analysis of box beams 7

Wing Box Beams

(Skin – Stringer Panels Design)

Transport Wing (Two ‐

cell box)

•Wing ,Empennage and Fuselage structures are essentially beams.

•Loaded in Bending Shear and Torsion.

•Cross section Typically Resembles a Box

Typical Wing and Body Loads

Wing Design Shear Envelope for static conditions


Wing bending moment envelope for static conditions


Wing Design Torsion envelope for Static Conditions


Body monocoque vertical shear envelope


Body monocoque vertical Bending Moment envelope


Body monocoque Torsion envelope


Body monocoque Lateral shear envelope


Body monocoque Lateral Bending Moment envelope


Purpose of Analysis

To obtain stresses due to bending

Stresses due to Shear

Stresses due to Torsion

Need for a Box Beam

• A simple beam such as ‘I’

Resists Bending.

• It is weak in torsion.

• A/C structure is Loaded in Torsion.

• Structure requires adequate Torsional Stiffness.

• Unstiffened Single Cell

• Two Stringer with Stiffened Cell(Light Aeroplanes)

• Three stringer Single Cell(Light Aeroplanes)

• Multistringer Single Cell( Transport‐Thick wings)

• Multistringer Multi Cell( Military‐Thin wings)

Types of Cross Section

• All Assumptions of beam theory are applicable mainly

– Plane sections remain Plane

– Closely placed rigid diaphragms

Assumptions

• Consists of wide skin‐stringer panel to resist bending

torsion and shear.

• Skins carry all the shear(Torsion and shear)

• Skins carry very little compression.

• Skins exhibit post buckled strength

• Box section generally not symmetric(unsymmetric bending)

• Box is tapered and must be considered.

General Characteristics of Box Beams

• Beam is not under pure bending

• Due to presence of shear stresses ,plane sections do not

remain plane

• This leads to a phenomenon called shear lag.

Limitations of Beam Theory.

Shear Lag effect in a wing Box Beam

• In general the Section is unsymmetric, hence bending stress is obtained from

• For symmetric section

Estimation of Bending stresses.

The principal stresses

• The moments of inertia of the section is computed considering the skin and the stringer.

• Need to account for skin buckling(depends on stress

• Requires an iterative procedure.

Section Properties

• Step1‐Assume skin fully effective(lump skin area with stringer)

• Step 2‐Compute Ix, Iz and Ixz

• Step 3‐Compute bending stress σb

• Step 4‐

Find effective width

• Step 5‐

Compute Ix,Iy and Ixz and σb

• Step 6‐

Repeat Steps 1 to 5 till convergence

Iterative Procedure

b

• Shear stress due to applied shear

• Shear stress due to applied torsion

• Shear flow =fs

x t

• Concept of shear flow is used extensively.

Estimation of Shear Stresses.

Basic equilibrium condition

• (qn+1

‐qn

)

Δy

=P2

‐P1

Equilibrium Equation

Axial Load and shear flows on a member.

Shear flow due to shear

Also Written as where

Example of an open section‐Shear flow two common idealizations

V

Sheet carries bending stress

V

Sheet carries only shear

Moment due to shear flow‐closed cell

Torque ‘My’

applied to a closed cell results in uniform shear flow q

Bredt Batho Eqn

Shear flow

Consider a unit length

Strain energy U=

From castiglianos theorem

Twist of a Closed Cell

Procedure to obtain q0

• Step 1 Starting at any point obtain

• Step 2 Calculate moment MI

due to qI about a point O

• Step 3 Moment due to q0

about the point is q0

.2A •• Step 4 obtain q0

from moment equilibriumMI

+ q0

.2A =VX0

• Make a cut to introduce unknown shear flow q0

Single cell Box subjected to shear

• If the shear is applied through shear center the twist is zero• Assume‐

the shear flow q=qI

+q0

is for shear through S.C

Twist

Hence

• The shear center may then be obtained by taking moments of

shear flow q about point O

Shear Center

• Shear center lies on an axis of symmetry

• Shear center may not coincide with centroid.

• Line joining shear centers is the axis of twist

• Aircraft wing structure is generally unsysmmteric

• Flexural axis is the neutral axis(principal axis)

• Shear flow is zero at an axis of symmetry when loading is

also symmetric about that axis.

Points to Remember

Section with Geometry and load is shown below, determine the

shear stress in the skin and web and the location of the shear

center for the cell

D‐cell with two stringers

D‐cell with two stringers

D‐Cell with 3 Stringer section

Calculate the line of action of the shear center for the three‐stringer cell as

shown below.

Single‐Cell Multi Stringer section

• In a multi cell box there are as many unknown shear flows

as there are cells

• Twist cell equations need to be used to obtain them

• Twist of each cell is equal to twist of the box

Multi Cell Box‐Basic Principle

• Let there be N cells• Make N cuts and obtain shear flow q1

,q2

…qn

• Let the unknown shear flows be q10

,q20

…qno

• Twist of any cell i=

• Equating the twist of each cell

Eqn θ1

=

θ2

=

θ3

………………………

[N‐1]Equations

• Taking moment of shear flows about a point ‘o’

Where ds‐panel length,P‐perpendicular distance from ‘o’

• With these equations q10

,q20

….qN0

are obtained

Equations for Twist

Two cell boxSame example considered except a middle spar to make the box

become a two cell (three spars)

Two cell box

analysis of box beams 7

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