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Unformatted text preview: 3.4 Sections Subjected to Torsion 3.4.1 Reinforced concrete members subjected to torsion
Reinforced concrete members are commonly subjected to bending
moments, to transverse shears associated with those bending moments
and, in the case of columns, to axial force often combined with bending
and shear. In addition, torsional forces may act, tending to twist a member about its longitudinal axis.
edge girder floor beam
column (a) Edge beams Fig. 3.41 Examples of concrete structures with significant torsion 140 3.4.2 Torsional behaviour
• A plain concrete beam subjected to pure torsion
o Torsional moment produces shear stresses, which result in principal tensile stresses inclined at approximately 45° to the
longitudinal axis.
o Diagonal cracks occur when these tensile stresses exceed the tensile strength of the concrete. These cracks will form a spiral
around the members as shown in Fig. 3.41. Fig. 3.41 Torsional cracks of plain concrete beams
• Torsional reinforcement
o Torsional reinforcement will carry the forces from increasing torsional moment after cracking in the form of closed links and
longitudinal bars by a special truss action, as shown in Fig.
3.42.
o Space truss analogy (Fig. 3.43) Steel reinforcement − tension members
Concrete – compressive struts between links
141 Closed links
Longitudinal bars Fig. 3.42 Torsional reinforcement ⎧ yielding of reinfocement
⎪
coupled with
o Failure will eventually occur by ⎨
.
⎪crushing of the concrete
⎩ Fig. 3.43 Truss action
o It is assumed that tension reinforcement in the form of closed links must be provided to resist the full torsional moment.
Then,
142 Torsional force in a twoleg link: F = Asv
× 0.95fyv
2 Torsional moment of resistance provided by one closed link T = F ⋅ x1 + F ⋅ y1
from
vertical legs (3.28)
from
horizontal legs o When links are provided at sv apart, the total torsional resistance T = F ⋅ x1 ⋅ y1
x
+ F ⋅ y1 ⋅ 1
sv
sv (3.29) where y1 / sv – number of vertical legs crossing a crack
x1 / sv – number of horizontal legs crossing a crack
Eq. (3.29) can be written as
T= Asv
(0.95 f yv )x1 y1 + Asv (0.95 f yv )y1 x1
sv
sv
2
2 (3.30) 3.4.3 Design formulas (BS 8110: Part 2, Clause 2.4.7)
• Closed links (closed stirrups)
o From Eq. (3.30), the total torsional resistance is expressed as T= Asv
x1 y1 (0.95 f yv )⋅ 0.8
sv (3.31) where x1 is smaller dimension of the link, y1 is larger dimension
of the link, Asv is sectional area of the two legs of a link, and sv
is stirrup spacing.
The efficiency factor of 0.8 is included to allow for errors in
assumptions made about the truss behaviour.
143 o Hence a twoleg closed link must be provided such that Asv
T
≥
sv 0.8 ⋅ x1 y1 (0.95 f yv ) (3.32) • Longitudinal bars
o Longitudinal bars resist the longitudinal component of the diagonal tension forces
○ Recall Fig. 3.42, Considering a length sv of the beam, the torsional longitudinal
reinforcement should be such that the total quantity is equal to
the same volume as the steel in the links: As sv = Asv
(2x1 +2y1).
2 The longitudinal bars and the links should fail simultaneously.
This is achieved by making the steel volume multiplied by the
characteristic strength the same for each set of bars. This gives
As sv fy = Asv (x1 + y1) fyv
or ⎛x y ⎞
As f y = Asv f yv ⎜ 1 + 1 ⎟
⎜s s ⎟
v⎠
⎝v ∴ The torsional longitudinal bars required
As ≥ Asv f yv
(x1 + y1 )
sv f y (3.33) 144 o Note that reinforcement in beams includes the reinforcement required for flexure and shear plus that required for torsion.
o To ensure the proper action of the links, longitudinal bars must be provided evenly distributed round the inside perimeter of the
links, as shown in Fig. 3.44. Fig. 3.44 Longitudinal bars evenly distributed round the links • Reinforcement detailing requirements (BS 8110: Part 2, clauses
2.4.8 and 2.4.9)
1. sv ⎧y
⎫
min ⎨ x1 , 1 , 200 mm ⎬ .
⎩2
⎭ 2. The links are to be of the closed type.
3. The clear distance between torsional bars 300 mm and at least four bars (one in each corner) should be used. Fig. 3.45 Detailing of torsional bars
145 ...
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 Spring '06
 JSKUANG

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