Reference #11 - Zia v49 No3 2004

7/26/2019 Reference #11 - Zia v49 No3 2004

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Paul Zia, Ph.D., FPCIDistinguished University ProfessorEmeritusDepartment of Civil EngineeringNorth Carolina State UniversityRaleigh, North Carolina

This paper presents a unified method for the torsion and sheardesign of prestressed and non-prestressed concrete flexuralmembers, and provides an alternative method to the provisions ofthe ACI Building Code. The method applies to rectangular, box andflanged sections such as L-beams. The approach depends onsubdividing the given section into component rectangles. Equationsare given for the shear and torsion web reinforcement of beams aswell as expressions for the minimum reinforcement and requiredamount of longitudinal steel. The design method is illustrated with a

fully worked numerical example of a spandrel beam. The shear andtorsion provisions in the Sixth Edition of the PCI Design Handbookare based on the principles outlined in this paper, which will also bereferenced in ACI 318-05. The design method has been shown to bereliable, accurate and easy to use.

F

or the past 30 years, a slightly

modified form of the Zia-McGee

method1 has been a widely used

method for the shear and torsion de-

sign of prestressed and non-pre-

stressed concrete flexural members.

The design equations in the original

Zia-McGee paper1 were derived from

a comprehensive set of test data and

correlated with existing design prac-

tice. Over the years, the design

method has proven to be reliable, ac-

curate and easy to use.

Design for Torsion and Shearin Prestressed ConcreteFlexural Members

34 PCI JOURNAL

Thomas T. C. Hsu, Ph.D.Moores ProfessorDepartment of Civil EngineeringUniversity of HoustonHouston, Texas

Prior to the publication of the Zia-

McGee paper in 1974,1 the only guid-

ance design engineers had for torsion

were the recommendations on torsion

design of reinforced concrete mem-

bers, developed by former ACI Com-

mittee 438 in 1969.2 These recommen-

dations formed the basis for the first

comprehensive torsion provisions to

be included in the 1971 ACI Building

Code (ACI 318-71).3 However, these

code provisions were confined only to

reinforced concrete members. There-

fore, a prime motivation for the Zia-

McGee study was to extend the ACI

Code provisions on torsion to pre-Note: This paper is being published through the cour-

tesy of the American Society of Civil Engineers.


2/9

stressed concrete. It should also be

mentioned that in 1971, the ACI Code

equations were expressed in terms of

nominal stresses.

For the 1977 ACI Code,4 a major

shift occurred whereby all the equa-

tions in the code were now expressed

in terms of forces and moments in-

stead of in nominal stresses. To ad-

dress this change, on October 18,1978, Paul Zia and Thomas T. C. Hsu

presented a paper on Design for Tor-

sion and Shear in Prestressed Con-

crete at an ASCE Convention in

Chicago, Illinois. This oral presenta-

tion (which was also distributed as a

preprint but was never published) es-

sentially updated the original Zia-

McGee paper to the 1977 ACI Code

while also adding a few refinements.

The major reason this paper is now

being published is that it forms thebasis for the shear and torsion provi-

sions in the Sixth Edition of the PCI

Design Handbook,5 and, therefore,

serves as an important background ref-

erence. The paper also serves as an al-

ternative design method to the current

ACI Code (ACI 318-02).6

The design method presented in this

paper essentially follows the Zia-

McGee procedure except for the fol-

lowing modifications or refinements:

1. All the design equations are ex-

pressed in terms of forces and mo-

ments instead of nominal stresses,

which is in conformance with the cur-

rent ACI Code (ACI 318-02).

2. For the sake of simplicity, the tor-

sion coefficient in the basic torsional

stress equation is taken as one-third

instead of a variable depending on the

aspect ratio as was proposed by Zia

and McGee.

3. New equations for minimum tor-

sional and shear web reinforcement

are presented based on research data.7

4. The torsion-shear interaction

curve is based on the concrete contri-

bution curve rather than on the

cracking curve, as implied in the

original Zia-McGee method.

5. The expression for maximum tor-

sional strength as originally proposed

by Zia and McGee is revised.

