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Chapter A13

Bending Stresses

Introduction:

The bar in fig ais subjected to an axial compressive loadP.If the compressive stresses are such that no buckling of

the bar takes place, then the bar sections such as 1-1 and

2-2 move parallel to each other as the bar shortens

under compressive stress.

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In fig bthe same bar is used as a simply supported beam

with two applied loads P as shown.

The shear and bending moment diagrams are shown. The

portion of the beam between section 1-1 and 2-2 under

the given loading is subjected to pure bending since the

shear is zero in this region

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Experimental evidence from a beam segment xtakenin this beam region under pure bending shows that

plane sections remain plane after bending but they

rotate w.r.t each other as illustrated in Fig.C. Here thedashed lines represent the unstressed beam element

and the solid line is the element after bending (pure)

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Fig d shows a cantilever beam subjected to a pure

moment at its free end. Under this bending, the

deformation of the beam is shown to an exaggerated

shape in fig d

A13.1 Location of Neutral axis

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The applied bending moment acts in a plane perpendicular

to the z-axisin the figureConsider a beam element of length Lshown in dotted lines

The distortion of this element due to bending where plane

sections remain plane is shown by solid lines in Fig d

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It will be assumed that the beam section is homogenous and thebeam stresses are below the proportional limit stress of the

material

In other words Hookslaw appliesFrom the geometry of similar triangle

c

c

c

ee ee2 2= or =y c y c

ee= y 1c

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The stress and strain at a

distance, y, from N.A are

related by Hookslaw:

c

c

LE or

e eL

e eEE. y

L L c

eEy

L c

y

y y

y

e

cee= y

c

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Similarly the compressive stress in theoutermost fiber can be expressed as

cc

c

EeL

eECL C

c

c

eEy

L cE y

ce

L

y

y

c c y

c

y

Substituting by in the equaE

e :L

y .. B

ti

on f

C

or

ory c

=y C

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For equilibrium of the bending stress perpendicular to

the beam cross section or in the x-direction we can write

However in this expression is not zero hence the term

yda must equal zero

This can only be true if the material axis coincides with

the centroidal axis of the beam cross section

x

c

c

F 0da 0

yda 0

yda 0

o

r

y

c

c

c

c

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y

-y

da

da

yda

-yda

Y

Y

However in this expression is not zero hence the term ydamust equal zero

This can only be true if the material axis coincides with the

centroidal axis of the beam cross section

c

c

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The neutral axis does not pass through the

beam section centroid when the beam is

not homogenous, that is, the modulus ofelasticity is not constant over the beam

section and also when the Hooks law does

not apply or the -relation is non-linear

These beam conditions are described laterin the chapter

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Stresses from moments, section properties and distancesreferred to any pair of rectangular axes through the

centroid of the section

A beam cross section subjected to bending moments Mxand My. The expression for bending stress bdue to Mxand Myis

A13.5 Method 3

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Where Ix, Iy and Ixy are moments of inertia about the

rectangular axes X and Y passing through the centroid

of the cross section

Thus the bending stresses can be found without resort

to principal axes or to the neutral axes

b 3 y 1 x 2 x 1 y

xy1 2

x y

y

2 2

x y

x3 2

x y

(K M K M )x (K M K M )y . 14

IK

I I I

IK

I I I

IK

I I I

wh ree

xy

xy

xy

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A13.10 Bending stresses of homogenous beams

stressed above the elastic limit stress range

In airplane structural design, the limit loads on the

airplane must be taken by the structure without

suffering permanent strain. This means that stresses

should fall within the elastic range

The airframe should also be able to take the ultimateload without collapse or rupture with no restriction on

permanent strain

Many airplane structures will not fail until the stresses

are considerably above the elastic stress range of the

materials

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The - curve of a

material in theinelastic range is

non linear.

Also in the inelasticrange the - curvein tension is

different than that incompression.

