stress transforamation 5

8/20/2019 Stress Transforamation 5

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1

P4 Stress and Strain Dr. A.B. ZavatskyHT08

Lecture 5

Plane Stress

Transformation Equations

Stress elements and plane stress.

Stresses on inclined sections.

Transformation equations.

Principal stresses, angles, and planes.

Maximum shear stress.



2

Normal and shear stresses on inclined sections

To obtain a complete picture of the stresses in a bar, we must

consider the stresses acting on an “inclined” (as opposed to a

“normal”) section through the bar.

PP

Normal sectionInclined section

Because the stresses are the same throughout the entire bar, the

stresses on the sections are uniformly distributed.

PInclinedsection

PNormalsection



3

P

V

N

P

θ

x

y

The force P can be resolved into components:Normal force N perpendicular to the inclined plane, N = P cos θShear force V tangential to the inclined plane V = P sin θ

If we know the areas on which the forces act, we can calculate

the associated stresses.

x

y

area A

area A( / cos )θ

σθ

x

y

τθ

area A

area A( / cos )θ



4

( ) 2cos12

cos

cos cos

cos

2

x

2

θ σ

θ σ σ

θ θ

θ σ

θ

θ

+==

====

x

A

P

A

P

Area

N

Area

Force

/

( ) 2sin2

cossin

cossin cos

sin

x θ σ

θ θ σ τ

θ θ θ

θ τ

θ

θ

x

A

P

A

P

Area

V

Area

Force

−=−=

−=−

=−

==/

x

yτθ

σθ

θ

σmax = σx occurs when θ = 0°

τmax = ± σx/2 occurs when θ = -/+ 45°



5

Introduction to stress elements

Stress elements are a useful way to represent stresses acting at some

point on a body. Isolate a small element and show stresses acting on all

faces. Dimensions are “infinitesimal”, but are drawn to a large scale.

x

y

z

σ =x

P A/σx

x

y

σxσ =x P A/

P σx= AP /x

y

Area A



6

Maximum stresses on a bar in tension

PP a

σx = σmax = P / Aσx

aMaximum normal stress,

Zero shear stress



7

PPab

Maximum stresses on a bar in tension

bσx/2

σx/2

τmax = σx/2

θ = 45°

Maximum shear stress,

Non-zero normal stress



8

Stress Elements and Plane Stress

When working with stress elements, keep in mind that only

one intrinsic state of stress exists at a point in a stressed

body, regardless of the orientation of the element used toportray the state of stress.

We are really just rotating axes to represent stresses in a

new coordinate system.

x

y

σx σx

θ

y1

x1



9

y

z

x

Normal stresses σx, σy, σz

(tension is positive)

σx σx

σy

σy

σz

τxy

τyx

τxzτzx

τzy

τyz

Shear stresses τxy

= τyx

,

τxz = τzx, τyz = τzy

Sign convention for τabSubscript a indicates the “face” on which the stress acts

(positive x “face” is perpendicular to the positive x direction)

Subscript b indicates the direction in which the stress acts

Strictly σx = σxx, σy = σyy, σz = σzz



10

When an element is in plane stress in the xy plane, only the x and

y faces are subjected to stresses (σz = 0 and τzx = τxz = τzy = τyz = 0).

Such an element could be located on the free surface of a body (no

stresses acting on the free surface).

σx σx

σy

σy

τxy

τyx

τxy

τyx

x

y

Plane stress element in 2D

σx, σy

τxy = τyx

σz = 0



11

Stresses on Inclined Sections

σx σx

σy

τxy

τyx

τxy

τyx

x

y

x

y

σx1

θ

y1

x1

σx1

σy1

σy1

τx y1 1

τy x1 1

τx y1 1

τy x1 1

The stress system is known in terms of coordinate system xy.

We want to find the stresses in terms of the rotated coordinate

system x1y1.

Why? A material may yield or fail at the maximum value of σ or τ. This

value may occur at some angle other than θ = 0. (Remember that for uni-

axial tension the maximum shear stress occurred when θ = 45 degrees. )



12

Transformation Equations

y

σx1

x

θ

y1

x1

σx

σy

τx y1 1

τxy

τyx

θ

Stresses

y

σ θx1 A sec

x

θ

y1

x1

σx A

σ θy A tan

τ θx y1 1 A sec

τxy A

τ θyx A tan

θ

Forces

Left face has area A.

Bottom face has area A tan θ.

Inclined face has area A sec θ.

Forces can be found from stresses ifthe area on which the stresses act is

known. Force components can then

be summed.



