MOM Chapter 09 New-edited

7/29/2019 MOM Chapter 09 New-edited

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Chapter Objectives

Navigate between rectilinear co-ordinate systems for

stress components

Determine principal stresses and maximum in-plane

shear stress Determine the absolute maximum shear stress in 2D

and 3D cases

Copyright 2011 Pearson Education South Asia Pte Ltd


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1. Reading Quiz

2. Applications

3. General equations of plane-stress transformation

4. Principal stresses and maximum in-plane shear stress5. Mohrs circle for plane stress

6. Absolute maximum shear stress

7. Concept Quiz

In-class Activities



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READING QUIZ

1) Which of the following statement is incorrect?

a) The principal stresses represent the maximum and minimum

normal stress at the point

b) When the state of stress is represented by the principalstresses, no shear stress will act on the element

c) When the state of stress is represented in terms of the maximum

in-plane shear stress, no normal stress will act on the element

d) For the state of stress at a point, the maximum in-plane shear

stress usually associated with the average normal stress.



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APPLICATIONS



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GENERAL EQUATIONS OF PLANE-STRESSTRANSFORMATION


The state of plane stress at a point is uniquely represented by three

components acting on an element that has a specific orientation atthe point.

Sign Convention:

Positive normal stress acts outward

from all faces

Positive shear stress acts upwards

on the right-hand face of the element


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GENERAL EQUATIONS OF PLANE-STRESSTRANSFORMATION (cont)


Sign convention (continued)

Both the x-y and x-y system follow the right-hand rule

The orientation of an inclined plane (on which the normal and shearstress components are to be determined) will be defined using theangle . The angle is measured from the positive x to the positive

x-axis. It is positive if it follows the curl of the right-hand fingers.



Normal and shear stress components:

Consider the free-body diagram of the segment



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+Fx= 0; xA(xyA sin ) cos (yA sin ) sin

( xyA cos ) sin (xA cos ) cos = 0

x = x cos2 + y sin

2 + xy (2 sin cos )

+Fy= 0; xyA + (xyA sin ) sin (yA sin ) cos

( xyA cos ) cos + (xA cos ) sin = 0

xy = (y x) sin cos + xy (cos2sin2)

x =x + y

2xy

2 cos 2 + xy sin 2 +

xy = x + y2 sin 2 + xy cos 2

y =x + y

2xy

2 cos 2xy sin 2


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VARIABLE SOLUTIONS


Please click the appropriate icon for your computer to access the

variable solutions
http://localhost/var/www/apps/conversion/tmp/scratch_10/variable_solutions_9_2_windows.exehttp://localhost/var/www/apps/conversion/tmp/scratch_10/show-variable_solutions_9_2.html


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EXAMPLE 1


The state of plane stress at a point on the surface of the

airplane fuselage is represented on the element oriented as

shown in Fig. 94a. Represent the state of stress at the point

on an element that is oriented 30 clockwise from the

position shown.


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EXAMPLE 1 (cont)


The free-body diagram of the segment is as shown.

The element is sectioned by the line a-a.

Solutions


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EXAMPLE 1 (cont)

Applying the equations of force equilibrium in the x and y direction,

Solutions

(Ans)MPa15.4

030cos30sin2530sin30sin80

30sin30cos2530cos30cos50;0

'

''

x

xx

AA

AAAF

(Ans)MPa8.68

030sin30cos2530cos30sin80

30cos30cos2530sin30cos50;0'

x' y'

x'y'y

AA

AAAF



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EXAMPLE 1 (cont)

Repeat the procedure to obtain the stress on theperpendicularplaneb

b.

Solutions


(Ans)MPa8.25

030sin30sin5030cos30cos25

30cos30cos8030sin30cos25;0

'

''

x

xx

AA

AAAF

(Ans)MPa8.68

030cos30sin5030sin30sin25

30sin30cos8030cos30cos25-;0'

x' y'

x' y'y

AA

AAAF

The state of stress at the point can be

represented by choosing an element

oriented.


