Chapter 9 Extension, Torsion and Flexure of Elastic Cylinders

Chapter 9 Extension, Torsion and Flexure

of Elastic CylindersPrismatic Bar Subjected to End Loadings

x

y

z P

M

ℓ

S

R

0 xyyx :Assume

Semi-Inverse Method

0

0

2

2

2

2

2

2

2

yxzyx

zz

zzzz

yzxz

Equations ityCompatibil

Equations mEquilibriu

654321 CyzCxzCzCyCxCz gIntegratin

Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island

Extension of CylindersAssumptions- Load Pz is applied at centroid of cross-

section so no bending effects- Using Saint-Venant Principle, exact end

tractions are replaced by statically equivalent uniform loading

- Thus assume stress z is uniform over any cross-section throughout the solid

0, yzxzz

z A

P

x

y

z Pz

ℓ

S

R

Using stress results into Hooke’s law and combining with the strain-displacement relations gives

0,0,0

,,

z

u

x

w

y

w

z

v

x

v

y

u

AE

P

z

w

AE

P

y

v

AE

P

x

u zzz Integrating and dropping rigid-body motion terms such that displacements vanish at origin

zAE

Pw

yAE

Pv

xAE

Pu

z

z

z

0 xyyx and


Torsion of Cylinders

x

y

z T

ℓ

S

R

Guided by Observations from Mechanics of Materials• projection of each section on x,y-plane rotates as

rigid-body about central axis• amount of projected section rotation is linear

function of axial coordinate• plane cross-sections will not remain plane after

deformation thus leading to a warping displacement


Torsional Deformations

x

y

O

P'P

r

R

S

xrv

yru

cos

sin

z

= angle of twist per unit length

),( yxww

xzv

yzu

w = warping displacement

Now must show assumed displacement form will satisfy all elasticity field equations


Stress Function Formulation

),( yxww

xzv

yzu

xy

we

yx

we

eeee

yz

xz

xyzyx

2

1

2

1

0

xy

w

yx

w

yz

xz

xyzyx 0

0

yxyzxz

2xyyzxz

Equilibrium Equations Compatibility Relation

Introduce Prandtl Stress Function = (x,y) :xy yzxz

,

Equilibrium will be identically satisfied and compatibility relation gives

22

2

2

22

yx

a Poisson equation that is amenable to several analytical solution techniques


Boundary ConditionsStress Function Formulation

x

y

z T

ℓ

S

R

000

0

000

000

constantds

d

ds

dy

yds

dx

x

nnnT

nnnT

nnnT

zzyyzxxzn

z

zzyyyxxyn

y

zzxyyxxxn

x

RR

nx

nyz

R

nzy

R

nzx

R

nzz

R

nyy

R

nxx

dxdyTTdxdyyTxTM

dxdyxTM

dxdyyTM

dxdyTP

dxdyTP

dxdyTP

2)(

0

0

0

0

0

On Lateral Side: S

On End: R (z = constant)

dn

dy

ds

dxn

dn

dx

ds

dyn

y

x

Normal Unit

n


Displacement Formulation

0

yxyzxz 0

2

2

2

2

y

w

x

w

Displacement component satisfies Laplace’s equation

)(0

0

yxyx

zzyyzxxzn

z

xnyndn

dwnx

y

wny

x

w

nnnT

or

On Lateral Side: S

R

R

nx

nyz

dxdyx

wy

y

wxyxT

TdxdyyTxTM

)(

)(

22

On End: R

JT Rigidity Torsional...22

Rdxdy

x

wy

y

wxyxJ


Formulation Comparison

S

Ryx

0

22

2

2

22

Snxy

wny

x

w

Ry

w

x

w

yx

0

02

2

2

2

x

y

O R

S

Stress Function Formulation Displacement Formulation

Relatively Simple Governing EquationVery Simple Boundary Condition

Very Simple Governing EquationComplicated Boundary Condition


Multiply Connected Cross-SectionsBoundary conditions of zero tractions on all lateral surfaces apply to external boundary So and all internal boundaries S1, . . . Stress function will be a constant and displacement be specified as per (9.3.20) or (9.3.21) on each boundary Si, i = 0, 1, . . .

x

y

C So

R

S1

iyxii Snxy

wny

x

wS

0 or

where i are constants. Value of i may be arbitrarily chosen only on one boundary, commonly taken as zero on So .

