Lecture 4: Saint Venant Problems

2101607 Advanced Mechanics of Materials

Lecture 4

Saint Venant Problems

2101607 Advanced Mechanics of Materials Saint Venant Problems


Problem Description

Consider a cylindrical body (W) of length L A body is cylindrical if there exists a direction, called the axis, in which all sections

normal to that direction are identical

L

AxisCross section

Boundary of the body consists of three parts: the side face , the left end , and

the right end

S lS

rS

rS

S

lS

Material properties The body is made of a homogeneous, isotropic, linear elastic material

Two material constants from the set are prescribed{ , , , , }E k



Problem Description

Choice of coordinate system

L

AxisCross section

rS

S

lS

Origin (O) is at the centroid of cross section at lS

O

x3-axis directs along the axis of the cylinder

3x

x1- and x1-axes directs along principal axes of the cross section at lS

2x

1x

Resulting properties: 1 2 1 2

, , ,

0

l r l r l rS S S S S S

x dA x dA x x dA

Definitions:2

1 2

,l rS S

x dA I

Moment of inertia about x2-axisMoment of inertia about x1-axis

Polar moment of inertia

to simplify calculations

2

2 1

,

;

l rS S

x dA I2 2

1 2

,

; ( )

l rS S

x x dA J



Problem Description

Loading conditions

L

AxisCross section

rS

S

lS

Body force vanishes throughout, i.e., 0b

O 3x

Side face S

2x

1x

Boundary conditions: pure traction BCs, i.e., ,u tS S W

t 0: traction vanishes, i.e.,

: t 0

Left endlS

0lt t: traction is prescribed, i.e.,

Left endrS

0rt t: traction is prescribed, i.e.,

0: rt t

0: lt t

Prescribed traction data must satisfy overall equilibrium0 0

l r

l r

S S

dA dA t t 00 0;

l r

l r

S S

dA dA r t r t 0

Position vector to any reference point



Saint Venant Principle

Consider following situations:

L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t

0: lt t

The length of cylinder (L) is sufficient large in comparison with the dimensions of

the cross section There exists an interior region far away from the end zones

Only solution within the interior region is of interest

Saint Venant Principle: The dependence of the elastic field on the distribution of

the prescribed tractions at both ends decays and becomes insignificant as we move

away from both ends, i.e., solution is strongly dependent on their resultants not how

they distribute.




