Singly reinforced beam ast - over reinforced
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The document outlines the analysis of singly reinforced beams in reinforced cement concrete according to IS 456:2000 and SP-16. It details the types of beam sections based on limiting moments, calculations for areas of steel reinforcement for under-reinforced and over-reinforced sections, and provides an example problem to determine the required steel for specific beam dimensions and materials. Additionally, it reinforces the need for proper design when sections become over-reinforced, suggesting the inclusion of compression reinforcement.
![CE8501 Design Of Reinforced Cement Concrete Elements
Unit 1-Introduction
Analysis of Singly reinforced beam (Ast)
[As per IS456:2000] & SP -16
Presentation by,
P.Selvakumar.,B.E.,M.E.
Assistant Professor,
Department Of Civil Engineering,
Knowledge Institute Of Technology, Salem.
1](https://image.slidesharecdn.com/singlyreinforcedbeam-ast-overreinforced-200804063450/85/Singly-reinforced-beam-ast-over-reinforced-1-320.jpg)
![Area of steel (Ast)
• For balanced and Under reinforced sections
4
[Refer IS456 Pg.96]
Ast derived from](https://image.slidesharecdn.com/singlyreinforcedbeam-ast-overreinforced-200804063450/85/Singly-reinforced-beam-ast-over-reinforced-4-320.jpg)
![Limiting moment of resistance
[SP -16] (Alternate Method)
5
For Fe 415 HYSD bars,
Mu,lim = 0.138 fck b d2 [SP -16, Pg:10, Table C]
(Ast unknown)
𝑥 𝑢
𝑑
=
0.87 𝑓𝑦 𝐴𝑠𝑡
0.36 𝑓𝑐𝑘 𝑏 𝑑
Not applicable](https://image.slidesharecdn.com/singlyreinforcedbeam-ast-overreinforced-200804063450/85/Singly-reinforced-beam-ast-over-reinforced-5-320.jpg)
![Area of tension reinforcement (Over reinforced section)
Ast= Ast1 + Ast2 • Mu1 derived from
• Mu1 = Mu,lim
8
Mu2 derived from
• Mu2 = 𝐌 𝐮 − 𝐌𝐮, 𝐥𝐢𝐦
• Mu – (Actual B.M at beam)
• Given in question
Ast formula (Ast1)
Mu1 = 0.87 fy Ast1 d [1 -
𝑨 𝒔𝒕𝟏
𝒇𝒚
𝒃 𝒅 𝒇𝒄𝒌
]
Ast =
𝑷𝒕 𝒃 𝒅
𝟏𝟎𝟎or
IS 456 SP 16
Ast formula (Ast2)
Ast2 =
𝑴 𝒖𝟐
𝟎.𝟖𝟕 𝒇𝒚 (𝒅−𝒅′)
Mu2 = 0.87 fy Ast2 (d-d’)
SP 16, pg:12
Mu1 Mu2](https://image.slidesharecdn.com/singlyreinforcedbeam-ast-overreinforced-200804063450/85/Singly-reinforced-beam-ast-over-reinforced-8-320.jpg)
![Step 1 : Limiting moment of resistance
[IS 456:2000]
12
For Fe 415 HYSD bars,
Mu,lim = 0.36
𝒙𝒖,𝒎𝒂𝒙
𝒅
[1 – 0.42
𝒙𝒖,𝒎𝒂𝒙
𝒅
] b d2 fck
Mu,lim = 0.36 * 0.48 [1 – 0.42 ∗ 0.48 ] * 200 * 4002 * 20
= 88.30 x 106 N.mm
= 88.30 kN.m < 120 kN.m [Mu>Mu,lim Over reinforced section]
(Ast unknown)
𝑥 𝑢
𝑑
=
0.87 𝑓𝑦 𝐴𝑠𝑡
0.36 𝑓𝑐𝑘 𝑏 𝑑
Not applicable](https://image.slidesharecdn.com/singlyreinforcedbeam-ast-overreinforced-200804063450/85/Singly-reinforced-beam-ast-over-reinforced-12-320.jpg)
![Step 3 : Area of steel at tension zone (Ast)
