Problem 1 – An experiment is set up in which a steel slab is shocked from both ends. The
pressure generated in the left going shock is 20 GPa. The pressure generated in the right going
shock is 10 GPa. Draw the p-u diagram and Determine
a) The particle velocity in the right going shock. – Answer:
=s
km
256.0
1
u
b) The particle velocity in the left going shock. – Answer:
−= s
km
479.0
2
u
c) The resultant particle velocity in the material. – Answer:
−= s
km
223.0
3
u
d) The resultant pressure generated. – Answer:
GPa872.32
3=p
The slab has the following properties:
Steel
490.1
s
km
569.4
cm
g
896.7
0
3
0
=
=
=
Steel
s
cSteel
Steel
Solution:
If we examine our diagram from the notes we see that we should be able to determine the answer
0850.0066.3 1
2
1=−+ uu
( ) ( )( )( )
( )
−=
−−−
=s
km
789.1533.1
12
850.014066.3066.3 2
1
u
=s
km
256.0
1
u
The particle velocity in the left going shock is found again by noting that we have the left-going
Hugoniot passing through the origin. Our equation for the left-going Hugoniot is
( ) ( )
2
20200
2
202002 00 suucusucp
+=−+−=
Inserting our values for steel we have
( ) ( ) ( ) ( )
2
2
2
3
2
3s
km
490.1
cm
g
896.7
s
km
s
km
569.4
cm
g
896.7GPa20
+
=uu
2
22 765.11077.36GPa20 uu +=
0700.1066.3 2
2
2=−+ uu
( ) ( )( )( )
( )
−=
−−−
=s
km
012.2533.1
12
700.114066.3066.3 2
2
u
084.3047.12765.11471.18077.36 3
2
333 +−++−= uuup
555.21124.48765.11 3
2
33 +−= uup
(A)
Now we need to examine the right going Hugoniot where we know points (u2,p2) and (2u2,0)
( ) ( )
2
23023003 22 uusuucp −+−=
( )( ) ( )( ) ( )( ) ( )( )
2
333 479.02490.1896.7479.02569.4896.7 −−+−−= uup
( )( ) ( )( )
2
333 958.0765.11958.0077.36 +++= uup
797.10542.22765.11562.34077.36 3
2
333 ++++= uuup
359.45619.58765.11 3
2
33 ++= uup
(B)
If we now subtract equation (A) from equation (B) we can solve for u3 so we have
0804.23743.106 3=+u
−= s
km
223.0
3
u
The pressure then can be found from either equation (A) or (B). We’ll use (B).
( ) ( )
359.45223.0619.58223.0765.11 2
3+−+−=p
GPa872.32
3=p
Problem 2 – A strange jeweler wants to make an earring by launching a quartz slab at a slab of
gold. His high tech instrumentation measures the induced velocity in the gold as 0.5 km/s.
Determine
0
b) The shock pressure at the interface.
1=p
c) The speed at which the shock wave travels in the gold.
km
Au
d) The speed at which the shock wave travels in the quartz.
km
Q
695.1
s
km
794.0
cm
g
204.2
0
3
0
=
=
=
Q
s
cQ
Q
572.1
s
km
056.3
cm
g
24.19
0
3
0
=
=
=
Au
s
cAu
Au
Solution:
We need to recall that the induced velocity in the gold is the velocity at the interface. We can
find the pressure at the interface if we write the p–u Hugoniot equations for the gold, again by
convention assume that the quartz plate is flying from left to right, then we need a right-going
0877.9532.0 0
2
0=−− uu
We solve this using
a
acbb
x2
4
2−−
=
and get
( ) ( )( )( )
( )
=
−−
=s
km
154.3266.0
12
877.914532.0532.0 2
0
u
=s
km
420.3
0
u
18.2 Rarefaction Waves
Problem 3 – An experiment is set up in which a tungsten penetrator is fired against a rigid target.
The impact velocity is 500 m/s. Determine the shock pressure, tensile stress and also if the
penetrator will break up.
