Ch1 Beams Bending

Founded 1905

NATIONAL UNIVERSITY OF SINGAPORE

Department of Mechanical Engineering

MECHANICS OF

MATERIALS II

ME2114

Course Lecturer: A/P CJ TAY

1

Founded 1905

SESSION 2014-15

Semester 2

ME2114 Mechanics of Materials II Modular Credits: 3

Part I Lecture Notes

A/P CJ TAY

2

Recommended Books

Basic Text:

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

(Chapter 4, 12 & 13 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. 5. A.L. Window ed., Strain gauge technology, London : Elsevier

Applied Science , 2nd ed., 1992.

6. J.W.Dally & W.F. Riley, Experimental Stress Analysis, McGraw-Hill, 3rd Ed., 1991.

3

TABLE OF CONTENTS Chapter 1

ELASTIC-PLASTIC BENDING OF BEAMS AND TORSION

OF CIRCULAR BARS 1.1 BENDING OF BEAMS STRESS CONCENTRATION 1.2 BENDING OF BEAMS INELASTIC BENDING 1.3 TORSION OF CIRCULAR BARS ASSUMPTIONS 1.4 FIRST YIELD TORQUE 1.5 ELASTIC-PLASTIC TORQUE 1.6 FULLY PLASTIC TORQUE 1.7 RESIDUAL SHEAR STRESS DISTRIBUTION

Chapter 2 BUCKLING OF COLUMNS

2.1 EULER BUCKLING OF SLENDER COLUMNS 2.2 SUMMARY 2.3 EFFECTIVE LENGTH 2.4 COLUMNS INITIALLY CURVED 2.5 COLUMNS WITH ECCENTRIC LOADS

Chapter 3 EXPERIMENTAL STRESS ANALYSIS

3.1. THEORY

3.2. TYPE OF BRIDGE CIRCUITS

3.3. APPLICATIONS OF STRAIN GAUGES

3.4. GRAPHICAL SOLUTION

3.5. TRANSDUCERS

3.6. TEMPERATURE EFFECT

3.7. THREE-LEAD-WIRE ARRANGEMENT

3.8. SPECIAL PURPOSES GAUGES

3.9. ADVANTAGES AND DISADVANTAGES OF STRIAN GAUGES

2

CHAPTER 1

BENDING OF BEAMS

1.1 BENDING OF BEAMS STRESS CONCENTRATION

Bending stress formula I

MCmax is used only for a constant

cross-sectional area. For cross-section that changes suddenly,

the stress-strain distributions become nonlinear.

Examples of changes in cross-sections

3

The stress distribution for case (a) is :

The max stress occurs at base of grooves, max stress is given

by:

I

MCkmax

k stress concentration factor The value of k and the stresses through the section are

determined by experiment or theory (sometime).

4

For a certain beam geometry, the stress concentration values

can be from the following Figs:

Beams with shoulder fillets (Fig. F1):

5

Beams with grooves (Fig. G1):

6

Example 1.1-1

Beams with shoulder fillets

A steel bar with shoulder fillets as shown in the following

figure is subjected to a bending moment of 5 kNm, determine

the maximum bending stress developed in the steel. Given

that r = 16 mm, h = 80 mm, w = 120 mm, t = 20 mm.

Solutions:

Given r = 16 mm, h = 80 mm, w = 120 mm

We have

r/h = 16/80 = 0.2 , w/h = 120/80 = 1.5

Second moment of area 4308.002.0

12

1mI

Using I

MCkmax

7

From Fig. F1, the value of k is given by 1.45

Using M = 5 kNm, C = 0.04 m,

MPa

x

x340

1053.8

04.0545.1

7max

Stress distribution below the fillets :

Stresses away from the fillets are not affected by the stress

concentration and the max stress is given by:

I

MCmax i.e. k = 1

Hence MPax

x234

1053.8

04.057max

Stress distribution away from the fillets :

9

Example 1.1-2

A simply supported beam with thickness of 10 mm is loaded

as shown in the Fig. Determine the length L of the center

portion of the beam so that the maximum bending stress at

section A, B, C is the same.

