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University of Liège

Aerospace & Mechanical Engineering

Aircraft Structures

Beams - Bending & Shearing

Ludovic Noels

Computational & Multiscale Mechanics of Materials – CM3

http://www.ltas-cm3.ulg.ac.be/

Chemin des Chevreuils 1, B4000 Liège

[email protected]

http://www.ltas-cm3.ulg.ac.be/

• Balance of body B– Momenta balance

• Linear

• Angular

– Boundary conditions• Neumann

• Dirichlet

• Small deformations with linear elastic, homogeneous & isotropic material

– (Small) Strain tensor , or

– Hooke’s law , or

with

– Inverse law

with

Aircraft Structures - Beams - Bending & Shearing

Elasticity

b

T

n

2ml = K - 2m/3

2013-2014 2

• Assumptions

– Symmetrical beams

– Cross-section remains plane and

perpendicular to fibers

• Bernoulli or Kirchhoff-Love theory

• Only for thin structures (h/L << 1)

• Limited bending: kL << 1

– Linear elasticity

• Small deformations

• Homogeneous material

• Hooke ‘s law

– For pure bending

• Vertical axis of symmetry

• Constant curvature of the

neutral plane


Pure bending of symmetrical beams

x

z

hL y

z

Mxx

Mxx

2013-2014 3

• Kinematics

•

• Linear elasticity

– Hooke’s law & stress-free edges

•



zx

z

hL

Mxx

Mxx

2013-2014 4

• Resultant forces

– Tension

• Should be equal to zero

(pure bending)

• So the neutral axis is defined

such that the first moment of

area

– The neutral axis passes through

the centroid of area of the cross-

section

– As, here, Oz is a symmetry axis

the neutral axis passes through

the centroid of the cross-section



x

z

hL y

z

Mxx

Mxx

y

z

y

z

2013-2014 5

• Resultant forces (2)

– Bending moment

• With the

second moment of area

• As



x

z

hL y

z

Mxx

Mxx

2013-2014 6

• Example

– I-beam subjected to a bending moment

• 100 kN.m

• Applied in a 30°- inclined plane

– Distribution of stress?

– Neutral axis?



My

z

tf = 20 mm

tf = 20 mm

tw = 25 mm

h = 300 mm

b = 200 mm

30°

2013-2014 7

• Second moments of area

– Doubly symmetrical cross-section

origin of the axes is the centroid

–

–



Transport theorem

My

z

tf = 20 mm

tf = 20 mm

tw = 25 mm

h = 300 mm

b = 200 mm

30°

2013-2014 8

• Bending moment components

–

–

• Stress distribution

– Linear elasticity superposition

–



My

z

tf = 20 mm

tf = 20 mm

tw = 25 mm

h = 300 mm

b = 200 mm

30°

2013-2014 9

• Stress distribution (2)

–

– For z = ± 0.15 m

– For y = 0

• Neutral axis

– Such that sxx = 0



My

z

h = 300 mm

b = 200 mm

30°

My

z

76.4°

2013-2014 10

• Assumptions

– Same as for symmetrical bending

• Bernoulli or Kirchhoff-Love theory

• Only for thin structures (h/L << 1)

• Limited bending: kL << 1

• Linear elasticity

• Bending moment in a

plane being q-inclined

– Let us assume

• The existence of a neutral axis

– Being a-inclined

– Including the axes origin


Pure bending of unsymmetrical beams

x

y

z

Mxx

q

Mxx

y

z

q

Mxx

y

zq

Mxx

a

2013-2014 11

• Bending moment components

–

–

• Pure bending

– Signed distance from neutral axis

– Plane section remains plane

– Pure bending

• As the neutral axis still passes through the cross-section centroid



y

z

a

My

Mz

d

y

z

q

Mxx

2013-2014 12

• Neutral axis

– As

– It yields

•

•

• In tensorial form

– So position of the neutral axis depends on

• The geometry

• The loading



y

z

a

My

Mz

d

y

z

q

Mxx

2013-2014 13

y

z

q

Mxx

• Principal axes

– As

– The referential can be changed such that

• We are in the principal axes of the section

– This referential is load-independent

• The neutral axis is then obtained by

• And the stresses are rewritten as

– So in the principal axes, everything happens as for symmetrical loading



y

z

a

My

Mz

d

2013-2014 14

• Example

– Beam subjected to a bending moment

• 1.5 kN.m

• In the vertical plane

– Neutral axis?