1x

2x

d3x

L

BVP1

BVP23x

BVP33x

P P

P P

P P

1x

2x

d3x

L

Simplified BVPP P

End zone

(2 3)to d

End zone

(2 3)to d

Interior region

(4 6)L to d

Solutions in this zone for all 3 BVPs are almost the same


Replace tractions by their resultants



Compute Resultants of Prescribed Tractions

LrS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

Left end boundarylS

0

2

lF

0

3

lF0

1

lF

0

0 1

1

0 0 0 0

2 2

0

03

3

l

l l

l

l

lS

l l l l

S Sl

l

S

t dAF

dA F t dA

Ft dA

F t

0

0 2 3

1

0 0 0 0

2 1 3

0

0 03

1 2 2 1

l

l l

l

l

lS

l l l l

S Sl

l l

S

x t dAM

dA M x t dA

Mx t x t dA

M r t

0

2

lM

0

3

lM

0

1

lM

Right end boundaryrS

0

2

rF

0

3

rF0

1

rF

0

0 1

1

0 0 0 0

2 2

0

03

3

r

r r

r

r

rS

r r r r

S Sr

r

S

t dAF

dA F t dA

Ft dA

F t

0

2

rM

0

3

rM

0

1

rM

0

0 2 3

1

0 0 0 0

2 1 3

0

0 03

1 2 2 1

r

r r

r

r

rS

r r r r

S Sr

r r

S

x t dAM

dA M x t dA

Mx t x t dA

M r t



Compute Resultants of End Tractions Generated by Stress

L

rS

S

lS

O 3x

2x

1x

: t 0

Left end boundarylS

2

lF

3

lF1

lF

2

lM

3

lM

1

lM

1 11 12 13 13

2 12 22 23 23

3 13 23 33 33

0

0

1

l

l l l

l

t

t

t

t σn

0

0

1

l

n

1 13

1

2 2 23

3

3 33

l l

l l l

l l

l

lS S

l l l l

S S Sl

l

S S

t dA dAF

dA F t dA dA

Ft dA dA

F t

2 3 2 33

1

2 1 3 1 33

3

1 2 2 1 1 23 2 13

l l

l l l

l l

l

lS S

l l l l

S S Sl

l l

S S

x t dA x dAM

dA M x t dA x dA

Mx t x t dA x x dA

M r t



Compute Resultants of End Tractions Generated by Stress

L

rS

S

lS

O 3x

2x

1x

: t 0

Right end boundaryrS

2

rF

3

rF1

rF

2

rM

3

rM

1

rM

1 11 12 13 13

2 12 22 23 23

3 13 23 33 33

0

0

1

r

r r r

r

t

t

t

t σn

0

0

1

r

n

1 13

1

2 2 23

3

333

l r

r l r

rl

r

rS S

r r r r

S S Sr

r

SS

t dA dAF

dA F t dA dA

FdAt dA

F t

2 3 2 33

1

2 1 3 1 33

3

1 2 2 1 1 23 2 13

r r

l r r

r r

r

rS S

r r r r

S S Sr

r r

S S

x t dA x dAM

dA M x t dA x dA

Mx t x t dA x x dA

M r t



Original versus Simplified BVPs

Original BVP

LrS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

Find the displacement field , strain field , stress field such that

u

ε σ

, 0ij j ib

, ,( )/2ij i j j iu u

2ij ij kk ij

for Wx

and0 l

ij j in t for lSx0 r

ij j in t for rSx

0ij jn for Sx

Simplified BVPFind the displacement field , strain field , stress field such that

u

ε σ

, 0ij j ib

, ,( )/2ij i j j iu u

2ij ij kk ij

for Wx

and0 0, l l l l F F M M for lSx0 0, r r r r F F M M for rSx

0ij jn for Sx

0

2

lF

0

3

lF0

1

lF

0

2

lM

0

3

lM

0

1

lM

0

2

rF

0

3

rF0

1

rF

0

2

rM

0

3

rM

0

1

rM



Uniaxial Tension Problem

Consider a cylindrical body of arbitrary cross section subjected to

0 0

0 0

0 , 0

0

l l

P

F M

0 0

0 0

0 , 0

0

r r

P

F M

Arbitrarily distributed traction at the left end with0lt

Arbitrarily distributed traction at the right end with0rt

L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

P P




Solution Procedure

0 0 0

0 0 0

0 0 /P A

Guessing stress field from knowledge in basic course of strength of materials

L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

P P

Inverted method by guessing form of stress field

is the cross sectional areaA

Check equilibrium equations

, 0ij j ib

Since the guessed stress field is constant and the body force vanishes b

are satisfied identically




Solution Procedure

Check compatibility of the guessed stress field

L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

P P

Check Beltrami-Mitchell equations: ,

10

1ij kk ij

(for = )b 0

Since the guessed stress field is constant

,

10

1ij kk ij


The guessed stress field is compatible

The corresponding strain field is compatible The displacement field exists




Solution Procedure

Check boundary conditions on the side face S

L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

P P

1

2

3

t

t

t

1

2

0 0 0

0 0 0

0 0 / 0

n

n

P A

0

0

0

The boundary conditions on the surface are satisfiedS




Solution Procedure

Check boundary conditions on the facelS

L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

P P

13

23

33

l

l

l

S

l

S

S

dA

dA

dA

F

0

0

( / )

l

l

l

S

S

S

dA

dA

P A dA

0

0

P

2 33

1 33

1 23 2 13

l

l

l

S

l

S

S

x dA

x dA

x x dA

M

2

1

1 2

( / )

( / )

(0) (0)

l

l

l

S

S

S

P A x dA

P A x dA

x x dA

0

0

0

0l l F F0, l lM M

The boundary conditions on the surface are satisfiedlS




Solution Procedure

Check boundary conditions on the facerS

L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

P P

13

23

33

r

r

r

S

r

S

S

dA

dA

dA

F

0

0

( / )

r

r

r

S

S

S

dA

dA

P A dA

0

0

P

2 33

1 33

1 23 2 13

r

r

r

S

r

S

S

x dA

x dA

x x dA

M

2

1

1 2

( / )

( / )

(0) (0)

r

r

r

S

S

S

P A x dA

P A x dA

x x dA

0

0

0

0r r F F0, r rM M

The boundary conditions on the surface are satisfiedrS




Solution Procedure Conclusion on the guessed stress field

0 0 0

0 0 0

0 0 /P A

L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

P P

- Satisfy equilibrium equations

- Satisfy compatibility conditions

- Satisfy traction boundary conditions on side face S

- Satisfy simplified boundary conditions on lS- Satisfy simplified boundary conditions on rS

