Strength Prediction of Continuous R/ C Deep Beams Using the Strut-and-Tie Method

Strength Prediction of Continuous R/ C Deep Beams Using the Strut-and-Tie Method Mohamed E. El-Zoughiby, Salah E. El- Metwally, Ahmed T. Al-Shora & Essam E. Agieb Arabian Journal for Science and Engineering ISSN 1319-8025 DOI 10.1007/s13369-013-0755-2 1 23

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DOI 10.1007/s13369-013-0755-2 RESEARCH ARTICLE - CIVIL ENGINEERING Strength Prediction of Continuous R/C Deep Beams Using the Strut-and-Tie Method Mohamed E. El-Zoughiby Salah E. El-Metwally Ahmed T. Al-Shora Essam E. Agieb Received: 22 March 2012 / Accepted: 12 July 2013 King Fahd University of Petroleum and Minerals 2013 Abstract In this paper, the current developments on the method of strut-and-tie model are reanalyzed and, then, adjusted for the application to continuous deep beams. Based on concrete strength (normal or high) and the shear spanto-depth ratio, the effectiveness factors of concrete struts are modified. The effectiveness factors for nodal zones are chosen based on the force boundary conditions. The proposed strut-and-tie approach is, then, applied to two different groups of top and bottom loaded continuous deep beams that had been tested experimentally. The obtained results are compared with the experimental results and the findings based on the effectiveness factors of two selected building codes (ACI Code-08 and the UK CIRIA Guide-2). From the comparison, the strut-and-tie method with the proposed modified effectiveness factors gives satisfactory results. Keywords Strut-and-tie Continuous deep beams Shear strength Effectiveness factors M. E. El-Zoughiby S. E. El-Metwally (B) A. T. Al-Shora E. E. Agieb Structural Engineering Department, Mansoura University, Mansoura, Egypt e-mail: selmetwally@yahoo.com 1 Introduction There have been extensive experimental investigations of simply supported deep beams [1 4] but very few tests on continuous reinforced concrete deep beams; therefore, the shear design of reinforced concrete continuous deep beams, in most building codes, is based on tests of simply supported deep beams. In simple deep beams, the region of high shear coincides with the region of low moment. On the other hand, in continuous deep beams, the regions of high shear and high moment coincide and failure usually occurs in these regions. Hence, the failure mechanisms of continuous deep beams are different from those of simply supported deep beams. Continuous deep beams are divided into two major groups according to their loading conditions; top and bottom loading. Top loaded continuous deep beams are commonly used in reinforced concrete buildings, while indirectly loaded or bottom loaded deep beams are widely used as cross-girders in concrete bridges, water tanks, etc. The two groups behave differently under the same applied loads [5 7]. Based on previous studies [3,8] and the findings of solving about 160 beams (simple and continuous deep and short

Table 1 Classification of simple beams according to the shear-span-to-depth ratio (a/d) [7,9] Model name Shear span-to-depth ratio Beam type Model type Type I a. (a/d) 0.9 Deep Direct (one-panel) model b. 0.90 <(a/d) 1.5 Short c. 1.50 <(a/d) 2.00 Type II 1.50 <(a/d) 2.00 Short Two-panel (fan- or arch-action) model Type III (a/d) >2.00 Slender Beam (fan or arch) action model (a)type I (Direct or simply One-Panel) Model (i) Type II arch-action (ii) Type II fan-action (b) Type II (Two-Panel) Model (i) Type III arch-action (ii) Type III fan-action (c) Type III (Beam Action) Model Fig. 1 The strut-and-tie approach for simple deep beams: a Type I direct model; b Type II: i arch-action and ii fan-action, and c Type III: i arch-action and ii fan-action beams subjected to different loadings and boundary conditions) using both the strut-and-tie method and a 3D nonlinear finite element analyses, three strut-and-tie models, on the basis of the shear span-to-depth ratio (a/d), are considered for simple deep beams, namely; Type I, Type II, and Type III [7,9], Table 1 and Fig. 1. Types I and II cover deep and short beams, respectively, and Type III covers slender beams. In Type I, direct strut-and-tie model is utilized, whereas a two-

panel (fan- or arch-action) model is used in Type II. The choice between the two types I and II, in some cases is controlled by the shear span-to-depth ratio (a/d), presence of vertical web reinforcement, and the concrete strength, Table 1 and Fig. 1a, b. The proposed general classification of simple beams is shown in Table 1. In Type I (direct model), the load is transferred from the point load to the support directly by only one compression strut, Fig. 1a. Type II may be divided into two subtypes, namely; arch- and fan-action. In Type II arch-action, Fig. 1b(i), the load is transferred by struts forming an arch mechanism with vertical web reinforcement acting as a single tie. In Type II fan-action, Fig. 1b(ii), the load is transferred to the support by a combination of major and minor compression struts. Herein, hanger reinforcement is required to return the vertical components of forces developed in the minor compression struts to the top of the member. In Type III (where (a/d) exceeds 2.0), the load is transferred across a strut-and-tie system which has an arch- or fan-action model as shown in Fig. 1c(i), c(ii). Considering the effect of negative moment at interior supports, continuous deep and short beams can be, simply, modeled utilizing Type I, Fig. 2, and Type II, Fig. 3, models [7,9]. In this paper, for simplicity sake and because Type II (a) Single top point load (a) Single top point load (b) Two top point loads Fig. 3 Type II strut-and-tie model for continuous deep beams subjected to single top point load; a arch-action model; and b fan-action model model (Fig. 3) requires more complex calculations, only Type I model will be used. Table 2 shows the proposed general classification of continuous beams according to the shearspan-to-depth ratio [7]. In order to assist designers and code writers, simple and general design procedures for modeling Type I of continuous deep beams are developed in this paper considering the recommendations adopted by the 2004 Canadian Code and following references [7,8,10 14]. These beams are modeled in terms of different design parameters such as the shear-spanto-depth ratio, concrete strength, and percentage of web reinforcement; in addition, the effectiveness factors for concrete struts and nodal zones are modified. To verify the reliability of the strut-and-tie method with the proposed effectiveness factors, the obtained results are compared with experimental tests on continuous deep beams, Ashour [6] and Adly [5]. For comparison purposes, the strut-and-tie models are reanalyzed utilizing the effectiveness factors of both the ACI Code-08, Appendix A [15] and the UK CIRIA Guide-2 [16]. (b) Two top point loads 2 Strength Limits of Strut-and-Tie Model Components For the proposed strut-and-tie model, the strength of ties, concrete struts, and nodal zones are as in the following: 2.1 Reinforced Ties (c) Bottom loads Fig. 2 Simplified strut-and-tie model for continuous deep beams, Type I, subjected to: a single top point load; b two top point loads: and c bottom loads In this study, the contribution of tensile strength of a concrete tie is ignored and normally tie forces are carried by reinforcement. The tie cross-section is constant along its length and is obtained from the tie force and the yield strength of the steel. The nominal strength of a tie F nt shall be taken as