– Maximum stress due to bending ?



M

y’

z’

t = 8 mm

t = 8 mm

h = 80 mm

b1 = 40 mm b2= 80 mm

2013-2014 15

• Position of centroid

– New frame Cyz linked to centroid C

•

•



M

y’

z’

t = 8 mm

t = 8 mm

h = 80 mm

b1 = 40 mm b2= 80 mm

y

z

C

2013-2014 16


–

–



M

t = 8 mm

t = 8 mm

h = 80 mm

b1 = 40 mm b2= 80 mm

y

z

C

17.6 mm

12 mm

2013-2014 17


• Second moments of area (2)

–

– As each part is symmetrical, Iyz with respect to C is equal to the sum of

• Area part times

• y-distance from C of section part to global section C times

• z-distance from C of section part to global section C


M

t = 8 mm

t = 8 mm

h = 80 mm

b1 = 40 mm b2= 80 mm

y

z

C

17.6 mm

12 mm

2013-2014 18

• Neutral axis

– As

with q = p/2, so

– Neutral axis in inclined by 14.7° as

• Stresses

–

– By inspection the maximum stress is reached



My

z

C

12 mm 14.7°

t = 8 mm

h = 80 mm

17.6 mm

2013-2014 19

• Bending deflection

– Let us consider

• Unsymmetrical beam

• Frame origin at the

cross-section centroid

• A bending moment in

a planed being q-inclined

• A resulting neutral-axis

being a-inclined



x

y

z

Mxx

q

Mxx

a

y

z

y

zq

Mxx

a

2013-2014 20

• Bending deflection (2)

– Due to the bending moments

• There is a deflection normal

to the neutral-axis

• The centroid is deflected by x

• The neutral surface has

a curvature k with

– Deflection components

•

•



y

zq

Mxx

a

x

y

z

a

q

a y

z

x

Mxx

2013-2014 21

• Bending deflection (3)

– As

•

•

– The neutral-axis is obtained by

•

&

• An unsymmetrical beam

– Deflects both vertically & horizontally even for a loading in the vertical plane

– Excepted in the principal axes (Iyz=0)



y

zq

Mxx

a

2013-2014 22

• Shearing-bending relationship

– Linear balance equation

– Momentum balance equation

as second order terms vanish

– Eventually

•

•


Bending of unsymmetrical beams

y

z

a

My

Mz

d

Tz+ ∂xTz dx

x

z

My

fz(x)

Tz dx

My+ ∂xMy dx

y

z

aMy

Mz

d

x

y

Mz

fy(x)

Ty dx

Mz+ ∂xMz dx

Ty+ ∂xTy dx

2013-2014 23

• Bending approximation: deflection equations

– As &

– Deflection equations read

• Euler-Bernoulli equations for x in [0; L]

• Low order boundary conditions

– Either on displacements &

– Or on shearing

• High order boundary conditions

– Either on rotations &

– Or on couple



x

z f(x) TzMxx

uz =0

duz /dx =0 M>0

L

2013-2014 24

• Bending approximation: deflection equations with energy method

– Same results can be obtained with an energy method

– Let us assume we are in the principal axes and Mz = 0

• As

• Internal energy variation

• Work variation of external forces



x

z f(x) TzMxx

uz =0

duz /dx =0 M>0

L

2013-2014 25

• Bending approximation: deflection equations with energy method (2)

– Energy conservation

• Integration by parts of the internal energy variation

• Work variation of external forces



x

z f(x) TzMxx

uz =0

duz /dx =0 M>0

L

2013-2014 26

• Bending approximation: deflection equations with energy method (3)

– Energy conservation (2)

• As duz is arbitrary:

