ASSHTO Example Strut and Tie

8/17/2019 ASSHTO Example Strut and Tie

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B

C

D

E

F

G

H

I

J

K

L

530 k 545 k 585 k

5.00'5.00' 7.57' 2.43'

0.5'

'

0.75'

522 k 843 k 1653 k1653 k 16531653k 1653k

-1653k-1653k

1.57'2.43'1.57' 2.43'

- 1 0 7 5

k

- 5 8 5

k

- 1 0 7 5

k

8.132'

8.132'

8.132'

- 5 8 5

k

6.291'

0 k 4

0 3

k

3 2 5

k

3.858'5.074'

13.7°

47.0° - 7 4 4 k

- 5 1 5 k - 1 1 8 9 k

0 k

0

k

0

k 0 k

- 5 3 8 k

- 8 6 7 k

- 1 1 8

stirrup

band

1= 5.00'

stirrup

band

2

= 5.00'

stirrup

band

3

= 5.00'

AASHTO LRFD Strut-and-Tie Model

Design Examples

Denis Mitchell, McGill University

Michael P. Collins, University of Toronto

Shrinivas B. Bhide and Basile G. Rabbat,

Portland Cement Association


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Direct all correspondence to:

Shrinivas B. BhidePortland Cement Association

5420 Old Orchard RoadSkokie, Illinois 60077-1083

Voice: 847.972.9100Fax: 847.972.9101E-mail: [email protected]


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ENGINEERING BULLETIN EB231

AASHTO LRFDStrut-and-Tie Model

Design Examples

Denis Mitchell, McGill University; Michael P. Collins, University of Toronto; andShrinivas B. Bhide and Basile G. Rabbat, Portland Cement Association

5420 Old Orchard RoadSkokie, Illinois 60077-1083

847.966.6200 Fax 847.966.9781www.cement.org

An organization of cement companies to improve andextend the uses of portland cement and concrete throughmarket development, engineering, research, education,and public affairs work.


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iii

© Portland Cement Association 2004

First edition

Printed in U.S.A.

ISBN 0-89312-241-6

All rights reserved. No part of this book may be reproduced inany form without permission in writing from the publisher,except by a reviewer who wishes to quote brief passages in areview written for inclusion in a magazine or newspaper.

EB231

WARNING: Contact with wet (unhardened) concrete,mortar, cement, or cement mixtures can cause SKINIRRITATION, SEVERE CHEMICAL BURNS (THIRDDEGREE), or SERIOUS EYE DAMAGE. Frequent expo-sure may be associated with irritant and/or allergic con-tact dermatitis. Wear waterproof gloves, a long-sleeved

shirt, full-length trousers, and proper eye protection whenworking with these materials. If you have to stand in wetconcrete, use waterproof boots that are high enough tokeep concrete from flowing into them. Wash wet con-crete, mortar, cement, or cement mixtures from your skinimmediately. Flush eyes with clean water immediatelyafter contact. Indirect contact through clothing can be asserious as direct contact, so promptly rinse out wet con-crete, mortar, cement, or cement mixtures from clothing.Seek immediate medical attention if you have persistentor severe discomfort.

Portland Cement Association (“PCA”) is a not-for-profitorganization and provides this publication solely for the con-tinuing education of qualified professionals. THIS PUBLICA-TION SHOULD ONLY BE USED BY QUALIFIED PRO-FESSIONALS who possess all required license(s), who arecompetent to evaluate the significance and limitations of theinformation provided herein, and who accept total responsibil-ity for the application of this information. OTHER READERSSHOULD OBTAIN ASSISTANCE FROM A QUALIFIEDPROFESSIONAL BEFORE PROCEEDING.

PCA AND ITS MEMBERS MAKE NO EXPRESS ORIMPLIED WARRANTY WITH RESPECT TO THIS PUB-LICATION OR ANY INFORMATION CONTAINED HERE-IN. IN PARTICULAR, NO WARRANTY IS MADE OFMERCHANTABILITY OR FITNESS FOR A PARTICULARPURPOSE. PCA AND ITS MEMBERS DISCLAIM ANYPRODUCT LIABILITY (INCLUDING WITHOUT LIMITA-TION ANY STRICT LIABILITY IN TORT) IN CONNEC-TION WITH THIS PUBLICATION OR ANY INFORMA-TION CONTAINED HEREIN.

Cover photos:I-25/I-40 “Big-I” Interchange, Albuquerque, New Mexico,winner of the 2002 PCA Bridge Design AwardsCompetition. (IMG15512)

KEYWORDS: AASHTO LRFD Specifications, anchorage, bearing stress, B-region, bridge, compressive strut, concentratedload, concrete, crack control reinforcement, dapped end, deep beam, development length, D-region, equilibrium, factored load,footing, geometric discontinuity, hammerhead, limiting compressive stress, moving load, nodal region, node, pier, pile cap, post-tensioned, prestressed concrete, prestressing steel, principal tensile strain, reinforcing steel, resistance factor, shear, St. Venant’sprinciple, stirrup, Strut-and-Tie Model, tension tie, transfer length, truss model, wall pier

ABSTRACT: The strut-and-tie method (STM) prescribed in the AASHTO LRFD Specifications is explained. Disturbed regions of

structures resulting from geometric or force discontinuities where STM must be used are identified. A step-by-step procedure for STMis provided. Five detailed design examples are also provided; they include: (1) Design of cap beam, (2) Design of footing, (3) Designof pile cap, (4) Design of dapped end region of girder, and (5) Design of hammerhead pier.

REFERENCE:Denis Mitchell, Michael P. Collins, Shrinivas B. Bhide, and Basile G. Rabbat, AASHTO LRFD Strut-and-Tie Model Design Examples, EB231, Portland Cement Association, Skokie, Illinois, USA, 2004, 58 pages.


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Preface

In an effort to advance the state of the art, the AASHTO LRFD Bridge Design Specifications have

introduced several new concepts and design methods. They include the use of limit states design

principles, reliability-based factored load combinations and, for concrete structures, new shear design

methods and the introduction of a general strut-and-tie design model. These new methods are signifi-

cantly different from those included in the Standard Bridge Design Specifications.

By 2007 all bridges receiving federal aid will have to be designed by the LRFD Bridge Design

Specifications. States have begun the transition from the Standard Specifications to the LRFD

Specifications. The purpose of this bulletin is to assist bridge engineers in the proper application of

the new strut-and-tie model for the design of disturbed regions of bridge components. The emphasis

is placed on “how to use” the specifications. This publication also should be a valuable aid to educa-

tors and students.

While every attempt has been made to ensure the accuracy of the design examples presented, PCA

would be grateful to any reader who brings any typographical or other errors to our attention. Other

suggestions for improvement are also most welcome.

Shri Bhide

Engineered Structures Department

iv


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Acknowledgement

The authors wish to thank the following individuals without whose help the publication of this

bulletin would not have been possible:

Jian Zhou and William Cook for preparing the drawings. Dale McFarlane, Diane Vanderlinde, Caron

Johnsen, and Cheryl Taylor, PCA, for the word processing, layout, and formatting of the document,

and David Bilow, PCA, for technical review.

v


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Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1. Visualize Flow of Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2. Sketch an Idealized Strut-and-Tie-Model. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

3. Select Area of Ties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

4. Check Nodal Zone Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

5. Check Strength of Struts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

6. Provide Adequate Anchorage for the Ties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

Example 1 – Design of Cap Beam. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-1

Example 2 – Design of Footing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-1

Example 3 – Design of Pile Cap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-1

Example 4 – Design of Dapped End Region of Girder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-1

Example 5 – Design of Hammerhead Pier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-1

Table of Contents

vi


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vii


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1

Introduction

The purpose of this bulletin is to provide bridge designers with representative examples illustrating the use of

strut-and-tie models in accordance with the AASHTO LRFD Specifications (AASHTO 2004). Although the

concept of modeling the flow of forces in structural concrete members using struts and ties was used in the early

1900’s (Ritter, 1899; Mörsch, 1909), design procedures using strut-and-tie models only recently have beencodified (CSA, 1984; AASHTO, 1994, and ACI, 2002).

