DIAB Sandwich Handbook - STM Boats · DIAB SANDWICH HANDBOOK 09.03- 2 SANDWICH Whenever new...

DIAB SANDWICH HANDBOOK

09.03- 1

EK

P

SANDWICH CONCEPT


09.03- 2

SANDWICHWhenever new materials or production methods appear there is a resistance touse them. Mostly the resistance originates in conservatism and ignorance. Theonly way to overcome the resistance is to try to teach and convince theopponents.

This handbook has been written to spread knowledge about sandwich and under-standing of its behaviour.

HISTORYHistorically, the principal of using two cooperating faces with a distance betweenthem was introduced by Delau about 1820. The first extensive use of sandwichpanels was during World War II. In the ”Mosquito” aircraft sandwich was used,mainly because of the shortage of other materials in England during the war.The faces were made of veneer and the core of balsa wood.

During World War II the first theoretical writings about sandwich appeared. Inthe 50’s the development was mainly concentrated on honeycomb materials.Honeycomb was mainly used as core material in the aircraft industry. However,it had some limitations, for example there were big problems with corrosion.

At the end of the 50’s and during the 60’s different cellular plastics were produced,suitable as core materials. In the beginning rather soft materials were usedbecause of their insulation properties, for example polystyrene and polyuret-hane.

Later it was possible to produce harder cellular plastics with higher densitiesand by that time sandwich became a very useful and flexible concept. Todaythere is an enormous number of different qualities of cellular plastics as corematerials.

Fig. 1.1 Sandwich

Every part has its specific function to make it work as a unit.

The aim is to use the material with a maximum of efficiency. The two faces areplaced at a distance from each other to increase the moment of inertia, andthereby the flexural rigidity, about the neutral axis of the structure. A comparisoncould be made with a solid beam. A Sandwich beam of the same width andweight as a solid beam has a remarkably higher stiffness because of its highermoment of inertia.

upper face

core

lower face

joints

DESIGN

MAIN PRINCIPLESSandwich is built up of three elements, see Fig. 1.1.

* two faces* core* joints


09.03- 3

DESIGN

weight flexural rigidity 1 1

1 (+ core) 12

Fig. 1.2 Comparison of stiffness in bending between solid beam and sandwich beam

An important difference in comparing an Ι-beam with a sandwich beam is thepossibility for each to bear transverse loads. For an Ι−beam the web is stiffenough to give Navier’s assumption validity, (i.e. plane cross sections remainplane). In a sandwich beam the core material is usually not rigid in shear andthe assumption is not fullfilled. In bending the shear deflection in the core is notnegligible in most cases. There is also shear deflection in the faces but this canbe ignored.

The effect of shear rigidity in the core is shown in fig. 1.3.

The upper case shows an ideal sandwich beam which is relatively stiff in shear.It is obvious how the faces cooperate without sliding over each other.

Fig 1.3 Comparison between cores that are rigid or weak in shear

Fig. 1.2 demonstrates, as a simple example, the difference in flexural rigidity fora solid beam versus a sandwich beam.

The lower case shows, as a comparison, a sandwich beam which is not veryrigid in shear. Here the faces do not cooperate and the faces work as plates inbending, independent of each other. The local flexural rigidities for the facescan in most cases be ignored. Accordingly, the result of a core that is weak inshear is a loss of the sandwich effect.

d/2

d/4

d/4

b

d

b

d


09.03- 4

DESIGN

a/ The facesThe faces carry the tensile and compressive stresses in the sandwich. Thelocal flexural rigidity is so small that it can often be ignored. Conventionalmaterials such as steel, stainless steel and aluminium are often used for facematerial. In many cases it is also suitable to choose fibre- or glass- reinforcedplastics as face materials. These materials are very easy to apply. Reinforcedplastics can be tailored to fulfill a range of demands like anisotrophic mechanicalproperties, freedom of design, excellent surface finish etc.

Faces also carry local pressure. When the local pressure is high the facesshould be dimensioned for the shear forces connected to it.

b/ The coreThe core has several important functions. It has to be stiff enough to keep thedistance between the faces constant. It must also be so rigid in shear that thefaces do not slide over each other. The shear rigidity forces the faces tocooperate with each other. If the core is weak in shear the faces do not cooperateand the sandwich will lose its stiffness. (See fig. 1.3).

This presentation demonstrates that it is the sandwich structure as a wholethat gives the positive effects. However, it should be mentioned that the corehas to fulfill the most complex demands. Strength in different directions andlow density are not the only properties that the core has to have. Often thereare special demands for buckling, insulation, absorption of moisture, ageingresistence, etc.

c/ Adhesive (Bonding layer)To keep the faces and the core co-operating with each other the adhesive betweenthe faces and the core,must be able to transfer the shear forces between thefaces and the core. The adhesive must be able to carry shear and tensile stresses.It is hard to specify the demands on the joints. A simple rule is that the adhesiveshould be able to take up the same shear stress as the core.

Each of the parts in sandwich have their particular functions and will be described.


09.03- 5

BEAMS AND STRUTS

ASSUMPTIONSIn this chapter it is assumed that the faces are thin and of the same thickness.Shear and bending strains in the faces are small and can be ignored. The shearstress is assumed to be constant throughout the thickness of the core at anygiven section. For a beam with faces on the sides, the shear and bending strainsin the side faces cannot be ignored.

In this chapter the beams are considered narrow. The conditions and directionsfor when a beam is to be considered narrow or wide are found in the chapter”Beams considered narrow or wide”.

SIGN CONVENTION FOR BENDING OF BEAMSThe sign conventions to be adopted for deflection, slope, curvature, bendingmoment and shear forces are illustrated in fig 2.1.

y

x

z

Fig. 2.1. Sign conventions. Left, positive deflection, slope and curvature; negative bendingmoment. Right, positive shear force, shear stress and shear strain.

Loads and deflections (w) are measured positive downwards, in the direction ofthe z-axis. As a result of the choice of sign convention it is necessary to introducenegative signs in some of the relationships between distributed load (q), shearforce (Q), bending moment (M), slope (dw/dx), and deflection (w).

For reference, the full set of relationships, with the correct signs, is given:

Deflection w

Slope + dw/dx = w’

Curvature + w’’(2.1)

– M + Dw’’

– Q + Dw’’’

+ q + Dw (4)


09.03- 6

OPEN BEAMS (FREE SIDES)

FLEXURAL RIGIDITYThe theory for engineering stresses in beams is easily adapted to sandwichbeams with some modifications. Effects caused by shear deflections in the coremust be added and certain terms may be neglected when calculating flexuralrigidity.

To use ordinary beam theory we should first find a simple way to calculate theflexural rigidity, here denoted D, of the beam. In an ordinary beam D would bethe product of the modulus of elasticity, (E) and the second moment of area (I).In a sandwich beam D is the sum of the flexural rigidities of the different parts,measured about the centroidal axis of the entire section:

12

3bccE2

2btdfE6

3btfED ++= (2.2)

Ef and E

c are the moduli of elasticity of the faces (index f) and the core (index c)

respectively. Dimensions according to fig. 2.1.

Fig. 2.2. Dimensions of sandwich beam. Section AA on right.

W

A

x

A z

z

t

t

yL /2 L /2

C C

b

c /2

c /2

d/2

d/2

h /2

h /2

The first term amounts to less than 1% of the second when:

5.77td> (2.3)

At a ratio of d/t > 11.55 the proportion is less than 0.25% and since we haveassumed that the faces are thin the first term can for the present be ignored.

The third term amounts to less than 1% of the second (and may consequentlybe ignored) when:

16.73c

2td

cEfE >⋅ (2.4)

In many practical sandwich beams this condition is fulfilled but, considering themany combination possibilities of Divinycell, this term must be checked. Theerror may be too big to be acceptable. With condition (2.3) the expression forthe flexural rigidity is:

12

3bccE2

2btdfED += (2.5)

If condition (2.4) is fulfilled this expression will be reduced to:

2

2btdfED = (2.6)

The first term in equation (2.2) is local flexural rigidity of the faces about theirown centroidal axes. The second term is the first term transposed for bendingabout the centroidal axis of the entire cross section. The third term is flexuralrigidity of the core about its own centroidal axis, which is the same as for theentire cross section.

STRESSESThe stresses in a sandwich beam may also be determined by the use of theoryfor engineering stresses in beams, with a few modifications. Due to assumptions(sections remain plane and perpendicular to the centroidal axis) the strain at apoint the distance z below the centroidal axis cc is Mz/D.

To obtain the bending stress at the same point the strain may be multiplied withthe appropriate modulus of elasticity. For instance, the stresses in the faces andcore are respectively:


09.03 7


−≤≤−≤≤=

2c

z2h

;2h

z2c

fEDMz

fσ (2.7a)

≤≤−=

2c

z2c

cEDMz

cσ (2.7b)

The maximum stresses are obtained with the maximum value of z within theinterval. The ratio of the maximum membrane stress in the faces and the maxi-mum core stress is (E

f /E

c ) . (h/c).

The assumptions of the theory of bending lead to the common expression forthe shear stress (τ) in a homogeneous beam at depth z, below the centroid ofthe cross section:

IbQS

=τ (2.8)

Here Q is the shear force at the section under consideration, I is the secondmoment of area of the entire section about the centroid, b is the width at level z

1

and S is the first moment of area of the part of the section for which z > z1. The

familiar distribution of such shear stress in an Ι-beam is illustrated in Fig 2.3.

Fig. 2.3. Shear stress distribution in an Ι-beam.

