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 , /
7/28/2019 08 Stresstrans Print
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
() Lecture : Analysis o f stress ( I) stress t ransformation
Oct , /
7/28/2019 08 Stresstrans Print
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
() Lecture : Analysis o f stress ( I) stress t ransformation
Oct , /
7/28/2019 08 Stresstrans Print
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 , /
7/28/2019 08 Stresstrans Print
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 , /
7/28/2019 08 Stresstrans Print
6/41
. Plane Stress State
()
. . . . . .
7/28/2019 08 Stresstrans Print
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 , /
7/28/2019 08 Stresstrans Print
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 ()
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
7/28/2019 08 Stresstrans Print
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
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
7/28/2019 08 Stresstrans Print
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 )
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
7/28/2019 08 Stresstrans Print
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
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
7/28/2019 08 Stresstrans Print
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 =
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
7/28/2019 08 Stresstrans Print
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
()
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
7/28/2019 08 Stresstrans Print
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,)
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
7/28/2019 08 Stresstrans Print
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
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
7/28/2019 08 Stresstrans Print
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?
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
7/28/2019 08 Stresstrans Print
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!
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
7/28/2019 08 Stresstrans Print
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
()
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
7/28/2019 08 Stresstrans Print
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
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
E l
7/28/2019 08 Stresstrans Print
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
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
E l ( )
7/28/2019 08 Stresstrans Print
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 , /
7/28/2019 08 Stresstrans Print
22/41
. Principal Stresses ()
. . . . . .
M i / i i i l l t
7/28/2019 08 Stresstrans Print
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.
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
Maximum/minimum in plane normal stress (continued)
7/28/2019 08 Stresstrans Print
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
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
Correspondence between and
7/28/2019 08 Stresstrans Print
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
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
Principal stresses and principal planes
7/28/2019 08 Stresstrans Print
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
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
Procedure for finding principal stresses and orientations
7/28/2019 08 Stresstrans Print
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 .
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
An alternative way way-
7/28/2019 08 Stresstrans Print
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.
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
Comparison between the two methods
7/28/2019 08 Stresstrans Print
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
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
Example - page (first part)
7/28/2019 08 Stresstrans Print
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
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
Example -, page (first part, in a different way)
7/28/2019 08 Stresstrans Print
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 .)
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
7/28/2019 08 Stresstrans Print
32/41
. Max. Shear Stresses
()
. . . . . .
Derivation for maximum in-plane shear stress
7/28/2019 08 Stresstrans Print
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
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
max and s
7/28/2019 08 Stresstrans Print
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
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
max/min in-plane shear stresses and the complementary
7/28/2019 08 Stresstrans Print
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).
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
Example -, page (nd part)
7/28/2019 08 Stresstrans Print
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
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
Example -, page , (nd part)
7/28/2019 08 Stresstrans Print
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 = . = .
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
Remarks on plotting in stress elements
7/28/2019 08 Stresstrans Print
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!
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
Remarks on the principal stresses and max. shear stresses
7/28/2019 08 Stresstrans Print
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)
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
Summary
7/28/2019 08 Stresstrans Print
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
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
Homework
7/28/2019 08 Stresstrans Print
41/41
. . . . . .
Chap. , , ,
Due: .. (Mon.)(to L)
() Lecture : Analysis o f stress ( I) stress t ransformation Oct , /
Top Related