1P4 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.

2Normal 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, thestresses on the sections are uniformly distributed.

P Inclined section PNormal section

3PV

N

P

x

y

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

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

x

y

areaA

area A( / cos )

x

y

area

A

area A( / cos )

4( ) 2 cos12

cos

cos cos cos

2x

2

+==

====x

AP

AP

AreaN

AreaForce

/

( ) 2 sin2

cos sin

cos sin cos sin

x

x

AP

AP

AreaV

AreaForce

==

====/

x

y

max = x occurs when = 0

max = x/2 occurs when = -/+ 45

5Introduction to stress elementsStress 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

= x P A/xx

y

x = x P A/

P x = AP / x

y

Area A

6Maximum stresses on a bar in tension

PPa

x = max = P / Ax

aMaximum normal stress,

Zero shear stress

7PPab

Maximum stresses on a bar in tension

bx/2

x/2

max = x/2

= 45

Maximum shear stress,Non-zero normal stress

8Stress 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 to portray the state of stress.

We are really just rotating axes to represent stresses in a new coordinate system.

x

y

x x

y1x1

9y

z

x

Normal stresses x, y, z(tension is positive)

x x

y

y

z

xy

yx

xzzxzyyz

Shear stresses xy = yx, xz = zx, yz = zy

Sign convention for abSubscript 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 actsStrictly x = xx, y = yy, z = zz

10

When an element is in plane stress in the xy plane, only the x and y faces are subjected to stresses (z = 0 and zx = xz = zy = yz = 0).Such an element could be located on the free surface of a body (no stresses acting on the free surface).

x x

y

y

xyyx

xyyx

xy

Plane stress element in 2D

x, yxy = yxz = 0

11

Stresses on Inclined Sections

x x

y

xyyx

xyyx

xy

x

y

x1

y1

x1

x1

y1

y1

x y1 1 y x1 1

x y1 1 y 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 or . This value may occur at some angle other than = 0. (Remember that for uni-axial tension the maximum shear stress occurred when = 45 degrees. )

12

Transformation Equations

y

x1 x

y1x1

x

y

x y1 1

xyyx

Stresses

y

x1 A sec x

y1x1

xA

y A tan

x y1 1 A sec

xy A yx A tan

Forces

Left face has area A.

Bottom face has area A tan .Inclined face has area A sec .

Forces can be found from stresses if the area on which the stresses act is known. Force components can then be summed.

13

y

x1 A sec x

y1x1

xA

y A tan

x y1 1 A sec

xy A yx A tan

( ) ( ) ( ) ( ) 0costansintansincossec :direction in the forces Sum1 1 = AAAAAx

yxyxyxx

( ) ( ) ( ) ( ) 0sintancostancossinsec :direction in the forces Sum11 1 =+ AAAAAy

yxyxyxyx

( ) ( )

2211

221

sincoscossin

cossin2sincos

:gives gsimplifyin and Using

+=++=

=

xyyxyx

xyyxx

yxxy

14

22sincossin

22cos1sin

22cos1cos

identities tric trigonomefollowing theUsing

22 ==+=

2sin2cos22

:for 90 substitute face, on the stressesFor

1

1

xyyxyx

y

y

+=+

yxyx

yx +=+ 11

11 :gives and for sexpression theSumming

( )

2cos2sin2

2sin2cos22

:stress planefor equationsation transform thegives

11

1

xyyx

yx

xyyxyx

x

+=

+++=HLT, page 108

Can be used to find y1, 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

xy

50 MPa

25 MPa

Define the stresses in terms of the established sign convention:x = -80 MPa y = 50 MPaxy = -25 MPaWe need to find x1, y1, and x1y1 when = -30.

Substitute numerical values into the transformation equations:

( ) ( ) ( ) MPa9.25302sin25302cos2

50802

5080

2sin2cos22

1

1

=+++=

+++=

x

xyyxyx

x

16

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

25080

2cos2sin2

11

11

=+=

+=

yx

xyyx

yx

( ) ( ) ( ) MPa15.4302sin25302cos

25080

25080

2sin2cos22

1

1

=+=

+=

y

xyyxyx

y

Note that y1 could also be obtained (a) by substituting +60 into the equation for x1 or (b) by using the equation x + y = x1 + y1+60

25.8 MPa

25.8 MPa

4.15 MPa

x

y

4.15 MPa68.8 MPa

x1

y1

-30o

(from Hibbeler, Ex. 15.2)

17

Principal StressesThe maximum and minimum normal stresses (1 and 2) are known as the principal stresses. To find the principal stresses, we must differentiate the transformation equations.

( ) ( )( )

yx

xyp

xyyxx

xyyxx

xyyxyx

x

dddd

=

=+=

=+=

+++=

22tan

02cos22sin

02cos22sin22

2sin2cos22

1

1

1

p are principal angles associated with the principal stresses (HLT, page 108)

There are two values of 2p in the range 0-360, with values differing by 180.There are two values of p in the range 0-180, with values differing by 90.So, the planes on which the principal stresses act are mutually perpendicular.

18

2sin2cos22

22tan

1 xyyxyx

x

yx

xyp

+++==

We can now solve for the principal stresses by substituting for pin the stress transformation equation for x1. This tells us which principal stress is associated with which principal angle.

