08 Stresstrans Print

7/28/2019 08 Stresstrans Print

1/41

. . . . . .

Lecture : Analysis of stress (I) stress transformation

Yubao Zhen

Oct ,

() Lecture : Analysis o f stress ( I) stress t ransformation

Oct , /


2/41

. . . . . .

Review: Angle of twist and statically indeterminate problems

Angle of twist

general: = L

T(x)GIp(x)dxprismatic (single): =

TL

GIpGIp : torsional rigidity

prismatic (multiple): =n

i=

TiLi

GiIpi

Torque diagram: construction; sign convention; jumps and ext. loads

statically indeterminate torsion problems

. compatibility + torque-displacement+ equilibrium.

. the force method.

torsion with non-circular cross-section warping

shear flow; average shear stress on thin walled tubes [optional]

f= avgt=T

Am, avg =

T

tAm


Oct , /


3/41

. . . . . .

Outline

. Plane stress state ()

characteristics; transformation equations; signs

. Principal stresses ()

definition; characteristics; procedures

. Maximum shear stresses ()

in-plane / out-of-plane (/); physical sense; relation to principal stressstate


Oct , /


4/41

. . . . . .

Review of the D stress state

IMPORTANT: stress state is physical and objective ()

can only be observed through its components in a certain coordinate system

the general d stress state

. notation of stress components a. normal

stresses: d (d= x,y, z)

b. shear stresses: dd (d, d = x,y, z)

d: plane normal (orientation);d: acting direction

. sign convention of stress components

()

judge by the physical loading(++,) positive(+,+) negative() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /


5/41

. . . . . .

Understanding stress state in a full scope

A A A

A

A

A

ob

server

(coo

rd.sys.)

orientation (relative)

differential element

stress components in different coord. sys. reflect the same objective state() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /


6/41

. Plane Stress State

()

. . . . . .


7/41

. . . . . .

The D stress state in a stress element ()

z= zx = zy=

y

x

z

O

ys

ys

xsxs

t

yx

tyx

txytxy

Plane stress () (d view)

(b)

y

xO

ys

ys

xs

xs

tyx

tyx

txy

txy

(d view)

() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /


8/41

. . . . . .

Engineering cases

engineering cases of plane stress

. skin () of any ( D ) stressed body (loading free)

. thin plates (shells, membranes, ...) ( D )

. rods ( D ) with special loading styles:

tension/compression; torsion; bending;

. ...

the reason to study D state

.

all previous formulas deal with only in D and on cross-sections. material failure ()maximum stress level (normal/shear). max. on inclined sections ()



9/41

. . . . . .

Features of plane stress state

characteristics:

.

all non-zero stresses are in a single plane. non-coplanar planes are needed to specify a unique state

General D case: x, y, xy Recall: xyacts on four sides

all possible cases:

uniaxial tension/compression

pure shear (). e.g. torsion

+ torsion + uniaxial load

x + y D stretching

e.g. tubular pressure tank

x + y+ general

x

y

xy

x

y

general D stress state



10/41

. . . . . .

Plane stress state at a point

*** The same state of stress ***

[,

)

x

y

xy

x

y

xoy(unprimed)

(knowns: x, y, xy)

x

y

x

y

y

x

x

oy

(primed)

(unknowns: x , y , xy )



11/41

. . . . . .

Stress components on an arbitrary section

x

y

xy

x

y

x

y

x

y

face

face

+

2

for x

-face

the procedure

. set cuts

. draw FBD

. apply eqns. of equil.

fory

-face



12/41

. . . . . .

A step-by-step derivation

Assuming unit length along the zdirection of the wedge (),

x

x

y

xyA sin

yA sin

xA cos

xyA cos

x

A

xyA

FBD

Fx = x

A

(xyA sin

)cos

(xyA cos

)sin

(yA sin ) sin (xA cos ) cos =

Fy = x

y

A + (xyA sin ) sin (xyA cos ) cos (yA sin ) cos +

(xA cos

)sin =



13/41

. . . . . .

Transformation equations for plane stress

{x

= x cos+ ysin

+ xy

(sin cos

)xy = (y x) sin cos + xy(cos sin )the trigonometric identities ():

cos =

( + cos

), sin =

( cos

), sin cos =

sin

results: for an inclined section x

(angle from xto x

):

x =

x + y

+

x y

cos+ xysin

xy =

x y

sin+ xycos

Transformation equations for plane stress

()



14/41

. . . . . .

The matrix () form

An easier form to memorize

x

xy

=

x + y

+ cos sin

sin cos

x y

xy

Determinant () of the matrix = angle involved: always

leading term(): average ofx, y for normal; zero for shear

A special case:

ifx = y= x

= + xysin, xy = xycos

further,

ifxy= x = , xy = (hydrostatic state,)



15/41

. . . . . .

