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.
1

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#03
• Determine the ultimate moment of resistance of an isolated T-
beam, having span of 6m and cross sectional dimensions as shown
in fig. Assuming fck = 20Mpa and grade 415 steel.
• Given:
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fck = 20 N/mm2
fy = 415 N/mm2
Ast = 2454 mm2
bf = 900mm
Df = 120mm
bw = 300mm
d = 600mm
900mm
120mm
300mm
600mm
25 mm dia

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 ∗2454
0.36 ∗20 ∗900
xu = 136.7mm > Df is 120mm [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 𝐷 𝑓
0.36 𝑓𝑐𝑘 𝑏𝑤
=
0.87 ∗415 ∗2454 −0.447 ∗20 900 −300 ∗120
0.36 ∗20 ∗300
xu = 112.2 mm
3/7 * 112.2 = 48.08 mm < Df is 120mm [Hence our assumption is wrong]
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8. Step 1: Neutral axis depth
• Assuming the depth of NA lies outside the flange (Category 2)
xu=
0.87 𝑓𝑦 𝐴𝑠𝑡 −0.447 𝑓𝑐𝑘 𝑏 𝑓
−𝑏𝑤 𝑦 𝑓
0.36 𝑓𝑐𝑘 𝑏𝑤
=
0.87 ∗415 ∗2454 −0.447 ∗20 900 −300 ∗(0.15 𝑥𝑢 +78)
0.36 ∗20 ∗300
2160 xu = 886016.7 – 5364 (0.15 xu + 78)
2160 xu = 886016.7 – 804.6 xu - 418392
2964.6 xu = 467.62 x 103
xu = 157.7mm
3/7 * 157.7 = 67.6 mm < Df is 120mm [Hence our assumption is true]
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𝑦 𝑓 = (0.15 xu + 0.65 Df)
= (0.15 xu + 0.65 * 120)
𝑦 𝑓 = 0.15 xu +78

9. Step 2: Compare xu with xu,max
For Fe415
𝑥𝑢,𝑚𝑎𝑥
𝑑
= 0.48
𝑥 𝑢, 𝑚𝑎𝑥 = 0.48 * 600
𝑥 𝑢, 𝑚𝑎𝑥 = 288 mm [𝑥 𝑢 < 𝑥 𝑢, 𝑚𝑎𝑥 , Hence the section is under reinforced]
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10. 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) yf (d-
𝑦 𝑓
2
)
= {0.36 * 0.26 [ 1- 0.42 ∗ 0.26 ] 20 * 300 * 6002 }+ {0.45 *20 (900 –300) 102 (600-
102
2
)}
= (180 x 106) + (302.38 x 106)
Mu = 482.38 x 106 N.mm
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𝑦 𝑓 = 0.15 xu +78
= 0.15* 157.7 + 78
𝑦 𝑓 = 101.65 mm
𝑋𝑢
𝑑
=
157.7
600
= 0.26

11. Assignment#2.1
• A T – Beam having 120mm thick slab has a flange width of
1100mm, Web width of 450mm & web depth of 550mm. Area of
steel in tension is 3041mm2 at a effective cover of 70mm from
bottom. Use M20 & Fe415.
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12. Thank You
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