3.6 Mean Rotation

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Sometimes a rigid body motion is just too restrictive. Consider a beam with square cross

section, fixed at one end. The other end is twisted by an angle γ . It is not known what

motion each node on the twisted surface makes, only the mean rotation γ is known. Hence,

the twisted surface can expand, contract, warp, and so on, which violates a rigid surface

condition.

To formulate an appropriate multiple point constraint, consider the motion of the nodes

on the twisted surface as a translation of the center of gravity of this set of nodes, followed

by a motion about it. The location of the center of gravity pcg of a set of N nodes at

locations pi satisfies

pcg = 1

N

N

 

j=1

pj . (3.93)

The relative location p

_

i of node i is

p

_

i

= pi

− 1

N

N

 

j=1

pj . (3.94)

The translation of the center of gravity is given by the mean of the displacements ui :

ucg = 1

N

N

 

j=1

uj (3.95)

168 GEOMETRIC NONLINEAR EFFECTS

and the relative displacement u

_

i of each nodes i satisfies

u

_

i

= ui − 1

N

N

 

j=1

uj . (3.96)

The rotation of each node i about the center of gravity is expressed by the angle αi in

Figure 3.19. This angle satisfies

| sin αi| =

_p

_

i

× (p

_

i

+ u

_

i )_

_p

_

i

_ ・ _p

_

i

+ u

_

i

_ (3.97)

(the underscore removes implicit summation). However, the plane defined by p

_

i , u

_

i in

Figure 3.19 will generally be different for each node i. Generally, we are interested in the

rotation about an axis. Let this axis be defined by a unit vector a. Then, the rotation γi of

node pi about this axis can be expressed as

γi = arcsin

a ・ [p

_

i

× (p

_

i

+ u

_

i )]

_p

_

i

_ ・ _p

_

i

+ u

_

i

_ (3.98)

and expressing that the mean angle amounts to γ leads to

1

N

 

i

γi = γ. (3.99)

Because of Equation (3.98), this is a nonlinear equation in the displacements. To linearize

this equation, we first focus on Equation (3.98) and define

λi := sin γi (3.100)

and use component notation in a rectangular coordinate system yielding

λi =

eknj akp

_

in(p

_

ij

+ u

_

ij )

_p

_

i

_ ・ _p

_

i

+ u

_

i

_ (3.101)

u

_

i

p

_

i

p

_

i

+ u

_

i

αi

Center of gravity

Figure 3.19 Rotation about the center of gravity

GEOMETRIC NONLINEAR EFFECTS 169

where p

_

ij and u

_

ij are the j components of pi and ui , respectively. The only terms in

Equation (3.101) depending on uk, k = 1, . . . ,N are u

_

ij and u

_

i through Equation (3.96).

Because of the term _p

_

i

+ u

_

i

_ in the denominator of Equation (3.101), λi is nonlinear in

uk. To linearize λi(uk), we first focus on the derivative of some simpler expressions:

∂u

_

ij

∂upq

= δipδjq − 1

N

N

 

k=1

δkpδjq

= δjq(δip − 1

N ). (3.102)

Now,

_p

_

i

+ u

_

i

_2 = p

_

ijp

_

ij

+ 2u

_

ijp

_

ij

+ u

_

iju

_

ij . (3.103)

Hence,

∂_p

_

i

+ u

_

i

_2

∂upq

= 2(δip − 1

N )(p

_

iq

+ u

_

iq ) (3.104)

and

∂_p

_

i

+ u

_

i

_

∂upq

=

(δip − 1

N )(p

_

iq

+ u

_

iq )

_p

_

i

+ u

_

i

_ . (3.105)

Using Equations (3.102) to (3.105), one obtains for the derivative of γi :

∂γi

∂upq

= 1

_1 − λ2

i

eknjakp

_

in(δip − 1

N )

_p

_

i

_ ・ _p

_

i

+ u

_

i

_3

_δjq_p

_

i

+ u

_

i

_2 − (p

_

iq

+ u

_

iq)(p

_

ij

+ u

_

ij )_ .

(3.106)

Defining

ξ i :=

(p

_

i

+ u

_

i )

_p

_

i

+ u

_

i

_ (3.107)

and

ηi :=

p

_

i

_p

_

i

_ (3.108)

Equation (3.106) can be transformed into

∂γi

∂upq

= 1

_1 − λ2

i

_δip − 1

N

_

_p

_

i

+ u

_

i

_ [eknqakηin − λiξiq ] (3.109)

where

λi = eknj akηinξij . (3.110)

170 GEOMETRIC NONLINEAR EFFECTS

h

8 h h

M

ν = 0.3

Figure 3.20 Cantilever beam with square cross section subject to torsion

The governing nonlinear equation, Equation (3.99), can finally be linearized at position 0

yielding

N

 

i=1

_

γi |

0 + ∂γi

∂upq

____

0

(upq − upq_

_0)

_ = Nγ (3.111)

or

_ N

 

i=1

∂γi

∂upq

____

0

_

(upq − upq_

_0) = Nγ −

N

 

i=1

γi |

0 . (3.112)

This is a linear scalar equation in the unknowns upq, p = 1, . . . ,N, q = 1, . . . , 3. Notice

that the coefficients of the linear terms can at times be zero. Since the dependent term in

an equation must have a nonzero coefficient, the selection of the dependent variable may

have to change from one iteration to the next.

The mean-rotation concept only makes sense if more than one node is involved. If one

of the nodes k happens to coincide with the center of gravity of the node set, the angle

γk is not determinate since p

_

k

= 0 and the contribution γk|

0 and ∂γk

∂upq

___

0

are left out in

the sums in Equation (3.112). Equation (3.112) is less restrictive than a rigid body motion.

Accordingly, less energy is needed for applying a mean rotation than for a rigid body

motion.

This can be nicely illustrated by the cantilever beam in Figure 3.20. A torque is applied

at the free end such that a rotation of 45◦ results. Three conditions are examined here:

1. The beam theory is applied and the torque is determined analytically by (Popov 1968)

M = 0.141ϕ(bc)3G

L

(3.113)

where, in the actual example, ϕ = π/4, b = c = h and L = 8 h.

2. The cross section at the free end of the beam is considered as a rigid body.

3. The mean-rotation condition is applied.

The torque required for each of these conditions is listed in Table 3.2. The analytical result

is close to the rigid body condition. The mean-rotation condition requires a torque that is

10% less due to the relaxed constraints.

GEOMETRIC NONLINEAR EFFECTS 171

Table 3.2 Torque needed

for a twist of 45◦.

Condition

M

Gh3

Beam theory 0.0138

Rigid body 0.0141

Mean rotation 0.0126

a

b

Figure 3.21 A straight-line kinematic constraint

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