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Unformatted text preview: Chapter Five
_____________________________________________________________________ Design of Reinforced Concrete Beams The design procedure of reinforced concrete members consists of a series
of interrelated steps and checks, which may generally be condensed into
the following three basic stages:
1. Preliminary analysis and member sizing
2. Design and detailing of steel reinforcement
3. Serviceability calculations 5.1 Preliminary Analysis and Member Sizing The layout and size of members are often controlled by architectural
details. The engineers must either check that the beam sizes are adequate
to carry the loading, or alternatively, decide on sizes that are adequate.
5.1.1 Preliminary analysis and basic dimension requirements
• The preliminary analysis of beams generally only provides the
maximum bending moments and shear forces in order to ascertain
reasonable dimensions.
• Basic requirements of beam dimensions 190 o overall depth (h) and effective depth (d): d = h – (50 ~ 60) mm 1
3 1
2 o breadth (b): b = ( ~ )h
o cover to reinforcement: (generally minimum cover = 25 mm) Fig. 5.11 Beam dimensions 5.1.2 Suitable dimensions for b and d
Suitable dimensions for b and d can be decided by a few trial
calculations.
(1) For singly reinforced section (no compression reinforcement), M
< 0.156
bd 2 f cu
For doubly reinforced section (with compression reinforcement if
the area of bending steel is not to the excessive),
0.156 < M
10
<
2
f cu
bd f cu 191 (2) In no case should v = V/(bvd) exceed or 0.8√fcu or 5 N/mm2 (Hong
Kong code: 0.8√fcu or 7 N/mm2) whichever is the lesser. To avoid
congested shear reinforcement.
The value of v should preferably be somewhat closer to half (or
less) of the maximum allowed.
(3) The span/effective depth ratio (for l ≤ 10 m) should be within the
basic values given below:
Cantilever: 7; Simply supported: 20; Continuous: 26
All calculations in design are generally based on the effective span:
o For simply supported beams – distance c/c of the supports, or clear distance between the supports + d (effective depth);
o For continuous beams – distance c/c of the supports;
o For cantilever beams – length to the face of the support + d/2, or distance to the centre of the support it the beam is continuous
5.1.3 Flow chart for beam design 192 5.2 Design for Bending 5.2.1 Singly reinforced rectangular section
• Summary of design calculations (Notes Section 2.3) (1) K = M
bd 2 f cu ( ) (2) Lever arm z = d 0.5 + 0.25 − K / 0.9 or from table shown (3) As = M
0.95 f y z Select suitable bar sizes (φ16, 20, 25, 32, 40…; Notes p.46)
(4) Check steel ratio
For highyield steel (fy = 460 N/mm2), 0.13% ≤ As/(bd) ≤ 4%;
For mild steel (fy = 250 N/mm2), 0.24% ≤ As/(bd) ≤ 4%. 193 • Using design charts in the code (Notes p.49) Steps (2) and (3) can be replaced using design charts in BS 8110:
Part 3, chart No. 1 and chart No. 2.
Text Example 7.2 (p.149): Design of tension steel for a singly
rectangular section 5.2.2 Doubly reinforced rectangular section
• βb ≥ 0.9 (Moment redistribution ≤ 10%) and d'/d > 0.185
(common cases in practice)
o If M > 0.156 bd2fcu , compression steel is required. M − 0.156 f cu bd 2
(1) As′ =
0.95 f y (d − d ') (2) As = 0.156 f cu bd 2
′
+ As , where z = 0.775d
0.95 f y z o If d'/d > 0.185, the stress in compression steel should be determined first as presented in Subsection 2.4.2 of the Notes.
• βb < 0.9 (Moment redistribution > 10%)
o The limiting depth of the neutral axis is x = (β b − 0.4d )
Compression steel is required if
s M > 0.45 f cu bs d − 2 where s = 0.9x.
194 o The design procedure is as follows. (1) K = M
bd 2 f cu (2) K ′ = 0.402(β b − 0.4) − 0.18(β b − 0.4) 2 If K < K', no compression steel is required.
If K > K', compression steel is required.