The original ASCE preprint was the

first attempt to provide a method that

unifies the shear and torsion design for

both prestressed and non-prestressed

May-June 2004 35

concrete flexural members.

To illustrate the application of the

design method, a fully worked numer-

ical example is included in Appendix

B at the end of the paper.

DESIGN CONSIDERATIONS

The design method applies to both

symmetric and unsymmetric sections.

Typical symmetric sections are mem-

bers with rectangular, box, I or T

shapes, whereas unsymmetric (flanged)

sections are members with L (ledger or

spandrel beams) or step shapes (sta-

dium seats). Implicit in the method is

that the section area can be divided into

component rectangles.

Fig. 1 shows a typical member sub-

ject to shear, torsion and applied load.

Such beams are frequently used in

buildings, parking structures, and

other types of structures.

When the torsional moment:

the torsional effect must be considered

in design.

The notation in Eq. (1) is:

= strength reduction factor = a pres tress factor =

= average prestress in a mem-

ber after losses

fc = specified compressive

strength of concrete

x,x = shorter dimension of rectan-gular component of cross

section

y,y = longer dimension of rectan-gular component of cross

section

In computing the sum x2y for a

flanged section, the section must be

divided into component rectanglessuch that the quantity x2y is a maxi-

mum. However, the overhanging

flange width used in design shall not

be greater than three times the flange

thickness.

When torsion design is required, the

member cross section must be propor-

tioned such that the factored torsional

moment:

Tu Tn (2)

where Tn is the nominal torsional mo-

ment strength of the member com-

posed of the strength provided by the

concrete, Tc, and the strength provided

by the torsion reinforcement, Ts. Thus:

Tn = Tc + Ts (3)

Torsion-Shear Interaction

The torsional moment strength of a

member contributed by concrete is af-

fected by the presence of shear and

1 10+ /fc

T f x yu c> ( ) 0 5 2. (1)

Fig. 1. Typical member subject to shear, torsion, and applied load.

P

P

P

Vu

Tu

y

x

Vu

Tu

x

y


3/9

36 PCI JOURNAL

similarly shear is affected by the pres-

ence of torsion. The interaction be-

tween torsion and shear may be repre-

sented by a circular curve. Thus, the

nominal torsional moment strength

provided by the concrete in combined

loading is:

and the nominal shear strength pro-

vided by the concrete in combined

loading is:

where

Tc = nominal torsional moment

strength provided by concrete

under pure torsion

Vc = nominal shear strength pro-

vided by concrete without tor-

sion

Tu = factored torsional moment at

section

Vu = factored shear force at section

Note that the torsional moment, Tc,

can be obtained from:

Also, the shear strength, Vc, is the

lesser of Vci and Vcw (the nominal

shear strength provided by the con-

crete when diagonal cracking results

from combined shear and moment,

and from excessive principal tensile

stress in the web, respectively), as

given by Eqs. (11-10) and (11-12), re-

= T f x yc c0 8 2 5 1 52. ( . . ) (6)

V V

V

T

T

V

cc

c

c

u

u

=

+

1

2 (5)

T T

TV

VT

cc

c

c

u

u

=

+

1

2 (4)

spectively, in ACI 318-02.

For the special case of a non-pre-

stressed concrete member (i.e., when

= 1), Vc = 2 bwd, in which bw isthe web width and d is the effective

depth of the member (see Notation in

Appendix A for a more complete defi-

nition of d).

A comparison of Eqs. (4) and (5)

with the corresponding equations pre-viously suggested by Zia and McGee1

is shown in Fig. 2. It is clear that Eqs.

(4) and (5) postulate that, under tor-

sion-shear interaction, concrete would

contribute more to the torsional

strength but less to the shear strength

than what is implied by the Zia-

McGee equations.