Euler-Bernoulli equations

• on [0, L] &

•

• Remark: if displacement or rotation constrained at x=0 or x=L

– duz =0 is no longer arbitrary at this point

– The boundary condition

» Becomes uz = value or/and uz,x = value

» Instead of being on shear or/and couple



x

z f(x) TzMxx

uz =0

duz /dx =0 M>0

L

2013-2014 27

• Bending approximation: shearing

– As

•

•

– There are no shearing loads only if the bending moment is constant

• Cross section remains plane only if constant bending moments

• Beam with cross-section dimensions << L

– If bending variation is reduced (Saint-Venant)

– Shear stress O(T/bh) ~ bending stress x h/L

– Shear stress can be neglected

• For thin-walled cross sections (as plates/shells)

– Shear stress O(T/th) cannot be neglected

– If bending variation is reduced (Saint-Venant)

Deflection is primarily due to bending strain



h

L

L

h t

2013-2014 28

• Second moments of area for thin-walled sections

– Thin walled section

• Thickness t << cross-section dimensions

• Example Iyy of a U-thin walled cross-section

• This result is obtained directly if the section is

represented as a single line


Bending for thin-walled sections

y

z

t

t

t

h

b

z

y

t

t

t

h

b

2013-2014 29

Bending for thin-walled sections

• Second moments of area for thin-walled sections (2)

– Centroid & z’C= 0

– Second moments of area

• &

• null as Cy is a symmetry axis


y

t

t

t

h

b

z

y’

z’

C

2013-2014 30

• Shear stress

– A naïve solution is a uniform stress

•

• By reciprocity this would lead to szx = t

• Impossible as the surface is stress-free

– So the shear stress has to be tangent to the contour

For thick cross section For thin-walled section

– How to compute these profiles?


Shearing of beams

h

L

x

z

TzTz

x

ty

z

t

h

L

x

z

TzTz t

z

y

z

H

H

T

2013-2014 31

• Assumptions

– Thick section

– Symmetrical beam

• My = ||Mxx||

• Mz = 0

• Shear stress

– From equilibrium (previous slide):

– Let us consider the lower part of the beam (cross section A*)

• Normal force acting on this lower part extremities

• With the reduced moment of area


Shearing of symmetrical beams

x

z

My

fz(x)

Tz dx Tz+ ∂xTz dx

N*x+ ∂xN

*x dxN*

x

My+ ∂xMy dx

y

z

A*

2013-2014 32

• Shear stress (2)

– We found

– But lower part of the beam is at equilibrium

– So the upper part should apply

• On the lower part

• A force Rdx

• With

– Shear stress

• On the cut surface, assuming uniform shear stress on b(z)



x

z

My

fz(x)

Tz dx Tz+ ∂xTz dx

N*x+ ∂xN

*x dxN*

x

My+ ∂xMy dxRdx

y

z

A*

x

t

t

b(z)

2013-2014 33

• Deformation

– Section shear strain in linear elasticity

•

• But shear stress is not uniform

– g = 0 on top and bottom surfaces

– at neutral axis

• The average shear strain g is defined as

– The angle of the deformed neutral axis

with its original direction



ht

z

y

z

t

b(z)

A*

t

h

x

z

Tz

dx

Tz+ ∂xTz dx

gmax

g

2013-2014 34

• Deformation (2)

– Average g obtained by energy method

• Work of external forces

–

– If the shear stress was uniform

one would have

– To account for the non-uniformity, a

corrected area A’ is used

• Internal energy variation

–

– As



x

z

Tz

dx

Tz+ ∂xTz dx

gmax

gg dx

2013-2014 35

• Deformation (3)

– Average g obtained by energy method (2)

• Work of external forces = Internal energy variation

– Eventually

• Average shear strain

• With

• Under the assumption of a constant shear stress on b(z)



x

z

Tz

dx

Tz+ ∂xTz dx

gmax

gg dx

2013-2014 36

• Deformation (4)

– Shearing equations

• Shearing is the difference between

– Neutral fiber deflection

– Orientation of the cross section

» Angle qy

• Curvature is then computed from the variation in cross-section direction

• Timoshenko equations on [0, L]

