AE114: Wednesday February 18, 2015

Shear Flow of Members in Torsion

Assignment 5 due todayAssignment 6 due Wednesday 2/25

Exam 1 Wednesday 3/4

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 to

pure torsion. There are no end restraints on the tube so the

tube ends and tube cross-sections are free to warp out of

their plane.

T

Shearing stresses are uniformly distributed through the

skin thickness.

Define: shear flow q = τ t

ds = differential length of the skin

dF = q ds = differential shearing force

dT = (dF) (h) = q h ds = 2q dA

h = moment arm of dF about O

dA = area of shaded triangle

Area of triangle= ½ 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):

2A

Tq =

and for shearing stress:

2At

Tτ =τ

Bruhn A6.9: Torsion of Multiple-cell Closed Sections

In a wing cross section, there will be spars which

divide the section into cells.

How can we relate applied torque to shear flow when

the wing section has more than one cell?

nx

ny

We can choose any axis (coming out of the page)

as our moment axis and sum moments caused by the

shear flow. Let us choose an axis through point O.

We know that: T = 2qA, so if we multiply the shear

flow by the area it encloses, we will know the torque

about 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’ – 2q3 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

nx

ny

For equilibrium of shear at the junction point

of the interior web and the outside wall:

q3 = q1 – q2

Substituting this equality gives us:

To = 2q1 (A1 + A2’) + 2q2 A3’ – 2(q1 – q2)A2’

= 2q1 A1 + 2q1 A2’ + 2q2 A3’ – 2q1 A2’ + 2q2 A2’

= 2q1 A1 + 2q2 (A2’ + A3’)

= 2q1 A1 + 2q2 A2

To = 2q1 (A1 + A2’) + 2q2 A3’ – 2q3 A2’

We can generalize for a multiple-cell wing structure:

∑=

=n

1i

iio 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 θ.

How can we define θ?

Define: A = cross sectional area enclosed by skin

θ = 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 γ

=

Also, we know that:2A

Tq =

Assuming small θ: 1

tanθθγ

≈≈

2AGt

T

tG

q

G(1)(θ ===≈

τ)γ

Defining Strain Energy:

( )

dsGt8A

T

ds2AGt

T

2A

T

2

1

ds q2

1dU

2

2

=

=

=

γ

We can find the total energy by integrating dU

over the entire skin boundary:

∫∫ ==t

ds

G8A

TdU U

2

2

From Castigliano’s Theorem:

dU = θ dT → =dU

θdT

In summation form:where: L = length of skin

section with

thickness t

θ = angle of twist of

the cross section

per unit length

∑=t

L

2AG

1θ q

)2A

Tq:(where

t

ds

2AG

1

t

ds

G4A

T

dT

dUθ

2

==

==

∫

∫

q

Example: Find the shear flow and angle of twist in a 3-cell

thin-wall closed section.

nx

ny

T

Let us use the symbol: ∑=t

La

∑=t

L

2AG

1θ q

∑=t

L

A

12Gθ q Convenient form

for constant G (shear modulus)

nx

ny

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

θ1 = θ2 and θ2 = θ3

So the equations can be solved simultaneously for

q1, q2, q3 and θ 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?

{ }

{ }

( ){ }3032332

3

3

23321221202

2

2

1221101

1

1

aqaqqA

12Gθ

a)q(qa)q(qaqA

12Gθ

a)q(qaqA

12Gθ

+−−=

−+−−=

−+=

How many unknowns?

We can solve for the q’s and θ

if we know applied torque and section geometry.

Or, we can find T if we know the resisting q’s and

resulting angle of twist, θ.

Numerical Example

of a simple two-cell wing section

Page A6.8 of Bruhn

The wing section shown is subjected to torque, T.

Find the shear flows and the angle of twist.

T

nx

ny

Loading and geometric conditions:

L10 = 26.9 in

L12 = 13.4 in

AB = 25.25 in

BC = 15.7 in

CD = 25.3 in

Total A1 + A2 = 493.2 in2

A1 = 105.8 in2

A2 = 387.4 in2T = 83,450 in lb

T

nx

ny

Solution

Find a10, a12, a20:

Find the angles of twist:nx

ny

nx

ny

Angle of twist: 2Gθ1 = 446.54 lb/in2Gθ2 = 446.5 lb/in

θ1 = θ2 = 0.032 deg

)( in

lb37.07qqq

in

lb55.48)

in

lb(92.550.5994q

in

lb92.55

901.6383,450

901.63Tq

2112

1

2

direction assumed of opposite

−=−=

==

===∴

55.48 lb/in

37.07 lb/in

92.55 lb/in

Assignment 6 – due Wednesday 2/25