composite strut and tie model for reinforced concrete deep

Journal of Advanced Concrete Technology Vol. 7, No. 1, 97-109, February 2009 / Copyright © 2009 Japan Concrete Institute 97

Scientific paper

Composite Strut and Tie Model for Reinforced Concrete Deep Beams Kil-Hee Kim1, Woo-Bum Kim2, Jin-Man Kim2 and Sang-Woo Kim3

Received 2 October 2008, accepted 30 January 2009

Abstract Monotonic shear loading tests were conducted on three half-scaled reinforced concrete deep beams with shear span-to-depth ratios of 0.5 to 0.75. The obtained test results were investigated in detail based on the experimental measurements and finite element analysis. From these investigations, a new macro model for deep beams was established. This model is composed of two crooked main struts formed between both beam end sections and branched-off sub struts. The com-pressive force introduced to main struts balances the flexural compression and the external shear force. The bond stress of the longitudinal reinforcement and the tensile force of the stirrup produce the diagonal compression in the sub strut. Theoretically predicted shear strengths of tested deep beams showed good agreement with experimentally observed shear strengths, where the effective strength of concrete was assumed to be 75% of the cylinder strength.

1. Introduction

Deep beams are widely used as perimeter beams of rein-forced concrete (RC) frames for functional and architec-tural reasons. In multi-story wall structures, deep beams are also used for coupling walls constructed side by side. Because the beams have small span-to-depth ratios, brit-tle failure in shear may be observed under seismic lat-eral loads for some of them. Examples of such failure due to earthquakes have been reported in the literature (Kim et al. 2001; Paulay 1981; Paulay and Binney 1974). To avoid such failure, a sufficient amount of transverse reinforcement should be provided to induce ductile flexural failure prior to shear failure. An alterna-tive method is the arrangement of diagonal reinforce-ment in a beam, which was proposed by Park and Pau-lay (1975).

When a deep beam is subjected to combined bending and shear with anti-symmetric moment distribution, the truss mechanism is hardly developed due to the overlap of tension shift regions of longitudinal reinforcement. That is, the main resisting mechanism is developed by a single strut formed between both end sections of the member. Thus the final shear failure may occur due to crushing of diagonally compressed concrete struts. Based on this basic mechanism, several shear design equations for deep beams have been proposed (ACI 2008; AIJ 1990, 1999; CEB-FIP 1995; Nielsen 1984). However the role of the transverse reinforcement is not necessarily clear.

This paper presents a new macro model to predict the shear strength of deep beams with anti-symmetric mo-ment distribution. Shear loading tests and a nonlinear finite element (FE) analysis for deep beams were con-ducted to derive the mechanical shear resisting model. The proposed model is composed of two crooked main struts and branched off sub struts, where the contribu-tion of the transverse reinforcement and the bond action of the longitudinal reinforcement are taken into account.

2. Experimental program

Three specimens with various shear span-to-depth ratios and amounts of transverse reinforcement, as indicated in Table 1, were cast and tested to verify the accuracy of the proposed model. The details of typical specimen S-0.75-50 are shown in Fig. 1. Each specimen has a cen-tral test region with a 200×300 mm sectional dimension and loading stubs at both ends of the specimen. The clear spans of S-0.5-50, S-0.75-50 and S-0.75-75 were 300, 450 and 450 mm, respectively.

The specimens were designed to fail in shear before yielding of the flexural reinforcement. All the specimens contained four deformed bars with a nominal diameter of 16 mm and a yield strength of 795 MPa as top and bottom reinforcements. Deformed and welded closed stirrups were used in this experiment. The transverse steel bars had the yield strength of 305 MPa and the nominal diameter of 6 mm. The spacing of stirrups of S-0.5-50, S-0.75-50 and S-0.75-75 was 50, 50 and 75 mm, respectively. Normal weight concrete was used in the construction of the test specimens. The average com-pressive strength of concrete at the time of the beam test was 30.6 MPa.

Figure 2 illustrates the loading equipment and the test setup of the specimens. This loading method was developed by Jirsa et al. (1978), where the rotation of the top stub of the specimen is restricted by two vertical jacks and results in the anti-symmetric moment distribu-tion of the central beam test region. In this system, the

1Assistant Professor, Department of Architectural Engineering, Kongju National University, South Korea.E-mail:[email protected] 2Professor, Department of Architectural Engineering,Kongju National University, South Korea. 3Research Assistant Professor, Department of Architectural Engineering, Kongju National University, South Korea.

98 K.-H. Kim, W.-B. Kim, J.-M. Kim and S.-W. Kim / Journal of Advanced Concrete Technology Vol. 7, No. 1, 97-109, 2009

vertical movement of the top stub is free, that is, no ad-ditional vertical load is applied to the specimen sub-jected to lateral shear force by two parallel horizontal jacks. During the test, the rotation angle (drift) of the specimen was measured by two electronic displacement transducers and recorded by a computer controlled data

acquisition system. Wire-type strain gauges were at-tached on the surface of the steel bars to measure the strain of longitudinal and transverse reinforcements. Readings from the strain gauges were used to estimate the tensile and bond stresses of the reinforcing bars.

Table 1 Details of test specimens.

