Shear strength criteria for design of RC beam–column joints in building codes

Abstract

The paper presents a comprehensive review on shear strength provisions of RC beam–column joint in various national codes viz. ACI 318-2014, NZS 3101-1:2006, EN 1998-1:2004, CSA A23.3:2004, AIJ:2010, and IS 13920:2016. The shear strength equation given in these codes are generic and simple in application, which is based on the contribution of only a few governing parameters. However, the effects of governing parameters in different codes are considered in different ways. As a result, the code prediction varies significantly among themselves as well as with experimental studies. Considering these differences, the influence of various governing parameters on the joint shear strength are evaluated. A database is compiled from 492 experimental results of beam–column joints from literature. To find the cause of variation between code prediction and experimental observations, different type of failure modes of beam–column joints is studied. Consequently, two parameters namely, aspect ratio of joint and area ratio of column to beam cross-section is observed to be affecting the code predictions considerably. The influence of these two parameters on the joint shear strength is validated with the compiled experimental results. Therefore, to ameliorate the code prediction, two approaches i.e. aspect ratio approach and area ratio approach are proposed. The first approach is based on the effect of variation of strut angle on joint shear strength, whereas, the second approach proposes various empirical modification factors based on area ratio of column to beam cross-section. By using these two approaches, it is observed that the difference between the code predictions and experimental results can be minimized considerably. These approaches make the code prediction suitable for design purpose.

Appendices

Appendix 1

See Tables 13 and 14.

Table 13 Experimental database of interior RC beam–column joints

Table 14 Experimental database of exterior RC beam–column joints

Appendix 2

2.1 Sample joint shear strength calculation for considered national codes

Sample calculations for the shear strength prediction of the codes are presented in this Appendix for interior and exterior beam–column joints. Five interior beam–column joint specimens viz. ‘O5’ of Hakuto et al. (2000); ‘IPB’ of Melo et al. (2015); ‘B11’ of Joh et al. (1991); ‘A2’ of Li et al. (2002) and ‘RA6’ of Hegger et al. (2003) and five exterior joint specimens viz. ‘T1-400’ of Hwang et al. (2014); ‘Test #1’ De Risi et al. (2016); ‘T5’ of Masi et al. (2008); ‘BS-L-450’ of Wong and Kuang (2008); and ‘A-1’ of Tsonos (2007) have been considered arbitrarily from the total database. Tables 15 and 16 shows the summary of actual code prediction and modified code prediction based on two approaches for the above considered specimens. Out of aforementioned specimens, the detail calculation of one specimen from each interior and exterior joints are presented in the following subsections.

Table 15 Comparison of shear strength prediction of considered codes based on two approaches with experimental results for five interior beam–column joints (number indicate the serial number in experimental database)

Table 16 Comparison of shear strength prediction of considered codes based on two approaches with experimental results for five exterior beam–column joints (number indicate the serial number in experimental database)

1. Interior beam–column joint Experimental Specimen: “O5” Hakuto et al. (2000) [Ref. “Appendix 1”, Table13, Sr. no. 118]

Specimen details:

1. Column section details
	Width of column (bc)	=	460 mm
	Depth of column (hc)	=	460 mm
	Longitudinal reinforcement	=	6–28 mm dia
	Yield strength of reinforcement (fyc)	=	321 MPa
	Column axial load (Nc)	=	0
	Transverse reinforcement	=	6 mm dia. @ 230 mm c/c
2. Beam section details
	Width of beam (bb)	=	300 mm
	Depth of beam (hb)	=	500 mm
	Longitudinal reinforcement	=	2–32 mm dia. at top and 2–32 mm dia. at bottom
	Yield strength of reinforcement (fyb)	=	306 MPa
	Transverse reinforcement	=	6 mm dia. @ 380 mm c/c
3. Joint details
	No shear reinforcement
4. Concrete compressive strength (fc)		=	33 MPa
5. Joint shear strength from experiment as reported in the paper (Vjexpt)		=	1069 kN

Based on above specimen details the shear strength calculation according to the code equations are presented below,

• 1. ACI 318-14

Joint Shear strength (Table 1); $V_{j}^{ACI} = \lambda \phi \sqrt {f_{c} } A_{j}$. where λ (represents the confinement effect of beams to the joint) = 1.20 (from Table 2) and ϕ (represents the factor for concrete type) = 1.00.

