Unformatted text preview: Chapter Three
_____________________________________________________________________ Shear, Bond and Torsion 3.1 Shear Failure of RC Beams without Shear Reinforcement 3.1.1 Shear in homogeneous beams and in RC beams
• In mechanics of materials, the shear stress of homogeneous, elastic and uncracked beams can be calculated by (average shear stress)
Ib where V − shear force
Q – first moment of area ( Q = A ⋅ y ) (a) (b) (c) Fig. 3.1-1 Normal, shear and principal stresses in homogeneous uncracked
beam. (a) Flexural and shear stresses acting on elements in shear span; (b)
distribution of shear stress; (c) principal stresses on elements in shear span
103 Fig. 3.1-2 Principal compressive stress trajectories in an uncracked beam
• Crack vs flexural and shear stresses for a RC beam (g) Fig. 3.1-3 Orientation of principal stresses. (a) Geometry and loading; (b)
axial stress at section x-x; (c) shear stress distribution at section x-x; (d)
segment A; (e) segment B; (f) segment C; (g) principal compressive stress
paths 104 o Crack patterns follow the principal compressive stress paths
o Stresses in different segments Segment A (at the bottom):
v = 0 and σ = σmax Vertical flexural crack. Segment B (at the neutral axis):
v = vmax and σ = 0 about 45°−diagonal crack. Segment C (between the N.A. and the bottom):
Combination of v and σ 45°−90° diagonal crack. 3.1.2 Types of shear failure
The type of shear failure of RC beams depends mainly on various factors:
Shear span-to-depth ratio
Quantity of longitudinal reinforcement ratio ρ = As /(bd )
Geometry of the beam, etc.
• One of the most significant factors is the shear span-to-depth ratio, av / d . Shear span av is defined as the distance between points of
zero and maximum moments. Fig. 3.1-4 Shear span. (a) Geometry and loading; (b) bending moment
105 The effective shear-span/depth ratio is defined as M / Vd , where
M is the bending moment and V the corresponding shear force.
This is true for both distributed loading or concentrated loading.
• Inclined cracks must develope before complete shear failure. Inclined cracks can form by shear-web cracking or, more
commonly, by shear-flexural cracking.
• The failure mode is strongly dependent on the shear-span/depth ratio a v / d .
The shear-span/depth ratio can be divided into four general
categories. For different members falling into the same category,
the sequence of events and the nature of the failure are
approximately the same.
Category Shear span/depth ratio Category I av
d Category II 1< av
d Mode of failure
Deep beam failure
Dowel failure Category III 2.5 < av
d Shear-tension failure
(Diagonal tension failure)
Shear-bond failure Category IV 6< av
d Flexural failure 106 (1) Category I ( av / d < 1 ): Deep beam failure The structural behaviour approaches that of deep beams.
The diagonal crack occurs approximately along a line joining
the loading and support points. It forms as a result of the
splitting action of the compression force transmitted directly
from the loading point to the support.
The shear-web crack initiates frequently at about d/3 about the
bottom face, and propagates simultaneously towards the loading
and support points. When the crack has penetrated sufficiently
deeply into the concrete zone at the loading/the support points,
crashing failure of concrete occurs.
The mode of failure may be an anchorage failure at the end of
the reinforcement due to the large tensile force. 107 (2) Category II: 1 < av / d < 2.5
Shear-compression failure The diagonal crack (shear-web crack) forms independently and
not as a development of a flexural crack. Eventually, the diagonal crack penetrates into the concrete compression zone at
the loading point. Crashing failure of the concrete occurs.
Dowel failure The member can fail owing to dowel failure of the longitudinal
reinforcement at the point of the inclined crack.
(3) Category III ( 2.5 < av / d < 6 ): Diagonal tension failure 108 Flexural crack (a-b) is developed before the compressive force
is great enough to develop shear-web cracks. Then the flexural
crack (a-b) nearest the support would propagate towards
loading point, becoming an inclined crack (a-b-c) known as a
If a v / d is relative low, the diagonal crack (a-b-c) spreads but
stops at j ; the crack widens and propagates along the level of
tension reinforcement (g-h), destroying the bond between the
reinforcement and the surrounding concrete. This is the shearbond failure.
