https://file.aiquickdraw.com/mathbot/file/1726140764243z7d0y06p.jpg

The problem in the image is about a lap joint with three 20-mm diameter rivets subjected to an axial load \( P = 60 \, \text{kN} \). We are tasked with finding:

1. **The shearing stress in each rivet**.
2. **The bearing stress in each plate**.
3. **The maximum average tensile stress in each plate**.

### Step 1: Shearing Stress in Each Rivet

The load is distributed equally among the three rivets, meaning each rivet carries one-third of the total load.

\[
P_{\text{rivet}} = \frac{P}{3} = \frac{60 \, \text{kN}}{3} = 20 \, \text{kN} = 20 \times 10^3 \, \text{N}
\]

The shearing stress, \( \tau \), is given by:

\[
\tau = \frac{P_{\text{rivet}}}{A_{\text{rivet}}}
\]

Where \( A_{\text{rivet}} \) is the cross-sectional area of the rivet, which can be calculated as:

\[
A_{\text{rivet}} = \frac{\pi d^2}{4} = \frac{\pi (20 \, \text{mm})^2}{4} = 314.16 \, \text{mm}^2 = 314.16 \times 10^{-6} \, \text{m}^2
\]

Substituting the values:

\[
\tau = \frac{20 \times 10^3 \, \text{N}}{314.16 \times 10^{-6} \, \text{m}^2} = 63.66 \, \text{MPa}
\]

### Step 2: Bearing Stress in Each Plate

The bearing stress, \( \sigma_{\text{bearing}} \), is given by:

\[
\sigma_{\text{bearing}} = \frac{P_{\text{rivet}}}{t \cdot d}
\]

Where \( t = 25 \, \text{mm} \) (the plate thickness) and \( d = 20 \, \text{mm} \) (rivet diameter). Substituting the values:

\[
\sigma_{\text{bearing}} = \frac{20 \times 10^3 \, \text{N}}{25 \, \text{mm} \times 20 \, \text{mm}} = \frac{20 \times 10^3}{500} = 40 \, \text{MPa}
\]

### Step 3: Maximum Average Tensile Stress in Each Plate

The maximum average tensile stress, \( \sigma_{\text{tensile}} \), is calculated by dividing the total load by the effective cross-sectional area of the plate. The width of the plate is given as 130 mm, but we must subtract the area removed by the three rivet holes.

The net width is:

\[
\text{Net width} = 130 \, \text{mm} - 3 \times 20 \, \text{mm} = 130 - 60 = 70 \, \text{mm}
\]

The cross-sectional area of the plate is:

\[
A_{\text{plate}} = \text{Net width} \times \text{thickness} = 70 \, \text{mm} \times 25 \, \text{mm} = 1750 \, \text{mm}^2 = 1750 \times 10^{-6} \, \text{m}^2
\]

The tensile stress is:

\[
\sigma_{\text{tensile}} = \frac{P}{A_{\text{plate}}} = \frac{60 \times 10^3 \, \text{N}}{1750 \times 10^{-6} \, \text{m}^2} = 34.29 \, \text{MPa}
\]

### Summary of Results:
- (a) Shearing stress in each rivet: \( 63.66 \, \text{MPa} \)
- (b) Bearing stress in each plate: \( 40 \, \text{MPa} \)
- (c) Maximum average tensile stress in each plate: \( 34.29 \, \text{MPa} \)

Would you like more details on any part? 

Here are 5 related questions:
1. How is the shearing stress in rivets affected by changing the number of rivets?
2. What happens to the bearing stress if the thickness of the plate is increased?
3. How does the rivet diameter impact both shearing and bearing stresses?
4. Why is the net width of the plate used for tensile stress calculations?
5. How would different materials with varying strengths influence the joint design?

**Tip:** When designing riveted joints, always consider both shearing and bearing stresses to ensure structural integrity.

The lap joint shown in the figure is fastened by three 20-mm diameter rivets. Assuming that P = 60 kN, determine (a) the shearing stress in each rivet, (b) the bearing stress in each plate, and (c) the maximum average tensile stress in each plate. Assume that the axial load P is distributed equally among the three rivets.

This problem involves calculating the shearing stress in rivets, the bearing stress in plates, and the tensile stress for a lap joint fastened with three 20-mm diameter rivets under a load of 60 kN. Step-by-step solutions include relevant formulas and concepts such as stress analysis and mechanical engineering principles.

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