Chapter 4 Equations of Elastic-Plastic Equilibrium,

Simplest Problems

4.1 Equations of Elastic-Plastic Equilibrium

4.1.1 Basic Equations

（1）Equilibrium equations

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

=+∂∂

+∂∂

+∂∂

=+∂∂

+∂∂

+∂∂

=+∂∂

+∂∂

+∂∂

0

0

0

zzzyzx

yyzyyx

xxzxyx

Fzyx

Fzyx

Fzyx

σττ

τστ

ττσ

（4-1）

or

0, =+ ijij Fσ （ 321,,, ，，orzyxji = ） （4-2）

（2）Geometrical equations

zu

xw

zw

yw

zv

yv

xv

yu

xu

zxz

yzy

xyx

∂∂

+∂∂

=∂∂

=

∂∂

+∂∂

=∂∂

=

∂∂

+∂∂

=∂∂

=

γε

γε

γε

,

,

,

（4-3）

or

)(21

,, ijjiij uu +=ε （ 321,,, ，，orzyxji = ） （4-4）

（3）Constitutive (Physical) equations （a）In elastic stage

kkijijij EEσδνσνε −

+=

1 （ 321,,,, ，，orzyxkji = ） （4-5）

or

eeEEijijijijij λδμεδ

νννε

νσ +=

−++

+= 2

)21)(1(1 （4-6）

（b）In plastic stage

σελλ

με

23,

21 ddsddsd ijijij =+= Incremental (flow) theory

ijij seσε

23

= Deformation theory （4-7）

（4）Boundary conditions

On stress boundary, ： TS

⎪⎩

⎪⎨

⎧

=++=++=++

zzzyzx

yyzyyx

xxzxyx

TlllTlllTlll

321

321

321

στττστττσ

， i.e. ijij Tn =σ （4-8）

On displacement boundary, ： uS

ii uu = （4-9）

4.1.2 The Problems Putting Forward The purpose of solving an elastic-plastic problem, often called a boundary value problem, is to find the distributions of stress, strain and displacement in a body. Therefore the way to put forward the boundary value problem is to find the stress field and displacement field in a solid body subject to external effect (including temperature, external loading, etc.) to the interior and whole surface of the body. More specifically, the stress, strain and displacement at every point of the body should satisfy the equilibrium equations, geometrical equations and constitutive equations, as well as all given boundary conditions. （1）Boundary value problem of the incremental elastic-plastic theory

A body V with stress boundary and displacement boundary . Find the

stress field

TS uS

ijσ , strain field ijε and displacement field of the body under

external loading: volume force , surface force i（on ）and given displacement

iu

iF T TS

iu （on ）. uS

According to loading path, the load is applied in stepwise：

ii FF Δ∑= ， ii TT Δ∑= ， ii uu Δ∑= （4-10）

the increments from time to t tt Δ+ are

tFF ii Δ=Δ ， ，tTT ii Δ=Δ tuu ii Δ=Δ （4-11）

where, , and iF iT iu are called the rate of volume force, the rate of surface force

and the rate of displacement (velocity) separately. The reason to introduce the forms of those rates is to simplify the expression to the relevant formulae.

The problem solution procedure is：

If displacement , strain iu ijε , stress ijσ and the history recording parameter

ξ at time t have been obtained and the present loading surface is 0),( =ξσ ijf ,

when the volume force rate and surface force rate on boundary and

displacement rate (velocity )

iF iT TS

iu on boundary have been given, we now need to

find velocity , strain rate

uS

iu ijε and stress rate ijσ .

The unknown ，iu ijε and ijσ should satisfy the following equations：

1）Equilibrium equations 0, =+ ijij Fσ

2）Geometrical equations )(21

,, ijjiij uu +=ε

3）Constitutive equations，such as Prandtl-Reuss equation ijijij sdλσμ

ε +=21

4）Boundary conditions

on ： ； on ：TS ijij Tn =σ uS ii uu =

Moreover, the evolution law of internal variable（history recording parameter）ξ

follows：

( )ξξσεξεξ ),,(),( ijijij gg == （4-12）

When ，iu ijε ， ijσ and at time have been solved, the displacements,

strains, stresses and recording parameter at time

ξ t

tt Δ+ will be

tuuu ititti Δ+=Δ+

tijtijttij Δ+=Δ+

εεε

tijtijttij Δ+=Δ+

σσσ

tttt Δ+=Δ+

ξξξ

and the new loading surface will be：

0),( =Δ+Δ+ ttf ijij ξξσσ （4-13）

Practically, the effect of the length of load increment on accuracy, convergence and stability of calculations should be considered, thus it is often necessary to solve

the problem numerically. （2）Boundary value problem of elastic-plastic deformation theory（omitted） 4.2 Elastic-Plastic Analysis of Thick-Wall Cylindrical Tube under Pressure This is an axial symmetric problem, so the cylindrical coordinate system

（ zr ,,θ ）is used, therefore 0=== zrzr γγγ θθ . Since the cylinder is very long, the

problem can be considered as plane strain problem, 0=zε , or generalized plane

stress problem where constz == 0εε . p

ab

Fig. 4-1

Equilibrium equation:

0=−

+rdr

d rr θσσσ （4-14）

Geometric equations:

drdu

r =ε ，ru

=θε —axial displacement（4-15） u

Compatibility equation:

0=−

+rdr

d rεεε θθ （4-16）

Boundary condition:

（4-17） ⎩⎨⎧

==−

=brarp

r ,0,

σ

∫= ba zrdrT σπ2 —Axial force （4-18）

（1）Elastic solution Substituting stress-strain relation

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

=+−=

+−=

+−=

0)]([1

)]([1

)]([1

εσσνσε

σσνσε

σσνσε

θ

θθ

θ

rzz

rz

zrr

E

E

E （4-19）

into Eq.（4-16）and utilizing equilibrium equations (4-14), we get the compatibility equation in terms of stresses：

0)(=−

+dr

ddr

d zr σνσσ θ

According to the last equation in Eq.（4-19）：

0)(=

+−=

drd

drd

drdE rzz θσσνσε

From the above two equations

0)(=

+dr

d r θσσ， 0=

drd zσ

or

constAr ==+ 2θσσ ， constz =σ

Introducing stress function ϕ：

drd

rrϕσ 1

= ， 2

2

drd ϕσθ =

which automatically satisfies equilibrium equation, then

Adrd

rdrd 21

2

2=+

ϕϕ

Solving the above equation we obtain

CrBrA+−= ln

22ϕ

2rBAr −=σ ， 2r

BA +=θσ

constants A and B can be determined by using boundary condition （4-17）as

01 2

2≤⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

rbprσ ， 01 2

2>⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

rbpθσ （4-20）

where, pab

ap 22

2

−= .

From the third equation in Eq.（4-19）and Eq.（4-20）we know

constEpErz =+=++= 00 2)( ενεσσνσ θ （4-21）

substituting into boundary condition (4-18), we obtain

)( 22 abT

z −=π

σ （4-22）

and

)(2)2(1

22

2

0 abEpaTp

E z −−

=−=π

νπνσε （4-23）

From the second equation in Eq.（4-19）and the second equation in Eq.（4-15）：

（ )]([1rzE

σσνσε θθ +−= ，ru

=θε ）

rr

brpEE

ru rz 0

2])21[(1)]([ νεννσσνσθ −+−

+=+−= （4-24）

Elastic limit pressure can be determined using Tresca yield criterion：

rσσθ > ， 0max, == brr σσ ， pb 2min, == θθ σσ

thus when pz 20 ≤≤σ , i.e. p20

zσθσrσ

Fig. 4-2

22

2

222

)(0

abpa

abT

−≤

−≤π

（4-25）

or

paT 220 π≤≤ （4-26）

zσ is the intermediate principal stress. If the ends are free, 0=T , and if closed,

, all lie in the range of Eq. (4-25). So in terms of Tresca yield criterion paT 2π=

2

22

rbpr =−σσθ

which reaches the maximum value at ar = and yielding firstly occurs at here

sr abp σσσθ ==− 2

2

max 2)(

so the elastic limit load (pressure) is obtained

2

2

2bap s

eσ

= ， ⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−== 2

2

2

221

2 bap

aabpp s

eeσ （4-27）

（2）Elastic-plastic solution

When , plastic region enlarges gradually. Supposing the border between

elastic region and plastic region is located at

epp >

cr = , bca << .

（a）Solution in plastic zone cra ≤≤

We still assume zσ be the intermediate

principal stress and notice Tresca yield criterion

sr σσσθ =− , then equilibrium equation（4-14）

becomes

Elastic zone

Plastic zone

cb

a

Fig. 4-3

0=−=−

+rdr

drdr

d srrr σσσσσ θ

i.e.

rdrd sr σσ

=

by integration of the above we know

Crsr ′+= lnσσ

by utilizing boundary condition parr −==

σ we get

apC s lnσ−−=′ , so

⎪⎩

⎪⎨

⎧

++−=+=

+−=−−=

)ln1(

lnlnln

arp

arpapr

srs

sssr

σσσσ

σσσσ

θ

（4-28）

the solution is static determinant and the geometric relation is not necessary. （b）Solution in elastic zone brc ≤≤ Elastic solution（4-20）still apply in elastic zone by replacing in the solution expressions of Eq. (4-20) with .

ac

scaec bcppp σ2

2

2===

→， ⎟⎟

⎠

⎞⎜⎜⎝

⎛−===

→ 2

21

2 bcppp s

caecσ

⎪⎪⎭

⎪⎪⎬

⎫

⎟⎟⎠

⎞⎜⎜⎝

⎛+=⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

2

2

2

2

2

2

2

2

2

2

2

2

12

1

12

1

rb

bc

rbp

rb

bc

rbp

s

sr

σσ

σσ

θ

（4-29）

Because ）（）（ elasticplastic crrcrr == =σσ , from Eq.（4-28） and Eq.（4-29）

we know

⎟⎟⎠

⎞⎜⎜⎝

⎛−=+− 2

2

2

21

2ln

cb

bc

acp s

sσσ

thus

⎟⎟⎠

⎞⎜⎜⎝

⎛−+= 2

21

21ln

bc

acp

sσ （4-30）

gives the relation between pressure p and elastic-plastic border . c

When , whole cylinder becomes yielding, bc = p is not able to increase, we

obtain the plastic limit pressure ： sp

abp ss lnσ= （4-31）

If zσ is still the intermediate principal stress, then yield criterion is

sr σσσθ =− ,

0=∂∂

=z

pz

fddσ

λε ， 0== constpzε

and

)]([10 θσσνσεεε +−=== rz

ezz E

so

0)( εσσνσ θ Erz ++=

Substituting into boundary condition and integrating it，we get ∫= ba zrdrT σπ2

)(2

22

2

0 abEpaT

−−

=π

νπε （4-32）

The verification of zσ being the intermediate principal stress is neglected.

