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Government Engineering College,
Bhavnagar.
Civil Engineering Department
Advanced Engineering Mathematics
Topic:-
Application of Ordinary Differential Equation
Elastic Beams.
Content
 Beams
 Elasticity
 Application of Beams
 Ordinary Differential Equations
 Methods to Solve Ordinary Differential Equation
 Euler–Bernoulli beam theory
 Static beam equation
 Derivation of bending moment equation
 Assumptions of Classical Beam Theory
 Second Order Method For Beam Deflections
 What we learned?
Beams
 Beams are the most common type of structural component,
particularly in Civil Engineering. Its length is comparatively more
than its cross sectional area.
 A beam is a bar like structural member whose primary function is to
support transverse loading and carry it to the supports. See Figure.
 A beam resists transverse loads mainly through bending action,
Bending produces compressive longitudinal stresses in one side of
the beam and tensile stresses in the other.
Elasticity
 It is the property of the material to regain its original shape after the
removal of external force.
 As per the Hooke’s Law:-
“ Stress is directly proportional to the strain.”
i.e.
σ ∝ ɛ,
σ = E ɛ,
Therefore we have,
E = σ/ɛ. ( Where , E = Young’s Modulus of Elasticity,
σ = Stress.
ɛ = Strain. )
Application of Beams
 They are used as structural members in a structure.
 They transfer load from the superstructure to the columns
and to the sub-soil foundation.
 They are used in bridges for the transportation of light as
well as heavy vehicles.
 They are also as a strap beam in strap footing in
foundations to connect and transfer loads from the
columns.
Ordinary Differential Equations
 The Differential equation containing a single
independent variable and the derivative with respect to
it is called Ordinary Differential Equation.
 Example:-
dy/dx = cx
dV/dx + q = 0
2x2 y’’ + x (2x + 1) y’ − y = 0.
Methods to Solve Ordinary Differential
Equation
 Variable Separable Method
 Homogeneous and Non - Homogeneous Differential Equation
Method
 Exact and Non-exact Differential Equation Method
 Linear and Non- Linear Differential Equation Method
 Able’s Formula to find second solution of a Differential Equation
 Wronskian method
 Methods to find Complementary Function and Particular Integral of a
Differential Equation.
Euler–Bernoulli beam theory
 Euler–Bernoulli beam theory (also known as engineer's beam
theory or classical beam theory) is a simplification of the
linear theory of elasticity which provides a means of calculating the
load-carrying and deflection characteristics of beams.
 It covers the case for small deflections of a beam that is subjected
to lateral loads only. It is thus a special case of Timoshenko beam
theory that accounts for shear deformation and is applicable for
thick beams.
 It was first enunciated circa 1750, but was not applied on a large
scale until the development of the Eiffel Tower and the Ferris
wheel in the late 19th century. Following these successful
demonstrations, it quickly became a cornerstone of engineering
and an enabler of the Second Industrial Revolution.
 The Euler-Bernoulli beam equation is a combination of
four different subsets of beam theory. These being the
Kinematics
Constitutive
Force resultant
Equilibrium theories
Euler–Bernoulli beam theory
Kinematics
 Kinematics describes the motion of the beam and its
deflection.
 Kirchoff’s Assumptions
 The normal remain straight
 The normal remain unstretched.
 The normal remain to the normal (they are always perpendicular
to the neutral plane)
Constitutive
 The next equation needed to explain the Euler-Bernoulli
equation is the constitutive equation.
 Constitutive equations relate two physical quantities and in
the case of beam theory the relation between the direct
stress s and direct strain e within a beam is being
investigated.
Force Resultant
 These force resultants allow us to study the important stresses
within the beam.
 They are useful to us as they are only considered functions of,
whereas the stresses found in a beam are functions.
 Consider a beam cut at a point It is possible to find the direct
stress and shear stress, and respectively, of that point. This is
due to the direct stress at the cut producing a moment
around the neutral plane. By summing all these individual
moments over the whole of the cross-section it is possible to
find the moment resultant M
Equilibrium
 To keep things simple it is sufficient to deal with the stresses
resultants rather than the stresses themselves.
 The reason, explained previously, is that the resultants are
functions of only.
 The equilibrium equations are used to describe how the
internal stresses of the beam are affected by the external
loads being applied.
Static beam equation
 Bending of an Euler–Bernoulli beam. Each cross-section of the beam
is at 90 degrees to the neutral axis.
