ch09 stress transformation

8/21/2019 Ch09 Stress Transformation

1/116

2005 Pearson Education South Asia Pte Ltd

9. Stress Transformation

1

Derive equations for

transforming stress

components between

coordinate systems of

different orientation Use derived equations to

obtain the maximum normal

and maximum shear stress

at a pt Determine the orientation of elements upon which

the maximum normal and maximum shear stress

acts

CHAPTER OBJECTIVES


2/116



2

Discuss a method for

determining the absolute

maximum shear stress at a

point when material is

subjected to plane and3-dimensional states of stress

CHAPTER OBJECTIVES


3/116



3

CHAPTER OUTLINE

1. Plane-Stress Transformation

2. General Equations of Plane Stress

Transformation

3. Principal Stresses and Maximum In-Plane

Shear Stress4. Mohrs Circle Plane Stress

5. Stress in Shafts Due to Axial Load and Torsion

6. Stress Variations Throughout a Prismatic Beam7. Absolute Maximum Shear Stress


4/116



4

9.1 PLANE-STRESS TRANSFORMATION

General state of stress at a pt is characterized bysix independent normal and shear stress

components.

In practice, approximations and simplifications are

done to reduce the stress components to a singleplane.


5/116



5

The material is then said to be

subjected to plane stress.

For general state of plane stress at a

pt, we represent it via normal-stress

components, x, yand shear-stresscomponent xy.

Thus, state of plane stress at the pt is

uniquely represented by three

components acting on an element

that has a specific orientation at

that pt.



6/116



6

Transforming stress components from one

orientation to the other is similar in concept to how

we transform force components from one system of

axes to the other.

Note that for stress-component transformation, weneed to account for

the magnitude and direction of each stress

component, and

the orientation of the area upon which each

component acts.



7/1162005 Pearson Education South Asia Pte Ltd


7

Procedure for Analysis

If state of stress at a pt is known for a given

orientation of an element of material, then state of

stress for another orientation can be determined





8


1. Section element as shown.

2. Assume that the sectioned area is A, then

adjacent areas of the segment will be Asinand

Acos.

3. Draw free-body diagram of segment,

showing the forces that act on the

element. (Tip: Multiply stresscomponents on each face by the

area upon which they act)





9


4. Apply equations of force equilibrium in thex andy

directions to obtain the two unknown stress

components x, and xy.

To determine y(that acts on the +y face of theelement), consider a segment of element shown

below.

1. Follow the same procedure asdescribed previously.

2. Shear stress xyneed not be

determined as it is complementary.





10

EXAMPLE 9.1

State of plane stress at a pt on surface of airplane

fuselage is represented on the element oriented as

shown. Represent the state of stress at the pt that is

oriented 30clockwise from the position shown.




11

EXAMPLE 9.1 (SOLN)

CASE A (a-asection)

Section element by line a-aand

remove bottom segment.

Assume sectioned (inclined)

plane has an area of A,horizontal and vertical planes

have area as shown.

Free-body diagram of

segment is also shown.




12

EXAMPLE 9.1 (SOLN)

Apply equations of force equilibrium

in thexandydirections (to avoid

simultaneous solution for the two

unknowns)

+ Fx= 0;

MPa15.4

030sin30sin25

30sin30sin8030sin30cos25

30cos30cos50

'

'

x

x

A

AA

AA




13

EXAMPLE 9.1 (SOLN)

+ Fy= 0;

Since xis negative, it acts

in the opposite directionwe initially assumed.

MPa8.68

030sin30sin25

30cos30sin8030cos30cos25

30sin30cos50

''

''

yx

yx

A

AA

AA




14

EXAMPLE 9.1 (SOLN)

CASE B (b-bsection)

Repeat the procedure to obtain

the stress on the perpendicular

plane b-b.

Section element as shownon the upper right.

Orientate the +x axis

outward, perpendicular to

the sectioned face, with

the free-body diagram

as shown.




