MECHANICS OF CHAPTER 7MATERIALS - Civil …naresh/teaching/ce221/L9... · · 2014-10-14MECHANICS...

MECHANICS OF MATERIALS

CHAPTER

7 Transformations of Stress and Strain


7 - 2

Transformations of Stress and Strain

IntroductionTransformation of Plane StressPrincipal StressesMaximum Shearing StressSample Problem 1Sample Problem 2Mohr’s Circle for Plane StressSample Problem 3Sample Problem 4Transformation of Plane StrainStresses in Thin-Walled Pressure Vessels


7 - 3

Introduction• General state of stress at a point represented by

6 components,

),, :(Note

stresses shearing,,

stresses normal,,

xzzxzyyzyxxy

zxyzxy

zyx

• If axis are rotated, the same state of stress is represented by a different set of components, i.e., the stress components get transformed.

• First we consider transformation of stress components, due to rotation of coordinate axes. Then we consider a similar transformation of strain components.

),, :(Notestresses shearing,,

stresses normal,,

''''''''''''

''''''

'''

zxxzyzzyxyyx

xzzyyx

zyx


7 - 4

Plane Stress

• Plane Stress - state of stress in which two faces of the cubic element are free of stress. Example,

.0nonzero, are,, xy zyzxzyx

• For example, state of plane stress occurs in thin plate subjected to forces acting in midplane of plate.

• Another example of plane stress is on free surface, i.e., unloaded point on surface.


7 - 5

Transformation of Plane Stress

sinsincossin

coscossincos0

cossinsinsin

sincoscoscos0

AA

AAAF

AA

AAAF

xyy

xyxyxy

xyy

xyxxx

• Consider equilibrium of prismatic element with faces perpendicular to x, y, and x’ axes.

cos 2 sin 22 2

cos 2 sin 22 2

sin 2 cos 22

x y x yx xy

x y x yy xy

x yx y xy

• Solving for transformed stress components,


7 - 6

Principal Stresses• Eliminating q , this yields equation of a

circle,

22

222

22

where

xyyxyx

ave

yxavex

R

R

• Principal stresses occur on principal planes on which there exist zero shearing stresses.

o

''

22

minmax,

90by separated angles twodefines :Note

0from,2

2tan

22

yxyx

xyp

xyyxyx


7 - 7

Maximum Shearing StressMaximum shearing stress occurs for avex

2

:are

45by fromoffset and 90by separated angles twodefines :Note

from,2

2tan

22

''

o

o

ave'

minmax22

max

yxaveyx

p

xxy

yxs

xyyx

stressesnormalingCorrespond

R


7 - 8

Mohr’s Circle for Plane Stress

• Used to graphically find principal stresses and planes and maximum shear stresses and planes

• For known plot points X and Yand construct circle centered at C.

xyyx ,,

22

22 xyyxyx

ave R

• Principal stresses obtained at A and B.

yx

xyp

ave R

22tan

minmax,

Direction of rotation of Ox to Oa (ie., in physical plane) is same as CX to CA(ie., in Mohr plane)


7 - 9

Mohr’s Circle for Plane Stress

• From Mohr’s circle we can find state of stress at other axes orientations.

• For state of stress at angle q with respect to the xy axes, construct a new diametral line X’Y’ at angle 2q with respect to XY.

• Coordinates of X’, Y’ are the transformed normal and shear stresses.

MECHANICS OF MATERIALSProof of Mohr’s Circle construction

7 - 10

2sin2cos2sin2

2cos

)2sin2cos2cos2(sin)22sin(

2sin2cos22

)22cos(

2sin2cos22

2sin2cos2

)2sin2sin2cos2(cos)22cos(

''

'

'

RRRR

RR

R

RRR

RR

yxxy

yxxy

pppyx

xyyxyx

pavey

xyyxyxxyyx

ave

ppavepavex

So we get same formulae as before


7 - 11

Mohr’s Circle for Plane Stress• Mohr’s circle for centric axial loading:

0, xyyx AP

(plane)element45on2

0

max A

Pxyyx

• Mohr’s circle for torsional loading:

JTc

xyyx 0(plane)element45on

0;

0

max xyyx JTc

MECHANICS OF MATERIALSPoints to note when drawing stress block.

7 - 12

plane.on which acts stress principalwhich identify not do they However,).,,(act they on which planes principal theand ve),–or ve( sense/sign

and magnitude their yield stresses principalfor Formulae

minmax p

plane.on which acts stress principalhich identify w torelations

ation transformin the angles put the , angles associated and, stresses principal findyou once So

p

minmax


7 - 13

. of ve)-or ve( sense/sign theyieldnot do they However, act.hey on which t planes and of

magnitudeonly yield stressshear maximumfor Formulae

max

smax

. of ve)-or ve( sense/signcorrect find to relationsation transformin theit put , findyou once So

max

s

sense/signcorrect with , , associated find torelations

rmation in transfo themuse and and findjust can you So

maxmaxmin

ps


7 - 14

act.hey on which t

, planes theand,,, of sense and magnitudescorrect gives which circleMohr usecan you y,Alternatel

spmaxminmax


7 - 15

Example1

For state of plane stress shown, find (a) principal planes, (b) principal stresses, (c) maximum shearing stress and corresponding normal stress.


