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Priit Põdra 8. Strength of Components under Combined Loading 1
STRENGTH OF MATERIALSSTRENGTH OF MATERIALS
8. Strength of Components under Combined Loading
8. Strength of Components under Combined Loading
8.1 Combination of Two Bendings
8.2 Combination of
Bending and Axial Load
8.3 Combined Stress
State Analysis
8.4 Combination of
Bending and Shear
8.5 Combination of
Torsion and Bending
Mehhanosüsteemide komponentide õppetool
Priit Põdra 8. Strength of Components under Combined Loading 2
STRENGTH OF MATERIALSSTRENGTH OF MATERIALS
8.1. Combination of Two Bendings8.1. Combination of Two Bendings
Priit Põdra 8. Strength of Components under Combined Loading 3
x
y
z
F
Mz diagram
My diagram
Cantilever Beam
zx Principal Plane
z
xy Principal Plane
y
x
Three-Dimensional Bending TasksThree-Dimensional Bending Tasks
Structure
Bending deformation is present in both component’ principal planes
F2z
z
x
FBz
My diagram
FAz
zx Principal Plane
Bending Moment My Diagram
Loads do not act at the same plane
Loads act at the same plane
F1
y
z
x
Shaft of a Belt Drive
F2
L
y
F1
L1
x
F2y
L2
A B
FAy FBy
Mz diagram
xy Principal Plane
Bending Moment Mz Diagram
But this plane is not a principal plane
Structure and its Bending Moment Diagrams
Priit Põdra 8. Strength of Components under Combined Loading 4
Reduction of Load(s) to the Principal Axes
F
z
y
y
z
Fz
FyCentroid
Principal Centroidal Axis
Principal Centroidal
Axis
Shear forces Qy and Qz, may also be present, but their action is neglected
Internal Forces for Two Bendings’ CaseInternal Forces for Two Bendings’ Case
Bending moments (My and Mz) act in both principal planes of the component
For the cases, when the load direction line
does not cross the centroid
Torque T is also acting at the
cross-section
Bending Moments at the Critical Section
)( LFM yz
)( LFM zy
x
y
z
F
Mz diagram
My diagram
Cantilever Beam
z
y
x
L
Critical Cross-Section
FyL
FzL
sin
cos
FF
FF
z
y
Priit Põdra 8. Strength of Components under Combined Loading 5
Bending stresses are normal stresses, i.e. they both act
perpendicular to the section – this is ONE-DIMENSIONAL stress
Bending Stress AnalysisBending Stress Analysis
zI
M
y
yMy
yI
M
z
zMz
Resultant of two bending stresses at some point (y; z) equals to their algebraic sum
yI
Mz
I
M
z
z
y
yMzMy
My diagram
z
MyMax
MyMin
y
Mz diagram
MzMax
MzMin
Cross-Section’ Bending Stresses Diagrams
Principal Centroidal
Axes
Centroid
Cross-Section Bending Moment
Cross-Section Moment of Inertia
CoordinateAccording to Hooke’ law,
the stresses of each cross-section point follow the
value of respective strain
MyMy E MzMz E
Resultant of the deformations acting in same direction at some point equals to the sum of these
values (considering the signs +/-)
Stress at a Cross-Section Certain Point
Tension
Ten
sio
nCompression
Co
mp
ress
ion
Priit Põdra 8. Strength of Components under Combined Loading 6
Location of Cross-Section Critical PointsLocation of Cross-Section Critical Points
Equation of NEUTRAL LINE
0 zI
My
I
M
y
y
z
z
Cross-Section NEUTRAL LINE = projection of the beam’ neutral layer
line at the beam cross-section, which’ points have no stress (bending stress equals to zero)
=
z
y
Centroid OT (y1;z1)
OS (y2;z2)
Maximum distance
Maximum distance
Maximum Tensile Stress
Maximum Compressive Stress
Cross-Section’ Critical Points
Cross-Section NEUTRAL LINE
Combination of two bendings
• Neutral Line crosses the centroid• Neutral Line is straight
Cross-section criticla points are those, that are located most distant from the neutal line
The signs (+/-) of both bending moments and coordinates must be considered
Neutral line divides the section in two: area of tension and area of compression
Priit Põdra 8. Strength of Components under Combined Loading 7
Strength ConditionsStrength Conditions
22OS
11OT
zI
My
I
M
zI
My
I
M
y
y
z
z
y
y
z
z
Normal Stresses at critical Points
Compr22OS
Tension11OT
zI
My
I
M
zI
My
I
M
y
y
z
z
y
y
z
z
Strength Conditions
Maximum Compressive Stress
at the point OS
Maximum Tensile Stress at the point OT
Design Stress for Tension
Design Stress for Compression
z
y
OT (y1;z1)
OS (y2;z2)
Maximum Tensile Stress
Maximum Compressive
Stress
Cross-Section’ Critical Points
Cross-Section NEUTRAL
LINE
z-coordinate of point OT
diagram
OT
OS
NB! The signs (+/-) of both bending moments and coordinates must be considered
y-coordinate of point OT
z-coordinate of point OS
y-coordinate of point OS
Priit Põdra 8. Strength of Components under Combined Loading 8
z
+ + + + + + +
++++++
Stresses of Rectangular Section
OS
OT
Mz diagram
My
diagramy
Mz
Wz
Mz
WzMy
Wy
My
Wy
Maximum Compressive Stress
Max
imu
m T
ensi
le S
tres
s
Special Case – Rectangular SectionSpecial Case – Rectangular Section
Critical points are always located at the
diagonal corners of the rectangle
For all cross-sections, which’s ouside contour is rectangle, both principal axes are the axes of symmetry
Diagonal corners are always critical
z
z
y
y
W
M
W
M MinMax
Value of Maximum Bending Stress
At the critically tensioned point OT
At the critically compressed point OS
I-profile Square