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EXAMPLE 2


The state of plane stress at a point is represented by the

element shown in Fig. 97a. Determine the state of stress at

the point on another element oriented 30 clockwise from

the position shown.


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EXAMPLE 2 (cont)


From the sign convention we have,

To obtain the stress components on plane CD,

Solutions

30MPa25MPa50MPa80 xyyx

(Ans)MPa8.682cos2sin2

(Ans)MPa8.252sin2cos22

''

'

xy

yx

yx

xy

yxyx

x


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EXAMPLE 2 (cont)


To obtain the stress components on plane BC,

The results are shown on the element as shown.

Solutions

60MPa25MPa50MPa80 xyyx

(Ans)MPa8.682cos2sin2

(Ans)MPa15.42sin2cos22

''

'

xy

yx

yx

xy

yxyx

x


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IN-PLANE PRINCIPAL STRESS


The principal stresses represent the maximum and

minimum normal stress at the point.

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

Solving this equation leads to = p

2

2

2,122

2/

2tan

xy

yxyx

yx

xy

p

2cos22sin2

2

'xy

yxx

d

d


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IN-PLANE PRINCIPAL STRESS (cont)


2cos22sin2

2

'xy

yxx

d

d


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IN-PLANE PRINCIPAL STRESS (cont)


Solving this equation leads to = p; i.e

2/

2tan

yx

xy

p

2

2

2,122

xy

yxyx


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MAXIMUM IN-PLANE PRINCIPAL STRESS


The state of stress can also be represented in terms of

the maximum in-plane shear stress. In this case, an

average stress will also act on the element.

Solving this equation leads to = s; i.e

And there is a normal stress on the

plane of maximum in-plane shear stress

2

2

plane-inmax2

xy

yx

02sin2cos22

''

xy

yxyx

d

d

xy

yx

s

2/2tan

2

yx

avg


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EXAMPLE 3


When the torsional loading Tis applied to the bar in Fig. 9

13a, it produces a state of pure shear stress in the material.

Determine (a) the maximum in-plane shear stress and the

associated average normal stress, and (b) the principal

stress.


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EXAMPLE 3 (cont)


From the sign convention we have,

a) Maximum in-plane shear stress is

b) For principal stress,

Solutions

xyyx 00

(Ans)02

2

2

2

plane-inmax

yxavgxy

yx

(Ans)22

135,452/

2tan

22

2,1

12

xy

yxyx

pp

yx

xy

p


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EXAMPLE 3 (cont)


If we use

Thus, acts at as shown in Fig. 913b, and acts on the

other face

Solutions

452p

90sin00

2sin2cos22

' xy

yxyx

x

2 452p 1 45

1p


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EXAMPLE 4


When the axial loading Pis applied to the bar in Fig. 914a,

it produces a tensile stress in the material. Determine (a) the

principal stress and (b) the maximum in-plane shear stress

and associated average normal stress.


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EXAMPLE 4 (cont)


From the established sign convention,

Principal Stress

Since no shear stress acts on this element,

Solutions

(Ans)0 21

000 xyyx


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EXAMPLE 4 (cont)


Maximum In-Plane Shear Stress

To determine the proper orientation

of the element,

Solutions

2090sin

2

02cos2sin

2''

xy

yx

yx

(Ans)22

0

2

(Ans)2

02

0

2

45,45;0

2/02/2tan

avg

2

2

2

2

planeinmax

21

yx

xxy

yx

ss

xy

yx

s


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MOHRS CIRCLE OF PLANE STRESS


A geometrical representation of equations 9.1 and 9.2;

i.e.

Sign Convention:

is positive to the right, and is positive downward.

2sin2sin2

2sin2cos22

''

'

xy

yx

yx

xy

yxyx

x


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EXAMPLE 5


Due to the applied loading, the element at pointA on the

solid shaft in Fig. 918a is subjected to the state of stress

shown. Determine the principal stresses acting at this point.


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EXAMPLE 5 (cont)


Construction of the Circle

From Fig. 918a,

The center of the circle is at

The reference point A(-12,-6) and the center C(-6, 0) are plotted in Fig. 9

18b.The circle is constructed having a radius of

Solutions

MPa62

012

avg

MPa60MPa,12 xyyx ,

MPa49.86612 22 R


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EXAMPLE 5 (cont)


Principal Stress

The principal stresses are indicated by

the coordinates of points B and D.

