handout castigliano

Download Handout Castigliano

Post on 26-Dec-2014

56 views

Category:

Documents

3 download

Embed Size (px)

TRANSCRIPT

Elastic Deflection

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

Complementary Energy U

Stored Elastic Energy U

Incremental: Deflection:

U = U' = Q 2 dU = dU' = dQ = dU dQ

Castiglinos Theorem:

= U Q

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

Elastic Deflection

Castiglianos MethodTable 5.3 (p193): Energy and Deflection Equations

Example: Axial Tension Stored Elastic Energy: Case 1 from Table 5.1: gives:

For varying E and A:

Elastic Deflection Castiglianos MethodTable 5.3 (p193): Energy and Deflection Equations

(1) Obtain expression for all components of energyTable 5.3

(2) Take partial derivative to obtain deflectionCastiglinos Theorem:

= U Q

Elastic Deflection: Castiglianos Method

first compute Energy, then Partial Derivative to get deflectionHere 2 types of loading: Table 5.3 magnitude Bending and Shear

@ x:

1. Energy: here it has two components:

(23=8)*3*4 = 96

2. Partial Derivatives for deflection:

Elastic Deflection: Castiglianos Method

Table 5.3

TWO METHODS

Differentiate after Integral

Differentiate under Integral

Elastic Deflection: Castiglianos Method

m

m

Transverse shear contributes only only 4 energy components: 1) BENDING portion a_b: Mab=Py 2) BENDING portion b_c: Mbc=Qx +Ph 3) TENSION portion a_b: Q 4) COMPRESSION portion b_c: P

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

:

Elastic Deflection: Castiglianos Method

Eccentrically Load Column No Buckling

500kg x 9.8m/s2 =4900 N

Redundant SupportGuy wire

Now Deflection known (=0)

Find necessary Tension Force F

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

(Nm)2

Nm

(Nm)2

m

Nm3

m3

Nm3

m3

finite value

=! 0 F=2940 N