lecture notes msc - bending-v1
TRANSCRIPT
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AEROSTRUCTURES -BENDING STRESS
Introduction:
In the following, the theory will be developed to enable an analysis to be made, of the bending stresses in beams. Initially, bending stress relationships will be derived and then the theory and practicalities of finding principal axis properties for various beam geometries will be illustrated. The notes then address the analysis of shear stress distribution in a beam section.
Bending Stresses
This section firstly derives the fundamental relationship between the bending moment, second moment of area, direct stress, Young's modulus, radius of curvature and position in the beam cross section for a beam. This is then taken a step further with the relationship made to the differential equation of the deflection curve; used to calculate gradients of and deflections in beams.
For the portion of beam shown above, when subject to bending, the 'top' will be
in compression and the bottom will be in tension. Consequently, it follows that there must be a portion in between bearing no stress. This is known as the neutral plane or in a longitudinal sense, the neutral axis, N.A., and is where the longitudinal deformation is zero
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Now, consider a fibre EF, between sections AC and BD, which is dx long. Initially, EF will be the same length as GH at the N.A. The beam bends and EF extends to E'F'; but GH stays unstrained to G'H'.
Let R be the radius of curvature of G'H', therefore
We can write the longitudinal strain as
but
so it follows that
thence
θR.d =
dx = HG = GH ′′
Also, it follows that
θy).d + (R = FE ′′
EF
EF - FE = x
′′ε
θR.d =
HG = GH = EF ′′
θ
θθε
R.d
R.d - y).d + (R = x
R
y = xε
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so
therefore
Stress-Strain Relationship
We may write
and therefore
So, by substituting in for equation (1)
or alternatively
Further more, as θR.d = dx
dx
d =
R
1 θ
.ydx
d = x
θε
ε
σ
x
x = E
E = x
x
σε
R
y =
E = x
x
σε
R
E =
y
xσ
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Now consider the force on an elemental area due to bending
therefore, the total force in the x-direction (longitudinal)
where A = the total area of the section.
The total force over the section must be ZERO (as no external force is acting over the section, i.e. ΣF = 0). Consequently we may write
Substituting equation (2) for σx gives
but as E/R is a constant it follows that
and hence
this is the first moment of area of the section about the N.A. The 1st moment of area about its centroid is zero, and therefore the N.A. and centroid are coincident.
Taking moments, the elemental force dFx acts at a distance y from the N.A.,
therefore
.dA = dF xx σ
.dA = F xAx σ∫
0 = .dA = F xAx σ∫
0 = y.dAR
EA∫
0 R
E≠
0 = y.dAA∫
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where dM = the moment of axial force @ the neutral surface.
Consequently, for the whole section
(as dFx = σxdA)
The externally applied moment M must balance this moment, i.e. ∫AdM = M and hence
Again substituting for σx from equation (2) (i.e. E/R = σx/y) gives
Now, ∫Ay2
dA is the second moment of area of the cross section and denoted I. Hence we may write that
or alternatively
So the final bending relationship may be written as
The relationship σx = My/I is commonly used.
dFy. = dM x
dAy. =
dFy.dM
xA
xAA
σ∫
∫=∫
.dAy = M xA σ∫
.dAyR
E = M
2
A∫
IR
E = M
EI
M =
R
1
y =
R
E =
I
M xσ
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The Curvature-Bending Moment Relationship.
The figure above shows a beam that is bent by a small amount. Now, it can be seen that
(as dθ is -negative for positive increments of ds)
It is true that ds ≈ dx, as the beam is subjected to a small deformation, and that θ
≈ tanθ. Additionally, tanθ = dv/dx where v are vertical deflections which are relatively small.
Hence we may write that θ = dv/dx and differentiate with respect to x giving
which upon substituting ds for dx
θR.d = ds and that
ds
d- =
R
1 θ
dx
vd =
dx
d2
2θ
dx
vd =
ds
d2
2θ
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Previously we showed that
Using the equation derived earlier
we may arrive at the relationship
or, rewriting in a more common form
We may use this equation to calculate displacements by integrating twice over.