Constant stress function values on each interior boundary are found by requiring displacements w to be single-valued, expressed by

1

0),(S

yxdw 121

AdsS

where A1 is area enclosed by S1

Value of 1 on inner boundary S1 must therefore be chosen so that relation is satisfied. If cross-section has more than one hole, relation must be satisfied for each hole.Boundary conditions on cylinder ends will be satisfied, and resultant torque condition will give

1122 AdxdyTR


Membrane AnalogyStress function equations are identical to those governing static deflection of an elastic membrane under uniform pressure. This creates an analogy between the two problems, and enables particular features from membrane problem to be used to aid solution of torsion problem. Generally used to providing insight into qualitative features and to aid in developing approximate solutions.

x

y

S

z

R

p

Deflected Membrane

Static Deflection of a Stretched Membrane

Ndy

Ndy

Ndx

Ndx

Membrane Element

dx dy

x

z Ndy

Ndy x

z

dx

x

z

x

z2

2

pdxdy

Equilibrium of Membrane Element

N

p

y

z

x

zFz

2

2

2

2

0

Sz on0

R

zdxdyV

Membrane Equations

22

2

2

2

yx

Son0

R

dxdyT 2

Torsion Equations

Equations are same with: = z , p/N = 2 , T = 2V

dn

dzs

z

z

zt

zn

00

constant :Line Contour

n t


Torsion Solutions Derived from Boundary Equation

x

y

R

S

If boundary is expressed by relation f(x,y) = 0, this suggests possible simple solution scheme of expressing stress function as = K f(x,y) where K is arbitrary constant. Form satisfies boundary condition on S, and for some simple geometric shapes it will also satisfy the governing equation with appropriate choice of K. Unfortunately this is not a general solution method and works only for special cross-sections of simple geometry.

S

Ryx

0

22

2

2

2

Problem Value-Boundary

0),( yxf


Example 9.1 Elliptical Section

x

y

ab

12

2

2

2

b

y

a

x

Look for Stress Function Solution

1

2

2

2

2

b

y

a

xK

satisfies boundary condition and will satisfy governing governing if 22

22

ba

baK

Since governing equation and boundary condition are satisfied, we have found solution


Elliptical Section Results

Loading Carrying CapacityAngle of Twist

RRRdxdydxdyy

bdxdyx

aba

baT 2

22

222

22 112

Stress Field

(Stress Function Contours)

(Displacement Contours)

322

2

322

2

22

22

ba

Txx

ba

b

ab

Tyy

ba

a

yz

xz

4

2

4

222 2

b

y

a

x

ab

Tyzxz

Displacement Field

xyba

abTw

33

22 )(

33

22

22

33 )(

ba

baT

ba

baT or

2max

2),0(

ab

Tb


Elliptical Section Results3-D Warping Displacement Contours

T


Example 9.2 Equilateral Triangular Section

x

y

a 2a

For stress function try product form of each boundary line equation

))(23)(23( axayxayxK

satisfies boundary condition and will satisfy governing governing if

Since governing equation and boundary condition are satisfied, we have found solution

aK

6


Equilateral Triangular Section Results

Loading Carrying CapacityAngle of Twist

Stress Field


(Displacement Contours)

Displacement Field

3max

22

18

35

2

3)0,(

)2(2

)(

a

Taa

yaxxa

yaxa

yz

yz

xz

pIaT 5

3

35

27 4

)3(6

22 yxya

w


Additional Examples That Allow Simple Solution Using Boundary Equation Scheme

Section with Higher Order Polynomial Boundary (Example 9-3)

Circular Shaft with Circular Keyway (Exercise 9-22/23)

r = b

r = 2acos

r

x

y

.