 The Euler–Bernoulli equation describes the relationship between the
beam's deflection and the applied load:
 The curve describes the deflection of the beam
in the z direction at some point x.
Static beam equation
Continued…
 q is a distributed load, in other words a force per unit length (analogous to
pressure being a force per area); it may be a function of or other
variables.
 Note that E is the elastic modulus and that I is the second moment of area of the
beam's cross-section. I must be calculated with respect to the axis which passes
through the centroid of the cross-section and which is perpendicular to the
applied loading. Explicitly, for a beam whose axis is oriented along x with a
loading along z, the beam's cross-section is in the yz plane, and the relevant
second moment of area is
,
Static beam equation
Continued…
 where it is assumed that the centroid of the cross-section occurs at
y = z = 0.
 Often, the product EI (known as the stiffness) is a constant, so that
 This equation, describing the deflection of a uniform, static beam, is
used widely in engineering practice.
Static beam equation
 dw/dx is the slope of the beam.
 , is the bending moment in the beam.
 , is the shear force in the beam.
 The stresses in a beam can be calculated from the above expressions after
the deflection due to a given load has been determined.
Continued….
Derivation of bending moment equation
 Here is a short derivation,
 The length of the neutral axis in the figure,
Bending of an Euler–Bernoulli beam, .
 The length of a fibre with a radial distance, e,
below the neutral axis is .
 Strain of this fibre is
Deflection of a beam deflected
symmetrically and principle of superposition
Derivation of bending moment
equation
 The differential force vector,
 The differential bending moment vector, dM , associated with dF is
given by
 This expression is valid for the fibres in the lower half of the beam.
Derivation of bending moment
equation
 the resulting bending moment vector will still be in the -y direction since
 we integrate over the entire cross section of the beam and get for M the
bending moment vector exerted on the right cross section of the beam the
expression
where I is the second moment of area
 We know that when dw/dx is small as it is for an Euler–Bernoulli beam,
( is the radius of curvature). Therefore

Assumptions of Classical Beam Theory
 1. Planar symmetry. The longitudinal axis is straight and the cross
section of the beam has a longitudinal plane of symmetry. The
resultant of the transverse loads acting on each section lies on that
plane. The support conditions are also symmetric about this plane.
 2. Cross section variation. The cross section is either constant or
varies smoothly.
 3. Normality. Plane sections originally normal to the longitudinal axis
of the beam remain plane and normal to the deformed longitudinal
axis upon bending.
Assumptions of Classical Beam Theory
 4. Strain energy. The internal strain energy of the member accounts
only for bending moment deformations. All other contributions,
notably transverse shear and axial force, are ignored.
 5. Linearization. Transverse deflections, rotations and deformations
are considered so small that the assumptions of infinitesimal
deformations apply.
 6. Material model. The material is assumed to be elastic and
isotropic. Heterogeneous beams fabricated with several isotropic
materials, such as reinforced concrete, are not excluded.
Second Order Method For Beam
Deflections
 Example 1: Cantilever Load Under Tip Point Load
 The problem is defined in Figure:-
Mz(x) = −P x
 For convenience we scale v(x) by E Izz
 so that the ODE linking bending moment to
 deflection is E Izz v (x) = Mz(x) = −P x.
 Integrating twice:- E Izz v’ (x) = −P x2/2 + C1
E Izz v’’(x) = −P x3/6 + C1 x + C2
Example 1:- Continued
 The kinematic BCs for the cantilever are v’(L) = 0 and v(L) = 0 at the fixed
end B. The first one gives
E Izz v’(L) = −PL2/2 + C1 = 0, whence C1 = PL2/2.
 The second one gives
E Izz v(L) = −PL3/6 + C1L + C2 = −PL3/6 + PL3/2 + C2 = 0 whence C2 = −PL3/3.
 Substituting into the expression for v(x) gives
v(x) = − P/6E Izz ( x3 − 3L2x + 2L3)= − P/6E Izz (L − x)2(2L + x)
 Of particular interest is the tip deflection at free end A, which is the largest
one. Setting x = 0 yields
 v(0) = vA = − PL3 /3E Izz ⇓ (The negative sign indicates that the beam
deflects downward if P > 0.)
What we learned?
Total amount of deflection which will occur.
Merits and Demerits of this theory.
Comparison Between the two theories.
(Timoshenko Beam Theory and Euler–Bernoulli
beam theory.)
Frequency of a Vibrating System Using a Vertical
beam.
Thank You For Bearing.