In the design of reinforced and prestressed concrete elements, there are two types of regions: flexural (bending)

regions (B-regions) and regions near discontinuities (D-regions) (Schlaich et al., 1987; Collins and Mitchell,

1986). Within flexural regions, it is accurate to assume that plane sections prior to loading remain plane after

loading and that the shear stresses are distributed in a reasonably uniform manner over the effective web area. For

the design of B-regions, a sectional design approach is used. In this approach it is not necessary to address how

the forces are introduced into the member. Strut-and-tie models are used primarily to design regions near

discontinuities or D-regions. These regions have a disturbed flow of stresses and hence plane sections do not

remain plane and the shear stresses are not uniformly distributed over the effective shear area. For the design of

these regions it is important to consider how the forces are introduced into the member. Discontinuities are caused by abrupt changes in cross-sectional dimensions (e.g., the presence of openings or dapped ends) or abrupt changes

in applied forces (e.g., support reactions, large concentrated loads and post-tensioned anchorage zones). For

flexural regions, the AASHTO LRFD Specifications (§5.8.1.1)* permit the use of either traditional sectional

models (§5.8.3) or the strut-and-tie model (§5.6.3). For regions near significant discontinuities, the use of the

strut-and tie model is required (§5.8.1.2). A concentrated load which causes more than 50% of the shear at the

face of the support and is closer than 2d from the support face is a situation that requires a strut-and-tie model

(§5.8.1.1).

If we consider the case of a simply supported beam of depth h, subjected to a concentrated load applied at

midspan, there will be three disturbed regions, one adjacent to each support and one centered at midspan. The

regions near the supports will be about h long while the disturbed region near midspan will be about 2h long(St. Venant’s principle). When the distance between the applied load and the support is less than about 2h, the

disturbed regions will overlap. For typical girders (without geometric discontinuities) where the span is often

about 18h, the presence of the disturbed regions has very little influence on the overall behavior of the member

and hence the localized influence of these regions normally is ignored in design. However, if the span of the beam

is less than 4h, all of the beam will be a disturbed region and hence the overall behavior will be influenced

strongly by the disturbed flow of stresses. For this case the strut-and-tie model is used for design.

Figure 1 illustrates some examples of disturbed regions with the flow of stresses modeled with concrete

compressive struts and the tension ties provided by reinforcement. The compressive struts are shown with dashed

lines while the tension ties are indicated with thick solid lines. Figure 1(a) shows a simply supported B-region in

which the principal compressive stress trajectories will be essentially parallel. This flow is modeled by diagonalstruts. In the D-region near the support, the principal compressive stress trajectories fan out from the support.

These fanning compressive stresses can be represented by a diagonal strut along the centerline of the fan. The

strut-and-tie model uses straight-line compressive struts that are assumed to act along the center of the flow of the

compressive stresses. Each vertical tension tie represents the tension forces in a number of stirrups over a certain

* Article numbers of the AASHTO LRFD Specifications are preceded by the symbol “§”


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2

length of the member. The length of this stirrup band usually is taken as between 0.8h to about 2h, depending on

the geometry and loading.

(a) Simply supported beam (b) Pier cap

(c) Deep beam (d) Wall pier with concentratedloads

D-regionwith fanningcompressive

stresses

B-regionwith uniformcompressive

stresses

tension tie

Figure 1. Examples of D-regions modeled with struts and ties.

Figure 1(b) shows a double-sided corbel, with the flow of the compressive stresses into the corbel modeled byconcrete compressive struts. The tension tie along the top of the corbel is required for the equilibrium of the truss

formed by the strut-and-tie model. Figure 1(c) shows a deep beam subjected to a concentrated load at midspan.

The entire beam is a D-region, in accordance with St. Venant’s principle. The flow of forces into the beam is

modeled using compressive struts from the applied load into the support reaction areas. The tension tie for such

beams typically consists of several layers of reinforcing bars or post-tensioned tendons. Figure 1(d) illustrates the

fanning compressive stresses resulting from concentrated forces being applied to the top of a wall pier and the

resulting tension ties required for equilibrium.


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3

The flow of compressive stresses can follow a curved path as shown in Figure 2(a) as the compressive struts

widen between the loading and reaction plates. This curving of the compressive stresses causes tension

perpendicular to the flow as shown by the inclined tension ties in Figure 2(b). Instead of using curved

compressive struts, the AASHTO LRFD Specifications prescribe simple straight-line struts to model the flow of

compression and additional uniformly distributed horizontal and vertical reinforcement (§5.6.3.6) to control

cracking in the disturbed region (Figure 2[c]). It is noted that slabs and footings are exempt from the crack control

reinforcement requirement (§5.6.3.6). For these types of members the diagonal compressive stresses are usuallyconsiderably lower in magnitude than those for regions such as dapped ended beams, hammerhead piers, or deep

beams.

(b) Tensions due to flow of compressive stresses

(a) Flow of compressivestresses

(c) Assumption of straightcompressive struts

crack controlreinforcement

Figure 2. Straight-line struts and required crack control reinforcement (adapted from Schlaich et al., 1987).

The main steps for design of a D-region can be summarized as follows:

1. Visualize flow of stresses

2. Sketch an idealized strut-and-tie model3. Select area of ties4. Check nodal zone stresses5. Check strength of struts6. Provide adequate anchorage for the ties

The steps required to design a D-region, such as the deep beam shown in Figure 3, are given below:

1. Visualize Flow of Stresses

Visualize the flow of the compressive stresses in the D-region, and idealize the flow with straight-line struts.

Determine the locations of tension ties required for equilibrium. In sketching the truss, made up of struts and ties,

make suitable assumptions for the positions of the centroids of the ties, allowing sufficient space for placement ofthe required reinforcement.

2. Sketch an Idealized Strut-and-Tie Model

Sketch the strut-and-tie model in the form of a truss idealization. In modeling bridge girders subjected to moving

loads, it is convenient to space the nodes of the truss such that the truck axle loads can be applied directly to the

nodes. Determine the forces required in the struts and ties due to the applied factored loads. For moving loads one


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4

should consider different positions of the axle loads to achieve the maximum design forces for a particular

member.

If the member being designed is statically indeterminate (e.g., a two-span continuous deep beam subjected to

point loads in each span), a realistic strut-and-tie model also will be statically indeterminate. The support

reactions for this indeterminate truss can be found from elastic analysis of the actual member, and the internal

forces in the truss model then can be determined from statics. Alternatively, stiffness values can be assigned to themembers of the truss and hence the forces in the members and the reactions of the truss can be determined by

analyzing the statically indeterminate truss. For this analysis the stiffness of the ties can be taken as the axial

stiffness of the reinforcing bars and prestressing steel (e.g., ps psts AEAE + ) that the tie represents. The stiffness

of a strut may be taken as the cross-sectional area of the strut multiplied by the concrete modulus and the area of

any compression steel multiplied by the steel modulus (i.e., ps psscsc AEAEAE ++ ).

nodal zone

tension tie

truss node

αstension tie

0.75 f ' φ c

develop tensiontie force over this length

ε1

f f c cu


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5

4. Check Nodal Zone Stresses

Compare the nodal zone stresses with the nodal zone stress limits. Because of the dimensions of the struts and the

reinforcement making up the ties, the truss joint, or node, represents a nodal zone with finite dimensions

(see Figure 3). The nodal zones serve to transfer the forces between the ties, the struts, the support reaction areas ,

(a) Deep beam (b) Dapped end

(c) Strut anchored by bearing plate and reinforcing bars (CCT node)

(d) Strut anchored by bearing plate and strut (CCC node)

(e) Strut anchored by two tension ties (CTT node)

(f) Continuous beam support (CCC node)

CCT node

CTT node

CCT node

CCC node

ha

ha

0.5 ha

0.5 ha

lb

lb lb

l sin + h cosb aθ θ

l sin + h cosb aθ θ

θ

θ

θ

lb

da

l sin + d cosb aθ θ

Stress limit = 0.75 f 'φ c Stress limit = 0. 5 f '8 φ c

Stress limit = 0. 5 f '8 φ cStress limit = 0. 5 f '6 φ c

Figure 4. Types of nodal zones and dimensions of struts.

and the loaded bearing surfaces. The nodal zones occur at the intersections of the truss elements and at the loading

points and support reaction areas. It is important to allow for the transfer of forces without overstressing the


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6

concrete in the nodal zones. In many practical cases it will be necessary to spread the tie reinforcement into

several layers so that the nodal zone stress limit is not exceeded in the effective anchorage area (see Figure 3[b]).

The integrity of the nodal zone is checked by comparing the normal stresses applied to the boundaries of the nodal

zone with the specified nodal zone stress limits. The compressive strength of the nodal zone depends on the

tensile straining from intersecting tension ties and on confinement due to the presence of transverse

reinforcement. The nodal zone stress limits in the AASHTO specifications (§5.6.3.5) depend on the number of

ties that are being anchored in the nodal zone. Figure 4 illustrates a number of typical nodal zone conditions andtheir corresponding stress limits depending on the presence of compressive struts (indicated by “C”) and tension

ties (indicated by “T”).

5. Check Strength of Struts

Compare the factored resistance of the struts with the calculated factored loads in the strut members (§5.6.3.2 and

§5.6.3.3). The nominal resistance of the strut is determined by multiplying the limiting compressive stress, cuf , by

the effective cross-sectional area of the strut, csA . The limiting compressive stress depends on the angle, sα ,

between the compressive strut and the tension tie and the tensile strain, sε , in the tie where it crosses the strut.