Where in particular th critical points OT and OS are located, depends on the loading
Priit Põdra 8. Strength of Components under Combined Loading 9
Special Case – Circular SectionSpecial Case – Circular Section
W
M MinMax
Value of Maximum Normal Stress
22zy MMM
Bending Moment
Neutral Line
++++
++++
++
++
Circle has an infinite number of principal
centroidal axes
Neutral Line is also a
principal axis
OS
OT
z
Stresses of Circular Section
Mz diagram
My diagramy
Mz
W
Mz
WMy
W
My
W
z
y
diagram MW
MW
Maximum Tensile Stress
Maximum Compressive Stress
Critical points are always located diametrally on the perimeter of the section
RESULTANT BENDING MOMENT –can be calculated for CIRCULAR
section only
Priit Põdra 8. Strength of Components under Combined Loading 10
STRENGTH OF MATERIALSSTRENGTH OF MATERIALS
8.2. Combination of Bending and Axial Loads
8.2. Combination of Bending and Axial Loads
Priit Põdra 8. Strength of Components under Combined Loading 11
Slender bars have may fail by buckling due to cmpression
Eccentric Tension/CompressionEccentric Tension/Compression
SHORT Bar = Cross-section dimensions and bar length have the same magnitude
Buckling will be
studied later
ECCENTRIC TENSION/COMPRESSION= • load is acting parallel to the component axis• load action line does not coincide with axis
Eccentrically Compressed Bar
xF
zy
Loacation of the Load
yz
Fx Component axisPrincipal
Centroidal Axes
Centroid
Eccentrically Tensioned Bar
x
y
z
F
Load eccentricity may exceed the bar
cross-section dimensions
Short Bar
Priit Põdra 8. Strength of Components under Combined Loading 12
Internal Forces due to Eccentric LoadInternal Forces due to Eccentric Load
x F
z
ez
N (-)
Section
Bending Moment
My (-)
F eyx
y
Mz (-)
Internal Forces at the Principal Planes
zx Principal Plane xy Principal Plane
N (-)Axial ForceCross-Section Internal Forces
FN
yz
zy
FeM
FeM
Axial Force N (compression) is shown at both principal planes
Signs of bending moments My ja Mz depend on the directions of
axes y ja z
TENSION (+) or COMPRESSION (--)
BENDING may be three-dimensional (My 0 AND Mz 0)
or planar (My = 0 OR Mz = 0)
Priit Põdra 8. Strength of Components under Combined Loading 13
Normal Stresses due to Eccentric LoadNormal Stresses due to Eccentric Load
My diagramMyMax
MyMin
y
Mz diagram
MzMax
Diagrams of Cross-Section’ Normal Stresses
Centroid
Tension
Compression
z
MzMin
Ten
sio
n
Co
mp
ress
ion
N diagram N
Compression
Load
zI
M
y
yMy
A
NN
yI
M
z
zMz
Cross-Section
Axial Force
Cross-Section Area
Axial Stress of Cross-section Points
Cross-Section Bending Moment
Cross-Section Moment of Inertia
Coordinate of a Point
Bending Stress at a Point
Axial Stress and Bending Stress are normal stresses, i.ethey both act perpendicular to the cross-section – this is
ONE-DIMENSIONAL stress state
Resultant of an axial and two bending stresses at some point (y; z) equals to their algebraic sum:
yI
Mz
I
M
A
N
z
z
y
yMzMyN
Priit Põdra 8. Strength of Components under Combined Loading 14
Cross-Section Critical PointsCross-Section Critical Points
Equation of NEUTRAL LINE
Eccentric Tension/Compression
• Neutral Line does not cross the centroid• Neutral Line is straight,
Cross-Section critical points are te most distant ones from the
neutral line
0 yI
Mz
I
M
A
N
z
z
y
y
yCentroid
z
OS (y2;z2)
Maximum Distance
Maximum Distance
Critical Points of the Cross-Section
Maximum Compressive Stress
OT (y1;z1)Maximum Tensile Stress
Cross-Section NEUTRAL
LINE
Cross-Section NEUTRAL LINE = projection of the beam’ neutral layer
Neutral line divides the section in two: area of tension and area of compression
line at the beam cross-section, which’ points have no stress (bending stress equals to zero)
=
The signs (+/-) of both internal forces and coordinates must be considered
Priit Põdra 8. Strength of Components under Combined Loading 15
2
1
=
y
z
Cross-Section Critical Points
Maximum Compressive Stress
OT (y1;z1)Maximum Tensile Stress
diagram
Cross-Section NEUTRAL LINE
O1
O2
OS (y2;z2)
Strength ConditionsStrength Conditions
22OS
11OT
zI
My
I
M
A
N
zI
My
I
M
A
N
y
y
z
z
y
y
z
z
Normal Stresses at the Critical Points
Compr22OS
Tension11OT
zI
My
I
M
A
N
zI
My
I
M
A
N
y
y
z
z
y
y
z
z
Strength Conditions for Eccentric Load
Maximum Tensile Stress at the Point OT
Maximum Compressive Stress at the
Point OS
DesignStress for Tension
Design Stress for Compression
y-coordinate of point OT
z-coordinate of point OT
z-coordinate of point OSy-coordinate of point OS
NB! The signs (+/-) of both internal forces and coordinates must be considered
Priit Põdra 8. Strength of Components under Combined Loading 16
y
z
OS
Stress Distribution of TWO Signs
Maximum Compressive Stress
OT
Maximum Tensile Stress
diagram
OT
OS
Centroid
Stress Distributions of One SignStress Distributions of One Sign
The load is applied FAR from the centroid
The load is applied NEAR to the centroid
O2y
z
Stress Distribution of ONE Sign
diagramOS
Cross-Section CORE
Cross-Section Centroid
Min
imu
m
Co
mp
ress
ive
Str
ess
OS Maximum Compressive Stress
Cross-section’ CORE = area around the centroid
When the load is applied insede the core
Stress distribution of one sign is formed
