zhang 2007

Engineering Structures 29 (2007) 2987–3001www.elsevier.com/locate/engstruct

Direct strut-and-tie model for single span and continuous deep beams

Ning Zhang, Kang-Hai Tan∗

School of Civil and Environmental Engineering, Nanyang Technological University, 50 Nanyang Avenue, Singapore 639798, Singapore

Received 8 October 2006; received in revised form 4 January 2007; accepted 2 February 2007Available online 27 March 2007

Abstract

A modified strut-and-tie model (STM) for determining the shear strength of reinforced deep beams is proposed in this paper. The modifiedmodel is based on Mohr Coulomb’s failure criterion. Several improvements have been made to the original model [Tan KH, Tong K, Tang CY.Direct strut-and-tie model for prestressed deep beams. J Struct Eng 2001;127(9):1076–84; Tan KH, Tang CY, Tong K. A direct method for deepbeams with web reinforcement. Mag Concr Res 2003;55(1):53–63]. The influence of the stress distribution factor k on the model prediction isstudied quantitatively, and a parametric study shows that the k value derived from a linear stress assumption is sufficiently accurate for predictingdeep beam shear strength. The use of an interactive failure criterion (Mohr’s failure equation) to model the tension–compression nodal zones inSTM means that no empirical stress limit is needed for calculating the ultimate strength, and that the softening effect of concrete compressivestrength due to transverse tensile strain is taken into consideration. Unlike the original model, the softening effect in the modified model isexplicitly derived and is comparable to MCFT and Belarbi and Hsu’s equations. The modified model for simply supported deep beams (SSDBs) isevaluated using 233 test results and it gives better agreement than the original model. The modified STM is further applied to concrete continuousdeep beams (CDBs) and is in good agreement with a total of 54 experimental results.c© 2007 Elsevier Ltd. All rights reserved.

Keywords: Continuous deep beams; Interactive; Reinforced concrete; Single span; Strut-and-tie; Ultimate strength

1. Introduction

While the strut-and-tie model (STM) is a conceptuallysimple design tool, common practices of STM involveassuming different levels of stress limits under different stressconditions on the STM components, i.e. nodal zones and struts.However, although stress limits are set in such a way as toinclude all the stress conditions, they are empirically definedand may be over- or under-conservative for some specialconditions. There has been considerable debate about the valueof concrete efficiency factors. Researchers in the past havetried to postulate various stress criteria for the concrete strutand nodal zones in the STM (Marti [1], Schlaich et al. [2],MacGregor [3], Bergmeister et al. [4], Alshegeir [5], Foster andMalik [6]). It should be noted that ACI 318-02 [7], CSA [8]and CEB-FIP model code 1990 [9] give different values onthe stress limits according to their respective investigations.Another school of researchers (Mau and Hsu [10], Hwang

∗ Corresponding author. Tel.: +65 679 052 97; fax: +65 679 166 97.E-mail address: [email protected] (K.-H. Tan).

0141-0296/$ - see front matter c© 2007 Elsevier Ltd. All rights reserved.doi:10.1016/j.engstruct.2007.02.004

et al. [11]) relate the stress limits with the transverse strainperpendicular to the stress direction so as to consider thesoftening effect of concrete compressive strength. However,it may not be appropriate to apply the smeared crack modelconcept to the D-Region, that is, the disturbed region wherethe conventional plane-section-remains-plane principle is notvalid [2].

Tan et al. [12,13] have developed a direct STM forsimply supported deep beams, which can consider theeffects of different web reinforcement configurations (vertical,horizontal, or inclined), and prestressing tendons. Other thandefining the stress limits for the STM components, the modelutilizes a failure criterion from the Mohr–Columb theory fornodal zones (tension–compression stress state) as below:

f1

ft+

f2

f ′c

= 1 (1)

where f1 and f2 are principal tensile and compressive stressesat the nodal zone respectively; f ′

c is the cylinder compressivestrength representing the maximum compressive strength inthe f2 direction; and ft represents the maximum tensile

http://www.elsevier.com/locate/engstruct

mailto:[email protected]

http://dx.doi.org/10.1016/j.engstruct.2007.02.004

2988 N. Zhang, K.-H. Tan / Engineering Structures 29 (2007) 2987–3001

Nomenclature

a Shear span measured between concentrated loadand the support centre

Ac Beam effective cross-sectional area, equal tobwdc

As1, As2 Area of top and bottom longitudinal reinforce-ment, respectively

Astr1, Astr2, Astr3 Cross-sectional areas of the concretediagonal struts at node A, B, C, respectively(Fig. 9)

Astr3, Astr4 Average cross-sectional areas of the outer andinner concrete struts, respectively (Fig. 8)

Asw Total area of web reinforcement that criss-crossesthe diagonal strut

Asw1 Area of single web steel bar that criss-crosses thediagonal strut

Ash, Asv Total areas of vertical and horizontal steel rein-forcement that criss-cross the strut, respectively

bw Width of deep beamc1, c2 Distances from the centroid of the top and bottom

longitudinal steel to the beam top and beam soffit,respectively

dc Effective beam depth (Figs. 1 and 7)dw Distance from the beam top to the intersection of

web reinforcement with the line connecting thesupport center and the load center.

Ec, Es Elastic modulus of concrete and steel, respec-tively

f1, f2 Principal tensile and compressive stresses, re-spectively

f ′c Concrete compressive strength

fct Tensile contribution of concrete to the compositetensile capacity

fst Tensile contribution of reinforcement to thecomposite tensile capacity

ft Combined tensile strength of reinforcement andconcrete

fy1, fy2 Yield strength of top and bottom longitudinalsteel reinforcement, respectively

fyh, fyv Yield strength of horizontal and vertical webreinforcement, respectively

fyw Yield strength of web reinforcementFc Compressive force in the diagonal strut of a

SSDBFc1, Fc2 Compressive force in the outer and inner

diagonal strut of a CDB,h Overall height of deep beamk Stress distribution factor at the bottom nodal zonek′ Stress distribution factor at top nodal zonela, lb, lf Widths of the load bearing, outer support, and

inner support plates, respectivelylc, ld Effective depths of the bottom and top nodal

zones, respectivelyle Effective span measured between center-to-

center of supports

m, n, p Ratios of axial stiffness of truss members in STMns Nos. of web steel bars evenly distributed along

the strutpt Average equivalent tensile stress across the

diagonal strutP Load applied on the CDBsPexp Experimental ultimate load of CDBsPn Predicted ultimate load of CDBsT Tensile force in the reinforcementT1, T2 Tension force in the top and bottom steel,

respectivelyT1 max, T2 max Yield strength of top and bottom steel,

respectivelyT1a, T2a Corresponding tension force in the top and

bottom steel at the yielding of bottom andtop steel, respectively. They are limited by therespective top and bottom steel yield strength.

Uc Total complementary energy in the trussVexp Experimental shear strength of SSDBsVn Predicted shear strength of SSDBsv Concrete softening coefficientX Reaction force of the inner support of CDBsε1, ε2 Principal tensile and compressive stress at the

tension–compression nodal zone, respectivelyεs The strain in the reinforcementθw Angle between the web reinforcement and the

horizontal axis of beams at the intersection of thereinforcement and the diagonal strut (Fig. 1)

θs Angle between the longitudinal tension reinforce-ment and the diagonal strut

ρa Total reinforcement ratio of beams, equal to(As1 + As2)/Ac

ρs Effective longitudinal reinforcement ratio of thebeam, equal to As/Ac

capacity in the f1 direction. It is worth noting that the termft is the combined tensile strength comprising contributionsfrom concrete and reinforcement (web and main). This failurecriterion is adopted because it offers a simple linear interactiverelationship between two limit failure modes, namely, brittleconcrete crushing and ductile reinforcement yielding. Byadopting the interactive stress-based failure criterion, i.e.Mohr’s failure criterion, no other empirically set stress limit isneeded in the model for calculating the ultimate strength. Casestudies have been carried out and the model gives consistent andconservative predictions of the ultimate strengths of normal andhigh strength concrete deep beams (Tan et al. [12,13]).