7 - 16

Example 1

• Find element orientation for principal stresses:

1.233,1.532

333.11050

40222tan

p

yx

xyp

6.116,6.26p

• Find principal stresses:

22

22

minmax,

403020

22

xy

yxyx

MPa30MPa70

min

max

MPa10

MPa40MPa50

y

xyx


7 - 17

Example 1

MPa10MPa40MPa50

y

xyx

21050

2''

yxaveyx

• Find corresponding normal stress:

MPa20

• Find maximum shearing stress:

22

22

max

4030

2

xy

yx

MPa50max

45 ps

6.71,4.18s


7- 18

Example 2

Horizontal force P = 600 N magnitude applied to end D of lever ABD. Find (a) normal and shearing stresses on element at H having sides parallel to x and y axes, (b) principal planes and principal stresses at H.


7- 19

Example 2

Find equivalent force-couple system at center of transverse section passing through H.

Nm150m25.0N600

Nm270m45.0N600N600

xMTP

Find normal and shearing stresses at H.

421

441

m015.0m015.0Nm270

m015.0m015.0Nm150

JTcI

Mc

xy

y

MPa9.50MPa6.560 yyx

Note: tyz due to bending is zero at H


7- 20

Example 2Find principal stresses and planes.

119,0.612

8.16.560

9.50222tan

p

yx

xyp

5.59,5.30p

22

22

minmax,

9.502

6.5602

6.560

22

xy

yxyx

MPa9.29MPa5.86

min

max


7 - 21

Example 3

For state of plane stress shown, (a) construct Mohr’s circle, find (b) principal planes, (c) principal stresses, (d) maximum shearing stress and corresponding normal stress.

• Construct Mohr’s circle

MPa504030

MPa40MPa302050

MPa202

10502

22

CXR

FXCF

yxave


7 - 22

Example 3• Principal planes and stresses

5020max CAOCOA

MPa70max

5020max BCOCOB

MPa30max

1.53230402tan

p

p CFFX

6.26p


7 - 23

Example 3

• Maximum shear stress

45ps

6.71s

Rmax

MPa 50max

ave

MPa 20


7 - 24

Example 4

For state of plane stress shown, find (a) principal planes and the principal stresses, (b) stress components on element obtained by rotating given element counterclockwise through 30 degrees. • Construct Mohr’s circle

MPa524820

MPa802

601002

2222

FXCFR

yxave


7 - 25

Example 4

• Principal planes and stresses

4.672

4.220482tan

p

p CFXF

clockwise7.33 p

5280max

CAOCOA5280

max

BCOCOA

MPa132max MPa28min


7 - 26

Example 4

6.52sin52

6.52cos52806.52cos5280

6.524.6760180

XK

CLOCOLKCOCOK

yx

y

x

• Stress components after counterclockwise rotation by 30o

Points X’ and Y’ on Mohr’s circle, that correspond to stress components on rotated element, are obtained by rotating XY ccw through 602

MPa3.41

MPa6.111MPa4.48

yx

y

x


7 - 27

Transformation of Plane Strain• Plane strain - deformations of the material

take place in parallel planes and are the same in each of those planes.

• Example: Long bar subjected to uniformly distributed transverse loads (ie., normal to z-axis). State of plane strain exists in any transverse section not located too close to the ends of the bar.

• Plane strain occurs in a plate subjected along its edges to a uniformly distributed (in z-direction) load and restrained from expanding or contracting laterally by smooth, rigid and fixed supports

0

:strain of components

x zyzxzxyy


7 - 28

Transformation of Plane Strain• State of strain at point Q results in

different strain components with respect to the xy and x’y’ coordinate systems.