Effective joint area, A_j = b_j * h_c.

Effective joint width, b_j = min (b_b+ 2x; b_b+ h_c; and b_c) = 460 mm.

Joint depth, h_c= 460 mm.

$$V_{j}^{ACI} = 1.20*1.00*\sqrt {33} *\left( {460*460} \right)$$

$$\varvec{V}_{\varvec{j}}^{{\varvec{ACI}}} = 1458.65\,{\text{kN}} > 1069\,{\text{kN}}$$

• 2. EN1998-1:2004

Joint Shear strength (Table 1); $V_{j}^{EN} = \eta f_{cd} \sqrt {1 - \frac{{\nu_{d} }}{\eta }} A_{j}$. Where η is the reduction factor on concrete compressive strength due to tensile strain in transverse direction.

$$\eta = 0.60\left( {1 - \frac{{f_{c} }}{250}} \right)$$

$$\eta \, = \,0.60\, * \,\left[ {1\, - \,\left( {35/250} \right)} \right]\, = \,0.52$$

ν_d is the column axial load ratio = $\frac{{N_{c} }}{{A_{c} f_{c} }}_{ } = \, 0$

Effective joint area, A_j = b_j * h_c.

Effective joint width, b_j = min (b_c; b_b+ 0.5h_c) = 460 mm.

Joint depth, h_c= 460 mm.

Design strength of concrete from Eurocode; f_cd= 0.566 * 33 = 19 MPa.

$$V_{j}^{EN} = 0.52*19*\sqrt {1 - \frac{0}{0.41}} *\left( {460*460} \right)$$

$$\varvec{V}_{\varvec{j}}^{{\varvec{EN}}} = 1853\,\varvec{ }{\text{kN}} > 1069\,{\text{kN}}$$

• 3. NZS 3101: 1-2006

Joint Shear strength (Table 1); $V_{j}^{NZS} = 0.2f_{c} A_{j}$.

Effective joint area, A_j = b_j * h_c.

Effective joint width, b_j = min (b_c; b_b+ 0.5h_c) = 460 mm.

Joint depth, h_c= 460 mm.

$$V_{j}^{NZS} = 0.2*33*\left( {460*460} \right)$$

$$\varvec{V}_{\varvec{j}}^{{\varvec{NZS}}} = 1397\,{\text{kN}} > 1069\,{\text{kN}}$$

• 4. CSA A23.3: 2004

Joint Shear strength (Table 1); $V_{j}^{CSA} = \lambda \phi \sqrt {f_{c} } A_{j}$.

ϕ represents the concrete factor = 1.00.

φ represents the factored concrete tensile strength = 0.65.

λ represents the confinement factor of beam = 1.60.

Effective joint area, A_j = b_j * h_c.

Effective joint width, b_j = min (b_c; 2b_b) = 460 mm.

Joint depth, h_c= 460 mm.

$$V_{j}^{CSA} = 1.6*0.65*1.00*\sqrt {33} *\left( {460*460} \right)$$

$$\varvec{V}_{\varvec{j}}^{{\varvec{CSA}}} = 1265\,{\text{kN}} > 1069\,{\text{kN}}$$

• 5. AIJ:2010

Joint Shear strength (Table 1); $V_{j}^{AIJ} = k\lambda F_{j} A_{j}$.

k represents the shape factor of joint = 1.00.

λ represents the confinement factor of beam = 0.85.

F_j represents the concrete factor = 0.8 f ^0.7 _c  = 0.8 * (33)^0.7 = 9.24.

Effective joint area, A_j = b_j * h_c.

Effective joint width, b_j = b_b+ b_a1+ b_a2.

b_b = 300 mm.

b_a1 and b_a2= Minimum of hc/4 and (b_c − b_b)/2 on either side.

hc/4 = 460/4 = 115 mm.

(b_c − b_b)/2 = (460 − 300)/2 = 80 mm.

b_j = 300 + 80 + 80 = 460 mm.

