s3.698s7.343s6.354 μμμtshock =+=
If we approximated it using twice the length divided by the longitudinal wave speed we would
( )( ) ( )( )
( )
fff uuu 323 941.5896.70.4232.2 −=
(E)
The way to find u2f is to note that the two Hugoniots for the steel are reflected about the velocity
u1. So we can write
ff uuuu 0112 −=−
ff uuu 012 2−=
ff uu 33 270.0910.46928.8 −=
=s
km
227.0
3f
u
( )
( )
in24.52mm327,1
s
mm
2
721.4
s2.562
2==
==
U
tl
Problem 10 – A 2 inch long steel bar impacts a 6 inch thick slab of 4340 steel at 1000 m/s and
bounces off. Assuming the impact is normal and using 1 dimensional hugoniot equations
determine a material type and thickness of a backing material that is required to keep the
interface pressure below 2.5 GPa for the duration of the impact event. For full credit state all of
your assumptions
Use these properties for the steel
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
Note: use the Cooper text for material hugoniot properties but you may need to research wave
speeds for some of the materials – looking on the internet is fine.
Solution: Since the target and penetrator are the same material, we know that
01
2uu =
Now we can simply write our right going Hugoniot in the target as
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
( )
010
uu
( )
( )( )
0.15.0896.7
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
s56.9
s
mm
314.5
mm8.50
=
== 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
mm
059.6
s
km
896.7
842.47
Rhead
U
2
202002 suucp
+=
(RH-8)
Inserting values for the steel we have
( )
( ) ( ) ( ) ( )
2
2
2
3
2
3s
km
490.1
cm
g
775.1
s
km
s
km
569.4
cm
g
896.7GPa5.2
+
=uu
( )
2
22 645.208.36GPa5.2 uu +=
(A)
05.208.36645.2 2
2
2=−+ uu
095.064.13 2
2
2=−+ uu
( ) ( )( )( )
( )
−=
−−−
=s
km
889.6820.6
12
95.01464.1364.13 2
2
u
p
u
Left going Hugoniot
for material “A”
u = 2u1A
p = p2
u = u2
p = 0
Right going Hugoniot for
material “B”
u = 0
Pressure behind original
shock
in material “A”
Particle velocities in both materials
behind generated shocks
Right going Hugoniot for
material “A”
u = u1A
p = p1
Pressure behind generated
Shocks in both materials
Particle velocity behind original
shock in material “A”
=s
km
069.0
2
u
Now this result may seem strange at first but we need to think it through a bit. we would expect
to see a velocity smaller than but very close to 1 km/s. If we note that we used the right going
707.1
s
km
cm
g
153.2
0
3
0
=
=
PTFE
PTFE
s
PTFE
So we write
( ) ( ) ( ) ( ) ( )( )
2
2
33 s
km
931.0707.1
cm
g
153.2
s
km
931.0
s
km
841.1
cm
g
153.2GPa
+
=
PTFE
p
GPa88.6=
PTFE
p
Let’s try rubber
865.1
s
km
852.0
cm
g
010.1
0
3
0
=
=
=
rubber
rubber
s
c
rubber
So we are good with rubber as the material – now for the thickness. Just to get the data correct
let’s simultaneously solve the equations for the steel and the rubber to find the correct interface
velocity.
( ) ( ) ( ) ( ) ( )( )
2
2
33 s
km
1490.1
cm
g
775.1
s
km
1
s
km
569.4
cm
g
896.7
−
+
−
=uup
(B)
( ) ( ) ( ) ( )
2
2
33 s
km
865.1
cm
g
010.1
s
km
s
km
852.0
cm
g
010.1
+
=uup
(C)
Rewriting equations (B) and (C) we have
Equation the two we have
Then the pressure is
For the rubber we have
rubber
Now we can make a simplifying assumption. We know the length of the pulse in the rod was
17.94 microseconds. I know that to get the thickness of the rubber correct I need to use the
shock speed to the free surface and the rarefaction wave speed back to the interface. Since the
s
mm
020.4
So the total pulse time would be
18.4 Detonation Physics
Problem 11 – Tetryl (C7H5N5O8) is detonated in standard sea level air. Assuming non-ideal
behavior and
+=
=
=gmol
kcal
67.4 and
gmol
g
213 ,
cm
g
86.0 0
3fmix hMW
a.) Determine the reaction equation assuming no dissociation
b.) Determine the temperature of the products behind the detonation wave, T2
2
c.) Determine the speed of the detonation wave, D
m
2=p
e.) Determine the induced velocity of the gas behind the wave, u2
m
2
Solution: Let’s first determine the chemical reaction. We form the hydrogen and oxygen into
water.
K
K 4000K 4000
vv cNc
Now we estimate the ratio of specific heats at our assumed temperature, T2* of 4000 K.
+=
K 4000
1
K 4000
v
c
Rn
( )
( )
( )
251.1
K
cal
8.94
Kgmol
cal
99.1gmol12
1
K 4000 =
+=
Now we can use this value to calculate the gas temperature. If we get 4000 K, we’re done, if not,
we try again using the value we do obtain (or one that’s close).
The equation for the temperature change is:
2
2
1
Q
TRn
++=
Q
TRn
2
2
1
Kgmol
K 3700
v
K
K 3700
v
CO 7.00 5.5 38.50
N2 6.97 2.5 17.43
==
K
cal
2.94
K 3700K 3700
vv cnc
( )
( )
( ) ( )
( )( )
( )
( )
+
+=
K
cal
2.94
cal200,287
K
cal
2.94253.1
K700,3
Kgmol
cal
99.1gmol12
2
1
K298
2
T
( )
( )
( )
K049,3K374K298
2
++=
T
K721,3
2
=
T
We can call it quits here. Even though this correlation is not exact, it is good enough (within
0.6%) for our purposes. We shall use 3,700 K as our ideal final temperature. So
K700,3
2
=
T
( )
( )
( ) ( )
( ) ( ) ( ) ( )
( )( ) ( )
8557 ONHC
2
*
gmol1
gmol
g
287253.1
kg
g
1000
N
s
mkg
1
J
mN
1
cal
J
18.4K700,3
Kgmol
cal
99.1gmol12
1253.1
+=
D
and
=s
m
283,2
*
D
( )
=
=g
cm
163.1
cm
g
86.0
13
3
v
We can now compute the parameter x1 for use in the correction tables from
mix
avMWT
k
x=
1
029.2
2
020.2039.2
11 =
+
=x
125.2
2
104.2126.2
12 =
+
=x
2
D
077.2=
D
D
and
( )( )
=
=s
m
742,4
s
m
283,2077.2D
Now that we have this we can estimate the pressure behind the wave directly from
−=
2
1
2
21x
x
Dp
( ) ( ) ( )
−
=734.1
26.1
1
g
kg
1000
1
m
cm
100
cm
g
86.0
s
m
742,4 3
3
3
32
2
2
2
p
2