jntuh 3-1 aerospace vehile stuctures -II

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7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.AEROSPACE VEHICLE STRUCTURES II( AVS II )

7/11/2014SYLLABUSTHIN PLATE THEORYBUCKLING OF THIN PLATES AND STIFFENED PANELSBENDING AND SHEAR OF THIN WALLED BEAMSTORSION OF THIN WALLED BEAMSSTRUCTURAL IDEALISATION OF THIN WALLED BEAMSSTRUCTURAL AND LOADING DISCONTINUITIES IN THIN WALLED BEAMSSTRESS ANALYSIS OF AIRCRAFT COMPONENTS- WINGSTRESS ANALYSIS OF AIRCRAFT COMPONENTS- FUSELAGE7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.SUGGESTED READINGSMegson, T.H.G., Aircraft Structures for Engineering Students, 4th edn., Elsevier, 2007, ISBN 0-750-667397.Peery, D.J. and Azar, J.J., Aircraft Structures, 2nd edn., McGra-Hill, 1982, ISBN 0-07-049196-8.Bruhn. E.H, Analysis and Design of Flight Vehicles Structures, Tri-state Off-set Company, USA, 1965.Rivello, R.M., Theory and Analysis of Flight Structures, McGraw Hill, 1993.Sechler.E.E. and Dunn, L.G., Airplane Structural Analysis and Design, John Wiley & Sons.7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.UNIT - 1Introduction To Pure Bending Theory.Thin Plate TheoryAnalysis Of Thin Rectangular Plates Subject To BendingAnalysis Of Thin Rectangular Plates Subject To Bending And TwistingAnalysis Of Thin Rectangular Plates Subject To Distributed Transverse Load, Combined Bending And InplaneLoading- Thin Plates Having Small Initial Curvature And Energy Methods Of Analysis.7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.BENDING STRESSES IN BEAMSBeams are subjected to bending moment and shearing forces which vary from section to section. To resist the bending moment and shearing force, the beam section develops stresses.Bending is usually associated with shear. However, for simplicity we neglect effect of shear and consider moment alone ( this is true when the maximum bending moment is considered---- shear is ZERO) to find the stresses due to bending. Such a theory wherein stresses due to bending alone is considered is known as PURE BENDING or SIMPLE BENDING theory.

7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.5Bending action:7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.

cNeutral Axist

Neutral layercNeutral layerAssumptions made in Pure bending theoryThe beam is initially straight and every layer is free to expand or contract.The material is homogenous and isotropic. Youngs modulus (E) is same in both tension and compression. Stresses are within the elastic limit. The radius of curvature of the beam is very large in comparison to the depth of the beam. A transverse section of the beam which is plane before bending will remain plane even after bending.Stress is purely longitudinal.

7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.Results of pure bending theoryThe neutral plane passes through the centroid of the cross section.The bending equation is given byWhere:E= Youngs modulus, R= Radius of curvature,M= Bending moment at the section,I= Moment of inertia about neutral axis, f or = Bending stressy = distance of the fibre from the neutral axis

7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.

Bending of thin plates in one dimension7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.

For bending moment


Bending of thin plates in two dimensionsThe thin rectangular plate of Fig (a) is subjected to pure bending moments of intensity Mx and My per unit length uniformly distributed along its edges.

According to simple beam theory, the middle plane of the plate does not deform during the bending and is therefore a neutral plane.

Let us consider an element of the plate of side x y and having a depth equal to the thickness t of the plate as shown in Fig (b).7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.

Fig (a) Plate subjected to pure bending

Fig (b) Direct stress on lamina of plate elementBending of thin plates in two dimensionsSuppose that the radii of curvature of the neutral plane n are x and y in the xz and yz planes respectively (Fig. (c)).Simple beam theory, the direct strains x and y corresponding to direct stresses x and y of an elemental lamina of thickness z a distance z below the neutral plane are given by

we have,

rearranging above eqs. and substiduding in x and y gives,7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.

Fig.(c ) radii of curvature of neutral plane

12Bending of thin plates in two dimensionsThe internal direct stress distribution on each vertical surface of the element must be in equilibrium with the applied bending moments. Thus,and

Substituting x and y in above,

Let,

ThenWhere, D is known as the flexural rigidity of the plate.7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.

Bending of thin plates in two dimensionsIf w is the deflection of any point on the plate in the z direction, then we may relate w to the curvature of the plate in the same manner as the well-known expression for beam curvature. Hence

So,

If either Mx or My is zero7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.

Bending of thin plates in two dimensionsThe case of My =0 is illustrated in Fig. (d)A surface possessing two curvatures of opposite sign is known as an anticlastic surface, as opposed to a synclastic surface which has curvatures of the same sign.if Mx =My =M

Therefore, the deformed shape of the plate is spherical and of curvature7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.

Fig(d) Anticlastic bending

Anticlastic Bending7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.