From vertical equilibrium

RD = RE = (350L)/2 = 175L

BM at section A or B

RD RE

D E

10

MA = 175L (0.3) = 52.5L

BM at section C

For r = 8 mm, h = 40 mm, w = 60 mm

r/h = 8/40 = 0.2 , w/h = 60/40 = 1.5

MC = 175L (0.3 + L/2)

- 350(L/2)(L/4)

= 52.5L + 43.75L2

350 N/m

11

From Fig. F1, the value of k is given by 1.45

Maximum bending stress at either section A or B

LxL

I

CMk AAat

7

3max 1085.2

04.001.012

1

02.05.5245.1

Maximum bending stress at section C

266

3

2

max

1029.71075.8

06.001.012

1

03.075.435.52

LL

LL

I

CMCCat

Problem requires that

CatAat maxmax i. e.

mL

LLL

72.2

1029.71075.81085.2 2667

12

1.2 BENDING OF BEAMS INELASTIC

BENDING

1.2.1 STRESS-STRAIN CURVE

In general, deformation of material under load can be divided into

four stages:

I: Linear elastic deformation

-small strains and displacements; Hookes law

II: Non-linear deformation

-permanent "set" after unloading

Str

ess

u

yu yl Rupture

Strain

13

III: Large deformation

-eg. in metal forming processes

IV: Rupture/Fracture

Up till now, mainly concerned with stage I. This section considers

stage II and transition from stage I to stage II (elastic-plastic

behaviour).

14

Why the need to study the elastic-plastic behaviour of structures?

It would be unwise if designers knew nothing of what would

happen to components that were grossly overloaded to the point

where marked yielding and plastic deformation had occurred;

One important aspect - in some circumstances, enhanced

performance can be achieved by prior plastic deformation resulting

in favourable residual stresses eg, in thick-walled pressure vessel.

In elastic-plastic analysis, usually the actual - curves for the real

strain-hardening material introduces some complications in

analyses.

Because of this, semi-idealised behaviour is often assumed in which

(a) strain-hardening occurs linearly from initial yield, or (b) strain-

hardening is ignored.

15

strain-hardening assumed linear from initial yield

( bi-linear stress-strain curve)

Strain-hardening is ignored

Y

16

1.2.2 ASSUMPTIONS

1. That any cross-section of the beam will remain plane during bending as in elastic bending.

2. That the fibres are in a condition of simple tension or

compression.

RECTANGULAR SECTION

(a) Limit of total elastic action

In elastic bending of a beam, there is a linear stress distributed over

its cross-section. The extreme fibres reach the yield stress when the

bending moment is

MY MY

Y

Y

d

b

17

6

2

12

1I 2

3

bd

d

bd

yM

I

YM

YYYY

Y

Y

(b) Partial elastic-plastic action

When the bending moment is increased further beyond MY, some of

the fibres near the top and bottom surfaces begin to yield and plastic

deformation penetrates deeper into the beam as shown above. The

Y

Y

c

c

b

d

Plastic zone

Mp > My Mp

(a)

18

strain at outer fibres of beam may increase, but stress will remain

at Y .

Bending moment Mp in beam is given by:

dyybM p

where b is width of beam at distance y from N.A.

For rectangular section beam (b = constant)

(Elastic portion):

Yc

y ; (Plastic portion): = Y

c

o

d

cY

Yp dyybdyyby

cM

2/2

=

2/2

0

3

232

d

c

Y

c

Y yby

c

b

=

343

41

4

22

2

22 cdb

d

cbdY

Y

(2.1)

When stresses at outermost fibres of beam just reach yield i.e. c =

d/2, the bending moment MY is

M b d d bd

YY Y

2 2 2

4 12 6 (2.2)

Same as Eq. (a)

19

(c) Total plastic action

When stresses throughout the beam section reach yield as in above,

i.e. c = 0, the "ultimate B.M." or fully plastic moment is

Y

Y

Mult Mult

d

b

From Eq. 2.1

The ratio is the "shape factor"; it depends only on the shape of the cross-section of beam.

ultYYPM

db

cdbM

434

222

5.1

6

42

2

d

b

db

M

M

Y

Y

Y

ult

20

EI

M

Rand

R

yEE xx

1

From beam bending equations (ME2113)

R is the radius of curvature of the beam neutral axis

When first yield has just occurred (at the extreme fibres) the value of c is d/2. The radius of curvature of the beam at first yield is

1

2R EY

Y

d

1

R EcY

o

R

cEYx

I.e.

