Chapter 6 Transformation

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Chapter 6

.

Materials for this chapter are Originally taken from :

Ferdinand P. Beer, E Russell Johnston, Jr, John T. Dewolf and David Mazurek: Mechanics of Materials ,th

FOR INTERNAL USE ONLY

Introduction

• The most general state of stress at a point may

be re resented b 6 com onents,

stressesshearing,,

stressesnormal,,

zx yz xy

z y x

τ τ τ

σ σ σ

),, :(Note xz zx zy yz yx xy τ τ τ τ τ τ ===

• ame state o stress s represente y a

different set of components if axes are rotated.

• The first part of the chapter is concerned with

how the components of stress are transformed

under a rotation of the coordinate axes. The

second part of the chapter is devoted to a

components of strain.

7- 2



Introduction

• Plane Stress - state of stress in which two faces of

t e cu c e ement are ree o stress. For t e

illustrated example, the state of stress is defined by

.,, xy zy zx z y x

• State of plane stress occurs in a thin plate subjected

to forces acting in the midplane of the plate.

• State of plane stress also occurs on the free surface

of a structural element or machine component, i.e.,

at any point of the surface not subjected to an

7- 3

.

Transformation of Plane Stress

• Consider the conditions for equilibrium of a

rismatic element with faces er endicular to

( ) ( ) θ θ τ θ θ σ σ sincoscoscos0 A A AF xy x x x Δ−Δ−Δ==∑ ′′

the x, y, and x’ axes.

( ) ( )

( ) ( ) θ θ τ θ θ σ τ

θ θ τ θ θ σ

coscossincos0

cossinsinsin

A A AF

A A

xy x y x y

xy y

Δ−Δ+Δ==∑

Δ−Δ−

′′′

( ) ( ) θ θ τ θ θ σ sinsincossin A A xy y Δ+Δ−

θ τ θ σ σ σ σ

σ 2sin2cos y x y x

+−

++

=′

• The equations may be rewritten to yield

θ τ θ σ σ σ σ

σ 2sin2cos

22

22

xy y x y x

y −−

−+

=′

θ τ θ σ σ

τ 2cos2sin2

xy y x

y x +−

−=′′

7- 4



Principal Stresses

• The previous equations are combined to

ield arametric e uations for a circle,

( ) 222 y xave x Rτ σ σ =+− ′′′

22

xy y x y x

ave R τ σ σ σ σ

σ +⎟ ⎞

⎜⎛ −

=+

=

•

planes of stress with zero shearing stresses.

2− 2

minmax,

2

22xy

y x y x

τ

τ σ +

⎟

⎟

⎠⎜

⎜

⎝

±=

o90 bse aratedan lestwodefines: Note

2tan y x

pσ σ

θ −

=

7- 5

Maximum Shearing Stress

Maximum shearing stress occurs for ave x σ σ =′

2

22

max xy y x

R τ σ σ

τ +⎟⎟ ⎞

⎜⎜⎛ −

==

22tan xy

y x

s τ

σ σ

θ

−−=

45 byfromoffset

and90 byseparatedanglestwodefines: Note

o

o

θ

2

y xave

σ σ σ σ

+==′

7- 6



Example 6.1

SOLUTION:

• n e e emen or en a on or e pr nc pa

stresses from

xτ 2

y x p σ σ −=an

• Determine the rinci al stresses from

22

minmax,22

xy y x y x

τ σ σ σ σ

σ +⎟⎟ ⎞

⎜⎜⎛ −

±+

=Fig. 7.13

,

determine (a) the principal planes,

(b) the principal stresses, (c) the• Calculate the maximum shearing stress with

maximum shearing stress and the

corresponding normal stress.

2max 2 xy

y x

τ

σ σ

τ +⎟⎟ ⎠⎜⎜⎝

−

=

2

y x σ σ σ

+=′

7- 7

Example 6.1

SOLUTION:

• n e e emen or en a on or e pr nc pa

stresses from

( )+ 4022 xτ

( )

°°=

=−−

=−

=

1.233,1.532

.1050

an

p

y x p

θ

σ σ

°°= 6.116,6.26 pθ MPa40MPa50 +=+= xy x τ σ

Fig. 7.13

• e erm ne e pr nc pa s resses rom

22

⎞⎛ −+ y x y x σ σ σ σ

MPa10−= xσ

( ) ( )22

m nmax,

403020

22

+±=

⎠⎝ xy

MPa30

MPa70max

−=

=

σ

σ

7- 8

Fig. 7.14



Example 6.1

• Calculate the maximum shearing stress with

22

max2

+⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −= xy

y xτ

σ σ τ

( ) ( )22 4030 +=

MPa50max =τ

45−= ps θ θ MPa40MPa50 +=+= xy x τ σ

Fig. 7.13

°°−= 6.71,4.18sθ MPa10−= xσ

1050 −=

+==′ y x

ave

σ σ σ σ

• e correspon ng norma s ress s

MPa20=′σ

7- 9

Fig. 7.16

Example 6.2

A plane stress system is as shown SOLUTION:

n t e agram e ow. Determ ne

the direct and shear stress acting

on the lane P-P.

fordirectandshearstress:

σ σ σ σ + −cos s n

2 2

10 2 10 2 cos2 75 8 sin 2 75

N xy

o o

σ τ = + +

− += + + −

( )

2 2

5.196 compressive MPa= −

sin 2 cos22

x y

xy

σ σ τ θ τ θ

−= − +

)sin 2 75 8 cos2 752

4.928

o o

MPa

= − + −

=

7- 10



Mohr’s Circle for Plane Stress

• With the physical significance of Mohr’s circle

for lane stress established, it ma be a lied

with simple geometric considerations. Critical

values are estimated graphically or calculated.

