Strut and Tie Modelling

12/10/2010

1

Design Using the Strut-and-Tie Method, Part 2(A)

ACI Spring 2010 Xtreme Concrete ConventionMarch 21 - 25, Chicago, IL

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ACI Conventions

12/10/2010

2

ACI Web Sessions

This ACI Web Session includes four speakers presenting at the ACI Xtreme Concrete convention held in Chicago, IL, March 21st through 25th, 2010.

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Design Using the Strut-and-Tie Method, Part 2(A)

ACI Spring 2010 Xtreme Concrete ConventionMarch 21 - 25, Chicago, IL

Daniel Kuchma holds a B.A.Sc., M.A.Sc., and Ph.D., all in civil engineering, from the University of Toronto. Since 1997, he has been an Associate Professor in the department of Civil and Environmental Engineering at the University of Illinois, and has taught courses in structural dynamics, statics, reinforced concrete, and pre-

stressed concrete. His work includes a variety of consulting projects involving offshore structures, hydroelectric dams, towers, buildings and specialty structures. Dr. Kuchma is an active member of ACI, and the Federation International de Beton(fib). He received a National Science Foundation CAREER Award on Tools and Research to Advance the Use of Strut-and-Tie Models in Education and Design. He is also a National Center for Supercomputing Applications Faculty Fellow and University of Illinois Collins Scholar.

10

Dan KuchmaSukit Yindeesuk

Tjen Tjhin

Propped Cantilever with Opening

University of Illinois

11

Design problem5625 kN

12

Selected truss model

5800 kN

3229.5 kN

3659 kN

418.8 kN

4136 kN

A B C D E

FG H I J

K L M N

O P Q R

535

mm

570

mm

2895

mm

1015 mm

1590 mm

4000 mm 2030 mm

1015 mm

1350 mm

S

T

U

49 degrees

Externally and internally indeterminate truss5625 kN

12/10/2010

3

13

ACI design; calculation of nominal capacity Calculated plastic truss capacity Calculated non-linear truss capacity Predicted capacity and behavior by non-

linear finite element analysis Measured capacity and behavior by

experimental testing Observations and Conclusions

Outline of Presentation

14

ACI design; calculation of nominal capacity

-6947

5615 -4090

-3860 -3835

-7500 -3796 -79

-3869

-254

-5360

-4863

2083

2662

-421

-949

2535

-4570

520

154

1243 821 3650

-305

167

0 1

582

341

0 3

716

-87

-4430

-821

2146

209

3860 3650

182

8 1

781

144

46

326

5 3

409

Member forces determined assuming equal stiffness of member forces

Reinforcement and strut/node dimensions selected to provide adequate capacity

15

0.392

0.520 0.188 0.185 0.343 0.227 1.01(O/S)

0.454

0.07

0.328

0.764

0.010.698

0.7740.798 0.0540.0

10.

890.

92

0.48

3

0.45

2

1.01

(O/S

)

0.586

0.496

0.0330.990 0.9420.078 0.633

0.0160.4960.525 0.04 0.0350.

524

0.49

50.

039

0.03

90.

429

0.92

40.

02

Nominal design strength taken as when first members reaches its capacity

Occurs at top right tie; Pn = 7500 kN Utilization rates shown in figure

ACI design; calculation of nominal capacity

16

Calculated plastic truss capacity Member stress-strain characteristics

17

Calculated plastic truss capacity

0.527

0.999

0.52

8

0.274

0.363

0.284 0.122 0.115 0.639

0.919 0.994 0.003

0.743 0.734 0.009 0.018

0.025 0.993 0.922

0.48

2

0.033

0.597

0.011

0.743

0.731 0.025

0.005 0.480

0.04

10.

997

1.00

9

0.52

10.

562

0.01

3

0.00

20.

526

0.48

9

1.00

91.

002

0.00

1

Very different distribution of demands Capacity reaches when a mechanism forms P = 9469 kN

18

Calculated non-linear truss capacity

0.520

0.481 0.267 0.271 0.012 0.047 0.714

0.511

0.031

0.369

0.985

0.0130.762

0.9880.959 0.0190.0

150.

981

0.97

3

0.52

0

0.47

6

1.00

00.613

0.605

0.070.988 0.9710.016

0.7380.006

0.7300.724 0.006 0.0080.52

1

0.50

60.

008

0.01

40.

468

0.97

30.

