Elastic DeflectionCastigliano’s Method

If deflection is not covered by simple cases in Table 5.1 (p186)

Stored Elastic Energy

U

Complementary Energy U’

2Q∆UU ' ⋅==dQ∆UdU ' ⋅== dIncremental:

dQdU∆ =Deflection:

Castiglino’s Theorem:

When a body is elastically deflected by any combination of loads, the deflection at any point and in any direction is equal to the partial derivative of strain energy (computed with all loads acting) with respect to a load located at that point and acting in that direction

QU∆ ∂∂=

Elastic DeflectionCastigliano’s Method

Table 5.3 (p193): Energy and Deflection Equations

Example: Axial TensionStored Elastic Energy:

Case 1 from Table 5.1:

gives:

For varying E and A:

Elastic DeflectionCastigliano’s Method

(1) Obtain expression for all components of energyTable 5.3

(2) Take partial derivative to obtain deflection

QU∆ ∂∂=Castiglino’s Theorem:

Table 5.3 (p193): Energy and Deflection Equations

Elastic Deflection: Castigliano’s Method

Table 5.3

Here 2 types of loading: Bending and Shear

magnitude @ x:

1. Energy: here it has two components:

first compute Energy, then Partial Derivative to get deflection

2. Partial Derivatives for deflection:(23=8)*3*4 = 96

Table 5.3

Elastic Deflection: Castigliano’s Method

TWO METHODS

Differentiate after Integral Differentiate under Integral

Elastic Deflection: Castigliano’s Method

Transverse shear contributes only <5% to deflection

m

m

Elastic Deflection: Castigliano’s Method

Use of “Dummy Load” Q=0

•90° bend cantilever beam•shear neglected

•Shear neglected => only 4 energy components:

1) BENDING portion a_b: Mab=Py2) BENDING portion b_c: Mbc=Qx +Ph

3) TENSION portion a_b: Q4) COMPRESSION portion b_c: P

(Tension and Compression mostly negligible if torsion and bending are present)

:

Elastic Deflection: Castigliano’s Method

•Eccentrically Load Column•No Buckling

Redundant Support

500kg x 9.8m/s2

=4900 N

•Now Deflection known (δ=0) •Find necessary Tension Force F

Guy wire

•Hence partial derivative of total elastic energy with respect to F must be zero

•Omit zero derivatives- all energy terms above a- compression term below a

•Only bending term is left: M= (4900 N)(1.2m)-Fy = 5880 Nm - Fy

=! 0 finite value F=2940 N

(Nm)2 m Nm3 m3

(Nm)2 Nm

Nm3 m3