(3) x = (β b − 0.4 )d
If d ' / x < 0.37 , compression steel will have yielded; thus
f s′ = 0.95 f y . If d ' / x > 0.37 , calculate εsc ε sc = ε cu x − d' and then f's .
x (4) As′ = ( K − K ') bd 2 fcu
f s′ ( d − d ') (5) As = K ′bd 2 f cu
f s′
+ As′
, where z = d – s/2
0.95 f y z
0.95 f y • Links should be provided as lateral restraint to the outer layer of compression steel:
o Link size ≥ 1/4 size of the largest compression bar;
o sv ≤ 12 times the size of the smallest compression bar. Text Examples 7.3 (βb > 0.9) and 7.4 (βb < 0.9) (pp.151153): Design of
tension and compression steel 195 5.2.3 Tbeams Fig. 5.21 Tbeam and Lbeam sections (1) Empirical rules for the effective width of flange
The lesser of the actual flange width or the width of web + 0.2 lz for Tsection
, where lz = 0.7× times the effective span. 0.1 lz for Lsection
(2) Design procedure ( ) 2
1. Calculate M /(bd f cu ) and z = d 0.5 + 0.25 − K / 0.9 . 2. If d – z < h f / 2 , the stress block falls within the flange depth,
design may proceed as for a rectangular section with breadth b.
3. Transverse reinforcement in flange, as shown in Fig. 5.21,
should be provided across the top of the flange to prevent
cracking:
Area = 0.15% hf × 1000 = 1.5hf mm2 per meter length of the
beam (0.15% of the longitudinal crosssection of the flange).
Text Examples 7.5 (p.155): Design of reinforcement for a Tsection 196 5.2.4 Anchorage
• Design anchorage length: l = 0.95 f y
4 f bu φ , where the ultimate anchorage bond stress fbu = β√fcu, in which values of β are
presented in Table 3.31 of Notes p.132; or l = KAφ, where values
of anchorage coefficient KA are given in Table 3.32 of Notes p.132.
• Lap length Minimum lap length = max {15φ, 300 mm, tension anchorage
length}; or Lap length = KLφ (value of KL are presented in Table
3.34 of Notes p.137)
5.2.5 Curtailment of reinforcing bars
The magnitude of the bending moment on a beam decreases along its
length, and then may the area of bending reinforcement be reduced by
curtailing bars as they are no longer required, as shown in Fig. 5.22. Fig. 5.22 Curtailment of reinforcement 197 • Each curtailed bar should extend beyond the point at which it is no longer needed so that it is well anchored into the concrete:
o The curtailment anchorage should not be less than 12φ or the effective depth d of the beam, whichever is the greater.
o Where a bar is stopped off in the tension zone, one of the following conditions must be satisfied.
1. The bar must extend to the point where shear capacity is
twice the design shear force.
2. The bars continuing past the actual cutoff point provide
double area to resist moment at that point.
3. The curtailment anchorage is increased to a full anchorage
length based on a stress of 0.95fy. • Simplified rules for curtailment of bars in beams:
12 φ d/2 + 12φ Fig. 5.23 Anchorage lengths at a simple support 198 (a) Simply supported beam (b) Continuous beam Fig. 5.24 Simplified rules for curtailment of bars in beams Text Examples 7.6 (pp.158159): Design a beam – bending reinforcement
5.3 Design for Shear and Torsion 199 5.3.1 Design for shear
Shear stress: v = V/ bd
(1) Types of links Fig. 5.31 Types of links Fig. 5.32 Bending of links (2) Design equations
o The size and spacing of stirrups Asv b(v − v c )
≥
sv
0.95 f yv
o If v < (vc + 0.4), minimum links should be provided for whole length of beam such that (unless v < 0.5vc and the beam is a very
minor one), Asv 0.4bv s = 0.95 f
yv v min
o The shear resistance of (concrete + minimum stirrups) is Vn = (vc + 0.4)bv d
Text Example 7.7 (pp.162163): Design of shear steel for a beam.
200 5.3.2 Design for torsion
(1) Torsional shear stress vt =
ο h 2
min 2T
( hmax − hmin 3) When vt > vt,min, where vt,min = 0.067√fcu , but not more than 0.4
N/mm2, torsional reinforcement is required; ο (v + vt ) ≤ vtu, where vtu ≤ 0.8√fcu or 5 N/mm2 (Hong Kong code:
vtu ≤ 0.8√fcu or 7 N/mm2), whichever is the lesser. ο For small sections where y1 < 550 mm,
vt ≤ vtu y1
550 (2) Torsional shear reinforcement
Asv
T
=
sv
0.8 x1 y1 (0.95 f yv ) where sv > min{x1 , y1
, 200 mm}.
2 (3) Torsional longitudinal reinforcement As = Asv f yv ( x1 + y1 )
sv f y The torsional longitudinal bars should be distributed evenly around
the inside perimeter of the links. At least four bars should be used
and the clear distance between bars 300 mm. Text Example 7.9 (pp.166170): Design of a continuous beam.1
1 This Example should carefully be studied again. 201 ...
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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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