MECHANICS OF MECHANICS OF MATERIALS MATERIALS

MECHANICS OFFourth Edition

MECHANICS OF MATERIALS

CHAPTER

MATERIALSFerdinand P. BeerE. Russell Johnston, Jr.John T DeWolf

Transformations of John T. DeWolf

Lecture Notes:Stress and Strain

J. Walt OlerTexas Tech University

© 2006 The McGraw-Hill Companies, Inc. All rights reserved.


FourthEdition

Beer • Johnston • DeWolf

Transformations of Stress and Strain

IntroductionTransformation of Plane StressTransformation of Plane StressPrincipal StressesMaximum Shearing StressExample 7.01Sample Problem 7.1Mohr’s Circle for Plane StressMohr s Circle for Plane StressExample 7.02Sample Problem 7.2G l St t f StGeneral State of StressApplication of Mohr’s Circle to the Three-Dimensional Analysis of StressYield Criteria for Ductile Materials Under Plane StressFracture Criteria for Brittle Materials Under Plane StressStresses in Thin-Walled Pressure Vessels

© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 7 - 2


FourthEdition


Introduction• The most general state of stress at a point may

be represented by 6 components,tl

),,:(Note

stresses shearing,,

stressesnormal,,

xzzxzyyzyxxy

zxyzxy

zyx

ττττττ

τττ

σσσ

=== ),, :(Note xzzxzyyzyxxy ττττττ

• Same state of stress is represented by a different set of components if axes are rotateddifferent set of components if axes are rotated.

• The first part of the chapter is concerned with h h f f dhow 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.



FourthEdition


Introduction

• 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

.0,, and xy === zyzxzyx ττστσσ

• State of plane stress occurs in a thin plate subjected• State of plane stress occurs in a thin plate subjected to forces acting in the midplane of the plate.

• State of plane stress also occurs on the free surface• 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

t l f


external force.


FourthEdition


Transformation of Plane Stress• Consider the conditions for equilibrium of a

prismatic element with faces perpendicular to the x y and x’ axes

( ) ( )( ) ( ) θθτθθσ

θθτθθσσ

cossinsinsin

sincoscoscos0

AA

AAAF

xyy

xyxxx

∆−∆−

∆−∆−∆==∑ ′′

the x, y, and x axes.

( ) ( )( ) ( )

( ) ( ) θθτθθσ

θθτθθστ

sinsincossin

coscossincos0

AA

AAAF

xyy

xyxyxy

xyy

∆+∆−

∆−∆+∆==∑ ′′′

σσσσ +

• The equations may be rewritten to yield

θτθσσσσ

σ

θτθσσσσ

σ

2sin2cos

2sin2cos22

xyyxyx

y

xyyxyx

x

−−

−+

=

+−

++

=

′

′

θτθσσ

τ

θτθσ

2cos2sin2

2sin2cos22

xyyx

yx

xyy

+−

−=′′



FourthEdition


Principal Stresses• The previous equations are combined to

yield parametric equations for a circle,

( )

2

222

whereyxavex Rτσσ =+− ′′′

22

22 xyyxyx

ave R τσσσσ

σ +⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

+=

• Principal stresses occur on the principal planes of stress with zero shearing stresses.

22

minmax, 22 xyyxyx τ

σσσσσ +⎟⎟

⎠

⎞⎜⎜⎝

⎛ −±

+=

o90byseparatedanglestwodefines:Note

22tan

yx

xyp σσ

τθ

−=


90by separatedanglestwodefines :Note


FourthEdition


Maximum Shearing StressMaximum shearing stress occurs for avex σσ =′

2

22

2

max xyyxR

σσ

τσσ

τ

−

+⎟⎟⎠

⎞⎜⎜⎝

⎛ −==

and 90by separated angles twodefines :Note

22tan

oxy

yxs τ

σσθ −=

45by fromoffset o

yxave

p

σσσσ

θ

+==′

2aveσσ



FourthEdition


Example 7.01SOLUTION:

• Find the element orientation for the principal stresses from

yx

xyp σσ

τθ

−=

22tan

yx σσ

• Determine the principal stresses from

22

yxyx σσσσ ⎞⎜⎛ −+

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

2minmax, 22 xy

yxyx τσσσσ

σ +⎟⎠

⎞⎜⎜⎝

⎛±

+=

• Calculate the maximum shearing stress with(b) the principal stresses, (c) the maximum shearing stress and the corresponding normal stress.

