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The given structure shows a portal frame subjected to a horizontal force of 50 kN at the top left corner of point \( D \). The frame consists of vertical columns \( AB \), \( BC \), and horizontal beams \( DE \), \( EF \) with fixed supports at points \( A \) and \( C \).

### Steps to Identify Forces Using the Portal Method:
The portal method is a simplified structural analysis technique primarily used for analyzing frames subjected to lateral loads, such as wind or earthquake loads. It assumes the following:
1. The columns are subject to axial forces, bending moments, and shear forces.
2. Internal hinges are assumed at midpoints of the horizontal beams (between \( D \) and \( E \), \( E \) and \( F \)).
3. The columns take on equal shear forces at each story level.

#### 1. **Shear Forces:**
The lateral load of 50 kN will be distributed equally between the two columns (\( AB \) and \( BC \)) since the portal method assumes equal distribution of shear in the columns.
- Shear force in \( AB \) and \( BC \):
  \[
  V_{AB} = V_{BC} = \frac{50 \, \text{kN}}{2} = 25 \, \text{kN}
  \]

#### 2. **Axial (Normal) Forces:**
The axial forces in the columns are determined by the fact that the sum of the axial forces in each column must resist the moments caused by the lateral loads.
- Assuming equal axial forces in \( AB \) and \( BC \), and the overall moment is zero at the base due to symmetry.

#### 3. **Bending Moments:**
- In the portal frame, the bending moment at the base of the columns (points \( A \) and \( C \)) and the top corners of the frame will resist the applied force of 50 kN. Assuming that the moment at the center of the beams \( DE \) and \( EF \) is zero due to internal hinges:
  - The moment at the base of the columns will balance the horizontal force.

Let’s calculate the moment at points \( A \) and \( C \) using:
\[
M_A = M_C = V_{AB} \times \frac{h}{2} = 25 \, \text{kN} \times \frac{6 \, \text{m}}{2} = 75 \, \text{kN} \cdot \text{m}
\]

#### Summary of Forces:
- **Shear Force** at columns \( AB \) and \( BC \): 25 kN each.
- **Axial Forces**: Balanced to prevent lateral sway.
- **Moment** at supports \( A \) and \( C \): 75 kN·m each.

Would you like to see more detailed calculations or ask anything else?

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Here are 5 related questions to expand your understanding:
1. How does the distribution of shear forces change if the frame was asymmetrical?
2. What happens to the moment if the columns had different heights?
3. How would axial forces vary in each column if the applied force was positioned off-center?
4. What assumptions does the portal method rely on for simplification?
5. How would internal hinges affect the moment distribution across the beams?

**Tip**: In portal method analysis, ensure that the load distribution assumptions are valid, especially for frames with non-uniform members.

Identifikasi gaya-gaya yang ada dalam struktur di bawah ini dgn metode portal (geser, normal, momen)

This analysis focuses on identifying shear forces, axial forces, and moments in a portal frame under a horizontal load of 50 kN. Key topics include distribution of shear forces, calculation of axial forces, and moment at the supports using the portal method, suitable for civil engineering students.

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