However, there are several limitations in Tan’s model, asfollows. Firstly, when deriving the stress distribution factork, only the force equilibrium was satisfied. Secondly, themodel was highly sensitive to the value of concrete tensilestrength, which is difficult to be determined accurately inpractice. Thirdly, the component force of bottom reinforcementin the direction of the concrete diagonal strut is neglectedfor simplicity. This term can be significant if the strut angle

N. Zhang, K.-H. Tan / Engineering Structures 29 (2007) 2987–3001 2989

Fig. 1. Determination of tensile stress factors at nodal zones.

is small. Fourthly, the softening factor which is implicitin the model was not derived explicitly. The authors re-examine Tan’s model [12,13] and its shortcomings and proposea modified STM. The stress distribution factor k is newlyderived from the consideration of both force and momentequilibrium, and its influence on model prediction of beamshear strength is discussed through a parametric study. Concretetension–stiffening properties are used instead of concretetensile strength to improve model prediction consistency. Thecomponent force of tension tie in the direction of the concretediagonal strut is also included in the model for completeness.The authors also illustrate that, by the use of interactive failurecriteria (such as Mohr’s failure criterion), the softening effectof concrete strength due to the presence of transverse tensilestrain is implicitly taken into consideration. The softeningeffect in the authors’ model is comparable to MCFT (Vecchioand Collins [14]) and Belarbi and Hsu [15]’s equations. Themodified model is verified against 233 test results of normal andhigh strength concrete simply supported deep beams (SSDBs).Comparison results show that the modified model yields moreaccurate and consistent predictions of beam ultimate shearstrength than the original one [12,13]. The modified STM isfurther extended to concrete continuous deep beams (CDBs)for the first time and is shown to be in good agreement with54 published test results. It is also interesting to see thatthe distribution of reaction forces among 3 supports can bereasonably well predicted by the proposed strut-and-tie model.

2. Modified strut-and-tie model

From the equilibrium of forces at the bottom nodal zone ofthe inclined strut (Fig. 1)

Fc =Vn

sin θs(2)

Ts =Vn

tan θs(3)

where Fc and Ts are the forces in the concrete strut and bottomreinforcement (tension tie), respectively. Vn is the shear strengthof the beam. The inclined angle of the diagonal strut θs can be

Fig. 2. Assumed tensile stress distribution arising from bottom steel.

computed from

tan θs =h −

lc2 −

ld2

a(4)

where h is the beam depth, lc and ld are the respective depths ofbottom and top nodal zones, and a is the shear span measuredfrom centre lines between the load and support bearing plates.

In the original model, the principal tensile stress f1at the tension–compression nodal zone arises from thecomponent force of longitudinal reinforcement in the directionperpendicular to the diagonal strut, namely, Ts sin θs. Thus, f1can be expressed by

f1 =kTs sin θs

Ac/sin θs(5)

where Ts sin θs/(Ac/sin θs) is the average equivalent tensilestress across the diagonal strut and Ac is the effective beamcross-sectional area, k in the numerator is a factor takingaccount of the non-uniformity of the stress distribution. Theoriginal model assumed a triangular stress distribution alongthe diagonal strut due to the presence of the bottom steel.According to force equilibrium, namely, by equating Ts sin θsto the force represented by the triangular stress block, Tanet al. [12] proposed k = 2. Although the original modelsatisfies force equilibrium, it is worth noting that the momentequilibrium is violated. To satisfy both moment and forceequilibrium, the stress distribution is shown in Fig. 2 and thefactor k can be determined accordingly.

First, consider one reinforcing bar that criss-crosses thediagonal strut and inclines at an angle θw from horizontal(Fig. 1). The effect of single reinforcement T is “smeared”across the entire strut length. For simplicity, the tensile stressdue to tension force T is assumed to distribute linearly alongthe diagonal strut. To achieve equivalence between the discreteforce T and the assumed tensile stress distribution, the lattermust satisfy both force and moment equilibrium. Applyingforce equilibrium in the f1 direction to tensile force T of thesteel bar and the idealized stress distribution along the diagonalstrut (Fig. 1), the following equation can be established:(

k + k′

2

)pt · bwdc/sin θs = T sin (θs + θw) (6)


where k and k′ are the stress distribution factors at the respectivebottom and top nodal zones, θs is the diagonal strut angle, bw isthe beam width, and dc is the beam effective depth taken as thevertical distance between the centroids of top and bottom nodalzones. It should be noted that pt = T sin(θs + θw)/(Ac/sin θs)

is the average equivalent tensile stress across the diagonal strutdue to the component of T in the principal tensile direction ofthe bottom nodal zone.

From moment equilibrium about the top node between Tand the idealized tensile stress distribution:(

k′

2+

k − k′

3

)(dc/sin θs)

2 bw pt = T sin (θs + θw)

· (dw/sin θs) . (7)

Comparing Eqs. (6) and (7), the following factors fordetermining the principal tensile stress at the respective top andbottom nodal zones can be obtained

k′= 4 − 6 ·

dw

dc

k = 6 ·dw

dc− 2.

(8)

For the case of bottom reinforcement (Fig. 2), dw = dc andθw = 0, the stress distribution factor is:{

k′= −2 (compression)

k = 4 (tension). (9)

Thus, the principal tensile stress f1 in the model can beexpressed as below:

f1 =4Ts sin θs

Ac/sin θs. (10)

The maximum tensile capacity in the f1 direction is a compositeterm and can be expressed by

ft = fst + fct. (11)

The term fct represents the contribution from concretetensile strength. The original model (Tan et al. [12,13]) usesthe concrete tensile strength fct = 0.5

√f ′c . However, at the

ultimate state, as concrete cracks extensively, concrete tensilestrength is significantly reduced. It would be un-conservative touse full concrete tensile strength as fct. Besides, the predictionsof the original model are highly dependent on the magnitudeof concrete tensile strength, which is to some extent difficult tobe determined accurately. This is one factor that has significantinfluence on the model accuracy and consistency. To improveon it, a concrete tension stiffening effect is used instead. Thefollowing equation for the tensile strength of cracked reinforcedconcrete proposed by Belarbi and Hsu [16] is adopted in themodified model:

fct = 0.31√

f ′c

(εcr

ε1

)0.4

(12)

where εcr is the strain of concrete at cracking, taken as 0.00008(Belarbi and Hsu [16]). ε1 is the principal tensile strain of

concrete strut and can be calculated from

ε1 = εs + (εs + ε2) cot2 θs (13)

where εs and ε2 are the tensile strain of longitudinal steeland peak compressive strain of the concrete strut at crushing,respectively. Usually ε2 is taken as 0.002 for normal strengthconcrete. In the modified model, the term fct is relativelysmall compared to the tensile contribution from reinforcementfst and can usually be neglected for conservatism. However,for improved accuracy in comparison, it is included in thisproposed model. Thus, through incorporating Eq. (12), themodified model is not so dependent on the magnitude ofconcrete tensile strength and is more robust in predictionconsistency.

The term fst represents the contribution from steelreinforcement. It consists of two parts: fsw from the webreinforcement and fss from the longitudinal reinforcement.

fst = fsw + fss. (14)

The presence of web reinforcement in the strut restricts thediagonal crack from quickly propagating to either end of thestrut. Although this “stitching” action only occurs at discretelocations of web reinforcement, the congregate sum effects areconsidered in the STM at the interface between the top-and-bottom nodal zones and the diagonal strut itself.

For web reinforcement, a composite factor can be obtainedfrom the summation of k factors of all individual web steel bars.Assume that there are ns web steel bars evenly distributed alongthe strut, the stress distribution factors for web steels can bewritten as below:

k′=

ns∑i=1

(4 − 6

dwi

dc

)= 4ns − 3ns = ns

k =

ns∑i=1

(6

dwi

dc− 2

)= 3ns − 2ns = ns.

