Example of Slenderness Ratio:

An IS flat is to be used as tie to carry a tensile force of 250 kN. The effective length is 1.1 m. It is to be connected by 16 mm dia hand driven rivets to the gusset plate. σ_y for MS flat is 250 MPa.

Solution

Permissible steel stress (σ_at) = 0.6 σ_y = 0.6 × 250 = 150 MPa

Net area of flat = A_n = P/σ _at = 250000/150 = 1667 mm²

Adding 25% for deductions A_g = 1.25 × 1667 = 2083 mm²

∴ Provide 300 × 8 mm IS flat A_g = 2400 mm²

Rivets: diameter of rivets by Unwin's formula

= 6.01 √t = 6.01   √8 = 17 mm

∴ Provide 18 mm dia rivets ⇒ dia of rivet hole = 18 +1.5 = 19.5 mm

Value of one rivet (i) in shear = π/4 d ² τ _vf = π/4 × (19.5)2 × 80 = 23, 892 N

 (ii) in bearing = dt σ _pt  = 19.5 × 8 × 250 = 39, 000 N

∴ Rivet value (R) = 23892 N

No. of rivets required in the connection = 250000/23,892 = 10.5 say 11 (rivets)

They are arranged in 3 rows of 4, 3, 4 as display in Figure.

Figure

For tearing of plates:

Considering line a-b-d-e:   l = 300 - 2 × 19.5 = 261 mm

a-b-c-f:   l = 300 - 2 × 19.5 + 40²/(4 × 100) = 265 mm

a-b-c-d-e :   l = 300 - 3 × 19.5 + 2 ×40²/(4 × 100) = 249.5 mm

Hence, adopting the smallest of the above, i.e. 249.5.

∴ Net area of plate (A_n) = 249.5 × 8 = 1996 mm² > 1667 mm²

∴ Load carrying capacity = 1996 × 150 = 299400 N > 250 000 N ∴ OK.