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MEE 3090
Design of Machine Elements
Module 4
Designing against plane stress
Designing against compound stress
Theories of static failure
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
1
Plane stress
Mohr’s circle for plane stress systems
Computing principal stresses and angles
Understanding complementary shear stresses
Special cases: pure tension, pure compression, pure shear
Explaining failure signatures using Mohr’s circle
Applications: thin pressure vessels, shaft under pure torsion, helical springs
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
2
Plane stress systems
• Stress element
• An infinitesimally small cube-shaped
region encompassing a point in the
stress field (or the stressed structure),
at which we are interested to compute
the stresses
• Plane stress systems
• May include both normal and shear
stresses
• All stress vectors on the element are
contained in the same plane
• There is no component of stress out of
the plane (normal to the image)
• Plane stress problems can be studied as
special case of 3-D stress systems
y
100
80
200
x
All stresses in MPa
Example of a plane
stress element
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
3
Understanding complementary shear stresses
Applied stress
• Questions
• Why do we draw the shear stresses on all four sides of the element?
• Why do they appear in sets of fours?
• How to determine their directions?
• Consider an element
• Apply a shear stress on any one side
• This vector is the stress applied externally
• Then try to establish static equilibrium
• First balance the forces (stresses)
• Then balance the moments
σ𝐹 = 0
Reaction stress
σ𝑀 = 0
• Note that for each “applied stress” you need three “reaction stresses”
• The shear stresses are all equal in magnitude, so just name them 𝜏𝑥𝑦
• In 3-D stress systems, each applied shear stress must be balanced in this manner,
which may produce a complex system of stresses
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
4
Mohr’s circle for plane stress systems
• Topics
• How to construct the circle to represent a given stress element
• How to interpret the circle
• What does a diameter represent?
• What does rotating a diameter through 2𝜃 mean physically?
• How to compute stresses (magnitude and directions)
• Principal stresses (𝜎, 𝜏) and their orientations
• 𝜎’s and 𝜏’s for any orientation – graphical and analytical
y
100
80
200
x
All stresses in MPa
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
5
2-d Mohr’s circle
• Sign convention
• Shear CW upward
• Normal tension right
• There is only one Mohr’s
circle for a given element
• Each diameter represents
a different orientation of
the same element.
• Rotation by 𝜃 on the
actual element
corresponds to a rotation
of 2𝜃 on the circle.
Demonstrate in class: Given a stress element,
1. Draw its Mohr’s circle
2. Determine principal stresses, max shear stress
3. Determine the principal stress angles (normal, shear)
4. Determine the stress element at a different orientation
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
6
How to draw a 2-D Mohr’s circle
• Identify the “x-faces” and “y-faces”
• Transfer the data to the plot separately for x and y faces
• Normal stress: tension to the right
• Shear stress: clockwise moment upward
• Join the x and y points
• Doing this creates a diameter
• This diameter represents the stresses, as given
• The diameter is bisected at the 𝜎 axis
• Draw a circle using this line as a diameter
• Center always on the 𝜎 axis
• Examine the followings
• By rotating the diameter 180°, x and y faces switch
• By rotating the diameter through 2𝜃, a physical element rotates through 𝜃
• When both 𝜎 and 𝜏 are present, the maximum stress experienced by the
element can be higher than any of the individual stresses applied
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
7
Special cases of Mohr’s circles
• Special cases
• Uniaxial pure compression (x or y), no shear
• Uniaxial pure tension (x or y), no shear
• Tension on x, equal or different tension on y, no shear
• Tension on x (or y), different compression on y (or x), no shear
• Tension on x (or y) and equal compression on y (or x), no shear
• Pure shear stresses on all four sides, no normal stress
• Equivalency of the previous two cases
• Uniaxial pure tension (x or y), no shear
• Is ductile failure limited by tensile strength or shear strength?
• Why are Leuder’s lines produced?
• Why does MSST predict ductile failure better than MNST?
• Why does MNST predict failure better in the case of brittle materials?
