ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311MECHANICAL DESIGN
Module 2Review of Solid Mechanics
Dulyachot CholaseukMechanical Engineering Department
Thammasat University
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 2
Contents
1. Stress Analysis2. Theory of Failure3. Deformation analysis
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 3
Stress Analysis
Normal Stress from Axial Load
Normal Stress from Bending
Transverse Shear Stress
Shear Stress from Torsion
Contact Stress
Basic Types of Loads and Stresses
1Stress is intensity of internal force.
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 4
Normal Stress from Axial Load
σx
σx
Tensile (+)
Compressive (-)
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 5
Normal stress from Bending
-max
xy
+max
σxσx
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 6
Transverse shear stress
max
τxy
xy
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 7
Shear stress from torsion
max
τxy
xy
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 8
Contact stress
Called Hertzian Stress
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 9
Stress concentrations
σavg = F/A
σmax > σavg
F
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 10
Application of stress concentration
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 11
Stress concentration factors
σmax > σavg
Let σmax = Kσavg
K = σmax / σavg
F
Value of K …•depends on geometry.•can be obtained by photoelastic experiment
or by computer simulation.
Stress concentration factor,
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 12
Photoelastic experiment
Computer simulation (FEA)
Photoelastic experiment
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 13
Stress concentration factor chart
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 14
Stress concentration factor chart
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 15
Design to avoid stress concentration factor
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 16
Exercise
Is point A or B the critical point of the beam?
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 17
Exercise
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 18
Combined Stress
Machine elements
are subjected to different
kind of stress simultaneously
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 19
Curved Beam
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 20
Example: Curved Beam
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 21
Example: Curved Beam
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 22
Example: Curved Beam
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 23
Example: Curved Beam
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 24
Example: Curved Beam
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 25
Drill bit
T
F
+ =
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 26
Shaft
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 27
State of Stress
Tri-axial Plane Stress
σx=100 MPa, σy= 80 MPa and τxy= 30 MPa,while Sy = 120 MPa. Is it safe?
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 28
Stress transformation
Values of stresses depend on directions
σy
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 29
Stress transformation
Values of stresses depend on directions
( ) ( ) ( ) ( ) 0coscossincossinsincossin;0 =−−−−=∑ ′ θθσθθτθθσθθτσθ AAAAAF xxyyxyx
( ) ( ) ( ) ( ) 0sincoscoscoscossinsinsin;0 =−−−+=∑ ′ θθσθθτθθσθθττθ AAAAAF xxyyxyy
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 30
Stress transformation
θτθσσσσ
σθ 2sin2cos22 xy
yxyx +
−+
+=
θτθσσ
τθ 2cos2sin2 xy
yx +
−−=
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 31
Principal stresses
02cos22sin2
2 =+
−−= θτθ
σσθσθ
xyyx
dd
yx
xy
σστ
φσ −=
22tan
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 32
Principal stresses
yx
xy
σστ
φσ −=
22tan
22
21 22, xy
yxyx τσσσσ
σσ +
−±
+=
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 33
Maximum shear stress
°±= 45στ φφ
22
max 2 xyyx τ
σστ +
−±=
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 34
Mohr circle
θτθσσσσ
σθ 2sin2cos22 xy
yxyx +
−+
+=
θτθσσ
τθ 2cos2sin2 xy
yx +
−−=
22
21 22, xy
yxyx τσσσσ
σσ +
−±
+=
22
max 2 xyyx τ
σστ +
−±=
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 35
Absolute maximum shear stress
03 =σ 1σ2σ σ
τ
maxτ
max,absτ
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 36
Exercise
Consider a 1-meter-long solid shaft of 15 mm diameter. Answer the
following questions:
(a) Locate the critical point.
(b) Find the state of stress at the critical point.
(c) Find principal stresses and the principal direction at the critical
point.
(d) Draw the corresponding Mohr’s circle.
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 37
Principal stress trajectories
φσ= 0°
φσ= 20°
φσ= 45°
φσ= 60°
φσ= 90°
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 38
Principal stress (σ1) field and trajectories
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 39
Principal stress trajectories in a cantilever beam
σ1
σ2
Principal stress trajectories derived from beam theory(consider both bending stress and tranverse shear)
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 40
Sketch of principal stresses trajectories in human femur(based on cantilever curved beam model)
Principal stress trajectories in human femur
Anterior-to-posterior roentgenogram of a thin-sectioned human proximal femur
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 41
Trabecular Microstructure
“Trabecular Microstructure Differs Greatly between Trabecular Groups in
Proximal Femurs of Postmenopausal Women”
Wang, J; Zhou, B; Guo, X Columbia University, New York, NY, USA
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 42
Stress distribution
High stress
High stress
Low stress
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 43
Maximum shear stress distribution
SHEAR"
6587866133745679625225504771384317263863153409032954912500792046671592561138446843223020.1
F
−
=2
,2
,2
max 2121max
σσσστ
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 44
Examples of stress distribution
High
Low
Stress magnitude
Principal stress (beam theory)
Von Mises stress (FEA)
Von Mises stress (FEA)
Von Mises stress (FEA)
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 45
Stress distribution and stress trajectories
SHEAR"
6587866133745679625225504771384317263863153409032954912500792046671592561138446843223020.1
F
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 46
Theory of Failure2STATIC LOAD
BRITTLE – MNST
DUCTILE – MSST
– DET
FATIGUE
SODERBERG
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 47
Tensile Testing
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 48
Brittle Failure
Brittle material fail when bonding
between molecules breaks.
