Lecture Note for Solid Mechanicscontents.kocw.net/document/Solid Mechanics - Stress... · 2014-02-21 · Advanced Materials & Smart Structures Lab. 금오공대기계공학과윤성호교수

Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수

Lecture Note for Solid Mechanics- Stress-Strain Relationships and Material Properties -

Prof. Sung Ho YoonDepartment of Mechanical EngineeringKumoh National Institute of Technology


Text book : Mechanics of Materials, 6th ed.,

W.F. Riley, L.D. Sturges, and D.H. Morris, 2007.

Prerequisite : Knowledge of Statics, Basic Physics, Mathematics, etc.


Stress-strain diagrams

It is necessary to relate loads and temperature changes on the structure to the deformations produced by the loads and temperature changes

Relationship between load and deformation depends on dimension of member and type of material

Stress-Strain Relations and Material Properties

Dependent of cross sectional area & length

Independent of cross sectional area

Independent of length cross sectional area and length


The tensile test

Data for stress-strain curves are obtained by applying an axial load to a specimen and measuring the load and deformation simultaneously

Stress is obtained by dividing the load by the initial area (nominal stress) or the actual area (true stress)

Extensometer

Loadcell

Grip or fixture



Strain measurement

Instruments for measuring strain : strain gages or extensometers Nominal strain is computed on the initial length of the specimen True strain is computed on the actual length of the specimen

Example of stress-strain curves



Modulus of elasticity (Young’s modulus)

The slope of straight line portion of stress-strain curve is used to measure the stiffness of material Young’s modulus (Modulus of elasticity) : Shear modulus (Modulus of rigidity) :

E

G

Proportional limit (A)The maximum stress for which stress and strain are proportional

Young’s modulus (E)The slope of straight line portion of stress-strain curve

Tangent modulus (Et)The slope of stress-strain curve at a particular point beyond the proportional limit

Secant modulus (Es)The slope of stress-strain curve at any point



Elastic limit Elastic if strain resulting from loading disappears when the load is removed Elastic limit (point D) is the maximum stress for which material acts elastically Line BC is parallel to line OA Precise value is difficult to obtain

Work hardening (or strain hardening)

If loading again, stress-strain diagram follows the unloading curve until it reaches a stress a little less than the maximum stress during initial loading

The proportional limit (point B) is greater than that for the initial loading



Yield point

The stress at which there is an appreciable increase in strain with no increase in stress

Low carbon steels

Yield strength

The stress that induce a specified permanent set with 0.2 percent (0.002)

Stress indicated by intersection of CB and stress-strain diagram

Ultimate strength

The maximum stress based on the original area developed in a material before rupture

The nominal stress decreases beyond the ultimate strength

The true stress continues to increase until rupture



Elastoplastic Materials

When the stress exceeds the proportional limit, empirical equations have been proposed relating between stress and strain

Plastic strain is 16-20 times the elastic strain at the proportional limit

Poisson’s ratio

Ratio of the lateral strain to longitudinal strain

//

a

l

long

lat

GE )( 12

Ductility

Capacity for plastic deformation

Controls the amount of cold forming to which a material may be subjected



(Example 4-1) A 100-kip axial load is applied to a 1x4x90-in rectangular bar. When loaded, the 4-in side measures 3.9986-in and the length has increased 0.09-in. determine Poisson’s ratio, the modulus of elasticity, and the modulus of rigidity of the material.

(Sol)

ksiEGrigidityofModulus

ksiEelasticityofModulus

ratiosPoisson

ksiAP

L

L

in

long

lat

longlong

latlat

lat

260935012

0002512

0002500100025

350001000000350

2514

100

00100090090

000350400140

00140499863

,).(

,)(

:

,.

:

...:'

..

....



Principle of superposition

If the stresses do not exceed the proportional limit and the deformations are small, the effects of separate loadings can be added algebraically

Generalized Hooke’s law



dyE

dyE

dyd

dxE

dxE

dxd

xyyy

yxxx

yxz

xyy

yxx

E

E

E

1

1

Extension to the state of tri-axial stresses

yxzz

xzyy

zyxx

E

E

E

1

1

1

yxzz

xzyy

zyxx

E

E

E

)())((

)())((

)())((

1211

1211

1211

For the case of plane stress

yxzzE

)(

))((1

2110 yxz

1



2

0)45cos()45sin(0

sincoscossin 22

x

x

xyyxnt

)1(2

EG

zxzxzx

yzyzyz

xyxyxy

EG

EG

EG

)(

)(

)(

12

12

12

1

0)45cos()45sin(2

sincoscossin2 22

x

xx

yx

yx

xyyxnt

xy

x

y

a

t

12

1

EG

GG

xx

xntnt



(Example 4-2) When the machine part of alloy steel with E=210GPa and =0.3 was subjected to a biaxial stress, the measured strains were x=+1394m/m, y=-660m/m, and xy=2054m/m. Determine (a) stress components at the point (b) principal stresses and maximum stress at the point. Locate the planes on which these stresses act.

