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Unformatted text preview: 3.2.4 Procedures for design of shear reinforcement
(1) Calculate the design shear stress at critical sections:
v = V/(bvd).
(2) Check v < 0.8 f cu or 5 N/mm2, whichever is the lesser.
(3) Determine the design concrete shear stress vc using Eq. (3.2)
or directly from Table 3.8 of BS 8110: Part 1 (Notes p.114).
(4) Determine Asv
using Eq. (3.11).
sv (5) Calculate the spacing of minimum links using Eq. (3.13).
(6) Determine the shear resistance of concrete plus minimum
links from Eq. (3.9); then plot the envelope of this shear
resistance.
(7) Space the web reinforcement (i.e. stirrups arrangement). Asv
for varying stirrup diameter and spacing
sv 123 Example 3.21: Design of shear reinforcement
Design load = 75.2 kN/m
d = 550 b = 300
2R10
2T25
300 fcu = 25 N/mm2
6000 214.3 kN
214.3 kN Vcriticalsection shear force diagram (1) Maximum shear stress
Shear force at the face of support
6
0.3
V = 75.2 × − 75.2 ×
= 214.3 kN
2
2 The maximum shear stress
v max V
214.3 × 10 3
= 1.30 N mm 2 < 0.8 f cu = 4 N mm 2
=
=
bv d
300 × 550 OK
(2) Design concrete shear stress vc
2T25: As = 982 mm 2
100 As 100 × 982
=
= 0.59
bv d
300 × 550 From BS8110: Part 1, Table 3.8 (Notes p. 114)
vc = 0.52 N mm 2 124 (3) Links (stirrups)
The shear force of the section at a distance d from the support face Vcriticalsection = 214.3 – 75.2d = 214.3 – 75.2 × 0.55 = 172.9 kN
Shear stress
v= Vcriticalsection 172.9 × 103
=
= 1.05 N mm 2
300 × 550
bv d Shear links (stirrups) required
Asv bv (v − vc ) 300 × (1.05 − 0.52)
=
=
= 0.67
0.95 f yv
0.95 × 250
sv For R10 links ( Asv = 157 mm2 )
sv = 157
= 234 mm < 0.75d = 413 mm
0.67 Provide R10 links at 230 mm centres (R10@230)
(4) Minimum links ⎛ Asv ⎞
0.4bv
0.4 × 300
=
= 0.51
⎜
⎟=
⎝ sv ⎠ min 0.95 f yv 0.95 × 250
For R10 links
sv = 157
= 308 mm < 0.75d = 413 mm
0.51 Provide R10@300 ( Asv
= 157/300 = 0.52)
sv (5) Shear resistance of concrete plus minimum links Vn = bv vc d + Asv
0.95 f yv d
sv = 300 × 0.52 × 550 + 0.52 × 0.95 × 250 × 550
= 153.7 kN
125 (6) Envelope of the shear resistance of concrete plus minimum links
214.3 kN Required links Minimum links 153.7 kN
153.7 kN
s = 806 mm The distance s = 214.3 − 153.7
= 806 mm
75.2 214.3 kN (7) Stirrup arrangement
5R@230 12R@300 5R@230 2T12 hanger
2T25
50 Note:
In design practice, the enhanced shear strength of sections near supports
may not be considered. The shear force at the centre line of support is
often taken as the design shear force. In this example, the design shear
force will be
V = 75.2 × 6/2 = 225.6 kN The values of V = 214.3 kN and V = Vcriticalsection = 172.9 kN are not used. 126 3.2.5 Shear resistance of bentup bars
• To resist the shearing force, bars may be bent up near the supports as shown in Fig. 3.23. The bentup bars and the concrete in
compression are considered to act as an analogous lattice girder. Fig. 3.23 Bentbar systems
• Single system − The diagonal crack only intercepts one bentup bar.
From Fig. 3.23(a), the spacing of the bentup bars is
sb = (d − d ′)(cot α + cot β ) (3.16) where α = angle between bentup bar and axis of the beam
β = angle between compression strut and axis of the beam The shear resistance of the single bar is Vb = 0.95 f yv Asb sin α (3.17) where Asb is the crosssection area of the bentup bar.
127 • Multiple system − The diagonal crack intercepts more than one bentup bar.
o From Fig. 3.23(b), the shear resistance is given by Vb = 0.95 f yv Asb sin α where (d − d ′)(cot α + cot β )
sb (3.18) (d − d ′)(cot α + cot β )
can be considered as the number
sb of bentup bars intercepting the diagonal crack.
o The truss should be arranged as follows: Both α and β are greater than or equal to 45°; and
Spacing sb has a maximum value of 1.5d [referring to Eq.
(3.16): When α = β = 45°, sb,max = 2(d − d'). Take sb,max =
1.5d]
o With α = β = 45° (this is a common practice) and sb = d – d’, Eq. (3.18) becomes V = 1.34 f yv Asb (3.19) This arrangement is commonly referred to as a double system.
This is because the number of bentup bars crossing the
diagonal crack is (d − d ′)(cot α + cot β ) (d − d ′)(cot 45° + cot 45°)
=
=2
sb
d − d′
• BS 8110: Part1, Clause 3.4.5.6 states that at least 50% of shear resistance provided by the steel should be in the form of links.
128 Example 3.22
Shear resistance of a beam section with both stirrups and bentup bars 100 As 100 x 982
=
= 0.43
bd
350 x 650
From BS 8110 Table 3.8 (Notes p.114), vc = 0.5 N/mm2 by interpolation. Crosssectional area of a size 12 bar = 113 mm2. Vs = Asv /sv·0.95fyvd + bvcd = 349 + 114 = 463 kN 303 kN V = Vs + Vb
= 463 + 303 = 766 kN 129 ...
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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
 Shear Stress

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