08

Non-linear stress-strain relationship Similar demands as by plastic truss model Capacity reaches when a mechanism forms P = 9301 kN

12/10/2010

4

19

Comparison of strength calculations

Pn

Pu

20

Predicted capacity and behavior by non-linear finite element analysis

Predicted state of cracking at ultimate P = 16622 kN

21


Predicted distribution of steel stress at ultimate

Stress (Steel): -truss at crack46.0664.5282.98

101.43

119.89138.35156.81175.27

193.73212.18230.64249.10

267.56286.02304.48322.93

341.39359.85378.31396.77

415.23433.68452.14470.60

22


Vital Signs: Fcm0.040.080.130.17

0.210.250.290.33

0.380.420.460.50

0.540.580.630.67

0.710.750.790.83

0.880.920.961.00

Compressive Demand: ratio of compressive stress to compressive capacity at failure

P = 16622 kN

23

Measured capacity and behavior by experimental testing

Reinforcing cage

24

Measured capacity and behavior by experimental testing Test Setup

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5

25

Measured capacity and behavior by experimental testing Mode of failure

26

Comparison of strength calculations

Pn

Pu

27

Observations and Conclusions

1. Truss member design forces in statically indeterminate strut-and-tie models depend on the relative stiffness of members

2. Plastic truss capacity can be modestly larger than when the first member reaches its capacity

3. Truss models cannot provide a good estimate of deformation; much softer than in reality

4. Non-linear finite element analysis can predict well the behavior of complex STM designed regions

28

Questions

Hakim Bouadi is a Senior Associate with Walter P Moore & Associates in Houston, Texas, which provides structural, structural diagnostics, civil, traffic and transportation engineering, and parking consulting services to clients worldwide.

STM Design of two Link Beams at a Medium-Rise Building

Hakim Bouadi, Ph.D., P.E.Asif Wahidi, Ph.D., P.E.

WALTER P MOORE

12/10/2010

6

STM Design of two Link Beams at a Medium-Rise Building WALTER P MOORE31

Outline

Building Overview Link Beam Overview Link Beam with Moderate Shear Link Beam with High Shear Conclusions

31 STM Design of two Link Beams at a Medium-Rise Building WALTER P MOORE32

Building Overview

32

256'-0"142'-0"

148'

-0"

351'

-0"

Hospital buildingLocation: Las Vegas, NevadaLateral design controlled by seismic forces

STM Design of two Link Beams at a Medium-Rise Building WALTER P MOORE33 33

18'-0

"

48'-7"

18'-0

"15

'-0"

15'-0

"15

'-0"

256'-0"142'-0"

148'

-0"

351'

-0"

Plan size: about 500 ft by 400 ftLateral resisting system: shear wallsControlling lateral loads: seismic forces

Building Overview


18'-0

"48'-7"

18'-0

"15

'-0"

15'-0

"15

'-0" Shear walls with link beams above openings

Link beam: deep beams per ACI 318 definition

Review link beam at roof and at level 3

Shear Wall Overview


Beam under constant shear and moment reversal Forces on nodes obtained from global lateral analysis Reduce forces to ends

Link Beam Overview


Roof Link Beam

External forces applied at nodes Shear force equal to about:

dbf wc '25.4

.

External nodes at location of wall reinforcement Horizontal tie at location of reinforcement Vertical tie at mid-span Improvement: Extend model into wall

12/10/2010

7


Forces on Members

37

Force resolved by analysis Check struts Design ties Check nodes Detailing


Design of Struts and Ties

38

Strut and tie dimensions from geometry

Struts: fan shaped Capacity of struts checked

at strut and at nodes Tie force resisted by

reinforcement


Design of Struts and Ties

39

215.46075.0

187 inf

FAy

us

Vertical tie: 8#5 stirrups

Horizontal tie: 4 #7218.2

6075.09.94 in

fFA

y

us

kipswbfF csu 215125.85.560.085.075.0'85.0 Strut

Develop beyond extended nodal zone


Node Capacity

40

Nodal dimensions from geometry

Node type CCT due to tension force in wall reinforcement

Check capacity on each face

kipsbwfcn 286125.85.58.085.075.085.0'


Node Capacity

Node type CTT due to tension force in wall reinforcement

Check capacity on each facekipsbwfcn 1761275.56.085.075.085.0

'


Design: Roof Link Beam,

42

12/10/2010

8


Link Beam at Level 3

43

18'-0

"

