AircrAft StructurAl AnAlySiS

Prof. Ravi kumarAM.Ae.SI, M.Tech(Aero)

SASTRA University, Thanjavur

Airframe

Function of Aircraft Structures

GeneralThe structures of most flight vehicles are thin walled structures (shells)

Resists applied loads (Aerodynamic loads acting on the wing structure)

Provides the aerodynamic shape

Protects the contents from the environment

Definitions

Primary structure:A critical load-bearing structure on an aircraft. If this structure is severely damaged, the aircraft cannot fly.

Secondary structure:Structural elements mainly to provide enhanced aerodynamics. Fairings, for instance, are found where the wing meets the body or at various locations on the leading or trailing edge of the wing.

Definitions…

Monocoque structures:Unstiffened shells. must be relatively thick to resist bending, compressive, and torsional loads.

Definitions…

Semi-monocoque Structures:Constructions with stiffening members that may also be required to diffuse concentrated loads into the cover.

More efficient type of construction that permits much thinner covering shell.

SPATER

WING LAYOUT

Function of Aircraft Structures:Part specific

Skin Reacts the applied torsion and shear forces . Transmits aerodynamic forces to the longitudinal and transverse supporting members . Acts with the longitudinal members in resisting the applied bending and axial loads . Acts with the transverse members in reacting the hoop, or circumferential load when the structure is pressurized.

Function of Aircraft Structures:Part specific

Ribs and Frames1. Structural integration of the wing and fuselage.2. Keep the wing in its aerodynamic profile.

Function of Aircraft Structures:Part specific

Spar1. resist bending and axial loads.2. form the wing box for stable torsion resistance.

Function of Aircraft Structures:Part specific

Stiffener or Stringers

1. resist bending and axial loads along with the skin.2. divide the skin into small panels and thereby increase its

buckling and failing stresses.3. act with the skin in resisting axial loads caused by

pressurization.

STRUCTURAL IDEALISATION

STRUCTURAL IDEALISATION1) The longitudinal stiffeners and spar flanges carry only axial stresses. 2) The web, skin and spars webs carry only shear stresses. 3) The axial stress is constant over the cross section of each longitudinal

stiffener .4) The shearing stress is uniform through the thickness of the webs .5) Transverse frames and ribs are rigid within their own planes and

have no rigidity normal to their plane.

•The stiffeners are represented by circles called booms, which have a concentrated mass in the plane of the skin. •The direct stresses are calculated at the centroid of these booms and are assumed to have constant stress through their cross-section.

Idealization Of a Panel

Suppose that we wish to idealize the panel as in Fig.a in to a combination of direct stress carrying booms and shear stress only carrying skin as shown in Fig.b.

Suppose also that direct stress distribution in actual panel varies linearly from unknown stress σ1 to unknown stress σ2.

The Boom area can be found as; - ( as loading produces same direct stress in both panels), taking moments about the Right-hand edge of each panel.

or

Similarly ; For axial load (σ1 / σ2 ) = 1, and for pure bending(σ1 / σ2 ) = -1 .

Thus different idealizations of the same structure are required for different loading conditions.

Shear of open section beams

.

For skin,

If skin carries only shear stress thn tD =0 .Now taking equilibrium of stresses at rth boom,

Or ….. ( A )

Shear of open section beams

So

Or,

This is the increment in shear flow in boom, which is already subjected to direct stress. So for n Booms shear flow at any point is:

Shear loading of closed section beams.

Solved examples :

Q.1: Part of the wing section is in the form of the two cell box as shown in Figure, in which vertical spars are connected to the wing skin through angle sections all having a cross sectional area 300 mm2. Idealize the section in to an arrangement of direct stress carrying booms and shear stress only carrying panels suitable for resisting bending moments in the vertical plane. Position the Booms at the spar/ skin junctions only.

Solution • The idealized section as in Figure, by symmetry,

Since section is required to resist bending in vertical plane. The direct stress at any point in the section is directly proportional to its distance from the horizontal axis of symmetry.

Boom Areas are:

or

or

Solution .

or Ans.

Solved examples :

Q.2. The fuselage section as shown in figure is subjected to a bending moment of 100 kN m applied in the vertical plane of symmetry. If the section has been completely idealized in to a combination of direct stress carrying booms and shear stress only carrying panels, determine direct stress in each boom.