13

y

σ θx1 A secx

θ

y1

x1

σx A

σ θy A tan

τ θx y1 1 A sec

τxy Aτ θyx A tan

θ

( ) ( ) ( ) ( ) 0costansintansincossec

:directionin theforcesSum

1

1

=−−−− θ θ τ θ θ σ θ τ θ σ θ A A A A Aσ

x

yx y xy x x

( ) ( ) ( ) ( ) 0sintancostancossinsec

:directionin theforcesSum

11

1

=−−−+ θ θ τ θ θ σ θ τ θ σ θ τ A A A A A

y

yx y xy x y x

( ) ( )θ θ τ θ θ σ σ τ

θ θ τ θ σ θ σ σ

τ τ

22

11

22

1

sincoscossin

cossin2sincos

:givesgsimplifyinand Using

−+−−=

++=

=

xy y x y x

xy y x x

yx xy



14

2

2sincossin

2

2cos1sin

2

2cos1cos

identitiestrictrigonomefollowingtheUsing22 θ

θ θ θ

θ θ

θ =−

=+

=

θ τ θ σ σ σ σ

σ

θ θ

2sin2cos

22

:for90substituteface,on thestressesFor

1

1

xy

y x y x

y

y

−−

−+

=

+ °

y x y x

y x

σ σ σ σ +=+ 11

11 :givesand forsexpressiontheSumming

( )θ τ θ

σ σ τ


σ

2cos2sin

2

2sin2cos22

:stress planeforequationsationtransformthegives

11

1

xy

y x

y x

xy

y x y x

x

+−

−=

+−

++

=

HLT, page 108

Can be used to find σy1, instead of eqn above.



15

Example: The state of plane stress at a point is represented by the

stress element below. Determine the stresses acting on an element

oriented 30° clockwise with respect to the original element.

80 MPa 80 MPa

50 MPa

x

y

50 MPa

25 MPa

Define the stresses in terms of the

established sign convention:σx = -80 MPa σy = 50 MPa

τxy = -25 MPa

We need to find σx1

, σy1

, and

τx1y1 when θ = -30°.

Substitute numerical values into the transformation equations:

( ) ( ) ( ) MPa9.25302sin25302cos2

5080

2

5080

2sin2cos22

1

1

−=°−−+°−−−

++−

=

+−++=

x

xy

y x y x

x

σ


σ



16

( )( ) ( ) ( ) MPa8.68302cos25302sin

2

5080

2cos2sin2

11

11

−=°−−+°−−−

−=

+

−

−=

y x

xy

y x

y x

τ

θ τ θ

σ σ

τ

( ) ( ) ( ) MPa15.4302sin25302cos2

5080

2

5080

2sin2cos22

1

1

−=°−−−°−−−

−+−

=

−

−

−

+

=

y

xy

y x y x

y

σ


σ

Note that σy1 could also be obtained

(a) by substituting +60° into the

equation for σx1 or (b) by using the

equation σx

+ σy

= σx1

+ σy1

+60°

25.8 MPa

25.8 MPa

4.15 MPa

x

y

4.15 MPa

68.8 MPa

x1

y1

-30o

(from Hibbeler, Ex. 15.2)



17

Principal StressesThe maximum and minimum normal stresses (σ1 and σ2) are

known as the principal stresses. To find the principal stresses,

we must differentiate the transformation equations.

( ) ( )

( )

y x

xy

p

xy y x x

xy

y x x

xy

y x y x

x

d

d

d

d

σ σ

τ θ

θ τ θ σ σ θ

σ

θ τ θ σ σ

θ

σ


σ

−=

=+−−=

=+−−

=

+−++=

22tan

02cos22sin

02cos22sin22

2sin2cos22

1

1

1

θp are principal angles associated with

the principal stresses (HLT, page 108)

There are two values of 2θp in the range 0-360°, with values differing by 180°.

There are two values of θp in the range 0-180°, with values differing by 90°.

So, the planes on which the principal stresses act are mutually perpendicular.



18


σ

σ σ

τ θ

2sin2cos22

22tan

1 xy

y x y x

x

y x

xy

p

+−

++

=

−=

We can now solve for the principal stresses by substituting for θp

in the stress transformation equation for σx1. This tells us whichprincipal stress is associated with which principal angle.

2θp

τxy

(σx – σy) / 2

R

R

R

xy

p

y x

p

xy

y x

τ θ

σ σ θ

τ σ σ

=

−=

+⎟⎟

⎠

⎞⎜⎜

⎝

⎛ −=

2sin

22cos

2

2

2

2

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−+

+=

R R

xy

xy

y x y x y x τ τ

σ σ σ σ σ σ σ

2221



19

Substituting for R and re-arranging gives the larger of the two

principal stresses:

2

2

122

xy

y x y xτ

σ σ σ σ σ +⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+

+=

To find the smaller principal stress, use σ1 + σ2 = σx + σy.

