Duhamel's Integral in SDOF Systems - Dynamic Response Analysis

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

4.1 General Dynamic Excitation

Figure 4.1.1 General Dynamic Excitation

In this section we will be exploring the response of a SDOF system under arbitrary dynamic loading. Any arbitrary dynamic loading can be expressed as a seried of impulse loads that can be solved taking the response of one impulse as the initial conditions for the next impulse and so on. The the total response can then be obtained as the summation of these impulse loads. A SDOF system subjected to an impulse load can be described by Duhamel’s Integral which, for simple cases, can be solved analytically. For more complex cases, Duhamel’s Integral can be solved using numerical methods. We will also show that for impulsive and low duration excitations, the effect of daming for many cases can be neglected.

4.1.1 Duhamel’s Integral for Undamped System

Let's assume that the undamped system is subjected to a general dynamic loading $F(t)$ , the equation of motion can be written as,

m \ddot{u} + ku = F(t)

Eq. 4.1.1

The solution of this equation can be obtained in the form,

u(t) = u_o \cos (\omega t) + \frac{v_o}{\omega} \sin (\omega t) \ +

\frac{1}{m \omega} \int_{0}^{t} F(\tau) \ \sin (\omega(t-\tau)) \ d \tau

Eq. 4.1.2

Knowing the boundary conditions and by evaluating the integral, we can obtain the response of the system for the dynamic excitation $F(t)$ .

Response to a Constant Force Applied Suddenly

Figure 4.1.2 Constant Load Applied Suddenly to SDOF

The equation defining a constant force applied suddenly at ${t=0}$ can be defined as,

F(t) = F_o

Eq. 4.1.3

The Duhamel's integral will now become,

u(t) = u_o \cos (\omega t) + \frac{v_o}{\omega} \sin (\omega t) \ +

\frac{1}{m \omega} \int_{0}^{t} F_o \ \sin (\omega(t-\tau)) \ d \tau

Eq. 4.1.4

Performing the integration,

u(t) = u_o \cos (\omega t) + \frac{v_o}{\omega} \sin (\omega t) \ +

\frac{F_o}{m \omega^2} | \cos (\omega(t-\tau))|_0^t

Eq. 4.1.4

u(t) = u_o \cos (\omega t) + \frac{v_o}{\omega} \sin (\omega t) \ +

\frac{F_o}{k} (\cos (0) - \cos (\omega t))

Eq. 4.1.5

u(t) = u_o \cos (\omega t) + \frac{v_o}{\omega} \sin (\omega t) \ +

u_{st} (1 - \cos (\omega t))

Eq. 4.1.6

The solution can be represented in the below chart, try to change the parameters to see how the solution changes.

u_o

v_o

F_o

Assuming that the initial conditions are zero, i.e. $u_o = 0$ and $v_o = 0$ ,the solution reduces to,

u(t) = u_{st} (1 - \cos (\omega t))

Eq. 4.1.7

Or,

\text{DAF} = \frac{u(t)}{u_{st}} = 1 - \cos (\omega t)

Eq. 4.1.8

The value of $\frac{u(t)}{u_{st}}$ is referred to as the Dynamic Amplification Factor (DAF) and represent the amplification in the response because of the dynamic nature of the excitation force.

Representing this Dynamic Amplification Factor (DAF) on the y-axis instead of the absolute response, we get,

u_o

v_o

F_o

To maximize the value of $\frac{u(t)}{u_{st}}$ , the value of $\cos (\omega t)$ must be equal to -1, this will make the maximum value of $\frac{u(t)}{u_{st}}$ equal to 2.

Observations

The maximum displacement is exactly equal to $2 u_{st}$ . This indicates that if a constant force is applied suddenly to a system, it will have double the effect as if it is applied in a slow static fashion. This amplified effect is usually referred to as the Dynamic Amplification Factor, $\text{DAF} = \frac{u(t)}{u_{st}}$

The amplification effect for the displacment is also valid for stresses and strains developed because of the effect of the dynamic load.