（3）Displacement analysis is omitted （4）Residual stress after unloading

If the internal pressure is increased to （ ）and the axial load

reaches

*p se ppp << *

*T , the pressure is allowed to unload until totally removed. The way to do

this is to superpose a reversal elastic solution on basis of the original stress

distribution and ： *rσ

*θσ

，*pp −=Δ *TT −=Δ

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−−=Δ 2

2

22

*21

rb

abpa

rσ ； ⎟⎟⎠

⎞⎜⎜⎝

⎛+

−−=Δ 2

2

22

*21

rb

abpa

θσ

)(

222

*

022

*2

abTE

abpa

z −−=−

−−=Δ

πενσ

so the residual stresses are：

⎪⎪⎩

⎪⎪⎨

⎧

≤≤Δ+⎟⎟⎠

⎞⎜⎜⎝

⎛−

≤≤Δ++−=Δ+=

)(,12

)(,ln

2

2

2

2

*

*0

brcrb

bc

craarp

rs

rs

rrrσσ

σσσσσ

⎪⎪⎩

⎪⎪⎨

⎧

≤≤Δ+⎟⎟⎠

⎞⎜⎜⎝

⎛+

≤≤Δ+⎟⎠⎞

⎜⎝⎛ ++−

=Δ+=)(,1

2

)(,ln1

2

2

2

2

*

*0

brcrb

bc

craarp

s

s

θ

θ

θθθσσ

σσσσσ

（4-33） （5）The elastic-plastic solution of the thick-wall cylinder made up of hardening material

Assumption:

① Plane strain state, so 0 0zε ε= = ;

② The volume is incompressible, therefore 0m r zθε ε ε ε= + + = ;

③ Simple loading, so that the deformation theory applies, and 23ij ijs eσε

= ;

④ Power function hardening law, that is nAσ ε= .

Solution procedure: From the assumption ③, we know

23z zs eσε

= 2 ( )3z m z mσσ σ εε

− = −ε

From the assumption ① and ②, we know

2 ( ) 03

z rz m

θσ σ σσ σ − +− = =

1 ( )2z r θσ σ σ= +

The equilibrium equation: 0rrddr r

θσ σσ −+ =

The geometric equation: rr

dudr

ε = , rurθε =

Considering the conditions of incompressible volume and plane strain, we know

( )z r θ 0ε ε ε= − + =

so we have

0r rdu udr r

+ = rCur

=

where C is the constant to be determined. Therefore

2r

rdu Cdr r

ε = = − , 2ru C

r rθε = = , 0zε =

The equivalent strain

( ) ( ) ( )2 2 22

2 23 3r r z z

Crθ θε ε ε ε ε ε ε= − + − + − =

Noticing that 1 (2z r )θσ σ σ= + , the equivalent stress is

2 2 21 3( ) ( ) ( ) ( )22 r z z rθ θ θ rσ σ σ σ σ σ σ σ σ= − + − + − = −

or

23rθσ σ σ− =

Substituting the above equation and nAσ ε= into the equilibrium equation

0rrddr r

θσ σσ −+ = , we obtain

2

2 2 2 2( )3 3 3 3

nn

r rdr dr dr C drd Ar r r rθσ σ σ σ ε ⎛ ⎞= − = = = ⎜ ⎟

⎝ ⎠A

r

Integration of the above expression in the range of r to the outer radius b

2

2 23 3

nbb

r r r

C drAr r

σ ⎛ ⎞= ⎜ ⎟⎝ ⎠∫

which yields

1 2 22

2 1

3

n nb

r n n nr

ACb rn

σ +⎛ ⎞= − −⎜ ⎟⎝ ⎠

1

Noticing that 0r r bσ

== , we get

1 2 22

2 1

3

n n

r n n n

ACb rn

σ +⎛ ⎞= −⎜ ⎟⎝ ⎠

1

Substituting r r apσ

== − into the above expression, we obtain

1 112 22

1 12 2

3

2 ( )

nn n

n

n nn n

n a bC pA b a

+

=−

therefore the stresses 2 2 2

1 2 2 2 2 22

2 1 1 ( )( )3

n n n n n

r n n n n n n

AC b r a pb r b a rn

σ +

−⎛ ⎞= − = −⎜ ⎟ −⎝ ⎠

2

23

nn CA A

rσ ε ⎛ ⎞= = ⎜ ⎟

⎝ ⎠

2 2 2 2

2 2 2

2 ( 2 )( )3

n n n n

r n n n

r b nb a pb a rθσ σ σ − +

= + =−

（4-34）

and the displacement

1 112 22

2 2

32( )

nn n

nr n n

C a b nur b a r A

+

⎛ ⎞= = ⎜ ⎟− ⎝ ⎠p （4-35）

4.3 The Free Torsion of the Perfect Plastic Prismatic Bar 4.3.1 Basic equations in the theory of elasticity

The equilibrium equations:

0

0

0

yxx zx

xy y zy

yzxz z

Xx y z

Yx y z

Zx y z

τσ τ

τ σ τ

ττ σ

∂∂ ∂+ + + =

∂ ∂ ∂∂ ∂ ∂

+ + + =∂ ∂ ∂

∂∂ ∂+ + + =

∂ ∂ ∂

（4-36）

The geometric equations:

xux

ε ∂=∂

； xyu vy x

γ ∂ ∂= +∂ ∂

yvy

ε ∂=∂

； yzv wz y

γ ∂ ∂= +∂ ∂

（4-37）

zwz

ε ∂=∂

； zxw ux z

γ ∂ ∂= +∂ ∂

The constitutive equations:

1 1[ ( )],

1 [ ( )],

1 1[ ( )],

1

x x y z xy

y y z x yz yz

z z x y zx zx

E G

E G

E G

xyε σ ν σ σ γ τ

ε σ ν σ σ γ τ

ε σ ν σ σ γ τ

= − + =

= − + =

= − + =

（4-38）

The stress boundary conditions:

x x yx y zx z

xy x y y zy z

xz x yz y z z

n n n

n n n

n n n

σ τ τ

τ σ τ

τ τ σ

+ + =

+ + =

+ + =

X

Y

Z

（4-39）

By elimination of the displacement components in the geometric equations, we obtain the compatibility equation denoted by the strain components:

2 2 2 2 22 2 22

2 2 2 2 2 2, ,y y yz yx zx xz

z y y z x z z x y x xxy

yε ε γ ε γε γ εε∂ ∂ ∂ ∂ ∂∂ ∂ ∂∂

+ = + = + =∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

（4-40）

Utilizing the equilibrium equations and the constitutive equations, the above compatibility equation equations can be transformed into the those denoted by the stress components, which when neglecting the body forces shows

2 22 21 1

2

2 22 21 1

2

2 22 21 1

2

(1 ) 0, (1 ) 0

(1 ) 0, (1 ) 0

(1 ) 0, (1 ) 0

x yz

y zx

z xy

I Ix yI I

y zI I

z x

ν σ ν τ

ν σ ν τ

ν σ ν τ

∂ ∂+ ∇ + = + ∇ + =

∂ ∂

∂ ∂+ ∇ + = + ∇ + =

∂ ∂

∂ ∂+ ∇ + = + ∇ + =

∂ ∂

z

x

y

∂

∂

∂

（4-41）

In order to find the solution for any problem in terms of stress, it should satisfy the equilibrium equations, compatibility equations as well as the stress boundary conditions.

4.3.2 The elastic solution

Assuming

0x y z xyσ σ σ τ= = = = （4-42）

then the equilibrium equations are

0, 0, 0zy yzzx xz

z z x yτ ττ τ∂ ∂∂ ∂

= = +∂ ∂ ∂ ∂

=

From the above first two equations, we

notice that zxτ and zyτ are only the functions of x and y. So the above third equation

is written as

x

y

z

O

zM

zM

( )xz yx y zτ τ∂ ∂= −

∂ ∂

We introduce the stress function ( , )x yϕ and denote the stress components as

,zx xz zy yzy xϕ ϕτ τ τ τ∂ ∂

= = = = −∂ ∂

（4-43）

then from the compatibility equations we know 2 20, 0yz xzτ τ∇ = ∇ =

and

2 20, 0x y

ϕ ϕ∂ ∂∇ = ∇

∂ ∂=

so 2ϕ∇ must be a constant:

2 Cϕ∇ = （4-44）

Applying the boundary conditions:

The lateral surface is a free surface, so 0X Y Z= = = , 0zn = . If we substitute

Eq. (4-42) into the boundary conditions (4-39), we will see that the first two equations in Eq. (4-39) are satisfied naturally, and the third equation requires

0xz x yz yn nτ τ+ =

Since ,xz yzy xϕ ϕτ τ∂

= = −∂

∂ ∂, and having ,x y

dy dxn nds ds

= = − on the boundary,

the above equation becomes

0dy dx dy ds x ds dsϕ ϕ ϕ∂ ∂

+ =∂ ∂

= , i.e. sϕ = const （4-45）

Because a constant difference between two stress functions makes no difference on the stress components, we will let

0sϕ = （4-45）

At bar ends, say the end 0x = , 0x yn n= = , 1zn = − , 0Z = , the third equation

in Eq. (4-39) is satisfied naturally, and the first two equations become

,zx zyX Yτ τ− = − = （a）

The equilibrium conditions on the end surface are

0, 0, ( ) zXdA YdA yX xY dA M= = −∫∫ ∫∫ ∫∫ = （b）

x

y

o

A

B

x

Because that

( )

zx

B A 0

XdA dxdy dxdyy

dx dy dxy

ϕτ

ϕ ϕ ϕ

∂= − = −

∂∂

= − = − − =∂

∫∫ ∫∫ ∫∫

∫ ∫ ∫

we can see the first equation in Eq. (b) is satisfied, and the same for its second equation.