Shear reinforcement Specimens Concrete

(f’c, MPa) Pitch (mm)

σwy (MPa)

ρwσwy

(MPa)

Longitudinal reinforcement

S-0.5-50 S-0.75-50

50 1.95

S-0.75-75 30.6

75 305

1.30

4-D16 fy = 795 MPa

Note: S-0.5-50 means S (specimen), 0.5 (shear span-to-depth ratio), and 50 (spacing of shear reinforcement, mm)

30 140 30

200

40

300

50 50

225

450 450 450

Wire Strain Gauge

30 140 30

200

40

300

50 50

225

450 450 450

Wire Strain Gauge

Fig. 1 Details of specimen (S-0.75-50).

Loading Frame

PIN

Load Cell JACK C

Parallelogramic Apparatus

Specimen

JACK A JACK B

Bearing Floor

Jack A Jack B

Specimen

Jack C

Pin

Fig. 2 Test setup of specimen.

K.-H. Kim, W.-B. Kim, J.-M. Kim and S.-W. Kim / Journal of Advanced Concrete Technology Vol. 7, No. 1, 97-109, 2009 99

3. Experimental results

3.1 Load-displacement response and crack pat-tern All test specimens failed in shear due to crushing of diagonally compressed concrete. Table 2 indicates the experimentally observed maximum strength of each specimen. From the comparison between S-0.75-50 and S-0.75-75, the contribution of the shear reinforcement seems to be small.

The shear strengths predicted by ACI 318-08 Code (ACI 2008) and AIJ (Architectural Institute of Japan) Code (AIJ 1990, 1999) are also indicated in Table 2. It can be seen from the table that all the code equations give very conservative shear strengths for test deep beams. More reliable design equations for the shear strength of deep beams with anti-symmetric moment distribution should be established. With respect to the contribution of the shear reinforcement, Fig. 3 shows the experimentally observed shear force versus drift angle responses of test specimens. While all specimens showed almost the same responses up to peak loads, the ductility of the specimen increased as the amount of transverse reinforcement grew higher in the post-peak behavior.

Crack propagation of each specimen for three loading stages is indicated in Fig. 4. After initial development of a diagonal crack, the crack width increased with in-creases in shear force up to the maximum load. Finally, each specimen reached the maximum capacity showing the crushing of diagonally compressed concrete near the beam end sections.

3.2 Strain of reinforcement Figure 5 shows the variation of the strain distribution of reinforcing bars. The x-axis of the figure represents the location of strain gauges along a beam length. In the figure, the thick solid line corresponds to the strain at the maximum capacity. After diagonal cracking, stirrup strain increased rapidly and most stirrups yielded at the maximum capacity. With respect to the strain of the longitudinal reinforcement, tension shift was observed after diagonal cracking, and then the strain was main-tained in tension. This means that the beam action dis-appeared over the entire beam length. It can be also seen from Fig. 5 that the longitudinal reinforcement did not

yield in any of the loading stages as intended in the de-sign of specimens.

The average bond stress along a beam length was computed from the experimentally obtained strain of the longitudinal reinforcing bars. The computed average bond stresses are indicated in Fig. 6 with the drift angle. The bond stresses were very small and ranged from 1.5 to 2.0 MPa though no splitting bond failure of the longi-tudinal reinforcement was observed. This may be due to the specific mechanical behavior of deep beams.

4. Analytical investigation

To investigate the shear resisting mechanism of deep beams, a nonlinear FE analysis was carried out for specimens S-0.5-50 and S-0.75-50, with shear span-to-depth ratios of 0.5 and 0.75, respectively. The FE analy-sis was performed using a two-dimensional FE analysis program, VecTor2, developed by Vecchio and his re-search group at the University of Toronto in Canada. The analytical model used in the VecTor2 was the Modified Compression Field Theory (MCFT) (Vecchio and Collins 1986) and the constitutive models that were used (Kent and Park 1971; Kupfer et al. 1969; Popovics 1970) are listed in Table 3. The details of each constitu-tive model and their implementation into the VecTor2

Table 2 Experimental results of test specimens.

Specimens Concrete (f’c, MPa)

ρwσwy

(MPa) VAIJ (kN)

VACI (kN)

Vexp (kN)

Failure mode

S-0.5-50 190 149 297 S-0.75-50

1.95 181 149 290

S-0.75-75 30.6

1.30 174 149 274

Shear failure(crushing)

VAIJ, VACI : shear force calculated by AIJ and ACI Code, respectively; Vexp : Experimentally obtained shear force.

0 1 2 3 4 5 60

50

100

150

200

250

300

350

Shea

r for

ce (k

N)

Drift angle (%)

S- 0.5 -50 S-0.75-50 S-0.75-75

Fig. 3 Shear force versus drift angle relationships of test specimens.


Diagonal crack

Peak

After peak

Diagonal crackDiagonal crack

Peak

After peak

Peak

After peak

Diagonal crack

Peak

After peak

Diagonal crackDiagonal crack

Peak

After peak

Peak

After peak

(a) S-0.5-50 (b) S-0.75-50 (c) S-0.75-75

Fig. 4 Crack propagation.

