Question
Two round shafts both of diameter 80 mm are to be joined by a fillet weld that is applied around the complete circumference.
A 5 mm weld size is used with no return welds. The permissible tensile stress in the joint must not exceed 60 MPa. Calculate a safe tensile load.
A 5 mm weld size is used with no return welds. The permissible tensile stress in the joint must not exceed 60 MPa. Calculate a safe tensile load.
Answers
GPT-4o mini
To calculate the safe tensile load for the fillet weld joining the two round shafts, we need to follow these steps:
1. **Calculate the effective throat of the weld**:
The effective throat of a fillet weld is given by the formula:
\[
t = \frac{a \cdot \sqrt{2}}{2}
\]
where \( a \) is the size of the weld. In this case, \( a = 5 \, \text{mm} \).
Plugging in the values:
\[
t = \frac{5 \cdot \sqrt{2}}{2} = \frac{5 \cdot 1.4142}{2} \approx 3.54 \, \text{mm}
\]
2. **Calculate the perimeter of the weld**:
Since the weld is applied around the complete circumference of the shaft, we first need to calculate the circumference of the shafts. The diameter of each shaft is \( 80 \, \text{mm} \), so the circumference \( C \) is given by:
\[
C = \pi \cdot d = \pi \cdot 80 \, \text{mm} \approx 251.33 \, \text{mm}
\]
3. **Calculate the weld cross-sectional area**:
The cross-sectional area \( A_w \) of the weld can be calculated as:
\[
A_w = \text{Effective Throat} \times \text{Length of weld}
\]
The length of the weld is equal to the circumference, thus:
\[
A_w = t \cdot C = 3.54 \, \text{mm} \cdot 251.33 \, \text{mm} \approx 891.63 \, \text{mm}^2
\]
4. **Calculate the safe tensile load**:
The safe tensile load \( P \) can now be calculated using the permissible tensile stress \( \sigma \) in the weld:
\[
P = \sigma \cdot A_w
\]
Given that \( \sigma = 60 \, \text{MPa} = 60 \, \text{N/mm}^2 \), we can calculate:
\[
P = 60 \, \text{N/mm}^2 \cdot 891.63 \, \text{mm}^2 \approx 53497.8 \, \text{N}
\]
5. **Convert to more practical units**:
To convert this to kilonewtons (kN):
\[
P \approx 53.5 \, \text{kN}
\]
**Final Result**:
The safe tensile load for the weld is approximately \( 53.5 \, \text{kN} \).
1. **Calculate the effective throat of the weld**:
The effective throat of a fillet weld is given by the formula:
\[
t = \frac{a \cdot \sqrt{2}}{2}
\]
where \( a \) is the size of the weld. In this case, \( a = 5 \, \text{mm} \).
Plugging in the values:
\[
t = \frac{5 \cdot \sqrt{2}}{2} = \frac{5 \cdot 1.4142}{2} \approx 3.54 \, \text{mm}
\]
2. **Calculate the perimeter of the weld**:
Since the weld is applied around the complete circumference of the shaft, we first need to calculate the circumference of the shafts. The diameter of each shaft is \( 80 \, \text{mm} \), so the circumference \( C \) is given by:
\[
C = \pi \cdot d = \pi \cdot 80 \, \text{mm} \approx 251.33 \, \text{mm}
\]
3. **Calculate the weld cross-sectional area**:
The cross-sectional area \( A_w \) of the weld can be calculated as:
\[
A_w = \text{Effective Throat} \times \text{Length of weld}
\]
The length of the weld is equal to the circumference, thus:
\[
A_w = t \cdot C = 3.54 \, \text{mm} \cdot 251.33 \, \text{mm} \approx 891.63 \, \text{mm}^2
\]
4. **Calculate the safe tensile load**:
The safe tensile load \( P \) can now be calculated using the permissible tensile stress \( \sigma \) in the weld:
\[
P = \sigma \cdot A_w
\]
Given that \( \sigma = 60 \, \text{MPa} = 60 \, \text{N/mm}^2 \), we can calculate:
\[
P = 60 \, \text{N/mm}^2 \cdot 891.63 \, \text{mm}^2 \approx 53497.8 \, \text{N}
\]
5. **Convert to more practical units**:
To convert this to kilonewtons (kN):
\[
P \approx 53.5 \, \text{kN}
\]
**Final Result**:
The safe tensile load for the weld is approximately \( 53.5 \, \text{kN} \).