The limiting compressive stress is given as:

'c

1

'c

cu f 85.01708.0

f f ≤

ε+=

in which the principle tensile strain in the concrete, 1ε , is taken as:

( ) s2

ss1 cot002.0 α+ε+ε=ε

where sα is the smallest angle between the compressive strut and the tension tie and sε is the average tensile

strain in the concrete in the direction of the tension tie. For a tension tie consisting of reinforcing bars, sε , can be

taken as the tensile strain due to factored loads in the reinforcing bars. As shown in Figure 5, the limitingcompressive stress, f cu, reduces significantly as the angle, sα , becomes smaller. A strain value of sε of 0.002

would correspond to a tie yielding in tension at the location where it crosses the centerline of the strut. If the

tension force in the tie changes as it crosses the strut, then sε may be taken as the value of the strain where the tie

meets the centerline of the strut. For example if the tie is developed within the width of the strut, the tensile strain

in the tie might change from 0.002 at the inner edge of the strut to zero at the outer edge of the strut, resulting in

an sε value of 0.001. For a tension tie consisting of prestressed steel, sε can be taken as zero until the

precompression in the concrete due to the prestress is overcome. Figure 5 illustrates the influence of sα and sε on

the limiting compressive stress, f cu. It can be seen that even struts at very shallow angles (e.g., 20°) can resist

significant stresses ( 'cf 297.0 ) if the tie is prestressed.


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7

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0 10 20 30 40 50 60 70 80 90

f c u

/ f ' c

αs

εs = 0

εs = 0.002

εs = 0.001

Figure 5. Influence of sα and sε on the limiting compressive stress in a strut.

Figure 6 shows two specific examples of calculating f cu for cases involving reinforcing bars as ties. For the two

cases shown, if the strain sε is 0.002 at the inner edge of the strut, and it is conservatively assumed that this strain

reduces to 0.001 at the centerline of the strut, then the limit on the compressive strength of the struts, f cu, changes

from 'cf 68.0 for the case where sα equals 45o to 'cf 30.0 for the case where sα equals 25

o (see § 5.6.3.3.3).

εs εs

ε1

ε1

αs = 45o

αs = 25o

(a) = 45αso

f = 0.68 f 'cu c

( ) = 25b αso

f = 0.30 f 'cu c

Figure 6. Examples of determining f cu..

If the strut is anchored by a bearing area, the cross-sectional dimensions of the strut will be influenced by the

length of the bearing area, the dimensions of the adjacent ties or struts, and the inclination of the strut (see Figures

4[c] and 4[d]). If the strut is anchored only by reinforcement, the effective dimensions of the strut are related to


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8

the reinforcement details as shown in Figure 7. The strut bears against the longitudinal reinforcing bars which in

turn are anchored by the stirrups. It is assumed that the effective width of the strut across the thickness of the

member can extend a distance of up to six times the diameter of the longitudinal bar anchored by the stirrups (i.e.,

bad6 ) (see § 5.6.3.3.2).

6dba 6dba

l a

s

dba

l a ssinθ

θs

dba


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9

References

AASHTO LRFD Bridge Design Specifications, first edition 1994, second edition 1998, third edition 2004,

American Association of State Highway and Transportation Officials, Washington, DC, USA.

ACI, 2002, Building Code Requirements for Structural Concrete (ACI 318-02) and Commentary (ACI 318R-02),

American Concrete Institute, Farmington Hills, MI, 443 pages.

Collins, M. P., and Mitchell, D., 1986, A Rational Approach to Shear Design – The 1984 Canadian Code

Provisions,” ACI Journal, Vol. 83, No.6, Nov.-Dec., pages 925 to 933.

CSA Standard A23.3, 1984, Design of Concrete Structures for Buildings, Canadian Standards Association,

Rexdale, Ontario, Canada, 281 pages.

Mörsch, E., 1909, Concrete-Steel Construction, McGraw Hill, New York, (English translation by E.P. Goodrich),

368 pages.

Ritter, W., 1899, Die bauweise hennebique,“ Schweizerische Bauzeitung ,” Vol. 33, No. 7, pages. 59 to 61.

Schlaich, J.; Shäfer, K., and Jennewein, M., 1987, Towards a Consistent Design of Reinforced Concrete

Structures,” PCI Journal, Vol. 32, No. 3, May-June, pages. 74 to 150.


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SpecificationExample 1 (cont’d) Calculations and Discussion Reference

1-2

Step 1 - Draw Idealized Truss Model and Solve for Member Forces

The idealized truss model shown in Figure 1.2 represents the flow of forces in the cap beam.

The dashed lines represent compressive struts and the solid lines represent tension ties. The

point loads at the locations of the bearings have been increased to account for the factored self-weight of the cap beam. For example, the tributary factored self-weight of the beam at Nodes B

and D is ( )[ ] kips3.251035.415.025.1 =×××× , plus an allowance for the bearing block and bearing, giving about 26 kips. Similarly the factored self-weight of the beam, bearing block, and

bearing at Node A is about 18 kips. For simplicity the moment in the column will be neglectedfor the design of the cap beam, i.e., the axial stresses will be considered uniform at the top of

the column pier.

In order to allow for the placement of the tension reinforcement and to account for the depth of

the concrete compressive struts, it has been assumed that the centroids of the top and bottomchords of the truss are located 4 in. from the top and bottom concrete surfaces. In establishing

the horizontal locations of the nodes of the truss, Nodes A, B, and D are located directly below

the applied loads, while node C is located halfway between nodes B and D. Node G is locateddirectly below node B, and Node H is located directly below Node C. Although a single node

could be used to represent the column support, it is somewhat more accurate to represent thecolumn support by two Nodes, E and F. Node E is responsible for supporting the 267 kip load

from the cantilever, while Node F is responsible for the 434 kip load from the loads in the span

(i.e., 2/298285+ ). The positions of the two resultant compressive forces in the column (Nodes

E and F) have been chosen such that a uniform compressive stress exists in the column. To

achieve this, the width of the column has been subdivided so that the width 48.02× ft is

carrying the 267 kip force and the width 77.02× ft is carrying the 434 kip force. The resulting

factored forces in the truss members are determined by satisfying equilibrium (see Figure 1.2).

Figure 1.2. Truss idealization and member forces.


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1-3

Step 2 – Check Size of Bearings

The limiting concrete stresses under the bearings depend on the conditions at the nodal zone.

Node D is a CCC node with a limiting stress of 'cf 85.0 φ and Node A is a CCT node with a

limiting stress of 'cf 75.0 φ . Because Node B anchors the horizontal Tie A-B and the vertical Tie

B-G (representing the uniformly distributed stirrups), it will be designed as a CTT node with a

limiting stress of 'cf 65.0 φ .

The critical bearing stress occurs at Node B caused by the girder load and hence the minimum

bearing area can be determined as:

2

'c

u in.142470.065.0

259

f 0.65

Prequiredarea bearing =

××=

φ=

With dimensions of 18 x 30 in., the bearing block has sufficient area )in.540( 2 .

Step 3 – Choose Tension Tie Reinforcement

(a) Top reinforcement over column

The required area of tension tie reinforcement, stA , in Tie AB is:

Use 6 No. 9 bars,2

s in.0.6A =

(b) Bottom reinforcement at midspan

The required area of tension tie reinforcement, stA , at midspan is:

Use 12 No. 9 bars, 2s in.0.12A =

(c) Stirrups

The vertical tension Ties, BG and CH, each must resist a factored tension force of 149 kips.

This tension force can be provided by stirrups within a certain length of the beam as indicated

by the stirrup bands in Figure 1.2.

Using No. 5 stirrups with 2 legs, the number of stirrups, n, required in each band is:

2

y

ust in.46.5

609.0

295

f

PA =

×=

φ=

2

y

ust in.20.11

609.0

605

f

PA =

×=

φ=

§5.6.3.4.1

§5.6.3.5


21/62


1-4

45.46031.029.0

149

f A

Pn

yst

u =×××

=φ

=

Hence, the required spacing, s, within the 5-ft band is:

in.5.1345.4

60s =≤

Try No. 5 double-legged stirrups at 12 in.

Step 4 – Check Capacity of Struts

Strut FB carries the highest compression force (671 kips, see Figure 1.2). Also, this strut is

anchored at Joint B which also anchors tension Tie AB and Tie BG. Hence this is the most

critical strut. The limiting compressive stress in the strut, cuf , usually is controlled by the

tensile strain in the tie which is at the smallest angle to the strut. From the geometry of the trussidealization, the angle between tension Tie AB and Strut FB is 40.3o. The tensile strain in Tie

AB is:

3

sst

us 10695.1

000,290.6

295

EA

P −×=×

==ε

671

kips

295

kips

285

kips

217kips

= 1.695×10-3

εs = 0.848×10-3

18" bearing

8"

ε ≈ 0

18 sin 40.3° + 8 cos 40.3° = 17.7”

εs

40.3o

εs = 1.657×10-3

149 kips

49.7o

Figure 1.3. Details of strut near Node B.