By,
 Nitin Charel (130210106011)
 Kartik Hingol (130210106030)
 Bhavik Shah (130210106049)
 Yash Shah (130210106052)
 Digvijay Solanki (130210106055)

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Elastic beams

  • 2. Advanced Engineering Mathematics Topic:- Application of Ordinary Differential Equation Elastic Beams.
  • 3. Content  Beams  Elasticity  Application of Beams  Ordinary Differential Equations  Methods to Solve Ordinary Differential Equation  Euler–Bernoulli beam theory  Static beam equation  Derivation of bending moment equation  Assumptions of Classical Beam Theory  Second Order Method For Beam Deflections  What we learned?
  • 4. Beams  Beams are the most common type of structural component, particularly in Civil Engineering. Its length is comparatively more than its cross sectional area.  A beam is a bar like structural member whose primary function is to support transverse loading and carry it to the supports. See Figure.  A beam resists transverse loads mainly through bending action, Bending produces compressive longitudinal stresses in one side of the beam and tensile stresses in the other.
  • 5. Elasticity  It is the property of the material to regain its original shape after the removal of external force.  As per the Hooke’s Law:- “ Stress is directly proportional to the strain.” i.e. σ ∝ ɛ, σ = E ɛ, Therefore we have, E = σ/ɛ. ( Where , E = Young’s Modulus of Elasticity, σ = Stress. ɛ = Strain. )
  • 6. Application of Beams  They are used as structural members in a structure.  They transfer load from the superstructure to the columns and to the sub-soil foundation.  They are used in bridges for the transportation of light as well as heavy vehicles.  They are also as a strap beam in strap footing in foundations to connect and transfer loads from the columns.
  • 7. Ordinary Differential Equations  The Differential equation containing a single independent variable and the derivative with respect to it is called Ordinary Differential Equation.  Example:- dy/dx = cx dV/dx + q = 0 2x2 y’’ + x (2x + 1) y’ − y = 0.
  • 8. Methods to Solve Ordinary Differential Equation  Variable Separable Method  Homogeneous and Non - Homogeneous Differential Equation Method  Exact and Non-exact Differential Equation Method  Linear and Non- Linear Differential Equation Method  Able’s Formula to find second solution of a Differential Equation  Wronskian method  Methods to find Complementary Function and Particular Integral of a Differential Equation.
  • 9. Euler–Bernoulli beam theory  Euler–Bernoulli beam theory (also known as engineer's beam theory or classical beam theory) is a simplification of the linear theory of elasticity which provides a means of calculating the load-carrying and deflection characteristics of beams.  It covers the case for small deflections of a beam that is subjected to lateral loads only. It is thus a special case of Timoshenko beam theory that accounts for shear deformation and is applicable for thick beams.  It was first enunciated circa 1750, but was not applied on a large scale until the development of the Eiffel Tower and the Ferris wheel in the late 19th century. Following these successful demonstrations, it quickly became a cornerstone of engineering and an enabler of the Second Industrial Revolution.
  • 10.  The Euler-Bernoulli beam equation is a combination of four different subsets of beam theory. These being the Kinematics Constitutive Force resultant Equilibrium theories Euler–Bernoulli beam theory
  • 11. Kinematics  Kinematics describes the motion of the beam and its deflection.  Kirchoff’s Assumptions  The normal remain straight  The normal remain unstretched.  The normal remain to the normal (they are always perpendicular to the neutral plane)
  • 12. Constitutive  The next equation needed to explain the Euler-Bernoulli equation is the constitutive equation.  Constitutive equations relate two physical quantities and in the case of beam theory the relation between the direct stress s and direct strain e within a beam is being investigated.
  • 13. Force Resultant  These force resultants allow us to study the important stresses within the beam.  They are useful to us as they are only considered functions of, whereas the stresses found in a beam are functions.  Consider a beam cut at a point It is possible to find the direct stress and shear stress, and respectively, of that point. This is due to the direct stress at the cut producing a moment around the neutral plane. By summing all these individual moments over the whole of the cross-section it is possible to find the moment resultant M
  • 14. Equilibrium  To keep things simple it is sufficient to deal with the stresses resultants rather than the stresses themselves.  The reason, explained previously, is that the resultants are functions of only.  The equilibrium equations are used to describe how the internal stresses of the beam are affected by the external loads being applied.
  • 15. Static beam equation  Bending of an Euler–Bernoulli beam. Each cross-section of the beam is at 90 degrees to the neutral axis.  The Euler–Bernoulli equation describes the relationship between the beam's deflection and the applied load:  The curve describes the deflection of the beam in the z direction at some point x.
