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To determine the elongation of a steel bar that tapers from a diameter of 100 mm at the top to 50 mm at the bottom over a length of 1 meter under its own weight, we'll follow a detailed, step-by-step solution. Given the unit weight of steel is 78.5 kN/m³ and the Young's modulus \( E \) is 200 GPa, let's proceed with the calculations.
### Step-by-Step Solution
#### Step 1: Parameters and Conversions
First, convert all necessary parameters into consistent SI units:
- Lower diameter (at the bottom), \( D_1 \): 50 mm = 50 × 10\(^{-3}\) m = 0.050 m
- Upper diameter (at the top), \( D_2 \): 100 mm = 100 × 10\(^{-3}\) m = 0.100 m
- Length of the bar, \( L \): 1 meter
- Unit weight of steel, \( \gamma \): 78.5 kN/m³ = 78.5 × 10³ N/m³
- Young's modulus, \( E \): 200 GPa = 200 × 10^9 Pa
#### Step 2: Radii Calculations
- Radius at the bottom, \( r_1 \): \( \frac{D_1}{2} = \frac{0.050}{2} \) m = 0.025 m
- Radius at the top, \( r_2 \): \( \frac{D_2}{2} = \frac{0.100}{2} \) m = 0.050 m
#### Step 3: Cross-sectional Area at Any Point \( x \)
To find the cross-sectional area at any point \( x \) along the length of the bar, we first find the radius at that point:
[tex]\[ r(x) = r_1 + \left(\frac{r_2 - r_1}{L}\right) x \][/tex]
The cross-sectional area \( A(x) \) at point \( x \) is given by:
[tex]\[ A(x) = \pi [r(x)]^2 = \pi \left[ r_1 + \left(\frac{r_2 - r_1}{L}\right) x \right]^2 \][/tex]
#### Step 4: Differential Element and Integrand
The weight of the bar is distributed along its length, thus the elongation \( \delta L \) due to a small segment \( dx \) can be described by:
[tex]\[ d(\delta L) = \frac{\gamma x \, dx}{E \cdot A(x)} \][/tex]
#### Step 5: Integration
Integrate from \( x = 0 \) to \( x = L \):
[tex]\[ \delta L = \int_0^L \frac{\gamma x \, dx}{E \cdot \pi \left[ r_1 + \left( \frac{r_2 - r_1}{L} \right) x \right]^2} \][/tex]
#### Step 6: Plug-In the Numbers
Solving this integral involves detailed calculus, but using the given parameters, after integration, the elongation of the bar can be found.
#### Step 7: Result
The resulting elongation of the bar, \(\delta L\), over its length \( L \) of 1 meter is:
[tex]\[ \delta L \approx 3.861 \times 10^{-5} \text{ meters} \][/tex]
Therefore, considering all derived steps and numerical integration:
- The lower radius \( r_1 = 0.025 \) m
- The upper radius \( r_2 = 0.050 \) m
- The length \( L = 1 \) m
- The elongation \( \delta L \approx 3.861 \times 10^{-5} \) meters
Hence, the elongation of the steel bar under its own weight is approximated to [tex]\( 3.861 \times 10^{-5} \)[/tex] meters.
### Step-by-Step Solution
#### Step 1: Parameters and Conversions
First, convert all necessary parameters into consistent SI units:
- Lower diameter (at the bottom), \( D_1 \): 50 mm = 50 × 10\(^{-3}\) m = 0.050 m
- Upper diameter (at the top), \( D_2 \): 100 mm = 100 × 10\(^{-3}\) m = 0.100 m
- Length of the bar, \( L \): 1 meter
- Unit weight of steel, \( \gamma \): 78.5 kN/m³ = 78.5 × 10³ N/m³
- Young's modulus, \( E \): 200 GPa = 200 × 10^9 Pa
#### Step 2: Radii Calculations
- Radius at the bottom, \( r_1 \): \( \frac{D_1}{2} = \frac{0.050}{2} \) m = 0.025 m
- Radius at the top, \( r_2 \): \( \frac{D_2}{2} = \frac{0.100}{2} \) m = 0.050 m
#### Step 3: Cross-sectional Area at Any Point \( x \)
To find the cross-sectional area at any point \( x \) along the length of the bar, we first find the radius at that point:
[tex]\[ r(x) = r_1 + \left(\frac{r_2 - r_1}{L}\right) x \][/tex]
The cross-sectional area \( A(x) \) at point \( x \) is given by:
[tex]\[ A(x) = \pi [r(x)]^2 = \pi \left[ r_1 + \left(\frac{r_2 - r_1}{L}\right) x \right]^2 \][/tex]
#### Step 4: Differential Element and Integrand
The weight of the bar is distributed along its length, thus the elongation \( \delta L \) due to a small segment \( dx \) can be described by:
[tex]\[ d(\delta L) = \frac{\gamma x \, dx}{E \cdot A(x)} \][/tex]
#### Step 5: Integration
Integrate from \( x = 0 \) to \( x = L \):
[tex]\[ \delta L = \int_0^L \frac{\gamma x \, dx}{E \cdot \pi \left[ r_1 + \left( \frac{r_2 - r_1}{L} \right) x \right]^2} \][/tex]
#### Step 6: Plug-In the Numbers
Solving this integral involves detailed calculus, but using the given parameters, after integration, the elongation of the bar can be found.
#### Step 7: Result
The resulting elongation of the bar, \(\delta L\), over its length \( L \) of 1 meter is:
[tex]\[ \delta L \approx 3.861 \times 10^{-5} \text{ meters} \][/tex]
Therefore, considering all derived steps and numerical integration:
- The lower radius \( r_1 = 0.025 \) m
- The upper radius \( r_2 = 0.050 \) m
- The length \( L = 1 \) m
- The elongation \( \delta L \approx 3.861 \times 10^{-5} \) meters
Hence, the elongation of the steel bar under its own weight is approximated to [tex]\( 3.861 \times 10^{-5} \)[/tex] meters.
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