2

2

1222

xy

y x y x

y x τ σ σ σ σ

σ σ σ σ +⎟⎟

⎠

⎞⎜⎜

⎝

⎛ −−

+=−+=

These equations can be combined to give:

2

2

2,1

22 xy

y x y xτ

σ σ σ σ σ +

⎟⎟ ⎠

⎞

⎜⎜⎝

⎛ −±

+=

Principal stresses

(HLT page 108)

To find out which principal stress goes with which principal angle,

we could use the equations for sin θp and cos θp or for σx1.



20

The planes on which the principal stresses act are called theprincipal planes. What shear stresses act on the principal planes?

Solving either equation gives the same expression for tan 2θp

Hence, the shear stresses are zero on the principal planes.

( )

( )

( ) 02cos22sin

02cos22sin

02cos2sin2

0and 0forequationstheCompare

1

11

111

=+−−=

=+−−

=+−−=

==

θ τ θ σ σ θ

σ

θ τ θ σ σ

θ τ θ σ σ

τ

θ σ τ

xy y x x

xy y x

xy

y x

y x

x y x

d

d

d d



21

Principal Stresses

2

2

2,122

xy

y x y xτ

σ σ σ σ σ +⎟⎟

⎠

⎞⎜⎜⎝

⎛ −±

+=

y x

xy

pσ σ

τ θ

−=

22tan

Principal Angles defining the Principal Planes

x

y

σ1

θp1

σ1

σ2

σ2

θp2

σy

σx σx

σy

τxy

τyx

τxy

τyx

x

y



22


stress element below. Determine the principal stresses and draw thecorresponding stress element.

80 MPa 80 MPa

50 MPa

x

y

50 MPa

25 MPa

Define the stresses in terms of theestablished sign convention:

σx = -80 MPa σy = 50 MPa

τxy = -25 MPa

( )

MPa6.84MPa6.54

6.6915252

5080

2

5080

22

21

2

2

2,1

2

2

2,1

−==

±−=−+⎟ ⎠

⎞⎜⎝

⎛ −−±

+−=

+⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −±

+=

σ σ

σ

τ σ σ σ σ

σ xy

y x y x



23

( )

°°=

°+°=

=−−

−=

−=

5.100,5.10

18021.0and 0.212

3846.05080

2522tan

2

2tan

p

p

p

y x

xy

p

θ

θ

θ

σ σ

τ θ

( ) ( ) ( ) MPa6.845.102sin255.102cos2

5080

2

5080

2sin2cos22

1

1

−=°−+°−−

++−

=

+−

++

=

x

xy

y x y x

x

σ


σ

But we must check which angle goes with which principal stress.

σ1 = 54.6 MPa with θp1 = 100.5°

σ2 = -84.6 MPa with θp2 = 10.5°

84.6 MPa

84.6 MPa

54.6 MPa

x

y

54.6 MPa

10.5o

100.5o



24

The two principal stresses determined so far are the principal

stresses in the xy plane. But … remember that the stress elementis 3D, so there are always three principal stresses.

y

xσxσx

σy

σy

ττ

τ

τ

σx, σy, τxy = τyx = τ

yp

zp

xp

σ1

σ1

σ2

σ2

σ3 = 0

σ1, σ2, σ3 = 0

Usually we take σ1 > σ2 > σ3. Since principal stresses can be com-

pressive as well as tensile, σ3 could be a negative (compressive)

stress, rather than the zero stress.



25

Maximum Shear Stress

( )

⎟⎟ ⎠ ⎞⎜⎜

⎝ ⎛ −−=

=−−−=

+

−

−=

xy

y x s

xy y x

y x

xy

y x

y x

d

d

τ

σ σ θ

θ τ θ σ σ θ

τ

θ τ θ

σ σ

τ

22tan

02sin22cos

2cos2sin2

11

11

To find the maximum shear stress, we must differentiate the trans-

formation equation for shear.

There are two values of 2θs in the range 0-360°, with values differing by 180°.

There are two values of θs in the range 0-180°, with values differing by 90°.

So, the planes on which the maximum shear stresses act are mutuallyperpendicular.

Because shear stresses on perpendicular planes have equal magnitudes,

the maximum positive and negative shear stresses differ only in sign.



