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Unformatted text preview: 2.3 Singly Reinforced Rectangular Section in Bending 2.3.1 Design formulae
The design formulae are derived using the BS 8110 simplified
stress block. z = la d Fst εs Fig. 2.31 Singly reinforced section with rectangular stress block (1) Design formulae
For equilibrium, the ultimate design moment M = Fcc ⋅ z = Fst ⋅ z (2.15) where z is the lever arm (the distance between Fcc and Fst)
o Concrete compressive force Fcc = 0.45fcu b(0.9x) (2.16a) Fst = 0.95fy As (2.16b) z = d – s/2 = d − 0.45x (2.16c) o Steel tensile force o Lever arm 43 Substituting Eqs (2.16a) and (2.16c) into Eq. (2.15) gives M = 0.9 f cu b(d − z ) z (2.17) Let
K= M
bd 2 f cu (2.18) Eq. (2.17) becomes ( z / d ) 2 − ( z / d ) + K / 0 .9 = 0 (2.19) Solving this quadratic equation gives z = d [0.5 + 0.25 − K / 0.9 ] (2.20) which is the equation presented in BS 8110 for determining the
lever arm of a singly reinforced section.
o By substituting Eq. (2.16b) into Eq. (2.15), the tension steel As = M
0.95 f y ⋅ z (2.21) o Both Eqs (2.20) and (2.21) are used to design the area of tension steel to resist an ultimate moment, M.
In BS 8110: Part 1,
o Upper limit of the lever arm (Clause 3.4.4.1): z max = 0.95d ( xmin = 0.11d )
for maintaining a minimum area of concrete in compression
o Lower limit of the lever arm (Clause 3.4.4.4): z min = 0.775d ( xmax = 0.5d )
44 i.e. 0.11d ≤ x ≤ 0.5d Fig. 2.32 Leverarm example
o With x = xmax = 0.5d (z = 0.775d), Eq. (2.17) becomes M = 0.9 f cu b(d − 0.775d ) × 0.775d (2.22) M = 0.156 bd 2 f cu (2.23) or This is the maximum moment of resistance for a singly
reinforced section. As compared to Eq. (2.18), K = 0.156. o When M
= K > 0.156
bd 2 f cu
compression reinforcement is required to supplement the
moment of resistance of concrete.
45 (2) Design procedure
(i) Calculate K value (Eq. 2.18) K = M /(bd 2 f cu )
(ii) Determine lever arm (Eq. 2.20) z = d [0.5 + 0.25 − K / 0.9 ]
(iii) Determine area of steel reinforcement (Eq. 2.21) As = M
0.95 f y ⋅ z (iv) Select suitable bar sizes
Table 2.31 Areas of groups of bars
Number of bars in groups Diameter (mm)
1 3 4 5 6 7 8 28
50
78
113
201
314
490
804 6
8
10
12
16
20
25
32 2
56
100
157
226
402
628
981
1608 84
150
235
339
603
942
1472
2412 113
201
314
452
804
1256
1963
3216 141
251
392
565
1005
1570
2454
4021 169
301
471
678
1206
1884
2945
4825 197
351
549
791
1407
2199
3436
5629 226
402
628
904
1608
2513
3927
6433 (v) Check reinforcement ratio ρ =
i. ρ max = 4% As
bd (In BS 8110: Part 1, Clause 3.12.6.1: As /Ac or As′ / Ac ≤ 4% ).
ii. ρ min = ⎧0.13% (highyield steel)
⎨
⎩0.24% (mild steel) 2
In Clause 3.12.5.3: As /Ac ≥ 0.13% (for f y = 460 N/mm ) ,
2
and As /Ac ≥ 0.24% (for f y = 250 N/mm ) . 46 Example 2.31
Design of a singly reinforced rectangular section
The ultimate design moment to be resisted by the section in the following
figure is 185 kNm. Determine the area of tension reinforcement (As) required, given that characteristic material strengths are fy = 460 N/mm2
and fcu = 30 N/mm2. K= M
bd 2 f cu 185 × 10 6
= 0.122
260 × 440 2 × 30
< 0.156
= Therefore compression steel is not required. Design example – singly
reinforced section Lever arm:
⎡
K⎤
z = d ⎢0.5 + 0.25 −
⎥
0.9 ⎦
⎣
⎡
0.122 ⎤
= 440⎢0.5 + 0.25 −
⎥
0.9 ⎦
⎣
= 369 mm < 0.95d = 418 mm. O.K. (Or alternatively, the value of z = la d may be obtained form the leverarm
diagram, Fig. 2.32.)
