chapter 7 combined stresses and strains.(som-201)

8/8/2019 Chapter 7 Combined Stresses and Strains.(SOM-201)

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MEC201Mechanics of

Solids


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Combined

Stresses and

Strains


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Combined stresses

P


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Wire over pulley

Section atAA

100 N

500

Dia 1A A

Bending stresses

+

Direct stresses=

Resultant stresses


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Spirally-woundcylindrical vessel

z

r

zz

nn

nt

Consider a spirally-wound cylindrical vessel of diameter 40 mm. Thevessel is made from a thin copper strip of width 30 mm wound spirally.The edges of the strip are brazed together. The diameter of thecylinder is 40 mm and the wall thickness is 2 mm. What is the shear

strength of the brazed joint if the internal pressure that this cylindercan withstand is 3MPa.

30 mm

(

)

40

mm

= 0.234 rad= 42.2o


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zz

nn

nt

nn and nt are obtained as 23.2 MPa and = 7.5

MPa, respectively


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Plain stresses

x

z

y

xxxx

yy

yy

yxxy

xyyx


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Two-Dimensional Stress

x

yxxxx

yy

yy

xy

xy

y

x

y

x

x

y

First convert stresses in toforces since m is of forcesnot stresses.

Take depth = 1


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Stresses on Plane

xx

yy

xy

x

y

n

nnnt

+


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Stresses on Plane...

xx

yy

xy

xy

nnnt+

+xy


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xx

yy

xy

x

y

nnnt


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xx

yy

xyx

y

nnnt


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Ifxy= 0

2

n

n'


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Ifxy 0

X

Y


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( nn , nt )


X

Y

N

x

y

x

yn

t

2

2


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( nn , nt )

X

Y

N2

2x

y

x

yn

t

Plot the stress components on the -plane using the convention thata positive shear stress on an x-like face is plotted downward while apositive stress on a y-like face is plotted upwards. Reverse for thenegative shear stresses


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X- and Y- Faces

A

B

B

A

A

B

A

B


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Principal Planes

x

y

X

Y

2

AB a

a,b =


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Uni-axial Tension

xx

XY

C

D

Ductile Failure

Brittle Failure

C

D


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Uni-axial Compression

yy

Y X

S1

S2

S1

S2


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Hydrostatic Loading

1

1

X,Y No shearCompressive stressindependent of direction


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Equal Tensile Loading

1

1

X,Y

No shear

Tensile stressindependent ofdirection


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Shear Loading

Z

2 1 Principal Directions

o

o

z

1

2


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Torsion Failure

ox

y 12

Principal Directions

Torsion failure of aductile material

Torsion failure of abrittle material


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An Example

x

y

40 kPa

110 kPa

50 kPa

X

Y

CP2

P1

C = (( + )/ , ) = ( , ),110 50 2 0 80 0 R = [{( )/ }110 50 2 2 + 402)]= 50P1= (( + ), ) = ( , ),80 50 0 130 0 P2= (( ), ) = ( , )80 50 0 30 0

P1

P2

12

(c)

2

63 7 MPa1


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Torsional Pendulum

.63 7 MPa

.41 8 MPa

z

50 N60o

=1 mm

L= 1 m

z

S2

1

Z

S1

S2

2

1


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Torsional Pendulum

z

50 N60o

z

S2

1

Z

S1

S2

2


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Displacement and strain

A

D

C

B

D

B

u

u+(u/x)x

v

v+(v/y)y

v+(v/x)x

x,u

y,v

u+(u/y)y

A


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Transformation of planestrains

y,v

x,uA

B

B

C

ED

X

y

B

A

The coordinates of point B arelcos and lsin


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Transformation of planestrains

y,v

x,uA

B

B

C

ED

X

y

B

A


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y

xC

( xx +yy)/2

Mohr circle of strain

X

Y

/2

O

2

Y

X

y'y

x'x

x'y /2

x'

y'

yy

xx

yx

yy

xx

xy


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An Example:

In a condition of plane stress in a steelplate, xx = 130 MPa, yy = -70 MPa, xy=

80 MPa, Find the principal strains.

y

x

Y

Radius = sqrt(1002

+ 802

)= 128

X

1 = 30 + 128 = 158 MPa

2 = 30 - 128 = -98 MPa

30


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Example, continued:

In the last example:1 = 158 MPa

2 = -98 MPa

1 = (1 2)/E = (158 0.25(-98))/200103 = 0.910-3

2 = (21)/E = (-98 0.25(158))/200103 = -

0.6910-3

Find principal strains


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Distortion in computergraphics

x

y1% stretch

5%

com

pression

/2

y

x

S2

S1

Radius = [0.01 (0.05)]/2 = 0.03.

Maximum angular strain S1S2equal to two

times the radius of the circle, i.e., 0.06

e a ons p e ween


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e a ons p e weenelastic properties E, G

and

12

x

y

y

x

o

x

y 12

X

Y

12

X

Y

12

=o/G 1 = 2 = o

e a ons p e ween


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e a ons p e weenelastic properties E, G

and yx

x

y

o

=o/G

d d i S i


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Bonded Resistance StrainGauges

Soldertabs

Alignmentmarks

A

ctive

gridlength

Backing sheet


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An application

Air Gauge

Cantilevered reed

Modelvocalchord

easur ng s ear


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easur ng s earstresses with strain

gauges o

z

/2

O

Z

o/2

o/2

P1P2

z

P1P2

= o/2

o/2

chapter 7 combined stresses and strains.(som-201)

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