The material has the following properties:
Tungsten
GPa0.2
s
km
541.5
237.1
s
km
029.4
cm
g
224.19
0
3
0
=
=
=
=
=
Dynamic
W
W
W
UTS
L
W
c
s
c
Solution:
The only piece of information we have is the impact velocity but we do know the equations for
1
2
00
0
u
p
cWW L−
−
=
→
102 ucp WW L
=
( )( )( )
GPa26.535.0541.5224.19
2==p
Since this value is greater than the dynamic tensile strength of the material, the part will spall
This is a Matlab plot of the actual curves
Problem 4 – A 4 inch long steel bar impacts a 12 inch thick slab of 4340 steel at 1000 m/s and
bounces off. Assuming the impact is normal and using 1 dimensional equations determine:
b) The pressure developed at the interface
1=p
4340 steel
u
p
Impact velocity
(500 m/s)
=
=
=
=
s
km
941.5
490.1
s
km
569.4
cm
g
896.7
0
3
0
Steel
Steel
L
Steel
Steel
c
s
c
Solution:
Since the target and penetrator are the same material, we know that
01
2uu =
111 490.1896.7569.4896.7 uup +=
0
2
01
2
11 039.18941.2077.36765.11 uuuup +=+=
This yields
1=+=p
( )
010
uu
( )
0980.20
−
=U
−=
−= s
mm
314.5
s
km
314.5
U
If the shock was moving toward the rear surface of our rod at 5.314 mm/microsecond
then it would reach the rear of the plate in
s12.19
s
mm
314.5
mm6.101
=
== U
l
t
Recall that the speed of the head of the rarefaction wave is the slope of the p-u Hugoniot curve at
the material pressure. In the notes, an example was written for a left going rarefaction wave. In
our case the slope is the negative of this value which we know.
( ) ( )
2
10010001 uusuucp −+−=
( )( )( ) ( )( )( )
2
10101 490.1896.7569.4896.7 uuuup −+−=
( )( )( ) ( )( )( )
2
111 0.1490.1896.70.1569.4896.7 uup −+−=
765.11530.23765.11077.36077.36 1
2
111 +−++−= uuup
s
896.7
Rhead
s
mm
059.6
== U
s87.35s77.16s12.19 μμμtshock =+=
Problem 5 – We would like to determine how fast a large fragment will be propelled by an
explosion using Hugoniots. Assume we have a 6 inch cube of steel that has a pressure of 5.0
GPa applied to one face for 3μs. Determine the velocity of the piece of steel by
a.) Using the Hugoniot curve
b.) Using impulse and momentum (It might help to reference a statics and dynamics
book)
c.) Compare the two methods and comment on the differences
d.) What can you do to the problem parameters to make the answers the same?
Assume that the material does not spall at all. The material has the following properties:
Steel
=
=
=
=
s
km
900.5
490.1
s
km
569.4
cm
g
896.7
0
3
0
Steel
Steel
Steel
L
Steel
c
s
c
Solution: We shall define the motion of the material as beginning when the surface away from
the explosion begins to move, that is the shock reflects from the free surface as a rarefaction. In
this case the material velocity will be the speed of the metal (assuming it holds together) and we
00.5077.36765.11 1
2
1=−+ uu
0425.0066.3 1
2
1=−+ uu
( ) ( )( )( )
( )
−=
−−−
=s
km
666.1530.1
12
425.014066.3066.3 2
1
u
=s
km
133.0
1
u
p
u
Left going Hugoniot
for material
u = 2u1A
p = p1
u = u1A
p = 0
Right going Hugoniot for
material
u = 0
Pressure behind the
shock
in material
Particle velocity behind original
shock in material
At first blush, the velocity of the bar is then
( )( )
=
== s
km
266.0
s
km
133.022 1
uu
Now let’s look at an impulse and momentum approach. Recall from statics and dynamics that
==
s
ftlbm
3.521,2FtFdt
t
The mass of the block is
V
0Steel
m
=
m
Fdt
Vt
=
or
( )
( )
=
=
=s
m
54.12
s
ft
14.41
lbm128.61
s
ftlbm
3.521,2
V
This is very different and we should recover the same answer – so what is wrong? Let’s look at
=
=s
mm
767.4
s
km
767.4
U
The plate is 6” thick so that is
( )
( )
mm4.152
in
mm
4.25in6=
=l
So the shock reaches the free surface in
s97.31
s
mm
767.4
mm4.152
=
== U
l
t
This says that the pulse width is such that the event is over well before the shock wave interacts
with the free surface. To make the calculations match we must hold the pressure just until the
shock reaches the free surface.
