Example 1
Check body structural strength for the following proposed EX rocket design:
Body material 6061-T6 aluminum alloy, 4.0 inch OD x 0.035 inch wall thickness
Nosecone shape = tangent ogive
Design Stability Margin = 2
Maximum expected velocity = mach 1 @ 2000 feet AGL (at motor burn out)
Maximum expected acceleration = 30 G’s
Drag coefficient, plotted in Figure 1 (ref. AeroLab)
Design angle-of-attack = 10 degrees
Liftoff weight = 30 pounds
Motor length = 16 in.
Launch site elevation 700 feet ASL
Figure 1 – Rocket drag coefficient as function of mach number
The first step is to calculate the maximum drag force acting on the rocket at maximum velocity.
Air density (r) at 2000 ft AGL corresponds to 2700 feet ASL. From this Table, using linear interpolation, the value of r is:
r = (2700 – 0)/(5000 – 0) × (20.48 – 23.77) + 23.77 = 22.0 slug(104)/ft3 = 0.00220 slug/ft3
The reference area, A, of the rocket is
A = ¼ π (4.0/12)2 = 0.0873 ft2
From this Table, using linear interpolation, mach 1 at 2700 feet ASL corresponds to a speed of:
V = (2700 - 2461)/(3281 - 2461) × (1103 – 1107) + 1107 = 1106 ft/sec.
Although the target maximum velocity of the rocket is mach 1, a glance at the drag coefficient versus mach number curve (Fig.1) shows a very steep rise of slope as mach 1 is crossed. If the rocket exceeds its target performance, it may approach mach 1.2 or 1.3, where the highest drag occurs. As such, to be conservative, use the highest drag coefficient:
CD = 0.67
This gives a total drag force on the rocket of:
The design (worst-case) angle-of-attack is 10 degrees:
a = 10/180 × π = 0.175 radians
The next step is to calculate the normal forces acting on the nosecone and the fins, given by:
NNOSE = q A a (CNa)N
NFINS = q A a (CNa)F
The dynamic pressure (q) acting on the rocket is:
q = ½ (0.00220) (1106)2 = 1346 lbf/ft2
The slope of the normal force coefficient with respect to angle of attack, at α = 0, for the nosecone and fins is obtained using the Barrowman method. For any nosecone shape:
(CNa)N = 2 per radian
For the fins, the Barrowman equation is:
where N = number of fins, and d = base diameter of the nosecone. The other parameters are defined by the following figure:
The mid-chord sweep angle theta is given by:
and mid-chord length, LF, by:
For the example rocket, the values of these parameters are:
N = 4
d = 4.0 in.
R = 2.0 in.
S = 4.5 in.
XR = 3.5 in.
CT = 3.0 in.
CR = 8.0 in .
This gives
LF = 4.5/cos(12.5) = 4.61 in.
The slope of the normal force coefficient for the fins may now be calculated.
As such, the normal forces acting on the nosecone and the fins are:
NNOSE = 1346 (0.0873) 0.175 (2.0) = 41.1 lbf.
NFINS = 1346 (0.0873) 0.175 (11.48) = 236 lbf.
Take a minute to check to see if units are consistent
The next step is to calculate the location of the nosecone centre of pressure (XN) and the fins centre of pressure (XF), illustrated below. This is where the normal force resultants acts. Note the ‘global’ coordinate system, denoted by ‘large X’, has its origin at the nosecone tip.
These are obtained using the Barrowman method. For a nosecone, the location of the CP is dependant upon the nosecone shape and the nosecone length. For the example rocket, which has a tangent ogive nosecone of length 8.0 inches:
XN = 0.466 (8.0) = 3.73 in.
For the fin set, the location of the CP is given by
where XB is the distance from the nosecone tip to the start of the finset.
XB = 76 in.
Plugging in the values gives:
The lateral acting inertia loads, w1 and w2, as shown in the figure below, are now calculated.
Note the introduction of a ‘local’ coordinate system, ‘small x’, with its origin at the nosecone CP. This coordinate system is for convenience with regard to calculating the structural loads on the rocket body.
The location of the CG of the rocket is determined from the design stability margin and location of the rocket’s CP. As this rocket will reach supersonic velocity, compressibility effects will affect the CP position that the Barrowman method neglects. Figure 2 (ref. AeroLab) shows the CP variation with mach number.