For a sandwich beam, equation (2.8) must be modified to take into account themoduli of elasticity of the different elements of the cross section:

( )SEDbQ

Σ=τ (2.9)

In this expression D is the flexural rigidity of the entire section and Σ (S;E;)represents the sum of the products of S and E of all parts of the section forwhich z < z

1. For example, if equation (2.9) is used to determine the shear

stress at a level z in the core of the sandwich in fig. 2.1,

+

−+==∑ z2

cz2c

2bE

2btdE)(SE cf

The shear stress in the core is therefore

−+= 2z

4

2c2cE

2td

fEDQτ (2.10)

An analogous expression may be obtained for the shear stress in the faces, andthe complete shear stress distribution across the depth of the sandwich isillustrated in fig. 2.4a.The maximum shear stress in the core is obtained byinserting z = 0 in (2.10).

+=

4

2c2c

E

2td

fEDQτ (2.11)

The ratio of the maximum core shear stress (at z = 0) to the minimum coreshear stress (at z = ± c/2) is

+

td

2c4t

fE

cE1

The second term amounts to less than 1% of the expression provided

100cd

ct

cEfE4 > (2.12)

b τ

z > z1

z1


09.03- 8


If condition (2.12) is satisfied, the shear stress can be assumed constant over thethickness of the core. Because d≈c, conditions (2.4) and (2.12) are similar ineffect.Therefore it may be concluded that where a core is too weak to provide asignificant contribution to the flexural rigidity of the sandwich, the shear stressmay be assumed constant over the depth of the core. For a weak core, it istherefore permissible to write E

c = 0 in equations (2.2) and (2.8); the constant

shear stress in the core is then given by:

(2.13)

The way the shear stresses are distributed across the section is illustrated in fig.2.4.b.

t

t

c

a) b) c)

d

DEj td

2

τ = τ =Q Q

bd

If, in addition, the flexural rigidities of the faces about their own separate axes issmall (i.e. if condition (2.3) is fulfilled), then the first term on the right-hand side ofequation (2.2) may be ignored as well as the third, leaving:

(2.14)

In this case equation (2.10) for the shear stress in the core is reduced to thesimplest possible form:

(2.15)

The corresponding shear stress distribution is illustrated in fig. 2.3c. The differencebetween fig. 2.3b and 2.3c is that in the latter the principle stress in each face isassumed to be uniform (because the local bending stress is ignored). It followsfrom this that the shear stress in the faces varies with depth in a linear fashion,not a parabolic one.

It is often convenient to invoke the concept of an ”antiplane” core (σx = σy = τxy =0). An antiplane core is an idealised core in which the modulus of elasticity inplanes parallel with the faces in zero but the shear modulus in planesperpendicular to the faces is finite. By this definition Ec = 0 and the antiplanecore makes no contribution to the flexural rigidity of the beam. Conditions (2.4)and (2.9) are automatically satisfied and the shear distribution is similar to thatshown in fig. 2.3b.

DEFLECTIONS

a/ Symmetrical loadsThe loads considered here are symmetrical, i.e. the load is symmetrical withrespect to the geometry of the beam and/or a relative horizontal displacement ofthe faces is prevented somewhere (for example at a clamped end).

In this case the flexural rigidity of the sandwich and the shear stress in the coreare defined by equations (2.14) and (2.15). The shear stress distribution appearsin fig. 2.4c.

Fig. 2.4. Shear stress distribution in a sandwich beam.(a) True shear stress distribution.(b) Effect of weak core (conditions (2.4) and (2.12) satisfied).(c) Effect of weak core, ignoring the local flexural rigidity of the faces (conditions(2.3), (2.4) and (2.12) satisfied).

2

2btdfED =

bdQ

=τ

2

tdE

DQ f=τ


09.03- 9


In the first instance the transverse displacements (w1) of the beam may be

calculated by the theory of bending, using the relationship (2.1). For example,fig. 2.5b shows the bending deformation of a simply supported beam with acentral point load W. The points a,b,c, ... lie on the centrelines of the faces andthe cross sections aa, bb, cc, ... rotate but nevertheless remain perpendicular tothe longitudinal axis of the deflected beam. It is obvious that the upper face iscompressed as the points a, b, c, ... move closer together, while the lower faceis loaded in tension.

The shear stress in the core at any section is τ = Q/bd (equation (2.15b) ). Thisis associated with a shear strain γ = Q/Gbd which like τ, is constant through thedepth of the core; G is the shear modulus of the core material. These shearstrains lead to a new kind of deformation illustrated in fig. 2.5c.

a) b)

a b c d e

a b c d eca b d e

a b d ex

W

L

w1

z

c) d)

c ea b d

a b d

W W

w2

w2'

w2'

Fig. 2.5. Deflection of sandwich beam.

On the centrelines of the faces lie the points a, b, c, ... . They are not movedhorizontally but in a vertical direction w2 due to shear strain. The faces and thelongitudinal centreline of the beam tilt, and the relationship between the slopeof the beam, dw2 /dx, and the core shear strain γ may be obtained from fig. 2.6.In this figure, which shows a deformation of a short length of the sandwich, thedistance d e is equal to d(dw2 /dx). It is also equal to c f, which in turn is equal toγ c.

.

.

.

...

d e

c

c d

a

b

f

ϒ

w2'

Fig. 2.6. Shear deformation of a beam.

Hence

dc

AGQ

dc

GbdQ

dc

dx

dw 2 === γ (2.16a)

Since the faces are assumed to be thin, c is about the same as d which meansthat w2' = γ and

VQ

AGQ

w2 ==′

(2.16b)

The product V is often referred to as the shear stiffness of the sandwich. (Theproduct also contains a factor called β but because of its rectangular shape, inthis case it is 1.) The displacement w2, associated with shear deformation on thecore, may be obtained by integration of equation (2.16a) in any particular pro-blem.

For example, in the simply supported beam with a central point load W, thetransverse force Q in the left-hand half of the beam is + W/2. Integration ofequation (2.16a) with Q = + W/2 provides the displacement:

L/2x 0constantx2VW

w2 ≤≤+=


09.03- 10


b/ Unsymmetrical loadIn the previous section it was assumed that during shear deformation all pointson the centrelines of the faces moved only in the vertical direction, as in fig. 2.5c.In general, it is possible for one face as a whole to move horizontally with respectto the other.

For example, a simply supported beam of span L with a uniformly distributedload q has a central bending deflection ∆

1 equal to + 5qL4/384 D. The bending

moment at the centre is + qL2/8 and the central shear deflection ∆2 is therefore +

qL2/8V. The total deflection ∆ at the centre is given by:

8VqL

D 3845qL 24

21 +=∆+∆=∆ (2.18)

In the same way expressions for total deflections are obtained for other cases.At the end of this chapter a few of the most usual load cases are presented. Themaximum values of bending moment and shear forces are presented and maybe used to give the stresses in the core and in the faces.

..

....

c'

a

d'

e'

f'c

d

c

d

e

f

b

γ

0γ w2

'

Fig. 2.7. Effect of γ0 on shear deformation.

The effect is illustrated in fig. 2.7, which is similar to fig. 2.6 in showing the axisof the beam at an angle w

2' to the horizontal as a result of pure shear deforma-

tion of the core. However, the upper face has also been displaced to the left, sothat the points cdef in figs. 2.6 and 2.7 now appear in new positions at c’d’ e’ f’.The angle cbc’ is denoted by γ

0 and the following relationships exist:

( ) dwdecccfccf 20 ⋅′==⋅−=′−′= γγHence

( )dcw 02 γγ −=′ (2.19a)

For other cases an elementary table of load cases can be used. Insert theappropriate value for Q in (2.16a) and use boundary conditions to integrate thewhole expression.

The constant vanishes because w2 = 0 at x = 0. The maximum value of w

2 occurs

at the centre of the beam, x = L/2, and is equal to:

4VWL

2 =∆

The total central deflection ∆ is therefore the ordinary bending displacement ∆1

with the displacement ∆2 superimposed:

4VWL

48DWL3

21 +=∆+∆=∆

In general the displacement of any symetrically loaded sandwich beam with anantiplane core and thin faces may be found by similarly superimposing the ben-ding and shear deflections w

1 and w

2. The bending deflections are found in the

usual way and the shear deflections by integrating equation (2.16a).It may beconvenient to integrate equation (2.16b) in general terms with the following result:

constantVMw2 += (2.17)

For a simply supported beam with the origin at one support the constant is alwayszero. Consequently the shear displacement diagram is the same as the bendingmoment diagram, with a factor 1/V applied to it.


09.03- 11


Fig. 2.8.

The boundary condition w2 = 0 at x = 0, L shows that the constant vanishes and

γ0 is equal to – M

0d/AGLc. Substitution for γ

0 in equation (2.19c) shows that the

transverse shear displacement w2 is zero everywhere. However, all the sections

through the core have rotated through an angle γ0 as in fig. 2.8. The shear strain

γ at all points in the core is given by equation (2.19a) as

c

d

AGL0

M

002w

c

d−==+

′= γγγ

The rotation γ0 is always zero when the beam is loaded in a symmetrical man-

ner, or when the relative horizontal displacement of the faces is prevented, forexample at a clamped end.

Consider, for example, a simply-supported beam with a moment M0 applied at

one end (fig. 2.8). The bending moment at x is – M0x/L, which value may be

inserted in equation (2.19c):

constantdcx0AGL

x0M2w +−−= γ

x

M0

L

M0L

M0

P P P P

β = 2 β = 1 β = 0,699 β = 0,5

Fig. 2.9. The Euler load for different cases

The Euler load represents the smallest value for an axial load P at which thestrut will not return to straight condition after being displaced in lateral direction.

In the case of a sandwich strut the occurring shear deformations reduce thestiffness of the strut and the buckling load will be smaller than the correspondingEuler load.

A pin-ended sandwich strut will be considered here. The flexural rigidity is givenby equation (2.5). When the axial thrust P reaches a critical value Pcr, the dis-placement consists of two superimposed displacements: w1 (bending displace-ment) and w

2 (displacement associated with shear deformation of core). The

buckled strut is shown in fig. 2.10. At a section x the bending moment M is,referring to equation (2.1):

( ) ″−=+=1

w1

D2

w1

wPM (2.20)

dc

0AGQ

2wOr γ−=′

(2.19b)

constantdc

x0AGM

2wOr +−= γ (2.19c)

Equations (2.16a) and (2.17) are merely special cases of (2.19b) and 2.19c).

