2πal½r(r4o−r4i) +rFILLr4i 2. T= 796,600[lbf-in.]
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 155 mm howitzer launch. Assume the explosive sticks completely to the interior wall. Thefiring conditions at the time of the failure were as follows:
1 in.
6.092 in.
1 in.
4 in.
10 in.
• Axial acceleration = 10,000g
• Angular acceleration = 300,000 rad/s2
• Angular velocity = 100 Hz The projectile was as shown earlier:
The wall is AISI 4140:
E= 30106 lbf in:2
n= 029, r= 0:283 lbf
in:3
The explosive is composition B:
rfill= 0:71 g cm3
Write the stress tensor for a point on the ID, 4 in. from the base.
Answer:
s =
srr trq trz trq sqq tqz
trz tqz szz 2
66 4
3 77 5=
−2265 21 −266 21 5564 −2864
−266 −2864−19,307 2
66 4
3 77 5 lbf
in:2
Problem 6
A 155 mm projectile isfired from a tube with a 1 in 20 twist. Its muzzle velocity is 1000 m/s.
What is the spin rate at the muzzle in hertz?
Answer: 322.6 [Hz]
Problem 7
It is requested that a brass slip ring be constructed for a spin testfixture to allow electrical signals to be passed (although really noisy) to some instrumentation. The design require- ments are for the ring to have an ID of 4 in. and a length of 2 in. 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/in.3
Answer: 1/4 in. thickness will work, but it can be thinner.
Problem 8
A 155 mm projectile is to be designed with a rotating band designed to discard as the projectile leaves the muzzle of the weapon. The rotating band isfixed 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, respectively. The projectile mass is 98 lbm and the axial moment of inertia is 41 lbm−in.2 Geometry constraints require the band to have an engraved OD of 6.2 in., and the outer diameter of the band seat (band ID) is 5.5 in. The band is 2.5 in. long. The 48 rifling lands are 0.05 in. high and are 0.2 in. wide. Consider both a copper and a soft iron band with the properties provided next and determine whether the bands will.
1. Withstand the shear at peak angular acceleration 2. Break up upon muzzle exit
3. 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?
Assume the following: The weapon has a 1 in 20 twist. At peak acceleration, only shear on the rotating band need be considered. Ignore the increase in shear area caused by the rifling helix. Ignore the stress concentration developed by the engraving for discard calculations.
Ignore the effect of the splines. For conservatism, on muzzle exit, ignore the mass of material above the rifling marks (i.e., use an OD of 6.1 in. for the band).
The properties for the copper and steel are as follows:
Copper Soft Iron Ultimate tensile strength (psi) 35,000 40,000
Shear strength (psi) 19,250 22,000
Density (lbm/in.3) 0.316 0.263
Poisson’s ratio 0.28 0.34
Problem 9
An experimental 40 mm gun has an average chamber inner diameter of 60 mm. The weapon is expected to develop a maximum breech pressure of 35,000 psi. If we would like the weapon to withstand 10,000 cycles at this pressure and given the properties of the steel given later, determine the OD of the chamber. Without proper design experience, an interferencefit can sometimes be catastrophic. If we were instead to design this chamber out of two tubes, each at half of this thickness but with the outer tube compressing the inner tube by 0.002 in. diametrally, what is the maximum pressure the design will accommodate and still function for the 10,000 cycles?
Assume AISI 4340 steel with a yield strength (SY) of 100,000 psi. The endurance stress (S0n) for 4340 is 0.875SYfor the amount of cycles desired. Assume the following factors from our cyclic loading discussion: CR = 0.93, CG = 0.95, and CS = 0.99. Assume the chamber is open ended. The modulus of elasticity and Poisson’s ratio are 30 × 106 psi and 0.3, respectively.
