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
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MECHANICS OF MATERIALS
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
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MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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.
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MECHANICS OF MATERIALS
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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
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external force.
MECHANICS OF MATERIALS
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Beer • Johnston • DeWolf
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
+−
−=′′
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MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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
−=
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90by separatedanglestwodefines :Note
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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
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MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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+
=′
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MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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
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MPa30min −=σ
MECHANICS OF MATERIALS
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Beer • Johnston • DeWolf
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
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MECHANICS OF MATERIALS
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Beer • Johnston • DeWolf
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
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, ( ) p p pprincipal stresses at the point H.
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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
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ksi96.7ksi84.80 ++ yyx τσσ
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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
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MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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
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The direction of rotation of Ox to Oa is the same as CX to CA.
MECHANICS OF MATERIALS
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Beer • Johnston • DeWolf
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’.
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MECHANICS OF MATERIALS
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Beer • Johnston • DeWolf
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
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JTc
xyyx === τσσ 0 0=== xyyx JTc τσσ
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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
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( ) ( ) MPa504030 22 =+== CXR
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
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
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MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Example 7.02
• Maximum shear stress
°+= 45ps θθ
°= 671θ
R=maxτ
MPa50max =τ
aveσσ =′
MPa20=′σ
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6.71sθ MPa50maxτ MPa20σ
MECHANICS OF MATERIALS
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Beer • Johnston • DeWolf
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
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( ) ( ) ( ) ( ) MPa524820 2222 =+=+= FXCFR
MECHANICS OF MATERIALS
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Beer • Johnston • DeWolf
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
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clockwise7.33 °=pθMPa132max +σ MPa28min +σ
MECHANICS OF MATERIALS
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Beer • Johnston • DeWolf
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
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MPa3.41=′′yx
y
τ
MECHANICS OF MATERIALS
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Beer • Johnston • DeWolf
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
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and the normal stresses are the principal stresses.
MECHANICS OF MATERIALS
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Beer • Johnston • DeWolf
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.
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(shearing stress is zero)minmaxmax 2
1 σστ −=
MECHANICS OF MATERIALS
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Beer • Johnston • DeWolf
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
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are at 45o to the principal planes.
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