(15)

Thus, the tensile contribution of web reinforcement at theinterface of the nodal zone can be calculated as

fsw =Asw fyw sin(θs + θw)

Ac/sin θs(16)

where Asw = ns Asw1 represents the total area of webreinforcement criss-crossing the concrete strut, where Asw1 isthe area of an individual web steel. For the common cases ofvertical or horizontal web reinforcement, Eq. (16) reduces toAsv fyv sin 2θs

2Acor Ash fyh sin2 θs

Ac, respectively, where, Asv and Ash

are the respective total areas of vertical and horizontal webreinforcement within the shear span.

The contribution of bottom longitudinal steel fss can beobtained in a similar fashion as Eq. (10):

fss =4As fy sin θs

Ac/sin θs. (17)

The principal compressive stress f2 in the direction of theleft diagonal strut at the bottom nodal zone is computed from:


f2 =Fc − T cos θs

Astr(18)

where Astr is the cross-sectional area at the bottom end ofthe diagonal strut. In the original STM, the component forceT cos θs of the main longitudinal reinforcement is omitted forsimplicity and conservatism. It was argued that for a small a/dratio of deep beams, this component would be insignificant,as cos θs approaches zero. However, this may not alwaysbe the case. In many instances the a/d ratio is relativelyhigh so that T cos θs becomes significant and its influenceon ultimate strength is no longer negligible. By omitting it,the original model underestimates the shear strength of deepbeams.

Substituting Eqs. (2) and (3) into Eqs. (10) and (18), andcombining with Eq. (1), the following expression for thenominal shear strength Vn is proposed:

Vn =1

4 sin θs cos θsAc ft

+sin θsAstr f ′

c

. (19)

It is noteworthy that the top nodal zone is subjected to a biaxialcompression–compression stress state. The authors suggest thatthe compressive stress along the diagonal strut at the interfaceof the top nodal zone should not exceed f ′

c . Therefore, due tothe interactive nature between tension and compression stressesin the failure criterion, Eq. (19) safeguards the compressionfailure of the top nodal zone, given that the width of the topnodal zone is comparable with that of the bottom supportregion. Thus, no further consideration is given to the top node.

The accurate value of angle θs needs to be determinedthrough an iterative procedure as documented in the originalmodel (Tan et al. [12,13]), because ld is initially unknown inEq. (4). In practice, the iterative procedure can be simplifiedby assuming the depth of the top nodal zone to be equal tothat of the bottom nodal zone, i.e. ld = lc in Fig. 1. Bydoing so, it is found that the error introduced in Vn is generallyless than 2%, which is in accordance with the findings ofprevious researchers (Tan et al. [12], Tang and Tan [17]). This isbecause ld is typically ten times smaller than the beam height h.Therefore, even a rough estimate of ld will have little effect onthe accuracy of the inclined strut angle θs. However, for betteragreement with test results, the iterative procedure is adoptedin this paper for model verifications. Usually, the convergenceis very rapid. An iteration of at most 3 or 4 rounds will besufficient to obtain a convergent result. The entire procedurecan be easily implemented using a spreadsheet. The iterativeprocedure for calculating the ultimate strength of deep beamsby the modified model is shown in Fig. 3 for the purpose ofimplementation.

In summary, the major modifications made to the originalSTM [12,13] are listed below. First, the stress distribution factork in the model is obtained considering both force and momentequilibrium. Second, the component of tension force of thereinforcement in the direction parallel to the concrete diagonalstrut is included in the calculation of principal compressivestress f2. Third, the modified STM incorporates the morerealistic tension stiffening effect rather than the full concrete

Fig. 3. Iteration procedure for computing the ultimate strength of SSDBs.

tensile strength when evaluating the composite concrete tensilestrength.

3. Verification of the modified STM

The shear strength predictions of 233 simply supporteddeep beams, shown in Table 1, have been evaluated by theoriginal STM and the proposed modified STM (Eq. (19)). Theconcrete strength of the deep beams varied from 16 to 120 MPa,including normal and high strength concrete. The beams hadan overall depth h ranging from 200 to 1750 mm, and an a/dratio from 0.28 to 2.0. The longitudinal main reinforcementratio ranged from 0.90% to 4.07%, and vertical and horizontalweb reinforcement ratios ranged from zero to 2.86%, andzero to 3.17%, respectively. The predicted shear strengths Vnversus experimental shear strengths Vexp are plotted in Fig. 4b,and c for both original and modified models. The average(AVG) and coefficient of variation (C.O.V.) Vn/Vexp ratios havealso been reported in the figure. The modified STM predictsthe ultimate shear strength more accurately and consistentlythan the original STM, because the C.O.V. obtained with themodified STM is 38% lower [(0.130 − 0.212)/0.212] while theAVG is 7% higher [(0.91 − 0.85)/0.85].

The results obtained in terms of shear strength from themodified model on the 212 specimens have also been comparedwith results obtained from ACI 318-99. Ten specimens from


Table 1SSDBs used for experimental comparison

Kong et al. [18]: S-30, S-25, S-20, S-15, S-10, D-30, D-25, D-20, D-15, D-10

Smith and Vantsiotis [21]: 0A0-44, 0A0-48, 1A1-10, 1A3-11, 1A4-12, 1A4-51, 1A6-37, 2A1-38, 2A3-39, 1A4-40, 2A6-41, 3A1-42, 3A3-43, 3A4-45, 3A6-46,0B0-49, 1B1-01, 1B3-29, 1B4-30, 1B6-31, 2B1-05, 2B3-06, 2B4-07, 2B4-52, 2B6-32, 3B1-08, 3B1-36, 3B3-33, 3B4-34, 3B6-35, 4B1-09, 0C0-50, 1C1-14,1C3-02, 1C4-15, 1C6-16, 2C1-17, 2C3-03, 2C3-27, 2C4-18, 2C6-19, 3C1-20, 3C3-21, 3C4-22, 3C6-23, 4C1-24, 4C3-04, 4C3-28, 4C4-25, 4C6-26, 0D0-47,4D1-13

Subedi et al. [24], Subedi [25]: 1A2, 1B1, 1B2, 1C2, 1D1, 1D2, 2A2, 2B2, 2D1, 2D2, 4G1, 4G2, 4G3, 4G4

Walraven and Lehwalter [26]: V711/4, V022/4, V511/4, V411/4, V711/3, V022/3, V511/3, V411/3, V711, V022, V511

Tan et al. [20]: A-0.27-2.15, A-027-3.23, A-0.27-4.30, A-0.27-5.38, B-0.54-2.15, B-0.54-3.23, B-0.54-4.30, B-0.54-5.38, C-0.81-2.15, C-0.81-3.23, D-1.08-2.15,D-1.08-3.23, D-1.08-4.30, D-1.08-5.35, E-1.62-3.23, E-1.62-4.30, E-1.62-5.38, E-2.16-4.30, G-2.70-5.38

Tan et al. [27]: I-1/0.75, I-3/0.75, I-4/0.75, I-6S/0.75, II-2N/1.00, II-3/1.00, II-5/1.00, II-6N/1.00, III-2N/1.50, III-2S/1.50, III-6N/1.50

Tan et al. [23]: 1-2.00/0.75, 1-2.00/1.00, 2-2.58/0.25, 2-2.58/0.50, 2-2.58/0.75, 3-4.08/0.25, 3-4.08/0.50, 3-4.08/0.75, 3-4.08/1.00

Foster and Gilbert [28]: B1.2-1, B1.2-2, B1.2-3, B1.2-4, B2.0-1, B2.0-2, B2.0-3, B2.0A-4, B2.0B-5, B2.0C-6, B2.0D-7, B3.0-1, B3.0-2, B3.0-3, B3.0A-4, B3.0B-5

Tan and Lu [29]: 1-500/0.50, 1-500/0.75, 1-500/1.00, 2-1000/0.50, 3-1400/0.50, 4-1750/0.50

Shin et al. [30]: MHB1.5-50, MHB1.5-75, MHB1.5-100, MHB2.0-50, MHB2.0-75, MHB2.0-100, MHB2.5-50, MHB2.5-75, MHB2.5-100, HB1.5-50, HB1.5-75,HB1.5-100, HB2.0-50, HB2.0-75, HB2.0-100, HB2.5-50, HB2.5-75, HB2.5-100

