A Proposed Model and Performance Study on Prefabricated Cage-Reinforced Self-compacting Concrete Deep Beams

Abstract

High-rise buildings, bridges, pile foundations, and offshore structures comprise deep beams as an important structural component for transferring heavy loads. The modern era of construction demands speedy construction which led to the need for a change in reinforcement system known as a prefabricated cage system (PFCS). This study focuses on the application of PCS in deep beam construction using self-compacting concrete (SCC). The experimental investigation has been carried out by testing twelve deep beams, out of which two deep beams have been constructed with conventional reinforcement and ten deep beams have been constructed with prefabricated cages. The experimental behaviour of deep beams has been examined with different web reinforcement configurations and shear–span to depth ratios of 0.5, 0.75, and 1. The findings showed that as the a/D ratio rises, the failure mode shifts to flexural shear. Prefabricated cage-reinforced deep beams, incorporating both vertical and horizontal web reinforcement, have demonstrated higher ultimate strength ranging from 7.1% to 10.6% compared to conventional deep beams. A reserve strength factor of 0.45 indicates good reserve strength efficiency. Moreover, an increasing trend in displacement ductility and a decreasing trend in energy absorption capacity have been observed with the increase in the a/D ratio. The energy absorption capacity of PCS-reinforced deep beams has been observed to be in the range of 20.11% to 33.38% higher than conventional ones. The proposed equation for predicting the ultimate strength of prefabricated cage-reinforced deep beams is conservative, while the ACI 318–2019 equation slightly overestimates the ultimate strength. Thus, PCS represents an efficient construction method for deep beams, offering both commendable ultimate strength and ductility.

Appendices

Appendix

Solved Numerical Example for Predicting Ultimate Strength Using The Proposed Model for the Specimen I-PCR-0.5-VH2

The details of reinforcement and the method to calculate ‘y’ is shown in Fig.

Fig. 26

Method to calculate ‘y’ for specimen I-PCR-0.5-VH2

26

$${\text{b}} = {13}0\;{\text{mm}};\;{\text{D}} = {5}00\;{\text{mm}};\;{\text{f}}_{{\text{t}}} = 0.{1}\;{\text{f}}_{{{\text{cu}}}} = {4}.{8}\;{\text{N}}/{\text{mm}}^{{2}} ;\;{\text{x}} = {25}0\;{\text{mm}};\;\frac{x}{D} = 0.{5};\;{\text{K}}_{{1}} = {1}.{44};\;{\text{K}}_{{2}} = {39}0\;{\text{N}}/{\text{mm}}^{{2}} .$$

The predicted shear strength equation for prefabricated cage-reinforced deep beams using the proposed model is given below

$${\text{V}}_{{\text{u}}}^{{\text{ pred}}} { = }K_{1} \left( {1 - 0.35\frac{{\text{x}}}{{\text{D}}}} \right){\text{f}}_{{\text{t}}} {\text{bD}} + { }K_{2} \mathop \sum \limits_{{\text{n}}} {\text{A}}\frac{{\text{y}}}{{\text{D}}}{\text{ Sin}}^{2} {\upalpha }$$

Referring to Table 5

A₁ = Area of longitudinal reinforcement = 600 mm², A₂ = A₃ = Area of horizontal web reinforcement = 272 mm², A₄ = Area of compression reinforcement = 216 mm², A₅ = Area of vertical web reinforcement = 136 mm.²

$${\rm A}_{{1}} = \alpha_{{2}} = \alpha_{{3}} = \alpha_{{4}} = {65}.{43};\alpha_{{5}} = {24}.{57}$$

By measurement from the drawing in Fig. 26

y₁ = 454.5 mm; y₂ = 301 mm; y₃ = 159 mm; y₄ = mm; y₅ = 280 mm

$$\begin{aligned} V_{u}^{pred} = & 1.4 \left( { 1 - 0.35*0.5} \right)4.05*130*500 + 390 \mathop \sum \limits_{5 } \frac{y}{500} sin^{2} \alpha \\ = & {3}0{4}0{53} + {266245} \\ \end{aligned}$$

$$V_{u}^{pred} = {57}0\;{\text{kN}}$$

2.1 A. 2 Solved Numerical Example for Predicting Ultimate Strength using ACI 318-2019 Shear Strength Equation for the Specimen I-PCR-0.5-VH2

Figure 

Fig. 27

Idealized STM model

27 represents the idealized STM model for the Specimen I-PCR-0.5-VH2.

a = 250 mm; D = 500 mm; F_u = 356 kN; f^’_c = 38.76 N/mm²;

2.1.1 Calculation of jd

As shown in Fig. 27, By equating forces in strut and tie, F_u BC = F_u AD, we get $w_{t}$ = 1.25 $w_{t}{\prime}$ (or) ${\text{w}}_{{\text{t}}}{\prime} = 0.8 {\text{w}}_{{\text{t}}}$.

From Fig. 27, jd = $D - 1.125w_{t}{\prime}$.

