RCC design, Analysis of flanged beam, T beam, anna university, CE8501, Moment of resistance, neutral axis depth, Civil Engineering, design of beams, limit state method, IS 456, SP 16

1. CE8501 Design Of Reinforced Cement Concrete Elements
Unit 2 – Design of Beams
Analysis of Flanged beams
[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. Neutral axis depth
• Case I: Neutral axis lies within the flange [xu<Df, if true]
xu =
0.87 𝑓𝑦 𝐴𝑠𝑡
0.36 𝑓𝑐𝑘 𝑏
• Case II: Neutral axis lies ouside the flange
Category 1: 3/7 xu ≥ Df
Category 2: 3/7 xu < Df
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3. Neutral axis depth
Case II: Neutral axis lies ouside the flange
Category 1 : [3/7 xu ≥ Df]
xu=
0.87 𝑓𝑦 𝐴𝑠𝑡
−0.447 𝑓𝑐𝑘 𝑏 𝑓
−𝑏𝑤 𝐷 𝑓
0.36 𝑓𝑐𝑘 𝑏𝑤
Category 2 : [3/7 xu < Df]
xu=
0.87 𝑓𝑦 𝐴𝑠𝑡
−0.447 𝑓𝑐𝑘 𝑏 𝑓
−𝑏𝑤 𝑦 𝑓
0.36 𝑓𝑐𝑘 𝑏𝑤
3
𝑦 𝑓 = (0.15 xu + 0.65 Df)
[Refer IS456 Pg.97]

4. Moment of resistance
• Case I
Mu = 0.87 fy Ast d [1 -
𝐴 𝑠𝑡
𝑓𝑦
𝑏 𝑑 𝑓𝑐𝑘
]
• Case II (Category 1) (Df)
Mu = 0.36
𝑥 𝑢
,
𝑚𝑎𝑥
𝑑
[ 1- 0.42
𝑥 𝑢
,
𝑚𝑎𝑥
𝑑
] fck bw d2 + 0.45 fck (bf –bw) Df (d-
𝐷 𝑓
2
)
• Case II (Category 2) (yf)
Mu = 0.36
𝑥 𝑢
,
𝑚𝑎𝑥
𝑑
[ 1- 0.42
𝑥 𝑢
,
𝑚𝑎𝑥
𝑑
] fck bw d2 + 0.45 fck (bf –bw) yf (d-
𝑦 𝑓
2
)
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5. Problem#02
• Determine the moment of resistance of T- beam having
bf=1000mm, bw= 300mm, overall depth is 550mm, 10 nos. of 20mm
bars are provided. Use M20 & Fe415.
• Given:
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fck = 20 N/mm2
fy = 415 N/mm2
Ast = 3141 mm2
Assume effective Cover
of 50mm
d = 550 -50 = 500mm
120mm
300mm
550 mm
1000mm

6. Step 1: Neutral axis depth
• Assuming the depth of NA lies within the flange
xu =
0.87 𝑓𝑦 𝐴𝑠𝑡
0.36 𝑓𝑐𝑘 𝑏𝑓
=
0.87 ∗415 ∗3141
0.36 ∗20 ∗1000
xu = 157.5 mm > Df is 100mm [Hence our assumption is Wrong]
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7. Step 1: Neutral axis depth
• Assuming the depth of NA lies outside the flange
xu=
0.87 𝑓𝑦 𝐴𝑠𝑡 −0.447 𝑓𝑐𝑘 𝑏 𝑓
−𝑏𝑤 𝐷𝑓
0.36 𝑓𝑐𝑘 𝑏𝑤
=
0.87 ∗415 ∗3141 −0.447 ∗20 1000 −300 ∗100
0.36 ∗20 ∗300
xu = 233.36 mm
3/7 * 233.36 = 100.01 = Df is 100mm [Hence our assumption is correct]
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8. Step 2: Compare xu with xu,max
For Fe415
𝑥𝑢,𝑚𝑎𝑥
𝑑
= 0.48
𝑥 𝑢, 𝑚𝑎𝑥 = 0.48 * 500
𝑥 𝑢, 𝑚𝑎𝑥 = 240 mm [𝑥 𝑢 < 𝑥 𝑢, 𝑚𝑎𝑥 , Hence the section is under reinforced]
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9. Step 3: Moment of resistance
Mu = 0.36
𝑥 𝑢
,
𝑚𝑎𝑥
𝑑
[ 1- 0.42
𝑥 𝑢
,
𝑚𝑎𝑥
𝑑
] fck bw d2 + 0.45 fck (bf –bw) Df (d-
𝐷 𝑓
2
)
= {0.36 * 0.46 [ 1- 0.42 ∗ 0.46 ] 20 * 300 * 5002 }+ {0.45 *20 (1000 –300) 100 (500-
100
2
)
= (200.4 x 106) + (283.5 x 106)
Mu = 483.9 x 106 N.mm
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𝑋𝑢
𝑑
=
233.6
500
= 0.46

10. Thank You
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