ch 1 stress strain 14 ivle

8/9/2019 CH 1 Stress Strain 14 Ivle

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National University of Singapore

SESSION

2014

-

15

Semester

1

ME2113

Mechanics of Materials

I (Part 1)

Department of Mechanical Engineering

Applied Mechanics Group

NUS ME Dept Web Page

Office EA-05-13

[email protected]

Tel: 65162557

A/P Tay C. J.Tay Cho Jui
mailto:[email protected]:[email protected]


2/83

SESSION

2014

-

15

Semester 1

ME2113

Mechanics of Materials I

Modular Credits

: 3

Pre

-

Requisites

:

EG1109

-

Statics and Mechanics of Materials

Workload:

Lecture hr: 26 Tutorial hr: 5 Lab.

hr: 6 Two Lab sessions

Assessment:

Final Examination: 80%

Continuous assessment (

Consisting of Labs and quiz)

20% Course Lecturers

A/P C J Tay Part I, A/P

Vincent Tan Part II ;

Brief description of module: Part I

Analysis of Stress and Strain

Bending of Beams

Stresses in Loaded Beams

Deflection of Beams


3/83

Basic Text:

A.C.

Ugural

,

Mechanics of Materials, McGraw

-

Hill,1993

(Chapters 4, 7, 9 & 10 for part I)

Supplementary Readings:

F. P. Beer and E. R. Johnston, Jr. and J.T.

DeWolf

,

Mechanics of Materials, McGraw

-

Hiil

, SI 3rd Ed., 2004. R. C.

Hibbeler,Mechanics

of Materials, Prentice Hall,

SI2nd Ed., 2005.

J. M.

Gere

, Mechanics of Materials, Thomson

Brooks/Cole Publishing Company, 6th ed., 2004.

R

. R. Craig, Jr., Mechanics of Materials, McGraw

-

Hill,2nd ed., 2000.



4/83

NUS- Web Page IVLE

Module ME2113

Workbin

Lecture notes, Tutorial sheets, Lab manuals

FAQ

Tutorial Solutions, Past Exams Solutions

Discussion Forum

Announcement


Anonymous feedback via IVLE


5/83

ME2113 Part I (6 weeks)

14 AugStresses and Strains

21 Aug

Stresses and Strains

28 Aug

Shear force and Bending Moment in Beams04 Sept

Stresses in Loaded Beams

11 Sep

Stresses in Loaded Beams/Deflection of Beams

18 SepDeflection of Beams

Mid-Semester Break

ME2113 Part II - Prof. Vincent Tan

LESSON PLAN

SESSION 2014-15


6/83

ME2113 (part 1)

Tutorial Session

Day: Every Friday 9-10 am

Venue: Tutorial Room E3-06-02Timetable: http://me.nus.edu

.sg/

Consultation time for ME2113 (part 1)Day: Every Thursday

Time: 5.00-6.30 pm Venue: Office EA-05-13
http://me.nus.edu.sg/http://me.nus.edu.sg/http://me.nus.edu.sg/


7/83

ME 2113 Part I Assessment (on-line)

14/10/2014

(3rd week after mid-sem break)

multiple-choice question quiz


8/83


9/83

Chapter 1


10/83

ANALYSIS OF STRESS AND STRAIN

n is normal to dA

S1 , S2 are tangential

(in plane)

Apply general force dF on dA

Defn:

As dA 0, stress state is at the point P.

dA

dF

SheardA

dF

NormaldA

dF

S

dAS

S

dAS

n

dAn

2

02

1

01

0

lim

lim

lim

Note: Stress values depend on magnitude of dF and

also the direction of dF.

STRESSES

dA

P

s1

n

dF

s2


11/83

y

x

z

yy

yx

yz

zx

zy

zz

xx

xy

xz

Stresses shown are

all positive on a

cube of 1 unit

length

For a small isolated element with planes perpendicular

to coordinate axes and surrounding a point P, there

exist 9 stress components.

They are

3 normal stresses

6 shear stresses

As size of parallelepiped reduces, in the limit, these 9 stress

components will define completely, the state of stress at the

point P.

