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1. CE8501 Design Of Reinforced Cement Concrete Elements
Unit 1-Introduction
Design of singly reinforced beam
[As per IS456:2000]
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. Design steps as per IS:456:2000
2. Reinforcement details
3. Example#06
4. Example#07
2

3. Design Steps – Singly reinforced concrete beam as
per IS:456:2000
1. Determination of required dimensions of beam
2. Classification of beam as Under reinforced or Balanced or Over
reinforced section.
3. Finding area of reinforcement in tension zone only (Ast).
4. Providing suitable rebar based on Ast.
5. Determination of depth of NA.
6. Check for Ast.
3

4. Area of steel (Ast)
• If Mu < Mu,lim then it is under reinforced section
4
[Refer IS456 Pg.96]
Ast derived from

5. Area of steel (Ast)
• If Mu = Mu,lim then it is balanced section
5
[Refer IS456 Pg.96]
Ast derived from

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. Problem#06
Deign of singly reinforced rectangular beam
• Design a singly reinforced concrete beam of width 250 mm, subjected to
an ultimate moment of 130 kNm. Assume fck =20 MPa and fy = 415 MPa.
Given :
b = 250mm d = ?
M-20 – fck = 20N/mm2 Fe415 – fy = 415N/mm2
Ast = ? Mu= 130 kNm = 130 x 106 N.mm
7
b= 250mm
d =?
Ast

8. Step 1 : Determination of required dimensions of beam
[IS 456:2000]
8
Mu,lim = 0.36
𝒙𝒖,𝒎𝒂𝒙
𝒅
[1 – 0.42
𝒙𝒖,𝒎𝒂𝒙
𝒅
] b d2 fck
130*106 = 0.36 * 0.48 [1 – 0.42 ∗ 0.48 ] * 250 * d2 * 20
d = 434.11 mm
d ≈ 460 mm
Assume cover of 40mm, Hence overall depth D = 500mm

9. Step 2 : Classification of beam
[IS 456:2000]
9
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 ] * 250 * 4602 * 20
= 146 x106 N.mm
= 146 kN.m > 130 kN.m [Mu<Mu,lim Under reinforced section]

10. Step 3 : Area of tensile reinforcement (Ast)
[IS 456:2000]
For under reinforced section, Ast derived from Mu
Mu = 0.87 fy Ast d [1 -
𝑨 𝒔𝒕
𝒇𝒚
𝒃 𝒅 𝒇𝒄𝒌
]
130 * 106 = 0.87 * 415 * Ast * 460 [1-
Ast∗415
250∗460∗20
]
= (- 29.97 Ast
2 )+ (166.08 * 103 Ast ) – (130 * 106)
Ast = 943.34 mm2
10

11. Step 4 : Selecting Rebar size
Area for tension zone
Ast = 943.3 mm2
Assume 20mm dia bar,
No.of bars =
943.3
314.2
= 3
Hence use 3 numbers of 20mm dia bar
Area provided,
Ast = 942.6 mm2
Hanger bar size
Use 8mm dia bar as hanger bar
11
Area of rebar
Area =
𝜋
4
(𝑑2)
= 50.3 mm2 (8mm ϕ)
= 78.5 mm2 (10mm ϕ)
= 113.1 mm2 (12mm ϕ)
= 201.1 mm2 (16mm ϕ)
= 314.2 mm2 (20mm ϕ)
= 490.9 mm2 (25mm ϕ)
= 804.2 mm2 (32mm ϕ)

12. Step 5 : Depth of neutral axis
12
𝑥 𝑢 =
0.87 𝑓𝑦 𝐴𝑠𝑡
0.36 𝑓𝑐𝑘 𝑏
𝑥 𝑢=
0.87 ∗415 ∗ 942.6
0.36 ∗20 ∗250
𝑥 𝑢= 189 mm

13. Step 6 : Check for Ast[IS 456:2000]
13
[Refer IS456 Pg.46,47]

14. Step 5 : Check for Ast[IS 456:2000]
• Minimum tension reinforcement required
𝐴 𝑠𝑡
𝑏𝑑
=
0.85
𝑓 𝑦
𝐴 𝑠𝑡
250∗460
=
0.85
415
Ast = 235.54 mm2 < 943.34 mm2 Hence safe
14
[Refer IS456 Pg.47]