– As

– As



z

g

dx

x

g

qy

qy

x

z f(x) TzMxx

uz =0

duz /dx =0 M>0

L

2013-2014 37

• Bernoulli assumption

– Timoshenko beam equations

– Euler-Bernoulli equations

– It has been assumed that

the cross-section remains planar

and perpendicular to the neutral axis

– uz,x = -qy

– Validity: (EI)/(L2A’m) << 1


Shear effect

z

x

g

qy

qy

x

z f(x) TzMxx

uz =0

duz /dx =0 M>0

L

2013-2014 38

• Example: shear stress for a rectangular section

– Using

• Reduced moment of area

• Shear stress

• Maximum shear stress reached at z=0:

– Exact solution (elasticity theory):

• Maximum shear stress reached at y=z=0

• Shear stress is not uniform on b, but this is a correct approximation for h>b



y

z

x

t

b(z)

A*

t

h

ht

z

h/b 2 1 1/2

a 1.033 1.126 1.396

2013-2014 39

• Example: shear strain for a rectangular section

– Reduced moment of area

– Corrected area

– So if shear stress is uniform on b (ok for h>b)



y

z

x

t

b(z)

A*

t

h

2013-2014 40

• Extending these equations to unsymmetrical beams

– The equations remain valid

• If we are in principal axes as Iyz = 0

as everything happens as if uncoupled

• In linear elasticity, as superposition

principle holds, the same equations

can be derived for uy

– But if the loads do not pass through a point called the shear center S there

will be a twist of the beam

• This point is difficult to obtained for unsymmetrical non-thin-walled sections

– This problem is not relevant for light aeronautic structures based on thin-

walled sections

• Only thin-walled sections will be considered


Shearing of unsymmetrical beams

x

z f(x) TzMxx

uz =0

duz /dx =0 M>0

L

2013-2014 41

• General relationships

– Consider thin-walled sections

• Symmetrical or not

• Closed or open

– Assumption

• Shear is uniform on the thickness t

Definition of the shear flow

– Balance equations

•

•


Shearing of thin-walled cross section beams

L

q + ∂sq ds

s

x

dx

ds

qq

q + ∂xq dx

ss

ss + ∂sss ds

sxx + ∂xsxx dx

sxx

L

tx

z

y

h

2013-2014 42

• Shear flow

– Assumption

• No twisting of the beam

Shear loads pass through

a particular point called

shear center S

• If this is not the case there is

a torsion combined to the shearing

– Equation

• Direct bending stress obtained from

• ! We use pure bending theory: section assumed to remain planar (not true)



x

z

y

Tz

Tz

Ty

Ty

y

z

S

Tz

TyC

2013-2014 43

• Shear flow (2)

– Equations (2)

•

– Assuming the variation of moment is larger than the variation of section

• But , &

• Eventually



y

z

S

Tz

TyC

q

s

2013-2014 44

• Shear flow for an open section beam

– For an open section q(s) = 0 as

on the boundary sxz(0) = szx(0) = 0

• In the principal axes

• To be compared to the thick cross section approximation

–

– Has been obtained for symmetrical beam

and assuming constant stress on b

– For unsymmetrical beams, the same approximation leads to

– Accurate only for h/b > 1


Shearing of thin-walled open section beams

y

z

S

Tz

TyC

q

s

y

z

x

t

b(z)

A*

t

h

2013-2014 45

• Shear center

– Definition:

• No twisting of the beam if shear loads pass through

a particular point called shear center S

– Practically

• Shear loads are represented by loads passing

through S and by a torque

– Location

• If axis of symmetry: S lies on it

• If cruciform or angle sections, as q is along these section, S is at the intersection

• Generally speaking: obtained by considering the moment equilibrium



y

t

t

t

h

b

z

C=S

t

y

t

t

h

b

z

C

t

S

y

t

t

h

b

z

C

t

S

y

z

S

Tz

TyC

q

s

2013-2014 46

• Example

– U-thin walled cross section

• From previous exercise

– Location of the shear center S?