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 49
Maximum Normal Stress Theory
utS=1σ
1σ
2σ
ucS
ucS
utS
utS
Failure occurs when
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 50
Failure of concrete beam
Principal stress trajectories
Maximum shear
stress trajectories
Insufficient bottom
reinforcement
Sufficient bottom
reinforcement
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 51
Ductile Failure
Ludwig line
Ductile material fail when the molecule
deforms permanently.
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 52
Maximum Shear Stress Theory
Ductile material fail when the
molecule deforms permanently. 2max,y
abs
S=τ
1σ
2σ
yS
yS
yS
yS
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 53
Distortion Energy Theory
Failure of ductile material occurs when
strain energy per unit volume of the
material exceeds the strain energy per
unit volume at the yield point of the same
material under tensile test. ye S=σ
1σ
2σ
yS
yS
yS
yS
xyyyxx
e
τσσσσ
σσσσσ
322
2221
21
++−=
+−=
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 54
Comparison
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 55
Chalk twisting experiment
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 56
Exercise
Which one is brittle?
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 57
Exercise
Consider a 1-meter-long solid shaft of 15 mm diameter. Answer the
following questions:
(e) If the material is gray cast iron (Su=125MPa), will it fail?
(f) If the material is medium carbon steel (Sy=300MPa), find Ns
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 58
Fatigue
Cumulative damage caused by alternated load.
Stress lower than the static threshold.
Source of more than 80% of mechanical part failure.
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 59
Alternated Load
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 60
Alternated load in machines
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 61
Rotating beam experiment
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 62
Universal Testing Machine
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 63
S-N Diagram
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 64
Endurance Limit
Endurance limit is maximum the value of fatigue stress that
will not cause failure of the parts within 107 cycles.
Aluminum does not have endurance limit
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 65
Low cycle fatigue (N<1000)
ul SS 9.0=′
ul SS 75.0=′
ul SS 72.0=′
Bending
Axial load
Torsion
Steel parts will last for about 1000 cycles
if fatigue stress is less than lS ′
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 66
High cycle fatigue (103<N<107)
( ) sbt
cf NS ′=′ 10
Steel parts will last for about N cycles
if fatigue stress is fS ′
′′
−=e
ls S
Sb log31
( )e
l
SSc′′
=2
log
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 67
Modified Endurance Limit
Actual conditions differ from the experiments,
adjustment factors are needed:
Surface finishing factor
Size factor
Temperature factor
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 68
Soderberg theory
1=+e
a
y
m
SSσσ
For non-zero mean stress
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 69
Exercise
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 70
Deformation Analysis3Other than stress compliance, a design has to comply with
deformation criteria.
Deformation = function of geometry,
load and Young's modulus
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 71
Strain Energy
The external work done on an elastic member in deforming
it is transformed into strain energy (like spring). If the
member is deformed a distance y, this energy is equal to
the product of the average force and the deflection,
for spring kFyFU
2
2==
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 72
Strain Energy in Various Load Types
Rod under tension or compression: AElFU
2
2
=
Rod under torsion: AG
lTU2
2
=
Element under shear:AG
lFU2
2
=
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 73
Strain Energy in Various Load Types
Beam under pure bending moment: ∫= dxEI
MU2
2
Beam under transverse shear: ∫= dxAG
CVU2
2
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 74
Strain Energy Density
Tension and compression Eu
2
2σ=
Direct shear Gu
2
2τ=
TorsionG
u4
2maxτ
=
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 75
Castigliano’s Theorem
When forces act on elastic systems subject to small
displacements, the displacement corresponding to any force,
collinear with the force, is equal to the partial derivative of the
total strain energy with respect to that force.
ii F
U∂∂
=δ
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 76
Statically Indeterminate Problems
Use compatibility conditions
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 77
Buckling
Failure of long members under compression occurs
before yield point. [More detail in MODULE 6:
POWER SCREWS]
ME TUMechanical Engineering DepartmentFaculty of Engineering, Thammasat University
ME311 Module 2: Review of Solid Mechanics 78
Buckling
Lateral deformation due to axial load