For a partially elastic-plastic beam at a distance c from the neutral axis, the stress in the fibres has just reached the value

Y

(note: C = d/2)

21

The moment-curvature relationship for the rectangular beam is

thus linear up to a value M = MY and beyond this point the

relationship is non-linear and asymptotic to Mult as shown in the

figure below.

Curvature

Ben

din

g

Mo

men

t

Mult

My

22

1.2.3 SYMMETRICAL I-SECTION

(a) When yielding is about to occur at extreme fibres (first

yield)

2

12

12

I

3

11

3

Y

d

dbbd

yM

I

YM

xx

YY

Y

Y

b1/2 b1/2

b

d1 d

Y

Y

23

(b) When top and bottom flanges have yielded

Plastic portion: = Y

Elastic portion:

2

1

Y y

d

Y

Y

b2

Y

Y

24

Recall M b y dy

2

23

4

2 4

2 2

2

2

1

22

1

0

3

2

1

2

1

0

2

2

1

2

2

1

2

1

0

2

2

12

1

d

dY

d

Y

d d

d Y

Y

d d

d Y

Y

yb

yb

d

dyybdyybd

dyybdyybyd

M

Substituting the value of y and simplifying we have

M b b d d

Y

+b d2

1 12

12

6 4

(c) Fully Plastic Condition

Y

Y

Y

Y

25

2

1

0

2

2

12Y 2

d d

d YultdyybdyybM =

Ybd b d2 1 1

2

4

Shape factor,

M

M

bd d

bd d

dult

Y

b

b

21 1

2

31 1

3

12

4 2

26

Example 1.2-1

The steel wide-flange beam has the dimensions as shown

in Fig. 3.1. If it is made of an elastic perfectly plastic

material having a tensile and compressive stress of

Y = 250 MPa. Determine the shape factor of the beam.

Fig. 3.1

27

a) Determine the maximum elastic moment MY :

I of X-section:

46

233

1044.82

75.1185.1220012

5.122002

12

2255.12

mm

I

A1 A2

N

A1

A2

A

118.75 mm

56.25 mm

mm

28

Applying the beam bending formula:

kNm

yM

I

YM

YY

Y

Y

88.164

125

1044.82250

I 6

29

b) Determine the fully plastic moment Mp :

kN

areaflangetopCForce Y

625

2005.12250

2

30

kN

areawebtopCForce Y

56.351

5.1125.12250

1

kNm

kNmm

webtopandflangetoptodueMoment

94

25.5656.35175.118625

Hence Mp = 2 x 94 = 188 kNm

The "shape factor" 14.188.164

188

Y

p

M

M

Note: the ultimate moment (Mp) is only 14% higher than the

moment at first yield (MY).

31

Example 1.2-2

The beam x-section as shown in the following figure is

made of an alloy of titanium that has a stress-strain

relationship as shown. If the material behaviour is the same

in both tension and compression, determine the moment that

is applied to the beam to cause a strain of 0.05 at the

extreme top and bottom fibre of the beam.

32

The material exhibits elastic-plastic behaviour with linear

strain hardening. The strain distribution is given by:

From similar triangles

We have

mmcmy

y

33.0

050.0

010.0

5.1

The stress and force distribution on the x-section are:

33

252000 N=

34

1.2.4 ASYMMETRICAL SECTION (Y-Y is only plane

of symmetry)

(a) When yielding is about to occur at top fibre

(a) Determine position of N.A. ie. "h" from bottom

h = 70 15 200 7 5 10 170 85 15 100 15 7 5

70 15 10 170 100 15

. .

