1) The document discusses the analysis and design of singly reinforced concrete beams according to Indian Standard Code IS 456:2000 and SP-16. It provides formulas and steps to calculate the limiting moment capacity, check if the section is under-reinforced, balanced or over-reinforced, and determine the required area of tension reinforcement.
2) Two example problems are presented to demonstrate calculating the area of steel for an under-reinforced beam section and determining the minimum depth and steel area required for a beam.
3) Key concepts covered include limiting moment capacity formulas, using equilibrium equations to calculate steel area for under-reinforced sections, and tables from SP-16 for determining steel percentages.
information on types of beams, different methods to calculate beam stress, design for shear, analysis for SRB flexure, design for flexure, Design procedure for doubly reinforced beam,
Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show.
Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.
information on types of beams, different methods to calculate beam stress, design for shear, analysis for SRB flexure, design for flexure, Design procedure for doubly reinforced beam,
Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show.
Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.
check it out: http://goo.gl/vqNk7m
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this slide will clear all the topics and problem related to singly reinforced beam by limit state method, things are explained with diagrams , easy to understand .
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Singly reinforced beam ast type 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.
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 (Ast) as per IS 456:2000
5. Area of steel (Ast) as per SP-16
6. Example problems #03,04
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)
• 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. Limiting moment of resistance
[SP -16] (Alternate Method)
7
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
9
b= 300mm
d =675 mm
Ast
10. Step 1 : Limiting moment of resistance
[IS 456:2000]
10
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
11
12. Step 1 : Limiting moment of resistance
[SP -16] (Alternate Method)
12
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
13
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
14
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
15
Alternate method
[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
17
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
18
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.
19
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.
20