Response to a Rectangular Pulse Force

Figure 4.1.3 Rectangular Pulse Load

The equation defining a rectangular pulse force can be defined as,

F(t) = \begin{cases} F_o & \text{if } t \le t_d \\ 0 & \text{if } t \ge t_d \end{cases}\

Eq. 4.1.9

Assuming zero initial conditions, $u_o = 0$ and $v_o = 0$ , the solution can be divided into two stages, the forced vibration response obtained from the constant force solution which is valid for $t ≤ t_d$ and the free vibration when $t ≥ t_d$ ,

Forced Vibration Stage

As before, the forced vibration response can be written as,

\frac{u(t)}{u_{st}} = 1 - \cos (\omega t) \space \space \space \space t \le t_d

Eq. 4.1.10

It is convenient to replace $\omega$ by $\frac{2 \pi}{T}$ ,

\frac{u(t)}{u_{st}} = 1 - \cos \Big(2 \pi \frac{t}{T} \Big) \space \space \space \space t \le t_d

Eq. 4.1.11

Free Vibration Stage

u(t) = u(t_d) \cos(\omega (t-t_d)) \space +

\frac{v(t_d)}{\omega } \sin(\omega (t-t_d)) \ \ \ \ \ \ t > t_d

Eq. 4.1.12

Since $u(t) = u_{st} (1-\cos(\omega t_d))$ and $v(t_d) = \dot{u}(t_d) = u_{st} \omega \sin (\omega t_d)$ ,

The free vibration stage response can be written as,

u(t) = u_{st} (1-\cos(\omega t_d)) \cos(\omega (t-t_d)) \space +

u_{st} \sin(\omega t_d)) \sin(\omega (t-t_d)) \ \ \ \ t > t_d

Eq. 4.1.13

\frac{u(t)}{u_{st}} = (1-\cos(\omega t_d)) \cos(\omega (t-t_d)) \space +

\sin(\omega t_d)) \sin(\omega (t-t_d)) \ \ \ \ t > t_d

Eq. 4.1.14

\frac{u(t)}{u_{st}} = \cos(\omega (t-t_d))- \cos(\omega (t-t_d)) \cos(\omega t_d)) \space +

\sin(\omega t_d)) \sin(\omega (t-t_d)) \ \ \ \ t > t_d

Eq. 4.1.15

Using the trigonometric identity,

\cos(A+B) = \cos(A) \cos(B) - \sin(A) \sin(B)

\frac{u(t)}{u_{st}} = \cos(\omega (t-t_d)) - \cos(\omega t)

Eq. 4.1.16

Similarly replace $\omega$ by $\frac{2 \pi}{T}$ ,

\frac{u(t)}{u_{st}} = \cos(2 \pi \frac{t-t_d}{T} ) - \cos(2 \pi \frac{t}{T} )

Eq. 4.1.17

\frac{u(t)}{u_{st}} = \cos \Big(2 \pi (\frac{t}{T} - \frac{t_d}{T}) \Big) - \cos\Big(2 \pi \frac{t}{T} \Big)

Eq. 4.1.18

The solution can be represented in the below chart, try to change the parameters to see how the solution changes.

t_d

t_d / T = 0.50

Maximum Response in the Forced Vibration Stage

During the forced vibration stage, the deformation is defined as,

\frac{u(t)}{u_{st}} = 1 - \cos \Big(2 \pi \frac{t}{T} \Big) \space \space \space \space t \le t_d

Eq. 4.1.10

The first maximum value occurs when $\cos \Big(2 \pi \frac{t}{T} \Big) = -1$

Or when $\frac{t}{T} = \frac{t_d}{T} = \frac{1}{2}$ . The first maximum value is equal to 2.

If $\frac{t_d}{T} \le 2$ , the response will build up toward the maximum value of $\frac{u(t)}{u_{st}}$ which is equal to 2 until $\frac{t}{T} = \frac{1}{2}$ .

If $t_d$ is long enough so that $\frac{t_d}{T} = \frac{3}{2}$ , a second maximum value of the response will develop, which is also equal to 2.