Problem 10
You are to design a fragment-throwing gun system for another organization to be used in fragment impact testing. The gun is to throw a 22 g, 0.500 in. diameter, cylindrical fragment at 2300 m/s. 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 its 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 × 106psi. 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 Rateb(in./s/psi)
Solid Density d(lbm/in.3)
Adiabatic Flame TemperatureT0(°R)
Propellant Force l(ft-lbf/lbm)
Specific Heat Ratiog
IMR 0.000132 0.0602 5103 327,000 1.2413
M12 0.000137 0.0600 5393 362,000 1.2326
Bullseye 0.000316 0.0590 6804 425,000 1.2523
Red dot 0.000153 0.0593 5774 375,000 1.2400
Navy pyro 0.000135 0.0566 4477 321,000 1.2454
6. Assume the cartridge case is brass and use a bilinear kinematic hardening model where the brass has a modulus of elasticity of 15 × 106psi, a local tangent modulus of 13 × 106psi, and a yield stress of 16,000 psi (yield occurs in this material ate= 0.002).
7. Weight is not a major concern; however, you should make the design light enough to be moved using reasonable test range equipment.
The design is to proceed as follows (not necessarily in the order given):
1. Interior ballistics design
a. Size the chamber length and diameter.
b. Determine the amount of propellant needed based on your choice of the aforementioned propellants and propellant geometry (make sure itfits in the chamber).
c. Determine a web thickness for the propellant.
d. Determine the length of the gun.
e. DetermineV,pB, andxfor the projectile at peak pressure.
f. DetermineVc,pBc, andxcfor the projectile at charge burnout.
g. Determine the muzzle velocity of the projectile.
2. Gun tube design
a. Based on the calculations of part 1 develop a pressure–distance curve to use as criteria for your gun design.
b. Determine the OD of the gun tube. To keep the design light as possible, use the design rules provided in this chapter and taper the tube toward the muzzle. If needed, over the chamber, you may shrinkfit cylinders to build up a composite tube.
c. Determine the weight of your gun and comment on if it is reasonable.
3. Cartridge case design
a. Determine a thickness and tolerance for your cartridge case.
b. Determine the OD and tolerance for the cartridge case.
c. Decide on a tolerance for your chamber ID.
Note that for these calculations show that the case may be easily extracted at the limits of the tolerance.
It is important that you write down all 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.
4.11 Buttress Thread Design
There is a variety of instances where a buttress thread form is the desired means of transmitting loads between mating components. In some instances, the thread form is not the usual continuous spiral associated with a normal thread but a series of discontinuous
grooves that exhibit the cross-sectional form of the buttress. In this section, we will discuss a true thread with lead-ins and partial thread shapes, but we will assume that the basic analysis will apply to buttress grooves as well.
Buttress threads are designed to maximize the load-carrying capability in one direction of a threaded joint. There are many variations on such threads, but on ammunition compo- nents, we predominantly use threads with a pressureflank angle (described later) of 7° as shown in Figure 4.18. Thread callouts on drawings usually appear, e.g.,
2. 750-4UNC-2A LH Buttress The meanings of these callouts are as follows:
• First number is the major diameter of the thread (here it is in inches).
• Second number is how many threads per inch.
• The letters are the thread form callout (UNC = Unified National Coarse).
• The last number is the class of fit of the thread related to clearances in the engagement (3 is the tightestfit, 1 the loosest).
• The last letter determines whether the thread is male (A) or female (B) (mnemonic− A = Adam = male).
• LH means left handed (there will be no callout if the threads are right-hand twist or if the thread is a groove and not a continuous spiral).
Thread nomenclature of relevance is as follows:
• The major diameter is the largest diameter of the thread form.
• The minor diameter is the smallest diameter of the thread form.
• The pitch diameter is the diameter where there is 1/2 metal and 1/2 air.
We use buttress threads for several reasons: most important is to improve the directional loading characteristics of the thread; also to allow for a more repeatable, controllable shear during an expulsion event, i.e., if we want the thread to intentionally and controllably fail allowing separation of the components; and to prevent thread slip in joints with fine threads or threads on thin shell walls. If thread slip occurs, the threads can either dilate or
Pitch of the thread
Pressure flank 45°
7°
Load carrying or shearing
FIGURE 4.18
Depiction of a standard buttress thread.
contract elastically and the joint can pop apart with little or no apparent damage to the threads.
When we design for strength, we typically calculate the strength based on the shear area at the pitch diameter in the weaker material. This, of course, translates to half the length of engagement of the threads. This is acceptable because we usually use conservative prop- erties and add a safety factor to account for material variations and tolerances. We must always base our calculations on the weaker material if the design is to be robust. When designing to actually fail the threads, however, we need to be more exact in our analysis and take everything such as actual material property variation and tolerancing into account or our answers will be wrong.
