Ch 1 Stress Strain 2015 Ivle

i

Founded 1905

NATIONAL UNIVERSITY OF SINGAPORE

Department of Mechanical Engineering

ME2113

MECHANICS OF

MATERIALS I

Course Lecturer: A/P CJ TAY

ii

Founded 1905

SESSION 2015-16

Semester 1

ME2113 – Mechanics of Materials I Modular Credits: 3

Part I Lecture Notes

A/P CJ TAY

iii

Recommended Books

Basic Text:

A. C. Ugural, Mechanics of Materials, McGraw-Hill, 1993

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

Supplementary Readings:

1. F. P. Beer and E. R. Johnston, Mechanics of Materials, McGraw-Hill,

3rd Ed., 2003.

2. R. C. Hibbeler, Mechanics of Materials, Prentice Hall, 4th Ed., 2000.

3. J. M. Gere and S. P. Timoshenko, Mechanics of Materials, PWS

Publishing Company, 4th ed., 1997.

4. R. R. Craig, Jr., Mechanics of Materials, McGraw-Hill, 2nd ed., 2000.

v

TABLE OF CONTENTS

Chapter 1

ANALYSIS OF STRESS AND STRAIN 1.1 STRESS

1.2 DEFORMATION AND STRAIN

Chapter 2

BENDING OF BEAMS

(SINGULARITY FUNCTIONS AND SIGN CONVENTION)

2.1 SINGULARITY FUNCTIONS

2.2 SIGN CONVENTION

Chapter 3

STRESSES IN LOADED BEAMS 3.1 PURE BENDING

3.2 SHEAR STRESSES IN BEAMS

3.3 RELATIVE MAGNITUDES OF BENDING

AND SHEAR STRESSES

Chapter 4 DEFLECTION OF BEAMS

4.1 STATICALLY DETERMINATE BEAMS

4.2 STATICALLY INDETERMINATE BEAMS

4.3 BEAM DEFLECTION BY MOMENT – AREA

1

Chapter 1

ANALYSIS OF STRESS AND STRAIN

1.1 STRESS

Apply general force dF on dA

Defn:

dA

dF

SheardA

dF

NormaldA

dF

S

dAS

S

dAS

n

dAn

2

02

1

01

0

lim

lim

lim

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

Note: Stress values depend on magnitude of dF

and also the direction of dF.

dA

P

S2

S1

n

dF

2

1.1.1 Stress state at a point

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.

y

x

z

yy

yx

yz

zx

zy

zz

xx

xy

xz

Stresses shown are

all positive on a

cube of 1 unit

length

P

3

From equilibrium (i.e. taking moment about any

axis), we can show that:

zyyz

zxxz

yxxy

Number of “unknown” stresses reduced to 6.

Take Moment about Z-axis

xy

· 1x1 - yx

· 1x1 = 0

i.e. xy

= yx

The cube is stationary (in equilibrium)

y

x

z

yy

yx

yz

zx

zy

zz

xx

xy

xz

4

1.1.2 Stress Component on Arbitrary plane (2

dimensional case)

z

y

yx x’

x’y’

x

y

xy

x

x

y

For an arbitrary plane whose normal makes an

angle with the horizontal, what are the values

of x’, x’y’ in terms of x, y and xy ?

5

Consider a cut section on the element

Resolving forces in the x-direction

We have 111 BCTABAC xyxx

Divide above Eq. by BC

sincos yxxyxxxBC

AB

BC

ACT

Note: from the geometry of triangle ABC,

sin,cos BC

AB

BC

AC

Ty

B A

C

X’

x

yx

y

Tx

y

x xy

6

Similarly, we can show that cossin xyyyT

Consider the equilibrium of forces in the x’ direction on

plane BC, we have

sincos..

sin1cos11

'

'

yxx

yxx

TTei

BCTBCTBC

Consider the equilibrium of forces in the y’ direction on

plane BC, we have

sincos..

