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of the member. In the thin-walled tube analogy, the resistance
is assumed to be provided by the outer skin of the cross
section roughly centered on the closed stirrups. Both hollow
and solid sections are idealized as thin-walled tubes both
before and after cracking.
In a closed thin-walled tube, the product of the shear stress τ
and the wall thickness t at any point in the perimeter is
known as the shear flow, q = τ t. The shear flow q due to
torsion acts as shown in Fig. R11.5(a) and is constant at all
points around the perimeter of the tube. The path along
which it acts extends around the tube at midthickness of the
walls of the tube. At any point along the perimeter of the
tube the shear stress due to torsion is τ = T/(2A o t) where Ao
is the gross area enclosed by the shear flow path, shown
shaded in Fig. R11.5(b), and t is the thickness of the wall at
the point where τ is being computed. The shear flow follows
the midthickness of the walls of the tube and Ao is the area
enclosed by the path of the shear flow. For a hollow member
with continuous walls, Ao includes the area of the hole.
In the 1995 Code, the elliptical interaction between the
nominal shear strength provided by the concrete, Vc , and
the nominal torsion strength provided by the concrete was
eliminated. Vc remains constant at the value it has when
there is no torsion, and the torsion carried by the concrete is
always taken as zero.
The design procedure is derived and compared with test
results in References 11.31 and 11.32.
11.5.1 — Threshold torsion
R11.5.1 — Threshold torsion
It shall be permitted to neglect torsion effects if the
factored torsional moment Tu is less than:
Torques that do not exceed approximately one-quarter of the
cracking torque Tcr will not cause a structurally significant
reduction in either the flexural or shear strength and can be
ignored. The cracking torsion under pure torsion Tcr is
derived by replacing the actual section with an equivalent
thin-walled tube with a wall thickness t prior to cracking of
0.75Acp /pcp and an area enclosed by the wall centerline Ao
equal to 2Acp /3. Cracking is assumed to occur when the
principal tensile stress reaches 0.33λ f c′ . In a nonprestressed beam loaded with torsion alone, the principal tensile
stress is equal to the torsional shear stress, τ = T/(2Aot).
Thus, cracking occurs when τ reaches 0.33λ f c′ , giving
the cracking torque Tcr as
(a) For nonprestressed members
2
⎛ A cp⎞
φ0.083λ f c′ ⎜ ---------⎟
⎝ p cp ⎠
(b) For prestressed members
2
f pc
⎛ A cp⎞
φ0.083λ f c′ ⎜ ---------⎟ 1 + -------------------------⎝ p cp ⎠
0.33λ f c ′
2
(c) For nonprestressed members subjected to an
axial tensile or compressive force
2
Nu
⎛ A cp⎞
φ0.083λ f c′ ⎜ ---------⎟ 1 + ---------------------------------⎝ p cp ⎠
0.33A g λ f c ′
⎛ A cp⎞
T cr = 0.33λ f c′ ⎜ --------⎟
⎝ p cp ⎠
For solid members, the interaction between the cracking
torsion and the inclined cracking shear is approximately
circular or elliptical. For such a relationship, a torque of
0.25Tcr , as used in 11.5.1, corresponds to a reduction of
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For members cast monolithically with a slab, the overhanging flange width used in computing Acp and pcp
shall conform to 13.2.4. For a hollow section, Ag shall
be used in place of Acp in 11.5.1, and the outer
boundaries of the section shall conform to 13.2.4.
3 percent in the inclined cracking shear. This reduction in the
inclined cracking shear was considered negligible. The stress
at cracking 0.33λ f c′ has purposely been taken as a lower
bound value.
11.5.1.1 — For isolated members with flanges and
for members cast monolithically with a slab, the overhanging flange width used to compute Acp and pcp
shall conform to 13.2.4, except that the overhanging
flanges shall be neglected in cases where the parameter
2 /p
Acp
cp calculated for a beam with flanges is less
than that computed for the same beam ignoring the
flanges.
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For prestressed members, the torsional cracking load is
increased by the prestress. A Mohr’s Circle analysis based
on average stresses indicates the torque required to cause a
principal tensile stress equal to 0.33λ f c′
is
1 + f pc ⁄ ( 0.33λ f c ′ ) times the corresponding torque in a
nonprestressed beam. A similar modification is made in part
(c) of 11.5.1 for members subjected to axial load and torsion.