Table 2 The proposed classification of continuous beams according to (a/d) [7] Model name Shear span-to-depth ratio Beam type Model type Type I a. (a/d) 1.00 Deep Direct (one-panel) model b. 1.00 <(a/d) 2.00 Short c. 2.00 <(a/d) 2.75 Type II 2.00 <(a/d) 2.75 Short Two-panel (fan- or arch-action) model Type III (a/d) >2.75 Slender Beam (fan or arch) action model F nt = A st f y (1) where A st is the area of the reinforcement and f y is the yield strength of the reinforcement. Depending on the distribution of the tie reinforcement, the effective tie width w t may vary between the following limits: 1. The minimum width for configurations where only one layer of reinforcement is provided in a tie, w t, can be taken as the diameter of the bars, bars, in the tie plus twice the concrete cover, 2c, to the surface of the bars, or simply w t = 2c + bars (2a) Should the tie be wider than this, the reinforcement shall be distributed evenly over the width and thus, w t = 2c + n bars + (n 1)s (2b) where n is the number of steel layers and s is the clear spacing between steel layers. 2. The upper limit of the width is established as the width in a hydrostatic nodal zone, calculated as w t,max = F nt /f n ce (3) where fce n is the applicable effective compressive strength of a nodal zone and is computed from [15,17,18] as fce n = 0.85 f c βn or 0.67 f cuβ n (4) the 0.85 f c stands for a cylinder concrete compressive strength, 0.67 f cu for a cube concrete compressive strength, and β n is the effectiveness factor for nodal zones. 2.2 Concrete Struts The strength of the concrete in a compression stress field depends to a great extent on the multi-axial state of stress and on the disturbance from cracks and reinforcement. The effective compressive strength of the concrete in a strut fce s is obtained from [15,17,18]: fce s = 0.85 f c β s or 0.67 f cu β s (5) where β s is the effectiveness factor for concrete struts. In this study and based on previous studies [20,14,19,21], new values for β s (which considers the effect of shear spanto-depth ratio and concrete strength) have been developed, Table 3. For instance, for Normal Strength Concrete NSC ( f cu 50 MPa), β s is taken equal to 0.75 for the case of Type I, (a/d 10) whereas, for high strength concrete, HSC ( f cu > 50 MPa), β s is taken equal to 0.65. The reduction in is reasonably assumed since it simulates the reduction in concrete strength as a result of transverse tensile strains and agrees with a conclusion suggested by Elzanaty et al. in 1986 [22], which states that, for the case of high strength concrete, the force carried by aggregate interlock is lower than that for conventional strength concretes. In this paper, as already stated in the introduction and as shown in Table 3, for continuous deep and short beams, only Type I is considered. For Type I, based on a strut-and-tie approach and a 3D nonlinear finite element analysis [7], the shear span-to-depth ratio is divided into three ranges; the first range may reasonably represent deep beams limitations, and the last two ranges may reasonably represent short beams limitations. The proposed limitations and the values of effectiveness factor (β s for concrete struts and β n for nodal zones, Tables 3 and 4, respectively) are estimated and verified as follows: Table 3 Proposed values for the β s factor of a concrete strut (continuous deep and short beams) Concrete struts Concrete strength NSC HSC Undisturbed and uniaxial state of stress 1.00 1.00 Type I (a/d) 1.0 0.75 0.65 1.0 <(a/d) 2.0 0.70 0.60 2.0 <(a/d) 2.75 0.65 0.55 Table 4 Proposed values for the β n factor of a nodal zone Nodal zone Compression compression compression, C C C 1.00 Compression compression tension, C C T 0.80 Compression tension tension, C T T a 0.60 Tension tension tension, T T T 0.40 Effectiveness factor a In nodal zones anchoring two or more ties with the presence of one strut