b/3b
h/3
h
Core Centroid
Rectangle’ core
D/4
D
Core
Centroid
Circle’ Core
Determination of the core is usually not needed in strength
analysis
Priit Põdra 8. Strength of Components under Combined Loading 17
Extremities of Eccentric Loading CasesExtremities of Eccentric Loading CasesIncreasing distance between the
load application point and the centroid
More resemblance of the case to that of bending (two bendings)
Nearer the load application point to the centroid
More resemblance of the case to that of axial loading
Neutral line is located nearer to the centroid
Increasing distance between the neutral line and the centroid
y
z
OS
Stress Distribution of Two Signs
OT
diagram
OT
OS
y
z
OS
Stress Distribution of Two Bendings
OT
diagram
OT
OS
y
z
Stress Distribution of ONE Sign
diagramOS
OS y
z
Stress Distribution of Axial Load
diagram
Load (Force) is applied infinitely distant from the centroid
Neutral Line crosses the Centroid
Neutral Line is infinitely distant from the Centoid
Load is applied at the centroid
Priit Põdra 8. Strength of Components under Combined Loading 18
Special Case – Rectangular SectionSpecial Case – Rectangular SectionCross-sections, which’ outside contour is rectangle
and both principal centroidal axes are axes of symmetry
Critical are the diagonal corners
At the critical point of tension OT ORcompression OS
At the critical point of compression OS OR tension OT
z
z
y
y
z
z
y
y
W
M
W
M
A
N
W
M
W
M
A
N
Max
Min
Min
Max
Maximum Normal Stress Valuesz
+ + + + + +
+++++
Stresses of Eccentrically Loaded Rectangular Section
OS
OT
Mz diagram
My diagramy
Mz
Wz
Mz
WzMy
Wy
My
Wy
Maximum Compressive Stress
Max
imu
m
ten
sile
Str
ess
NA
N diagram
Critical points are always located at the diagonal corners (or in one corner)
I-profile Square
Priit Põdra 8. Strength of Components under Combined Loading 19
Special Case – Circular SectionSpecial Case – Circular Section
22zy MMM
Bending Moment
Resultant bending
moment is calculated only
for circular sections
Neutral Line
+++
++
++
++
++
+Circle has an infinite number of principal
centroidal axes
OS
OT
z
y
Mdiagram M
W
MW
Maximum Tensile Stress
Maximum Compr. Stress
Ndiagram
NA
This axis is also a principal
centroidal axisz
Stresses of Eccentrically Loaded circular Section
Mz diagram
My diagramy
Mz
W
Mz
WMy
W
My
W
NA
Ndiagram
W
M
A
N
W
M
A
N
Max
Min
Min
Max
Maximum Normal Stress Values
Critical points are always located
diametrally on the perimeter
At the critical point of tension OT
ORcompression OS
At the critical point of compression OS
ORtension OT
Priit Põdra 8. Strength of Components under Combined Loading 20
STRENGTH OF MATERIALSSTRENGTH OF MATERIALS
8.3. Combined Stress State Analysis8.3. Combined Stress State Analysis
Priit Põdra 8. Strength of Components under Combined Loading 21
Strength Analysis Problem of a Combined Stress StateStrength Analysis Problem of a Combined Stress State
STRENGTH CONDITION compares the value of actual stress to that of design stress
Is dependent on the material limit
state stress
DESIGN STRESS
Limit state stress is determined by the tensile test
ONE-DIMENSIONALStress is acting in the test specimen
ACTUAL STRESS
Is dependent on the component geometry and
loading
ONE-DIMENSIONAL,
PLANE orTHREE-
DIMENSIONALstress is acting
in the component
Classical Strength Condition
ACTUAL STRESS DESIGN STRESS
THREE_DIMENSIONAL and ONE-DIMENSIONAL
stresses CAN NOT be compared
ONE-DIMENSIONAL stress CAN be compared with another ONE_DIMENSIONAL stress
PLANE and ONE-DIMENSIONAL stresses CAN NOT be compared
Priit Põdra 8. Strength of Components under Combined Loading 22
Equivalent Stress of the Combined Stress StateEquivalent Stress of the Combined Stress State
KNOWN ARE the strength conditions for ONE-DIMENSIONAL stress state
Strength of components, that have PLANE or THREE-DIMENSIONAL
stress state NEED TO BE ANALYSED
COMBINED stress state must be reduced to the ONE-DIMENSIONAL one
EQUIVALENT STRESS = ONE-DIMENSIONAL stress, that has equal criticality with the given COMBINED
stress state
Strength Condition for COMBINED Stress State
);;( 321Eq f Equivalent one-dimensional stress of a combined stressstate
Function of principal stresses
Design stress (Permissible stress) for one-dimensional case
Equivalent Stress of Three-Dimensional Stress State
);;( 321Eq f
Eq
K
ONE-DIMENSIONAL Stress at the Point K
THREE-DIMENSIONAL Stress at the Point K
K
12
3
Reduction of THREE-DIMENSIONAL Stress State
Stress states of equal criticality
Priit Põdra 8. Strength of Components under Combined Loading 23
General Principles of Strength TheoriesGeneral Principles of Strength Theories
STRENGTH THEORIES
Criterial Theories
They give theoretical hypothesis about the general cause of the limit state occurance (limit state criterion)
Are based on the mathematical processing results of test data, with no further analysis of
physical phenomena
Older
STRENGTH THEORY (or Limit State Theory) = Theoretical concepts for the stress state criticality analysis
STRENGTH THEORY • is based on some hypthesis for limit state appearance: “What kind of stress state causes the material limit state to occur?”