For the left hand side of Eq. (b)

( ) ( ) ( )

( )zx zyyX xY dA yX xY dxdy y x dxdy

y x dxdy dx y dy dy x dxy x y x

τ τϕ ϕ ϕ ϕ

− = − = − −∂ ∂ ∂ ∂

= − + = − −∂ ∂ ∂ ∂

∫∫ ∫∫ ∫∫∫∫ ∫ ∫ ∫ ∫

where

[ ]BAdx y dy dx y dy dxdy

yϕ ϕ ϕ ϕ∂

− = − − =∂∫ ∫ ∫ ∫ ∫∫

and also

dy x dx dxdyxϕ ϕ∂

− =∂∫ ∫ ∫∫

so the third equation in Eq. (b) becomes

2 zdxdy Mϕ =∫∫ （4-46）

To determine the constant C in Eq. (4-44), we substitute Eq. (4-42) and (4-43) into the compatibility equation (4-38), and then obtain

1 10, 0, 0, , , 0x y z yz zx xyG x G yϕ ϕε ε ε γ γ γ∂ ∂

= = = = − =∂ ∂

=

substitution of the above expressions into the geometric equation (4-37) yields

0, 0, 0,

1 1, ,

u v wx y zw v u w v uy z G x z x G y x y

ϕ ϕ

∂ ∂ ∂= = =

∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

+ = − + = + =∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

0

By means of the integration and derivation, we obtain

0 0,y z z xu u z y Kyz v v x z Kxzω ω ω ω= + − − = + − +

where 0 0, , , ,x yu v zω ω ω are the integration constants interpreting the rigid

displacement, K is the constant representing the torsional deformation. We keep only the deformation terms, then

,u Kyz v Kxz= − = （4-47）

they are denoted in the cylindrical coordinate system as 0,ru u rθ Kz= = .

The rotation angle of the cross-section is Kzα = , and d Kdzα= , so that K is the

torsional angle per unit length of the bar. Substituting Eq. (4-47) into the fourth and fifth equations of the geometric

equations, we obtain

1 1,w wKy Kxx G y y G x

ϕ ϕ∂ ∂ ∂ ∂= + = − −

∂ ∂ ∂ ∂ （4-48）

taking the derivatives of the two equations about y and x respectively, and then make the extraction of them, we obtain

2 2GKϕ∇ = − （4-49）

2C GK= − （4-50） Membrane analogy Assuming that a stretched membrane is held along a closed boundary being the same as the outline of the cross-section of the torsional bar. The membrane is blown

up from its bottom surface by an uniform pressure . The membrane is subjected

deflection and uniform in-plane tension force （the tension force per unit

length）. So the torsional bar problem can be solved with the help of the solution of the membrane under uniform pressure.

q

( , )z x y T

The value of the membrane deflection on the boundary is

0sz = (4-51)

Considering a micro element dxdy taking from the membrane, the equilibrium of all the forces acting on the element along z direction yields

2 qzT

∇ = − （4-52）

If the volume between the deformed membrane and its boundary plane is V, then

T

T

T

T

o

o x

y

z

dy

dx

q

x

zdxdy V=∫∫ （4-53）

so the membrane analogy can be summarized in the table below.

物理问题 弹性扭转 薄膜问题

2 22

2 2 2GKx yϕ ϕϕ ∂ ∂

∇ = + = −∂ ∂

2 2

22 2z zz

Tx y∂ ∂ q

∇ = + = −∂ ∂

0sϕ = 0sz = 基本方程

2 zdxdy Mϕ =∫∫ 2 2zdxdy V=∫∫

比拟条件 2 qGKT

=

ϕ z

zM 2V 对应的 物理量

,xz yzy xϕ ϕτ τ∂ ∂

= = −∂ ∂

,y xz zK Ky x∂ ∂

= − = −∂ ∂

The shear stresses at any point on the cross-section of the torsional bar are equal to the slopes of the membrane at the corresponding point but in perpendicular directions. The maximum shear stress corresponds to the maximum slope of the membrane, but perpendicular to each other.

x

y

a

b

A

B

oM

Example. 1: An elliptical section bar under torsion. Solution:

From 0sϕ = , we can assume

2 2

2 2 1x yma b

ϕ⎛ ⎞

= + −⎜ ⎟⎜ ⎟⎝ ⎠

From : 2 Cϕ∇ =2 2

2 22( )a bm Ca b

=+

2 2 2 2

2 2 2 2 12( )

a b x yCa b a b

ϕ⎛ ⎞

= + −⎜ ⎟⎜ ⎟+ ⎝ ⎠

From 2 dxdy Mϕ =∫∫ : 2 2

2 22( )a bC M

a bπ+

= − 2 2

2 2 1M x yab a b

ϕπ

⎛ ⎞= − +⎜ ⎟⎜ ⎟

⎝ ⎠−

2 22 2

3 3

2 2 2, ,xz yz zx zy 4 4

M M My xy ab x a b ab a bϕ ϕτ τ τ τ τ

π π π∂ ∂

= = − = − = = + = +∂ ∂

x y

If the stress results are transformed into the polar coordinate system,

cosx r θ= , siny r θ= , we will easily find the resultant shear stress distributes linearly

along the radial coordinate. According to the membrane analogy, the maximum membrane slopes occur at points A and B on the boundary and perpendicular to it. Therefore, the maximum shear stresses on the section will occur at the same points and parallel to the boundary.