0 50 100 150 200 250 3000

500

1000

1500

2000

2500

3000

Yielding Strain

Diagonal Crack

PEAK

Hoo

p St

rain

(μ)

0.0 112.5 225.0 337.5 450.00

500

1000

1500

2000

2500

3000

Diagonal Crack

Yield Strain

PEAK

Hoo

p St

rain

(μ)

0.0 112.5 225.0 337.5 450.00

500

1000

1500

2000

2500

3000

Diagonal Crack

PEAK

Yield Strain

Hoo

p St

rain

(μ)

Stirr

up st

rain

(μ)

Stirr

up st

rain

(μ)

Stirr

up st

rain

(μ)

Clear length (mm) Clear length (mm) Clear length (mm)

0 50 100 150 200 250 300-500

0

500

1000

1500

2000

2500

3000

Diagonal Crack

PEAK

Reba

r stra

in (μ

)

0.0 112.5 225.0 337.5 450.0-500

0

500

1000

1500

2000

2500

3000

Diagonal Crack

PEAK

Reb

ar S

train

(μ)

0.0 112.5 225.0 337.5 450.0-500

0

500

1000

1500

2000

2500

3000

Diagonal Crack

PEAK

Reba

r Stra

in (

μ)

Long

itudi

nal r

einf

orce

men

t stra

in (μ

)


Long

itudi

nal r

einf

orce

men

t stra

in (μ

)

Long

itudi

nal r

einf

orce

men

t stra

in (μ

)

0 50 100 150 200 250 3000

500

1000

1500

2000

2500

3000

Yielding Strain

Diagonal Crack

PEAK

Hoo

p St

rain

(μ)

0.0 112.5 225.0 337.5 450.00

500

1000

1500

2000

2500

3000

Diagonal Crack

Yield Strain

PEAK

Hoo

p St

rain

(μ)

0.0 112.5 225.0 337.5 450.00

500

1000

1500

2000

2500

3000

Diagonal Crack

PEAK

Yield Strain

Hoo

p St

rain

(μ)

Stirr

up st

rain

(μ)

Stirr

up st

rain

(μ)

Stirr

up st

rain

(μ)


0 50 100 150 200 250 300-500

0

500

1000

1500

2000

2500

3000

Diagonal Crack

PEAK

Reba

r stra

in (μ

)

0.0 112.5 225.0 337.5 450.0-500

0

500

1000

1500

2000

2500

3000

Diagonal Crack

PEAK

Reb

ar S

train

(μ)

0.0 112.5 225.0 337.5 450.0-500

0

500

1000

1500

2000

2500

3000

Diagonal Crack

PEAK

Reba

r Stra

in (

μ)

Long

itudi

nal r

einf

orce

men

t stra

in (μ

)


Long

itudi

nal r

einf

orce

men

t stra

in (μ

)

Long

itudi

nal r

einf

orce

men

t stra

in (μ

)

(a) S-0.5-50 (b) S-0.75-50 (c) S-0.75-75

Fig. 5 Strain of reinforcements along length of specimen.


program have been reported elsewhere (Wong and Vec-chio 2002). The longitudinal and transverse reinforce-ments were modeled as truss elements and connected to nodal points of concrete elements by the contact ele-ments.

A comparison of the analytical and experimental re-sults for the shear force versus displacement response of test specimens is shown in Fig. 7. In S-0.5-50, the maxi-mum load was reached at step 5 in the computational process, and in S-0.75-50, at step 12. These predicted maximum shear strengths were almost the same as the experimentally observed ones. After the peak load, however, the predicted curve showed a somewhat pre-mature reduction of load carrying capacity compared to the experimental one.

Figure 8 illustrates the crack patterns of S-0.5-50 and S-0.75-50 at loading steps 8 and 19, respectively. It can be seen from this figure that the theoretically predicted crack propagation agreed well with experimental obser-vation (see Fig. 4). Based on the comparison results, the applied FE analysis method in this paper can be used for

predicting the internal stress flow of specimens. Figure 9 shows the magnitude and direction of prin-

cipal tensile and compressive stresses at three loading stages, i.e. pre-peak, peak and post-peak loads. From the figure for the pre-peak stage, it can be seen that one main compressive strut developed on a beam diagonal. At peak and post-peak loads, however, this diagonal strut seems to separate into two main struts as the width of the diagonal crack increases. These two main struts are located at both sides of a beam diagonal. During this process, the truss action produced by the stirrup tension and the diagonal compression of the concrete existed outside of the main strut. However, this truss action was small and most of the shear force could be resisted by the diagonal main strut.

0.0 0.5 1.0 1.5 2.0 2.5 3.00.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

Bon

d st

ress

(M

Pa)

Drift angle (%)

S- 0.5 -50 S-0.75-50 S-0.75-75

Fig. 6 Average bond stress versus drift angle responses of specimens.

0

50

100

150

200

250

300

350

0 5 10 15 20Displacement (mm)

Forc

e (k

N)

AnalysisTest

Step 4 Step 5

Step 8

0

50

100

150

200

250

300

350

0 5 10 15Displacement (mm)

Forc

e (k

N)

Analysis

TestStep 8 Step 12

Step 19

0

50

100

150

200

250

300

350


Forc

e (k

N)

AnalysisTest

Step 4 Step 5

Step 8

0

50

100

150

200

250

300

350


Forc

e (k

N)

AnalysisTest

Step 4 Step 5

Step 8

0

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Forc

e (k

N)

Analysis

TestStep 8 Step 12

Step 19

0

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350


Forc

e (k

N)

Analysis

TestStep 8 Step 12

Step 19

(a) S-0.5-50 (b) S-0.75-50 Fig. 7 Comparison of analytical and experimental results for shear force versus displacement relationship of specimens.

