SD Lecture05 Elastic Stress Waves
-
Upload
ronaldo-hertez -
Category
Documents
-
view
217 -
download
0
Transcript of SD Lecture05 Elastic Stress Waves
-
7/28/2019 SD Lecture05 Elastic Stress Waves
1/33
Soil Dynamics
Lecture 05
Elastic Stress Waves in Bars
Luis A. Prieto-Portar, August 2006.
-
7/28/2019 SD Lecture05 Elastic Stress Waves
2/33
The propagation of stress in an elastic medium.
When stress is applied to a body, that stress will propagate away from the point of
application via stress waves. Different materials will propagate the stress at different
speeds.
For example, in sands the stress will propagate at about 1,000 feet/sec. In sandstones,the stress will propagate at 14,000 feet/sec.
This lecture deals with the propagation of stress waves in elastic media in the form of
rods, bars or beams.
An example of this type of stress is that induced by a pile driving hammer striking
the head of a pre-cast concrete pile (shown on the next three slides).
The propagation of stress is central to the understanding of how dynamic loads
propagate in soils, whether the loads come from gravity, wind, explosions, etc., orfrom earthquakes.
-
7/28/2019 SD Lecture05 Elastic Stress Waves
3/33
-
7/28/2019 SD Lecture05 Elastic Stress Waves
4/33
-
7/28/2019 SD Lecture05 Elastic Stress Waves
5/33
-
7/28/2019 SD Lecture05 Elastic Stress Waves
6/33
Stress and strain in elastic media.
The notation for the normal and shear stresses in an idealized very small element of amuch larger elastic body is shown below.
-
7/28/2019 SD Lecture05 Elastic Stress Waves
7/33
1
21
2
1
2
x
y
z
xy yx
yz zy
zx xz
x
y
z
xy x
yz y
zx z
Normal stresses ,
Shear stresses
uNormal strains
x
v
yw
z
v u w vShear strains and
x y y zw v u w
y z z x
u w v u
z x x y
=
=
=
=
=
=
= + =
= + =
= + =
where is the components
of rotation about the x, yand the z axes.
-
7/28/2019 SD Lecture05 Elastic Stress Waves
8/33
Constitutive Relations (Hookes law).
In elastic and isotropic media, the stresses and the strains are related to each otherthrough relationships called constitutive relationships. These are:
( )
( )
( )
1
2
12
1
2
2 1
x x y z
y y z x
z z x y
xy xy
yz yz
zx zx
where E is Young' s elastic mod ulus and is Poisson' s ratio.
Shear stresses and shear strains are related via the shear mod ulus G,
EG where G
( )G
G
= +
= +
= +
= =+
=
=
-
7/28/2019 SD Lecture05 Elastic Stress Waves
9/33
( ) ( )
( )
22
2
1 1 2
2
x x
y y
z z
x y z
Normal stresses can also be exp ressed in t erms of strains,
GG
G
where,E
and called the volumetric strain
Notice that ,
G
= += +
= +
= = + ++
=+
-
7/28/2019 SD Lecture05 Elastic Stress Waves
10/33
Longitudinal Elastic Stress Waves in a Bar.
The rod shown below is experiencing the propagation of a stress wave from left to
right. Using Newtons second law of motion (F = ma) we can find the differential
equation of motion.
-
7/28/2019 SD Lecture05 Elastic Stress Waves
11/33
( )
( )
2
2
2
2
2
2
2 22
2 2
1
C
A xW V uA x A ma a a
g g g t
Since g ,
ux A x Assume A , and
x t
usince E E , replace in the above equation,
x
u uEx x t
u uor v whe
t x
+ + = = = =
=
= =
= =
=
=
C
Cv is the longitudinal stress wav
Ere v
w e velohe ce yr .it
=
-
7/28/2019 SD Lecture05 Elastic Stress Waves
12/33
( ) ( )( )
( )
1
2
C C
P
t C
solution to find the displacement due to the stress wave could be ,
u F v t x G v t x The function F v t x is the propagation going along x ( Block # ) at time t ,
u F v t x
At time t t the function is represented by Block # shown below.
= + + +
= +
+
-
7/28/2019 SD Lecture05 Elastic Stress Waves
13/33
( ) ( )
( ) ( ) ( )
1 2
t t C
t t t
C P
C
C
Therefore,at time t t,
u F v t t x x If the block moves left without changing its shape from position# to position# ,
u u
orF v t x F v t t x x
or
v t x
Therefore, the longitudinal stress wave velocity v
+
+
+
= + +
=
+ = + +
=
=
( )C
x
t
Similarly, the functionG v t x represents a wave traveling to the right.