)cos2

1)((2

22

r

arb

2 ofion Concentrat Stress

22

)(

)(0/ As

max

max

a

aab

shaftsolid

keyway

x

y

22 cxay

22 cyax

22 cyax

22 cxay

a

a

))(( 222222 ycxacyxaK

83,)21(4 2

c

aK

aaa 2),0()0,(max


Examples That Do Not Allow Simple Solution Using Boundary Equation Scheme

x

y

a

b

x = a

y = m1x

y = -m2x

x

y

General Triangular Section Rectangular Section


Example 9.4 Rectangular SectionFourier Method Solution

Previous boundary equation scheme will not create a stress function that satisfies the governing equation. Thus we must use a more fundamental solution technique - Fourier method. Thus look for stress function solution of the standard form

withph )(),( 22 xayxp

)(),(,0),(,0 222 xabxya hhh

homogeneous solution must then satisfy

)()(),( yYxXyxh Separation of Variables Method

22

2

2

2

yx

a

yn

a

xnByx

nnh 2

cosh2

cos),(1

a

bnnaB n

n 2cosh/)1(32 332/)1(2

x

y

a

b

a

yn

a

xn

a

bnn

axa

n

n

2cosh

2cos

2cosh

)1(32)(

5,3,1 3

2/)1(

3

222


Rectangular Section Results

a

yn

a

xn

a

bnn

ax

x

a

yn

a

xn

a

bnn

a

y

n

n

yz

n

n

xz

2cosh

2sin

2cosh

)1(162

2sinh

2cos

2cosh

)1(16

5,3,1 2

2/)1(

2

5,3,1 2

2/)1(

2

a

bn

n

abaT

n 2tanh

11024

3

16

5,3,155

43

5,3,1 2

2max

2cosh

1162)0,(

nyz

a

bnn

aaa

a

yn

a

xn

a

bnn

axyw

n

n

2sinh

2sin

2cosh

)1(32

5,3,1 3

2/)1(

3

2

Stress Field

Displacement Field

Loading Carrying Capacity/Angle of Twist


Rectangular Section Results



(Displacement Contours, a/b = 1.0)



Torsion of Thin Rectangular Sections (a<<b)

x

y

a

b

Investigate results for special case of a very thin rectangle with a << b. Under conditions of b/a >> 1

12

tanh2

cosh

a

bn

a

bnand

baT

a

xa

3

max

22

3

16

2

)(

x

y

1

2

3

(Stress Function Contours) (Composite Section)

Composite Sections

N

iii baT

1

3

3

16

Torsion of sections composed of thin rectangles. Neglecting local regions where rectangles are joined, we can use thin rectangular solution over each section. Stress function contours shown justify these assumptions. Thus load carrying torque for such composite section will be given by


Example 9.5 Hollow Elliptical Section

x

y

12

2

2

2

b

y

a

x1

)()( 2

2

2

2

kb

y

ka

x For this case lines of constant shear stress coincide with both inner and outer boundaries, and so no stress will act on these lateral surfaces. Therefore, hollow section solution is found by simply removing inner core from solid solution. This gives same stress function and stress distribution in remaining material.

1222

22

kba

bai

12

2

2

2

22

22

b

y

a

x

ba

ba

Constant value of stress function on inner boundary is

Load carrying capacity is determined by subtracting load carried by the removed inner cylinder from the torque relation for solid section

)1()()(

)()( 4332222

33

22

33

kbabakbka

kbka

ba

baT

Maximum stress still occurs at x = 0 and y = b 42max 1

12

kab

T


Hollow Thin-Walled Tube Sections

aa

A B

Tube Centerline

t(Section aa)

A B

o

MembraneC

With t<<1 implies little variation in membrane slope, and BC can be approximated by a straight line. Since membrane slope equals resultant shear stress

to

Load carrying relation: coioo

ioR

AAAAdxdyT

222

222

where A = section area, Ai = area enclosed by inner boundary, Ac = area enclosed by centerline

Combining relations tA

T

c2

Angle of twist: tA

TSAds

c

cc

Sc 24

2 where Sc = length of tube centerline


Cut Thin-Walled Tube Sections

Cut creates an open tube and produces significant changes to stress function, stress field and load carrying capacity. Open tube solution can be approximately determined using results from thin rectangular solution. Stresses for open and closed tubes can be compared and for identical applied torques, the following relation can be established (see Exercise 9-24)

