Aero Structures

45
Reminders • HW #3 due Thursday 10/6

description

Notes

Transcript of Aero Structures

Page 1: Aero Structures

Reminders

• HW #3 due Thursday 10/6

Page 2: Aero Structures

Thin-walled Sections- Review

Page 3: Aero Structures

Thin-walled Sections

• Aircraft structures are characterized by thin sections- we can use this to simplify analysis

Assumptions:

– Thickness is small compared to the cross-section

– Stress is constant across the thickness

– Higher order terms of t can be neglected

1.0max b

t

Page 4: Aero Structures

Thin-walled Sections

• In calculating the second moment of inertia, we can neglect higher order terms of t to get:

NOTE:

Thin-walled assumptions in the case of bending only affects the cross-sectional properties. The formulation is unchanged.

12

22

32 h

tbthI xx

Page 5: Aero Structures

Thin-walled Sections

• We also assumed that the stress does not change through the thickness.

A B C D A B C D

τzx distribution in the flange

Page 6: Aero Structures

Thin-walled Sections

• Most of the shear stress is carried by the web

• This means we can approximate the beam with a web and two connected (lumped) areas for the flange

• In this simplified model:–Web carries transverse shear

– Flanges carry bending loads

Page 7: Aero Structures

Shear of Beams

Page 8: Aero Structures

Shear of Beams

• This theory applies to:1. Open section under shear load

2. Closed section under shear and/or torsion

NOTE: Torsion of open sections requires a different formulation

Page 9: Aero Structures

Shear of Beams

• Assumptions:1. Axial constraints are negligible

2. Shear stresses normal to the surface are negligibleThey must be zero at the free surface and the thickness must be small

1. Direct and shear stresses on planes perpendicular to the surface are constant across the thickness

Page 10: Aero Structures

Shear of Beams

• Assumptions:4. The beam is of uniform cross-section so that the

thickness may vary around the section (s-coordinate) but is constant along the length of the beam

5. In the section properties calculation, neglect higher order terms of the thickness t

Page 11: Aero Structures

Shear of Beams

• Consider an element of the wall of a loaded beam of size δs x δz x t

Figure 1: Shear of closed section beam

Figure 2: Stress on element of beam (fig 17.1a).

Page 12: Aero Structures

Shear of Beams

• From elasticity (eqn 1.4) we deduce that:

• In terms of shear flow, q, can be written as:

szzs Complimentary Shear Stresses

tq Shear Flow

Figure 2: Stress on element of beam (fig 17.1a).

Page 13: Aero Structures

Shear of Beams

• Replacing the shear stress values with shear flow, the elemental diagram can be written as:

Figure 3: Stress on element of beam (fig 17.1b).

Page 14: Aero Structures

Shear of Beams

• Equilibrium in the z-direction without body forces:

• Dividing by δz and in the limit as δz0:

0

zqzss

qqststz

zF z

zzz

Figure 3: Stress on element of beam (fig 17.1b).

0

zt

s

q z

Page 15: Aero Structures

Shear of Beams

• Using equilibrium in the s-direction gives:

Figure 3: Stress on element of beam (fig 17.1b).

0

st

z

q s

Page 16: Aero Structures

Shear of Beams

Figure 4: Axial, tangential and normal components of displacement(fig 17.2).

• We can define:w = displacement in the z axisvt = tangential displacement (+ with increasing s)

vn = normal displacement (+ outwards)

• From elasticity (eqn 1.18):

z

wz

Page 17: Aero Structures

Shear of Beams

• Shear strain is then defined as the addition of the two angles of rotation of the sides:

• As both δs and δz

tend to zero:

Figure 5: Element distorted due to shear (fig 17.3).

Distorted shape of element due to shear

21

z

v

s

w t

Page 18: Aero Structures

Shear of Beams

• It is now necessary to define the term vt as a function of displacements u and v (in x & y) and angle of twist of the section θ

• Further assumptions:– The shape of the beam cross-section is maintained

– There is no resistance to axial displacement

Page 19: Aero Structures

Shear of Beams

Figure 6: Rotated beam cross-section.

