Mechanics of Materials（甲） of... · · 2017-03-14Mechanics of Materials ... 泊松比（Poisson’s ratio ... 2 1.039 3.039mm sin 30 tan 30 1 2 3 3 4 ' ' $ l l G y AA A

Hongtao WANG王宏涛

浙江大学应用力学所

玉泉校区教12-223

Email htwzjueducn

Mechanics of Materials（甲）I

Chapter 2 Tension and Compression

Axial tension compression and the internal force and

stresses on both cross section and inclined section

Mechanical behavior of materials subjected to tension

and compression as well as Hookersquos law

Deformation and strain energy under axial tension and

compression stress concentration

Statistically indeterminate problems with tension and

compression (including thermal stress and assembly

stress)

21 Axial Tension and Compression


Force feature

The resultant force of the external forces must be along the axis of the member


Deformation feature

The member will be elongated or shortened along its axis

Which one is uniaxial tension


F FBefore tension

After tension

(A)

(B)

Plane Cross-section Assumption

Plane cross-section assumption

If the cross section is plane at first it still remains plane

after deformed and perpendicular to the axial line


F Fl0

l

Along the axial direction

Elongation

Strain


Perpendicular to the axial direction

Elongation

Strain

l0

l

d d0

对同一种材料为常数称之为泊松比（Poissonrsquos ratio）

Poissonrsquos ratio

Simeacuteon Denis Poisson (21 June 1781 ndash

25 April 1840) was a French

mathematician geometer and physicist


Cross-section

Before tension

After tension

d1

d2

d01

d02

1205761 =1198891minus 1198890111988901

1205762 =1198892minus 1198890211988902

1205760 =119897 minus 11989701198970

Tensile strain

1205921 = minus12057611205760

1205922 = minus12057621205760

Poissonrsquos ratio

Poissonrsquos Ratio

Estimate the value of Poissonrsquos ratio

Material property Poissonrsquos ratio

l0

l

d d0Poissonrsquos ratio

109Pa无量纲

Poissonrsquos ratio

Can Poissonrsquos ratio be negative

What is the Poissonrsquos ratio for non-compressible material

Range for Poissonrsquos ratio

22 Internal force amp Stress under Axial Tension amp Compression

PP

m

m

Internal force on cross section

Three steps

1 Cut

2 Replace with Resultant force and resultant moment

3 Equilibrium equations

F1

F3

F2

Fn

F1 FR

F3M

0

0

F

M

Review Cross-sectional Method

Internal force amp Stress Axial Tension amp Compression

PP

m

m

Sign convention for resultant force

N axial forceP

m

m

SIGN CONVENSION

N gt 0 ( tension ) N lt 0 ( compression )

P2P1

m

m

P3 PP

m

m

If a bar is subjected to external forces P1 P1 hellip Pn and

ngt2 then N=N(x)

Axial force diagram (轴力图)


Problem A bar is subjected to the force system as shown in the following Fig Plot the axial force diagram of the bar

P1=2kNP2=3kN

P3=1kNA BC

N2P1 =2kNP2 =3kN

AC

A BC2kN

1kN

Axial force diagram

sumX=0 N1-P1=0

sumX=0 N2+P2-P1=0

1

1

2

2

P1=2kN

A

N1

1

1N1=P1=2kN

N2=P1-P2= -1kN

Example 21

P1=2kNP2=3kN

P3=1kNA BC

1

1

2

2

Is there a more simple way to obtain the axial force on 2-2

section

Example 21

P A

C

m

m

C

m

m

p

m

Pp

A

0 0

lim limmA A

Pp p

A

normal stress

shear stress

Review Stress

PP

m

m

PN

22 Stress on cross section

Uniform stress distribution But why

PN

A

N dA

A

N dA A N

A

+ (tension stress)

- (compression stress)

Unit Pa=Nm2

1MPa=106Nm2

22 Stress on Cross Section

Uniform stress distribution Advantage


Assumption Same cross-section area


Every part of the material has the same efficiency

Good for equal-life design

( )( )

( )

N xx

A x

1 Bar with varying cross-sections

Fixed PA(x)

2 Bar with varying axial force

Fixed Pp(x)

Axial force diagram

Stress on Cross Section

Is this correct


Is this correct ( )( )

( )

N xx

A x

Correct Good approximation Poor approximation

N N N


Real stress at point P is

A(x)

N

Approximate stress at point P is a NA(x)

P

Compare and a

Comparison

Saint-Venantrsquos Principle for axial tensioncompression

The difference between the

effects of two different but

statically equivalent loads

becomes very small at

sufficiently large distances

from load

Adheacutemar Jean Claude Barreacute de Saint-Venant (August 23

1797 Villiers-en-Biegravere Seine-et-Marne ndash January 1886)

Example 22

Given F=20kN calculate the stress in rod AB

and CB The cross-section of rod AB is a circle

with diameter of 20mm while it is a square

（15times15mm）for rod BC

F

A

B

C

0yF

kN3281 NF

Solution

1 Calculate the axial force

kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF

2 Calculate the stress

MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k

Internal force on inclined section

It is observed from experiments that during tensile or

compressive loading a bar may break along inclined

section instead of cross-section

P

A Normal stress on cross section

PPk

k

k

k

P P

k

k

PP

p

Internal Force on Inclined Section

k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion

1 If =0 takes the greatest value

2 If =45ordm takes the greatest value

max

max2

3 If =90ordmmax max 0

23 Hookersquos Law

F Fl0

l

Review

External load Material response

F

Global

Local

k 包含了材料和几何尺寸所有的性质

23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量

弹性模量E 是材料属性

Material property Elastic Modulus

l0

l

d d0

109Pa无量纲

23 Hookersquos Law

Robert Hooke FRS (hʊk 28 July [OS 18 July]

