Problem 1 – For the state of stress below, find the principal stresses and the maximum shear
stress
MPa
900
0415
01520
−
=σ
2
2
1
3
Here we go
222
2zxyzxyxxzzzzyyyyxx
I
−−−++=
zxyzxyxyzzzxyyyzxxzzyyxx
I
2
222
3+−−−=
2
1
I−+=
32
1
2
+−+= 3
4
2
1
I
( )
−
23
2
2
1
32
3
II
MPa159420
1=−+=I
2
max =−−=
4.2 Failure Criteria
Problem 2 – A component has principal stress values of 20,000, 56,000 and –220,000 psi (note
that negative means compressive stress), if the yield strength in a simple tension test of the
material was found to be 180,000 psi will the part survive based on the von Mises failure
criteria?
2222
( ) ( )( ) ( )
rad096.0
305,127361159152
cos
3
1
23
2
1=
+−−
=−
( ) ( ) ( )
MPa25
3
4
096.0cos3613152
3
15 2
3−=
+−−+=
( ) ( )( ) ( )
2
2
10
222
2
in
lbf
105.13000,56000,220000,220000,20000,20000,562
=−−+−−+−
Y
Since this value is greater, the part will fail.
4.8 Cartridge Case Design
Problem 3 – A design for a 105mm weapon is being considered. The chamber is stated to
withstand the desired 35,000 psi and is essentially a steel cylinder of 4.5 inch I.D. and 7”
O.D.(Etube = 30106 psi, = 0.3). We have decided to use brass with an outer diameter of 4.490
inches. If we use a bi-linear kinematic hardening model where the brass has a modulus of
elasticity of 15106 psi, a local tangent modulus of 12.5106 psi, a yield stress of 15,000 psi
(yield occurs in this material at
= 0.001) and an ultimate tensile strength of 45,000 psi. based
on the information given what is the radial clearance between the case and the chamber after
firing neglecting thermal effects?
tube
( )
in
lbf
1015
2
6
fR
We can convert this to a diameter as follows
case
Thus the case has almost an 0.008 inch diametral clearance and can be extracted. It is worthy
of note here that the cartridge case thickness did not come into play at all. This is because we
assumed the case to be a thin wall cylinder (i.e. with no gradient in stress in the radial direction)
rucase
=
where we note that the radial displacement is a function of the radial location
Y
f
R
4.10 Shell Structural Analysis
Problem 4 – A High Explosive Projectile is to be designed for a 155mm cannon using a ½ inch
thick steel wall with TNT as the filler material. Assume the shell and filler are a cylinder 0.75
meters in length. It is to be capable of surviving a worn-tube Torsional Impulse (angular
acceleration) of 440,000 radians/sec2.
a) Derive the expression to calculate the torque on the projectile that achieves this
acceleration if the torque is applied at the O.D. of the shell
b) Calculate the value of the torque assuming the density of steel is 0.283 lbm/in3 and
TNT is 0.060 lbm/in3
Hint: start from
TT maF=
444
1
332
3
2
2 io
r
r
tWALL rrldrrlF
o
i
−==
(B)
Performing the same operation on the filler yields
3
0
2
3
2
2 iFILLER
r
FILLERtFILLER rldrrlF
i
==
(C)
Thus the total tangential force is
( )
333
3
2
iFILLERiotFILLERtWALLt rrrlFFF
+−=+=
(D)
Problem 5 – To participate in a failure investigation of an explosive someone asks you to look at
their design of a cylinder that was supposed to hold the explosive during a 155mm Howitzer
launch. Assume the explosive sticks completely to the interior wall. The firing conditions at the
time of the failure were as follows
Axial Acceleration = 10,000 g’s
Angular acceleration = 300,000 rad/s2
Angular Velocity = 100 Hz
The projectile was as shown below
The wall is AISI 4140
= 2
6
in
lbf
1030E
29.0=
=3
in
lbm
283.0
The explosive is Composition B
=3
cm
g
71.0
fill
Write the Stress tensor for a point on
the inside diameter, 4” from the base
1”
6.092”
1”
10”
4”
1”
6.092”
1”
10”
4”
zz
( ) ( ) ( )
( ) ( )
( )
22
2
22
22
2222
V
io
disk
io
io
ioio
axial
zz rr
tar
rr
larr
rr
a
rr
am
A
F
−
+
−
−
=
−
=
−
==
( )
22
2
io
disk
zz rr
tar
la −
+=
( ) ( )
( ) ( )
( )
( )
−
−
+
−
=lbm
lbf
000,10
2
092.4
2
092.6
in1in
2
2092.6
in
lbm
283.0
lbm
lbf
000,10in410
in
lbm
283.0 22
2
2
3
3
zz
−= 2
in
lbf
307,19
zz
I have inserted the negative sign here because the loading is compressive. We have also