Joint depth, h_c= 460 mm.

$$V_{j}^{AIJ} = 1.00*0.85*9.24*\left( {460*460} \right)$$

$$\varvec{V}_{\varvec{j}}^{{\varvec{AIJ}}} = 1661\,{\text{kN}} > 1069\,{\text{kN}}$$

• 6. IS 13920:2016

Joint Shear strength (Table 1); $V_{j}^{IS} = \lambda \sqrt {f_{c} } A_{j}$. where λ (represents the confinement effect of beams to the joint) = 1.20 (from Table 2) (Compressive strength of concrete f_c= 33 MPa is based on strength of cube specimen but to compare with other codes it has been converted into strength of cylindrical specimen by multiplying with the factor of 0.8).

Hence, concrete compressive strength f_c = 0.8 * 33 = 26.4 MPa.

Effective joint area, A_j = b_j * h_c.

Effective joint width, b_j = min (b_b+ 2x; b_b+ h_c; and b_c) = 460 mm.

Joint depth, h_c = 460 mm.

$$V_{j}^{IS} = 1.20*\sqrt {26.4} *\left( {460*460} \right)$$

$$\varvec{V}_{\varvec{j}}^{{\varvec{IS}}} = 1304\,{\text{kN}} > 1069.88\,{\text{kN}}$$

From the above calculations, it is observed that the codes overestimate the joint shear strength for same specimen. Now based on two proposed approaches (i.e. strut angle approach and area ratio approach) the modified code prediction is presented as,

2.2 Approach-1: Strut angle modifier

Depth ratio (α) = $\frac{{h_{b} }}{{h_{c} }}$ = $\frac{500}{460}$ = 1.08 and Width ratio (β) = $\frac{{b_{b} }}{{b_{c} }}$ = $\frac{300}{460}$ = 0.65.

Table 8 gives the modification factors based on the width and depth ratio as,

For h_b> h_c and b_c> b_b, the modifier coefficient is (β/α) as given below,

$$\left( {\upbeta/\upalpha} \right)\, = \,0.65/1.08\, = \,0.60.$$

The modification in the code predicted joint shear strength is shown as,

Code	Proposed modified strength using Approach-1
ACI 318-14	0.60 × 1458 = 875 kN
EN1998-1:2004	0.60 × 1853 = 1112 kN
NZS 3101: 1-2006	0.60 × 1397 = 838 kN
CSA A23.3: 2004	0.60 × 1265 = 759 kN
AIJ: 2010	0.60 × 1661 = 997 kN
IS13920: 2016	0.60 × 1304 = 782 kN

2.3 Approach-2: Area ratio modifier

Cross sectional area of column (A_c) = b_c× h_c = 460 × 460 = 211,600 mm².

Cross sectional area of beam (A_b) = b_b× h_b = 300 × 500 = 150,000 mm².

Area ratio of column to beam = (A_c/A_b) = (211,600/150,000) = 1.41.

For the considered specimen the column to beam area ratio is in between the range of 1.40–1.59. Therefore, from Table 11, by using the strength modification factors (ψ), the modified code predicted joint shear strength is shown as,

Code	strength modification factors (ψ) for the range of 1.40–1.59	Proposed modified strength using Approach-1
ACI 318-14	0.65	0.65 × 1458 = 948 kN
EN1998-1:2004	0.55	0.55 × 1853 = 1019 kN
NZS 3101: 1-2006	0.60	0.60 × 1397 = 838 kN
CSA A23.3: 2004	0.70	0.70 × 1265 = 886 kN
AIJ: 2010	0.55	0.55 × 1661 = 913 kN
IS13920: 2016	0.60	0.60 × 1304 = 782 kN

2. Exterior beam–column joint Experimental Specimen: “Test #1” of De Risi et al. (2016) [Ref. “Appendix 1”, Table14, Sr. no. 267].