• For a known state of plane stress

lot the oints X and Y and construct the xy y x τ σ σ ,,

circle centered at C .

22

y x y x σ σ σ σ ⎞⎛ −+

22xyave

⎠⎝ ==

• e pr nc pa s resses are o a ne a an .

ave Rσ σ ±=minmax,

y x

xy p

σ σ θ

−=2tan

The direction of rotation of Ox to Oa is

7- 11

the same as CX to CA.


• With Mohr’s circle uniquely defined, the state

o stress at ot er axes or entat ons may e

depicted.

• For the state of stress at an angle θ with

res ect to the xy axes, construct a new

diameter X’Y’ at an angle 2θ with respect to

XY .

• Normal and shear stresses are obtained

’ ’ .

7- 12




• Mohr’s circle for centric axial loading:

0, === xy y x A

τ σ σ A

xy y x2

=== τ σ σ

• ’

Tc Tc

7- 13

J xy y x === τ σ σ === xy y x

J τ σ σ

Example 6.3

Fig. 7.13

,

(a) construct Mohr’s circle, determine

(b) the principal planes, (c) the SOLUTION:

principal stresses, (d) the maximum

shearing stress and the corresponding• Construction of Mohr’s circle

( ) ( )1050−++

y xσ σ

.

MPa40MPa302050

22

==−= FX CF

ave

7- 14

( ) ( ) MPa504030 22 =+== CX R



Example 6.3

• Principal planes and stresses

5020max +=+== CAOC OAσ

MPa70max =σ

5020min −=−== BC OC OBσ

MPa30min −=σ

==40

2tan pFX

θ

°= 1.532 pθ

°= 6.26θ

7- 15

Example 6.3

• Maximum shear stress

°+= 45 ps θ θ

°= 6.71sθ

R=maxτ

MPa50max =τ

aveσ σ =′

MPa20=′σ

7- 16



Sample Problem 6.4

For the state of stress shown,

e erm ne a e pr nc pa p anes

and the principal stresses, (b) the

stress com onents exerted on the

element obtained by rotating thegiven element counterclockwise

SOLUTION:

• Construct Mohr’s circlet roug egrees.

MPa802

60100

2=

+=

+=

y xave

σ σ σ

7- 17

( ) ( ) ( ) ( ) MPa524820 =+=+= FX CF R

Sample Problem 6.4

• Principal planes and stresses

48 XF max +== CAOC OAσ max −== BC OC OAσ

°=

===

4.672

.20

an

p

pCF

θ 5280 += 5280 −=

MPa132max +=σ MPa28min +=σ

7- 18

clockwise7.33 °= pθ



Sample Problem 6.4

°−=−==

°=°−°−°=

′ 6.52cos5280

6.524.6760180

KC OC OK xσ

φ • Stress components after rotation by 30o

’ ’ ’

°=′=

°+=+==

′′

′

6.52sin52

6.52cos5280

X K

CLOC OL

y x

y

τ

σ o n s an on o r s c rc e a

correspond to stress components on therotated element are obtained b rotatin

XY counterclockwise through °= 602θ

MPa6.111

MPa4.48

+=

+=

′

′

y

x

σ

σ

7- 19

MPa3.41=′′ y xτ

Supplementary Problem 6.1

1. At a point in the structural member, the stresses

are represented as in figure below. Find ;

(a) the magnitude and orientation of the

principal stresses and

(b) the magnitude and orientation of the

normal stresses.

2. For the state of plane stress shown, determine (a)

the principal planes and the principal stresses, (b)

the stress components exerted on the element

obtained by rotating the given element

counterclockwise via 30o.

7- 20



Supplementary Problem 6.1

3. A structural member is sub ected to two set of loadin s. Each se aratel roduce the

.

stress conditions at a point A as shown in Figure Q1(a) and FigureQ1(b).

a) Determining σx, σy and τxy for the stress conditions.

b) By continuing the effect of both stress conditions Q1 (a) and Q2(b), determine

the principal stresses and their orientations with respect to the x-axis and also

e magn u e o e max mum s ear ng s ress.

7- 21

General State of Stress

• Consider the general 3D state of stress at a point and

the transformation of stress from element rotation

• State of stress at Q defined by: zx yz xy z y x τ τ τ σ σ σ ,,,,,

• Consider tetrahedron with face perpendicular to the

line QN with direction cosines: z y x λ λ λ ,,

• The requirement leads to,∑ = 0nF

222=

x z zx z y yz y x xy

z z y y x xn

λ λ τ λ λ τ λ λ τ 222 +++

• Form of equation guarantees that an element

orientation can be found such that

ccbbaan λ σ λ σ λ σ σ ++=

These are the principal axes and principal planes

7- 22

.



Application of Mohr’s Circle to the ThreeDimensional Analysis of Stress

• The three circles represent the• Transformation of stress for an element

normal and shearing stresses for

rotation around each principal axis.

rotated around a principal axis may be

represented by Mohr’s circle.

• Points A, B, and C represent the

principal stresses on the principal planes• Radius of the largest circle yields the

maximum shearing stress.

7- 23

minmaxmax

2σ σ τ −=

Application of Mohr’s Circle to the Three-Dimensional Analysis of Stress

,

perpendicular to the plane of stress is a

principal axis (shearing stress equal zero).

• If the points A and B (representing the

principal planes) are on opposite sides of

a) the corresponding principal stresses

the origin, then

are the maximum and minimum

normal stresses for the element

b) the maximum shearing stress for the

element is equal to the maximum “in-

”

c) planes of maximum shearing stresso .

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Chapter 6 Transformation