A welded plate shown in Fig. 3.3, is subjected to force (eccentric) of 50 KN in the plane of the welds. Determine the size of the welds, if the permissible shear stress for the weld is 120 N/mm2. Assume static conditions.

Subject: Machine Design -I

Topic: Design against static Loads, Bolted and welded joints under eccentric loading. Power Screw

Difficulty: Medium

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 $t=14.91 mm$

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