Lecture7 Bending Stresses in Beams

7/30/2019 Lecture7 Bending Stresses in Beams

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Unit 2- Stresses in Beams

Lecture -1 Review of shear force and bendingmoment diagram

Lecture -2 Bending stresses in beams Lecture -3 Shear stresses in beams Lecture -4- Deflection in beams Lecture -5 Torsion in solid and hollow shafts.

Topics Covered


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Theory of simple

bending (assumptions) Material of beam is homogenous and isotropic => constant E in alldirection

Youngs modulus is constant in compression and tension => to simplifyanalysis

Transverse section which are plane before bending before bending remainplain after bending. => Eliminate effects of strains in other direction (nextslide)

Beam is initially straight and all longitudinal filaments bend in circular arcs=> simplify calculations

Radius of curvature is large compared with dimension of cross sections =>simplify calculations

Each layer of the beam is free to expand or contract => Otherwise they willgenerate additional internal stresses.


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Key Points:

1. Internal bending moment causes beam to deform.2. For this case, top fibers in compression, bottom in

tension.

Bending in beams


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Key Points:

1. Neutral surface no change in length.2. Neutral Axis Line of intersection of neutral surface

with the transverse section.

3. All cross-sections remain plane and perpendicular tolongitudinal axis.

Bending in beams


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Key Points:

1. Bending momentcauses beam to deform.

2. X = longitudinal axis3. Y = axis of symmetry4. Neutral surface does

not undergo a changein length

Bending in beams


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P

AB

RBRAM M

Radius of Curvature, R

Deflected

Shape

Consider the simply supported beam below:

M M

What stresses are generated

within, due to bending?

Bending Stress in beams

Neutral Surface


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M=Bending Moment

M M

Beam

x (Tension)

x

(Compression)

x=0

(i) Bending Moment, M

(ii) Geometry of Cross-section

x is NOT UNIFORM through

the section depth

xDEPENDS ON:

Axial Stress Due to Bending:

stress generated due to bending:

Neutral Surface


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Neutral axis

N A

y

dy

dAforce on the layer=stress on layer*area of layer

= dA

=

E

R y dA

Total force on the beam section

=

E

R y dA

=

E

R y dAFor equilibrium forces should be 0

y dA = 0

Neutral axis coincides with the geometrical

axis

Stress diagram

x

x

M M

x


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Moment of resistance

N A

y

dy

dAforce on the layer=stress on layer*area of layer

= dA

=

E

R y dA

Moment of this force about NA

=

E

R y dA y

=

E

R

y

2

dA

Total moment M=E

R y 2 dA =

E

R y 2 dA

Stress diagram

y 2 dA = I

M =E

RI

M

I=

E

R

x

x

M M

x


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Flexure Formula

M

I=

E

R=

y


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Beam subjected to 2 BM

In this case beam is subjected to

moments in two directions y and z.The total moment will be a resultant

of these 2 moments.

You can apply principle of superposition

to calculate stresses. (topic covered inunit 1).

Resultant moments and stresses


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Section Modulus

Section modulus is defined as ratio of moment of inertia about the neutral axis to

the distance of the outermost layer from the neutral axis

Z=

I

ymax

M

I=

y

M

I=

max

ymax

M =max

I

ymax

M =maxZ


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Section Modulus of

symmetrical sections

Source:-http://en.wikipedia.org/wiki/Section_modulus


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Section Modulus of

unsymmetrical sections

In case of symmetrical section neutral axis passes through geometrical center of

the section. But in case of unsymmetrical section such as L and T neutral axisdoes not pass through geometrical center.

The value of y for the outermost layer of the section from neutral axis will not be

same.


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Composite beamsComposite beams consisting of layers with fibers, or rods strategically placed to

increase stiffness and strength can be designed to resist bending.


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Composite beams

b

t t

y

d

1

E1

=

2

E2

1=

E1

E2

2

= m2 m=modular ratio

M=

yI

M= M1+M

2

=

1

y I1+

2

y I2

=

2

ymI

1+I

2[ ]

Equivalent I (moment of inertia)= mI1+ I

2

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Lecture7 Bending Stresses in Beams