!"#$ %

& &(!)

* +&

Wing Example: Draw the approximate shear and bending moment diagrams for the following wooden wing beam due to the lift force as

shown. Also show the location and magnitude of the maximum bending moment.

Use Modulus of Elasticity: E = 1.3 x 106 lb/in2 (pine) and moment of inertia: I = 17 in4.

Fuselage

30

25 lb/in

WoodenWing

Cable

96 in 40 in

Fuselage

30

25 lb/in

WoodenWing

Cable

96 in 40 in

Fuselage

30

25 lb/in

WoodenWing

Cable

96 in 40 in

Fuselage

30

25 lb/in

WoodenWing

Cable

96 in 40 in

30

25 lb/in

96 in 40 in

TVT

TH

AV

AH

A 30

25 lb/in

96 in 40 in96 in 40 in

TVT

TH

AV

AH

A

Solution

= 0MA

{ }

2410lbT

0)2

136((25)(136)(96)T

V

V

=

=-

Also:

\ AH = 4180lb

AV = (25)(136) - 2410 = 990 lb

HHH TA0F -==

\

4180lb0.5772410

tan30T

T VH ===

(25)(136)TA0F VVV +--==

= 0MA

{ }

2410lbT

0)2

136((25)(136)(96)T

V

V

=

=-

Also:

\ AH = 4180lb

AV = (25)(136) - 2410 = 990 lb

HHH TA0F -==

\

4180lb0.5772410

tan30T

T VH ===

(25)(136)TA0F VVV +--==

30

25 lb/in

96 in 40 in

TVT

TH

AV

AH

A 30

25 lb/in

96 in 40 in96 in 40 in

TVT

TH

AV

AH

A

25 lb/in

96 in

TV = 2410lb

P = TH = 4180 lb

AV = 990lb

P = AH

M2 = 20,000 lb-inM1 = 0

1000 lb

25 lb/in

96 in

TV = 2410lb

P = TH = 4180 lb

AV = 990lb

P = AH

M2 = 20,000 lb-inM1 = 0

1000 lb

52804180

17)10(1.3PEI

j6

2 =

==

0.968jL

sin

0.246jL

cos

=

=

75.751.32rad72.796

jL

72.7inj

2 -=-=-=

===

=

52804180

17)10(1.3PEI

j6

2 =

==

0.968jL

sin

0.246jL

cos

=

=

75.751.32rad72.796

jL

72.7inj

2 -=-=-=

===

=

.968)132,000)(0(.246)132,000)(0(112,000

jL

sinD

jL

cosDD

CC

jx

tan

1

12

2

1

----

=-

==

0.622127,80079,500

127,80032,500112,000

=--

=-

+-=

40.4in.556)(72.7in)(0x

0.556radj

x

max

max

==

=

max 1

2

x Ctan

j C= =

lb-in 23,500132,000155,000132,0000.849132,000

wjj

xcos

DM 21max -=+-=+

-=+=

96 in 40 in

40.4 in

Mmax=-23,500 in-lb

20,000

96 in 40 in

40.4 in

Mmax=-23,500 in-lb

20,000

Mmax = -19,602 in lb

Without axial force:xmax = 39.6 in

V

990

1410

-1000

39.6

2410

30

25 lb/in

96 in 40 in

TVT

TH

AV

AH

A 30

25 lb/in

96 in 40 in96 in 40 in

TVT

TH

AV

AH

A

= 990 lb= 2410 lb

What is the difference if we neglect Axial Force?

This figure shows a straight cylindrical bar subjected to a torque, T.

As the bar twists, each section is subjected to a shearing stress. Assuming the left end is stationary relative to the rest of the beam, a line AB will move to AB under these shearing stresses.

The longitudinal lines become twisted.