Oh and Shin [31]: N4200, N42A2, N42B2, N42C2, H4100, H41A2(1), H41B2, H41C2, H4200, H42A2(1), H42B2(1), H42C2(1), H4300, H43A2(1), H43B2,H43C2, H45A2, H45B2, H45C2, H41A0, H41A1, H41A2(2), H41A3, H42A2(2), H42B2(2), H42C2(2), H43A0, H43A1, H43A2(2), H43A3, H45A2(2), U41A0,U41A1, U41A2, U41A3, U42A2, U42B2, U42C2, U43A0, U43A1, U43A2, U43A3, U45A2, N33A2, N43A2, N53A2, H31A2, H32A2, H33A2, H51A2, H52A2,H53A2

Yang et al. [32]: L5-40, L5-60, L5-60R, L5-75, L5-100, L10-40R, L10-60, L10-100, UH5-40, UH5-60, UH5-75, UH5-100, UH10-40, UH10-40R, UH10-60,UH10-100

Kong et al. [18] have been excluded from the case study becauseACI 318-99 does not include the design of deep beams withinclined web reinforcement. The predicted shear strength Vnversus experimental shear strength Vexp are plotted in Fig. 4(a)for the ACI code. It can be observed that while ACI 318-99 [19]gives AVG = 0.72 and C.O.V. = 0.526, the modified modelprovides AVG = 0.91 and C.O.V. = 0.130, highlighting itsaccuracy and consistency.

4. Influence of factor k on the prediction of ultimatestrength Vn

The actual tensile stress distribution along the diagonal strutis highly nonlinear and difficult to be determined mechanically.For simplicity, it is assumed that the tensile stress transverseto the diagonal strut is kpt at the bottom nodal zone. Thus themaximum tensile capacity in the f1 direction is:

ft =k fy As sin θs

Ac/sin θs+

fyw Asw sin(θs + θw)

Ac/sin θs+ fct. (20)

The beam ultimate shear strength can be expressed fromEq. (19)

Vn =1

k sin θs cos θsft Ac

+sin θsf ′c Astr

=1

cos θs

fy As sin θs+Asw fyw sin(θs+θw)

k +fct Ac

k sin θs

+sin θsf ′c Astr

. (21)

From the mathematical form of Eq. (21), it is noted that if thedeep beam is not provided with any web reinforcement Asw and

the concrete tensile contribution fct is ignored, the predictionof ultimate shear strength Vn is completely independent ofthe k factor. That is to say, whatever the form of stressdistribution is along the diagonal strut, the model predicts thesame value of Vn for the beam. However, as mentioned inthe previous section, for optimal predictions, fct is considered.Thus, from Eq. (21), it can be deduced that with an increasein k factor, Vn will slightly decrease. This is because as kincreases, the contributions from web steel and concrete (whichare independent of k) to the composite tensile strength ftbecome relatively insignificant. Thus the f1/ ft ratio increasesas k increases, resulting in a relatively smaller failure load. Todemonstrate how the value of k influences the ultimate shearstrength quantitatively, a parametric study is carried out. Two-hundred-and-sixteen fictitious beams with the same a/d ratioof 0.75 and a main reinforcement ratio of 1.5% are analyzedusing the proposed STM. The variables are web reinforcementratio (ρw) ranging from 0% to 1.0% and concrete strength ( f ′

c)from 20 to 80 MPa. The influence of the factor k on the modelprediction of beam shear strength Vn can be represented by thepercentage of Vn, expressed by (Vn − Vk=4)/Vk=4, where Vk=4represents the beam shear strength obtained by the proposedSTM with k = 4. The plotted curves are shown in Fig. 5.For higher web reinforcement ratios, the model predictionsdecrease more quickly as the factor k increases. Similarly,for higher concrete strengths, the model predictions decreasemore quickly as k increases. Generally, there is about 3%reduction in shear strength when ρw goes from 0% to 1.0%and about 2% reduction when f ′

c goes from 20 to 80 MPa. For2 ≤ k ≤ 4, the predictions are greater than Vk=4, but when4 ≤ k ≤ 10, the predictions are smaller than Vk=4. However,


Fig. 4. Ultimate strength predictions by means of: (a) ACI 318-99 (212 beams);(b) Original STM; (c) Modified STM for 233 reinforced concrete tested deepbeams.

as shown in Fig. 5 the reduction is generally less than 5% ofthe predicted beam shear strength when k is greater than 4.Thus, it can be concluded that k = 4 derived from the linearstress distribution assumption is sufficiently accurate for themodel predictions. Besides, the modified model now satisfiesboth force and moment equilibrium.

5. Softening effect in the STM

Experimental studies have provided strong evidence that theability of diagonally cracked concrete to resist compressiondecreases as the amount of tensile straining increases. Anumber of analytical models (Vecchio and Collins [14,22],Belarbi and Hsu [15]) have been proposed to represent thecompression softening effect observed in cracked reinforcedconcrete in tension–compression stress states. According to theproposed STM, the equation that yields the concrete softeningcoefficient v can be expressed as:

Fc

Astr= v f ′

c . (22)

Substituting Eqs. (10), (18) and (22) into Eq. (1), thefollowing equation can be obtained:

v = 1 +Ts cos θs

Astr f ′c

−4Ts sin2 θs

Ac ft(23)

where Ts is the tension force in the bottom reinforcementand can be expressed as εs Es As. It is noteworthy that thesecond term in Eq. (23) represents the contribution of thetension force of the bottom steel resolved in the directionof the diagonal strut, which negates a certain amount ofcompression force in the strut itself. The third term in Eq. (23)represents the contribution from transverse tensile stress dueto the bottom steel, which is the main factor for concretesoftening. Substituting Eq. (20) into Eq. (23) and utilizing therelationship between Ts and εs, the following expression for theconcrete softening coefficient is established,

v = 1 +Es Asεs cos θs

f ′c Astr

−4Esεs

4 fy +Asw sin(θs+θw)

As sin θsfyw +

fctρs sin2 θs

(24)

where Es and As are respectively the elastic modulus and cross-sectional area of longitudinal reinforcement; εs is the strain inthe longitudinal reinforcement; Asw is the cross-section area ofweb reinforcement crossing the diagonal strut; θs is the inclinedangle of the strut; fy and fyw are the yield strengths of thelongitudinal and web reinforcement, respectively; Astr is thecross-sectional area of the diagonal strut; ρs = As/Ac is theeffective longitudinal reinforcement ratio.

From Eq. (24), it is worth noting that the softening factorv in the STM is not only influenced by the strut angle θs butalso by other variables such as concrete compressive strength,yield strength and reinforcing ratios of steel reinforcement(main and web). However, the strut angle θs is the mainvariable that affects the softening coefficient v significantly.For a larger reinforcement ratio or a greater yield strength ofsteel, the slope becomes gentler, i.e. the decreased portion ofconcrete compressive strength is less than lightly reinforcedbeams. Web reinforcement seems to reduce the softening slope,but the effect is minimal. A typical deep beam specimen,say, 3–4.08/0.75 is chosen and the softening relationship bythe STM is plotted in Fig. 6 together with Vecchio and


Fig. 5. Influence of factor k on ultimate shear strength predictions.

Fig. 6. Typical maximum concrete compressive stress as function of tensilestrain in reinforcement.