For under reinforced section, Ast1 derived from Mu
Mu1 = 0.87 fy Ast1 d [1 -
𝑨 𝒔𝒕𝟏
𝒇𝒚
𝒃 𝒅 𝒇𝒄𝒌
]
88.30 * 106 = 0.87 * 415 * Ast1 * 400 [1-
Ast1
∗
415
200∗400∗20
]
= (- 37.46 Ast1
2 )+ (144.42 * 103 Ast1 ) – (88.30 * 106)
Ast1 = 762 ≈ 770 mm2
14
Ast1](https://image.slidesharecdn.com/singlyreinforcedbeam-ast-overreinforced-200804063450/85/Singly-reinforced-beam-ast-over-reinforced-14-320.jpg)
Singly reinforced beam ast - over reinforced
- 1. CE8501 Design Of Reinforced Cement Concrete Elements Unit 1-Introduction Analysis of Singly reinforced beam (Ast) [As per IS456:2000] & SP -16 Presentation by, P.Selvakumar.,B.E.,M.E. Assistant Professor, Department Of Civil Engineering, Knowledge Institute Of Technology, Salem. 1
- 2. Singly Reinforced Rectangular beam – Analysis 1. Types of section based on limiting moment 2. Moment of resistance as per IS 456:2000 3. Limiting moment of resistance as per SP-16 4. Area of steel for over reinforced beam as per IS 456:2000 5. Example#05 2
- 3. Types of section based on limiting moment • If Mu < Mu,lim then it is under reinforced section • If Mu = Mu,lim then it is balanced section • If Mu > Mu,lim then it is over reinforced section 3
- 4. Area of steel (Ast) • For balanced and Under reinforced sections 4 [Refer IS456 Pg.96] Ast derived from
- 5. Limiting moment of resistance [SP -16] (Alternate Method) 5 For Fe 415 HYSD bars, Mu,lim = 0.138 fck b d2 [SP -16, Pg:10, Table C] (Ast unknown) 𝑥 𝑢 𝑑 = 0.87 𝑓𝑦 𝐴𝑠𝑡 0.36 𝑓𝑐𝑘 𝑏 𝑑 Not applicable
- 6. Limiting Moment of resistance • If Mu > Mu,lim then it is over reinforced section • The section must be designed as doubly reinforced section. 6
- 7. Area of steel (Over reinforced section) 7 Mu>Mu,lim Over reinforced section. Hence it is recommended to design as doubly reinforced section For a doubly reinforced section, reinforcement steel is necessary to provide both in tension and compression zone. • Ast = (Ast1 + Ast2 ) Area of steel at tension zone • Asc – Area of steel at compression zone Ast Asc
- 8. Area of tension reinforcement (Over reinforced section) Ast= Ast1 + Ast2 • Mu1 derived from • Mu1 = Mu,lim 8 Mu2 derived from • Mu2 = 𝐌 𝐮 − 𝐌𝐮, 𝐥𝐢𝐦 • Mu – (Actual B.M at beam) • Given in question Ast formula (Ast1) Mu1 = 0.87 fy Ast1 d [1 - 𝑨 𝒔𝒕𝟏 𝒇𝒚 𝒃 𝒅 𝒇𝒄𝒌 ] Ast = 𝑷𝒕 𝒃 𝒅 𝟏𝟎𝟎or IS 456 SP 16 Ast formula (Ast2) Ast2 = 𝑴 𝒖𝟐 𝟎.𝟖𝟕 𝒇𝒚 (𝒅−𝒅′) Mu2 = 0.87 fy Ast2 (d-d’) SP 16, pg:12 Mu1 Mu2
- 10. Area of Compression reinforcement (Asc) Mu2 = 𝐴 𝑠𝑐 𝑓 𝑠𝑐 − 𝑓𝑐𝑐 𝑑 − 𝑑′ Asc = 𝑀 𝑢2 𝑓 𝑠𝑐 −𝑓𝑐𝑐 𝑑 −𝑑′ 10 SP 16, pg:12 SP 16, pg:13 𝑓𝑐𝑐 = 0.446 𝑓𝑐𝑘 Asc = 𝑀 𝑢2 𝑓 𝑠𝑐 −0.446 𝑓𝑐𝑘 𝑑 −𝑑′ d’ is the distance from the extreme compression fibre to the centroid of compression reinforcement.