Note that since Member BC is in compression (see Figure 1.2 and Figure 1.3), it will have a

small compressive strain. Because there is a large tensile strain in Tie AB and a relatively small


22/62


1-5

compressive strain (say zero strain) in Member BC, it will be assumed that at the center of Strut

FB, near Node B, that

( ) 33s 10848.02/010695.1 −− ×=+×=ε .

Therefore, the principal strain, 1ε , can be determined as:

( ) ( ) 30233s2ss1 1081.43.40cot002.010848.010848.0cot002.0 −−− ×=+×+×=α+ε+ε=ε

and the limiting compressive stress, cuf , in the strut is:

ksi3.4040.85ksi47.21081.41708.0

4f 85.0

1708.0

f f

3

'c

1

'c

cu =×≤=××+

=≤ε+

=−

Note that the tension strain in Tie BG is:

3

sst

us 10657.1

000,29)12/60(31.02

149

EA

P −×=×××

==ε

Hence from this tie 1ε would need to be:

( ) ( ) 30233s2ss1 1029.47.49cot002.010657.110657.1cot002.0 −−− ×=+×+×=α+ε+ε=ε

As this is less than 31081.4 −× this tie does not govern the compressive capacity of the strut.

The nominal resistance of the strut is based on the limiting stress, cuf , and the strut dimensions.The strut width is computed in Figure 1.3 at 17.7 in. and the strut thickness can be

conservatively taken as the bearing block dimension, namely 30 in. Hence,

kips1312307.1747.2Af P cscun =××==

The factored resistance of the strut is:

requiredkips671kips918131270.0PP nr ≥=×=φ=

Therefore, the strut capacity is adequate.

Step 5 – Check Anchorage of Tension Tie

The top No. 9 longitudinal bars must be developed at the inner edge of the bearing at Node A.

An embedment length of 36 + 9 – 2 in. cover = 43 in. is available to develop the bars (see

Figure 1.1). The basic development length for a straight No. 9 bar is 34 in., but with a top bar

factor of 1.4, the required length is 48 in. Therefore, it is necessary to provide hooks at the end

of the bars.

§5.6.3.3.3

§5.6.3.3.3

§5.6.3.3.1

§5.6.3.3.3

§5.6.3.3.1

§5.6.3.2

§5.11.2.1.1

§5.6.3.4.1


23/62


1-6

It is also necessary to check if the tension ties are spread out sufficiently in the effective

anchorage area, which is 2 x 4 = 8 in. in depth. The nodal zone stress to anchor the tension tie

is:

ksi024.13642

295f c =××

=

For this nodal zone, the limiting nodal zone stress is:

ksi1.2470.075.0f 75.0f 'cc =××=φ=

Therefore, the nodal zone stress in the anchorage area is acceptable.

Step 6 – Provide Crack Control Reinforcement

(a) D-region (region near discontinuity)

Due to the presence of the concentrated loads at A and B, which are within a distance less than

the member depth from the faces of the column, the zone between Nodes A and B will beconsidered as a D-region (St. Venant’s principle).

For “disturbed regions” (D-regions) the AASHTO LRFD Specifications require that crack

control reinforcement in the form of an orthogonal grid of reinforcement on both faces be

provided. The minimum ratio of reinforcement to gross concrete area is 0.003 in each direction.For a spacing of 12 in. this requires:

2s in.30.13612003.0A =××=

Therefore, use 4 No. 5 horizontal bars at 12 in. spacing and 4 legs of No. 5 stirrups at 12 in.

spacing, giving 2s in.24.131.04A =×= in the D-region (this is within 5% of the required

area).

(b) B-region (flexural region)

The region between Nodes B and D is considered to be a B-region because the concentrated

load being transmitted in this region is more than twice the member depth from the supporting

column. As such this flexural region can be designed using the strut-and-tie model but the

detailing requirements of the sectional design procedures for flexural regions must be satisfied.

Hence it is appropriate to use the minimum transverse reinforcement requirements of §5.8.2.5.

Although two-legged No. 5 stirrups at 12 in. spacing are required for strength (Step 3[c]), it is

necessary to ensure that minimum shear reinforcement is provided to control inclined cracking.

Assuming a stirrup spacing of 12 in., the minimum area of stirrups shall not be less than:

2

y

v'cv in.46.0

60

123640316.0

f

s bf 0316.0A =

×××==

§5.6.3.5

§5.6.3.6

§5.8.1.1

§5.8.2

§5.6.3.1

§5.8.2.5

§5.8.2.5


24/62


1-7

Since 2v in. 62.031.02A =×= , an amount greater than minimum has been provided.

Use double-legged No. 5 stirrups at a spacing of 12 in. in the B-region.

Step 7 – Sketch the Required Reinforcement

The resulting reinforcement of the cap beam is shown in Figure 1.4.

D - region B - region

6 – No.94 legged No.5stirrups at 12"

2 legged No.5stirrups at 12"

4 – No.5 2 – No.5 12 – No.9

4 – No.5 typ. 2 – No.5 typ.each face

6 – No.9 top

4 legs of No.5

closed stirrups

@ 12"

12 – No.9 bot 12 – No.9 bot

2 No.5@12"

2 – No.9 top

Figure 1.4. Reinforcement details of cap beam.


25/62

Example 2 – Design of Footing

2-1

Design the footing supporting the column shown in Example 1 (see Figures 1.1 and 2.1). The square spread

footing is 9 x 9 x 3 ft. thick. The specified concrete compressive strength, 'cf , is 4 ksi and the specified yield

strength of the reinforcing steel is 60 ksi.

Design the footing using the AASHTO LRFD strut-and-tie method.

Figure 2.1. Footing dimensions.

SpecificationCalculations and Discussion Reference


Traditionally, footings are designed as flexural regions by checking the moment capacity at the

face of the column and checking the one-way and two-way shear capacities at the appropriate

critical sections. While this procedure leads to satisfactory designs, a more accurate model forthe flow of forces in the footing results if the strut-and-tie method is used (see Figure 2.2).

The idealized truss model shown in Figure 2.2 represents the flow of forces in the footing. The

dashed lines represent compressive struts and the solid line represents the tension tie.

In order to allow for the minimum 3-in. concrete cover and an assumed bar size of 1 in. for the

tension reinforcement, the centroid of the tension tie has been taken as 4 in. from the bottom

surface of the footing. It is further assumed that the centroid of the top compressive strut islocated 2 in. below the top surface of the footing.

In determining the loading on the footing it is necessary to account for the factored self-weight

of the column. The factored dead load of the column is ( )[ ] kips.9.12115.25.215.025.1 =×××× It is noted that footings are designed for the factored net pressure due to the applied column

loads only. The weight of the footing and pressure from the soil surcharge are transmitted

directly through the footing thickness and are resisted by equal and opposite soils pressures.


26/62


27/62


2-3

Solving the statics of the truss, the member forces shown in Figure 2.2 are determined.


Figure 2.2 illustrates the required tension tie forces in the bottom reinforcement from this strut-and-tie model. Figure 2.2 also shows the variation of tie force if it is assumed that the tie force

results from flexure only, that is a tie force equal to the moment calculated at each section

divided by a lever arm of 2.5 It is important to note that, except for the location directly under

the column, the tie forces required by the strut-and-tie model considerably exceed those

required for flexure alone. The increased tie force is due to the fact that the inclined

compression, which is carrying the shear, causes tension in the reinforcement.

The required area of tie reinforcement, stA , which is a maximum in Ties DE and EF, is:

In components subjected to flexure, the amount of reinforcement provided must be adequate to

develop a factored flexural resistance, Mr , at least 1.2 times the cracking moment unless anamount of reinforcement capable of carrying 1.33 times the factored moment is supplied. The

factored flexural resistance required to resist 1.2Mcr , using the modulus of rupture, f r , of

'cf 24.0 ksi, specified in §5.4.2.6 is:

ft-kip1120in-kip437,13424.06

361292.124.0

62.12.1

2'

2

==××

×=×=≥ ccr r f bh

M M

The corresponding minimum area of flexural reinforcement can be determined by dividing thisfactored flexural resistance by the product of flexural lever arm (2.5 ft), and the factored yield

strength of the reinforcement giving:

2min,s in.30.8

609.05.2

1120A =

××=

The AASHTO specifications indicate that this minimum amount need not be supplied if 1.33times the required resistance is provided. Hence, increase the tension tie reinforcement to:

Use 8 No. 8 bottom bars in each direction (

2

in.32.6 provided).