BUCKLING OF SANDWICH STRUTSStandard analysis of uniform beams and struts has shown that instability appearswhen the axial load p reaches the value of the Euler load PE. The Euler load ishere presented in four different cases, the elementary Euler cases.

( )2LD

EPβ

2π=


09.03- 12


P

P

P

M

P

x

c c

a)

b)

w2

w1

′+′ 2w1w

Boundary conditions provide that C2 = 0 and if (w1 + w2) = 0 for x = 0 and x = L.This yields:

(2.25)

Equation (2.22b) now yields:

(2.26)

Where P represents the critical load Pcr of the sandwich strut. The expression isoften given in this equal form:

(2.27)

In which is easily seen (V = AG):

* when G is finite, Pcr is less than the Euler load* when G is infinite, Pcr is equal to the Euler load* when G is small, Pcr approaches the value of AG.

These formulas can be used for all cases in fig. (2.9) with the appropriate Eulerload inserted.

By differentiating (2.23) and inserting in the right-hand term of (2.21) the totaldeflection w1 + w2 will be obtained from

[ ]

( )P/V1x cosCxsinC

xcosCx sincP

Dww

21

22

21

121

+

++=

−−−=+

αα

αααα

(2.24)

21

2

EE

E

L

DP:where

/VP1

PP

π=

+=

32,1,nn(PI)αL ==

V

1

P

1

P

1

Ecr

+=

Fig. 2.10 Buckled strut with hinged ends.

Fig. 2.10 shows that P has a component P(w1' + w2') acting perpendicular to theaxis of the strut. This represents the transverse force. Corresponding to equation(2.16a) the shear force is related to w2 by:

(2.21)

The term w2' may be eliminated from equations (2.20) (differentiated once) and

(2.21) to yield a differential equation for w1.

0wαw 12

1 =+′″′ (2.22a)

where

(2.22b)

(2.22a) has a solution in the form:

(2.23)

( )[ ]P/V1DP

1

2

−=α

3211 CxcosCx sinCw ++= αα

V

wwPw

21

2

+

=′

′′


09.03- 13

BEAMS WITH FACES ON FOUR SIDESBOXED BEAMS

b

hdc

e

Fig. 2.11. Boxed beam

The expression for flexural rigidity in this case is:

12

3ecc

E12

)3ec3(bhf

ED +−

= (2.28)

This is the flexural rigidity for bending about the centroidal axis of the crosssection. Terms are the flexural rigidity of the box and the core respectively. If thesecond term amounts to less than 1% of the first it can be ignored. This meansif:

10013ec

3bh

cEf

E>−

(2.29)

The second term is of no importance. Practically this is usually the case.

STRESSES IN BEAMS WITH FACES ON ALL FOUR SIDESThe stresses will in this case be calculated in the same way as before. However,first the bending moment distribution between faces and core must be foundout. To evaluate an expression for the distribution of the bending moment ashort beam section can be studied.

R

Fig. 2.12.

The curvature is the same for the core and the face throughout the whole beam.Due to the theory of engineering stress in beams the curvature (κ) is given bythe expression:

DM

=κ (2.30)

HenceM

f = κE

fIf

Mc = κE

cIc

It is easy to see that Mc amounts to less than 1% of M

f when condition (2.29) is

fullfilled. In practical cases the ratio will be even smaller, a fact that leads to theassumption that the bending moment is taken up in the face material only.

The normal stress in the core is then approximately zero and in the faces thestresses are calculated by:

fE

D

Mzfσ ⋅= (2.31)

R1

=κ

FLEXURAL RIGIDITYTo get a really strong sandwich beam a ”boxed” section can be chosen. Withfaces on all four sides the shear stiffness will be higher and the shear deflectionwill be smaller, though not negligible.


09.03- 14


t

yTp

z

b'

Q

1τ

1τ

1τ

1τ

2τ2τ

2τ 2τ

3τ 3τ t

d

Fig 2.13 Shear stress distribution in a ”box”γ⋅= ff VQ

γ⋅= cc VQ

where the total shear force is:

cf QQQ +=

which means that Qc amounts to less than 1% of Q

f if

DEFLECTIONS OF BEAMS WITH FACES ON ALL FOUR SIDESAlso in this kind of beam the deflection consists of two parts, bending deflectionand shear deflection. In some case the shear deflection can be neglected, butthe following example shows that this is not always the case.

Example:A simple supported beam with faces on four sides is loaded by a concentratedload on the mid point of the beam. Face material is FRP (Fibre Reinforced Plastic)and the core is of Divinycell H 60. The deflection is given by expression (2.18).The first term is the bending deflection.

The flexural rigidity is given by (2.28) and with the length L = 1 m the bendingdeflection BD amounts to 3.1.10–6.P.

The maximum values for σf, i.e on the top and bottom, are obtained with maxi-

mum values for z.

The shear stresses depend on transverse forces. They are, in the same way asthe bending moment, taken up in both the core and the faces. A study of a thinboxed beam section show that the ability for the core to take up shear forces canbe ignored.

The shear deformation γ is the same for face and core, and is given by:

i

ii V

Q=γ (2.32)

where V contains a form factor β (see 2.35) , related to each part of the beam.For the core, β is assumed to be 1 and for the faces it is given by:

web

f

A

Aβ = (2.33)

where Aweb

is the cross-section area of the sides. This gives

1001

V

V

f

c < (2.34)

The condition will in most practical cases be fulfilled and Qc can be ignored. Thus

we have expressions for the different shear stresses in the faces. (See fig. 2.13)

( )

⋅

⋅+′

=

=⋅′

=

⋅′

=

Q8l

ddb2

Q4l

db

Q4l

db

y3

1y

2

y1

τ

ττ

τ


09.03- 15


The deflection caused by shear deformation is given by the second term. Herethe shear stiffness AG is denoted V. To calculate V we have to consider theshear stiffness of both the core and the side faces. When this is done the sheardeflection SD amounts to 1.94.10–7.P.

The ratio SD/BD is 0.064, which means that the shear deflection is about 6% ofthe bending deflection and should not be ignored.

With a very long and slender beam the shear deflection can be ignored but inother cases the shear deflection must be considered. The shear stiffness wascalculated:

f

fAfG

c

cAcGfVcVV ββ

+=+= (2.35)

where β is a factor mentioned earlier by (2.33). For the core βc is assumed to be

1 but for the faces βf is given by (2.33). Due to equation (2.34) the first term in

(2.35) can be ignored, leaving:

f

fAfGfVV β== (2.36)

BUCKLING BEAMS WITH FACES ON ALL FOUR SIDESIn the case of sandwich struts with faces on all four sides (”boxed struts”) thecalculations will be made in the same way as for ordinary sandwich struts. Theformulas (2.26) and (2.27) can be used, but with D calculating according to(2.28) and the shear stiffness AG, here denoted V, according to (2.36). Sincethe shear stiffness is highly increased compared with an open strut, the criticalload is higher, usually close to Euler load.


09.03- 16

BEAMS WITH ODD PROPERTIES

centroidd

d

c

b

a

t1

t2

Fig. 2.14. Dimensions of sandwich with faces of unequal thickness.

2t2E1t1E2t1t2E1E

2bdD

+= (2.38)

It is useful to note that equation (2.15) for the core shear stress is unaltered. drepresents as usual the distance between the centroids of the upper and thelower faces.

2t1t1dt

+

BEAMS CONSIDERED NARROW OR WIDEA beam is considered narrow when the width b is less than the core dept c.Then the lateral expansions and contractions of the faces in the y-direction,associated with the membrane stress in the x-direction, may take place freelywithout causing large shear strains in the core in the yz-plane. The stresses inthe faces are therefore mainly in one direction, and the ratio of stress to strain isequal to E. This has been assumed in the analysis of beams in this chapter.

The same argument does not apply to the local bending stresses in the faces.Each face is a thin plate in bending and the ratio of stress to strain is strictly E/(1–ν2). However, these stresses and strains are of secondary importance and itseems reasonable to adopt E throughout in order to avoid complications.

A beam is considered wide when the width b >> the core depth c. Then lateralexpansions and contractions of the faces in the y-direction are restricted by theinability of the core to undergo indefinitely large shear deformations in the yz-plane. In this case it is more reasonable to assume that the strains in the y-direction are zero. The ratio of stress to strain in the x-direction is therefore E/(1–ν2) for both the membrane stresses and the local bending stresses. Thisvalue should be used in place of E in all equations of this chapter when a beamis considered wide. Note that if a wide beam can curve freely in the yz-plane, forinstance if it is permitted to lift off its support, then E should be used in preferenceto E/(1–ν2).

BEAMS WITH DISSIMILAR FACESIf the faces are not of the same material or of unequal thickness the results inchapters ”Flexural Rigidity” and ”Stresses” have to be modified. The principalbeam equations are unchanged provided that the flexural rigidity is written asfollows:

( )32t2E31t1E12b

2t2E1t1E

2t1t2E1E2bd

D +⋅++

= (2.37)

where the suffixes 1 and 2 refer to the upper and lower faces respectively. It ishere assumed that condition (2.4) is fulfilled and the contribution from the coreto the flexural rigidity is negligible. If the local flexural rigidities for the faces arenegligible, i.e. if the condition (2.3) is fulfilled for each of the faces, the secondterm in (2.37) can also be ignored.

Then (which is assumed in the analysis in chapter ”Analysis method for sand-wich beams”) the flexural rigidity should be written as follows:

BEAMS IN WHICH THE CONTRIBUTION TO THE FLEXURAL RIGIDITYFROM THE CORE IS NOT SMALLWhen E

c is not small, i.e. when condition (2.12) is not fulfilled, some modification

must be made to use chapter ”Open beams (free sides)”. For example theexpression (2.2) must be used fully for flexural rigidity D. Since condition (2.12)is not satisfied, the shear stress τ and the shear strain γ are not to be consideredconstant throughout the depth of the core. The means equation (2.10) is validbut (2.13) is not.

