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Let's break down the problem step by step using the given dataLet's break down the problem step by step using the given 

### Given:
- Young's modulus \( E = 150 \, \text{GPa} = 150,000 \, \text{MPa} \)
- Force \( F_1 = 50 \, \text{kN} \) (acting on the top section BC)
- Force \( F_2 = 20 \, \text{kN} \) (acting on the lower section AB)
- Lengths:
  - \( L_1 = 220 \, \text{mm} \) (BC segment)
  - \( L_2 = 250 \, \text{mm} \) (AB segment)
- Cross-sectional areas:
  - \( A_1 = 140 \, \text{mm}^2 \) (BC segment)
  - \( A_2 = 410 \, \text{mm}^2 \) (AB segment)

### To Calculate:
1. **Internal forces**:
   - \( F_{BC} \) in kN (force in segment BC)
   - \( F_{AB} \) in kN (force in segment AB)
2. **Stresses**:
   - \( \sigma_{BC} \) in MPa (stress in segment BC)
   - \( \sigma_{AB} \) in MPa (stress in segment AB)
3. **Total displacement** at point C (\( \delta_C \)) in mm.

---

### Step 1: Calculate the Internal Forces

Since the bar is prismatic and subjected to axial forces, the internal forces are balanced by the external forces applied.

- \( F_{BC} = F_1 = 50 \, \text{kN} \) (since the external force \( F_1 \) is applied to segment BC)
- \( F_{AB} = F_1 + F_2 = 50 + 20 = 70 \, \text{kN} \) (since both \( F_1 \) and \( F_2 \) contribute to the force in AB)

---

### Step 2: Calculate the Stresses

Stress is given by the formula:
\[
\sigma = \frac{F}{A}
\]
Where:
- \( F \) is the internal force in the segment (in kN)
- \( A \) is the cross-sectional area (in mm²)

#### For BC:
\[
\sigma_{BC} = \frac{F_{BC}}{A_1} = \frac{50 \, \text{kN}}{140 \, \text{mm}^2} = \frac{50,000 \, \text{N}}{140 \, \text{mm}^2} = 357.14 \, \text{MPa}
\]

#### For AB:
\[
\sigma_{AB} = \frac{F_{AB}}{A_2} = \frac{70 \, \text{kN}}{410 \, \text{mm}^2} = \frac{70,000 \, \text{N}}{410 \, \text{mm}^2} = 170.73 \, \text{MPa}
\]

---

### Step 3: Calculate the Total Displacement at Point C

The displacement \( \delta \) of a bar under axial load is given by:
\[
\delta = \frac{F L}{A E}
\]
Where:
- \( F \) is the internal force
- \( L \) is the length of the segment
- \( A \) is the cross-sectional area
- \( E \) is Young's modulus

#### For BC:
\[
\delta_{BC} = \frac{F_{BC} L_1}{A_1 E} = \frac{50,000 \times 220}{140 \times 150,000} = \frac{11,000,000}{21,000,000} = 0.5238 \, \text{mm}
\]

#### For AB:
\[
\delta_{AB} = \frac{F_{AB} L_2}{A_2 E} = \frac{70,000 \times 250}{410 \times 150,000} = \frac{17,500,000}{61,500,000} = 0.2846 \, \text{mm}
\]

#### Total Displacement at Point C:
\[
\delta_C = \delta_{BC} + \delta_{AB} = 0.5238 + 0.2846 = 0.8084 \, \text{mm}
\]

---

### Final Results:
- \( F_{BC} = 50 \, \text{kN} \)
- \( F_{AB} = 70 \, \text{kN} \)
- \( \sigma_{BC} = 357.14 \, \text{MPa} \)
- \( \sigma_{AB} = 170.73 \, \text{MPa} \)
- \( \delta_C = 0.81 \, \text{mm} \)

Let me know if you'd like further details or have any questions!

---

### Related Questions:
1. How would the stress values change if \( A_1 \) and \( A_2 \) were swapped?
2. How does changing Young's modulus affect the total displacement?
3. What would be the effect if the lengths \( L_1 \) and \( L_2 \) were interchanged?
4. How would the system behave if \( F_2 \) was a tensile force instead of a compressive one?
5. What happens to the displacement if only \( F_2 \) was applied and \( F_1 = 0 \)?

### Tip:
Remember, stress is a function of both force and cross-sectional area, so increasing the area reduces the stress for a given force.

The figure shows a prismatic bar subjected to axial forces F1 and F2. Given: Young's modulus E = 150 GPa, forces F1 = 50 kN, F2 = 20 kN, lengths L1 = 220 mm and L2 = 250 mm, cross-section areas A1 = 140 mm² and A2 = 410 mm², calculate the internal forces FBC and FAB, stresses σBC and σAB, and total displacement δC.

Solve for internal forces, stresses, and displacement in a prismatic bar subjected to axial forces. Using Young's modulus, force values, lengths, and cross-sectional areas, this problem illustrates key concepts in mechanics of materials. Suitable for undergraduate engineering students.

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