The guessed stress field is the solution of the simplified BVP




Post-process for other quantities Strain field

L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

P P

1ij ij kk ij

E E

0 0 0

0 0 0

0 0 /P A

/ 0 0

0 / 0

0 0 /

P EA

P EA

P EA

ε




Post-process for other quantities Displacement field u

L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

P P

, ,

1

2ij i j j iu u

/ 0 0

0 / 0

0 0 /

P EA

P EA

P EA

ε

1

1 2 3 3 2 1

22 1 3 3 1 2

3 1 2 2 1 33

Px

EAu B x B x CPx

u B x B x CEA

u B x B x CPx

EA

uIntegration

Rig

id r

ota

tio

n B

1, B

2, B

3

Rig

id t

ran

slat

ion

C1, C

2, C

3

Rigid body motion




Remarks

L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

P P

Tractions generated by on the surfaces σ ,l rS S

0 0 0 0

0 0 0 0

0 0 / 1

l

P A

t

0

0

/P A

0 0 0 0

0 0 0 0

0 0 / 1

r

P A

t

0

0

/P A

What if and 0l lt t0r rt t

Solutions of simplified and original BVPs are identical for the whole body



Pure Bending Problem

Consider a cylindrical body of arbitrary cross section subjected to

0 0

0 0

0 ,

0 0

l l M

F M

0 0

0 0

0 ,

0 0

r r M

F M



L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

M

M

0 0

0 0

0 ,

0 0

l l M

F M 0 0

0 0

0 ,

0 0

r r M

F M




L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

M

M

Solution Procedure

1

0 0 0

0 0 0

0 0 Cx

Guessing form of the stress field from beam theory

Inverted method by guessing form of stress field

is unknown constant to be determinedC

Check equilibrium equations

, 0ij j ib

Since the guessed stress field is constant and the body force vanishes b


(one can try with / )C M I

0 0

0 0

0 ,

0 0

l l M

F M 0 0

0 0

0 ,

0 0

r r M

F M




L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

M

M

Solution Procedure

Check compatibility of the guessed stress field

Check Beltrami-Mitchell equations: ,

10

1ij kk ij

(for = )b 0

Since the guessed stress field is constant

,

10

1ij kk ij


The guessed stress field is compatible

The corresponding strain field is compatible The displacement field exists

0 0

0 0

0 ,

0 0

l l M

F M 0 0

0 0

0 ,

0 0

r r M

F M




L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

M

M

Solution Procedure


1

2

3

t

t

t

1

2

1

0 0 0

0 0 0

0 0 0

n

n

Cx

0

0

0

The boundary conditions on the surface are satisfied for any value of the constant C

S

0 0

0 0

0 ,

0 0

l l M

F M 0 0

0 0

0 ,

0 0

r r M

F M




L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

M

M

Solution Procedure


13

23

33

l

l

l

S

l

S

S

dA

dA

dA

F

1

0

0

l

l

l

S

S

S

dA

dA

C x dA

0

0

0

2 33

1 33

1 23 2 13

l

l

l

S

l

S

S

x dA

x dA

x x dA

M

1 2

2

1

1 2(0) (0)

l

l

l

S

S

S

C x x dA

C x dA

x x dA

2

0

0

CI

0l l F F is satisfied for any value of the constant C0l l M M is satisfied if the constant C = - M/I2

0 0

0 0

0 ,

0 0

l l M

F M 0 0

0 0

0 ,

0 0

r r M

F M




L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

M

M

Solution Procedure


13

23

33

l

l

l

S

r

S

S

dA

dA

dA

F

1

0

0

r

r

r

S

S

S

dA

dA

C x dA

0

0

0

2 33

1 33

1 23 2 13

r

r

r

S

r

S

S

x dA

x dA

x x dA

M

1 2

2

1

1 2(0) (0)

r

r

r

S

S

S

C x x dA

C x dA

x x dA

2

0

0

CI

0r r F F is satisfied for any value of the constant C0r r M M is satisfied if the constant C = - M/I2

0 0

0 0

0 ,

0 0

l l M

F M 0 0

0 0

0 ,

0 0

r r M

F M




L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

M

M

Solution Procedure

0 0

0 0

0 ,

0 0

l l M

F M 0 0

0 0

0 ,

0 0

r r M

F M

Conclusion on the guessed stress field

1

0 0 0

0 0 0

0 0 Cx

- Satisfy equilibrium equations

- Satisfy compatibility conditions

- Satisfy traction boundary conditions on side face S

- Satisfy simplified boundary conditions on if C = - M/I2lS- Satisfy simplified boundary conditions on if C = - M/I2rS