15. Step 6 : Reinforcement details of Singly
reinforced beam
15
N A
Compression Zone
Tension Zone
d= 460mm
Effective cover = 40mm
b= 250mm
20mm dia main bar
8mm dia hanger bar
Cross section
xu= 189 mm
D= 500mm

16. Assignment#05
• Design a beam to carry a factored moment of 145kNm using grade
of M25 and Fe415. (Assume b= 230mm)
16

17. Problem#07
Deign of singly reinforced rectangular beam
• Design a singly reinforced concrete beam subjected to an ultimate moment of
315 kNm. Assume fck = 25 N/mm2 and fy= 415 N/mm2. In this beam, due to
architectural considerations, the width has to be restricted to 230 mm.
Given :
b = 230mm d = ?
M-25 – fck = 25N/mm2 Fe415 – fy = 415N/mm2
Ast = ? Mu= 315 kNm = 315 x 106 N.mm
17
b= 230mm
d =?
Ast

18. Step 1 : Determination of required dimensions of beam
[IS 456:2000]
18
Mu,lim = 0.36
𝒙𝒖,𝒎𝒂𝒙
𝒅
[1 – 0.42
𝒙𝒖,𝒎𝒂𝒙
𝒅
] b d2 fck
315*106 = 0.36 * 0.48 [1 – 0.42 ∗ 0.48 ] * 230 * d2 * 25
d = 630.14 mm
d ≈ 650mm
Assume cover of 50mm, Hence overall depth D = 700mm

19. Step 2 : Classification of beam
[IS 456:2000]
19
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 ] * 230 * 6502 * 25
= 335.17 x106 N.mm
= 335.17 kN.m > 315 kN.m [Mu<Mu,lim Under reinforced section]

20. Step 3 : Area of tensile reinforcement (Ast)
[IS 456:2000]
For under reinforced section, Ast derived from Mu
Mu = 0.87 fy Ast d [1 -
𝑨 𝒔𝒕
𝒇𝒚
𝒃 𝒅 𝒇𝒄𝒌
]
315 * 106 = 0.87 * 415 * Ast * 650 [1-
Ast∗415
230∗650∗25
]
= (- 26.06 Ast
2 )+ (243.68 * 103 Ast ) – (315 * 106)
Ast = 1549 mm2
Ast = 1550 mm2
20

21. Step 4 : Selecting Rebar size
Area for tension zone
Ast = 1550 mm2
Assume 20mm dia bar,
No.of bars =
1550
314.2
= 4.9
Hence use 6 numbers of 20mm dia bar
Area provided,
Ast = 1885.2 mm2 > 1550 mm2 Hence safe
21
Area of rebar
Area =
𝜋
4
(𝑑2)
= 50.3 mm2 (8mm ϕ)
= 78.5 mm2 (10mm ϕ)
= 113.1 mm2 (12mm ϕ)
= 201.1 mm2 (16mm ϕ)
= 314.2 mm2 (20mm ϕ)
= 490.9 mm2 (25mm ϕ)
= 804.2 mm2 (32mm ϕ)

22. Step 5 : Depth of neutral axis
22
𝑥 𝑢 =
0.87 𝑓𝑦 𝐴𝑠𝑡
0.36 𝑓𝑐𝑘 𝑏
𝑥 𝑢=
0.87 ∗415 ∗1885.2
0.36 ∗25 ∗230
𝑥 𝑢= 328.8 mm

23. Step 5 : Reinforcement details of Singly
reinforced beam
23
N A
d= 640mm
Clear cover = 30mm
b= 230mm
20mm dia main bar
D= 700mm
Cross section
xu= 328.8 mm
Spacing 20 mm
• 6 bars cannot be provided
at single layer
• Hence has to provide at 2
layers
• Assume
clear cover of 30mm
Spacing of 20mm
• Effective depth
d = 700-30-20-10
d= 640 mm

24. Step 6 : Check for Ast , d, Mu
Mu = 0.87 fy Ast d [1 -
𝑨 𝒔𝒕
𝒇𝒚
𝒃 𝒅 𝒇𝒄𝒌
]
= 0.87 * 415 * 1885 * 640 [1 -
1885 ∗ 415
25 ∗ 230 ∗640
]
Mu = 344.4 x 106 N.mm.
Mu = 344.4 kN.m > 315 kN.m
Hence safe
24

25. Thank You
25