Special cases of plane stress. Uniaxial stress state ()

x , y= , xy=

x =

x

( + cos), xy = x sin

. Pure shear ()

x = , y= , xy

x = xysin, xy = xycos

. Biaxial stress state ()

x , y , xy=

x =

x + y

+

x y

cos

xy =

x y

sin



16/41

. . . . . .

The missing components: stresses on plane y

x

y

x

y

y

x

Apply (+ )stresses at planey

cos sin

sin cos

cos sin

sin cos

y =x + y

x y

cos xysin

y?= +

x y

sin xycos= xy !!!

whats wrong with the shear stress?



17/41

. . . . . .

Sense of shear stress and the local coordinate system

x

x

y

x

x

y

y

x

+ 2

positiv

epos

itive

+

2

physical

positive directions for shear in

and

system are opposite

No wonder!



18/41

. . . . . .

The first invariant of stress tensor

Recall

x =

x + y

+

x y

cos+ xysin

y =

x + y

x y

cos xysin

Observations: x + y = x + y

generalization:

I = x + y+ z is independent of coordinate system

sum of the diagonal elements in stress matrix the st invariant of stress tensor

()



19/41

. . . . . .

Remarks

. applicable to any plane stress state, regardless of materials

. general plane stress =pure shear + biaxial stress

. periodic function of inclined angle (

/ period of)

(originally) [, ), values repeat at + ( face of an element). isconfined to a domain extent of. e.g. [, ), [

,

)

each inclined section (orientation) has its

OWN local coordinate system for transformation


E l


20/41

. . . . . .

Example -, page Given: Plane stress state as shown.

Solve: state of stress with an element oriented clockwise

y

xO

12 MPa

46 MPa

19 MPa


E l ( )


21/41

. . . . . .

Example -, page (cont.)

Solution:y

xO

12 MPa

46 MPa

19 MPa

read from fig.:

x = MPa,

y= MPa,

xy= MPa

y

xO

1.4 MPa

32.6 MPa

31.0 MPa

y1

x1

u= 15

substitute = into the formula

x =x + y

+x y

cos+ xysin= . MPa

xy=

x y

sin+ xycos= . MPa

Similarly, for plane = : x = . MPa, xy = . MPa

check: for = , y

= x + y x = (.) = . MPa() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /


22/41

. Principal Stresses ()

. . . . . .

M i / i i i l l t


23/41

. . . . . .

Maximum/minimum in-plane normal stress

Local extrema off(x): dfdx

= .

dx

d=

x

y (sin) + xycos=

Extrema occur at p where

tanp =xy

x

y

, x y

Observations:. ifx = y= xycos=

. ifxy= hydrostatic. max/mininplane

(

)=

.

ifxy

=

,

max/mininplane=

xy

. further, if = (pure shear), =

,

, max/min

inplane= xy

. ifxy= but x y (biaxial stress state) sin=

= ,

/ identifying x, y themselves as max/min.


Maximum/minimum in plane normal stress (continued)


24/41

. . . . . .

Maximum/minimum in-plane normal stress (continued)

tanp =xy

x y, x y

. Two roots: p , p .

. p and p are apart p and p are

apart.

. For p, one is in

[,

]; the other is in

[

,

).

.

The extrema of normal stresses occur on two mutually perpendicular planes


Correspondence between and


25/41

. . . . . .

Correspondence between p , p and max/mintanp =

xy

x y, x y

set set sinp xy/R xy/Rcosp x y

/R x y

/R

where R =x y + xy. Checking againstd

x

d=

x y

(cos) xysin; d

x

d(set ) < ; d

x

d(set ) >

Conclusions:

maxinplane

pwhere sinp = xy/R and cosp =

x y

/R simultaneously.

mininplane

p= p


Principal stresses and principal planes


26/41

. . . . . .

Principal stresses and principal planesSubstituting p and p into

x

=x + y

+

x y

cos+ xysin

xy = x y

sin+ xycos

. The max/min for in-plane normal stress

, =x + y

x y

+ xy, ( , algebraically)

=x + y

R , in-plane principal stresses

p, p : principal orientations/angles/planes (/)

. xy p = (direct substitution). Or, by: dxd p = xydx

d

p= xy = . principal planes are shear free


Procedure for finding principal stresses and orientations


27/41

. . . . . .

Procedure for finding principal stresses and orientations

Way-: the direct way

. evaluate R =x y

+ xy, then , = x + y

R;

. determine the unique angle

(p

)in

[,

)that satisfies SIMULTANEOUSLY

cosp =x y

R , sinp =

xy

R

This p corresponds to .

(Remark: one should understand why this is valid.)

. find p by plus/minus

to p : p = p

This p corresponds to .


An alternative way way-


28/41

. . . . . .