Synclastic Bending

Plates subjected to bending and twistingThe perpendicular components are seen to be Mx and My as before, while the tangential components Mxy and Myx (again these are moments per unit length) produce twisting of the plate about axes parallel to the x and y axes.From a consideration of complementary shear stresses Mxy=Myx, so that we may represent a general moment application to the plate in terms of Mx, My and Mxy as shown in Fig(b). These moments produce tangential and normal moments, Mt and Mn, on an arbitrarily chosen diagonal plane FD.7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.

Fig (a).Plate subjected to bending and twisting.Fig (b).Plate subjected to bending and twistingPlates subjected to bending and twistingFor equilibrium of the triangular element ABC of Fig.(c). in a plane perpendicular to AC

Similarly for equilibrium in a plane parallel to CA


Fig (c)tangential and normal moments on an arbitrary plane.

For Mt =0, leaving normal moments of intensity Mn on two mutually perpendicular planes. These moments are termed principal moments and their corresponding curvatures principal curvatures.

Plates subjected to bending and twistingNow consider an element in a platesubjected to twisting moment asshown in the fig(d).on the face ABCD,

and on the face ADFE

Giving

or in terms of the shear strain xy and modulus of rigidity G


Fig(d).Complementary shear stresses due to twisting moments Mxy.

Plates subjected to bending and twistingThe shear strain xy

An element taken through thethickness of the plate will sufferrotations equal to w/x and w/yin the xz and yz planes respectively.Considering the rotation of such an element in the xz plane, as shown in Fig.(e).The displacement u in the x direction of a point a distance z below the neutral plane is


Fig(e). Determination of shear strain xyPlates subjected to bending and twistingHence, substituting for u and v in the expression for xy we have

Replacing G by the expression E/2(1+)

Multiplying the numerator and denominator of this equation by the factor (1) yields


Plates subjected to a distributed transverse loadConsider a distributed transverse load of intensity q per unit area such thatdistributed load may, in general, vary overthe surface of the plate and is therefore afunction of x &y as shown in the Fig(a).Assume middle plane as a neutral plane.Assume that although xz =xz/G andyz =yz/G are negligible the correspondingshear forces are of the same order of magnitude as the applied load q and the moments Mx, My and Mxy.For the Fig(b) the vertical shear forces Qx and Qy per unit length on faces perpendicular to the x and y axes, respectively.The variation of shear stresses xz and yz along the small edges x, y of the element is neglected and the resultant shear forces Qxy and Qyx are assumed to act through the centroid of the faces of the element.7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.

Fig(a).Plate supporting a distributed transverse load.Plates subjected to a distributed transverse loadWe have7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.

Fig(b).Plate element subjected to bending, twisting and transverse loads.

In a similar fashion,

For equilibrium of the element parallel to Oz and assuming that the weight of the plate is included in q

Taking moments about the x axis


Simplifying this equation and neglecting small quantities of a higher order than those retained gives

Similarly taking moments about the y axis we have

Substituting Qx and Qy from above eqs. To below eq.7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.

or7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.

Boundary ConditionsBefore discussing the solution of Laplace Eq. for particular cases we shall establish boundary conditions for various types of edge support.The simply supported edgeLet us suppose that the edge x =0 of the thin plate shown in Fig. (c) is free to rotate but not to deflect. The edge is then said to be simply supported. The bending moment along this edge must be zero and also the deflection w=0. Thus

The condition that w=0 along the edge x =0 also means that

Along this edge. The above boundary conditions therefore reduce to7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.

Fig (c).Plate of dimensions ab.

Boundary ConditionsThe built-in edgeIf the edge x =0 is built-in or firmly clamped so that it can neither rotate nor deflect, then, in addition to w, the slope of the middle plane of the plate normal to this edge must be zero. That is

The free edgeAlong a free edge there are no bending moments, twisting moments or vertical shearing forces, so that if x =0 is the free edge then

Consider two adjacent elements y1 and y2 along the edge of the thin plate of Fig.(d),7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.

Boundary ConditionsThe twisting moment Mxyy1 onthe element y1 may be replacedby forces Mxy a distance y1apart.The twisting moment on the adjacent element y2 is[Mxy +(Mxy/y)y]y2.Again this may be replaced by forces Mxy +(Mxy/y)y.At the common surface of the two adjacent elements there is now a resultant force (Mxy/y)y or a vertical force per unit length of Mxy/y.For a statically equivalent vertical force per unit length of (Qx Mxy/y).The separate conditions for a free edge of (Mxy)x=0 =0


Fig (d).Equivalent vertical force system

Boundary ConditionsIn terms of deflection,

and


Sample caseThe solution for the simple case of a thin rectangular plate of dimensions ab.Consider a thin rectangular plate of dimensions ab, simply supported along each of its four edges and carrying a distributed load q(x, y)We havethe governing eq.,

The boundary conditions,Navier (1820) showed that these conditions are satisfied byrepresenting the deflection w as an infinite trigonometrical or Fourier series.Where,m represents the number of half waves in the x direction and n the corresponding number in the y direction.7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.

Sample caseWe may also represent the load q(x, y) by a Fourier series, thus

A particular coefficient amn is calculated by first multiplying both sides of above Eq. by sin(mx/a) sin(ny/b) and integrating with respect to x from 0 to a and with respect to y from 0 to b. Thus

Since,7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.

