6-Transformation of Stress & Strain [Jan2013]

8/17/2019 6-Transformation of Stress & Strain [Jan2013]

1/49

Faculty of Mechanical Engineering

Engineering Mechanics Centre of Studies

V{t àxÜ I

Transformations of

Stress & Strain

Ch 6- 1

1. Ferdinand P. Beer, E. Russell Johnston,Jr, John T. Dewolf, David F. Mazurek “ Mechanics of Materials” 5th Edition in SI units

2. R.C.Hibbeler “ Mechanics of Materials “ Seventh Edition

Materials for this chapter are taken from :


2/49



The most general state of stress at a point may be

represented by 6 components,

Introduction

),, :(Note

stressesshearing,,

,,

xz zx zy yz yx xy

zx yz xy

z y x

τ τ τ τ τ τ

τ τ τ

===

Same state of stress is represented by a different set

of components if axes are rotated.

Ch 6- 2

the components of stress are transformed under arotation of the coordinate axes. The second part of

the chapter is devoted to a similar analysis of the

transformation of the components of strain.


3/49



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

cubic element are free of stress. For the illustrated

example, the state of stress is defined by

Plane Stress

.0,, and xy === zy zx z y x τ τ σ τ σ σ

State of plane stress occurs in a thin plate subjected to

forces acting in the midplane of the plate.

Ch 6- 3

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.


4/49



Consider the conditions for equilibrium of a prismatic

element with faces perpendicular to the x , y , and x’

axes.

Transformation of Plane Stress

( ) ( )

( ) ( )

( ) ( )

( ) ( ) θ θ τ θ θ σ

θ θ τ θ θ σ τ

θ θ τ θ θ σ

θ θ τ θ θ σ σ

sinsincossin

coscossincos0

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

∆+∆−

∆−∆+∆==∑

∆−∆−

∆−∆−∆==∑

′′′

′′

The equations may be rewritten to yield

Ch 6- 4

θ τ θ

σ σ

τ

θ τ θ σ σ σ σ

σ

θ τ θ σ

2cos2sin2

2sin2cos22

2sin2cos22

xy

y x

y x

xy y x y x

y

xy

y x y x

x

+

−

−=

−−

−+

=

+

−

+=

′′

′

′


5/49



The previous equations are combined to yield

parametric equations for a circle,

222 τ σ σ =+− ′′′

Principal Stresses

22

22

where

xy y x y x

ave R τ σ σ σ σ

σ +

−=

+=

Principal stresses occur on the principal planes

of stress with zero shearing stresses.

Ch 6- 5

o

2minmax,

90 byseparatedanglestwodefines : Note

22tan

22

y x

xy p

xy y x y x

σ σ

τ θ

τ σ σ σ σ

σ

−=

+

−±

+=


6/49



Maximum shearing stress occurs for ave x σ σ =′

Maximum Shearing Stress

45 byfromoffset

and90 byseparatedanglestwodefines: Note

2

2tan

2

o

o

22

max

p

xy

y x s

xy y x

R

θ

τ

σ σ θ

τ σ σ

τ

−−=

+

−==

Ch 6- 6

2 y xave σ σ σ σ

+==′


7/49



For the state of plane stress

shown, determine (a) the

principal planes, (b) the principal

Find the element orientation for the principal

stresses from

Example 6.1

Solut ion

stresses, (c) the maximum

shearing stress and the

corresponding normal stress.

y x

xy p

σ σ

τ θ

−=

22tan

Determine the principal stresses from

22

minmax,22

xy y x y x

τ σ σ σ σ

σ +

−±+=

Calculate the maximum shearin stress with

Ch 6- 7

22

max2

xy y x

τ σ σ

τ +

−=

2

y x σ σ

σ

+

=′

F lt f M h i l E i i


8/49



Find the element orientation for the principal

stresses from:

( )=

+== 333.1

40222tan

xyτ θ

Example 6.1

°°=−−−

1.233,1.5321050

p

y x

θ σ σ

°°= 6.116,6.26 pθ

Determine the principal stresses from:

2 −+ x σ σ σ σ

50MPa 40MPa

10MPa

x xy

y

σ τ

σ

= + = +

= −

Decl are !

