Flanged beams design for t beam

CE8501 Design Of Reinforced Cement Concrete Elements
Unit 2 – 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.
1

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
2

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
3
𝑦 𝑓 = (0.15 xu + 0.65 Df)
[Refer IS456 Pg.97]

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
)
4

Area of steel (Approximate)
Approximate Ast can be calculated from following expression
Ast =
𝑀 𝑢
0.87 𝑓𝑦 𝑧
Approximate Lever arm can be calculated from following expression
z = d -
𝐷 𝑓
2
5

Problem#04
• 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.
• Given:
6
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

Step 1: Approximate Lever arm distance
z = d -
𝐷 𝑓
2
= 315 -
125
2
z = 252.5 mm
7

Step 2 : Approximate Ast
Ast =
𝑀 𝑢
0.87 𝑓𝑦 𝑧
=
(240 ∗106
)
0.87 ∗415 ∗252.5
Ast= 2632.4 mm2
• Assume 25mm dia bars
No. of Bars =
2632.4
490.9
= 5.36
Hence provide 6#25mm dia bars
Ast= 2945.4 mm2
8

Step 3: Neutral axis depth
Case - I
• Assuming the depth of NA lies within the flange
xu =
0.87 𝑓𝑦 𝐴𝑠𝑡
0.36 𝑓𝑐𝑘 𝑏𝑓
=
0.87 ∗415 ∗2945.4
0.36 ∗20 ∗960
xu = 153.85mm > Df is 125mm [Hence our assumption is Wrong]
9

Step 3: Neutral axis depth
Case II:
• Assuming the depth of NA lies outside the flange
𝐷 𝑓
𝑑
=
125
315
= 0.39 > 0.2
Hence it comes under category 2
10

Step 3 : 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
=
0.87 ∗415 ∗2945.4−0.447 ∗20 960 −200 ∗(0.15 𝑥𝑢 +81.25)
0.36 ∗20 ∗200
1440 xu = (1.06 * 106) – 6794.4 (0.15 xu + 81.25)
1440 xu = 950428 – 1019.6 xu - 552045
2459.6 xu = 511.4 x 103
xu = 207.9 mm
3/7 * 162 = 89.1 mm < Df is 125mm [Hence our assumption is true]
11
𝑦 𝑓 = (0.15 xu + 0.65 Df)
= (0.15 xu + 0.65 * 125)
𝑦 𝑓 = 0.15 xu +81.25

Step 4: Check for 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 * 0.66 [ 1- 0.42 ∗ 0.66 ] 20 * 200 * 3152 }+ {0.45 *20 (960 –200) 112.45 (315 -
112.45
2
)}
= (68 x 106) + (199 x 106)
Mu = 267 x 106 N.mm > 240 kN.m
Hence safe
12
𝑦 𝑓 = 0.15 xu +81.25
= 0.15* 208 + 81.25
𝑦 𝑓 = 112.45 mm
𝑋𝑢
𝑑
=
208
315
= 0.66

Reinforcement details
13
d = 315 mm
xu = 207.9 mm
6 nos. of
25mm dia main bar
bf = 960 mm
Df = 125 mm
Spacing 25mm
bw = 200 mm

Flanged beams design for t beam

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Flanged beams design for t beam