The guessed stress field is the solution of the simplified BVP if C = - M/I2

1 2

0 0 0

0 0 0

0 0 /Mx I

The solution is the same as that from the beam theory




L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

M

M

0 0

0 0

0 ,

0 0

l l M

F M 0 0

0 0

0 ,

0 0

r r M

F M

Post-process for other quantities Strain field

1ij ij kk ij

E E

1 2

0 0 0

0 0 0

0 0 /Mx I

1 2

1 2

1 2

/ 0 0

0 / 0

0 0 /

Mx EI

Mx EI

Mx EI

ε




L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

M

M

0 0

0 0

0 ,

0 0

l l M

F M 0 0

0 0

0 ,

0 0

r r M

F M

Post-process for other quantities Displacement field u

, ,

1

2ij i j j iu u

1 2

1 2

1 2

/ 0 0

0 / 0

0 0 /

Mx EI

Mx EI

Mx EI

ε

2 2 2

3 1 2

2

1 2 3 3 2 1

1 22 1 3 3 1 2

2

3 1 2 2 1 3

1 3

2

( )

2

M x x x

EIu B x B x C

Mx xu B x B x C

EIu B x B x C

Mx x

EI

u

Rig

id r

ota

tio

n B

1, B

2, B

3

Rig

id t

ran

slat

ion

C1, C

2, C

3

Rigid body motion




L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

M

M

0 0

0 0

0 ,

0 0

l l M

F M 0 0

0 0

0 ,

0 0

r r M

F M

Remarks Tractions generated by on the surfaces σ ,l rS S

1 2

0 0 0 0

0 0 0 0

0 0 / 1

l

Mx I

t

1 2

0

0

/Mx I

1 2

0 0 0 0

0 0 0 0

0 0 / 1

r

Mx I

t

1 2

0

0

/Mx I





Pure Torsion Problem

Consider a cylindrical body of circular cross section subjected to

0 0

0 0

0 , 0

0

l l

T

F M

0 0

0 0

0 , 0

0

r r

T

F M



0 0

0 0

0 , 0

0

l l

T

F M0 0

0 0

0 , 0

0

r r

T

F M

T

L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

T




0 0

0 0

0 , 0

0

l l

T

F M0 0

0 0

0 , 0

0

r r

T

F M

T

L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

T

Solution Procedure

1 2 3

2 1 3

0

C x x

C x x

u

Guessing form of the displacement field u

Inverted method by guessing displacement field

1 2, are unknown constants to be determinedC C

- Assume no movement of cross section along the axis (no warping displacement)

- Displacement should be linear in x3 (why ???)- Displacement u1 should be linear in x2 (why ???)- Displacement u2 should be linear in x1 (why ???)




0 0

0 0

0 , 0

0

l l

T

F M0 0

0 0

0 , 0

0

r r

T

F M

T

L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

T

Solution Procedure Generate strain field from , , / 2ij i j j iu u

1 2 3

2 1 3

0

C x x

C x x

u

Generate stress field from 2ij ij kk ij

1 2 3 1 2

1 2 3 2 1

1 2 2 1

0 ( ) /2 /2

( ) /2 0 /2

/2 /2 0

C C x C x

C C x C x

C x C x

ε

1 2 3 1 2

1 2 3 2 1

1 2 2 1

0 ( ) /2 /2

( ) /2 0 /2

/2 /2 0

C C x C x

C C x C x

C x C x

ε

1 2 3 1 2

1 2 3 2 1

1 2 2 1

0 ( )

( ) 0

0

C C x C x

C C x C x

C x C x

σ




0 0

0 0

0 , 0

0

l l

T

F M0 0

0 0

0 , 0

0

r r

T

F M

T

L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

T

Solution Procedure Check equilibrium equations:

1 2 3 1 2

1 2 3 2 1

1 2 2 1

0 ( )

