CONTINUUM MECHANICS ( TORSION as BOUNDARY VALUE PROBLEM - BVP )

1/17M.Chrzanowski: Strength of Materials

SM1-12: Continuum Mechanics: Torsion as BVP

CONTINUUM MECHANICS(TORSION

as BOUNDARY VALUE PROBLEM - BVP)



Body shape: straight prismatic bar with end surfaces perpendicular to the bar axis, cross-section of arbitrary shape.

Loading: distributed loading over end surfaces yielding torque as only cross-sectional force, side surface free of loading, no volume forces.

Kinematics boundary conditions: Bar fixed at one end (all displacements and their derivatives vanish there).

MS

Problem formulation

In a further analysis we shall adopt the assumption of replacing kinematics conditions by statics ones (reaction torque); the bar is considered as being in the equilibrium but free to be twisted (free torsion).

M

We will make also use of de Sain-Venant principle replacing distributed loading with a torque

MS=M



x3

x1

x2

cossinsincoscoscoscos1 rrru

x1

x2

A

A’

r

r’

Assume: r’ = r sincossincossinsinsin2 rrru

Assume: 3x

0,, 21

sin1cos

21 sin xru

12 cos xru

Twist angle per unit length (unit angle)

1,0,0

321 xxu 312 xxu

213 , xxu ?

AA’

1

1

Total twist angle

Distortion function

u



i

j

j

iij x

u

x

u

2

1321 xxu 312 xxu

213 , xxu

2

12

3

3

223 2

1

2

1

xx

x

u

x

u

02

222

x

u 03

333

x

u01

111

x

u

02

1

2

133

1

2

2

112

xxx

u

x

u

1

21

3

3

113 2

1

2

1

xx

x

u

x

u

0

00

00

T

ijkkijij G 20kk

0

00

00

2GT

132123 ' xG

311213 ' xG

- distortion function



0

j

iji x

P

311213 ' xG

0

0

0

3

33

2

32

1

31

3

23

2

22

1

21

3

13

2

12

1

11

xxx

xxx

xxx

00 0

0 0 0

022

2

21

2

xx

0

0''0''0 21 GG

322123 ' xG

The governing equation of torsion boundary value

problem

02 or:

Laplacian



Statics boundary conditions

On a bar surface 0,0,0q 0,, 21 jijiq

0'' 221112 xGxG

On bar ends: 1,0,0 0,, 21 qqq

121 ' xGq 212 ' xGq

x1

x2

qν1

qν2

S

A

MdAxqxq 2112

S

A

MdAxxxxG 212121 '' +

MS

S

S

GJ

M

JS Torsion inertia moment

This is boundary value condition for distortion function differential equation

By de Saint Venant hypothesis



21

,x

f

x

fn

022

2

21

2

xx

2R

x1

x2

n 0, 22

22121 Rxxxxf

21

21 44 xxn

21 2,2 xxn

RxRxn

n21 ,

02

21

1

12

R

x

xx

R

x

xx

02

2

1

1

R

x

xR

x

x

0Governing equation and boundary condition are

homogeneous

Solid circular shaft

Contour equation:

No distortion!

Rxxn 22 21

21

022

111

2

x

xx

x



2

123 2 xx

GJ

M

S

S

GTT 2

dAxxxxJA

S 212121 ''

S

S

GJ

M

2

123 2 xx

1

213 2 xx

1

213 2 xx

GJ

M

S

S

2

123 xx

J

M

S

S

1

213 xx

J

M

S

S

321 xxu

312 xxu

213 , xxu

022

2

21

2

xx

02

211

12

x

xx

x 00,0,0



dAxxxxJA

S 212121 ''

S

S

GJ

M

2

123 xx

J

M

S

S

1

213 xx

J

M

S

S

0 0

222

21 JdArdAxxJ

AA

S 2

013 x

J

M S

10

23 xJ

M S

rJ

Mxx

J

M SS

0

21

21

0

223

213

r

00max W

MR

J

M SS 3

03 x

GJ

Mx S

Twist angle

Unit twist angle

Total twist angle for a shaft of length llGJ

M S

0max

)(r

Solid circular shaft

1x

2x

Torsion section modulus

Polar inertia moment



S

S

GJ

M0

S

S

W

Mmax

ss JJ ss WW

For hbb

h/2 max

hbJ s3 hbWs

2

h/b 1 1,5 2 2,5 3 4 6 8

0,208 0,231 0,246 0,258 0,267 0,282 0.299 0,307 0,333

0,141 0,196 0,229 0,249 0,263 0,281 0.299 0,307 0,333

Rectangular bar

),( bh h/2



Bars of open cross-sectionsBars of closed cross-sections

The behaviour of the above types of bars differs significantly when subjected to the action of a torque. One can make a simple experiment cutting a tube:

Thin-walled bars



1

2

1

Assume: constant distribution of shear stress across of tube thickness

2

Assume: prismatic tube of varying wall thickness

From equilibrium condition:

const 2211

02211

minmax const

Closed thin-walled cross-sections



ds

dA

r(s)

s

dssrdA )(2

1

dAdssrMss

s 2)(

const

SdAMs

s 22

S – area of the figure embedded

within central curve s

S

M s

2

s

s

W

Mmax SW s min2

minmax 2

S

M s




b1

h1

b2

h2

b3

h3

n

isis MM

1

i

ii bh

Solutions for torsion of rectangular bars obey:

s

si

GJ

M

si

sii W

Mmax

iiisi hbJ 3 iiisi hbW 2

A2

iii

Sii hbG

M3 iiisi hbGM 3

A3

n

iiiis hbGM

1

3 ss

n

iiiis JGMhbGM

1

3

Assumptions:Cross-section partitioning:

A1

A2

A3

A1

Open thin-walled cross-sections



si

sii W

Mmax

iiisi hbW 2

iiisi hbGM 3

ss

n

iiiis JGMhbGM

1

3

iiis

ssi hb

GJ

MGM 3

ii

i

s

s

iii

iii

s

si b

J

M

hb

hb

J

M

2

3

max

is

si b

J

Mmaxmax max

i

i

i

s

si b

J

M

maxmax max

For hi/bi >6 i =i




is

s bJ

Mmaxmax

maxh/2

h/2

max




ds

dA

r(s)

s

S – area of the figure embedded

within central curve s

minmax 2

S

M s




stop

CONTINUUM MECHANICS ( TORSION as BOUNDARY VALUE PROBLEM - BVP )

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