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= δ_M + δ_V

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 ‘δ_i’ within elastic limits. Express strain energy of an elastic structure as a function of generalized displacement δ_i; then find the partial derivative of the strain energy with respect to generalized displacement, δ_i. This partial derivative gives the force F_i. In the mathematical form, it is given as

F_i = ∂U/∂δ_i

Where U is the total strain energy due to the forces/moments/torques etc.

_δi is the displacement at any point ‘i’

F_i 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

δ_i = ∂U/∂F_i

Where

δ_i is the displacement of the point under the force F_i in the direction of F_i

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

θ_i = ∂U/∂M_i

Where U is the total strain energy,

θ_i is the slope of the moment M_i 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

2. In Simple Compression

Extension in compression δL= PL/AE

Strain energy = F_av δL = (F/2) (PL/AE) = =F²L/2EA

3. In Simple Bending

Strain energy = Average Bending moment x deflection

= (M/2) (ML/EI) = M²L/2EI

4. 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 y_f in the direction of a real force F_f 

1. Prepare an expression for the total strain energy.

2. Obtain the linear deflection y_f from the relationship y_f = ∂U /∂F_f

3. If the force is fictitious, set F_f = 0 and solve the resulting equation)

 

       (b) To determine the slope (an angular displacement) θ_f in the direction of a fictitious moment M_f

1. Write the  expression for the total strain energy.

2. Determine the angular deflection from the relationship θ_f= ∂U /∂M_f

3. If the moment is fictitious, then set M_f = 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_i (W in this case) will be found from the partial derivative.

yi = ∂U /∂Fi = ∂U /∂W

y_i =∫ [2W x²/4EI] dx + ∫ (6 W/10GA) dx

=  ∫(Wx²/2EI) dx +∫ (3W/5GA) dx                                           

        integration limits are from 0 to L/2

Deflection y_i = 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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