Ductile, axial tension
Brittle, axial tension
• Explain the failure signatures with Mohr’s circle
Pure tension
Pure torsion
Ductile
45
90
Brittle
90
45
Brittle (above), ductile (below)
in torsion
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
8
2-d Mohr circle problem types
• Given an element with stresses…
• Draw the Mohr’s circle and label it
• Represent the principal stress element on the circle
• Represent the maximum shear stress element on the circle
• Represent the element that is 𝜃 = 30° rotated in CW or CCW direction
• Compute the principal stresses
• Draw the principal stress element with respect to the given element
• Compute the max. shear stress
• Draw the max. shear stress element with respect to the given element
y
• Compute the stresses on the element rotated by 𝜃 = 30°
100
• Draw the rotated element with respect to the given element
80
200
x
All stresses in MPa
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
9
Design applications of plane stress
Thin wall pressure vessels
Line shafts for power transmission
Helical springs
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
10
Thin-wall pressure vessels
• Condition for thin wall : t ≤ d/10
• Thin wall: Assume uniform stress distribution on the thickness
• Thick wall: stress is higher on the inner wall than outer wall
These are examples of THIN wall pressure vessels
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
11
Thin-wall pressure vessels
• Design problem
• A cylindrical pressure vessel of mean wall diameter of 1 m and wall
thickness of 10 mm carries an internal pressure of 1000 kPa. Determine
the following for a stress element on the wall.
• The longitudinal stress
• The hoop stress
• The maximum shear stress
• The maximum normal stress
• The Mohr’s circle for an element on the pressure vessel
• The material has a yield strength of 300 MPa. What is the factor of
safety guarding the pressure vessel against failure?
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
12
Hoop stress and longitudinal stress
• Burst test results of thin-wall cylinders
• Why do pipes burst parallel to the axis?
• Follow in-class derivations and discussion
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
13
Design of line shafts
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
14
Design of line shafts
• A shaft is loaded and supported as shown. Compute the following:
• Critical section where failure is most likely to happen
• Max normal and shear stresses at the critical section
• Mohr’s circle for the critical section
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
15
Design of line shafts
• Derive the free body diagram
• Driver and driven pulleys
• Direction of rotation
• Draw bending moment diagrams
• Looking from +z axis
• Looking from +y axis
• Add them as vectors
• At all critical points
• Choose the section with max M as critical
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
16
Design of line shafts
8000 lb-in
8246 lb-in
β
2000 lb-in
CAUTION:
The vector diagram on the left is at
a skewed perspective. Refer to the
diagram on the top for a proper
perspective.
Read it just as a usual vector
addition, with the 2000 and 8000
lb-in moments being mutually
perpendicular.
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
17
Design of line shafts
• Now, do the following:
• Compute the bending normal stress at the critical section – tension, compression
• Draw the orientation of the max bending stress (tens, comp), using 𝛽
• Add the torsional shear stresses
• Not in all spans, but uniform “all around” and “all along” when present
• Draw the elements with stresses for the max bending stress points (T, C)
• Draw Mohr’s Circles and compute the max normal and shear stresses
• Then use the failure theories to determine the likelihood of failure
• Critical section: Failure is likely to happen at point B (in this case)
• Point C has a lower value for the resultant bending moment, and same 𝜏
• At mid-span, in this case, the resultant bending moment is less than B
• At bearings, both 𝜎 and 𝜏 are zero, no little chance of failure
• Anywhere else is unlikely, as follows from the trends of bending moment
• In general, for any other shaft problem:
• Find the likely locations for failure (e.g., B, C, mid-span, bearings, etc.)
• Compute the stresses (bending, torsion) for each
• Choose the worst stressed location only for analysis
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
18
Design of helical springs
• Follow in-class discussions
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
19
Compound stress
Computing stress at a point – stress tables
Mohr’s circle for planar (2d) stress system
Mohr’s circle for compound (3d) stress system
Computing principal stresses and angles
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
20
Compound stress
Bending moment, M
(Normal, trans. shear)
Twisting moment, T
(Torsional shear)
Applied force, P
b
Local stresses may be
different from the global
ones. Consider the contact
points of the spanner head.
a
Reaction, R
(Direct shear)
Applied force, P
Applied force, P
Applied force, P
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
21
Compound stress
• Compute the state of stress (3-D Mohr’s circle) at points A, A', B, B'
• Steps:
1. Free body diagrams
2. Stress tables
3. Stress element w/ values
4. 3-D stress equation
a) Identify 𝜎’s and 𝜏’s
b) Put in equation
c) Solve iteratively / prog.
d) Order 𝜎1 ≥ 𝜎2 ≥ 𝜎3
5.
Draw 3-d Mohr’s circle
• Follow in-class discussion.
