So there are two pressure calculations; the chamber assuming 5,000 psi and the muzzle
assuming 2,500 psi with a linear taper beginning at the peak pressure location.
We shall assume that there actually is an axial force (assuming the projectile gets stuck) so for
each location we can write the following equations for stress assuming no shear:
( ) ( ) ( )
2
222
zzrrrrzz
Y
−+−+−
=
( ) ( )
2222
44 io
c
io
axial
zz
dd
pA
dd
F
−
=
−
=
( )
( )
22
22
io
io
dd
dd
p−
+
=
p
rr −=
I put these equations onto MathCAD and iterated. A reasonable solution is for a 2 inch OD
( )
( )
( )
( ) ( )
−
+
+−
−
−
=21
2
22
2
2
2
1
22
1
1pp
r
ba
bpap
abE
r
u
tube
i
tube
422
=2
in
lbf
000,16
We now need to see what the strain level is with the cartridge case O.D. expanded to the
chambers pressurized I.D. We do this by first calculating the increase in the diameters assuming
the cartridge case does not provide any strength. We shall assume a gap of 0.0005 inches
Y=16,000
4.12 Sabot Design
Problem 11 – You are to design a rifled 20mm gun system for an anti-vehicle application. The
gun is to throw a 0.25 lbm, 0.5-inch diameter, cylindrical sub projectile at 4,500 ft/s. A sabot
will be placed on the outside of the sub projectile. The sabot can be any material you like and
can find properties for. The sabot shall be made of 3 pieces which are at least 3 inches in
length. Your design must use a brass cartridge case to assist with obturation of the breech.
Other assumptions and information are:
1.) You do not need to design the breech – assume it will hold the cartridge case in
properly (in reality we can always add more threads to the design).
2.) Even though there will be a slight taper on the chamber (which must be larger than
the bore diameter for seating purposes) assume, for calculation purposes, that the chamber is
cylindrical at it’s maximum diameter.
3.) The tube is to be steel and assume that the yield strength is 60,000 psi (this accounts
for the effect of cyclic loading). The modulus of elasticity is 29 × 106 psi. Poisson’s ratio is 0.29.
4.) Assume the propellant is either cylindrical or single perforated (and state your
assumption)
5.) Choose from the following propellants
Propellant
Linearized
burn rate, β
(in/s/psi)
Solid
density, δ
(lbm/in3)
Adiabatic
flame
temperature,
T0 (°R)
Propellant
force, λ (ft-
lbf/lbm)
Specific
Heat
ratio, γ
IMR
0.000132
0.0602
5,103
327,000
1.2413
M12
0.000137
0.0600
5,393
362,000
1.2326
Bullseye
0.000316
0.0590
6,804
425,000
1.2523
Red Dot
0.000153
0.0593
5,774
375,000
1.2400
Navy Pyro
0.000135
0.0566
4,477
321,000
1.2454
6.) Assume the cartridge case is brass and use a bi-linear kinematic hardening model
where the brass has a modulus of elasticity of 15106 psi, a local tangent modulus of 13106
psi, a yield stress of 16,000 psi (yield occurs in this material at
= 0.002)
7.) The weapon shall be as light as possible.
The design is to proceed as follows (not necessarily in the order given):
A.) Interior Ballistics Design
1.) Size the chamber length and diameter
3.) Determine a web thickness for the propellant
5.) Determine V, pB and x for the projectile at peak pressure
7.) Determine the muzzle velocity of the projectile
B.) Gun Tube Design
1.) Based on the calculations of part A.) develop a pressure-distance curve to use as
criteria for your gun design
2.) Determine the outside diameter of the gun tube. To keep the design light as
possible use the design rules provided in the text and taper the tube towards the muzzle. If
C.) Cartridge Case Design
1.) Determine a thickness and tolerance for your cartridge case
Note that for these calculations you must show that the case may be easily extracted at
the limits of the tolerance.
D.) Sabot Design
1.) Based on a twist rate of 1 in 30, calculate the thickness of a restraining band, 1” wide
located over the petal C.G., required to reliably break at shot exit. This band can be
considered as a thin cylinder evenly loaded by the three sabot petals (i.e. no
moment). “Reliable” breakage can be considered as when the hoop stress is 1.25
times the allowable hoop stress. Neglect the dynamic pressure cause by the forward
velocity of the projectile.
It is important that you write down all of your assumptions. It is also highly likely that as you
proceed further along with your design you may come upon a situation that requires you to
revisit an assumption you made earlier – this is to be expected and it is part of the design
process.
( )
3
petal
The total throw mass is then
I will then set up the equations and vary the tube length and the charge quantity until I obtain a
reasonable solution. The steps I followed are below. The result (certainly NOT optimized) was
a gun with a projectile travel of 75 inches, a chamber of 23.27 inches length which uses 250
grams of IMR propellant. The procedure was as follows.
