Question

A 200 mm x 20 mm plate arrangement is shown in figure below. The weld along the outer edges of the plate is of 10 mm size while the weld in the slot is of 6 mm size. The maximum pull that can be applied on the plate is ____ kN. Plates are of grade Fe410 steel.

Answer 1

Tensile strength of plate is minimum of

(a) Gross yielding strength

$=\frac{A_g{f}_y}{{\gamma }_{m_0}}=\frac{200\times 20\times 250}{1.1\times 1000}\text{kN}=909.1\text{kN}$

(b) Net section rupture strength

$=\frac{0.9A_n{f}_u}{{\gamma }_{m_1}}=\frac{0.9\times 200\times 20\times 410}{1.25\times 1000}\text{kN}=1180.8\text{kN}$

$\therefore$ Tensile strength of plate is 909.1 kN

Strength of 10 mm fillet weld

$=\frac{f_u{L}_w{t}_c}{\sqrt{3}{\gamma }_{mw}}=\frac{410\times 360\times (0.7\times 10)}{\sqrt{3}\times 1.25\times 1000}\text{kN}=477.2\text{kN}$

Strength of 6 mm fillet weld

$=\frac{200\times (0.7\times 6)\times 410}{\sqrt{3}\times 1.25\times 1000}\text{kN}=159.1\text{kN}$

Total strength of weld = 477.2 + 159.1
= 636.3 kN

$\therefore \text{Strength of weld}\le \text{Strength of plate}$

$\therefore \text{Safe pull = Strength of the weld}$
$=636.3kN$