• gives a formula for equivalent stress value calculationHypothesis of limit state occurance depends on the type of limit state and the stress state in question
• Maximum Normal Stress Theory• Maximum Strain Theory• Maximum Shear Stress Theory• Strain Energy Theory
Phenomenological Theories
• Mohr’ Theory, etc
Newer
Priit Põdra 8. Strength of Components under Combined Loading 24
Material Ductility and BrittlenessMaterial Ductility and Brittleness
Limit State of DUCTILE Material = YIELDING
Limit State of BRITTLE Material = FRACTURE
Yielding
Elastic Elongation Elstic Elongation
Fracture
Material Ductility and Brittleness
depend on TEMPERATURE
Du
ctili
ty
Temperature
Ductile
Brittle
Malleable Steel
Dependence of Steel Ductility on the Temperature
Ductility of many steels start to decrease at the temperatures (0…-10)ºC
Stress-Strain Diagram
Stress-Strain Diagram
Priit Põdra 8. Strength of Components under Combined Loading 25
Verified in practice by tensile and pure torsion tests
A.K.A. I (first) Strength Theory
Theory of Maximum Normal StressTheory of Maximum Normal Stress
G. Galilei 1564...1642
Brittle material fractures, when the principal stress of maximum absolute value exceeds a certain limit value, independent on the values of other principal
stresses at that point
HYPOTHESIS:
Value of Equivalent Stress
313
311IEq if
if
Cast Iron, Concrete, Rock, ...
Applicable for BRITTLE MATERIALS under TENSION
W.J.M. Rankine 1820…1872
Ultimate Strength of that material, from the ordinary tensile test
According to I (first)
Strength Theory
Maximum Tensile StressMaximum Compressive Stress
Priit Põdra 8. Strength of Components under Combined Loading 26
Theory of Maximum StrainTheory of Maximum Strain
A.K.A. II (second)
Strength TheoryApplicable for BRITTLE MATERUIALS under COMPRESSION
J.V. Poncelet 1788…1867
B. de Saint-Venant 1797…1886
Brittle material fractures, when the strain of maximum absolute value exceeds a certain limit value, independent on the values of other strains of that point
HYPOTHESIS:
Strain, that corresponds to the Ultimate Strength of that material, from the ordinary tensile test
2133
3211
1
1
E
E
Maximum Strains of the Stress State
Values of Equivalent Stress
31213
31321IIEq i
if
f
According to II (second)
Strength Theory
Cast Iron, Concrete, Rock, ...
Priit Põdra 8. Strength of Components under Combined Loading 27
Theory of Maximum Shear StressTheory of Maximum Shear Stress
A.K.A. III (third) Strength Theory
Steels
Applicable for DUCTILE MATERIALS, if the limit state is yielding
H. Tresca, 1868
Yielding = Shear of material layers
Ductile material starts to yield, when the maximum shear stress of the stress state exceeds a certain limit value, independent of the values of principal
stresses at that
HYPOTHESIS:
Maximum shear stress, corresponding to the tensile yield strength of that material, from the ordinary tensile test
Maximum Shear Stress of Three-Dimensional
Stress State
231
Max
Maximum Shear Stress of Equivalent
Stress State
2Eq
Max
Value of Equivalent Stress
31IIIEq
According to III (third) Strength Theory
J.J.Guest, 1900
Priit Põdra 8. Strength of Components under Combined Loading 28
H. Hencky, 1925
Strain Energy TheoryStrain Energy Theory
A.K.A. IV (fourth)
Strength Theory
Metals, incl. Steels
Applicable for DUCTILE MATERIALS, if the limit state is yielding or plasticity
R. von Mises, 1913
Ductile material starts to deform plastically or yield, when the strain energy density of the stress state exceeds a certain limit value
Strain energy density, that corresponds to the elastic limit stress of that material, from the ordinary tensile test
HYPOTHESIS:
M.T. Huber, 1904E. Beltrami, 1903
Strain Energy of a Three-Dimensional Stress State
13322123
22
21D 3
1
Eu
Strain Energy of Equivalent Stress State
2EqD 3
1 E
u
Value of Equivalent Stress
13322123
22
21
IVEq
According to IV (fourth) Strength Theory
Priit Põdra 8. Strength of Components under Combined Loading 29
Phenomenological Theory
Mohr’ Strength TheoryMohr’ Strength Theory
From tests with brittle materials
In the cases of combined stress state, the most critical, for the limit state to occur,
are the principal stresse of extremal value1 and 3 (influence of 2 is negligible)
Each three-dimensional stress state can be regarded as a plane
stress state
TensionU1Lim
When the Tensile and Compressive Strengths are Equal
CU
TU 3Lim1Lim
Applicable for both DUCTILE and BRITTLE materials
C.O. Mohr, 1835…1918 Does not have theoretical hypothesis
Material Limit State Stresses from Ordinary Tests
ComprU3Lim
Maximum possible value of 1
Material tensile Strength form the tensile test
Maximum possible value of 3
Material Compressive Strength form the compression test
Value of Equivalent Stress
33Lim
1Lim1
MEq
According to Mohr’ Strength Theory
IIIEq
MEq
Mohr’ Strength Theory
III Strength
Theory
Priit Põdra
COMBINED STRESS STATE = any PLANE and THREE-DIMENSIONAL Stress State
8. Strength of Components under Combined Loading 30
Stresses of different directions are acting at the same point
General Metod for Combined Stress State Strength AnalysisGeneral Metod for Combined Stress State Strength Analysis
1. Location of critical section is determined using internal force diagrams
2. Location of cross-section critical points is determined
using stress diagramsCross-sections, where the maximum internal force values are acting
Point, where the equivalent stress maximum value is
actingCross-sections with decreased
surface area
3. Application of strength condition at the critical point(s) of critical section(s)
);;( 321Eq f
Equivalent one-dimensional stress of the given stress state
Function of stess state’ principal
stresses
One-dimensional design stress SS
Eq
