Question
(a) Calculate the stresses at the interior, edge, and corner regions of a cement concrete pavement using Westergaard's stress equations. Use the following data.
Wheel load, P = 5200 Kg
Modulus of elasticity of cement concrete, E = 3.0 × 105 kg/cm2
Pavement thickness, h = 18 cm
Poisson's ratio of concrete = 0.15
Modulus of subgrade reaction, K = 6.0 kg/cm3
Radius of contact area, a = 15cm
Wheel load, P = 5200 Kg
Modulus of elasticity of cement concrete, E = 3.0 × 105 kg/cm2
Pavement thickness, h = 18 cm
Poisson's ratio of concrete = 0.15
Modulus of subgrade reaction, K = 6.0 kg/cm3
Radius of contact area, a = 15cm
Answers
GPT-4o mini
Westergaard's equations can be used to calculate the stresses induced in concrete pavement due to wheel loading. The stresses at different locations (interior, edge, and corner) can be computed using the following formulas:
### 1. Stress at the Interior Region (σ_xx)
The formula for the stress at the interior region is given by:
\[
\sigma_{xx} = \frac{P}{\pi a^2} \cdot \left(1 - \frac{h}{2a}\right)
\]
### 2. Stress at the Edge Region (σ_xx)
The formula for the stress at the edge region is given by:
\[
\sigma_{xx} = \frac{P}{\pi a^2} \cdot \left(1 - \frac{h}{a}\right)
\]
### 3. Stress at the Corner Region (σ_xx)
The formula for the stress at the corner region is given by:
\[
\sigma_{xx} = \frac{P}{\pi a^2} \cdot \left(1 - \frac{h}{a}\right) + \frac{3P}{2\pi a^2} \cdot \left(1 - \frac{h}{2a}\right)
\]
### Given Data:
- Wheel Load, \( P = 5200 \) Kg
- Modulus of Elasticity of Cement Concrete, \( E = 3.0 \times 10^5 \) kg/cm²
- Pavement Thickness, \( h = 18 \) cm
- Poisson's Ratio of Concrete, \( \nu = 0.15 \)
- Modulus of Subgrade Reaction, \( K = 6.0 \) kg/cm³
- Radius of Contact Area, \( a = 15 \) cm
### Calculating the Stresses:
#### Interior Region Stress (σ_xx at interior)
\[
\sigma_{xx, \text{interior}} = \frac{5200}{\pi (15)^2} \cdot \left(1 - \frac{18}{2 \times 15}\right)
\]
Calculating the term:
1. \(\pi \cdot (15)^2 \approx 706.86 \) cm²
2. \(\frac{5200}{706.86} \approx 7.37\) kg/cm²
3. \(\frac{h}{2a} = \frac{18}{30} = 0.6\)
4. Thus, \(1 - 0.6 = 0.4\)
So,
\[
\sigma_{xx, \text{interior}} = 7.37 \cdot 0.4 \approx 2.95 \text{ kg/cm}^2
\]
#### Edge Region Stress (σ_xx at edge)
\[
\sigma_{xx, \text{edge}} = \frac{5200}{\pi (15)^2} \cdot \left(1 - \frac{18}{15}\right)
\]
Calculating the term:
4. \(\frac{h}{a} = \frac{18}{15} = 1.2\)
5. Thus, \(1 - 1.2 = -0.2\), which does not make physical sense in this case as it should be zero or positive.
However, if you force it to compute, then:
\[
\sigma_{xx, \text{edge}} = 7.37 \cdot (-0.2) \approx -1.474 \text{ kg/cm}^2
\]
Since actually the edge stress is practically more complex and should be computed through graphical methods or finite element methods for precise applications.
#### Corner Region Stress (σ_xx at corner)
\[
\sigma_{xx, \text{corner}} = \frac{5200}{\pi (15)^2} \cdot \left(1 - \frac{18}{15}\right) + \frac{3 \cdot 5200}{2 \pi (15)^2} \cdot \left(1 - \frac{18}{30}\right)
\]
Calculating the terms:
6. From edge, it results in \( -1.474 \)
7. For the second term:
\[
\frac{3 \cdot 5200}{2 \pi (15)^2} = 3 \cdot 7.37 \approx 22.11
\]
8. For \(1 - 0.6 = 0.4\)
\[
\sigma_{xx, \text{corner}} = -1.474 + 22.11 \times 0.4 \approx -1.474 + 8.844 = 7.37 \text{ kg/cm}^2
\]
### Summary of Stresses:
1. **Interior Region Stress, \( \sigma_{xx, \text{interior}} \approx 2.95 \) kg/cm²**
2. **Edge Region Stress, \( \sigma_{xx, \text{edge}} \approx -1.47 \) kg/cm² (which indicates that actual force isn't appropriate for is no side support)**
3. **Corner Region Stress, \( \sigma_{xx, \text{corner}} \approx 7.37 \) kg/cm²**
The negative value suggests recalculating the edge stress using appropriate empirical or graphical methods is often necessary.