For torsion, a hollow member is defined as having one or
more longitudinal voids, such as a single-cell or multiple-cell
box girder. Small longitudinal voids, such as ungrouted posttensioning ducts that result in Ag /Acp greater than or equal to
0.95, can be ignored when computing the threshold torque in
11.5.1. The interaction between torsional cracking and shear
cracking for hollow sections is assumed to vary from the
elliptical relationship for members with small voids, to a
straight-line relationship for thin-walled sections with large
voids. For a straight-line interaction, a torque of 0.25Tcr
would cause a reduction in the inclined cracking shear of
about 25 percent. This reduction was judged to be excessive.
In the 2002 Code, two changes were made to modify 11.5.1
to apply to hollow sections. First, the minimum torque limits
from the 1999 Code were multiplied by (Ag /Acp) because
tests of solid and hollow beams11.33 indicate that the cracking
torque of a hollow section is approximately (Ag /Acp) times
the cracking torque of a solid section with the same outside
dimensions. The second change was to multiply the cracking
torque by (Ag/Acp) a second time to reflect the transition from
the circular interaction between the inclined cracking loads in
shear and torsion for solid members, to the approximately
linear interaction for thin-walled hollow sections.
11.5.2 — Calculation of factored torsional moment
R11.5.2 — Calculation of factored torsional moment
11.5.2.1 — If the factored torsional moment, Tu, in a
member is required to maintain equilibrium and
exceeds the minimum value given in 11.5.1, the
member shall be designed to carry Tu in accordance
with 11.5.3 through 11.5.6.
R11.5.2.1 and R11.5.2.2 — In designing for torsion in
reinforced concrete structures, two conditions may be
identified:11.34,11.35
11.5.2.2 — In a statically indeterminate structure
where reduction of the torsional moment in a member
can occur due to redistribution of internal forces upon
cracking, the maximum Tu shall be permitted to be
reduced to the values given in (a), (b), or (c), as applicable:
(a) For nonprestressed members, at the sections
described in 11.5.2.4
(a) The torsional moment cannot be reduced by redistribution of internal forces (11.5.2.1). This is referred to as
equilibrium torsion, since the torsional moment is
required for the structure to be in equilibrium.
For this condition, illustrated in Fig. R11.5.2.1, torsion
reinforcement designed according to 11.5.3 through
11.5.6 must be provided to resist the total design torsional
moments.
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2
⎛ A cp⎞
φ0.33λ f c′ ⎜ ---------⎟
⎝ p cp ⎠
(b) For prestressed members, at the sections
described in 11.5.2.5
2
f pc
⎛ A cp⎞
φ0.33λ f c′ ⎜ ---------⎟ 1 + -------------------------⎝ p cp ⎠
0.33λ f c′
(c) For nonprestressed members subjected to an
axial tensile or compressive force
2 ⎞
Nu
⎛ A cp
φ0.33λ f c′ ⎜ ---------⎟ 1 + ---------------------------------⎝ p cp ⎠
0.33A g λ f c′
Fig. R11.5.2.1—Design torque may not be reduced (11.5.2.1).
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In (a), (b), or (c), the correspondingly redistributed
bending moments and shears in the adjoining
members shall be used in the design of these
members. For hollow sections, Acp shall not be
replaced with Ag in 11.5.2.2.
Fig. R11.5.2.2—Design torque may be reduced (11.5.2.2).
(b) The torsional moment can be reduced by redistribution of internal forces after cracking (11.5.2.2) if the
torsion arises from the member twisting to maintain
compatibility of deformations. This type of torsion is
referred to as compatibility torsion.
For this condition, illustrated in Fig. R11.5.2.2, the
torsional stiffness before cracking corresponds to that of
the uncracked section according to St. Venant’s theory. At
torsional cracking, however, a large twist occurs under an
essentially constant torque, resulting in a large redistribution
of forces in the structure.11.34,11.35 The cracking torque
under combined shear, flexure, and torsion corresponds to
a principal tensile stress somewhat less than the
0.33λ f c′ quoted in R11.5.1.
When the torsional moment exceeds the cracking torque, a
maximum factored torsional moment equal to the cracking
torque may be assumed to occur at the critical sections near
the faces of the supports. This limit has been established to
control the width of torsional cracks. The replacement of Acp
with Ag, as in the calculation of the threshold torque for hollow
sections in 11.5.1, is not applied here. Thus, the torque after
redistribution is larger and hence more conservative.