15

EXAMPLE 9.1 (SOLN)

+ Fx

= 0;

MPa8.25

030sin30sin50

30cos30sin2530cos30cos80

30sin30cos25

'

'

x

x

A

AA

AA




16

EXAMPLE 9.1 (SOLN)

+ Fy

= 0;

Since xis negative, it acts

opposite to its directionshown.

MPa8.68

030cos30sin50

30sin30sin2530sin30cos80

30cos30cos25

''

''

yx

yx

A

AA

AA

S f




17

EXAMPLE 9.1 (SOLN)

The transformed stress

components are as shown.

From this analysis, we conclude

that the state of stress at the pt can

be represented by choosing anelement oriented as shown in the

Case A or by choosing a different

orientation in the Case B.

Stated simply, states of stress are equivalent.

9 St T f ti




18

9.2 GENERAL EQNS OF PLANE-STRESS TRANSFORMATION

Sign Convention

We will adopt the same sign convention as

discussed in chapter 1.3.

Positive normal stresses, xand y, acts outward

from all faces Positive shear stress xyacts

upward on the right-hand

face of the element.

9 St T f ti




19


Sign Convention

The orientation of the inclined plane is determined

using the angle .

Establish a positivexandyaxes using the right-

hand rule. Angle is positive if it

moves counterclockwise

from the +xaxis to

the +xaxis.

9 St T f ti




20


Normal and shear stress components

Section element as shown.

Assume sectioned area is A.

Free-body diagram of element

is shown.

9 St T f ti




21



Apply equations of force

equilibrium to determine

unknown stress components:

+ Fx= 0;

cossin2sincos

0coscos

sincossinsin

cossin

22'

'

xyyxx

x

xyy

xyx

A

AA

AA

9 St T f ti




22



+ Fy= 0;

Simplify the above two equations usingtrigonometric identities sin2= 2 sincos,sin2= (1 cos2)/2, and cos2=(1+cos2)/2.

22''

''

sincoscossin

0sincoscoscoscossin

sinsin

xyyxyx

x

xyy

xyyx

AAA

AA

9 St T f ti




23



If yis needed, substitute (= + 90) for into

Eqn 9-1.

292cos2sin2'' -

xyyx

yx

192sin2cos22

' -

xyyxyx

x

392sin2cos22

' -

xyyxyx

y

9 Stress Transformation


24/116



24



To apply equations 9-1 and 9-2, just substitute the

known data for x, y, xy, andaccording to

established sign convention.

Ifx andxyare calculated as positive quantities,then these stresses act in the positive direction of

thexandyaxes.

Tip: For your convenience, equations 9-1 to 9-3 can

be programmed on your pocket calculator.



25/116



25

EXAMPLE 9.2

State of stress at a pt is represented by the element

shown. Determine the state of stress at the pt onanother element orientated 30clockwise from the

position shown.



26/116



26

EXAMPLE 9.2 (SOLN)

This problem was solved in Example 9.1 using

basic principles. Here we apply Eqns. 9-1 and 9-2.

From established sign convention,

Plane CD +xaxis is directed outward,

perpendicular to CD,

and +y axis directed along CD.

Angle measured

is = 30(clockwise).

MPaMPaMPa 255080 xyyx



27/116



27

EXAMPLE 9.2 (SOLN)

Plane CD

Apply Eqns 9-1 and 9-2:

The negative signs indicate that xand xyact in

the negativexandydirections.

MPa8.25

302sin25302cos2

5080

2

5080

'

'

x

x

MPa8.68

302cos25302sin2

5080

''

''

yx

yx



28/116



28

EXAMPLE 9.2 (SOLN)

Plane BC

Similarly, stress components

acting on faceBCare

obtained using = 60.

MPa15.4

602sin25602cos2

5080

2

5080

'

'

x

x

MPa8.68

602cos25602sin2

5080

''

''

yx

yx



29/116



29

EXAMPLE 9.2 (SOLN)

As shown, shear stress xywas computed twice to

provide a check.

Negative sign for xindicates that stress acts in the

negativexdirection.

The results are shown below.