48'-7"

18'-0

"15

'-0"

15'-0

"15

'-0"

Shear force equal to about: dbf wc '10 Design using STM Follow also Chapter 21 of ACI:

Seismic Design/Detailing


Forces for Level 3 Link Beam

Shear force equal to about:dbf wc '10

Design using STM Follow also Chapter 21 of ACI:

Seismic Design/Detailing

External nodes at location of wall reinforcement Horizontal tie at location of reinforcement Vertical tie at mid-span Improvement: Extend model into wall


External nodes at location of wall reinforcement

Transfer forces through X configuration

Model for Level 3 Link Beam


Forces resolved through geometry Symmetric design due to load

reversal Tie force resisted by reinforcement

(4#11 and 2#9) Strut force resisted by concrete

and reinforcement

Design for Level 3 Link Beam


development length of beyond extended nodal zone extended by 25%

Minimum web reinforcement Appendix A ACI Chapter 21 (controls)

Enclose Tie reinforcement with stirrups

Detailing for Level 3 Link Beam


Summary for Level 3 Link Beam

12/10/2010

9


Conclusions

49

STM use for link beam design Different models are possible Model can be extended into the wall to follow force

transfer Check detailing (in addition to design)

STM Design of two Link Beams at a Medium-Rise Building

Hakim Bouadi, Ph.D., P.E.Asif Wahidi, Ph.D., P.E.

WALTER P MOORE

Thank you

Richard Beaupre received his Bachelor of Science and Engineering from the University of Florida and his Master of Science from the University of Texas at Austin. While at the University of Texas he was involved in research pertaining to deviation saddle behavior and

design for externally post-tensioned segmental concrete girder bridges. He is currently a senior bridge engineer for URS Corporation in Tampa, Florida, where he is responsible for design of steel and concrete bridges, ship impact designs, structural modeling, and quality control. He is experienced in design of major bridge structures, including cable-stayed, post-tensioned segmental concrete and movable.

MIC-Earlington Heights Connector Metrorail

ACI 2010, Chicago, IL

Richard Beaupre, PE

Robert (Bob) Anderson, PE

Velvet Bridges, PE

URS Corporation

Tampa, Florida

Diaphragm for a Segmental Concrete Bridge



Introduction

Many Areas of a Concrete Segmental Bridge can be Classified as a D Region Pier Diaphragms Interior Segment Diaphragms at Deviation

Points for External Tendons Openings in Flanges and Webs Pile Caps



Project Overview

12/10/2010

10



Miami Intermodal Center

MIC-EARLINGTON HEIGHTS METRORAIL

EXTENSION



Superstructure Requirements

130 Spans in Miami Intermodal Center 225 Span to Clear the Miami River with a 40

Vertical Clearance 180 Span for the South Florida Railroad Corridor 256 Span to Cross to SR112 and the Future

Dade Expressway Height Restrictions Set by FAA Airspace near

MIA



Superstructure Types

72 Florida Prestressed U-Beams Segmental Concrete Boxes 30 Cast-In-Place Concrete Slabs Single Track and Dual Track Cross-

Sections



Guideway Structures Overview 72 Florida U-Beams - Single Track Guideway



Guideway Structures Overview 72 Florida U-Beams - Dual Track Guideway



Guideway Structures Overview Single Track Guideway (Units 1 thru 4 & 14)

12/10/2010

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Guideway Structures Overview Dual Track Guideway (Units 5 thru 9 & 11 thru 13)



Diaphragm Example

Layout Function Boundary Forces Strut-Tie Model



Diaphragm Example

Unit 8



Layout



LayoutMIC-Earlington Heights Connector Metrorail


Pier Diaphragm Function

Transfer Loads from the Webs to the Support around Access Openings

Distribute Tendon Anchorage Forces to the Cross-section

12/10/2010

12



Rapid Transit Live Load Vehicle

Full Live Load Weight of 120 kips Train 2 to 8 Vehicles



Diaphragm Loadings

Factored Loading Case

Shear Torsion

Kips/Box (kN/Box) Kip-Ft/Box

(kN*m/Box)

Strength I (Maximum Shear) 3,291 (14,638) 21 (28)

Extreme Event III (Maximum Torsion)

2,634 (11,716) 5,355 (7,260)