Ch 6- 8

( ) ( )22

minmax,

403020

22

+±=

+

= xyτ σ

MPa30

MPa70

min

max

−=

=

σ

σ



9/49



Calculate the maximum shearing stress with

22

+

−

= y x

τ σ σ

τ

Example 6.1

( ) ( )22 4030 +=

MPa50max =τ

45−= p s θ θ

°°−= 6.71,4.18 sθ

50MPa 40MPa

10MPa

x xy

y

σ τ

σ = + = += −

Ch 6- 9

2

1050

2

−=

+==′ y x

ave

σ σ σ σ

The corresponding normal stress is

MPa20=′σ



10/49



A plane stress system is as shown in the

diagram below. Determine the direct and

Example 6.2

SOLUTION: METHOD OF EQUILIBRIUM

=- .

10 MPa

8 MPa

15°

P

'

'

10 sin15cos75 8 sin15cos75

2 cos15cos15 8 cos15cos75 0

.

0;

10 sin15sin75 8 sin15sin15

x

y

A A A

A A

Ans

F

A A A

σ

σ

τ

∆ − ∆ + ∆

+ ∆ + ∆ =

=

Σ =

∆ + ∆ + ∆

-5.916MPa

տ

Ch 6- 10

2 MPa

P 2 cos15cos75 8 cos15cos15 0

.

A A

Ansτ + ∆ − ∆ == 3.93MPa



11/49



With the physical significance of Mohr’s circle for

plane stress established, it may be applied with

simple geometric considerations. Critical values

Mohr’s Circle for Plane Stress

.

For a known state of plane stress

plot the points X and Y and construct the circle

centered at C .

22

22 xy

y x y xave R τ

σ σ σ σ σ +

−=

+=

Ch 6- 11

.

y x

xy p

ave R

σ σ

τ θ

σ σ

−=

±=

22tan

minmax,

The direction of rotation of Ox to Oa is the sameas CX to CA.



12/49

y g g


With Mohr’s circle uniquely defined, the state of

stress at other axes orientations may be depicted.


For the state of stress at an angle q with respect to

the xy axes, construct a new diameter X’Y’ at an

angle 2q with respect to XY .

Normal and shear stresses are obtained from the

Ch 6- 12

coordinates X’Y’.



13/49


Mohr’s circle for centric axial loading:


0, === xy y x P

τ σ σ P

xy y x2

=== τ σ σ

Mohr’s circle for torsional loading:

Ch 6- 13

Tc xy y x === τ σ σ 0 0=== xy y x Tc τ σ σ



14/49


Example 6.3

Solut ion

For the state of plane stress shown, (a)

’

Ch 6- 14

,

principal planes, (c) the principal stresses,

(d) the maximum shearing stress and the

corresponding normal stress.

Construction of Mohr’s circle

( ) ( )

( ) ( ) MPa504030

MPa40MPa302050

MPa202

1050

2

22 =+==

==−=

=−+

=+

=

CX R

FX CF

y xave

σ σ σ


E i i M h i C t f St di


15/49


Principal planes and stresses

5020max +=+== CAOC OAσ

Example 6.3

MPa70max =σ

5020min −=−== BC OC OBσ

MPa30min −=σ

°=

==

1.532

30

402tan p

CP

FX

θ

θ

Ch 6- 15

°= 6.26 pθ




16/49


Example 6.3

Ch 6- 16

Maximum shear stress

°+= 45 p s θ θ

°=6.71 sθ

R=maxτ

MPa50max =

τ

aveσ σ =′

MPa20=′

σ




17/49


Example 6.4

Solut ion

For the state of stress shown,determine (a) the principal planes

and the principal stresses, (b) the

Ch 6- 17

stress components exerted on the

element obtained by rotating the

given element counterclockwise

through 30 degrees.

Construct Mohr’s circle

( ) ( ) ( ) ( ) MPa524820

MPa802

60100

2

2222 =+=+=

=+

=+

=

FX CF R

y xave

σ σ σ




18/49


Example 6.4

Ch 6- 18

Principal planes and stresses

°=

===

4.672

4.220

482tan

p

pCF

XF

θ

θ

clockwise7.33 °= pθ

5280

max

+=

+== CAOC OAσ

5280

max

−=

−== BC OC OAσ

MPa132max +=σ MPa28min +=σ




19/49

g g

Example 6.4

180 60 67.4 52.6

80 52cos52.6

80 52cos52.6

x

y

OK OC KC

OL OC CL

ϕ

σ

σ

′

′

= ° − °− ° = °

= = − = − °

= = + = + °Stress components after rotation by 30o

Ch 6- 19

s n . x yτ ′ ′ = = −

Points X’ and Y’ on Mohr’s circle thatcorrespond to stress components on the

rotated element are obtained by rotating XY

counterclockwise through °= 602θ

48.4MPa

111.6MPa

41.3MPa

x

y

x y

σ

σ

τ

′

′

′ ′

= +

= +

= −

S l P bl 6 1


20/49

Supplementary Problems 6.1

1) At a point in the structural member, the stresses

are represented as in figure below. Find;

a) the magnitude and orientation of theprincipal stresses and,

[96.1MPa;23.94 MPa;28.15o]

a) the magnitude and orientation of the

maximum shearing stresses andassociated normal stresses.

