ce270 stress mohrs s10 purdue university

7/25/2019 CE270 Stress Mohrs S10 Purdue University

1/24

J. Liu CE270 Spring 2010

CE270 Spring 2010

Stress Transformations

Often interested in maximum normal and

shear stresses at a point and orientation on

the element


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Plane Stress

Stress can be analyzed in a single plane (no

load on surface)

Typically on thin sections, such as a beamweb

x

y

z

x

y


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x

x

y

y

xy

State of planestress for elementis unique to

orientation


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How do we transform the

stress components when we

change orientation?

x

xy

Use equilibrium of

FORCES!


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First, Sign Convention

Positive (+) if pointing in (+) direction on (+)

face

Positive (+) if pointing in (-) direction on (-)face

x

y

OR normal stress

(outward)

shear stress (up on

right-hand side)


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x

xy

A x face

Asin

Acos

y

x

xy

yx

)cos)(cos(

)cos)(sin()sin)(sin()sin)(cos(0

''

A

AAAAF

x

xyy

xyxx

4/23/10 Fix! (had them backwards on

Wednesday)


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cossin22sin

sincos2cos

)cossin2(sincos

22

22' xyyxx

Repeat for other stresses


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Stress Transformations

2sin2cos22

2cos2sin2

2sin2cos22

'

''

'

xyyxyx

y

xyyx

yx

xyyxyx

x

Hibbeler Chp. 9


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Example

x

z

A

L

T=294.5 N-m

wood grainGiven:

Wooden shaft

dia. = 100 mm

Find:

stresses parallel

and perpendicular

to the grain at

point A ( to z

axis)A


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x

z

A

L

T=294.5 N-m

wood grain

JTcmax

4)05.0(

2

)05.0)(5.294(

m

mNm

Pax 6105.1

BUT! (+) on (+) face, therefore (+) shear stress

x

y

max


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x

z

A

L

T=294.5 N-m

wood grain

x

y

max

72

72


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x

z

A

L

T=294.5 N-m

wood grain

72

Perpendicular to grain (could cause splitting)

2sin2cos

22

' xyyxyx

y

MPaPaxy 88.0))72(2(sin)105.1( 6'

Compression perpendicular to grain


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x

z

A

L

T=294.5 N-m

wood grain

72

Parallel to grain (could shear along grain?)

2cos2sin

2

'' xyyx

yx

MPaPaxyx 2.1144cos)105.1( 6

''


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Parallel to grain (pulling on grain)

2sin2cos22

' xyyxyx

x

MPax 88.0'

0.88 MPa

0.88 MPa

1.2 MPa

Final

State ofStress


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Principal Stresses

Maximum and minimum normal stresses

acting at a point

Can differentiate with respect to and set=0,

etc., or can use Mohrs Circle

Equations for plane stress transformation can

be put into a graphical format


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Mohrs Circle

22

22''

2'

2

2

xyyx

yxavg

yxavgx

R

R


17/24

Sign Convention Mohrs Circle

positive to the right

clockwise above,counterclockwise below

In the kitchen, the clock is above, and the counter is below.


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General Procedure

1. Plot C = avg

2. Plot (x

,xy

), (y

,yx

)

3. Determine R

4. Draw circle

C

( x, xy)

( y, yx)x

y

avg

Reference state


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on element = 2 on Mohrs Circle

( x, xy)

( y, yx)

x

y

( y, yx)

( x, xy)

x

y

90

180


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Principal Stresses

( x, xy)

( y, yx)

p from

reference

state

12

p

R

R

avg

avg

2

1

1

2


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Maximum Shear Stress

Ryx max''

( x, xy)

( y, yx)

s from

referencestate

xymax

s

45

9022

ps

ps


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Example

x

z

A

L

T=294.5 N-m

wood grain

Given:

Wooden shaft

dia. = 100 mm

Find:stresses parallel

and perpendicular

to the grain at

point A ( to zaxis) using Mohrs

Circle


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Reference

state

x

y

max =1.5MPa = R (0, yx)

(0, xy)

2 =144

144 -90 =54

( y, yx)

( x, xy)

MPaRx 88.054cos'

MPaRy 88.054cos'

MPaRyx 2.154sin''

C


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0.88 MPa

0.88 MPa

1.2 MPa

Final

State of

Stress

MPaRx 88.054cos'

MPaRy 88.054cos' MPaR

yx

2.154sin''

ce270 stress mohrs s10 purdue university

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