2t1t2dt

+


09.03- 17

BEAMS WITH ODD PROPERTIES

.

..

. .

.

centroid

A

B

c

p

z

u

A'

B'

Fig. 2.15. Shear deformation of sandwich with stiff core

Now suppose that the core is replaced by a true antiplane (σx = σ

y = τ

xy = 0)

core with a shear modulus G’, different from Gx, but keep the former D. Thevalue of G’ is chosen so that the section ACB is deformed to the straight lineA’CB’. As the core is antiplane, E

c vanishes and the horizontal displacement

becomes:

′

=′4

tdcf

E

DG

QBB (2.43)

Since G’ has been chosen so that equations (2.39) and (2.41) give the sameresults for BB’, the antiplane core (G’) is exactly equivalent to the real corepermitting us to use the analyses in chapter ”Analysis method for sandwichbeams”. These analyses deal only with the core-edge displacements AA’, BB’and do not depend on the shape of the distorted section A’CB’. Therefore theequivalent antiplane core has a shear modulus as follows:

( )tct2c

f6E

cE1

GG

+⋅+

=′

(2.44)

The procedure is now to use the analysis in chapter ”Analysis method for sand-wich beams”, except that:

* D should be written as in equation (2.2)* G is replaced with G’.

This procedure yields the correct deflections and stresses in the faces. Toobtain the shear stress in the core equation (2.10) ((or (2.40)) should be used.

In fig. 2.15. a short length of a sandwich beam is shown undergoing shear defor-mation of the core.The section ACB has distorted into the curve A’CB’. Thetypical point p has moved a distance u to the right. At p the strain is γ = du/dzwhich gives the stress τ:

dzdu

G=τ (2.39)

Equation (2.10) and (2.39) may be combined and integrated to yield andexpression for u.

−+=

3

3z4

z2c2cE

2

tdzfE

GDQ

u (2.40)

For example the displacements AA’ and BB’ are obtained by writing z = ± c/2.

+=′

24

3ccEtdc4

fE

GDQBB (2.41)

The maximum shear stress is obtained by writing z = 0 in equation (2.10).

+=

8

2ccE

2

tdfE

DQ

maxτ (2.42)


09.03- 18

CALCULATIONS

1 Check out in chapter ”Beams considered narrow or wide” if the beamis to be considered wide or narrow. In the following calculations thevalid stress to strain ratio must be used.

2 If the faces are of unequal thickness the flexural rigidity is to be writtenas in equation (2.37).

3 Open beam Check if condition (2.12) is fulfilled for the core tobe considered weak, and if not,goto chapter ”Beamsin which the contribution to the flexural rigidity fromthe core is not small” for proper adjustments.

Boxed beam In most cases Gf >> G

c and the shear stresses are

taken up in the faces and are notconstant through-out the beam.

4 Flexural rigidityOpen beam:use (2.2) and check condition (2.3) and (2.4) forignoring terms.

Boxed beam:use (2.28) and check condition (2.29) forignoring the second term.

5 Shear stiffness Open beam: use (2.16)Boxed beam: use (2.35)

6 Stresses a. Look in the elementary table for the moment andthe transverse force.

b. Open beam: σf from (2.7a)

σc from (2.7b)

τc from (2.10)

and check condition(2.12)for ignoring the second term.

Boxed beam: σf from (2.31)

τf from Fig. 2.14

7 Deflections Open beams: see elementary table or use(2.16b) and (2.17)

Boxed beams: use elementary table or (2.16)and (2.17) with current rigiditiesaccording to point 4 and 5 above.

ANALYSIS METHOD FOR SANDWICH BEAMS


09.03- 19

SANDWICH PANELS

ASSUMPTIONSThe faces are assumed to be thin and of equal thickness.

y

z

b

a

x

t

t

0

c/2

c/2

Fig. 3.1. Dimensions of sandwich panel with equal faces

Because the faces are thin compared to the core it is assumed that c≈ d andthat the local flexural rigidity of the faces is negligible. This means that the nor-mal stress is constant throughout the faces. It is assumed that there are nostresses worth considering in the z-direction. The faces and the core are isotropic.The faces are assumed to be rigid in shear in yz- and zx-planes.

For the flexural rigidity of the panel, the core is assumed to be considerably lessstiff than the faces. Consequently Ec ~ 0 in the xy-plane which leads to the factthat they do not contribute to the flexural rigidity. The core shear stresses areassumed to be constant throughout the depth of the core.

Further, the deflections are assumed to be small. Accordingly, ordinary bendingtheory is valid and there is no strain in the middle plane of the panel formtransverse displacements, i.e.σ

y = ε

y = σ

x = ε

x = 0 for z = 0.

SIGN CONVENTIONSThe sign convention that will be used for plates is shown in fig. 3.2:

Myx

My

Mx Mxy

Nyx Nxy

Nx

Ny

Qy

Qz

q

x

z

y

dx = 1dy = 1

Fig. 3.2. Sign conventions for plates

The figure shows positive directions of bending and torsion moment (Mx, My,M

xy, M

yx), shear forces (Q

x, Q

y) and membrane forces (N

x, N

y, N

xy, N

yx).


09.03- 20

BENDING AND BUCKLING OFSANDWICH PANELS

BENDING AND BUCKLING OF SANDWICH PANELS SUPPORTED ON TWOSIDESFor sandwich panels supported on two opposite sides the theory and the formu-las are the same as for open sandwich beams provided the load is a uniformpressure. However, it must be noted that the panel is considered as a widebeam due to chapter ”Beams considered narrow or wide”. Therefore, in theanalysis E should be replaced by E/(1–ν2). Assumptions are made similarly.The conditions used in chapter ”Open beams (free sides)” for ignoring termswhen calculating flexural rigidity are the same.

From this it follows that in case of a sandwich panel supported on only two sidesthe reader is recommended to use chapter ”Open beams (free sides)” and thetheory for open beams with E replaced by E/(1–ν2).

BENDING AND BUCKLING OF PANELS SUPPORTED ON FOUR SIDESFor obtaining useful formulas, energy methods are applied to sandwich panelssupported on all four edges.

The method aims to find expressions for the total potential energy in the materialas a function of assumed displacements. The energy consists of two main parts:the strain energy U because of strain in core and faces of the deformed mate-rial, and the potential energy H because of movement of loads when deformingthe panel.

The method is also based on the fact that the total energy (U + H) will have aminimum value when the deflected plate is in equilibrium. Accordingly the totalenergy (U + H) will be minimized with respect to deflection due to bending andshear to find the critical load, stresses and deflections. In fig. 3.3 a part of adeflected panel is shown.

The centre line AG and the normal AE have both rotated an angle ðw/ðx. Becauseof shear deformation the line AF has rotated a smaller angle λðw/ðx, where λmay take any value between + 1 and 0. From this is obtained the shear strain inthe section (the angle EAF). λ = 1 means that the panel is rigid in shear and λ =0 that there is no shear stiffness in the panel.

. . .

x

z

z

a

A

∂w∂x

∂w∂xE

F D d2

c2

=

d2

c2

=

λ

γzx

Fig. 3.3. Section through deflected sandwich panel in zx-plane

( )xwλ1zx ∂∂

−=γ (3.1)

Since deformations are assumed to be small the displacement of F in the x-direction is:

xw

zu∂∂

−= λ (3.2)


09.03- 21


The shear strain in the xy-plane is:

( )yxw

z -xv

yu 2

xy ∂∂∂

+=∂∂

+∂∂

= µλγ (3.7)

It must be added that λ and µ are treated as being independent of x and y duringdifferentiation.

STRAIN ENERGYThe strain energy of an isotropic solid is given by integrating the strain energyover the volume:

dV)(2G

dV)ee2e(e2gEU

2zx

2yz

v

2xy

vyx

2yx

2

γγγ

ν

++

+++=

∫

∫

(3.8)

In the same way

( )yw

1yz ∂∂

−= µγ (3.3)

yw

zv∂∂

−= µ (3.4)

Where µ is the term corresponding to λ and v is the displacement in y-direction.

The strains in x- and y-direction are given by the displacements:

2

2

xx

wz

xu

e∂

∂−=

∂∂

= λ (3.5)

2

2y

ywz

yve

∂∂−=

∂∂= µ (3.6)

where g = 1–ν2.

This expression is to be used for core and faces respectively.

STRAIN ENERGY OF CORE, UCAccording to the assumptions, the strain energy from terms containing e

x, e

y

and γxy

will be zero. This leaves only shear strains γzx

and γyz

from (3.1) and (3.3)to be inserted in (3.8).

( )

( ) dxdyxw)(1

yw1

2dG

Vdxw)(1

yw1

2G

U

22

22

b

0

a

0

c

22

22

v

cc

∂∂

−+

∂∂

−=

=

∂∂

−+

∂∂

−=

∫∫

∫

λµ

λµ

(3.9)

where dV = dx dy dz.

STRAIN ENERGY OF FACES, UfAccording to assumptions γ

yz and γ

zx are zero. This leaves terms e

x, e

y and γ

xy to

be inserted in (3.8) For the lower face z is + d/2 and the strain energy here is:

dVyx

w)(4

d2

G

Vdyw

xw

4d2

xw

4d

xw

4d

2gEU

222

2

v

f

2

2

2

222

2

22

22

2

22

2

vlower

∂∂∂

++

+

∂

∂⋅

∂∂

+

∂

∂+

∂

∂=

∫

∫

µλ

λµνµλ

(3.10)


09.03- 22


POTENTIAL ENERGY OF APPLIED LOADSWhen a beam of length L is given a transverse deformation w, the ends of thebeam approach each other by an amount δ.