[36.08MPa; -16.85o ; 60 MPa]

80 MPa

40 MPa

30 MPa

y

10 MPa

Ch 6 - 20

2) For the state of plane stress shown, determinethe normal and shearing stresses on the planes

whose normal are at +60o and + 150o with the x-

axis. Show stresses on a sketch of the element.

[for +60: σ N = 7.32MPa; τ N = -10MPa]

[for +150: σ N = -27.32MPa; τ N = 10MPa]

x

20 MPa

10 MPa

S l P bl 6 1


21/49


3) A structural member is subjected to two set of loadings. Each separately produce

stress conditions at a point A as shown in Figure 3(a) and Figure 3(b).

x, y xy .

[37.5 MPa; 22.5 MPa; -14.33 MPa]

a) By continuing the effect of both stress conditions Q3(a), determine the

principal stresses and their orientations with respect to the x -axis and also the

magnitude of the maximum shearing stress.

10 MPa

20 MPa30 MPa

Ch 6 - 21

Figure 3(a) Figure 3(b)

AA

15 MPa

5 MPa

30o

x

+




22/49

State of stress at Q defined by: zx yz xy z y x τ τ τ σ σ σ ,,,,,

Consider the general 3D state of stress at a point and thetransformation of stress from element rotation

General State of Stress

Consider tetrahedron with face perpendicular to the line QN

with direction cosines: z y x λ λ λ ,,

The requirement leads to,∑ = 0n F

x z zx z y yz y x y

z z y y x xn

λ λ τ λ λ τ λ λ τ

λ σ λ σ λ σ σ

222

222

+++

++=

Ch 6- 22

Form of equation guarantees that an element orientationcan be found such that

222ccbbaan λ σ λ σ λ σ σ ++=

These are the principal axes and principal planes and the

normal stresses are the principal stresses.




23/49

Application of Mohr’s Circle to the Three- Dimensional Analysis of Stress

Transformation of stress for an element The three circles represent the normal

Ch 6- 23

ro a e aroun a pr nc pa ax s may e

represented by Mohr’s circle.

an s ear ng s resses or ro a on aroun

each principal axis.

Points A, B, and C represent the principal

stresses on the principal planes (shearing

stress is zero)minmaxmax

21 σ σ τ −=

Radius of the largest circle yields the

maximum shearing stress.




24/49

In the case of plane stress, the axis

perpendicular to the plane of stress is a principal

axis (shearing stress equal zero).

Application of Mohr’s Circle to the Three-Dimensional Analysis of Stress

a) the corresponding principal stresses arethe maximum and minimum normal

stresses for the element

If the points A and B (representing the principal

planes) are on opposite sides of the origin, then

Ch 6- 24

b) the maximum shearing stress for the

element is equal to the maximum “in-plane”shearing stress

c) planes of maximum shearing stress are at

45o to the principal planes.




25/49

Application of Mohr’s Circle to the Three-Dimensional Analysis of Stress

If A and B are on the same side of the origin (i.e.,

have the same sign), then

b) maximum shearing stress for the element is

equal to half of the maximum stress

a) the circle defining smax, smin, andt max for the element is not the circle

corresponding to transformations within

the plane of stress

Ch 6- 25

c) planes of maximum shearing stress are at

45 degrees to the plane of stress




26/49

Transformation of Plane Strain

Plane strain - deformations of the material takeplace in parallel planes and are the same in

each of those planes.

Plane strain occurs in a plate subjected along

its edges to a uniformly distributed load and

restrained from expanding or contracting

laterally by smooth, rigid and fixed supports

( )0 :strainof components

x === zy zx z xy y γ γ ε γ ε ε

Ch 6- 26

Example: Consider a long bar subjected to

uniformly distributed transverse loads. State

of plane stress exists in any transverse

section not located too close to the ends of

the bar.