More maximum values will develop as $t_d$ is lengthened with respect to $T.$

The complete solution in the forced vibration stage can then be written as,

R_d = \begin{cases} 1-cos(2 \pi \ t_d / T) & t_d/T \le 1/2 \\ 2 & t_d/T \ge 1/2 \end{cases}\

Eq. 4.1.19

t_d

t_d / T = 3.00

Observations

It is clear that the Dynamic Amplification Factor, $R_d = \frac{U}{u_{st}}$ only depends on the ratio $\frac{t_d}{T}$ .

Try to move the $t_d$ slider and observe the value of $\frac{t_d}{T}$ , you will realize that as long as $\frac{t_d}{T}$ is less than 0.5, $R_d$ is always $\lt 2$ . The first peak value of 2 will be reached when $\frac{t_d}{T} = 0.5$ .

As $t_d$ is lengthened, more maximum values can develop in the forced vibration stage at values of $\frac{t_d}{T} = 0.5,1.5,2.5,3.5,.....$ .

Maximum Response in the Free Vibration Stage

As we have shown before, in the free vibration stage, the system oscilates with displacement defined as,

u(t) = u(t_d) \cos(\omega (t-t_d)) \space +

\frac{v(t_d)}{\omega } \sin(\omega (t-t_d)) \ \ \ \ \ \ t > t_d

Eq. 4.1.12

Which has an amplitude defined as,

U = \sqrt{[u(t_d)]^2 + \Big[\frac{v(t_d)}{\omega} \Big]^2}

Eq. 4.1.20

Since $u(t_d) = u_{st} (1 - \cos (\omega t_d))$ ,

And $v(t_d) = \dot{u}(t_d) = u_{st} \omega \sin (\omega t_d)$ ,

U = \sqrt{[u_{st} (1 - \cos (\omega t_d))]^2 + \Big[\frac{u_{st} \omega \sin (\omega t_d)}{\omega} \Big]^2}

Eq. 4.1.21

U = u_{st} \sqrt{[ (1 - \cos (\omega t_d))]^2 + \sin^2 (\omega t_d) }

Eq. 4.1.22

U = u_{st} \sqrt{ 1 - 2 \cos (\omega t_d) + \cos^2 (\omega t_d) + \sin^2 (\omega t_d)}

Eq. 4.1.23

U = u_{st} \sqrt{ 1 - 2 \cos (\omega t_d) + 1}

Eq. 4.1.24

U = u_{st} \sqrt{ 2 - 2 \cos (\omega t_d) }

Eq. 4.1.25

U = u_{st} \sqrt{ 2 - 2 \cos (\frac{2 \pi t_d}{T} ) }

Eq. 4.1.26

Since $\cos(\frac{2 \pi t_d}{T}) = 1-2 \sin^2(\frac{\pi t_d}{T})$ , then,

U = u_{st} \sqrt{ 4 \sin^2 (\frac{\pi t_d}{T}) }

Eq. 4.1.27

Since we are only interested in the maximum values,

U = 2 u_{st} \Big| \sin( \frac{\pi t_d}{T}) \Big|

Eq. 4.1.28

Finally,

R_d = \frac{U}{u_{st}} = 2 \Big| \sin( \frac{\pi t_d}{T}) \Big|

Eq. 4.1.29

t_d

t_d / T = 3.00

The overall deformation response factor can then be defined as,

R_d = \begin{cases} 2 \ \sin(\frac{\pi t_d}{T} ) & \frac{ t_d}{T} \le \frac{ 1}{2} \\ 2 & \frac{ t_d}{T} \ge \frac{ 1}{2} \end{cases}\

Eq. 4.1.30

t_d

t_d / T = 3.00

Response to a Triangular Pulse Force

Figure 4.1.4 Triangular Pulse Load

The equation defining a triangular pulse force can be defined as,

F(t) = \begin{cases} F_o(1-t / t_d) & \text{if } t \le t_d \\ 0 & \text{if } t \ge t_d \end{cases}\

Eq. 4.1.31

Assuming zero initial conditions, $u_o = v_o = 0$ , the solution can be divided into two stages, the forced vibration response when $t \le t_d$ ,and the free vibration response when $t \gt t_d$ .

