RCC design, design 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 β Design of Beams
Design 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
β’ Case I: Neutral axis lies within the flange [xu<Df, if true]
xu =
0.87 ππ¦ π΄π π‘
0.36 πππ π
β’ Case II: Neutral axis lies ouside the flange
Category 1: 3/7 xu β₯ Df
Category 2: 3/7 xu < Df
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4. Moment of resistance
β’ Case I
Mu = 0.87 fy Ast d [1 -
π΄ π π‘
ππ¦
π π πππ
]
β’ Case II (Category 1) (Df)
Mu = 0.36
π₯ π’
,
πππ₯
π
[ 1- 0.42
π₯ π’
,
πππ₯
π
] fck bw d2 + 0.45 fck (bf βbw) Df (d-
π· π
2
)
β’ Case II (Category 2) (yf)
Mu = 0.36
π₯ π’
,
πππ₯
π
[ 1- 0.42
π₯ π’
,
πππ₯
π
] fck bw d2 + 0.45 fck (bf βbw) yf (d-
π¦ π
2
)
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5. Area of steel (Approximate)
Approximate Ast can be calculated from following expression
Ast =
π π’
0.87 ππ¦ π§
Approximate Lever arm can be calculated from following expression
z = d -
π· π
2
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6. Problem#04
β’ In a flanged beam, bf= 960mm, bw=200mm, Df=125mm, d= 315mm
and factored moment = 240 kNm. Check the capacity of the beam
to carry the load and if it is safe, design the steel required. Assume
Fe 415 steel and M20 concrete.
β’ Given:
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fck = 20 N/mm2
fy = 415 N/mm2
Ast = ?
bf= 960mm
bw=200mm
Df=125mm
d= 315mm
Mu = 240 kNm.
125 mm
200 mm
315 mm
960 mm
7. Step 1: Approximate Lever arm distance
z = d -
π· π
2
= 315 -
125
2
z = 252.5 mm
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9. Step 3: Neutral axis depth
Case - I
β’ Assuming the depth of NA lies within the flange
xu =
0.87 ππ¦ π΄π π‘
0.36 πππ ππ
=
0.87 β415 β2945.4
0.36 β20 β960
xu = 153.85mm > Df is 125mm [Hence our assumption is Wrong]
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10. Step 3: Neutral axis depth
Case II:
β’ Assuming the depth of NA lies outside the flange
π· π
π
=
125
315
= 0.39 > 0.2
Hence it comes under category 2
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