Elasticity/Rotating rectangular beam

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Example : Rotating Rectangular Beam

File:Elastic rotating beam.png
A rotating rectangular beam

The body force potential is given by

(52)V=ρθ˙22(x12+x22)

Hence,

(53)2V=V,11+V,22=ρθ˙22(2+2)

or,

(54)2V=2ρθ˙2

The compatibility condition (in terms of stress) is

(55)4φ+(21α)2V=0

Plug V in to get

(56)4φ(21α)2ρθ˙2=0

Since V is even in x1 and x2 and BCs are homogeneous, assume

(57)φ=Ax14+Bx12x22+Cx24+Dx12+Ex22

Hence,

(58)σ11=φ,22+V=2Bx12+12Cx22+2Eρθ˙22(x12+x22)(59)σ22=φ,11+V=12Ax12+2Bx22+2Dρθ˙22(x12+x22)(60)σ12=φ,12=4Bx1x2

The traction BCs are

(61)atx1=±at1=t2=0σ11=σ12=0(62)atx2=±bt1=t2=0σ12=σ22=0

Apply BCs at x2=±b.

(63)σ22=0=12Ax12+2Bb2+2Dρθ˙22(x12+b2)(64)σ12=0=4Bbx1

Therefore,

(65)B=0(66)A=ρθ˙224(67)D=ρθ˙2b24

We then have,

(68)φ=ρθ˙224x14+Cx24+ρθ˙2b24x12+Ex22

Plug into compatibility equation

(69)φ,1111+2φ,1122+φ,2222(21α)2ρθ˙2=0

to get

(70)24(ρθ˙224+C)(21α)2ρθ˙2=0

or,

C=(21α)ρθ˙212ρθ˙224=(32α)ρθ˙224(71)

Apply BCs at x1=±a.

(72)σ11=0=2Ba2+12Cx22+2Eρθ˙22(a2+x22)=(32α)ρθ˙22x22+2Eρθ˙22(a2+x22)(73)

Strong BCs imply that

(74)(32α)ρθ˙22ρθ˙22=0

which cannot be true. So weak BCs on σ11 need to be applied at x1=±a.

(75)atx1=±abbσ11dx2=0

Hence,

(76)(32α)ρθ˙222b33+4Ebρθ˙22(2a2b+2b33)=0

or,

(77)(22α)ρθ˙2b23+4Eρθ˙2a2=0

Hence,

(78)E=ρθ˙24[a2(22α)b23]

The stress field is, therefore,

(79)σ11=(32α)ρθ˙22x22+ρθ˙22[a2(22α)b23]ρθ˙22(x12+x22)(80)σ22=ρθ˙22x12+ρθ˙2b22ρθ˙22(x12+x22)σ12=0

or,

(81)σ11=ρθ˙22[(a2x12)+2(11α)(x22b23)](82)σ22=ρθ˙22(b2x22)σ12=0

The displacements can be found in the standard manner.

File:Elastic rotating beam sigxx.png
Stresses (σ11) in a rotating rectangular beam
File:Elastic rotating beam sigxx.png
Stresses (σ22) in a rotating rectangular beam

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