘ fy ’. Minimum tensile reinforcement Ast should not be less than obtained by the given equation as per IS 456-2000 ( cl. 26.5.1.1a )

4. VAKA/RCC DESIGN/KR/2023-24 4. Maximum reinforcement in Beams 5. Selection of diameters of bar of tension reinforcement Reinforcement bars are available in different diameters such as 6, 8, 10, 12, 14, 16, 18, 20, 22, 25, 28, 30, 32, 36 and 40 mm. Some of these bars are less available. The selection of the diameter of bars depends on its availability, minimum stiffness to resist while persons walk over them during construction, bond requirement etc. Normally, the diameters of main tensile bars are chosen from 12, 16, 20, 22, 25 and 32 mm. For Lintel beams over doors & windows ( 10mm, 12mm) For stirrups- 8mm, 10mm, 12mm 6. Selection of grade of concrete & grade of steel

5. VAKA/RCC DESIGN/KR/2023-24 7. Minimum clear cover to reinforcement depends on exposure conditions. Should not be less than 25mm for beams 8. Reinforcement in the beams should be places symmetrically about the vertical axis of the beam and when its difficult to accommodate all the rebars in one layer , should be placed vertically over one other in a symmetric way. This is important so that the flow of concrete takes place without any obstruction caused by rebars 9. Distance between the reinforcement in a layer should not be less than the maximum among the following- Diameter of the bars when all are of same diameter. Diameter of the larger bar if the diameter are unequal. 5 mm more than the nominal maximum size of coarse aggregate 10. Each layer shall have minimum 2 no. of rebars . Minimum vertical distance between the bars shall be greater of following :

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7. VAKA/RCC DESIGN/KR/2023-24 Design Of RCC beams can be done by 3 methods The three methods are: Direct computation method, Use of charts of SP-16 and Use of tables of SP-16 Design of beams mainly involves Finding area of tensile reinforcement Dimension of the beam section ( width and depth)

8. VAKA/RCC DESIGN/KR/2023-24 Singly Reinforced Beams- Design Steps using SP16 Design Aids Given data : UDL on a beam ( w KN/m) , Span (l m) ( Effective/ Clear) Calculate Max BM due to udl . For a simply supported beam subjected to UDL, BM is max at the center and can be calculated using expression 3. Factored Moment Mu is obtained by multiplying with 1.5 4. If the dimensions of the beam are not given, suitable dimensions can be assumed using thumb rules. Assume width b= 230mm Over all depth = effective span/10 to effective span/15 Effective span= Clear span+ c/c supports

9. VAKA/RCC DESIGN/KR/2023-24 5. Assume suitable effective cover ( say 30 to 40 mm), effective depth may be calculated. 6. Effective depth= Overall depth- effective cover 7. Calculate Moment Factor Mu/bd^2 8. Calculate Mu max/ Mu lim using formulae 9. Compare Mu & Mu max. If Mu < Mumax ; Beam would be designed as a Singly Reinforced Beam . Which means steel would be provided only in tension zone. ( 2 Anchor bars would be provided on top side/compression zone to hold stirrups) 10. Calculate ‘ Pt’ , percentage of Tensile steel. Use SP16-1980 Design aids. Use Table 2, 3, 4 Page 47. Using Pt value, Area of steel reinforcement can be calculated using Ast = pt x b x d 11. Check the Percentage steel should satisfy the minimum reinforcement criteria as per IS-456-2000 100

10. VAKA/RCC DESIGN/KR/2023-24 12. Compare Ast values calculated in step 10 & 11. 13. If Ast in step 10 > Ast step 11, then the Ast in step 10 can be used for calculating the No. of bars by assuming a suitable diameter of bar. 14. If Ast in step 10 < Ast step 11, The Ast obtained as in step 11 should be used. As it’s the minimum tensile reinforcement required in the beam section as per IS456:2000. 15. A suitable diameter of bar is assumed ( Say 16mm dia ) and the no. of bars can be calculated by No.of bars = Ast ( Area of steel reinforcement provided) Area of 1 bar 16. Always round of the no. of bars obtained on the higher side. For example if Np . Of bars are 3.5, consider 4 No. bars . 17 Finally a reinforcement detail sketch is prepared for the beam section, showing following details. Over all depth, Effective depth, Cover, width of beam, No. of bars and diameter. Stirrups should always be shown in the sketch supported on anchor bars.

