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Unformatted text preview: 2.5 Flanged Section in Bending (Tbeam and Lbeam sections) Fig. 2.51 Tbeams and Lbeams in building systems Neutral axis in flange Section at span Section at support Neutral axis in web Section at span Fig. 2.52 Location of neutral axis in Tbeams Actual stress block Equivalent
stress block Actual N.A. Fig. 2.53 Stress blocks of Tbeams. (a) Actual stress block; (b) equivalent
stress block
84 2.5.1 Effective width
• The Tbeam and Lbeam in Fig. 2.54 are examples of flanged
beams. In practice the flange is often the floor slab and the question arises of what width of the slab is to be taken as the
effective width; that is, the width b in Fig. 2.54. Fig. 2.54 Effective width of T and Lbeam sections
• BS 8110 (clause 3.4.1.5) gives the following recommendations:
ο For a Tbeam: the effective width b should be taken as (1) bw +
0.2lz or (2) the actual flange width, whichever is less;
ο For a Lbeam the effective width b should be taken as (1) bw +
0.1lz or (2) the actual flange width, whichever is less; where bw
is the web width (Fig. 2.54);
Tsection slab beam column 85 ο lz is the distance between points if zero moment along the span of
the beam (Clause 3.4.1.5).
lz may be determined from the bending moment diagram; or
lz may be taken as 0.7 times the effective span for a continuous
beam.
Definitions of effective span and effective length:
(1) (Clause 3.4.1.2) Effective span of a simply supported beam
may be taken as the smaller of
the distance between centres of supports; or
the clear distance between supports plus the effective depth.
(2) (Clause 3.4.1.3) Effective span of a continuous beam is the
distance between centres of supports.
(3) (Clause 3.4.1.4) Effective length of a cantilever is:
its length to the face of the support plus half its effective
depth; or, the length to the centre of the support where it
forms the end of a continuous beam.
• Approach of analysis
ο If the depth of stress block is within or equal to the flange thickness,
a flanged section may be analysed and designed as a rectangular
section of the same width b and effective depth d.
ο If the depth of stress block is beyond the flange thickness, the
equivalent rectangular section should not be used.
86 2.5.2 Section with the depth of stress block lying within the flange
( s ≤ hf ) Fig. 2.55 Tsection, stress block within the flange, s < hf
• The beam section can be considered as an equivalent rectangular section of width b equal to the flange width. This is because the section below the neutral axis is in tension and is, therefore,
considered to be cracked and inactive.
• Design − same as that of a singlyreinforced rectangular beam section:
K= M
M
and z = d (0.5 + 0.25 − K / 0.9) , then As =
2
0.95 f y z
bd f cu • Checking Assume that tension steel has yielded at failure, so that fst = 0.95fy .
From 0.45 f cu bs = 0.95 f y As s (or x) z=d− s
2 The moment of resistance is M = Fcc z = 0.45 f cu bs ⋅ z
or M = Fst z = 0.95 f y As ⋅ z
87 Example 2.51: Analysis of a flanged section
Determine the ultimate moment of resistance of the Tbeam section
shown in the figure, given that fy = 460 N/mm2 and fcu = 30 N/mm2. Fst Assume initially that the stress block depth lies within the flange and the
reinforcement is strained to the yield, so that fs = 0.95fy .
For equilibrium: Fcc = Fst therefore
0.45 f cu b f s = 0.95 f y As and solving for the depth of stress block
0.95 × 460 × 1470
0.45 × 30 × 800
= 59 mm < h f = 150 mm s= Hence the stress block does lie within the flange and with this depth of
neutral axis the steel will have yielded as assumed.
z = d – s/2
= 420 – 59/2 = 390 mm
Taking moments about the centroid of As , the moment of resistance is
M = Fcc × z
= 0.45 f cu b f s ⋅ z
= 0.45 × 30 × 800 × 59 × 390 × 10 −6
= 249 kNm
88 2.5.3 Section with the depth of stress block extending to the web (s > hf)
The moment of resistance of the flange is
M f = 0.45 f cu bh f (d − h f / 2) (2.51) If the design ultimate moment
M > Mf
the stress block must extend blow the flange, i.e.
s > hf
(1) Design by an exact method
A T−section (or L−section) can be divided into flange component
and web component, as shown in Fig. 2.56. mf M − mf Fig. 2.56 Flange and web components of a Tsection o Moment of resistance of the flange component m f = 0.45 f cu (b − bw )h f (d − 0.5h f ) (2.52) Steel area corresponding to the flange component
Asf = mf
0.95 f y (d − 0.5h f ) (2.53) 89 o Moment of resistance of the web component mw = M − m f (2.54) Steel area corresponding to the web component
Asw = where M − mf
mw
=
0.95 f y z 0.95 f y z ( z = d 0.5 + 0.25 − K w / 0.9 Kw = (2.55) ) M − mf (2.56) bw d 2 f cu o Total reinforcement area As = Asf + Asw (2.57) Note that when Kw > K’ (K’ can be determined from Eq. 2.41),
compression steel is required.