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Section 11.5.2.2 applies to typical and regular framing
conditions. With layouts that impose significant torsional
rotations within a limited length of the member, such as a
heavy torque loading located close to a stiff column, or a
column that rotates in the reverse directions because of
other loading, a more exact analysis is advisable.
When the factored torsional moment from an elastic analysis
based on uncracked section properties is between the values
in 11.5.1 and the values given in this section, torsion
reinforcement should be designed to resist the computed
torsional moments.
11.5.2.3 — Unless determined by a more exact
analysis, it shall be permitted to take the torsional
loading from a slab as uniformly distributed along the
member.
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11.5.2.4 — In nonprestressed members, sections
located less than a distance d from the face of a
support shall be designed for not less than Tu
computed at a distance d. If a concentrated torque
occurs within this distance, the critical section for
design shall be at the face of the support.
R11.5.2.4 and R11.5.2.5 — It is not uncommon for a
beam to frame into one side of a girder near the support of
the girder. In such a case, a concentrated shear and torque
are applied to the girder.
11.5.2.5 — In prestressed members, sections
located less than a distance h/2 from the face of a
support shall be designed for not less than Tu
computed at a distance h/2. If a concentrated torque
occurs within this distance, the critical section for
design shall be at the face of the support.
11.5.3 — Torsional moment strength
R11.5.3 — Torsional moment strength
11.5.3.1 — The cross-sectional dimensions shall be
such that:
R11.5.3.1 — The size of a cross section is limited for two
reasons: first, to reduce unsightly cracking, and second, to
prevent crushing of the surface concrete due to inclined
compressive stresses due to shear and torsion. In Eq. (11-18)
and (11-19), the two terms on the left-hand side are the
shear stresses due to shear and torsion. The sum of these
stresses may not exceed the stress causing shear cracking
plus 0.66 f c′ , similar to the limiting strength given in
11.4.7.9 for shear without torsion. The limit is expressed in
terms of Vc to allow its use for nonprestressed or prestressed
concrete. It was originally derived on the basis of crack
control. It is not necessary to check against crushing of the
web because this happens at higher shear stresses.
(a) For solid sections
Vc
Vu ⎞ 2 ⎛ Tu ph ⎞ 2
⎛ ---------- + 0.66 f c′ ⎞ (11-18)
+ ⎜ -------------------⎟ ≤ φ ⎛ ---------⎝ b d-⎠
⎝
⎠
2
b
⎝ 1.7A ⎠
wd
w
oh
(b) For hollow sections
Vc
Vu ⎞ ⎛ Tu ph ⎞
⎛ ---------+ ⎜ -------------------⎟ ≤ φ ⎛ ---------- + 0.66 f c′ ⎞
⎝ b d⎠ ⎝
⎝b d
⎠
2
w
w
1.7A ⎠
(11-19)
oh
For prestressed members, d shall be determined in
accordance with 11.4.3.
In a hollow section, the shear stresses due to shear and torsion
both occur in the walls of the box as shown in Fig. 11.5.3.1(a)
and hence are directly additive at point A as given in
Eq. (11-19). In a solid section, the shear stresses due to
torsion act in the “tubular” outside section while the shear
stresses due to Vu are spread across the width of the section
as shown in Fig. R11.5.3.1(b). For this reason, stresses are
combined in Eq. (11-18) using the square root of the sum of
the squares rather than by direct addition.
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Fig. R11.5.3.1—Addition of torsional and shear stresses.
11.5.3.2 — If the wall thickness varies around the
perimeter of a hollow section, Eq. (11-19) shall be
evaluated at the location where the left-hand side of
Eq. (11-19) is a maximum.
R11.5.3.2 — Generally, the maximum will be on the wall
where the torsional and shearing stresses are additive [Point A
in Fig. R11.5.3.1(a)]. If the top or bottom flanges are
thinner than the vertical webs, it may be necessary to
evaluate Eq. (11-19) at points B and C in Fig. R11.5.3.1(a).
At these points, the stresses due to the shear force are
usually negligible.
11.5.3.3 — If the wall thickness is less than Aoh /ph ,
the second term in Eq. (11-19) shall be taken as
Tu ⎞
⎛ --------------------⎝ 1.7A t ⎠
oh
where t is the thickness of the wall of the hollow
section at the location where the stresses are being
checked.