Thence
dx
vd- =
R
12
2
dx
vdy- =
R
y =
2
2
xε
EI
M =
R
1
EI
M- =
dx
vd2
2
M- = dx
vdEI
2
2
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Principal Axes
In this section we will look at the analysis of stresses developed in symmetrical and then unsymmetrical beam sections. For this analysis we need to sum or integrate expressions for the total area in question. The Integrals which need to be evaluated are:
1. the first moment of area 2. the second moment of area 3. the product moment of area
The transformation of axes to principal axes is then evaluated with examples of the practical application of this theory. First Moment of Area
Used to find the position of the centre of area, or centroid of a shape.
where C.A. is the Centre of Area
For Example
from which we can find the centroidal distance, i.e. y .
y.dA = yAA∫
y.A + y.A = y).A + A(221121
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Second Moment of Area
If the first moment of area of an element dA is again multiplied by its co-ordinate then the second moment of area results, namely
For Example
giving
And similarly
Parellel Axis Theorem
There are numerious cercumstances when it is most useful to be able to determine the second moment of area about an axis parallel to the centroidal axis.
Taking the second moment of area of the element dA about the x' axis gives (y+b)
2.dA. So for the entire section
.dA)b +(y = I2
Ax ∫′
.dAy2
A∫
∫
3
by =b.dy y = I
3d/2
d/2-
2d/2+
d/2-x
12
bd = I
3
x
12
db = I
3
y
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and multiplying out
Now, as ∫A y dA = 0 (the 1st moment of area) about its centroidal axis, it follows that
And similarly
Product Moment of Area
So far in lectures we have assumed that the beam sections are symmetrical about the plane of bending
.dAb + y.dA2b + .dAy = I2
AA
2
Ax ∫∫∫′
Ab + I = I2
xx′
Aa + I = I2
yy′
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Now, if a bending moment is applied to a beam about the x-x axis such that no bending occurs about the y-y axis then
as the sum of the internal forces must be ZERO
Now we know that
and therefore it follows that
However, as M and Ix are constant then
This last integral is known as the product moment of area or, sometimes, the
product of inertia and is measured about the x-x and the y-y axes. In this case, it was shown for pure bending: when the product moment of area is taken about the centroidal axes and perpendicular to the plane of bending it is ZERO.
When the product moment of area is ZERO about a set of mutually perpendicular axes, then these axes will be principal axes; and will be the principal axes of the section in question. The second moments of area about a set of principal axes are called the principal second moments of area; being maximum and minimum values about any set of axes passing through the centroid. Moments of Area About Inclined Axes
So, for the unsymmetrical bending of beams it can be necessary to consider bending about a rotated set of axes; that is relative to our reference axes used previously.
0 = x x .dAAσ∫
I
My =
x
xσ
0 = x.dAI
My
x
A∫
0 = xy.dAA∫
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Let us consider the figure above which shows a set of reference axes denoted x and y with second moments of area Ix, Iy and Ixy. Also shown are a set of axes equivalent to the previous axes rotated anti-clockwise through and angle θ and denoted x', y'; with their respective second moments of area Ix', Iy' and Ix'y'.
Now it follows that x' = xcosθ + ysinθ
y' = ycosθ – xsinθ
and so as Ix' = ∫A y2.dA we can write
Expanding and rearranging
In a similar manner the procedure may be repeated for Iy',
arriving at the following result
θθ 2I + 2)I - I(2
1 - )I + I(
2
1 = I xyyxyxy sincos′
Finally, we need to calculate a corresponding result for the product moment of
area. So
.dA)x - (y = I2
Ax θθ sincos∫′
θθθθ cossinsincos I2 - I + I = I xy2
y2
xx′
θθ 2I - 2)I - I(2
1 + )I + I(
2
1 = I xyyxyxx sincos′
.dA)y + (x = .dAx = I2
A
2
Ay θθ sincos∫∫′
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axes for which the product moment of area is zero are the principal axes, and so by setting the final equation for Ix'y' = ZERO we may determine the angle of the axes about which the maximum and minimum principal second moments of area occur.
Hence
which upon rearranging yields
The maximum and minimum principal second moments of area are given the
notation Iu and Iv. It should be noted that Iu + Iv = Ix + Iy.
In Summary, so far.
• The axes for which Ixy is zero are called the principal axes of a section and are the axes about which bending takes place.
• For a symmetrical section the axes of symmetry are also principal axes.
• For an unsymmetrical section, e.g. an angle section, the position of the principal axes must be determined.
).dAx - )(yy + (x = dAyx = I AAyx θθθθ sincossincos∫′′∫′′
).dAxy - y + x - (xy =222
Aθθθθθθ 2
sinsincossincoscos∫
).dAx - y( + )xy.dA - ( = 22
A
22
Aθθθθ sincossincos ∫∫
.dA]x - .dAy.[22
1 + xy.dA.2 = 2
A
2
AA ∫∫∫ θθ sincos
)I - I.(22
1 + I.2 = I yxxyyx θθ sincos′′
)I - I.(22
1 = I.2 xyxy θθ sincos
I - I
I2 = 2
xy
xyθtan
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Asymmetrical Bending Introduction
So far, our analysis has concerned itself with finding first and second moments of area, product moments of area, and then applying these properties of a shape to identifying the principal axes. Now, it is also important to be able to analyse the bending in a component which either does not have any axes of symmetry or for a symmetrical section about an asymmetrical axis. We will now, thus, consider a beam with unsymmetrical bending moment or skew bending. Beam With Unsymmetrical Bending Moment
Consider an arbitrary section, shown above, in which a bending moment M acts in the direction of the y-axis, i.e. the bending is about x-x. The principal axes, u-u and v-v, are inclined at an angle α to x-x and y-y. Now, the bending moment may be resolved onto the principal axes thus
and
From the engineer's theory of bending the direct stress along the axis of the
beam, σz, is given by
αcos M.= M u
αsin M.= M v
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Now, if we consider an elemental area dA having ordinates u and v relative to the principal axes, and also given that the total stress axial to the beam may be formed by a superposition of the effects of the two moments Mu and Mv, then it follows that
Or, simplifying
When there is a moment inclined to the principal axes then the neutral plane will
no longer be perpendicular to the plane of bending. However, the neutral plane can be simply determined by setting the last equation equal to ZERO, i.e. σz = 0 which is the
case at the neutral plane. Thus
and comparing with the equation of a straight line y = mx + c
i.e. a line with a gradient -(Iuu/Ivv.tanα) Consequently we may give the inclination of the neutral axis as a result of this unsymmetrical bending moment. Denoting this inclination φ and referencing it from the principal axis u-u
Alternative Treatment Asymetrical pure bending of a beam is considered for positive external moments of My and Mx applied about centroidal axes. Considering the zy-plane (where the z-axis runs down the beam axis) equations of equilibrium may be written, thus
I
My = zσ
.uI
M. + .v
I
M. =
vvuu
z
αασ
sincos
.u
I + .v
I M.=
vvuu
z
αασ
sincos
I
u.- =
I
v.
vvuu
αα sincos
.uI
I- = v
vv
uu
αtan
αφ tantanI
I- =
uu
vv
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0=∫A
zdAσ
x
A
z MdAy =∫ .σ and y
A
z MdAx =∫ .σ
However, we can recall that
y
zR
yE=σ
where Ry is the radius of curvature of the zy-plane.
Consequently we may write the following
∫=Ay
x dAyR
EM
2
y
x
R
EI=
Similarly for My in the zx-plane
∫=Ay
y dAyxR
EM .
y
yx
R
EI=
For bending only in the yx-plane may be treated in a similar manner
∫=Ax
y dAxR
EM
2
x
y
R
EI=
Similarly for Mx in the zx-plane
∫=Ax
x dAyxR
EM .
x
yx
R
EI=
For simultaneous bending the above results may be superimposed to yield the following
y
yx
x
y
yR
EI
R
EIM =+=
x
yx
y
x
xR
EI
R
EIM =+=
The above equations can be solved to give radii of curvature in y and x
)(
12
yxxy
yxyyx
y IIIE
IMIM
R −
−=
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PAGE 17.