Table 3 Constitutive models used in FE analysis.

Analytical parameters Models Compression Pre-peak curve Popovics

Compression post-peak curve Modified Park-Kent

Compression softening

Vecchio 1992-A

Tension stiffening Modified Bentz Tension softening Linear/no residual Tension splitting Not considered Confined strength Kupfer/Richart Dilation Variable-Kupfer Cracking criterion Mohr-coulomb (Stress)Crack width check Crack limit (Agg/2)

Concrete

Concrete hysteresis Linear w/Plastic offsetsSteel hysteresis Sekin model Steel bar Dowel action Tassios et al.

Bond Concrete bond Fujii model


The above-mentioned phenomena can be confirmed by the shear stress and principal compressive stress dis-tributions along a section height, as shown in Fig. 10. At the critical section of the beam, Section A-A’, the location of the peak stress shifted from the section edge to the center of the section as the progress of loading steps. This peak shift was observed in both beams and may be due to the local crushing of the concrete at the extreme compression fiber. A similar phenomenon was observed at Section B-B’ of the S-0.5-50 beam. The other specific behavior was the separation of the diago-nal main strut at and after peak load. At Section C-C’ of S-0.5-50, and at Sections B-B’ and C-C’ of S-0.75-50, the stress distributions had two peaks at and after peak load. This suggests that the main diagonal strut is going to separate into two struts due to the change of internal mechanism. Final failure of the beam occurred due to crushing of the diagonally compressed concrete, and average principal compressive stress at peak load was from 0.7 to 0.8 times the compressive strength of con-crete.

5. Curved dual strut model (CDS-M)

When a deep beam with a shear span-to-depth ratio of one or less is subjected to bending moment and shear, uniform compression stress field (in other word, truss mechanism) is hardly developed due to the tension shift of the flexural reinforcement. This means that the main resisting mechanism for shear should be a diagonal strut developed between both beam ends. This has been be-ing pointed out experimentally and theoretically in past research works. Therefore design equations for deep beams have been developed based on the strut and tie model. However, the bond action of longitudinal rein-forcement and the force of the transverse reinforcing bar cannot be properly evaluated using the strut and tie model, because the model is able to control the shear strength of deep beams only by the compressive strength of diagonal strut.

In this chapter, a new shear resisting model incorpo-rating dual curved struts is proposed, where the bond action of longitudinal reinforcement and the force of the stirrup are taken into account. Based on the numerical

deformed undeformed

Wcr＜2㎜

Wcr ≥ 2㎜

deformed undeformed

Wcr＜2㎜

Wcr ≥ 2㎜

(a) S-0.5-50 (Step 8) (b) S-0.75-50 (Step 19)

Fig. 8 deformation and crack pattern.

Step 4 Step 5 Step 8

SteelTensionstressesCompressionstresses



Step 4Step 4 Step 5Step 5 Step 8Step 8







(a) S-0.5-50

Step 19Step 8 Step 12




Step 19Step 8 Step 12 Step 19Step 19Step 8Step 8 Step 12Step 12







(b) S-0.75-50

Fig. 9 Principal concrete stresses.

Wcr: crack width


simulations and experimental observations described in the previous chapter, a new macro model to predict the shear strength of deep beams was developed.

5.1 Modeling and basic assumptions When a deep beam is subjected to a reverse symmetric moment under constant shear (e.g. a coupling beam), the shear resisting mechanism is generally expressed by a diagonal direct strut and supplemental fan shaped mechanism (AIJ 1999). However, when the theory of plasticity is applied for the modeling, the stress condi-tion of the concrete near the pivot of a fan is not clear

and hardly modeled because of the changing width of each fan bone.

In this study, the shear resisting mechanism is mod-eled using the lower bound of the theory of plasticity. The feature of this modeling is that two curved diagonal main struts are introduced as the main shear resisting mechanism in conjunction with supplemental sub struts. This curved dual strut model was devised from the ex-perimental observations and the theoretical analysis using the FE method.

The new model is indicated in Fig. 11 for two beams with different shear span-to-depth ratios. All of the main

(a) Shear stresses in concrete (S-0.5-50)

(b) Principal stresses (S-0.5-50)

(c) Shear stresses in concrete (S-0.75-50)

(d) Principal stresses (S-0.75-50)

Fig. 10 Distribution of shear and principal stresses.

A

A’

B

B’

C

C’

A

A’

B

B’

C

C’

0

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0 3 6 9 12 15 18

Equivalent Shear Stress (MPa)

Loca

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on v

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irect

ion

(mm

) Step 4 Step 5 Step 8

Section A-A’0

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ertic

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irect

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(mm

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and sub struts are subjected to an axial compressive stress, cσ . Two main struts are changing their angle along a beam length due to the external compression given by the sub struts. At an intersection point between the main and sub struts, a triangular hydro pressure stress field is assumed. The axial compressive forces of sub struts balance the horizontal bond forces of the lon-gitudinal reinforcement and the vertical tension forces of the stirrup at the intersection points of the sub strut and the longitudinal reinforcement. The basic assump-tions used in the proposed model are as follow: (a) At the maximum shear resistance, all of the shear

reinforcement is to be yielded. This comes from the experimentally observed strain distributions of stir-rups indicated in Fig. 5.

(b) Based on the experimental results described in Sec-tion 3.2, the distribution of the bond stress of longi-tudinal reinforcement is assumed to have a triangu-lar shape as recognized from Fig. 11.

