Aero Structures
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Transcript of Aero Structures
Reminders
• HW #3 due Thursday 10/6
Thin-walled Sections- Review
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
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
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
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
Shear of Beams
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
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
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
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).
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).
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).
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
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
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
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
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
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
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
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 ,
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.
• 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
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
• 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
• 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
• 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).
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).
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
• 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
• 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
• 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
• 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
• Shear flow in z-section:
Shear of Beams Example 1
Figure 9: Shear flow in the z-section (fig 17.7).
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.
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.
Shear of Beams Shear Center
• Shear Center
SC
SC
SC
Figure 10: Shear center position for open section beams (fig 17.8).
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
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).
• 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
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
• 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
• 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
• 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 ,
Reminders
• HW #3 due Thursday 10/6