Introduction

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Corbel is a short structural element that cantilevers out from column/wall to support load. Generally, the corbel is casted monolithically with column/wall.

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There are several typical modes of failure in the corbel. The most common of which are yielding of the tension tie, failure of the end anchorages of the tension tie, either under the load point or in the column, failure of the compression strut by crushing or shearing, and local failures under the bearing plate.

The following figures shows the failure mode of corbel

Design of Corbel

The corbel must be designed to resist simultaneously Vu, a factored moment Mu and a factored horizontal tensile force Nuc. ACI Code Section 11.8 requires corbels having a/d between 1 and 2 to designed using Appendix A, strut-and-tie models, where a is the distance from the load to the face of column and d is the depth of the corbel below the tie, measured at the face of the column. Corbels having a/d less than or equal 1 may be designed using either strut-and-tie models or traditional ACI designed method, Section 11.8. This paper presents Corbels design according to traditional ACI method.

ACI Design Method

Shear Design of Corbel

To avoid the crack that occurs in the interface of the corbel and the column we must provide the shear friction reinforcement perpendicular with the cracks direction. We use coefficient of friction μ to transform the horizontal resisting force into vertical resisting force.

The nominal shear strength of shear reinforcement can be determined using equations below

V_n = A_vf f_y μ for vertical shear friction reinforcement, and

V_n = A_vf f_y (μ sin α_f + cos α_f) for inclined shear reinforcement

where

V_n : nominal shear strength of shear friction reinforcement

A_vf : area of shear friction reinforcement

f_y : yield strength of shear friction reinforcement

μ : coefficient of friction

Method	Coefficient of Friction, μ
Concrete cast monolithic	1.4λ
Concrete placed against roughened hardened concrete	1.0λ
Concrete placed against unroughened hardened concrete	0.6λ
Concrete anchored to structural steel	0.7λ

The value of λ is 1.0 for normal weight concrete, 0.85 for sand light weight concrete and 0.75 for all light weight concrete.

The maximum nominal shear force, Vn shall not exceed the smallest of 0.2 fc’ b_w d and 5.5 b_w d, where

fc’ : compression strength of concrete (MPa)

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b_w : width of corbel section (mm)

d : effective depth of corbel (mm)

Flexural Design of Corbel

The corbel is designed to resist ultimate flexural moment result from the supported beam reaction, Vu and horizontal force from creep and shrinkage, Nuc. The minimum value of Nuc is 0.2 Vu and not greater than Vu.

Tension Reinforcement

The ultimate horizontal force, Nuc shall be resisted by tension reinforcement as follow

A_n = N_uc / ϕf_y

Where:

A_n : area of tension reinforcement

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N_uc : ultimate horizontal force at corbel

f_y : yield strength of shear friction reinforcement

ϕ : strength reduction factor

Flexural Reinforcement

The ultimate flexure moment, Mu is

Mu = Vu a_v + Nuc (h – d)

Where:

Mu : ultimate flexure moment

Vu : ultimate shear force

a_v : distance from Vu to the face of column

Nuc : ultimate horizontal force at corbel

h : height of corbel

d : effective depth of corbel

Mu ≤ ϕ A_f f_y (d-a/2) where

a = A_f f_y / 0.85f’_c b

From the equation above, area of flexural reinforcement, A_fcan be determined using trial and error. As first trial, (d – a/2) can be assumed 0.9d so that

A_f ≥ Mu / ϕ f_y (0.9d)

For practical reason, the value of (d – a/2) can be used 0.85d

A_f ≥ Mu / ϕ f_y (0.85d)

After finding A_vf, A_n, and A_f, we must then calculate the primary tension reinforcement A_sc from the larger of A_f + A_n and 2A_vf/3 + A_n

Reinforcement Limits

The primary steel reinforcement at corbel design,

A_sc shall not be less than 0.04 f_c‘ bw d /f_y

The horizontal closed stirrups,

A_h shall not be less than 0.5 (A_sc – A_n)

Distribution of Corbel Reinforcement

The horizontal closed stirrups, Ah shall be distributed uniformly within (2/3) d adjacent to primary tension reinforcement.

