ch1-Beam-Bending

8/10/2019 ch1-Beam-Bending

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Founded 1905

NATIONAL UNIVERSITY OF SINGAPORE

Department of Mechanical Engineering

MECHANICS OF

MATERIALS II

ME2114

Course Lecturer: A/P CJ TAY


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Founded 1905

SESSION 2013-14

Semester 2

ME2114 Mechanics of Mater ials II

Modular Credits: 3

Part I Lecture Notes

A/P CJ TAY


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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.
http://linc.nus.edu.sg/search/aKoch%2C+Jacobus+Johannes./akoch+jacobus+johannes/-5,-1,0,B/browsehttp://linc.nus.edu.sg/search/aKoch%2C+Jacobus+Johannes./akoch+jacobus+johannes/-5,-1,0,B/browse


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3.7. THREE-LEAD-WIRE ARRANGEMENT

3.8. SPECIAL PURPOSES GAUGES

3.9. ADVANTAGES AND DISADVANTAGES OF STRIAN

GAUGES


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CHAPTER 1

BENDING OF BEAMS

1.1 BENDING OF BEAMSSTRESS

CONCENTRATION

Bending stress formulaI

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


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For a certain beam geometry, the stress concentration values

can be from the following Figs:

Beams with shoulder fillets (Fig. F1):


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Beams with grooves (Fig. G1):


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Example 1.1-1

Beams with shoulder fillets

A steel bar with shoulder fillets as shown in the followingfigure 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 area4308.002.0

12

1mI


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Using I

MCkmax

From Fig. F1, the value of kis given by 1.45

Using M = 5 kNm, C = 0.04 m,

MPax

x340

1053.8

04.0545.1

7max

Stress distribution below the fillets :

Stresses away from the fillets are not affected by the stressconcentration and the max stress is given by:

I

MCmax i.e. k= 1

Hence MPaxx 234

1053.804.05 7max


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Stress distribution away from the fillets :


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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 thecenter 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

RDRE

D E


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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


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From Fig. F1, the value of kis 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


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1.2 BENDING OF BEAMSINELASTIC

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

u

yu

yl Rupture

Strain


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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).


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strain-hardening assumed linear from initial yield

( bi-linear stress-strain curve)

Strain-hardening is ignored

Y


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1.2.2 ASSUMPTIONS

1.That any cross-section of the beam will remain plane duringbending 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

MYMY

Y

Y

d

b


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above. The strain at outer fibres of beam may increase, but stress

will remain at Y.

Bending momentMpin beam is given by:

dyybMp

where bis width of beam at distanceyfromN.A.

For rectangular section beam (b = constant)

(Elastic portion):

Y

c y; (Plastic portion): = Y

c

o

d

c Y

Yp dyybdyyby

cM

2/2

=

2/2

0

3

232

d

c

Y

c

Y yby

c

b

=

343

41

4

22

2

22cd

bd

cbdY

Y

(2.1)

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

d/2, the bending momentMYis

M bd d bd

YY Y

2 22

4 12 6 (2.2)

Same as Eq. (a)


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(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.

ultYYP M

db

cdbM

434

222

5.1

6

42

2

d

b

db

M

M

Y

Y

Y

ult


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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

12R EY

Yd

1

R Ec

Y

R

cE

Yx I.e.

For a partially elastic-plastic beam at a distance cfrom the neutral axis, the stress

in the fibres has just reached the value Y

(note: C = d/2)


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The moment-curvature relationship for the rectangular beam is

thus linear up to a valueM = MYand beyond this point the

relationship is non-linear and asymptotic toMult as shown in thefigure below.

Curvature

Bending

Mult

My


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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


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(b) When top and bottom flanges have yielded

Plastic portion: = Y

Elastic portion:

2

1

Y y

d

Y

Y

b2

Y

Y


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Recall M b y dy

22

34

24

22

2

2

1

22

1

0

3

2

1

2

1

0

2

2

1

2

2

1

2

1

0

2

212

1

d

dY

d

Y

d d

d Y

Y

d d

d Y

Y

yb

yb

d

dyybdyybd

dyybdyybydM

Substituting the value of y and simplifying we have

M

b b d dY

+b d2

1 1

212

6 4

(c) Fully Plastic Condition

Y

Y

Y

Y

flange(plastic portion)stem


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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


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a)Determine the maximum elastic moment MY :

I of X-section:

46

233

1044.82

75.1185.12200

12

5.122002

12

2255.12

mm

I

A1A2

N

A1

A2

A

118.75 mm

56.25 mm

parallel axis theo, Ad^2


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b)Determine the fully plastic moment Mp :

kN

areaflangetopCForce Y

625

2005.122502


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kN

areawebtopCForce Y

56.351

5.1125.12250

1

kNm

kNmm

webtopandflangetoptodueMoment

94

25.5656.35175.118625

HenceMp= 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).


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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.


