Pure bending, SOM | Strength of Materials (SOM) - Mechanical Engineering PDF Download

 Page 1


Simple Bending or Pure Bending
Consider a simply supported beam AB of length L subjected to moment M0 at its ends
as shown in ?gures.
Consider an element of length dx at distance x from hinged support A. Due to applied
moment, the beam will sag in a manner.
          
Page 2


Simple Bending or Pure Bending
Consider a simply supported beam AB of length L subjected to moment M0 at its ends
as shown in ?gures.
Consider an element of length dx at distance x from hinged support A. Due to applied
moment, the beam will sag in a manner.
          
It can be seen in ?gure the element dx will be converted into an curve. Let the normal
tangents at end point of element intersect a point O . This point O is known as centre
of curvature. The distance Ox and Ox is known as radius of curvature. Let the angle
made at centre of curvature by both the normal is denoted by d?.
Curvature, C is de?ned as reciprocal of radius of curvature R i.e., C = 1/R
As from ?gure,  ds = Rd?
R = ds/d?
Here, ds is length of arch x1x2
As, curvature is reciprocal of radius of curvature, so, from eq. (i)
Curvature, C = d?/ds
For small de?ection, as is usually case with beam, ds = dx
Hence, curvature C = d?/dx
These equations are used to calculate normal or bending stress and strains in beam.
Bending moment diagram for beam.
As clear from bending moment diagram, the beam is free from shear force. This type
of bending in which there is no shear force is called as simple bending or pure
bending. Bending in presence of shear force is known as non-uniform bending.
In pure bending, radius of curvature is constant at di?erent sections of beam. For
example, as the beam in ?gure is the case of pure bending, radius of curvature is R at
section x3 also. As circular of curvature is constant, de?ection curve of beam is
circular.
Another case of pure bending can be a simply supported beam of length L subjected
to concentrated loads as shown in ?gure
1 2
          
Page 3


Simple Bending or Pure Bending
Consider a simply supported beam AB of length L subjected to moment M0 at its ends
as shown in ?gures.
Consider an element of length dx at distance x from hinged support A. Due to applied
moment, the beam will sag in a manner.
          
It can be seen in ?gure the element dx will be converted into an curve. Let the normal
tangents at end point of element intersect a point O . This point O is known as centre
of curvature. The distance Ox and Ox is known as radius of curvature. Let the angle
made at centre of curvature by both the normal is denoted by d?.
Curvature, C is de?ned as reciprocal of radius of curvature R i.e., C = 1/R
As from ?gure,  ds = Rd?
R = ds/d?
Here, ds is length of arch x1x2
As, curvature is reciprocal of radius of curvature, so, from eq. (i)
Curvature, C = d?/ds
For small de?ection, as is usually case with beam, ds = dx
Hence, curvature C = d?/dx
These equations are used to calculate normal or bending stress and strains in beam.
Bending moment diagram for beam.
As clear from bending moment diagram, the beam is free from shear force. This type
of bending in which there is no shear force is called as simple bending or pure
bending. Bending in presence of shear force is known as non-uniform bending.
In pure bending, radius of curvature is constant at di?erent sections of beam. For
example, as the beam in ?gure is the case of pure bending, radius of curvature is R at
section x3 also. As circular of curvature is constant, de?ection curve of beam is
circular.
Another case of pure bending can be a simply supported beam of length L subjected
to concentrated loads as shown in ?gure
1 2
          
Its shear force diagram and bending moment diagram is shown in ?gure above.
As from shear force diagram, it is cleared that there is no shear force in region CD of
beam. So in region CD, pure bending will occur due to which beam will be converted
into an arc of circle.
Assumptions in Theory of Pure Bending
1. Material of the beam is homogeneous, isotropic and linear elastic in which Hooke’s
law is valid.
2. The beam is straight before loading.
3. Cross-section of beam is prismatic throughout the length.
4. The plane section before bending remains plane after bending, it means, the
longitudinal strain vary linearly from zero at neutral axis to maximum at the
surface and longitudinal strain at any distance y is directly proportional to its
distance (y) from neutral axis.
5. Every layer of material is free to expand or contract longitudinally and laterally
under stress and do not exert pressure upon each other. Thus, the Poisson’s e?ect
at the interface of the adjoining di?erently stressed ?bres are ignored.
6. The value of Young’s modulus (E) for the material is same in tension and in
compression.
          
Page 4


Simple Bending or Pure Bending
Consider a simply supported beam AB of length L subjected to moment M0 at its ends
as shown in ?gures.
Consider an element of length dx at distance x from hinged support A. Due to applied
moment, the beam will sag in a manner.
          
It can be seen in ?gure the element dx will be converted into an curve. Let the normal
tangents at end point of element intersect a point O . This point O is known as centre
of curvature. The distance Ox and Ox is known as radius of curvature. Let the angle
made at centre of curvature by both the normal is denoted by d?.
Curvature, C is de?ned as reciprocal of radius of curvature R i.e., C = 1/R
As from ?gure,  ds = Rd?
R = ds/d?
Here, ds is length of arch x1x2
As, curvature is reciprocal of radius of curvature, so, from eq. (i)
Curvature, C = d?/ds
For small de?ection, as is usually case with beam, ds = dx
Hence, curvature C = d?/dx
These equations are used to calculate normal or bending stress and strains in beam.
Bending moment diagram for beam.
As clear from bending moment diagram, the beam is free from shear force. This type
of bending in which there is no shear force is called as simple bending or pure
bending. Bending in presence of shear force is known as non-uniform bending.
In pure bending, radius of curvature is constant at di?erent sections of beam. For
example, as the beam in ?gure is the case of pure bending, radius of curvature is R at
section x3 also. As circular of curvature is constant, de?ection curve of beam is
circular.
Another case of pure bending can be a simply supported beam of length L subjected
to concentrated loads as shown in ?gure
1 2
          