Design Procedure

Step 1. Find factored shear V_u and tensile force N_uc

If N_uc is not specified, use a minimum value of N_uc = 0.2 V_u(ACI 11.9.3.4)

Compute nominal values of shear and tensile force

V_n = V_u / 0.75 ; N_nc = N_uc / 0.75

If V_n > 0.2 fc’ b d OR

V_n > 5.5 b d then section size is inadequate (ACI 11.9.3.2)

Step 2. Compute shear-friction reinforcement (ACI 11.7.4.1)

A_vf = V_n /μ f_y

Step 3. Calculate required flexural reinforcement (11.9.3.3)

Mu = Vu a_v + Nuc (h – d)

A_f = Mu / ϕ f_y (jd) (assume jd = 0.85d)

Step 4. Reinforcement to carry tensile force (ACI 11.9.3.4)

A_n = N_uc / ϕf_y

Step 5. Required main flexural steel (A_sc) is given by (ACI 11.9.3.5 and 11.9.5)

the larger of

A_f + A_n and 2A_vf/3 + A_n

Step 6. Provide closed horizontal stirrups (ACI 11.9.4):

A_h = 0.5 (A_sc – A_n)

Ensure adequate detailing (ACI 11.9.6 & 11.9.7)

Example

Corbel Geometri

Width of corbel, b = 300 mm

Total thickness of corbel, h = 500 mm

Depth to main reinforcement, d = 450 mm

Material Properties

Yield strength of reinforcement, fy = 415 Mpa

Compressive strength of concrete, fc’ = 35 Mpa

Normal weight concrete, λ = 1

Coefficient friction, μ = 1.4 x λ = 1.4

Design Load Data

Factored vertical load, Vu = 370 kN

Distance from face to column, a = 100 mm

Horizontal force, Nu = 75 kN

Strength reduction factor, ϕ = 0.75

Design Procedure

Step 1. Find factored shear V_u and tensile force N_uc

V_u = 370 kN

N_{uc_min} = 0.2 x 370 = 74 kN

N_{uc_act} = the larger of 74 kN and 75 kN

= 75 kN

Compute nominal value of shear and tensile force

V_n = 370 / 0.75 = 493.33 kN

N_nc = N_uc / 0.75 = 100 kN

Check section

0.2 x fc’x b x d = 945.0 kN > 493.33 kN

5.5 x b x d = 742.5 kN > 493.33 kN

Section is OK

Step 2. Compute shear-friction reinforcement (ACI 11.7.4.1)

A_vf = V_n /μ f_{y =} 493.33 / (1.4×415) = 849.11 mm²

Step 3. Calculate required flexural reinforcement (11.9.3.3)

Mu = Vu a_v + Nuc (h – d) = 370 x 1000 x 100 + 75 x 1000 (500-450)

A_f = Mu / ϕ f_y (0.85d) =342.28 mm²

Step 4. Reinforcement to carry tensile force (ACI 11.9.3.4)

A_n = N_uc / ϕf_{y =} 75×1000 / 0.75×415 =240.96 mm²

Step 5. Required main flexural steel (A_sc) is given by (ACI 11.9.3.5 and 11.9.5)

A_f + A_n = 342.28 + 240.96 = 583.25 mm²

2A_vf/3 + A_n = 2/3 849.11 + 240.96 = 807.04 mm²

A_sc = 807.04 mm²

Use 3 dia 20 bars

A_{sc_prov} = 3 x 314 = 942 mm²

Step 6. Provide closed horizontal stirrups (ACI 11.9.4):

A_h = 0.5 (A_sc – A_n)= 0.5(807.04 – 240.96) =283.04 mm²

Use 3 dia 8 bars

A_{h_prov}=2×3×50 = 300 mm²

S_h = (2×d/3)/3 = (2×450/3)/3 = 100 mm

Detailed reinforcement