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This problem involves the calculation of rivet dimensions and plate width for a plate connection subjected to a tensile load of 10 kN.

### Given Data:
- **Load (P)**: 10 kN = 10,000 N
- **Permissible tensile stress of plate material** (\( \sigma_t \)): 137.3 N/mm²
- **Permissible shear stress of rivet material** (\( \tau \)): 109.8 N/mm²
- **Thickness of plate (t)**: 5 mm
- **Number of rivets (n)**: 2

We need to determine:
1. **Diameter of rivet (d)**
2. **Distance between rivets**
3. **Required width of the plate**

### Step-by-Step Solution:

#### a. **Determine the appropriate rivet diameter:**

The rivet is subjected to shear. To find the diameter, we first calculate the load each rivet must carry in shear:

\[
P_{rivet} = \frac{P}{n} = \frac{10,000}{2} = 5,000 \text{ N}
\]

The shear capacity of each rivet can be determined using the formula for shear:

\[
P_{rivet} = \tau \times A
\]

where:
- \( A \) is the cross-sectional area of the rivet in shear, which for a circular rivet is \( \frac{\pi d^2}{4} \)
- \( \tau \) is the allowable shear stress

Rearranging for diameter \( d \):

\[
5000 = 109.8 \times \frac{\pi d^2}{4}
\]

\[
d^2 = \frac{5000 \times 4}{109.8 \times \pi}
\]

\[
d = \sqrt{\frac{20,000}{344.91}} \approx \sqrt{58.00} \approx 7.62 \text{ mm}
\]

Thus, the **diameter of the rivet** is approximately 7.62 mm.

#### b. **Determine the spacing between the rivets:**

The minimum spacing (pitch) between rivets in a single row should be approximately 3 to 4 times the diameter of the rivet to avoid stress concentration and ensure proper load distribution. Therefore:

\[
\text{Spacing} \approx 3 \times d = 3 \times 7.62 \approx 22.86 \text{ mm}
\]

#### c. **Determine the required width of the plate:**

To calculate the required width of the plate, we need to ensure that the plate can carry the tensile load safely. The tensile strength of the plate is given by:

\[
P = \sigma_t \times t \times w
\]

Where:
- \( \sigma_t \) is the allowable tensile stress
- \( t \) is the thickness of the plate
- \( w \) is the required width of the plate

Rearranging for \( w \):

\[
w = \frac{P}{\sigma_t \times t}
\]

Substituting the values:

\[
w = \frac{10,000}{137.3 \times 5} \approx \frac{10,000}{686.5} \approx 14.57 \text{ mm}
\]

Thus, the **required width of the plate** is approximately 14.57 mm.

### Final Answers:
- **a. Diameter of rivet**: 7.62 mm
- **b. Spacing between rivets**: 22.86 mm
- **c. Required plate width**: 14.57 mm

Let me know if you'd like more details or further clarification. Here are 5 related questions to expand your understanding:

1. How would the results change if the number of rivets were increased to 4?
2. What if the allowable tensile stress of the plate material were lower?
3. How do we calculate the maximum load the rivet system can carry?
4. How would the results change if the plate thickness were increased to 10 mm?
5. Can we use the same rivet diameter for different materials with lower shear stress?

**Tip:** Always ensure that the material properties (shear and tensile stress) are compatible with the load and design requirements to prevent failure in structural connections.

Dua buah plat akan disambung dengan kampuh bilah tunggal dikeliling tunggal satu baris, direncanakan menerima beban sebesar 10 kN. Bila bahan plat mempunyai tegangan tarik izin 137,3 N/mm² dan bahan paku dengan tegangan geser izinnya 109,8 N/mm², tebal plat 5 mm dan jumlah paku yang digunakan sebanyak 2 buah. Tentukan: a. Diameter paku keling yang sesuai. b. Jarak antar paku keling. c. Lebar plat yang dibutuhkan.

Solve a problem involving tensile and shear stress in a plate-rivet connection, where two plates are joined with rivets and subjected to a 10 kN load. Learn how to calculate the diameter of rivets, spacing, and required plate width. Given tensile stress of 137.3 N/mm², shear stress of 109.8 N/mm², and plate thickness of 5 mm.

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