Fig. 3.4. Deformed beam and deformed element with length ds

ds in fig. 3.4 can be written:

( )2dw/dx1dx2dw2dxds +=+=

and then developed in a series. Serial development yields:

( )

+

+=+ ....

2

dxdw

21

1dx2dw/dx1dx

For a beam element with the length ds the ends approach each other by anamount ds – dx. The equation above gives:

dx2

dxdw

21

dxds

=−

The total approach δ will be obtained by integrating over the length of the beam.

dx2L

0 dxdw

21∫

=∂

Consider now a narrow strip of the plate in fig. (3.1) parallel with the x-axis andof width dy. In the same way the ends of this strip approach each other as theplate bends by an amount:

dx2a

0 xw

21

∫

∂∂

If a compressive force Nx is applied at the edge (x = 0 and x = a) in the plane of

the plate, then the force on the strip Nxdy and the change in potential energy as

the plate bends is:

∫

∂−

a

0dx

2

dxw

2

dyxN

P P

δ

dx

ds

ds - dx

dw

The total strain energy of both faces, Uf, is obtained by integrating over the thickness

t and doubling. It is also convenient to write G = E/{2(1+ν)}.

( )dxdy

yxw

21

yw

xw2

yw

xw

4gtEdU

222f

2

2

2

2

f

2

2

22

2

2

22

a

0

b

0

2

f

∂∂∂

+

−+

+∂

∂⋅

∂

∂+

∂

∂+

∂

∂

= ∫ ∫µλ

ν

λµνµλ

(3.11)


09.03- 23


The total decrease in potential energy for the force Nx, is obtained by integrating

from y = 0 to y = b.

dydxxw

2N

V2a

0

b

0

x1 ∫ ∫

∂∂

−=

If the plate also supports a uniform transverse pressure q in the z-direction, thedecrease in potential energy V

2 for the load is:

dxdywgVa

0

b

02 ∫ ∫−=

The displacement w for a simply-supported rectangular plate may be expressedby sums of trigonometric functions:

bynsin

axmsinaW

1nmn

1m

ππ∑∑∞

=

∞

=

= (3.12)

Where amn

is the amplitude of the (m, n)th mode of deformation. This expressionsatisfies the boundary conditions of a simply-supported plate.

The total energy of the system, (U + V), is obtained by adding the expressionsfor U and V respectively and substituting w by the series (3.12). Consider forexample the first term of U

c.

( ) dxdyxw1

2Gd1U

b

0

2a

0

a

c ∫∫

∂∂

−= µ (3.13)

Substituting w according to (3.12) gives:

( ) dxdyb

yncosa

xmsinb

na12

Gd1U2b

0 1m 1nmnmn

a

0c ∫ ∑ ∑∫

−=

∞

=

∞

=

πππµ

(3.14)

When the series is squared the integrals of the cross-product terms vanish becauseof the ortogonal properties of the chosen functions for w. Only the squared termsare left.

( ) dxdyb

yncosa

xmsinb

na12

Gd1Ub

0 1m 1n

222

mn2

a

0c ∫ ∑ ∑∫

−=

∞

=

∞

=

πππµ

(3.15)

The series may be integrated term by term. The following integral is equal to ab/4 for all values of m and n.

4abdxdy

byncos

axmsin 2

a

0

b

0

2 =∫ ∫ππ

(3.16)

Hence

( )4ab

bna1

2GdU

2

222mn

2

1nmn

1m1c ⋅−= ∑∑

∞

=

∞

=

πµ (3.17)

All the energy terms are to be treated in the same way. In this process it is usefulto have knowledge about the following relationships, derived from equation (5.12).

4ab

amadxdy

xw

4

442mn

2a

0

b

02

2⋅=

∂

∂∑∑∫ ∫

π

4ab

anadxdy

yw

4

44

mn

2a

0

b

02

2⋅=

∂

∂∑∑∫ ∫

π

4ab

banmadxdy

yw

xw

22

2222

mn2

2a

0

b

02

2⋅=

∂∂

∂

∂∑∑∫ ∫

π


09.03- 24


4ab

ba2madxdy

yxw

22

22n2

mn

2a

0

b

0

2⋅=

∂∂

∂∑∑∫ ∫

π

( ) ( ) ( ) 0VUa

VUµ

VUλ mn

=+∂∂

=+∂∂

=+∂∂

(3.19)

These equations can be used to determine the values of amn

, λ and µ. Since the(m, n)tn value of a

mn, λ and µ only appears in the (m, n)th mode, (U + V) in

equation (3.19) could be replaced by (U +V)mn

only.

It is easier to see the connections and follow the line of equations if the totalenergy is written in the form:

0yxxy2

yy2

xxmn B2B2B2BBBV)(U +++++=+ µλλµµλ (3.20)Where:

2mn22

22

4

4

22

2

1xx abanm

21

amEA

amGAB

⋅

−++=

ν(3.21a)

(m, n both odd)

= 0 (otherwise)

By substituting these values in the former expressions for Uc, U

f, V

1 and V

2 the

following expressions are obtained for the energy terms.

( ) ( ) ( ) 2mn22

22

22

1mnc abn1

am1GAU

−+−= µλ (3.18a)

( ) ( ) 22222

222

22

4

42

4

42

2mnf mnabanm

21

banm2

bn

amEAU

−

−+++= µλννλµµλ

(3.18b)

( )2

22mn

2xmn1 a

m4

aba2

NV ⋅= π (3.18c)

( )mnaba4qV 2

mn2 ⋅−=

π(3.18d)

Where:

ab16gtdAandab

8dA 4

2

22

1 ππ ==

For simplicity, only the (m, n)th mode is shown above and there are no suffixes onλ and µ. There are different values for each mode m, n.

2mn22

22

4

4

22

2

1γγ abanm

21

bnEA

bnGAB

⋅

−++=

ν(3.21b)

2mn22

22

2xy abanm

21EAB

+=

ν(3.21c)

2mn2

2

12mn2

2

1x abnGABya

amGAB −=−= (3.21d, e)

4ab

bnadxdy

yw

2

222

mn

2a

0

b

0

⋅=

∂∂

∑∑∫ ∫π

mnab4adxdyw 2

mna

0

b

0

⋅=∑∑∫ ∫ π

Evidently (U + V) is a function of amn

, λ and µ. If the plate is to be in equilibrium, (U+ V) has to be stationary with respect to each of these variables. From this itfollows that for each mode the following conditions must be fulfilled.


09.03- 25


( )mn212mn22

2

2

10 VVabn

amGAB ++

+= (3.21f)

2

2

btd

GE

2gπρ = (3.25c)

The variables λ, Ω and ρ are non-dimensional. λ and Ω take different values fordifferent modes. ρ is constant and represents the ratio of the flexural rigidity Etd2/2g and the shear stiffness Gd.

When the expression for λ is inserted in (3.24) an expression for (U + V) asfunction of a

mn is given.

mnaba4q

am

4aba

2N

1baa

8GdV)(U 2

mn2

22mn

2x2

2mn

2mn

ππ

ρρ

π −−Ω+

Ω=+

(3.26)For equilibrium (3.26) must be stationary with respect to a

mn.

0mnab4qa

am

4abN

ρΩ1ρΩ

ba

4GdV)(U

a 2mn22

2x

22

mnmn

=−

−

+=+

∂∂

πππ

(3.27)

Equation (3.19) then gives:

( ) ( ) 0BBλBλVU

21VU

λ21

xxyxx =++=∂+∂

=+∂∂ µ (3.22a)

( ) ( ) 0BλBBµVU

21VU

µ21

yxyyy =++=∂+∂

=+∂∂

µ (3.22b)

If equations 3.22 are multiplied by λ and µ respectively and then added, then:

0BB2BBB yxxy2

yy2

xx =++++ µλλµµλ ( 3.23)

(3.23) inserted in (3.20) leaves only:

0yxmn BBBV)(U ++=+ µλ (3.24)

By solving (3.22) and (3.22b) it is possible to show that in this particular problemthe solution of equations (3.22) is such that µ = λ.

Substitution for µ by λ in equation (3.22) gives the following result for λ:

(3.25a)

where

(3.25b)

EDGE LOAD, NxSuppose that the transverse pressure q is zero. The critical load is then thevalue of N

x which causes the panel to buckle.

If the panel buckles in the (m, n)th mode amn

is non-zero. Equation (3.27) is thensatisfied only when:

(3.28)

Pxmn

is defined as the critical edge load per length unit which causes buckling inthe (m, n)th mode. For any given m the lowest critical load is obtained for n = 1.Equation (3.28) can then be written as follows:

Ω++=

+=

ρλ

11

BxBxB

yx

x

22

22n

abm

+=Ω

xmn

22

2xP

1ba

mGdN =

Ω+Ω

=

ρρ


09.03- 26


1Kb

DP

22

2

xmnπ= (3.29)

where:

( ) ( )[ ]( )

++

+=1mb/a 1

a/mbmb/aK2

2

1ρ (3.30)

And D2 is the flexural rigidity of the sandwich:

−

==2

22f

212

Eftd2gtdED

ν (3.31)

Notice the factor g = (1–ν2) in the expression for the flexural rigidity of the plate.It originates from the conditions for a beam to be considered narrow or wide in”Beams with odd properties” (see chapter ”Beams considered narrow or wide”and is always to be in the expressions in this chapter because panels are naturallyconsidered wide.

It is observed that if shear rigidity is infinite, ρ vanishes and equation (5.30) isidentical with the result for buckling of a plate not subjected to shear deformations.

Fig. 3.5 shows the value of K1 plotted against a/b for m = 1,...,4 and four different

values of ρ (0, 0.1, 0.2 and 0.4). Since only the lowest value of K1 is of interest,

only the lower envelopes of K1 for m = 1,...,4 are to be used. Since the figure

only shows four curves (m = 1,...4) it should be noticed that when a/b >> 1 thelower envelope of the curves get close to a straight horizontal line. The diagramsare valid for 0 < a/b < ~ 3.5 and for higher values the lowest value in the diagramcan be used.