P


12/83

y

x

z

yy

yx

yz

zx

zy

zz

xx

xy

xz

Take Moment about Z-axis

xy 1x1 -yx 1x1 = 0

i.e. xy = yx

The cube is stationary (in equilibrium)


13/83

From equilibrium (i.e. taking moment about any axis), we can show

that:

zyyz

zxxz

yxxy

Number of unknown stresses reduced to 6.

i.e. xx ,yy , zz,


yzxzxy ,,

y

x

z

yy

yx

yz

zx

zy

zz

xx

xy

xzP


14/83

Stress Component on Arbitrary plane (2 dimensional case)

z

y

yx x

xy

x

y

xy

x

x

y

Arbitrary plane whose normal makes an angle with horizontal

Q. What are the values of x, xy in terms of x, y and xy ?


15/83


16/83

Substitute for Tx and Ty

cossin)()sin(cos

22

'' xyxyyx

y can be found by substituting + /2 for

in expression for x,

i.e.

cossin2cossin 22 xyyxy


Using the relations

cossin2sincos' 22 xyyxx

z

y

yx x

xy

x

y

xy

x

x

y

etc 2222

sin211cos2sincos2cos

cossin22 Sin

x

y x


17/83

Rewriting,

(i)

(ii)

(iii)

2cos2sin2

)(,, xy

yx

yx

Element A

y

y

xy

xx

y

y'

xy

x x'

x

Stresses on element A

2sin2cos2

)(

2

)(xy

yxyx

x

2sin2cos2

)(

2

)(xy

yxyx

y


18/83

When ,

)(

2tan

2

1

)(

22tan..

02cos2sin2

)(

1

yx

xy

yx

xy

xy

yx

or

ei

0,, yx

x' which is denoted as 1 is known as the maximum

principal stress

y' which is denoted as 2 is known as the minimum principalstress and

, which is denoted as is known as the principal angle

Eq. a


19/83

y

x

z

yy

yxyz

zx

zy

zz

xx

xy

xz

Convention for denoting stress

1. Normal stress, ijiindicates the direction of a normal to the plane on

which the stress component acts;

jindicates the direction of the stress.

Usually denoted by ij , e.g. xx

For simplicity

xx is written as xyy is written as y


20/83

2. Shear stress

iindicates the direction of a normal to the plane on

which the stress component acts;

jindicates the direction of the stress. y

x

z

yy

yx

yz

zx

zy

zz

xx

xy

xz

e.g. xy

ij


21/83

x

y

x

y

Positive Shear Negative Shear

+xy -xy

Sign convention

Sign convention- Normal stresses

Stress is positive Tension (e.g. xx)

negative Compression (e.g. -xx)

Sign convention- Shear stresses

M h Ci l f 2 D St


22/83

2sin2cos2

)(

2

)(xy

yxyx

x

(i)

22

2sin2cos2

)(

2

)(

xy

yxyx

x (a)

2sin2cos2

)(2

)(xy

yxyx

y (ii)

22

2sin2cos2

)(

2

)(

xy

yxyx

y (b)

2cos2sin2

)(,, xy

yx

yx

(iii)

22 2cos2sin

2

)(,,

xy

yx

yx(c)

We haveMohrs Circle for 2-D Stresses


23/83

22

2sin2cos2

)(

2

)(

xy

yxyx

x (a)

22

2sin2cos2

)(

2

)(

xy

yxyx

y(b)

22 2cos2sin2

)(,,

xy

yx

yx(c)

Eqs. a) + c) gives

222

''

2

)2()2(xy

yx

yx

yx

x

Eqs. b) + c) gives

222

''

2 )2

()2

(xy

yx

yx

yx

y

(iv)

(v)


24/83

Eqns (iv) and (v) represent a circle in the -planewith center

)0,2

( yx

2

2

2 xyyxR

and radius R given by:



25/83

)0,2

( yx

2

2

2 xyyxR

x-axis represents normal stress

y-axis represents shear stress

1

2

max

min

222

''

2 )2

()2

(xy

yx

yx

yx

x

222)( Ryax

x

xyA point on the circumference is

given by itsx andy coordinates

ie it represents the stresses ona plane in the element

Q: What does a point on the

circumference of the circle represent?
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Y


