8_Strength_of_Components_under_Combined_Loading

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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’ 
Case
Internal 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
z
I
M
y
y
My 
y
I
M
z
z
Mz 
Resultant of two bending stresses at some 
point (y; z) equals to their algebraic sum
y
I
M
z
I
M
z
z
y
y
MzMy  
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
Coordinate
According 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
Te
n
si
o
n
Compression
C
o
m
p
re
ss
io
n
Priit Põdra 8. Strength of Components under Combined Loading 6
Location of Cross-Section Critical 
Points
Location of Cross-Section Critical 
Points
Equation of NEUTRAL LINE
0 z
I
M
y
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
z
I
M
y
I
M
z
I
M
y
I
M
y
y
z
z
y
y
z
z


Normal Stresses at critical Points
 
 









Compr22OS
Tension11OT


z
I
M
y
I
M
z
I
M
y
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
diagram
y
Mz
Wz
Mz
WzMy
Wy
My
Wy
Maximum Compressive Stress
M
ax
im
u
m
 T
en
si
le
 S
tr
es
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
22
zy 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 diagram
y
Mz
W
Mz
WMy
W
My
W
z
y
 diagram M
W
M
W
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
x
F
zy
Loacation of the Load
y
z
F
x 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 ey
x
y
Mz (-)
Internal Forces at the Principal Planes
zx Principal Plane xy Principal Plane
N (-)Axial Force
Cross-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 EccentricLoad
My diagramMyMax
MyMin
y
Mz diagram
MzMax
Diagrams of Cross-Section’ 
Normal Stresses
Centroid
Tension
Compression
z
MzMin
Te
n
si
o
n
C
o
m
p
re
ss
io
n
N diagram N
Compression
Load
z
I
M
y
y
My 
A
N
N 
y
I
M
z
z
Mz 
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.e
they 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:
y
I
M
z
I
M
A
N
z
z
y
y
MzMyN  
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 y
I
M
z
I
M
A
N
z
z
y
y
y
Centroid
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
z
I
M
y
I
M
A
N
z
I
M
y
I
M
A
N
y
y
z
z
y
y
z
z


Normal Stresses at the Critical Points
 
 









Compr22OS
Tension11OT


z
I
M
y
I
M
A
N
z
I
M
y
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
 diagram
OS
Cross-Section 
CORE
Cross-Section 
Centroid
M
in
im
u
m
 
C
o
m
p
re
ss
iv
e 
S
tr
es
s
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/3
b
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 Cases
Increasing 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
 diagramOS
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 Section
Cross-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 OR
compression 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 diagram
y
Mz
Wz
Mz
WzMy
Wy
My
Wy
Maximum 
Compressive Stress
M
ax
im
u
m
 
te
n
si
le
 S
tr
es
s
N
A
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
22
zy 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
M diagram M
W
M
W
Maximum Tensile Stress
Maximum 
Compr. 
Stress
N diagram
N
A
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
N
A
N diagram















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
OR
compression OS
At the critical point of compression OS
OR
tension 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 State
Strength 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-DIMENSIONAL
Stress is acting in 
the test specimen
ACTUAL STRESS
Is dependent on 
the component 
geometry and 
loading
ONE-
DIMENSIONAL, 
PLANE or
THREE-
DIMENSIONAL
stress 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 State
Equivalent 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
1 2
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
D
u
ct
ili
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
311I
Eq 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 Theory
Applicable 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
31321II
Eq 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
2
31
Max
 
Maximum Shear 
Stress of Equivalent 
Stress State
2
Eq
Max

 
Value of Equivalent 
Stress
31
III
Eq  
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
 133221232221D 3
1  
E
u
Strain Energy of Equivalent Stress State
2
EqD 3
1 
E
u


Value of Equivalent Stress
 133221232221IVEq  
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 value
1 and 3 (influence of 2 is negligible)
Each three-dimensional 
stress state can be 
regarded as a plane 
stress state
Tension
U1Lim  
When the Tensile and 
Compressive Strengths are Equal
C
U
T
U   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
Compr
U3Lim  
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
3
3Lim
1Lim
1
M
Eq 

 
According to Mohr’ Strength Theory
III
Eq
M
Eq  
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 Analysis
General 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 diagrams
Cross-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 
State
Equivalent Stress of a Plane Stress 
State
Stress Theory













 








 