Shear strain, g= surface) (at L r

ABBB

=

Torsion: Stresses and Deflections

{ }

Shear strain is a function of: angle of twist, q distance from center of beam, r

distance)arbitrary an (at L

=

r

=max

g

r

r

g

gratio of (shear strain) : (max shear strain) = ratio of (distance from center) : (radius of beam)

For linearly elastic material, Hookes Law is:

t = G g where G = shear modulus

r

=max

tt

For a cross-section:

T = dT = r (t dA) = r (t max dA) area area area

dAr

T 2max =t

r

t = dF / dA(force / area)

So: t dA = force (dF)

area

J (Polar Momentof Inertia)

Jr

T maxt

=

JTr

max = tJ

Trt =

To find polar moment of inertia (J):

1 Solid shaft:

rrprprrr dddAJcc

===0

3

0

22 2)2(

change from dA to dr

44r

r2

04r

2p

pr

p =

-== 24

0

4

2 Tubular shaft: same as previous slide, just change limitsof integration to:

radius) outsider radius, inside(r oi ==o

i

r

r

Relating Shear Stress and Torque to Angle of Twist

distance)arbitrary an (at L

=gRecall:

dxd =

Writing this in differential form:

q varies along the length of the beam also varies wi th applied torque.

g is a function of q.

L=

Shear strain:

dxJGTdx

GJT

d =

==

dx

GJT

GJ

T

G

===

Rewriting Hookes Law:

Substituting for g, the differential angle of twist is:

Integrating along the length:

JGTL

dx JGT

d === L L

0 0

Bruhn A6.8: Torsion of Thin-Walled Closed Sections

Aircraft wings, fuselages, tails, control surfaces are thin-walled tubes of one or more cells.

Flight and landing loads produce torques (twisting moments) on these structural elements.Designers must determine the torsional stresses and deformations of these structures.

T

This figure shows a portion of a wing which is subjected topure torsion. There are no end restraints on the tube so thetube ends and tube cross-sections are free to warp out oftheir plane.

T

Shearing stresses are uniformly distributed through theskin thickness.

Define: shear flow q = t t

ds = differential length of the skin

dF = q ds = differential shearing force

dT = (dF) (h) = q h ds = 2q dAh = moment arm of dF about OdA = area of shaded triangle

Area of triangle= h ds = dA

q: force/unit lengthds: length

T

How is Shear flow, q, related to applied torque, T?

The total torque due to the shear flow over the entire wing section can be obtained by integrating dT

over the entire boundary curve of the section.

section wingthe in enclosed area totalA:where

2qAT

dA2qdTT

=

=

==

Solving for shear flow (shear force intensity):

2AT

q =

and for shearing stress:

2AtT

=t

Bruhn A6.9: Torsion of Multiple-cell Closed Section s

In a wing cross section, there will be spars which divide the section into cells.

How can we relate applied torque to shear flow whenthe wing section has more than one cell?

*

We can choose any axis (coming out of the page)as our moment axis and sum moments caused by theshear flow. Let us choose an axis through point O.

We know that: T = 2qA, so if we multiply the shearflow by the area it encloses, we will know the torqueabout that part of the wing.

q1 causes a positive moment about point O q2 also causes a positive moment about point O q3 causes a negative moment about point O

To = 2q1 (A1 + A2) + 2q2 A3 2q 3 A2

(To = resisting moment of the shear flow about an arbitrary point O)

q1 causes a positive moment about point O q2 also causes a positive moment about point O q3 causes a negative moment about point O

*

For equilibrium of shear at the junction point of the interior web and the outside wall:

q3 = q1 q 2

Substituting this equality gives us:

To = 2q1 (A1 + A2) + 2q2 A3 2(q 1 q 2)A2

= 2q1 A1 + 2q1 A2 + 2q2 A3 2q 1 A2 + 2q2 A2

= 2q1 A1 + 2q2 (A2 + A3)

= 2q1 A1 + 2q2 A2

To = 2q1 (A1 + A2) + 2q2 A3 2q 3 A2

We can generalize for a multiple-cell wing structure:

=

=n

1iiio A2qT

A single cell wing is fixed at one end and subjected to torsion on the other end.

The torsion produces a shear flow, q, on the cross section.