Lim
Strength Condition by Safety
Points, where some stress maximum value is acting
Value of Design Factor
Real Value of Safety Factor
Stresses of different types are acting at the same pointThere could be many of them
There could be many of them
Priit Põdra 8. Strength of Components under Combined Loading 31
Equivalent Stress of a Plane Stress StateEquivalent Stress of a Plane Stress State
Stress Theory
2
2
3
2
2
1
22
22
xyyxyx
xyyxyx
Principal Stresses of Planar Stress State
from previous
According to Maximum Shear Stress’ Strength Theory (Tresca)
31IIIEq 22III
Eq 4 xyyx
According to Strain Energy’ Strength Theory (von Mises)
222IVEq 3 xyyxyx 31
23
21
IVEq
Cross-Section Normal Stress
Longitudinal Section Normal Stress
Cross-Section Shear Stress
Priit Põdra 8. Strength of Components under Combined Loading 32
STRENGTH OF MATERIALSSTRENGTH OF MATERIALS
8.4. Combination of Bending and Shear8.4. Combination of Bending and Shear
Priit Põdra 8. Strength of Components under Combined Loading 33
MaxM
yz
yK
M diagram Q diagramMaxM
MaxQ
Cross-Section Stresses
x
y
z
F
Põikpaindes konsool
Ohtlik ristlõige
Q epüür
M epüür
Internal Forces and Cross-Section StressesInternal Forces and Cross-Section Stresses
Combination of Bending and Shear = Interaction of Bending moment M and Shear Force Q
yI
MM
Bending Stress
y
yQ Ib
QS
Shear Stress
Bending MomentShear Force
Combination of Bending and Shear is Important:
Component is relatively short Shear force Q has high value compared with bending moment M
Component has thin wallsMaximums of normal and shear stresses act in close vicinity
Stresses of IPE-Section
Thinwalled Section
Cantilever beam
Critical Section
M diagram
Q diagram
Priit Põdra 8. Strength of Components under Combined Loading 34
Equivalent StressEquivalent Stress
Stresses
Qxy Mx
0yAccording to the maximum Shear Stress Theory (Tresca)
31IIIEq 22III
Eq 4 QM
22
2
22
1
22
22
QMM
QMM
Principal Stresses at a Point
According to the Strain Energy Theory (von Mises)
22IVEq 3 QM 31
23
21
IVEq
Priit Põdra
6
20
FA = 20 kN
300
x
y300
A BC
F = 40 kN
FB = 20 kN
20
Q diagram, kN
M diagram, kNm
Transversely Loaded Beam
Critical Section
IPE - Beam
8. Strength of Components under Combined Loading 35
Example: Strength Analysis of a IPE-Beam (1)
Example: Strength Analysis of a IPE-Beam (1)
Determine the apropriate IPE-profile!
Solution
1. Dimensioning for BENDING
Maximum bending stress is a function of W, [m3]
Complex cubic equation is to be solved
Maximum shear stress is a function of A, [m2]
The relation of A and W is not apparent
2. Check for adequate strength under the combination of bending and shear
Material: Steel S235 DIN 17100
y = ReH = 235 MPa
[S] = 2,5
Yield Strength
Design Factor
kN20
kNm6
C
C
Q
M
Critical Section
For a Combination of Bending and Shear
Dimensioning for the combination of bending and shear is not easy
Priit Põdra 8. Strength of Components under Combined Loading 36
Example: Strength Analysis of a IPE-Beam (2)
Example: Strength Analysis of a IPE-Beam (2)
Dimensioning for BENDING
Strength Condition for Bending
SW
M yMax
For Moment of Resistance
SM
WWy
3366
3
y
cm64m1063.82.510235
106
SM
W
Table of IPE Steel Profiles
Profilele IPE 140 fulfills the strength condition for bending
33 cm64cm77.3 WWx
2cm16.4 A 4cm541 Ix
Priit Põdra 8. Strength of Components under Combined Loading 37
diagram
Stress Distributions of the IPE 140 Cross-Section
diagram
y
z
Critical Points
O1
O3
O2
O2
O2
Critical Points
Critical Points
Max
Max
Max
Example: Strength Calculation (3)Example: Strength Calculation (3)
Check for adequate strength for the COMBINATION OF BENDING AND SHEAR at the critical points of IPE 140 cross-section
0O1
MaxO1
Critical points O1
MaxO3
O3 0
Critical Points O3
0
0
O2
O2
Critical points O2
Critical Points of the Cross-Section
Priit Põdra 8. Strength of Components under Combined Loading 38
diagram, MPa
Stress Distributions of the IPE 140 Cross-Section
MaxQ
diagram
y O1
z
78
78
Shear stress is absent
The stress with the same value, but opposite sign is acting at the symmetrically located points
Example: Strength Analysis of a IPE-Beam (4)
Example: Strength Analysis of a IPE-Beam (4)
Maximum bending stress is acting at the critical points O1
Strength is adequate at the critical points O1
Check for adequate strength under bending
Check for adequate strength for BENDING at the critical points O1 of the IPE 140 cross-section
MPa78Pa1077.61077.3
106 66
3
MaxO1
W
M
Maximum Bending Stress at the Points O1
2.53.03.0178
235
Max
y SSFactor of Safety at the Points O1
Priit Põdra 8. Strength of Components under Combined Loading 39
Normal stress is absent
Example: Strength Analysis of a IPE-Beam (5)
Example: Strength Analysis of a IPE-Beam (5)
Maximum shear stress is acting at the critical
points O3
Strength is adequate at the critical points O3
Check for adequate strength under shear
MPa34Pa1033.760.004710541
1042.91020
6
8
-630.5
MaxO3
Is
QS
Maximum Shear Stress at the Points O3
diagram, MPa
Stress Distributions of the IPE 140 Cross-Section
diagram, MPa
y
O3 z
78
140
(h)
73 (b)
4.7 (s)
6.9
(t)
34
78
32
2
0.5
cm42.942.870.69140.697.30.692
140.47
2
1
22
1
thbtt
hsS
Statical Moment of the Half-Section about z-axis
Check for adequate strength for SHEAR at the critical points O3
of the IPE 140 cross-section
2.53.43.4534
2350.5
Max
y
SSFactor of Safety at the Points O3
Priit Põdra 8. Strength of Components under Combined Loading 40
Example: Strength Analysis of a IPE-Beam (6)
Example: Strength Analysis of a IPE-Beam (6)
3
Flange
cm33.533.52
0.69140.697.32
1
2
1
thbtS
Flange’ Statical Moment about z-axis
At the critical points O2 is acting:- bending stress value, close to the maximum;
- shear stress value, close to the maximum.