( ) 0

0

C C x C x

C C x C x

C x C x

σ

, 0ij j ib

0

0

0

b

11,1 12,2 13,3 1b 0 0 0 0 0

21,1 22,2 23,3 2b 0 0 0 0 0

31,1 32,2 33,3 3b 0 0 0 0 0

Generated stress field is in equilbrium




0 0

0 0

0 , 0

0

l l

T

F M0 0

0 0

0 , 0

0

r r

T

F M

T

L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

T

Solution Procedure


1

2

3

t

t

t

1 2 3 1 2 1

1 2 3 2 1 2

1 2 2 1

0 ( ) /

( ) 0 /

0 0

C C x C x x R

C C x C x x R

C x C x

1 2 2 3

1 2 1 3

1 2 1 2

( ) /

( ) /

( ) /

C C x x R

C C x x R

C C x x R

The boundary conditions on the surface are satisfied ifS

1 2 0C C 1 2C C

1

2

/

/

0

x R

x R

n

R1 2( , )x x




0 0

0 0

0 , 0

0

l l

T

F M0 0

0 0

0 , 0

0

r r

T

F M

T

L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

T

Solution Procedure

By choosing , then1 2C C C

- The boundary conditions on the surface are satisfied andS

2 3

1 3

0

Cx x

Cx x

u

2

1

2 1

0 0 /2

0 0 /2

/2 /2 0

Cx

Cx

Cx Cx

ε

2

1

2 1

0 0

0 0

0

Cx

Cx

Cx Cx




0 0

0 0

0 , 0

0

l l

T

F M0 0

0 0

0 , 0

0

r r

T

F M

T

L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

T

Solution Procedure


13

23

33

l

l

l

S

l

S

S

dA

dA

dA

F

2

1

0

l

l

S

S

C x dA

C x dA

0

0

0

2 33

1 33

1 23 2 13

l

l

l

S

l

S

S

x dA

x dA

x x dA

M

2 2

1 2

0

0

lSC x x dA

0

0

CJ

0l l F F is satisfied for any value of the constant C0l l M M is satisfied if the constant C = T/J




0 0

0 0

0 , 0

0

l l

T

F M0 0

0 0

0 , 0

0

r r

T

F M

T

L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

T

Solution Procedure


13

23

33

r

r

r

S

r

S

S

dA

dA

dA

F

2

1

0

r

r

S

S

C x dA

C x dA

0

0

0

2 33

1 33

1 23 2 13

l

l

l

S

l

S

S

x dA

x dA

x x dA

M

2 2

1 2

0

0

lSC x x dA

0

0

CJ

0r r F F is satisfied for any value of the constant C0r r M M is satisfied if the constant C = T/J




0 0

0 0

0 , 0

0

l l

T

F M0 0

0 0

0 , 0

0

r r

T

F M

T

L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

T

Solution Procedure Conclusion on the guessed displacement field

1 2 3

2 1 3

0

C x x

C x x

u

- Generate stress field that is in equilibrium

The guessed displacement field is the solution of the simplified BVP ifu

- Satisfy traction boundary conditions on side face if S 1 2C C

- Satisfy simplified boundary conditions on iflS1 2 /C C C T J

- Satisfy simplified boundary conditions on ifrS1 2 /C C C T J

1 2 /C C C T J




0 0

0 0

0 , 0

0

l l

T

F M0 0

0 0

0 , 0

0

r r

T

F M

T

L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

T

Solution Procedure Final solutions of all fields

2 3

1 3

/

/

0

Tx x J

Tx x J

u

2

1

2 1

0 0 /2

0 0 /2

/2 /2 0

Tx J

Tx J

Tx J Tx J

ε

2

1

2 1

0 0 /

0 0 /

/ / 0

Tx J

Tx J

Tx J Tx J

σ




0 0

0 0

0 , 0

0

l l

T

F M0 0

0 0

0 , 0

0

r r

T

F M

T

L

rS

S

lS

O 3x

2x

1x

: t 0

0: rt t0: lt t

T

Remarks Tractions generated by on the surfaces σ ,l rS S

2

1

2 1

0 0 / 0

0 0 / 0

/ / 0 1

l

Tx J

Tx J

Tx J Tx J

t

2

1

/

/

0

Tx J

Tx J

2

1

2 1

0 0 / 0

0 0 / 0

/ / 0 1

r

Tx J

Tx J

Tx J Tx J

t

2

1

/

/

0

Tx J

Tx J



2 2 2 2

1 2 1 2|| ( ) ( ) /l l lt t T x x J Tr J || t2 2 2 2 2

1 2 1 2|| ( ) ( ) /r rt t T x x J Tr J || t

Lecture 4: Saint Venant Problems

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Transcript of Lecture 4: Saint Venant Problems