Question

A circular plate 100 mm diameter is welded to another plate by means of 6 mm size fillet weld. If the permissible shear stress in the weld is 100 n/mm2, then the greatest twisting moment that can be resisted by the weld is

Answer 1

The greatest twisting moment that can be resisted by the weld is $\bold{214.2\pi \ kg.m}$

Given:

Diameter of the plate = d = 100 mm

Size of another plate = 6 mm

Shear Stress = 100 $\bold{N/mm^2}$ = 10.2 $\bold{kg/mm^2}$

To find:

Greatest twisting moment = ?

Formula used:

$\bold{Twisting \ moment = P \times \frac{d}{2}}$

Where,

P = Permissible stress

$\bold{P = Strength \ of \ weld \times Length \ of \ weld}$

Solution:

From the question, fillet weld is used to join the two plates. Thus, the two plates are perpendicular to each other.

The allowable force is less than 70% when the welds were placed perpendicular to the applied load.

Strength of weld:

$\bold{Strength \ of \ weld = 0.7 \times Shear \ Stress \times Size \ of \ another \ plate}$

$\bold{Strength \ of \ weld = 0.7 \times 10.2 \times 6}$

$\bold{\therefore Strength \ of \ weld = 42.84 \ kg/mm}$

Length of weld:

$\bold{Length \ of \ weld = \pi \times Diameter}$

$\bold{Length \ of \ weld = \pi \times 100 \ mm}$

$\bold{\therefore Length \ of \ weld = 100\pi \ mm}$

Twisting moment:

$\bold{Twisting \ moment = P \times \frac{d}{2}}$

$\bold{Twisting \ moment = (Strength \ of \ weld \times Length \ of \ weld) \times \frac{d}{2}}$

$\bold{Twisting \ moment = (42.84 \times 100\pi) \times \frac{100}{2}}$

$\bold{Twisting \ moment = (42.84 \times 100\pi) \times 50}$

$\bold{Twisting \ moment = 214200\pi \ kg.mm}$

$\bold{\therefore Twisting \ moment = 214.2\pi \ kg.m}$