1. Based on previous studies [14,19 21], new values for β s and β n (considering the effects of (a/d), concrete strength, and force boundary conditions) were first reasonably estimated. 2. At second, about 160 tested beams [1,3 6,8] (simple and continuous deep and short beams subjected to different loadings and boundary conditions) were reanalyzed using a 3D nonlinear finite element analysis using ANSYS program [7], in which the effects of (a/d), (L/d), loading and boundary conditions, the amount and arrangement of the main and web reinforcement, and the concrete strength were considered. In the 3D nonlinear finite element analysis, nonlinear constitutive models for concrete and reinforcement were introduced. Concrete was modeled using a 3D reinforced concrete element SOLID65 (8-nodes having 3-translational degrees of freedom at each node), which is capable of cracking in tension and crushing in compression. The main and web reinforcement were modeled using bar elements within the concrete element SOLID65. The obtained results were compared with the experimental results and the comparison was very satisfactory. Some of the basic assumptions and general considerations of the 3D nonlinear analyses were: i. The concrete material was assumed to be initially isotropic. ii. Cracking is permitted in three-orthogonal directions at each integration point. iii. Loads were applied slowly (utilizing the arc-length method) to prevent possible fictituos crushing of the concrete before proper load can occur through a closed crack. 3. At third, the output results of the 3D nonlinear finite element analysis were utilized to construct a strut-and-tie model for each case. The failure load and failure mode could be obtained using a proposed numerical scheme utilizing the values of β s and β n from step #1. 4. At fourth, the output of the strut-and-tie model (failure load and failure mode) was compared with the available test results. The comparison may reveal that the assumed values for β s and β n are acceptable; otherwise, refined estimates are assumed and repeat from step #3. As shown in Table 3, for Type I model, the increase in the shear span-to-depth ratio (a/d) results in a reduction in β s value. The reduction is caused by the decrease of the strut angle which is associated with skew cracks (this justifies why Type I is divided into three sub-types, namely; Type Ia, Ib, and Ic). Also, the increase in concrete strength results in a reduction in β s value. The proposed β s values for continuous deep beams are smaller than that for simple deep beams [7,9]. This is mainly because their failure mechanisms are different. The nominal compressive strength of a strut without longitudinal reinforcement F ns shall be taken as the smaller value of: F ns = fce s A cs (6) at the two ends of the strut, where A cs is the cross-sectional area at one end of the strut, and fce s is the smaller of: 1. The effective compressive strength of the concrete in the strut; β s, based on Table 3. 2. The effective compressive strength of the concrete in the nodal zone; β n, based on Table 4. Design of struts shall be based on F ns F us (7) or in other form (0.85 f c β s)a cs F us (8) where F us is the largest factored force acting in a strut and obtained from the applicable load combinations and the factor is 0.75 for ties, struts, and nodes, ACI-08 [15]. It is noted that the strut width changes linearly between the two nodes. However, in this paper the cross-sectional area of a strut is assumed constant along its length. 2.3 Nodal Zones In this paper, the effective compressive strength of concrete in a nodal zone can be obtained from [15,17,18]: fce n = 0.85 f c β n or 0.67 f cu β n (9) where β n is the effectiveness factor of a nodal zone and it is assumed as given in Table 4. For safety purposes, for compression compression compression (C C C) nodes, the value 1.10 suggested by Schliach et al. in 1991 [14] is reduced to 1.0. Also, in order to appropriately consider the effect of the tensile strain on the concrete compressive strength, the value of 0.80 taken by Schliach et al. [14] for other nodes is replaced by 0.80 for compression compression tension (C C T) nodes, 0.60 for compression tension tension (C T T) nodes (two or more ties), and 0.40 for tension tension tension (T T T) nodes. The nominal compressive strength of a nodal zone, F nn, shall be F nn = fce n A nz (10) where fce n is the effective compressive strength of the concrete in the nodal zone (Table 4) and A nz is the smaller of: 1. The area of the face of the nodal zone on which F u acts, taken perpendicular to the line of action of the strut force.f u.

2. The area of a section through the nodal zone, taken perpendicular to the line of action of the resultant force on the section. In smeared nodes, where the deviation of forces may be smeared or spread over some length, the check of stress is often not critical and it is only required to check the anchorage of the reinforcing bars. On the other hand, singular or concentrated nodes have to be carefully checked. 3. Inclination of struts S 1 and S 2 1 = tan 1 L d L 1 (13) and, 2 = tan 1 If L 1 = a, then L d 2a L 1 (14) 3 Application of Type I Model to Top Loaded Continuous Deep Beams 1 = 2 = tan 1 L d a (15) Figure 4 shows a continuous deep beam (two bays) subjected to a single top load along with the proposed strut-andtie model. From symmetry, the model has two compression struts S 1 ands 2, two tension ties; T 1 to represent the main bottom longitudinal reinforcement, and T 2 to represent the main upper longitudinal reinforcement, and three nodes N 1, N 2, and N 3. The load is transferred directly from the point load at node N 2 to the external and intermediate supports by struts S 1 ands 2, respectively. With reference to Fig. 4, the next notations may be introduced: h, b, and d are the beam height, breadth, and depth, respectively; L 1 (or a when L 1 = a) and L are the shear span and the beam effective length; ai, bi, Ld, and j are the height and length of node i, the internal lever arm, and the angle of strut S j and; w 1 andw 2,are the widths of struts S 1 and S 2 measured perpendicular to the strut centerline, respectively. The width of a strut is always measured perpendicular to its center line. 3.1 Numerical Scheme 1. Input data The terms of beam size (h, b, b 1, b 2, and b 3 ), shear span-to-depth ratio (L 1 /d) or(a/d) when (L 1 = a) and the used concrete and reinforcement strengths ( f c and f y) are defined as input data. 2. The internal lever arm, L d The heights of nodes N 1 and N 2 (a i, i = 1,2) are given by: a i = 2(c + str ) + n bars + (n 1)s (11) where c is the clear concrete cover, str is the stirrup diameter, bars is the longitudinal steel diameter, n is the number of steel layers, and s is the clear spacing between steel layers. Thus, L d = h 0.5(a 1 + a 2 ) (12) 4. Width of struts (w 1 and w 2 ) From the details of Node N 1, shown in Fig. 4c, w 1 may be obtained from: w 1 = a 1 cos 1 + b 1 sin 1 (16) The geometry at node N 2 is more complicated than that at node N 1 and, thus, it can be traced as shown in Fig. 4c. With reference to Fig. 4a, from equilibrium of the model forces we have: S 2n = T 1n + T 2n cos 2 (17) In the strut-and-tie analogy, the tension tie member framing into Node N 1 is extended to the opposite face of that node, Fig. 4c, such that the anchored tension force becomes the compression force on that vertical face. Based on this assumption, for tension tie T 1 : T 1n = (0.85 f c )β n 1 a 1 b = (0.85 f c )(0.80)a 1b (18) the 0.80 represents the β n1 factor of Node N 1 (C C T node). For concrete strut S 2 : S 2n = (0.85 f w 2 b = (0.85 f c )(0.80)w 2b (19) the 0.80 represents the β n2 factor of Node N 2 (C C T node). Similarly, for the tension tie T 2, T 2n = (0.85 f a 2 b = (0.85 f c )(0.80)a 2b (20) and thus, w 2 = a 1 + a 2 cos 2 (21)

