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The given problem is related to calculating the axial force \( N_c \) acting on a member based on its cross-sectional pressure-bearing capacity. Let's break it down step by step:

### Given:
1. **Upper beam cross-section**:
   - Width \( b = 160 \, \text{mm} \)
   - Height \( h = 180 \, \text{mm} \)

2. **Inclined beam cross-section**:
   - Width \( b = 160 \, \text{mm} \)
   - Height \( h = 100 \, \text{mm} \)

3. **Inclined force \( D_1 \)**:
   - Magnitude \( D_1 = 13.94 \, \text{kN} \)

4. **Angle between inclined beam and upper beam**:
   - \( \alpha = 60^\circ \)

5. **Compressive strength of the concrete**:
   - \( f_{c60} = 4.1 \, \text{N/mm}^2 \)

### Required:
The axial force \( N_c \) based on the pressure-bearing capacity of the compressive area.

### Formula:
To find the compressive force \( N_c \), we generally use:

\[
N_c = f_c \times A
\]

Where:
- \( f_c \) is the compressive strength of the material.
- \( A \) is the area of the cross-section resisting the load, which depends on the geometry and applied force.

### Steps:
1. **Determine the compressive area**:
   - The compressive area in this case would be the cross-sectional area of the inclined beam.
   - Area \( A = b \times h \) where \( b = 160 \, \text{mm} \) and \( h = 100 \, \text{mm} \).

2. **Resolve the force component**:
   - Since the inclined force \( D_1 = 13.94 \, \text{kN} \) acts at an angle of \( 60^\circ \), we need to resolve this into its vertical component.
   - Vertical force \( F_v = D_1 \times \sin(\alpha) \).

3. **Calculate the compressive stress**:
   - From the stress formula \( \sigma = \frac{F_v}{A} \), we compare it with the allowable compressive stress \( f_{c60} \) to ensure the structure's safety.

Let me go ahead and calculate these values for you.### Calculations:
1. **Area of the inclined beam cross-section**:  
   \[
   A_{\text{inclined}} = 160 \, \text{mm} \times 100 \, \text{mm} = 16000 \, \text{mm}^2
   \]

2. **Vertical component of the inclined force**:
   \[
   F_{\text{vertical}} = 13.94 \, \text{kN} \times \sin(60^\circ) = 12072.39 \, \text{N}
   \]

3. **Compressive force \( N_c \)** (based on the pressure-bearing capacity):
   \[
   N_c = \min\left(F_{\text{vertical}}, f_c \times A_{\text{inclined}}\right) = \min(12072.39 \, \text{N}, 4.1 \times 16000) = 12072.39 \, \text{N}
   \]

Thus, the axial force \( N_c \) is approximately **12072.39 N**.

Would you like more details on any specific step? Here are five related questions that could help expand your understanding:

1. How do we determine the horizontal component of a force acting at an angle?
2. What is the significance of limiting force by the compressive strength?
3. How would the calculation change if the angle \( \alpha \) were different?
4. How can we ensure the structure remains safe under increasing loads?
5. What role does the cross-sectional area play in structural strength?

**Tip**: Always ensure that the resolved components of forces are correctly aligned with the structural members they affect to avoid errors in calculations.

This problem focuses on calculating the axial force Nc acting on a member using its cross-sectional pressure-bearing capacity. Learn the step-by-step solution involving force resolution, compressive area determination, and stress calculation. Suitable for civil engineering students.

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