27/49

Transformation of Plane Strain

State of strain at the point Q results indifferent strain components with respect to

the xy and x’y’ reference frames.

22

( ) ( )

( ) y xOB xy

xy y xOB

y y x

ε ε ε γ

γ ε ε ε ε

γ ε ε ε

+−=

++=°=

=

2

45

coss ns ncos

21

Applying the trigonometric relations used

for the transformation of stress,

Ch 6- 27

cos2 sin 22 2 2

cos2 sin 22 2 2

sin 2 cos22 2 2

x y x y xy

x

x y x y xy

y

x y x y xy

ε ε ε ε γ

ε θ θ

ε ε ε ε γ ε θ θ

γ ε ε γ θ θ

′

′

′ ′

+ −

= + ++ −

= − −

−= − +




28/49

Mohr’s Circle for Plane Strain

The equations for the transformation of plane

strain are of the same form as the equations

for the transformation of plane stress - Mohr’s

circle techni ues a l .

Abscissa for the center C and radius R ,

22

222

+

−=

+=

xy y x y xave R

γ ε ε ε ε ε

Principal axes of strain and principal strains,

Ch 6- 28

R R aveave

y x

xy

p

−=+=

−=

ε ε ε ε

ε ε

γ

θ

minmax

2tan

( ) 22max 2 xy y x R γ ε ε γ +−==

Maximum in-plane shearing strain,




29/49

Example 6.5

The strain at a point in a body are as

follows:

800 200 600ε ε = = =

Solut ion

Using Mohr’s circle

800 xε µ =

Determine:

a) The principal strains and directions,

and

b) The strain εa in the direction of 60º

with the x -axis and the strain εbperpendicular to εa and the shearing

x y xy

Setting the coordinates

( )

( )

800, 300

200,300

X

Y

µ

µ

−

2001

3002

y

xy

ε µ

γ µ

=

=

Ch 6- 29

strain γab.

Centre OC800 200

5002

OC +

= =

Radius, R

2 2300 300 424.3 R µ = + =


30/49




31/49

Absolute Maximum Shear Strain

Absolute maximum shear strain is found from the circle having the largest radius.

It occurs on the element oriented 45° about the axis from the element shown in its

original position.

minmax

minmaxmaxabs

ε ε ε

ε ε γ

+=

−=

Ch 6- 31

Faculty of Mechanical EngineeringEngineering Mechanics Centre of Studies


32/49

Absolute Maximum Shear Strain

For plane strain we have,

This value represents the absolute maximum shear strain for the material.

Ch 6- 32

( ) maxmax''maxabs ε γ γ ==

z x minmaxmax''maxabs ε ε γ γ −==

y x



33/49

Example 6.6

The state of plane strain at a point is represented by the strain components,

( ) ( ) ( )6 6 6400 10 , 200 10 , 150 10 x y xyε ε γ − − −= − = =

e erm ne e max mum n-p ane s ear s ra n an e a so u e max mum s ear s ra n.

From the strain components, the centre of the circle is on the ε axis at:

( ) ( )66 10100102

200400 −− −−=+−

=avg ε

−γ −−

Solut ion

Ch 6- 33

nce , e re erence po n as coor na es

2

= ,−

Thus the radius of the circle is

( ) ( ) ( )9622 103091075100400 −− =

+−= R



34/49

Example 6.6

Computing the in-plane principal strains, we have

( )66

66

max 1020910309100

−−

−−

−=−−=

=+−=ε

From the circle, the maximum in-plane shear strain is

m n

( )[ ] (Ans) 1061810409209 66minmax planeinmax−− =−−=−= ε ε γ

10409,0, 10209 6minint6

max

−− −=== ε ε ε From the above results, we have

Thus the Mohr’s circle is as follows,

Ch 6- 34



35/49

Three-Dimensional Analysis of Strain

Previously demonstrated that three principal

axes exist such that the perpendicular

element faces are free of shearing stresses.

By Hooke’s Law, it follows that the shearing

strains are zero as well and that the principal

planes of stress are also the principal planes

of strain.

Ch 6- 35

represented by Mohr’s circles.



36/49


For the case of plane strain where the x and y

axes are in the plane of strain,

the z axis is also a principal axis

the corres ondin rinci al normal

strain is represented by the point Z = 0 or the origin.

