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Unformatted text preview: Finding Second Moments of Area for Thin Walled Sections Most of the beam crosssections found in the aircraft industry are made of a thin skin of material.
This might be a complex fabricated skin as in the wing crosssection or it might be just a simple folded shape as in a channel or zsection stiffener. In all of these cases the skin thickness, t, is very small and in ﬁnding the second moments of
area terms of order t2 and higher are ignored. Only terms of order are retained, In all cases lengths of sides composing the section are taken as the mid thickness lengths For
the thin skins considered here this will be essentially the same as the inner and outer lengths of the side.
b=t —e4—— Second Moments of Area of a Single inclined Locating the origin at the centroid the distance 3 is Face
measured locally along the member length and at some distance 5 there is an element of area dA as shown in FigureThe second moment of areas are then In =13?sz =13: sin.2 aid/i
[I .II I. I.
t 3.1 3.2
a  .rtsm o: 2 II. sm a
= Is‘stn‘ards: _.._ :_
r. 3 r 12 —2‘
id
12  sin of)2 = _1 his 1 A 2
  =— , . =—L
Ix dA 12 (l coscr) 12 x IL A
‘7“ = —Lsina.ﬂcosa = "Lrl’.
I” 12 12 ’ A
L: = I. cos 0: = projected t'ength on 3? axis
L}. = Lsin a projected length on )7 axis Example — Second Moments of Area 3 all members
Of lhiCki'lESS t X
— 1 2 I‘T—H It is desired to ﬁnd the second moments of area of the inclined Z—section shown in the
above Figure . The process for doing this can be broken down into a series of steps. Step 1 — Break the section into a series of straightline segments.
Segments #2, 23, 34 Step 2 — Find the area and centre of area for each segment.
12 Area = Sat. Centre of gravityr xcoord = 1.53 y—coord = 0
23 Area = Sat. xcoord = 1.5a ycoord = 2a 34 Area = 3st xcoord = 1.5a ycoord = 4a 30 Step 3 — Find the centre of area {centre of gravity) of the complete section.
By inspection xcoord = 1.5a ycoord = 2a Step 4 Find the second moments of area for each straig ht—Iine segment with respect
to its own local centroid. 1—2
23 34 Step 5 — Use the parallel axis theorem to ﬁnd the second moments of area of each
segment with respect to the centroid of the complete section. 12 23
34 Step 6 — Sum the results of step 5 to give the ﬁnal second moments of area. Resolving Shear forces There are occasions when it is necessary to resolve shear forces from Global direction i? Y ’ 5y
to Local direction i' y' In this case. the shear force transformation is given as x cos or — sin or 3 xi Figure 6 (13) sin or cos on S y 3" Inverse transformation is as follows cos 0L sin or
— sin 0: cos (1 Principal Stress lithe principal axes of the crosssection are used to calculate the section
properties then, by deﬁnition. Ix}. = 0 and the equation for the direct stress
simpliﬁes to Where all of the parameters are
transformed to the principal axis
direction. i.e. 5?; icos¢+§sin¢ G ~ _    21 i
y x 51“ q) + y cos 4) ( Direction of the Principal Axes 33 The crossproduct second moment of area is then — 21x}. Ixx _],v\ (23) For the principie axes 1 £9 = 0 giving tan 291: = This equation gives two angles ninety degrees apart and either one can be used.
Having chosen one then it follows using algebra very similar to equation ( 22) that
the principal second moments of area are: =1“ c052$~21wsin¢cos¢i+lyﬁsinzcb (24) 1“. =1,“ sin2 d) + 2]“. sin ¢cos¢ + 1).}. cos2 it (25} Having found the orientation of the principal axes the bending moments
can be transformed to the principal axes. .1 The Neutral Axis The stress varies linearly over the beam crosssection. For the bending
component, with any end load component removed. it will have values that are
both positive and negative over the section. This implies that there will be
a line on the section where the stress is zero. This is called the neutral
axis. The peak stress on the section will then occur at the point on the section
that is at the greatest perpendicular distance from the neutral axis If the stress is zero then from equation ( 15} Noting that for the deﬁnition of the angle w in Figure then 1E1” My!“ —Mx I”
tanWZ—TZ————
M I Mg” —M 1 x xv y 90’ Where Wis measured anticlockwise from the Cm axis. Alternatively Alternatively a : Mylii
Mxlﬁ, tanw Where 1?! is measured ciockwise from the GS: principal axis. Principal Axes of a Z—Section The angle between the original Cartesian axes and the principal axes is, from equaﬁon
s___4 all members
DI‘ Inlckness t the principal second moments of area of the Z—section are 2   2 _ 3
In =1“ cos d)—2IXY smqicoquIW sm ¢—3.17313at I“, :1xx sin2 d) + 21x), sin¢cos¢+ IW cos2 ¢=0.71853a3t
3? Y} Modification to Ctassicst Beam Theory for Thin Walled Beams and Tubes The aircraft structure is a beam but it is special in that it is essentially a thin
walled hollow tube form of beam. This means that extra assumptions can be made and some of the assumptions in
classical beam theory to be dropped, The modiﬁcations to classical beam
theory for thin walled tubes are: e. The stress over the complete cross~section varies linearly as in d but the stress
is constant through the thickness of the skin. f. The shear loads give rise to shear stresses within the thin skin and these shear
stresses are not insignificant with respect to the direct stresses. g. The shear stresses do not distort the crosssection but they can contribute to
the overall deﬂections ofthe beam. h. In addition to shear and bending loads the tube can also experience signiﬁcant
torsional loads which give rise to shear stresses in the skin. i. The torsional loads can cause out of plane distortions of the cross~section, i.e.