2p

xy

(x y) / 2

R

R

R

R

xyp

yxp

xyyx

=

=

+

=

2sin

22cos

22

22

+

++=

RRxy

xyyxyxyx

2221

19

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

22

1 22 xyyxyx +

++=

To find the smaller principal stress, use 1 + 2 = x + y.

22

12 22 xyyxyx

yx +

+=+=

These equations can be combined to give:

22

2,1 22 xyyxyx +

+= Principal stresses(HLT page 108)

To find out which principal stress goes with which principal angle, we could use the equations for sin p and cos p or for x1.

20

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

Solving either equation gives the same expression for tan 2pHence, the shear stresses are zero on the principal planes.

( )( )( ) 02cos22sin

02cos22sin

02cos2sin2

0 and 0for equations theCompare

1

11

111

=+==+

=+===

xyyxx

xyyx

xyyx

yx

xyx

dd

dd

21

Principal Stresses

22

2,1 22 xyyxyx +

+=

yx

xyp

=2

2tan

Principal Angles defining the Principal Planes

x

y

1 p1

1

2

2

p2

y

x x

y

xyyx

xyyx

xy

22

Example: The state of plane stress at a point is represented by the stress element below. Determine the principal stresses and draw the corresponding stress element.

80 MPa 80 MPa

50 MPa

xy

50 MPa

25 MPa

Define the stresses in terms of the established sign convention:x = -80 MPa y = 50 MPaxy = -25 MPa

( )MPa6.84MPa6.54

6.6915252

50802

5080

22

21

22

2,1

22

2,1

===+

+=

+

+=

xyyxyx

23

( )

=+=

===

5.100,5.10

18021.0and0.212

3846.05080

2522tan

22tan

p

p

p

yx

xyp

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

50802

5080

2sin2cos22

1

1

=+++=

+++=

x

xyyxyx

x

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

1 = 54.6 MPa with p1 = 100.52 = -84.6 MPa with p2 = 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 element is 3D, so there are always three principal stresses.

y

xxx

y

y

x, y, xy = yx =

yp

zp

xp1

1

2

2

3 = 0

1, 2, 3 = 0Usually we take 1 > 2 > 3. Since principal stresses can be com-pressive as well as tensile, 3 could be a negative (compressive) stress, rather than the zero stress.

25

Maximum Shear Stress

( )( )

=

==

+=

xy

yxs

xyyxyx

xyyx

yx

dd

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 2s in the range 0-360, with values differing by 180.There are two values of s in the range 0-180, with values differing by 90.So, the planes on which the maximum shear stresses act are mutually perpendicular.

Because shear stresses on perpendicular planes have equal magnitudes, the maximum positive and negative shear stresses differ only in sign.

26

2sxy

(

x

y

)

/

2

R

( )

2cos2sin2

22tan

11 xyyx

yx

xy

yxs

+=

=

We can now solve for the maximum shear stress by substituting for s in the stress transformation equation for x1y1.

R

R

R

yxs

xys

xyyx

22sin

2cos

22

22

=

=

+

=

maxmin2

2

max 2 =+

= xyyx

27

Use equations for sin s and cos s or x1y1 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 s in the equations for x1 and y1 to get

syx

yx =+== 211

x

y

s s

s

s

s

max

max

max

maxy

x x

y

xyyx

xyyx

xy

28

Example: The state of plane stress at a point is represented by the stress element below. Determine the maximum shear stresses and draw the corresponding stress element.

80 MPa 80 MPa

50 MPa

xy

50 MPa

25 MPa

Define the stresses in terms of the established sign convention:x = -80 MPa y = 50 MPaxy = -25 MPa

( ) MPa6.69252

5080

2

22

max

22

max

=+

=

+

=

xyyx

MPa152

50802

=+=

+=

s

yxs

29

( )=

+==

=

=

5.55,5.341800.69and0.692

6.2252

50802

2tan

s

s

xy

yxs

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

25080

2cos2sin2

11

11

=+=

+=

yx

xyyx

yx

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

max = 69.6 MPa with smax = 55.5min = -69.6 MPa with smin = -34.5

15 MPa

15 MPa

15 MPa

x

y

15 MPa

69.6 MPa

-34.5o55.5o

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

22

21

22

2,1

==

+

=

+

+=

xyyx

xyyxyx

pp

s

yx

xyp

xy

yxs

2cot2tan12tan

22tan

22tan

===

=

31

( )

==

==

=+=+

=+

45

45

9022

022cos

02cos2cos2sin2sin

02sin2cos

2cos2sin

02cot2tan

ps

ps

ps

ps

psps

p

p

s

s

ps

So, the planes of maximum shear stress (s) occur at 45 to the principal planes (p).

32

80 MPa 80 MPa

50 MPa

xy

50 MPa

25 MPa

Original Problem

x = -80, y = 50, xy = 25

84.6 MPa

84.6 MPa

54.6 MPa

x

y

54.6 MPa

10.5o

100.5o

Principal Stresses

1 = 54.6, 2 = 0, 3 = -84.6

15 MPa

15 MPa

15 MPa

x

y

15 MPa

69.6 MPa

-34.5o55.5o

Maximum Shear

max = 69.6, s = -15

==

=

5.55,5.34455.10

45

s

s

ps

( )

MPa6.692

6.846.542

max

max

21max

=

=

=