If the points A and B lie on opposite sides of

the origin, the maximum shearing strain is themaximum in-plane shearing strain, D and E .

Ch 6- 36

If the points A and B lie on the same side of theorigin, the maximum shearing strain is out of the

plane of strain and is represented by the points

D’ and E’ .


Th D l A l f S


37/49


Corresponding normal strains,

Consider the case of plane stress,

0=== z b ya x σ σ σ σ σ

( ) ( )babac

bab

baa

E

E E

E E

ε ε ν

ν σ σ

ν ε

σ σ ν ε

σ ν σ ε

+−

−=+−=

+−=

−=

1

Ch 6- 37

If B is located between A and C on the Mohr-

circle diagram, the maximum shearing strain is

equal to the diameter CA.

not zero.



38/49

1) For the given state of plane strain, determine (a) the orientation and magnitude of the

principal strains, (b) the maximum in-plane strain, and (c) the absolute maximum

shearing strain.a 37.53o -32.4e-6 -187.6e-6 b 155.2e-6 c 187.6e-6

ε x = - 90 µ ; ε y = - 130 µ ; γ xy = 150 µ

2) For the given state of plane strain, determine the state of strain associated with axes

x’ and y’ rotated via the angle θ .[-653e-6; 303e-6; -829e-6]

ε x = - 800 µ ; ε y = 450 µ ; γ xy = 200 µ ; θ = -25o

Ch 6 - 38


G li d H k ’ L


39/49

For a triaxial state of stress, the general form for Hooke’s law is as follow:

( ) ( ) ( ) x z x z x x vvv σ σ σ ε σ σ σ ε σ σ σ ε +−=+−=+−=1

, 1

, 1

Generalized Hooke’s Law

They are valid only for a linear–elastic materials.

Hooke’s law for shear stress and shear strain is written as

And relationship between G, E and v is;

1 1 1 xy xy yz yz xz xz

G G Gγ τ γ τ γ τ = = =

Ch 6- 39

( )v E G+= 12


G li d H k ’ L


40/49

Generalized Hooke’s Law

Stresses in terms of strains

2 E

v or G eσ ε ε ε ε σ ε λ = + + − = +

( ) ( ) ( )( )

( ) ( ) ( )( )

1 1 2

21 1 2

21 1 2

: ;

y y x z y y y

z z x y z z z

x y z

v v

E v or G e

v v

E

or G ev v

vE ote and e

σ ε ε ε ε σ ε λ

σ ε υ ε ε ε σ ε λ

λ ε ε ε

+ −

= + + − = ++ −

= + + − = ++ −

= = + +

Ch 6- 40

−

( ) ( ) ( ); ;

2 1 2 1 2 1 xy xz yz y xz xz

E E E

v v vτ τ τ γ γ γ = = =

+ + +


E am le 6 7


41/49

Example 6.7

The copper bar is subjected to a uniform loading along its edges. If it has a = 300 mm, b = 50mm, and t = 20 mm before load is applied, find its new length, width, and thickness after

application of the load. Take E CU = 120 GPa, v = 0.34.

From the loading we have

The associated normal strains are determined from the generalized Hooke’s law,

800 MPa , 500 MPa , 0 , 0 x y xy z

σ σ τ σ = = − = =

Ch 6- 41

( )

( )

( )

0.00808

0.00643

0.000850

x x y z

y

y x z

z z x y

v E E

v

E E

v E E

σ ε σ σ

σ ε σ σ

σ ε σ σ

= − + =

= − + = −

= − + = −


42/49


Example 6 8


43/49

Example 6.8

At a point in a stresses body, the strains related to the coordinate set xyz are

given by:

Determine the complete stress components for this body. Assume Young

Modulus, E = 100 GPa and Poisson’s ratio, v = 0.33.

6300 100 500 10

200 500 400

− − × −

Ch 6- 43


Example 6 8


44/49

Example 6.8

( )( ) xx xx yy zz xx E

σ ε υ ε ε ε = + + −−

Normal stresses on each axis:

( ) ( )( ) ( )( )

( )

( ) ( ) ( )( )

6

6 6 6 6

11

11

100 10200 10 0.33 100 10 400 10 200 10

1 0.33 1 2 0.33

2.2114 10 0.000031 .