under torsion plane sections do not remain plane. This distortion is called the 33
warping of the crosssection. Shear Stress and Sheer Strains in Thin Walled
Tube Bending Theory “W Direction
of the
normal to
the
surface There is no component of loading in the x and y directions and. by deﬁnition: when compared
to its length the beam is short in these directions. These two conditions, that there is no load in
the x and y directions and there is no signiﬁcant distance for any internal stresses to build up,
means that the assumption; 0' = (T)? = O (23) 39 XX Significant Stress Components In the Thin Walled Skin
3 C On the thin skin the stresses in the through thickness tdtrectlon, 0 M Gt 6 m 0
can all be assumed to be insignificant. They all have to be zero on the top and bottom
surfaces of the skin and. since the skin is thin, they cannot develop signiﬁcant values
within the skin. 0' ts st 40 The Shear Flow The shear stress GSZ will have a signiﬁcant value since it is equilibrating the applied shear
loads. it is varying in some manner through the
skin thickness typically as shown in the Figure q : Canaverage ) t (29)
Typical a”. Distribuliou Equilibrium of Skin Stresses For moment equilibrium about the centre of the element ignoring second order terms gives Considering equiiibrium of forces in the
zdirection then Substituting for q = 0'3! gives Which simpliﬁes to Recalling Differentiating with respect to 2
as: _
6—2 _ Muitiplying by t gives since tag“ =w§ﬂ 6?. ('35 Now recalling s and S _ 3M
x .. y _ We can write (31) where (32) Equation ( 31} then becomes
_ “[5303: (32)
' as Integrating this to some distance 5 around the periphery of the tube gives the shear ﬂow
at that point as q = (33) where Dx(s)=—It?ds and Dy(s)=~ftids
D D (10 is the value of the shear ﬂow at S = 0 lfthis corresponds to a free edge. as in the open tube then go = 0 if s is measured from the free edge. Hence, the shear ﬂow (and the average shear stress q/‘L‘ ) at any
point s on the tube periphery for an OPEN TUBE is given by S g 34)
s=—5’D+*D ‘
c101 x I y XX Example  The Shear Flow in an Open Channel Section Subject to 3 Vertical
Shear Force Sy The channel section is symmetric about the
y axis and. from symmetry arguments
alone, the centre of area lies on the line of
symmetry and the line of symmetry is always
a principal axis. From earlier calculations
the shear flow on the _
vertical web is given by 10 Example ~ The Shear Flow in an Open Channel Section Subject to 8 Vertical
Sheer Force 8y h
To simplify the algebra put [3 = a Step 1 — Calculation of the Centroid
Taking the origin of the coordinate axes at the centre of the vertical web‘ then Step 2 — Calculation of Second Moments ofAree th3 6 2d3t 2
1..=— 1+“ ; I‘..——q 2+5 +2 ; IU=0 12[ a] * 3i2+m~( B B) ' Step 3  Calculation of the D x diStriblﬁion 5 Dx(s)=—It§ds Cl 11 Comments upon the Shear Flow Distribution in an Open Section At any free edge the shear ﬂow has to be zero. In the example this is used at
point 1 to start the integration prooess. However, it is also true at point 4. the other free end. if the calcuiation of the shear ﬂow does not result in a zero value at a freeend
then the calculation is in error. Integrating D x around the complete section gives as Dx = —SJ.tYds = —SjydA
ﬂ 0 This is the ﬁrst moment of area about the Z centroidal axis. which, by deﬁnition. is zero. Hence the ﬁnal shear ﬂow at a free end must be zero. 12 ...
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 Winter '09
 S.PinhoandM.Aliabadi

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