= 90.2 mm

Fibre furthest away from N.A. will yield first.

ymax = 200 - 90.2 = 109.8 mm (i.e. at top fibre)

First yield moment, My

YY NA

I

max

d = 200

h

Y

N A

10

b1/2=30 b1/2=30

di=170

b=100

35

47

2

3

2

3

2

3

10556.2

7.821510012

151008.917010

12

170103.1021570

12

1570

mm

INA

Y

Y

YM

57

10328.28.109

102.556

36

(b) Bending moment is increased until bottom fibre yields:

Assume Y is the same in tension and compression.

Let y = depth of plastic flow.

From the condition that nett axial force on section is zero:

F and F b dy 0

0152

110015

2

200

210

2

200

2

11015101570

1

1

Y

YYY

y

yy

Y

Y

37

2

100

285

2

200

152

200

2

200

152

200

1

1

y

y

y

y

y

y

YY

Y

Only unknown is y , so can be solved.

0151

2y100

2y85

2100

2y85

2y100

2y85

210

2

y200

2

11015y101570

Y

Y

YYY

y y2 245 6975 0 y = 32.9 mm

38

Note: The position of zero stress is now (200 - y )/2 = 83.6 mm

from the bottom (compared to 90.2 mm for elastic case)

In elastic-plastic loading mode, the N.A. moves to such a position

that the total force over the x-sectional area subjected to the tensile

stress system is equal to that experiencing compressive stresses and

the state of equilibrium remains undisturbed.

The corresponding B.M. is given by M b y dy

39

Bending Moment calculation

152

100153

2

215100

5.72

10015100

152

1003

215

2100

210

2100

3

2

2100

210

2100

2

151510

5.72

1001570

1

1

1

y

y

yy

yy

yyy

yy

M

Y

Y

Y

Y

Y

Y- l

40

(c) Fully Plastic Case

Since there can be no longitudinal resultant force in the beam,

A Y Y1 2 A

where A1 and A2 are the areas of the c/s above and below N.A.

respectively.

That is A1 = A2 = 1

2A where A is the total area of the cross-section.

Thus for the fully plastic state, the neutral axis divides the cross-

section into two equal areas and the stress diagram is shown above.

n

41

Thus

7015 + 10n = 10(170 - n) + 10015

n = 107.5 mm

Note: The position of zero stress is now:

(170-n) + 15 = (170-107.5) + 15 = 77.5 mm from the bottom

(compare to 90.2 mm for elastic case).

M n nn

nn

ult Y Y

Y Y

70 15 7 5 102

10170

2100 15 170 7 5

2

.

.

Hence "shape factor M

Mult

Y

1.3

n line

42

1.2.5 RESIDUAL STRESSES

When a beam that has undergone plastic deformation is unloaded,

the unloading is assumed to be linearly elastic.

If applied moment (causing plastic deformation) is Mp and the

unloading moment is Me then for equilibrium,

Mp - Me = 0 (5.1)

(Me is negative as it is applied in the opposite direction)

Mp + Me = 0 (5.1)

43

Note: Compressive where it was tensile (and vice versa).

Y e

Mp Mp

e Y

+

+

-

-

e - Y

Me Me

44

e is calculated from the equilibrium condition ie.

Mp -Me = 0 (5.1)

For rectangular section, b x d

From Eq. 2.1

3

4

22 cdbM Yp

6 by given is moment recovery elastic The

12

12 Using

2

3

bdMM

bd

dM

eee

e

e

Hence Eq. 5.1 becomes

)2.5(2

2

3

3

4

6

0 6

- 3

4

2

2

222

222

d

c

cdb

bd

bdcdb

Y

e

Ye

eY

Hence we can find e Note: The maximum possible value of e is

when beam is fully plastic ie. c = 0. Then 2

3

Y

e

45

Example 1.2-3

The steel wide-flange beam shown in the following figure is

subjected to a fully plastic moment of Mp. If this moment is

removed, determine the residual stress distribution in the beam. The

material is elastic perfectly plastic and has a yield stress of y = 250

MPa.

From example 3.1

I = 82.44 x 106 mm4

Mp = 188 kNm

Applying the beam bending formula:

46

MPa

I

YM ee

1.285

1044.82

125101886

6

From similar triangles, we have

mmy

y

61.109

1.285

250

125

Superposition of the above stresses give the residual

stress distribution as shown below:

e = 285.1 MPa

e = 285.1 MPa

47

1.2.6 RESIDUAL CURVATURE

For a beam which has reached yield, unloading will NOT result in

the beam returning to its original (straight) condition.