The procedure is to read K1 from fig. 3.5 and then insert the value in equation

(3.29) to determine the buckling load. If a value of ρ is obtained that it does notfit with the diagrams in fig. 3.5, the value for K

1 has to be calculated with (3.30).

0

1

2

3

4

5

0 1 2 3 4 5a/b

Fig. 3.5a. Buckling coefficient K1 plotted against a/b for m = 1..4 and ρ = 0.Simply supported isotropic sandwich with thin faces.

0

1

2

3

4

5

0 1 2 3 4 5a/b

Fig. 3.5b. Buckling coefficient K1 plotted against a/b for m = 1..4 and ρ = 0.1.Simply supported isotropic sandwich with thin faces.


09.03- 27


Fig. 3.5d. Buckling coefficient K1 plotted against a/b for m = 1..4 and ρ = 0.4. Simply supported isotropic sandwich with thin faces.

UNIFORM PRESSURE, gNow suppose that the edge load N

x is zero and that the load is q, a uniform

pressure.

Substitution of ρ from equation (3.25c) provides an expression for the amplitudeof the (m, n) th mode.

21

2mnD6

416qbmna

ΩΩ+⋅= ρ

π if m, n are both odd

(3.32)= 0 , otherwise.

In this case too, an infinite shear rigidity in the core causes ρ to vanish and thenequation (3.32) corresponds to the standard result for bending of a plate notsubjected to shear deformation.

Equation (3.32) may also be written in the form:

Ω⋅

Ω=

mnGd4

216qb2

2mnD6

416qbmna

ππ , if m, n are both odd (3.33)

= 0 , otherwise

The terms on the right side represent bending and shear deformations respec-tively. The ratio of the shear deformation to the bending deformation is

Ω⋅−

ΩGd2b

2D2

or,1

or,π

λλρ

To obtain the deflection w the value amn

must be inserted in equation (3.12). Themaximum deflection w

max is at the centre of the panel, x = a/2, y = b/2, and is

obtained by summation:

( ) ( )

+

⋅

−−

−−

=2

1

mn2

1n12

1m1

2D6

416qbmaxW

Ω

ρΩΣΣπ (3.34)

0

1

2

3

4

5

0 1 2 3 4 5a/b

0

1

2

3

4

5

0 1 2 3 4 5a/b

Fig. 3.5c. Buckling coefficient K1 plotted against a/b for m = 1..4 and ρ = 0.2.

Simply supported isotropic sandwich with thin faces.


09.03- 28


For practical use this expression may be written:

( )212

4

max Dqb

w ββ ρ+= (3.35)

where:

( ) ( )2

2

1n

2

1m

61 mn

1116βΩ

−−ΣΣ=

−−

π, m, n odd

( ) ( )Ω

βmn

11162

1n2

1m

62

−−

−−ΣΣ=

π, m, n odd

β1 and β2 can be read from fig. 3.6 for panels with various a/b ratios.

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

0 1 2 3 4 5a/b

ß2

ß1

Fig. 3.6. Coefficients β1 and β2. Simply supported isotropic sandwichwith thin faces.

The stresses in the faces and the core may also be obtained from (3.32). Forexample the normal stresses in the face (in x-direction) are equal to (E/g).(e

x+νe

y).

The strains ex and e

y are defined in (3.5) and (3.6). By inserting z = ± d/2 the

stresses in the x- and y-direction are

∂

∂+

∂

∂±=

2

2

f2

2

xw

x

w2g

Ed

yσ νλ (3.36a)

∂

∂+

∂

∂±=

2

2

f2

2

yx

w

y

w2g

Ed νλσ (3.36b)

The shear stress τxy

in the faces is equal to[E/(2(1+ν))] gxy

where the strain γxy

isgiven by (3.7). For z = ± d/2 the shear stress is:

( ) yxw

ν12Ed 2

xy ∂∂∂⋅

+±=

λτ (3.37)

The shear stress τzx

in the core is equal to Gγzx

. When the strain γzx

is given by(3.1).

( )xw

1Gzx ∂∂

−= λτ (3.38a)

and similarly

( )xw

1Gyz ∂∂

−= λτ (3.38b)

Usually the maximum stresses are of interest. For practical use it is convenientto write the expressions in the same way as equation (3.35).

.10-2

It can be shown that the normal stresses in the faces are maximum at the centreof the panel (x = a/2, y = b/2). The shear stress in the faces is the highest at acorner (x = 0, y = 0), the core shear stress τ

zx is highest in the middle of the sides

of length b (x = 0, y = b/2) and the core shear stress τyz

is highest in the middleof the sides of length a (x = a/2, y = 0). The results may be summarized in thefollowing forms:


09.03- 29


( )4f32

x dtqb ββσ ν+= (3.39a)

( )3f42

y dtqb ββσ ν+= (3.39b)

( ) 5f2

xy 1dtqb βντ −= (3.39c)

6zx dqb β=τ (3.39d)

7yz dqb β=τ (3.39e)

where

( ) ( )2

2

2

21m

43 a

bnm11

16

⋅⋅Ω

−−ΣΣ=

−−2

πβ

1n

(3.40a)

( ) ( )mn11

16

2

21n

21m

44⋅

Ω

−−ΣΣ=

−−

πβ (3.40b)

245 a

b16

Ωπβ ⋅ΣΣ= (3.40c)

( )ab

n 1-16 2

1n-

36⋅⋅ΣΣ=

Ωπβ (3.40d)

( )Ω

⋅ΣΣ=m 1-16 2

1-m

37 πβ (3.40e)

Fig. 3.7 shows β3 – β

7 plotted against a/b. All the stresses are independent of

shear stiffness of the core and it is possible to show that the results in equation(3.39) and (3.40) are the same as when the core shear deformation is ignored.

Results for a simply-supported rectangular panel can therefore also be used tocalculate the stresses in a sandwich panel.

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

0 1 2 3 4 5a/b

ß4

ß5

ß6

Fig. 3.7a. Constants β3 – β5. Simply supported isotropic sandwich with thin faces.


09.03- 30


ß7

ß6

0.0

0.1

0.2

0.3

0.4

0.5

0 1 2 3 4 5a/b

Fig. 3.7b. Constants β6 and β7. Simply supported isotropic sandwich with thin faces.

EDGE LOAD AND UNIFORM PRESSURE ACTING SIMULTANEOUSLY

When the uniform transverse pressure and the compressive edge load ρ perunit length act simultaneously, the value of a

mn can again be obtained from

equation (3.27). The expression for amn

is:

( )xmnP/P1

0mnamna

−= (3.41)

where (amn

)0 is the amplitude when P is zero, given by equation (3.32) and P

xmn

the critical load given by equation (3.29). Of course the expression is for the (m,n)th critical load.

The practical effect of this load arrangement is to multiply each term in the seriesfor the β -functions by a factor (1 – P/P

xmn)–1. Because P

xmn depends on the ratio

ρ, stresses in the panel are no longer independent of the shear stiffness.


09.03- 31

PANELS WITH ODD PROPERTIES

PANELS WITH DISSIMILAR FACES

When the panels have faces of unequal thickness, or are of different materials,a few modifications have to be made. The buckling and bending equations (3.29)and (3.35) are unchanged provided both faces have the same Poisson’s ratio ν

f

and the following alterations are made:

( ) ( )22112f

22121

2tEtE1

dttEED

+⋅−=

ν (3.42)

( ) ( )22112121

2f

2

2

tEtEG

cttEE

1b +⋅

−=

νπρ (3.43)

CALCULATIONS

ANALYSIS METHOD FOR PANELS SIMPLY SUPPORTED ON FOUR SIDES

1: If the faces are of unequal thickness then modifications according to chapter”Panels with dissimilar faces” have to be made.

2: Edge load, Nx. (buckling load)

The buckling load is given by (3.29) with flexural rigidity from (3.31), K1

from fig. 3.5 as a function of a/b and ρ from (3.25c).

3: Uniform pressure, q. (Deflection and stresses)

Maximum deflection wmax

is given by (3.35) with flexural rigidity from (3.31),the constants β

1 and β

2 from fig. 3.6 as a function of a/b and r from (3.25c).

The stresses are given by (3.39a-e) with the constants β3 – β

7 from fig.

3.7 as a function of a/b.

4: Edge load and uniform pressure acting simultaneously.

Instructions written in chapter ”Edge loading and uniform pressure actingsimultaneously”.


09.03- 32

EXAMPLES

INTRODUCTIONIn this section a few calculations are made to exemplify the use of the analysismethods presented in chapters ”Analysis method for sandwich beams” andAnalysis method for panels simply supported on four sides”. The expressionsreferred to are easy to find due to their numbering.

Example 1: Beam with concentrated load and simply supported endsAn open beam have the following measurements:

L = 0.5 mb = 0.05 mh = 0.054 mtf = 2 mm

The faces are of equal thickness and are made of aluminium 4054-7 with Ef =

61 GPa. The core is made of DIVINYCELL H 45 with the following properties:E

c = 45 MPa

Gc = 18 MPa

The beam is simply supported at both ends and the load is a point load W = 25kg at a = 0.25 . L = 0.125 m (see fig. 4.1). L≥a.

a L - a

W

Fig. 4.1.Find out σ

f, τ

c and the deflection at L/2 = 0.25 m. Solution according to chapter

3: Condition (2.12) is checked:

The condition is fulfilled Ec is considered small.

4: Flexural rigidity.Conditions (2.3) and (2.4) are checked:

⇒ condition (2.3) is fulfilled.

⇒ condition (2.4) is fulfilled.

This leaves from equation (2.2) only the expression (2.6) fothe flexural rigidity D.

5: Shear stiffness.

6: a/Elementary table gives:

”Analysis method for sandwich beams”.