Forced Vibration Stage

The response for the forced vibration stage can be obtained using the Duhamel's integral. Substituting the forcing function into Duhamel's integral for $0 < \tau \le t_d$ ,

u(t) = \frac{1}{m \omega} \int_{0}^{t} F_o \Big(1-\frac{\tau}{t_d} \Big) \ \sin \omega(t-\tau) \space \space d \tau

Eq. 4.1.32

The response for the forced vibration stage can Peforming the intergation, the forced vibration stage solution is,

u(t) = \frac{F_o}{k} \Big(\frac{\sin(\omega t)}{\omega t_d} -\cos(\omega t) - \frac{t}{t_d} +1 \Big)

Eq. 4.1.33

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

u(t) = u_{st}\Big(\frac{\sin(\omega t)}{\omega t_d} -\cos(\omega t) - \frac{t}{t_d} +1 \Big)

Eq. 4.1.34

\frac{u(t)}{u_{st}} = \Big(\frac{\sin(\omega t)}{\omega t_d} -\cos(\omega t) - \frac{t}{t_d} +1 \Big) , t \le t_d

Eq. 4.1.35

Free Vibration Stage

The free vibration stage response can be obtained from the well-known free vibration response with initial conditions at the end of the loading stage, (i.e. at $t = t_d$ ) ,

u(t) = u(t_d) \cos(\omega (t-t_d)) +

\frac{v(t_d)}{\omega} \sin(\omega (t-t_d)) , \space \space t \gt t_d

Eq. 4.1.36

Since,

u(t) = u_{st}\Big(\frac{\sin(\omega t)}{\omega t_d} -\cos(\omega t) - \frac{t}{t_d} +1 \Big)

Eq. 4.1.37

Then,

u(t) = u_{st}\Big(\frac{\sin(\omega t_d)}{\omega t_d} -\cos(\omega t_d) \Big)

Eq. 4.1.38

Differentiating Eq. 4.1.34, we can have $v(t)$ defines as,

v(t) =\dot{u}(t) = u_{st}\Big(\frac{\cos(\omega t)}{t_d} + \omega \sin(\omega t) -\frac{1}{t_d} \Big)

Eq. 4.1.39

And,

v(t_d) =\dot{u}(t_d) = u_{st}\Big(\frac{\cos(\omega t_d)}{t_d} + \omega \sin(\omega t_d) -\frac{1}{t_d} \Big)

Eq. 4.1.40

And the solution for the free vibration stage becomes,

u(t) = u_{st} \Big(\frac{\sin(\omega t_d)}{\omega t_d} -\cos(\omega t_d)\Big) \cos(\omega (t-t_d) ) +

\frac{u_{st}}{\omega} \Big( \frac{\cos(\omega t_d)}{t_d} + \omega \sin(\omega t_d) - \frac{1}{t_d} \Big) \sin(\omega (t-t_d))

, t \gt t_d

Eq. 4.1.41

This equation can be written as,

\frac{u(t)}{u_{st}} = \Big(\frac{\sin(\omega t_d)}{\omega t_d} -\cos(\omega t_d)\Big) \cos(\omega (t-t_d) ) +

\frac{1}{\omega} \Big( \frac{\cos(\omega t_d)}{t_d} + \omega \sin(\omega t_d) - \frac{1}{t_d} \Big) \sin(\omega (t-t_d))

, t \gt t_d

Eq. 4.1.42

t_d

t_d / T = 1.50

u(t)/u_st = 1.69

Observations

The maximum will approach 2 for very high $\frac{t_d}{T}$ values.

Try to move the $t_d$ slider and observe the value of $\frac{t_d}{T}$ and it's corresponding $\frac{u(t)}{u_{st}}$ , after the end of the force, the system oscillates around zero.

If the excitation is a single pulse load, the effect of damping usually can be neglected, this is because the energy dissipated by damping is small for systems subjected to short impulsive loads unless the system is highly damped.