Solution As section has symmetry about y axis and resisting B.M. Mx = 100 kN m.

therefore,

The origin of axes Cxy coincides with the position of the centroid of the direct stress carrying area, i.e, the centroid of the boom areas. Thus taking moments of the area about the boom 9.

Which gives,

and Moment of inertia of boom is

So results can be tabulated as :

Solution

.

Solved examples :

Q.3. Idealize the box section as shown in Figure in to an arrangement of direct stress carrying booms positioned at the four corners and panels which are assumed to carry only shear stresses. Hence determine the position of shear centre from the left hand-Web.

Solution

Idealized section is :

So boom areas are:-

So,

Solution Shear flow distribution is:

Here,

Cut the section in the wall 12, Then

Since shear load is applied through the shear centre , the rate of twist is zero. And is given as

Solution Hence,

Which gives,

The complete shear flow distribution is then

Solution

Position of shear centre: taking moments about the intersection of the horizontal axis of symmetry and the left hand web,

Which gives, Ans.

Wing spar and Box beams

Wing spar and Box beamsTapered wing spar :-Consider a wing spar positionedin YZ plane and comprises two Flanges and one web. At the sectionz the beam is subjected to a positiveMoment Mx and a positive shearForce Sy.

In which, in the direction of taper shown,δy2 is negative. So axial load in flange 1 is :…….. ( 1)

substituting Py,1 in equation (1), we have

, Similarly

Tapered wing spar :-• Internal shear force Sy comprises the resultant Sy,w of the web shear flows together

with the vertical components of P1 and P2. thus

Or,

• As δy2 is negative, the above equation may be assumed to calculate the shear flow distribution in the web. For a completely idealized beam the web shear flow is constant through the depth and is Sy,w / h.

• For a beam in which web is fully effective in resisting direct stresses, the web shear flow is given by:

Or,

Solved Example:

Q1. Determine the shear flow distribution in the web of the tapered beam as shown, at a section midway along the length. The web of the beam has a thickness of 2 mm and is fully effective in resisting the direct stress. The beam tapers symmetrically about its central axis and the cross- sectional area of each flange is 400 mm2.

Solution • Internal bending moment and shear load at the section A-A are:

Direct stress parallel to z axis is:

and Hence,

and axial loads:Shear load resisted by web is;

So from Fig,

Hence

Solution

• Shear flow in the web is:

i.e,

So shear flow distribution is parabolic.Shear flow distribution: ( N / mm)

Open and closed section beams

,

….. (1)

Where,

Eqn (1) can be directly applied to a tapered beam subjected to to forces positioned in relation to the moment centre.

Solved Example: Q.2. The cantilever wing as shown in Figure, is uniformly tapered along its

length in both x and y directions and carries a load of 100 kN at its free end. Calculate the forces in the booms and the shear flow distribution in the walls at a section 2 m from the built – in end if the booms resist all the direct stresses while the walls are effective only in shear. Each corner boom has a cross – sectional area of 900 mm2 while both central booms have cross –sectional areas of 1200 mm2.

Solution

• The internal force system at a section 2 m from built –in end is:

Beam is doubly symmetric so Ixy = 0. so

So direct stress at the rth boom is :

Or &

Solution From column (6)

From column (10)

From column (11)

So shear loads:

Hence

Shear flow distribution in the walls of the beam;

….. ( A )

SolutionWe now ‘cut ‘ one of the walls, say 16 . The resulting open section shear flow is

given as:

Thus &

Taking moments about center of symmetry, we have

Which gives,

Complete shear flow distribution can be found by adding qs,0 to qb shear flow distribution.

SolutionShear flow distribution of open section in N /mm ;

Total shear flow distribution in N / mm :

Solved Example: Q3. A wing spar has the dimensions as shown in Figure, and carries a uniformly

distributed load of 15 kN/m along its complete length. Each flange has the cross–sectional area 500 mm2 with top flange being horizontal. If the flanges are assumed to resist all direct loads while the spar web is effective only in shear, determine the flange loads and the shear flows in web at the sections 1 and 2 m from the free end.

Solution

Bending moment at section 1 :

Thus

Also

As

Then And

Shear force at section 1 is 15 kN and this is resisted by the shear force in web ( Py,L).

Hence shear flow

Solution

At section 2 :

Hence

Also Then

and

The Shear force at the section 2 is 15* 2 = 30 kN. Hence shear force in the web is 30 – 7.5 = 22.5 kN, which gives shear flow as;

FUSELAGES

Fuselages

Aircraft fuselage consists of thin sheets of material stiffened by large numbers of longitudinal stringers together with transverse frames.