• We get strain transformation relations similar to those for stress transformation (see details in next two slides)

MECHANICS OF MATERIALSTransformation of Plane Strain

7 - 29

2sin2

2cos22

cossinsincos)(

'

22'

xyyxyxx

xyyxx

,sin/,cos/,)()()(Use

)(2)21()()21()()21()(

)2/cos()1()1(2

)1()()1()()1()(strains,inquadraticsneglectingrule,cosineUse

222

22'

2

22222'

2

sysxyxs

yxyxs

yxyxs

xyyxx

xyyx

yxx

2sin2

2cos22'

xyyxyxy

,in2/put 'x

MECHANICS OF MATERIALSTransformation of Plane Strain

7 - 30

2cos2

2sin22

'' xyyxyx

xyyxxOB

21

' )4/(,4/For

OyOxOB

yxOBxy

andofbisectoriswhere

2

2cos2

2sin22

)4/(

Thus,'.and'ofbisectoris'where

22system,inThus,

''

''''''

xyyxyxxOB

yxOByxOByx

OyOxOB

x'y'

''''

'''' 2but;;:Note yx

yxyyxx


7 - 31

Mohr’s Circle for Plane Strain• Since strain transformation relations are of

same form as stress transformation, for plane problems, Mohr’s circle techniques apply.

• Abscissa for center C, and radius R , are22

222

xyyxyx

ave R

• Principal axes of strain and principal strains,

RR aveave

yx

xyp

minmax

2tan

22max 2 xyyxR

• Maximum in-plane shearing strain,

MECHANICS OF MATERIALSDrawing strain block.

7 - 32

MECHANICS OF MATERIALSDrawing strain block.

7 - 33


7 - 34

Measurements of Strain: Strain Rosette

• Strain gages indicate normal strain through changes in resistance.

yxOBxy 2

• With a 45o rosette, ex and ey are measured directly. gxy is obtained indirectly with,

3332

32

3

2222

22

2

1112

12

1

cossinsincos

cossinsincos

cossinsincos

xyyx

xyyx

xyyx

• Normal and shearing strains may be obtained from normal strains in any three directions,


7 - 35

Stresses in Thin-Walled Cylindrical Pressure Vessels

• Cylindrical vessel with principal stressessq=s1 = hoop stresssx = s2 = longitudinal stress

tpr

xrpxtFz

1

1 220

• Hoop stress:

21

2

22

22

20

tpr

rprtF

x

x

• Longitudinal stress:


7- 36

Stresses in Thin-Walled Cylindrical Pressure Vessels

Maximum in-plane shearing stress:

tpr42

12)planeinmax(

Maximum out-of-plane shearing stress corresponds to a 45o rotation of the plane stress element around a longitudinal axis

tpr22max

Points A and B correspond to hoop stress, s1, and longitudinal stress, s2

Note: Plane stress Mohr’s circle is only ADBE. The remaining is due to 3-D state of stress with third principal stress being zero. 3-D state of stress is also used in finding maximum out-of-plane shearing stress by using third principal stress as zero. This is not covered in this course. Only maximum in-plane shearing stress is covered in this course.


7- 37

Stresses in Thin-Walled Spherical Pressure Vessels

Spherical pressure vessel:

tpr

rptr

2

)2(

21

22

Maximum out-of-plane shearing stress

tpr412

1max

Mohr’s circle for in-plane transformations reduces to a point

0constant

plane)-max(in

21

Note: Plane stress Mohr’s circle is only AB, i.e., circle with zero radius, i.e., a point. The dotted circle is due to 3-D state of stress with third principal stress being zero. 3-D state of stress is also used in finding maximum out-of-plane shearing stress, by using third principal stress as zero. This is not covered in this course. Only maximum in-plane shearing stress is covered in this course.


8- 38

Principle Stresses in a Beam

For beam subjected to transverse loading

ItVQ

ItVQ

IMc

IMy

mxy

mx

Can the maximum normal stress within the cross-section be larger than

IMc

m


8- 39

Principle Stresses in a Rectangular section Beam


8- 40

Principle Stresses in a Beam

Cross-section shape results in large values of txy near the surface where sx is also large.

smax may be greater than sm (since tb is large)


8- 41

Example 5

160-kN force applied at tip of W200x52 rolled-steel beam.

Neglect effects stress concentrations at fillets, determine whether normal stresses at section A-A’ satisfy sall =150 MPa


8- 42

Example 5Determine shear and bending moment in

Section A-A’

kN160

m-kN60m375.0kN160

A

AVM

Calculate normal stress at top surface and at flange-web junction.

MPa9.102mm103mm4.90MPa2.117

MPa2.117m10512

mkN6036

cyσ

SM

bab

Aa


8- 43

Example 5

Calculate principal stresses at flange-web junction

MPa 150MPa9.159

5.952

9.1022

9.102 22

2221

21

max

bbb

Design specification not satisfied.

Calculate shear stress at flange-web junction.

MPa5.95m0079.0m107.52

m106.248kN160

m106.248

mm106.2487.966.12204

46

36

36

33

ItQV

Q

Ab


8- 44

Example 6

sall = 165 Mpa, tall = 100 Mpa. Select wide-flange beam to be used.


