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8/10/2019 teknik kekuatan bahan

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1

Transformations of Stress



2

• The most general state of stress at a point may

be represented by 6 components,

),, :(Note

stressesshearing,,

stressesnormal,,

xz zx zy yz yx xy

zx yz xy

z y x

• Same state of stress is represented by adifferent set of components if axes are rotated.

• The first part of the chapter is concerned with

how the components of stress are transformed

under a rotation of the coordinate axes. The

second part of the chapter is devoted to a

similar analysis of the transformation of the

components of strain.



3

• Plane Stress - state of stress in which two faces of

the cubic element are free of stress. For theillustrated example, the state of stress is defined by

.0,, and xy zy zx z y x

• State of plane stress also occurs on the free surface

of a structural element or machine component, i.e.,

at any point of the surface not subjected to an

external force.



4



5

MPa

y F

MPa

x F

3.43

030sin000.2002310

0

0.75

030cos000.2002310

0



6



7

Transformation of Plane Stress

• Consider the conditions for equilibrium of a

prismatic element with faces perpendicular to

the x, y, and x’ axes.

2cos2sin

2

2sin2cos22

2sin2cos22

xy

y x

y x

xy

y x y x

y

xy

y x y x

x

• The equations may be rewritten to yield



8

sinsincossin

coscossincos0

cossinsinsinsincoscoscos0

A A

A A A F

A A A A A F

xy y

xy x y x y

xy y

xy x x x



9

sinsincossincoscossincos0

cossinsinsin

sincoscoscos0

A A A A A F

A A

A A A F

xy y

xy x y x y

xy y

xy x x x

2

2

2

2

sincossin

cossincos0

cossinsin

sincoscos0

xy y

xy x y x y

xy y

xy x x x

F

F

2

2cos1

sin

cossin22sinand 2

2cos1cos

2

2



10

2cos2

2sin

2cos2

2sin0

2sin2cos22

2sin2cos22

0

2

2sin

2

2cos1

2

2sin

2

2cos1

0

xy y x y x

xy y x y x

xy

y x y x

x

xy

y x y x

x

xy y xy x x



11

2cos2sin2

2cos2sin2

2cos2sin2

2sin2cos2

2sin2cos22

2sin2cos22

2cos2sin2

2sin2cos22

2222

2

2

2

222

22

22

xy

y x

xy

y x

y x

xy

y x

y x

xy

y x

xy

y x y x

x

xy

y x y x

x

xy y x

y x

xy

y x y x

x



12

2

2

2

2

22222

2

2

2

222

2

2

222

22

22

2sin2cos2sin2cos22

2cos2sin22cos2sin2

2sin2cos2

2sin2cos22

xy

y x

y x

y x

x

xy

y x

y x

y x

x

xy

y x

xy

y x

y x

xy

y x

xy

y x y x

x

22

222

22

where

xy y x y x

ave

y xave x

R

R



13

Principal Stresses

• The previous equations are combined to

yield parametric equations for a circle,

22

222

22

where

xy y x y x

ave

y xave x

R

R

• Principal stresses occur on the principal

planes of stress with zero shearing stresses.

o

22

minmax,

90 byseparatedanglestwodefines : Note

22tan

22

y x

xy p

xy

y x y x



14

Maximum Shearing Stress

Maximum shearing stress occurs for ave x

2

45 byfromoffset

and90 byseparatedanglestwodefines: Note

2

2tan

2

o

o

22

max

y xave

p

xy

y x s

xy y x

R



15

Example 1

For the state of plane

stress shown, determine

(a) the principal planes,

(b) the principal stresses,(c) the maximum shearing

stress and the

corresponding normal

stress.