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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:

anything below 0,01 strain is in elastic range


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252000 N=

T= stress x area


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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 ofN.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 fromN.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=30b1/2=30

di=170

b=100

symmetrical abt y axis

A1 y1 A2 y2 A3 y3

A1 A2 A3

A1

A3

A2


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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


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

Assume Yis 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 dy0

0152

110015

2

200

210

2

200

2

11015101570

1

1

Y

YYY

y

yy

Y

Ycongruent triangle

A1 A2

A3

A4 A5


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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

2

y100

2y85

2100

2y85

2y100

2y85

210

2

y200

2

11015y101570

Y

Y

YYY

y y2 245 6975 0 y= 32.9 mm


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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 correspondingB.M.is given by M b y dy


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(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

force above N.A = force below N.A


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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

n

n

ult Y Y

Y Y

70 15 7 5 10 2

10

170

2 100 15 170 7 5

2

.

.

Hence "shape factorM

Mult

Y

1.3

n line


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

Y e

Mp

Mp

e Y

+

+

-

-

e- Y

Me Me


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eis calculated from the equilibrium condition ie.

Mp -Me= 0 (5.1)

For rectangular section, bx dFrom Eq. 2.1

3

4

22 cdbM Yp

6bygivenismomentrecoveryelasticThe

12

1 2Using

2

3

bdMM

bd

dM

eee

e

e

Hence Eq. 5.1 becomes

)2.5(2

2

3

3

4

6

06

-3

4

2

2

222

222

d

c

cdb

bd

bdcdb

Y

e

Ye

e

Y

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

when beam is fully plastic ie. c = 0. Then2

3

Y

e


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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 106mm4

Mp= 188 kNm

Applying the beam bending formula:


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MPa

I

YMee

1.285

1044.82

12510188

6

6

From similar triangles, we have

mmy

y

61.1091.285

250

125

e = 285.1 MPa

e = 285.1 MPa


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Superposition of the above stresses give the residual

stress distribution as shown below:

1.2.6 RESIDUAL CURVATURE

For a beam which has reached yield, unloading will NOTresult in

the beam returning to its original (straight) condition.

We have,

Plastic curvature -Unloading curvature = Residual curvature

ie.re

111p


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For an elastic case,Ey

1 where is the radius of theN.A.

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

stresses ie.p

1=

Ec

Y

During unloading, stresses behave elastically:

MpMp

c

+ Y


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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

1

3

2

dd

c

E

Y

r

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

SPRINGBACK

MeMed/2

- e

e


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Example 1.2-4

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

is formed into part of a circle using a round mandrel. Whatmandrel 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 103MPa.

(a) On loading: recall for the elastic caseEyEI

M

1

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

1

315625.0

008.0200x10

250=

1

mEc

Y

p

Mandrel

Beam

25

25

16

+ Y


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p= 6.4 m

Hence mandrel diameter = 12.8 m

What is the elastic-plastic moment?

Mp 2 25 45 250 225 8. .

83

2250825

2

1+

= 843.2 Nm

(b) On unloading: stresses are wholly elastic

2dMp

e

Sectional area of bar

+Y

Sectional area of bar


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=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 curvaturer

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 aty= 12.5mm is

250-323.8 = -73.8 MPa

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

e aty= 8mm


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e =12

4

3

1012

25

1082.843

= -207.2 MPa

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

Residual stress aty= 8mm is 250 - 207.2 = 42.8 MPa

(a) Loading (b) Unloading

25

25

16

BeamMandrel

-73.8

+42.8

+y - e

(c) Residual Stress

+Y


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1.3TORSION 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


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

torque is given by:

RR drrrdrrT 0

20 22 (3.1)


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The torque in the bar is then :

T= 2 2 2r dr r r dr o

c

Yc

R

(Elastic) (Plastic)

T= Y

r

cr dr r dr

o

c

Yc

R

2 22 2

1232

3

2

2c

2c

2T

33

334

2

0

3

cR

cRc

drrdrr

Y

YY

R

cYc

Y

(5.1)


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1.6 FULLY PLASTIC TORQUE

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

The torque TPis 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


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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.


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First yield plastic torque:

We have

kNmT

T

JcT

Y

Y

YY

42.3

03.005.02

05.01020

44

6

Fully plastic torque:

From Eq. (6.1)

kNm

r

drr

drrT R

YP

1.4

31066.125

10202

2

05.0

03.0

36

05.0

03.0

26

0

2


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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

50 mm 30 mm

Plastic shear strain distribution


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We have

rad

LR

008.0

5.102.06.0


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mm

m

r

R

r

4

004.0

02.0008.0

0016.0008.0

0016.0


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Using Eq. (5.1)

kNm

cRT Y

25.1

12

0.004

3

0.0210752

12

32

336

33

c


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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


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For elastic unloading we have:

J

RTPR

From Eq. (6.1)

3

3

2RT YP

Y

Y

R

R

RR

34

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


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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 0253

0 0112

3 3

. .

= 4.83 kNm

(b) Elastic unloading gives

205.032

05.0104.83

2

D

4

3

e

J

Tp

= 196.8 MN/m2


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rat r= 10mm is4.83 10 3

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 MPae- Y


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(c) Residual Twist:

Recall

GrLor

GrL

GJ

L

r

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/mResidual twist = Initial twistTwist due to elastic unloading

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