stronger is tubeclosed thusand open tubein higher are Stresses

1sincebut 6

2

23

ClosedTubeOpenTubeClosedTube

OpenTubesc

s

c

c

s

ClosedTube

OpenTube τττ

τA A,

A

A

tATaAT

τ

τ

Cut


Torsion of Circular Shafts of Variable Diameterx

y

z

r

.Displacement Assumption

ur = uz = 0 u = u (r,z)

z

ue

r

u

r

ue

eeee

zr

rzzr

2

1,

2

1

0

z

u

r

u

r

uzr

rzzr

,

0

Equilibrium Equations 033

r

u

zr

zr

u

rr

r

Stress Function Approach

z

r

r

r

u

zr

r

r

r

u

rr

z

23

23

03

2

2

2

2

zrrr

constant

002 ds

d

ds

dz

zds

dr

rr

Boundary Condition

)],0()),(([2 zzzRT Load Carrying Torque


Conical Shaft Example 9-7

z

2

r

Stress Function Solution

2/322

3

22 )(3

1

zr

z

zr

zC

)cos3

1cos

3

2(2 3

T

C

Stresses Displacement

2/522

2/522

2

)(

)(

zr

rzC

zr

rC

z

r

r

zr

Cru

2/322 )(3

r is rigid-body rotation about z-axis and can be determined by specifying shaft rotation at specific z-location

boundaryon constant cos22

zr

z


Conical Shaft Example 9-7 = 30o

Comparison with Mechanics of Materials

4 5 6 7 8 9 100

0.01

0.02

0.03

0.04

0.05

0.06

z

( z

) max

/ T

Max Shear Stress Comparison

Mechanics of Materials

Elasticity Theory


Numerical FEA Torsion Solutions

(4224 Elements, 2193 Nodes)






A


Flexure of Cylinders

x

y

z

ℓ

S

R Px

Py

.(xo,yo)

Consider flexure of cantilever beam of arbitrary section with fixed end at z = 0 and transverse end loadings Px and Py at z = ℓ. Problem is solved in Saint-Venant sense, so only resultant end loadings Px and Py will be used to formulate boundary conditions at z = ℓ.

0 xyyx

))(( zlCyBxz From general formulation , and motivated from strength of materials choose , where B and C are constants. Stresses xz and yz will be determined to satisfy equilibrium and compatibility relations and all boundary conditions.

Remaining equilibrium equation will be identically

satisfied if we introduce stress function F(x,y) such that

0)(

CyBxyxyzxz

2

2

2

1

2

1

Cyx

F

Bxy

F

yz

xz


Flexure FormulationRemaining Beltrami-Michell Compatibility Relations

01

)(

01

)(

2

2

CF

x

BF

y

2)(

12 ByCxF

Zero Loading Boundary Condition on Lateral Surface S

0 yyzxxz nn )(2

1 22

ds

dxCy

ds

dyBx

ds

dF

Separate Stress Function F into Torsional Part and Flexural Part

),(),(),( yxyxyxF

Sds

d

R

on0

in22

Sds

dxCy

ds

dyBx

ds

d

RByCx

on)(2

1

in)(1

22

2


Flexure FormulationGeneral solution to )(

12 ByCx

)(16

1),(),( 33 ByCxyxfyx

where 2f = 0

Boundary Conditions on end z = ℓ

R yyz

R xxz

Pdxdy

Pdxdyxxyy PCIBI

yxxy PCIBI 2

2

xyyx

xyxyy

xyyx

xyyxx

III

IPIPC

III

IPIPB

xoyoRPyPxdxdy

yy

xxyBxCxyJ

)()(2

1 22

xoyoR xzyz PyPxdxdyyx ][

where x, y and xy are the area moments of inertia of section R

where J is the torsional rigidity – final relation determines angle of twist


Flexure Example - Circular Section with No Twist

x

y

za

P

ℓ

cos

12 r

I

P

x

araI

P

a x

onsin2

11 32

Polar Coordinate Formulation

Solution:

322

)1(24

21

)1(8

21

)1(8

23xxyxa

I

P

x

Stress Solution:

)(

]23

21[

)1(8

23

1

21

4

222

zlyI

P

xyaI

P

xyI

P

xz

xyz

xxz

)1(2

23)0,0(

2max

a

Pyz

Strength of Materials: 2max 3

4

a

P


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Chapter 9 Extension, Torsion and Flexure of Elastic Cylinders

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