(Sol)

(a)

MPamN

E

CMPamN

E

TMPamN

E

xyxy

xyy

yxx

9165109165

1020543012

1021012

85510855

1013943066030110210

1

27610270

1066030139430110210

1

26

69

26

62

9

2

26

62

9

2

.)(.

))(().()(

)(.)(.

)())(.().(

)()(

)())(.().(



(b)

0

51245124

3457344

62341110

91652

8552762

855276

22

3

2

1

22

221

zp

p

p

xyyxyx

pp

CMPaMPaTMPaMPa

)(..)(.

..

).(..

,

522

018555276

9165222

62345124734421

21

21

.

.).(.

).(tan

.).(.

max

p

yx

xyp

ppp

MPa



(Example 4-3) When the machine part of alloy steel with E=30,000ksi and =0.3 was subjected to a biaxial stress, the measured strains were a=+650in/in, b=+475in/in, and c=-250in/in. Determine (a) stress components at the point (b) principal strains and maximum strain at the point. Locate the planes on which these stresses act (c) principal stresses and maximum stress at the point. Locate the planes on which these stresses act.

(Sol) (a)

ksiksiE

CksiksiE

TksiksiE

xyxy

xyy

yxx

xyxyyxb

ycbxa

356356105503012

0003012

813181311065030250301

000301

961896181025030650301

000301

550475

250475650

6

622

622

22

..))(().(

,)(

)(..)())(.().(

,

)(..)())(.().(

,

cossinsincos



(b)

711543312

611110250650

5502

1055

41711

3274527200

72745272004527200

222

21

3

2

1

22

21

..

.)(

tan

.)(

.

..

,

max

pp

yx

xyp

ppp

yxzp

p

p

xyyxyxpp

ksiE 171210810543012

0003012

6 .))(.().(

,)( maxmax

(c)

06035993

104727304327301

000301

72074220

104327304727301

000301

3

62

1222

62

2121

zp

ppp

ppp

Cksiksi

ETksiksi

E

)(..

).(...

,

)(..

).(...

,



Thermal strain

Most materials when unconstrained expand under heating and contract under cooling

Coefficient of thermal expansion : thermal strain due to one degree change in temperature

TT

Total strains

Strains due to temperature change and strains due to applied loads are independent

Most materials expand uniformly in all directions when heated

Neither the shape of the body nor the shear stresses and shear strains are affected by temperature changes

TETtotal



(Example) The aluminum [E=70GPa and =22.5(10-6)/oC ] block rests on a smooth, horizontal surface. When the body is subjected to a temperature change of ∆T=20 oC. Determine (a) The thermal strains (b) The deformations in the coordinate directions (c) The shearing strain

(Sol) (a)

mmmm

TTzTyTxT

/450/)10(450

)20)(10(5.226

6

Thermal strain :

(b)

TTtotal

),,( TzTyTx

),,( zyx

)( xy

Since the block is not constrained, there are no applied forces and there are no stresses. The load portion of the strain is zero, .

mmL

mmL

mmL

Tzz

Tyy

Txx

0338.0)75)(10(450

0225.0)50)(10(450

0450.0)100)(10(450

6

6

6

(c) Since the block is not constrained, there are no stresses and the total strains the same as the thermal strain. The thermal strain is the same in all directions. The original angle do not change. That is, there are no shearing strain, . 0xy



(Example) Consider a 1.5m long brass [E=100GPa and =17.6(10-6)/oC ] strap. Determine (a) how much the strap would have to be heated to lengthen the strap enough to insert the pin through the eyelets. (b) stress existing in the strap after the strap cooed back to room temperature.

(Sol) (a) TEL

Total strain :

If no stress is applied to the strap and the temperature is raised to stretch the strap by 1mm (0.001m)

CT

TLTE

o937

001051106170 6

.

.).())(.(

(b)

)(.)(.

.).()(

TMPamN

LTE

766106766

001051010100

26

9

If the strap stays stretched by 1mm after it cools down



(Homework)

(4.5), (4.16), (4.19), (4.27), (4.31)

Documents

Lecture Note for Solid Mechanicscontents.kocw.net/document/Solid Mechanics - Stress... · 2014-02-21 · Advanced Materials & Smart Structures Lab. 금오공대기계공학과윤성호교수