max 22

A BMab

τ τ τπ

= = =

then we know

2 2

3 3( )

2C a b MKG a b Gπ

+= − =

2 2 2 2

3 3 3 3( ) ( ),a b M a b Mu Kyz yz v Kxz xz

a b G a b Gπ π+ +

= − = − = =

2 2

3 32 2

3 3

1 1 ( )

1 1 (

zx

zy

w aKy Ky yx G y G a b Gw aKx Kx xy G x G a b G

ϕ τπ

ϕ τπ

∂ ∂ −= + = + = −

∂ ∂∂ ∂ −

= − − = − = −∂ ∂

)

b M

b M

2 2 2 2

1 23 3 3 3( ) ( )( ), ( )a b M a b Mw xy f y w xy

a b G a b Gπ π− −

= − + = − + f x

0

From the above expression, we can deduce that 1 2( ) ( )f y f x w= = , where w0 is

indicates the rigid displacement and can be ignored. So the distortional displacement is obtained

2 2

3 3( )a b Mw x

a b Gπ−

= − y

************************************************************ If the bar section is round, and the radius of the section is R=a=b, by using the

radial coordinates, we have

cos , sinx r y rθ θ= =

2 22 2

4 4 4 4

2 2 2 ;2p

p

M x y M Mr Mr 41x y Iab a b R R I

Rτ ππ π π

= + = + = = =

max 32

r RMR

τ τπ== =

the rotation angle per unit length is 2 2

3 3 4( ) 2

p

a b M M MKGIa b G R Gπ π

+= = =

and the distortion is 2 2

3 3( ) 0a b Mw x

a b Gπ−

= − =y

Example 2: A narrow rectangular section bar under torsion. Solution: a b>>

We can estimate from the membrane

analogy that the stress function ϕ does not

change with x on the section except for the areas near both ends of the section, so we have

x

y

a

b o M

22

20, ,d d Cx y d y d yϕ ϕ ϕ ϕϕ∂ ∂= = ∇ =

∂ ∂=

Conducting the integration of the above equations and noticing that 2 0y bϕ =± = ,

we get 2

2

2 4C byϕ⎛ ⎞

= −⎜ ⎟⎜ ⎟⎝ ⎠

Since , we obtain 2 dxdy Mϕ =∫∫2

23 3 3

6 3 6, , ,4 zx zy

M M b MC y yy xab ab abϕ ϕϕ τ τ

⎛ ⎞ ∂ ∂= − = − = = − = − =⎜ ⎟⎜ ⎟ ∂ ∂⎝ ⎠

0

Using the membrane analogy we can deduce that the maximum shear stresses occur at the mid-points of the long edges of the rectangular section:

max 223

zx y bM

abτ τ =±= =

4.3.3 The elastic-plastic solution （1）The elastic-plastic torsion of the perfect plastic round bar

According to the above example, the resultant shear stress and the torsional angle per unit length of the round sectional bar are

41; ;2p

p p

Mr MK II GI

Rτ π= = = (4-54)

The maximum shear stress occurs at the boundary of the section and tangent to it

max 32

r RMR

τ τπ== = (4-55)

the yielding condition is

max sτ τ= (4-56)

Substituting Eq. (4-55) into (4-56), we get the elastic limit torsion moment 31

2e sM Rπ τ=

When eM M> , the region for the shear

stress reaches sτ is extending. When the inner

radius of the plastic region arrives rp, the shear stress distribution can be denoted by

eMprM pM

Rpr

sτ τ= sτ sτ

, 0

,

s pp

s p

r r rr

r r R

ττ

τ

⎧ ≤ ≤⎪= ⎨⎪ ≤ ≤⎩

Based on the force moment equilibrium, the relation of the torsional moment M and the elastic-plastic boundary rp is

32 2 3

0

2 12 2 13 4

p

p

r Rp

s s sp r

M rdA

rr r dr r dr Rr R

τ

τ π τ π π τ

=⎡ ⎤⎛ ⎞⎢ ⎥= ⋅ + ⋅ = − ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

∫∫

∫ ∫

When , whole section becomes totally plastic, the

present torsional moment is called the plastic limit moment

0pr =

323s sM Rπ τ=

If the torsional moment is applied till ( )e sM M M M< < and then unloaded to

zero, the residual stress and residual torsional angle will be produced in the round bar. Assuming that τΔ is the elastically removed stress in the unloading process (do not

consider the reversal yielding), then 3

4 13

ps

p

rMr rI R R

τ τ⎡ ⎤⎛ ⎞⎢ ⎥Δ = = −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

and the residual stress is 3

3

3

3

1 4 1 , 03 4

1 4 1 ,3 4

ps p

pr

ps p

rr r

r R R

rr r

r R R

τ

τ τ τ

τ

⎧ ⎡ ⎤⎛ ⎞⎪ ⎢ ⎥⎜ ⎟ r

r R

− − ≤⎜ ⎟⎪ ⎢ ⎥⎪ ⎝ ⎠⎣ ⎦= − Δ = ⎨ ⎡ ⎤⎛ ⎞⎪ ⎢ ⎥⎜ ⎟

≤

− − ≤⎪ ⎜ ⎟⎢ ⎥⎝ ⎠⎪ ⎣ ⎦⎩≤

Let us now calculate the torsional angle per unit length θ when

e sM M M< < :

