CE8501 DESIGN OF REINFORCED CEMENT CONCRETE ELEMENTS, Singly reinforced beam, ast problems,

1. CE8501 Design Of Reinforced Cement Concrete Elements
Unit 1-Introduction
Analysis 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.
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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 (Ast) as per IS 456:2000
5. Area of steel (Ast) as per SP-16
6. Example problems #03,04
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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
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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.
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7. Limiting moment of resistance
[SP -16] (Alternate Method)
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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

8. Area of tensile reinforcement (Ast)
[SP -16] (Alternate Method)
Ast =
𝑷𝒕
𝒃 𝒅
𝟏𝟎𝟎
[SP -16, Pg:17]
8
𝑀𝑢
𝑏𝑑2 value, Corresponding
pt for 415
Percentage of steel
Pt
[SP -16, Pg:48, Table – 2]

9. Problem#03 Area of steel (Ast) type
• Determine the area of reinforcement required for a singly reinforced concrete
section having a breadth of 300mm and effective depth of 675 mm to support a
factored moment of 185 kNm. Adopt M-20 grade concrete and Fe-415 grade
HYSD bars.
• Given :
b = 300mm d = 675mm
M-20 – fck = 20N/mm2 Fe415 – fy = 415N/mm2
Ast = ? Mu= 185 kNm = 185 x 106 N.mm
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b= 300mm
d =675 mm
Ast

10. Step 1 : Limiting moment of resistance
[IS 456:2000]
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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 ] * 300 * 6752 * 20
= 377.15 x 106 N.mm
= 377.15 kN.m > 185 kN.m [Mu<Mu,lim Under reinforced section]
(Ast unknown)
𝑥 𝑢
𝑑
=
0.87 𝑓𝑦 𝐴𝑠𝑡
0.36 𝑓𝑐𝑘 𝑏 𝑑
Not applicable

11. Step 2 : 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
]
185 * 106 = 0.87 * 415 * Ast * 675 [1-
Ast∗415
300∗675∗20
]
= (- 24.9 Ast
2 )+ (243.71 * 103 Ast ) – (185 * 106)
Ast = 829.37 ≈ 830 mm2
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12. Step 1 : Limiting moment of resistance
[SP -16] (Alternate Method)
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For Fe 415 HYSD bars,
Mu,lim = 0.138 fck b d2 [SP -16, Pg:10, Table C]
Mu,lim = 0.138 * 20 * 300 * 6752
= 377.25 x 106 N.mm
= 377.25 kN.m > 185 kN.m [ Under reinforced section]
(Ast unknown)
𝑥 𝑢
𝑑
=
0.87 𝑓𝑦 𝐴𝑠𝑡
0.36 𝑓𝑐𝑘 𝑏 𝑑
Not applicable

13. Step 2 : Area of tensile reinforcement (Ast)
[SP -16] (Alternate Method)
For under reinforced section, Ast derived from Mu
Ast =
𝑷𝒕
𝒃 𝒅
𝟏𝟎𝟎
[SP -16, Pg:17]
Ast =
0.409 ∗300 ∗675
100
Ast = 828.2 ≈ 830 mm2
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𝑀𝑢
𝑏𝑑2
=
185 ∗ 106
300 ∗ 6752
𝑀𝑢
𝑏𝑑2 = 1.35
Corresponding pt for 415
Percentage of steel
Pt = 0.409
[SP -16, Pg:48, Table – 2]

14. Problem#04 Area of steel (Ast) and depth of beam
• Design the minimum effective depth required and the area of reinforcement
for a rectangular beam having a width of 300mm to resist an ultimate moment
of 200 kNm, using M20 grade concrete and Fe-415 HYSD bars.
Given :
b = 300mm d = ?
M-20 – fck = 20N/mm2 Fe415 – fy = 415N/mm2
Ast = ? Mu= 200 kNm = 200 x 106 N.mm
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b= 300mm
d =?
Ast

15. Step 1 : Depth of the section
[IS 456:2000]
Mu,lim = 0.36
𝒙𝒖,𝒎𝒂𝒙
𝒅
[1 – 0.42
𝒙𝒖,𝒎𝒂𝒙
𝒅
] b d2 fck
200*106 = 0.36 * 0.48 [1 – 0.42 ∗ 0.48 ] * 300 * d2 * 20
d = 491.5 mm
d ≈ 500 mm
For Fe 415 HYSD bars,
Mu,lim = 0.138 fck b d2 [SP -16, Pg:10, Table C]
Mu,lim = 0.138 * 20 * 300 * d2
d=
𝑀 𝑢
,
𝑙𝑖𝑚
0.138 𝑓𝑐𝑘 𝑏
=
200∗106
0.138 ∗20 ∗300
= 491.47 𝑚𝑚
d ≈ 500 𝑚𝑚
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Alternate method
[SP -16]

16. Step 2 : Check for Mu,lim
[IS 456:2000]
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Mu,lim = 0.36
𝒙𝒖,𝒎𝒂𝒙
𝒅
[1 – 0.42
𝒙𝒖,𝒎𝒂𝒙
𝒅
] b d2 fck or
Mu,lim = 0.36 * 0.48 [1 – 0.42 ∗ 0.48 ] * 300 * 5002 * 20
= 207 x 106 N.mm
Mu,lim = 207 kN.m > 200 kN.m [Mu<Mu,lim Under reinforced section]
[SP-16]

17. Step 2 : 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
]
200 * 106 = 0.87 * 415 * Ast * 492 [1-
Ast∗415
300∗492∗20
]
= (- 24.97 Ast
2 )+ (177.64 * 103 Ast ) – (200 * 106)
Ast = 1402 mm2 ≈ 1410 mm2
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18. Step 2 : Area of tensile reinforcement (Ast)
[SP -16] (Alternate Method)
For under reinforced section, Ast derived from Mu
Ast =
𝑷𝒕
𝒃 𝒅
𝟏𝟎𝟎
[SP -16, Pg:17]
Ast =
0.995 ∗300 ∗492
100
Ast = 1409.5 ≈ 1410 mm2
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𝑀𝑢
𝑏𝑑2
=
200 ∗ 106
300 ∗ 4922
𝑀𝑢
𝑏𝑑2 = 2.75
Corresponding pt for 415
Pt = 0.995
[SP -16, Pg:48, Table – 2]

19. Assignment#02
• Determine the area of reinforcement required for a singly
reinforced concrete section having a breadth of 300 mm and an
effective depth of 500 mm to support a factored moment of
230 kN.m. Assume M25 grade concrete and Fe 145 steel.
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20. Assignment#03
• Determine the minimum effective depth required and the
correspond ing area of tension reinforcement for a rectangular
beam having a width of 300 mm to resist an ultimate moment of
250 kN.m, Using M-30 Grade concrete and Fe-415 HYSD bars.
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21. Thank You
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