We will proceed in this analysis in meticulous detail, initially, as a cantilevered beam subjected to compressive and tensile stresses caused by contact forces and bending moments. This technique was first developed during the US Army’s sense and destroy armor program by Dan Pangburn of Aerojet Corporation [5] and has been used by the US Army.
We consider the thread form as a short, tapered, cantilever beam and assume that failure will occur as a result of a combination of stresses and that combined bending and com- pressive stress precipitate the failure. This is depicted in Figure 4.19. If we examine this figure, we see that the distributed forceFcauses our beam to bend in the classical sense with the loadedflank in tension and the unloadedflank in compression about the neutral axis.
We have separated an element of material out from point A in thefigure. The free-body diagram of this element shows that the bending of the beam puts it in tension, while the loading on the pressureflank puts it in compression. It is this combined load that will cause failure of the material.
If we were analyzing this in afinite element code, the bending and compression would cause combined stresses and the part would fail by one of the failure criteria that were discussed earlier. However, in this case, we will use the maximum shear criteria to check for failure at some radius in the thread and will check the load at which failure occurs with the von Mises criterion at the thread roots,dion the male thread anddoon the female thread.
These are the diameters of the loading (i.e., the mating thread contact areas) as depicted in Figure 4.20.
For simplicity, we shall call the male thread the“bolt”(subscript 1) and the female thread the“nut”(subscript 2). The loading is further described by Figure 4.21. In thisfigure, the radiusris the plane at which the threads will shear.
r
di F
Element at point A
Tensile stress from Compressive stress
from F
I σ=Mc
I σ=Mc
Neutral axis
FIGURE 4.19
Depiction of a standard buttress thread.
If we assume the contact is frictionless, the average normal stress is simply the total axial forceFdivided by the projected areaA. We have assumed that the normal stress is constant over the contact area. This gives us a negative value because the stress is compressive.
Figure 4.22 shows the configuration where the normal force has been termed F4and the thread area isA4. Since an axial loading is what shears the threads, we need to project the components of this force along the axis of the projectile (i.e., rotate through the anglef1).
This allows us to express the stress as
sN=−F4 A4 =
− F cosf1
A cosf1
=−F
A=sv (4.112)
r
di
F do
Bolt (1)
The location where these stresses are the greatest is here along the contact
surface FIGURE 4.20
Definition of load radii.
r do
di
t2
t1
F2
F1 Nut (2)
Bolt (1)
di= Inner diameter
1
r= Shear radius do= Outer diameter
2
FIGURE 4.21
Loading diagram of buttress threads.
WheresNandsvare the normal and axial stresses, respectively. By substituting the areaA, we get
sv=− F π
4 d2o−d2i (4.113)
If we assume that failure takes place at a radiusr, yet to be determined, the bearing force on the external thread (bolt) that produces bending in the thread is
F d
1 r
2
= − ⎛ 2
⎝⎜ ⎞
⎠⎟ −
⎡
⎣⎢ ⎤
⎦⎥ πσv o
2 (4.114)
Similarly, the force that produces bending in the internal thread (nut) is
F r d
2 2 2
= − − ⎛
⎝⎜ ⎞
⎠⎟
⎡
⎣⎢ ⎤
⎦⎥
πσv i
2 (4.115)
Now the pitch diameter is defined as the location where the thickness of the thread is one- half the thread pitch. Since thread failure occurs at an assumed radiusr, we need to define the thicknesses of both the male and female threads at this location.
First, recall that the thread pitch is pand then definedpfas the internal (female) thread pitch diameter anddpmas the external (male) thread pitch diameter. Then,t1andt2from our earlier diagram can be expressed as follows:
t p
r d
1= −2 ⎛ − 1 2
⎝⎜ ⎞
⎠⎟
(
+)
pm
2 tanφ tanφ (4.116)
t p d
2 f 1 2
2 2
= −⎛ −
⎝⎜ ⎞
⎠⎟ +
pf (tanφ tanφ ) (4.117)
Then the bending stress can be calculated from simple beam theory as s =Mc
I = Mt 1 2 12ð2πrÞt3
= 3M
πrt2 (4.118)
A4
F
F4
1
A FIGURE 4.22
Loading of a thread surface.
