Lecture Mechanics of materials (Third edition) - Chapter 7: Transformations of stress and strain

Third Edition

CHAPTER

MECHANICS OF
MATERIALS
Ferdinand P. Beer
E. Russell Johnston, Jr.
John T. DeWolf

Transformations of
Stress and Strain

Lecture Notes:
J. Walt Oler
Texas Tech University

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Third
Edition

MECHANICS OF MATERIALS

Beer • Johnston • DeWolf

Transformations of Stress and Strain
Introduction
Transformation of Plane Stress
Principal Stresses

Maximum Shearing Stress
Example 7.01
Sample Problem 7.1
Mohr’s Circle for Plane Stress
Example 7.02
Sample Problem 7.2
General State of Stress
Application of Mohr’s Circle to the Three- Dimensional Analysis of Stress
Yield Criteria for Ductile Materials Under Plane Stress
Fracture Criteria for Brittle Materials Under Plane Stress
Stresses in Thin-Walled Pressure Vessels

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Introduction
• The most general state of stress at a point may
be represented by 6 components,
σ x ,σ y ,σ z

normal stresses

τ xy , τ yz , τ zx shearing stresses
(Note : τ xy = τ yx , τ yz = τ zy , τ zx = τ xz )

• Same state of stress is represented by a
different 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.

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

• 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
of a structural element or machine component, i.e.,
at any point of the surface not subjected to an
external force.
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Transformation of Plane Stress
• Consider the conditions for equilibrium of a
prismatic element with faces perpendicular to
the x, y, and x’ axes.
∑ Fx′ = 0 = σ x′∆A − σ x (∆A cosθ ) cosθ − τ xy (∆A cosθ )sin θ
− σ y (∆A sin θ )sin θ − τ xy (∆A sin θ ) cosθ

∑ Fy ′ = 0 = τ x′y ′∆A + σ x (∆A cosθ )sin θ − τ xy (∆A cosθ ) cosθ
− σ y (∆A sin θ ) cosθ + τ xy (∆A sin θ )sin θ

• The equations may be rewritten to yield
σ
σ
τ

σ
σ

σ
2

σ

σ

2

σ

σ
σ

2

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σ
2
2

σ

cos 2θ τ

sin 2θ

cos 2θ τ

sin 2θ

sin 2θ τ

cos 2θ

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MECHANICS OF MATERIALS

Beer • Johnston • DeWolf

Principal Stresses
• The previous equations are combined to
yield parametric equations for a circle,

(σ x′ − σ ave )2 + τ x2′y′ = R 2
where

σ ave =

2

σ x +σ y

⎛σ x −σ y ⎞
2
⎟⎟ + τ xy
R = ⎜⎜
2
⎝
⎠

2

• Principal stresses occur on the principal
planes of stress with zero shearing stresses.
σ max,min =
tan 2θ p =

σ x +σ y
2

2

⎛σ x −σ y ⎞
2
⎟⎟ + τ xy

± ⎜⎜
2
⎝
⎠

2τ xy

σ x −σ y

Note : defines two angles separated by 90o
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Maximum Shearing Stress
Maximum shearing stress occurs for

σ x′ = σ ave

2

⎛σ x −σ y ⎞

2
⎟⎟ + τ xy
τ max = R = ⎜⎜
2
⎝
⎠
σ x −σ y
tan 2θ s = −
2τ xy
Note : defines two angles separated by 90o and
offset from θ p by 45o

σ ′ = σ ave =

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σ x +σ y
2

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Example 7.01

SOLUTION:
• Find the element orientation for the principal
stresses from
2τ xy
tan 2θ p =
σ x −σ y
• Determine the principal stresses from
σ max,min =

σx +σ y

2

⎛σ x − σ y ⎞
2
⎟⎟ + τ xy
± ⎜⎜
2
⎝
⎠

2
For the state of plane stress shown,
determine (a) the principal panes, • Calculate the maximum shearing stress with
(b) the principal stresses, (c) the
2
σ
σ
−
⎛

⎞
x
y
maximum shearing stress and the
2
⎟⎟ + τ xy
τ max = ⎜⎜
2
corresponding normal stress.
⎝
⎠

σx +σ y
′
σ =
2

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Example 7.01

SOLUTION:
• Find the element orientation for the principal
stresses from
2τ xy
2(+ 40 )
= 1.333
=
tan 2θ p =
σ x − σ y 50 − (− 10 )
2θ p = 53.1°, 233.1°

σ x = +50 MPa
σ x = −10 MPa

θ p = 26.6°, 116.6°

τ xy = +40 MPa

• Determine the principal stresses from
σ max,min =

σx +σ y
2

= 20 ±

2

⎛σ x − σ y ⎞
2

⎟⎟ + τ xy
± ⎜⎜
2
⎠
⎝

(30)2 + (40)2

σ max = 70 MPa
σ min = −30 MPa
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Example 7.01
• Calculate the maximum shearing stress with
2

⎛σ x −σ y ⎞
2
⎟⎟ + τ xy
τ max = ⎜⎜

2
⎠
⎝
=

(30)2 + (40)2

τ max = 50 MPa
σ x = +50 MPa
σ x = −10 MPa

τ xy = +40 MPa

θ s = θ p − 45
θ s = −18.4°, 71.6°

• The corresponding normal stress is
σ x + σ y 50 − 10
σ ′ = σ ave =
=
2

2

σ ′ = 20 MPa

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Sample Problem 7.1
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.
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.
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MECHANICS OF MATERIALS