Fig. 4 Symbols and details of strut-and-tie model for a top loaded continuous deep beam subjected to single point load: a strut-and-tie model, b details of the strut-and-tie model, c details of nodes N 1 and N 2, and d details of half node N 3 (a) Strut-and-tie model Φ Φ Φ (b) Details of the strut-and-tie model (c) Details of nodes N 1 and N 2 (d) Details of half node N 3 5. STM forces Assuming that the steel bars will reach their yield strength, the tensile forces in the STM elements are as follows: T 1n = A st1 f y1 (22) T 2n = A st2 f y2 (23) From equilibrium of Node 1 S 1n = T 1n cos 1 (24) where A st1 and A st2 are the areas of the bottom and top reinforcement, respectively, f y1 and f y2 are the yield

strengths of the bottom and top reinforcement, respectively. 6. Checking of stress limits (a) Concrete struts Utilizing Table 3, one can simply check if: S 1n (0.85 f c )β s 1 w 1 b or f s 1 ce w 1b (25) S 2n (0.85 f c )β s 2 w 2 b or fce s2 2b (26) If the above checks are not satisfied, assume new adjusted values of either S 1n or S 2n and repeat from Step 5. (b) Nodes Upon the utilization of Table 4, Node N 1 : S 1n (0.85 f c )β n 1 w 1 b or f n 1 ce w 1b (27) V 1n (0.85 f c )β n 1 b 1 b or f n 1 ce b 1b (28) T 1n (0.85 f c )β n 1 a 1 b or f n 1 ce a 1b (29) Node N 2 : S 1n (0.85 f w 1 b or f n 2 ce w 1b (30) S 2n (0.85 f w 2 b or f n 2 ce w 2b (31) T 2n (0.85 f a 2 b or f n 2 ce a 2b (32) P n (0.85 f b 2 b or f n 2 ce b 2b (33) Node N 3 : S 2n (0.85 f c )β n 3 w 2 b or f n 3 ce w 2b (34) V 2n (0.85 f c )β n 3 b 3 b or f n 3 ce b 3b (35) T 1n (0.85 f c )β n3a 1 b or f n3 ce a 1b (36) If the above mentioned checks are satisfied, the required nominal shear capacity can be obtained. Otherwise, assume new adjusted values and repeat from Step 5. Finally, the nominal shear capacity P STM can be obtained from: P STM = 2P n = 2(S 1n sin 1 + S 2n sin 2 ) (37) 3.2 Verification Examples To illustrate how to model and analyze reinforced concrete continuous deep beams, using Type I model, 8-top loaded reinforced concrete continuous deep beams are chosen and examined. These beams were tested by Ashour [6], Table 5. The studied parameters are: (1) shear span-to-depth ratio and (2) the amount of longitudinal reinforcement. These beams are modeled using the strut-and-tie model shown in Fig. 4.To verify the accuracy and consistency of the proposed effectiveness factors (β s for concrete struts and β n for nodal zones), the deep beams are reanalyzed using the strut-and-tie model utilizing the previous numerical scheme considering both the proposed effectiveness factors and those from the ACI Code- 08 and the CIRIA Guide-2. Finally, the obtained results are compared with the experimental results. Case Study 1: Normal Strength Concrete NSC continuous deep beams subjected to single top point loads [6]: 3.3 Numerical Scheme for Beam CDB1 1. Input data Beam size: h = 625 mm, d = 585 mm, b = 120 mm, b 1 = 120 mm, and b 2 = b 3 = 250 mm. Shear span-to-depth ratio: L 1 = a = 660 mm, and (a/d) = (660/585) = 1.12 Materials: f c = 30.6 MPa, f y 1 = f y2 = 480 MPa, A s1 = 4 12, and A s2 = 4 12+2 10. 2. The internal lever arm, L d Utilizing the following relation: a i = 2(c + str ) + n bars + (n 1)s Table 5 The basic characteristics of the tested top loaded deep beams [6,7] Beam f c, MPa h, mm b, mm a, mm L, mm (a/d) CDB1 30.60 625 120 660 1,320 1.12 CDB2 33.70 625 120 660 1,320 1.12 CDB3 22.40 625 120 660 1,320 1.12 CDB4 28.50 625 120 660 1,320 1.12 CDB5 29.20 625 120 660 1,320 1.10 CDB6 23.00 425 120 660 1,320 1.70 CDB7 27.20 425 120 660 1,320 1.70 CDB8 24.00 425 120 660 1,320 1.65