S

Tz

Ty

y

t

t

t

h

b

z

y’

z’

C

2013-2014 47

• Location of the shear center

– By symmetry:

• On Cy

– Horizontal location?

• Consider a shearing Tz (Ty is useless

as we know the vertical location)

• Ensure moment equilibrium

– Shear flow

•

• As

• With



q

s

S

Tz

y

t

t

t

h

b

z

y’

z’

C

q

q

2013-2014 48

• Shear flow

– Shear flow on bottom flange

• As

• Maximum shear flow

• Moment around O’ induced by the shearing of bottom flange



q

s

S

Tz

y

t

t

t

h

b

z

y’

z’

C

q

q

O’

+

2013-2014 49

• Shear flow (2)

– Shear flow on the web

• As

• “Boundary” shear flow

• Maximum shear flow

• Moment around O’ induced by the shearing of the web



q

s

S

Tz

y

t

t

t

h

b

z

y’

z’

C

q

q

O’

+

2013-2014 50

• Shear flow (3)

– Shear flow on top flange

• As

• Moment around O’ induced by the

shearing of the top flange



q

s

S

Tz

y

t

t

t

h

b

z

y’

z’

C

q

q

O’

+

2013-2014 51

• Shear center

– Moment due to the shear flow:

– It has to be balanced by the shearing



q

s

S

Tz

y

t

t

t

h

b

z

y’

z’

C

q

q

O’

+

2013-2014 52

• Shear flow for a closed section beam

– Equation

•

still holds

• But q(s=0) is now ≠ 0

– Method

• The cross section is virtually cut at s=0

• With the open section contribution

• q(s=0) is computed to balance the momentum

– Shear loads pass through a given point T (not necessarily shear center S)

– We will see that, for closed sections, shear and torsion stresses have the

same form, so we do not really have to pass through the shear center S as it

becomes useless to decompose shearing and torque


Shearing of thin-walled closed section beams

y

z

T

Tz

Ty

C

q s

2013-2014 53

• Shear flow for a closed section beam (2)

– Evaluation of q(s=0)

• Momentum balance (if q & s anticlockwise)

with p the distance from the wall tangent to C

• If Ah is the area enclosed by the section mid-line

–

• At the end of the day



y

z

T

Tz

Ty

C

q s

p

y

z

T

Tz

Ty

C

q s

pds

dAh

2013-2014 54

• Shear flow for a closed section beam (2)

– Final expression

•

• qo(s) is computed as for an open section

•

– Remarks

• The completely around integrals are related to the

closed part of the section, but if there are open parts,

their contributions has be taken when computing qo(s)

• This last expression assumes q, s anticlockwise

• If momentum and distance p are related to the lines of action

of the shear load T, this simplifies into

• By cutting the section we are actually substituting the shearing by

– A shear load passing through the shear center of the cut section

» Which depends on the cut location

– A torque leading to constant shear flow q(s=0)

» Which depends on the cut location



y

z

T

Tz

Ty

C

q s

2013-2014 55

• Shear center

– As we have already used the momentum balance,

how to determine the shear center S ?

– As loads passing through this point do not lead to

section twisting, we have to evaluate this twist



y

z

T

Tz

Ty

C

q s

2013-2014 56


• Shear deformations

– As loads passing through shear center do not lead to

section twisting, we have to evaluate the twist in the

general case

• Shear strain in the local axes

• Remark

y

z

T

Tz

Ty

C

q s

s

x

dx

dsdus

dux

n

2013-2014 Aircraft Structures - Beams - Bending & Shearing 57

• Twisting and warping

– Closely Spaced Rigid Diaphragm (CSRD)

assumption

• The cross section can deform along Cx

• The shape of the cross-section in its own

plane remains constant

• These 2 motions correspond to

warping & twisting

– So in the Cyz plane only a rigid rotation is seen

• Aeronautical structures

– You do not want to change the airfoil shape

– So structures are rigid enough for this

assumption to hold

– Let us call R the center of twist

• When warping is constrained the center of

twist is different from the shear center

(see lectures on structural discontinuities)


x

z

y

y

z

R

C

pR

us

q


• Center of twist

– Equations

• pR is the distance from the wall tangent to R

• p is the distance from the wall tangent to C


y

z

R

C p

pRY

us

q

y

z

R

C

pR

us

q


• Center of twist (2)