Example 2.52
The Tsection beam shown in the figure is required to resist an
ultimate design moment of 180 kNm. The characteristic material
strengths are fy = 460 N/mm2 and fcu = 30 N/mm2. Calculate the
area of reinforcement required. FC1
FC2 90 o Moment of resistance of the whole flange: M f = 0.45 f cu b f h f (d − h f / 2)
= 0.45 × 30 × 400 × 100(350 − 100 / 2) × 10−6
= 162 kNm < 180 kNm (the design moment) ∴ The stress block must extend below the flange, i.e. s > h f .
o Moment of resistance of the flange component: m f = 0.45 f cu (b − bw )h f (d − 0.5h f )
= 0.45 × 30 × (400 − 200) × 100(350 − 0.5 × 100)
= 81 kNm
Asf =
= mf
0.95 f y (d − 0.5h f )
81 × 106
= 618 mm 2
0.95 × 460(350 − 0.5 × 100) o Factor for the moment of resistance of the web component: Kw =
= M −Mf
bw d 2 f cu
(180 − 81) × 10 6
= 0.135 < 0.156
200 × 350 2 × 30 z = d (0.5 + 0.25 − K w / 0.9 )
= 350(0.5 + 0.25 − 0.135 / 0.9 )
= 286 mm > 0.775d = 271 mm, O.K. Asw = M − mf
0.95 f y z = (180 − 81) × 106
= 792 mm 2
0.95 × 460 × 286 o Total area of tension steel As = Asf + Asw = 618 + 792 = 1410 mm2
Provide 3T25 (As = 1472 mm2); then ρ = As / ( bw d ) = 1472 / ( 200 × 350 ) = 2.1% < 4% , O.K.
91 (2) Design by the code method (BS 8110: Part 1, clause 3.4.4.5)
o A safe but conservative design for a flanged section with s > hf can be achieved by setting x = d/2. Fst Fig. 2.57 Flange section with x = d/2 The compression forces developed by Area 1 and Area 2 are
Fc1 = 0.45 f cu bw ⋅ s = 0.45 f cu bw ⋅ 0.9d / 2 = 0.2fcubwd
Fc 2 = 0.45 f cu (b − bw ) h f Lever arms are
z1 = d – s/2 = 0.775d and z2 = d – 0.5hf Taking moment about the centroid of the flange
M = Fst ( d − 0.5h f ) − Fc1 ( s / 2 − 0.5h f )
= 0.95 f y As ( d − 0.5h f ) − 0.1 f cu bw d (0.45d − h f ) Therefore,
As = M + 0.1 f cu bw d (0.45d − h f )
0.95 f y ( d − 0.5h f ) (2.58) Eq. (2.58) is only applicable when hf < 0.45d.
92 o With x = d
, the maximum moment of resistance of the
2 concrete is M c = Fc1 z1 + Fc 2 z 2
= 0.156 f cu bw d 2 + 0.45 f cu (b − bw )h f (d − 0.5h f ) or
M c = β f f cu bd 2 (2.59) where β f = 0.45 hf
d (1 − hf
bw
b
)(1 − ) + 0.156 w
2d
b
b (2.60) Table 2.51 Values of the factor βf (Clause 3.4.4.4) o When using Eq. (2.58) to calculate the area of tension steel As, it is necessary to confirm that
the design ultimate moment should be ≤ M c = β f f cu bd 2
(compression steel is not required);
not more than 10% of redistribution has been carried out. 93 2.5.4 Flanged section with compression reinforcement (not more than
10% moment redistribution)
• When the design ultimate moment M > M c = β f f cu bd 2 compression steel is required.
• The moment [M – Mc] will be resisted by compression steel. Hence,
′
M − M c = 0.95 f y As ( d − d ′) (2.61) then
′
As = M − Mc
0.95 f y (d − d ′) ′
As = M − β f f cu bd 2
0.95 f y (d − d ′) or
(2.62) • By considering the equilibrium condition of forces on the section (see Fig. 2.57),
0.95fyAs = compression of flange component
+ compression of web component
+ compression of steel As′
′
= 0.45 f cu (b − bw )h f + 0.45 f cu bw (0.9d / 2) + 0.95 f y As Therefore,
As = 0.45 f cu (b − bw ) h f + 0.202 f cu bw d
0.95 f y ′
+ As (2.63) 94 ...
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This note was uploaded on 12/24/2011 for the course CIVL 232 taught by Professor Jskuang during the Spring '06 term at HKUST.
 Spring '06
 JSKUANG

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