1635 ndash 3 March 1703) was an English natural

philosopher architect and polymath

The linear relation between

stress and strain for a bar in a

simple tension or compression

is expressed by = Ee

E is defined as modulus of

elasticity

Note Hookersquos law is a first-order approximation

Example 23

C

Plot the enlargement sketch of the small deformation

Accurate method to plot diagram of deformation the arc line as shown in the figure

CCrsquo is the REAL displacement

Determine deformation Li of each rod as shown in Fig l

Approximate method to plot the diagram of deformation the tangent of the arc line shown in the figure

CCrsquorsquo is the APPROXIMATE displacement

A B

C

L1 L2

P1L

2L

C

Determine the displacement of Point C

Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23

The length of bar AB is 2m the area 200mm2 The

area of AC is 250mm2 E=200GPa F=10kN Determine

the displacement of A

0yFkN202sin1 FFFN

Solution 1 Calculate axial force AB is taken as 1

while AC 2 At point A

kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN

2 Calculate the deformation of two bars

1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23

3 Calculate displacement at A

A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA

Changes in lengths of nonuniform bars - bar consisting of several prismatic segments

Fixed PA(x)

Fixed Pp(x)

Bars with continuously varying loads or dimensions

ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy

The energy which is stored within an deformable body due to its deformation

The strain energy of a straight bar under tension

P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl

The work of external force = strain energy

l

lEA

EA

lPlPWU

2

)(

22

1 22

The strain energy density

EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理

应变能 =应变能密度的体积积分=外力对边界做的功

偶然必然数学上有无类似的定理

观察

F

l

Δl系统应变能对作用在该点的力的偏导数等于力作用点沿力作用方向的位移

这种方法叫能量法EA

lFV

2

2

e

能量法求位移

Calculate the displacement of A

Difference

Axial force can be obtained from

equilibrium equationsAxial force can NOT be obtained from

equilibrium equations

Statistically determinate problem Statistically indeterminate problem

25 Statistically Indeterminate Structures

约束反力不能由平衡方程求得

静不定度（次）数约束反力多于

独立平衡方程的数

独立平衡方程数平面任意力系

3个平衡方程平面共点力系

2个平衡方程

平面平行力系2个平衡方程共线力系1个平衡方程


Compatibility equation

(geometric equation kinematic

equation consistent deformation

equation)

of unknown axial forces gt of equilibrium equations

Equilibrium equation

Constitutive equation


Example 25

1 Equilibrium equation

Solution

210 NNx FFF

FFFF NNy 31 cos20

2 Geometric equation

cos321 lll

3 Constitutive relation

cos

11

EA

lFl N

EA

lFl N 33

4 Complementary equation

coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF

5 Solving equilibrium amp complementary equation

3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25

A bar-AB subjected to axial external force P is rigidly fixed at two

ends Area of cross-section is AElastic modulus is ETry to

determine the support reaction

（1）equilibrium equation

00 PRRY BA

（2） Deformation geometric

relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC

（4） Complementary equation

EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）

（5） solving equilibrium equation amp

complementary equation

Example 25

3 bars are made of the same materials Area of AB is

200mm2 area of AC 300 mm2 and area of AD 400 mm2

F=30kN Determine the axial force of each bar

3

2lll ADAB

(1) Equilibrium equation

0 xF0

32

0

1 30cos30cos NNN FFF

FFFF NNy 0

3

0

1 30sin30sin0

that is 1323 321 NNN FFF

2231 FFF NN

(2) Compatibility equation

Solution (0) Assuming the length of bar AB is l

lengths of AB amp AD are therefore

F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

The components of the

displacement at A are projected to

each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll

Deformation relation

213 3 lll

Substituting them into constitutive relation

2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN

322 213 NNN FFF We obtain

Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF

From ①②③ we obtain

kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps

Equilibrium equations

Geometric equationmdashcompatibility equation of

deformation

Physical equation mdash Constitutive law

Complementary equationfrom geometric equation and

physical equation

Solving equilibrium equations and the complementary

equation

Method and steps for solving statically indeterminate problems

Source of Stress

Is there any stress if NO external load

2010年10月26日沪杭高铁开通时速最高可达4166公里

1979年台风ldquo泰培rdquo时速306公里这是地球纪录上最强的热带气旋飞机的起飞速度300kmh

沪杭高铁轨道（无砟轨道）

无缝钢轨（404根小钢轨焊接形成一条完全无缝隙的ldquo大钢条rdquo长202公里）

轨道板

填充层（厚2-4厘米和Ⅱ型板一样宽水泥乳化沥青砂浆它的作用是缓冲力道调平）

底座板

滑动层（5毫米左右厚其实是两层布加一层膜）

五层结构

26 Thermal Stress

The bar-AB is shown in figure Try to

determine the thermal stress after increasing

T

（1）Equilibrium equationSolution

（2）Geometric equation

T Rl l

（3）Physical relation

lTlT

（4）Complementary equation

EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA

Thermal stress Stresses due to temperature change

No thermal stress in statistically determined structure

Thermal stress exists in statistically indetermined structure with partial or total

deformation constrained

（5）Solving complementary equation amp

equilibrium equation

26 Thermal Stress


Geometric equationmdashcompatibility equation of deformation

(including thermal item)