We have,

Plastic curvature - Unloading curvature = Residual curvature

ie. re

111

p

For an elastic case, Ey

1 where is the radius of the N.A.

48

For the elastic-plastic beam, use the extremity of the elastic stresses

ie. p

1 =

Ec

Y

During unloading, stresses behave elastically:

Mp Mp

c

+ Y

Me Me d/2

- e

e

49

2/

1

dE

e

e

2/ -

1

dEEc

eY

r

From Eq. (5.2)

2

2

3

2

2

d

cYe

Hence

3

4

c

1

13

2

dd

c

E

Y

r

Note: Amount of elastic (unloading) movement is known as

SPRINGBACK

50

Example 1.2-4

A square steel bar (25 x 25 mm) of an elastic-perfectly plastic

material is formed into part of a circle using a round mandrel. What

mandrel diameter would be required so that an elastic zone of 16 x

25 mm is attained? Determine the final curvature after springback

and the residual stress distribution in the bar. Assume Y = 250

MPa, E = 200 x 103 MPa.

(a) On loading: recall for the elastic case EyEI

M

1

For elastic-plastic beam, elastic core of 16mm x 25mm is obtained.

1

3 15625.0

008.0200x10

250=

1

mEc

Y

p

p = 6.4 m

Mandrel

Beam

25

25

16

+ Y

51

Hence mandrel diameter = 12.8 m

What is the elastic-plastic moment?

Mp 2 25 45 250 2 25 8. .

8

3

2 250 8 25

2

1 +

= 843.2 Nm

(b) On unloading: stresses are wholly elastic

2

dM p

e

Sectional area of bar

+Y

Sectional area of bar

52

= 12

4

3

1012

25

105.122.843

= 323.8 MN/m2

Unloading curvature 0125.010200

8.323

2/

1

3

dE

e

e

= 0.12952 m-1

Residual curvature r

1 = 0.15625 - 0.12952

= 0.02673 m-1

Hence r = 37.4 m

After springback, final diameter of formed curve is 74.8m

(c) Residual stress at y = 12.5mm is

250-323.8 = -73.8 MPa

(stress recovery is in the opposite direction, hence substraction)

e at y = 8mm

53

e = 12

4

3

1012

25

1082.843

= -207.2 MPa

(stress recovery is in the opposite direction, hence -ve)

Residual stress at y = 8mm is 250 - 207.2 = 42.8 MPa

(a) Loading (b) Unloading

25

25

16

Beam Mandrel

-73.8

+42.8

+y - e

(c) Residual Stress

+Y

54

1.3 TORSION OF CIRCULAR BARS

ASSUMPTIONS An ideal stress-strain relationship for the material as

shown below

A plane cross-section of the shaft remains plane when in

the plastic state;

The radius remains straight after torque is applied.

YY

Y

55

Recall that when a torque T is applied on a circular bar, the

torque is given by:

RR

drrrdrrT0

2

022 (3.1)

56

The shear stress distribution is given by:

From similar triangles,

R

r

R

rwhen

R

r

Y

YYR

R

,

57

If the torque is further increased, the yielded region will

extend inwards resulting in an outer plastic annulus and an

inner elastic core.

If the torque is increased, yielding will first occur at the outer most fibres.

Let TY be the torque at which the bar first reaches

yield: From Eq. (3.1)

1.4 FIRST YIELD TORQUE

3

0

4

0 0

22

2

4222

R

R

rdrr

R

rdrrT

Y

R

YR R Y

Y

(4.1)

1.5 ELASTIC-PLASTIC TORQUE

58

The torque in the bar is then :

T = 2 2 2r dr r r dro

c

Yc

R

(Elastic) (Plastic)

T = Y

r

cr dr r dr

o

c

Yc

R

2 2

2 2

1232

3

2

2c

2c

2T

33

334

2

0

3

cR

cRc

drrdrr

Y

YY

R

cYc

Y

(5.1)

59

1.6 FULLY PLASTIC TORQUE

If the torque is further increased until the bar is fully plastic

The torque TP is given by:

3

0

2

3

2

2

R

drrT

Y

RYP

(6.1)

Comparing with the first yield torque (Eq. 4.1) we have

i.e. the plastic torque is 33% greater than the first yield torque.