30/116



30

9.2 PRINCIPAL STRESSES AND MAXIMUM IN-PLANE SHEAR STRESS

In-plane principal stresses

Differentiate Eqn. 9-1 w.r.t. and equate to zero:

Solving the equation and let= P, we get

Solution has two roots, p1, and p2.

02cos22sin22

'

xy

yxx

d

d

492/)(

2tan -yx

xyP



31/116



31



For p1,

For p2,

2

2

1

22

1

222cos

22sin

xyyxyxp

xyyx

xyp

22

2

22

2

22

2cos

22sin

xyyxyx

p

xyyx

xyp



32/116



32



Substituting either of the two sets of trigonometric

relations into Eqn 9-1, we get

The Eqn gives the maximum/minimum in-plane

normal stress acting at a pt, where 12. The values obtained are the principal in-plane

principal stresses, and the related planes are the

principal planes of stress.

59222

2

2,1 -xyyxyx



33/116



33



If the trigonometric relations for p1and p2are

substituted into Eqn 9-2, it can be seen that

xy= 0.

No shear stress acts on the principal planes.Maximum in-plane shear stress

Differentiate Eqn. 9-2 w.r.t. and equate to zero:

692/)(2tan -xy

yxS



34/116



34


Maximum in-plane shear stress

The two roots of this equation, s1and s2can be

determined using the shaded triangles as shown.

The planes for maximum

shear stress can bedetermined by orienting

an element 45from the

position of an element

that defines the planeof principal stress.



35/116



35



Using either one of the roots

s1and s2, and taking trigo

values of sin 2sand cos 2s

and substitute into Eqn 9-2:

Value calculated in Eqn 9-7 is referred to as the

maximum in-plane shear stress.

792

)( 22

-plane-in

max xyyx



36/116



36


Substitute values for sin 2sand cos 2sinto

Eqn 9-1, we get a normal stress acting on the

planes of maximum in-plane shear stress:

You can also program the above equations onyour pocket calculator.


892

-yx

avg



37/116



37

IMPORTANT

Principals stresses represent the maximum andminimum normal stresses at the pt.

When state of stress is represented by principalstresses, no shear stress will act on element.

State of stress at the pt can also be represented interms of the maximum in-plane shear stress. Anaverage normal stress will also act on the element.

Element representing the maximum in-plane shearstress with associated average normal stresses isoriented 45from element represented principalstresses.




38/116



38

EXAMPLE 9.3

When torsional loading T is applied to bar, it produces

a state of pure shear stress in the material. Determine(a) the maximum in-plane shear stress and

associated average normal stress, and (b) the

principal stress.



39/116



39

EXAMPLE 9.3 (SOLN)

From established sign convention:


Apply Eqns 9-7 and 9-8,

xyyx 00

220

2

)( 22

xyyx

plane-in

max

02

00

2

yxavg



40/116



40

EXAMPLE 9.3 (SOLN)


As expected, maximum in-plane shear stress

represented by element shown initially.

Experimental results show that materials that are

ductile will fail due to shear stress. Thus, with atorque applied to a bar

made from mild steel,

the maximum in-plane

shear stress will causefailure as shown.



41/116



41

EXAMPLE 9.3 (SOLN)

Principal stress

Apply Eqns 9-4 and 9-5,

13545

;2/)00(2/)(

2tan

12 pp

yx

xyP

22

222,1

00

22 xyyxyx



42/116



42

EXAMPLE 9.3 (SOLN)

Principal stress

Apply Eqn 9-1 with p2= 45

Thus, if 2= acts at p2= 45

as shown, and 1= acts onthe other face, p1= 135.

90sin00

2sin2cos22

2,1 xyyxyx



43/116

2005 Pearson Education South Asia Pte Ltd 43

EXAMPLE 9.3 (SOLN)

Principal stress

Materials that are brittle fail due to normal stress. An

example is cast iron when subjected to torsion, fails

in tension at 45inclination as shown below.



44/116


EXAMPLE 9.6

State of plane stress at a pt on a body is represented

on the element shown. Represent this stress state interms of the maximum in-plane shear stress and

associated average normal stress.