The terms a 1 and a 2 are as follows: a 1 = 2(15 + 8) + 2 12 + 1 10 = 80 mm, a 2 = 2(10 + 8) + (2 12 + 1 10) + 2 10 = 90 mm, and thus L d = h 0.5(a 1 + a 2 ) = 540 mm 3. Inclination of struts S 1 and S 2 1 = 2 = tan 1 L d a = 39.2 4. Width of struts w 1 = a 1 cos 1 + b 1 sin 1 = 137 mm, and w 2 = a 1 + a 2 = 219 mm cos 2 5. STM forces Assuming that: T 1n = A st1 f y1 = 4 113.14 480 = 217.10 kn, and T 2n = A st2 f y2 = (4 113.14) + (2 78.5) 480 = 292.50 kn The forces S 1n and S 2n are thus: S 1n = T 1n cos 1 = cos 217.1 39.2 = 280.10 kn, and S 2n = T 1n+T 2n cos 2 = 217.1+292.5 cos 39.2 = 658 kn 6. Checking of stress limits Utilizing Tables 3 and 4 and knowing that f c = 30.6 MPa, the term ( f ce = 0.85 f c β) will be: f s 1 ce = 0.85 f c β s 1 = 0.85 30.6 0.70 = 18.20 MPa for Strut S 1 fce s2 = 0.85 f c β s2 = 0.85 30.6 0.70 = 18.20 MPa for Strut S 2 f n 1 ce = 0.85 f = 0.85 30.6 0.80 = 20.80 MPa for Node N 1 f n 2 ce = 0.85 f = 0.85 30.6 0.80 = 20.80 MPa for Node N 2 fce n3 = 0.85 f c β n3 = 0.85 30.6 0.80 = 20.80 MPa for Node N 3. Concrete struts, S 1 and S 2 : S 1n,max = 299.20 kn which is the smaller value of: 18.20 137 120 = 299.2 kn (for the strut itself) and 20.80 137 120 = 341.9 kn (at the weaker end, both Nodes N 1 and N 2 have the same strength) S 2n,max = 478.40 kn which is the smaller value of: 18.20 219 120 = 478.4 kn (for the strut itself) and 20.80 219 120 = 546.6 kn (at the weaker end, both Nodes N 2 and N 3 have the same strength) Since S 2n > S 2n,max, which is unsafe, try S 2n = S 2n,max = 478.40 kn and upon applying S 2n = T 1n + T 2n cos 2 = T 1n + T 2n cos 39.2 = 478.40 kn which gives T 1n + T 2n = 370.70 kn and knowing that: T 1n = A st 1 = 0.74 T 2n A st2 Thus, T 1n = 157.60 kn, T 2n = 213.0 kn,s 1n = 203.40 kn, and the nominal shear capacity P STM is P STM = 2P n = 2(S 1n sin 1 + S 2n sin 2 ) = 2(203.40 + 478.40) sin 39.2 = 861.80 kn P STM /P EXP = 861.80 1100 = 0.78 where P EXP is the experimental failure load. Nodes N 1, N 2 and N 3 : Node 1 Based on the recommendations of past researchers [23], it was determined that it is unnecessary to apply the bonding stresses from a developed bar to the back face of a CCT node. Therefore, only directly applied stresses, such as those due to bearing of a plate or to an external indeterminacy, are applied to the back face of CCT nodes and checked with the 0.80 effectiveness factor. Knowing that: 0.85 f = 20.80 MPa (at Node N 1 ) S 1n = 203.40 1000 = (0.85 f )w 1 b = f n 1 ce 137 120 which results in f n 1 ce =12.37 < 20.80 MPa, Okay. Similarly, V 1n = 203.40 sin 39.2 1000 = (0.85 f )b 1 b = f n 1 ce 120 120 which results in f n 1 ce =8.93 < 20.80 MPa, Okay, and T 1n = 157.6 1000 = (0.85 f )a 1 b = f n 1 ce 80 120 which results in f n 1 ce = 16.42 < 20.80 MPa, Okay. Node 2 Knowing that: 0.85 f = 20.80 MPa (at Node N 2 ) S 1n = 203.40 1000 = (0.85 f )w 1 b = f n 2 ce 137 120 which results in f n 2 ce = 12.37 < 20.80 MPa, Okay. Similarly, S 2n = 478.4 1000 = (0.85 f )w 2 b = f n 2 ce 219 120 which results in f n 2 ce =18.20 < 20.80 MPa, Okay,

T 2n = 213.0 1000 = (0.85 f )a 2 b = f n 2 ce 90 120 which results in f n 2 ce =19.72 < 20.80 MPa, Okay, and P n = (213.40 + 478.40) sin 39.2 1000 = (0.85 f c β n 2 )b 2 b = f n 2 ce 250 120 which results in f n 2 ce = 14.36 < 20.80 MPa, Okay. Node 3 Knowing that: 0.85 f c β n3 = 20.80 MPa (at Node N 3 ) S 2n = 478.4 1000 = (0.85 f c β n3)w 2 b = fce n3 219 120 which results in f n3 ce = 18.20 < 20.80 MPa, Okay, V 2n = 2 478.40 sin 39.2 1000 = (0.85 f c β n3)b 3 b = fce n3 2 250 120 which results in f n3 ce = 20.15 < 20.80 MPa, Okay, and T 1n = 157.6 1000 = (0.85 f c β n3)a 1 b = f n3 ce 80 120 which results in f n3 ce = 16.42 < 20.80 MPa, Okay, and, thus, P STM /P EXP = 861.80 1100 = 0.78. In this paper, for a strut-and-tie model, two types of failure mode are considered, namely; Shear failure (the crushing of concrete in a strut or at the face of a node) Tension failure (yielding of a tie or anchorage failure of a tie). Herein, because failure had occurred due to the crushing of the concrete strut S 2 it is a diagonal shear, or simply, shears failure. Upon following the previous numerical scheme, the failure load and the corresponding failure mode of all other cited top loaded continuous deep beams are determined and presented in Table 6. As shown in the table, the strut-and-tie method gives a mean value of 88 % of the ultimate experimental value and a standard deviation of 0.14. Obviously, these values show the accuracy and consistency of the method. Considering the effectiveness factors of the ACI Code-08 gives a mean value of 0.86 and a higher standard deviation SD of 0.23. In addition, the predictions are unsafe for two beams. Considering the effectiveness factors of the CIRIA Guide-2 gives unsafe predictions for six beams. Neglecting the high ratio of web reinforcement in beams CDB1 and CDB6, the strut-and-tie model gives lower ultimate loads compared with the available test results. Ignoring the horizontal web reinforcement in beam CDB3 slightly affects the ultimate load obtained from the strut-and-tie method. 4 Application of Type I Model to Bottom Loaded Continuous Deep Beams The behavior of a continuous deep beam depends greatly on its loading and boundary conditions. The indirectly or bottom loaded continuous deep beams behave in a manner different from that of top loaded beams. Indirect application of loads changes the mechanism of load transfer, the failure mode, and the role of reinforcement in continuous deep beams. For bottom loaded continuous beams (based on an experimental work [5], a 3D nonlinear finite element analysis [7], and demonstrated by strut-and-tie models), it is found that: The beams exhibit lower ultimate loads than that for top loaded beams. Introductory bars (compression struts S in Fig. 5) are needed to transmit the external bottom load to the top of the model. This justifies the appearance of compression forces Table 6 The strut-and-tie method results compared with selected building codes [15,16] for top loaded deep beams No. Beam f c, MPa P EXP, kn Failure mode, tests P STM / P EXP Failure mode, STM Proposed STM ACI-08 App. A CIRIA Guide 2 1 CDB1 30.60 1100 Shear 0.78 1.11 0.96 Shear 2 CDB2 33.70 950 Shear 0.99 0.85 1.05 Shear 3 CDB3 22.40 570 Shear 1.10 1.23 1.57 Shear 4 CDB4 28.50 885 Shear 0.90 0.57 1.05 Shear 5 CDB5 29.20 820 Shear 0.68 0.97 0.98 Tension 6 CDB6 23.00 495 Shear 0.77 0.86 1.09 Shear 7 CDB7 27.20 445 Shear 1.02 0.62 1.14 Shear 8 CDB8 24.00 385 Shear 0.80 0.70 1.14 Shear Mean 0.88 0.86 1.12 SD 0.14 0.23 0.19