Collins [14] and Belarbi and Hsu [15] softening curves. Detailsof the specimen are reported in Tan et al. [23]’s experimentalprogramme. In the figure, εs is the strain in the bottom mainsteel and the strut angle θs is 45◦. It is shown that the concretecompressive strength has a nearly linear descending trend withan increase of strain in the bottom steel. This is due to theadoption of a linear interactive failure criterion in the STM.Therefore, although the failure criterion adopted in the STMis based on stress values (Eq. (19)), the softening effect ofconcrete compressive strength due to transverse tensile strain is

implicitly taken into account through Eq. (24). It is comparableto the softening curves proposed by Vecchio and Collins [14]and Belarbi and Hsu [15]. Furthermore, the descending trendwith an increase of steel strain is determined not only by thestrut angle θs but also by the web and main reinforcementdetails. This seems more reasonable for deep beam specimens.As can be seen from Fig. 6, the softening coefficient of STMis slightly larger than the other two curves when steel strainεs is relatively small. This is the situation when the beam isover-reinforced so that the strain in longitudinal reinforcementis small when the beam approaches failure. When εs approachesthe yield strain εy of the longitudinal reinforcement (for fy =

499 MPa, E = 20 000 MPa, εy = 0.0025), the STMsoftening coefficient v is comparable to the other softeningcurves (Fig. 6), and when εs exceeds the yield strain, the STMcoefficient is less than those determined by the other threeequations, indicating a greater softening effect.

6. Modelling continuous deep beams

A STM for two-span continuous deep beams (CDBs) witha top point load at each mid-span is given in Fig. 7. It canbe idealized as a statically indeterminate truss as shown inFig. 8. Assuming perfect elastic–plastic material properties forconcrete and steel bars, the internal forces of the truss are solved(detailed derivation can be found in Appendix A). They aredenoted as Fc1 = A · P , T1 = B · P , T2 = C · P , Fc2 = D · P ,where Fc1 and Fc2 represent the respective forces in the outer


Fig. 7. Two-span continuous deep beam under two-point loading.

Fig. 8. Truss model for two-span CDB under two-point loading.

and interior concrete strut, T1 and T2 represent the respectivetension forces in the top and bottom longitudinal reinforcement,and P represents the point load acting on the beam (Fig. 8).Again, Mohr’s linear interaction failure criterion is adopted atthe interface between the strut and tension–compression nodalzones:

f1

ft+

f2

f ′c

= 1. (25)

Three tension–compression nodal zones are identified in theSTM for CDBs as shown in Fig. 7. Similar to the modellingof SSDBs, Morh’s failure criterion is applied to the three nodalzones, as follows:

6.1. Nodal zone A

The principal tensile stress f1A across the diagonal strut andthe principal compressive stress f2A in the diagonal strut can beobtained in a similar manner as the SSDB:

f1A =4T2 sin θs

Ac/sin θs= 4C sin2 θs ·

PAc

(26)

f2A =Fc1 − T2 cos θs

Astr1= (A − C cos θs) ·

PAstr1

(27)

where θs is the inclined angle of diagonal strut with respect tothe horizon, Astr1 is the cross-sectional area at the bottom of theouter concrete strut (Fig. 9).

From (25), (26) and (27), the following expression can bederived for the ultimate force PnA:

PnA =1

4C sin2 θsftA Ac

+(A−C cos θs)

f ′c Astr1

(28)

where ftA is the maximum tensile capacity of nodal zone A inthe f1 direction and can be similarly expressed by:

ftA =4As2 fy sin θs

Ac/sin θs+

Asw fyw sin(θs + θw)

Ac/sin θs+ fct.

6.2. Nodal zone B

The principal tensile stress f1B across the diagonal strut atnodal zone B consists of two components: the contributionsfrom the top and bottom reinforcement. As the k factor is −2(compression) for top steel and 4 (tension) for bottom steel, thecombined f1B can be expressed as below

f1B =4T2 sin θs − 2T1 sin θs

Ac/sin θs= (4C − 2B) sin2 θs ·

PAc

. (29)

The principal compressive stress at nodal zone B is obtainedsimilarly as nodal zone A

f2B =Fc2 − T2 cos θs

Astr2= (D − C cos θs) ·

PAstr2

(30)

where Astr2 is the cross-sectional area at the bottom of theinterior concrete strut (Fig. 9).

From Eqs. (25), (29) and (30), the following expression canbe derived for the ultimate load PnB:

PnB =1

(4C−2B) sin2 θsftB Ac

+(D−C cos θs)

f ′c Astr2

. (31)

It is noteworthy that the maximum tensile capacity of nodalzone B ftB is expressed by

ftB =4T2 max − 2T1a

Ac/sin2θs+


Ac/sin θs+ fct


Fig. 9. Details of nodal zones in CDB.

where T2 max is the yield strength of bottom steel and T1a is thecorresponding tension force in the top steel at the yielding ofbottom steel (= B

C T2 max). The term T1a should not exceed theyield strength of top steel.

6.3. Nodal zone C

f1C =4T1 sin θs − 2T2 sin θs

Ac/sin θs= (4B − 2C) sin2 θs ·

PAc

(32)

f2C =Fc2 − T1 cos θs

Astr3= (D − B cos θs) ·

PAstr3

. (33)

From Eqs. (25), (32) and (33), the following expression can bederived for the ultimate load PnC:

PnC =1

(4B−2C) sin2 θsftC Ac

+(D−B cos θs)

f ′c Astr3

. (34)

The term ftC in Eq. (34) is the maximum tensile capacity ofnodal zone C:

ftC =4T1 max − 2T2a

Ac/sin2θs+


Ac/sin θs+ fct.

Similarly, T1 max is the yield strength of top steel and T2ais the corresponding tension force in the bottom steel at theyielding of top steel (=C

B T1 max). T2a should not exceed theyield strength of bottom steel.

Thus the predicted ultimate load P will be the minimumamong Eqs. (28), (31) and (34), denoted as Pn.

Pn = Min(PnA, PnB, PnC). (35)

7. Reliability of STM for CDBs

The computing of STM for CDBs can be easily implementedby hand calculations or a spreadsheet. An example of handcalculations is shown in the appendix for illustrative purposes.

Fig. 10. Ultimate strength predictions for 54 continuous deep beams.

Fifty-four concrete continuous deep beams reported by otherresearchers have been evaluated by the authors’ model.The details of the specimens and the predicted-versus-actualultimate strength ratios are summarized in Table 2. The beamshad an overall depth h ranging from 400 to 1000 mm, and anle/d ratio from 0.95 to 4.49. The top-and-bottom longitudinalmain reinforcement ratios ranged from 0.07% to 1.88% and0.32% to 1.88%, respectively. The vertical and horizontal webreinforcement ratios ranged from zero to 0.90% and zero to1.71%, respectively. The STM for CDBs generally performswell in predicting the ultimate strengths with an overall averageprediction to test result of 0.95 and a C.O.V of 0.130. Theultimate strength Pn versus the experimental strength Pexp areplotted in Fig. 10. It shows the comparison of model predictionswith 54 test results. Generally speaking, the proposed model ison the safe side and gives consistent predictions. For simplicity,the model assumes linear material properties for steel andconcrete when solving the statically-indeterminate truss. It


Table 2Predictions of ultimate load for CDBs

No. Reference f ′c (MPa) h (mm) bw (mm) le (mm) fy As1/ fy As2

(kN/kN)fyv Av/ fyh Ah(kN/kN)

Pexp (kN) Pn (kN) STM Pn/Pexp

1 CD-0.5/V 40.4 960 47 860 163/163 43/0 400 427 1.072 CD-0.5/H 40.6 960 47 860 163/163 0/93 425 454 1.073 CD-0.5/I 45.1 960 47 860 163/163 (93) 500 485 0.974 CD-1.0/H 42.2 960 47 860 163/163 0/185 425 385 0.915 CD-1.0/I 41.0 960 47 860 163/163 (162) 510 485 0.956 CD-0.0 50.4 960 47 860 163/163 0/0 400 466 1.167 MC2-200 50.9 1000 40 860 15/314 23/42 598 610 1.028 MC2-150 48.4 1000 40 860 15/314 23/53 590 593 1.009 MC2-100 51.7 1000 40 860 15/314 23/87 599 598 1.00