(c) The resultant vector of the force of the sub-strut located in the C region in tension indicates point T. Equation (1) shows the vector equation, and the an-gles of the other two sub-struts can be computed from Eq. (2) based on the assumptions of (a) and (b) above.

32tan da

=θ (1)

2 3tan (1/ 3) tan=θ θ , 1 3tan (1/ 5) tan=θ θ (2)

where d is the distance between the centers of the strut and the bottom re-bar; a is a third of the clear span of a beam; and θ is the angle between the sub-strut and the longitudinal axis.

(d) As shown in Fig. 11, the rim of crooked main strut A in the A region is bent by the influence of the sub-struts, and thus it is assumed that the Crooked Main Strut starts from the center of the main re-bar.

(e) The width of crooked main strut A is set below one tenth of the effective depth of the member. This imitation is based on the result of the FE analysis for the reason that the region traversing the diago-nal shear crack becomes quite large if the width of strut A is too large, as can be seen in Fig. 11.

5.2 Computation of basic values for analysis First of all, the angles of the sub-struts are determined based on the experimental and analytical results. Figure 11 shows that the shear resistance mechanism is gov-erned by a geometric condition. Based on the assump-tion that the stress of each strut is the same, the shear force can be obtained by computing the width of the strut that satisfies the geometric constraint.

Figure 11 can be drawn in more detail as Fig. 12(a). The force per unit length of strut A and its coordinate can be expressed in Eqs. (3) and (4), respectively.

1STA cC b x= ⋅ ⋅σ (3)

(tan )y x d= − +β (4)

where cσ is the stress of concrete on the strut; b is the width of the cross section; 1x is the width of strut A; β is the angle between the strut A and the longitudinal axis; x and y are the x- and y-axes in the x-y coordinate system, respectively.

Similarly, the force of strut E and its coordinate can be obtained as

1STE cC b s= ⋅ ⋅σ (5)

1 1(tan ) ( / 2) tany x a= − −θ θ (a = L/3) (6)

where 1s is the width of strut E; θ1 is an angle between the strut E and the longitudinal axis; and L is the clear span of the member.

The coordinates for the nodal point II of struts A and E can be given by the following Eqs. (7) and (8) based on Eqs. (4) and (6).

1

1

( / 2) tantan tan

d ax

⎡ ⎤+= =⎢ ⎥−⎣ ⎦

θα

β θ (7)

1 1(tan ) ( / 2) tany a= − ⋅ −θ α θ (8)

The B region of Fig. 11 is illustrated in detail in Fig. 12(b). The force per unit length and the coordinate of strut B can be given by Eqs. (9) and (10), respectively.

2STB cC b x= ⋅ ⋅σ (9)

1 1 1 1(tan ) (tan tan ) ( / 2) tany x a= − + − −φ α φ θ θ (10)

1 1 11

1 1 1

sin sintan

cos cosx sx s

⎛ ⎞⋅ + ⋅= ⎜ ⎟⋅ + ⋅⎝ ⎠

β θφ

β θ (11)

( ) ( )1

2 2 22 1 1 1 1 1 1sin sin cos cosx x s x s⎡ ⎤= + + +⎣ ⎦β θ β θ (12)

where 2x is the width of strut B and φ1 is the angle between the strut B and the longitudinal axis.

Sub-strut F can be expressed as the following equa-tions.

2STF cC b s= ⋅ ⋅σ (13)

2 2(tan ) (2 / 3) tany x a= − ⋅ −θ θ (14)

where s2 is the width of strut F and θ2 is the angle be-tween the strut F and the longitudinal axis.

The coordinates for the nodal point III of struts B and F can be given by the following Eqs. (15) and (16) based on Eqs. (10) and (14).

1 2 1 1

2 1

( / 2) tan (3 / 2) tan (tan tan )(tan tan )

a ax

− − −= =

−

⎡ ⎤⎢ ⎥⎣ ⎦

θ θ α φ θγ

θ φ

(15)


Bond Force τb

Shear Reinforcement Force ρwσwy

a aa/2

3a

Strut A (main)

Strut B (main)

Strut C (main)Strut D (main)

Strut E (sub) Strut F (sub) Strut G (sub)

Strut A (main)

Strut B (main)


Strut E (sub)

Strut F (sub)Strut G (sub)

Ⅰ

Ⅱ

Ⅲ

Ⅳ

Ⅴ

Ⅰ

Ⅱ

Ⅲ

Ⅳ

Ⅴ

σc

σc

σc

σc

VM V M

A regionB regionC region D regionD region

Strut H

Strut H

θ3 θ2

Bond Force τb

Shear Reinforcement Force ρwσwy

a aa/2

3a

Strut A (main)

Strut B (main)


Strut E (sub) Strut F (sub) Strut G (sub)

Strut A (main)

Strut B (main)


Strut E (sub)

Strut F (sub)Strut G (sub)

Ⅰ

Ⅱ

Ⅲ

Ⅳ

Ⅴ

Ⅰ

Ⅱ

Ⅲ

Ⅳ

Ⅴ

σc

σc

σc

σc

VM V M

A regionB regionC region D regionD region

Strut H

Strut H

θ3 θ2

(a) S-0.5-50

M

VM V

σc

σc

σc

σc

Crooked Main Strut

Crooked Main Strut

Sub-Strut

Sub-Strut

a aa

θ1

θ2

ab ⋅⋅ΣΨ 1τ

abwyw ⋅⋅σρbsc ⋅⋅ 1σ

a

θ3

M

VM V

σc

σc

σc

σc

Crooked Main Strut

Crooked Main Strut

Sub-Strut

Sub-Strut

a aa

θ1

θ2

ab ⋅⋅ΣΨ 1τ

abwyw ⋅⋅σρbsc ⋅⋅ 1σ

a

θ3

(b) S-0.75-50

Fig. 11 Crooked dual strut model.


