Question

A single story building model is shown in the figure. The rigid bar of mass \( m \) is supported by three massless elastic columns whose ends are fixed against rotation. For each of the columns, the applied lateral force (P) and corresponding moment (M) are also shown in the figure. The lateral deflection (\( \delta \)) of the bar is given by \( \delta = \dfrac{PL^3}{12EI} \) where \( L \) is the effective length of the column, \( E \) is the Young's modulus of elasticity and \( I \) is the area moment of inertia of the column cross-section with respect to its neutral axis. 
For the lateral deflection profile of the columns as shown in the figure, the natural frequency of the system for horizontal oscillation is 

• A. \( 6 \sqrt{\frac{EI}{mL^3}} \) rad/s
• B. \( \frac{1}{L} \sqrt{\frac{2EI}{m}} \) rad/s
• C. \( 2 \sqrt{\frac{6EI}{mL^3}} \) rad/s
• D. \( \frac{2}{L} \sqrt{\frac{EI}{m}} \) rad/s

Hint:

For the natural frequency of a system with lateral deflection, use the relationship between the stiffness of the system and the deflection, and apply the formula \( \omega_n = \sqrt{\frac{k}{m}} \).

Answer 1

The Correct Option is A

To find the natural frequency of the system, we need to use the formula for the natural frequency of a system with lateral deflection. The relationship between the deflection \( \delta \) and the applied force \( P \) for a column is given by: \[ \delta = \frac{PL^3}{12EI} \] This describes the deflection at the column under the applied lateral load \( P \). The equation for the natural frequency \( f \) of the system is derived from the balance of forces and moments. The formula for the natural frequency \( \omega_n \) for lateral oscillation is: \[ \omega_n = \sqrt{\frac{k}{m}} \] where \( k \) is the stiffness of the system, and \( m \) is the mass. The stiffness \( k \) of a column under lateral deflection is inversely proportional to the deflection \( \delta \), which can be expressed as: \[ k = \frac{12EI}{L^3} \] Substituting this into the equation for the natural frequency: \[ \omega_n = \sqrt{\frac{12EI}{mL^3}} \] The natural frequency is then: \[ \omega_n = 6 \sqrt{\frac{EI}{mL^3}} \] Thus, the correct answer is option (A).
Final Answer: (A) \( 6 \sqrt{\frac{EI}{mL^3}} \) rad/s