10 MC2-75 53.1 1000 40 860 15/314 23/118 596 594 1.0011 MC2-50 56.0 1000 40 860 15/314 23/182 590 624 1.0612 MC2-75-2 48.8 1000 40 860 29/314 23/237 570 652 1.1413 CDB1 30.0 625 120 1340 305/226 262/149 550 492 0.9014 CDB2 33.1 625 120 1340 305/226 135 /74 475 457 0.9615 CDB3 22.0 625 120 1340 305/226 0/74 285 314 1.1016 CDB4 28.0 625 120 1340 305/226 135/0 443 394 0.8917 CDB5 28.7 625 120 1340 113/113 135/74 410 329 0.8018 CDB6 22.5 425 120 1340 192/192 139/39 248 249 1.0019 CDB7 26.7 425 120 1340 192/192 72/20 223 239 1.0820 CDB8 23.6 425 120 1340 113/113 72/20 193 186 0.9721 BM 7/1.0(T1) 34.5 1000 200 2100 484/363 0/0 705 840 1.1922 BM 7/1.0(T2) 34.5 1000 200 2100 484/363 0/0 1085 840 0.7723 BM 6/1.0(T1) 35.8 1000 200 2100 484/363 0/194 1095 995 0.9124 BM 6/1.0(T2) 35.8 1000 200 2100 484/363 0/194 1059 995 0.9425 BM 5/1.0(T1) 36.9 1000 200 2100 484/363 518/0 1280 1188 0.9326 BM 5/1.0(T2) 36.9 1000 200 2100 484/363 518/0 1147 1188 1.0427 BM 4/1.0(T1) 28.5 1000 200 2100 456/342 0/65 1083 798 0.7428 BM 4/1.0(T2) 28.5 1000 200 2100 456/342 0/65 988 798 0.8129 BM 3/1.0(T1) 28.9 1000 200 2100 456/342 130/0 1084 856 0.7930 BM 3/1.0(T2) 28.9 1000 200 2100 456/342 130/0 1163 856 0.7431 HW 51.6 700 100 1400 338/225 0/247 646 522 0.8132 VW 49.5 700 100 1400 338/225 330/0 580 563 0.9733 N2-1.0-WO 36.9 600 150 1200 735/735 0/0 650 663 1.0234 N2-1.0-WV 37.2 600 150 1200 735/735 162/0 750 678 0.9035 N2-1.0-WVH 37.3 600 150 1200 735/735 162/216 923 694 0.7536 N2-1.5-WO 33.9 600 150 1800 735/735 0/0 500 597 1.1937 N2-1.5-WV 35.1 600 150 1800 735/735 206/0 725 667 0.9238 N2-1.5-WVH 35.4 600 150 1800 735/735 206/165 650 673 1.0439 2CB4 38.0 600 75 1680 116/323 182/114 420 361 0.8640 2CB3 38.0 600 75 1680 323/323 182/114 425 333 0.7841 1CB2 48.0 400 50 1000 99/99 48/29 180 175 0.9742 1CB1 48.0 400 50 500 99/99 29/29 330 259 0.7943 BM 1.0/1/1(r) 29.6 1000 150 2300 356/265 354/0 800 671 0.8444 BM 1.0/1/2 27.5 1000 150 2300 356/265 257/0 754 621 0.8245 BM 1.0/1/3 25.8 1000 150 2300 356/265 161/0 595 573 0.9646 BM 1.0/2/1 25.2 1000 150 2300 265/356 354/0 914 728 0.8047 BM 1.0/2/2 28.2 1000 150 2300 265/356 257/0 743 722 0.9748 BM 1.0/2/3 30.4 1000 150 2300 265/356 161/0 686 697 1.0249 BM 1.5/1/1 29.6 600 150 2300 445/356 354/0 595 514 0.8650 BM 1.5/1/2 28.9 600 150 2300 445/356 257/0 519 479 0.9251 BM 1.5/1/3 25.9 600 150 2300 445/356 161/0 398 425 1.0752 BM 1.5/2/1 27.8 600 150 2300 356/445 354/0 580 550 0.9553 BM 1.5/2/2 26.0 600 150 2300 356/445 257/0 535 500 0.9354 BM 1.5/2/3 28.5 600 150 2300 356/445 161/0 384 480 1.25

Mean 0.95S.D. 0.123C.O.V. 0.130

Note: Source—Specimens no. 1–6, Chemrouk [33]; 7–12, Maimba [34]; 13–20, Ashour [35]; 21–30, Rogowsky et al. [36]; 31–32, Choo and Lim [37]; 33–38,Poh [38]; 39–42, Subedi [39]; 43–54, Asin [40].


Table 3Comparison bewteen calculated and measured interior support reaction

Reference Predicted interiorsupport reactionXn (kN)

Tested interiorsupport reactionXexp (kN)

Xn/Xexp

CDB1 672.2 702.0 0.96CDB2 624.2 612.0 1.02CDB3 427.1 360.4 1.19CDB4 537.4 567.8 0.95CDB5 433.0 516.0 0.84CDB6 325.6 312.2 1.04CDB7 313.7 281.0 1.12CDB8 244.7 247.4 0.99BM 7/1.0(T1) 1145.0 842.0 1.36BM 7/1.0(T2) 1145.0 1376.0 0.83BM 6/1.0(T1) 1357.4 1281.0 1.06BM 6/1.0(T2) 1357.4 1186.0 1.14BM 5/1.0(T1) 1621.5 1741.0 0.93BM 5/1.0(T2) 1621.5 1491.0 1.09BM 4/1.0(T1) 1085.1 1330.0 0.82BM 4/1.0(T2) 1085.1 1228.0 0.88BM 3/1.0(T1) 1164.3 1374.0 0.85BM 3/1.0(T2) 1164.3 1540.0 0.76HW 719.6 828.0 0.87VW 775.4 726.0 1.07N2-1.0-WO 1026.2 844.0 1.22N2-1.0-WV 1081.3 936.0 1.16N2-1.0-WVH 1133.5 1156.0 0.98N2-1.5-WO 738.6 548.0 1.35N2-1.5-WV 826.6 887.0 0.93N2-1.5-WVH 858.0 856.0 1.00BM 1.0/1/1(r) 935.3 1053.0 0.89BM 1.0/1/2 862.0 989.0 0.87BM 1.0/1/3 794.1 776.0 1.02BM 1.0/2/1 913.4 1171.0 0.78BM 1.0/2/2 905.8 937.0 0.97BM 1.0/2/3 875.4 845.0 1.04BM 1.5/1/1 699.9 798.0 0.88BM 1.5/1/2 652.3 694.0 0.94BM 1.5/1/3 578.0 691.0 0.84BM 1.5/2/1 699.9 745.0 0.94BM 1.5/2/2 635.7 675.0 0.94BM 1.5/2/3 610.5 499.0 1.22

Mean 0.99S.D. 0.147C.O.V. 0.149

would be interesting to see how the model-predicted supportreactions compared with the test results. Table 3 shows thecomparison between model predictions and test results ofreaction forces at the interior support. Clearly, it shows thatthe STM for continuous deep beams can predict the reactionforces reasonably well. The fifty-four Pn/Pexp ratios obtainedfrom the proposed model are plotted versus le/d ratio, concretestrength f ′

c , and total reinforcement ratio ρa in Fig. 11. It isshown that the scatter is very low and uniform for the entire setof three variables. It can be concluded that the application ofthe modified STM to CDBs is feasible and successful. It alsoshows the validity and versatility of the proposed STM fromsimply-supported to continuous deep beams.

It is also noteworthy that the overall average prediction-to-test result ratios for the 233 SSDBs and 54 CDBs are very close:0.91 and 0.95, respectively, and the C.O.V. of both studies are

Fig. 11. Effect of: (a) concrete strength ( f ′c); (b) effective span to overall depth

ratio (le/d); and (c) total reinforcement ratio on ultimate strength predictions(ρa).

same at 0.130. This shows that the predictions are consistentand accurate for both SSDBs and CDBs with differentgeometrical properties, concrete strengths and reinforcementconfigurations.


