Interface Shear Transfer
CSA A23.3
v1.0.0
File:
Project:
Designer:
Date:

Concrete Conditions

Case Description c (MPa) μ
a For concrete placed against hardened concrete with the surface clean but not intentionally roughened 0.25 0.60
b For concrete placed against hardened concrete with the surface clean and intentionally roughened to a full amplitude of at least 5 mm 0.50 1.00
c For concrete placed monolithically 1.00 1.40
d For concrete anchored to as-rolled structural steel by headed studs or by reinforcing bars 0.00 0.60

Concrete condition case:
MPa

Material Properties

Concrete
Compressive Strength f'c :
MPa

Reinforcing Steel
Yield Strength, Fy :
MPa
Lambda, λ:
MPa

Load

Normal force, N:
kN
(Unfactored permanent load perpendicular to shear plane. Positive for compression, negative for tension)

Section

Ag:
mm2 (Gross area of section)
Avf:
mm2 (Area of shear friction reinforcement)

Acv:
mm2 (Area of concrete section resisting shear transfer)

Angle, αf:
degrees (Angle between shear friction reinforcement and shear plane)

Values

Condition case = c
c = 1 MPa
μ = 1.4
f'c = 25 MPa
fy = 400 MPa
λ = 1
N = 0 N
Ag = 1000000 mm2
Avf = 0 mm2
Acv = 1000000 mm2
α = 90 °
Φc = 0.65
Φs = 0.85

Design & Solution

[11.5.1]

General Method

Effective normal stress: σ = ρv * fy * sin(α) + (N / Ag) = 0 MPa
Ratio of shear friction reinforcement: ρv = Avf / Acv = 0
vr = λ * Φc * (c + μ * σ) + Φs * fy * cos(α) = 0.65 + 0
Concrete stress: λ * Φc * (c + μ * σ) = 0.65 MPa
Concrete stress limit = 0.25 * Φc * f'c = 4.06 MPa
Steel stress = Φs * fy * cos(α) = 0 MPa
vr = 0.65 + 0 = 0.65 MPa
Vr = vr * Acv = 650000 kN

Vr = 650000 kN


[11.5.3]

Alternative Equation Method

vr = k * Φc * (σ * f'c )0.5 + Φs * fy * cos(α)= 0 + 0
k = 0.6
Concrete stress: λ * Φc * (c + μ * σ) = 0 MPa
Concrete stress limit = 0.25 * Φc * f'c = 4.06 MPa
Steel stress = Φs * fy * cos(α) = 0 MPa
vr = 0 + 0 = 0.65 MPa
Vr = vr * Acv = 0 kN

Vr = 0 kN