wherecis the distance from the point of interestrto the neutral (bending) axis andIis the area moment of inertia of the cross-section. The bending stress in the external (male) thread is then
σ1 π
1 12
=3 ⎛ −
⎝⎜ ⎞
⎠⎟ F d
r rt
o
2 2
(4.119) Similarly, we can show that the bending stress in the internal (female) thread is
σ2 π
2 22
=3
⎛ −
⎝⎜ ⎞
⎠⎟ F r d
rt
i
2 2
(4.120)
In considering the failure criteria, we shall assume that the maximum shear stress in the material must not exceed 0.6 times the material strength in a tensile test. We will use the yield strength as the strength of this material because at that point in failure, the geometry of the part is changing. Experience has shown that once this begins to happen, the part is in the process of failing anyway and will not recover.
In a state of combined loading, the maximum shear stress can be found from tmax=1
2jsmax+sminj (4.121)
This averaging can be shown to be
tmax=s−sN
2 = 0:6Y (4.122)
Here we are reminded thatsNandsvare therefore compressive negative numbers andY is the yield stress in tension. The equivalent stress at failure in the male thread is then
Y1 =s1−sv
1:2 (4.123)
and in the female thread, it is
Y2 =s2−sv
1:2 (4.124)
In these equations, Y1 and Y2 are the yield stress in the male and female threads, respectively.
We will now combine Equations 4.123 and 4.119 as well as Equations 4.124 and 4.120 to eliminates1ands2, respectively. This yields
Y1= 1:25F1 1 2do−r
πrt21 − sv
1:2 (4.125)
and
Y2= 1:25F2 r−1
2di πrt22 − sv
1:2 (4.126)
We now combine Equations 4.125 and 4.116 as well as Equations 4.126 and 4.117 to eliminate the thicknessest1andt2, respectively. This yields
Y F
d r
r p r d
1 1
1 2
1 25 2
1 2 1
2
1 2
= −
−⎛ −
⎝⎜ ⎞
⎠⎟
(
+)
⎡
⎣⎢
⎤
⎦⎥
. o −
pm
v
tan tan 1
π φ φ
σ
..2 (4.127)
and
Y F
r d
r p d r
2 2
1 2
1 25 2
1 2 1
2 1 2
= −
−⎛ −
⎝⎜ ⎞
⎠⎟
(
+)
⎡
⎣⎢
⎤
⎦⎥
. −
tan tan
i
pf
v
π φ φ 1
σ
..2 (4.128)
We will now insert Equation 4.114 into Equation 4.127 and Equation 4.115 into Equation 4.128 to eliminateF1andF2, respectively. This yields
Y d r
d r
r p r d
1 2 2
1 2
0 3125 4
1 2 1
2
1 2
= −
(
−)
−−⎛ −
⎝⎜ ⎞
⎠⎟
(
+)
. σ
φ φ
v o
o
pm tan tan
⎡⎡
⎣⎢
⎤
⎦⎥
2− σv
1.2 (4.129)
and
Y r d
r d
r p d r
2 2 2
1 2
0 3125 4
1 2 1
2 1 2
= −
(
−)
−−⎛ −
⎝⎜ ⎞
⎠⎟
(
+)
. σ
φ φ
v i
i
pf tan tan
⎡⎡
⎣⎢
⎤
⎦⎥
2− σv
1.2 (4.130)
Now we must solve Equations 4.129 and 4.130 in terms ofsv. Thefirst of these is sv= −Y1
G3+G2+G1+G0+ 1 1:2
(4.131)
where
G3 = 0:15625d3o
r0:5p−rtanf1−rtanf2+ 0:5dpmtanf1+ 0:5dpmtanf22 (4.132)
G2= −0:3125d2o
0:5p−rtanf1−rtanf2+ 0:5dpmtanf1+ 0:5dpmtanf2
2 (4.133)
G1= −0:625rdo
0:5p−rtanf1−rtanf2+ 0:5dpmtanf1+ 0:5dpmtanf2
2 (4.134)
G0= 1:25r2
0:5p−rtanf1−rtanf2+ 0:5dpmtanf1+ 0:5dpmtanf2
2 (4.135)
The second equation is
sv= −Y2
H3+H2+H1+H0+ 1 1:2
(4.136)
where
H3 = 0:15625d3i
r0:5p+rtanf1+rtanf2−0:5dpftanf1−0:5dpftanf22 (4.137)
H2= −0:3125d2i
0:5p+rtanf1+rtanf2−0:5dpftanf1−0:5dpftanf2
2 (4.138)
H1= −0:625rdi
0:5p+rtanf1+rtanf2−0:5dpftanf1−0:5dpftanf2
2 (4.139)
H0= 1:25r2
0:5p+rtanf1+rtanf2−0:5dpftanf1−0:5dpftanf2
2 (4.140)
We now solve Equation 4.113 forF, and we get F=π
4sv d2o−d2i
(4.141) Substitution of Equation 4.131 forsv yields (for a full thread on the bolt)
F=π
4 d2o−d2i −Y1
G3+G2+G1+G0+ 1 1:2
(4.142)
We perform a similar operation with Equation 4.136, giving us (for a full thread on the nut) F=π
4 d2o−d2i −Y2
H3+H2+H1+H0+ 1 1:2
(4.143)
Equations 4.142 and 4.143 now contain only two unknowns,randF. The procedure now involves solving both Equations 4.142 and 4.143 and plotting the forceFvs.r. The lowest value in either equation is then the force (and location) at which the joint will fail. It is recommended that these solutions be performed with the aid of a computerized numerical calculation program such as MathCAD.