Beer • Johnston • DeWolf

Sample Problem 7.1
SOLUTION:
• Determine an equivalent force-couple
system at the center of the transverse
section passing through H.
P = 150 lb
T = (150 lb )(18 in ) = 2.7 kip ⋅ in
M x = (150 lb )(10 in ) = 1.5 kip ⋅ in

• Evaluate the normal and shearing stresses
at H.
σy =+

(1.5 kip ⋅ in )(0.6 in )
Mc
=+
1 π (0.6 in )4
I
4

τ xy = +

(2.7 kip ⋅ in )(0.6 in )
Tc

=+
1 π (0.6 in )4
J
2

σ x = 0 σ y = +8.84 ksi τ y = +7.96 ksi
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Beer • Johnston • DeWolf

Sample Problem 7.1
• Determine the principal planes and
calculate the principal stresses.
2τ xy
2(7.96 )
tan 2θ p =
=
= −1.8
σ x − σ y 0 − 8.84
2θ p = −61.0°,119°

θ p = −30.5°, 59.5°

σ max,min =

σx +σ y
2

2

⎛σ x −σ y ⎞
2
⎟⎟ + τ xy
± ⎜⎜
2
⎠
⎝
2

0 + 8.84
⎛ 0 − 8.84 ⎞
2
=
± ⎜
⎟ + (7.96 )
2
⎝ 2 ⎠

σ max = +13.52 ksi
σ min = −4.68 ksi

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MECHANICS OF MATERIALS

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. Critical
values are estimated graphically or calculated.
• For a known state of plane stress σ x , σ y ,τ xy
plot the points X and Y and construct the
circle centered at C.
σ ave =

σ x +σ y
2

2

⎛σ x −σ y ⎞
2
⎟⎟ + τ xy
R = ⎜⎜
2

⎝
⎠

• The principal stresses are obtained at A and B.
σ max,min = σ ave ± R
tan 2θ p =

2τ xy

σ x −σ y

The direction of rotation of Ox to Oa is
the same as CX to CA.
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Mohr’s Circle for Plane Stress
• 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 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 obtained
from the coordinates X’Y’.

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

σx =

P
, σ y = τ xy = 0
A

σ x = σ y = τ xy =

P

2A

• Mohr’s circle for torsional loading:

σ x = σ y = 0 τ xy =

Tc
J

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σx =σy =

Tc
τ xy = 0
J
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Example 7.02

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

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

SOLUTION:
• Construction of Mohr’s circle
σ x + σ y (50 ) + (− 10 )
=
= 20 MPa
σ ave =
2
2
CF = 50 − 20 = 30 MPa FX = 40 MPa
R = CX =

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(30)2 + (40)2 = 50 MPa
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Example 7.02

• Principal planes and stresses
σ max = OA = OC + CA = 20 + 50
σ max = 70 MPa
σ max = OB = OC − BC = 20 − 50

σ max = −30 MPa
FX 40
=
CP 30
2θ p = 53.1°

tan 2θ p =

θ p = 26.6°

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Example 7.02

• Maximum shear stress

θ s = θ p + 45°

τ max = R

σ ′ = σ ave

θ s = 71.6°

τ max = 50 MPa

σ ′ = 20 MPa

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Sample Problem 7.2

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 the SOLUTION:
given element counterclockwise • Construct Mohr’s circle
through 30 degrees.
σ x + σ y 100 + 60
=
= 80 MPa
σ ave =
2

2

R=

(CF )2 + (FX )2 = (20)2 + (48)2 = 52 MPa

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Sample Problem 7.2

• Principal planes and stresses

XF 48
=
= 2.4
CF 20
2θ p = 67.4°

tan 2θ p =

θ p = 33.7° clockwise

σ max = OA = OC + CA
= 80 + 52

σ max = +132 MPa

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σ max = OA = OC − BC
= 80 − 52

σ min = +28 MPa
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Sample Problem 7.2

• Stress components after rotation by

30o

Points X’ and Y’ on Mohr’s circle that
correspond to stress components on the
rotated element are obtained by rotating
XY counterclockwise through 2θ = 60°

φ = 180° − 60° − 67.4° = 52.6°
σ x′ = OK = OC − KC = 80 − 52 cos 52.6°
σ y′ = OL = OC + CL = 80 + 52 cos 52.6°
τ x′y′ = KX ′ = 52 sin 52.6°
σ x′ = +48.4 MPa
σ y′ = +111.6 MPa
τ x′y′ = 41.3 MPa

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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: σ x ,σ y ,σ z ,τ xy ,τ yz ,τ zx
• Consider tetrahedron with face perpendicular to the
line QN with direction cosines: λx , λ y , λz
• The requirement ∑ Fn = 0 leads to,
σ n = σ x λ2x + σ y λ2y + σ z λ2z
+ 2τ xy λx λ y + 2τ yz λ y λz + 2τ zx λz λx

• Form of equation guarantees that an element
orientation can be found such that
σ n = σ a λ2a + σ bλb2 + σ cλc2

These are the principal axes and principal planes
and the normal stresses are the principal stresses.
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Application of Mohr’s Circle to the ThreeDimensional 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.

• Points A, B, and C represent the
• Radius of the largest circle yields the
principal stresses on the principal planes
maximum shearing stress.
1
(shearing stress is zero)
τ max = σ max − σ min
2

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Application of Mohr’s Circle to the ThreeDimensional 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
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 “inplane” shearing stress
c) planes of maximum shearing stress
are at 45o to the principal planes.
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