This set of Earthquake Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Force Vibration Analysis of SDOF System”.

1. What do we call the ratio of deflection due to dynamic load and deflection due to static load?

a) Logarithmic decrement

b) Damping ratio

c) Magnification factor

d) Logarithmic increment

View Answer

Explanation: The value of deflection under the dynamic load divided by the deflection under the static load is called as the magnification factor. The dynamic magnification factor value varies depending on the type of loading conditions. The ratio of amplitude of motion caused by external forces and deflection due to static vibration is denoted using this factor.

2. What is the critical damping coefficient (C_{cr}), if circular natural frequency is 40 rad/sec and mass is 1 kg?

a) 40 N.s/m

b) 100 N.s/m

c) 60 N.s/m

d) 80 N.s/m

View Answer

Explanation: Given,

Circular natural frequency = 40 rad/sec

Mass = 1 kg

Critical damping coefficient (C

_{cr}) = 2mω

= 2*1*40

= 80 N.s/m

3. What is the logarithmic decrement, if the amplitude of a vibrating body reduces to 1/3^{rd} in 2 cycles?

a) -0.5493

b) 0.5493

c) 1.0986

d) -1.0986

View Answer

Explanation: Given,

No. of cycles (n) = 2

A2 = A1/3 or A1 = 3*A2

Logarithmic decrement = 1/n ln (A1/A2)

= 1/2 ln (3*A2 / A2)

= 1/2 ln (3)

= 0.5493

4. What is the value of natural frequency if stiffness is 1000 KN/m and mass is 2 ton?

a) 22.36 rad/sec

b) 0.707 rad/sec

c) 0.045 rad/sec

d) 44.72 rad/sec

View Answer

Explanation: Given,

Stiffness (k) = 1000 KN/m

Mass = 2 ton

Natural circular frequency (ω) = \(\sqrt{\frac{k}{m}}\)

\(\sqrt{\frac{1000}{2}}\)

= 22.36 rad/sec

5. How many cycles are required to reduce the amplitude to 1/5th, if logarithmic decrement is 0.23?

a) 6

b) 7

c) 5

d) 2

View Answer

Explanation: Given,

A2 = A1/5 or A1 = 5*A2

Logarithmic decrement = 0.23

No. of cycles (n) =?

Logarithmic decrement = 1/n ln (A1/A2)

0.23 = 1/n ln (5*A2 / A2)

0.23 = 1/n ln (5)

n = 7

6. Which of the following statement is true for numerically computed response and analytical solution?

a) The numerically computed response is a poor approximation for the analytical solution for large time intervals

b) The numerically computed response is an accurate approximation for the analytical solution for large time intervals

c) The numerically computed response is an accurate approximation for the analytical solution for smaller time intervals

d) Both numerically computed response and analytical solutions are inaccurate for any time intervals

View Answer

Explanation: For large time intervals (∆t) the numerically computed response gives a poor approximation for the analytical solution. Especially in the free vibration era, the numerically computed responses are not acceptable. For small ∆t also, the numerically computed response lags behind the true response and the amount of lag is proportional to the choice of ∆t.

7. Inertial forces are considered in both static and dynamic analysis.

a) True

b) False

View Answer

Explanation: Inertial forces are considered only in dynamic analysis. No inertia forces are developed in case of static analysis. Static response is a function of type of force and stiffness of structure. Whereas, dynamic response is a function of type of force, mass of structure and stiffness of structure.

8. What should be the value of time interval (∆t) for reasonable accuracy of the computed response, where T_{n} is the natural period of the oscillator?

a) ∆t = T_{n}

b) ∆t ≥ 0.5 T_{n}

c) ∆t ≤ 0.5 T_{n}

d) ∆t ≤ 0.1T_{n}

View Answer

Explanation: For maintain the accuracy of the computed response, the value of ∆t should be less than equal to 0.1T

_{n}. For large ∆t, the numerically computed response is very poor approximation of the true response and as the time interval becomes smaller the numerically computed response starts getting closer to the true response. Therefore, ∆t ≤ 0.1T

_{n}.

9. What is the value of the natural circular frequency (ω) if natural time period (T) is 0.5 sec?

a) 1.256 rad/sec

b) 3.545 rad/sec

c) 12.57 rad/sec

d) 0.08 rad/sec

View Answer

Explanation: Given,

Natural time period (T) = 0.5 sec

natural circular frequency (ω) = 2π / T

= 2π / 0.5

= 12.57 rad/sec

10. What is the other name of body-friction damping?

a) Euler damping

b) Coulomb damping

c) Newton damping

d) Thomson damping

View Answer

Explanation: Body-friction damping is also known as coulomb damping. It occurs because of the friction generated at the support points or connections. It is constant regardless of the amount of displacement or velocity. This type of damping is large in infilled masonry walls when it cracks and provides effective seismic resistance.

**Sanfoundry Global Education & Learning Series – Earthquake Engineering.**

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