At node B,

$${\text{F}}_{{\text{u}}} {\text{BC}} = {\text{f}}_{{{\text{ce}}}} {\text{A}}_{{{\text{cs}}}} = \phi \left( {0.{{85\beta }}_{{\text{s }}} {\text{f}}_{{\text{c}}}{\prime} { }} \right){\text{b}}w_{t}{\prime} = 0.{85}*{1}*{38}.{76}*{13}0*w_{t}{\prime}$$

where ${\upbeta }_{{\text{s }}}$ = 1 (C–C-C node) and ϕ is the strength reduction factor = 0.75.

From Fig. 

Fig. 28

Free body diagram

28, By taking moment about equilibrium at point A

$$\begin{aligned} & {25}0\;{\text{F}}_{{\text{u}}} {-}{\text{F}}_{\text{u}} {\text{BC}}\left( {{\text{jd}}} \right) = 0 \\ & 500\left( {{356}} \right){-}0.{75}*0.{85}*{1}*{38}.{76}*{13}0*w_{t}^{\prime} \left( {500{-}{1}.{125}w_{t}^{\prime} } \right) \\ & {3613}.{76}w_{t}^{{\prime}2} - {1606117.5}w_{t}^{\prime} + {89}0*{1}0^{{5}} \\ \end{aligned}$$

Solving this equation, $w_{t}{\prime}$ = 64.89 mm and $w_{t}$ = 81.11 mm.

2.1.2 Calculation of ${{\varvec{\uptheta}}}$.

From the Values of jd, the value of ${\uptheta }$ is obtained as

$$\begin{aligned} & \theta = \tan^{ - 1} \frac{jd}{a} \\ & \theta = 59.65{^\circ } \\ \end{aligned}$$

2.1.3 Calculation of width of strut

$$\begin{aligned} & w_{st } = \frac{{\left( {{\text{w}}_{{\text{t}}}^{\prime} + {\text{w}}_{{\text{t}}} } \right){{cos\theta }} + \left( {{\text{l}}_{{{\text{ps}}}} + {\text{l}}_{{{{pl}}}} } \right){{sin\theta }}}}{2} \\ & \quad \quad \quad = \frac{{1. *81.11*cos 59.65 + \left( {130 + 130} \right)sin 59.65}}{2} \\ & w_{st } = {149}.0{7}\;{\text{mm}} \\ \end{aligned}$$

$${\text{For}}\;{\text{bottle}}\;{\text{shaped}}\;{\text{struts}},\;{\upbeta }_{{\text{s }}} = 0.75$$

$$\begin{aligned} {\text{Strength}}\;{\text{of}}\;{\text{strut}},{\text{F}}_{{{\text{ns}}}} & = (0.{{85\beta }}_{{\text{s }}} {\text{f}}_{{\text{c}}}^{\prime} { })\;{\text{b}}w_{st} \\ & = 0.{85}*0.{75}*{38}.{76}*{13}0*{149}.0{7} = {478}.{85}\;{\text{kN}} \\ \end{aligned}$$

2.1.4 Calculation of V_u ^ACI

To calculate the shear load in the strut model AB, the node B is considered and by applying the hydrostatic condition the actual load taken by the strut AB is calculated. Consider the Fig. 

Fig. 29

Resolving of forces in node B

29 in which the forces acting on the node B has been shown along with the dimensional details on the node and strut AB.

The load acting on the strut AB is 478.85 kN. F_AB is acting on the face q-s^’, which is used to find the stress on q-s

$${\text{Considering}}\;{\text{triangle}}\;{\text{qrs}},{\text{qs}} = \sqrt {130^{2 } + 81.11^{2} } = {145}.{29}\;{\text{mm}}$$

Also, qs^’ = l_pl Sin θ + w_t^’ Cos θ = 130 * sin 59.65 + 64.89 * cos 59.65 = 144.97 mm.

By knowing qs and qs^’, from triangle qss^’, cos α^’ = 144.97 / 145.29.

α^’ = 3˚49’.

Strength of strut, F_ns = F_AB = 478.85 kN.

F^’_AB = F_AB cos α^’ = 478.85 * cos 3° 49’ = 477.78 kN.

Since the forces on all faces of the node should be equal, equating the stress in the face q-s and q-r.

$$\begin{aligned} & {\text{F}}_{{\text{u}}} ^{{{\text{ACI}}}} /{\text{13}}0 = {\text{F}}^{{\prime }} _{{{\text{AB}}}} /{\text{145}}.{\text{29}} \\ & {\text{F}}_{{\text{u}}} ^{{{\text{ACI}}}} = \left( {{\text{477}}.{\text{78}}*{\text{13}}0} \right)/{\text{145}}.{\text{29}} \\ & \quad \quad = {\text{427}}.{\text{48}}\;{\text{kN}} \\ & {\text{V}}_{{\text{u}}} ^{{{\text{ACI}}}} = {\text{2}}*{\text{F}}_{{\text{u}}} ^{{{\text{ACI}}}} \\ & {\text{V}}_{{\text{u}}} ^{{{\text{ACI}}}} = {\text{854}}.{\text{975}}\;{\text{kN}} \\ \end{aligned}$$