3 — Shear strength provided by concrete for prestressed members
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CHAPTER 11
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COMMENTARY
11.3.3.1 — Vci shall be computed by
V i M cre
V ci = 0.05λ f c′ b w d p + V d + -----------------M max
(11-10)
where dp need not be taken less than 0.80h and
Mcre = (I /yt )(0.5λ f c′ + fpe – fd )
(11-11)
and values of Mmax and Vi shall be computed from the
load combination causing maximum factored moment
to occur at the section.Vci need not be taken less than
0.17λ f c′ bwd.
11.3.3.2 — Vcw shall be computed by
Vcw = (0.29λ f c′ + 0.3fpc )bwdp + Vp
(11-12)
Fig. R11.3.2—Application of Eq. (11-9) to uniformly loaded
prestressed members.
where dp need not be taken less than 0.80h.
Alternatively, Vcw shall be computed as the shear force
corresponding to dead load plus live load that results in a
principal tensile stress of 0.33λ f c′ at the centroidal axis
of member, or at the intersection of flange and web
when the centroidal axis is in the flange. In composite
members, the principal tensile stress shall be
computed using the cross section that resists live load.
Fig. R11.3.3—Types of cracking in concrete beams.
Web-shear cracking begins from an interior point in a
member when the principal tensile stresses exceed the
tensile strength of the concrete. Flexure-shear cracking is
initiated by flexural cracking. When flexural cracking
occurs, the shear stresses in the concrete above the crack are
increased. The flexure-shear crack develops when the
combined shear and tensile stress exceeds the tensile
strength of the concrete.
Equations (11-10) and (11-12) may be used to determine the
shear forces causing flexure-shear and web-shear cracking,
respectively. The nominal shear strength provided by the
concrete Vc is assumed equal to the lesser of Vci and Vcw.
The derivations of Eq. (11-10) and (11-12) are summarized
in Reference 11.17.
In deriving Eq. (11-10) it was assumed that Vci is the sum of
the shear required to cause a flexural crack at the point in
question given by
V i M cre
V = -----------------M max
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plus an additional increment of shear required to change the
flexural crack to a flexure-shear crack. The externally
applied factored loads, from which Vi and Mmax are determined, include superimposed dead load, earth pressure, and
live load. In computing Mcre for substitution into Eq. (11-10), I
and γt are the properties of the section resisting the externally
applied loads.
For a composite member, where part of the dead load is
resisted by only a part of the section, appropriate section
properties should be used to compute fd . The shear due to
dead loads, Vd , and that due to other loads, Vi , are separated
in this case. Vd is then the total shear force due to unfactored
dead load acting on that part of the section carrying the dead
loads acting prior to composite action plus the unfactored
superimposed dead load acting on the composite member.
The terms Vi and Mmax may be taken as
Vi = Vu – Vd
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Mmax = Mu – Md
where Vu and Mu are the factored shear and moment due to
the total factored loads, and Md is the moment due to unfactored dead load (the moment corresponding to fd).
For noncomposite, uniformly loaded beams, the total cross
section resists all the shear and the live and dead load shear
force diagrams are similar. In this case, Eq. (11-10) reduces to
V u M ct
V ci = 0.05λ f c′ b w d + --------------Mu
where
Mct = (I/yt)(0.5λ f c′ + fpe)
The symbol Mct in the two preceding equations represents
the total moment, including dead load, required to cause
cracking at the extreme fiber in tension. This is not the same
as Mcre in Code Eq. (11-10) where the cracking moment is
that due to all loads except the dead load. In Eq. (11-10), the
dead load shear is added as a separate term.
Mu is the factored moment on the beam at the section under
consideration, and Vu is the factored shear force occurring
simultaneously with Mu. Since the same section properties
apply to both dead and live load stresses, there is no need to
compute dead load stresses and shears separately. The
cracking moment Mct reflects the total stress change from
effective prestress to a tension of 0.5λ f c′ , assumed to
cause flexural cracking.
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Equation (11-12) is based on the assumption that web-shear
cracking occurs due to the shear causing a principal tensile
stress of approximately 0.33λ f c′ at the centroidal axis of
the cross section. Vp is calculated from the effective
prestress force without load factors.
11.3.4 — In a pretensioned member in which the
section at a distance h/2 from face of support is closer
to the end of member than the transfer length of the
prestressing steel, the reduced prestress shall be
considered when computing Vcw . This value of Vcw
shall also be taken as the maximum limit for Eq. (11-9).
The prestress force shall be assumed to vary linearly
from zero at end of the prestressing steel, to a
maximum at a distance from end of the prestressing
steel equal to the transfer length, assumed to be 50
diameters for strand and 100 diameters for single wire.
R11.3.4 and R11.3.5 — The effect of the reduced prestress
near the ends of pretensioned beams on the shear strength
should be taken into account. Section 11.3.4 relates to the
shear strength at sections within the transfer length of
prestressing steel when bonding of prestressing steel
extends to the end of the member.
