Question

An L-shaped elastic member ABC with slender arms AB and BC of uniform cross-section is clamped at end A and connected to a pin at end C. The pin remains in continuous contact with and is constrained to move in a smooth horizontal slot. The section modulus of the member is same in both the arms. The end C is subjected to a horizontal force \( P \) and all the deflections are in the plane of the figure. Given the length AB is \( 4a \) and length BC is \( a \), the magnitude and direction of the normal force on the pin from the slot, respectively, are.

• A. \( \frac{3P}{8}, \) and downwards
• B. \( \frac{5P}{8}, \) and upwards
• C. \( \frac{P}{4}, \) and downwards
• D. \( \frac{3P}{4}, \) and upwards

Hint:

For an L-shaped member with forces applied at the end, use the principles of equilibrium (force balance and moment balance) to calculate the reactions and deflections.

Answer 1

The Correct Option is A

In this question, we are given an L-shaped elastic member subjected to a horizontal force \( P \) at end C. We need to find the magnitude and direction of the normal force on the pin from the slot. Step 1: Determine the equilibrium equations:
To analyze this system, we first need to apply the principles of equilibrium. The system has two unknowns, namely the normal force at pin C and the deflection caused by the applied force. We can write equilibrium equations based on force and moment balance. Step 2: Calculate the deflection and normal force at pin C:
The deflection in each arm of the L-shaped member will depend on the force applied and the section modulus. By applying the appropriate relationships for beam deflections and using the length ratios \( 4a \) and \( a \), we calculate the normal force on the pin and its direction. Step 3: Conclusion:
After solving the equilibrium equations and calculating the normal force, we find that the magnitude of the normal force is \( \frac{3P}{8} \), and the direction is downwards. Thus, the correct answer is (A) \( \frac{3P}{8}, \) and downwards.