At pr r= , the shear strain p prγ θ= and

/p s Gγ τ= , thus pr

θ

1

pγ

s

pGrτθ =

Let θΔ be the range of torsional angle per unit length removed during the elastic unloading process, i.e.

3 3

3 34 41 1

3 34 4p ps

p

r r rMGI R G R

p

R Rτθ θ

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟Δ = = − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

then the residual torsional angle 3

341 13 4

p pr e r rR R

θ θ θ θ⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟= − Δ = − −

⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

（2）Full plastic torsion and the sandpile analogy The basic equations

If the bar of any shape under torsion reaches the full plastic state, its stress components still need to satisfy the equilibrium equation

0yzxz

x yττ ∂∂

+ =∂ ∂

(4-57)

We introduce the stress function for the full plastic torsion, pϕ , and have

,pzx zy

p

y xϕ ϕ

τ τ∂ ∂

= − =∂ ∂

(4-58)

In full plastic state, the combination of the stress components must satisfies the yield condition

2 2 2zx zy sτ τ τ+ = (4-59)

Substituting Eq. (4-58) into (4-59), we obtain 2 2

2p psx y

ϕ ϕτ

∂ ∂⎛ ⎞ ⎛ ⎞+ =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

grad p sϕ τ= (4-60)

which shows that the slope of pϕ on the section remains constant.

For the solid sectional bar, we have

( ) 0p sϕ = (4-61)

and the boundary condition on the bar end is

2 p sdxdy Mϕ =∫∫ (4-62)

We can solve pϕ from Eqs. (4-60)～(4-62), and then obtain the solution of

stress components in full plastic state according to Eq. (4-58). Nadai’s sandpile analogy Make a plate with the same shape as the cross section of a torsional bar, and pile

the dry and fine sand on it until the maximum amount. The surface of the sandpile will have the constant slope about the horizontal surface. If w is the height of the pile at any location and α the internal frictional angle, the following relation will be satisfied everywhere

grad tanw α= (4-63)

The height of the sandpile at the plate boundary is zero:

0sw = (4-64)

If V is the volume of the pile, then we have

2 wdxdy V= 2∫∫ (4-65)

The sandpile analogy of the full plastic torsion

Physical problem Full plastic torsion Sandpile problem

grad p sϕ τ= grad tanw α=

( ) 0p sϕ = 0sw = Basic equations

2 p sdxdy Mϕ =∫∫ 2 2wdxdy V=∫∫

Analogical condition tansτ α=

wpϕ Corresponding physical terms

pM 2V

Example 1：The full plastic torsion of round sectional bar. Solution： The sand pile is a cone and the height of which is

α α

R

h

2A Rπ=

tanh R α= . The analogical condition is tan sα τ= , then we

know sh Rτ= , and

31 22 23 3s sM V Ah Rπ τ= = × =

which is the same as that obtained by the elastic-plastic solution. Example 2: The rectangular sectional bar under torsion

A B

a

b

A B

A B

45

α α

45

/ 2a/ 2a b a−

h

a

b

Solution: The sand pile is shown in the right figure. The angles

of the four declining surfaces about the bottom face are all α . The volume of the sand pile is

21 1( )2 3

V b a ah a= − + h

tan2 2 sa ah α τ= = , The analogical condition is:

so the plastic limit torsional moment is: 2

2 (36

ss

a )M V bτ= = − a

For the square section, , we have a b=

3

3s saM τ=

（3）The elastic-plastic torsion and the membrane-house roof analogy

When e sM M M< < , the cross section of a torsional bar is divided into plastic

region and elastic region.

satisfies: In the elastic region, the stress function eϕ

2 2e Gϕ θ∇ = −

In the elastic region, the stress function satisfies: pϕ

grad p sϕ τ=

At the boundary of the cross section:

0, 0p eϕ ϕ= =

At the border of the elastic region and the plastic region, the stress components must continue, thus

,p pe e

x x y yϕ ϕϕ ϕ∂ ∂∂ ∂

= =∂ ∂ ∂ ∂

so we have , or directly p eϕ ϕ= +constant

p eϕ ϕ=

the border is changing with the increase of torsional moment. The solution of the above equations is usually difficult to get, so the elastic-plastic torsion problem can be solved by the use of the membrane-house roof analogy.

the membrane-house roof analogy Cutting a hole which is the same as the cross section of a torsional bar on a flat

plate, and putting up a membrane covering the hole and fixed along the hole border. Above the hole a “house-roof” is installed which has the same shape as the “sandpile” surfaces. The membrane is blown up with uniform pressure. If the pressure is small, the deformed membrane does not contact with the roof and the whole section is in elastic. With the increase of the pressure, part of the membrane contacts with and confined by the roof, indicating that the related part of the cross-section is already in plastic region, while the other part of the cross-section is still in elastic. This is the situation of elastic-plastic torsion and, in this way, the border of elastic region and plastic region can be located. The border is changing with the increase of pressure and if the pressure is large enough, the membrane will get fully in touch with the roof, which means that the whole section has been in plastic state.