Specimen details:

1. Column Section details
	Width of column (bc)	=	300 mm
	Depth of column (hc)	=	300 mm
	Longitudinal reinforcement	=	8-20 mm dia
	Yield strength of reinforcement (fyc)	=	450 MPa
	Column axial load (Nc)	=	260
	Transverse reinforcement	=	8 mm dia. @ 100 mm c/c
2. Beam section details
	Width of beam (bb)	=	300 mm
	Depth of beam (hb)	=	500 mm
	Longitudinal reinforcement	=	4-20 mm dia. at top and 4-20 mm dia. at bottom
	Yield strength of reinforcement (fyb)	=	450 MPa
	Transverse reinforcement	=	8 mm dia. @ 100 mm c/c
3. Joint details
	No Shear reinforcement
4. Concrete compressive strength (fc)		=	28.8 MPa
5. Joint shear strength from experiment as reported in the paper (Vjexpt)		=	256 kN

The detailed calculation as per the considered six codes is presented below;

• 1. ACI 318-14

Joint Shear strength (Table 1); $V_{j}^{ACI} = \lambda \varphi \sqrt {f_{c} } A_{j}$

where λ (represents the confinement effect of beams to the joint) = 1.00 (from Table 2).

and ϕ (represents the factor for concrete type) = 1.00.

Effective joint area, A_j = b_j * h_c.

Effective joint width, b_j = min (b_b+ 2x; b_b+ h_c; and b_c) = 300 mm.

Joint depth, h_c= 300 mm.

$$V_{j}^{ACI} = 1.00*1.00*\sqrt {28.8} *\left( {300*300} \right)$$

$$\varvec{V}_{\varvec{j}}^{{\varvec{ACI}}} = 483\,{\text{kN}} > 256\,{\text{kN}}$$

• 2. EN1998-1:2004

Joint Shear strength (Table 1); $V_{j}^{EN} = \eta f_{cd} \sqrt {1 - \frac{{\nu_{d} }}{\eta }} A_{j}$ where η is the reduction factor on concrete compressive strength due to tensile strain in transverse direction.

$$\eta = 0.48\left( {1 - \frac{{f_{c} }}{250}} \right)$$

η = 0.48 * [1 − (28.8/250)] = 0.42.

ν_d is the column axial load ratio = $\frac{{N_{c} }}{{A_{c} f_{c} }}$. = $\frac{260000}{{\left( {300*300*28.8} \right)}}$ = 0.10.

Effective joint area, A_j = b_j * h_c.

Effective joint width, b_j = min (b_c; b_b+ 0.5h_c) = 300 mm.

Joint depth, h_c= 300 mm.

Design strength of concrete from Eurocode; f_cd= 0.566 * 28.8 = 16 MPa.

$$V_{j}^{EN} = 0.42*16*\sqrt {1 - \frac{0.10}{0.42}} *\left( {300*300} \right)$$

$$\varvec{V}_{\varvec{j}}^{{\varvec{EN}}} = 545\,\varvec{ }{\text{kN}} > 256\,{\text{kN}}$$

• 3. NZS 3101: 1-2006

Joint Shear strength (Table 1); $V_{j}^{NZS} = 0.2f_{c} A_{j}$.

Effective joint area, A_j = b_j * h_c.

Effective joint width, b_j = min (b_c; b_b+ 0.5h_c) = 300 mm.

Joint depth, h_c= 300 mm.

$$V_{j}^{NZS} = 0.2*28.8*\left( {300*300} \right)$$

.

$$\varvec{V}_{\varvec{j}}^{{\varvec{NZS}}} = 518\,{\text{kN}} > 256\,{\text{kN}}$$

.

• 4. CSA A23.3: 2004

Joint Shear strength (Table 1); $V_{j}^{CSA} = \lambda \varphi \phi \sqrt {f_{c} } A_{j}$.

ϕ represents the concrete factor = 1.00.

φ represents the factored concrete tensile strength = 0.65.

λ represents the confinement factor of beam = 1.30.

Effective joint area, A_j = b_j * h_c.

Effective joint width, b_j = min (b_c; 2b_b) = 300 mm.

Joint depth, h_c= 300 mm.

$$V_{j}^{CSA} = 1.3*0.65*1.00*\sqrt {28.8} *\left( {300*300} \right)$$

.

$$\varvec{V}_{\varvec{j}}^{{\varvec{CSA}}} = 408\,{\text{kN}} > 256\,{\text{kN}}$$

.

• 5. AIJ:2010

Joint Shear strength (Table 1); $V_{j}^{AIJ} = k\lambda F_{j} A_{j}$.

k represents the shape factor of joint = 0.70.