Strut AJ carries the highest compression force (128 kips, see Figure 2.2) and has the smallest

angle of inclination from the tension tie. Figure 2.3 shows the details of the strut near Node A.

2

y

ust in.41.4

609.0

0.238

f

PA =

×=

φ=

2st in.87.541.433.1A =×=

§5.6.3.4.1

§5.7.3.3.2

§5.7.3.3.2

§5.4.2.6


28/62


2-4

Figure 2.3. Details of strut near Node A.

From the geometry of the truss idealization, the angle between tension Tie AB and the Strut AJis 38.3o. The tensile strain in Tie AB is:

3

sst

us 10548.0

000,2932.6

5.100

EA

P −×=×

==ε

The strain, sε , in the tension tie crossing the strut varies from a maximum at the inside face of

the strut to zero at the outside edge of the strut. Hence, it will be assumed that at the center of

the strut, at Node A, ( ) 33s 10274.02/010548.0 −− ×=+×=ε .

The principal strain, 1ε , can be determined as:

( ) ( ) 30233s2ss1 1092.33.38cot002.010274.010274.0cot002.0 −−− ×=+×+×=α+ε+ε=ε

and the limiting compressive stress, cuf , in the strut is then:

ksi3.4040.85ksi73.21092.31708.0

4f 85.0

1708.0

f f

3

'c

1

'c

cu =×≤=××+

=≤ε+

=−

Multiplying this stress by the area of the strut, the nominal resistance of the strut is thus:

kips21821294.773.2Af P cscun =×××==

With a capacity reduction factor for compression in strut-and-tie models of 0.70, the factored

resistance of the strut is:

requiredkips281kips1527218270.0 ≥=×== nr P P φ

§5.6.3.3.3

§5.6.3.3.3

§5.6.3.3.1

§5.5.4.2.1


29/62


2-5

It can be seen that the factored resistance of the strut greatly exceeds the required factored force.

This will often be the case in large, lightly reinforced members.

It is noted that Nodes J, K and L represent the CCC node where the column frames into the

footing. The compressive struts in this region are not crossed by any tension ties. In addition,there will be some vertical reinforcement passing through the column-footing interface. This

node and the struts in this region are not critical.

Step 4 – Check Anchorage of Tension Tie

The No. 8 bars are required to develop a force of 100.5 kips at a distance of only 12 in. from the

edge of the footing (see Figure 2.3). Note that the stress in the 8 No. 8 bars at this location is

only:

ksi9.1579.08

5.100f s =

×=

With the provision of a 180o hook, as often used for reinforcing bars in footings, the distance

from the critical section to the end of the hook is in.9(cover)312 =− This detail will be

sufficient to develop the relatively small stress in the bars.

It is necessary to check if the tension ties are spread out sufficiently in the effective anchorage

area, which is 2 x 4 = 8 in. in depth. The nodal zone stress to anchor the tension tie is:

ksi12.012942

5.100f c =

×××=

For this CCT nodal zone, the limiting nodal zone stress is:

ksi1.2470.075.0f 75.0f 'cc =××=φ=

Therefore, the nodal zone stress limit in the anchorage area is considerably below the limit.


It is noted that slabs and footings are exempt from the crack control requirements of §5.6.3.6.

The requirements for minimum flexural reinforcement of §5.7.3.3.2 have been satisfied in

Step 2.

§5.6.3.4.1

§5.6.3.5

§5.6.3.6

§5.7.3.3.2


30/62


2-6


The resulting reinforcement of the footing is shown in Figure 2.4.

Figure 2.4. Reinforcement details of footing.


31/62

Example 3 – Design of Pile Cap

3-1

Design the pile cap supporting the pier shown in Figure 3.1. The square pile cap is 12 x 12 x 4 ft. thick. The 653

kip pier load acting on the top surface of the pile cap includes factored dead loads, factored truck and lane

loadings, and an allowance for impact. The specified concrete compressive strength, 'cf , is 4 ksi and the specified

yield strength of the reinforcing steel is 60 ksi. There is a 2-ft overburden on the pile cap with a unit weight of 110

pcf.

Design the pile cap using the AASHTO LRFD strut-and-tie method.

Figure 3.1. Pile cap dimensions.



Traditionally, pile caps are designed as flexural regions by checking the moment capacity at the

face of the pier and checking the one-way and two-way shear capacities at various critical

sections. For deep pile caps special investigations typically are conducted to justify higher shear

stresses on the critical sections. A more accurate model for the flow of forces in the pile capresults if the strut-and-tie method is used (see Figure 3.2).

Figure 3.2 shows the idealized truss model for the pile cap. For the design of the pile cap it is

important to appreciate the three-dimensional flow of forces from the pier to the piles. The load

applied to the top of the pile cap includes not only the factored pier load but also the factored

weight of the pile cap and the soil overburden. Hence the total factored load applied to the top

of the pile cap is 653 + 1.25 (0.150) (12 x 12 x 4) + 1.3 (0.110) (12 x 12 – 3 x 3) 2.0 = 800 kips.


32/62


3-2

¼ × 3' = 0.75'

3'-0"

3'-0"

6"4"

2"

38.7°

9'-0"

200 kips

200 kips

200 kips

200 kips200 kips

200 kips200 kips

250 kips

2 5 0 k i

p s

force in strut

= -406 kips

200 kips

250 kips

Figure 3.2. Truss idealization and member forces.

The total factored load at the top of the pile cap has been applied at four points (E, F, G, and H),

located at distances equal to the pier dimension divided by four from the pier faces, to simulateuniform compressive stress in the pier. It is assumed that these four loads will be transferred to

the support reaction areas of the four piles by four inclined compressive struts, indicated by


33/62


3-3

dashed lines in Figure 3.2. The inclined compressive struts, spreading from Nodes E, F, G, and

H, to Nodes A, B, C, and D, respectively, just above the piles, cause tension in the Ties AB, BC,

CD, and AD indicated by solid lines. To define the geometry of the truss, it is assumed that the

centroid of the tension tie reinforcement is located 4 in. above the top of the piles to allow for a

3 in. concrete cover and a bar size of 1 in. Nodes E, F, G, and H are assumed to be 2 in. belowthe top surface of the pile cap to allow for the dimensions of Struts EF, FG, GH, and EH. Nodes

A, B, C, and D are located directly above the centers of the piles.

Solving the statics of the truss, the member forces shown in Figure 3.2 are determined. The

elevation view is useful in determining the tension tie force required in Tie AB. The isometric

view is useful in visualizing the complete three-dimensional truss and for determining the forces

in the compressive struts.

Step 2 – Check Nodal Zone Stress Limits

The nodal zone at the pier-pile cap interface has a stress of:

ksi62.03636

800f c =

×=

This is considerably below the nodal zone stress limit for this CCC node of

ksi38.2470.085.0f 85.0 'c =××=φ

Furthermore, the pier vertical bars would extend some distance into the pile cap along with the

pier ties. This would increase the resistance of this nodal zone.

The stress in the nodal zones immediately above the piles is:

ksi78.01616

200f c =

×=

These nodal zones immediately above the pile caps have tension ties in two directions passing

through the nodal zones (CTT node) and hence have a reduced nodal zone stress limit of

ksi82.1470.065.0f 65.0 'c =××=φ . Therefore, the dimensions of these nodal zones are

adequate.


All four tension ties have a factored tension force of 250 kips. Hence, the required area of tiereinforcement, stA , is:

In components subjected to flexure, the amount of reinforcement provided must be adequate to

develop a factored flexural resistance, Mr , at least 1.2 times the cracking moment unless an

2

y

ust in.63.4

609.0

250

f

PA =

×=

φ=

§5.6.3.5

§5.6.3.5

§5.6.3.2

§5.6.3.4.1


34/62


3-4

amount of reinforcement capable of carrying 1.33 times the factored moment is supplied. The

factored flexural resistance required to resist 1.2Mcr , using the modulus of rupture, f r , of

'cf 24.0 ksi, specified in §5.4.2.6 is:

ft-kip2654in.-kip850,31424.06

4812122.124.0

62.1

2'

2

==××

×=×≥ cr f bh

M

The corresponding minimum area of flexural reinforcement can be determined by dividing this

factored flexural resistance by the product of flexural lever arm (3.0 ft), and the factored yield

strength of the reinforcement giving:

2min,s in.4.16

609.00.3

654,2A =

××=

It should be noted that the total area of this flexural tension reinforcement,s

A , will be provided

by the reinforcement in the two parallel tension ties, e.g., Ties AB and CD (in Figure 3.2).