Its shear force diagram and bending moment diagram is shown in ?gure above.
As from shear force diagram, it is cleared that there is no shear force in region CD of
beam. So in region CD, pure bending will occur due to which beam will be converted
into an arc of circle.
Assumptions in Theory of Pure Bending
1. Material of the beam is homogeneous, isotropic and linear elastic in which Hooke’s
law is valid.
2. The beam is straight before loading.
3. Cross-section of beam is prismatic throughout the length.
4. The plane section before bending remains plane after bending, it means, the
longitudinal strain vary linearly from zero at neutral axis to maximum at the
surface and longitudinal strain at any distance y is directly proportional to its
distance (y) from neutral axis.
5. Every layer of material is free to expand or contract longitudinally and laterally
under stress and do not exert pressure upon each other. Thus, the Poisson’s e?ect
at the interface of the adjoining di?erently stressed ?bres are ignored.
6. The value of Young’s modulus (E) for the material is same in tension and in
compression.
          
7. The section of the beam is symmetrical in the loading plane. If section is non
symmetrical then twisting and warping may occur apart from bending.
Analysis of stress and strain in pure bending
a) Normal strain in beams
Consider a simply supported beam of length L subjected to moments at ends A and B
as shown in ?gure.
Take two section x -x and x -x which are dx length apart and section x -x is at
distance x from end A, as shown
Due to applied moments M , the beam will sag in a manner as shown.
 
       
     
  
    
1 1 2 2 1 1
0
          
Page 5


Simple Bending or Pure Bending
Consider a simply supported beam AB of length L subjected to moment M0 at its ends
as shown in ?gures.
Consider an element of length dx at distance x from hinged support A. Due to applied
moment, the beam will sag in a manner.
          
It can be seen in ?gure the element dx will be converted into an curve. Let the normal
tangents at end point of element intersect a point O . This point O is known as centre
of curvature. The distance Ox and Ox is known as radius of curvature. Let the angle
made at centre of curvature by both the normal is denoted by d?.
Curvature, C is de?ned as reciprocal of radius of curvature R i.e., C = 1/R
As from ?gure,  ds = Rd?
R = ds/d?
Here, ds is length of arch x1x2
As, curvature is reciprocal of radius of curvature, so, from eq. (i)
Curvature, C = d?/ds
For small de?ection, as is usually case with beam, ds = dx
Hence, curvature C = d?/dx
These equations are used to calculate normal or bending stress and strains in beam.
Bending moment diagram for beam.
As clear from bending moment diagram, the beam is free from shear force. This type
of bending in which there is no shear force is called as simple bending or pure
bending. Bending in presence of shear force is known as non-uniform bending.
In pure bending, radius of curvature is constant at di?erent sections of beam. For
example, as the beam in ?gure is the case of pure bending, radius of curvature is R at
section x3 also. As circular of curvature is constant, de?ection curve of beam is
circular.
Another case of pure bending can be a simply supported beam of length L subjected
to concentrated loads as shown in ?gure
1 2
          
Its shear force diagram and bending moment diagram is shown in ?gure above.
As from shear force diagram, it is cleared that there is no shear force in region CD of
beam. So in region CD, pure bending will occur due to which beam will be converted
into an arc of circle.
Assumptions in Theory of Pure Bending
1. Material of the beam is homogeneous, isotropic and linear elastic in which Hooke’s
law is valid.
2. The beam is straight before loading.
3. Cross-section of beam is prismatic throughout the length.
4. The plane section before bending remains plane after bending, it means, the
longitudinal strain vary linearly from zero at neutral axis to maximum at the
surface and longitudinal strain at any distance y is directly proportional to its
distance (y) from neutral axis.
5. Every layer of material is free to expand or contract longitudinally and laterally
under stress and do not exert pressure upon each other. Thus, the Poisson’s e?ect
at the interface of the adjoining di?erently stressed ?bres are ignored.
6. The value of Young’s modulus (E) for the material is same in tension and in
compression.
          
7. The section of the beam is symmetrical in the loading plane. If section is non
symmetrical then twisting and warping may occur apart from bending.
Analysis of stress and strain in pure bending
a) Normal strain in beams
Consider a simply supported beam of length L subjected to moments at ends A and B
as shown in ?gure.
Take two section x -x and x -x which are dx length apart and section x -x is at
distance x from end A, as shown
Due to applied moments M , the beam will sag in a manner as shown.
 
       
     
  
    
1 1 2 2 1 1
0
          
As it is case of pure bending, the elastic curve will be circular in shape.
Point O is the centre of curvature. Cross-section x -x and x -x will remain plane and
normal to longitudinal ?bres of beam so as to keep radius of curvature constant.
As it is evident , longitudinal ?bres EF are shortened while ?bres AB are elongated due
to bending.
Thus top ?bres are in compression while bottom ?bres are in tension. Somewhere
between top and bottom ?bres, a surface of layer exist at which there is no change in
length known as neutral layer as represented by CD. The intersection of neutral
surface with any cross-sectional plane is called neutral axis.
Radius of curvature R is taken as distance from centre of curvature O to neutral layer.
As derived earlier,
Radius of curvature,
R = dx/d?
Curvature,
C = d?/dx
To calculate longitudinal strain, take a ?bre GH at distance y from neutral layer CD as
shown.
The length L of ?bre GH = (R + y) d?
Here, y is the distance of ?bre from neutral axis. Strains are positive below neutral axis
and negative above neutral axis.
(b) Normal Stresses in beams
1 1 2 2
1