Let's break down the problem step-by-step. --- ## **1. System Parameters** - **Building mass, \( m = 10^5 \) kg** - **Stiffness of each wall, \( k_{\text{each}} = 3.519 \times 10^6 \) N/m** - **Total stiffness, \( k = 2 \times k_{\text{each}} = 7.038 \times 10^6 \) N/m** - **Ground motion amplitude, \( Y_ = .1 \) m** - **Ground motion frequency, \( \omega = 7.5 \) rad/s** - **Maximum allowed steady-state amplitude, \( X_{\text{max}} = .15 \) m** --- ## **2. Equation of Motion** The building acts as a **single degree of freedom (SDOF) system** subject to a base excitation: \[ m\ddot{x} + c\dot{x} + kx = k y(t) \] where \( y(t) \) is the ground displacement. Let: - \( x(t) \): relative displacement of mass wrt ground - \( y(t) = Y_ \sin(\omega t) \): ground displacement The steady-state amplitude of the relative displacement is given by: \[ X = \frac{Y_}{\sqrt{\left[1 - \left(\frac{\omega}{\omega_n}\right)^2\right]^2 + \left[2\zeta \left(\frac{\omega}{\omega_n}\right)\right]^2}} \] Where: - \( \omega_n = \sqrt{\frac{k}{m}} \): natural frequency - \( \zeta \): damping ratio --- ## **3. Calculate System Frequencies** \[ \omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{7.038 \times 10^6}{10^5}} = \sqrt{70.38} \approx 8.397 \, \text{rad/s} \] \[ r = \frac{\omega}{\omega_n} = \frac{7.5}{8.397} \approx .893 \] --- ## **4. Steady-State Amplitude Equation** \[ X = \frac{Y_}{\sqrt{(1 - r^2)^2 + (2 \zeta r)^2}} \] Plug in the numbers: - \( Y_ = .1 \) m - \( X_{\text{max}} = .15 \) m - \( r = .893 \) Set \( X \leq .15 \): \[ .15 = \frac{.1}{\sqrt{(1 - .893^2)^2 + [2 \zeta (.893)]^2}} \] --- ## **5. Solve for Damping Ratio \( \zeta \)** First, calculate \( .893^2 = .797 \): \[ 1 - .797 = .203 \] \[ (.203)^2 = .0412 \] \[ .15 = \frac{.1}{\sqrt{.0412 + (2 \zeta \times .893)^2}} \] \[ .15 = \frac{.1}{\sqrt{.0412 + (1.786 \zeta)^2}} \] \[ \sqrt{.0412 + (1.786 \zeta)^2} = \frac{.1}{.15} = .6667 \] \[ .0412 + (1.786 \zeta)^2 = (.6667)^2 = .4444 \] \[ (1.786 \zeta)^2 = .4444 - .0412 = .4032 \] \[ 1.786 \zeta = \sqrt{.4032} = .6352 \] \[ \zeta = \frac{.6352}{1.786} \approx .356 \] --- ## **6. **Final Answer** ### **The damping ratio required is:** \[ \boxed{.36} \] (Rounded to two decimal places) --- ## **7. **Summary Table** | Parameter | Value | |-------------------|-------------------------------| | Mass, \( m \) | \( 10^5 \) kg | | Stiffness, \( k \)| \( 7.038 \times 10^6 \) N/m | | Natural freq., \( \omega_n \) | \( 8.40 \) rad/s | | Ground amp., \( Y_ \) | .1 m | | Ground freq., \( \omega \) | \( 7.5 \) rad/s | | Max allowed amp., \( X \) | .15 m | | Required damping ratio, \( \zeta \) | **.36** | --- ## **Key Steps Recap** 1. Calculate total stiffness and natural frequency. 2. Use the forced vibration formula for base excitation. 3. Substitute values and solve for the damping ratio. If you need further clarification or steps, let me know!

# Explanation of Damping Ratio Calculation for Building Model ## 1. System Parameters - **Building mass, \( m = 10^5 \) kg** - **Stiffness of each wall, \( k_{\text{each}} = 3.519 \times 10^6 \) N/m** - **Total stiffness, \( k = 2 \times k_{\text{each}} = 7.038 \times 10^6 \) N/m** - **Ground motion amplitude, \( Y = 0.1 \) m** - **Ground motion frequency, \( \omega = 7.5 \) rad/s** - **Maximum allowed steady-state amplitude, \( X_{\text{max}} = 0.15 \) m** ## 2. Equation of Motion The equation of motion for the system is: \[ m\ddot{x} + c\dot{x} + kx = -m\ddot{y(t)} \] where \( y(t) \) is the ground displacement, modeled as: \[ y(t) = Y \sin(\omega t) \] The steady-state relative amplitude \( X \) can be expressed as: \[ X = \frac{r^2}{\sqrt{(1 - r^2)^2 + (2\zeta r)^2}} Y \] where \( r = \frac{\omega}{\omega_n} \). ## 3. Calculate Natural Frequency The natural frequency \( \omega_n \) is calculated as follows: \[ \omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{7.038 \times 10^6}{10^5}} = \sqrt{70.38} \approx 8.39 \, \text{rad/s} \] ## 4. Frequency Ratio Calculate the frequency ratio: \[ r = \frac{\omega}{\omega_n} = \frac{7.5}{8.39} \approx 0.894 \] Now, compute \( r^2 \): \[ r^2 = (0.894)^2 \approx 0.799 \] ## 5. Steady-State Amplitude Equation Rearranging the steady-state amplitude equation for \( \zeta \): \[ \sqrt{(1 - r^2)^2 + (2\zeta r)^2} = \frac{r^2 Y}{X} \] Substituting the values: 1. Calculate \( (1 - r^2) \): \[ 1 - r^2 = 1 - 0.799 \approx 0.201 \] \[ (1 - r^2)^2 = (0.201)^2 \approx 0.0404 \] 2. Compute the right-hand side: \[ \frac{r^2 Y}{X} = \frac{0.799 \times 0.1}{0.15} \approx 0.5327 \] 3. Square both sides: \[ (1 - r^2)^2 + (2\zeta r)^2 = (0.5327)^2 \approx 0.2845 \] 4. Solve for \( (2\zeta r)^2 \): \[ (2\zeta r)^2 = 0.2845 - 0.0404 \approx 0.2441 \] 5. Take the square root: \[ 2\zeta r = \sqrt{0.2441} \approx 0.4941 \] ## 6. Solve for Damping Ratio \( \zeta \) Now, solve for \( \zeta \): \[ \zeta = \frac{0.4941}{2r} = \frac{0.4941}{2 \times 0.894} \approx 0.276 \] --- ## Final Answer ### The required damping ratio is: \[ \boxed{0.276} \] --- ## Summary Table | Parameter | Value | |-------------------|-------------------------------| | Mass, \( m \) | \( 10^5 \) kg | | Stiffness, \( k \)| \( 7.038 \times 10^6 \) N/m | | Natural freq., \( \omega_n \) | \( 8.39 \) rad/s | | Ground amp., \( Y \) | 0.1 m | | Ground freq., \( \omega \) | \( 7.5 \) rad/s | | Max allowed amp., \( X \) | 0.15 m | | Required damping ratio, \( \zeta \) | **0.276** | --- ## Key Steps Recap 1. Calculate total stiffness and natural frequency. 2. Use the correct steady-state amplitude formula. 3. Substitute values and solve for the damping ratio. If you need further assistance or clarification, feel free to ask!