Torsion Reinforcement

When the factored torsional mo-

ment, Tu, exceeds the torsional mo-

ment strength, Tc, torsion reinforce-ment in the form of closed stirrups

must be provided to satisfy Eqs. (2)

and (3), where the torsional strength

contributed by the reinforcement is:

in which

At = area of one leg of a closed

stirrup resisting torsion within

a distance s

fy = specif ied yield strength of

mild steel reinforcement

x1 = shorter center-to-center di-

mension of closed rectangular

stirrup

y1 = longer center-to-center dimen-

sion of closed rectangular stir-

rup

s = spacing of shear or torsion re-

inforcement in direction paral-

lel to longitudinal reinforce-ment

t = torsion coefficient as a func-tion ofy1/x1

Note that the torsion coefficient

must be:

To avoid brittle failure, a minimum

amount of web reinforcement must be

provided to resist both shear and tor-

sion. For lack of sufficient research

t y x= + 0 66 0 33 1. . ( / ) 1.5 (8)1

TA x y f

ss

t t y= 1 1

(7)

fc

Fig. 2. Comparison of Eqs. (4) and (5) with Zia and McGee equations (Ref. 1).


4/9

May-June 2004 37

data, Zia and McGee1 had suggested

that a beam should be reinforced for

no less than its cracking torque. How-

ever, a subsequent study made by Hsu

and Myers7 indicated that the mini-

mum area of closed stirrups for shear

and torsion may be determined from:

where

Av = area of shear reinforcement in

section

At = area of torsion reinforcement

in section

Fig. 3 shows the comparison of Eq.

(9) with the experimental data.7 Note

that the web reinforcement index r inFig. 3 is (Av + 2At)/bws.

Longitudinal bars distributed around

the perimeter of the closed stirrups,At,

must be provided to resist the longitu-

dinal component of the diagonal ten-

sion induced by torsion. This longitu-

dinal reinforcement, A l, should be

approximately of equal volume as that

of stirrups for torsion.

Therefore:

or

whichever is greater. In Eq. (11):

The values of Al computed by Eq.

(11) need not exceed that obtained by

substituting:

501 12 2

b s

f f

b s

fAw

y c

w

y

t+ 200 for

(13)

C b d

x yt

w= 2

(12)

A

xs

f

T

T V

C

A x y

s

l

y

u

uu

t

t

=

+

+

400

3

2 1 1

(11)

A A x y

sl t=

+ 2

1 1

(10)

A A

b s

f f

b s

f

v t

w

y c

w

y

+ =

+

2

50 1 12 002

(9)

It should be emphasized that pre-

stressed concrete beams without longi-

tudinal mild steel reinforcement would

fail abruptly under high torsion.

Therefore, it is essential that a reason-

able amount of longitudinal mild rein-

forcing steel be provided even if there

is an excess of prestressing steel (i.e.,

more than what is required for flex-

ure). This reinforcement must be prop-

erly distributed around the perimeter

of the closed stirrups.

Fig. 3. Comparison of Eq. (9) with test data (Ref. 1).

Fig. 4.Comparison ofEq. (15) withexperimentaldata (Ref. 1).

(9)

Maximum Torsional Moment

In proportioning a member to resist

torsion, consideration should be given

to the possible danger of over-rein-

forcing the beam so that a compres-

sive failure of the concrete might

occur before the reinforcement yields.

To avoid this type of failure, a maxi-

mum limit of factored torsional mo-

ment to be carried by a member must

be established.


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38 PCI JOURNAL

Based on the study by Zia and

McGee,1 and incorporating the torsion

coefficient of one-third rather than in the basic torsion stress equation,1 it

can be shown that:

in which the coefficient, C, is:

C= 12 10(/fc) (15)

A comparison of Eq. (15) with the

experimental data is shown in Fig. 4.

It can be seen that the data points rep-

resenting those members which devel-

oped compression failure all lie well

above the line for Eq. (15). Thus, the

equation represents a fairly conserva-tive limit.

Note that and fpc (average com-pressive stress in concrete after pre-

stress losses) are interchangeable as

shown in the design example of Ap-

pendix B.

Torsional Strength of Box Sections

The torsional strength of a rectangu-

lar box section may be taken as a solid

section provided its wall thickness, h,

is at leastx/4. A box section with wallthickness less thanx/4 but greater than

x/10 may be taken as a solid section

except that the quantity x2y shall be

multiplied by 4h/x. Similarly, the mul-

tiplying factor, 4h/x, shall be applied

to Eq. (7).