11. VAKA/RCC DESIGN/KR/2023-24 Example: Determine the main tension reinforcement required in a rectangular beam of size 30x60 cm. Used M15, Fe 415 steel. Factored moment = 170KN-m Solution: b= 300mm D= 600mm Lets assume effective cover= 40mm Effective depth d= 600-40= 560mm Mu= 170 KN-M Calculate Moment factor- Mu/bd^2= 170x 10^6/( 300 x 560 x560)= 1.806 Pt= 0.993 % Calculate Ast = 0.993 x 300 x 560/ 100 =1668.24 mm ^2 Tables in SP16

12. VAKA/RCC DESIGN/KR/2023-24 Check for Min reinforcement Ast = 0.85 x 300 x 560/(415)= 344. mm^2 Area of steel required = 1664.8 sqmm Lets consider 20mm dia bars. Area of 1 bar= 3.14 x 20 x 20/ 4= 314 mm^2 No. of bars= 1664.8/ 314= 5.3 Nos. Consider 6 Nos 20mm dia bars. 6 No. 20mm Dia bars 300mm 600mm 2 Nos 8mm dia Anchor bars 8mm dia stirrups 20mm L section of beam Main steel/ Longitudinal bars/ Tension steel/ Ast Stirrups/Transverse Reinforcement

13. VAKA/RCC DESIGN/KR/2023-24 DOUBLY REINFORCED BEAMS

14. VAKA/RCC DESIGN/KR/2023-24 A beam in which reinforcement is provided in both the zones i.e. Tension and compression zone.

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17. VAKA/RCC DESIGN/KR/2023-24 Doubly reinforced beams, therefore, have moment of resistance more than the singly reinforced beams of the same depth for particular grades of steel and concrete. In many practical situations, architectural or functional requirements may restrict the overall depth of the beams. For example where head room restrictions are there. Hence doubly reinforced beam (DRB) would be suitable in such cases. Merits: Stronger than SRB Enables to provide small depth/size beam; hence taking care of architectural requirements/ head room/ aesthetics. Demerit: 1. Its costlier than SRB

18. VAKA/RCC DESIGN/KR/2023-24 Reinforcement in compression zone ( above NA in case of a SSB) is called compression reinforcement/ Steel provided at top/ Asc Effective cover for top reinforcement is represented as d’

19. VAKA/RCC DESIGN/KR/2023-24 STEPS TO BE FOLLOWED FOR Ast & Asc CALCULATION Check if Mu > Mu lim. Then beam to be designed as a DRB. Such a beam needs reinforcement at top ( Asc ) and bottom( Ast ) Calculate d’/d where d is effective depth of the beam d’ is effective cover for compression steel ( use 40mm if data not given) Based on d’/d valve, appropriate tables to be referred from SP16-1980 to find out Pt and Pc. Where Pt- %age of tensile reinforcement Pc- %age of compression reinforcement Note: Refer tables 50, 51, 52 Page 81 onwards 4.If d’/d value is different from the values provided in tables of SP16-1980, linear interpolation may be done to exact %age of steel. For quick calculation however approx value may be considered to avoid interpolation. 5. Calculation of Area of tensile steel Ast & area of compression steel Asc is given by Ast = pt x b x d Asc = pc x b x d 100 100

20. VAKA/RCC DESIGN/KR/2023-24 6. Always provide steel bars with area greater than whats required as calculated in step 5 7. Make a detailed sketch of reinforcement detail, showing Diameter & No. of bars, along with the dimensions of beam, cover on top and bottom, stirrups etc complete in all respects. Reinforcement should be symmetrically placed. 8. In DRB, the total reinforcement area in compression & tension should not exceed 4% of gross cross section of beam section i.e. 4%( bD ). This is important to ensure and avoid congestion . It avoids difficulty in placing & compacting the concrete.

21. VAKA/RCC DESIGN/KR/2023-24 Example 1 : Calculate the dimensions of the given RCC beam based on the thumb rules. Assume its supported over walls 300mm thick and the clear span is 5m. Solution: The width of the beam may be assumed to be equal to the support width i.e. 300mm. So consider b = 300mm According the thumb rule For a SSB, depth should lie between Leff /10 to leff /15 Clean span= 5m = 5000mm Effective span L eff = 5000+ 300/2 + 300/2= 5300 mm or 5.3m So the depth can be calculated by assuming a value between Leff /10 to leff / 15 Leff / 10= 5300/10= 530 mm Leff /15= 5300/15 = 353mm So we can consider the depth value D= 400mm