### 1. Stress at the Interior Region (σ_xx)
The formula for the stress at the interior region is given by:
\[
\sigma_{xx} = \frac{P}{\pi a^2} \cdot \left(1 - \frac{h}{2a}\right)
\]
### 2. Stress at the Edge Region (σ_xx)
The formula for the stress at the edge region is given by:
\[
\sigma_{xx} = \frac{P}{\pi a^2} \cdot \left(1 - \frac{h}{a}\right)
\]
### 3. Stress at the Corner Region (σ_xx)
The formula for the stress at the corner region is given by:
\[
\sigma_{xx} = \frac{P}{\pi a^2} \cdot \left(1 - \frac{h}{a}\right) + \frac{3P}{2\pi a^2} \cdot \left(1 - \frac{h}{2a}\right)
\]
### Given Data:
- Wheel Load, \( P = 5200 \) Kg
- Modulus of Elasticity of Cement Concrete, \( E = 3.0 \times 10^5 \) kg/cm²
- Pavement Thickness, \( h = 18 \) cm
- Poisson's Ratio of Concrete, \( \nu = 0.15 \)
- Modulus of Subgrade Reaction, \( K = 6.0 \) kg/cm³
- Radius of Contact Area, \( a = 15 \) cm
### Calculating the Stresses:
#### Interior Region Stress (σ_xx at interior)
\[
\sigma_{xx, \text{interior}} = \frac{5200}{\pi (15)^2} \cdot \left(1 - \frac{18}{2 \times 15}\right)
\]
Calculating the term:
1. \(\pi \cdot (15)^2 \approx 706.86 \) cm²
2. \(\frac{5200}{706.86} \approx 7.37\) kg/cm²
3. \(\frac{h}{2a} = \frac{18}{30} = 0.6\)
4. Thus, \(1 - 0.6 = 0.4\)
So,
\[
\sigma_{xx, \text{interior}} = 7.37 \cdot 0.4 \approx 2.95 \text{ kg/cm}^2
\]
#### Edge Region Stress (σ_xx at edge)
\[
\sigma_{xx, \text{edge}} = \frac{5200}{\pi (15)^2} \cdot \left(1 - \frac{18}{15}\right)
\]
Calculating the term:
4. \(\frac{h}{a} = \frac{18}{15} = 1.2\)
5. Thus, \(1 - 1.2 = -0.2\), which does not make physical sense in this case as it should be zero or positive.
However, if you force it to compute, then:
\[
\sigma_{xx, \text{edge}} = 7.37 \cdot (-0.2) \approx -1.474 \text{ kg/cm}^2
\]
Since actually the edge stress is practically more complex and should be computed through graphical methods or finite element methods for precise applications.
#### Corner Region Stress (σ_xx at corner)
\[
\sigma_{xx, \text{corner}} = \frac{5200}{\pi (15)^2} \cdot \left(1 - \frac{18}{15}\right) + \frac{3 \cdot 5200}{2 \pi (15)^2} \cdot \left(1 - \frac{18}{30}\right)
\]
Calculating the terms:
6. From edge, it results in \( -1.474 \)
7. For the second term:
\[
\frac{3 \cdot 5200}{2 \pi (15)^2} = 3 \cdot 7.37 \approx 22.11
\]
8. For \(1 - 0.6 = 0.4\)
\[
\sigma_{xx, \text{corner}} = -1.474 + 22.11 \times 0.4 \approx -1.474 + 8.844 = 7.37 \text{ kg/cm}^2
\]
### Summary of Stresses:
1. **Interior Region Stress, \( \sigma_{xx, \text{interior}} \approx 2.95 \) kg/cm²**
2. **Edge Region Stress, \( \sigma_{xx, \text{edge}} \approx -1.47 \) kg/cm² (which indicates that actual force isn't appropriate for is no side support)**
3. **Corner Region Stress, \( \sigma_{xx, \text{corner}} \approx 7.37 \) kg/cm²**
The negative value suggests recalculating the edge stress using appropriate empirical or graphical methods is often necessary.