SOLUTION:

• Find the element orientation for the principalstresses from

y x

xy p

2

2tan

• Determine the principal stresses from

22

minmax,22 xy

y x y x

• Calculate the maximum shearing stress with

22

max 2 xy

y x

2

y x



16

Example 1

SOLUTION:

• Find the element orientation for the principalstresses from

1.233,1.532

333.11050

40222tan

p

y x

xy p

6.116,6.26 p

• Determine the principal stresses from

22

22

minmax,

403020

22

xy

y x y x

MPa30

MPa70

min

max

MPa10

MPa40MPa50

x

xy x



17

Example 1

MPa10

MPa40MPa50

x

xy x

2

1050

2

y x

ave

• The corresponding normal stress is

MPa20

• Calculate the maximum shearing stress with

22

22

max

4030

2

xy

y x

MPa50max

45 p s

6.71,4.18 s



18

Sample Problem 2

A single horizontal force P of 150 lb

magnitude is applied to end D of lever

ABD. Determine (a) the normal and

shearing stresses on an element at

point H having sides parallel to the x

and y axes, (b) the principal planes

and principal stresses at the point H .

SOLUTION:

• Determine an equivalent force-couple

system at the center of the transverse

section passing through H .

• Evaluate the normal and shearing

stresses at H .• Determine the principal planes and

calculate the principal stresses.



19

Sample Problem 2

SOLUTION:

• Determine an equivalent force-couplesystem at the center of the transverse

section passing through H .

inkip5.1in10lb150

inkip7.2in18lb150

lb150

x M

T

P

• Evaluate the normal and shearing stresses

at H .

421

4

4

1

in6.0

in6.0inkip7.2

in6.0

in6.0inkip5.1

J

Tc

I

Mc

xy

y

ksi96.7ksi84.80 y y x



20

Sample Problem 2

• Determine the principal planes and

calculate the principal stresses.

119,0.612

8.184.80

96.7222tan

p

y x

xy p

5.59,5.30 p

22

22

minmax,

96.72

84.80

2

84.80

22

xy

y x y x

ksi68.4

ksi52.13

min

max



21

Mohr’s Circle for Plane Stress

• With the physical significance of Mohr’s circle

for plane stress established, it may be applied

with simple geometric considerations. Criticalvalues are estimated graphically or calculated.

• For a known state of plane stress

plot the points X and Y and construct the

circle centered at C .

xy y x ,,

22

22 xy

y x y xave R

• The principal stresses are obtained at A and B.

y x

xy p

ave R

22tan

minmax,

The direction of rotation of Ox to Oa is

the same as CX to CA.



22


• With Mohr’s circle uniquely defined, the state

of stress at other axes orientations may bedepicted.

• For the state of stress at an angle with

respect to the xy axes, construct a new

diameter X’Y’ at an angle 2 with respect to XY .

• Normal and shear stresses are obtained

from the coordinates X’Y’.



23


• Mohr’s circle for centric axial loading:

0, xy y x

A

P

A

P xy y x

2

• Mohr’s circle for torsional loading:

J

Tc xy y x 0 0 xy y x

J

Tc



24

Example 3

For the state of plane stress shown,

(a) construct Mohr’s circle, determine

(b) the principal planes, (c) the

principal stresses, (d) the maximumshearing stress and the corresponding

normal stress.

SOLUTION:

• Construction of Mohr’s circle

MPa504030

MPa40MPa302050

MPa202

1050

2

22

CX R

FX CF

y xave



25

Example 3

• Principal planes and stresses

5020max CAOC OA MPa70max

5020max BC OC OB

MPa30max

1.532

30

402tan

p

pCP

FX

6.26 p



26

Example 3

• Maximum shear stress

45 p s

6.71 s

Rmax

MPa50max

ave

MPa20



27

Sample Problem 4

For the state of stress shown,

determine (a) the principal planes

and the principal stresses, (b) the

stress components exerted on the

element obtained by rotating thegiven element counterclockwise

through 30 degrees.

SOLUTION:

• Construct Mohr’s circle

MPa524820

MPa802

60100

2

2222

FX CF R

y xave



28

Sample Problem 4

• Principal planes and stresses

4.672

4.220

482tan

p

pCF

XF

clockwise7.33 p

5280

max

CAOC OA

5280

max

BC OC OA

MPa132max MPa28min



29

Sample Problem 4

6.52sin52

6.52cos5280

6.52cos5280

6.524.6760180

X K

CLOC OL

KC OC OK

y x

y

x

• Stress components after rotation by 30o

Points X’ and Y’ on Mohr’s circle that

correspond to stress components on therotated element are obtained by rotating

XY counterclockwise through 602

MPa341

MPa6.111

MPa4.48

y x

y

x

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