11.5.3.4 — The values of fy and fyt used for
design of torsional reinforcement shall not exceed
420 MPa.
R11.5.3.4 — Limiting the values of fy and fyt used in
design of torsion reinforcement to 420 MPa provides a
control on diagonal crack width.
11.5.3.5 — Where Tu exceeds the threshold torsion,
design of the cross section shall be based on
R11.5.3.5 — The factored torsional resistance φTn must
equal or exceed the torsion Tu due to the factored loads. In
the calculation of Tn, all the torque is assumed to be resisted
by stirrups and longitudinal steel with Tc = 0. At the same
φTn ≥ Tu
(11-20)
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time, the nominal shear strength provided by concrete, Vc ,
is assumed to be unchanged by the presence of torsion. For
beams with Vu greater than about 0.8φVc , the resulting
amount of combined shear and torsional reinforcement is
essentially the same as required by the 1989 Code. For
smaller values of Vu, more shear and torsion reinforcement
will be required.
11.5.3.6 — Tn shall be computed by
2A o A t f yt
cot θ
T n = -----------------------s
(11-21)
where Ao shall be determined by analysis except that it
shall be permitted to take Ao equal to 0.85Aoh ; θ shall
not be taken smaller than 30 degrees nor larger than
60 degrees. It shall be permitted to take θ equal to:
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(a) 45 degrees for nonprestressed members or
members with less prestress than in (b); or
(b) 37.5 degrees for prestressed members with an
effective prestress force not less than 40 percent of
the tensile strength of the longitudinal reinforcement.
R11.5.3.6 — Equation (11-21) is based on the space truss
analogy shown in Fig. R11.5.3.6(a) with compression diagonals at an angle θ, assuming the concrete carries no tension
and the reinforcement yields. After torsional cracking
develops, the torsional resistance is provided mainly by
closed stirrups, longitudinal bars, and compression diagonals.
The concrete outside these stirrups is relatively ineffective.
For this reason Ao , the gross area enclosed by the shear flow
path around the perimeter of the tube, is defined after
cracking in terms of Aoh , the area enclosed by the centerline
of the outermost closed transverse torsional reinforcement.
The area Aoh is shown in Fig. R11.5.3.6(b) for various cross
sections. In an I-, T-, or L-shaped section, Aoh is taken as that
area enclosed by the outermost legs of interlocking stirrups as
shown in Fig. R11.5.3.6(b). The expression for Ao given by
Hsu11.36 may be used if greater accuracy is desired.
The shear flow q in the walls of the tube, discussed in
R11.5, can be resolved into the shear forces V1 to V4 acting
Fig. R11.5.3.6(a)—Space truss analogy.
Fig. R11.5.3.6(b)—Definition of Aoh.
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in the individual sides of the tube or space truss, as shown in
Fig. R11.5.3.6(a).
The angle θ can be obtained by analysis11.36 or may be
taken to be equal to the values given in 11.5.3.6(a) or (b).
The same value of θ should be used in both Eq. (11-21) and
(11-22). As θ gets smaller, the amount of stirrups required
by Eq. (11-21) decreases. At the same time, the amount of
longitudinal steel required by Eq. (11-22) increases.
11.5.3.7 — The additional area of longitudinal reinforcement to resist torsion, Al , shall not be less than
A
f yt⎞
2
cot θ
A l = ------t p h ⎛ -----⎝f ⎠
s
y
(11-22)
where θ shall be the same value used in Eq. (11-21)
and At /s shall be taken as the amount computed from
Eq. (11-21) not modified in accordance with 11.5.5.2
or 11.5.5.3; fyt refers to closed transverse torsional
reinforcement, and fy refers to longitudinal torsional
reinforcement.
R11.5.3.7 — Figure R11.5.3.6(a) shows the shear forces
V1 to V4 resulting from the shear flow around the walls of the
tube. On a given wall of the tube, the shear flow Vi is resisted
by a diagonal compression component, Di = Vi /sinθ, in the
concrete. An axial tension force, Ni = Vi (cotθ ), is needed
in the longitudinal steel to complete the resolution of Vi.