Generally they carry bending moments, shear forces and torsion loads which induces axial stresses in the stringers and skin together with shear stresses in the skin.

Shear flow is constant between adjacent stringers ( distance is very small).

The analysis of the fuselages thus involves the calculation of the direct stresses in the stringers and the shear stress distributions in the skin.

Fuselage bending example:

Q.1. The fuselage of a light passenger aircraft has the circular cross-section shown in Fig(a). The cross-sectional area of each stringer is 100 mm2 and the vertical distances given in Fig (a) are to the mid-line of the section wall at the corresponding stringer position. If the fuselage is subjected to a bending moment of 200 kN m applied in the vertical plane of symmetry, at this section. Calculate the direct stress distribution.

Solution

• The section is first idealized, as an approximation we shall assume that the skin between the adjacent stringers is flat.

From symmetry, Boom areas :

i.e,

Solution

Similarly

Stringers 5 and 13 lie on the neutral axis of the section and therefore are unstressed. and the calculation of boom areas for 5 and 13th stringers are not required. Since centroid is on an axis of symmetry. So Ixy = 0. My= 0.

So direct stress is ….. ( 1)

Here

So direct stresses for complete fuselage is tabulated as:

Solution

Results are :

Fuselage Shear flow example:

Q.2. the fuselage of Example 1 is subjected to a vertical shear load of 100 kN applied at a distance of 150 mm from the vertical axis of symmetry as shown, for the idealized section as shown in Figure. Calculate the shear flow distribution in the section.

Solution

As fuselage is symmetric, so Ixy = 0, and Sx=0, hence

Or,

i.e,

The first term of above equation is the ‘ open section’ shear flow . We therefore ‘cut ‘ one of the skin panels, say 12, and calculate .

Results are presented in the table ( in next slide);

The shear flow in the panel 12 is now found by taking moments about a convenient moment centre, say C.

Solution

.

Solution

• Taking moments about the point C,

Where, , and since shear flows are constant between the booms, the above equation may be written as;

In which A12, A23,……, A161 are the areas subtended by the skin panels 12, 23,…, 161 at the centre C and anticlockwise moments are taken positive.

So,

Or

SolutionComplete shear flow distribution can be get by adding qb and qs,0, as shown in figure:

Fuselage Torsion example:

Q.3. The fuselage of example 1 is subjected to a pure torque of 15 MN mm and a shear load of 100 kN applied through the shear centre as shown in figure. Find the shear flow distribution.

Solution• As shear centre coincides with the centre of symmetry and a shear force of

100 kN is applied through it . which produces pure torque.• A fuselage is basically a single cell closed section beam and hence

Then from symmetry, and using the results of the table of example 2.

The resultant of these shear flows will be statically equal to applied shear load.so

Which gives,

Solution

SoShear force due to applied torque:

• Acting in the anticlockwise sense completely around the section. This value of the shear flow is now superimposed on the shear flows produced by the shear load.

Complete shear flow distribution is shown in Figure ( next slide)

Solution

• .

Tension field beam ( Wagner’s Beam )

Tension field beams (Wagner’s beam)

The spars of aircraft wings usually comprise an upper and a lower flange connected by thin stiffened webs. These webs are often of such a thickness that they buckle under shear stresses at a fraction of their ultimate load. The form of the buckle is shown in Fig. (a),

where the web of the beam buckles under the action of internal diagonal compressive stresses produced by shear, leaving a wrinkled web capable of supporting diagonal tension only in a direction perpendicular to that of the buckle; the beam is then said to be a complete tension field beam.

Incomplete diagonal tension field beams

In modern aircraft structures, beams having extremely thin webs are rare.

They retain, after buckling, some of their ability to support loads so that even near failure they are in a state of stress somewhere between that of pure diagonal tension and the pre-buckling stress.

Such a beam is know as an incomplete diagonal tension field beam.

Shear lag :

• Shear lag is caused when a framing member is connected to another member by only a portion of its cross-section.

• An example would be connecting steel stringer beams to girders. The top and bottom flanges of the stringer are not connected to the girder. Shear lag occurs because the forces cannot be transmitted directly into the entire cross-section of the stringer. This means that the area of the member that is effective in resisting the force is something less than the total area.

Shear lag could occur within an aircraft structure in cases where only portions of the cross-section of a framing member are used in a connection.

Example : in spars web, diagonal strut in wing etc.

ANY QUERY ??