DESIGN PROCEDURE

The following design procedure

may be used to determine the shear

and torsional reinforcement in pre-

stressed concrete flexural flanged

members:

1. Determine the deck load to be

supported by the beam.

2. Calculate the factored shear force

and factored torsional moment.

3. Divide the section of the member

into separate rectangular sections.

4. Calculate the required sectional

properties of the member.

5. Calculate the shear and torsional

constant Ctshown in Eq. (12).

6. Perform the design for the right

(or left) support of the member.

7. Check minimum torsion.

8. Check maximum torsion.

9. Calculate nominal shear strength

provided by concrete.

10. Calculate nominal torsional

strength provided by concrete.

11. Compute web reinforcement for

torsion.

12. Compute web reinforcement for

shear.

13. Check minimum reinforcement.

14. Determine longitudinal steel re-inforcement.

15. Repeat the same design procedure

for the other support of the member.

CONCLUDING REMARKS

The equations presented in this

paper provide a unified approach for

the design of shear and torsion in both

prestressed and non-prestressed con-

crete flexural members. Expressed in

terms of forces and moments rather

than nominal stresses, these equations

are similar to the shear and torsion de-

sign provisions in the current ACI

Code (ACI 318-02).

While the general design approach

follows that of the Zia-McGee

method, new expressions are proposed

for torsion-shear interaction and mini-

mum torsion reinforcement and longi-

tudinal steel. Application of these

equations is illustrated by a fully

worked numerical design example.

Experience has shown that the

method outlined in this paper is accu-

rate, safe and easy to use.

The shear and torsion provisions in

the Sixth Edition of the PCI Design

Handbook are based on the principles

outlined in this article, which will also

be referenced in the next edition of the

ACI Building Code (ACI 318-05).

ACKNOWLEDGMENT

The authors would like to express

their gratitude to George Nasser, edi-

tor emeritus, PCI JOURNAL, for his

helpful comments and assistance in

modifying the original preprint.

TC f x y

C V

C T

u

c

u

t u

+

( / )1 3

1

30

2

2

(14)

1. Zia, Paul, and McGee, W. Denis, Torsion Design of Pre-

stressed Concrete, PCI JOURNAL, V. 19, No. 2, March-April

1974, pp. 46-65.

2. ACI Committee 438, Tentative Recommendations for the De-

sign of Reinforced Concrete Members to Resist Torsion,ACI

Journal, V 66, No. 1, January 1969, pp. 1-8.3. ACI Committee 318, Building Code Requirements for Rein-

forced Concrete (ACI 318-71), American Concrete Institute,

Detroit, MI, 1971, 144 pp.

4. ACI Committee 318, Building Code Requirements for Rein-

forced Concrete (ACI 318-77), American Concrete Institute,

Detroit, MI, 1977, 102 pp.

5. PCI Design Handbook Precast and Prestressed Concrete,

Sixth Edition, Precast/Prestressed Concrete Institute, Chicago,

IL, 2004.

6. ACI Committee 318, Building Code Requirements for Struc-

tural Concrete (ACI 318-02), American Concrete Institute,Farmington Hills, MI, 2002.

7. Myers, G., Minimum Torsional Web Reinforcement for Pre-

stressed Concrete, M.S. Thesis under the supervision of

Thomas T. C. Hsu, Department of Civil Engineering, Univer-

sity of Miami, Coral Gables, FL, 1978.

REFERENCES


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May-June 2004 39

Ag = gross area of section, sq in.

Al = total area of longitudinal reinforcement to resist

torsion, sq in.

Aps = area of prestressed reinforcement in tension zone,

sq in.

As = area of non-prestressed tension reinforcement, sq in.

Av = area of shear reinforcement in section

At

= area of one leg of torsion reinforcement in sec-

tion, sq in.

b = width of compression face of member, in.

bw = width of web, in.