An alternative way way-

. evaluate , as in the previous method (R first, then ,);

. solve two ps from tanp formula.no association between p, and , yet;

. pick one p, substitute backinto formula for x for matching;

. adjust the convention ofp, to make angle-principals correspondence

. Possible mistakes:

wrong formula: anything other than tanp =xy

x y

(sign, numerator/denominator, etc.) mistakenly calculated ps from tanp or wrong units

correct ps, but no back-substitution for association:p , p .

caution: dont simply choose one randomly.


Comparison between the two methods


29/41

. . . . . .

Comparison between the two methods

Way-: the direct way (to get p by two criteria)

. needs two calculations on p ;

. faster

Way-: the trial-and-error way (to get p

by tanp formula)

. testing on two values;

. post-adjusting on convention ;

. slower;

. more traditional (less to memorize)

can use both to double checkagainst each other


Example - page (first part)


30/41

. . . . . .

Example , page (first part)

Given: plane stress state with x = MPa, y= MPa, xy= MPa

Solve: principal stresses and its sketch

x

y

84 MPa

32 MPa

30 MPa


Example -, page (first part, in a different way)


31/41

. . . . . .

Example , page (first part, in a different way)Solution:

x

y

84 MPa

32 MPa

30 MPa

p1 = 14.7

1 = 92.4 MPa

2 = 38.4 MPa

principal stresses

. R =x y + xy= . MPa;, =

x + y

R = . MPa

= . MPa, = . MPa

. Determine p :

cosp =x y

R= .,

sinp =xy

R= .

For p

[, ),p = arcsin(.) = . rad,p

= . rad = .;

. Hence p = p + = . (or .)



32/41

. Max. Shear Stresses

()

. . . . . .

Derivation for maximum in-plane shear stress


33/41

. . . . . .

pSimilarly, for maximum shear stresses and the acting planes

dxy

d= (x y)(cos ) xysin=

Extrema occur at s where

tans = x y

xy, xy

observation:. Two roots: s , s they are

apart.

One in [, ]; the other in [

, ).

. p and s are related by

tans =

tanp= cotp = tan(p )

s p =

s = p

planes of max. shear occur at to the principal planes


max and s


34/41

. . . . . .

max s

xy = x y

sin+ xycos; tans =

x y

xyObservation: one of the two sets of solution to tans eqn.

coss =xy

R, sins =

x y

R

gives xy = R > . Recall that

cosp =x y

R , sinp =

xy

R

direct verification s = p (rotate clockwise from p)

Check:

maxinplane

=x y

+ xy= R =

,

mininplane

= R, s = s

, avg =

x + y


max/min in-plane shear stresses and the complementary


35/41

. . . . . .

p p y

property

min/max have the same magnitude;

Shear stresses are evaluated in the faces associated local coordinate system;Algebraic sense is in accordance with the complementary property;

Double check the arrows (head to head, tail to tail).


Example -, page (nd part)


36/41

. . . . . .

p p g p

Given: plane stress state with x = MPa, y= MPa, xy= MPa

Solve: Maximum in-plane shear stresses and its sketch

x

y

84 MPa

32 MPa

30 MPa


Example -, page , (nd part)


37/41

. . . . . .

Solution:

p1 = 14.7

1 = 92.4 MPa

2

= 38.4 MPa

principal stresses

s1 = 59.7

s2 = 30.3

27 MPa

65.4 MPa

maximum shear

stresses

solution-: maxinplane

= R = . MPa cos s =xy

R= .

sins = x y

R= .

For s [,

),

s = + arcsin(.) = . rad,s = . rad = .

Then s = s + = .;

avg =x + y

= MPa

solution-:

or simply s = p = . = .


Remarks on plotting in stress elements


38/41

. . . . . .

Key points: sign conventions ofnormal and shear stresses

. all data from transformation eqn. are for the local coordinate system

. two angles apart indicate the same plane

(one faces () sides; complementary properties). convert algebraic value to physical sense while plotting. dont forget the avg for max/min shear stress plot

. dont forget to mark the angles

.

double check!


Remarks on the principal stresses and max. shear stresses


39/41

. . . . . .

. Maximum/minimum normal/shear stresses and

. the associated orientations of the planes

are of practical engineering interests, since

different materials (brittle, ductile) fail on different types of stressesbrittle (): fails on normal stress;ductile (): fails on shear stress

need ofsafety checkagainst allowable stress

consideration of orientation (to strengthen on a specific direction)


Summary


40/41

. . . . . .

plane stress state

transformation equations for plane stresses

matrix form; local coordinate system for each inclined section

principal stresses/planes

max/min normal stresses; shear free;

two perpendicular directions p,

, = x + y

R, R =x y + xy

way-: direct; way-: trial-and-error

max. in-plane shear stress/planes

max/min (algebraically only); two perpendicular directions

max = R =

; avg =

x + y

relations between principal angles/planes

s = p ; s = s


Homework


41/41

. . . . . .

Chap. , , ,

Due: .. (Mon.)(to L)


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