Module #2

Transformation of stresses in 2-D

READING LISTDIETER: Ch. 2, Pages 20-27

Ch. 2 in McClintock and ArgonCh. 7 in Edelglass

HOMEWORKFrom Dieter2-1, 2-2, 2-5

Objectives

• Develop equations for transformation of axes.

• Apply equations to determine principal normal and shear stresses.

• Perform transformation graphically using Mohr’s circle.

Why bother?

• Engineering structures are subjected to many different types of applied loads:

– Tension

– Compression

– Bending

– Torsion

– Pressure

– Combinations of the above

• Complex states of normal and shear stress occur.

Why bother?

• These stress components vary in magnitude and direction relative to their location in the structure and the coordinate system with which they are related to.

• As engineers we need to make sure that the structures we design don’t fail as a result of the applied stresses.

• Thus, we need to identify locations where stresses are the most severe. Then we can do more detailed analysis if needed.

Principal Stresses and Axes

• Principal stresses:

– The most severe stresses (i.e., the maximum and minimum stresses) in a stress state.

• Principal axes:

– Directions in which principal stresses act.

The Stress Tensor

x

y

z

xx

xy

xz

zz

zyzx

yy

yx

yz

xy

zO

xx xy xz

yy yzij yx

zx z zzy

If we limit ourselves to two dimensions, we achieve a condition called plane stress.

This is where all stresses in one dimension become zero.[see the next page]

PLANE STRESSAll stresses are zero in one primary direction

(a) (c)(b)

xy

z

00

0 0 0

xx xy

xy yy

00 0 0

0

xx xz

zx zz

0 0 000

yy yz

zy zz

xy

z

xy

z

yx yy

xyxx

xz

xx

zz

zx

zz

yyyz

zy

xy yx xz zx yz zy

0 00 0

0 0 0 0 0 0

xx xy x x x y

xy yy x y y y

A*

x

yyy

xxxy

yx x x

x

yy y y x

x y

Transformation is accomplished via a force balance

Let’s first balance forces parallel to x′followed by y′

A*

y′

yy

xx

xy

yx

x x

x

yy y y x

x y

x x x y

A

AcosθAsinθ

x′

0 sin cos cos sin

cos cos sin sin

0 cos cos sin sin

cos sin sin cos

x x x yx xy

xx yy

y x y xy yx

xx yy

F A A A

A A

F A A A

A A

xy yx

2 2

2 2

cos sin 2 sin cos

( )sin cos (cos sin )

x x xx yy xy

x y yy xx xy

Solving for σx′x′ and τx′y′ yields:

THESE THREE BOXED EQUATIONS ARE KNOWN AS THE TRANSFORMATION OF STRESS EQUATIONS

2 2

2 2

cos sin 2 sin cos2 2 2 2

sin cos 2 sin cos

y y xx yy xy

y y xx yy xy

Since σy′y′ is 90° away from σx′x′, σy′y′ is:

We can simplify the Transformation of Stress equations by invoking the double-angle identities

2

2

1cos 1 cos 22

1sin 1 cos 22

2sin cos sin 2

cos 2 sin 22 2

cos 2 sin 22 2

sin 2 cos 22

xx yy xx yyx x xy

xx yy xx yyy y xy

yy xxx y xy

They provide us with the stresses re-oriented to a new coordinate system

Transformation of Stress Equations(for plane strain)

00

0 0 0

x x x y

x y y y

cos2 sin22 2

cos2 sin22 2

xx yy xx yyxy

xx yyx x y yxx yy xx yy

xy

This quantity is invariant! It doesn’t change regardless of the coordinate system.

xx yyx x y y

An interesting observation

2 stress invariants for plane strain2 stress invariants for plane strain• I1 is the sum of the main diagonal of the

stress tensor.

• I2 is the sum of the principal minors.

1 xx yy x x y yI

2 22 xx yy xy x x y y x yI

0 20 40 60 80 100 120 140 160 180

-400

-200

0

200

400

600

800

1000max

max

x'y'

x'x'

Compression

S

tress

(MP

a)

(degrees)

xx = 800 MPayy = 200 MPaxy = -300 MPa

Tension

90o

45o

min

min

'

'

'

'

cos2 s

sin 2 c

i

os22

n 22 2

xx yy x

yy xxx

x yyx x x

y y

y

x

Principal planes

cos 2 sin 22 2

from previous viewgraph, normal stress is maximum (or minimum) when:

0 2sin 2 2 cos 22

Solving for yields the plane where maximum andminimum nor

xx yy xx yyx x xy

xx yyx xxy

dd

mal stress occur.

tan22

xyprincipal

xx yy

By definition this is also the plane where shear stress vanishes

Maximum in-plane shear stress

sin 2 cos 22

Shear stress is maximum/minimum when:

0 cos 2 2 sin 22

Solving for yields the plane of maximum/minimum shear stress

2tan2

An average norm

yy xxx y xy

x y xx yyxy

xx yyshear

xy

dd

al stress is superimposed on these planes

2xx yy

average

MAXIMUM & MINUMUM PRINCIPAL STRESSES FOR A 2-D STATE OF STRESS

2max 1 2

min 2

22

max

2 2

2

xx yy xx yyxy

xx yyxy

5 minute break

Mohr’s Circle for Stress• Developed in 1882; a graphical way to represent the

transformation of stress equations.

