STRENGTH OF MATERIALS - data.vk.edu.eedata.vk.edu.ee/RDER/RDER32/Tugevusopetus I/MHE0011 Strength of... · Priit Põdra 8. Strength of Components under Combined Loading 1 STRENGTH

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



8.1. Combination of Two Bendings8.1. Combination of Two Bendings


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


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


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


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


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


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 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


z

+ + + + + + +

++++++

Stresses of Rectangular Section

OS

OT

Mz diagram

My

diagramy

Mz

Wz

Mz

WzMy

Wy

My

Wy


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


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



Critical points are always located diametrally on the perimeter of the section

RESULTANT BENDING MOMENT –can be calculated for CIRCULAR

section only



8.2. Combination of Bending and Axial Loads

8.2. Combination of Bending and Axial Loads


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


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)


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


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


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


2

1

=

y

z

Cross-Section Critical Points


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


y

z

OS

Stress Distribution of TWO Signs


OT


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


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


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


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


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 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



8.3. Combined Stress State Analysis8.3. Combined Stress State Analysis


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


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


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


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


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


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, ...


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


31IIIEq

According to III (third) Strength Theory

J.J.Guest, 1900


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


13322123

22

21

IVEq

According to IV (fourth) Strength Theory


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


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


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



8.4. Combination of Bending and Shear8.4. Combination of Bending and Shear


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


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


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




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


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


diagram, MPa


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



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


Normal stress is absent



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


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




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


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




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



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


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


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


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


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


Causes the sfaft to TWIST



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


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



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


Shear forces Q are neglected



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


Maximum Shear Stress Theory

222IIIEq TMMM zy



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




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


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


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


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



8.6. Practical Strength Analysis of a Complex Mechanical Structure

8.6. Practical Strength Analysis of a Complex Mechanical Structure


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


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


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


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


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.


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



THANK YOU!THANK YOU!

Questions, please?

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Documents

STRENGTH OF MATERIALS - data.vk.edu.eedata.vk.edu.ee/RDER/RDER32/Tugevusopetus I/MHE0011 Strength of... · Priit Põdra 8. Strength of Components under Combined Loading 1 STRENGTH