- 11. Problem#05 Area of steel (Ast) type (over reinforced beam) • A rectangular beam has b = 200 mm, d = 400 mm. If steel used is Fe 415 and grade of concrete is M20, find the steel required to carry a factored moment of 120 kNm. Given : b = 200mm d = 400mm M-20 – fck = 20N/mm2 Fe415 – fy = 415N/mm2 Ast = ? Mu= 120 kNm = 120 x 106 N.mm 11 b= 200mm d =400 mm Ast
- 12. Step 1 : Limiting moment of resistance [IS 456:2000] 12 For Fe 415 HYSD bars, Mu,lim = 0.36 𝒙𝒖,𝒎𝒂𝒙 𝒅 [1 – 0.42 𝒙𝒖,𝒎𝒂𝒙 𝒅 ] b d2 fck Mu,lim = 0.36 * 0.48 [1 – 0.42 ∗ 0.48 ] * 200 * 4002 * 20 = 88.30 x 106 N.mm = 88.30 kN.m < 120 kN.m [Mu>Mu,lim Over reinforced section] (Ast unknown) 𝑥 𝑢 𝑑 = 0.87 𝑓𝑦 𝐴𝑠𝑡 0.36 𝑓𝑐𝑘 𝑏 𝑑 Not applicable
- 13. Step:2 Calculation of Mu1 & Mu2 13 Mu2 derived from • Mu2 = 𝐌 𝐮 − 𝐌𝐮, 𝐥𝐢𝐦 Mu2 = 120 − 88.30 Mu2 = 𝟑𝟏. 𝟕 𝐤𝐍. 𝐦 • Mu1 derived from • Mu1 = Mu,lim Mu1 = 88.30 kN.m
- 14. Step 3 : Area of steel at tension zone (Ast) For under reinforced section, Ast1 derived from Mu Mu1 = 0.87 fy Ast1 d [1 - 𝑨 𝒔𝒕𝟏 𝒇𝒚 𝒃 𝒅 𝒇𝒄𝒌 ] 88.30 * 106 = 0.87 * 415 * Ast1 * 400 [1- Ast1 ∗ 415 200∗400∗20 ] = (- 37.46 Ast1 2 )+ (144.42 * 103 Ast1 ) – (88.30 * 106) Ast1 = 762 ≈ 770 mm2 14 Ast1
- 15. Step 3 : Area of steel at tension zone (Ast) For under reinforced section, Ast2 derived from Mu Ast2 = 𝑴 𝒖𝟐 𝟎.𝟖𝟕 𝒇𝒚 (𝒅−𝒅′) = 31.7 ∗106 0.87 ∗415 (400−40) Ast2 = 243.88 mm2 ≈ 250 mm2 15 Ast2 𝑑′ 𝑑 = 0.10 𝑑′ = 0.10 * 400 = 40 mm 𝑓𝑠𝑐 = 353 N/mm2
- 16. Step 3 : Area of steel at tension zone (Ast) Area of tension reinforcement Ast1 = 770 mm2 Ast2 = 250 mm2 Ast= Ast1 + Ast2 Ast= 1020 mm2 16 Ast
- 17. Step 2 : Area of steel at Compression zone (Asc) Asc = 𝑀 𝑢2 𝑓 𝑠𝑐 −0.44 𝑓𝑐𝑘 𝑑 −𝑑′ Asc = 31.7 ∗106 353 −0.446 ∗20 400−40 Asc = 255.91 mm2 ≈ 260 mm2 17 𝑑′ 𝑑 = 0.10 𝑑′ = 0.10 * 400 = 40 mm 𝑓𝑠𝑐 = 353 N/mm2 SP 16, pg:13 Asc
- 18. Area of Steel At tension zone, Ast = 1020 mm2 At compression zone, Asc = 260 mm2 18 Ast AscAst Asc
- 19. Assignment#04 • A reinforced concrete beam of rectangular section of size 250 x 550 mm overall is to be designed for a factored moment of 225 kNm. Compute the reinforcement required at the effective cover of 50 mm. The concrete mix to be used is M20 and the grade of steel is Fe415. 19
- 20. Thank You 20