2 2(tan ) (3 / 2) tany a= − ⋅ −θ γ θ (16)

Figure 12(c) depicts the C region of Fig. 11 in detail. The force per unit length and the coordinate of strut C can be given by Eqs. (17) and (18), respectively.

3STC cC b x= ⋅ ⋅σ (17)

2 2 2 2(tan ) (tan tan ) (3 / 2) tany x a= − + − −φ γ φ θ θ (18)

1 1 1 2 22

1 1 1 2 2

sin sin sintan

cos cos cosx s sx s s

⎛ ⎞⋅ + ⋅ + ⋅= ⎜ ⎟⋅ + ⋅ + ⋅⎝ ⎠

β θ θφ

β θ θ (19)

( )( )

12 2

1 1 1 2 223

1 1 1 2 2

sin sin sincos cos cos

x s sxx s s

⎡ ⎤+ += ⎢ ⎥+ + +⎢ ⎥⎣ ⎦

β θ θβ θ θ

(20)

where 3x is the width of strut C, and φ2 is the angle between the strut C and the longitudinal axis.

Sub-strut G can be expressed as the following equa-tions.

3STG cC b s= ⋅ ⋅σ (21)

3 3(tan ) (5 / 2) tany x a= − ⋅ −θ θ (22)

where 3s is the width of strut G and θ3 is the angle between the strut G and the longitudinal axis.

x1

d

s1

βθ1

φ1

x2

ρwσwy

τb1

(Στb1ψ)a

ρwσwybaShear Reinforcement Force

Bond Force

Strut E (sub)

Strut A (main)

Strut B (main)

a

point Ⅰ

point Ⅱ

σc

a

σc

τb2

ρwσwy

ρwσwyba

(Στb2ψ)a

Strut C (main)

Strut B (main)

Strut F (sub)

Bond Force

Shear Reinforcement Force

x2

x3

s2

θ2

φ1φ2

point Ⅲ

point Ⅱ

σc

x1

d

s1

βθ1

φ1

x2

ρwσwy

τb1

(Στb1ψ)a


Bond Force

Strut E (sub)

Strut A (main)

Strut B (main)

a

point Ⅰ

point Ⅱ

σc

x1

d

s1

βθ1

φ1

x2

ρwσwy

τb1

(Στb1ψ)a


Bond Force

Strut E (sub)

Strut A (main)

Strut B (main)

a

point Ⅰ

point Ⅱ

σc

a

σc

τb2

ρwσwy

ρwσwyba

(Στb2ψ)a

Strut C (main)

Strut B (main)

Strut F (sub)

Bond Force


x2

x3

s2

θ2

φ1φ2

point Ⅲ

point Ⅱ

σc

a

σc

τb2

ρwσwy

ρwσwyba

(Στb2ψ)a

Strut C (main)

Strut B (main)

Strut F (sub)

Bond Force


x2

x3

s2

θ2

φ1φ2

point Ⅲ

point Ⅱ

σc

(a) A region (b) B region

x4

x3

s2

φ2φ3

θ3

σc

ρwσwy


τb3(Στb3ψ)aBond Force

point Ⅲ

point Ⅳ

Strut C (main)

Strut G (sub)

Strut D (main)a

x4・cos φ3

φ3

β

x4

x1

x4・sin φ3

x4/(2cosφ3)

x1･sinβ･tanφ3

x1・sin β

point Ⅰ

Strut A (main)

Strut D (main)point Ⅴ

point Ⅳ

x1・cos β

σc

σc

c

Strut H

x4

x3

s2

φ2φ3

θ3

σc

ρwσwy



point Ⅲ

point Ⅳ

Strut C (main)

Strut G (sub)

Strut D (main)a

x4

x3

s2

φ2φ3

θ3

σc

ρwσwy



point Ⅲ

point Ⅳ

Strut C (main)

Strut G (sub)

Strut D (main)a

x4・cos φ3

φ3

β

x4

x1

x4・sin φ3

x4/(2cosφ3)

x1･sinβ･tanφ3

x1・sin β

point Ⅰ

Strut A (main)


point Ⅳ

x1・cos β

σc

σc

c

Strut H

x4・cos φ3

φ3

β

x4

x1

x4・sin φ3

x4/(2cosφ3)

x1･sinβ･tanφ3

x1・sin β

point Ⅰ

Strut A (main)


point Ⅳ

x1・cos β

σc

σc

c

x4・cos φ3

φ3

β

x4

x1

x4・sin φ3

x4/(2cosφ3)

x1･sinβ･tanφ3

x1・sin β

point Ⅰ

Strut A (main)


point Ⅳ

x1・cos β

σc

σc

c

Strut H

(c) C region (d) D region

Fig. 12 Details of crooked dual strut model.