The shear-bond failure may also occur in the Category II
members: If a v / d is relative high, the diagonal crack (a-b-c) spreads to
e, splitting the beam into two pieces along an inclined crack (ab-c-e). This is typical diagonal tension failure.
(4) Category IV ( av / d > 6 ): Flexural failure Beams with such a high value of a v / d usually fail in bending.
109 • Summary of types of failure: Fig. 3.1-5 Summary of types of failure for beams without shear reinforcement 3.1.3 Mechanisms of shear transfer
Consider a typical beam in bending and shear shown below, which is
reinforced with longitudinal steel against bending. Fig. 3.1-6 A typical beam in bending and shear
• The shear force is transmitted through the cracked beam by a combination of three mechanisms (Fig. 3.1-7):
(1) Dowel action of longitudinal reinforcement
(2) Aggregate interlock
(3) Shear stress in uncracked concrete
110 Fig. 3.1-7 Shear transfer mechanisms. (a) Dowel action; (b) aggregate
interlock; (c) shear stresses in uncracked concrete • The external shear force V is considered to be resisted by the combined action of Vcz, Va and Vd . Fig. 3.1-8 Transmitted shear forces Thus, V = Vcz + Va + Vd
111 The shear force V is carried in the approximate proportions:
Compression zone shear Vcz = 20 − 40%; Aggregate interlock
Va = 35 − 50%; Dowel action Vd = 15 − 25% .
• The shear capacity is strongly dependant upon the shear span-depth ratio a v / d , but may also be affected by:
o Tension reinforcement ratio ρ = As / (bd) ρ ↑, probably the aggregate-interlock and also dowel action ↑ Fig. 3.1-9 Effect ρ and fcu on
nominal ultimate shear stress v o Concrete strength fcu fcu ↑, possibly the shear capacities due to aggregate-interlock
and dowel action and in compression zone generally ↑; fcu ↑,
concrete tensile strength ↑, then inclined cracking load ↑.
o Aggregate type − effect on aggregate interlock capacity For lightweight concrete, the shear stress is equal to 80% of that
of normal concrete (BS 8110).
o Beam size (particularly the beam depth) − size effect: larger beams are proportionately weaker than smaller beams.
112 3.1.4 Design concrete shear stress
• The shear force carried by the concrete (incorrectly referred as) Vc = Vcz + Va + Vd
The design concrete shear stress (nominal concrete shear stress) vc = Vc
bv d (3.1) where bv = the beam width (bv = b for rectangular sections; bv=
bw for flanged sections).
• An accurate analysis for the shear strength of the concrete is not possible. The problem is solved by establishing the strength of
concrete in shear from test results.
In BS 8110: Part 1, clause 184.108.40.206, values for the design concrete
shear stress vc can be calculated by vc =
where 100 As 1/ 3 400 1/ 4
d γm (3.2) ○ γm = 1.25.
○ 100As /(bvd) should not be taken as greater than 3.
○ 400/d should not be taken less than 1.
○ For fcu > 25 N/mm2, the results may be multiplied by
( f cu / 25)1/ 3 , in which the values of fcu should not be taken as greater than 40. 113 Values of the design concrete shear stress, vc, are derived from Eq. (3.2)
and given in the following table (BS 8110: Part 1, Table 3.8): 3.1.5 Design shear stress (BS 8110: Part 1, clause 220.127.116.11)
• The design shear stress at any cross-section should be calculated
bv d (3.3) Definitions of bv can be found in Eq. (3.1).
• For a beam with web reinforcement, the shear resistance may be
regarded as being made up of the sum of the concrete resistance
and the web steel resistance: v = vc + v s (3.4) where vc is the design concrete shear stress and vs is the nominal
shear stress of web reinforcement. 114 ...
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