( )
( ) ( )
s00003197.0
slbf
ftlbm
2.32lbf000,100,26 2
==
FtFdt
t
==
s
ftlbm
868,26FtFdt
t
( )
( )
=
=
=s
m
133
s
ft
4.438
lbm128.61
s
ftlbm
868,26
V
So we see that with round off error the Hugoniot curve will give us the correct material velocity
behind the shock – ONLY if the pressure is held until the shock reaches the free surface. But
1
=
du
uu
=
=s
mm
965.4
s
km
896.7
203.39
Rhead
U
So the time it takes the rarefaction to move from the rear surface of the flyer to the interface is
Our equation for the left-going Hugoniot is
( ) ( )
2
10010001 uusuucp ff −+−=
(RH-10)
( )( )( ) ( )( )( )
2
111 461.0460.1350.11461.0051.2350.11 uup −+−=
( )( ) ( )( )
2
111 461.0571.16461.0279.23 uup −+−=
639.7278.15571.16279.23018.11 1
2
111 +−+−= uuup
657.18557.38571.16 1
2
11 +−= uup
(A)
Here we assume the units are correct and we know the answer will be in GPa. Also for our right
going Hugoniot in the steel we can write
2
101001 suucp
+=
(RH-8)
( )( ) ( )( )
2
111 490.1896.7569.4896.7 uup +=
1
2
11 077.36765.11 uup +=
(B)
Equating equations (A) and (B) yields
p
u
Left going Hugoniot
For flyer plate
u = u0f
u = u1
p = 0
Right going Hugoniot for
flyer plate
(used for rarefaction wave)
u = 0
Pressure behind shock
in both flyer and target
p1f = p1t
Particle velocities in both materials
behind generated shocks
Right going Hugoniot for
target
p = p1
Initial velocity of flyer plate
u = u2f
A
B
Material velocity behind
rarefaction wave in flyer
0657.18634.74806.4 1
2
1=+− uu
( )
806.42
1
=s
km
254.0
1
u
Based on our diagram it had to be positive. The pressure now comes from inserting this value in
either equation (A) or (B). We’ll use (B).
( )( ) ( )( )
GPa923.9254.0077.36254.0765.11 2
1=+=p
p
u
Left going
Hugoniot
u = 2u1
p = p1
u = u1
p = 0
Right going Hugoniot for
incipient shock
u = 0
Shock
Slope =
0URhead
Head of rarefaction
Slope =
0URtail
Tail of rarefaction
18.3 Stress Waves in Solids
p
u
Left going
Hugoniot
for rarefaction
u = 2u1
p = p1
u = u1
p = 0
Right going Hugoniot for
incident shock
u = 0
Shock
Slope =
0CL
Problem 7 – A 4 inch long steel bar impacts a 12 inch thick slab of 4340 steel at 1000 m/s and
bounces off. Assuming the impact is normal and using 1 dimensional equations determine:
a) The duration of the impact event (use Hugoniots)
b) The pressure developed at the interface (use Hugniots)
1=p
c) The thickness of the first spalled piece (if any) assuming the input pulse is a constant
square wave pulse throughout the impact event
d) Illustrate your answer to b) above
e) Illustrate your answer to c) above
4340 steel
Modulus of Elasticity (× 106 psi) 30.0
Modulus of Rigidity (shear) (× 106 psi) 11.5
Poisson’s Ratio 0.29
Ultimate tensile stress (lbf/in2) 250,000
=
=
=
=
s
km
941.5
490.1
s
km
569.4
cm
g
896.7
0
3
0
Steel
Steel
L
Steel
Steel
c
s
c
Solution:
Since the target and penetrator are the same material, we know that
01
2uu =
2
00000
2
101001 4
1
2
1usucsuucp
+=+=
( )( ) ( )( )
2
111 490.1896.7569.4896.7 uup +=
0
2
01
2
11 039.18941.2077.36765.11 uuuup +=+=
This yields
( ) ( )
GPa980.201039.181941.2 2
1=+=p
s
If the shock was moving toward the rear surface of our rod at 5.314 mm/microsecond
then it would reach the rear of the plate in
Recall that the speed of the head of the rarefaction wave is the slope of the p-u Hugoniot curve at
10010001 uusuucp −+−=