C = a coefficient defined in Eq. (15) = 12 10(/fc)Ct = a factor relating shear and torsional stress proper-

ties = bwd/x2y

d = distance from extreme compression fiber to cen-

troid of longitudinal tension reinforcement, but

need not be less than 0.80h for prestressed con-

crete members, in.

e = eccentricity of prestressing force

F = effective prestress force, kips

fc = specified compressive strength of concrete, psi

fd = stress due to unfactored dead load at extreme

fiber of section where tensile stress is caused by

externally applied loads, psi

fpc = average compressive stress in concrete due to ef-

fective prestress force only, psi (same as )fpe = compressive stress in concrete due to effective

prestress force (after allowance for prestress

losses) at extreme fiber of section where tensile

stress is caused by applied loads, psi

fy = specified yield strength of non-prestressed rein-

forcement, psi

h = wall thickness of box section, in., or height ofbeam

I = moment of inertia of section, in.4

Mcr = moment causing flexural cracking at section due

to externally applied loads, in.-lb

MD = service dead load moment, in.-lb

Mmax = maximum factored moment at section due to ex-

ternally applied loads, in.-lb

P = applied load on member, kips

s = spacing of shear or torsion reinforcement in direc-

tion parallel to longitudinal reinforcement, in.

Tc = nominal torsional moment strength provided by

concrete in combined loading, in.-kipsTc = nominal torsional moment strength provided by

concrete under pure torsion, in.-kips

Tn = nominal torsional moment strength, in.-kips

Ts = nominal torsional moment strength provided by

torsion reinforcement, in.-kips

Tu = factored torsional moment at section, in.-kips

Tmax = maximum torsional moment, in.-kips

Tmin = minimum torsional moment, in.-kips

Vc = nominal shear strength provided by concrete in

combined loading, kips

Vc = nominal shear strength provided by concrete

without torsion, kips

Vci = nominal shear strength provided by concrete

when diagonal cracking results from combined

shear and moment, kips

Vcw = nominal shear strength provided by concrete

when diagonal cracking results from excessive

principal tensile stress in web, kips

Vd = shear force at section due to unfactored dead load,

lb

Vi = factored shear force at section due to externally

applied loads occurring simultaneously with

Mmax, lb

Vn = nominal shear strength, kips

Vp = vertical component of effective prestress force at

section, lb

Vs = nominal shear strength provided by shear rein-

forcement, kips

Vu = factored shear force at section, kips

wd = dead load, lb per ft

x,x = shorter dimension of rectangular component of

cross section, in.

y,y = longer dimension of rectangular component of

cross section, in.

x2y = torsional section properties, in.3

yb = distance form neutral axis to bottom fiber of

member, in.yt = distance from neutral axis to top fiber of member,

in.

x1 = shorter center-to-center dimension of closed rect-

angular stirrup

y1 = longer center-to-center dimension of closed rect-

angular stirrup

= torsion coefficient in original Zia-McGee equa-tion1

t = coefficient as a function ofy1/x1 = strength reduction factor = a prestress factor =

= correction factor related to unit weight of con-crete

= average prestress in member after losses, psi(same asfpc)

= summation symbol

1 10+ /fc

APPENDIX A NOTATION


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40 PCI JOURNAL

Given: A prestressed concrete spandrel beam with dimen-

sions shown in Fig. B1. The beam is loaded (transmitted

through the deck) with concentrated loads, P, as shown in

Fig. B2. The span is 30 ft (9.14 m) long with a 12 in. (305

mm) wide bearing at each end.

Ag = 696 sq in. (449,031 mm2)

yb = 33.15 in. (842 mm)

yt = 41.85 in. (1063 mm)I = 360,400 in.4 (1.5 1011 mm4)

Six 1/2 in. (12.7 mm) diameter 270 ksi (1862 MPa) strands

Aps = 0.918 sq in. (592 mm2)

d= 69 in. (1753 mm) throughout

e = 69 41.85 = 27.15 in. (690 mm)

fc = 5000 psi (normal weight concrete)

fy = 60,000 psi

Note that = 1 for normal weight concrete.wD = 725 lb per ft (1080 kg/m)

Torsion arm (assumed) = 8 in. (203 mm)

Deck to be supported by spandrel beam:

Span = 60 ft (18.3 m)Spacing between stems = 4 ft (1.22 m)

Dead load = 89.5 psf (4.29 kPa)

Live load = 50.0 psf (2.39 kPa)

Find: Required shear and torsional reinforcement of

spandrel beam.