1: b < c ⇒ The beam is considered to be narrow.

2: The faces are of equal thickness.

1002541050

1052

1050

102

1040

10614

3

3

3

3

6

9>=

⋅

⋅⋅

⋅

⋅⋅

⋅

⋅⋅

−

−

−

−

5.7726102

10523

3>=

⋅

⋅−

−

( )( )

16.766.01050

1052 10 2

1040

106123

233-

6

9>=

⋅

⋅⋅⋅⋅

⋅

⋅−

−

( ) 223 3-3-9 Nm8247.22

1052 10 210501061D =

⋅⋅⋅⋅⋅⋅=

−

N10391015105210 50 bdG V(2.16) 363-3 ⋅=⋅⋅⋅⋅⋅==⇒ −

( ) ( )Nm15.33

0.50.250.50.125245.25

L

L/2)LWa(L/2)M =

−⋅⋅=

−=


09.03- 33

EXAMPLES

4: Flexural rigidity.D = 8247.2 Nm2. (see example 1)

5: Shear stiffness.V = 39.103 N (see example 1)

6: a/Elementary table gives: ( ) Nm10.4224

20.5310L/2M =

⋅=

Q(L/2) = 0

b/Stresses.σ

f, max from (2.7a) with z = h/2.

( )MPa2.0491061

8247.2

0.054/210.24fσ =⋅⋅

⋅=

(2.13) gives τc(L/2) = 0 as Q(L/2) = 0.

( ) ( ) 61.31N-0.5

0.125245.25

L

Wa(L/2)Q =

⋅−=−=

b/Stressesσ

f,max is obtained with z = h/2

( )MPa3.0691061

8247.2

/23105415.33fσ:(2.7a) =⋅⋅

−⋅⋅=

τc is constant throughout the core as the condition (2.12)

is fulfilled.

( )kPa23.58

2

31052310291061

8247.261.31-

c:(2.13) =−⋅⋅−⋅⋅⋅

⋅=τ

7: Deflection.Elementary table 1 gives:

( ) ( )( )

( ) mm0.393310392

0.125245.25

2V

WaL/2

2w

mm0.05320.125420.538247.248

0.125245.25

24a23L48D

WaL/21w

=⋅⋅

⋅==

=⋅−⋅⋅⋅

⋅=

=−⋅=

Total deflection w

mm0.452w1ww =+=

Example 2: Beam with uniform pressure and clamped ends

The same beam as in example 1 is here loaded by a uniform load q = 1 kN/mand the ends are clamped.

Find out σf, τ

c and deflection at L/2 = 0.25 m.

Solution according to chapter ”Analysis method for sandwich beams”.

1: ⇒≤ cb The beam is considered to be narrow.


3: Condition (2.12) is fulfilled (Ec is small). (see example 1)


09.03- 34

EXAMPLES

At the ends N2502

0.5310

2

qLQ =

⋅=±= , which with

( ) kPa96.152

0.0520.00291061

8247.2

2500Lc

gives(2.13)

=⋅⋅⋅

⋅==τ

Conclusion: it is important to study the shear force.

7: Deflection.Elementary table gives:

( ) mm0.0208247.2384

20.5310L/21w =⋅

⋅=

( ) mm0.801310398

20.5310L/2 2w =

⋅⋅

⋅=

Total deflection w.

mm0.222w1ww =+=

Example 3: Boxed beam with concentrated load and simply supportedendsA boxed beam has the following measurements:

L = 0.5 mb = 0.05 mh = 0.054 me = 0.05 mc = 0.046 mtf = 2 mm

The faces are of equal thickness and are made of FRP with the followingproperties:

Ef = 12 GPaGf = 4.8 GPa

Load and support are the same as in example 1.

Find out σf, max

, τc, max

and the deflection at L/2 = 0.25 m.

Solution according to chapter ”Flexural rigidity”.



3: Gf >> G

c.

4: Flexural rigidity.Condition (2.29) is checked:

100111130.050.046

30.0540.0561040

91012>=−

⋅

⋅⋅

⋅

⋅

2Nm2123.212

30.050.04630.0540.0591012D =

⋅−⋅⋅=

5: Shear stiffness.

(2.33) gives: 20.0020.052

0.0020.054cβ =⋅⋅

⋅⋅=

then (2.35) gives:

N31034.51

0.0500.04661015

cβcAcG

cV ⋅=⋅⋅⋅

==

( )N310960

2

0.04620.05420.0029104.8

fβfAfG

fV ⋅=⋅+⋅⋅⋅⋅

==

Which means that the expression (2.28) with the second termignored can be used for the flexural rigidity.


09.03- 35

EXAMPLES

Example 4: Panel with uniform pressure and simply supported edgesA panel has the following dimensions:

a = b = 3 m h = 70 mmt = 5 mm c = 60 mm

The faces are of equal thickness and are made of FRP with Ef = 12.0 GPa and

νf = 0.25. The core has the following properties: E

c = 200 MPa, G

c = 80 MPa and

νc = 0.30. The panel is simply supported at all edges and the load is a uniform

pressure q = 10 kPa.

Find out the maximum deflection and σf. Solution according to chapter ”Analysis

method for panels simply supported on four sides”.


3: Flexural rigidity from (3.31).

( )2Nm310135.2

20.2512

0.0650.00591012.02D ⋅=

−⋅

⋅⋅⋅=

ρ from (3.25c) 0.0285

23 6108020.2512

0.0650.005910122=

⋅⋅⋅

−⋅

⋅⋅⋅⋅=πρ

β1 and b

2 are given in fig. 3.6 with a/b = 1

β1 = 0.42.10–2 β

2 = 0.73.10–2

Then (3.35) gives

( ) mm25.62100.740.0292100.405310135.2

4331010maxw =

−⋅⋅+−⋅⋅⋅

⋅⋅=

σf from (3.39a) with β

3 and β

4 from fig. 3.7.a.

β3 = 0.0371 β

4 = 0.0385 which gives

( ) MPa12.940.03850.250.03710.0050.065

2331010yσxσ =⋅+

⋅

⋅⋅=

(2.34): 100

10.035

310960

31034.5

fV

cV >=⋅

⋅=

The condition is not fulfilled, which means that the whole of expression(2.35) must be used.V = V

c + V

f = 994.5.103

6: Stressesa/ M(L/2) = 15.33 Nm Q(L/2) = -61.31 Nb/ (2.31) gives with z = h/2 and I = I

y = D/E

f = 0.1769.10–6

MPa2.346100.1769

2/0.05415.33fσ =−⋅

⋅=

τmax

is given by fig. 2.13.

( )

( )MPa0.3361.310.052

6100.17698

0.0520.0482

Qdy81

db23max

=−⋅−⋅⋅

+⋅=

=⋅+′

==ττ

7: DeflectionElementary table 1 gives:

( )

mm0.20720.125420.532123.248

0.125245.25

24a23L48DWa

L/21w

=

⋅−⋅⋅

⋅⋅

=

=

−⋅=

( ) mm0.015310994.52

0.125245.25

2V

WaL/22w =

⋅⋅

⋅==

Total deflection w:

mm0.222w1ww =+=


09.03- 36

Example 6: Beam with concentrated load, simply supported ends and stiffcoreThe same beam as in example 1 and with the same dimensions. The face ma-terial is the same but the core is made of Divinycell H 130.

L = 0.5 m Ef = 61 GPa E

c = 130 MPa

b = 0.05 m Gc = 50 MPa

h = 0.054 mtf = 2 mm

The load is a concentrated load W = 25 kg (245.24N) at a = 0.25 L = 0.125 m.

Find out σ f, σ

c, τ

c and deflection at L/2 = 0 .25 m.Solution according to chapter

”Flexural rigidity”.


Example 5: Panel with edge load Nx and simply supported edgesThe same panel as in example 4 but in this case the load is an edge load. Findout the buckling load.

Solution according to chapter ”Analysis method for panels simply supported onfour sides”.


2: Flexural rigidity from (3.31)D2 = 135.2

.103 Nm2 (see example 4)ρ is given from (3.25c)ρ = 0.0285 (see example 4)

Then K1 is obtained from an interpolation between fig. 3.5a and 3.5b.a/b = 1 ⇒ K1 = 3.78

The critical load is then given by (2.29):

kN/m5603.7823

310135.2xmnP

2

=⋅⋅⋅

=π

EXAMPLES


3: Condition (2.12) is checked.

10078.080.05

0.052

0.05

0.002610130

910614


09.03- 37

EXAMPLES

( ) mm 0.05320.125420.538319.048

0.125245.25L/21w =

⋅−⋅⋅

⋅⋅

=

( ) mm 0.1193101292

0.125245.25L/22w =

⋅⋅

⋅=

Total deflection: w = w1 + w

2 = 0.17 mm

Example 7: Beam with concentrated load, simply supported ends and facesof unequal thicknessesThe same load case as in example 1, but now the faces are of unequalthicknesses. The thickness of the upper face is still 2 mm but the lower is 4 mm.The faces are made of aluminium 4054-7 and the core is made of Divinycell H40. Measurements and properties:

L = 0.5 m Ef = 61 GPa E

c = 40 MPa

b = 0.05 m tf1 = 2 mm G

c = 15 MPa

h = 0.056 m tf2 = 4 mm

c = 0.05 m d = 0.053 m

Here the suffixes 1 and 2 represent the upper and the lower face respectively.

The beam is simply supported at both ends and the load is a concentrated loadW = 25 kg (245.5 N) at a = L/4 = 0.125 m.

Find out tc and deflection at L/2 = 0.25 m. Solution according to chapter ”Flexural

Rigidity”.

1: ⇒≤ cb The beam is considered to be narrow.2: The faces are of unequal thicknesses and therefore the flexural

rigidity shall be written as in (2.37).

3: Condition (2.12) is checked (for the thinner face) in example 1and is fulfilled.

4: Flexural rigidity.Condition (2.3) is checked for both of the faces.