27/83

Hence the state of stress on Plane A or Plane B as shown

above can be represented by a point on the

circumference of a Mohrs circle

of center

and radius

)0,2

( yx

2

2

2 xy

yxR

xy

y

Y

X

xx

xy

y

Plane A

Plane B

A

B


28/83

2

2

1

2

2

2 xy

yxR

X

Y

xy

x

yPlane B

Plane A

y

x

2

yx

CD direction of 1

(2 ) (1 )P

Q

2

yx

1 = max prin stress

2 = min prin stress

max = max shear

stress


29/83


30/83


31/83

O C

xy

For a plane inclined at , x,

y,

xycan be obtained.

x

y

2 1

)22cos('

ROCx

22

Plane B

Plane A

x

center )0,2

( yx

2

2

2 xy

yx

R

)22cos( R

PQ

Principal angle =2

2

(x, -xy)A

(y , xy)B

2

2

( x, xy - )D

( y xy , )E

xy x

y

x

Plane E

Plane D


32/83

Stress components at any arbitrary plane can be determined from Mohrs

circle:

)22cos(

ROCx

2

yxOC

2

2

2 xy

yxR

On substituting and simplifying,

(cf. eqn(i))

2sin2cos2

)(

2

)(xy

yxyx

x

(i)

2sin2cos2

)(

2

)(xy

yxyx

x

2sin2sin2cos2cos)22cos(: Note

2sin2cos2

)(

2

)(xy

yxyx

y

and

2cos2sin2

)(,, xy

yx

yx

Similarly,

)(

2

2tanyx

xy

and


33/83

O C

xy

x

y

2 1

Plane B

Plane A

x

)22cos( R

PQ

2

(x, -xy)A

(y , xy)B

2

2

( x, xy - )D

( y xy , )E

xy x

y

x

Plane E

Plane D

Note that rotation on Mohrs circle is twice that of element and sense of

direction of rotation of axes is the same for Mohrs circle and element.

Sign convention for stresses when constructing and analyzing Mohrs circle:


34/83

2

2

1

2

2

2 xy

yxR

X

Y

xyx

yPlane B

Plane A

y

x

2

yx

CD direction of 1

(2 ) (1 )

Sign convention for stresses when constructing and analyzing Mohr s circle:

B?

A?
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35/83

Sign convention for stresses when constructing and analyzing Mohrs circle:

Shear stressesif the shear stresses on opposite faces of the element

produce forces that result in a clockwise couple, these stresses are taken as

positive.


Plane A clockwise ,

+ve on Mohrs circle

Plane B anticlockwise,

-ve on Mohrs circle


36/83


Sign convention for normal stresses

Positive is tensile and plotted along positive x-axis

- negative is compressive and plotted along negative x-axis

x

y

x

y

Positive Shear Negative Shear

+xy -xy

IMPORTANT:

Do not confuse Mohrs circle sign convention with that of an entire element


37/83

Please be punctual

Lectures will start on time

not puncture!

ME 2113 Lectures will be webcast

Power point slides are available on webcast


38/83

Applications of Mohrs circle

i) Spherical Pressure Vessel

Consider a spherical pressure

vessel with radius rand wall

thickness tsubjected to an

internal gage pressurep.



39/83

P


40/83

The normal stresses can berelated to the pressurep by

inspecting a free bodydiagram of the pressurevessel. To simplify theanalysis, we cut the vesselin half as illustrated.

From equilibrium, the stressaround the wall must have anet resultant to balance theinternal pressure across thecross-section.


41/83


42/83

Applications of Mohrs circle

Stresses in SphericalPressure Vessel


43/83


44/83


45/83


46/83


47/83


48/83


49/83


50/83


51/83

ii) Cylindrical

Pressure Vessel

Consider a cylindrical

pressure vessel with

radius rand wall

thickness tsubjected to

an internal gage

pressurep.



52/83

P

1


53/83

To determine the

longitudinal stress, we

make a cut across the

cylinder similar toanalyzing the spherical

pressure vessel.

From equilibrium,


54/83

dx

rP

h

h


55/83

To determine the hoop

stress, we make a cut

along the longitudinal

axis.

From equilibrium, the

hoop stress yields,


56/83

NOTE:

The hoop stress is twice as much as the longitudinal stress for thecylindrical pressure vessel.

The above formulas are good for thin-walled pressure vessels. ieradius ris larger than 5 times its wall thickness t(r> 5 t).

When a pressure vessel is subjected to externalpressure, thestresses are negative since the wall is now in compression insteadof tension.



