Dynamics of Structures

Structural Engineering Simplified

Chapter 4. Response of SDOF to General Dynamic Loading

Chapter 4

4.1 General Dynamic Excitation 4.2 Numerical Solution of the Equation of Motion 4.3 Examples

4.3 Examples

Figure 4.3.1 Example 4.3.1

Example 4.3.1

An undamped SDOF system with frequency $\omega$ is suddenly subjected to a forcing function that decays exponentially with time and has a frequency $\bar{\omega}$ . If the system was initially at rest, derive an expression for the Dynamic Amplification Factor.

The system equation of motion can be written as,

m \ddot{u} + ku = F_o e^{-\bar{\omega}t}

Since the system was at rest, we can write the Duhamel's integral solution as follows,

u(t) = \frac{F_o}{m \omega} \int_{0}^{t} e^{-\bar{\omega}\tau} \ \sin (\omega(t-\tau)) \ d \tau

Let's do the integration by parts, assume that $v = \sin (\omega(t-\tau))$ and $dy = e^{-\bar{\omega}\tau} d \tau$ ,

dv = \omega \cos (\omega(t-\tau))

And,

y = -\frac{1}{\bar{\omega}} e^{-\bar{\omega}\tau}

We can write,

u(t) = vy - \int_{0}^{t} ydv

Or,

u(t) =-\frac{F_o}{m \omega \bar{\omega}} \Big[\Big(\sin (\omega(t-\tau)) e^{-\bar{\omega}\tau} \Big)|_0^t

- \int_{0}^{t} \omega e^{-\bar{\omega}\tau} \cos (\omega(t-\tau)) d\tau \Big]

Integrating the second part again by parts using $v = \cos (\omega(t-\tau))$ and $dy = e^{-\bar{\omega}\tau} d \tau$ ,

u(t) =\frac{F_o}{m \omega (\bar{\omega}^2 + \omega^2)} \Big[\bar{\omega} \sin (\omega t) - \omega \cos(\omega t) + \omega e^{-\bar{\omega}t} \Big]

Rearranging the terms,

u(t) =\frac{F_o}{m (\bar{\omega}^2 + \omega^2)} \Big[\frac{\bar{\omega}}{\omega} \sin (\omega t) - \cos(\omega t) + e^{-\bar{\omega}t} \Big]

u(t) =\frac{F_o}{m \omega^2 (\frac{\bar{\omega}^2}{\omega^2} + 1)} \Big[\frac{\bar{\omega}}{\omega} \sin (\omega t) - \cos(\omega t) + e^{-\bar{\omega}t} \Big]

u(t) =\frac{F_o}{k (\frac{\bar{\omega}^2}{\omega^2} + 1)} \Big[\frac{\bar{\omega}}{\omega} \sin (\omega t) - \cos(\omega t) + e^{-\bar{\omega}t} \Big]

Knowing that $u_{st} = \frac{F_o}{k}$ ,

u(t) =\frac{u_{st}}{ (\frac{\bar{\omega}^2}{\omega^2} + 1)} \Big[\frac{\bar{\omega}}{\omega} \sin (\omega t) - \cos(\omega t) + e^{-\bar{\omega}t} \Big]

\frac{u(t)}{u_{st}} =\frac{1}{ (\frac{\bar{\omega}^2}{\omega^2} + 1)} \Big[\frac{\bar{\omega}}{\omega} \sin (\omega t) - \cos(\omega t) + e^{-\bar{\omega}t} \Big]

The Dynamic Amplification Factor can be obtained as,

DAF =\frac{1}{ (\frac{\bar{\omega}^2}{\omega^2} + 1)} \Big[\frac{\bar{\omega}}{\omega} \sin (\omega t) - \cos(\omega t) + e^{-\bar{\omega}t} \Big]

The following graph plots the DAF againist time and calculates the frequency ratio and the first peak value of the DAF. Notice that after some time, the forcing function decays leaving the system oscillating under the steady state condition. Try to change the values of ${\omega}$ and $\bar{\omega}$ and visualize the solution.

Tip: to micro change the slider values, place the mouse cursor on the slider, then use your keyboard right and left arrows to adjust the slider value.

$\bar{\omega}$

${\omega}$

$t$

$\frac{\bar{\omega}}{\omega} = 1.00$

$\frac{u(t)}{u_{st}} = 0.76$

Figure 4.3.2 Example 4.3.2

Example 4.3.2

The shown beam is supporting a concentrated weight of 10 kips at its mid span and is subjeted to the triangular pulse shown with $F_o = 20 \text{ kips}$ and $t_d = 0.15 \text{ s}$ . Neglecting the weight of the beam, determine the maximum dynamic bending stress that will develop in the beam $E = 29000 \text{ ksi}$ .

We first need to calculate the equivalent stiffness of the system $k,$

k = \frac{48EI}{L^3}

k = \frac{48 \times 29000 \times 800.6}{(20 \times 12)^3}

k = 80.62 \text{ kip/in}

The mass of the system $m,$

m = \frac{W}{g}=\frac{10}{ 386 }

m = 0.0259 \space \text{kip. s}^2 \text{/in}

Calculate the system period $T,$

T = 2 \pi \sqrt{\frac{m}{k}}= 2 \pi \sqrt{\frac{0.0259}{80.62}}

T = 0.113 \space \text{s}

\frac{t_d}{T} = \frac{0.15}{0.113} = 1.33

The solution for the triangular pulse was derived in equations 4.1.35 and 4.1.42 and can be plotted as a function of time. From the graph, the maximum $DAF$ is 1.65 at t= 0.053 s.

t_d / T = 1.33

u(t)/u_st = 1.65

The maximum static bending stress in the beam can be calculated as,

\sigma_{st} = \frac{F_o L/4}{S}

\sigma_{st} = \frac{20 \times (20 \times 12)/4}{89}

\sigma_{st} = 13.65 \text{ ksi}

The maximum dynamic bending stress in the beam can be calculated as,

\sigma_{max} = \sigma_{st} \times DAF

\sigma_{max} = 13.65 \times 1.65

\sigma_{max} = 22.25 \text{ ksi}