sin1cos11

''

''

xyyx

xyyx

TTei

BCTBCTBC

Substituting for Tx and Ty into the Eqs for 'x and '' yx

and note that yxxy , we have

cossin2sincos 22

' xyyxx

and

cossin)()sin(cos 22'' xyxyyx

y’ can be found by substituting + /2 for in the

expression for x’ , i.e.

cossin2cossin 22

' xyyxy

7

Rewriting,

2sin2cos2

)(

2

)(xy

yxyx

x

(i)

2sin2cos2

)(

2

)(xy

yxyx

y

(ii)

2cos2sin2

)(,, xy

yx

yx

(iii)

Element A

y

y

xy

x x

y’

y'

x’y’ x’ x'

x

Stresses on element A inclined at counter-clockwise to x-axis

8

Note that

When ,

)(

2tan

2

1

)(

22tan..

02cos2sin2

)(

1

yx

xy

yx

xy

xy

yx

or

ei

0,, yx

σx’ which is denoted as σ1 is known as the maximum

principal stress

σy’ which is denoted as σ2 is known as the minimum

principal stress and

θ , which is denoted as α is known as the principal angle and

Eq. a

(a')

x

y

θ which is denoted as α is known as the principal

angle

The angle is positive for counter-clockwise rotation.

(Usually denoted as α)

9

y

x

z

yy

yx

yz

zx

zy

zz

xx

xy

xz

Convention for denoting stress

1. Normal stress, ij

i – indicates the direction of a normal to the plane on

which the stress component acts;

j – indicates the direction of the stress.

Usually denoted by ij , e.g. xx

Sign convention

Stress is positive Tension (e.g. xx)

negative Compression (e.g. -xx)

For simplicity

xx is written as x

yy is written as y

10

2. Shear stress

i – indicates the direction of a normal to the plane on

which the stress component acts;

j – indicates the direction of the stress. y

x

z

yy

yx

yz

zx

zy

zz

xx

xy

xz

e.g. xy

Denoting stresses

ij

11

x

y

x

y

Positive Shear Negative Shear

+ xy - xy

Sign convention

12

2sin2cos2

)(

2

)('xy

yxyxx

(i)

22' 2sin2cos

2

)(

2

)(

xy

yxyxx (a)

2sin2cos2

)(

2

)('xy

yxyxy

(ii)

22' 2sin2cos

2

)(

2

)(

xy

yxyxy (b)

2cos2sin2

)(,, xy

yx

yx

(iii)

22 2cos2sin

2

)(,,

xy

yxyx

(c)

We haveMohr’s Circle for 2-D Stresses

)(2sin2cos2

)(

2

)(22

' bxyyxyx

y

x =

y =

)(2sin2cos2

)(

2

)(22

bxy

yxyx

y

)(2sin2cos2

)(

2

)(22

axy

yxyx

x

13

22' 2sin2cos

2

)(

2

)(

xy

yxyxx (a)

22' 2sin2cos

2

)(

2

)(

xy

yxyxy (b)

22 2cos2sin

2

)(,,

xy

yxyx

(c)

Eqs. a) + c) gives

222''

2' )2

()2

( xyyx

yxyx

x

Eqs. b) + c) gives

222''

2' )2

()2

( xyyx

yxyx

y

(iv)

(v)

22

' 2sin2cos2

)(

2

)(

xy

yxyxy (b)

)(2sin2cos2

)(

2

)(22

axy

yxyx

x

)(2sin2cos2

)(

2

)(22

bxy

yxyx

y

222''

2 )2

()2

( xy

yx

yx

yx

x

222''

2 )2

()2

( xy

yx

yx

yx

y

14

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

with center )0,

2(

yx

2

2

2xy

yxR

and radius R given by:

National University of SingaporeNational University of SingaporeNational University of Singapore

15

)0,2

(yx

2

2

2xy

yxR

National University of SingaporeNational University of SingaporeNational University of Singapore

