Question

A lap joint is used to connect two flat plates of width 200 mm and 12 mm thickness with chain pattern bolts having bolt hole diameter 18 mm for M16 bolts as shown in figure. The yield stress and ultimate tensile stress of plate are 250 N/mm${}^2$ and 410 N/mm${}^2$ respectively. Partial safety factor for material governed by yield stress and ultimate tensile stress are respectively
${\gamma }_{mo}=1.10and$ ${\gamma }_{m1}=1.25$.
Determine design tensile strength of plate by considering net section rupture.

• A. 315.79 kN
• B. 453.43 kN
• C. 646.48 kN
• D. 394.2 kN

Answer 1

The correct option is B 453.43 kN

Design tensile strength of plate based on net section rupture
$T_d=0.9A_n{f}_u/{\gamma }_{m1}$

Diameter of bolt hole
$=d_h=18mm$

No. of bolt holes, n = 4

Net section area at any critical section
$=(B-n{d}_h)t$

$=(200-4\times 18)\times 12=1536{mm}^2$

$T_d=\frac{0.9A_n{f}_u}{{\gamma }_{m1}}$

$=\frac{0.9\times 1536\times 410}{1.25}$

$=453427.2N=453.43kN$