-
7/28/2019 SD Lecture05 Elastic Stress Waves
14/33
The velocity of the particles in the stressed zone.
There is a difference between the longitudinal stress wave propagation vc and thevelocity of the particles vc in the stressed zone.
Consider a compression stress pulse of intensity xand duration tapplied to the rod.
-
7/28/2019 SD Lecture05 Elastic Stress Waves
15/33
The compressive stress pulse could be the result of a striking hammer upon a pre-cast
concrete pile. When this stress pulse is initially applied, that zone of the rod or pile
will be under compression. This compression wave will soon move to an adjacentzone, such that, during the time interval t the stress will have traveled a distance x,
( )
C
C
x xC
x v t
A t an y time later on, that is , when t t' a segment of the pile of length x willconstitute the comressed zone. Hence,
x v t'
The elastic shortening u of the rod in t he wave packet x is thus,
u x v t' where u is the disE E
=
>
=
= =
x C
placement of the end of the rod .
The velocity of the particles u is therefore,
vuu
t' E
= =
-
7/28/2019 SD Lecture05 Elastic Stress Waves
16/33
Therefore, theparticle velocity is a function of the intensity of stress x, whereas the
longitudinal stress wave propagates at a velocity is a function of the material properties
only, that is,
The particle and the stress wave velocities are in the same direction when a
compression stress is applied to the end of the pile (or rod).
However, when a tensile stress is applied at the end of the rod, the particle and the
stress wave velocities are in opposite directions.
C
Ev
=
-
7/28/2019 SD Lecture05 Elastic Stress Waves
17/33
The reflection of elastic stress waves at the end of a rod or bar.
What happens when the stress wave reaches the opposite end of the pile (or rod)?
Notice the compression wave moving towards the right along the bar (or pile) in the
figure (a) below, and the tensile wave moving left.
-
7/28/2019 SD Lecture05 Elastic Stress Waves
18/33
Notice that when the two stress waves meet at section a-a they cancel each others
stress, but the particle velocity doubles!
After the waves pass each other the stress and velocity go back to zero.
-
7/28/2019 SD Lecture05 Elastic Stress Waves
19/33
Notice that the section a-a is essentially like a free end. Therefore, in figure (d) below
notice that a compression wave is reflected back as a tension wave. Both have the
same magnitude and shape. Obviously, a tension wave approaching the pile (or rod)end will be reflected back as a compression wave.
-
7/28/2019 SD Lecture05 Elastic Stress Waves
20/33
In contrast to the compression versus tension example seen before, the figure below
shows the effect of two identical compression waves traveling in opposite directions.
When they cross each other (shown below) at section a-a, the stress is doubled but
the particle velocity becomes zero.
-
7/28/2019 SD Lecture05 Elastic Stress Waves
21/33
After the two compression waves pass each other, the stress and particle velocity
return to zero at section a-a.
Notice again, that section a-a remains stationary and hence behaves as the fixed end
of the rod or pile. Notice (below) that at the end, the compression wave is reflectedback also as a compression wave. At the end the stress is doubled.
-
7/28/2019 SD Lecture05 Elastic Stress Waves
22/33
Torsional waves in a rod (bar or pile).
In the figure below, the rod is experiencing a torsional force or torque T; the bar willbe rotated an angle at a distancex.
T
-
7/28/2019 SD Lecture05 Elastic Stress Waves
23/33
2
2
2 2 2 22
2 2 2 2s
s
TT T x J
x t
where J is the polar moment of inertia of the bar cross sec tion.
The torque T itself is given by,
T JGx
Re placing T in t he force equation yields,
Gor v
t x t x
where,
Gv Notice th
+ + =
=
= =
= C
sv is the velo
Ee similarity t
city of torsion
o v
wh al wa sere .ve
=
The differential equation for the torsional stress wave in a bar is found again,
starting with Newtons second law (F = ma),
-
7/28/2019 SD Lecture05 Elastic Stress Waves
24/33
Former Secretary of State Henry Kissinger listens to a lecture on soil dynamics.
-
7/28/2019 SD Lecture05 Elastic Stress Waves
25/33
Longitudinal vibration of short bars.