C g

22

max 2 xyyx τ

σστ +⎟⎟

⎠

⎞⎜⎜⎝

⎛ −=

p g ⎠⎝

2yx σσ

σ+

=′



FourthEdition


Example 7.01SOLUTION:

• Find the element orientation for the principal stresses from

( )( ) =−−+

=−

= 333.11050

40222tan

yx

xyp σσ

τθ ( )

°°= 1.233,1.532 p

yx

θ

°°= 6.116,6.26pθ p

• Determine the principal stresses from2⎞⎛

MPa10

MPa40MPa50

−=

+=+=

x

xyx

σ

τσ

( ) ( )22

22

minmax,

403020

22

±

+⎟⎟⎠

⎞⎜⎜⎝

⎛ −±

+= xy

yxyx τσσσσ

σ

( ) ( )22 403020 +±=

MPa30MPa70max

==

σσ


MPa30min −=σ


FourthEdition


Example 7.01

• Calculate the maximum shearing stress with2⎞⎛ σσ

( ) ( )22

2max

4030

2

+=

+⎟⎟⎠

⎞⎜⎜⎝

⎛ −= xy

yx τσσ

τ

( ) ( )4030 +=

MPa50max =τ

MPa10

MPa40MPa50

−=

+=+=

x

xyx

σ

τσ 45−= ps θθ

°°−= 6.71,4.18sθ

1050 −=

+==′ yx σσ

σσ

• The corresponding normal stress is

22=== aveσσ

MPa20=′σ



FourthEdition


Sample Problem 7.1

SOLUTION:

• Determine an equivalent force couple• Determine an equivalent force-couple system at the center of the transverse section passing through H.

• Evaluate the normal and shearing stresses at H.

A i l h i l f P f 150 lb

• Determine the principal planes and calculate the principal stresses.

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 yaxes, (b) the principal planes and


, ( ) p p pprincipal stresses at the point H.


FourthEdition


Sample Problem 7.1SOLUTION:

• Determine an equivalent force-couple q psystem at the center of the transverse section passing through H.

lb150=P( )( )( )( ) inkip5.1in10lb150

inkip7.2in18lb150lb150

⋅==⋅==

xMTP

• Evaluate the normal and shearing stresses at H.

( )( )in60inkip51Mc ( )( )( )

( )( )

441

in6.0inkip7.2

in6.0in6.0inkip5.1

πσ

⋅

⋅+=+=

Tc

IMc

y

( )( )( )42

1 in6.0in6.0inkip7.2

πτ +=+=

JTc

xy

ksi96.7ksi84.80 +=+== yyx τσσ


ksi96.7ksi84.80 ++ yyx τσσ


FourthEdition


Sample Problem 7.1• Determine the principal planes and

calculate the principal stresses.( )

°°

−=−

=−

=

1190612

8.184.80

96.7222tan

yx

xyp

θ

σστ

θ

°°−= 119,0.612 pθ

°°−= 5.59,5.30pθ

22

minmax, 22+⎟⎟

⎠

⎞⎜⎜⎝

⎛ −±

+= xy

yxyx τσσσσ

σ

( )22

96.72

84.802

84.80+⎟

⎠⎞

⎜⎝⎛ −

±+

=

ksi68.4ksi52.13

min

max−=+=

σσ



FourthEdition


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 Criticalwith simple geometric considerations. Critical values are estimated graphically or calculated.

F k t t f l t• For a known state of plane stressplot the points X and Y and construct the circle centered at C.

xyyx τσσ ,,

22

22 xyyxyx

ave R τσσσσ

σ +⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

+=

• The principal stresses are obtained at A and B.ave Rσσ ±=minmax,

yx

xyp σσ

τθ

−=

22tan

The direction of rotation of Ox to Oa is


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


FourthEdition


Mohr’s Circle for Plane Stress

• With Mohr’s circle uniquely defined, the state of stress at other axes orientations may be ydepicted.

F th t t f t t l θ ith• For the state of stress at an angle θ with respect to the xy axes, construct a new diameter X’Y’ at an angle 2θ with respect to XY.

• Normal and shear stresses are obtainedNormal and shear stresses are obtained from the coordinates X’Y’.