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
22
1. Free body diagram
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
23
2. Stress tables (all stresses in kPa)
@A
Axial
BN
𝑀𝑐
𝐼
2 kN
3 kN
4 kN
0
0.8 × 0.025
= 𝜋
0.025 4
4
0
0
(neutral axis)
𝐹
𝐴
𝑀𝑐
+ 0 (𝑁. 𝐴. )
𝐼
4
=
𝜋 0.025 2
1.2 × 0.025
= 𝜋
0.025 4
4
BTS
DS
TS
𝑉
𝐴
0
(outer fiber)
=
0
2
𝜋 0.025 2
4 𝑉
∙
3 𝐴
𝑉
𝐴
𝑇𝑟
𝐽
4
3
=
3 𝜋 0.025 2
3
=
𝜋 0.025 2
0.9 × 0.025
= 𝜋
0.025 4
2
0+0
(outer fiber,
purely axial)
0
0
𝑇𝑟
𝐽
3 kN-m
0
0
0
0
3 × 0.0.25
= 𝜋
0.025 4
2
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
24
2. Stress tables (all stresses in MPa)
@ “A”
2 kN
Axial
BN
BTS
0
𝑀𝑐
𝐼 a
0
(outer fiber)
= 65 MPa
3 kN
0
(neutral axis)
0
4 kN
3 kN-m
f
𝑀𝑐
+ 0 (N. A. )
𝐼
= 2 MPa
= 98 MPa
0
g
0
TS
𝑉
𝐴 b
0
= 1 MPa
4 𝑉
∙ c
3 𝐴
= 2 MPa
𝐹
𝐴
DS
0+0
(outer fiber,
purely axial)
0
𝑉 d
𝐴
= 1.5 MPa
𝑇𝑟
𝐽
e
= 37 MPa
0
0
𝑇𝑟
𝐽
0
h
= 122 MPa
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
25
3. Stress element w/ values
Caution:
Axis directions have changed from the last slide.
Caution:
Only one stress vector
is shown for each
action-reaction pair, for
clarity. Complimentary
stresses are not shown.
𝜎𝑥 = 0
𝜎𝑦 = 𝑎 + 𝑓 + 𝑔
𝜎𝑧 = 0
𝜏𝑥𝑦 = 𝑐 + 𝑑 + 𝑒 + ℎ
𝜏𝑦𝑧 = 𝑏
𝜏𝑧𝑥 = 0
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
26
4. 3-d stress equation
Examine the equation first:
What if it was reduced to 2-d?
𝜎𝑧 = 0, 𝜏𝑦𝑧 = 0, 𝜏𝑧𝑥 = 0
𝜎 3 − 𝜎𝑥 + 𝜎𝑦 + 𝜎𝑧 𝜎 2
+ 𝜎𝑥 𝜎𝑦 + 𝜎𝑦 𝜎𝑧 + 𝜎𝑧 𝜎𝑥 − 𝜏𝑥𝑦 2 − 𝜏𝑦𝑧 2 − 𝜏𝑧𝑥 2 𝜎
− 𝜎𝑥 𝜎𝑦 𝜎𝑧 + 2𝜏𝑥𝑦 𝜏𝑦𝑧 𝜏𝑧𝑥 − 𝜎𝑥 𝜏𝑦𝑧 2 − 𝜎𝑦 𝜏𝑧𝑥 2 − 𝜎𝑧 𝜏𝑥𝑦 2 = 0
𝜎𝑥 = 0
𝜎𝑦 = 𝑎 + 𝑓 + 𝑔 = 65 + 2 + 98 𝑀𝑃𝑎 = 165 𝑀𝑃𝑎
𝜎𝑧 = 0
𝜏𝑥𝑦 = 𝑐 + 𝑑 + 𝑒 + ℎ = 2 + 1.5 + 37 + 122 𝑀𝑃𝑎 = 162.5 𝑀𝑃𝑎
𝜏𝑦𝑧 = 𝑏 = 1 𝑀𝑃𝑎
http://www.graniteng.com/mohr-3d?lang=en
𝜏𝑧𝑥 = 0
𝜎 = 264.7, 0, −99.75
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
27
3-d Mohr’s circle
• Recall: biaxial stress. The max. normal and shear stresses
occur at different orientations of the element than the
orientation at which the external stresses are applied.
• The same happens in 3-d.
• In general, the max-stress element could be three rotations
away from the given orientation: 𝑥, 𝑦, 𝑧. The max-stress
element, in general, looks crooked in space due to these
rotations.
• However, all three principal stresses happen on the same
element at this orientation – on three different sides.
• The three roots of the cubic stress equation are the three
principal stresses in 3-d.
𝜎1 > 𝜎2 > 𝜎3
𝝈𝟏 − 𝝈𝟐
𝝉𝟏,𝟐 =
𝟐
• Then draw tangentially touching Mohr circles, for
𝜎1 , 𝜎2 , 𝜎2 , 𝜎3 , 𝜎3 , 𝜎1 .
𝝈𝟐 − 𝝈𝟑
𝝉𝟐,𝟑 =
𝟐
• Order them as:
• 𝜏𝑚𝑎𝑥 occurs between the “farthest” sigmas.