2
2
3
2
2
1
22
22
xy
yxyx
xy
yxyx






Principal Stresses of Planar Stress State
from previous
According to Maximum Shear Stress’ Strength Theory (Tresca)
31
III
Eq     22IIIEq 4 xyyx  
According to Strain Energy’ Strength Theory (von Mises)
222IV
Eq 3 xyyxyx   312321IVEq  
Cross-Section 
Normal Stress 
Longitudinal Section 
Normal Stress
Cross-Section 
Shear StressPriit 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
MaxM
yz
y
K
M diagram Q diagramMaxM
MaxQ
Cross-Section Stresses
x
y
z
F
Põikpaindes konsool
Ohtlik ristlõige
Q epüür
M epüür
Internal Forces and Cross-Section 
Stresses
Internal Forces and Cross-Section 
Stresses
Combination of Bending and Shear = Interaction of 
Bending moment M and Shear Force Q
y
I
M
M 
Bending Stress
y
y
Q Ib
QS

Shear Stress
Bending Moment
Shear 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 walls
Maximums 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  
0y
According to the maximum Shear Stress Theory (Tresca)
31
III
Eq   22IIIEq 4 QM   


















2
2
2
2
2
1
22
22
Q
MM
Q
MM


Principal Stresses at a Point
According to the Strain Energy Theory (von Mises)
22IV
Eq 3 QM   312321IVEq  
Priit Põdra
6
20
FA = 20 kN
300
x
y
300
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 y
Max

 
For Moment of Resistance
   SMWW
y

    336
6
3
y
cm64m1063.82.5
10235
106



 S
M
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
Max
Q
 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
MPa78Pa1077.6
1077.3
106 6
6
3
MaxO1 

 W
M
Maximum Bending Stress at the Points O1
  2.53.03.01
78
235
Max
y  SS


Factor 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
MPa34Pa1033.76
0.004710541
1042.91020
6
8
-63
0.5
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
14
0 
(h
)
73 (b)
4.7 (s)
6.
9 
(t
)
34
78
 
  3
2
2
0.5
cm42.942.870.69140.697.30.69
2
14
0.47
2
1
22
1














 














  thbtt
h
sS
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
  2.53.43.45
34
2350.5
Max
y 

 SS


Factor 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
0.69140.697.3
2
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
MPa27Pa1026.34
0.004710541
1033.51020 6
8
-63
Flange
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
14
0 
(h
)
73 (b)
4.7 (s)
6.
9 
(t
)
34
78
Flange
63
.1
 (
y)
70
27
27
MPa70Pa1069.90.0631
10541
106 6
8
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
2
O2
III
O2Eq,   4
Equivalent Stress at the Points O2
  2.52.6642
89
235
III
Eq
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 epüür
y
Mz
W
Mz
WMy
W
My
W
M
W
M
W
T
W0
T
W0
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-Section
Combination 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
22
zy 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
T
T 20
Max 
Maximum 
Torsional Stress For a Circular Section
32
3D
W


16
3
0
D
W


Moment of Resistance
Polar Moment of Resistance
Priit Põdra8. Strength of Components under Combined Loading 44
Maximum Equivalent Stress for 
CIRCULAR Section
Maximum Equivalent Stress for 
CIRCULAR Section
At the points
O1 and O2
According to the Maximum Shear Stress Theory (Tresca)
  22IIIEq 4 xyyx  
   
W
TMM zy
TΜ
222
2Max2MaxIII
Eq 4

 
According to the Strain Energy Theory (von Mises)
   
W
TMM zy
TΜ
222
2Max2MaxIV
Eq
0.75
3

 
222IV
Eq 3 xyyxyx  
Stresses
Max
Τxy  
Max
Μx  
0y
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
MEq
Eq 
NB! Apliccable only for CIRCULAR sections!!!
W
TMM zy
222
III
Eq


Maximum Equivalent Stress for a Circular Section
W
TMM zy
222
IV
Eq
0.75

Maximum Bending Stress for a 
Circular Section
W
MM zy
22 
W
M
 or
Maximum Shear Stress Theory (Tresca)
222III
Eq TMMM zy 
Strain Energy Theory (von Mises)
222IV
Eq 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 mm
D2 = 100 mm
c = F/f = 2,1
Material: steel S355 DIN 17100
Y = ReH = 355MPa
[S] = 2,5
Yield Strength
Design Factor
Power 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
rad
9.49.42
60
2
90
60
2



 n
Moment Transmitted by Shaft
Nm3231.9
9.4
300


PM
Shaft BENDING deformations are caused by:
• Belts’ tensioning forces,
• Transmittable loads, that act transverselyLoads of the Belts’ Sides
F1 f1
F2
f2D
2
D
1
Belt’ Sides’ Tensioning 
Forces
 






cfF
Dc
f
1-
2
M
Relation of Shaft Twisting Moment and Belts’ Forces
 
 







2
2
2
22
1
11
D
fF
D
fF
M    
2
1
2
c
D
-cf
D
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
D
2
D
1
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
N970969.6
0.0612.1
32
2
1-
2
11
1
1
cfF
Dc
f
M
   