Check for adequate strength under the combination of bending and shear
MPa27Pa1026.340.004710541
1033.51020 68
-63Flange
O2
Is
QS
Shear Stress at the Points O2
This is the combination of bending and shear
diagram, MPa
Stress Distributions of the IPE 140 Cross-Section
diagram, MPa
y
O2
z
78
140
(h)
73 (b)
4.7 (s)
6.9
(t)
34
78
Flange
63.1
(y)
7027
27
MPa70Pa1069.90.063110541
106 68
3
O2
yI
M
Bending Stress at the Points O2
Check for adequate strength for the COMBINATION of BENDING and SHEAR at the critical points O2 of the IPE 140 cross-section
Priit Põdra 8. Strength of Components under Combined Loading 41
Example: Strength Analysis of a IPE-Beam (7)
Example: Strength Analysis of a IPE-Beam (7)
MPa8988.427470 222O2
2O2
IIIO2Eq, 4
Equivalent Stress at the Points O2
2.52.664289
235IIIEq
y SS .
Factor of Safety at the Poinys O2
Strength is adequate at the points O2
ALL strength conditions are fulfilled
IPE 140 profile is adequate for the given application
Check for adequate strength for the COMBINATION of BENDING and SHEAR at the critical points O2 of the IPE 140 cross-section
Priit Põdra 8. Strength of Components under Combined Loading 42
STRENGTH OF MATERIALSSTRENGTH OF MATERIALS
8.5. Combination of Torsion and Bending8.5. Combination of Torsion and Bending
Priit Põdra
z
Mz epüür
My epüüry
Mz
W
Mz
WMy
W
My
W
MW
MW
TW0
TW0
M diagram
T diagram
Stress Distributions
Neutral Line
O1
O2
Mz diagram
My diagram
8. Strength of Components under Combined Loading 43
Combination of Torsion and Bending for the Circular Cross-SectionCombination of Torsion and Bending for the Circular Cross-Section
Maximum torsional stresses are acting at the edge of the cross-section
Maximum bending stresses are acting at the edge of the cross-section
22zy MMM
Resultant Bending moment for a CIRCULAR Section
Combination of bending and torsional stress MAXIMUMS is acting at the points O1 ja O2
W
MM
W
M zy
M
22
Max
Maximum Bending Stress
This is a plane stress state
W
T
W
TT 20
Max
Maximum Torsional Stress For a Circular Section
32
3DW
16
3
0
DW
Moment of Resistance
Polar Moment of Resistance
Priit Põdra 8. Strength of Components under Combined Loading 44
Maximum Equivalent Stress for CIRCULAR SectionMaximum Equivalent Stress for CIRCULAR Section
At the pointsO1 and O2
According to the Maximum Shear Stress Theory (Tresca)
22IIIEq 4 xyyx
W
TMM zy
TΜ
2222Max2MaxIII
Eq 4
According to the Strain Energy Theory (von Mises)
W
TMM zy
TΜ
2222Max2MaxIV
Eq
0.753
222IVEq 3 xyyxyx
Stresses
MaxΤxy
MaxΜx
0y
Priit Põdra 8. Strength of Components under Combined Loading 45
Equivalent Bending MomentEquivalent Bending Moment
Critical section of a uniform circular bar is located, where the equivalent bending moment has maximum value
W
MEqEq
NB! Apliccable only for CIRCULAR sections!!!
W
TMM zy222
IIIEq
Maximum Equivalent Stress for a Circular Section
W
TMM zy222
IVEq
0.75
Maximum Bending Stress for a Circular Section
W
MM zy22
W
M or
Maximum Shear Stress Theory (Tresca)
222IIIEq TMMM zy
Strain Energy Theory (von Mises)
222IVEq 0.75TMMM zy
Priit Põdra
F1
Driving Pulley
Driven PulleyRadial-Thrust
Bearing
Radial Bearing
Shaft
f1
F2
D1
f2
D2
Shaft of a Belt Drive
Direction of Rotation
8. Strength of Components under Combined Loading 46
Example: Strength Analysis of a Circular Uniform Shaft (1)
Example: Strength Analysis of a Circular Uniform Shaft (1)
Calculate the diameter of this uniform shaft
Ratio of Belt Sides’ Loads –
calculated using Euler’ formula
D1 = 60 mmD2 = 100 mmc = F/f = 2,1
Material: steel S355 DIN 17100
Y = ReH = 355MPa
[S] = 2,5
Yield Strength
Design FactorPower to be transmitted: P = 300 W
Frequency of rotation : n = 90 rev/min
All relevant dimensions are shown on the structural models
General Scheme of Shaft Strength Analysis
1. Bending and Twisting Loads
2. Internal Forces at the Principal Planes
3. Equivalent Bending Moment
4. Strength Calculation in Critical Section(s)
5. Check of Adequate Strength
Priit Põdra 8. Strength of Components under Combined Loading 47
Causes the sfaft to TWIST
Example: Strength Analysis of a Circular Uniform Shaft (2)
Example: Strength Analysis of a Circular Uniform Shaft (2)
1. Bending and Twisting Loads
Shaft Angular Velocity
s
rad9.49.42
60
290
60
2
n
Moment Transmitted by Shaft
Nm3231.99.4
300
PM
Shaft BENDING deformations are caused by:• Belts’ tensioning forces,
• Transmittable loads, that act transverselyLoads of the Belts’ Sides
F1 f1
F2
f2D2
D1
Belt’ Sides’ Tensioning Forces
cfFDc
f1-
2M
Relation of Shaft Twisting Moment and Belts’ Forces
2
22
22
111
DfF
DfF
M 2
12
cD
-cfD
ff M
Twisting moment is Caused by the Difference of Belt’ Sides’ Tensioning Forces