Diaphragm Unit Loads for ShearMIC-Earlington Heights Connector Metrorail


Diaphragm Unit Loads for Torsion



Strut-Tie Model Steps Step 1: Determine strut-and-tie arrangement

based on boundary forces Step 2: Solve for the member forces Step 3: Determine the amount of steel for ties Step 4: Arrange tie steel Step 5: Check anchorage zone for the ties Step 6: Check diagonal struts Step 7: Check nodal zones



Pier DiaphragmsReference: Schlaich et. al., Towards a Consistent Design of Structural

Concrete, PCI Journal, Vol. 32, No. 3, May-June 1987

12/10/2010

13



Model Members with Shear Unit Loads



Model Members with Torsion Unit Loads



Material Properties

Concrete: fc=8,500 psi (58.7 MPa) Reinforcement: fy= 60,000 psi (414 MPa)



Results for Shear Only

Strength I V=3291 k (14,638 kN)

Extreme Event III V=2634 k (11,716 kN) Member Unit Force Kips (kN) Kips (kN)

1 0.450 740.9 (3295.3) 593.0 (2637.5) 2 0.267 439.4 (1954.4) 351.7 (1564.2) 3 0.267 439.4 (1954.4) 351.7 (1564.2) 4 0.450 740.9 (3295.3) 593.0 (2637.5) 5 -0.792 -1302.7 (-5794.4) -1042.6 (-4637.7) 6 -0.467 -768.5 (-3418.1) -615.0 (-2735.7) 7 -0.467 -768.5 (-3418.1) -615.0 (-2735.7) 8 -0.792 -1302.7 (-5794.4) -1042.6 (-4637.7) 9 -1.000 -1645.5 (-7319.2) -1317.0 (-5858.0) 10 -1.000 -1645.5 (-7319.2) -1317.0 (-5858.0) 11 -0.363 -596.5 (-2653.2) -477.4 (-2123.5) 12 -0.363 -596.5 (-2653.2) -477.4 (-2123.5) 13 0.000 0.0 (0.0) 0.0 (0.0) 14 0.000 0.0 (0.0) 0.0 (0.0) 15 1.020 1678.7 (7466.9) 1343.6 (5976.2) 16 0.001 0.9 (4.2) 0.7 (3.3) 17 1.020 1678.7 (7466.9) 1343.6 (5976.2) 18 0.001 0.9 (4.2) 0.7 (3.3)



Results for Torsion Only

Strength I T=21 k-ft (28 kN-m)

Extreme Event III T=5354 k-ft (7258 kN-m) Member Unit Force

Kips (kN) Kips (kN) 1 -0.015 -0.3 (-1.4) -81.1 (-360.8) 2 -0.009 -0.2 (-0.8) -48.1 (-214.0) 3 0.009 0.2 (0.8) 48.1 (214.0) 4 0.015 0.3 (1.4) 81.1 (360.8) 5 -0.027 -0.6 (-2.5) -142.6 (-634.4) 6 0.031 0.7 (2.9) 168.5 (749.3) 7 -0.031 -0.7 (-2.9) -168.5 (-749.3) 8 0.027 0.6 (2.5) 142.6 (634.4) 9 -0.105 -2.2 (-9.8) -562.3 (-2501.2) 10 0.105 2.2 (9.8) 562.3 (2501.2) 11 0.012 0.3 (1.1) 65.3 (290.5) 12 -0.012 -0.3 (-1.1) -65.3 (-290.5) 13 0.086 1.8 (8.0) 458.5 (2039.3) 14 -0.086 -1.8 (-8.0) -458.5 (-2039.3) 15 0.700 0.7 3.2 183.8 (817.5) 16 0.0 0.0 0.0 0.0 (0.1) 17 -0.700 -0.7 -3.2 -183.8 (-817.5) 18 -0.0 0.0 0.0 0.0 (-0.1)



Results for Shear and Torsion Combined

Strength I Extreme Event III Member Kips (kN) Kips (kN)

1 740.5 (3293.9) 511.8 (2276.7) 2 439.2 (1953.5) 303.6 (1350.2) 3 439.2 (1955.2) 399.8 (1778.2) 4 741.2 (3296.8) 674.1 (2998.3) 5 -1303.3 (-5796.8 -1185.3 (-5272.1) 6 -767.8 (-3415.1) -446.6 (-1986.4) 7 -769.1 (-3421.0) -783.5 (-3485.0) 8 -1302.1 (-5791.9) -900.0 (-4003.3) 9 -1647.7 (-7329.0 -1879.3 (-8359.2)