N12301222.25822.1
N582581.8
0.1012.1
32
2
1-
2
22
2
2
cfF
Dc
f
M
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
FBz
FCz FDz
90
Bending Moment Diagram on x-z Plane
Mz diagram, Nm
FA
y
x
25050 60
A BC D
FByFCy
FDy
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
222III
Eq 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
2222
B
2
B
2
B
III
B Eq,
2222
D
2
D
2
D
III
D Eq,
2222
C
2
C
2
C
III
C Eq,
2222
A
2
A
2
A
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
155
115MEq diagram, Nm
III
Maximum equivalent 
bending moment is acting 
at the cross-section C
Strength Condition
 SD
M
W
M y
3
III
Eq
III
EqIII
Eq
32 

    mm25m0.02232.510355
1553232
3
6
3
y
III
Eq 




S
M
D
Shaft Diameter for Cross-Section C
4. 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
MPa101Pa10101.0
0250
1553232 6
33
Max
Ο1 




D
M
W
M zz
M
Maximum Torsional Stress for Cross-Section C
MPa11Pa1010.4
0250
321616 6
33
0
Max
Ο1 




D
T
W
T
T
Maximum Equivalent Stress for Cross-Section C
    MPa104Pa10103.31141014 6222O12O1IIIEq  
Check for Adequate Strength via 
Equivalent Stress
  2.53.43.41
104
355
III
Eq
Y  SS


Check for Adequate strength via 
Normal Stress
  2.53.53.51
101
355
Max
Y  SS
M

Check for Adequate Strength 
via Shear Stress
  2.51616.1
11
3550.5
Max
Y 

 SS
T

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-Section
Combination of Torsion and Bending 
for the Rectangular Cross-Section
z
Rectangular Cross-Section Stresses
Mzdiagram
My diagram
y Mz
Max
Mz
Max
My
Max
My
Max
Thdiagram
Th
Max
Tb
Max
h
b
O1
O2
O3
Three Critical Points:
This is ONE-
DIMENSIONAL stress
- Combination of maximum 
bending stresses
MaxMax
1 ;:O MzMy 
This is PLANE stress
MaxMax
2 ;:O ThMz  - Combination of maximum 
bending stress and 
maximum torsional stress
This is PLANE stress
MaxMax
3 ;: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
TbdiagramPriit Põdra 8. Strength of Components under Combined Loading 53
Strength Analysis for Rectangular 
Section
Strength Analysis for Rectangular 
Section
  MaxMax1O MzMy  
Combination of Normal Stresses at thePoint O1
     
     





2Max
b
2Max
3
III
Ekv
2Max
h
2Max
2
III
Ekv
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
Min
Y
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 
Methodics
Options of Strength Analysis 
Methodics
Strength analysis general 
methodics depend on the shape of 
cross-section
Strength analysis using 
equivalent bending moment W
MEq
Eq 
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
Deformeerunud
raam/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 raam
ehk vaheraam
Veokast
algasendis
Tõstesilinder
Tõstesilinder
Tõstesilindri telg
on külgsihis
vertikaalne
Kallutatav
veokast
Tipper 
BoxHydraulic cylinder 
axis must be 
strictly vertical
Hydraulic 
Cylinder for 
Tipper Box 
Lifting
Tipper box 
initial 
position
Superstructure 
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 
Failure
Non-Rigid Sub-Frame Caused the 
Failure
FTõste
Pealisehituse raam
ehk vaheraam
Tõstesilinder
Tõstesilinder
FKülg
Külgjõud
Tõstejõud
Veokast liigend
raami tagaosas
Tõstesilindri telg on
külgsihis vertikaali
suhtes kaldu
Veokast
algasendis
Kallutatav
veokast
Vaheraami
esiosa
Vaheraami
tagaosa on
pöördunud
esiosa suhtes
Hydraulic cylinder 
axis is inclined 
about vertical
Tipper 
Box
Hydraulic 
Cylinder
Tipper box 
initial 
position
Superstructure 
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 Violated
Vehicle Body Builder Instructions 
were Misunderstood and Violated
Volvo nõuded
pealisehituse
konstruktsioonile
Tagumiste diagonaal-
sidemete nõutav ulatus
Tagumiste diagonaal-
sidemete tegelik ulatus
Kasti stabilisaatori
kinnituse nõutav asukoht
Kasti stabilisaatori
kinnituse tegelik asukoht
Põiksideme
nõutav asukoht
Eesmine
esitelg (F)
Tagumine
esitelg (G)
Eesmine
tagatelg (H)
Pealisehituse
raam ehk
vaheraam
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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