 # Structural Analysis Using Simple Beam Theory

It is relatively straightforward to use simple Euler-Bernoulli beam and torsion theory to analyze a spar that has uniform properties along its length. Since we almost always use uniform section CFRP tubes, this represents the best place to begin the preliminary structural analysis. The key results that are needed are for point and uniform forces and moments applied to cantilever beams with built-in (encastre) ends. These can be obtained from any standard undergraduate text: the key results are summarized in Table 14.1 for convenience. We then use the principle of superposition to sum up the effects of loads along the length of any given spar. If a simple end-point force and moment pair or a completely uniform load is to be applied, this is very straightforward. For example, if we want to place a 4g maneuver load on a main spar of 3 m length (wing tip to wing tip), which we assume is clamped into the fuselage at the center plane and ignoring the mass of the wings, then our UDL for a 15-kg aircraft is just w = 4 x 15 X 9.81/3 = 196.2 N/m. Table 14.1 then shows that for this case the maximum

Table 14.1 Shear forces (Q), bending moments (M), slopes (в, in radians), and deflections (<5) for Euler-Bernoulli analysis of uniform encastre cantilever beams.

 Model Distribution Max. value Concentrated tip force P Q(x) = P Qmax = P M(x) = P(x — L) Mroot = PL d(x) = P(2Lx-x2) v 7 2EI в = PL2 °lip 2EI S(x) = Px2(3L-x) V 7 6EI S = PL3 °tip 3EI Concentrated tip moment q Q(x) = о Qmax = 0 M(x) = q Mmax = Е 0(x) = Ex II № S(x) = Ex2 2EI S. = qL2 tlp 2EI Uniformly distributed load w Q(x) = w(L — x) Qroot = wL M(x) = w(L2—2Lx+x2) M = — Mroot 2 wx( 3L2—3Lx+x2) 0(x) = 6EI wL3 etip = 6Ё7 wx2 (6L2 —4Lx+x2) (x) = 24EI S wL4 Stip = "8E7

Note that for a hollow cylinder of outer diameter d and wall thickness t, the second moment of area I = Kid4 — (d — 2t)4)/64, while for a rectangular cross-section of breadth b, height h, and wall thickness t, the second moment of area I = bh3/12 — (b — 2t)(h — 2t)3/12.

bending moment occurring at the root (mid-span) of the spar is 196.2 x 1.52/2 = 220.7 N m. If the spar is made of CFRP with a Young’s modulus of 70 GPa and ultimate tensile strength of 570 MPa (see Table 18.1) and inner and outer diameters of 25 and 20 mm, respectively, M/I =

^bending/У tells us that the peak bending stress wih be ^bending max = ж((2002574-0з0224)/б4 = 243.7

MPa, that is, 43% of the ultimate tensile strength of the material. The equivalent tip deflection is &ip = 4xi5x9.8i/3xi.5 4 = 0.1567 m, an acceptable fraction of the total wing span.

If the lift force was not acting directly through the spar but some distance in front or behind it, there would be an additional torque load to be dealt with. If, for example, the lift was acting 20 mm behind the spar centerline, the total torque would be T = 0.02 x 4 x 15 x 9.81/2 = 5.886 N m. The maximum stress caused by such a torque rmax = J = ^d'—^—= 13 MPa, a negligible extra amount.

For a nonuniform loading, some form of discretization is needed, but a simple hand calculation is still possible. If the load distribution is to be taken from XFLR5, the code conveniently gives section bending moments along each lifting surface that can be applied directly to the spars, station by station along the spans. For a uniform spar, one simply takes the highest bending moment tabulated by XFLR5 for a given spar and again uses M/I = o/y to get the relevant maximum spar stresses. If the tip deflection is also required, one must sum up the deflections section by section along the spar from the root, taking care to allow for the cumulative effect of the slope changes caused by each local section bending moment. To make this clear, consider again the previous case but now assume each spar half is analyzed in just two parts and where the local section lift and drag coefficients as given by XFLR5 (lift coefficient = Cl, viscous drag coefficient = PCd, and induced drag coefficient = ICd) are outer half 0.176423,

0.007079,0.003350, and inner half 0.303362,0.007320,0.000282, and the mean local section chords are 0.2641 and 0.3727, respectively. We take the velocity to be 30m/s and p to be 1.211 kg/m2. We first convert the coefficients to an equivalent uniformly distributed force by noting that the drag and lift are perpendicular to each other and also allow for our maneuver load factor of 4 so that the force per unit length are w = 4 x 0.5pV2c/Cl2 + (PCd + ICd)2 = 101.7 and 246.5 N/m, respectively. (Here, c is the local mean section chord and we ignore the slightly different angles these resultant forces have with respect to each other and the airframe.) The deflection and slope at the root are of course zero. At the junction between the two parts, the deflection is due to a combination of the UDL on the inner half of the wing and the force and moment applied at the inner half’s tip by the outer half. These quantities are w = 246.5 N/m, P = 0.75 x 101.7 = 76.3 N, and q = 0.5 x 101.7 x 0.75 x 0.375 = 28.6 N m. Their effects on the inner half of the wing may be deduced from Table 13.1 as a deflection

at the junction of function = 24x70x109x^(0.0254 -0.024)/64 = °.°36 m and a slope at the

junction of <9junction = 0-75(246-5x09752+3x7643x0-754+6x28-6) = 0.076 rad. The tip deflection at the end of the outer part is then that for the outer half plus that for the inner half plus that caused by the slope there times the length of the outer half, that is, (tip = шл^л5 4 +

0.036 + 0.0076 x 0.75 = 0.0981 m, which is somewhat less than predicted from a fully uniform loading, as would be expected. Clearly, such an approach can be extended to multiple subdivisions with the use of a simple spreadsheet.

Following this logic, Figure 14.4 shows how deflection and slope vary for the Decode-1 main spar when flying at 30 m/s and an angle of attack of 2.53° using loading results from XFLR5, again with a load factor of 4. For this 15-kg aircraft, the spar has a total length Figure 14.4 Deflection and slope variations for the Decode-1 main spar when flying at 30m/s and an angle of attack of 2.53° using loading taken from XFLR5, a load factor of 4, and simple beam theory analysis. The spar is assumed to be made from a circular CFRP section of outer diameter 20 mm, wall thickness 2mm, Young’s modulus of 70 GPa, and extending the full span of the aircraft, being clamped on the center plane.

of 2.94 m, an outer diameter of 20 mm and a wall thickness of 2 mm. The resulting tip deflection using simple beam theory is predicted to be 0.292 m. XFLR5 reports a maximum bending moment for this condition of 48.77 N m, which when combined with a load factor of 4 leads to a peak stress in the spar of 420.7 MPa, which is 74% of the yield stress of the CFRP material, and an acceptable margin, bearing in mind that the spar is actually supported by the fuselage over a significant length rather than just at the center plane; this will reduce the stresses and deflections significantly as we will see shortly. To gain further insights, it is simpler to make use of one of the many standard FEA packages for such analyses, an approach we consider next. 