Physical equation mdash Constitutive law (including thermal

item)


physical equation


equation

Method and steps for solving thermal stress problems

Tl T l

26 Thermal Stress

Materials and dimensions of rod 1and2 are all

the same as shown in the figure If temperature

of the structure changes from T1 to T2

determine the thermal internal forces of each

rod（linear thermal expansion coefficient of

each rod i T= T2 -T1)

C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation

Solution Equilibrium equations

1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Complementary equation

cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN

Solving equilibrium equations and the

complementary equation we get

cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2

The upper and lower ends of a ladder-like steel shaft are fixed

at temperature T1=5 as shown in the figure Areas of the

upper and lower segments are 1=cm2 and 2=cm2

respectively When its temperature reaches T2=25

determine the temperature stress of each rod(Linear thermal

expansion coefficient = 125times10 -6 1oCElastic Modulus

E=200GPa)

Geometric equation

Solution Equilibrium equation

021 NNY

0 NT LLL

26 Thermal Stress

Physical equation


complementary equation we get kN 33321 NN

Complementary equation 2

2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT

Temperature stresses

MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress

Assembly stress Stresses due to manufacturing error

No Assembly stress in statistically determined structure

Assembly stress exists in statistically indetermined structure with manufacturing error

A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL

The dimension error of rod 3 is as shown in the figure

Determine the assemble internal force of each rod

A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN

Physical equation and complementary equation

Solving equilibrium equations and

the complementary equation we get

cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress



(including manufacturing error)



physical equation


equation

Method and steps for solving assembly stress problems

The End

Chapter 2 Tension and Compression

Axial tension compression and the internal force and

stresses on both cross section and inclined section

Mechanical behavior of materials subjected to tension

and compression as well as Hookersquos law

Deformation and strain energy under axial tension and

compression stress concentration

Statistically indeterminate problems with tension and

compression (including thermal stress and assembly

stress)



Force feature



Deformation feature




F FBefore tension

After tension

(A)

(B)






F Fl0

l


Elongation

Strain



Elongation

Strain

l0

l

d d0


Poissonrsquos ratio





Cross-section

Before tension

After tension

d1

d2

d01

d02

1205761 =1198891minus 1198890111988901

1205762 =1198892minus 1198890211988902

1205760 =119897 minus 11989701198970

Tensile strain

1205921 = minus12057611205760

1205922 = minus12057621205760

Poissonrsquos ratio

Poissonrsquos Ratio



l0

l


109Pa无量纲

Poissonrsquos ratio





PP

m

m


Three steps

1 Cut



F1

F3

F2

Fn

F1 FR

F3M

0

0

F

M



PP

m

m


N axial forceP

m

m

SIGN CONVENSION


P2P1

m

m

P3 PP

m

m


ngt2 then N=N(x)




P1=2kNP2=3kN

P3=1kNA BC

N2P1 =2kNP2 =3kN

AC

A BC2kN

1kN

Axial force diagram

sumX=0 N1-P1=0

sumX=0 N2+P2-P1=0

1

1

2

2

P1=2kN

A

N1

1

1N1=P1=2kN

N2=P1-P2= -1kN

Example 21

P1=2kNP2=3kN

P3=1kNA BC

1

1

2

2


section

Example 21

P A

C

m

m

C

m

m

p

m

Pp

A

0 0

lim limmA A

Pp p

A

normal stress

shear stress

Review Stress

PP

m

m

PN



PN

A

N dA

A

N dA A N

A

+ (tension stress)


Unit Pa=Nm2

1MPa=106Nm2








( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End



Force feature



Deformation feature




F FBefore tension

After tension

(A)

(B)






F Fl0

l


Elongation

Strain



Elongation

Strain

l0

l

d d0


Poissonrsquos ratio





Cross-section

Before tension

After tension

d1

d2

d01

d02

1205761 =1198891minus 1198890111988901

1205762 =1198892minus 1198890211988902

1205760 =119897 minus 11989701198970

Tensile strain

1205921 = minus12057611205760

1205922 = minus12057621205760

Poissonrsquos ratio

Poissonrsquos Ratio



l0

l


109Pa无量纲

Poissonrsquos ratio





PP

m

m


Three steps

1 Cut



F1

F3

F2

Fn

F1 FR

F3M

0

0

F

M



PP

m

m


N axial forceP

m

m

SIGN CONVENSION


P2P1

m

m

P3 PP

m

m


ngt2 then N=N(x)




P1=2kNP2=3kN

P3=1kNA BC

N2P1 =2kNP2 =3kN

AC

A BC2kN

1kN

Axial force diagram

sumX=0 N1-P1=0

sumX=0 N2+P2-P1=0

1

1

2

2

P1=2kN

A

N1

1

1N1=P1=2kN

N2=P1-P2= -1kN

Example 21

P1=2kNP2=3kN

P3=1kNA BC

1

1

2

2


section

Example 21

P A

C

m

m

C

m

m

p

m

Pp

A

0 0

lim limmA A

Pp p

A

normal stress

shear stress

Review Stress

PP

m

m

PN



PN

A

N dA

A

N dA A N

A

+ (tension stress)


Unit Pa=Nm2

1MPa=106Nm2








( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End


Force feature



Deformation feature




F FBefore tension

After tension

(A)

(B)






F Fl0

l


Elongation

Strain



Elongation

Strain

l0

l

d d0


Poissonrsquos ratio





Cross-section

Before tension

After tension

d1

d2

d01

d02

1205761 =1198891minus 1198890111988901

1205762 =1198892minus 1198890211988902

1205760 =119897 minus 11989701198970

Tensile strain

1205921 = minus12057611205760

1205922 = minus12057621205760

Poissonrsquos ratio

Poissonrsquos Ratio



l0

l


109Pa无量纲

Poissonrsquos ratio





PP

m

m


Three steps

1 Cut



F1

F3

F2

Fn

F1 FR

F3M

0

0

F

M



PP

m

m


N axial forceP

m

m

SIGN CONVENSION


P2P1

m

m

P3 PP

m

m


ngt2 then N=N(x)