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For these two

reasons the beambending stress

formula

does not apply inthe inelastic range

since its derivation is

based on the linear

relation between

-

b

My

I

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Tests, however, have shown that even in the

inelasticstress range, plane sections remain

plane during bending. This implies thatacross the section (on either side of N.A) the

strain still varies linearlywith distance from

N.A even in the inelastic range

y= y,

Strain variation

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Because of non-linear -relationship in the inelastic range thestress variation is not linear across the section. Nevertheless,

the inelastic stresses can still be found from the -relationship

y= y,

Strain variation

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A general method of approach to

solving beams that are stressed abovethe elastic range can be explained by

the solution of a problem

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Example Problem-7

Portion (a) of Fig A13.20 shows a solid round bar made from

24ST aluminum alloy material.

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Let it be assumed that maximumfailing compressive stress occurs

at a strain of 0.01 in/in

The problem is to determine theultimate resisting bending

moment developed by this

round bar and then compare the

result with that obtained by

using the beam bending formula

based on linear -relation

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Solution:

Since the stress-strain

diagram in tension and

compression are differentthe neutral axis will not

coincide with the

centroidal axis of the

round bar regardless of

the fact that it is

symmetrical in shape

Thus the method of

solution is a trial and error

one since the location of

neutral axis cannot be

solved for directly

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Solution:

Since the stress-strain diagram in tension and compression are

different the neutral axis will not coincide with the centroidal axis

of the round bar regardless of the fact that it is symmetrical in

shape

Thus the method of solution is a trial and error one since the

location of neutral axis cannot be solved for directly

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Solution:

In fig b of Fig13.20 the N.A has been assumed to be

0.0375inch above the centroidal axis

Solution:

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Solution:

It was assumed

towards the

tension side

because for a

given inelastic

strain the tensile

stress is higher

than thecompressive

stress

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Solution:

This higher tensile stress has to be

kept in mind in ensuring that the total

tensile force must equal the totalcompressive force on the cross

section

As stated earlier the problem assumes that a compressive unit

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As stated earlier, the problem assumes that a compressive unit

strain of 0.01will cause failure of the cross section

Fig (b) shows the linear strain variation with most remote fiber

compressive strain = 0.01passing through zero strain through theassumed N.A which is 0.0375above the centroidal axis

This will be the strain distribution on the cross section at the

assumed failure compressive strain =0.01

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We need to estimate what

internal resisting bending

moment will correspond to

this ultimate strain

distribution

Table A13.2 gives the detailed

calculations for the ultimate

internal resisting bending

moment

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Column-1

The cross section of the rod is divided

in to 20strips of 0.1thick as shown in

Fig A13.20 (a)

Column-2

gives the areas of each strip

Column-3

gives the distance of the mid point of

the strip to the center line of the crosssection

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Column-4Based on the ultimatestrain variation in fig (b) the

strain in each strip is calculated

as

on the compression side of NA

0.01

y 0.0375 1 0.0375

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Column-4Based on the ultimatestrain variation in fig (b) the

strain in each strip is calculated

as

on the compression side of NA

0.01

y 0.0375 1 0.0375

Column 5

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Column-5

for each strip-strain values in column-

4, the corresponding stress values for

the strip is obtained from -curve in

Fig A13.19

C l 6

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Column-6

The forces (compressive

and tensile) on each strip =

stress times strip area arecalculated

Column-7

gives the moments of thestrip-force about N.A

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The total internal resistingmoment = 56735 in lb

(column 7)

If we assume that stress also varies linearly

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y

with distance from N.A then the maximum

compressive stress corresponding to

= 0.01will be 48,500 PSI

If this stress is used as failing stress in the

beam formula then the bending momentwill be

I IM 48500 0.785

c c

38,000 in lb

3

for round I

bar4

r

c

I I

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Therefore, the linear beam formula will

predict an ultimate M =38000 in lb where

as the true Munder elasto plasticstress =

56735 in lb

The linear M is 67% Mtrue

I IM 48500 0.785 38,000 in lb

c c

3

for round I

bar

4

r

c

Fi A13 20 ( ) h h di ib i h

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Fig A13.20 (c) shows the true stress distribution on the

cross section which explains why the resisting moment is

higher

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