  • 16. Static beam equation Continued…  q is a distributed load, in other words a force per unit length (analogous to pressure being a force per area); it may be a function of or other variables.  Note that E is the elastic modulus and that I is the second moment of area of the beam's cross-section. I must be calculated with respect to the axis which passes through the centroid of the cross-section and which is perpendicular to the applied loading. Explicitly, for a beam whose axis is oriented along x with a loading along z, the beam's cross-section is in the yz plane, and the relevant second moment of area is ,
  • 17. Static beam equation Continued…  where it is assumed that the centroid of the cross-section occurs at y = z = 0.  Often, the product EI (known as the stiffness) is a constant, so that  This equation, describing the deflection of a uniform, static beam, is used widely in engineering practice.
  • 18. Static beam equation  dw/dx is the slope of the beam.  , is the bending moment in the beam.  , is the shear force in the beam.  The stresses in a beam can be calculated from the above expressions after the deflection due to a given load has been determined. Continued….
  • 19. Derivation of bending moment equation  Here is a short derivation,  The length of the neutral axis in the figure, Bending of an Euler–Bernoulli beam, .  The length of a fibre with a radial distance, e, below the neutral axis is .  Strain of this fibre is Deflection of a beam deflected symmetrically and principle of superposition
  • 20. Derivation of bending moment equation  The differential force vector,  The differential bending moment vector, dM , associated with dF is given by  This expression is valid for the fibres in the lower half of the beam.
  • 21. Derivation of bending moment equation  the resulting bending moment vector will still be in the -y direction since  we integrate over the entire cross section of the beam and get for M the bending moment vector exerted on the right cross section of the beam the expression where I is the second moment of area  We know that when dw/dx is small as it is for an Euler–Bernoulli beam, ( is the radius of curvature). Therefore 
  • 22. Assumptions of Classical Beam Theory  1. Planar symmetry. The longitudinal axis is straight and the cross section of the beam has a longitudinal plane of symmetry. The resultant of the transverse loads acting on each section lies on that plane. The support conditions are also symmetric about this plane.  2. Cross section variation. The cross section is either constant or varies smoothly.  3. Normality. Plane sections originally normal to the longitudinal axis of the beam remain plane and normal to the deformed longitudinal axis upon bending.
  • 23. Assumptions of Classical Beam Theory  4. Strain energy. The internal strain energy of the member accounts only for bending moment deformations. All other contributions, notably transverse shear and axial force, are ignored.  5. Linearization. Transverse deflections, rotations and deformations are considered so small that the assumptions of infinitesimal deformations apply.  6. Material model. The material is assumed to be elastic and isotropic. Heterogeneous beams fabricated with several isotropic materials, such as reinforced concrete, are not excluded.
  • 24. Second Order Method For Beam Deflections  Example 1: Cantilever Load Under Tip Point Load  The problem is defined in Figure:- Mz(x) = −P x  For convenience we scale v(x) by E Izz  so that the ODE linking bending moment to  deflection is E Izz v (x) = Mz(x) = −P x.  Integrating twice:- E Izz v’ (x) = −P x2/2 + C1 E Izz v’’(x) = −P x3/6 + C1 x + C2
  • 25. Example 1:- Continued  The kinematic BCs for the cantilever are v’(L) = 0 and v(L) = 0 at the fixed end B. The first one gives E Izz v’(L) = −PL2/2 + C1 = 0, whence C1 = PL2/2.  The second one gives E Izz v(L) = −PL3/6 + C1L + C2 = −PL3/6 + PL3/2 + C2 = 0 whence C2 = −PL3/3.  Substituting into the expression for v(x) gives v(x) = − P/6E Izz ( x3 − 3L2x + 2L3)= − P/6E Izz (L − x)2(2L + x)  Of particular interest is the tip deflection at free end A, which is the largest one. Setting x = 0 yields  v(0) = vA = − PL3 /3E Izz ⇓ (The negative sign indicates that the beam deflects downward if P > 0.)
  • 26. What we learned? Total amount of deflection which will occur. Merits and Demerits of this theory. Comparison Between the two theories. (Timoshenko Beam Theory and Euler–Bernoulli beam theory.) Frequency of a Vibrating System Using a Vertical beam.
  • 27. Thank You For Bearing. By,  Nitin Charel (130210106011)  Kartik Hingol (130210106030)  Bhavik Shah (130210106049)  Yash Shah (130210106052)  Digvijay Solanki (130210106055)