# Shear strength in concrete joints

RESULT

Design value shear stress | Design shear resistance | Requirement |

v_{Edi} (MPa) |
v_{Rdi} (MPa) |
v_{Edi} ≤ v_{Rdi} |

0.89 | 1.06 | OK |

DETAILS OF CALCULATION

Notation and methodology according to clause 6.2.5 of EC2

**v _{Edi} (design value of the shear stress) = 0.89 MPa**

v_{Edi} = V_{Ed} / (z·b_{i}) = 70000 / (225·350) = 0.89 MPa

where:

- z (inner lever arm) = 0.9 · d = 0.9 · 250 = 225 mm

**v _{Rdi} (design shear resistance at the interface) = 1.06 MPa**

v_{Rdi} = min [v_{Rdi}(1) ; v_{Rdi}(2)] = min[1.06 ; 4.5] MPa

where:

- v
_{Rdi}(1) = c·f_{ctd}+ ρ·f_{yd}·(μ·sinα + cosα)

v_{Rdi}(1) = 0.4·1.2 + 0.0019·434.78·(0.7·sin90º+cos90º) = 1.06 MPa

with:- f
_{ctd}(design tensile strength) = α_{ct}· f_{ctk,0.05}/ γ_{c}

f_{ctd}= 1 · 1.8 / 1.5 = 1.2 MPa

with:

f_{ctk,0.05}= 0.7 · f_{ctm}= 0.7 · 2.56 = 1.8 MPa

f_{ctm}= 0,30 × f_{ck}^{(2/3)}= 0,30 × 25^{(2/3)}= 2.56 MPa - ρ = A
_{sw}/ (s·b_{i}) = 1.01 / (15·35) = 0.0019 - f
_{yd}= f_{yk}/ γs = 500 /1.15 = 434.78 MPa - c = 0.4 ; μ = 0.7 (interface Rough, No dynamic loads)

- f
- v
_{Rdi}(2) = 0.5·v·f_{cd}= 0.5 · 0.54 · 16.67 = 4.5 MPa

with:- v = 0.6 ·(1-f
_{ck}/250) = 0.54 - f
_{cd}= α_{cc}· f_{ck}/ γc = 1 · 25 / 1.5 = 16.67 MPa

- v = 0.6 ·(1-f