33.13

4

2

3

2

3

3

r

R

T

T

Y

Y

Y

P

60

Example 1.6-1

The tubular shaft as shown in the figure is made of an aluminum

alloy that is assumed to have a an elastic plastic stress-strain -

relationship as shown in the figure.

Determine

(a) the torque applied to the shaft when first yield occurs,

(b) the fully plastic torque that can be applied to the shaft,

(c) the shear strain at the outer radius when yield first occurs at

the inner radius.

61

First yield plastic torque:

We have

kNmT

T

J

cT

Y

Y

YY

42.3

03.005.02

05.01020

44

6

Fully plastic torque:

From Eq. (6.1)

kNm

r

drr

drrTR

YP

1.4

31066.125

10202

2

05.0

03.0

36

05.0

03.0

26

0

2

62

Shear strain at the outer radius:

Since the shear strain remains linearly distributed over the

x-section, the plastic strain at the outer fibres is:


33 10477.030

5010286.0

30

50

30

50

mm

mm

mm

mm

mm

mm

rR

r

R

Plastic shear stress distribution Plastic shear strain distribution

50 mm 30 mm

Plastic shear strain distribution

63

Example 1.6-2

A solid circular shaft has a radius of 20 mm and length of 1.5

m. The material has an elastic plastic stress-strain -

relationship as shown in the figure. Determine the torque

needed to twist the shaft by an angle of 0.6 rad.

Given that:

angle of twist = 0.6, R = 0.02 m, L = 1.5 m

64

We have

rad

LR

008.0

5.102.06.0

L

65


mm

m

r

R

r

4

004.0

02.0008.0

0016.0

008.0

0016.0

66

Using Eq. (5.1)

kNm

cRT Y

25.1

12

0.004

3

0.02 10752

12

3 2

336

33

c

67

1.7 RESIDUAL SHEAR STRESS DISTRIBUTION

Case (a)

Consider a bar subjected to a fully plastic torque TP

Case (b)

If the bar in case (a) is unloaded, the bar unloads elastically.

R

68

For elastic unloading we have:

J

RTPR

From Eq. (6.1)

3

3

2RT YP

Y

Y

R

R

RR

3

4

2

3

2

4

3

Unloading from fully plastic torque is equivalent to superposing

case (a) and case (b).

The resultant is:

+ =

R

R- Y

69

Example 1.7-1

A solid shaft diameter 50 mm is twisted so that an elastic

core of diameter 20 mm remains. Assuming elastic-

perfectly plastic behaviour, calculate: (a) applied torque,

(b) residual stress distribution on unloading, (c) residual

twist.

Assume y = 150 MN/m2 G = 77x103 MN/m2

(a) Torque TR c

p y

2

3 12

3 3

2 150

0 025

3

0 01

12

3 3

. .

= 4.83 kNm

(b) Elastic unloading gives

205.0

32

05.0104.83

2

D

4

3

e

J

Tp

= 196.8 MN/m2

70

r at r = 10mm is 4.83 103

0 01

320 054

.

.

= 78.7 MPa

Residual at r = 25 mm is 150 - 196.8 = -46.8 MPa

Residual at r = 10 mm is 150 - 78.7 = 71.3 MPa

Residual Shear Stress Distribution

At r = 10 mm, = 71.3 MPa

At r = 25 mm, = -46.8 MPa e- Y

71

(c) Residual Twist:

Recall

GrLor

Gr

L

GJ

Lr

J

r

JTei

Tr

G

LT

..J

andJ

Initial twist:

01.01077

10150

01.0

9

6

GL

y

= 0.195 rad/m

Elastic unloading:

025.01077

10196.8

025.0

9

6

GL

e

= 0.102 rad/m

Residual twist = Initial twist Twist due to elastic unloading

= 0.195 - 0.102 = 0.093 rad/m

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Ch1 Beams Bending