P1=2kNP2=3kN

P3=1kNA BC

N2P1 =2kNP2 =3kN

AC

A BC2kN

1kN

Axial force diagram

sumX=0 N1-P1=0

sumX=0 N2+P2-P1=0

1

1

2

2

P1=2kN

A

N1

1

1N1=P1=2kN

N2=P1-P2= -1kN

Example 21

P1=2kNP2=3kN

P3=1kNA BC

1

1

2

2


section

Example 21

P A

C

m

m

C

m

m

p

m

Pp

A

0 0

lim limmA A

Pp p

A

normal stress

shear stress

Review Stress

PP

m

m

PN



PN

A

N dA

A

N dA A N

A

+ (tension stress)


Unit Pa=Nm2

1MPa=106Nm2








( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

Force feature



Deformation feature




F FBefore tension

After tension

(A)

(B)






F Fl0

l


Elongation

Strain



Elongation

Strain

l0

l

d d0


Poissonrsquos ratio





Cross-section

Before tension

After tension

d1

d2

d01

d02

1205761 =1198891minus 1198890111988901

1205762 =1198892minus 1198890211988902

1205760 =119897 minus 11989701198970

Tensile strain

1205921 = minus12057611205760

1205922 = minus12057621205760

Poissonrsquos ratio

Poissonrsquos Ratio



l0

l


109Pa无量纲

Poissonrsquos ratio





PP

m

m


Three steps

1 Cut



F1

F3

F2

Fn

F1 FR

F3M

0

0

F

M



PP

m

m


N axial forceP

m

m

SIGN CONVENSION


P2P1

m

m

P3 PP

m

m


ngt2 then N=N(x)




P1=2kNP2=3kN

P3=1kNA BC

N2P1 =2kNP2 =3kN

AC

A BC2kN

1kN

Axial force diagram

sumX=0 N1-P1=0

sumX=0 N2+P2-P1=0

1

1

2

2

P1=2kN

A

N1

1

1N1=P1=2kN

N2=P1-P2= -1kN

Example 21

P1=2kNP2=3kN

P3=1kNA BC

1

1

2

2


section

Example 21

P A

C

m

m

C

m

m

p

m

Pp

A

0 0

lim limmA A

Pp p

A

normal stress

shear stress

Review Stress

PP

m

m

PN



PN

A

N dA

A

N dA A N

A

+ (tension stress)


Unit Pa=Nm2

1MPa=106Nm2








( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End



F FBefore tension

After tension

(A)

(B)






F Fl0

l


Elongation

Strain



Elongation

Strain

l0

l

d d0


Poissonrsquos ratio





Cross-section

Before tension

After tension

d1

d2

d01

d02

1205761 =1198891minus 1198890111988901

1205762 =1198892minus 1198890211988902

1205760 =119897 minus 11989701198970

Tensile strain

1205921 = minus12057611205760

1205922 = minus12057621205760

Poissonrsquos ratio

Poissonrsquos Ratio



l0

l


109Pa无量纲

Poissonrsquos ratio





PP

m

m


Three steps

1 Cut



F1

F3

F2

Fn

F1 FR

F3M

0

0

F

M



PP

m

m


N axial forceP

m

m

SIGN CONVENSION


P2P1

m

m

P3 PP

m

m


ngt2 then N=N(x)




P1=2kNP2=3kN

P3=1kNA BC

N2P1 =2kNP2 =3kN

AC

A BC2kN

1kN

Axial force diagram

sumX=0 N1-P1=0

sumX=0 N2+P2-P1=0

1

1

2

2

P1=2kN

A

N1

1

1N1=P1=2kN

N2=P1-P2= -1kN

Example 21

P1=2kNP2=3kN

P3=1kNA BC

1

1

2

2


section

Example 21

P A

C

m

m

C

m

m

p

m

Pp

A

0 0

lim limmA A

Pp p

A

normal stress

shear stress

Review Stress

PP

m

m

PN



PN

A

N dA

A

N dA A N

A

+ (tension stress)


Unit Pa=Nm2

1MPa=106Nm2








( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End


F FBefore tension

After tension

(A)

(B)






F Fl0

l


Elongation

Strain



Elongation

Strain

l0

l

d d0


Poissonrsquos ratio





Cross-section

Before tension

After tension

d1

d2

d01

d02

1205761 =1198891minus 1198890111988901

1205762 =1198892minus 1198890211988902

1205760 =119897 minus 11989701198970

Tensile strain

1205921 = minus12057611205760

1205922 = minus12057621205760

Poissonrsquos ratio

Poissonrsquos Ratio



l0

l


109Pa无量纲

Poissonrsquos ratio





PP

m

m


Three steps

1 Cut



F1

F3

F2

Fn

F1 FR

F3M

0

0

F

M



PP

m

m


N axial forceP

m

m

SIGN CONVENSION


P2P1

m

m

P3 PP

m

m


ngt2 then N=N(x)




P1=2kNP2=3kN

P3=1kNA BC

N2P1 =2kNP2 =3kN

AC

A BC2kN

1kN

Axial force diagram

sumX=0 N1-P1=0

sumX=0 N2+P2-P1=0

1

1

2

2

P1=2kN

A

N1

1

1N1=P1=2kN

N2=P1-P2= -1kN

Example 21

P1=2kNP2=3kN

P3=1kNA BC

1

1

2

2


section

Example 21

P A

C

m

m

C

m

m

p

m

Pp

A

0 0

lim limmA A

Pp p

A

normal stress

shear stress

Review Stress

PP

m

m

PN



PN

A

N dA

A

N dA A N

A

+ (tension stress)