– Twist

•

• Let & be respectively the components

along Cy and Cz of the displacement of C due

to the twist dq

–

–

• As

– Remarks

• Valid for closed and open sections

• uzC, uy

C & q depend on x

• us depends on x & s


y

z

R

C p

pRY

us

q

y

z

R

C

pRY

us

q

uyC

q -uzC


• Center of twist (3)

– Location

• As

• Eventually

• Remarks

– Remains valid for a point other than the centroid C

– Equations valid for open thin-walled sections but what about CSRD

assumptions for such sections ?


y

z

R

C p

pRY

us

q


• Warping

– In linear elasticity

• Shear flow

• Shear strain

– Shear flow

• As


y

z

R

C p

pRY

us

q

s

x

dx

dsdus

dux

n


• Warping (2)

– Shear flow integral

• As is the area

swept by p with the

area enclosed by the section mid-line


y

z

R

C p

pRY

us

q

y

z

T

Tz

Ty

C

q s

pds

dAh


• Warping (3)

– As

• Performing the integral all around the section

• The wrapping displacement reads

• If axes origin corresponds to twist center R

– With


y

z

R

C p

pRY

us

q


• Warping (4)

– If axes origin corresponds to twist center R

•

– ux(s=0) ?

• Symmetrical sections

– If origin of s lies on a symmetry axis ux(0)=0

• Unsymmetrical sections

– Linear response sxx(s) ÷ ux(s)-ux(0)

– Axial load due to shear or torsion is equal to zero


y

z

R

C p

pRY

us

q


• Shear center

– To compute yS

• Assume a shear load Tz passing through yS

• So there is no twist

• As

• Eventually

– With

• Shear center (Point T = S here) position using


y

z

S

Tz

C

q s

p ds


• Example

– Simply symmetrical thin-walled closed section

• Constant t and m on all walls

– Shear center ?


z’

y’t

8a

A

6a9a

B

C

D



– Simply symmetrical

• Centroid lies on Dy’

• Iyz = 0

• Shear center lies on Dy’

– So we have only to consider

• A vertical shear load

• Iyy = Iy’y’


z’

y’t

8a

A

6a9a

B

C

D

Tz

S


• Shear flow

– Origin of s on D

•

• With

– Open shear flow on line DA

•

•


z’

y’t

8a

A

6a9a

B

C

D

Tz

sS

z’

y’

A

B

C

D

Tz

sS

qo


• Shear flow (2)

– Open shear flow on line AB

•

•

– Open shear on BC & CD by symmetry

•

•


z’

y’

A

B

C

D

Tz

sS

qo

z’

y’

A

B

C

D

Tz

sS

qo


• Shear flow (3)

– Open shear flow

– Constant shear flow for zero twist

•

• Using symmetry properties & as mt is constant

–

–


z’

y’

A

B

C

D

Tz

sS

qo


• Shear flow (4)

– Constant shear flow for zero twist (2)

• As

&

• Eventually


z’

y’

A

B

C

D

Tz

sS

q


• Shear center

– General expression

• Shear flow balances the applied torque (Point T = S here)

• With

•

• &

• But one has also


z’

y’

A

B

C

D

Tz

sS

q

z’

y’t

8a

A

6a9a

B

C

D

p

17a



• Shear center (2)

– Torque due to open shear flow

– Shear center location

z’

y’t

8a

A

6a9a

B

C

D

p

17a



References

• Lecture notes

– Aircraft Structures for engineering students, T. H. G. Megson, Butterworth-

Heinemann, An imprint of Elsevier Science, 2003, ISBN 0 340 70588 4

• Other references

– Books

• Mécanique des matériaux, C. Massonet & S. Cescotto, De boek Université, 1994,

ISBN 2-8041-2021-X

2013-2014 75

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