This document discusses the analysis of flanged beams. Flanged beams are reinforced concrete beams where a portion of the integrated slab acts as a flange to resist loads. The document outlines the assumptions and equations used to calculate the neutral axis depth and moment of resistance for flanged beams. It then provides an example problem calculating these values for a T-beam with given dimensions, reinforcement, and material properties. The neutral axis depth is found to lie within the flange. The moment of resistance is then calculated accordingly.

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. Analysis of Flanged beams
1. Introduction to Flanged beams.
2. Neutral axis depth
3. Assumptions for Neutral axis depth
4. Compare Xu and Df
5. Compare Xu and Xu,max
6. Determine Moment of resistance
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3. Flanged beams
• In RCC construction, slabs and beams are cast monolithic-ally.
• In such construction, a portion of the slab act integrally with the
beam and bends along with the beam under the loads.
• Such beams are termed as flanged beams.
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4. Flanged beams
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5. Flanged beams
• The portion of the slab which acts integrally with the beam to resist
loads is called as Flange.
• The portion of the beam below the flange is called as Web or Rib of the
beam.
• The intermediate beams supporting the slab are called as T-beams and
the end beams are called as L-beams.
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6. Assumptions for Neutral axis depth
• Case I: Neutral axis lies within the flange [xu<Df, if true]
• Case II: Neutral axis lies ouside the flange
a) 3/7 xu ≥ Df
b) 3/7 xu < Df
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7. 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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8. 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 𝑏𝑓−𝑏𝑤 𝐷 𝑓
0.36 𝑓𝑐𝑘 𝑏𝑤
Category 2 : [3/7 xu < Df]
xu=
0.87 𝑓𝑦 𝐴𝑠𝑡 −0.447 𝑓𝑐𝑘 𝑏𝑓−𝑏𝑤 𝑦 𝑓
0.36 𝑓𝑐𝑘 𝑏𝑤
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𝑦 𝑓 = (0.15 xu + 0.65 Df)
[Refer IS456 Pg.97]

9. 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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10. Problem#01
• Find the moment of resistance of an existing T – Beam having the
following data bf= 740mm, d= 400mm, bw= 240mm, Df= 100mm,
Ast = 5 nos. of 20mm dia bars. Concrete grade is M15 & use mild
steel.
• Given:
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fck = 15 N/mm2
fy = 250 N/mm2
Ast = 1570 mm2

11. Step 1: Neutral axis depth
• Assuming the depth of NA lies within the flange
xu =
0.87 𝑓𝑦 𝐴𝑠 𝑡
0.36 𝑓𝑐𝑘 𝑏𝑓
=
0.87 ∗250 ∗1570
0.36 ∗15 ∗740
xu = 85.45mm < Df [Hence our assumption is true]
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12. Step 2: Compare xu with xu,max
For Fe250
𝑥𝑢,𝑚𝑎𝑥
𝑑
= 0.53
𝑥 𝑢, 𝑚𝑎𝑥 = 0.53 * 400
𝑥 𝑢, 𝑚𝑎𝑥 = 212 mm [𝑥 𝑢 < 𝑥 𝑢, 𝑚𝑎𝑥 , Hence the section is under reinforced]
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13. Step 3: Moment of resistance
Mu = 0.87 fy Ast d [1 -
𝐴 𝑠𝑡
𝑓𝑦
𝑏 𝑓
𝑑 𝑓𝑐𝑘
]
= 0.87 * 250 * 1570 * 400 [1- (
1570∗250
740 ∗ 400∗15
)]
Mu = 124.52 x 106 N.mm
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14. Thank You
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