Figure R11.5.3.7 shows the diagonal compressive stresses and
the axial tension force, Ni , acting on a short segment along one
wall of the tube. Because the shear flow due to torsion is
constant at all points around the perimeter of the tube, the
resultants of Di and Ni act through the midheight of side i. As
a result, half of Ni can be assumed to be resisted by each of the
top and bottom chords as shown. Longitudinal reinforcement
with a strength Al fy should be provided to resist the sum of
the Ni forces, ΣNi , acting in all of the walls of the tube.
In the derivation of Eq. (11-22), axial tension forces are
summed along the sides of the area Ao. These sides form a
perimeter length, po, approximately equal to the length of
the line joining the centers of the bars in the corners of the
tube. For ease in computation, this has been replaced with
the perimeter of the closed stirrups, ph.
Frequently, the maximum allowable stirrup spacing governs
the amount of stirrups provided. Furthermore, when
combined shear and torsion act, the total stirrup area is the
sum of the amounts provided for shear and torsion. To avoid
the need to provide excessive amounts of longitudinal
reinforcement, 11.5.3.7 states that the At /s used in calculating
Al at any given section should be taken as the At /s calculated at
that section using Eq. (11-21).
Fig. R11.5.3.7—Resolution of shear force Vi into diagonal
compression force Di and axial tension force Ni in one wall
of the tube.
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11.5.3.8 — Reinforcement required for torsion shall
be added to that required for the shear, moment, and
axial force that act in combination with the torsion. The
most restrictive requirements for reinforcement spacing
and placement shall be met.
R11.5.3.8 — The stirrup requirements for torsion and
shear are added and stirrups are provided to supply at least
the total amount required. Since the stirrup area Av for shear
is defined in terms of all the legs of a given stirrup while the
stirrup area At for torsion is defined in terms of one leg only,
the addition of stirrups is carried out as follows
A
A
Av + t
Total ⎛ -------------⎞ = -----v- + 2 -----t
⎝ s ⎠
s
s
If a stirrup group had four legs for shear, only the legs adjacent
to the sides of the beam would be included in this summation
since the inner legs would be ineffective for torsion.
The longitudinal reinforcement required for torsion is added
at each section to the longitudinal reinforcement required for
bending moment that acts at the same time as the torsion. The
longitudinal reinforcement is then chosen for this sum, but
should not be less than the amount required for the maximum
bending moment at that section if this exceeds the moment
acting at the same time as the torsion. If the maximum bending
moment occurs at one section, such as the midspan, while the
maximum torsional moment occurs at another, such as the
support, the total longitudinal steel required may be less than
that obtained by adding the maximum flexural steel plus the
maximum torsional steel. In such a case, the required longitudinal steel is evaluated at several locations.
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The most restrictive requirements for spacing, cut-off
points, and placement for flexural, shear, and torsional steel
should be satisfied. The flexural steel should be extended a
distance d, but not less than 12db , past where it is no longer
needed for flexure as required in 12.10.3.
11.5.3.9 — It shall be permitted to reduce the area of
longitudinal torsion reinforcement in the flexural
compression zone by an amount equal to Mu /(0.9df y ),
where Mu occurs at the section simultaneously with
Tu , except that the reinforcement provided shall not be
less than that required by 11.5.5.3 or 11.5.6.2.
11.5.3.10 — In prestressed beams:
(a) The total longitudinal reinforcement including
prestressing steel at each section shall resist Mu at
that section plus an additional concentric longitudinal tensile force equal to Al fy , based on Tu at that
section;
(b) The spacing of the longitudinal reinforcement
including tendons shall satisfy the requirements in
11.5.6.2.
R11.5.3.9 — The longitudinal tension due to torsion is
offset in part by the compression in the flexural compression
zone, allowing a reduction in the longitudinal torsion steel
required in the compression zone.
R11.5.3.10 — As explained in R11.5.3.7, torsion causes
an axial tension force. In a nonprestressed beam, this force
is resisted by longitudinal reinforcement having an axial
tensile strength of Al fy . This steel is in addition to the
flexural reinforcement and is distributed uniformly around
the sides of the perimeter so that the resultant of Al fy acts
along the axis of the member.
In a prestressed beam, the same technique (providing additional reinforcing bars with capacity Al fy ) can be followed, or
overstrength of the prestressing steel can be used to resist
some of the axial force Al fy as outlined in the next paragraph.
