To compute the vertical displacement of the roller at C, we need to approach the problem by analyzing the structure and forces acting on it. Here's a step-by-step breakdown of the method we will follow:

Given Data:

• Young's Modulus, E = 200 GPa = $200 \times 10^9 \, \text{Pa}$
• Cross-sectional area of the rod at B, A = 300 mm² = $300 \times 10^{-6} \, \text{m}^2$
• Length of the rod at B, L = 4 m
• Load, F = 70 kN = $70 \times 10^3 \, \text{N}$
• Geometry:
  • Distances: AB = 5 m, BC = 2 m, CD = 2.5 m (left and right)

Approach:

1. Static equilibrium analysis: The first step is to analyze the forces and moments acting on the rigid bar to determine the internal forces.

2. Compatibility condition: Use the fact that the bar is pinned at points A and D, and the steel rod at B supports the vertical load, while the roller at C allows vertical displacement.

3. Displacement calculation: Using the force in the steel rod at B, we can compute the elongation in the rod (due to axial deformation), which will help us find the vertical displacement at C.

Step 1: Equilibrium Equations

• Let the vertical reactions at A and D be $R_A$ and $R_D$, respectively.
• The force in the steel rod at B is denoted by $F_B$.

Using moment equilibrium around point D (sum of moments = 0): $\sum M_D = 0: \quad R_A \cdot (10 \, \text{m}) - 70 \, \text{kN} \cdot (2.5 \, \text{m}) = 0$ Solving for $R_A$: $R_A = \frac{70 \times 2.5}{10} = 17.5 \, \text{kN}$

Next, use the vertical force equilibrium (sum of vertical forces = 0): $\sum F_y = 0: \quad R_A + R_D = 70 \, \text{kN}$ Substituting $R_A = 17.5 \, \text{kN}$: $R_D = 70 - 17.5 = 52.5 \, \text{kN}$

Step 2: Force in the Rod at B

The vertical displacement of point C is affected by the deformation of the steel rod at B. The force in the rod $F_B$ is related to the displacement via Hooke's law for axial deformation: $\delta_B = \frac{F_B L}{A E}$

We need to determine $F_B$, which will be based on the deflections of the structure. Let me calculate the next steps to compute the deformation.The elongation of the steel rod at point B is approximately $1.17 \, \text{mm}$.

Step 3: Vertical Displacement at C

The vertical displacement at C ($\delta_C$) will be influenced by the deformation at B, since point C is constrained to move due to the roller. Given that the displacement at C is related to the geometry of the rigid bars, and point C is located between B and D, we can compute it based on the relative distances.

The proportion of vertical displacement at C due to the displacement at B is given by: $\delta_C = \frac{\text{distance from C to D}}{\text{distance from B to D}} \cdot \delta_B$ Substituting the values: $\delta_C = \frac{2.5 \, \text{m}}{5 \, \text{m}} \cdot \delta_B$

Let me calculate this next.The vertical displacement of the roller at point C is approximately $0.583 \, \text{mm}$.

Would you like a more detailed explanation of any specific step, or have any other questions?

Here are 5 related questions to explore further:

1. How would the displacement change if the cross-sectional area of the steel rod at B were doubled?
2. How does the distance between points B and D affect the vertical displacement at C?
3. What effect would a higher Young's modulus have on the system's behavior?
4. How would the forces in the system change if the load at C were 100 kN instead of 70 kN?
5. Could you apply the same method to find the horizontal displacement, if applicable?

Tip: Always check the units when calculating deformations, as working with consistent SI units (meters, Newtons, Pascals) prevents calculation errors.