What is the rotation of the cross-section due to torsion?

T

Consider differential segment abcd as a free body. Due to the shear flow, the lines ad and bc rotate by an angle q.

How can we define q?

Define: A = cross sectional area enclosed by skin

q = angle of twist in radians per unit length of the wing

dU = strain energy of the element abcd(area = ds x 1) due to the twisting deformation

Strain energy = energy stored by structure during deformation.

2(q)(ds)dU=

Also, we know that:2AT

q =

Assuming small q: 1

tan

2AGtT

tGq

G(1)( ===

)

Defining Strain Energy

( )

dsGt8A

T

ds2AGt

T2AT

21

ds q21

dU

2

2

=

=

=

We can find the total energy by integrating dUover the entire skin boundary:

== tds

G8A

TdU U 2

2

From Castiglianos Theorem:

dU = q dT

In summation form:where: L = length of skin

section with thickness t

q = angle of twist of the cross section per unit length

= tL

2AG1

)2AT

q:(where t

ds2AG

1

tds

G4AT

dTdU

2

==

==

Example: Find the shear flow and angle of twist in a 3-cellthin-wall closed section.

Let us use the symbol: = tL

a

= tL

2AG1

= tL

A1

2G

{ }

{ }

( ){ }30323323

3

233212212022

2

12211011

1

aqaqqA1

2G

)aq(q)aq(qaqA1

2G

)aq(qaqA1

2G

+--=

-+--=

-+=

During torsion of the wing, each common junction point(between a pair of adjacent cells)

must be displaced by the same amount.

This is a compatibility condition which requires q1 = q2 and q2 = q3

So the equations can be solved simultaneously for q1, q2, q3 and q in terms of applied torque, T,

and the geometric constraints.

Compatibility Condition

T = 2q1 A1 + 2q2 A2 + 2q3 A3

How many equations do we have?

{ }

{ }

( ){ }30323323

3

233212212022

2

12211011

1

aqaqqA1

2G

a)q(qa)q(qaqA1

2G

a)q(qaqA1

2G

+--=

-+--=

-+=

How many unknowns?

We can solve for the qs and qif we know applied torque and section geometry.

Or, we can find T if we know the resisting qs and resulting angle of twist, q.

Numerical Exampleof a simple two-cell wing section

Page A6.8 of Bruhn

The wing section shown is subjected to torque, T (clockwise). Find the shear flows and the angle of twist.

T

Loading and geometric conditions:

L10 = 26.9 inL12 = 13.4 in

AB = 25.25 inBC = 15.7 inCD = 25.3 in

Total A1 + A2 = 493.2 in2

A1 = 105.8 in2

A2 = 387.4 in2T = 83,450 in lbCW

T

Solution

Find a10, a12, a20:

17350.032in25.3in

0.05in15.7in

0.04in25.25in

a

3350.04in13.4in

a

10750.025in26.9in

a

20

12

10

=++=

==

==

Find the angles of twist:

{ }

{ }

21

211

12211011

1

q 3.166 q 13.33

)q(q 335q 1075105.8

1

)aq(qaq A1

2G

+-=

---=

---=

{ }

{ }

21

212

12212022

2

q 5.343q 0.865

)q335(qq 1735 387.4

1

)aq(qaq A1

2G

-=

-+-=

-+-=

21

2121

q 0.5994q

q 5.343q 0.865q 3.156q 13.33

=

-=+-

2

22

21

2211

q 901.63

q 774.8)q (0.5994 211.6

q 774.8q 211.6

q2Aq2AT:Also

=

+=

+=

+= =

Angle of twist: 2Gq1 = 446.54 lb/in2Gq2 = 446.5 lb/in

q1 = q2 = 0.032 deg

)( inlb

37.07qqq

inlb

55.48)inlb

(92.550.5994q

inlb

92.55901.6383,450

901.63Tq

2112

1

2

direction assumed of opposite

-=-=

==

===\

55.48 lb/in

37.07 lb/in

92.55 lb/in