HW1. CEE 247. Structural Dynamics. UCI. Fall 2006

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HW1. CEE 247. Structural Dynamics. UCI. Fall 2006

Nasser Abbasi

June 17, 2014

1 Problem 1.3
2 Problem 1.6
3 Problem 2.4
4 Problem 2.6

1 Problem 1.3

Solution

From steel manual, we obtain the following values for $Ixx$ for the diﬀerent $W$ sections:

$4 4 W 8× 24→ Ixx = 82.5 in ,W 10× 33 → Ixx = 170 in$ , and $6 E = 30× 10 psi$ .

Weight of girder $W = 50 kips$ or $50,000$ lbs.

Gravity Acceleration $g = 386$ $in∕ sec2$

$L= 12 × 12= 244′′$

We start by ﬁnding the eﬀective stiﬀness $ke$

But $ωn = 2πf$ hence

2 Problem 1.6

Solution

The equation of motion is derived in 2 methods. One based on the force method and the second is based on energy method. In both method we assume there is no damping in the system and no friction nor air resistance.

First method

First draw a free body diagram showing forces acting on pendulum mass, which are the weight $W$ and the tension $T$ in the cord.

Next resolve the forces parallel and perpendicular to the direction of motion as follows

Where $at$ is the tangential acceleration and $m$ is the mass of the pendulum bob. Since $at = d2d(t2s)$ where $s$ is an arc length, and $s= L𝜃$ when $𝜃$ is in radians, hence $a = Ld2𝜃-= L𝜃¨ t dt2$

So applying Newton second law of motion along the tangential direction we obtain

Where the minus sign is due to the fact that force acts in the direction opposite to increasing $𝜃$ .

But $W = mg$ hence

This is the second order diﬀerential equation of motion we need to solve. This ODE is non-linear in $𝜃$ . Assuming $𝜃$ is small, and since $3 sin𝜃 = 𝜃 − 𝜃3! + ⋅⋅⋅$ , hence we see that for small $𝜃$ , $sin 𝜃 ≈ 𝜃$ , we the ODE becomes

𝜃¨+ g-𝜃 = 0 L

This is free vibration undamped motion. Assume the solution is $𝜃 (t)= Aeλt$ , we obtain the characteristic equation

λ2+ g-= 0 L

which has solution $∘ g- λ = ±i L$ , Hence the solution is

This has the solution

𝜃(t)= A cosλt+ Bsinλt

Now we ﬁnd $A,B$ from initial conditions. at $t = 0,$ $𝜃0 = A$ , and at $t = 0,$ $_𝜃0 = B λ ⇒$ $B = _𝜃0 λ$

Hence the natural frequency

|-----------| | ∘ -- | | ωn = gL | -------------

Hence the general solution is

|----------------------------| | 𝜃 (t) = 𝜃 cosω t+ _𝜃0sin ω t | | 0 n ωn n | -----------------------------|

Let $√ -2----2 C = A + B$ , and $A ϕ = arctan B,$ the solution can also be written as

𝜃 (t)= C sin (ωnt+ ϕ)

We see that the natural frequency $ωn$ of the bob is

∘ -- ωn = g- L

And it does not depend on the mass of the bob.

Second method

Here is another derivation. Since there is no damping in the system then the energy of the system $E = P E+ KE$ is constant.

But $P E = mgh$ , where $h$ is the height above the reference level. Taking the reference level when the bob is at the lowest point, we see that at any instance of time $h = L− L cos𝜃 = L(1− cos𝜃)$

And the $KE$ any that moment is given by $1mv2 2$ , but $v = L_𝜃$ , hence $KE = 1m (l _𝜃)2 2$ , Hence we obtain the energy as

1 ( )2 E = mgL (1 − cos 𝜃)+ 2m L𝜃_

Since $E$ is constant, then $dE dt = 0$ , hence

We have 2 solution. Ignoring the solution that $_𝜃 = 0$ since this is trivial. We obtain the same ODE as above which is

g ¨𝜃 + -sin𝜃 = 0 L

The advantage of the derivation based on energy is that one is working with scalar quantities, hence one does not need to worry about sign of forces and direction of motion as one would with the force method.

3 Problem 2.4

Solution

Since

( ) -ui- ln ui+j = j2πξ

And since $j = 10$ in this case (10 cycles), hence

( ) ln -ui-- ξ = ----ui+10- 20π

But $ui = 1′′$ and $ui+10 = .4′′$ , hence

Therefore

|-----------| | ξ ≈ 1.5% | -------------

4 Problem 2.6

$k = 30,000 lb∕in$

$ωn = 25$ rad/sec

Since viscous damping force is proportional to speed, hence $Fd v = c$ , then

a) $ξ = cccr$ , but

Hence

|--------------| | | | ξ = 0.41667 | ---------------

b) Since

T TD = ∘------- 1− ξ 2

But $T = 2πω-= 22π5 = 0.25133 sec n$

Hence

---0.25133--- TD = √ 1− 0.416672

Then

|------------------| | | | TD = 0.27647 sec | -------------------

c)The logarithmic decrement $u δ = ln uii+1$ , But $u 2πξ ln uii+1 = √1-−ξ2$ , hence

2π× 0.41667 δ = √------------ 1− 0.416672

Hence

|------------| | | | δ = 2.8799 | --------------

c)Since

ui ln u---= δ i+1

Then

Hence

u --i-= 17.812 ui+1

$ˉn= ˉer = {cos𝜃 ,sin𝜃 ,0}$

$ˉn= − ˉez = {0,0,− 1}$

$vcyl = {r,𝜃 ,z}$

$vcartesian = {x,y,z}$

$da= dz d 𝜃 z$

$x= rcos𝜃$

$y= rsin𝜃$

$z= z$

Given a point $pt$ coordinates in cylindrical $vcyl = {r,𝜃,z}$ , and if we wish to obtain its coordinates in cartesian $vcartesian = {x,y,z}$ , then use the following transformation rules

Example using Mathematica:

Given a point $pt$ coordinates in cartesian $v = {x,y,z} cartesian$ , and if we wish to obtain its coordinates cylindrical $vcyl = {r,𝜃 ,z}$ , then use the following transformation rules

Example using Mathematica: