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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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PAGE 15.

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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PAGE 16.

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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PAGE 18.

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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PAGE 19.

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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PAGE 20.

A.y.x - I +Axy = I xy(Self))xy(Section ΣΣΣ