57/83

Q: What would happen to the sausages if they were

overcooked?

A: Cracks along the longitudinal direction appear first.

= 2

This is because the internal pressure generated by the

steam inside the sausage causes the skin to fail from hoop

stress since the hoop stress is twice as much as the

longitudinal stress in the sausage (cylindrical pressurevessel .


58/83

EXAMPLE 1

The state of plane stress at a point (in a pressurevessel) is represented by the figure shown.Determine the stresses on an element oriented at300 counterclockwise from the position shown.Illustrate your answer on a diagram.

6 MPa

y

x

12 MPa

8 MPa 300


From Eqns (i) to (iii) we have


59/83

From Eqns. (i) to (iii), we have

(i)

(ii)

(iii)

Substituting = 300 , we have

To construct the Mohrs circle,

2sin2cos2

)(

2

)(' xy

yxyx

x

2sin2cos2

)(

2

)(

' xy

yxyx

y

2cos2sin2

)(,,

xy

yx

yx

MPax 2.8'

MPay 2.12'

MPayx 66.5,,

circleofcenterMPayxavg 22128

2

circleofradiusR xyyx

66.116102

222

2

Question: Where should point A be ?
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60/83

2

R = 11.66

8

6 MPa

y

x

12 MPa

8 MPa

Plane A

Plane B

Point A ?

Point A ?

x

Clockwise,

+ve shear

6

600

x

y

12

y

6

300

X'

y

Q p
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61/83

From the Mohrs circle,

MPax

2.8'

MPay 2.12'

MPayx 66.5,,

5.66 MPa

y'

x

12.2 MPa

8.2 MPa

300

300


2

R = 11.66

xClockwise,

+ve shear 6

600

x

12

y

6

-ve shear stressat this point

y

8

Hence

anticlockwise


62/83

Deformation and Strain


63/83

Deformation and Strain

Deformation is a physical phenomenon it can be measured.

Strain is a mathematical concept.

Basic modes of deformation (displacement)

Rigid body and Non rigid body

Rigid body Translation, Rotation

Non rigid body Elongation, Angular Distortion

O Angular Distortion

A

A'

B B'

P P'A

B'

A'

B

P

ORotation

OElongation

B

B

PP A A'

O

B

AP

P' A'

B'

Translation

Definition of Strain


64/83

Definition of Strain

Line element (direct strain)

Engineering strain of line element PQ.

O

P

Q

Q'

P'

PQ

PQQP

,,

Rotation between two line elements (shearing strain)


65/83

Rotation between two line elements (shearing strain)

Shearing strain

When is small

,

)tan( ,,


1.2.2 StrainDisplacement Relationships


66/83

Normal Strains

P

x

y

z

x

u

x

y

v

y

z

w

z

where u,v and w represent the displacements in thex, y and

zdirections respectively.

1.2.2 StrainDisplacement Relationships


67/83

Shear Strains

(ii)

y

u

x

vxy

y

w

z

v

yz

x

w

z

u

xz

Assumptions


68/83

Assumptions

A

A' B B'

D

D'

C

C'

ii indicates the direction in which the elongation or contraction is required. (the

sides are of an undeformed element)

Positive Strain for elongation (e.g. x)

Negative Strain for contraction (e.g. y)

1. Deformations are infinitesimally small

2. Displacement of a point on the element is continuous, i.e.

no cracks, overlapping, slippage, etc. and also body from

which element is isolated, is continuous throughout.3. Element is small, i.e. surroundings within close

neighbourhood of point P.

Convention for Strains

Normal Strains


69/83

Shear Strains

A A' B

B'

C

C'

D D'

x

y

iji, j indicate directions of two mutually perpendicular sides of an

undeformed element whose extent of angular deformation is required

(e.g. xy

)


70/83

Strains at a point

The strains x, y, z, xy, yz and xz are those of a cube elementsurrounding the point of interest, P(x, y, z). If the size of the cube is

allowed to become infinitesimally small, the strains can be regarded as

being the strains at point P(x, y, z) within a body.

Thus, 6 components of strains are required to define completely the state of

strain at point.

Transformation of Axes.

Often the strain components at a point referred to a set of axes are different

from the original axes.