Hence the area of tension tie reinforcement required to ensure that the factored flexural

resistance is at least cr M2.1 is equal to2in.2.82/4.16 =

On the other hand, the amount of reinforcement required to resist 1.33 times the factored loads

is:

As the amount of reinforcement required to resist 1.33 times the factored loads is less than the

amount required to resist cr M2.1 , this smaller amount will be provided. Choose 8 No.8 bottom

bars for each of the Ties AB, BC, CD, and AD shown in Figure 3.2 ( 2in.32.6 provided).


Struts AE, BF, CG and DH each carry a compression force of 406 kips (see Fig. 3.2). In order to

determine the nominal compressive resistances of the diagonal struts it is necessary to

determine the effective cross-sectional area of these struts in the critical region near the

intersection of the tension ties and the pile reactions (e.g., Joint A in Figure 3.2) and the limiting

compressive stress that these struts can carry. In making these calculations it should be

appreciated that the actual flow of forces from the pile cap into the bearing areas of the piles is

complex and that the ability of these highly stressed struts to carry the loads is considerably

enhanced by the confinement provided by the surrounding mass of concrete that has been

presumed to be unstressed. Because of this confinement it is appropriate to make simplifying

assumptions when estimating the cross-sectional area of the struts.

2st in.16.663.433.1A =×=

§5.7.3.3.2

§5.4.2.6


35/62


3-5

As shown in Figure 3.3, it will be assumed that the compressive stresses from the head of the

pile disperse into the pile cap at a 45° angle. This results in the effective bearing area at the

middle plane of the tension tie reinforcement being a square with side dimensions of 24 x 24 in.

(see Figure 3.3).

Figure 3.3. Dispersion of compressive stresses above pile.

The vertical rise of the diagonal strut from A to E is 3.0 ft, while the horizontal distance from A

to the point directly below E is ft30.5275.3 = (see Figure 3.2). Hence, the angle between the

centerline of the strut and the horizontal square bearing area is 29.5° (see Figure 3.4).

The cross-sectional area of the strut can be estimated by first looking at the section passing

through the vertical plane (Section 1-1) shown in Figure 3.4. From geometry, it can be

determined that the area of this vertical plane is 411 in.2. It will be assumed that the cross-

sectional area of the strut, which is inclined at 29.5o from the vertical plane, is thus2o in.3585.29cos411 =


36/62


3-6

Figure 3.4. Details of strut near Node A.

The limiting compressive stress, cuf , in the strut depends on the principal tensile strain, 1ε , in

the concrete surrounding the tension ties.

The tensile strain in Tie AB is:

3

sst

us 10364.1

000,2932.6

250

EA

P −×=×

==ε

§5.6.3.3.3


37/62


38/62


3-8

and to factors which may reduce bond strength. Alternatively, headed bars could be used to

ensure positive anchorage and direct bearing on the nodal zone.

It is also necessary to check if the tension ties are spread out sufficiently in the effective

anchorage area. The standard hooks have an inside bend diameter of 6 in., giving out-to-outdimensions of 8 in. for the hooks. Considering the beneficial effects of the hooks, the effective

depth of the nodal zone anchoring the ties will be taken as the distance from the bottom of the

nodal zone to the centroid of the ties (4 in.) plus the hook dimension (8 in.). The nodal zone

stress due to anchorage of the tension tie is:

ksi87.02412

250f c =

×=

For this CTT nodal zone, the limiting nodal zone stress is:

ksi82.1470.065.0f 65.0f '

cc

=××=φ=

Therefore, the tension ties with hooks are spread out sufficiently in the nodal zone.

Step 6 – Check Minimum Reinforcement Requirements

It is noted that footing-type elements are exempt from the crack control requirements of

§5.6.3.6.

The requirements for minimum flexural reinforcement of §5.7.3.3.2 have been satisfied

in Step 3.


The resulting reinforcement of the pile cap is shown in Figures 3.5 and 3.6.

§5.6.3.6

§5.7.3.3.2

§5.11.2.4


39/62


3-9

square column

3'-0"×3'-0"

16"×16"

precastpile

3'-6"

6"

9'-0"

8 No. 8 bars

8 No. 8 bars 3" clear

4'-6" 4'-6"

24”

Note: hooks to be placed in vertical plane

Figure 3.5. Reinforcement details of pile cap using 180° hooks.

square column

3'-0"×3'-0"

16"×16"

precastpile

3'-6"

6"

9'-0"

8 No. 8 bars

8 No. 8 bars 3" clear

4'-6" 4'-6"

24”

Figure 3.6. Reinforcement details of pile cap using headed bars.


40/62

Example 4 – Design of Dapped End Region of Girder

4-1

The dapped end region of an interior girder of a single lane ramp structure is shown in Figure 4.1. The precast,

pretensioned girder has a total length of 100 ft and spans 97 ft. 2in., center-to-center of bearings. The girder

consists of a 54-in. deep AASHTO IV girder and an 8-in. thick deck slab. The girders are spaced at 10 The end

region, 56 in. in length, is solid with a width of 26 in., that is, equal to the width of the bottom flange of the girder.

As shown in Figure 4.2, the girder is pretensioned with forty-two 0.5-in. diameter low-relaxation strands (36 with

a straight strand profile and 6 draped strands). The harping points are located 33.33 ft from each end of the beam.

The specified concrete compressive strength, 'cf , is 8 ksi for the girder and 6 ksi for the deck slab, and the

specified yield strength of the reinforcing steel is 60 ksi.

In addition to the self-weight of the girder and deck slab, the girder must carry a 2-in. concrete wearing surface, a

superimposed dead load (including utilities) of 0.334 kips/ft, the Design Lane load, and the Design truck load

(HL93).

Design the dapped end region using the AASHTO LRFD strut-and-tie method.

Figure 4.1. Details of dapped end.


41/62


4-2

Figure 4.2. Details of pretensioning strand.


42/62


4-3

Step 1 – Determine Factored Loading

A concrete unit weight of 0.15 kcf has been assumed. Table 4.1 summarizes the loading:

Table 4.1 Girder Loadings

Component Loading

(klf)

Strength-I

Load

Factor

(§3.4.1)

Factored Loading

(klf)

Precast girder 0.822 1.25 1.0275

Deck slab 1 1.25 1.25

Solid block (26 in. x 54

in.)

1.463 (over solid

block region only)

1.25 1.829 (over solid

block region only)

Dap region (26 in. x 27in.)

0.731 (over dapregion only)

1.25 0.9138 (over dapregion only)

Superimposed dead load

(wearing surface)

0.25 1.5 0.375

Superimposed dead load

(utilities, etc.)

0.334 1.5 0.501

The Design Lane Load is 0.64 klf and the Design Truck has axle loads of 32 kips, 32 kips, and 8

kips, with axle spacings of 14 ft. The tandem axle load consists of two axle loads of 25 kips

each, spaced 4 ft apart. For this case the tandem axle load will produce a smaller shear at the

support than the design truck and hence does not control the design.

For a girder spacing, S, of 10 ft, the live load distribution factor for shear in interior beams is:

76.025

1036.0

25

S36.0 =+=+

Using a load factor of 1.75 results in a factored lane loading for an interior girder of:

klf 8512.076.064.075.1 =××

Using a load factor of 1.75 and a dynamic load allowance factor, IM, of 1.33 results in factored

axle loadings of:

kips6.5633.176.03275.1 =×××

andkips15.1433.176.0875.1 =×××

With the truck positioned as indicated in Figure 4.3, the maximum shear (reaction) is

determined to be 319 kips.

§3.6.1.2.4

§3.6.1.2.2

§3.6.1.2.3

§4.6.2.2.3

§3.4.1

§3.6.2.1

§3.4.1


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4-5

The resulting factored forces in the truss members determined from equilibrium are listed in

Table 4.2.

2"

4"

29"

27"

39°

2"

56"

4"

33.5" 45.5" 56"

21.9"

11.0 kips 15.7 kips73.5 kips

18.7 kips

310 kips

16.5"

Figure 4.4. Truss idealization.

Table 4.2 Forces in Truss Members (Tension is Positive, Compression is Negative)

Member Force, kips

AB -456.8

BC 503.8AD 345.4

CD -649.0

DE -365.0

BD -276.3

BE -178.9

CF 409.1

EF 209.8

EG -409.1

FH 618.9


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4-7

force varies linearly from zero at the end of the beam to a maximum at the transfer length is

therefore:

kips8.143153.0616230

29P

strands

=×××=

Additional reinforcing bars will be required to carry the difference between the required force of

351.8 kips and the force carried by the strands, that is, kips0.2088.1438.351 =− . Therefore,

the area of reinforcing bars to make up this difference is:

Therefore, use 5 No. 8 bars,2

s in.95.3A = . To provide adequate anchorage of this

reinforcement, provide either heads on the bars or use 90o bend hooks at each end. The bars

must be capable of developing their yield force at the inner edge of the bearing area at Node Aand must be anchored beyond Node D so that the tension force can be transferred to Struts DB

and DC (see Figure 4.4).

