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
$f_s=90.0N/m{m}^{2}and{f}_{br}=250N/m{m}^2.$

Maximum tensile stress in the tie in is _______$N/m{m}^2$ .

Answer 1

Diameter of rivet,
d = 18 mm
Gross diameter of rivet,
d' = d + 1.5 = 19.5mm
In truss tie maximum tensile stress will be at the section where net effective area is minimum. It is minimum at the location of rivet.



For 2 ISA 75 x 75 x 8 mm angles,
Area of connected leg of each angle,

$A_1=(75-\frac{8}{2}-19.5)\times 8=412m{m}^2$

Area of outstanding leg,

$A_2=(75-\frac{8}{2})\times 8=568m{m}^2$

As the angles are not tacked along the length so they will behave individually, hence net area would be twice the area corresponding to single angle connected to gusset plate.
Net effective area of each angle,

$A=A_1+k{A}_2$

$Where,k=\frac{3A_1}{3A_1+A_2}=\frac{3\times 412}{3\times 412+568}$

$=0.685$

$A=412+0.685\times 568$
$=801.16m{m}^2$

So, effective area of tie
$=2A$
$=2\times 801.16=1602.32m{m}^2$

Maximum tensile stress
$\frac{Load}{Area}=\frac{150\times {10}^3}{1602.32}=93.6N/m{m}^2$