𝝈1 − 𝝈3
𝝉𝒎𝒂𝒙 = 𝝉𝟑,𝟏 =
𝟐
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
28
HW: Compound stress problems
Carry them through the 3-d Mohr’s circle step
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
29
Theories of static failure
Theories for ductile materials
Theories for brittle materials
Computing FoS guarding against failure
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
30
Theories of failure
• Typical problem / question:
• Given a structure and loads, will the structure fail?
• If it does not fail, what is the factor of safety guarding it against failure?
• If is does, at which value of the load(s) will failure begin?
• Or, what is maximum value of the load that the structure will be able to
carry?
• Assumptions: At this point, you have already learned how to …
• 3-d Mohr Circles (Load, geometry, FBD, Stress tables, element stresses)
• Stress concentration effects and their impact on the factor of safety
• Factor of safety needed for the design (including stress concentration)
• Material properties, i.e., ductility, strengths (yielding, ultimate)
• Failure criteria (Yielding? Rupture? Crushing? Buckling?)
• Failure strength
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
31
Theories of failure
• The theories of static failure predict if failure will occur for a given static load
and geometry
• Brittle materials
• Maximum normal stress theory (MNST)
• Brittle Coulomb Mohr theory (BCMT)
• Modified Mohr Theory (MMT)
• Ductile materials
• Maximum shear stress theory (MSST)
• Ductile Coulomb-Mohr theory (DCMT)
• Distortion energy theory (DET)
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
32
Max. normal stress theory (MNST – brittle)
• Statement: “Failure will occur whenever one of the three principal stresses
equals or exceeds the strength”
1. The square is called the failure
envelope – represents material
strength values
• FIPTOI: 𝜎1 ≥ 𝑆𝑢𝑡 or 𝜎3 ≤ −𝑆𝑢𝑐
𝑆
FoS, 𝑛 = 𝜎𝑢𝑡 in tension, 𝑛 =
1
−𝑆𝑢𝑐
in comp.
𝜎3
𝜎𝐵
3. The region inside the envelope
represents stresses that are
SAFE
𝑆𝑢𝑡
A"
O
𝜎𝐴
−𝑆𝑢𝑐
−𝑆𝑢𝑐
4. If point A falls outside the
envelope, the structure will fail
𝑆𝑢𝑡
5. The ratio OB/OA gives the
factor of safety guarding against
failure
B
6. Test that with similar triangles
for each case shown
A
A'
2. The red point (A) is the plot of
the two extreme stresses
(largest Mohr’s circle)
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
33
MNST Problems
• Draw the given failure envelope and plot the stress
for each element shown.
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
34
Max. shear stress theory (MSST – ductile)
• Statement: “Yielding begins whenever the maximum shear stress produced in an
element equals or exceeds the max. shear stress that would be produced in a
uniaxial tensile test specimen of the same material at yielding.”
• FIPTOI:
𝜏𝑚𝑎𝑥 =
𝑆𝑦
𝜎1 −𝜎3
≥
2
2
i.e., when
𝜎1 − 𝜎3 ≥ 𝑆𝑦
FoS, 𝑛 =
• Concepts to understand
𝑆𝑦
𝜎1 −𝜎3
𝜎1 −𝜎3
, occurs between the “farthest” sigmas
2
𝑆𝑦
𝜎
Under pure tension, 𝜏 = 2 . Therefore at yielding, 𝜏𝑚𝑎𝑥 = 2
• 𝜏𝑚𝑎𝑥 =
•
• Under pure tension, ductile materials mostly fail under shear
• MSST failure is governed not by a stress, but by the largest difference between
the three principal stresses – the largest Mohr’s circle diameter
• This means that you could prevent failure by increasing the lowest stress!
• Does it mean that 𝜎1 could be even larger than 𝑆𝑦 , as long as 𝜎1 − 𝜎3 ≤ 𝑆𝑦 ?
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
35
MSST envelop (biaxial only)
𝜎𝐵
𝑆𝑦
Compare with MNST
A
−𝑆𝑦
• Failure criterion is changed from
rupture (Sut, Suc) to yielding (Sy),
because the material is ductile.
• The envelope limits are symmetric
about the axes, since Sy has the
same value in tension and
compression.
• Careful about plane stress: must
account for the ZERO stress, which
could be one of the extreme
stresses.
