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3 — Shear strength provided by concrete for prestressed members

3 — Shear strength provided by concrete for prestressed members

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CHAPTER 11



CODE



161



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



ACI 318 Building Code and Commentary



11



162



CHAPTER 11



CODE



COMMENTARY

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



11



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.



ACI 318 Building Code and Commentary



CHAPTER 11



CODE



163



COMMENTARY

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



11



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;



ACI 318 Building Code and Commentary



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CHAPTER 11



CODE



COMMENTARY



(c) Combinations of stirrups and bent longitudinal

reinforcement.



11



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



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