CASTIGLIANO’S THEOREMS CLASS NOTES FOR MECHANICAL ENGINEERING

CASTIGLIANO’S  THEOREMS CLASS

NOTES FOR MECHANICAL ENGINEERING

Castiglioni’s Method

This method deals with displacements with respect to strain energy. Strain energy is also called potential energy. It is also named as Internal energy. Deflection is found using strain energy. Strain energy is recoverable as mechanical energy. Displacement in Castilian’s method means
  1. Extension under an axial tensile force, δL= PL/AE
  2. Contraction under an axial compressive force, δL= — PL/AE
  3. Deflection due to Bending Moment and shear force, y= δM + δV
  4. Angle of twist due torque (torsional moment), ϴ = TL/GJ
  5. Extension in a spring due to a force, δ =W/k
  6. Contraction in a spring due to a force, δ =W/k and etc.

CASTIGLIANO’S  Method determines the displacement in a linearly elastic system at any point. It  uses the partial derivatives of the strain energy. There are two Castiglioni’s theorems.

(ii) Castiglioni’s First Theorem OR Castigliano’s Force Finding Method or Castigliano’s Force Theorem

Determines the force at a certain point of known displacement ‘δi’ within elastic limits. Express strain energy of an elastic structure as a function of generalized displacement δi; then find the partial derivative of the strain energy with respect to generalized displacement, δi. This partial derivative gives the force Fi. In the mathematical form, it is given as
Fi = ∂U/∂δi
Where U is the total strain energy due to the forces/moments/torques etc.
δi is the displacement at any point ‘i’
Fi is the force at the point ‘i’

(ii) Castigliano’s Second Theorem or Castigliano’s Displacement Finding Method or Castigliano’s Displacement  Theorem

It determines Displacement at a point of the loaded beam. It uses  the partial derivatives of the strain energy with respect to the force acting at that location. If the Displacement is positive, then it is in the direction of the force and vice versa. In the mathematical form, it is given as
δi = ∂U/∂Fi
Where
δi is the displacement of the point under the force Fi in the direction of Fi
U is the total strain energy
COR: Castigliano’s theorem
It also determines the slope at a point of the loaded beam. Here it uses the partial derivatives of the strain energy with respect to the bending moment acting at that location. In the mathematical form, it is given as
θi = ∂U/∂Mi
Where U is the total strain energy,
θi is the slope of the moment Mi in the direction of M (Clockwise or anticlockwise)

NOTE: THIS METHOD IS VERY USEFUL FOR OBTAINING THE DEFLECTION OF A POINT WHERE THERE IS NO FORCE.  IN THIS CASE A FORCE/ MOMENT (FICTITIOUS) IS ASSUMED TO ACT.  THEN MAKE force ZERO AT THAT POINT IN THE PARTIAL DERIVATIVE OF THE STRAIN ENERGY.

MEANING OF GENERALIZED DISPLACEMENTS

  1. In Simple Tension
Extension in tension δL= PL/AE
Strain energy = Average Force x δL = (F/2) (PL/AE) = =F2L/2EA
  1. In Simple Compression
Extension in compression δL= PL/AE
Strain energy = Fav δL = (F/2) (PL/AE) = =F2L/2EA
  1. In Simple Bending
Strain energy = Average Bending moment x deflection
= (M/2) (ML/EI) = M2L/2EI
  1. Angle of twist in simple torsion
Strain energy= Average torque x displacement =average torque x angle of twist
= (T/2) (TL/GJ) = T2 L/2GJ

STRAIN ENERGY IN A MEMBER FOR ANY ONE TYPE OF LOADING

Sr. No.

Type of loading

Strain energy

with variable loading

Strain energy with

constant loading

1.
Axial load(tensile or compressive), F
U =∫ F2dx/2EA
U =F2L/2EA
2.
Bending Moment, M
U = ∫ M2dx/2EA
U =M2L/2EI
3.
Torque T
U = ∫ T2dx/2GJ
U =T2L/2GJ
4.
Direct shear force, F
U = ∫ F2dx/2GA
U =F2L/2GA
5.
Transverse shear force, V
U = ∫ V2dx/2GA
U =K V2L/2GA
INTEGRATION LIMITS ARE FROM 0 to L in the tabular equations.
Where K is a correction factor depending upon the shape of the area
rectangular section K =1.2
solid circular section K =1.11
thin walled tube          K =2.0

PROCEDURE FOR APPLICATION OF CASTIGLIANO’S THEOREM

a) To determine deflection yf in the direction of a real force F
  1. Prepare an expression for the total strain energy.
  2. Obtain the linear deflection yf from the relationship yf = ∂U /∂Ff
  3. If the force is fictitious, set Ff = 0 and solve the resulting equation)

 

       (b) To determine the slope (an angular displacement) θf in the direction of a fictitious moment Mf
  1. Write the  expression for the total strain energy.
  2. Determine the angular deflection from the relationship θf= ∂U /∂Mf
  3. If the moment is fictitious, then set Mf = 0 and solve the resulting equation.
    EXAMPLE ON CASTIGLIANO’S THEOREM
    For a given rectangular beam under a central load W, find the deflection under the load.
Given:
L = 2m,
b=0.1m,
h= 0.05m,
W=40 000 N,
E=200 GPa,
G = 80 GPa
I = 417×10-6m4
For this example, the beam is of rectangular section, width b and depth h.   Total strain energy will be due to bending moment and due to traverse shear. Because the beam is symmetrical, Calculation is made for L/2 and then made 2 times  for the entire beam.
Consider a point at distance x from the left hand support.
Moment M = (W x/2)  and Transverse Shear Force V = W/2
1) The expression for the total strain energy in length L = 2 times that in length L/2
U = ∫M2dx/2EI +2 V2dx/2GA
M=W x / 2  V = W/2
k rect =1.2
(Integration limits are from 0 to L/2)
= [(W/2) x)2/2EI] dx + 2 ∫ ( 3V2/10GA) dx
=∫ [W2x2/4EI] dx + ∫  (3W2/10GA) dx                                     ( Integration limits are from 0 to L/2)
2) From Castigliano’s method
the deflection of the point ‘ i ‘ in the direction of the force F(W in this case) will be found from the partial derivative.
yi = ∂U /∂Fi = ∂U /∂W
yi =∫ [2W x2/4EI] dx + ∫ (6 W/10GA) dx
=  ∫(Wx2/2EI) dx +∫ (3W/5GA) dx                                           
        integration limits are from 0 to L/2
Deflection yi = WL3/48EI + 3WL/10GA
= [40000 x 23/48x 200 x 1000 x417x10-6] +3 x 40000/10 x 80 x 1000 x (0.1 x 0.05)
=7.8 + 0.060 = 7.860 mm
It is to be noted that deflection due to M is 7.8 mm. Deflection  due to shear force is only 0.060 mm. Also finding of strain energy due to transverse shear is complex. Therefore deflection due to transverse shear is quite small and hence normally not included in the calculations. So in a beam, find strain energy due to bending moment only and then find the deflection only due to the bending moment. Thus
First Theorem Castigliano’s ——For finding force in an elastic structure
Castigliano’s Second Theorem——For finding displacements in an elastic structure
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