)(
12
yxxy
yxxxy
x IIIE
IMIM
R −
−=
It should be noted that the last but one pair of equations can be re-written in
matrix form
=
y
x
xyx
yxy
x
y
R
R
EIEI
EIEI
M
M
/1
/1
So by taking the inverse matrix the radii of curvature can be determined
=
−
x
y
xyx
yxy
y
x
M
M
II
II
ER
R1
1
/1
/1
−
−
−=
x
y
yyx
yxx
yxxyy
x
M
M
II
II
IIIER
R
)(
1
/1
/12
The resulting bending stress is thus the summation of the components for
bending in each of the zy and zx-plane.
xy
zR
xE
R
yE+=σ
Leading to
yIII
IMIMx
III
IMIM
xyyx
xyyyx
xyyx
xyxxy
z
−
−+
−
−=
22σ
Alternatively, the last equation can be rearranged in the following way
−
−+
−
−=
22
..()..(
xyyx
xyxxy
xyyx
xyyx
zIII
yIxIM
III
xIyIMσ
It should be noted that if either My = 0 or Mx = 0 a stress will still be produced
that varies in both x and y.
Also when Ixy = 0 the bending stress relationship can be simplified to the following
+
=
y
y
x
x
zI
xM
I
yMσ
This condition will occur when either or both the centroidal axes lie as axes of
symmetry and are thus principal axes. The position of the neutral axis is such that it always passes through the
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centroid of the section and can be determined by setting the value of σz equal to zero.
0)()( =−+− yIMIMxIMIM xyyyxxyxxy
Tabular Method: First, Second & Product Moment of Area Calculation
A useful method of calculating the properties of areas for a chosen section is by the use of a table. The method will be shown by way of an example. Example
Determine the position of the centroid and the centroid second moments of area for the following section
All dimensions in Centimeters
Firstly a table should be draw up with the following format taking values with a x-y co-ordinate datum at the bottom left hand corner of the section. (In order to keep numbers small centimetres have been used; however this is not a preferred unit.)
(Self) Item
b
d
A
x
y
Ax
Ay
Ax
2
Ay
2
Axy
Ix
Iy
Ixy
1
5
2
10
7.5
9
75
90
565
810
675
3.3
20.8
0
2
2
6
12
6
5
72
60
432
300
360
36
4
0
3
10
2
20
5
1
100
20
500
20
100
6.7
167
0
Σ
42
247
170
1495
1130
1135
46
191.5
0
Then after calculating the summation values and placing them in their relavent
column in the last row the following calculations can be performed.
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Co-ordinates of Centroid
Second Moment of Area About reference axes (R.A.)
About centroidal axes
n.b. The calculation steps for the second moment of area can be condensed by using the three equations below, which are simply an amalgamation of the two previous steps
5.88cm = 42
247 =
A
Ax = x
Σ
Σ
4.05cm = 42
170 =
A
Ay = y
Σ
Σ
cm1176 = 46 + 1130 = I + yA = I4
x(Self)
2
x(RA) ΣΣ
cm1686 = 191.5 + 1495 = I + xA = I4
y(Self)2
y(RA) ΣΣ
cm1135 = 0 + 1135 = I +Axy = I4
xy(Self)xy(RA) ΣΣ
cm486 = x42)4.05( - 1176 = Ay - I = I422
x(RA)x(Section) Σ
cm233 = x42)5.88( - 1686 = Ax - I = I422
y(RA)y(Section) Σ
cm135 = x42)5.88x4.05( - 1135 = Ay.x - I = I4
xy(RA))xy(Section Σ
A.y - I + yA = I2
x(Self)
2
x(Section) ΣΣΣ
A.x - I + xA = I2
y(Self)2
y(Section) ΣΣΣ
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A.y.x - I +Axy = I xy(Self))xy(Section ΣΣΣ