Partial threads can have a significant effect on the failure strength of a joint. If the joint were designed to survive, it is generally best to ignore the additional strength afforded by partial threads and base the design margin on the calculation method earlier. When a joint is designed to fail, however, the lead-in and run out must be accounted for unless sufficient margin is available in the expulsion system such that two additional threads may be added to the calculation, yet still allow the joint to be overcome with ease.
4.12 Sabot Design
Sabots (French for“wooden shoe”) are used in both rifled and smoothbore guns to allow a standard weapon tofire a high-density, streamlined subprojectile whose diameter is much smaller than the bore, at a velocity higher than would normally be possible if the gun were sized to the diameter of the subprojectile. Discarding sabots have been in general use since the Second World War and are still popular (in fact, an artist’s rendition of a discarding sabot is illustrated on the cover of this book). They are called “discarding sabots”since they are shed from the subprojectile at the muzzle allowing it tofly unencumbered to the target.
As previously stated, velocity is proportional to the square root of the pressure achieved in the tube, the area of the bore, and the length of travel and inversely proportional to the square root of the projectile weight. In mathematical terms,
V∽ ffiffiffiffiffiffiffiffiffi pAL wp s
(4.144) We can see that if the area over which the pressure is applied is much greater than the area presented at the rear of the subprojectile, a larger force would be applied to accelerate it than if it werefired at the same pressure from a bore of its own diameter. Furthermore, decreasing the launch weight of the as-fired assembly also increases the velocity. Therefore, we must design as light a sabot as feasible so that we can maintain a very dense, small diameter subprojectile (usually an armor penetrator). The combination of the full bore area; a dense, streamlined subprojectile; and a lightweight sabot has the overall effect of generating unusually high velocities, a characteristic essential for kinetic energy armor penetration.
There are many requirements for a successful sabot:
• It must seal the propellant gases behind the projectile (obturate).
• It must support the subprojectile during travel in the bore to provide stable motion (called providing a suitable wheelbase).
• It must transfer the pressure load from the propellant gases to the subprojectile.
• It must completely discard at the muzzle of the weapon without interfering with theflight of the subprojectile.
• The discarded sabot parts must also reliably fall within a danger area in front of the weapon so as not to injure troops nearby.
• It must be minimally parasitic, i.e., it must be as light as possible and remove as little energy from the subprojectile as possible.
These are formidable requirements that necessitate great ingenuity on the part of the designers.
The problem has been solved in a variety of ways. In the 1950s, designers, chiefly British, used cup- or pot-type sabots to launch APDS subprojectiles (Figure 4.23). The guns from which these munitions were fired were rifled to launch conventional full caliber, spin- stabilized rounds, and so the subprojectiles of the APDS rounds were spin-stabilized too.
Such armor-defeating munitions were highly effective against the tank armor of the times, and pot-type, saboted, kinetic energy penetrators were adopted in tank cannon around the world.