A

P

l

an

e

B

B

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

above can be represented by a point (point A and point

B) on the circumference of a Mohr’s circle

With center at

and radius R

)0,2

(yx

2

2

2xy

yxR

16

Hence the state of stress along the x- and y-plane as

shown above can be represented by a point on the

circumference of a Mohr’s circle

of center

and radius

)0,2

(yx

2

2

2xy

yxR

τxy

σy

Y

Xσxσx

τxy

σy

Plane A

Plane B

Hence the state of stress at any plane on the above element can be

represented by a point on the circumference of a Mohr’s circle:

I.e. the stresses on plane A is represented by a point on the

circumference and the stresses on plane B by another point on the

circumference.

P

la

n

e

B

P

la

n

e

B

17

on plane A or plane B as

The normal stresses at this plane are the maximum

(1) and minimum (

2) principal stresses. From

Mohr’s circle

2

2

122

)(xy

yxyx

2

2

222

)(xy

yxyx

2

2

2xy

yxR

)(

2tan

2

1 1

yx

xy

Eq. a’

Compare with earlier derivations the

angles are similar

2

yx

xy

yx

xy

yx

xy

yx

xy

2tan

2

1

2tan2

2

2

2tan

1

1

18

Y

X

Plane A

Plane B

• On the above element, the stresses on plane A and B

are represented by points A and B on the Mohr ’ s circle

• Point D and E represent the maximum ( 1 ) and

minimum ( 2 ) principal stresses.

• Line CA represents the direction of x - axis

• Line CD represents the direction of

2 • θ represents twice the maximum principal angle

(also the direction of ) of the element x

principal stress the maximum

19

x

''' ,, yxyx

maximum principal angle = α

Plane E

Plane D

20

Stress components at any arbitrary plane can be determined from Mohr’s

circle:

)22cos(' ROCx

2

yxOC

2

2

2xy

yxR

On substituting and simplifying,

(cf. eqn(i))

Similarly,

2sin2cos2

)(

2

)('

xyyxyx

x

2sin2cos2

)(

2

)('xy

yxyxy

and

2cos2sin2

)(,, xy

yx

yx

Note that sense of direction of rotation of axes is same for Mohr’s circle and

element, but rotation for element is half that of Mohr’s circle.

2sin2cos2

)(

2

)(xy

yxyx

x

2sin2cos2

)(

2

)(

,

xy

yxyx

y

Similarly

and

but rotation on Mohr’s circle is twice that of element.

21

Sign convention for stresses when constructing and analyzing

Mohr’s circle:

Shear stresses – if the shear stresses on opposite faces of the

element produce forces that result in a clockwise couple, these

stresses are taken as positive.

on Plane B the shear stress

direction is anticlockwise,

hence it is -ve shear stress

on Mohr’s circle

on Plane A the shear stress

direction is clockwise, hence

it is a +ve shear stress on

Mohr’s circle

22

x

y

x

y

Positive Shear Negative Shear

+xy -xy

IMPORTANT:

Do not confuse Mohr’s circle sign convention with that of an entire element

Sign convention for normal stresses on Mohr’s circle

– Positive is tensile and plotted along positive x-axis

- negative is compressive and plotted along negative x-axis

23

1.1.4 Applications of Mohr’s circle

i) Spherical Pressure Vessel

Consider a spherical pressure vessel with radius

r and wall thickness t subjected to an internal

gage pressure p.

The normal stresses can be related to the

pressure p by inspecting a free body diagram of

the pressure vessel. To simplify the analysis, we

cut the vessel in half as illustrated.

24

From equilibrium, the stress around the wall

must have a net resultant to balance the internal

pressure across the cross-section.

ii) Cylindrical Pressure Vessel

25

Consider a cylindrical pressure vessel with

radius r and wall thickness t subjected to an

internal gage pressure p.

To determine the longitudinal stress, we make a

cut across the cylinder similar to analyzing the

spherical pressure vessel.

