1. Determine the efficiency of the following riveted joints assuming the given permissible

Question

stresses:
• Tensile stress in the plates: 130 MPa.
• Shearing stress in the rivets: 80 MPa.
• Crushing stress in the rivets: 200 MPa.
1.1 A double-riveted lap joint made of 8 mm plates using 22 mm diameter rivets and a pitch of 70 mm.
1.2
A single-riveted lap joint made of 8 mm plates using 22 mm diameter rivets and a pitch of 60 mm.

Answer 1

To determine the efficiency of the riveted joints, we need to first calculate the capacities of the rivets and the plates based on the given permissible stresses for each type of joint. The efficiency of a riveted joint is calculated using the following formula:

\[ \text{Efficiency} (\eta) = \frac{\text{Strength of joint}}{\text{Strength of connected parts}} \times 100% \]

Given Data

• Tensile stress in plates: \(\sigma_t = 130 , \text{MPa}\)
• Shearing stress in rivets: \(\tau_s = 80 , \text{MPa}\)
• Crushing stress in rivets: \(\sigma_c = 200 , \text{MPa}\)
• Thickness of plates: \(t = 8 , \text{mm}\)
• Diameter of rivets: \(d = 22 , \text{mm}\)

1.1 Double-Riveted Lap Joint

• Number of rivets (n): 2 (double riveted)
• Pitch (p): 70 mm

Strength of the Plates:

The area of one plate (per unit length):

\[ \text{Area}_{\text{plate}} = \text{Width} \times \text{Thickness} = W \times t \]

Assuming the width \(W\) of the joint is large enough for simplicity, we will just analyze based on the unit length.

The tensile strength of the plate per unit length:

\[ \text{Strength of plates} = t \times \sigma_t = 8 \times 130 = 1040 , \text{N/mm} \]

Strength of the Rivets:

For each rivet, the shear area \(A_s\) per rivet:

\[ A_s = \frac{\pi d^2}{4} = \frac{\pi (22^2)}{4} = 380.13 , \text{mm}^2 \]

Total shear area for 2 rivets:

\[ \text{Total shear area} = 2A_s = 2 \times 380.13 = 760.26 , \text{mm}^2 \]

The shear strength of the rivets:

\[ \text{Strength of rivets} = 2 \times \tau_s \times A_s = 2 \times 80 \times 380.13 = 60820.8 , \text{N} \]

Check for Crushing:

The area resisting crushing is the projected area of the rivets:

\[ A_c = d \times t = 22 \times 8 = 176 , \text{mm}^2 \]

The crushing strength of the rivets:

\[ \text{Strength against crushing} = 2 \times \sigma_c \times A_c = 2 \times 200 \times 176 = 70400 , \text{N} \]

Strength of the Joint:

The strength of the joint is the minimum of tensile strength of the plates, the shear strength, and crushing strength of the rivets.

\[ \text{Strength of joint} = \min(1040, 60820.8, 70400) = 1040 , \text{N} \]

Efficiency:

\[ \eta = \frac{\text{Strength of joint}}{\text{Strength of plates}} \times 100% \] \[ \eta = \frac{1040}{1040} \times 100% = 100% \]

1.2 Single-Riveted Lap Joint

• Number of rivets (n): 1 (single riveted)
• Pitch (p): 60 mm

Strength of the Plates:

The tensile strength of the plate remains the same:

\[ \text{Strength of plates} = 1040 , \text{N/mm} \]

Strength of the Rivets:

The shear strength for a single rivet:

\[ A_s = \frac{\pi d^2}{4} = 380.13 , \text{mm}^2 \]

The shear strength of the rivet:

\[ \text{Strength of rivet} = \tau_s \times A_s = 80 \times 380.13 = 30410.4 , \text{N} \]

Check for Crushing:

The area resisting crushing remains the same:

\[ A_c = 176 , \text{mm}^2 \]

The crushing strength of the rivet:

\[ \text{Strength against crushing} = \sigma_c \times A_c = 200 \times 176 = 35200 , \text{N} \]

Strength of the Joint:

The strength of the joint is:

\[ \text{Strength of joint} = \min(1040, 30410.4, 35200) = 1040 , \text{N} \]

Efficiency:

\[ \eta = \frac{\text{Strength of joint}}{\text{Strength of plates}} \times 100% \] \[ \eta = \frac{1040}{1040} \times 100% = 100% \]

Conclusion

• The efficiency of the double-riveted lap joint is 100%.
• The efficiency of the single-riveted lap joint is 100%.

Both joints perform at their optimal capacity based on the given parameters.