Interface Shear Transfer
CSA A23.3
v1.0.0
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
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
Vr = vr * Acv = 650000 kN
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 MPaConcrete stress limit = 0.25 * Φc * f'c = 4.06 MPa
Steel stress = Φs * fy * cos(α) = 0 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
Vr = vr * Acv = 0 kN
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 MPaConcrete 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 = vr * Acv = 0 kN