26

From equilibrium,

To determine the hoop stress, we make a cut

along the longitudinal axis.

27

From equilibrium, the hoop stress yields,

NOTE:

• The above formulas are good for thin-walled

pressure vessels. ie radius r is larger than 5

times its wall thickness t (r > 5 · t).

• When a pressure vessel is subjected to

external pressure, the stresses are negative

since the wall is now in compression instead

of tension.

• The hoop stress is twice as much as the

longitudinal stress for the cylindrical

pressure vessel. This is why an overcooked

hotdog usually cracks along the longitudinal

direction first (i.e. its skin fails from hoop

stress, generated by internal steam pressure).

28

Example 1

The state of plane stress at a point is represented

by the figure shown. Determine the stresses on

an element oriented at 300 counterclockwise

from the position shown. Illustrate your answer

on a diagram.

6 MPa

y

x

12 MPa

8 MPa

29

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

2sin2cos2

)(

2

)('

xyyxyx

x

(i)

2sin2cos2

)(

2

)('xy

yxyxy

(ii)

2cos2sin2

)(,, xy

yx

yx

(iii)

Substituting θ = 300 , we have

MPax 2.8'

MPay 2.12'

MPayx 66.5,,

To construct the Mohr’s circle,

MPayx

avg 22

128

2

66.116102

2222

xy

yxR

2sin2cos2

)(

2

)(xy

yxyx

x

2sin2cos2

)(

2

)(xy

yxyx

y

x

y

30

σ

Clockwise,

+ve shear

12.2 MPa 2

12.2 8

τ

6 R = 11.66

600

12 8.2

σx

σy

σy’

σx’

τx’y’ = 5.66

31

From the Mohr’s circle,

MPax 2.8'

MPay 2.12'

MPayx 66.5,,

5.66 MPa

y'

x’

12.2 MPa

8.2 MPa

300

300

32

1.2 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

33

1.2.1 Definition of Strain

O

P

Q

Q'

P'

Line element (direct strain)

Engineering strain of line element PQ.

PQ

PQQP

,,

34

Rotation between two line elements (shearing

strain)

Shearing strain ,

When , is small

)tan( ,,

35

(i) Normal Strains

The normal strains are given by:

x

ux

y

vy

z

wz

(i)

where u,v and w represent the displacements in

the x, y and z directions respectively.

(ii) Shear Strains

The shear strains are given by:

y

u

x

vxy

y

w

z

vyz

(ii)

x

w

z

uxz

1.2.2 Strain – Displacement Relationships

36

1.2.3 Assumptions

Convention for Strains

Normal Strains

A

A' B B'

D

D'

C

C'

i – i 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. -x)

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.

37

Shear Strains

AA' B

B'

C

C'

DD'

x

y

ij – i, j indicate directions of two mutually

perpendicular sides of an undeformed element

whose extent of angular deformation is required

(e.g. xy)

38

1.2.4 Strains at a point

The strains x, y, z, xy, yz and xz are those of a

cube element surrounding 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 the equivalent strains x’, y’

and x’y’ referred to x’ and y’ axes that make an

angle with the x-y axes

39

The three strain components referred to x-y axes which

are at an angle to x-y axes are:

2sin2

2cos22

'xyyxyx

x

(d)

2sin2

2cos22

'xyyxyx

y

(e)

2cos2

2sin22

'' xyyxyx

(f)

1.2.5 Principal Strains

There will be a plane in the element which does not

experience any shear strains,

i.e. x’y’=0 .

From eqn(f)

0

2cos2sin''

xyyxyx

40

i.e.

)(2tan

yx

xy

or

)(

tan2

1 1

yx

xy

(g)

For this value of ,

22

'222

)(

xyyxyxx

(h)

22

'222

)(

xyyxyx

y

(i)

'x and 'y are called the maximum and minimum

principal strains respectively.