Figure 2 – CP location as a function of mach number
As seen in Figure 2, the Barrowman CP (denoted by ×) is further forward than the CP shown by the red line. Therefore, to be conservative, the Barrowman CP will be used in the analysis. The CG location is now calculated based on the design Stability Margin.
XCG = 67.79 – 2 (4.0) = 59.79 in.
From this, distances x1 and x2 may be calculated.
x1 = 59.79 – 3.73 = 56.1 in.
x2 = 79.0 – 59.79= 19.21 in.
The values of distributed inertia loads w1 and w2 are next calculated.
The value of w2 is calculated first, then w1.
The lateral shear loading, as a function of x, is given by
Inserting the values for the first range of x:
lbf
The shear force at x1 is needed in order to calculate V(x) over the 2nd range of x.
For the range x1 <x ≤ L, where the length L represents the body length from nose CP to the fin CP (i.e. L = x1 + x2), or L = 56.1 + 19.21 = 75.31 in. This gives the shear force over this range of
lbf
The maximum bending moment occurs where V = 0. Since this location may be anywhere along x, both equations for V(x) are solved in terms of x.
applicable to range of x = 0 to 56.1
As this value of x is outside the range of x = 0 to 56.1, then the true location of V= 0 must lie in the range x = 56.1 to 75.31 in., and therefore
The maximum bending moment is therefore given by:
Plugging in the values gives
= 2013 lbf-in.
As well as maximum bending moment, it will be useful to have a shear-bending moment diagram in case joints are introduced into the design. Moment as a function of x is given by:
The M(x) and V(x) curves are most conveniently plotted using a spreadsheet app such as Excel.
The largest shear force magnitude, Vmax = 236 lbf, is at the location of the fins CP (XF = 79.0). The maximum shear stress is calculated at this location.
where As is the tube cross-sectional area.
giving
The maximum bending moment (2013 lbf-in.) is at x = 58.2 inches. In terms of the global coordinate system, the maximum bending moment is at X = 58.2 + 3.73 = 61.9 in.
The locations of the maximum shear and bending moment on the example rocket are illustrated below.
Now that the maximum axial load and bending load on the rocket have been determined, the combined axial and bending stress is calculated. Note that axial stress due to mass inertia will be neglected as the maximum expected acceleration is relatively low (30 G’s).
The compressive stress on the rocket body tube due to aerodynamic drag , fca, is given by
where FD = 79 lbf, determined earlier, and body tube inside diameter, d = 4.0 – 2 (0.035) = 3.93 in.
Giving a maximum compressive stress of
The maximum bending stress on the rocket body is given by
where Mmax = 2013 lbf-in. and Z is the section modulus of the body tube, given by
giving
Therefore the combined compressive stress due to axial and bending components is
In order to determine if the rocket body tube will fail by compressive instability (buckling), the D/t ratio is calculated, where t is the wall thickness of the body tube.
D/t = 4.0/0.035 = 114
As the D/t ratio is greater than 70, the critical bending stress is obtained from the graph of compressive instability curves for aluminum. In order to use the graph, the ratios r/t and L/r are needed:
r/t = (4.0/2)/0.035 = 57
At this point in the design, the location of supports within the body tube (such as couplers) is unknown other than the location of the motor thrust bulkhead. Conservatively the length of the rocket body between the nosecone and the motor bulkhead will be considered to be the unsupported length L.
L = 85 – 8 – 16 = 61 inches.
L/r = 61/(4.0/2) = 30.5
The graph below is used to determine the critical buckling stress (Fbcr).
Fbcr = 23,500 psi
This results in a compression Safety Factor of
S.F. = 23,500/4884 = 4.81
The maximum applied shear stress, calculated earlier, is compared to the shear strength of the body tube material.
The allowable shear stress for 6061-T6 aluminum alloy is taken from Table 3.6.2.0(c1) of MMPDS.
Su = 27 ksi, or 27,000 psi.
This results in a shear Safety Factor of
S.F. = 27,000/1083 = 24.9
Conclusion
The Safety Margins calculated for both body tube compression and shear are healthy (greater than 2) and as such, the proposed rocket body design is sound, based on conservatively assumed loading.
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