Castigliano-Lecture14

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Castigliano¶s 2nd

Theorem Energy method derived by Italian engineer

Alberto Castigliano in 1879.

Allows the computation of a defection atany point in a structure

Previously, we have shown how to use

energy methods to compute deflection onlyat the point where real external force didwork.



Derivation

Let the body be linear elastic; and let (i be

the component of deflection in the direction

of the i

th

force, Fi. The total work done is then:

U =½F1(1+ ½F2(2 ½F3(3+«.½Fn(n

F1

Consider a solid object acted upon by n

forces, Fi=1 to n, as shown in the figure.

F2F3

Fn



Increase force Fn by an amount dF

This changes the state of deformation and

increases the total strain energy slightly:

Hence, the total strain energy after the

increase in the nth force is:

nn

dF F

U dU

x

x

!

n

n

dF F

U U

x

x



Now suppose, the order of this process is reversed;

i.e., Apply a small force dFn to this same body and observe a deformation d(n; thenapply the forces, Fi=1 to n.

As these forces are being applied, dFn

goesthrough displacement (n.( Note dFnisconstant) and does work:

dU = dFn(n

Hence the total work done is:U+ dFn(n



The end results are equal

Since the body is linear elastic, all work is

recoverable, and the two systems are

identical and contain the same stored

energy:

n

n

nnn

n

F

U

dF dF F

U U

x

x!@

!x

x U



Castigliano¶s Second Theorem

The term ³force´ may be used in its most

fundamental sense and can refer for

example to a Moment, M

, producing arotation, U, in the body.

M

Un

n M

U

x

x

!U



Applications:

Castigliano¶s 2nd theorem can be usedto determine the deflections instructures (eg, trusses, beams, frames,

shells) and we are not limited toapplications in which only 1 externalforce or moment acts.

Furthermore, we can determine thedeflection or rotation at any point, evenwhere no force or moment is appliedexternally.



Try it:

The truss shown is subjected to a force F at point B; UseCastigliano¶s 2nd theorem to determine the vertical deflection

at point C.



OK!«

There is no external force at C, but we candetermine the vertical displacement (c, by

applying a vertical force, F¶, at C and then apply

Castigliano's 2nd theorem, and then setting F¶ =0 in

the resulting expression for the displacement.



Solve for internal forces:

Forces Length of

Members

1 - 1 0

2 F 2 2 02 L

3 - 3 0

4 -2( ¶) 4 0

5F (2 ¶) 5 02 L

6 6 0

7 0 7 0

um

§!

!7

1

2

2i

ii

EA

L P U Internal Strain Energy



Castigliano¶s 2nd:

? A0

2

0

2

0

2

0

2

0

2

0

2 222222

1 L F L F F L F F L F L F L F EA

U !

(C

? A00 '24'222

1

' L F F L F F

EA F

U

C !

x

x!

Now, since there is no actual force, F¶; set F¶=0

? A00 2442

1 L F L F

EA

C !

EA

F Lc

0212 !(



With Numbers«

A lightweight aluminum truss,E=70000 MPa, has a height,

Lo=1m, and is made of tubular

stock with a cross sectional area

of 250 mm2.Determine the

deflection at C when the load at

B, F=20k N.

EA

F Lc

0212 !(

mm

c

03.11)250(70000

2000)20000(212

!

!(

(Recall, (B was 14.4 mm)

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Castigliano-Lecture14