Aircraft Structures

In general, my intention was to give some indication of the role and limitations of each method of analysis. Part 11, 'Analysis of Aircraft Structures', Chapters 7-1 1 , contains an analysis of the thin-walled cellular structure typical of aircraft.

I. units are used throughout

In addition, Chapter 7 includes a discussion of structural materials, fabrication and function of structural components, and an introduction to structural idealization. An introduction to computational methods for structural analysis is presented in Chapter 11, which also includes some elementary work with the relatively modern finite element method for continuum structures.

Preface to Second Edition

Only minor changes have been made to Chapter 11, while Chapter 12 now includes a detailed study of fatigue, fatigue strength of components, fatigue life prediction and crack propagation. Finally, Chapter 13 now contains a much more detailed study of flutter and the determination of critical flutter speed.

Preface to Third Edition

Elasticity

Basic elasticity

In general, the direction of SP is not perpendicular to the area SA, in which case it is common to split SP into two components: one, SP, perpendicular to the plane, and the other SPs, acting in the plane itself (see Figure 1). 1.2).

Fig. 1.1 Internal force at a point in an arbitrarily shaped body.

Boundary conditions 9
Determination of stresses on inclined planes
Prir icipal stresses
Mohr’s circle of stress 13
Mohr's circle of stress 15

Also determine the value of the second principal stress and the maximum value of the shear stress at the point. The change in length of element OA is (O'A' - OA), so the direct strain at 0 in the x direction is given by Eq.

Fig. 1.4 Components of stress at a point in a body.

Determination of strains on inclined planes 21

Suppose that the arm a of the rosette is inclined at some unknown angle 8 to the maximum principal strain as in Fig. .

Fig. 1.14 Mohr's circle of strain for Example 1.3.

Problems 33

Derive, using the geometry of Mohr's stress circle, expressions for the maximum values of direct stress that can be applied to the x and y planes in terms of the three parameters given above. If the diameter of the shaft is 150 mm, calculate the values of the principal stresses and their directions at a point on the surface of the shaft.

Problems 35

Two-dimensional problems 37 -_
Two-dimensional problems
Inverse and semi-inverse methods 39 The English mathematician Airy proposed a stress function 4 defined by the
Bending of an end-loaded cantilever 43

The final form of the stress function is then determined by the boundary conditions relating to the problem at hand. By assuming second or third degree polynomials for the stress function, we ensure that the compatibility equation is identically satisfied regardless of the values of the coefficients.

Fig. 2.2 (a) Required loading conditions on rectangular sheet in Example 2.2 for A = B = C = 0; (b) as in (a) butA = C = D = 0

E”i&3acements

Bending of an end-loaded cantilever 45

Venant's principle we can assume that the solution is accurate for areas of the beam away from the built-in end and the applied load. If we now assume that the inclination of the neutral plane is zero at the built-in end then.

Bending of an end-loaded cantilever 47

The cross-section, if permitted, would therefore have the shape of a shallow inverted S shown in the figure. Distortion of the cross-section occurs due to the variation of shear stress with the depth of the beam.

Fig. 2.7 (a) Distortion of cross-section due to shear; (b) effect on distortion of rotation due to shear

Show that q5 satisfies the internal compatibility conditions and determine the distribution of the stresses within the plate. The threshold voltage function values of T~~ do not correspond to the assumed constant equilibrium scales at x = 0 and 1.

Torsion of solid sections

Prandtl stress function solution 53
St. Venant warping function solution 59
St. Venant warping function-solution
Torsion of a narrow rectangular strip 63

On the cylindrical surface of the rod there are no external applied forces, so that. Each longitudinal fiber in the rod therefore remains unstressed, as we have actually assumed. So if the membrane has the same external shape as the cross-section of the rod, then

Fig. 3.2 Formation of the direction cosines I and rn of the normal to the surface of the bar

Stress concentrations are made clear by the fact that the contour lines are dense where the slope of the membrane is large. These equations represent exact solutions when the assumed shape of the deflected membrane is the actual shape. We should not close this chapter without mentioning alternative methods of solving the torsion problem.

Problems

Determine the shear stress distribution, the twist rate, and the twist of the cross section.

Strain energy and complementary energy 69

4.1 (a) Strain energy of an element subjected to simple tension; (b) load-deflection curve for a nonlinear elastic element. To be mathematically correct, however, it is the complementary energy differential C that must be equated to the deviation (compare Eqs. (4.3) and (4.4)). 4.l(a) and assuming that the potential energy of the system is zero in the discharged state, then the potential energy loss of the charge P after producing a deflection y is Py.