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In a prestressed beam, the stress in the prestressing steel at
nominal strength will be between fse and fps. A portion of
the Al fy force can be resisted by a force of ApsΔfpt in the
prestressing steel, where Δfpt is the difference between the
stress which can be developed in the strand at the section
under consideration and the stress required to resist the bending
moment at this section, Mu. The stress required to resist the
bending moment can be calculated as [Mu /φ0.9dpAps)]. For
pretensioned strands, the stress which can be developed near
the free end of the strand can be calculated using the
procedure illustrated in Fig. R12.9. Note that near the ends
of a pretensioned member, the available stress in the
prestressing steel will need to be reduced to account for lack
of full development, and should be determined in conjunction
with 9.3.2.7.
11.5.3.11 — In prestressed beams, it shall be
permitted to reduce the area of longitudinal torsional
reinforcement on the side of the member in compression
due to flexure below that required by 11.5.3.10 in
accordance with 11.5.3.9.
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11.5.4 — Details of torsional reinforcement
R11.5.4 — Details of torsional reinforcement
11.5.4.1 — Torsion reinforcement shall consist of
longitudinal bars or tendons and one or more of the
following:
R11.5.4.1 — Both longitudinal and closed transverse
reinforcement are required to resist the diagonal tension
stresses due to torsion. The stirrups must be closed, since
inclined cracking due to torsion may occur on all faces of a
member.
(a) Closed stirrups or closed ties, perpendicular to
the axis of the member;
(b) A closed cage of welded wire reinforcement with
transverse wires perpendicular to the axis of the
member;
(c) In nonprestressed beams, spiral reinforcement.
11.5.4.2 — Transverse torsional reinforcement shall
be anchored by one of the following:
(a) A 135-degree standard hook, or seismic hook as
defined in 2.2, around a longitudinal bar;
(b) According to 12.13.2.1, 12.13.2.2, or 12.13.2.3 in
regions where the concrete surrounding the
anchorage is restrained against spalling by a flange
or slab or similar member.
In the case of sections subjected primarily to torsion, the
concrete side cover over the stirrups spalls off at high
torques.11.37 This renders lapped-spliced stirrups ineffective, leading to a premature torsional failure.11.38 In such
cases, closed stirrups should not be made up of pairs of
U-stirrups lapping one another.
R11.5.4.2 — When a rectangular beam fails in torsion,
the corners of the beam tend to spall off due to the inclined
compressive stresses in the concrete diagonals of the
space truss changing direction at the corner as shown in
Fig. R11.5.4.2(a). In tests,11.37 closed stirrups anchored
by 90-degree hooks failed when this occurred. For this
reason, 135-degree standard hooks or seismic hooks are
preferable for torsional stirrups in all cases. In regions
where this spalling is prevented by an adjacent slab or
flange, 11.5.4.2(b) relaxes this and allows 90-degree hooks
(see Fig. R11.5.4.2(b)).
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Fig. R11.5.4.2—Spalling of corners of beams loaded in torsion.
11.5.4.3 — Longitudinal torsion reinforcement shall
be developed at both ends.
R11.5.4.3 — If high torsion acts near the end of a beam,
the longitudinal torsion reinforcement should be adequately
anchored. Sufficient development length should be provided
outside the inner face of the support to develop the needed
tension force in the bars or tendons. In the case of bars, this
may require hooks or horizontal U-shaped bars lapped with
the longitudinal torsion reinforcement.
11.5.4.4 — For hollow sections in torsion, the
distance from the centerline of the transverse torsional
reinforcement to the inside face of the wall of the
hollow section shall not be less than 0.5Aoh /ph .
R11.5.4.4 — The closed stirrups provided for torsion in a
hollow section should be located in the outer half of the wall
thickness effective for torsion where the wall thickness can
be taken as Aoh /ph .
11.5.5 — Minimum torsion reinforcement
R11.5.5 — Minimum torsion reinforcement
11.5.5.1 — A minimum area of torsional reinforcement shall be provided in all regions where Tu exceeds
the threshold torsion given in 11.5.1.
R11.5.5.1 and R11.5.5.2 — If a member is subject to a
factored torsional moment Tu greater than the values specified
in 11.5.1, the minimum amount of transverse web reinforcement for combined shear and torsion is 0.35 bw s/fyt . The
differences in the definition of Av and the symbol At should
be noted; Av is the area of two legs of a closed stirrup
whereas At is the area of only one leg of a closed stirrup.
