Question

A truss tie consisting of 2 ISA 75 x 75 x 8 mm carries a pull of 150 kN. At ends the two angles are connected, one each on either side of a 10 mm thick gusset plate, by 18 mm diameter rivets arranged in one row. The allowable stresses in rivet are
$f_s$ = 90.0 N/$m{m}^{2}and{f}_{br}$ = 250 N/$m{m}^2$.

Minimum number of rivets required at each end is ________.

• A. 3
• B. 5
• C. 2
• D. 4

Answer 1

The correct option is D 4

Gross dia of rivet,
d' = 19.5 mm
Allowable shear stress in rivet,
$f_s=90N/m{m}^2$
Allowable bearing stress in rivet,
$f_{br}=250N/m{m}^2$

Rivet value = Min {shearing strength, Bearing strength}

$=Min\{2\times (\frac{\pi d^{\prime 2}}{4}\times f_s),(d^{\prime }\times t\times f_{br}\left)\right\}$

$=Min\{\frac{2\times \pi \times {19.5}^2}{4}\times 90,19.5\times 10\times 250\}$

= Min {53729.33 N, 48750 N}
= 48.75 kN
Number of rivet required

$=\frac{Load}{Rivet.Value}=\frac{150}{48.75}=3.077$

= 4 Rivets