Question
Two hollow shafts 120 mm in diameter are joined using a fillet weld. If a force of 600 kN is applied and permissible shear stress in the weld is 68 MPa, calculate the weld size.
Answers
GPT-4o mini
To calculate the required weld size for the given hollow shafts connected by a fillet weld, we can use the following steps:
### Given Data:
- **Diameter of the hollow shaft (D)** = 120 mm
- **Force (F)** = 600 kN = \( 600 \times 10^3 \) N
- **Permissible shear stress in the weld (\( \tau \))** = 68 MPa = \( 68 \times 10^6 \) Pa
### Step 1: Calculate the required shear area
The shear area of the weld can be calculated using the formula:
\[
A_s = \frac{F}{\tau}
\]
where:
- \( A_s \) is the shear area of the weld,
- \( F \) is the applied force,
- \( \tau \) is the permissible shear stress.
### Step 2: Substitute in the values
Substitute the known values into the formula:
\[
A_s = \frac{600 \times 10^3}{68 \times 10^6}
\]
Calculating \( A_s \):
\[
A_s = \frac{600000}{68000000} = 0.008823529 \, \text{m}^2 = 8823.53 \, \text{mm}^2
\]
### Step 3: Relate shear area to weld size
For a fillet weld, the shear area (A_s) can be calculated as:
\[
A_s = L \cdot t
\]
where:
- \( L \) is the length of the weld,
- \( t \) is the size of the weld.
Assuming that the weld is applied all around the perimeter of the shaft, the length \( L \) can be approximated as the circumference of the shaft:
\[
L = \pi \cdot D
\]
Substituting the diameter \( D \):
\[
L = \pi \cdot 120 \, \text{mm} \approx 376.99 \, \text{mm}
\]
### Step 4: Substitute and solve for weld size \( t \)
Substituting \( L \) back into the shear area equation:
\[
A_s = \pi \cdot D \cdot t
\]
\[
8823.53 = \pi \cdot 120 \cdot t
\]
Now we can solve for \( t \):
\[
t = \frac{8823.53}{\pi \cdot 120}
\]
Calculating \( t \):
\[
t = \frac{8823.53}{376.99} \approx 23.36 \, \text{mm}
\]
### Conclusion
The required size of the fillet weld should be approximately **23.36 mm**.
For practical applications, typically, we round it to the nearest standard weld size or consider any other design factors, so you might choose a weld size based on standard sizes available (e.g. 25 mm).
Thus, the calculated need is about **23.36 mm** for non-standard, and **25 mm** for practical applications.
### Given Data:
- **Diameter of the hollow shaft (D)** = 120 mm
- **Force (F)** = 600 kN = \( 600 \times 10^3 \) N
- **Permissible shear stress in the weld (\( \tau \))** = 68 MPa = \( 68 \times 10^6 \) Pa
### Step 1: Calculate the required shear area
The shear area of the weld can be calculated using the formula:
\[
A_s = \frac{F}{\tau}
\]
where:
- \( A_s \) is the shear area of the weld,
- \( F \) is the applied force,
- \( \tau \) is the permissible shear stress.
### Step 2: Substitute in the values
Substitute the known values into the formula:
\[
A_s = \frac{600 \times 10^3}{68 \times 10^6}
\]
Calculating \( A_s \):
\[
A_s = \frac{600000}{68000000} = 0.008823529 \, \text{m}^2 = 8823.53 \, \text{mm}^2
\]
### Step 3: Relate shear area to weld size
For a fillet weld, the shear area (A_s) can be calculated as:
\[
A_s = L \cdot t
\]
where:
- \( L \) is the length of the weld,
- \( t \) is the size of the weld.
Assuming that the weld is applied all around the perimeter of the shaft, the length \( L \) can be approximated as the circumference of the shaft:
\[
L = \pi \cdot D
\]
Substituting the diameter \( D \):
\[
L = \pi \cdot 120 \, \text{mm} \approx 376.99 \, \text{mm}
\]
### Step 4: Substitute and solve for weld size \( t \)
Substituting \( L \) back into the shear area equation:
\[
A_s = \pi \cdot D \cdot t
\]
\[
8823.53 = \pi \cdot 120 \cdot t
\]
Now we can solve for \( t \):
\[
t = \frac{8823.53}{\pi \cdot 120}
\]
Calculating \( t \):
\[
t = \frac{8823.53}{376.99} \approx 23.36 \, \text{mm}
\]
### Conclusion
The required size of the fillet weld should be approximately **23.36 mm**.
For practical applications, typically, we round it to the nearest standard weld size or consider any other design factors, so you might choose a weld size based on standard sizes available (e.g. 25 mm).
Thus, the calculated need is about **23.36 mm** for non-standard, and **25 mm** for practical applications.