• Re-write the equations as:

' ' average

' '

cos 2 sin 22

sin 2 cos 22

xx yyx x xy

yy xxx y xy

' '

' '

cos 2 sin 22 2

sin 2 cos 22

xx yy xx yyx x xy

yy xxx y xy

Mohr’s Circle

• Square both sides of each equation and add them together, which yields:

• Equation for a circle expressed in (,) coordinates with a center at (average, 0).

22 2 2

' ' average ' '

2 2 2' ' average ' '

2or

xx yyx x x y xy

x x x y R

• Recall the equation for a circle:

where h is the location of the center on the x axis (i.e., the center of the circle), and R is its radius.

• The center of Mohr’s circle lies at (average,0).

• The radius of Mohr’s circle is:

• max = 1 = average + R

• min = 2 = average - R

Mohr’s Circle – cont’d

2 2 2x h y R

average 2xx yy

22

max 32xx yy

xyR

The steps for construction are provided on the next 2

pages of the handout and/or in your mechanics

of materials book.

Steps in Construction of Mohr’s Circle1. Show the stresses xx, yy, and xy on a cube. Label the vertical plane V and

the horizontal plane H.

2. Write the coordinates of points V and H as V(xx, -xy) and H(yy, yx). A positive value for ij produces a CW moment about the center of the cube (i.e. CW rotation of the cube).

3. Draw the horizontal axis with the tensile normal stress to the right (i.e., positive) and the compressive normal stress to the left (i.e., negative). Draw the vertical axis with the clockwise (CW) direction of shear stress (i.e., positive) up and the counterclockwise (CCW) direction of rotation down.

4. Locate points V and H and join the points by drawing a line. Label the point where line VH intersects the horizontal axis as C, the center of the circle. The center has coordinates C(average, 0).

5. Draw Mohr’s circle with point C as the center and a radius, R of2

2max 32

xx yyxyR

Steps in Construction of Mohr’s Circle – cont’d6. The angle between lines CV and C1 is labeled 2 because the angles on

Mohr’s circle are double the actual angle between planes.

To determine the direction of rotation (i.e., the sign) we first record the direction in which we move from point V(σxx, -xy) to point (σ1, 0) on Mohr’s circle.

If the direction of rotation is CCW (i.e., towards the positive shear direction), then the sign of is positive. If the rotation is CW then the sign of is negative.

This is illustrated on the next viewgraph.

Intersection with the -axis is min=2

Intersection withthe -axis ismax = 1

max= 3

C(average, 0)

2

CW

CCW

V(xx, -xy)

H(yy, yx)

(1, )(2, )

• max = 1 = average + R

• min = 2 = average - R

xx

yy

xy

yx

VH

x

y

1

2

x

y

x′

y′

VERY IMPORTANT:

• Many engineering texts (and practicing engineers) construct Mohr’s circle with shear stress pointing downwards as is illustrated to the right. In this case, the rotations between the principal stress axes and the state of stress on the volume element will be opposite of that on Mohr’s circle.

• Be careful and know the system that your employer or the texts that you are referring to are using.

Intersection with the-axis is min=2

2

Coordinate is yy, xy

Coordinate is xx, - xy

Intersection with the -axis is max = 1

max= 3

Coordinate is average, 0

ProducesCW moment

Produces CCW moment

+

xx

yy

xy

yx

VH

x

y

1

2

x

y

x′

y′

2θ

Example Problem

1. Consider a point in a solid that is subjected to the following state of stress: σxx = 90 MPa; σyy = -50 MPa; τxy = -60 MPa.

a. Draw a free body diagram representing the stress state.

b. Determine the principal stresses, the maximum in-plane shear stress acting on the point, and the orientation of the principalplanes using Mohr’s circle.

c. Show the stresses no an appropriate diagram.

Example Problem - solution

a. Draw a free body diagram representing the stress state.

x

y

90 MPaxx

50 MPayy

60 MPaxy

Example Problem – solution cont’db. Determine the principal stresses, the maximum in-plane

shear stress acting on the point, and the orientation of the principal planes using Mohr’s circle.

xx

yy

average

22

max

, 90,60

, 50, 60

90 50,0 ,0 20,0

2

90 5060 850 92.2

2

xy

xy

V

H

C

R

V(90,60)

H(-50,-60)

C(20, 0)

60tan 270

2 40.6

xy

xx avg

1 112.2 MPaavg R 2 72.2 MPaavg R

2

2max

90 5060 850 92.2 MPa

2R

60

90 - 20 = 70

A*

solution cont’d

H*

20

20

V(90,60)

H(-50,-60)

C(20, 0)

60 .tan 270 .

2 40.6

xy

xx avg

oppadj

1 112.2 MPaavg R 2 72.2 MPaavg R

2

2max

90 5060 850 92.2 MPa

2R

60

90 - 20 = 70

B*

solution cont’d

Example Problem - solution

c. Show the stresses on an appropriate diagram.

x

y

90 MPaxx

50 MPayy 60 MPaxy

20.3p

P

1 112.2 MPa

2 72.2 MPa

1

2

Note division of p by 2