Quick reviewSt St i d th i l ti hi

Stress: (Sections 1.3~1.6)

Stress, Strain and their relationship

AF

AF

:stressShear ,:Stress Normal

Strain: (Sections 2.1~2.2)

AngleL xy :strainShear ,:Strain Normal

Relating stress and strain—Hooke’s Law: (Sections 3.1, 3.2, 3.4, 3.7, 4.2)

EFL

G EA G

Lecture 2

Chapter 4: Statically Indeterminate Problems (Sections 4.4 &

)4.5)Statically Indeterminate Problems

Parallel Statically Indeterminate Problems

Chapter 5: Thermal Strain (Section 4 6)Chapter 5: Thermal Strain (Section 4.6)Thermal strain

Chapter 6: Poisson’s Ratio (Section 3.6)Poisson’s ratioPoisson s ratio

Relationship between Young’s Modulus (E), Shear Modulus (G) and

Poisson’s Ratio (ν)Poisson s Ratio (ν)

Chapter 4 Statically Indeterminate ProblemsT fi d th t l ti d thi

2m 1mTo find the external reactions under this beam, we drew a free body diagram and used the equilibrium equations.

20N

W=60N R2

20N 3 equilibrium equations:A

R1 R3

60N

R4 0 ,0 ,0 MFF VH

NRRFveH 200200 22

0600 431

RRRFveV

3 unknown reactions

2 equations:0605.132 0 43

RRM

veA

2 equations:

equationsofnumberreactionsunknownofno equationsofnumber reactionsunknown ofno. Statically Indeterminate

Statically Indeterminate ColumnsR1

A

veVF 0A

R1

B B 021 FRR

C F C F One equilibrium equation

R2 Two unknowns (R1 and R2)

equations ofnumber reactionsunknown of no.

Statically Indeterminate

Solution: R1

A

VF 0A

B

ve

B

021 FRR

C FC

R2

F

R

Equivalent problem

0 A

R1 0Top 0 BCABTop

B )0( 21

BCBC

BC

ABAB

AB

EALR

EALR

C FBCBCABAB

Example 1: Example 1: Determine the reactions at the floor and ceilingDetermine the reactions at the floor and ceilingR1

A

AluminiumA=100mm2 A

F 0B

20kN

A=200mm2 B veVF

0

)0( FRRC

20kNSteel

C

20 )0( 21 FRR

Lengths: AB=BC=100mm Statically IndeterminateR2

R1 0

Equivalent problemg Statically Indeterminate

A0Top 0Top 0 BCAB

B

)0( 21

BCAB

EALR

EALR

CF

)( BCBCABAB EAEA

Step 1: Use Statics to relate the unknown reactions:Step 1: Use Statics to relate the unknown reactions:

R1

VF 0A

B

veV

020 RRB

20

02021 RR

Two unknowns (R1 and R2)One equationC

R2

Two unknowns (R1 and R2)One equation

Statically Indeterminate

Step 2: Equivalent problemStep 2: Equivalent problem

A

AluminiumA=100mm2

This will normally involve replacing one of the supports by a force, and

i l i di l hiB

20kNA=200mm2

stipulating a displacement at this point.

C 20kN

Steel

R1 0

Equivalent problem Additional Equation:

A

1 0Top 0Top

B

F

BCABTop

CF

Step 3: Relate Step 3: Relate δδTOPTOP to the applied loads (including Rto the applied loads (including R11))

A

R1 0Top Find the force in each section:R1

B 1RFAB A

1

For section AB:

C 20kN FAB

R1

A

For section BC: 20RFB

20kN

For section BC: 201 RFBC20kN

FBC

Step 3: Relate Step 3: Relate δδTOPTOP to the applied loads (including Rto the applied loads (including R11))

Fi d th l ti f h ti i

A

R1 0TopFind the elongation of each section in terms of (R1):

B

For section AB:

LF ABABC 20kN

)84291()3100(1

ERER

AE ABAB

ABABAB

)8429.1()6100()970(

)(1

1

EREE

For section BC:LF BCBC

)3100())320(( 1

EERAE BCBC

BCBCBC

)95.2())320(()6100()9200(

)3100())320((1

1

EER

EEEER

Step 3: Relate Step 3: Relate δδTOPTOP to the applied loads (including Rto the applied loads (including R11))

Fi d th di l t f th t

A

R1 0TopFind the displacement of the top:

B )8429.1(1

ERBCABTop

C 20kN )95.2())320(( 1 EER

Step 4: Use the additional equation to determine the Step 4: Use the additional equation to determine the applied loads applied loads

0T 0Top

0)95.2())320(()8429.1( 11 EERERNR 29781

Step 5: Return to the equilibrium equations to find other Step 5: Return to the equilibrium equations to find other reactionsreactions

Equilibrium equation:

0200002978 2 R

NR 17021NR 170212

Parallel Statically Indeterminate Problems

250mmP

PP

PA

5mm

5mm

P

PA

PB

5mm

Aluminium

Brass

Aluminium P veHF 0

30mm 0 PPPP ABA

F 0 00 02 PPP BA

00

veVF 0 00

0)35(0)35(0 EPPEPM ABAB

02 PPP BA

One equilibrium equation

00

Statically Indeterminate

0)35(0)35( 0

EPPEPM ABAve

B

Two unknowns (PA and PB)q q

BA

Example 2: to find stress in brass and stress in aluminium

250mmPkNP 30

250mm

5mm GPaEGPaE

B

A

10570

5mm

5mm

Aluminium

B

30mm

Brass

Aluminium P

Step 1: Use Statics to get the equilibrium equations

0)330( EPPP ABAveHF

0

0)330(2 EPP BA

Step 2: Relate the deformation of each component to unknown force

P

unknown forceFor the brass

250mm

5

P

BB

BB AE

LP

5mm

5mm

5mm

AluminiumFor the aluminium

30mm

Brass

Aluminium P

AA

AA AE

LP

AA

Step 3: Additional equationLPLP

AB

105)9105( EAELAA

A

BB

B

AELP

AELP

70105

)970()9105(

AAA

BAB P

EEP

AA

EE

LLPP

Step 4: Solve these two equations, to obtain values for PAand Pand PB

Equilibrium equation

Additional equation

0)330(2 EPP BA

0)330(70

1052 EPP AA

70105

AB PP NEPA 357.8

q 70

NEEEPB 386.12)357.8(2)330( 70

Step 5: Calculate the corresponding stresses

MPEPA 1357)357.8( MPaEEAA

AA 13.57

)330()35()(

EP )38612( MPaEE

EAP

B

BB 73.85

)330()35()386.12(

Chapter 5: Thermal StrainChapter 5: Thermal Strain

Wh thi t h tt it d Th it d f thWhen something gets hotter, it expands. The magnitude of the expansion is proportional to the original size of the object and to the increase in temperature

TL THERMAL

to the increase in temperature.

α is the constant of proportionality, called the coefficient of thermal expansion. It is property of the material. L is the original length.∆T is the increase in temperature.

Th l St i 1)(ssdimensinleofunits:

C

ofUnits

o

Thermal Strain:

T

THERMAL

1)(Tof units

CC

oo

TL

THERMALTHERMAL

Typical values of α are measured in 10-6(oC)-1Typical values of α are measured in 10 6(oC) 1. For structural steel, 1))(612( CE o

Example 3:A brass bar and an aluminium bar are held between two rigid supports with a gap of

B

Aluminium

A brass bar and an aluminium bar are held between two rigid supports with a gap of 0.5 mm between their ends. The temperature is raised by the 60oC. Determine the stress in each bar and the elongation of the brass bar.

Brass

For the brass bar:L=300mm

For the aluminium bar:L=250mm

0.5mm

B

Aluminium

E=105GPaDiameter=50mmα=(18E-6)(oC)-1

E=70GPaDiameter=75mmα=(23E-6)(oC)-1

Brass

Touch? ?50 mm

0.5mm

Touch? ?5.0min mmGapiumTAluTBrass

Brass

Aluminium Touch! Load ExternalThermalTotal

Brass

PPiumPAluiumTAluiumAlu

PBrassTBrassBrass

minminmin

GapiumAluBrass min P

Example 3:Step 1: Determine if the bars will touch

Brass

Aluminium mmETLTBrass 324.060300)618(

Step 1: Determine if the bars will touch.

BrassmmETLiumTAlu 345.060250)623(min

iumTAluTBrass 345.0324.0min

0.5mmmmmm Gap

iumTAluTBrass

5.0669.0min

Touch!Touch!

Step 2: After the bars touch, a compressive force P will act between the bars. Calculate the combined elongation due to the temperature rise and to P.

Elongation due to temperature rise,

mmiumTAluTBrasseTemperatur 669.0min

Elongation due to the compressive force P,

AlAl

Al

BrassBrass

BrassiumPAluPBrassP AE

PLAE

PL min

Example 3: Brass

Aluminium

5.0 GapPeTemperatur

Step 3: To find the value of P by δTotal=δGap.

0.5mmGapPeTemperatur

5.0)6444418()970(

250)6751963()9105(

300669.0

EE

PEE

P)644.4418()970()675.1963()9105( EEEE

kNP 67.74

Step 4: Use the value of P to determine the stress in each bar

MPaEA

PBrass 03.38

751963)367.74(

Stress in the brass bar,

ABrassBrass 75.1963

Stress in the aluminium bar,

MPaEA

P 9.1644.4418

)367.74(

AluminiumAluminium

Example 3: Brass

Aluminium

S 5 U h li i d i hStep 5: Use the earlier equation to determine the elongation in each bar

El ti i th b b 0.5mmElongation in the brass bar,

EPBTBB

300)367.74(324.0

mmEEPBrassTBrassBrass

215.0109.0324.0)675.1963()9105(

324.0

Elongation in the aluminium bar,

E 250)36774(

mmEE

EiumPAluiumTAluiumAlu

285.0060.0345.0)644.4418()970(

250)367.74(345.0minminmin

Chapter 6: Chapter 6: Poisson’s Ratio (Section 3.6)Poisson’s Ratio (Section 3.6)P Poisson’s Ratio

lateral

P Poisson’s Ratio

B l t

v is a material property.

allongitudinP Bulge out

essdimensionlssdimensinlessdimensinle

of units of units ofUnits

Becomes thin

A block has a stress in the X direction only and the strain

Typical values of v are 0.25 to 0.4. For structural steel, v=0.32.Becomes thin

x

A block has a stress in the X-direction only, and the strain in the X direction will be:

σx

Ex z

yThe strain in the Y and Z directions:

x σx E

xxy

Ex

xz

MultiMulti--axial loadingaxial loading

I l σ

zyx

In general,z

x y

σy σx σz

EEEzyx

x σx

σy

σz EEEzyx

y

EEEzyx

z

EEE

Relationships between Young’s Modulus (E), Shear Relationships between Young’s Modulus (E), Shear M d l (G) d P i ’ R ti ( )M d l (G) d P i ’ R ti ( )

EG

Modulus (G) and Poisson’s Ratio (v)Modulus (G) and Poisson’s Ratio (v)

)1(2 G