(c) Tension Tie CF

The force required in Tie CF is 409.1 kips (see Table 4.2). The distance from the ends of the

straight strands to the inner edge of the nodal zone at Node C is 33.5 in. (see Figure 4.5). As this

distance exceeds 60 strand diameters, the strands are capable of developing a stress of 162 ksi.

The bottom three rows of strands (30 strands) are capable of providing a force of:

kips P strands 6.743153.030162 =××=

Because this force exceeds the force required, no additional reinforcing steel is required.

2

y

ust in.85.3

609.0

0.208

f

PA =

×=

φ=

§5.8.3.2

§5.6.3.4.1

§5.6.3.4.1


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4-8

7@4"2.5"

2.25"

51°

39°

503.8 kips649 kips

lb = 33.5"6dba = 3"

409.1 kips

4"

ha

= 8"

width of strut

= lb sin θ + ha cos θ

= 33.5 sin 51° + 8 cos 51° = 31.1"

14"

Figure 4.5. Details of strut near Node C.

(d) Tension Tie EF

The vertical tension Tie EF must resist a factored tension force of 209.8 kips (see Table 4.2).This tension force can be provided by stirrups within a stirrup band that is assumed to start at

the edge of the closed stirrups making up Tie BC and ending halfway between Nodes F and H.

The distance from the last stirrup of Tie BC to Node F is in.5.31145.45 =− (see Figures 4.4

and 4.5). Hence the length of the stirrup band is in.5.59285.31 =+

Using No. 4 stirrups with 4 legs, the number of sets of stirrups, n, required in this band is:

86.46020.049.0

8.209

f A

Pn

yst

u =×××

=φ

=

Hence the required spacing, s, within the 59.5-in. band length is:

in.2.1286.4

5.59s =≤

Use 4-legged No. 4 closed stirrups at 12 in.

§5.6.3.4.1


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4-9

Step 5 –Check Capacity of Struts

Strut CD carries the highest compression force (649 kips, see Table 4.2). Also, this strut is

anchored at Joint C that also anchors tension Ties BC and CF. Hence, this is the most critical

strut. From the geometry of the truss, the smallest angle between the strut and the adjoiningtension ties is 39o . The tensile strain in Tie BC is:

3

sst

us 10751.1

000,2992.9

8.503

EA

P −×=×

==ε


( ) ( ) 30233s2ss1 10471.739cot002.010751.110751.1cot002.0 −−− ×=+×+×=α+ε+ε=ε

and the limiting compressive stress, cuf , in the strut is:

5.1ksiksi60.85ksi90.210471.71708.0

685.0

1708.0 3'

1

'

=×≤=××+

=≤+

=−c

ccu f

f f

ε

6db6db6db 6db 6db 6db 1.69"1.69"

effective thickness of strut = 2 × 4.69 + 2 × 6 = 21.4"db = 0.5"

clear cover on stirrups= 1 3/8"

Figure 4.6. Effective thickness of diagonal struts.

From Figure 4.5, the cross-sectional dimension of the strut in the plane of the beam is 31.1 in.

and from Figure 4.6 the effective thickness of the strut is 21.4 in. Hence, the nominal resistanceof the strut is:

§5.6.3.3.3

§5.6.3.3.3

§5.6.3.3.3

§5.6.3.3.2


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4-10

kips19304.211.3190.2Af P cscun =××==


kips1351193070.0PP nr =×=φ=

As the factored resistance of the strut exceeds the factored load in the strut, kips649 , the strut

capacity is adequate.

Step 6 – Check Anchorage of Tension Ties

The details of the anchorage of the stirrups near the top of the beam must be such that they are

effective in transferring their tension to the diagonal struts. At these locations the effective

bearing area will be enhanced if closed stirrups rather than open stirrups are provided. Closed

stirrups also provide superior anchorage. The horizontal reinforcing bars which constitute Tie

AD have either hooks or heads and have sufficient anchorage (see Step 4). The pretensioned

strand was used for tension ties and the variation of force over the transfer length was accounted

for in assessing the force capability of this reinforcement.

It is necessary to check that the tension ties are spread out sufficiently in the effective anchorage

area. Node C is a CTT node, which is 8 in. in depth (see Figure 4.5). The nodal zone stress to

anchor the tension tie force in CF of 409.1 kips is:

ksi39.24.218

1.409f c =

×=

For this CTT nodal zone, the limiting nodal zone stress is:

ksi73.2670.065.0f 65.0f 'cc =××=φ=

Therefore, the nodal zone stress limit in the anchorage area is acceptable.


It is assumed that the B-region has been designed using sectional design procedures and that at

least a minimum amount of stirrups has been provided in the girder. The D-region is assumed to

be in the solid block region, with the B-region starting 62 in. from the end of the full-depth (62-in.-deep) solid block (see Figure 4.7).

For “disturbed regions” (D-regions) the AASHTO LRFD Specifications require that crack

control reinforcement in the form of an orthogonal grid of reinforcement on both faces be

provided. The minimum ratio of reinforcement to gross concrete area is 0.003 in each direction.

For a spacing of 9 in. this requires:

2s in.70.0269003.0A =××=

§5.6.3.3.1

§5.6.3.2

§5.6.3.5

§5.6.3.6


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4-11

Therefore, use 4 No. 4 horizontal bars at 9 in. spacing and 4 legs of No. 4 stirrups at 9 in.

spacing, giving 2in.80.020.04 =×= s A in the D-region.


The resulting reinforcement of the dapped end beam is shown in Figure 4.7.


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4-12

5 No.8headedbars

D-region B-region

No. 4

long. bars

@ 9"

No. 4

long. bars@ 9"

8 – 4 legged

No. 5closed stirrups@ 4"

4 – 4 legged


4 – 4 legged


4 legged No. 4stirrups @ 9"

4 legged No. 5

closed stirrups

@ 4"

Figure 4.7. Reinforcement details of D-region of dapped end beam.


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Example 5 – Design of Hammerhead Pier

5-1

The hammerhead pier shown in Figure 5.1 consists of a rectangular pier and a variable depth cap beam that

supports 5 lines of precast, pretensioned girders. The girders sit on neoprene pads, which in turn are supported by

concrete bearing blocks having dimensions of 18 x 36 in. The Strength I factored loads acting on the 5 bearing

blocks include allowances for the factored self-weight of the cap beam.

The specified concrete compressive strength, '

c

f , is 4 ksi and the specified yield strength of the reinforcing steel is

60 ksi.

Design the hammerhead pier using the AASHTO LRFD Specifications.

Pu = 530 k Pu = 545 k Pu = 585 k Pu = 545 k Pu = 530 k

10' - 0"2'-6" 2'-6"10' - 0" 10' - 0" 10' - 0"

4'-6"

4'-6"

20'-0"

18"

8'-0"

rectangularcolumn

8'-0" × 3'-6"

cap beam3'-6" wide

1.000

0.2432

f c' = 4 ksif y = 60 ksi

cover = 2.5" main bars= 2.0" stirrups

Figure 5. Details of hammerhead pier.


The three central loads are located at a distance which is less than twice the member depth from

the supporting reaction. Hence the central 20 ft of the hammerhead pier is a D-Region and will

be designed using the strut-and-tie method. The outer portions of the hammerhead pier are

flexural regions (B-Regions) which can be designed for shear using either the sectional model

or the strut-and-tie model. For this example, the strut-and-tie model will be used.

§5.6.3

§5.8.1.1


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5-2


The idealized truss model shown in Figure 5.2 represents the flow of forces in the hammerhead

pier. The dashed lines coincide with the centerlines of the compressive struts that represent

compressive stresses in different areas of the concrete. The solid lines coincide with thecentroids of tension ties, which represent tension forces in different groups of reinforcing bars.

Under the action of the girder loads the ends of the cap beam will bend down causing tension

near the top face of the hammerhead pier and compression near the sloping bottom faces. To

allow appropriate room for placement of the longitudinal reinforcement, it has been assumed

that the centroid of the tension tie near the top face is located 6 in. below the top face. To

provide an appropriate space for the concrete compression zone, it has been assumed that the

centerline of the bottom compression strut is located 9 in. above the sloping bottom face and is

parallel to this face. The compression force in the pier is represented by 3 vertical struts. The

central strut carries the 585 kip load, while the outer two struts carry 1075 kips each. Assuming

that the pier is subjected to uniform compressive stresses, the width of each outer strut must be:

ft14.3810752585

1075=×

×+

Hence the centerline of the outer struts will be ft57.114.350.0 =× from the outer faces of the

pier.