8. Conclusions

In this paper, the authors proposed a modified strut-and-tie model (STM) based on previous work of Tan et al. [12,13]. The modified STM for SSDBs is evaluated using 233 testresults collected from the literature. The modified model yieldsbetter agreement than the original STM [12,13] in terms ofaccuracy and consistency. The influence of stress distributionfactor k on model prediction is studied numerically. It is shownthat a k value of 4 derived from a linear stress distributionassumption is sufficiently accurate for predicting the deep beamshear strength. The modified model also implies a softeningeffect in concrete compressive strength due to transverse tensilestrain, by adopting an interactive failure criterion, i.e. Mohr’sfailure criterion. The softening effect given by the model isshown comparable to that obtained from MCFT and Belarbiand Hsu’s equations.

The modified STM is further extended to predict the ultimatestrengths of concrete continuous deep beams (CDBs) for thefirst time. The predictions of the STM for CDBs including beamstrengths and support reactions are in good agreement with atotal of 54 test results of concrete CDBs.

Acknowledgements

The authors would like to express their gratitude tothe Nanyang Technological University for research accountRG9/99. The first author also acknowledges a scholarship fromthe Nanyang Technological University, Singapore.

Appendix A. Derivation of internal forces of strut-and-tiemodel

The stress–strain relationship of concrete under compressionis taken as linear:

σc = Ecεc

where Ec is the modulus of elasticity of concrete incompression and can be taken from the ACI code (Ec =

4730√

f ′c , MPa) for normal strength concrete.

The stress–strain relationship of steel in tension is given by:

σs = Esεs.

The truss members in Fig. 8 are internally determinate butexternally indeterminate to 1 degree. Releasing the interiorreaction force X and applying the Crotti–Engesser theorem, thetotal complementary energy in the truss is:

Uc =

∑ ∫V

(∫εdσ

)dV

=

(P−

X2

)2le

2 sin2 θs cos θs

Ec Astr4+

X2le8 sin2 θs cos θs

Ec Astr5+

(X−P)2le2 tan2 θs

Es As1+

(P−

X2

)2le

tan2 θs

Es As2(A1)

where le is the effective span measured between centre-to-centre of supports; θs is the inclined angle of the diagonal

strut; Ec and Es are the elastic modulus of concrete undercompression and steel bar under tension, respectively. Theterms Astr1, Astr2 and Astr3 are the cross-sectional areas at theends of the tapered concrete struts (Fig. 9), while Astr4 andAstr5 represent the average cross-sectional areas of the outerand inner tapered concrete struts, respectively (Fig. 8). Theyare expressed as follows:

Astr1 = bw (lc cos θs + lb sin θs) (A2)Astr2 = bw(lc cos θs + lf sin θs) (A3)Astr3 = bw (ld cos θs + la sin θs) (A4)

Astr4 =Astr1 + Astr3

2(A5)


2(A6)

θs = arctan2(h − c1 − c2)

le(A7)

where

la, lb and lf are the widths of the support and load bearingplates (Fig. 7);

lc and ld are the respective effective depths of the bottom andtop nodal zones, lc = 2c2, ld = 2c1 (Fig. 9);

c1 and c2 are the distances from the centroid of the top andbottom longitudinal steel bars to the beam top and beamsoffit respectively (Fig. 7).

From the Theorem of Minimum Complementary Potential(TMCP), at interior support B:

∆B =∂Uc

∂ X= 0. (A8)

Thus, from Eqs. (A1) and (A8), an expression for the reactionforce X can be obtained as follows:

X =2(m + 2n cos3 θs + 2p cos3 θs)

1 + m + 4n cos3 θs + 2p cos3 θsP (↑) (A9)

where m, n, p are the ratios of axial stiffness of truss members:

m =Astr5

Astr4; (A10)

n =Ec Astr5

Es As1; (A11)

p =Ec Astr5

Es As2. (A12)

After solving the reaction force X , the internal forces of trussmembers (Fig. 8) can be obtained as follows:

Fc1 =1 + 2n cos3 θs

sin θs(1 + m + 4n cos3 θs + 2p cos3 θs)P

= A · P (compression) (A13)

T1 =cos θs(m + 2p cos3 θs − 1)


= B · P (tension) (A14)


T2 =cos θs(1 + 2n cos3 θs)


= C · P (tension) (A15)

Fc2 =m + 2n cos3 θs + 2p cos3 θs


= D · P (compression) . (A16)

Appendix B. Worked example on continuous deep beam

The notation of the beam is BM 5/1.0 (Rogowsky et al. [36]),details of specimens are as follows:Geometry: le = 2100 mm, c1 = 50 mm, c2 = 25 mm,lc = 50 mm, ld = 100 mm, h = 1000 mm, bw = 200 mm,lb = 200 mm, la = 300 mm, lf = 400 mm, Ac = 185 000 mm2;Concrete strength: fc′ = 36.9 N/mm2, Ec = 3694

√f ′c =

22 439 MPa, fsp = 0.461√

f ′c = 2.80 MPa;

Longitudinal reinforcement: As1 = 1195 mm2, As2 =

896 mm2, fy As1 = 484 kN, fy As2 = 363 kN;Vertical web reinforcement: Asv fyv = 518 kN, θw = 90◦.Step 1: Determine θs from Eq. (A7);

θs = arctan2 (h − c1 − c2)

le= arctan

2 × (1000 − 50 − 25)

2100= 41.4◦.

Step 2: Determine m, n, p, from Eqs. (A10) to (A12);

Astr1 = bw (lc cos θs + lb sin θs) = 200 × (50 × 0.750+ 200 × 0.661) = 33 945 mm2

Astr2 = bw (lc cos θs + lf sin θs) = 200 × (50 × 0.750+ 400 × 0.661) = 60 386 mm2

Astr3 = bw (ld cos θs + la sin θs) = 200 × (100 × 0.750+ 300 × 0.661) = 54 669 mm2


2=

33 945 + 54 6692

= 44 307 mm2


2=

60 386 + 54 6692

= 57 528 mm2

m =Astr5

Astr4=

57 52844 307

= 1.298

n =Ec Astr5

Es As1=

22 439 × 57 528200 000 × 1195

= 5.401

p =Ec Astr5

Es As2=

22 439 × 57 528200 000 × 896

= 7.204.

Step 3: Determine A, B, C, D from Eqs. (A13) to (A16);

A =1 + 2n cos3 θs

sin θs(1 + m + 4n cos3 θs + 2p cos3 θs)

=1 + 2 × 5.401 × 0.422

0.661 × (1 + 1.298 + 4 × 5.401 × 0.422 + 2 × 7.204 × 0.422)

= 0.481

B =cos θs(m + 2p cos3 θs − 1)


=0.750 × (1.298 + 2 × 7.204 × 0.422 − 1)

0.661 × (1 + 1.298 + 4 × 5.401 × 0.422 + 2 × 7.204 × 0.422)

= 0.414

C =cos θs(1 + 2n cos3 θs)


=0.750 × (1 + 2 × 5.401 × 0.422)

0.661 × (1 + 1.298 + 4 × 5.401 × 0.422 + 2 × 7.204 × 0.422)

= 0.360

D =m + 2n cos3 θs + 2p cos3 θs


=1.298 + 2 × 5.401 × 0.422 + 2 × 7.204 × 0.422

0.661 × (1 + 1.298 + 4 × 5.401 × 0.422 + 2 × 7.204 × 0.422)

= 1.032.

Step 4: Determine PnA, PnB, and PnC from Eqs. (28), (31) and(34), respectively;

fct = 0.31√

f ′c

(εcr

ε1

)0.4

= 0.31 ×√

36.9

×

(0.00008

0.002 + 0.004/ tan2 41.38◦

)0.4

= 0.31 MPa

T1 max = fy As1 = 484 kN

T2 max = fy As2 = 363 kN

T1a = min(

fy As1,BC

T2 max

)= 416 kN

T2a = min(

fy As2,CB

T1 max

)= 363 kN.

Nodal zone A:

ftA =4T2 max

Ac/sin2θs+

Asv fyv sin 2θs

2Ac+ fct

=4 × 363 × 0.437 × 103

185 000+

518 × 0.992 × 103

2 × 185 000+ 0.31

= 3.43 + 1.39 + 0.31 = 5.13 MPa

PnA =1

4C sin2 θsftA Ac

+(A−C cos θs)

f ′c Astr1

=1/1000

4×0.360×0.4375.13×185 000 +

0.481−0.360×0.75036.9×33 945

= 1202 kN.

