Question 15.13: Design of a Welded Joint under Out-of-Plane Eccentric Loadin...

By Table 15.8, for E6010, S_{y} = 345 MPa. The centroid lies at the intersection of the two axes of symmetry of the area enclosed by the weld group. The moment of inertia is

I_x=2(60) t(45)^2+\frac{2(90)^3 t}{12}=364,500 t  mm ^4

The total weld area equals A=2(60 t+90 t)=300 t  mm ^2 . Moment is M = 15(50) = 750 kN mm.

The maximum shear stress, using Equation 15.49, is

\tau=\left(\tau_d^2+\tau_m^2\right)^{1 / 2}         (15.49)

\tau=\left[\left\lgroup \frac{15,000}{300 t} \right\rgroup ^2+\left\lgroup \frac{750,000 \times 45}{364,500 t} \right\rgroup ^2\right]^{1 / 2}=\frac{105.2}{t}  N / mm ^2

Applying Equation 15.44, we have

n=\frac{S_{y s}}{\tau}=\frac{0.5 S_y}{\tau}         (15.44)

n \tau=0.5 S_y ; \quad 3\left(\frac{105.2}{t}\right)=0.5(345) \quad \text { or } \quad t=1.83  mm

Hence,

h=\frac{t}{0.707}=\frac{1.83}{0.707}=2.59  mm

Comment: A nominal size of 3 mm fillet welds should be used throughout.