strut and time method

8/13/2019 Strut and Time Method

1/30

Development of the Strut-and-Tie Method for A endix A of

James K. Wight

the Building Code (ACI 318-08)

University of Michigan

Strut and Tie Modeling

be designed by idealizing the concrete

and reinforcement as an assembly of

axially loaded members, inter-

connected at nodes, to form a truss

capable of carrying loads across aregion or member.


2/30

Components of Strut and Tie Models

Steps to Build Strut & Tie Model

-

Compute forces or distribution of

stresses on boundary

Represent stress distributions as forces

forces across the member or D-region


3/30

Concept of D-Regions

(force discontinuities)

h

h 2h

Concept of D-Regions(geometric discontinuities)

h2h1

1 2


4/30

Basic Requirements

Model approximates stress flow

Define component dimensions and

strengths

Define and factors

Select reinforcement details

Modeling Stress Flow


5/30

Modeling Stress Flow in D-Regions(Dapped Beam)

How to Select the CorrectStrut-and-Tie Model

Some researchers suggest using a finite

element model to determine stress

trajectories, then selecting a STM to

model the stress flow.

,required amount or reinforcement is

close to an ideal model.


6/30

Required Definitions for Code

Geometric rules to follow when creating

a strut-and-tie model.

Component strengths for determining

members sizes and final geometry of

.

Other Codes with Rules for Useof Strut-and-Tie Models

AASHTO LRFD Specification

Canadian Code for Design of Concrete

Structures (CSA Standard, 2004)

ecommen a ons orPractical Design of Structural Concrete


7/30

Strength of Compression Struts

(what to consider)

Longitudinal cracking due to transverse

tension strain

Transverse tension forces

Sustained loads

Reinforcement grid crossing strut

Confinement by concrete or steel

Lateral Expansion of Strut

(bottle-shaped strut)


8/30

Effective Compressive Strength of Struts

ns uF F

0.75 =

ns cu cF f A=

min. cross-sectional area of strutcA =

s .cu c=

AASHTO and CSA Evaluation ofEffective Concrete Strength

Trans. tension strain,1= (s + 0.002) cot2s

1

0.850.8 170

cs c c

ff f

=

+

s = angle between strut and tie (steel)

s = strain in steel tie


9/30

FIP Recommendations fors in

Cracked Struts

0.80 - struts with longitudinal cracking

(splitting), but crossed by minimum

reinforcement grid

0.75 - struts crossed by normal width

0.60 - struts crossed by wide cracks

ACI Recommended s Values for Struts

1.0 prismatic shape (constant width) over its,

in a B-region

0.75 inclined (bottle-shaped) strut crossedby minimum reinforcement grid

0.60 inclined bottle-shaped strut notcrossed by minimum reinforcement grid;

where accounts for lightweight concrete 0.40 struts in flexural tension zones


10/30

Struts in a Flexural Tension Zone

A

A

Sect. A-A

Minimum Reinforcement Grid (fc

6000 psi)

Strut centerline2

12

1

sii

i

A in 0.003b s

s (A 4)


11/30

Reinforcement Grid

(only horizontal bars)

Min. Reinf. Grid in other Codes

AASHTO 0.003 EW

CSA 0.002 EW

FIP 0.001 EF, EW


12/30

Modification for Higher

Strength Concrete

(41 MPa), must calculate required

amount of transverse reinforcement

crossing strut.

This procedure can result in a

significant difference when compared to

requirements of ACI Eq. (A-4).

Modification for HigherStrength Concrete

=

2 longitudinal/1 transverse

equ re ransversetie capacity = 50%

of strut strength


13/30

Tie Dimensions

Full width (out or plane) of member

Width (in plane) of tie is function ofeffective compression strength ofconcrete in nodes where tie is anchored

dimensions

Tie - Dimensions and Strength

Asfy


14/30

Strength of Ties

Strength = As fy, where = 0.75*

Anchorage of ties at nodes is a major

concern

*A constant value of = 0.75 is to be used for

sizing the strut-and-tie model, but the use of

= 0.75 to also select the reinforcement mayneed further examination within the appropriate

Code subcommittee.

Anchorage Check

Definition of Extended Nodal Zone


15/30

Nodal Zone Shape and Dimensions

Width of compression face is same as

width of strut connecting to nodal zone

(smaller allowable strength governs)

Height (width of face perpendicular to

tie force) of nodal zone is equal to tie

force divided by effective compressivestrength of the concrete in the node

Node Shape and Size


16/30

Effective Strength of Nodal Zones

Function of type of members connected

to the node

Possible combinations are CCC, CCT

and CTT

Strength can be enhanced by addition

of confinement reinforcement

Examples of CCC and CCT Nodes


17/30

Examples of CCT and CTT Nodes

Examples of CCT and CTT Nodes


18/30

Strength of Nodes

nn uF F

0 75 0.75 =

nn cu nF f A=

n uA area of node face to force F=

n0.85

cu cf f =

Recommended n Values (Nodes)

1.0 - CCC node

. -

0.6 - CTT node


19/30

Use of STM in the ACI Code

- .However,

Is listed as alternate procedure inseveral sections of the code (e.g.corbels, short shear walls).