Unit Pa=Nm2

1MPa=106Nm2








( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End





F Fl0

l


Elongation

Strain



Elongation

Strain

l0

l

d d0


Poissonrsquos ratio





Cross-section

Before tension

After tension

d1

d2

d01

d02

1205761 =1198891minus 1198890111988901

1205762 =1198892minus 1198890211988902

1205760 =119897 minus 11989701198970

Tensile strain

1205921 = minus12057611205760

1205922 = minus12057621205760

Poissonrsquos ratio

Poissonrsquos Ratio



l0

l


109Pa无量纲

Poissonrsquos ratio





PP

m

m


Three steps

1 Cut



F1

F3

F2

Fn

F1 FR

F3M

0

0

F

M



PP

m

m


N axial forceP

m

m

SIGN CONVENSION


P2P1

m

m

P3 PP

m

m


ngt2 then N=N(x)




P1=2kNP2=3kN

P3=1kNA BC

N2P1 =2kNP2 =3kN

AC

A BC2kN

1kN

Axial force diagram

sumX=0 N1-P1=0

sumX=0 N2+P2-P1=0

1

1

2

2

P1=2kN

A

N1

1

1N1=P1=2kN

N2=P1-P2= -1kN

Example 21

P1=2kNP2=3kN

P3=1kNA BC

1

1

2

2


section

Example 21

P A

C

m

m

C

m

m

p

m

Pp

A

0 0

lim limmA A

Pp p

A

normal stress

shear stress

Review Stress

PP

m

m

PN



PN

A

N dA

A

N dA A N

A

+ (tension stress)


Unit Pa=Nm2

1MPa=106Nm2








( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End


F Fl0

l


Elongation

Strain



Elongation

Strain

l0

l

d d0


Poissonrsquos ratio





Cross-section

Before tension

After tension

d1

d2

d01

d02

1205761 =1198891minus 1198890111988901

1205762 =1198892minus 1198890211988902

1205760 =119897 minus 11989701198970

Tensile strain

1205921 = minus12057611205760

1205922 = minus12057621205760

Poissonrsquos ratio

Poissonrsquos Ratio



l0

l


109Pa无量纲

Poissonrsquos ratio





PP

m

m


Three steps

1 Cut



F1

F3

F2

Fn

F1 FR

F3M

0

0

F

M



PP

m

m


N axial forceP

m

m

SIGN CONVENSION


P2P1

m

m

P3 PP

m

m


ngt2 then N=N(x)




P1=2kNP2=3kN

P3=1kNA BC

N2P1 =2kNP2 =3kN

AC

A BC2kN

1kN

Axial force diagram

sumX=0 N1-P1=0

sumX=0 N2+P2-P1=0

1

1

2

2

P1=2kN

A

N1

1

1N1=P1=2kN

N2=P1-P2= -1kN

Example 21

P1=2kNP2=3kN

P3=1kNA BC

1

1

2

2


section

Example 21

P A

C

m

m

C

m

m

p

m

Pp

A

0 0

lim limmA A

Pp p

A

normal stress

shear stress

Review Stress

PP

m

m

PN



PN

A

N dA

A

N dA A N

A

+ (tension stress)


Unit Pa=Nm2

1MPa=106Nm2








( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End



Elongation

Strain

l0

l

d d0


Poissonrsquos ratio





Cross-section

Before tension

After tension

d1

d2

d01

d02

1205761 =1198891minus 1198890111988901

1205762 =1198892minus 1198890211988902

1205760 =119897 minus 11989701198970

Tensile strain

1205921 = minus12057611205760

1205922 = minus12057621205760

Poissonrsquos ratio

Poissonrsquos Ratio



l0

l


109Pa无量纲

Poissonrsquos ratio





PP

m

m


Three steps

1 Cut



F1

F3

F2

Fn

F1 FR

F3M

0

0

F

M



PP

m

m


N axial forceP

m

m

SIGN CONVENSION


P2P1

m

m

P3 PP

m

m


ngt2 then N=N(x)




P1=2kNP2=3kN

P3=1kNA BC

N2P1 =2kNP2 =3kN

AC

A BC2kN

1kN

Axial force diagram

sumX=0 N1-P1=0

sumX=0 N2+P2-P1=0

1

1

2

2

P1=2kN

A

N1

1

1N1=P1=2kN

N2=P1-P2= -1kN

Example 21

P1=2kNP2=3kN

P3=1kNA BC

1

1

2

2


section

Example 21

P A

C

m

m

C

m

m

p

m

Pp

A

0 0

lim limmA A

Pp p

A

normal stress

shear stress

Review Stress

PP

m

m

PN



PN

A

N dA

A

N dA A N

A

+ (tension stress)


Unit Pa=Nm2

1MPa=106Nm2








( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End


Cross-section

Before tension

After tension

d1

d2

d01

d02

1205761 =1198891minus 1198890111988901

1205762 =1198892minus 1198890211988902

1205760 =119897 minus 11989701198970

Tensile strain

1205921 = minus12057611205760

1205922 = minus12057621205760

Poissonrsquos ratio

Poissonrsquos Ratio



l0

l


109Pa无量纲

Poissonrsquos ratio





PP

m

m


Three steps

1 Cut



F1

F3

F2

Fn

F1 FR

F3M

0

0

F

M



PP

m

m


N axial forceP

m

m

SIGN CONVENSION


P2P1

m

m

P3 PP

m

m


ngt2 then N=N(x)