Sample caseIt follows that,

Substituting now for w and q(x, y)in governing equation,

This equation is valid for all values of x and y so that

in alternative form,giving

Hence,7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.

Combined bending and in-plane loading of a thin rectangular plateThe elevation and plan of a small element xy of the middle plane of a thin deflected plate are shown in Fig(a).

Direct and shear forces per unitlength produced by the in-planeloads are given the notation Nx,Ny and Nxy and are assumed to be acting in positive senses inthe directions shown.

Since there are no resultant forces in the x or y directions fromthe transverse loads, we needonly include the in-plane loads shown in Fig(a).7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.

Fig(a). In-plane forces on plate elementCombined bending and in-plane loading of a thin rectangular plateconsidering the equilibrium of the element in these directions. For equilibrium parallel to Ox

For small deflections w/x and (w/x)+(2w/x2)x are small and the cosines of these angles are therefore approximately equal to one. The equilibrium equation thus simplifies to

Similarly for equilibrium in the y direction,

The determination of the contribution of the shear loads to the equilibrium of the element in the z direction is complicated by the fact that the element possesses curvature in both xz and yz planes.7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.

Combined bending and in-plane loading of a thin rectangular plateTherefore, from Fig(a). the component in the z direction due to the Nxy shear loads only is

or

Similarly, the contribution of Nyx is

The components arising from the direct forces per unit length are readily obtained from Fig.(a)

orsimilarly from y direction


Combined bending and in-plane loading of a thin rectangular plateThe total force in the z direction is found from the summation of these expressions and is

Nyx is equal to and is replaced by Nxy

the in-plane forces do not produce moments along the edges of the element then following Eqs. remain unaffected.7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.

Combined bending and in-plane loading of a thin rectangular plateWe have the governing eq. for the thin plate under transverse loads,

The above eq. may be modified simply by the addition of the above vertical component of the in-plane loads to qxy.Therefore, the governing differential equation for a thin plate supporting transverse and in-plane loads is7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, IARE.

Bending of thin plates having a small initial curvatureSuppose that a thin plate has an initial curvature so that the deflection of any point in its middle plane is w0 and w0 is small compared to the thickness of the plate.The application of transverse and in-plane loads will cause the plate to deflect a further amount w1 so that the total deflection is then w=w0 +w1.From the governing eq.,

The effect of an initial curvature on deflection is therefore equivalent to the application of a transverse load of intensity

Thus, in-plane loads alone produce bending provided there is an initial curvature.7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, ASTI.

Bending of thin plates having a small initial curvatureAssuming that the initial form of the deflected plate is

and

where 7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, ASTI.

Energy method for the bending of thin platesStrain energy produced by bending and twisting:

The bending strain energy due to Mx is

The contribution of My to the bending strain energy is

7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, ASTI.

Energy method for the bending of thin platesThe strain energy due to the twisting moment per unit length, Mxy, applied to the y edges of the element, is obtained from Fig. 7.14(b).

The contribution of the twisting moment Mxy on the x edges is

The total strain energy of the element from bending and twisting is thus

Substitution for Mx, My and Mxy gives the total strain energy of the element as

On rearranging,7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, ASTI.

Hence the total strain energy U of the rectangular plate ab is

Note that if the plate is subject to pure bending only, then Mxy =0, giving

Potential energy of a transverse load:An element x y of the transversely loaded plate supports a load qxy. If the displacement of the element normal to the plate is w then the potential energy v of the load on the element referred to the undeflected plate position is V = w q x yThe potential energy V of the total load on the plate is given by

The potential energy of complete in-plane loading system:7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, ASTI.

43Test ProblemConsidering the rectangular plate as shown in fig., simply supported along all four edges and subjected to a uniformly distributed transverse load of intensity q0.we know that its deflected shape isgiven by Eq.

The total potential energy of the plate is

Substituding w in above eq.

7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, ASTI.

Test ProblemThe term multiplied by 2(1) integrates to zero and the mean value of sin2 or cos2 over a complete number of half waves is 1/2 , thus integration of the above expression yields

From the principle of the stationary value of the total potential energy

so that,

The deflected form is given by,7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, ASTI.

Assignmenmt - 1Derive the equation (1/) = [ D (1+ )] of thin plate subjected to pure bending.Derive the equation Mxy = D (1-) 2w/xy for a thin plate subjected to bending and twisting.Derive the Governing differential equation for a simply supported thin rectangular plate subjected to distributed transverse load of intensity q per unit area.A thin rectangular plate a b is simply supported along its edges and carries a uniformly distributed load of intensity q0. Determine the deflected form of the plate and the distribution of bending moment. Here a is length and b is width of the plate.Determine the deflected form of the thin rectangular plate a b is simply supported along its edges and carrying a uniformly distributed load of intensity q0 . In addition to that it supports an in-plane tensile force Nx per unit length. Here a is length and b is width of the plate.Examples 7.1, 7.2, 7.3 & 7.4.7/11/2014Y.SHARATH CHANDRA MOULI, Asst. Professor, AE Dept, ASTI.

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jntuh 3-1 aerospace vehile stuctures -II