Section 11.3.5 relates to the shear strength at sections within
the length over which some of the prestressing steel is not
bonded to the concrete, or within the transfer length of the
prestressing steel for which bonding does not extend to the
end of the beam.
11.3.5 — In a pretensioned member where bonding of
some tendons does not extend to the end of member,
a reduced prestress shall be considered when
computing Vc in accordance with 11.3.2 or 11.3.3. The
value of Vcw calculated using the reduced prestress
shall also be taken as the maximum limit for Eq. (11-9).
The prestress force due to tendons for which bonding
does not extend to the end of member shall be
assumed to vary linearly from zero at the point at which
bonding commences to a maximum at a distance from
this point equal to the transfer length, assumed to be 50
diameters for strand and 100 diameters for single wire.
11.4 — Shear strength provided by shear
reinforcement
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R11.4 — Shear strength provided by shear
reinforcement
11.4.1 — Types of shear reinforcement
11.4.1.1 — Shear reinforcement consisting of the
following shall be permitted:
(a) Stirrups perpendicular to axis of member;
(b) Welded wire reinforcement with wires located
perpendicular to axis of member;
(c) Spirals, circular ties, or hoops.
11.4.1.2 — For nonprestressed members, shear
reinforcement shall be permitted to also consist of:
(a) Stirrups making an angle of 45 degrees or more
with longitudinal tension reinforcement;
(b) Longitudinal reinforcement with bent portion
making an angle of 30 degrees or more with the
longitudinal tension reinforcement;
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(c) Combinations of stirrups and bent longitudinal
reinforcement.
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11.4.2 — The values of fy and fyt used in design of
shear reinforcement shall not exceed 420 MPa, except
the value shall not exceed 550 MPa for welded
deformed wire reinforcement.
R11.4.2 — Limiting the values of fy and fyt used in design of
shear reinforcement to 420 MPa provides a control on
diagonal crack width. In the 1995 Code, the limitation of
420 MPa for shear reinforcement was raised to 550 MPa for
welded deformed wire reinforcement. Research11.18-11.20 has
indicated that the performance of higher-strength steels as
shear reinforcement has been satisfactory. In particular,
full-scale beam tests described in Reference 11.19 indicated
that the widths of inclined shear cracks at service load levels
were less for beams reinforced with smaller-diameter
welded deformed wire reinforcement cages designed on the
basis of a yield strength of 520 MPa than beams reinforced
with deformed Grade 420 stirrups.
11.4.3 — Where the provisions of 11.4 are applied to
prestressed members, d shall be taken as the
distance from extreme compression fiber to centroid of
the prestressed and nonprestressed longitudinal
tension reinforcement, if any, but need not be taken
less than 0.80h.
R11.4.3 — Although the value of d may vary along the span
of a prestressed beam, studies11.2 have shown that, for
prestressed concrete members, d need not be taken less than
0.80h. The beams considered had some straight tendons or
reinforcing bars at the bottom of the section and had stirrups
that enclosed the steel.
11.4.4 — Stirrups and other bars or wires used as
shear reinforcement shall extend to a distance d from
extreme compression fiber and shall be developed at
both ends according to 12.13.
R11.4.4 — It is essential that shear (and torsion) reinforcement
be adequately anchored at both ends to be fully effective on
either side of any potential inclined crack. This generally
requires a hook or bend at the end of the reinforcement as
provided by 12.13.
11.4.5 — Spacing limits for shear reinforcement
11.4.5.1 — Spacing of shear reinforcement placed
perpendicular to axis of member shall not exceed d /2
in nonprestressed members or 0.75h in prestressed
members, nor 600 mm.
11.4.5.2 — Inclined stirrups and bent longitudinal
reinforcement shall be so spaced that every 45-degree
line, extending toward the reaction from mid-depth of
member d / 2 to longitudinal tension reinforcement, shall
be crossed by at least one line of shear reinforcement.
11.4.5.3 — Where Vs exceeds 0.33 f c′ bwd,
maximum spacings given in 11.4.5.1 and 11.4.5.2 shall
be reduced by one-half.
11.4.6 — Minimum shear reinforcement
R11.4.6 — Minimum shear reinforcement
11.4.6.1 — A minimum area of shear reinforcement,
Av,min , shall be provided in all reinforced concrete
flexural members (prestressed and nonprestressed)
where Vu exceeds 0.5φVc , except in members satisfying one or more of (a) through (f):
R11.4.6.1 — Shear reinforcement restrains the growth of
inclined cracking. Ductility is increased and a warning of
failure is provided. In an unreinforced web, the sudden
formation of inclined cracking might lead directly to failure
without warning. Such reinforcement is of great value if a
member is subjected to an unexpected tensile force or an
overload. Accordingly, a minimum area of shear reinforcement
(a) Footings and solid slabs;
ACI 318 Building Code and Commentary