1 1 2

2.2114 10 0.000133 .

yy yy xx zz yy

Ans

E

ns

σ ε υ ε ε ε υ υ

−

− − − −×= × + − × − × − ×

+ −

= × − =

= + + −+ −

= × − =

-6.86MPa

-29.4MPa

Ch 6- 44

( ) ( )1 1 2 zz z

E σ ε υ υ

=+ − ( )( )

( )112.2114 10 0.000235 .

z xx yy zz

Ans

υ ε ε ε + + −

= × − = -51.97MPa


Example 6 7


45/49

Example 6.7

Continue.

( )9

6100 10300 10 .

1 1 0.33 xy xy

E ns

vτ ε

−×= = × =

+ +22.6MPa

( )

( )

9

6

9

6

100 10 200 10 .1 1 0.33

100 10500 10 .

1 1 0.33

xz xz

yz yz

E nsv

E ns

v

τ ε

τ ε

−

−

×= = × =

+ +

×= = × =

+ +

15.04MPa

37.6MPa

Thus, complete stress component is:

−

Ch 6- 45

. . .

22.6 29.4 37.6 .

15.04 37.6 51.97

Pa Ans

−

= −−

σɶ



46/49

1) Determine the complete stress component of this material. Use E = 200 GPa and G

= 80 GPa. 200 100 0

100 300 400 µ

2) A square ABCD of 20 mm side is scribed on the surface

of a thin plate while the plate is unloaded, as shown in

Figure Q2. After the plate is loaded, the lengths of sides

AB and AD are observed to have increased by 4 x10-6 m and 6 x10-6 m, respectively while angle DAB is

observed to have decreased by 160 x10-6 rad.

Determine;

D C

y

Ch 6 - 46

i. the principal strains and its orientation,[344.34µ,155.66µ,29o]

ii. the maximum in-plane shear strain, and [188.68µ]

iii. the average normal strain. [250µ]

iv. The principal stresses using E = 200 GPa and v =

0.3. [85.94 MPa,56.92 MPa]

A B

x

Figure Q2

Previous Exam Questions


47/49

OCT2009/MEC411/KJM454 Q2(b)

An element is subjected to two mutually perpendicular strains, εx= 400µ and εy= 200µ,

together with an unknown shear strain γxy. If the maximum principal strain in the

i. the magnitude of the shear strain, [346.4µ]

ii. the other principal strain and the direction of the principal strain, [100µ,30o]

iii. the strain εa in a direction of 30° with the x-axis, and [500µ]

iv. If E = 200 GPa and v = 0.03, determine the principal stresses.

APR2011/MEC411 Q5(b)

Ch 6 - 47

e s a e o p ane s ra n a a po n n a e orma e o y s e ne y εx = – µ, εy =

– 140 µ, and γxy = 150 µ. Calculate :

i. the orientation and magnitude of the principal strains, [37.53o,-42.38µ,-197.62µ]

ii. the maximum in-plane strain, and [150 µ]

iii. the new state of εx , i.e., εx’, when the element is rotated + 30o with respect to the

x-axis. [-45.05 µ]



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JUN2011/MEC411 Q5(d)

The principal stresses (at a point on a certain material) are given as 90 MPa and 30

MPa, both is in tension. Determine the normal and shear stresses on a plane making an

angle of tan-10.25 (ccw) with the principal directions. [86.47 MPa,14.1 MPa]

JAN2012/MEC411 Q5(b)

A state of plane stress consists of a tensile stress 60 MPa exerted on vertical surfaces

and of unknown shearing stresses. If the largest normal stress is 100 MPa, calculate;

i. the average normal stress, [30 MPa]

Ch 6 - 48

ii. the magnitude of the shearing stress, [63.25 MPa]

iii. the maximum shearing stress, and [70 MPa]iv. the minimum principal stresses. [-40 MPa]



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JUN2012/MEC411 Q5(a)

A steel bar is subjected to an axial force of P . The state of stress at a point on the bar

caused by the axial loading is shown in Figure Q5(a). Knowing that the stresses on

plane a-a are σy

’ = -100 MPa and τx

’y

’ = 35 MPa, determine using the Mohr’s circle;

i) the normal stress, σX’ of the stress component on plane a-a, [-12.25 MPa]

ii) The angle α that plane a-a forms with the horizontal, and [19.3o]

iii) The maximum compressive stress in the bar. [-112.25 MPa]

P

σy

Ch 6 - 49

State of stress

P

α

a

a

σy

y

x

z

Figure Q5 (a)

6-Transformation of Stress & Strain [Jan2013]

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Transcript of 6-Transformation of Stress & Strain [Jan2013]