Priit Põdra 8. Strength of Components under Combined Loading 48
Sircle has an infinite nuber of principal centroidal axes
Shaft Loads and the Principal Axes
FA
D2
D1
M
M FB
y
z
Prinsipal Centroidal
Axis
It is assumed, that the belt sides
are parallel
Example: Strength Analysis of a Circular Uniform Shaft (3)
Example: Strength Analysis of a Circular Uniform Shaft (3)
1. Bending and Twisting Loads (2)
Tensile Loads of Belts’ Sides
N204020379702.1
N970969.60.0612.1
322
1-2
11
11
cfFDc
fM
N12301222.25822.1
N582581.80.1012.1
322
1-2
22
22
cfFDc
fM
Shaft’ bending loads are the resultants of belts’ tensile loads
They cause the shaft’ bending deformation
N182018125821230
N30109702040
22B
11A
fFF
fFF
Shaft Transverse Loads
The choice of the principal centroidal axes is free for
circular shaftsComponents of Bending Loads about the Principal Centroidal Axes
0
N3010
Az
AAy
F
FF
N14901490.8sin551820sin
N10501043.9cos551820cos
BBz
BBy
FF
FF
Prinsipal Centroidal
Axis
Priit Põdra 8. Strength of Components under Combined Loading 49
Shear forces Q are neglected
Example: Strength Analysis of a Circular Uniform Shaft (4)
Example: Strength Analysis of a Circular Uniform Shaft (4)
2. Internal Forces Diagrams about Principal Centroidal Axes
Nm 630.061050DB
Nm 151150.50.053010AC
BD
AC
yz
z
FM
FM
Nm 9089.40.061490DBFBD zyM
Internal Forces (by the Method of Sections)
Nm 32BA MTT
My diagram, Nm
z
x
A BC D
FBzFCz
FDz
90
Bending Moment Diagram on x-z Plane
Mz diagram, Nm
FA
y
x
25050 60
A BC D
FByFCyFDy
63
151
Bending Moment Diagram on x-y Plane
NB! In this problem, the calculation of
bearing reactions is not needed!!!
Reactions at the bearings can be
determined using the equations of equilibrium
T diagram, Nm
x
A C D
32
Torque Diagram
B
MM
Priit Põdra 8. Strength of Components under Combined Loading 50
Maximum Shear Stress Theory
222IIIEq TMMM zy
Example: Strength Analysis of a Circular Uniform Shaft (5)
Example: Strength Analysis of a Circular Uniform Shaft (5)
3. Diagram of Equivalent Bending Moment
Nm 323200
Nm 115114.4326390
Nm 155154.3321510
Nm 323200
2222B
2B
2B
IIIB Eq,
2222D
2D
2D
IIID Eq,
2222C
2C
2C
IIIC Eq,
2222A
2A
2A
III AEq,
TMMM
TMMM
TMMM
TMMM
zy
zy
zy
zy
Values of Equivalent Bending Moment
This is the critical section of UNIFORM shaft
x
A C D
32
Diagram of Equivalent Bending Moment
B
32
155115MEq diagram, Nm
III
Maximum equivalent bending moment is acting
at the cross-section C
Strength Condition
SD
M
W
M y
3
IIIEq
IIIEqIII
Eq
32
mm25m0.02232.510355
15532323
63
y
IIIEq
SM
D
Shaft Diameter for Cross-Section C4. Strength Calculation for Cross-Section C
Priit Põdra 8. Strength of Components under Combined Loading 51
Example: Strength Analysis of a Circular Uniform Shaft (6)
Example: Strength Analysis of a Circular Uniform Shaft (6)
Maximum Bending Stress for Cross-Section C
MPa101Pa10101.00250
1553232 633
MaxΟ1
D
M
W
M zzM
Maximum Torsional Stress for Cross-Section C
MPa11Pa1010.40250
321616 633
0
MaxΟ1
D
T
W
TT
Maximum Equivalent Stress for Cross-Section C
MPa104Pa10103.31141014 6222
O1
2
O1IIIEq
Check for Adequate Strength via Equivalent Stress
2.53.43.41104
355IIIEq
Y SS
Check for Adequate strength via Normal Stress
2.53.53.51101
355Max
Y SSM
Check for Adequate Strength via Shear Stress
2.51616.111
3550.5Max
Y
SST
All strength conditins are fulfilled
Answer: Diameter of that shaft must be 25 mm
z
y
Mz diagram, MPa
Stresses at the Cross-Section C
Neutral Line
O1
O2
T diagram, MPa
101
101
11
11
Ø 25
5. Check of Adequate Strength in Cross-Section C
Priit Põdra 8. Strength of Components under Combined Loading 52
Middle Point of Shorter Edge
Middle Point of Longer Edge
Corner of a Rectangle
Combination of Torsion and Bending for the Rectangular Cross-SectionCombination of Torsion and Bending for the Rectangular Cross-Section
z
Rectangular Cross-Section Stresses
Mzdiagram
My diagram
y MzMax
MzMax
MyMax
MyMax
Thdiagram
ThMax
TbMax
h
b
O1
O2
O3
Three Critical Points:
This is ONE-DIMENSIONAL stress
- Combination of maximum bending stresses
MaxMax1 ;:O MzMy
This is PLANE stress
MaxMax2 ;:O ThMz - Combination of maximum
bending stress and maximum torsional stress
This is PLANE stress
MaxMax3 ;:O TbMy - Combination of maximum
bending stress and maximum torsional stress
A point of equal criticality is located symmetrically to the
critical points above
Critical points are always located at the
corners and at the middle of edges
Tbdiagram
Priit Põdra 8. Strength of Components under Combined Loading 53
Strength Analysis for Rectangular SectionStrength Analysis for Rectangular Section
MaxMax1O MzMy
Combination of Normal Stresses at thePoint O1
2Max
b
2Max3