26

2θs

τxy

( σ x

–

σ y )

/ 2

R

( ) θ τ θ σ σ

τ

τ

σ σ θ

2cos2sin2

22tan

11 xy

y x

y x

xy

y x

s

+−−=

⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−=

We can now solve for the maximum shear stress by substituting for

θs in the stress transformation equation for τx1y1.

R

R

y x

s

xy

s

xy

y x

22sin

2cos

2

2

2

2

σ σ θ

τ θ

τ σ σ

−−=

=

+⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −=

maxmin

2

2

max2

τ τ τ σ σ

τ −=+⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −= xy

y x



27

Use equations for sin θs and cos θs or τx1y1 to find out which face

has the positive shear stress and which the negative.

What normal stresses act on the planes with maximum shear stress?

Substitute for θs in the equations for σx1 and σy1 to get

s

y x

y x σ σ σ σ σ =+==2

11

x

y

σs

θs

σs

σs

σs

τmax

τmax

τmax

τmax

σy

σx σx

σy

τxy

τyx

τxy

τyx

x

y



28


stress element below. Determine the maximum shear stresses anddraw the corresponding stress element.

80 MPa 80 MPa

50 MPa

x

y

50 MPa

25 MPa

Define the stresses in terms of theestablished sign convention:

σx = -80 MPa σy = 50 MPa

τxy = -25 MPa

( ) MPa6.69252

5080

2

2

2

max

2

2

max

=−+⎟ ⎠

⎞⎜⎝

⎛ −−=

+⎟⎟

⎠

⎞⎜⎜

⎝

⎛ −=

τ

τ σ σ

τ xy

y x

MPa152

5080

2

−=+−

=

+=

s

y x

s

σ

σ σ σ



29

( )

°°−=

°+−°−=−=⎟

⎟ ⎠

⎞

⎜⎜⎝

⎛

−

−−

−=⎟⎟

⎠

⎞

⎜⎜⎝

⎛ −

−=

5.55,5.34

1800.69and 0.692

6.2252

5080

22tan

s

s

xy

y x

s

θ

θ

τ

σ σ θ

( )( ) ( ) ( ) MPa6.695.342cos255.342sin

2

5080

2cos2sin2

11

11

−=−−+−−−

−=

+

−

−=

y x

xy

y x

y x

τ

θ τ θ σ σ

τ

But we must check which angle goes

with which shear stress.

τmax = 69.6 MPa with θsmax = 55.5°τmin = -69.6 MPa with θsmin = -34.5°

15 MPa

15 MPa

15 MPa

x

y

15 MPa

69.6 MPa

-34.5o

55.5o



30

Finally, we can ask how the principal stresses and maximum shear

stresses are related and how the principal angles and maximum

shear angles are related.

2

2

22

22

21max

max21

2

2

21

2

2

2,1

σ σ τ

τ σ σ

τ σ σ

σ σ

τ σ σ σ σ

σ

−=

=−

+⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −=−

+⎟⎟

⎠

⎞⎜⎜

⎝

⎛ −±

+=

xy

y x

xy

y x y x

p

p

s

y x

xy p

xy

y x s

θ θ

θ

σ σ

τ θ

τ

σ σ θ

2cot2tan

12tan

22tan2

2tan

−=−

=

−=⎟⎟ ⎠ ⎞⎜⎜

⎝ ⎛ −−=



31

( )

°±=°±=−

°±=−

=−

=+

=+

=+

45

45

9022

022cos

02cos2cos2sin2sin

02sin

2cos

2cos

2sin

02cot2tan

p s

p s

p s

p s

p s p s

p

p

s

s

p s

θ θ

θ θ

θ θ

θ θ

θ θ θ θ

θ

θ

θ

θ

θ θ

So, the planes of maximum shear stress (θs) occur at 45° tothe principal planes (θp).



32

80 MPa 80 MPa

50 MPa

x

y

50 MPa

25 MPa

Original Problem

σx = -80, σy = 50, τxy = 25

84.6 MPa

84.6 MPa

54.6 MPa

x

y

54.6 MPa

10.5o

100.5o

Principal Stresses

σ1 = 54.6, σ2 = 0, σ3 = -84.6

15 MPa

15 MPa

15 MPa

x

y

15 MPa

69.6 MPa

-34.5o

55.5o

Maximum Shear

τmax

= 69.6, σs= -15

°°−=

°±°=

°±=

5.55,5.34

455.10

45

s

s

p s

θ

θ

θ θ

( )

MPa6.69

26.846.54

2

max

max

21max

=

−−=

−=

τ

τ

σ σ τ

stress transforamation 5

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