The solution to the differential equation for the longitudinal stress waves of short barsvibrating in their natural mode is given by,
( )
( )
1 2
1 2
2 22
2 2
2
n n
n
C
u( x ,t ) U( x ) A sin t A cos t
where,U( x ) is the amplitude of the displacement along the length of the rod ,
A and A are cons tan ts and is the natural circular requency of vibration.
u uBacksubstituting int o v
t x
u x , t
= +
=
( )
( )
( ) ( )( )
( )
2
2 2
2
2
2
1 2
0
0n
n n
c c
u x , t x E t
or
U x , t U x
x E
The solution for the amplitude U x can be of the form,
x xU x B sin B cos
v v
=
+ =
= +
-
7/28/2019 SD Lecture05 Elastic Stress Waves
26/33
1) End conditions: free and free.
At the ends, the stress (and strain) at the ends are zero. In other words, at x = 0
dU(x)/dx = 0 and at x = L dU(x)/dx = 0. Differentiating our proposed solution U(x)(previous slide),
1 2
11
22
0 0
0
n n n n
c c c c
n
c
n n
c c
n
c
B x B xdU ( x )cos cos
dx v v v v
U sin g the first boundary condition yields ,
Btherefore B
v
U sin g the sec ond boundary condition yields ,B x
( ) sin and sin ce B is not zero ,v v
L
n orv
=
= =
=
=
21 2 3
c n
n c
n v L
therefore vL n
Thu s , the equation for the am plitude U ( x ) is ,
n xU ( x ) B cos where n , , ...
L
= =
= =
-
7/28/2019 SD Lecture05 Elastic Stress Waves
27/33
The free-free end condition for the vibration of a longitudinal wave in a short bar.
-
7/28/2019 SD Lecture05 Elastic Stress Waves
28/33
2) End conditions: fixed and fixed.
At x = 0 U(x) = 0 (that is, no displacement) and also at x = L U(x) = 0.
2
1 1
1
0
0 0n n
c c
c nn c
The first boundary condition yields,
B
and the sec ond boundary condition,
L LB sin and since B yields n
v v
n v Lor therefore v
L n
Thus, the equation for the displacement amplitude U( x ) is,
n xU( x ) B sin
=
= =
= =
= 1 2 3where n , , ...L
=
-
7/28/2019 SD Lecture05 Elastic Stress Waves
29/33
The fixed-fixed end condition for the vibration of a longitudinal short bar.
-
7/28/2019 SD Lecture05 Elastic Stress Waves
30/33
3) End conditions: fixed and free.
The boundary conditions for this case are,atx= 0 (fixed end) U(x) = 0
atx=L (free end) dU(x)/dx= 0
2
1
1
0
0
2 1 1 2 32
2
n
n n
c c
c
The first boundary condition yields,
U ( x ) B
and the sec ond boundary condition ,
B LdU ( x )cos
dx v vL
or ( n ) where n , , ...v
Thus , the equation for the displacement am plitude U ( x ) is ,
U ( x ) B sin
= =
= =
= =
=
( )2 1n x
L
-
7/28/2019 SD Lecture05 Elastic Stress Waves
31/33
The fixed-free end condition for the vibration of a longitudinal short bar.
-
7/28/2019 SD Lecture05 Elastic Stress Waves
32/33
The torsional vibration of short bars.
Torsional vibration is similar to longitudinal vibrations. The equation for the naturalmodes of vibration is given as,
[ ]1 2
2 22
2 2
n n
s
sn
( x,t ) ( x ) A sin t A cos t
where is the amplitude of the angular distorsion.The solution of the differential equation,
v with the solution above result s in,
t xn v
for the free free and the fixed fixed end conL
= +
=
=
( )2 12 sn
ditions, and
n vfor the fixed free end condition.
L
=
-
7/28/2019 SD Lecture05 Elastic Stress Waves
33/33
References.
Dowding C.H., Construction Vibrations, Prentice-Hall, New Jersey, 1996;
Das, B., Principles of Soil Dynamics, PWS-Kent Publishing Co., Boston, 1993;
Richart F.E., Hall J.R., Woods R.D., Vibrations of Soils and Foundations, Prentice-Hall Inc., New Jersey, 1970;
Humar J.L., Dynamics of Structures, Prentice-Hall, New Jersey, 1990;
Prakash S., Soil Dynamics, McGraw-Hill, New York, 1981;
Timoshenko S.P., Goodier J.N., Theory of Elasticity, McGraw-Hill Book Company,
New York, 1970;