10 -1643.4 (-7309.4) -754.7 (-3356.9) 11 -596.8 (-2652.1) -412.1 (-1833.0) 12 -596.8 (-2654.4) -542.7 (-2414.1) 13 1.8 (8.0) 458.5 (2039.3) 14 -1.8 (-8.0) -458.5 (-2039.3) 15 1679.4 (7470.1) 1527.4 (6793.8) 16 0.9 (4.2) 0.8 (3.5) 17 1678.0 (7463.7) 1159.8 (5158.7) 18 0.9 (4.2) 0.7 (3.2)

12/10/2010

14



Ties 1 to 4 Top TieAccording to ACI 318 equation A-1 Fnt>Fut. Where: Fut = Factored Design Force = 741 k (3,297 kN) = 0.75 (Section 9.3.2.6) Fnt = Nominal Strength of a Tie Where no prestressing steel is used, Fnt = Atsfy (Section A.4.1) Using 1 row of 11 # 11 diameter reinforcing bars.

)mm (10,064 in 17.2in 1.56bars 11rows 1 222 tsA kN) (3297k 741 kN) (3443k 774ksi 60in 17.275.0 2 ntF



Tie 13 Diagonal TieSimilarly for Tie 14 (depending on direction of torsion), Fut = Factored Design Force = 459 k (2,039 kN) = 0.75 (Section 9.3.2.6) Using 10 # 9 diameter reinforcing diagonal bars. )mm (6,452 in 10.0in 1.00bars 10 222 tsA kN) (2,039k 459 kN) 669(2,k 450ksi 06in 0.1075.0 2 ntF



Ties 15 and 17 Hanging Up Tie Fut = Factored Design Force = 1679 k (7,470 kN) = 0.75 (Section 9.3.2.6) Using 28 # 8 diameter reinforcing web bars plus 11 - # 11 bars (continue Tie 1 to 4 reinforcement) )mm (25,368 in 39.3in 2.17in 0.79bars 28 2222 tsA kN) (7,470k 1,679 kN) 864(7,k 1,768ksi 06in 39.375.0 2 ntF



Main Tie Reinforcing



Nodal Zone Detail at Bearing Support



Strut 14According to ACI 318 equation A-1 Fns>Fus Where: Fus = Factored Design Force = -459 k (-2,039 kN) = 0.75 (Section 9.3.2.6) Fns = Nominal Strength of a Strut = fceAcs Further, csce ff '85.0 Where: 60.0s (Section A.3.2.2 (b) bottle shaped struts without reinforcing of A.3.3.1) Therefore, MPa) 9.9(2 psi 4335psi 85000.600.85 cef Multiply the allowable compressive stress of a strut by the area of concrete available to carry the stress which is limited by the access opening (width is 4.9 in). )m (0.09 in 141.6in 9.28in 4.9 22csA kN) (2,039k 459kN) (2,048k 4601000/in 6.141 psi 43350.75 2 nsF

12/10/2010

15



Nodal Check at Member 9According to ACI 318 equation A-1 Fnn>Fun Where: Fun = Factored Design Force = 0.75 (Section 9.3.2.6) Fnn = Nominal Strength of a Node = fceAnz Further, cnce ff '85.0 Where: 0.1n (Section A.5.2.2 Nodes bounded by struts and bearing area) Therefore,

MPa) (49.9 psi 7225psi 85001.00.85 cef Multiply the allowable compressive stress on a face of a nodal zone by the area of concrete based on the geometry of the node. Fun = Factored Design Force = 1,879 k (8,359 kN) )m (0.74 in 1142in 39.5in 28.9 22nzA (Area of Bearing) kN) (8,359k 1,879kN) (27,525k 6188000 1/in 1421psi 72250.75 2 nnF



Total Diaphragm Reinforcement



Diaphragm Cracking



Summary

Strut-Tie Procedures can be Effectively Utilized for Diaphragm Design

Shear and Torsion Forces are Redirected into Support through the Diaphragm around the Access Opening

After Solving the Truss Forces, Ties can be Designed and Detailed

Struts and Nodes need to be Checked

Daniel Kuchma holds a B.A.Sc., M.A.Sc., and Ph.D., all in civil engineering, from the University of Toronto. Since 1997, he has been an Associate Professor in the department of Civil and Environmental Engineering at the University of Illinois, and has taught courses in structural dynamics, statics, reinforced concrete, and pre-

stressed concrete. His work includes a variety of consulting projects involving offshore structures, hydroelectric dams, towers, buildings and specialty structures. Dr. Kuchma is an active member of ACI, and the Federation International de Beton(fib). He received a National Science Foundation CAREER Award on Tools and Research to Advance the Use of Strut-and-Tie Models in Education and Design. He is also a National Center for Supercomputing Applications Faculty Fellow and University of Illinois Collins Scholar.