Fig. 5 Symbols and details of strut-and-tie model for a bottom loaded continuous deep beam: a strut-and-tie model, b details of the strut-and-tie model, and c details of nodes N 1 and N 2 (a) Strut-and-tie model (b) Details of the strut-and-tie (c) Details of nodes N 1 and N 2 (needed for equilibrium) in both failure crack pattern and strain and stress distributions from the 3D nonlinear finite element analysis [7]. Figure 5 shows a continuous deep beam (two bays) with two point bottom loads at nodes N 3 and N 4 along with the proposed strut-and-tie model. The model has three compression struts (S 1, S 2, and S 3 ), five tension ties (T 1, T 2, T 3, T 4, and T 5 ), and six nodes (N 1, N 2, N 3, N 4, N 5 and N 6 ). An additional concrete strut S(which is required for equilibrium of the model forces and appears in both failure crack pattern and strain and stress distribution from the 3D nonlinear finite element analysis for bottom loaded continuous deep beams [7]) is introduced. This proposed compression strut is also recommended by Adly [5]. The main bottom longitudinal reinforcement is represented by the tension ties T 1 and T 3 and the main top longitudinal reinforcement is represented by the tension tie T 2. The introductory bars are represented by the tension ties T 4 and T 5. The load is transferred indirectly from the points of load application at nodes N 3 and N 4 by

these introductory bars to the external and the intermediate supports through the two struts S 1 and S 2, respectively. 4.1 Numerical Scheme 1. Input data The terms of beam size (h, b, b 1, and b 2 ), shear span-to-depth ratio (a/d) and the used concrete and reinforcement strengths ( f c and f y) are defined as input data. 2. The internal lever arm, L d The heights a 1 and a 2 are given by: a i = 2(c + str ) + n bars + (n 1)s (11) where c is the clear concrete cover, str is the stirrup diameter, bars is the longitudinal steel diameter, n is the number of steel layers, and s is the clear spacing between steel layers. Thus, L d = h 0.5(a 1 + a 2 ) (12) 3. Inclination of struts S 1,S 2, and S: 1 = tan 1 L d L 1 (38) 2 = tan 1 L d L 2 (39) and = tan 1 L d L (L 1 + L 2 ) If L 1 = L 2 = a, then 1 = 2 = tan 1 L d a and = tan 1 L d L (2a) (40) (41) (42) 4. Width of struts (w 1, w 2, and w 3 ) From the details of nodes, the following widths can be obtained: w 1 = a 1 cos 1 + b 1 sin 1 (43) w 2 = a 1 cos 2 + 0.50b 2 sin 2 (44) w 3 = a 2 (45) 5. STM forces Assuming that the steel bars will reach their yield strength, the tensile forces in the STM elements are as follows: T in = A sti f yi (46) where A sti and f yi are the area of the reinforcement and the yield strength in the element i, (i = 1 5). Upon applying equilibrium of STM forces, we have P 1n = P 2n = P n (47) P 1n = T 4n (48) T 4n = S 1n sin 1 + S n sin (49) P 2n = T 5n S n sin (50) T 1n S 1n = (51) cos 1 P 1n = T 4n = T 1n sin 1 + S n sin 1 cos 1 = T 1n tan 1 + S n sin (52) Since, P 1n = P 2n (53) Then, 1 S n = (2sin) (T 5n T 1n tan 1 ) (54) S 2n = T 5n sin 2 (55) S 3n = S 1n cos 1 S n cos (56) T 2n = S 2n cos 2 S 3n (57) T 3n = T 1n S n cos (58) where P 1 and P 2 are the external bottom point loads at nodes N 3 andn 4, respectively. 6. Checking of stress limits a. Concrete struts Utilizing Table 3, one can simply check if: S 1n (0.85 f c )β s 1 w 1 b or f s 1 ce w 1b (59) S 2n (0.85 f c )β s2w 2 b or fce s2 2b (60) If the above checks are not satisfied, assume new adjusted values of either S 1n or S 2n and repeat from Step 5. b. Nodes Upon the utilization of Table 4, Node N 1 S 1n (0.85 f c )β n 1 w 1 b or f n 1 ce w 1b (61) V 1n (0.85 f c )β n 1 b 1 b or f n 1 ce b 1b (62) T 1n (0.85 f c )β n 1 a 1 b or f n 1 ce a 1b (63) Node N 2 S 2n (0.85 f w 2 b or f n 2 ce w 2b (64) T 3n (0.85 f a 1 b or f n 2 ce a 1b (65) V 2n (0.85 f b 2 b or f n 2 ce b 2b (66)

