Two long metallic strips are joined together by two rivets each of radius `2 mm`. Each rivet can withstand a maximum shearing stress of `1.5 xx 10^(9) N//m^(2)`. Assuming that each rivet shares the stretching load equally, the maximum tensile force the strip can exert without rupture is

Options

A. `1.88xx10^(4)N`
B. `3.8xx10^(4)N`
C. `6xx10^(7)N`
D. `3xx10^(4)N`

Explanation

To find the maximum tensile force that the metallic strips can exert without rupture, we need to follow these steps:

### Step 1: Calculate the area of one rivet
The area \( A \) of a circular rivet can be calculated using the formula:
\[ A = \pi r^2 \]
where \( r \) is the radius of the rivet.

Given that the radius \( r = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \):
\[ A = \pi (2 \times 10^{-3})^2 = \pi (4 \times 10^{-6}) \approx 1.25664 \times 10^{-5} \, \text{m}^2 \]

### Step 2: Calculate the total area of both rivets
Since there are two rivets, the total area \( A_{total} \) is:
\[ A_{total} = 2 \times A = 2 \times 1.25664 \times 10^{-5} \approx 2.51328 \times 10^{-5} \, \text{m}^2 \]

### Step 3: Calculate the maximum shear force for one rivet
The maximum shear force \( F_{max} \) that one rivet can withstand is given by:
\[ F_{max} = \tau_{max} \times A \]
where \( \tau_{max} \) is the maximum shearing stress.

Given \( \tau_{max} = 1.5 \times 10^9 \, \text{N/m}^2 \):
\[ F_{max, one} = (1.5 \times 10^9) \times (1.25664 \times 10^{-5}) \approx 18899.2 \, \text{N} \]

### Step 4: Calculate the total maximum tensile force
Since there are two rivets sharing the load equally, the total maximum tensile force \( F_{total} \) is:
\[ F_{total} = 2 \times F_{max, one} = 2 \times 18899.2 \approx 37798.4 \, \text{N} \]

### Final Answer
The maximum tensile force that the strip can exert without rupture is approximately:
\[ F_{total} \approx 37798.4 \, \text{N} \]

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