The coordinates for the nodal point IV of struts C and G can be given by the following Eqs. (23) and (24) based on Eqs. (18) and (22).

2 2 3 2

2 3

(tan tan ) (5 / 2) tan (3 / 2) tan(tan tan )

a ax

− + −= =

−

⎡ ⎤⎢ ⎥⎣ ⎦

γ φ θ θ θλ

φ θ (23)

3 3(tan ) (5 / 2) tany a= − ⋅ −θ λ θ (24)

The force per unit length and the coordinate of strut D can be given by Eqs. (25) and (26), respectively.

4STD cC b x= ⋅ ⋅σ (25)

3 3 3 3(tan ) (tan tan ) (5 / 2) tany x a= − + − −φ λ φ θ θ (26)

1 1 1 2 2 3 33

1 1 1 2 2 3 3

sin sin sin sintan

cos cos cos cosx s s s

x s s s⎛ ⎞⋅ + ⋅ + ⋅ + ⋅

= ⎜ ⎟⋅ + ⋅ + ⋅ + ⋅⎝ ⎠

β θ θ θφ

β θ θ θ (27)

( )( )

12 2

1 1 1 2 2 3 324

1 1 1 2 2 3 3

sin sin sin sincos cos cos cos

x s s sxx s s s

⎡ ⎤+ + += ⎢ ⎥+ + + +⎢ ⎥⎣ ⎦

β θ θ θβ θ θ θ

(28)

where 4x is the width of strut D and φ3 is the angle between the strut D and the longitudinal axis.

The coordinates for the point V of strut D in Fig. 12(d) can be given by the following equations.

(3 )x a= − (29)

ty j d= − −η (30)

[ ]1 4 3 1 3 1cos ( ) /(2 sin ) sin tan /(2 cos )x x x x= ⋅ + + ⋅ ⋅ −η β φ β φ β

where tj is the distance between the centers of the main re-bars.

Substituting Eqs. (29) and (30) for Eq. (26), Eq. (31) is obtained. Then, the value of the concrete stress in the strut is used to solve the equation to determine the width of each strut.

3 3 3 3(5 / 2) tan 3 tan (tan tan ) 0tj d a a− − η + θ − ⋅ φ − λ φ − θ =

(31)

In addition, the width c of strut H in the longitudinal axis can be expressed as in the following equation.

1 4 3sin cosc x x= ⋅ + ⋅β φ (32)

Thus, the desired shear force of the member is given by the following equation.

1 4 3( sin cos )c cV b c b x x= ⋅ ⋅ = ⋅ ⋅ + ⋅σ σ β φ (33)

As has been shown so far, the shear force can be computed by solving a geometric problem. The flow of the analysis is shown in Fig. 13. It involves several processes, namely (1) determining the angles θ1, θ2, and

θ3 of the three sub-struts that are formed by the sum of the resultant forces of the bond stress of the main re-bar and the tensile force of shear reinforcement; (2) assum-ing the compressive stress (σc) of the strut and the angle β, which is formed by main strut A and the longitudinal axis; and (3) calculating the width of the crooked main strut, which satisfies the geometric constraints illus-trated in Fig. 11.

6. Analysis results and discussion

Experiments and FE analysis with the shear-span ratio as the variable of interest were carried out to propose a new shear resistance model for the crooked main- and sub-struts and an analytical method to predict the shear force, as illustrated in Fig. 11(a) and Fig. 11(b) for the shear-span ratios of 0.5 and 0.75, respectively. From the figures, it can be seen that the main strut A inclines (is crooked) more as the shear-span ratio gets shorter, and the main strut exhibits the tendency to bend a little more due to the sub strut under the influence of the bond stress of the main re-bar, which gets larger as the shear-span ratio becomes greater.

Additionally, the characteristic mechanism of the proposed model in this study is that the shear force is

Calculate tanφ1 , tanφ2 , tanφ3

0 ≦ β < 90

Output V

END

Yes

No

No

Yes

Cha

nge

σC

Chan

ge β

Given θ1, θ2, θ3

Calculate s1, s2, s3

cbV ci ⋅⋅= σ

V = max?

Calculate c

Calculate x1, x2, x3, x4

Assume β (β = 0)

Assume σC (0< σC <f’c)

Specify geometric and material properties

START

Calculate tanφ1 , tanφ2 , tanφ3

0 ≦ β < 90

Output V

END

Yes

No

No

Yes

Cha

nge

σC

Chan

ge β

Given θ1, θ2, θ3

Calculate s1, s2, s3

cbV ci ⋅⋅= σ

V = max?

Calculate c

Calculate x1, x2, x3, x4

Assume β (β = 0)

Assume σC (0< σC <f’c)

Specify geometric and material properties

START

Fig. 13 Calculation procedure.


transmitted through a by-passing route to avoid the di-agonal shear cracks, which connect both ends of the beam, as the width of the diagonal cracks increases after the occurrence of the diagonal shear crack. This allows the rather free use of the effective factor of concrete compressive strength inside the strut.

Shown in Fig. 14 obtained from analytical simula-tions is the relationship between the shear-span ratio and the effective strength of concrete inside the strut. There is good agreement between the model prediction and the test results of the two cases. Additionally, the crooked main-strut and sub-strut mechanism, based on the ex-perimental and FE analytical results, are most consistent with each other when the effective compressive strength of concrete is set to 0.75 times the cylinder strength.