Nodal zone B:

ftB =4T2 max − 2T1a

Ac/sin2θs+

Asv fyv sin 2θs

2Ac+ fct

=(4 × 363 − 2 × 416) × 103

185 000/0.437

+518 × 0.992 × 103

2 × 185 000+ 0.31

= 1.46 + 1.39 + 0.31 = 3.16 MPa

PnB =1

4C−2BftB Ac/sin2θs

+(D−C cos θs)

f ′c Astr2

=1/1000

4×0.360−2×0.4143.16×185 000/0.437 +

1.032−0.360×0.75036.9×60 386

= 1251 kN.

Nodal zone C:


ftC =4T1 max − 2T2a

Ac/sin2θs+

Asv fyv sin 2θs

2Ac+ fct

=(4 × 484 − 2 × 363) × 103

185 000/0.437

+518 × 0.992 × 103

2 × 185 000+ 0.31

= 2.86 + 1.39 + 0.31 = 4.56 MPa

PnC =1

4B−2CftC Ac/sin2θs

+D−B cos θs

f ′c Astr3

=1/1000

4×0.414−2×0.3604.56×185 000/0.437 +

1.032−0.414×0.75036.9×54 669

= 1187 kN.

Step 5: Determine the ultimate load Pn from Eq. (35).

∴ Pn = Min(PnA, PnB, PnC) = 1187 kN.

Thus the predicted ultimate load of Pn is 1187 kN. Totalultimate load is 2Pn = 2374 kN. In this example, the ultimateload 2Pexp from experiment was 2559 kN (T1). Thus, the ratioof Pn/Pexp is 0.93.

References

[1] Marti P. Basic tools of reinforced concrete beam design. ACI J 1985;82(1):46–56.

[2] Schlaich J, Schaefer K, Jennewein M. Toward a consistent design ofstructural concrete. PCI J 1987;32(3):74–150.

[3] MacGregor JG. Reinforced concrete: Mechanics and design. PrenticeHall; 1988. p. 848.

[4] Bergmeister K, Breen JE, Jirsa JO. Dimensioning of the nodes anddevelopment of reinforcement. In: Rep IABSE colloquium on structuralconcrete. p. 551–6.

[5] Alshegeir A. Analysis and design of disturbed regions with strut-tiemodels. Ph.D. thesis. West Lafayette (IN): Purdue University; 1992.

[6] Foster SJ, Malik AR. Evaluation of efficiency factor models used instrut-and-tie modeling of nonflexural members. J Struct Eng 2002;128(5):569–77.

[7] ACI Committee 318. Building code requirements for structural concrete(ACI 318-02) and commentary (ACI 318R-02). Farmington Hills (MI):American Concrete Institute; 2002.

[8] Canadian Standards Association. Design of concrete structures: Structuresdesign. Rexdale (ON): Canadian Standards Association; 1994.

[9] CEB-FIP model code 1990. Design code. London: Thomas Telford Ltd;1993.

[10] Mau ST, Hsu TTC. Formula for the shear strength of deep beams. ACIStruct J 1989;86(5):516–23.

[11] Hwang S-J, Lu W-Y, Lee H-J. Shear strength prediction for deep beams.ACI Struct J 2000;97(3):367–76.

[12] Tan KH, Tong K, Tang CY. Direct strut-and-tie model for prestressed deepbeams. J Struct Eng 2001;127(9):1076–84.

[13] Tan KH, Tang CY, Tong K. A direct method for deep beams with webreinforcement. Mag Concr Res 2003;55(1):53–63.

[14] Vecchio FJ, Collins MP. The modified compression-field theory forreinforced concrete elements subjected to shear. ACI J 1986;83(2):219–31.

[15] Belarbi A, Hsu TTC. Constitutive laws of softened concrete in biaxialtension–compression. ACI Struct J 1995;92(5):562–73.

[16] Belarbi A, Hsu TTC. Constitutive laws of concrete in tension andreinforcing bars stiffened by concrete. ACI Struct J 1994;91(4):465–74.

[17] Tang CY, Tan KH. Interactive mechanical model for shear strength ofdeep beams. J Struct Eng 2004;130(10):1534–44.

[18] Kong FK, Robins PJ, Kirby DP, Short DR. Deep beams with inclined webreinforcement. ACI J 1972;69(3):172–6.

[19] ACI Committee 318. Building code requirements for structural concrete(ACI 318-99) and commentary (ACI 318R-99). Farmington Hills (MI):American Concrete Institute; 1999.

[20] Tan KH, Kong FK, Teng S, Guan L. High-strength concrete deep beamswith effective span and shear span variations. ACI Struct J 1995;92(4):395–405.

[21] Smith KN, Vantsiotis AS. Shear strength of deep beams. ACI J 1982;79(3):201–13.

[22] Vecchio FJ, Collins MP. Compression response of cracked reinforcedconcrete. J Struct Eng 1993;119(12):3590–610.

[23] Tan KH, Teng S, Kong FK, Lu HY. Main tension steel in high strengthconcrete deep and short beams. ACI Struct J 1997;94(6):752–68.

[24] Subedi NK, Vardy AE, Kubota N. Reinforced concrete deep beams—some test results. Mag Concr Res 1986;38(137):206–19.

[25] Subedi NK. Reinforced concrete deep beams: A method of analysis. ProcICE J 1988;85:1–30. [Part 2].

[26] Walraven J, Lehwalter N. Size effects in short beams loaded in shear. ACIStruct J 1994;91(5):585–93.

[27] Tan KH, Kong FK, Teng S, Weng LW. Effect of web reinforcement onhigh-strength concrete deep beams. ACI Struct J 1997;94(5):572–82.

[28] Foster SJ, Gilbert RI. Experimental studies on high-strength concrete deepbeams. ACI Struct J 1998;95(4):382–90.

[29] Tan KH, Lu HY. Shear behavior of large reinforced concrete deep beamsand code comparisons. ACI Struct J 1999;96(5):836–45.

[30] Shin S-W, Lee K-S, Moon J-I, Ghosh SK. Shear strength of reinforcedhigh-strength concrete beams with shear span-to-depth ratios between 1.5and 2.5. ACI Struct J 1999;96(4):549–56.

[31] Oh J-K, Shin S-W. Shear strength of reinforced high-strength concretedeep beams. ACI Struct J 2001;98(2):164–73.

[32] Yang K-H, Chung H-S, Lee E-T, Eun H-C. Shear characteristics of high-strength concrete deep beams without shear reinforcements. Eng Struct2003;25(10):1343–52.

[33] Chemrouk M. Slender concrete deep beams: Behaviour, serviceabilityand strength. Ph.D. thesis. Newcastle: The University of Newcastle uponTyne; 1988.

[34] Maimba PP. Effects of support continuity on the behaviour ofslender reinforced concrete deep beams. M.Eng. thesis. Newcastle: TheUniversity of Newcastle upon Tyne; 1989.

[35] Ashour AF. Tests of reinforced concrete continuous deep beams. ACIStruct J 1997;94(1):3–12.

[36] Rogowsky DM, MacGregor JG, Ong SY. Tests of reinforced concretedeep beams. ACI J 1986;83(4):614–23.

[37] Choo QL, Lim HL. Design and behaviour of continuous concretedeep beams. BEng. Final year project report. Singapore: NanyangTechnological University; 1993.

[38] Poh SP. Strength and serviceability of prestressed concrete deep beams.M.Eng. thesis. Singapore: Nanyang Technological University; 1995.

[39] Subedi NK. Reinforced concrete two-span continuous deep beams. ProcInstn Civ Engrs Struct & Bldgs 1998;128(1):12–25.

[40] Asin M. The behaviour of reinforced concrete continuous deep beams.Delft University Press; 1999. p. 168.

zhang 2007

Documents

tie model stm

modified model

stress direction

stress conditions

modified stm

original model

stm components

direct stm