Question

The design tensile strength of the plate $250\text{mm}\times 10\text{mm}$ with the holes as shown below is ____$\text{kN}$. The yield strength and the ultimate strength of the steel used are $250\text{MPa}$ and $410\text{MPa}$ respectively.

• A. 543.17

Answer 1

The correct option is A 543.17

For 20 mm diameter bolt, diameter of bolt hole = 22 mm

$A_{\text{net}}\text{calculation}$

$A_{1-1^{{}^{\prime }}}=(250-1\times 22)\times 10=2280{\text{mm}}^2$
$A_{2-2^{{}^{\prime }}}=(250-3\times 22)\times 10=1840{\text{mm}}^2$
$A_{3-3^{{}^{\prime }}}=(250-3\times 22+\frac{{50}^2}{4\times 30})\times 10$
$=2048.33{\text{mm}}^2$

$\therefore A_{\text{net}}=1840{\text{mm}}^2$

Tensile strength calculation
$\text{1. Gross yielding strength}=\frac{A_g{f}_y}{{\gamma }_{m_0}}$

$=\frac{250\times 10\times 250}{1.1\times 1000}\text{kN}=568.182\text{kN}$

$\text{2. Net section rupture strength}=\frac{0.9A_n{f}_u}{\gamma m_1}$

$=\frac{0.9\times 1840\times 410}{1.25\times 1000}\text{kN}=543.168\text{kN}$

$\therefore \text{The design tensile strength is}543.17\text{kN}$