• Deffine:ψ =  Angle between the tangent to the surface of the

beam's cross section and the x-axis.

• Tangential displacement at any point N:

sincos vupvt

Page 20: Aero Structures

Shear of Beams

• Relating these displacements about a point R, which is the centre of twist, we can rewrite the displacement as:

Figure 6: Rotated of beam section around the center of twist (fig 17.4).

cossin

cossin

RRt

RRR

Rt

yxpv

yxpp

pv

Page 21: Aero Structures

Shear of Beams

• Differentiating with respect to z gives:

Figure 6: Rotated of beam section around the center of twist (fig 17.4).

sincos

cossin

dz

dv

dz

du

dz

dp

dz

dv

dz

dy

dz

dx

dz

dp

dz

dv

t

RRt

dzd

dzdu

y

dzd

dzdv

x RR ,

Page 22: Aero Structures

Shear of Beams Open Sections

• Look at a beam with an open section, with applied shear forces Sx and Sy about a point which produces no twisting of the cross section.

Figure 7: Open beam section loaded with two shear loads (fig 17.5).

This point is the shear center.

Page 23: Aero Structures

• The equation for direct stress is (eqn 16.18):

Shear of Beams Open Sections

Figure 7: Open beam section loaded with two shear loads (fig 17.5).

yIII

IzM

IzM

xIII

IzMIz

M

z xyyyxx

xyy

yyx

xyyyxx

xyx

xxy

z22

Page 24: Aero Structures

Shear of Beams Open Sections

• We can use the relationships below to simplify the previous equation (eqns 16.23&24):

*Shear is now indicated with S*

Figure 7: Open beam section loaded with two shear loads (fig 17.5).

yIII

ISISx

III

ISIS

z

SVz

M

SVz

M

xyyyxx

xyxyyy

xyyyxx

xyyxxxz

yyx

xxy

22

Page 25: Aero Structures

• Recall the relationship between shear flow and axial stress:

• This gives:

Shear of Beams Open Sections

Figure 7: Open beam section loaded with two shear loads (fig 17.5).

zt

s

q z

tyIII

ISIStx

III

ISIS

s

q

xyyyxx

xyxyyy

xyyyxx

xyyxxx

22

Page 26: Aero Structures

• Integrating, we get:

• If the origin is considered to be at the open edge of the cross-section, then when

Shear of Beams Open Sections

s

xyyyxx

xyxyyys

xyyyxx

xyyxxxs

tydsIII

ISIStxds

III

ISISds

s

q

02

02

0

s

xyyyxx

xyxyyys

xyyyxx

xyyxxxs tyds

III

ISIStxds

III

ISISq

02

02

0q 0s

Shear Flow Equation:General Case

Page 27: Aero Structures

• If Cx or Cy is an axis of symmetry, then

Shear of Beams Open Sections

x

xy

y

yx

s

xx

ys

yy

xs

I

QS

I

QS

tydsI

Stxds

I

Sq

Q Q xy

00

0xyI

Shear Flow Equation:Symmetrical Case

Figure 7: Open beam section loaded with two shear loads (fig 17.5).

Page 28: Aero Structures

Shear of Beams Example 1

Determine the shear flow distribution in the thin walled channel section loaded by a single vertical force applied through the shear centre.

Figure 8: Shear loaded z-section (fig 17.6).

