( )( )( ) ( )( )( )
2
10101 490.1896.7569.4896.7 uuuup −+−=
765.11530.23765.11077.36077.36 1
2
111 +−++−= uuup
s
896.7
Rhead
( )
( )
in75.3mm31.95
s
mm
2
314.5
s87.35
2==
==
U
tl
The illustration clearly shows why. When the compressive pulse gets reflected as a tensile pulse,
the net stress where the two waves overlap is zero. If we assume that the reflected tensile wave
Problem 8 – A Japanese 20mm projectile with the properties below impacts a 7” thick concrete
wall at 0° obliquity. The concrete has a 1,500 psi unconfined compressive strength and density
of 0.080 lbm/in3. The concrete dynamic tensile strength is 1,000 psi. If the projectile has the
properties below:
a.) Determine the duration of the impact event using the assumption of non-penetration –
b.) Determine whether the concrete will spall and if so determine the extent (in inches of
thickness) of the total spallation – list all assumptions!
1=t
c.) Determine if the projectile perforates the concrete accounting for the spallation
d.) Using your ability to determine the timing of the penetration events explain why or
why not the above model is valid – i.e. prove it using the numbers
Estimated Penetrator Information
mm60
mm20
mm40
=
=
=
L
d
s
=
=
s
m
550
g128
s
V
m
=3
in
lbm
283.0
p
Steel Concrete
=
=
=
=
s
km
941.5
490.1
s
km
569.4
cm
g
896.7
0
3
0
Steel
Steel
Steel
L
Steel
c
s
c
)estimate(
s
km
0.4
)estimate(4.1
)estimate(
s
km
0.4
cm
g
232.2
0
3
0
=
=
=
=
Concrete
Concrete
Concrete
L
Concrete
c
s
c
Solution: We must use Hugoniots in part a.) To determine the pulse duration so we begin as we
have been doing all along. Our diagram tells us that to get the pressure generated at the interface
56.394.1276.1108.3684.19 1
2
111 +−+−= uuup
40.2302.4976.11 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 concrete we can write
2
101001 suucp
+=
( )( ) ( )( )
2
111 4.1232.20.4232.2 uup +=
1
2
11 928.8125.3 uup +=
(B)
Equating equations (A) and (B) yields
040.2395.57635.8 1
2
1=+− uu
0710.271.6 1
2
1=+− uu
( )
12
1
=s
km
432.0
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 (A).
1=+−=p
( )
010
uu
( )
( )( )
55.0432.0896.7
0418.4
−
−
=U
−=
−= s
mm
741.4
s
km
741.4
U
s66.12
s
mm
741.4
mm60
=
== U
l
t
To determine the speed of the leading edge of the rarefaction wave we need to recall that the
40.2302.4976.11 1
2
11 +−= uup
(A)
First check is to ask – is this faster than the shock? – the answer is yes so we’re doing well. The
time it takes the rarefaction to reach the front face is then
Then the total time for the shock pulse is the time between impact and the rarefaction wave
reaching the interface or
If we approximated it using twice the length divided by the longitudinal wave speed we would
have gotten
( )
s
mm
741.4
Which is different but fairly close – but is was not what was requested in the problem.
Even though we are pretty sure the pulse strength is sufficient to fail the concrete, we will check
the level using the Hugoniot’s. We have to theoretically determine the velocity at which the
projectile will rebound, even though it will not. We shall ignore this velocity other than using it
u1. So we can write
ff uuuu 0112 −=−
ff uuu 012 2−=
ff uu 33 364.0910.46928.8 −=
=s
km
305.0
3f
u
The tensile stress is then
303 ==== fL ucp ConcreteConcrete
This will spall the concrete and, if we assume a square pulse, there will be one big spalled chunk
whose thickness is given by
s
2
2==
if we account for the spallation then the projectile will only actually be required to penetrate 7-
2.32 or 4.68” of concrete in order to perforate. We now turn to this calculation. As in the
midterm and homework, the first thing we need to determine is the parameter, N. We shall
assume that B = 1 as was stated in the notes. Before we can determine N we need to find ψ so
d
s
=
2
mm20
mm40 ==
2
24
18
−
=N
( )
( )
156.0
224
128
2=
−
=N
Now we need the empirical parameter S but to get it we need to convert our unconfined
compressive stress, fc’ to MPa.
in
lbf
05.145
in
2
2=
c
Now we use our curve-fitted equation
5603.0
48.93 −
=c
fS
( )( )
251.2534.1048.93 5603.0 == −
S
Now we need to find V1, c and t1 (assuming the projectile gets to the end of the crater region).
Let’s get all of our units consistent first.
mm
1000
The rest of the units are consistent so we can write
Ndm
fSdmV
Vcs
3
32
12
2
+
−
=
( )
( ) ( )
( )
( )
( )
( )
( )( )
+
−
=
3
3
3
2
63
3
2
2
2
1
m
kg
216,2156.0m020.02kg128.0
m
N
1034.10251.25m020.02
s
m
550kg128.0
V
=s
m
420
1
V
( )
2
1
2
4VV
d
m
cs−=
( )
( )( )
( ) ( )
( )
=
−= 22
2
22
2
2s
kg
000,088,10
s
m
420550
m020.04
kg128.0
c
Finally we can get t1 from
=−
s
V
V
c
m
t1
1
1cos
( )
( )
s800s00008.0
550
420
cos
s
kg
000,088,10
kg128.0 1
2
1
==
=−
t
Now we have everything we need to calculate the penetration depth
d
fS
VN
Nd
m
P
c
21ln
22
1
2+
+=
( )( )
( )
( )( )
( )( ) ( )
( )
( )
( )( )
m020.02
m
N
1034.10251.25
s
m
420
m
kg
216,2156.0
1ln
m
kg
216,2156.0m020.0
kg128.02
2
6
2
2
2
3
3
2
2
+
+
=
P
in445.6m163.0 ==P
Since this is greater that the 4.68 inches of unspalled material, the projectile will perforate.
The model appears to be valid because it takes 800 microseconds for the projectile to emerge
from the crater region and the spallation takes approximately 23 microseconds to create the pulse
and then for the pulse to travel 7 inches into the concrete requires us to examine the U-u
Hugoniot
sucU += 0
s
s
s
Concrete
we need to calculate the left-going Hugoniot in the flyer and solve it for the pressure since we
065.1911.56635.8 1
2
1=+− uu
0275.250.6 1
2
1=+− uu
( )
12
1
=s
km
371.0
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 (A).
( )
010
uu
( )
( )( )
472.0371.0896.7
0765.3
−
−
=U
−=
−= s
mm
721.4
s
km
721.4
U
Why didn’t we use equation (RH-3) or (RH-1)? The reason is that if we used (RH-3) we would
s6.354
s
mm
721.4
mm674,1
=
== U
l
t
To determine the speed of the leading edge of the rarefaction wave we need to recall our diagram
from the notes below
Here we see 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, this example was written for a left going rarefaction
wave. In our case the slope is the negative of this value which we know.
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
65.1918.4776.11 1
2
11 +−= uup
(A)
First check is to ask – is this faster than the shock? – the answer is yes so we’re doing well. The
time it takes the rarefaction to reach the front face is then
s
mm
870.4
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
Slope is the velocity of the
rarefaction wave in the flyer
=
=s
870.4
s
896.7
Rhead
U