• Consider the following plane-stress
elements (assume 𝑆𝑦 = 100 𝑀𝑃𝑎):
B
𝑆𝑦
O
𝜎𝐴
−𝑆𝑦
40
40
40
Case 1
Case 2
Case 3
80
80
80
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
36
(MPa)
𝝈𝟏
𝝈𝟐
𝝈𝟑
MSST envelop (biaxial only)
Case 1
80
40
0
𝜎𝐵
Case 2
80
0
- 40
Case 3
0
- 40
- 80
𝑆𝑦
Case 1
B
𝑆𝑦
A
−𝑆𝑦
O
• Case 1: 𝜎1 − 𝜎3 = 𝜎1 = 80 𝑀𝑃𝑎
• Case 1 FIPTOI: 𝝈𝟏 ≥ 𝑺𝒚
• Case 1: What happens if the 40 MPa
stress increases to 50, 60, 70, … ?
𝜎𝐴
• Case 3: 𝜎1 − 𝜎3 = −𝜎3 = 80 𝑀𝑃𝑎
• Case 3 FIPTOI: −𝝈𝟑 ≥ 𝑺𝒚
• Case 3: What happens if the - 40 MPa
stress increases to - 50, - 60, - 70, … ?
Case 2
Case 3
−𝑆𝑦
40
40
40
1
2
3
80
80
80
• Case 2: 𝜎1 − 𝜎3 = 80 − (−40) =
120 𝑀𝑃𝑎
• Case 2 FIPTOI: 𝝈𝟏 −𝝈𝟑 ≥ 𝑺𝒚
• Case 1: What happens if all three
circles moved together to the left or
right on the Mohr’s circle plane?
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
37
MSST envelop (tri-axial)
Image source: Hamrock, et. al., 2005
• The biaxial envelop is a special case of the tri-axial envelop
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
38
Brittle Coulomb Mohr theory (BCMT – brittle)
• An approximation to better fit the empirical data
• A cross between MNST and MSST
In MSST:
𝜎𝐵
𝑛=
𝑆𝑢𝑡
∴
O
𝜎𝐴
A
−𝑆𝑢𝑐
𝑆𝑢𝑡
B
𝑆𝑦
𝑂𝐵
=
𝑂𝐴
𝜎1 − 𝜎3
1
𝜎1 − 𝜎3
𝜎1
𝜎3
=
=
−
𝑛
𝑆𝑦
𝑆𝑦
𝑆𝑦
In BCMT:
1. Yield stress is replaced with ultimate
2. Different limits for tension and compression
1
𝜎1
𝜎3
=
−
𝑛
𝑆𝑢𝑡
𝑆𝑢𝑐
−𝑆𝑢𝑐
Now think about Case 1, Case 2, and Case 3 for
planar stress - They all work out similar to MSST
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
39
Modified Mohr theory (MMT – brittle)
• A further modification on BCMT – to better fit the empirical data
𝜎𝐵
𝑆𝑢𝑡
O
𝜎𝐴
A
−𝑆𝑢𝑐
−𝑆𝑢𝑡
B
−𝑆𝑢𝑐
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
40
Distortion energy theory (DET – ductile)
• Statement: “Yielding will begin whenever the distortion strain energy per
unit volume reaches or exceeds the distortion strain energy per unit volume
of the same material during yielding, under simple tension or compression.”
FIPTOI: 𝜎 ′ =
1
2
2
2
𝜎1 − 𝜎2 + 𝜎2 − 𝜎3 + 𝜎3 − 𝜎1 2
2
≥ 𝑆𝑦
• Concepts to understand
• Ductile materials offer much higher strengths under hydrostatic stresses – in
tension or compression – than brittle materials.
• Difference between scaling and distortion
• Geometric distortion (change of angles) consumes energy.
• Hydrostatic stresses cause volume change, but no distortion.
• Distortion stresses cause only distortion, no volume change.
• Von Mises stress is not an actual stress experienced by the material. It is a
conceptual equivalent of the uniaxial stress that would produce the same
amount of distortion energy as the applied 3-d stress system.
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
41
DET continued
• Concepts to understand
• Distortion energy in 3-d stress system
1 + 𝜈 𝜎1 − 𝜎2 2 + 𝜎2 − 𝜎3 2 + 𝜎3 − 𝜎1 2
𝑢𝑑 =
3𝐸
2
• Distortion energy in uniaxial tension
1+𝜈 2
𝑢𝑑 =
𝑆
3𝐸 𝑦
• Equate these two terms, and you get the FIPTOI condition.
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
42
DET failure envelope
𝜎𝐵
𝑆𝑦
B
A
−𝑆𝑦
O
𝜎𝐴
𝑆𝑦
−𝑆𝑦
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
43
End of module
Copyright © Chiradeep Sen, csen@fit.edu, and Florida Institute of Technology. Do not redistribute.
44

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