45/116


EXAMPLE 9.6 (SOLN)

Orientation of element

Since x= 20 MPa, y= 90 MPa, and

xy= 60 MPa and applying Eqn 9-6,

Note that the angles are 45

away from principal planes

of stress.

3.11121802

3.215.422

60

2/90202/

2tan

121

22

sss

ss

s xy

yx



46/116


EXAMPLE 9.6 (SOLN)


Applying Eqn 9-7,

Thus acts in the +ydirection on this

face (= 21.3).

MPa4.81

602

9020

2

)( 22

22

xy

yx

maxlanep-in

''yxmaxlanep-in



47/116


EXAMPLE 9.6 (SOLN)

Average normal stress

Besides the maximum shear stress, the element is

also subjected to an average normal stress

determined from Eqn. 9-8:

This is a tensile stress.

MPa352

90202

yxavg



48/116


Equations for plane stress transformation have a

graphical solution that is easy to remember anduse.

This approach will help

you to visualize how thenormal and shear stress

components vary as the

plane acted on is oriented

in different directions.

9.4 MOHRS CIRCLE: PLANE STRESS



49/116


Eqns 9-1 and 9-2 are rewritten as

Parameter can be eliminated by squaring each

eqn and adding them together.


1092cos2sin2'' -

xyyx

yx

992sin2cos22

' -

xyyxyx

x

xyyx

yxyx

x2

2

''2

2

'22



50/116


If x, y, xyare known constants, thus we compact

the Eqn as,


1292

2

119

22

2''

22'

-

where

-

xyyx

yxavg

yxavgx

R

R



51/116


Establish coordinate axes; positive to the right

and positive downward, Eqn 9-11 represents acircle having radiusRand center on the axis at

pt C(avg, 0). This is called the Mohrs Circle.




52/116


To draw the Mohrs circle, we must establish

the and axes.

Center of circle C(avg, 0) is plotted from the

known stress components (x, y,xy).

We need to know at least one pt on the circle toget the radius of circle.




53/116


Case 1 (xaxis coincident withxaxis)

1. = 0

2. x= x

3. xy= xy.

Consider this as reference ptA, andplot its coordinatesA (x, xy).

Apply Pythagoras theorem to shaded triangle to

determine radiusR. Using pts CandA,

the circle can now

be drawn.




54/116


Case 2 (xaxis rotated 90counterclockwise)

1. = 90

2. x= y

3. xy= xy.

Its coordinates are G(y, xy).

Hence radial line CG

is 180

counterclockwisefrom reference

line CA.




55/116



Construction of the circle

1. Establish coordinate

system where abscissa

represents the normalstress , (+ve to the

right), and the ordinate

represents shear

stress , (+ve downward).2. Use positive sign convention for x, y, xy, plot the

center of the circle C, located on the axis at a

distance avg

= (x

+ y

)/2 from the origin.




56/116




3. Plot reference ptA(x, xy). This pt represents the

normal and shear stress components on the

elements right-hand vertical face. Sincexaxis

coincides withxaxis, = 0.




57/116




4. Connect ptAwith center Cof the circle and

determine CAby trigonometry. The distance

represents the radiusRof the circle.5. OnceRhas been

determined, sketch

the circle.




58/116



Principal stress

Principal stresses 1and 2(12) are

represented by two ptsBandDwhere the circle

intersects the -axis.




59/116



Principal stress

These stresses act on planes

defined by angles p1and p2.

They are represented on thecircle by angles 2p1and 2p2

and measured from radial

reference line CAto lines CBand CDrespectively.




60/116



Principal stress

Using trigonometry, only one of

these angles needs to be

calculated from the circle,since p1and p2are 90apart.

Remember that direction of

rotation 2pon the circle represents the same

direction of rotation pfrom reference axis (+x) toprincipal plane (+x).




61/116




The average normal stress

and maximum in-plane shear

stress components aredetermined from the circle as

the coordinates of either ptE

orF.




62/116




The angles s1and s2give

the orientation of the planes

that contain thesecomponents. The angle 2s

can be determined using

trigonometry. Here rotation is

clockwise, and so s1must beclockwise on the element.