For 2- D analysis, if x, y and xy are strains in x-y plane, what are theequivalent strains x, y and xy referred to x and y axes that make an

angle with the x-y axes.


71/83

The three strain components referred to x-y axes which are at an angle to

x-y axes:

(d)

(e)

(f)

2sin2

2cos22

'xyyxyx

x

2sin2

2cos22

'xyyxyx

y

2cos2

2sin22

'' xyyxyx


Principal Strains


72/83

There will be a plane in the element which does not experience any shear

strains,

i.e. xy=0 .

From eqn(f)

0

2cos2sin''

xyyxyx

i.e.

)(2tan yx

xy

or

(g)

For this value of ,

(h)

(i)

)(

tan2

1 1

yx

xy

22

'222

)(

xyyxyx

x

22

' 222

)(

xyyxyx

y


73/83

and are called the maximum and minimum principal strains

respectively.

(Usually denoted by 1 and 2)

The particular angle x or y denotes the direction of the principal axes, and

is denoted by . The rotation is defined as positive for counter-clockwise

rotation.

'x 'y



74/83

(i)

(ii)

(iii)

2sin2cos2

)(

2

)(' xy

yxyx

x

2sin2cos2

)(2

)(' xyyxyxy

2cos2sin2

)(,, xy

yx

yx

The three strain components referred to x-y axes which are at an angle to

x-y axes:

(d)

(e)

(f)

2sin2

2cos22

'

xyyxyx

x

2sin2

2cos22

'xyyxyx

y

2cos2

2sin22

'' xyyxyx

Mohrs Circle of Strain


75/83

Re-examining eqns (d), (e) & (f) and compare

with eqns (i), (ii) & (iii) for x , y & xy

The equations are similar in form.

Therefore a Mohrs circle for strain can similarly be constructed

with center at in the x , plane. The radius of the

circle is

0,

2

yx

2

xy

22

22

xyyxR

However for Mohrs strain circle,

x axis normal strainyaxis half shear stain

2

.


76/83


77/83

y

x

y

x

+ve Shear stresses +ve Shear strains

Gi th t i t t

Mohrs Circle of Strain


78/83

Given the strain state

x , y , xy

the corresponding Mohrs strain circle can be plotted.

Hence the strain components x , y and xy at any orientation can be determined.

x

y

x

xy

x

y

xy

2

2,

''

'

yx

y

2,

''

'

yx

x

2

2

2

2,

xy

y

y

2,

xy

x

2

yx

22

22

xyyxR

x 1

StressStrain Relationships

St d t i l t d th h th i i ti f th t i l


79/83

Stress and strain are related through the engineering properties of the material

of the body.

Assumptions

All the stresses / strains are within the elastic range of the material

Material is homogeneous (i.e. properties uniform throughout)

Material is isotropic (i.e. properties independent of direction)

Hookes law

(Uniaxial tensile test)E

xx

E Youngs modulus

Lateral contraction in y direction =

- Poissons ratio.

Application of yproduces strains

E

x

)(

)(

directionxE

directionyE

y

yy

Application of and simultaneously


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EE

EE

xyy

yxx

)(1

)(1

)(1

yxzz

zxyy

zyxx

E

E

E

GGG

xzxz

yzyz

xyxy

;;

(i)(a)

(b)

(a) (a)

Application of x and y simultaneously,

For threedimensions,

For shear strains,

G shear modulus of elasticity.

Eqns (ii) and (iii) represent the generalized Hookes law

(for isotropic, homogeneous materials).

(b)

(c)

(ii)

(iii)

The Hookes law is applicable to any orthogonal stress system


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The Hooke s law is applicable to any orthogonal stress system.

e.g. )(1

zrE

)(1 3211 E

Substituting the values of y in Eq. (i)b into Eq. (i)a and express the stress

in terms of strains, we have

yxx E

2

1

Likewise we can obtain

xyyE

21

EE

EE

xyy

yxx

(i)(a)

(b)

EXAMPLE 2


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EXAMPLE 2

The strain components at a point in a machinemember are given by

x = 900 , y = -100 , xy = 600 .

Using Mohrs circle, determine the principal strainsand the maximum shearing strains.

Centre of circle: (x+ y )/2 = (900 -100)/2 = 400

Radius of circle

5832

600

2

100900

22

2222

xyyxR


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ch 1 stress strain 14 ivle

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