If the above mentioned checks are satisfied, the required nominal shear capacity can be obtained. Otherwise, assume new adjusted values and repeat from Step 5. Finally, the nominal shear capacity can be obtained from: P STM = P 1n = P 2n = P n (67) 4.2 Verification Examples Five bottom loaded continuous deep beams had been tested by Adly [5]. In which, the considered main parameters are the shear span-to-depth ratio and the web reinforcement ratio. The main input data for these beams are as given in Table 7 [5,7]. These beams are used to verify the strut-and-tie model shown in Fig. 5. Case Study 2: Normal Strength Concrete NSC continuous deep beams subjected to two point bottom loads [5]: 4.3 Numerical Scheme for Beam S1 1. Input data Beam size: h = 1000 mm, d (top steel) = 931 mm, d (bottom steel) = 940 mm, d (average) = 935.5 mm, b = 100 mm, b 1 = 150 mm, b 2 = 400 mm, andl = 2000 mm. Shear span-to-depth ratio: L 1 = L 2 = a = 600 mm, and ((a/d)) = 0.644 Materials: f c = 22.0 MPa, f y( 8 ) = 312.7 MPa, f y ( 10 ) 438.1 MPa, and f y ( 12 ) = 483.0 MPa, A s1 = A s3 = 2 8 + 2 10, A s2 = 2 8 + 3 10, and A s4 = A s5 = (1 12 + 1 10) per side. 2. The internal lever arm, L d Utilizing the following relation: a i = 2(c + φ str ) + nφ bars + (n 1)s The terms a 1 and a 2 are as follows: a 1 = a 2 = 2(25 + 6) + (8 + 10) + 1 10 = 90 mm, and thus, L d = h 0.5(a 1 + a 2 ) = 1000 0.50(90 + 90) = 910 mm 3. Inclination of struts S 1,S 2, and S 1 = 2 = tan 1 Ld 910 a = tan 1 600 = 56.6 and inclination of strut S or is given by: = tan 1 910 2000 2 600 = 48.6 4. Width of struts w 1 = a 1 cos 1 +b 1 sin 1 = 90 cos 56.6+150 sin 56.6 = 174.8 mm w 2 = a 1 cos 2 + 0.50b 2 sin 2 = 90 cos 56.6 + 0.50 (400) sin 56.6 = 216.5 mm, and w 3 = a 2 = 90 mm 5. STM forces Assuming that: T 1n = A st1 f y1 = 2 50 312.7 + 2 78.57 438.1 = 100.0 kn and T 5n = A st5 f y5 = (2 113.14) 483 + (2 78.57) 438.1 = 178 kn, the STM forces are thus: S n = T 4n = 1 2sin (T 5n T 1n tan 1 ) 1 = (178 100.0 tan 56.6) = 17.55 kn, 2sin48.6 S 1n = T 1n = 100.0 = 181.65 kn, cos 1 cos 56.6 S 2n = T 5n = 178.0 213.2 kn, sin 2 sin 56.6 S 3n = S 1n cos 1 S n cos = 181.65 cos 56.6 17.55 cos 48.6 = 88.4 kn, and P 1n = P 2n = P n = T 1n tan 1 + S n sin = 100 tan 56.6 + 17.55 sin 48.6 = 164.8 kn 6. Checking of stress limits Utilizing Tables 3 and 4 and knowing that f c = 22.0 MPa, the term ( f ce = 0.85 f c β) will be: f s 1 ce = 0.85 f c β s 1 = 0.85 22.0 0.75 = 14.025 MPa for Strut S 1 fce s2 = 0.85 f c β s2 = 0.85 22.0 0.75 = 14.025 MPa for Strut S 2 fce s3 = 0.85 f c β s3 = 0.85 22.0 1.00 = 18.70 MPa for Strut S 3 Table 7 The basic characteristics of the tested bottom loaded deep beams [5,7] Beam f c, MPa h, mm b, mm a, mm L, mm (a/d) S1 22.00 1000 100 600 2000 0.644 S3 22.00 1000 100 400 2000 0.428 S4 20.30 1000 100 800 2000 0.854 S5 23.00 1000 100 600 2000 0.644 S6 20.20 1000 100 600 2000 0.644

f n 1 ce = 0.85 f = 0.85 22.0 0.80 = 14.96 MPa for Node N 1 f n 2 ce = 0.85 f = 0.85 22.0 0.80 = 14.96 MPa for Node N 2 Concrete struts S 1 to S 3 : S 1n,max = 174.8 100 14.025 = 245.60 kn this is greater than 181.65 kn, Okay. S 2n,max = 216.5 100 14.025 = 303.64 kn this is greater than 213.20 kn, Okay. S 3n,max = 90 100 18.70 = 168.30 kn this is greater than 88.40 kn, Okay, and thus, P STM /P EXP = 1634.8 = 0.66 250 Nodes N 1 and N 2 : Node 1 Knowing that: 0.85 f = 14.96 MPa (at Node N 1 ) S 1n = 181.65 1000 = (0.85 f )w 1 b = f n 1 ce 174.8 100 which results in f n 1 ce = 10.39 < 14.96 MPa, Okay. Similarly, V 1n = S 1n sin = 181.65 sin 56.6 1000 = (0.85 f )b 1 b = f n 1 ce 150 100 which results in f n 1 ce =10.11 < 14.96 MPa, Okay, and T 1n = 1000 1000= (0.85 f )a 1 b = f n 1 ce 90 100 which results in f n 1 ce =11.11 < 14.96 MPa, Okay. Node 2 Knowing that: 0.85 f = 14.96 MPa (at Node N 2 ) S 2n = 213.2 1000 = (0.85 f )w 2 b = f n 2 ce 216.5 100 which results in f n 2 ce = 9.85 < 14.96 MPa, Okay. Similarly, V 2n = 2S 2n sin = 2 213.2sin56.6 1000 = (0.85 f )b 2 b = f n 2 ce 400 100 which results in f n 2 ce = 8.89 < 14.96 MPa, Okay, and T 3n = T 1n S n cos 1 = 100 17.55 cos 48.6 = 88.4 kn this is less than T 3n,max = 100 kn, Okay. T 3n = 88.40 1000 = (0.85 f )a 1 b = f n 2 ce 90 100 which results in f n 2 ce = 9.8 < 14.96 MPa, Okay. To sum up, upon following the previous numerical scheme, the failure load and failure mode for all other cited beams were obtained and the results are shown in Table 8. As Table 8 illustrates, the strut-and-tie method gives a mean value of 0.68 and an average SD of 0.091. Considering the effectiveness factors of the ACI Code-08 [15] gives a lower mean value of 0.55 and a higher SD of 0.077. Considering the effectiveness factors of the CIRIA Guide-2 [16] gives a mean value of 0.615 and a SD of 0.01. This shows that the strutand-tie model is more accurate and consistent than the other approaches in comparison. 5 Conclusions From this study, the following conclusions can be drawn: 1. Considering the effect of negative moment at interior supports, continuous deep beams can be simply modeled utilizing Type I and Type II models of simple deep beams. For simplicity sake, only type I (direct, in which the web reinforcement is ignored as a safety margin) is considered in this paper. Type I is divided into three sub-types, namely; Type Ia, Ib, and Ic. Type Ia covers deep beams and Types Ib and Ic cover short beams. The choice between the three sub-types depends on the shear span-to-depth ratio. 2. Based on previous studies, the findings of solving about 160 beams, and a 3D nonlinear finite element analysis, effectiveness factors for concrete struts are proposed for Types Ia, Ib, and Ic for continuous deep beams as a function of the shear span-to-depth ratio and the concrete Table 8 The strut-and-tie method results compared with selected building codes [15,16] for bottom loaded deep beams No. Beam f c, MPa P EXP,kN Failure mode, tests P STM / P EXP Failure mode, STM Proposed STM ACI-08 App. A CIRIA Guide 2 1 S1 22.00 250 Shear 0.655 0.536 0.636 Shear 2 S3 22.00 270 Shear 0.540 0.463 0.626 Shear 3 S4 20.30 220 Shear 0.72 0.582 0.623 Shear 4 S5 23.00 275 Shear 0.77 0.665 0.589 Shear 5 V6 20.20 260 Shear 0.737 0.507 0.604 Shear Mean 0.68 0.550 0.615 SD 0.091 0.077 0.018