63/116



Stresses on arbitrary plane

Normal and shear stress

components xand xy

acting on a specified planedefined by the angle , can

be obtained from the circle

by using trigonometry to

determine the coordinatesof ptP.




64/116



Stresses on arbitrary plane

To locate ptP, known angle

for the plane (in this case

counterclockwise) must bemeasured on the circle in

the same direction 2

(counterclockwise), from the

radial reference line CAto theradial line CP.




65/116


EXAMPLE 9.9

Due to applied loading, element at ptAon solid

cylinder as shown is subjected to the state of stress.Determine the principal stresses acting at this pt.



66/116


EXAMPLE 9.9 (SOLN)


Center of the circle is at

Initial ptA(2, 6) and the

center C(6, 0) are plotted

as shown. The circle havinga radius of

MPaMPa 6012 xyyavg

MPa62

012

avg

MPa49.86612 22 R



67/116


EXAMPLE 9.9 (SOLN)

Principal stresses

Principal stresses indicated atptsBandD. For 1> 2,

Obtain orientation of element by

calculating counterclockwise angle 2p2, which

defines the direction of p2and 2and its associated

principal plane.

MPa

MPa

5.1449.86

49.2649.8

2

1

5.22

0.45612

6tan2 1

2

2

p

p



68/116


EXAMPLE 9.9 (SOLN)

Principal stresses

The element is orientated such thatxaxis or 2isdirected 22.5counterclockwise from the horizontal

x-axis.



69/116


EXAMPLE 9.10

State of plane stress at a pt is shown on the element.

Determine the maximum in-plane shear stresses andthe orientation of the element upon which they act.



70/116


EXAMPLE 9.10 (SOLN)

Construction of circle

Establish the , axes as shown below. Center of

circle Clocated on the -axis, at the pt:

MPaMPaMPa 609020 xyyx

MPa352

9020

avg



71/116


EXAMPLE 9.10 (SOLN)


Pt Cand reference ptA(20, 60) are plotted. ApplyPythagoras theorem to shaded triangle to get

circles radius CA,

MPa4.81

5560 22

R

R


(SO )


72/116


EXAMPLE 9.10 (SOLN)


Maximum in-plane shear stress and average normal

stress are identified by ptE or Fon the circle. In

particular, coordinates of ptE(35, 81.4) gives

MPa

MPaplane-in

max

35

4.81

avg


EXAMPLE 9 10 (SOLN)


73/116


EXAMPLE 9.10 (SOLN)


Counterclockwise angle s1can be found from the

circle, identified as 2s1.

3.21

5.4260

3520

tan2

1

1

1

s

s


EXAMPLE 9 10 (SOLN)


74/116


EXAMPLE 9.10 (SOLN)


This counterclockwise angle defines the direction of

thexaxis. Since ptEhas positive coordinates, then

the average normal stress and maximum in-plane

shear stress both act in the positivexandydirections as shown.


EXAMPLE 9 11


75/116


EXAMPLE 9.11

State of plane stress at a pt is shown on the element.

Represent this state of stress on an element oriented30counterclockwise from position shown.


EXAMPLE 9 11 (SOLN)


76/116


EXAMPLE 9.11 (SOLN)


Establish the , axes

as shown.Center of circle C

located on the

-axis, at the pt:

MPaMPaMPa 6128 xyyx

MPa22

128

avg


EXAMPLE 9 11 (SOLN)


77/116


EXAMPLE 9.11 (SOLN)


Initial pt for = 0has coordinatesA(8, 6) areplotted. Apply

Pythagoras theorem

to shaded triangle

to get circles

radius CA,

MPa66.11

610

22

RR


EXAMPLE 9 11 (SOLN)


78/116


EXAMPLE 9.11 (SOLN)

Stresses on 30element

Since element is rotated 30counterclockwise, we

must construct a radial line CP, 2(30) = 60

counterclockwise, measured

from CA(= 0).

Coordinates of ptP(x, xy)

must be obtained. From

geometry of circle,

04.2996.3060

96.30106tan 1


EXAMPLE 9 11 (SOLN)


79/116


EXAMPLE 9.11 (SOLN)


The two stress components act onfaceBDof element shown, sincethexaxis for this face if oriented 30counterclockwise from thex-axis.

