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**Check the shear connection for the IPE450-S275 for the design shear V _{Ed}=500KN. Bolts M20-8.8, plate thickness t=20mm.**

**Solution**

e 1=60mm > 1.2*do=1.2*22=26.4mm

e 2=50mm > 1.2*do=1.2*22=26.4mm

e 1, e 2< 4*t+40=120mm

ρ1>2.2*do=2.2*22=48.4mm

ρ2>2.4*do=2.4*22=52.8mm

ρ1 and ρ2< min(14t,200)=200mm

F_{V,Rd}=a_{v}*A_{S}*f_{ub}/γ_{Μ}_{2}=(2*0.6*245mm^{2} *800 N/ mm^{2 })/1.25=188.2KN

α=min(e1/ 3*do, ρ1/ 3*do-1/4, f_{ub}/f_{u},1)=

= min(60/ 3*22, 100/ 3*22-1/4, 800/430,1)=0.9

k_{1} = min(2.8*e_{2}/do-1.7,1.4 *ρ2/ do-1.7,2.5)=

=min(4.664,4.663,2.5)=2.5

F_{b,Rd}=k_{1}* α *d*t* f_{u}/γ_{Μ}_{b} =(2.5*0.9*20mm* 20mm*430 N/ mm^{2})/1.25=309.6KN

Every bold gets vertical force 500/3=166.6KN and the torsional moment is 500KN*50mm=2500KN*cm and it is assigned at the farthest bolts as horizontal force 2500KN*cm/20cm=125KN.

At the farthest bolts the maximum force is

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