5.7726.500.0020.053

5.7713.250.0040.053

>=>=

The condition is fulfilled.Condition (2.4) is checked.

16.768.5430.05

20.0530.00261040

91061>=

⋅⋅

⋅

⋅

which means that the condition is fulfilled for both of the faces.Then the flexural rigidity is given by (2.38):

2Nm31011.420.004910610.00291061

0.0040.00229106120.0530.05

D ⋅=⋅⋅+⋅⋅

⋅⋅

⋅⋅⋅

=

5: Shear stiffness.(2.16) gives:

N31039.75610150.0530.05cbdGV ⋅=⋅⋅⋅==

( )MPa3.0491061

8319.00.054/215.33

f =⋅⋅⋅

=σ

(2.7b) gives with z = c/2:

( )kPa6.00610130

8319.00.05/215.33

cσ =⋅⋅⋅

=

(2.10) gives with z = 0:

kPa23.684

20.052

61013020.0520.00291061

8319.061.31

c =

⋅

⋅+

⋅⋅⋅⋅

−=τ

7: DeflectionElementary table gives:


09.03- 38

EXAMPLES

6: a/Elementary table gives:Q(L/2) = – 61.31 N (see example 1)

b/τc is according to chapter ”Beams with dissimilar faces” given by (2.15):

kPa23.140.0530.05

61.31c −=

⋅

−=τ

7: Deflection.Elementary table gives:

( ) mm0.038)20.125420.5(331011.4248

0.125245.25L/21w =⋅−⋅⋅

⋅⋅

⋅=

( ) mm 0.38631039.752

0.125245.25L/22w =

⋅⋅

⋅=

Total deflection w: w = w1 + w

2 = 0.42 mm.


09.03- 39

FINITE ELEMENT METHOD FORANALYSIS OF SANDWICH

ANALYSIS OF SANDWICH BY FEMIn this chapter a metod to analyse sandwich stryctures by the Finite ElementMethod (FEM) is presented. The method is general and can be peformed usingany of the commercially available FE-codes.

To obtain the right stiffness in shear in the FEM-model the material properties ofthe core have been modified according to the geometry of the sandwich. Themethod has been verified by comparing the results of a FEM-model of a panel toanalytical calculated displacements, stresses and buckling loads. There are alsoverifying examples corresponding to the analytical examples in the chapter”Examples”.

Sandwich panels have been treated in detail in chapter ”Sandwich panels”. Therethe following assumptions are made:

1. The stresses perpendicular to the plane of the panel are negligible both inthe core and in the faces.

2. The material in both the core and the faces is isotropic.

3. In most cases the modulus of elasticity in the core is so low that the cont-ribution to bending stiffness is negligible.

4. The displacements are small, meaning that the theory of bending is valid.

5. The faces are thin compared to the core. This means that the local flexuralrigidity can be ignored and that c≈d (see fig. 5.1).

PRESENTATION OF METHODIn principle a FEM-model of a sandwich structure can be built up in three differentways.

a/ Both faces and core are modelled of solid elements.b/ The faces are modelled of shell elements and the core of solid elements.

These are shown in fig. 5.1.c/ Shell element with math model off shore.

Fig. 5.1. Different ways of modelling sandwich by FEM. To the left with solid elements andto the right with a combination of solid elements and shell elements.

If only solid elements are used the elements will have the same thickness as thefaces and the core respectively. The elements modelling the faces will be veryextended, as the faces are thin. However, the solid elements give a better resultif they are cubic, i.e. all sides have about the same length. Consequently a verylarge number of elements is required if the sandwich has thin faces. Consequentlythe FE-model will have a great number of degrees of freedom and the calculationtime will be unacceptable.

t

c d

t


09.03 40

FINITE ELEMENT METHOD FORANALYSIS OF SANDWICH

To avoid this problem and to reduce the number of degrees of freedom the modelcan be built up of shell elements representing the faces and of solids representingthe core. As the nodes of a shell element are located in a plane in the middle ofthe element they should be placed at distance (c + d)/2 from each other. This isin accordance with assumption no 5 in chapter ”Analysis of sandwich by FEM”.The shear stiffness of the core will be reduced as the thickness of the core hasincreased by (d – c)/2 in the model. To compensate for this the shear modulus ofthe core can be increased to obtain the right stiffness in shear. The correctedshear modulus is given by the following expression:

(5.1)cG2ccd

cGcc)/2(dc

corrc,G ⋅+

=⋅−+

=

The stiffness in tension will be increased when the thickness of the core isincreased; the modulus of elasticity should be corrected in the same way whichgives:

cEc d

2ccorrc,E ⋅

+= (5.2)

In reality the modulus of elasticity has a very small influence on the deflectionsof a sandwich structure. However, it must be noted that if the modulus of elasticityand the shear modulus are corrected, the core material has to be modelled asorthotropic ((the expression G = E/(2 (1+ν)) is not valid)). Poisson’s ratios forfaces and core respectively are assumed to be unaffected.


09.03- 41

ELEMENTARY TABLES

BEAM WITH CONCENTRATED LOAD AND SIMPLY SUPPORTED ENDS

Wa b

0

1

2

x

Q

M

+

-

+

( )L

WbxQ 10 =− ( )

LWbx

xM 10 =−L

WabMM 1max ==

( )L

WaxQ 21 −=− ( ) ( )

L x- L Wa

xM 21 =−

( )

−−=−

2

2

2

210

1L

x

L

b1

6DWLbx

xw ( )LV

Wbxxw 102 =

−

( ) ( )

−−=−

2

2

2

221

1L

x

L

a

L

2x6D

x- L WLaxw ( ) ( )

LV x- L Wa1xw 22 =

−

( )3DL

bWaaw

22

1 = ( ) LVWba

aw 2 =

( ) ( )221 4a3L48DWa

2 /Lw −= ( )2VWa

2 /Lw 2 =


09.03- 42

ELEMENTARY TABLES

BEAM WITH UNIFORM PRESSURE AND SIMPLY SUPPORTED ENDS

x

q

x

Q

M

+

+

-

( )

−= x

2L

qxQ ( )2

qx2

qLxxM

2−=

( )8

qLM2L/M

2

max ==

( )

+−=

3

3

2

23

1L

x

L

x21

24DxqL

xw( ) ( )22 xLx2V

qxw −=

384D5qL

w4

max1, = 8VqL

w2

max2, =


09.03- 43

ELEMENTARY TABLES

x

x

Q

M

+

+

-

BEAM WITH TRIANGLE LOAD AND SIMPLY SUPPORTED ENDS

( )2L

qx6

qLxQ

2−= ( )

−=

2

2

L

x1

6qLx

xM

( )16qL

/21M2

= L0.577xwhenqL0.064M 2max ==

( )

+−=

4

4

2

23

1L

x3

L

x107

360DxqL

xw ( )

−=

2

2

2L

x1

6VqLx

xw

( )768D5qL

/21w4

1 = L0.577xatappearsw max,2 =

L0.519xatD

qL0.00652w

4

max,1 ==


09.03- 44

BEAM WITH CONCENTRATED LOAD, ONE SIMPLY SUPPORTED AND ONE CLAMPED END

ELEMENTARY TABLES

aW

b

0

12

Q

M

x

+

+

-

-

( )

−=−

Lb

32L2Wb

xQ2

10 ( )

+=

La

22L

aWbaM

2

2

( )

−−=−

2

221

L

a3

2LWa

xQ ( )

−−=

2

2

L

a1

2Wa

LM

( )

+−=−

2

2210

1L

xLa

2La

312D

xWbxw

( )

−=−

Lb

3V2L

xWbxw

2

210

2

( ) ( )

−

−−

−=−

Lx

1L

a3

L

a13

12D

x-L Waxw

2

2

2

2221

1

(a)ww 2max,2 =


09.03- 45

BEAM WITH UNIFORM PRESSURE, ONE SIMPLY SUPPORTED AND ONE CLAMPED END

ELEMENTARY TABLES

x

x

q

Q

M

0 1

+

+

-

-

( ) qx8

3qLxQ −= ( )

−=

Lx

43

2gLx

xM

L0.375xatqL128

9M 2max ==+ ( ) 8

qLLMM

2

max −==−

( )

+−=

3

3

2

23

1L

x2

L

x31

48DxqL

xw ( )2Vqx

8V3qLx

xw2

2 −=

L0.42xat185DqL

w4

max,1 ==

192DqL

2) / (L w4

1 =


09.03- 46

BEAM WITH TRIANGLE LOAD, ONE SIMPLY SUPPORTED AND ONE CLAMPED END

ELEMENTARY TABLES

x

x

q

Q

M

0 1

+

+

-

-

( )

−=

2

2

L

x51

2qL

xQ ( )

−=

2

2

3L

x51

2qLx

xM

L0.447xatqL0.0298M 2max ==+

( )15qL

LMM2

max −==−

( )

+−=

4

4

2

23

1L

x

L

2x1

120DxqL

xw ( )

+=

2

3

23L

x

5

x2VqL

xw

L0.447xatw max1, =


09.03- 47

BEAM WITH CONCENTRATED LOAD AND CLAMPED ENDS

x

x

0

Wa b

12

Q

M

+

+

-

-

-

( )

+=−

L2a

1L

WbxQ

2

210 ( )

++−=−

L2a

1L

xWb

L

WabxM

2

2

2

210

( )

+−=−

L2b

1L

WaxQ

2

221

( ) ( )axWL2a

1L

xWb

L

WabxM

2

2

2

221 −−

++−=−

( )

+−−+−=−

3

3

2

2

2

2210

1L

a2

L

a31

DIAB Sandwich Handbook - STM Boats · DIAB SANDWICH HANDBOOK 09.03- 2 SANDWICH Whenever new...

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Transcript of DIAB Sandwich Handbook - STM Boats · DIAB SANDWICH HANDBOOK 09.03- 2 SANDWICH Whenever new...