Currently required for shear strength

design of deep beams.

640 k (includes member weight)

All member widths = 20 in.

Example: ACI Concrete International

Magazine, May 2003

dv60 in.

.

53 in. 107 in. 212 k428 k

16 in. 16 in.


20/30

640 k

All member widths = 20 in.

Beam Dimensions and Initial Truss Model

dv60 in.

.

53 in. 107 in. 212 k428 k

16 in. 16 in.

Check Max. Allowable Shear Force

=.

Max. All. Shear Force = 10

4000(max) 0.75 10 20 54 512 k (o.k.)

1000

c w

n

f b d

V

= =

Assumed thatd0.9h


21/30

212 k428 k

Splitting of Node 2

dv

=

?

1

2

3

4 51 = 42.2

o2 = ? 2

49.7 in. 50.2 in. 50.2 in.

10 in.428 k 212 k

Establish Truss Geometry: Start withleft portion of the beam

Assumes heights of Nodes 1 & 2 = 15 in.

Thus, dv= 60 in. 2(7.5 in.)

= 45 in. (1140 mm)

From eometr :

tan 1 = (45in./49.7in.),

And thus 1 = 42.2 deg.


22/30

212 k428 k

Initial Truss Geometry

dv

=

45

in.

1

2

3

4 51 = 42.2

o2 = ? 2

49.7 in. 50.2 in. 50.2 in.

10 in.428 k 212 k

Establish Equilibrium at Node 1

12

F14Fy = 428 - F12sin 1 = 0

F12 = 637 k

428 kFx = F14 F12cos 1 = 0

F14 = 472 k


23/30

Geometry and dimensions of Node 1

and Strut 1-2

14

b1

1

Establish/Check Dimensions at Node 1

fcu(1) = (0.85) n fc'= 0.750.850.804

1

428( ) 1.34 ksi (o.k.)

20 16w b

Rf base

b= = =

l

l

= 2.04 ksi

1414

472 11.6 12 in.(1) 20 0.75 2.72w cu

Fwb f

= = =


24/30

Establish/Check Width and Strength of

Strut 1-2

( ) ( )12 14 1 1 1

12

cos sin

8.89 10.8 19.7 in.

bw w

w

= +

= + =

l

(1-2) 0.85 0.85 0.75 4 2.55 ksicu s cf f = = =

12(1-2) (1-2)(1-2) 0.75 2.55 19.6 20 757 k 637 k (o.k.)

ns cu w

ns

F f w bF

= = =

Modified Truss Geometryand Member Forces

Thus, height of Node 1 = 12 in.

Because Node 2 is a CCC node, assume

it has a total height of 10 in.

, v= = .

Reestablish truss geometry and member

forces!


25/30

212 k428 k

Final truss geometry and member forces

dv

=

49

in.

434 k 217 k

217 k

1

2

3

4 51 = 44.6

o 2 = 44.3o

2

212

k

49.7 in. 50.2 in. 50.2 in.

10 in.428 k 212 k

Select Reinforcement for Tie 1-4

214434

( .) 9.64 in.sF

A req d = = =.

y

Select 13 No. 8 bars,

As = 10.3 in.2

3 in.

20 in.

3 in.

3 in.


26/30

Check Anchorage at Node 1

Tie 1-4

a = 6 in./tan1a = 6.09 in.

b1 = 16in. a

Critical section

1 6in.

Length 22.1 in.

Check Anchorage at Node 1

0.02 0.02 1 60, 0001.0 in.

e yf

d

= = l

b

1 4000

19.0 in. (> 8d and > 6 in.)

c

dh

f

=l

Although dh is less than 22 in., this would be a tight fit if

only 90 hooks were used. In-plane 180 hooks could be

used for some bars to partially relieve this rebar detailingproblem. The use of mechanical anchorage devices could

also be considered.


27/30

Minimum Reinforcement Crossing Strut 1-2

Strut centerline2

12

1

sii

i

A in 0.003b s

s

Minimum Reinforcement CrossingStrut 1-2 for s = 0.75

and width of cracks crossing the strut.

Because this is a deep beam, I recommend

that you also satisfy ACI Code Sections

11.7.4 and 11.7.5.

, ,the minimum skin reinforcement requirements

of ACI Code Section 10.6.7 must be satisfied.


28/30

2 #4

per layer

3 in.Final design

in left span

Four #4 legs6 at 8 in.

3 at 3 in.

20 in.

13 #8

212 k428 k

Comments on right half of truss

dv

=

49

in.

434 k 217 k

217 k

1

2

3

4 51 = 44.6

o 2 = 44.3o

2

212

k

49.7 in. 50.2 in. 50.2 in.

10 in.428 k212 k


29/30

22in.60in.

Analysis of fan-shaped struts 2-4

and 3-5, and tie 3-4

2

49in.

25o

25o

3

5

104 in.

22in. six stirrups at s = 10 in.

4

Section A

Final design of longitudinal andtransverse steel

5 at 10 in. 15 at 6 in.

2 in. 2 in.


30/30

Gracias

Preguntas?

strut and time method

Documents