90

Dan Kuchma

Future of ACI STM Provisions and Guidelines

University of Illinois at Urbana-Champaign

12/10/2010

16

91

Location of STM Provisions in ACI318-14?

Appendix A of ACI318-08 Location of Provisions in ACI318-14 Available Guideline Documents Challenges to Design by the STM ACI 445 Committee Document

92

Appendix A of ACI318-08: Basic Rules

T

C

T

C

C C

P

P2

> A f Ts yP2

>

Af

Cc

cu

> A f Ts y

>

Af

Cc

cu

93

Appendix A of ACI318-08: Basic RulesDesign Strength of Struts = Fns where Fns = fce Area of Strut and fce = 0.85sfc

s = 1.00 for prismatic struts in uncracked compression zones s = 0.40 for struts in tension members s = 0.75 when struts may be bottle shaped and crack control reinforcement* is included s = 0.60 when struts may be bottle shaped and crack control reinforcement* is not included s = 0.60 for all other cases *crack control reinforcement requirement is visini 0.003

Design Strength of Ties = Fnt where Fnt = Astfy + Atp(fse + fp)

Note that the tie reinforcement must be spread over a large enough area such that the tie force divided by the anchorage area (where the height is twice the distance from the edge of the region to the centroid of the reinforcement) is less than the limiting stress for that nodal zone.

Design Strength of Each Nodal Zone Face = Fnn where Fnn = fce Area on Face of Nodal Zone (perpendicular to the line of action of the associated strut or t ie force) Again fce = 0.85nfc

n = 1.00 in nodes bounded by struts and bearing areas n = 0.80 in nodes anchoring a tie in one direction only n = 0.60 in nodes anchoring a tie in more than one direction

94

Appendix A of ACI318-08: Explanatory Materials32 Figures

95

Location of STM Provisions in ACI318-14?

Separate appendix like in ACI318-08

Separate 318 referenced document

Basic rules put into main body of code and application guidelines in a separate document

96

Available Guideline Documents

Design Examples SP-208 Second SP

Textbook Materials Journal Papers fib Bulletin 3

12/10/2010

17

97

Selection of Shape of the STM Model Determination of Member Forces in Indeterminate Models Design for Multiple Load Cases Uncertainty in Nodal Zones Dimensions Time Consuming Geometric Calculations Selecting What Needs to be Checked and Not Checked Designing for Good Performance Under Service Loads Validity of Design in Complex Models Performance under Overloads

Challenges to Design by the STM

98

Selection of Shape of the STM Model


99

Challenges to Design by the STM Selection of Shape of the STM Model

100



101



102



12/10/2010

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103

Selection of Shape of the STM Model Determination of Member Forces in Indeterminate Models


5800 kN

3229.5 kN

3659 kN

418.8 kN

4136 kN

A B C D E

FG H I J

K L M N

O P Q R

535

mm

570

mm

2895

mm

1015 mm

1590 mm

4000 mm 2030 mm

1015 mm

1350 mm

S

T

U

49 degrees

5625 kN

104



105



106



107

Designing for Good Performance Under Service Loads Validity of Design in Complex Models Performance under Overloads


108



12/10/2010

19

109



110



111



45% of Pn

112



68% of Pn

113

Selection of model shape Examples of good strut-and-tie model shapes for a large number of

common design situations Guidance for complex shapes Use of predictions of stress trajectories and topology optimization

Selecting relative member stiffness in indeterminate situations Design for multiple load cases and load reversals Determination of nodal zone geometries Determination of what to check and not to check Evaluation of performance under service loads; minimum

reinforcement recommendations Validation of ACI code-calculated capacity Other design requirements

Content of Potential ACI Committee 445 DocumentClick on the text below to go to the web page.

Seminar Schedule Bookstore Web Sessions Conventions

Online CEU Program ACI eLearning Concrete Knowledge Center

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