Question

Two steel plates each of width 150 mm and thickness 10 mm are connected with three 20 mm diameter rivets placed in a zig-zag pattern. The pitch of the rivets is 75 mm and gauge is 60 mm. If the allowable tensile stress is 150 MPa, the maximum tensile force that the joint can withstand is _______.

• A. 192.75 kN
• B. 195.66 kN
• C. 225.00 kN
• D. 195.00 kN

Answer 1

The correct option is A 192.75 kN

Maximum tensile force that the joint can withstand = Rupture strength or tearing strength of plate.
($T_{dn}$)


$T_{dn}=A_{net}\times f_t$
$f_t=\text{allowable tensile stress}=150MPa$
$\text{Thickness of plate (t)}=10mm$
$\text{Nominal dia of rivet}\left(\varphi \right)=20mm$

$\therefore \text{dia. of hole (}{d}_0\text{)}=20+1.5=21.5mm$

Consider the following failure sections,

$\text{(i) Section 1 - 2 - 3}$
$A_{net}=(150-1\times 21.5)\times 10$
$=1285m{m}^2$

$\text{(ii) Section 4 - 5 - 2 - 3}$
$A_{net}=(150-2\times 21.5)\times 10+\frac{{75}^2\times 10}{4\times 60}$
$=1304.375{mm}^2$

$\text{(iii) Section 4 - 5 - 2 - 6 - 7}$
$A_{net}=(150-3\times 21.5)\times 10+2\times \frac{{75}^2\times 10}{4\times 60}$
$=1323.75m{m}^2$

$\therefore$ Minimum net sectional area available is along section 1- 2 -3
$\therefore$ Maximum tensile force of joint
$T_{dn}=1285\times 150\times {10}^{-3}$
$T_{dn}=192.75kN$