P1=2kNP2=3kN

P3=1kNA BC

N2P1 =2kNP2 =3kN

AC

A BC2kN

1kN

Axial force diagram

sumX=0 N1-P1=0

sumX=0 N2+P2-P1=0

1

1

2

2

P1=2kN

A

N1

1

1N1=P1=2kN

N2=P1-P2= -1kN

Example 21

P1=2kNP2=3kN

P3=1kNA BC

1

1

2

2


section

Example 21

P A

C

m

m

C

m

m

p

m

Pp

A

0 0

lim limmA A

Pp p

A

normal stress

shear stress

Review Stress

PP

m

m

PN



PN

A

N dA

A

N dA A N

A

+ (tension stress)


Unit Pa=Nm2

1MPa=106Nm2








( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

Poissonrsquos Ratio



l0

l


109Pa无量纲

Poissonrsquos ratio





PP

m

m


Three steps

1 Cut



F1

F3

F2

Fn

F1 FR

F3M

0

0

F

M



PP

m

m


N axial forceP

m

m

SIGN CONVENSION


P2P1

m

m

P3 PP

m

m


ngt2 then N=N(x)




P1=2kNP2=3kN

P3=1kNA BC

N2P1 =2kNP2 =3kN

AC

A BC2kN

1kN

Axial force diagram

sumX=0 N1-P1=0

sumX=0 N2+P2-P1=0

1

1

2

2

P1=2kN

A

N1

1

1N1=P1=2kN

N2=P1-P2= -1kN

Example 21

P1=2kNP2=3kN

P3=1kNA BC

1

1

2

2


section

Example 21

P A

C

m

m

C

m

m

p

m

Pp

A

0 0

lim limmA A

Pp p

A

normal stress

shear stress

Review Stress

PP

m

m

PN



PN

A

N dA

A

N dA A N

A

+ (tension stress)


Unit Pa=Nm2

1MPa=106Nm2








( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End


l0

l


109Pa无量纲

Poissonrsquos ratio





PP

m

m


Three steps

1 Cut



F1

F3

F2

Fn

F1 FR

F3M

0

0

F

M



PP

m

m


N axial forceP

m

m

SIGN CONVENSION


P2P1

m

m

P3 PP

m

m


ngt2 then N=N(x)




P1=2kNP2=3kN

P3=1kNA BC

N2P1 =2kNP2 =3kN

AC

A BC2kN

1kN

Axial force diagram

sumX=0 N1-P1=0

sumX=0 N2+P2-P1=0

1

1

2

2

P1=2kN

A

N1

1

1N1=P1=2kN

N2=P1-P2= -1kN

Example 21

P1=2kNP2=3kN

P3=1kNA BC

1

1

2

2


section

Example 21

P A

C

m

m

C

m

m

p

m

Pp

A

0 0

lim limmA A

Pp p

A

normal stress

shear stress

Review Stress

PP

m

m

PN



PN

A

N dA

A

N dA A N

A

+ (tension stress)


Unit Pa=Nm2

1MPa=106Nm2








( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

Poissonrsquos ratio





PP

m

m


Three steps

1 Cut



F1

F3

F2

Fn

F1 FR

F3M

0

0

F

M



PP

m

m


N axial forceP

m

m

SIGN CONVENSION


P2P1

m

m

P3 PP

m

m


ngt2 then N=N(x)




P1=2kNP2=3kN

P3=1kNA BC

N2P1 =2kNP2 =3kN

AC

A BC2kN

1kN

Axial force diagram

sumX=0 N1-P1=0

sumX=0 N2+P2-P1=0

1

1

2

2

P1=2kN

A

N1

1

1N1=P1=2kN

N2=P1-P2= -1kN

Example 21

P1=2kNP2=3kN

P3=1kNA BC

1

1

2

2


section

Example 21

P A

C

m

m

C

m

m

p

m

Pp

A

0 0

lim limmA A

Pp p

A

normal stress

shear stress

Review Stress

PP

m

m

PN



PN

A

N dA

A

N dA A N

A

+ (tension stress)


Unit Pa=Nm2

1MPa=106Nm2








( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End


PP

m

m


Three steps

1 Cut



F1

F3

F2

Fn

F1 FR

F3M

0

0

F

M



PP

m

m


N axial forceP

m

m

SIGN CONVENSION


P2P1

m

m

P3 PP

m

m


ngt2 then N=N(x)




P1=2kNP2=3kN

P3=1kNA BC

N2P1 =2kNP2 =3kN

AC

A BC2kN

1kN

Axial force diagram

sumX=0 N1-P1=0

sumX=0 N2+P2-P1=0

1

1

2

2

P1=2kN

A

N1

1

1N1=P1=2kN

N2=P1-P2= -1kN

Example 21

P1=2kNP2=3kN

P3=1kNA BC

1

1

2

2


section

Example 21

P A

C

m

m

C

m

m

p

m

Pp

A

0 0

lim limmA A

Pp p

A

normal stress

shear stress

Review Stress

PP

m

m

PN



PN

A

N dA

A

N dA A N

A

+ (tension stress)


Unit Pa=Nm2

1MPa=106Nm2








( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

Three steps

1 Cut



F1

F3

F2

Fn

F1 FR

F3M

0

0

F

M



PP

m

m


N axial forceP

m

m

SIGN CONVENSION


P2P1

m

m

P3 PP

m

m


ngt2 then N=N(x)




