Unformatted text preview: 3.2 Shear Strength of Beams with Shear Reinforcement
• The complementary shear stress gives rise to diagonal tensile and diagonal compressive stresses.
• Taking a simplified view, concrete is weak in tension, and shear failure is then caused by a failure in diagonal tension with cracks
running at about 45°.
• Shear reinforcement should be provided by bars that cross the cracks. Theoretically, either vertical links (stirrups) or inclined
bars will serve this purpose.
3.2.1 Types of shear reinforcement in beams
(1) Links or vertical stirrups (2) Bend-up bars (inclined bars) (3) Combination system (links + bend-up bars) 115 The shear force V is resisted by shear in concrete and shear in steel: V = Vc + Vs (3.5) where the shear resistance of the concrete Vc = vcbv d (3.6) 3.2.2 Shear resistance with links (stirrups) − The 45 o truss model
• The design and analysis for shear are based on the concept of an analogous truss. In the truss shown,
d tension d
~ 45º Links
T b (stirrups) compression
V Fig. 3.2-1 The 45° truss analogy
o The links are spaced at a distance equal to d, so that the diagonal concrete compression members are at an angle of
about 45°, which generally agrees with the experimental
observations of the cracking of reinforced concrete beams close
to the supports.
o The web bars are considered to form the tension members of the imaginary truss, while the trusts in the concrete constitute the
116 o The shear resistance of the vertical links (stirrups) is Vs = 0.95 f yv Asv (3.7) where Asv is the cross-sectional area of the stirrups.
o When the stirrup spacing is less than d, the number of stirrups crossing a 45°−crack is about d / s v , where sv is the stirrup
spacing. Fig. 3.2-2 Stirrups crossing a 45° diagonal crack
o The shear resistance of the stirrups is Vs = 0.95 f yv Asv ⋅ d ⎛ Asv ⎞
⎟ 0.95 f yv d
sv ⎝ sv ⎠ (3.8) • The ultimate shear strength (shear resistance) of the cross-section
with vertical stirrups is V = Vc + Vs
= vcbv d + ⎜ sv ⎟ 0.95 f yv d
⎝ sv ⎠
resistance (3.9) web reinforcement
resistance 117 Q The design shear stress is
bv d The shear resistance of the cross-section presented by Eq. (3.9) can
then be written as (v − vc )bv d = ( Asv / sv ) 0.95 f yv d (3.10) o Eq. (3.10) is expressed in the BS 8110’s form, Asv (v − vc )bv
0.95 f yv (3.11) ⎛A ⎞
where ⎜ sv ⎟ can be considered as a relative web reinforcement
⎝ sv ⎠
o The actual stirrup ratio is defined by ρv = Asv
sv bv (3.12) Total area = Asv bv sv /2 sv sv /2
• Stirrup sizes are usually from 8-mm to 16-mm diameter, and they are often of mild steel to minimise the radii of bends.
o Minimum links should be provided as (Clause 126.96.36.199) ⎛ Asv ⎞
⎝ sv ⎠ min 0.95 f yv (3.13) 118 By comparing Eqs (3.11) and (3.13), it can be seen that (v – vc) =
0.4 N/mm2, indicating minimum links should at least provide a
design shear resistance of 0.4 N/mm2 (i.e. 0.4 N/mm2 is the minimum design shear resistance).
Reason: where the web reinforcement is used, it remains
unstressed until diagonal cracking occurs. If the amount of web
steel is too small, the sudden stress increase may cause the instant
yielding of the web bars.
o Maximum shear stress (Clause 188.8.131.52) The design shear stress, v = V/(bv d), at any cross-section must in
no case exceed 0.8√fcu or 5 N / mm 2 , whichever is the lesser,
whatever shear reinforcement is provided (This limit has included
an allowance for γm = 1.25), i.e. v= V
≤ 0.8 f cu or 5 N/mm 2
bv d (3.14) whichever is the lesser.
This upper limit prevents crashing of the concrete in the
direction of the maximum principal compressive stresses (in
diagonal compression in general).
When the design shear stress v = V
is exceeded, the crossbv d section sizes should be increased. 119 o The maximum shear stress presented in Clause 184.108.40.206 of the Hong Kong code (Buildings Department, Code of Practice for Structural Use of Concrete, 2004, HKSAR):
≤ 0.8 f cu or 7 N/mm 2
bv d whichever is the lesser, whatever shear reinforcement is provided
(this limit has included an allowance for γm = 1.25).
o Maximum spacing of links (Clause 220.127.116.11) In the direction of span, sv,max = 0.75d. Thus,
sv ≤ 0.75d (3.15) This condition ensures that at least one stirrup intercepts one
• Form and area of shear reinforcement (BS 8110: Part 1, Table 3.7) 120 3.2.3 Enhanced shear strength of sections close to supports (Clause
• Shear failure at sections of beams without shear reinforcement will normally occur on a plane inclined at an angle of about 30° to the
horizontal. If the angle of failure plane is forced to be inclined
more steeply than this (due to the load or because the section where
the shear is to be checked, such as X-X in Fig. 3.1-9, is close to a
support), the shear forced required to produce failure is increased,
i.e. the shear capacity of these sections is increased. NOTE: The shear
causing failure is that
acting on section X-X Fig. 3.1-9 Shear failure near supports
• This enhancement of shear strength may be taken into account in the design of sections within a distance of 2d from the face of a
support by increasing the design concrete shear stress, from vc to
(2d/av)⋅vc, provided that the shear stress v at the face of the support
remains the lesser of 0.8√fcu or 5 N/mm2. (The enhanced shear
resistance is partly because the concrete in diagonal compression
resists shear. In this case, the shear span ratio is small.)
121 • Simplified approach (Clause 18.104.22.168)
o Take the critical section for design at a distance d form the face of the support using the code specified value vc . d d
Maximum shear stress
at face of support Critical section for
simplified approach Fig. 3.1-10 Critical section for design o Calculate the shear reinforcement required at this critical section; then provide this reinforcement at all sections up to the
face of the support.
In other words, the shear stress at the critical section is used for
design to replace the maximum shear stress at the face of the
support. 122 ...
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