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General Aerospace Shell Beam Structures-Material and Structures-Lecture Handout, Exercises of Structures and Materials

This lecture note is part of Material and Structures course. It was provided by Prof. Aparijita Singh at Andhra University. It includes: General, Aerospace, Shell, Beam, Structures, Axial, Load, Bending, Moments, Torsional, Torque, Monocoque, Construction, Shear

Typology: Exercises

2011/2012

Uploaded on 07/26/2012

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Thus far, we have concentrated on the bending of shell
beams. However, in the general case a beam is subjected to:
axial load, F
bending moments, M
shear forces, S
torque (torsional moments), T
Figure 15.1 Examples of general aerospace shell beam structures
Aircraft Wing Space Habitat Shell
connecting nodes
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beams. However, in the general case a beam is subjected to:Thus far, we have concentrated on the bending of shell

  • torque (torsional moments), • shear forces, S • bending moments, M• axial load, F

T

Figure 15.

Examples of general aerospace shell beam structures

Aircraft Wing

Space Habitat Shell

connecting nodes

Idealize the cross-section of the shell beam into two parts:

Parts that carry extensional stress,

σ

xx

(and thus the bending and

axial loads)

Parts that carry shear stress

σ

xs

(and thus the shear loads and

torques)

Two

examples

again…

high aspect ratio wing with

semi-monocoque

construction

Notes

monocoque

construction

  • all in one piece without internal framing
  • from French “coque” meaning “eggshell”
  • “mono” = one piece

semi-monocoque

  • stressed skin construction

with

internal framework

  • still have “eggshell” to carry shear stresses,

σ

xs

  • internal framework to carry axial stress,

σ

xx

Skins and webs are assumed to carry

only

shear stress

σ

xs

Flanges and stringers are assumed to carry

only

axial

stress

σ

xx

Space habitat

Figure 15.

Representation of space habitat semi-monocoque

construction

wall

wall

stiffeners

spar

flanges

Idealize as for wing:

Figure 15.

Idealization of space habitat semi-monocoque construction

Outer skin and walls are assumed to carry

only

shear stress

σ

xs

Flanges and stiffeners are assumed to carry

only

axial stress

σ

xx

and torsion. Utilize St. Venant assumptions: Analyze these cross-sections as a beam under combined bending, shear,

do such)of the cross-section (or enough stiffness in the internal bracing toThere are enough closely spaced rigid ribs to preserve the shape

The cross-sections are free to warp out-of-plane

Start to develop the basic equations by looking at the most basic case:

leave at zero for now The axial stress is due only to bending (and axial force if that exists --

) and is therefore independent of the twisting since the

* Find M, S,wing is free to warp (except near root -- St. Venant assumptions)

T

from statics at any cross-section x of the beam

Figure 15.7 Consider the cross-section:

Representation of cross-section of box beam

flange/stiffener i = A Area associated with

i

Find the modulus-weighted centroid (

Note

: flange/stiffeners may be

made from different materials)

use a “corner” of the beam)Choose some axis system y, z (convenience says one might

Find the modulus-weighted centroid location:

A

y i

i

y

A

i

A

z i

i

z

A

i

n

= sum over number of flanges/stiffeners)

i

1

number = n

Note

: If flanges/stiffeners are made of the

same material, remove the asterisks)

  • Find the moments of inertia with reference to the coordinate

system with origin at the modulus-weighted centroid

2

I y = ∑ A i z i

2

I

z

A

i

y i

I

yz

= ∑ A i y i z i

Fall, 2002

Figure 15.

Representation of joints, skins, and shear flows in

cross-section of box beam

Figure 15.

Look at the equilibrium of joint 1

Representation of skins and stringer and associated loads

and shear flows at joint

stringer

skin

skin

Notes

The stringer only carries axial load

The skin carries only shear flow

( cross-section cut). This is due to equilibrium“cut”) must be the same as at the edge (theThe shear flow at the “end” of the skin (where it is σ

xy

σ

yx

)

Apply equilibrium:

F

x

dP

⇒ − P + P +

dx

q 1 dx

q 6 dx

dx dP

q 1 − q 6 = −

dx

More generally note that:

Fall, 2002

If shear flows “

into

” joint, its contributions is in the

negative

x-

direction; if shear flows “

out of

” joint, its contribution is in the

positive

x-direction

Figure 15.

Representation of shear flowing (left)

into

and (right)

out of

joint

Adding all this up:

dP

q in

q out

dx

dP

q out

q in

dx

Use this in general

Unit 15 - 14

Fall, 2002

For a more complicated joint, use superposition

Figure 15.

Representation of joint with multiple skins

--> Need an expression for P -- start with:

P

A

xx^

differentiating:

d P

A

d

xx^

d A

xx

d x

d x

d x = 0

since are considering stringers

with a uniform cross-section

Most general case

q

out

q

in

A

d

xx^

Joint Equilibrium

d x

Q S

q

out

q

in

=

y

z

Symmetric section

I y

M

z

= 0

Shear stresses arise due to two reasons: Now have an equation for the equilibrium of shear stresses at the joints.

Shear resultant

Twisting

It is convenient to break up the problem into two separate problems:In general have at any cross-section:

Unit 15 - 17

(1) “Pure Shear”

(2) “Pure Twist”

there is no twisting at shear center so shear resultant acts

superposition)--> Solve each problem separately, then add the results (use

Condition

: The two force systems (S

z ,

T

and S

z ’ ,

T

’ ) must be

equipollent

Figure 15.

Demonstration of equipollence of force systems

Fall, 2002

Solution procedure

Given:

section properties

loading [T(x), S

z (x)]

Find:

shear stresses (flows)

(n joints)

shear center

(n + 1) variables

Consider

Pure Shear

case

a) Apply joint equilibrium equation at each joint

Note

: n joints will yield n-

independent

section is closed) equations. (one is dependent since the

b) Use Torque Boundary Condition

T

internal

T

applied

This is torque

equivalence

not

equilibrium

Do this about the line of action of S

z

Then:

T

applied

S

z

d

T

internal

q i ( moment arm

i ( skin length

i

c)

Specified no twist (Pure Shear Case), so apply the

No

Twist Condition

Recall from Torsion Theory:

ds

AG

d

d x

d

α

Here:

d x

and:

t q

q

ds

t