Additionally, when the effective factor of concrete compressive strength of Fig. 14 is set to 0.75, the crooked main-strut and sub-strut mechanism works only for the case of the member with shear-span ratio below 0.75. When the geometric behavior for the member with shear-span ratio below 0.75 is determined by the analy-sis results so far discussed, the shear resistance of the member can be predicted with much improved accuracy by the analytical computing method proposed and ex-plained in this study. However, to increase the precision of the macro model proposed in this study, it is neces-sary to determine the bond behavior of the main re-bars in more detail. Thus, additional studies in this area are expected in near future.

7. Conclusions

An experimental investigation was carried out for the RC beams with a main variable of the shear-span ratio. The test specimen was subjected to monotonic anti-symmetric moment. An analytical method to compute the shear force of the deep beams and a new macro model for the shear resistance mechanism were pro-posed based on the results of the experiments and nu-merical analysis. From the experimental and analytical investigations, the following conclusions can be de-

duced: (1) FE analysis was carried out on the test specimens

with shear-span ratios of 0.5 and 0.75 to delineate the stress flow and the failure process, and the nu-merical results showed good correlation with the experimental results.

(2) An analytical method for computing the shear force and a new macro model, which considered the bond action of the main re-bar and used the geometric property of the crooked dual strut model, was pro-posed.

(3) The analytical results revealed that the proposed shear resistance action was observed at a shear-span ratio of less than 0.75. In addition, when the effec-tive factor of concrete compressive strength was 0.75, the results of the analysis and experiment were the most consistent with each other.

Notations The following symbols are used in this paper: a = a third of the clear span of member b = width of beam section c = length of strut H in the longitudinal axis CST = force per unit length of strut d = distance between the centers of the strut and

bottom re-bar f ’c = cylinder compressive strength of concrete jt = distance between centers of main re-bars L = clear span of member si = width of sub-strut V = shear force of member Wcr = crack width x, y = x- and y-axes in the x-y coordinate system,

respectively xi = width of main strut β = angle between main strut A and longitudinal

axis φ = angle between main strut B~D and longitudinal

axis θ = angle between sub-strut and longitudinal axis ρw = shear reinforcement ratio

0.6 0.7 0.8 0.9 1.0240

280

320

360

S-0.5-50 (test)

Effectiveness factor ν (σc / fc)

Van

a (kN

)

Shear Span Ratio = 0.5

0.6 0.7 0.8 0.9 1.0240

280

320

360

Effectiveness factor ν (σc / fc)

Van

a (kN

)

S-0.75-50 (test)

Shear Span Ratio = 0.75

(a) S-0.5-50 (b) S-0.75-50

Fig. 14 Shear force versus effectiveness factor of concrete relationships.


σc = stress of concrete on the strut σwy = yield strength of shear reinforcement References American Concrete Institute, (2008). “Building code

requirements for structural concrete (ACI 318M-08) and commentary.” Farmington Hills, MI: American Concrete Institute.

Architectural Institute of Japan, (1990). “Design guidelines for earthquake resistant reinforced concrete buildings based on ultimate strength concept.” Tokyo: Architectural Institute of Japan, 104-150.

Architectural Institute of Japan, (1999). “Design guidelines for earthquake resistant reinforced concrete buildings based on inelastic displacement concept.” Tokyo: Architectural Institute of Japan, 138-162.

CEB-FIP, (1995). “Model code for concrete structures.” 3rd ed. CEB-FIP International Recommendations, 212-218.

Jirsa, J. O., Maruyama, K. and Ramirez, H. (1978). “Development of loading system and initial tests short columns under bidirectional loading.” CESRL Report No. 78-2.

Kent, D. C. and Park, R. (1971). “Flexural members with confined concrete.” Journal of the Structural Division, ASCE, 97(ST7), 1969-1990.

Kim, K.-H., Yoshida, A., Sato, Y. and Fujii, S. (2001).

“Effect of shear span ratio and reinforcing arrangement on shear and bond capacities of RC beam.” Proceedings of the Japan Concrete Institute, 23, 307-312. (in Japanese)

Kupfer, H., Hilsdorf, H. K. and Rüsch, H. (1969). “Behavior of Concrete under Biaxial Stress.” ACI Journal, 66(8), 656-666.

Nielsen, M. P. (1984). “Limit analysis and concrete plasticity.” Prentice Hall.

Park, R. and Paulay, T. (1975). “Reinforced concrete structures.” New York: John Wiley & Sons, 645-660.

Paulay, T. (1981). “Simulated seismic loading of spandrel beams.” Journal of the Structural Division, ASCE, 97(ST9), 2407-2419.

Paulay, T. and Binney, J. R. (1974). “Diagonally reinforced coupling beams of shear walls.” ACI Special Publication, 42, 579-598.

Popovics, S. (1970). “A review of stress-strain relationships for concrete.” ACI Journal, 67(3), 243-248.

Vecchio, F. J. and Collins, M. P. (1986). “The modified compression field theory for reinforced concrete elements subjected to shear.” ACI Journal, 83(2), 219-231.

Wong, P. S. and Vecchio, F. J. (2002). “VecTor2 and FormWorks user’s manual.” Technical Report, Dept. of Civil Engineering, Univ. of Toronto, Toronto, Canada, http://www.civ.utoronto.ca/vector/.

composite strut and tie model for reinforced concrete deep

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