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23. VAKA/RCC DESIGN/KR/2023-24 Example 3: Design a SS beam of span 6m with a load of 40KN/m. Assume the width of the beam to be 300mm. Concrete mix M20, Steel grade Fe415 HYSD bars. Solution: 40KN/m 6m Max BM (M) for a SS beam is at its center & is given by = 40x 6 x 6 = 180KN-m 8 Step 1: Assume 6m to be effective span Step 2: Calculate Mu, factored moment = 1.5 M= 1.5 x 180= 270KN-m Step 3: Size of beam: Given b= 300mm For depth (using thumb rule) given in slide 8 Considering depth = Eff span/12= 6000/12= 500mm Lets say over all depth D=500mm Step 4: Calculate effective depth ‘d’- Assume effective cover 40mm

24. VAKA/RCC DESIGN/KR/2023-24 So effective depth d= 500-40= 460mm Step 5: Calculate Mu lim = As the given grade here is Fe415, Mu lim = 0.138 Fck b d^2 = 0.138 x 20 x 300 x 460 x460 = 175.2 Kn -m < 270KNm Step 6: Mulim < Mu, so design the beam as a DRB Step 7- Assume Effective over for compression reinforcement d’ = 40mm Calculate d’/d= 40/460= 0.0869 0r 0.09= 0.1 ( rounded off for quick calculation) Calculate Mu/bd^2= 270 x 10^6/( 300x 460 x 460) = 4.25 Step 8: Refer page 86, table no. 50 ( Fck 20, Fe415) Pt = 1.414 Pc= 0.482 Step 9 calculate Ast & Asc Ast = 1.414 x 300 x 460 = 1951.32 Sqmm Asc = 0.482 x 300 x 460 = 665.16mm^2 100 100 415

25. VAKA/RCC DESIGN/KR/2023-24 Step 10: Calculation of bars No. & diameter For compression reinforcement Assume 16mm dia bars Area of 1 bar of diameter 16mm= 3.14 x 16 x 16/ 4= 200.96 mm^2 Area of compression steel required = 665.16 mm ^2 No. of bars needed= 665.16/ 200.96 = 3.3099 . Provide 4 No. 16mm dia bars on top ( Compression zone) For tensile reinforcement assume 25mm dia bars Area of 1 bar 25mm dia = 3.14 x 25 x 25/ 4= 490.625 mm ^2 Area of tensile reinforcement required = 1951.32 mm^2 No. of bars needed = 1951.32/ 490.625 = 3.977 Provide 4 no. 25mm dia bars at bottom( Tension zone) Tensile steel 4-25mm dia Comp steel 4-16mm dia 500mm 300mm

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27. VAKA/RCC DESIGN/KR/2023-24 4.2--------------- 1.399 4.25-------------- x 4.3 ---------------1.429 Pt =1.414 4.2----------------0.466 4.25---------------x 4.3-------------------0.498 Pc =0.482

28. VAKA/RCC DESIGN/KR/2023-24 Complete this using SP16- 1980 Self weight = volume of member x density of material RCC DENSITY = 25KN/M^3

29. VAKA/RCC DESIGN/KR/2023-24 PRACTICE QUESTIONS

30. Practice Exercise Q1: What is a balanced, Under reinforced and Over reinforced beam section. Which section would you design and why? Make sketches. Q2 What do you mean by partial safety factor. Write down the Partial FOS values for steel and concrete in LSM. Q3: What is Limit state method of collapse and limit state method of serviceability. Explain. Q4: Why is LSM method globally used for design of RCC structures? Why had working stress method been discarded? Q5: With the help of diagram show compression zone and tension zone in a simply supported beam and a cantilever beam. Where would you provide the reinforcement and why? Q6. Define terms Neutral Axis , Compression zone, Tension Zone and Neutral Axis. Q7Find the moment of resistance of a singly reinforced concrete beam of 200 mm width 400mm effective depth, reinforced with 3-16 mm diameter bars of Fe 415 steel. Take M20 grade of concrete. Q8 VAKA/RCC DESIGN/KR/2023-24

31. VAKA/RCC DESIGN/KR/2023-24 Q1 Differentiate between singly and doubly reinforced beam. Make sketches Q2 What are the merits & demerits of a DRB. Q3 Under what conditions do we prefer a DRB. Q4: A simply supported beam carries a UDL of 20KN/m over a span of 6m. The cross section of the beam is 300x500mm. Calculate %age and amount of steel required. Use M20 grade & Fe415 steel. Use SP16 tables and make a sketch of reinforcement detail. Q5 Repeat all the above calculation for Q4, considering M25 grade concrete. UDL= 45KN/m Q6