1. Calculate Factored Shear Force and Torsional Moment

Dead load of beam = 1.2 0.725

= 0.87 kips per ft (12.70 kN/m)

Dead load of deck = 1.2 0.0895 30 4

= 12.90 kips per stem (57.33 kN/stem)

Live load = 1.6 0.050 30 4

= 9.60 kips per stem (42.70 kN/stem)

Vu at right support

h/2 from edge of bearing

= (75/2 + 6)/12

= 3.62 ft (1.10 m)

Vu at h/2 = 86.55 3.62 0.87

= 83.4 kips (371 kN)Vu at left support

Vu at h/2 = 97.05 0.87 3.62 (12.9 + 9.6)

= 71.4 kips (318 kN)

Right support torque:

Left support torque:

2. Calculate Ct (see Fig. B3)

x2y Area x y x2y

A 8 63 4032

B 12 16 2304

x2y = 6336 in.3

bwd= 8 69 = 552 sq in. (356,128 mm2)

Ct= bwd/x2y = 552/6336 = 0.0871 per in.

DESIGN OF RIGHT SUPPORT

3. Check Minimum Torsional Moment

Effective prestress (assume a 22 percent loss)

F = 0.918(0.7 270)(0.78)

= 135.33 kips (602 kN)

=fpc = F/Ag= 135.33/696

= 0.194 ksi (1.34 MPa)

/fc = 0.194/5 = 0.04

= 198 in.-kips (22.4 kN-m)

198 in.-kips (22.4 kN-m) < 588 in.-kips (66.4 kN-m)

Therefore, torsion design is required.

4. Check Maximum Torsional Moment

C= 12 10(/fc ) = 11.6

= 1640.84 in.-kips (185.4 kN-m)

Therefore, 1640.84 in.-kips is greater than 588 in.-kips.

T C f x y

C V

C T

max

c

u

t u

=

+

=

+

( / )

( / )( . )( . ) ( )

( . )( . )( . )

( )( . )( )

1 3

130

1 3 11 6 1 185000

10006336

111 6 1 18 83 4

30 0 0871 588

2

2

2

T f x ymin c= ( )=

0 50 75 0 5 5000 1 18 6336 1000

2.

. ( . ) ( . )( ) /

= + =1 10 1 18/ .fc

Tu= +

=

( . . )( )

( )12 9 9 67 16

301 8

492 in.-kips (55.6 kN-m)

Tu= +

=

( . . )( )

( )12 9 9 67 14

308

588 in.-kips (66.4 kN-m)

= + +

=

0 87 157 16

3012 9 9 6

97 05

.( )

( . . )

. kips (432 kN)

= + +

=

0 87 157 14

3012 9 9 6

86 55

.( )

( . . )

. kips (385 kN)

APPENDIX B DESIGN EXAMPLE


8/9

May-June 2004 41

5. Calculate Nominal Shear Strength andTorsional Moment Strength Providedby Concrete

From Eq. (11-12), ACI 318-02:

Vcw = (3.5 + 0.3fpc)bwd+ Vp

= [3.5 + 0.3(194)](8)(69)/1000

= 168.74 kips (750.6 kN)

Note that Vp is zero.

= 0.194 + 135.33(27.15)(33.15)/360,400

= 0.532 ksi (3.67 MPa)

MD = 0.725(15)(3.62) 0.725(3.62)2/2

= 34.62 ft-kips (46.9 kN-m)

From Eq. (11-11), ACI 318-02:

M I

yf f f

V

cr

b

c pe d

i

= + ( )

= + ( )=

= +

=

,

./

( )( . . )

6

360 400

33 156 5000 532 38 1000

7 14

3012 9 9 6

9983 in.-kips (1128 kN-m)

73.5 kips (326.9 kN)

f M y

Id

D b=

=

=

( . )( . )( )

,

.

34 62 33 15 12

360 400

0 038 ksi (0.262 MPa)

f F

A

Fey

Ipe

g

b= +

5000

fc

Fig. B2. Loads and reaction forces acting on spandrel beam.

Fig. B3. Subdivisionof beam section intorectangles.

Fig. B1. Spandrel beam section.

