Joints - The Behaviour and Design of Steel Structures to EC3 - page 426

and the average shear stress is

τ yz = Q × 10 3 / { ( 280 − 4 × 22 ) × 10 } N / mm 2 = 0.521 Q N / mm 2 .

Using the equivalent of equation 9.15 for fracture with f u = 510N / mm 2
(EN10025-2),

( 2.170 Q ) 2 + 3 × ( 0.521 Q ) 2 ≤ 510 2

so that Q vtu ≤ 228.5 > 215.0 ( ≥ Q vty ) .
However, these calculations are conservative, since the maximum bending
stressesoccuratthetopandbottomoftheplate,wheretheshearstressesarezero.
If the shear stresses are ignored, then Q ty ≤ 355 / 1.531 = 231.9 < 254.3 ( ≥ Q v )
and

Q tu ≤ 510 / 2.170 = 235.0 > 231.9 ( ≥ Q ty ) .

9.10.5Example - fillet weld resistance

Problem. Determine the resistance of the two fillet welds shown in Figure 9.22a.
Weld forces per unit length.
Atthewelds,thedesignactionsconsistofaverticalshearof Q kNandamomentof

( − 0.2 Q − Q × ( 25 + 30 + 70 )/ 1000 ) kNm =− 0.325 Q kNm.
l eff = 280 − 2 × 8 / √ 2 = 268.7mm EC3-1-8 4.5.1(1)

The average shear force per unit weld length can be determined as

F L , Ed = ( Q × 10 3 )/( 2 × 268.7 ) N / mm = 1.861 Q N / mm,

and the maximum bending force per unit weld length from equation 9.50b as

F Ty , Ed = ( − 0.325 Q × 10 6 ) × ( 268.7 / 2 )
2 × 268.7 3 / 12

N / mm =− 13.51 Q N / mm,

and using equation 9.30, the resultant of these forces is
F w , Ed = √ [ ( 1.861 Q ) 2 + ( 13.51 Q ) 2 ]= 13.63 Q N / mm.

Simplified method of EC3-1-8.

For Grade S355 steel, β w = 0.9 EC3-1-8T4.1

√

f vw , d = 510 /

3
0.9 × 1.25 = 261.7N / mm 2

EC3-1-8 4.5.3.3

F w , Rd = 261.7 × ( 8 / √ 2 ) = 1481N / mm

EC3-1-8 4.5.3.3

Hence 13.63 Q ≤ 1481