Chapter 7 Transformations of Stress and Strain

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Chapter 7 Transformations of Stress and Strain. 7.1 Introduction. 3 normal stresses. --  x ,  y , and  z. General State of Stress. 3 shearing stresses --  xy ,  yz , and  zx. Plane Stress condition. 2-D State of Stress. Plane Strain condition. - PowerPoint PPT Presentation

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Chapter 7

Transformations of Stress and Strain

7.1 Introduction

General State of Stress

3 normal stresses

3 shearing stresses -- xy, yz, and zx

-- x, y, and z

2-D State of Stress

Plane Stress condition

Plane Strain condition

A. Plane Stress State:

B. Plane Stress State:

z = 0, yz = xz = yz = xz = 0z 0, xy 0

z = 0, yz = xz = yz = xz = 0z 0, xy 0

Examples of Plane-Stress Condition:

Thin-walled Vessels

Max. x & y

Max. xy

(Principal stresses)

7.2 Transformation of Plane Stress

0

0

' ': ( cos )cos ( cos )sin

( sin )sin ( sin )cos

x xyx x

y xy

F A A A

A A

0

0

' ' ': ( cos )sin ( cos )cos

( sin )cos ( sin )sin

x xyy x y

y xy

F A A A

A A

2 2 2' cos sin sin cosx y xyx

2 2' ' ( )sin cos (cos sin )x y xyx y

2 22 2 2sin sin cos , cos cos sin

After rearrangement:

(7.1)

(7.2)

2 21 2 1 22 2

cos coscos , sin

Knowing

2 22 2

' cos sinx y x yxyx

2 22

' ' sin cosx yxyx y

2 22 2

' cos sinx y x yxyy

Eqs. (7.1) and (7.2) can be simplified as:

(7.5)

(7.6)

'y Can be obtained by replacing with ( + 90o) in Eq. (7.5)

(7.7)

1. max and min occur at = 0

2. max and min are 90o apart. max and min are 90o apart.

3. max and min occur half way between max and min

7.3 Principal Stresses: Maximum Shearing Stress

Since max and min occur at x’y’ = 0, one can set Eq. (7.6) = 0

2 2 02

' ' sin cosx yxyx y

22tan xy

x y

1 22 2

22

4/

( ) /cos

( ) /

x y

x y xy

(7.6)

It follows,

Hence, 1 22 22

4/sin

( ) /

xy

x y xy

(a)

(b)

Substituting Eqs. (a) and (b) into Eq. (7.5) results in max and min :

2 2

2 2max, min ( )x y x yxy

2 x y

ave

2

2( )x y

xyR

This is a formula of a circle with the center at:

and the radius of the circle as:

(7.14)

(7.10)

Mohr’s Circle

The max can be obtained from the Mohr’s circle:

Since max is the radius of the Mohr’s circle,

2

2max ( )x yxyR

Since max occurs at 2 = 90o CCW from max,

Hence, in the physical plane max is = 45o CCW from max.

In the Mohr’s circle, all angles have been doubled.

7.4 Mohr’s Circle for Plane Stress

Sign conventions for shear stresses:

CW shear stress = and is plotted above the -axis,

CCW shear stress = ⊝ and is plotted below the -axis

7.5 General State of Stress – 3-D cases

Definition of Direction Cosines:

cos , cos , cosx x y y z z

with2 2 2 1x y z

0

0

: ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

n n x x x xy x y xz x z

yx y z y y y yz y z

zx z x zy z y z z z

F A A A A

A A A

A A A

Dividing through by A and solving for n, we have

2 2 2 2 2 2n x x y y z z xy x y yz y z zx z x

2 2 2 n a a b b c c

(7.20)

We can select the coordinate axes such that the RHS of Eq. *7.20) contains only the squares of the ’s.

(7.21)

Since shear stress ij = o, a, b, and c are the three principal stresses.

7.6 Application of Mohr’s Circle to the 3-D Analysis of Stress

A > B > C

1 12 2max max min A C = radius of the Mohr’s circle

7.9 Stresses in Thin-Walled Pressure Vessels

1

prhoop stress

t

0 2 2 0: ( ) ( )z lF t x p r x

(7.30)

Hoop Stress 1

Longitudinal Stress 2

220 2 0: ( ) ( )xF rt p r

2 2prt

Solving for 2 (7.31)

Hence 1 22

Using the Mohr’s circle to solve for max

2

12 4max( )in plane

prt

2 2max( )out of plane

prt

1 2

1 2 2

pr

t

max 1

1

2 4

pr

t

max 1

1( ) (1 )

2 2 t

r

prp

t

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