What will be the efficiency of single riveted lap joint of 10 mm thick plate with rivet diameter of 20 mm having the pitch of 50 mm? [Given: Permissible tensile stress in plate = 150 MPa; Permissible shear stress in rivet = 100 MPa; Permissible crushing stress in rivets = 200 MPa; π = 3.14]

Concept:

The efficiency of a riveted joint is the ratio of the strength of the joint (minimum of tearing, shearing, or crushing strength) to the strength of the solid plate without holes.

Given:

Plate thickness, t = 10 mm

Rivet diameter, d = 20 mm

Pitch of rivets, p = 50 mm

Permissible tensile stress in plate = 150 MPa

Permissible shear stress in rivet = 100 MPa

Permissible crushing stress in rivet = 200 MPa

π = 3.14

Calculation:

Tearing strength of plate, \( P_t = (p - d) \times t \times \sigma_t = (50 - 20) \times 10 \times 150 = 45000~N \)

Shearing strength of rivet, \( P_s = \frac{\pi}{4} \times d^2 \times \tau = \frac{3.14}{4} \times 20^2 \times 100 = 31400~N \)

Crushing strength of rivet, \( P_c = d \times t \times \sigma_c = 20 \times 10 \times 200 = 40000~N \)

Strength of joint is minimum of tearing, shearing and crushing: \( P = \min(45000, 31400, 40000) = 31400~N \)

Strength of solid plate (without rivet hole), \( P_{solid} = p \times t \times \sigma_t = 50 \times 10 \times 150 = 75000~N \)

Efficiency of the joint, \( \eta = \frac{P}{P_{solid}} \times 100 = \frac{31400}{75000} \times 100 = 41.86\% \)