The orientation of the element can be

determined by calculating the angle

Solutions

(Ans)MPa5.1449.86

(Ans)MPa49.2649.8

,haveWe

2

1

21

5.22

0.45612

6tan2

2

2

1

p

p


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EXAMPLE 6


The state of plane stress at a point is shown on the element

in Fig. 919a. Determine the maximum in-plane shear stress

at this point.

EXAMPLE 6 ( t)


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EXAMPLE 6 (cont)



We first construct of the circle,

The center of the circle Cis on the axis at

From point Cand the A(-20, 60) are plotted, we have

Solutions

60and90,20 xyyx

MPa4.815560 22 R

MPa352

9020

avg

EXAMPLE 6 ( t)


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EXAMPLE 6 (cont)


Maximum In-Plane Shear Stress.

Max in-plane shear stress and average normal stress are

The counter-clockwise angle is

Solutions

(Ans)MPa35,MPa81.4plane-inmax avg

(Ans)3.2160

3520tan2 1

1

s

EXAMPLE 7


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EXAMPLE 7


The state of plane stress at a point is shown on the element

in Fig. 920a. Represent this state of stress on an element

oriented 30counterclockwise from the position shown.

EXAMPLE 7 ( t)


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EXAMPLE 7 (cont)



We first construct of the circle,

The center of the circle Cis on the axis at

From point Cand the A(-8, -6) are plotted, we have

Solutions

6and12,8 xyyx

66.11610 22 R

MPa22

128

avg

EXAMPLE 7 ( t)


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EXAMPLE 7 (cont)


Stresses on 30 Element

From the geometry of the circle,

The stress components acting on the adjacent face DEof the element,

which is 60 clockwise from the positivexaxis, Fig. 920c, are represented

by the coordinates of point Q on the circle.

Solutions

(Ans)MPa66.504.29cos66.11

(Ans)MPa20.804.29cos66.112

04.2996.306096.3010

6tan

''

'

1

yx

x

(Ans)(check)MPa66.504.29sin66.11

(Ans)MPa22.1204.29cos66.112

y 'x'

'

x


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ABSOLUTE MAXIMUM SHEAR STRESS


State of stress in 3-dimensional space:

2

2

minmaxavg

minmaxmaxabs


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ABSOLUTE MAXIMUM SHEAR STRESS (cont)




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ABSOLUTE MAXIMUM SHEAR STRESS (cont)



EXAMPLE 8


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EXAMPLE 8


The point on the surface of the cylindrical pressure vessel in

Fig. 924a is subjected to the state of plane stress.

Determine the absolute maximum shear stress at this point.

EXAMPLE 8 (cont)


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EXAMPLE 8 (cont)


An orientation of an element 45 within this plane yields the state of absolute

maximum shear stress and the associated average normal stress, namely,

Same result for can be obtained from direct application of

Solutions

(Ans)MPa16,MPa16 avgmaxabs

(Ans)MPa16

2

032

MPa162

32

2

avg

1maxabs

EXAMPLE 8 (cont)


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EXAMPLE 8 (cont)


By comparison, the maximum in-plane shear stress can be determined from

the Mohrs circle,

Solutions

(Ans)MPa242

1632

MPa82

1632

avg

maxabs


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CONCEPT QUIZ

1) Which of the following statement is untrue?

a) In 2-D state of stress, the orientation of the element representing

the maximum in-plane shear stress can be obtained by rotating

the element 45 from the element representing the principle

stresses.b) In 3-D state of stress, the orientation of the element representing

the absolute maximum shear stress can be obtained by rotating

the element 45 about the axis defining the direction ofint.

c) If the in-plane principal stresses are ofopposite sign, then the

absolute maximum shear stress equals the maximum in-planestress, that is, abs max = (maxmin)/2

d) Same as (c) but the principal stresses are of the same sign.

MOM Chapter 09 New-edited

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