8- 45

Example 6

Calculate required section modulus, select appropriate beam section.

section beam 920select W53

mm1959Pa10651Nm10323 3

6

3max

min

all

MS

Maximum shear and bending moment from SFD, BMD.

kN4.193

kN 9.54withkNm2.323

max

max

V

VM

Reactions at A and D.

kN5.1840

kN5.2650

AD

DA

RM

RM

323.2


8- 46

Example 6Find maximum shearing stress.

Assume uniform shearing stress in web (conservative, see shallow parabolic variation, slide 10, shear stresses chapter)

MPa100MPa 6.35m 105436.6N 104.193

26-

3max

max

webA

V

Find maximum normal stress.

MPa1.10m106.4365

N 54900

MPa147mm5.266mm9.250MPa1.156

MPa1.156m1007.2

Nm323200

26b

36max

web

bab

a

AV

cyσ

SM

MPa165MPa1.147

MPa1.102MPa147

2MPa147 2

2

max


8- 47

Design of Transmission Shaft

If power is transferred to and from shaft by gears or sprocket wheels, the shaft is subjected to transverse loading (due to forces in mating gears) as well as shear loading (due to torque from these forces).

Normal stresses due to transverse loads may be large, should be included in determination of maximum shearing stress.

Shearing stresses due to transverse loads are usually small and their contribution to maximum shear stress may be neglected.


8- 48

Design of a Transmission Shaft

Shaft section requirement,

all

TM

cJ

max

22

min

Maximum shearing stress,

22max

222

2

max

2 section,-crossannular or circular afor

22

TMJc

JI

JTc

IMc

mm

At any section,

JTc

MMMI

Mc

m

zym

222where


8- 49

Example 7

Shaft rotates at 480 rpm, transmits 30 kW from motor to gears G and H; 20 kW is taken off at gear G and 10 kW at gear H. sall = 50 Mpa. Find smallest permissible diameter for shaft.


8- 50

Example 7Find gear torques and corresponding tangential

forces. Use,

kN49.2mN199Hz82

kW10

kN63.6mN398Hz82

kW20

kN73.3m0.16

mN597

mN597Hz82

kW302

DD

CC

E

EE

E

FT

FT

rTF

fPT

Find reactions at A and B.

kN90.2kN80.2

kN22.6kN932.0

zy

zy

BB

AA

HHHDDD

GGGCCC

MMMEEE

TPPTTPPTTPPT


8- 51

Example 7Identify critical shaft section from torque and bending

moment diagrams. D comes out as critical section.

mN1357

5973731160 222max

222

TMM zy

At C it is 1319 N.m. At E it is 1003 N.m


8- 52

Example 7Find minimum allowable shaft diameter.

36

222

m1014.27MPa50

mN 1357

all

zy TMMcJ

mm 7.512 cd

m25.85m02585.0

m1014.272

363

c

ccJ

For solid circular shaft,


8- 53

Stresses Under Combined Loadings

Wish to find stresses in slender structural members subjected to arbitrary loadings.

Pass section through points of interest. Determine force-couple system at centroid of section required to maintain equilibrium.

System of internal forces consist of three force components and three couple vectors.

Determine stress distribution by applying the superposition principle.


8- 54

Stresses Under Combined LoadingsAxial force and bending moments yield

normal stresses.Shear forces and twisting couple yield

shearing stresses.

Find principal stresses, maximum shearing stress.


8- 55

Example 8

Find principle stresses, principal planes, maximum shearing stress, at H.


8- 56

Example 8Internal forces in Section EFG.

mkN3m100.0kN300

mkN5.8

m200.0kN75m130.0kN50

kN75kN50kN 30

zy

x

zx

MM

M

VPV

Section properties,

463121

463121

23

m10747.0m040.0m140.0

m1015.9m140.0m040.0

m106.5m140.0m040.0

z

x

I

I

A


8- 57

Example 8Normal stress at H.

MPa66.0MPa2.233.8093.8m1015.9

m025.0mkN5.8m10747.0

m020.0mkN3m105.6

kN50

46

4623-

x

x

z

zy I

bMI

aMAP

Shearing stresses at H.

0MPa52.17

m040.0m1015.9m105.85kN75

m105.85

m0475.0m045.0m040.0

46

36

3611

yx

x

zyz tI

QV

yAQ

Note: 2-D (i.e., plane) state of stress only at H. In interior it is 3-D state of stress


8- 58

Example 8Principal stresses and maximum, shearing

stress, principal planes.

98.13

96.2720.33

52.172tan

MPa4.74.370.33

MPa4.704.370.33

MPa4.3752.170.33

pp

min

max

22max

p

CDCY

ROC

ROC

R

98.13

MPa4.7

MPa4.70

MPa4.37

min

max

max

p

MECHANICS OF CHAPTER 7MATERIALS - Civil …naresh/teaching/ce221/L9... · · 2014-10-14MECHANICS...

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