P1=2kNP2=3kN

P3=1kNA BC

N2P1 =2kNP2 =3kN

AC

A BC2kN

1kN

Axial force diagram

sumX=0 N1-P1=0

sumX=0 N2+P2-P1=0

1

1

2

2

P1=2kN

A

N1

1

1N1=P1=2kN

N2=P1-P2= -1kN

Example 21

P1=2kNP2=3kN

P3=1kNA BC

1

1

2

2


section

Example 21

P A

C

m

m

C

m

m

p

m

Pp

A

0 0

lim limmA A

Pp p

A

normal stress

shear stress

Review Stress

PP

m

m

PN



PN

A

N dA

A

N dA A N

A

+ (tension stress)


Unit Pa=Nm2

1MPa=106Nm2








( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End


PP

m

m


N axial forceP

m

m

SIGN CONVENSION


P2P1

m

m

P3 PP

m

m


ngt2 then N=N(x)




P1=2kNP2=3kN

P3=1kNA BC

N2P1 =2kNP2 =3kN

AC

A BC2kN

1kN

Axial force diagram

sumX=0 N1-P1=0

sumX=0 N2+P2-P1=0

1

1

2

2

P1=2kN

A

N1

1

1N1=P1=2kN

N2=P1-P2= -1kN

Example 21

P1=2kNP2=3kN

P3=1kNA BC

1

1

2

2


section

Example 21

P A

C

m

m

C

m

m

p

m

Pp

A

0 0

lim limmA A

Pp p

A

normal stress

shear stress

Review Stress

PP

m

m

PN



PN

A

N dA

A

N dA A N

A

+ (tension stress)


Unit Pa=Nm2

1MPa=106Nm2








( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

P2P1

m

m

P3 PP

m

m


ngt2 then N=N(x)




P1=2kNP2=3kN

P3=1kNA BC

N2P1 =2kNP2 =3kN

AC

A BC2kN

1kN

Axial force diagram

sumX=0 N1-P1=0

sumX=0 N2+P2-P1=0

1

1

2

2

P1=2kN

A

N1

1

1N1=P1=2kN

N2=P1-P2= -1kN

Example 21

P1=2kNP2=3kN

P3=1kNA BC

1

1

2

2


section

Example 21

P A

C

m

m

C

m

m

p

m

Pp

A

0 0

lim limmA A

Pp p

A

normal stress

shear stress

Review Stress

PP

m

m

PN



PN

A

N dA

A

N dA A N

A

+ (tension stress)


Unit Pa=Nm2

1MPa=106Nm2








( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End


P1=2kNP2=3kN

P3=1kNA BC

N2P1 =2kNP2 =3kN

AC

A BC2kN

1kN

Axial force diagram

sumX=0 N1-P1=0

sumX=0 N2+P2-P1=0

1

1

2

2

P1=2kN

A

N1

1

1N1=P1=2kN

N2=P1-P2= -1kN

Example 21

P1=2kNP2=3kN

P3=1kNA BC

1

1

2

2


section

Example 21

P A

C

m

m

C

m

m

p

m

Pp

A

0 0

lim limmA A

Pp p

A

normal stress

shear stress

Review Stress

PP

m

m

PN



PN

A

N dA

A

N dA A N

A

+ (tension stress)


Unit Pa=Nm2

1MPa=106Nm2








( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

P1=2kNP2=3kN

P3=1kNA BC

1

1

2

2


section

Example 21

P A

C

m

m

C

m

m

p

m

Pp

A

0 0

lim limmA A

Pp p

A

normal stress

shear stress

Review Stress

PP

m

m

PN



PN

A

N dA

A

N dA A N

A

+ (tension stress)


Unit Pa=Nm2

1MPa=106Nm2








( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

P A

C

m

m

C

m

m

p

m

Pp

A

0 0

lim limmA A

Pp p

A

normal stress

shear stress

Review Stress

PP

m

m

PN



PN

A

N dA

A

N dA A N

A

+ (tension stress)


Unit Pa=Nm2

1MPa=106Nm2








( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

PP

m

m

PN



PN

A

N dA

A

N dA A N

A

+ (tension stress)


Unit Pa=Nm2

1MPa=106Nm2








( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

PN

A

N dA

A

N dA A N

A

+ (tension stress)


Unit Pa=Nm2

1MPa=106Nm2








( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End







( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End






( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End




( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

( )( )

( )

N xx

A x


Fixed PA(x)


Fixed Pp(x)

Axial force diagram


Is this correct



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End



( )

N xx

A x


N N N



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End



A(x)