strength. The increase in the shear span-to-depth ratio results in a reduction in the effectiveness factor value.this is due to the decrease in the strut angle. Also, the increase in the concrete strength results in a reduction in the effectiveness factor value, since the force carried by aggregate interlock is lower than that for normal strength concrete. 3. The effectiveness factors for nodal zones are chosen based on the force boundary conditions. 4. It should be noted that the proposed methodology is limited to top and bottom loaded continuous deep and short beams Type I Model only. In addition, the validation of the model is made considering a limited number of test results. It generally leads to more accurate results for top loaded deep beams than for bottom loaded beams. Experimental work is needed, especially for bottom loaded beams, to obtain more reliable values of the effectiveness factors. 5. Increasing the concrete strength results in an increase in the ultimate shear strength especially when the shear spanto-depth ratio is less than 1.0. 6. Loading direction has a major effect on the behavior and the ultimate capacity of continuous deep beams. Bottom loaded beams exhibit lower ultimate loads than top loaded beams. 7. Introductory bars are needed to transmit the external bottom load to the top of the model. This is justified by the appearance of compression forces in both failure crack pattern and strain and stress distributions using a 3D nonlinear finite element analyses. 8. In comparison with the findings (utilizing only the effectiveness factors) of the ACI Code-08 and the CIRIA Guide 2, the proposed effectiveness factors lead to more refined results. References 1. Foster, S.J.; Gilbert, R.I.: Experimental studies on high-strength concrete deep beams. ACI Struct. J. 95(4), 382 390 (1998) 2. Smith, K.N.; Vantsiotis, A.S.: Shear strength of deep beams. ACI Struct. J. 79(3) 201 213 (1982) 3. Tan, K.H.; Weng, L.W.; Teng, S.: A strut-and-tie model for deep beams subjected to combined top-and-bottom loading. Struct. Eng. 75(13), 215 222 (1997) 4. Tan, K.H.; Lu, H.Y.; Teng, S.: Shear behavior of large reinforced concrete deep beams and code comparisons. ACI Struct. J. 96(5), 836 845 (1999) 5. Adly, A.F.; Behavior of bottom loaded continuous deep beams. Ph. D. Thesis, Cairo University, Faculty of Engineering, Egypt (1999) 6. Ashour, A.F.: Tests of reinforced concrete continuous deep beams. ACI Struct. J. 94(1), 3 12 (1997) 7. El-Shora, A.T.: Design and detailing of deep beams. PhD Thesis, Mansoura University, Egypt (2005) 8. Rogowsky, D.M.; MacGregor, J.G.; Ong, S.Y.: Tests of reinforced concrete deep beams. ACI J. Proc. 83(4), 614 623 (1986) 9. El-Zoughiby, M.E.; El-Metwally, S.E.; El-Shora, A.T.; Agieb, E.E.: Strength prediction of simply supported r/c deep beams using the strut-and-tie method. Arab. J. Sci. Eng. AJSE. 38(8), 1973 1991 (2013) 10. Arabzadeh, A.; Rahaie, A.R.; Aghayari A.: A simple strut-and-tie model for prediction of ultimate shear strength of rc deep beams. Int. J. Civ. Eng. 7(3), 141 153 (2009) 11. Collins, M.P.; Mitchell, D.: Shear and torsion design of prestressed and non-prestressed concrete beams. PCI J. 25(4), (1980) 12. Macgregor, J.G.; Wight, J.K.: Reinforced Concrete: Mechanics and Design, 6th edn. Pearson Prentice Hall, Upper Saddle River, NJ (2012) 13. Ramirez, J.A.; Alshegeir, A.: Strut-tie approach in pretensioned deep beams. ACI J. Proc. 89(3), 296 304 (1992) 14. Schlaich, J.; Schäfer, K.: Design and detailing of structural concrete using strut-and-tie models. Struct. Eng. 96(6), 113 125 (1991) 15. ACI Committee 318: Building Code Requirements for Structural Concrete (318-2008) and Commentary (318R-2008). American Concrete Institute, Framington Hills (2008) 16. CIRIA Guide 2: The design of deep beams in reinforced concrete. Over arup and partners and construction industry research and information association, London (1977, reprinted 1984) 17. Egyptian Code: The Design and Construction of RC Structures, Egypt (2006) 18. European Committee for Standardization Euro code 2: Design of concrete structures (2004) 19. Ashour, A.F.; Morley, C.T.; Effectiveness factor of concrete in continuous deep beams. ASCE J. Struct. Eng. 122(2), 169 178 (1996) 20. Schlaich, J.; Schäfer, K.; Jennewein, M.: Toward a consistent design of structural concrete. J. Prestress. Concrete Inst. 32(3), 74 150 (1987) 21. Schlaich, J.; Schäfer, K.: The design of structural concrete. IABSE Workshop, New Delhi (1993) 22. Elzanaty, A.H.; Nilson, A.H.; Slate, F.O.: Shear capacity of reinforced concrete using high-strength concrete. ACI J. Proc. 83(2), 290 296 (1986) 23. Thompson, M.K.; Young, M.J.; Jirsa, J.O.; Breen, J.E.; Klingner, R.E.: Anchorage of headed reinforcement in cct nodes. Research Report 1855-2. Center for Transportation Research, University of Texas at Austin, Austin (2003)