The area of tension reinforcement: As = M
0.95 f y z 185 ×106
0.95 × 460 × 369
= 1147 mm 2
= Provide 4T20, As = 1256 mm2 (from Table 2.31)
47 Example 2.32
A simply supported rectangular beam of 8 m span carries a uniformly
distributed dead load including an allowance for selfweight of 7kN/m
and an imposed load of 5 kN/m. The beam dimensions are breadth b =
250 mm and effective depth d = 400 mm. Use grade 30 concrete and
highyield steel reinforcement.
i) Design load = (1.4×7) + (1.6×5) = 17.8 kN/m
Design moment = 17.8 × 82/8 = 142.4 kNm
ii) K = 142.4 x 10 6
M
=
= 0.119
bd 2 f cu 250 x 400 2 x 30 iii) Lever arm
z = 400 [0.5 + (0.25  0.119/0.9)1/2 ]
= 337 mm
< 0.95d = 380 mm. O.K. iv) Steel area
As = M
0.95 f y z 142.4 × 10 6
0.95 × 460 × 337
= 967 mm 2
= Provide 2T25, As = 981 mm2
v) Reinforcement ratio ρ=
= As
bd
981
= 0.98%
250 × 400 ρmin = 0.13% < 0.98% < ρmax = 4%. O.K.
48 (3) Design charts
ο Design charts for singly reinforced beams in BS 8110: Part 3. ο Design procedure:
First, calculate M/(bd2); then directly determine 100As /(bd)
from the charts with the given value of fcu .
49 ο In Example 2.31,
185 ×10 6
M
= 3.67
=
bd 2 260 × 440 2 Using the Chart No. 2, 100 As
= 1.07
bd
∴ As = 1.07 / 100 × 260 × 440 = 1220 mm 2 (compare to the result As = 1147 mm2)
ο In Example 2.32,
M 142.4 × 10 6
= 3.56
=
250 × 40 2
bd 2 Using the Chart No. 2, 100 As
= 1.05
bd
∴ As = 1.05 / 100 × 250 × 400 = 1 050 mm 2 (compare to the result As = 967 mm2) 50 2.3.2 Checking (Section analysis)
• Aim at determining the moment of resistance of a given section
with a known area of steel reinforcement. • Moment of resistance
The tension reinforcement is assumed to have yielded; then z = la d εs Q Fcc = Fst or Fs t 0.45fcu bs = 0.95fy As ∴ The depth of stress block
s= and 0.95 f y As (2.24) 0.45 f cu b x = s / 0 .9 (2.25) ∴ The moment of resistance of the section is
M R = Fst ⋅ z
= 0.95 f y As (d − s / 2) or M R = 0.95 f y As (d − 0.95 f y As
0.9 f cu b ) (2.26) Question: If the tension steel has not yielded, how to determine
the moment of resistance of the section?
51 Example 2.33
Analysis of singly reinforced rectangular section in bending
Determine the ultimate moment of resistance of the crosssection shown
in the following figure, given that the characteristic strengths are fy = 460
N/mm2 for the reinforcement and fcu = 30 N/mm2 for the concrete. Analysis example
– singly reinforced
section Fst For equilibrium of the compressive and tensile forces on the section,
Fcc = Fst
therefore 0.45fcubs = 0.95fyAs i.e. 0.45×30×300×s = 0.95×460×1470 Then s = 159 mm and x = s / 0.9 = 159/0.9 = 177 mm Moment of resistance of the given crosssection is M = Fst × z
= 0.95 f y As (d − s / 2)
= 0.95 × 460 × 1470(520 − 145 / 2) × 10 −6
= 287 kNm 52 ...
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 Spring '06
 JSKUANG

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