doubly reinforced beam ,ast ,ast problems ,area of steel ,is456 ,sp16 ,beam ,mu ,mulim ,steel area ,anna university ,civil engineering

1. CE8501 Design Of Reinforced Cement Concrete Elements
Unit 1-Introduction
Design of Doubly reinforced beam
[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.
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2. Problem#09
• Design the doubly reinforced beam with b= 300mm, D= 600mm.
Factored moment = 320 kN.m. Assume fck= 20 N/mm2, fy= 415 N/mm2.
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b= 350mm
D = 600 mm
Ast
Given :
b = 300mm
D = 600mm
fck = 20N/mm2
fy = 415N/mm2
M = 320 kNm
To find
Ast = ?
Asc = ?
d’ = ?
Asc

3. Step 1 : Determination of required dimensions of beam
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Assume effective cover of 40mm, Hence overall depth D = 600mm
d = 600- 40
d = 560 mm

4. Step 2: Limiting moment of resistance (Mu,lim)
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 ] * 300 * 5602 * 20
= 259.59 x 106 N.mm
Mu,lim = 259.6 kN.m [Mu>Mu,lim Over reinforced section]
[Consider as Mu1]
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5. Step 3 : Area of steel at tension zone (Ast1)
For under reinforced section, Ast1 derived from Mu
Mu1 = 0.87 fy Ast1 d [1 -
𝑨 𝒔𝒕𝟏
𝒇𝒚
𝒃 𝒅 𝒇𝒄𝒌
]
259.6 * 106 = 0.87 * 415 * Ast1 * 560 [1-
Ast1
∗
415
300∗560∗20
]
= (- 25 Ast1
2 )+ (202.19 * 103 Ast1 ) – (259.6 * 106)
Ast1 = 1600 mm2
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Ast1

6. Step 3 : Area of steel at tension zone (Ast2)
Mut= Mu1 + Mu2
Mu2 = Mut - Mu1
= 320 - 259.6
Mu2 = 60.4 kN.m
60.4 x 106 = Ast2 (0.87 * 415) (560-56)
Ast2 = 331.9 ≈ 332 mm2
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Mu2 derived from
• Mu2 = Ast2 (0.87 fy) (d-d’)
Ast2
𝑑′
𝑑
= 0.10
𝑑′
= 0.10 * 560
= 56 mm

7. Step 3 : Area of steel at tension zone (Ast2)
7
Ast2
Ast = 1600 + 332
Ast = 1932 mm2
Ast1Ast

8. Step 4 : Selecting Rebar size (Tension Zone)
Area for tension zone
Ast = 1932 mm2
Assume 25mm dia bar,
No.of bars =
1932
490.9
= 3.93 ≈ 4
Hence use 4 numbers of 25mm dia bar
Area provided,
Ast = 1963.6 mm2 > 1932 mm2
Hence ok
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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)

9. Step 5 : Area of steel at Compression zone (Asc2)
Asc =
𝑀 𝑢2
𝑓 𝑠𝑐 −0.44 𝑓𝑐𝑘 𝑑 −𝑑′
Asc =
60.4 ∗106
353 −0.446 ∗20 560−56
Asc = 348.29 mm2 ≈ 350 mm2
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𝑑′
𝑑
= 0.10
𝑑′ = 0.10 * 560
= 56 mm
𝑓𝑠𝑐 = 353 N/mm2
SP 16, pg:13
Asc

10. Step 6 : Selecting Rebar size (Compression Zone)
Area for compression zone
Asc = 350 mm2
Assume 12mm dia bar,
No.of bars =
350
113.1
= 3.1 ≈ 4
Hence use 4 numbers of 12mm dia bar
Area provided,
Asc = 452.4 mm2 > 350 mm2
Hence ok
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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)

11. Area of Steel
At tension zone,
Ast = 2049 mm2
At compression zone,
Asc = 452 mm2
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Ast
AscAst
Asc

12. Step 7 : Check for depth of neutral axis
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xu =
0.87 ∗415 ∗1963 − 353−13.38 ∗452
0.36 ∗20 ∗300
xu = 257.05 mm
xu =
0.87 𝑓𝑦 𝐴𝑠𝑡 − 𝑓 𝑠𝑐
−𝑓𝑐𝑐 𝐴 𝑠𝑐
0.36 𝑓𝑐𝑘 𝑏
Xu,max = 0.48 * d
Xu,max = 0.48 * 560
Xu,max = 268.8 mm
[xu<xu,max Hence it is under reinforced section]

13. Step 6 : Reinforcement details of Singly
reinforced beam
13
N A
Compression Zone
Tension Zone
d= 560mm
Effective cover = 40mm
b= 300mm
25mm dia main bar
12mm dia main bar
Cross section
xu= 257 mm
D= 600mm

14. Assignment#07
• Design a rectangular beam as doubly reinforced section subjected to a
moment of 200kN.m. Consider concrete of grade M20 and steel of grade
Fe415. Assume b = 250mm.
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15. Thank You
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