Stress components acting on adjacent faceDEof

element, which is 60

clockwise from +x-axis, arerepresented by the coordinates of pt Qon the circle.

This pt lies on the radial line CQ, which is 180fromCP.

MPa

MPa

66.504.29sin66.11

20.804.29cos66.112

''

'

yx

x


EXAMPLE 9 11 (SOLN)


80/116


EXAMPLE 9.11 (SOLN)


The coordinates of pt Qare

Note that here xyacts in

the ydirection.

)(Check!MPa

MPa

66.504.29sin66.11

2.1204.29cos66.112

''

'

yx

x


9 5 STRESS IN SHAFTS DUE TO AXIAL LOAD AND TORSION


81/116


Occasionally, circular shafts are subjected to

combined effects of both an axial load and torsion. Provided materials remain linear elastic, and

subjected to small deformations, we use principle

of superposition to obtain resultant stress in shaft

due to both loadings.

Principal stress can be determined using either

stress transformation equations or Mohrs circle.

9.5 STRESS IN SHAFTS DUE TO AXIAL LOAD AND TORSION


EXAMPLE 9 12


82/116


EXAMPLE 9.12

Axial force of 900 N and torque of 2.50 Nm are

applied to shaft. If shaft has a diameter of 40 mm,determine the principal stresses at a ptPon its

surface.


EXAMPLE 9 12 (SOLN)


83/116


EXAMPLE 9.12 (SOLN)

Internal loadings

Consist of torque of 2.50 Nm andaxial load of 900 N.

Stress components

Stresses produced at ptParetherefore

kPa

m

mmN9.198

02.0

2

02.050.24

J

Tc

kPa

m

N2.716

02.0

9004

A

P


EXAMPLE 9 12 (SOLN)


84/116


EXAMPLE 9.12 (SOLN)

Principal stresses

Using Mohrs circle, center of circle Cat the pt is

Plotting C(358.1, 0) and

reference ptA(0, 198.9),

the radius found was

R= 409.7 kPA. Principalstresses represented by

ptsBandD.

kPaavg 1.358

2

2.7160


EXAMPLE 9 12 (SOLN)


85/116


EXAMPLE 9.12 (SOLN)

Principal stresses

Clockwise angle 2p2can be

determined from the circle.It is 2p2= 29.1. The element

is oriented such that thexaxis

or 2is directed clockwise

p1= 14.5with thexaxisas shown.

kPakPa

2

1

6.517.4091.358

8.7677.4091.358


9 6 STRESS VARIATIONS THROUGHOUT A PRISMATIC BEAM


86/116


The shear and flexure formulas are applied to a

cantilevered beam that has a rectangular x-sectionand supports a load Pat its end.

At arbitrary section a-aalong

beams axis, internal shear V

and moment Mare developed

from a parabolic shear-stress

distribution,

and a linearnormal-stress

distribution.

9.6 STRESS VARIATIONS THROUGHOUT A PRISMATIC BEAM




87/116


The stresses acting on elements at

pts 1 through 5 along the section. In each case, the state of stress

can be transformed into principal

stresses, using either

stress-transformation equations

or Mohrs circle.

Maximum tensile stress acting on

vertical faces of element 1becomes smaller on corresponding

faces of successive elements, until its zero on

element 5.





88/116


Similarly, maximum compressive stress of vertical

faces of element 5 reduces to zero on that ofelement 1.

By extending this analysis to many vertical sections

along the beam, a profile of the results can be

represented by curves called stress trajectories.

Each curve indicate the

direction of a principal

stress having a constantmagnitude.



EXAMPLE 9 13


89/116


EXAMPLE 9.13

Beam is subjected to the distributed loading of

= 120kN/m. Determine the principal stresses in thebeam at ptP, which lies at the top of the web.

Neglect the size of the fillets and stress

concentrations at this pt.I= 67.1(10-6) m4.


