Let's go through the problem step-by-step.

Given Data:

• Thickness of plates, $t = 10 \, \text{mm}$
• Width of plates, $w = 120 \, \text{mm}$
• Allowable bearing stress for the plate, $\sigma_b = 100 \, \text{MPa}$
• Allowable shearing stress for the rivet, $\tau = 50 \, \text{MPa}$

Problem Requirements:

1. Determine the diameter of the rivet, $d$.
2. Determine the maximum tensile stress in the plates, $\sigma_t$.

Solution:

Part (a): Diameter of the Rivet

The bearing stress $\sigma_b$ is given by: $\sigma_b = \frac{P}{d \times t}$ where:

• $P$ is the applied force,
• $d$ is the diameter of the rivet,
• $t$ is the thickness of the plate.

Rearrange to solve for $d$: $d = \frac{P}{\sigma_b \times t}$

Since $P$ is not given directly, let's proceed with the shearing stress to express $P$.

The shearing stress $\tau$ is given by: $\tau = \frac{P}{A_s}$ where:

• $A_s = \frac{\pi d^2}{4}$ is the cross-sectional area of the rivet under shear.

Rearrange to solve for $d$: $P = \tau \times A_s = \tau \times \frac{\pi d^2}{4}$

Using this, substitute into the bearing stress equation to find $d$: $\sigma_b \times d \times t = \tau \times \frac{\pi d^2}{4}$ Simplify and solve for $d$: $d = \frac{4 \sigma_b \times t}{\pi \tau}$

Now substitute the given values: $d = \frac{4 \times 100 \, \text{MPa} \times 10 \, \text{mm}}{\pi \times 50 \, \text{MPa}}$

Calculating: $d = \frac{4000}{157.08} \approx 25.46 \, \text{mm}$

Therefore, the diameter of the rivet, $d \approx 25.46 \, \text{mm}$.

Part (b): Maximum Tensile Stress in the Plates

The tensile stress $\sigma_t$ in the plates can be calculated by: $\sigma_t = \frac{P}{A_t}$ where:

• $A_t = w \times t$ is the cross-sectional area of the plate.

Using $P$ derived from the shearing stress of the rivet: $P = \tau \times \frac{\pi d^2}{4}$

Substitute $P$ and $A_t$: $\sigma_t = \frac{\tau \times \frac{\pi d^2}{4}}{w \times t}$

Plugging in the values: $\sigma_t = \frac{50 \times \frac{\pi (25.46)^2}{4}}{120 \times 10}$

Calculating $P$ first: $P = 50 \times \frac{\pi (25.46)^2}{4} \approx 25465.57 \, \text{N}$

Now, find $\sigma_t$: $\sigma_t = \frac{25465.57}{1200} \approx 21.22 \, \text{MPa}$

Therefore, the maximum tensile stress in the plates, $\sigma_t \approx 21.22 \, \text{MPa}$.

Filling in the Blanks:

1. $\sigma_b = 100 \, \text{MPa}$
2. $t = 10 \, \text{mm}$
3. $A_b = w \times t = 120 \times 10 = 1200 \, \text{mm}^2$
4. $\sigma_t = 21.22 \, \text{MPa}$
5. $\tau = 50 \, \text{MPa}$
6. $\sigma (\text{tensile stress}) = 21.22 \, \text{MPa}$
7. $d_{\text{rivet}} = 25.46 \, \text{mm}$

Would you like more details or have any questions?

Here are five related questions you might find helpful:

1. What are other failure modes that can occur in riveted joints?
2. How does increasing the plate thickness affect the required rivet diameter?
3. How would you determine the tensile strength of a rivet itself?
4. How can you reduce the shear stress on the rivet in this setup?
5. What factors influence the choice of rivet material in engineering applications?

Tip: When designing joints with rivets, always check for both shear and bearing stresses, as they often have different allowable limits based on material properties.