λ represents the confinement factor of beam = 0.85.

F_j represents the concrete factor = 0.8 f ^0.7 _c  = 0.8 *(28.8)^0.7 = 8.41.

Effective joint area, A_j = b_j * h_c.

Effective joint width, b_j = b_b+ b_a1+ b_a2.

b_b = 300 mm.

b_a1 and b_a2 = Minimum of hc/4 and (b_c − b_b)/2 on either side.

hc/4 = 300/4 = 75 mm.

(b_c − b_b)/2 = (300 − 300)/2 = 0.

b_j = 300 + 0+0 = 300 mm.

Joint depth, h_c= 300 mm.

$$V_{j}^{AIJ} = 0.70*0.85*8.41*\left( {300*300} \right)$$

.

$$\varvec{V}_{\varvec{j}}^{{\varvec{AIJ}}} = 450 {\text{kN}} > 256 {\text{kN}}$$

.

• 6. IS 13920: 2016

Joint Shear strength (Table 1); $V_{j}^{IS} = \lambda \sqrt {f_{c} } A_{j}$. where λ (represents the confinement effect of beams to the joint) = 1.00 (from Table 2) (Compressive strength of concrete f_c = 67.3 MPa is based on strength of cube specimen but to compare with other codes it has been converted into strength of cylindrical specimen by multiplying with the factor of 0.8).

Hence, concrete compressive strength f_c = 0.8 * 28.8 = 23.4 MPa.

Effective joint area, A_j = b_j * h_c.

Effective joint width, b_j = min (b_b+ 2x; b_b+ h_c; and b_c) = 300 mm.

Joint depth, h_c= 300 mm.

$$V_{j}^{IS} = 1.00*\sqrt {23.4} *\left( {300*300} \right)$$

.

$$\varvec{V}_{\varvec{j}}^{{\varvec{IS}}} = 432 {\text{kN}} > 256 {\text{kN}}$$

.

From the above calculations, it is observed that the codes overestimate the joint shear strength for same specimen. Now based on two proposed approaches (i.e. strut angle approach and area ratio approach) the modified code prediction is presented as,

2.4 Approach-1: Strut angle modifier

Depth ratio (α) = $\frac{{h_{b} }}{{h_{c} }}$ = $\frac{500}{300}$ = 1.66 and Width ratio (β) = $\frac{{b_{b} }}{{b_{c} }}$ = $\frac{300}{300}$ = 1.00.

Table 8 gives the modification factors based on the width and depth ratio as,

For h_b> h_c and b_c> b_b, the modifier coefficient is (β/α) as given below,

(β/α) = 1.00/1.66 = 0.60.

The modification in the code predicted joint shear strength is shown as,

Code	Proposed modified strength using Approach-1
ACI 318-14	0.60 × 483 = 290 kN
EN1998-1:2004	0.60 × 545 = 326 kN
NZS 3101: 1-2006	0.60 × 518 = 311 kN
CSA A23.3: 2004	0.60 × 408 = 245 kN
AIJ: 2010	0.60 × 450 = 270 kN
IS13920: 2016	0.60 × 432 = 260 kN

2.5 Approach-2: Area ratio modifier

Cross sectional area of column (A_c) = b_c× h_c = 300 × 300 = 90000 mm².

Cross sectional area of beam (A_b) = b_b× h_b = 300 × 500 = 150,000 mm².

Area ratio of column to beam = (A_c/A_b) = (90,000/150,000) = 0.60.

For the considered specimen the column to beam area ratio is in between the range of < 0.70. Therefore, from Table 10, by using the strength modification factors (ψ), the modified code predicted joint shear strength is shown as,

Code	strength modification factors (ψ) for the range of 1.40–1.59	Proposed modified strength using Approach-1
ACI 318-14	0.70	0.70 × 483 = 338 kN
EN1998-1:2004	0.55	0.55 × 545 = 300 kN
NZS 3101: 1-2006	0.50	0.50 × 518 = 260 kN
CSA A23.3: 2004	0.70	0.70 × 408 = 285 kN
AIJ: 2010	0.65	0.65 × 450 = 293 kN
IS13920: 2016	0.75	0.75 × 432 = 324 kN