Shear Lag

19/04/23 Prof. Carlos Montestruque 1

Shear Lag

19/04/23 2

SHEAR LAG Thin sheet structures under loading conditions that produce

characteristically large and non-uniform axial (stringer) stress.

More pronounced in shells of shallow section than in shells in

deep section.

Much more important in wings than in fuselage ( if the basic method

of construction is similar)

The effect of sheet panel shear strains is to cause some stringers to

resist less or more axial load than those calculated by beam

theory I/My

In general, the shear lag effect in skin-stringer box beam is not

appreciable except for the following situations:

Thin or soft (i.e., aluminum) skin

Cutouts which cause one or more stringers

to be discontinued

Large abrupt changes in external load applications

Abrupt changes in stringer areas

Shear Lag

19/04/23 3

Example:Axial constraint stresses in a doubly symmetrical, single cell,six boom beam subject to shear.

• The bending stress in box beams do not always conform very closely

to the predictions of the simple beam bending theory.• The deviations from the theory are caused primary by the shear

deformations in the skin panels of the box that constitutes the

flanges of the beam.• The problem of analyzing these deviations from the simple beam

bending theory become known as the SHEAR LAG EFFECT

Shear Lag

19/04/23 4

0zh2

SzqPz

z

PP y

BB

B

0h2

Sq

z

P yB

Top cover of beam

Equilibrium of an edge boom element

Loads on web and corner booms of beam

Solution:

Shear Lag

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Similarly for an element of the central boom

0q2z

PA

Now considering the overall equilibrium of a length z of the cover, we have

0zh

SPP2 y

AB

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We now consider the compatibility condition which exists in the displacement of

elements of the boom and adjacent elements of the panel.

zz

dz1z1 AB

in which and are the normal strains in the elements of boomA B

ABd

1

z

or

BE

PBB

Now

AE

PAA

Gt

q

A

P

B

P

dE

Gt

dz

dq AB

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Choosing , say, the unknown to be determined initially.AP

0zh

SPP2 y

AB

A

P

B

P

dE

Gt

dz

dq AB 0q2z

PA

A

P

Bh2

zS

B2

P

dE

Gt

z

P

2

1 AyA2A

2

From these equations, we have

dEBh

zGtSP

dEAB

AB2Gt

z

P yA2

A2

Rearranging we obtain

dEBh

zGtSP

z

P yA

22A

2

or

dEAB

AB2Gt2 Where is the shear lag constant

zAB2h

ASzDsinhzcoshCP y

A

The differential equation solution is

The arbitrary constant C and D are determinate from the boundary conditions of the

cover of the beam.

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when z = 0 ;0PB 0

Gt2z

P

Gt

qA

when z = L

From the first of these C = 0 and from the second

LcoshAB2h

ASD y

Thus

Lcosh

zsinhz

AB2h

ASP y

A

Lcosh

zsinhz

AB2h

SyA

The normal stress distribution follows APAA

The distribution of load in the edge boom is

Lcosh

zsinh

B2

Az

AB2h

BSP y

B

0z

h

SPP2 y

AB

Lcosh

zsinh

B2

Az

AB2h

SyB

whence

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Lcosh

zcosh1

AB2h2

ASq y

whence

0q2z

PA

The shear flow

Lcosh

zcosh1

AB2ht2

ASy

The shear stress t

q

Elementary theory gives

AB2h2

ASq y

AB2h

zSyBA

Shear Lag

19/04/23 10

Lsinh

zsinhLz

AB2h

ASP y

A

Lsinh

zsinh

B2

ALz

AB2h

BSP y

B

Lsinh

zcoshL1

AB2h2

ASq y

dEAB

AB2Gt2 Where is the shear lag constant

Rectangular section beam supported at corner booms onlyThe analysis is carried out in an identical manner to that in the previous case except

that the boundary conditions for the central stringer are when z = 0 and z = L. 0PA

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Beam subjected to combined bending and axial load

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1BP is load in boom 1

2BP is load in boom 2

PPPP A2B1B

Longitudinal equilibrium

d2PdPd2P A1B

Moment equilibrium about boom 2

22B q

z

P

Equilibrium of boom 2

11B q

z

P

Equilibrium of boom 2

21A qq

z

P

Equilibrium of central stringer A

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The compatibility condition now includes the effect of bending in addition to extension

as shown in figure below

zdz

vd

dz

ddz1z1

2

21

1BA

Where and z are function of z only1Thus

2

2

1BA1

dz

vd

d

1

dz

d

Similarly for an element of the

lower panel

2

2

A2B2

dz

vd

d

1

dz

d

Subtraction these equation

2B1BA21 2

d

1

dz

d

dz

d

or, as before

B

P

B

P

A

P2

dE

Gt

dz

dq

dz

dq 2B1BA21

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19/04/23 14

B

P

B

P

A

P2

dE

Gt

dz

dq

dz

dq 2B1BA21

21A qq

dz

dP

Choose as the unknown, and using these equationsAP

d2PdPd2P A1B PPPP A2B1B

dEAB

AB2Gt2 Where is the shear lag constant

dEB

PGtP

AB

AB2

dE

Gt

dz

PdA2

A2

we obtain

dEB

PGtP

dz

PdA

22A

2

or

AB2

PAzDsinhzcoshCPA

The differential equation solution is

The arbitrary constant C and D are determinate from the boundary conditions of the

cover of the beam.

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when z = 0 ;0PA z

P0qq A

21

when z = L

or, rearranging

Lcosh

zLcosh1

AB2

PAPA

Lcosh

zLcosh

A

AB4

AB22

PAP 1B

Lcosh

zLcosh1

AB22

PAP 2B

The distribution of load in the edge booms 1 and 2

Lcosh

zLsinh

AB22

PA

z

Pq 1B

1

Lcosh

zLsinh

AB22

PA

z

Pq 2B

2

Finally the shear flow distribution are

The shear flow and are self-equilibrating and are entirely produced by shear lag

effect ( since no shear loads are applied).1q 2q

zsinh Ltanhzcosh1AB2

PAPA

we have the distribution load in the central stringer