ignored the effect of the explosive adhering to the wall and assume that this mass will be
So the equivalent internal pressure is given by
in
in
eq
The stress on the internal wall is then
in
rr
Again we used the negative sign due to compression. Since there is no external pressure we
used the above. In the hoop direction we have to use
( )( )
222 inlbm58.152inlbm45.11inlbm55.609 −+−+−=
zz
I
2
inlbm16.652 −=
zz
I
Then it follows that the torque required to spin up the entire projectile is
( )
( )
−
−
−= in
ft
12
1
ftlbm
slbf
2.32
1
s
rad
000,300inlbm16.652
2
2
2
T
inlbf335,506 −=T
This is the torque acting at the rotating band location. The torque at the location of interest
uses the same method of calculation except that we only use the moment of inertia of all the
material (in this case) forward of the location of interest. This is given for the fill by:
2
rmI zz
=
l
d
rl
d
Ifillzz fill
=
=
164
4
2
2
( ) ( )
( )
2
4
4
3inlbm55.9in510
16
in092.4
in
lbm
026.0 −=−
=
fill
zz
I
For the cylinder we have
22
iozz rmrmI
−
=
( ) ( )
16
44
44 io
iozz
dd
lrrlI cyl
−
=−
=
( ) ( )
( ) ( )
( )
2
4
44
3inlbm73.365
16
in092.4092.6
in6
in
lbm
283.0 −=
−
=
cyl
zz
I
If we sum all of this up we have
( )( )
222 inlbm58.151inlbm55.9inlbm73.365 −+−+−=
zz
I
2
inlbm85.390 −=
zz
I
inlbf890,8 −=
fill
T
lbf355,27=
base
h
F
“smearing” this we get
( )
( )
−== 22 in
lbf
266
in84.102
lbf355,27
rz
We can now write our stress tensor
−−−
−
−−
=
=2
in
lbf
307,19766,5266
766,5564,543
26643265,2
zzzrz
zr
rzrrr
σ
Problem 6 – A 155 mm projectile is fired from a tube with a 1 in 20 twist. It’s muzzle velocity is
1000 m/s. What is the spin rate at the muzzle in Hz?
( )
( )
s
s
m155.0
calibers20
Problem 7 – It is requested that a brass slip ring be constructed for a spin test fixture to allow
electrical signals to be passed (although real noisy) to some instrumentation. The design
requirements are for the ring to have an I.D. of 4 inches, a length of 2 inches and be capable of
supporting itself during a 150 Hz. spin test. How thick does the ring have to be? The
properties of brass are as follows: Yield strength of 15,000 psi and density of approximately
0.32 lbm/in3.
Problem 8 – A 155mm projectile is to be designed with a rotating band designed to discard as
the projectile leaves the muzzle of the weapon. The rotating band is fixed to the projectile (so
it transmits the proper torque to the projectile) with splines that prevent rotational motion
relative to the projectile while allowing the band to expand in the radial direction for proper
discard. This function must occur at the highest as well as the lowest spin rates. The two
extreme muzzle velocities are 250 and 800 m/s, respectively, with corresponding peak axial
accelerations of 2,000 and 15,500 g’s, respectively. The projectile mass is 98 lbm and the axial
moment of inertia is 41 lbm-in2. Geometry constraints require the band to have an engraved
outside diameter of 6.2 inches and the outer diameter of the band seat (band inside diameter)
is 5.5 inches. The band is 2.5 inches long. The 48 rifling lands are 0.05 inches high and are 0.2
inches wide. Consider both a copper and a soft iron band with the properties provided below,
and determine whether the bands will:
a.) Withstand the shear at peak angular acceleration
b.) Break up upon muzzle exit.
c.) If one (or both) designs fail to work properly what can be done with the analysis
and/or design to make it work? Is there anything we must be careful of?
a
nd
2
=
( ) ( )
( )
( )
=
=22
1s
rad
766,39
s
m
81.9g000,2
m155.0
rev
cal
20
rev
rad
2
For the high zone firing we have
( ) ( )
( )
( )
=
=22
2s
rad
190,308
s
m
81.9g500,15
m155.0
rev
cal
20
rev
rad
2
Now we can determine the torque applied to the rotating band at each zone using equation
(PD-36)
zz
IT =
(PD-36)
( )
( )
−
−
−= 2
2
2
1s
rad
766,39
in
ft
12
1
ftlbm
slbf
2.32
1
inlbm41T
inlbf220,4
1−=T
( )
( )
lbf383,1
in3
inlbf220,4
1
1=
−
== R
T
F
( )
( )
lbf722,10
in3
inlbf701,32
2
2=
−
== R
T
F
4
V
22
io dd
lAlmeff −
===
( ) ( )
( )
2
22
5.51.6
lbm
2