IIIEkv
2Maxh
2Max2
IIIEkv
4O
4O
TMz
TMy
Combination of Normal and Shear Stress at the Points O2 and O3
Plane Stress States
According to the maximums hear
stress theory
Min3Eq2Eq1 O;O;Omax
Strength Condition for the Combination of Bending and Torsion
Strength Calculation for the Most Critical Point
Minimum Value of the Design
Stresses
SS 3Eq2Eq1
MinY
Min O;O;Omax
Strength Condition via Safety FactorsMinimum Safety Factor
Value
Minimum Yield Strength Value
One-Dimensional Stress State
Priit Põdra 8. Strength of Components under Combined Loading 54
Options of Strength Analysis MethodicsOptions of Strength Analysis Methodics
Strength analysis general methodics depend on the shape of
cross-section
Strength analysis using equivalent bending moment W
MEqEq
Strength analysis in all critical points
Critical points are located somewhere at the circle’ perimeter
Critical points are located at the corners and middle of the rectangle
edges
Priit Põdra 8. Strength of Components under Combined Loading 55
STRENGTH OF MATERIALSSTRENGTH OF MATERIALS
8.6. Practical Strength Analysis of a Complex Mechanical Structure
8.6. Practical Strength Analysis of a Complex Mechanical Structure
Priit Põdra 8. Strength of Components under Combined Loading 56
Design a Sub-Frame for aTipper!Design a Sub-Frame for aTipper!
Problems
• Complex ladder frame
• Complex loading pattern
• Real loads are not known
• Should be strong enough
• Should be rigid enough
• Should be ligthweighth enough
• Should be cheap enough
NB! Many requiremenst are contradictory
Priit Põdra 8. Strength of Components under Combined Loading 57
New Machine after Some Period of Use:New Machine after Some Period of Use:
Viltu paiknev kast
Viltu paiknev kast
Deformeerunudraam/vaheraam
The box is skewed
The box is skewed
The sub-frame is deformed
Priit Põdra 8. Strength of Components under Combined Loading 58
Rigid Sub-Frame is Reliable and SafeRigid Sub-Frame is Reliable and Safe
Tõstejõud FTõste
Pealisehituse raamehk vaheraam
Veokastalgasendis
Tõstesilinder
Tõstesilinder
Tõstesilindri telgon külgsihisvertikaalne
Kallutatavveokast
Tipper BoxHydraulic cylinder
axis must be strictly vertical
Hydraulic Cylinder for Tipper Box Lifting
Tipper box initial
positionSuperstructure
Frame or Sub-Frame
Hydraulic Cylinder
Lifting Force FLift
Priit Põdra 8. Strength of Components under Combined Loading 59
Non-Rigid Sub-Frame Caused the FailureNon-Rigid Sub-Frame Caused the Failure
FTõste
Pealisehituse raamehk vaheraam
Tõstesilinder
Tõstesilinder
FKülg
Külgjõud
Tõstejõud
Veokast liigendraami tagaosas
Tõstesilindri telg onkülgsihis vertikaali
suhtes kaldu
Veokastalgasendis
Kallutatavveokast
Vaheraamiesiosa
Vaheraamitagaosa onpöördunud
esiosa suhtes
Hydraulic cylinder axis is inclined
about vertical
Tipper Box
Hydraulic Cylinder
Tipper box initial
positionSuperstructure Frame or Sub-Frame
Lifting Force FLift
Tipper Box Hinge at the Rear
of Sub-Frame
Front Part of Sub-Frame
Rear Part of Sub-Frame –Twisted relative to the front
Hydraulic Cylinder
Lateral Force FLat
Priit Põdra 8. Strength of Components under Combined Loading 60
Where Was the Mistake Made?Where Was the Mistake Made?
In the Cases of:
• Complexity of structure and
• Complexity of loading
Classical approach of strength analysis is difficult to apply safely
Empirical (based on engineering experience) knowledge must be applied
• General engineering handbooks
• Specific handbooks
• Product catalogues
• Standards (ISO, GOST, DIN, ASME, etc)
• Company standards
• Special methods, etc.
Priit Põdra 8. Strength of Components under Combined Loading 61
Vehicle Body Builder Instructions were Misunderstood and ViolatedVehicle Body Builder Instructions were Misunderstood and Violated
Volvo nõudedpealisehitusekonstruktsioonile
Tagumiste diagonaal-sidemete nõutav ulatus
Tagumiste diagonaal-sidemete tegelik ulatus
Kasti stabilisaatorikinnituse nõutav asukoht
Kasti stabilisaatorikinnituse tegelik asukoht
Põiksidemenõutav asukoht
Eesmineesitelg (F)
Tagumineesitelg (G)
Eesminetagatelg (H)
Pealisehituseraam ehkvaheraam
Requirements of vehicle manufacturer for superstructure design
Front Axle (F)
Second Axle (G)
Third Axle (H)
Superstucture Frame of Sub-
Frame
Required position of box stabilizer fixing
Actual position of box stabilizer fixing
Required position of cross-bar
Cross-bar is absent
Required length and design of diagonal reinforcement
Actual length and design of reinforcement
Priit Põdra 8. Strength of Components under Combined Loading 62
STRENGTH OF MATERIALSSTRENGTH OF MATERIALS
THANK YOU!THANK YOU!
Questions, please?
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