(Usually denoted by 1 and 2)

The particular angle x’θ

or y’θ

denotes the direction of the principal

axes, and is denoted by . The rotation is defined as positive for

counter-clockwise rotation.

41

1.2.6 Mohr’s Circle of Strain

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

(i), (ii) & (iii) for x’ , y’ & x’y’

The equations are similar in form. Therefore similar to

Mohr’s circle of stress, a Mohr’s circle of strain can

similarly be constructed with center at

0,

2

yx in the

x’ , 2

xy plane. The radius of the circle is

22

22

xyyxR

However for Mohr’s strain circle,

x – axis normal strain

y – axis half shear stain 2

.

42

Mohr’s Circle of Strain

The convention for constructing and reading shear strain values from

Mohr’s strain circle is similar to that used for shear stress in Mohr’s

stress circle.

However for Mohr’s strain circle,

x – axis normal strain

y – axis half shear strain

Sign convention for Mohr’s strain circle

Normal strain: elongation +ve strain

contraction -ve strain

Shear strain: Positive shear stress results in positive shear strain

i.e. A positive shear strain corresponds to a clockwise shear stress

couple.

A negative shear strain corresponds to an anti-clockwise shear stress

couple.

43

Shear strain: Positive shear stress results in positive

shear strain

y

x

y

x

+ve shear stress +ve shear strain

44

45

1.2.7 Stress – Strain Relationships

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)

Hooke’s law

If only x is applied in the x-direction, the strain

in the x-direction is:

E

xx

(such as in an uniaxial tensile test)

E is the Young’s modulus

The lateral contraction in the y direction = E

x

is the Poisson’s ratio.

46

If only y is applied in the y-direction, the strain

in the y-direction is:

E

yy

The lateral contraction in the x- direction = E

y

If x and y are applied simultaneously, we have

EE

EE

xyy

yxx

For three –dimensions,

)(1

)(1

)(1

yxzz

zxyy

zyxx

E

E

E

(ii)

For shear strains,

GGG

xzxz

yzyz

xyxy

;; (iii)

G – shear modulus of elasticity.

Eqns (ii) and (iii) represent the generalized

Hooke’s law (for isotropic, homogeneous

materials).

(i) (a)

(b)

(a)

(b)

(c)

47

where the directions , r, and z are perpendicular to

each other

where the directions 1, 2, and 3 are perpendicular to

each other

Substituting the values of σy in Eq. (i)b into Eq.

(i)a and express the stress in terms of strains, we

have

Likewise we can obtain σy by substituting the value of σx

in Eq (i)a into Eq (i)b and obtain :

)(1

3211 E

)(1

zrE

yxx

E

21

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

(stresses which are inclined at 900 to each other).

For example we can write the normal strains as

xyy

E

21

48

EXAMPLE 2

The strain components at a point in a machine member are given by

x = 900 μ,

y = -100 μ,

xy = 600 μ .

Using Mohr’s circle, determine the principal strain and the maximum shearing strains.

Centre of circle: (x+

y )/2 = (900 -100)/2 = 400 μ

Radius of circle

5832

600

2

100900

22

2222

xyyxR

49

2

C 1

x

y

2

400

900 , -300

-100 , 300

Max shearing strain

+600μ

Plane B

x = 900 μ

y = -100 μ

xy

=

Plane A

R=583

A

B

x

From the Mohr’s circle

1 = 983 x 10

-6 ,

2 = -183 x 10

-6

Maximum shearing strain γmax

= 1166 x 10

-6

On plane A, x = 900 μ, on plane A the shear

direction is anticlockwise, hence it is indicated as a -ve shear stress on the Mohr’s circle

-(xy

)/2= -600/2 = -300 μ

On plane B, y = -100 μ, on plane B the shear

direction is clockwise, hence it is indicated as a +ve shear stress on the Mohr’s circle

clockwise shear on plane B

(xy

)/2= 600/2 = 300 μ