The distributed stirrups in the cap beam are represented by the vertical tension Ties AB, CD,

EF, and GH. To solve the statics of the truss model it is convenient to know the lengths of these

4 truss members. As can be seen from Figure 5.1 and Figure 5.2, the vertical distance between

the top tie, ACEG, and the bottom strut, BDFH, increases by 0.2432 ft for every additional foot

travelled away from the free end of the cantilever. As shown in Figure 5.2, the resulting lengths

of the 4 vertical ties are 3.858 ft, 5.074 ft, 6.29 ft., and 8.132 ft.

The member forces shown in Figure 5.2 were determined by the method of joints. Thus at Joint

A, the vertical component from Member AD must push the joint upwards with 530 kips. The

member must also push the joint to the left with a force of kips522074.5/00.5530 =× . The

square root of the sum of the squares of these two components is the force in Member AD,

namely a compression of 744 kips. Member AC must have a tension force of 522 kips to

balance the horizontal component of Member AD. Considering horizontal and vertical

equilibrium for Joints D, C, F, E, H, and G enables all of the member forces to be computed.


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5-4

Use 20 No. 11 bars, 2st in.2.3156.120A =×=

Figure 5.3. Layout of 20 – No. 11 top bars near pier.

As shown in Figure 5.3, the required 20 No. 11 bars can be provided in 2 layers of 10 bars. If

No. 5 stirrups are used the centroid of the 20 No. 11 bars will be about 4.7 in. from the top face.Hence the assumption that the centroid of the tension tie would be 6 in. below the top face was

conservative.

(b) At lowest tension location, AC

The required area of tension tie reinforcement is:

Therefore, use 8 No. 11 bars,2

stin.48.1256.18A =×= .

(c) Development of bars

The development length for a straight top horizontal No. 11 bar with ksi60f y = and

ksi4f 'c = is 82 in. If 90o hooks with at least 2.5 in. of side cover are used the development

length is reduced to 19 in. Hence terminate the 10 bars in the lower layer at a location 19 in.

2

y

ust in.61.30

609.0

1653

f

PA =

×=

φ=

2

y

ust in.67.9

609.0

522

f

PA =

×=

φ=

§5.6.3.4.1

§5.6.3.4.1

§5.11.2.1

§5.11.2.4


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5-5

beyond point E. Terminate the remaining 10 bars with 90o hooks at a location 27 in. beyond

point A.

Step 4 – Design Tension Ties Representing Stirrups

Try using No. 5 stirrups with 4 legs (see Figure 5.3).

(a) Stirrup spacing required for Tie CD

Vertical Tie CD has the highest tension. Hence the number of stirrups required in stirrup band 2

(see Figure 5.2), is:

02.66031.049.0

403

f A

Pn

yst

u =×××

=φ

=


in.97.902.6

60s =≤

Try a spacing of 9 in.

In the flexural region between A and E the minimum transverse reinforcement, assuming a

stirrup spacing of 9 in., is:

2

y

v'cv in.39.0

60

94240316.0

f

s bf 0316.0A =

×××==

Since 2v in.24.131.04A =×= , an amount greater than minimum has been provided in stirrup

band 2 (see Figure 5.2). While No. 5 stirrups with 2 legs could be used in stirrup band 1, which

will be governed by the minimum area requirement, it would be more practical to continue the

4-legged No. 5 stirrups at a spacing of 9 in. throughout this region.

(b) Stirrup spacing required for Tie EF

Vertical Tie EF must resist a tension of 325 kips. Hence the number of stirrups required in

stirrup band 3 (see Figure 5.2), is:

85.46031.049.0

325f A

Pnyst

u =×××

=φ

=


in.37.1285.4

60=≤ s

§5.6.3.4.1

§5.8.2.5

§5.6.3.4.1


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5-6

Try a spacing of 12 in.

For crack control in this disturbed region, the ratio of reinforcement area to cross-sectional area

shall not be less than 0.003 in both the vertical and horizontal directions. Hence:

003.0 bs

Ast ≥

Therefore:

.in84.942003.0

31.04

b003.0

As st =

×

×=≤

Thus use No. 5 stirrups with 4 legs spaced at 9 in. throughout the length of the beam.

Step 5 – Check Capacity of Bottom Strut BDFH

The highest compressive force in the bottom Strut BDFH is 867 kips in Member FH

(see Figure 5.2).

As this strut will be crossed by vertical stirrups, the compressive capacity of this strut may need

to be reduced. The area of Tie EF is 2in.27.831.04)9/60( =×× . Hence the strain in this stirrup

under the 325 kip tension is:

3

sst

us 1036.1

000,2927.8

325

EA

P −×=×

==ε

As the smallest angle between the strut and the tension tie is 90 - 13.7 = 76.3o, the principal

strain, 1ε , can be determined as:

( ) ( ) 30233s2ss1 1056.13.76cot002.01036.11036.1cot002.0 −−− ×=+×+×=α+ε+ε=ε

And, the limiting compressive stress, cuf , in the strut is:

ksi3.4040.85ksi76.31056.11708.0

4f 85.0

1708.0

f f

3

'c

1

'c

cu =×≤=××+

=≤ε+

=−

The centroid of the strut was assumed to be at 9 in. vertically from the bottom face(see Figure 5.2); hence the thickness of the strut perpendicular to the sloping bottom face is

in.5.177.13cos92 o =×× The nominal resistance of the strut is:

kips 24995.174240.3Af P cscun =××==

§5.6.3.6

§5.6.3.3.3

§5.6.3.3.3

§5.6.3.3.1


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5-7


kips1749249970.0PP nr =×=φ=

As the factored resistance exceeds the 867 kip compression due to factored loads, the strutcapacity is adequate.

While the truss geometry could be adjusted by reducing the thickness of the bottom strut and the

member forces recalculated, the changes in forces will be rather small, resulting in perhaps the

saving of only one bar in the main tension tie. Thus the original conservative assumptions areacceptable.

Step 6 – Check Capacity of Diagonal Struts of AD, CF, and EH

Of the three diagonal struts crossing the web, AD, CF, and EH, Member EH has the highest

compression. The details of the member at end E, where it crosses the tension ties, are shown in

Figure 5.4.

Figure 5.4. Details of Strut EH near Node E.

The strains in Ties CE and EG due to factored loads are shown in Figure 5.3. For determining

the strut capacity, the average value of these two strains has been assumed,

giving 31085.1 −×= sε .

§5.6.3.3.3

§5.6.3.2


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5-8


( ) ( ) 3023321 1020.50.47cot002.01085.11085.1cot002.0 −−− ×=+×+×=++= s s s α ε ε ε

and the limiting compressive stress, cu f , in the strut is:

ksi3.4040.85ksi38.21020.51708.0

485.0

1708.0 3'

1

'

=×≤=××+

=≤+

=−c

ccu f

f f

ε

The cross-sectional dimension of strut EH in the plane of the pier is 19.6 in. (see Figure 5.4),

while the effective thickness of the strut at end E could be conservatively taken as 36 in. which

is the width of the bearing block. However, the good anchorage conditions provided by the No.

11 bars in the corner of the stirrups enable the effective thickness of the strut to be increased.

As can be seen from Figure 5.3, the center-to-center distance of the vertical stirrups across the

42-in. width of the hammerhead pier is 12.5 in. As this distance is less than

in.9.16410.162d62 ba =××=× , the full 42-in. width of the pier cap is effective. Hence the

nominal resistance of the strut is:

kips19596.194238.2 =××== cscun A f P


requiredkips1189kips1357195970.0 ≥=×== nr P P φ

Therefore, the strut capacity is adequate.


In Step 4, the stirrup spacing was adjusted to satisfy the crack control requirements for

reinforcement in the vertical direction, but crack control reinforcement also must be provided in

the horizontal direction. The vertical spacing between these horizontal bars must not exceed 12

in. If this maximum spacing is used, the area of horizontal bars in each layer needs to be:

2st in.51.14212003.0 bs003.0A =××==

Therefore, use 4 No. 6 horizontal bars at 12 in. spacing ( 2.in76.144.04 =× provided), arranged

as shown in Figure 5.5.

§5.6.3.3.3

§5.6.3.3.1

§5.6.3.6

§5.6.3.3.2


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5-9


The resulting reinforcement for the hammerhead pier is shown in Figure 5.5. For clarity the pier

reinforcement is not shown.

Figure 5.5. Reinforcement details for hammerhead pier.

1 0 – No. 11 10 – No. 11

4 legged No.5 stirrups at 9" cc

4 – No. 6 bars at 12 ” cc 4 – No.6 bars

20 – No. 11


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5-10


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ASSHTO Example Strut and Tie

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Transcript of ASSHTO Example Strut and Tie