FourthEdition


Mohr’s Circle for Plane Stress• Mohr’s circle for centric axial loading:

0, === xyyx AP τσσ

AP

xyyx 2=== τσσ

• Mohr’s circle for torsional loading:

Tc Tc


JTc

xyyx === τσσ 0 0=== xyyx JTc τσσ


FourthEdition


Example 7.02

For the state of plane stress shown, (a) construct Mohr’s circle, determine

SOLUTION(b) the principal planes, (c) the principal stresses, (d) the maximum shearing stress and the corresponding

SOLUTION:

• Construction of Mohr’s circle( ) ( )1050+σσg p g

normal stress. ( ) ( )

MPa40MPa302050

MPa202

10502

==−=

=−+

=+

=

FXCF

yxave

σσσ


( ) ( ) MPa504030 22 =+== CXR


FourthEdition


Example 7.02• Principal planes and stresses

5020max +=+== CAOCOAσmax

MPa70max =σ

5020min −=−== BCOCOBσ

MPa30min −=σ

40FX

°=

==

1.53230402tan

p

p CPFX

θ

θ

°= 6.26pθ



FourthEdition


Example 7.02

• Maximum shear stress

°+= 45ps θθ

°= 671θ

R=maxτ

MPa50max =τ

aveσσ =′

MPa20=′σ


6.71sθ MPa50maxτ MPa20σ


FourthEdition


Sample Problem 7.2

For the state of stress shown, determine (a) the principal planes and the principal stresses, (b) the p p , ( )stress components exerted on the element obtained by rotating the given element counterclockwise

SOLUTION:

• Construct Mohr’s circlegiven element counterclockwise through 30 degrees.

• Construct Mohr s circle

MPa802

601002

=+

=+

= yxave

σσσ


( ) ( ) ( ) ( ) MPa524820 2222 =+=+= FXCFR


FourthEdition


Sample Problem 7.2

• Principal planes and stresses48XF CAOCOA BCOCOA

°=

===

4.672

4.220482tan

p

p CFXF

θ

θ5280

max+=

+== CAOCOAσ5280

max−=

−== BCOCOAσ

MPa132max +=σ MPa28min +=σ


clockwise7.33 °=pθMPa132max +σ MPa28min +σ


FourthEdition


Sample Problem 7.2

°°=°−°−°=

65252806.524.6760180

KCOCOKφ• Stress components after rotation by 30o

°=′=

°+=+==°−=−==

′′

′

′

6.52sin52

6.52cos52806.52cos5280

XK

CLOCOLKCOCOK

yx

y

x

τ

σσ

p y

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

d l b i d b i 6.52sin52XKyxτrotated element are obtained by rotating XY counterclockwise through °= 602θ

MPa6.111MPa4.48

+=+=

′

′

y

xσσ


MPa3.41=′′yx

y

τ


FourthEdition


General State of Stress• Consider the general 3D state of stress at a point and

the transformation of stress from element rotation

• State of stress at Q defined by: zxyzxyzyx τττσσσ ,,,,,

• Consider tetrahedron with face perpendicular to the Co s de tet a ed o w t ace pe pe d cu a to t eline QN with direction cosines: zyx λλλ ,,

• The requirement leads to,∑ = 0nFq ,∑ n

xzzxzyyzyxxy

zzyyxxn

λλτλλτλλτ

λσλσλσσ

222

222

+++

++=

yyyy

• Form of equation guarantees that an element orientation can be found such that

222ccbbaan λσλσλσσ ++=

These are the principal axes and principal planes


and the normal stresses are the principal stresses.


FourthEdition


Application of Mohr’s Circle to the Three-Di i l A l i f StDimensional Analysis of Stress

• Transformation of stress for an element rotated around a principal axis may be represented by Mohr’s circle.

• The three circles represent the normal and shearing stresses for rotation around each principal axis.p y p p

• Points A, B, and C represent the principal stresses on the principal planes ( h i t i ) 1

• Radius of the largest circle yields the maximum shearing stress.


(shearing stress is zero)minmaxmax 2

1 σστ −=


FourthEdition


Application of Mohr’s Circle to the Three-Di i l A l i f StDimensional Analysis of Stress

• In the case of plane stress, the axis perpendicular to the plane of stress is a principal axis (shearing stress equal zero).

• If the points A and B (representing the• If the points A and B (representing the principal planes) are on opposite sides of the origin, then

a) the corresponding principal stresses are the maximum and minimum normal stresses for the element

b) the maximum shearing stress for the element is equal to the maximum “in-

normal stresses for the element

qplane” shearing stress

c) planes of maximum shearing stress 45o h i i l l


are at 45o to the principal planes.

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