EXAMPLE 9 13 (SOLN)


90/116


EXAMPLE 9.13 (SOLN)

Internal loadings

Support reaction on the beam B is determined, and

equilibrium of sectioned beam yields

Stress components At ptP,

mkNkN 6.3084 MV

MPa

m

m

4.45

1074.6

100.0106.30

46

3

mN

I

My


EXAMPLE 9 13 (SOLN)


91/116


EXAMPLE 9.13 (SOLN)

Stress components

At ptP,

Principal stresses

Using Mohrs circle, the principalstresses atPcan be determined.

MPa

mm

mmmN

2.35

010.01074.6

015.0175.01075.0108446

3

It

VQ


EXAMPLE 9 13 (SOLN)


92/116


EXAMPLE 9.13 (SOLN)

Principal stresses

As shown, the center of thecircle is at (45.4 + 0)/2 = 22.7,

and ptA(45.4, 35.2). We find

that radiusR= 41.9, therefore

The counterclockwise angle

2p2= 57.2, so that

MPa

MPa

2

1

6.649.417.22

2.197.229.41

6.282p


9 7 ABSOLUTE MAXIMUM SHEAR STRESS


93/116


A pt in a body subjected to a general

3-D state of stress will have a normalstress and 2 shear-stress components

acting on each of its faces.

We can develop stress-transformation

equations to determine the

normal and shear stress

components acting on

ANY skewed plane ofthe element.

9.7 ABSOLUTE MAXIMUM SHEAR STRESS




94/116


These principal stresses are assumed

to have maximum, intermediate andminimum intensity: maxintmin.

Assume that orientation of the element

and principal stress are known, thus

we have a condition known as triaxial

stress.





95/116


Viewing the element in 2D (y-z, x-z,x-y) we then

use Mohrs circle to determine the maximumin-plane shear stress for each case.





96/116


As shown, the element have a

45orientation and is subjectedto maximum in-plane shear

and average normal stress

components.





97/116



Comparing the 3 circles,

we see that the absolutemaximum shear stress

is defined by the circle

having the largest radius.

This condition can also

be determined directly by choosing the maximum

and minimum principal stresses:

max

abs

1392

minmax -max

abs




98/116


Associated average normal stress

We can show that regardless of the orientation of

the plane, specific values of shear stress on theplane is always less than absolute maximum shear

stress found from Eqn 9-13.

The normal stress acting on any plane will have a

value lying between maximum and minimumprincipal stresses, maxmin.


1492

minmax -avg




99/116


Plane stress

Consider a material subjected to planestress such that the in-plane principal

stresses are represented as maxand

int, in thexandydirections respectively;

while the out-of-plane principal stress in thez

direction is min= 0.

By Mohrs circle and Eqn. 9-13,


1592max

max'' -

maxabs

zx




100/116


Plane stress

If one of the principal stresses hasan opposite sign of the other, then

these stresses are represented as

maxand min, and out-of-plane

principal stress int= 0.

By Mohrs circle and Eqn. 9-13,


1692

minmax

max''

-

max

abs

yx




101/116



IMPORTANT

The general 3-D state of stress at a pt can berepresented by an element oriented so that onlythree principal stresses act on it.

From this orientation, orientation of element

representing the absolute maximum shear stresscan be obtained by rotating element 45about theaxis defining the direction of int.

If in-plane principal stresses both have the same

sign, the absolute maximum shear stress occursout of the plane, and has a value of 2max

max

abs




102/116


IMPORTANT

If in-plane principal stresses are of opposite signs,the absolute maximum shear stress equals themaximum in-plane shear stress; that is

2minmax max

abs


EXAMPLE 9.14


103/116


Due to applied loading,

element at the pt on theframe is subjected to the

state of plane stress shown.

Determine the principal

stresses and absolute

maximum shear stress

at the pt.