Page 29: Aero Structures

Shear of Beams Example 1

• Since the applied vertical load passes through the shear center, no torsion exists, so shear flow equation applies.

or

s

yy

s

xyxyyyxx

ys

s

xyyyxx

yyys

xyyyxx

xyys

tydsItxdsIIII

Sq

tydsIII

IStxds

III

ISq

002

02

02

Page 30: Aero Structures

• The second moments of area of this section are:

• Substituting these into our shear flow equation:

s

ys dsyx

h

Sq

03

84.632.10

8 ,

12 ,

3

333 thI

thI

thI xyyyxx

Shear of Beams Example 1

Page 31: Aero Structures

• Evaluating the bottom flange-12 for

we have:

This gives:

121312

0

11312

74.116.5

74.132.101

hssh

Sq

dshsh

Sq

y

sy

12

2

shx

hy

20 1hs

Shear of Beams Example 1

S1=0 S1=h/2

Page 32: Aero Structures

• In flange-12:

121312 74.116.5 hss

h

Sq y

Shear of Beams Example 1

S1=0 S1=h/2

1

0)0(74.1)0(16.5 231 h

h

Sq y

h

S

hh

h

h

Sq

y

y

42.0

274.1

216.5

2

32

2

Page 33: Aero Structures

• Evaluating the web-23 for

we have:

This gives:

22 shy

hs 20

222

2323

0

222323

42.342.342.0

84.642.31

shshh

Sq

qdsshh

Sq

y

sy

RECALL:

Shear of Beams Example 1

hS

q y42.02

S2=0

S2=h

Page 34: Aero Structures

• Shear flow in z-section:

Shear of Beams Example 1

Figure 9: Shear flow in the z-section (fig 17.7).

Page 35: Aero Structures

Shear of Beams Shear Center

• Shear Center– Point in the cross section through which the shear

loads produce no twisting.

– Center of twist of sections when torsional loads are applied.

– As a rule, if a cross-section has an axis of symmetry, then the shear center must lie on that axis and in cruciform or angle sections, the shear centre is located at the intersections.

Page 36: Aero Structures

Shear of Beams Shear Center

• Shear Center– It is important to define the position of the shear

center because although most wings are not loaded at this point, if we know its location, we can represent the shear loads applied as combinations of shear loads through the shear center and a torque.

Page 37: Aero Structures

Shear of Beams Shear Center

• Shear Center

SC

SC

SC

Figure 10: Shear center position for open section beams (fig 17.8).

Page 38: Aero Structures

Shear of Beams Shear Center

• Calculating Shear Center– Determine the moment generated by the shear flow

about an appropriate point in the cross section.

– This moment is equal to the moment generated by the applied shear force about this same point.

shearflowapplied MM

Page 39: Aero Structures

For the open beam section shown determine the position of the shear center

Shear of Beams Example 2

Figure 11: Determination of shear center (fig 17.9).

Page 40: Aero Structures

• Because shape is symmetrical the shear centre must lie on the x-axis, a distance ξs from theweb

• The steps to determine the shear center are:1. Determine the equations which describe the shear

flow in the cross section

2. Find an appropriate point in the cross section and take moments about it.

Shear of Beams Example 2

Page 41: Aero Structures

Determine the equations which describe the shear flow in the cross section:

•Recall the general equation for shear flow:

•For Cx, and so this simplifies to:

Shear of Beams Example 2

s

xyyyxx

xyxyyys

xyyyxx

xyyxxxs tyds

III

ISIStxds

III

ISISq

02

02

0xyI 0xS

h

bththhbtItyds

I

Sq xx

s

xx

ys

61

121222 where,

332

0

Page 42: Aero Structures

• Substituting for gives:

• On the bottom flange:

Shear of Beams Example 2

xxI

sy

s yds

hbh

Sq

03 61

12

121261

6

2

s

hbh

Sq

hy

y

Page 43: Aero Structures

• On the bottom flange:

• Substituting in for q12:

Shear of Beams Example 2

b

sy dsh

qS0

112 22

1

0

12 261

62 dss

h

hbh

SS

by

sy

Page 44: Aero Structures

• Solving for :

• In the case of unsymmetrical sections, you must find : 1. Apply a vertical shear load, , and find

2. Apply a horizontal force, , and find

Shear of Beams Example 2

hbh

bs

61

3 2

s

syS

sxS

ss ,

Page 45: Aero Structures

Reminders

• HW #3 due Thursday 10/6