Purely elastic Elastic-plastic Totally plastic

4.4 The Pure Elastic-Plastic Bending of Rectangular Cross-sectional Beams Basic hypothesis： （1）Only the normal stress on cross-section of the beam is considered, and the

lateral bearing stress is neglected； （2）During bending, the cross section of the beam retains as plane and is normal to

the deformed beam axis.

x

y

z M

h

b

y

M

Due to the flat cross-section hypothesis：

h

b

y

dy

neutral axis 2

2

dxwdK −≈K0εε += Ky — Curvature，，

0),(22 == ∫−

hh dyyxbN σ （Pure bending）

∫−= 22 ),(h

h dyyxybM σ

1、Elastic stage

)( 0εεσ +== KyEE

0)(),( 02

2 02

2==+== ∫∫ −−

hEdyKyEbdyyxbNh

h

h

hεεσ ， 00 =ε

EKJbhEKdyKyyEbdyyxybMh

h

h

h==+== ∫∫ −− 12

)(),(32

2 02

2εσ

JMEK = ，

12

3bhJ =yJMKyE =+=∴ )( 0εσ ， where

Top and bottom layer of the beam reach yielding first， shy σσ =±= 2 ，so as to get

elastic limit load (bending moment)：

sebhM σ

6

2=

corresponding curvature is： eK

EhhEK s

eσσ 2

2=

⋅=

then from we know： EKJM =

ee KK

MM

=

2、Elastic-plastic stage（for ） eMM >

Assuming that material is perfectly plastic.

sσ−sσ− sσ−

2h

h

sσ sσ

2h

20 hy ζ=

sσ

eMM < sMM =se MMM <<eMM = Distribution of normal bending stress σ

20 hy ζ= 10 ≤≤ ζ At elastic-plastic border, ） ，（

⎪⎩

⎪⎨

⎧

−≤≤−−≤≤

≤=

0

0

0

2for,2for,

for,

yyhhyy

yyEKy

s

sσ

σσ

and bending moment is

⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛=⎥

⎦

⎤⎢⎣

⎡+⎟⎟

⎠

⎞⎜⎜⎝

⎛= ∫∫ 2

0

22

00

0 23

32)(

0

0 yhbdyydyyyybyM sh

y sy

sσσσ

or

es MhbM )3(

21)3(

12)( 22

2ζζσζ −=−=

2hEKsζσ ⋅=20 hyy ζ== , then we know When , , sσσ =

0

1

2 1eKK

eMM

1.5

3

ζζσ es K

EhK ==

2

therefore

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−=−=

−22 3

21)3(

21

ee KK

MM ζ

Or

)(231

ee MMKK

−=

When 0=ζ ，i.e. ，plastic limit bending moment is reached： 00 =y

ess MbhMM23

4

2

0 ====

σς

Example：A simply supported beam subjected to point force

When the mid-cross section becomes total yielding, the bending moment of it will be the plastic limit bending

moment, , i.e.

b

hx

:M

P

4pl

2l 2l

x

z

yy

sM

44

2 PlbhM ss == σ

then we get the corresponding point force

ss lbhPP σ

2==

and the bending moment cannot be further increased. At this time the plastic zone in

the beam is illustrated as the shaded area. The location of the cross section where yielding has just occurred can be determined as below：

x Let be the distance of the section from left beam end, so the bending moment

on it is ss lbhPP σ

2==xPM

2= where the point force now is and the bending

moment should be the elastic limit bending moment, eMM = , i.e.

ss bhxP σ

62

2=

we then get

33

2 lP

bhx ss

== σ

At mid-cross section, upper and lower plastic zone link up. Two parts at both sides of the section can then rotate about each other, similar to the ordinary structural hinge. Therefore such kind of hinge is called the plastic hinge.

3、Residual curvature and residual stress during unloading

Have the bending moment exceed the elastic limit, . The relation of eMM >*

KM − during unloading follows law of elastic behavior

ee KK

MM Δ

=Δ

the stress change is

yJMΔ

=Δσ

*MM −=Δ 0K, the residual curvature If loading is reduced to zero, i.e. can be

expressed as

eeeeeee MM

MMMM

KK

KK

KK

KK *

*

***0

23

1−

−=−=

Δ+=

And the residual stress as

yJ

MyEK*

**0 −=Δ+= σσσ

The differences between plastic hinge and structural hinge：

（1）Structural hinge does not carry bending moment, while for the plastic hinge,

. sMM =

（2）Beam sections at both sides of a structural hinge are able to rotate relatively at two directions. Nevertheless, a plastic hinge is a single-directional hinge since its rotation in opposite direction induces unloading and vanishing of the hinge.