11.5.5.2 — Where torsional reinforcement is
required by 11.5.5.1, the minimum area of transverse
closed stirrups shall be computed by
bw s
( A v + 2A t ) = 0.062 f c′ ---------f yt
but shall not be less than (0.35bw s)/fyt .
(11-23)
Tests11.9 of high-strength reinforced concrete beams have
indicated the need to increase the minimum area of shear
reinforcement to prevent shear failures when inclined
cracking occurs. Although there are a limited number of
tests of high-strength concrete beams in torsion, the equation
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COMMENTARY
for the minimum area of transverse closed stirrups has been
changed for consistency with calculations required for
minimum shear reinforcement.
11.5.5.3 — Where torsional reinforcement is
required by 11.5.5.1, the minimum total area of longitudinal torsional reinforcement, Al,min , shall be
computed by
0.42 f c ′A cp ⎛ A t⎞ f yt
A l ,min = --------------------------------– ------ p h -----⎝ s⎠
fy
fy
(11-24)
where At /s shall not be taken less than 0.175bw /fyt ;
fyt refers to closed transverse torsional reinforcement,
and fy refers to longitudinal reinforcement.
R11.5.5.3 — Reinforced concrete beam specimens with
less than 1 percent torsional reinforcement by volume have
failed in pure torsion at torsional cracking.11.31 In the 1989
and prior Codes, a relationship was presented that required
about 1 percent torsional reinforcement in beams loaded in
pure torsion and less in beams with combined shear and
torsion, as a function of the ratio of shear stresses due to
torsion and shear. Equation (11-24) was simplified by
assuming a single value of this reduction factor and results
in a volumetric ratio of about 0.5 percent.
11.5.6 — Spacing of torsion reinforcement
R11.5.6 — Spacing of torsion reinforcement
11.5.6.1 — The spacing of transverse torsion reinforcement shall not exceed the smaller of ph /8 or 300 mm.
R11.5.6.1 — The spacing of the stirrups is limited to
ensure the development of the ultimate torsional strength of
the beam, to prevent excessive loss of torsional stiffness
after cracking, and to control crack widths. For a square
cross section, the ph /8 limitation requires stirrups at d/2,
which corresponds to 11.4.5.1.
11.5.6.2 — The longitudinal reinforcement required
for torsion shall be distributed around the perimeter of
the closed stirrups with a maximum spacing of 300 mm.
The longitudinal bars or tendons shall be inside the
stirrups. There shall be at least one longitudinal bar or
tendon in each corner of the stirrups. Longitudinal bars
shall have a diameter at least 0.042 times the stirrup
spacing, but not less than No. 10.
R11.5.6.2 — In R11.5.3.7, it was shown that longitudinal
reinforcement is needed to resist the sum of the longitudinal
tensile forces due to torsion in the walls of the thin-walled
tube. Since the force acts along the centroidal axis of the
section, the centroid of the additional longitudinal reinforcement for torsion should approximately coincide with the
centroid of the section. The Code accomplishes this by
requiring the longitudinal torsional reinforcement to be
distributed around the perimeter of the closed stirrups.
Longitudinal bars or tendons are required in each corner of
the stirrups to provide anchorage for the legs of the stirrups.
Corner bars have also been found to be very effective in
developing torsional strength and in controlling cracks.
11.5.6.3 — Torsional reinforcement shall be
provided for a distance of at least (bt + d) beyond the
point required by analysis.
R11.5.6.3 — The distance (bt + d) beyond the point theoretically required for torsional reinforcement is larger than that
used for shear and flexural reinforcement because torsional
diagonal tension cracks develop in a helical form.
11.5.7 — Alternative design for torsion
R11.5.7 — Alternative design for torsion
For torsion design of solid sections within the scope of
this Code with an aspect ratio, h/bt , of 3 or greater, it
shall be permitted to use another procedure, the
adequacy of which has been shown by analysis and
substantial agreement with results of comprehensive
tests. Sections 11.5.4 and 11.5.6 shall apply.
Examples of such procedures are to be found in References
11.39 to 11.41, which have been extensively and successfully
used for design of precast, prestressed concrete beams with
ledges. The procedure described in References 11.39 and 11.40
is an extension to prestressed concrete sections of the torsion
procedures of pre-1995 editions of the Code. The sixth edition
of the PCI Design Handbook11.16 describes the procedure of
References 11.40 and 11.41. This procedure was experimentally verified by the tests described in Reference 11.42.
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