Fig. 4.2 Load-deflection curve for a linearly elastic member.

Principle of virtual work 71

For the case when the particle is in equilibrium, the resultant P R of the forces must be zero and Eq. A particle is in equilibrium under the action of a system of forces if the total apparent work done by the system of forces is zero for a small apparent displacement. An alternative formulation of the principle of apparent work forms the basis of using the total complementary energy (Section 4.5) to determine the deflections of structures.

Fig. 4.4 (a) Principle of virtual displacements; (b) principle of virtual forces

The principle of the stationary value of the total potential energy

The total potential energy of the particle in each of its three positions is proportional to its height h above some arbitrary datum, since we consider a. The assumed displaced shape of the beam must satisfy the boundary conditions for the beam. Also dvldz = 0 when z = L / 2 so that the displacement function satisfies the boundary conditions of the beam.

Fig. 4.5 States of equilibrium of a particle.

The principle of the stationary value of the total complementary energy

Furthermore, the approximate displacement is less than the exact displacement, since by assuming a displaced shape we have effectively forced the beam into that shape by imposing a constraint; the shaft is therefore harder.

Application to deflection problems 77

Application to deflection problems

At this point, before A can be evaluated, the load-displacement characteristics of the members must be known. On the other hand, if the load-displacement relationship is of a non-linear form, say. The calculation of A is best accomplished in tabular form, but before the procedure is illustrated by an example, some aspects of the solution deserve discussion.

Fig. 4.7 Determination of the deflection of a point on a framework by the method of complementary energy

Application to deflection problems 79

This element subtends an angle Sf3 at its center of curvature due to the application of the bending moment M. We therefore choose a linear relationship M - 0 as this is the case in most of the problems we consider. The fictitious load method of the frame example can be used in solving beam deflection problems where we require deflections at positions in the beam other than the concentrated load points.

Application to deflection problems 83

Calculate the quarter and mid-span vertical displacements B and C of a simply supported beam of length L and flexural stiffness EI loaded as shown in the figure.

Fig. 4.1 1 Deflection of a simply supported beam by the method of complementary energy

Solution of statically indeterminate systems 85

In this case, no advantage is gained by selecting any single member, although in some cases careful selection may result in a reduction in the amount of arithmetic work. Taking BD as a redundant member, we assume that it sustains the tensile force R due to the external load. The solution is now completed in Table 4.2, where, as in Table 4.1, the positive signs indicate tension.

Fig. 4.12 Analysis of a statically indeterminate framework by the method of complementary energy

Solution of statically indeterminate systems 87

Calculate the loads in the members of the single-redundant butt-jointed frame shown in Fig. A planar, rigidly assembled framework consists of six bars forming a rectangle ABCD 4000 mm by 3000 mm with two diagonals, as shown in fig. the sectional area of each bar is 200 mm2 and the frame is unstressed when the temperature of each element is the same.

Solution of statically indeterminate systems 89

In problems like this, AB is usually zero for a rigid support, or a known quantity (sometimes in terms of RB) for a sinking support. The remainder of this section is therefore concerned with the resolution of frames and rings that have varying degrees of redundancy. The frameworks we discussed in the first part of this section and in section 4.6 included members that could only withstand direct forces.

Fig. 4.15 Analysis of a propped cantilever by the method of complementary energy

Solution of statically indeterminate systems 91

The two terms in Eq. iii) can be evaluated separately, taking into account that only the bar ABC contributes to the first term, while the entire structure contributes to the second.

Solution of statically indeterminate systems 93

In the above expression for C, A is the displacement of the top, A, of the ring relative to the bottom, B.

Solution of statically indeterminate systems 95

The bending stiffness of the lower half of the frame is 2EI, while that of the upper half and also the straight part is EI. Calculate the distribution of bending moment in each part of the frame for the loading system shown in figure. Illustrate your answer using a sketch and clearly show the bending moment carried by each part of the frame at the intersection with the straight element.

Fig. 4.20 Determination of bending moment distribution in a shear and direct loaded ring

Solution of statically indeterminate systems 99

However, many structures are linearly elastic and have unique properties that in some cases allow solutions to be obtained more easily. In section 4.6 we discussed the method of virtual or fictitious loads for obtaining the deflections of structures. Following the procedure of Section 4.6, we would place a vertical apparent load Pf on C and write down the total complementary energy of the frame, i.e.

