The areas of three welds are as follows:

$A_1 = (60 t) mm^2 $

$A_2 = (60 t) mm^2 $

$A_3 = (90 t) mm^2$

$A = A_1 + A_2 + A_3 = (210 t mm^2)$

The primary shear stress in the weld is given $\tau=\frac{P}{A}=\frac{50000}{210*t}=\frac{238.1}{t}N/mm^2 (i) $

Step 2: Secondary shear stress:

As seen in Fig. 8.26, A is the farthest point from the Centre of gravity G and its distance r is given by, $r = \sqrt{(60 - 17.14)^2 + 60^2} = 73.73 mm $

Also, $tan \theta=\frac{60}{60-17.14} , \theta = 54.46° $

Therefore, the secondary shear stress is inclined at 54.46° with horizontal.

$e= (60 -\bar{x} )+ 100 = (60 - 17.14) + 100 = 142.86 mm $

$M= P x e= (50 * 10^3) (142.86) =7143 * 10^3 N-mm$

G1, G2 and G3 are the centers of gravity of the three welds and their distances from the common Centre of gravity G are as follows,

$G_IG= G_2G= \sqrt{(30-17.14)^2+ 60^2} = 61.36 mm$

$ r_1= r_2 = 61.36 mm $
$ r_3= G_3G = \bar{x} =17.14mm$

$J_1 =J_2 = A_1 \frac{l_1^2}{12}+r_1^2=60t \frac{60^2}{12}+61.36^2=243902.9 t mm^4$

$J_3 = A_3 \frac{l_3^2}{12}+r_3^2=90t \frac{90^2}{12}+17.14^2=331093.06 t mm^4$

The secondary shear stress at the point A is given by,
$\tau=\frac{Mr}{J}=\frac{ 7143 * 103 * 73.73}{331093.06t}=\frac{ 1590.65}{t} N/mm^2 $ (ii)

Step 3: Resultant shear stress:

The secondary shear stress is inclined at $54.45^\circ$ with the horizontal. It is resolved into vertical and horizontal components as shown

Vertical component= $\tau_2sin\phi$

$=\frac{1590.65}{t}sin54.56=\frac{1294.32}{t}N/mm^2$

Horizontal component=$\tau_2cos\phi$

$=\frac{1590.65}{t}cos54.56=\frac{924.59}{t}N/mm^2$

The resultant shear stress is given by,

$\tau=\sqrt{(\frac{1294.32}{t}+\frac{238.1}{t})^2+\frac{924.59}{t}}^2$

$=\frac{1789.74}{t} N/mm^2$

Step 4: Size of weld

The permissible shear stress for the weld material is $120 N/mm^2$. Therefore,

$120=frac{1789.74}{t}$ or $t=14.91 mm$

$h=\frac{t}{0.707}=\frac{14.91}{0.707}=21.09 mm$