bending theory 2013x

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Bending theoryBending theory

CE2182

07/02/13

Lecture OutlineLecture Outline

� Section properties: centroids

� Second moment of area

� Theory of bending

◦ Assumptions

◦ Second moment of area

◦ Elastic section modulus

Section propertiesSection properties

� Area of a section

� Centre of gravity and centroid

total iA A=∑

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CentroidCentroid

� Position of centroid?

Second moment of area ISecond moment of area I

� If

determine the second moments of area for the shapes

3 3

and 12 12yy zz

bd dbI I= =

Bending stressBending stress

� When a beam is loaded, it will deflect. At every cross section internal strains and stresses are introduced.

� Their distribution across the depth of the cross section is not uniform

� Their distribution along the length of the beam is not uniform

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� Let’s assume that the beam is made up of different layers

� If there is no bond/ adhesion between the layers they will deform independently

� A real beam will deflect without any relative slipping between layers


� Basic kinematic assumption

◦ Plane sections though a beam taken normal to its axis remain plane after the beam is subjected to bending

◦ The top fibres are in compression and the bottom fibres are in tension.

◦ The distribution of strains and stresses is changes across the depth of the beam, therefore there is a layer which is neither in compression nor in tension!


� The deformed shape of a loaded beam is assumed to be a circular arc with a centre of curvature O and radius R to the neutral level of the beam.

O

R

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� We will take a small segment of our beam ABCD

� Before loading AB=A’B’ and CD=C’D’

� After loading

AB<A’B’ and CD>C’D’

Therefore we can draw a diagram of the change in length (ds) of any layer


� If we express the change in length/ original length then that will give the strain distribution across the depth of the beam


� One of the important theoretical assumptions is that the strain distribution is linear. Also, the strain distribution in a layer is proportional to the distance from the neutral axis

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� For elastic analysis

� Or the stress is directly proportional to strain

Eσ ε= ×

Bending stressBending stress� The resultants of the stress blocks are

thus C and T, which act through the centroids of the blocks (triangles)

Bending stressBending stress� The resultant forces are:

max max

max max

average stress x area=2 2 4

average stress x area=2 2 4

, T=C

bdb dC

bdb dT

Therefore

σ σ

σ σ

××= × =

××= × =

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Bending stressBending stress� T=C, acting in opposite directions,

therefore they form a couple

Example: Minimum height of a rectangular beam with width 150mm if M=30kNm and max bending stress is 30MPa?

2max max2 2

3 4 3 6

bd bdd dM C

σ σ× ×= × = × =

2 26 2max 30 150

; 30 10 ; 400006 6

bd dM d

σ × ×= × = =

Bending stress : IBending stress : I� Let’s have a look at an irregular cross

section

Bending stress : IBending stress : I

� Force in the thin strip

� Moment of that force about NA

� Total moment of the forces

� The total moment M

because

Here,

b yσ δ=

b y yσ δ= ×1

2

y

y

bydyσ−

= ∫

12max

1 2

y

y

by dyy

σ−

= ∫

max

1y y

σ σ=

12

2

y

y

I by dy−

= ∫

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I for rectangular sectionI for rectangular section

� For a rectangular cross section

/21 /2 3 3 3 32 2

2 /2 /23 24 24 12

dy d

y d d

y d d bdI by dy by dy b b

− − −

= = = = + =

∫ ∫

Elastic section modulusElastic section modulus� Therefore to calculate the bending stress:

� In structural design an elastic section modulus Z is used where

max

; topi bi top b

i

MyMy My

I I II

My

σ σ σ

σ

= = =

=

max

max

thus I

Z My Z

σ= =

bending theory 2013x

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