From Eq. (11-10), ACI 318-02:

Now, 1.7bwd = 66.35 kips (295.1 kN) < 261.5 kips

Since 261.5 kips is greater than 168.74 kips, Vcw governs.

=

V V

cc

1

++

=

+

=

V

T

T

V

c

c

u

u

2

2

168 74

1168 74 588

519 71 83 4

.

( . )( )

( . )( . )

67.55 kips (300.5 kN)

=

=

=

=

+

=

+

=

V

T f x y

T T

T

V

V

T

c

c c

cc

c

c

u

u

168 7

0 8 2 5 1 5 1000

519 71

1

519 71

1519 71 83 4

168 74 588

476 25

2

2

2

.

. ( . . ) /

.

.

( . )( . )

( . )( )

.

4 kips (750.6 kN)

in.-kips (58.72 kN-m)

in.-kips (53.82 kN-m)

fc

V b d f V V M

Mci w c d i

cr

max

= + +

= + +=

0 6

0 6 0 552 5000 8 25 73 5 9983 3193

261 5

.

( . )( . ) . . ( / )

. kips (1163 kN)

Vd=

00 725 15 3 62

8 25

. ( . )

.

= kips (36.7 kN)

M Vmax i=

=

( . )( )3 62 12

193

3 in.-kips (360.8 kN-m)


9/9

42 PCI JOURNAL

6. Calculate Web Reinforcement for Torsion

Assume 11/4 in. (32 mm) cover and No. 4 reinforcing bars.

y1 = 75 2(1.5) = 72 in. (1829 mm)

x1 = 8.0 2(1.5) = 5 in. (127 mm)

Ts = Tu/ Tc= 588/0.75 476.25

= 307.8 in.-kips (34.77 kN-m)

t = 0.66 + 0.33(y1/x1)= 0.66 + 0.33(72/5)

= 5.41 > 1.5 (Use 1.5)

7. Calculate Web Reinforcement for Shear

8. Check Minimum Reinforcement andDesign Longitudinal Steel

Since 0.010 is less than 0.030, minimum reinforcement

does not govern.

Since 19.25 in. is greater than 12 in., use bars with 12 in.

(305 mm) maximum spacing.

Use No. 4 bars at 12 in. (305 mm) spacing.

Equal volume governs. Therefore, Al = 1.46 sq in. (942

mm2). Use sixteen No. 4 bars (at each end of member)

spaced around stirrups at 12 in. maximum spacing.

9. Repeat the Same Design Procedure for Left Support

A similar analysis for the left support shows that the same

web reinforcement for torsion and shear as well as longitu-

dinal torsion reinforcement should be used even though the

factored shear and factored torsion at the left support are

less than those at the right support.

10. Repeat the same design at other sections, especiallyat the load points where Vci may govern.

A A x y

s

A

xs

f

T

T V

C

A

x y

s

l t

ly

u

uu

t

t

= +

= +

=

= +

+

=+

2

2 0 0095 5 72

1 46

400

3

2

400 8 12

60 000

588

58883 4

3 0 0871

1 1

1 1

( )( . )( )

.

( )( )( )

, .

( )( . )

sq in. (942 mm )2

(( )( . )

.

2 0 205 72

12

0 1

+

= sq in. (64.5 mm )2

A

s

A

s

t v+ = + =2

0 0095 0 0053 0 0148. . .

s x y

min= +

= +

=

1 1

4

5 72

4

19 25. in. (489 mm)

A

s

A

s

A

s

A

s f

b

f

v t

v t

min c

w

y

+ = +

=

+

= +

= +

=

20 0105 2 0 0095

0 030

250 1 12

50 1 12194

5000

8

60 000

0 010

. ( . )

.

,

.

sq in. per in. (0.76 mm /mm)2

As

V Vd f

v u c

y

=

=

=

/

( . / . ) .

( )( )

.

83 4 0 75 67 55

69 60

0 0105 sq in. per in. (0.27 mm /mm)2

A

s

T

x y f

t s

t y

=

=

=

1 1

307 8

1 5 5 0 72 0 60

0 0095

.

( . )( . )( . )( )

. sq in. per in. (0.24 mm /mm)2

Documents

Reference #11 - Zia v49 No3 2004