EXAMPLE 9.14 (SOLN)


104/116


( )

Principal stresses

The in-plane principal stresses can be determinedfrom Mohrs circle. Center of circle is on the axis at

avg= (20 + 20)/2 = 10 kPa. Plotting controlling pt

A(20, 40), circle can be drawn as shown. The

radius is

kPa2.41

401020 22

R


EXAMPLE 9.14 (SOLN)


105/116


( )

Principal stresses

The principal stresses at the pt where the circleintersects the -axis:

From the circle, counterclockwise angle 2, measured

from the CAto the axis is,

kPa

kPa

2.512.4110

2.312.4110

min

max

0.38

0.761020

40tan2 1

Thus,


EXAMPLE 9.14 (SOLN)


106/116


( )

Principal stresses

This counterclockwise rotation definesthe direction of thexaxis or min and

its associated principal plane. Since

there is no principal stress on the

element in thezdirection, we have

kPa

kPa

2.51

0

2.31

min

int

max


EXAMPLE 9.14 (SOLN)


107/116


( )

Absolute maximum shear stress

Applying Eqns. 9-13 and 9-14,

kPa

avg

102

2.512.31

2minmax

kPa

maxabs

2.412

)2.512.31

2minmax


EXAMPLE 9.14 (SOLN)


108/116


( )

Absolute maximum shear stress

These same results can be obtained by drawingMohrs circle for each orientation of an element about

thex,y, andzaxes. Since maxand minare of

opposite signs, then the absolute maximum shear

stress equals the maximum in-planeshear stress. This results from a 45

rotation of the element about thez

axis, so that the properly oriented

element is shown.


EXAMPLE 9.15


109/116


The pt on the surface of the cylindrical pressure

vessel is subjected to the state of plane stress.Determine the absolute maximum shear stress at this

pt.


EXAMPLE 9.15 (SOLN)


110/116


( )

Principal stresses are max= 32 MPa, int= 16 MPa,

and min= 0. If these stresses are plotted along theaxis, the 3 Mohrs circles can be constructed that

describe the stress state viewed in each of the three

perpendicular planes.

The largest circle has a radius of 16 MPa anddescribes the state of stress in the plane containing

max= 32 MPa and min= 0.

An orientation of an element 45within this planeyields the state of absolute maximum shear stress

and the associated average normal stress, namely,


EXAMPLE 9.15 (SOLN)


111/116


An orientation of an element 45within this plane

yields the state of absolute maximum shear stress andthe associated average normal stress, namely,

Or we can apply Eqns 9-13 and 9-14:

MPaMPa avgmaxabs 1616

MPa

MPa

avg

maxabs

162

032

2

162

032

2

minmax

minmax


EXAMPLE 9.15 (SOLN)


112/116


By comparison, maximum in-plane shear stress can

be determined from the Mohrs circle drawn betweenmax= 32 MPa and int= 16 MPa, which gives a value

of

MPa

MPa

avg

maxabs

24

2

163216

821632


CHAPTER REVIEW


113/116


Plane stress occurs when the material at a pt is

subjected to two normal stress components xand yand a shear stress xy.

Provided these components are known, then

the stress components acting on an element

having a different orientation can bedetermined using the two force equations of

equilibrium or the equations of stress

transformation.


CHAPTER REVIEW


114/116


For design, it is important to determine the

orientations of the element that produces themaximum principal normal stresses and the

maximum in-plane shear stress.

Using the stress transformation equations, we

find that no shear stress acts on the planes ofprincipal stress.

The planes of maximum in-plane shear stress

are oriented 45from this orientation, and onthese shear planes there is an associated

average normal stress (x+ y)/2.


CHAPTER REVIEW


115/116


Mohrs circle provides a semi-graphical aid for

finding the stress on any plane, the principal normalstresses, and the maximum in-plane shear stress.

To draw the circle, the and axes are

established, the center of the circle [(x+ y)/2, 0],

and the controlling pt (x,xy) are plotted.

The radius of the circle extends between these two

points and is determined from trigonometry.


CHAPTER REVIEW


116/116

The absolute maximum shear stress will be

equal to the maximum in-plane shear stress,provided the in-plane principal stresses have

the opposite sign.

If they are of the same sign, then the absolute

maximum shear stress will lie out of plane. Itsvalue is .2/0max

max

abs

ch09 stress transformation

Documents