Determine the safe load that a single riveted butt joint:

Determine the safe load that a single riveted butt joint with two cover plates as shown in Figure 8 can carry if the pitch of the power-driven field rivets is 7.5 cm centers. (Nominal dia of rivets is 20 mm and f_y = 150 MPa.) Determine the efficiency of the joint.

Figure

Solution

Gross diameter = 20 + 1.5 = 21.5 mm

Minimum edge distance = 32 mm (35 mm say) (Table 1)

Minimum pitch = 2.5 × 21.5 = 53.75 m

Maximum pitch = 32 × 9 = 288 mm

∴          The given pitch of 75 mm is OK.

Number of rivets = 5

Maximum permissible stress in power-driven field rivets (Table 2)

Shear (τ_vf) = 90 MPa

Bearing on rivet (σ _pt) = 270 MPa

(a) Strength of rivets in double shear

 = [ π/4 × 2 × (21.5) ²× 90] × 5 = 326745

(b) Strength of rivets in bearing with respect to main plate

= [21.5 × 16 × 270] × 5 = 464400

(c) Strength of rivets in bearing with respect to cover plates

= [21.5 × 18 × 270] × 5 = 522450

(d) Tearing strength of main plate = (370 - 5 × 21.5) × 16 × 150

= 630,000 N

(e) Tearing strength of cover plate

= (370 - 5 × 21.5) × 18 × 150 = 708,750 N

The safe load is the minimum of the above i.e. 326.7 kN.

The load that the plate without rivet holes can carry

= 370 × 16 × 150 = 888000 N = 888 kN

∴ Efficiency of the joint (η) = 326.7 /888×100 = 36.80% .