N


P

Compare and a

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

Comparison







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End







from load



Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

Example 22





F

A

B

C

0yF

kN3281 NF

Solution


kN202 NF

0xF

45deg

045cos 21 NN FF

045sin1 FFN

1

2F

B

F

1NF

2NF x

y

45deg

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

Example 22

kN3281 NF kN202 NF


MPa90Pa1090

10204

10328

6

62

3

1

11

A

FN

MPa89Pa1089

1015

1020

6

62

3

2

22

A

FN

F

A

B

C

45deg

1

2F

B

F

1NF

2NF x

y

45deg


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End


PP

k

k





P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

P


PPk

k

k

k

P P

k

k

PP

p


k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

k

k

P P

k

k

PP

p

cos

AA

P P

cos cosP P P

pA A A

2cos cosp

sin 2sin

2p



k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End


k

k

P P

2cos cosp

sin 2sin

2p

Discussion



max

max2


23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

23 Hookersquos Law

F Fl0

l

Review


F

Global

Local


23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

23 Hookersquos Law

宏观物理量

载荷F

几何

微观物理量



l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End


l0

l

d d0

109Pa无量纲

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

23 Hookersquos Law









elasticity


Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

Example 23

C







A B

C

L1 L2

P1L

2L

C


Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

Example 23

C

A B

C

L1 L2

P1L

2L

C


What is the error

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

Example 23




0yFkN202sin1 FFFN



kN32173cos12 FFF NN

0xF 0cos 21 NN FF

0sin1 FFN


1mmm1011020010200

21020 3

69

3

11

111

AE

lFl N

A

F

1NF

2NF x

y

300

mm60m10601025010200

7321103217 3

69

3

22

222

AE

lFl N

AB tensed by

AC compressed by

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

Example 23


A

F

1NF

2NF x

y

300

1mm11

111

AE

lFl N

mm6022

222

AE

lFl N

A

A

1A2A

A

A

1A2A

mm111 lAA

mm6022 lAA

mm602 lx

mm039303912

30tan30sin

21433

llAAAAy

mm13

039360 2222

yxAA

3A

4A

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

Example 23

i ii

N ll l

EA


Fixed PA(x)

Fixed Pp(x)


ii

iii

AE

lNll

ii

i

Nll l

EA

( )( )

( )

N xx

A x

( )d

( )L

N x xl

EA x

( )d( )

( )

N x xd l

EA x

dx

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

Example 23

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

24 Strain Energy

Strain energy



P

l

Δl

EA

Pll

Δl

P

P

Δl

dP1

dΔl1

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

24 Strain Energy

0 0

1d d

2

l l EA lW P l l P l

l

Δl

P

P

Δl

P

dΔl


l

lEA

EA

lPlPWU

2

)(

22

1 22


EE

Al

lP

V

Uu

22

1

2

122

2 ee

Notes

Unit Jm3

dW=Pdl

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

能量原理



观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

观察

F

l



lFV

2

2

e

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

能量法求位移


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End


Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

Difference











2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End







2个平衡方程






equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End





equation)





Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

Example 25


Solution

210 NNx FFF

FFFF NNy 31 cos20


cos321 lll


cos

11

EA

lFl N

EA

lFl N 33


coscos

31

EA

lF

EA

lF NN

2

31 cosNN FF


3

2

21cos21

cos

FFF NN 33

cos21

FFN

1l 2l

3l

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

Example 25





00 PRRY BA


relationship

lll CBAC

（b）

Solution


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End


EA

aRl

EA

a2Rl

BCB

AAC


EA

aR

EA

a2R BA

BA RR2

3

P2R

3

PR BA

（c）



Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

Example 25




3

2lll ADAB


0 xF0

32

0


FFFF NNy 0

3

0

1 30sin30sin0


2231 FFF NN




F

30A

B

C30

D

1

2

3

F

A x

y1NF

2NF

3NF

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

Example 25

1323 321 NNN FFF

2231 FFF NN


F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y



each bar

cossin1 xyl

xl 2

cossin3 xyl

cos2 213 lll


213 3 lll


2

2

1

1

3

3 3

3

2

3

2

EA

lF

EA

lF

EA

lF NNN


Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

Example 25

F

30A

B

C30

D

1

2

3

x

y

F

A

1NF

2NF

3NF

x

y

A

Ax

y

1323 321 NNN FFF

2231 FFF NN

322 213 NNN FFF


kN6343

23 FFN

MPa6863

MPa8262

kN048232 FFN

(tension)

MPa1271

kN4253

221

FFN

(tension)

(compression)

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

Method and Steps



deformation



physical equation


equation


Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

Source of Stress






轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End





轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End



轨道板


底座板


五层结构

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

26 Thermal Stress



T



T Rl l


lTlT


EA

RllT

EA

RllR

TEAR TEA

R

0RR0X BA RRRSo BA







26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

26 Thermal Stress





item)


physical equation


equation


Tl T l

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

26 Thermal Stress







C

A

B D

1 2

3

A1

1L2L

3L

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L

Geometric equation


1 2sin sin 0X N N

0coscos 321 NNNY

cos31 LL

ii

ii

iii LT

AE

LNL

Physical equation

A

N1

N3

N2

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

26 Thermal Stress

C

A

B D

1 2

3

A1

1L2L

3L


cos)( 33

33

3311

11

11 LTAE

LNLT

AE

LN



cos21

)cos(

3311

3

2

311121

AEAE

TAENN

cos21

cos)cos(2

3311

3

2

31113

AEAE

TAEN

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

26 Thermal Stress

aa

aa

N1

N2







E=200GPa)

Geometric equation


021 NNY

0 NT LLL

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

26 Thermal Stress

Physical equation




2

1

1 2EA

aN

EA

aNLTaL NT

2

2

1

12EA

N

EA

NT


MPa 7661

11

A

N MPa 333

2

22

A

N

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

27 Assembly Stress




A

B C

1 2

A

B C

1 2

D

A1

3

Compatible

Incompatible

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

27 Assembly Stress

Geometric equation


0sinsin 21 NNX

0coscos 321 NNNY

13 cos)( LL



A

B C

1 2

D

A1

3

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

27 Assembly Stress

cos)(33

33

11

11

AE

LN

AE

LN




cos21

cos

3311

3

2

11

3

21AEAE

AE

LNN

cos21

cos2

3311

3

3

11

3

3AEAE

AE

LN

d

A1

N1

N2

N3

A

A13L

2L1L

27 Assembly Stress






physical equation


equation


The End

27 Assembly Stress






physical equation


equation


The End

The End

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Mechanics of Materials（甲） of... · · 2017-03-14Mechanics of Materials ... 泊松比（Poisson’s ratio ... 2 1.039 3.039mm sin 30 tan 30 1 2 3 3 4 ' ' $ l l G y AA A