The volume the charge takes up will be
Let’s double this volume for insurance so we have an empty chamber volume of
3
in272.18U=
We shall assume that the chamber is 1 inch in diameter
in0.1=
c
d
Then
( )
22
2
in785.0in
4
0.1 ==
c
A
and
( )
( )
in265.23
in
4
0.1
in272.18U
2
2
3
===
c
cA
L
Now let’s start the ballistics calculations. The bore area is
( )
22
2
in487.0in
4
787.0 ==
A
+
1
2
1
2
1
w
c
cw
( )
( )
( )
( )( )
( )
( )( )
( )
( )
( ) ( )
−
−
−
+
+
=
2
2
2
2
2
2
2
4
2
lbf
s
in
000132.0
ftlbm
slbf
2.32
1
lbm
lbfft
000,327lbm55.0lbm337.0
337.02
55.0
1
337.03
55.0
1
in02.0in487.0
M
354.1=M
The volume initially available for gas generation is
( )
( )
333
base in136.90in136.9in272.18VUV =−−=−−=
c
i
in24.32
max =
p
x
( ) ( ) ( )
( )
( )
( )( )
( )
( )( )
( )
( )( )
261.01354.1
3261.01
337.03
55.0
1
337.02
55.0
1
in9.136
lbm55.0
ft
in
12
lbm
lbfft
000,327
max
−−
−
+
+
−
=epB
=2
in
lbf
292,6
max
B
p
( )
( )
( )
( ) ( )
( )
( )( )
+
−
−
−
=
337.02
55.0
1
ftlbm
slbf
2.32
1
lbm337.0
in
lbf
s
in
000132.0
261.01in02.0in487.0
2
2
2
max
p
V
=s
ft
867,2
max
p
V
Now for the values at charge burnout
c
( ) ( ) ( )
( )
( )
( )( )
( )
( )( )
( )
354.1
3
337.03
55.0
1
337.02
55.0
1
in9.136
lbm55.0
ft
in
12
lbm
lbfft
000,327
−
+
+
−
=ep c
B
=2
in
lbf
980,5
c
B
p
( )
( )
( ) ( )
( )
( )( )
+
−
−
=
337.02
55.0
1
ftlbm
slbf
2.32
1
lbm337.0
in
lbf
s
in
000132.0
in02.0in487.0
2
2
2
c
V
=s
ft
881,3
c
V
2103
4103
6103
8103
6.5 103
p.B x( )
p.c x.1
( )
p.t x.2
( )
p.s x( )
p.sc x.1
( )
We shall assume that there actually is an axial force (assuming the projectile gets stuck) so for
each location we can write the following equations for stress assuming no shear:
( ) ( ) ( )
2
222
zzrrrrzz
Y
−+−+−
=
( ) ( )
2222
44 io
c
io
axial
zz
dd
pA
dd
F
−
=
−
=
( )
( )
22
22
io
io
dd
dd
p−
+
=
p
rr −=
I put these equations onto MathCAD and iterated. A reasonable solution is for a 2 inch OD
section from the chamber to the point of projectile travel at maximum pressure tapering to a 1
inch OD at the muzzle
for the barrel so the weight can be readily estimated as 96 lb.
C.) Cartridge Case Design:
We first determine how much radial expansion we get in the gun tube at 6,500 psi. Thus we
have.
( )
( )
( )
( ) ( )
−
+
+−
−
−
=21
2
22
2
2
2
1
22
1
1pp
r
ba
bpap
abE
r
u
tube
i
tube
422
Y=16,000
Thus the case will just barely stick. Dropping the tolerance on the chamber to +0.0005/-0.000
allows it to be extracted.
D.) Sabot Design: In this case, since we are neglecting dynamic pressure, and since the band is
located over the C.G., we can assume that there is no opening moment and determine an
( )
( )
lbm029.0lbm
3
087.0 ==
petal
m
The C.G. of each petal was located at the centroid of the petal which is found through
( ) ( )
( )( )
( )
( )
in322.0
mm
in
4.25
1
mm
2
7.12
mm
22
7.1220 =
+
−
=
CG
r
The projectile spin rate is
( ) ( ) ( ) ( )
( )
=
=s
rad
363,14
cal
mm
20
s
ft
500,4
ft
in
12
in
mm
4.25
cal
rev
30
1
rev
rad
2
The force exerted by the three petals is then
( )( )
( )
( )
( ) ( )
lbf981,14
slbf
ftlbm
2.32
ft
in
12
s
rad
363,14in322.0lbm029.03
2
2
2
=
=
tot
F
t
pr
=
(MM-41)
Based on the constraints of the problem statement his must be greater than 1.25 times the
allowable stress – let’s assume 200,000 psi so
( ) ( )
( )( ) ( )
in0095.0
in
mm
4.25
in
lbf
000,2001.25
mm10
in
lbf
056,6
25.1
2
2=
==
pr
t