Fig. 4.22 Distribution of bending moment in frame of Example 4.7.

For example, a load P I applied at a point 1 in a linear elastic body will produce a deflection A I at the point given by. Thus, the corresponding deflection A, at point 1 (i.e. the total deflection in the direction of P i produced by all the charges) is then. where aI2 is the deflection at point 1 in the direction of P I produced by a unit load at point 2 in the direction of the load P2 and so on. The deflection at a point 1 in a given direction due to a unit load at point 2 in a second direction is equal to the deflection at point 2 in the second direction due to a unit load at point 1 in the first direction.

Fig. 4.24 Linearly elastic body subjected to loads f , , f z , f 3 , . . . , f,,

The reciprocal theorem 105

A uniform temperature applied across a section of beam causes the beam to expand as shown in the figure. The load intensity is zero at both ends of the beam and wo at its middle. Assuming that the deflected beam shape can be represented by the series im.

Fig. 4.26 Model analysis of a fixed beam.

A bracket BAC is composed of a circular tube AB, whose second moment of area is 1.51, and a beam AC, whose second moment of area is I and which has

The load on the frame consists of the hydrostatic pressure due to the fuel and the vertical support reaction P, which is equal to the weight of fuel carried by the frame, shown in Fig. Calculate the distribution taking into account only loads due to bending of the bending moment around the frame in terms of the force P, the radius r of the frame and the angle 0. Determine the distribution of the bending moment in the frame and sketch the bending moment diagram.

P.4.13 Figure P.4.13 shows a plan view of two beams, AB 9150mm long and DE 6100mm long

Full consideration must be given to deformations due to shear and direct loads as well as bending. Find the extension of the spring when it is replaced and the weight W is transferred to B. The center line of the frame is an arc of a circle, and the section is uniform, of bending stiffness EI and depth d.

The centerline of the frame is a circular arc and the section is uniform, with flexural stiffness EI and depth d. Find an expression for the maximum stress produced by a uniform temperature gradient through depth, the temperatures at the outer and inner surfaces respectively increasing and decreasing by the amount T.

Bending of thin plates

Pure bending of thin plates 123
Plates subjected to bending and twisting 127
Distributed transverse load 129
Distributed transverse load 131 Taking moments about the x axis

The simply supported edge

Distributed transverse load 133

The built-in edge
The free edge

Distributed transverse load 135 Navier (1820) showed that these conditions are satisfied by representing the deflection
Combined bending and in-plane loading 139
Thin plates having a small initial curvature 141

We have shown that the deformed shape of the plate must satisfy the differential equation. In these cases, we assumed that the middle or neutral plane of the plate remained unstressed. These conditions can be satisfied assuming a deformed plate shape given by.

Fig. 5.1 Plate subjected to pure bending.

Strain energy produced by bending and twisting
Energy method for the bending of thin plates 143 rotation, of the ends of the element is negative as the slope decreases with increasing x

Potential energy of a transverse load
Potential energy of in-plane loads

Energy method for the bending of thin plates 145
Energy method for the bending of thin plates 147 We are now in a position to solve a wide range of thin plate problems provided that
A simply supported square plate a x a carries a distributed load according to the formula

In thin sheet analysis, we are concerned with deflections perpendicular to the loaded surface of the sheet. It immediately follows that the potential energy of the Nxy loads is a w a w. 5.6 Energy Method for Bending Thin Sheets 147. It is proposed to determine the deflected shape of the sheet using the Rayleigh-Ritz method, using a 'guessed' shape for the deflection.

Fig. 5.14 (a) Strain energy of element due to bending; (b) strain energy due to twisting

Structural instability

Euler buckling of columns 153
Inelastic buckling 157
Inelastic buckling 159
Effect of initial imperfections 161
Beams under transverse and axial loads 163
Energy method 167
Buckling of thin plates 169

We have shown that the critical voltage, Eq. 6.8), depends only on the elastic modulus of the material of the column and the slenderness ratio l/r. The final deflected shape, v, of the column depends on the shape of its unloaded shape, vo. So a good approximation for deflection when the axial load is in the area of the critical load.

Fig. 6.1 Definition of buckling load for a perfect column.

Loaded edges clamped

Instability of stiffened panels 175

In local instability, flanges and webs buckle as plates with a resulting change in column cross-section. It is clear that the minimum value of the above critical stresses is the critical stress for the panel taken as a whole. In fact, the ultimate load is not reached until the stress in the bulk of the plate exceeds the elastic limit.