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5 — Design for torsion

5 — Design for torsion

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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



ACI 318 Building Code and Commentary



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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.



ACI 318 Building Code and Commentary



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