STRENGTH OF MATERIALS - data.vk.edu.eedata.vk.edu.ee/RDER/RDER32/Tugevusopetus I/MHE0011 Strength of...STRENGTH OF MATERIALS 2.1. Structural Model. Priit Põdra 2. Strength of Components

Priit Põdra 2. Strength of Components under Axial Loading 1

STRENGTH OF MATERIALSSTRENGTH OF MATERIALS

2. Strength of Components under Axial Loading

2. Strength of Components under Axial Loading

2.1 Structural Model

2.2 Influences of Axial Loading

2.3 Internal Forces

Analysis

2.4 Normal Stress

Analysis

2.6 Strength

Calculations

2.5 Shear Stress

Analysis

Mehhanosüsteemide komponentide õppetool



2.1. Structural Model2.1. Structural Model


Structural Model

Real Structure and Calculation SchemeReal Structure and Calculation Scheme

Real Structure

The bar is deformable

Simplified mechanical system

Ideal mechanical system

There are many different influencers and phenomena

Some of them are important — some

of them are not

Bar

Joints

Puller

Vibration

Wind

Bracket

Force

Swing

strength of that

component must be analysed

Negligible influencers eliminated

Important influencer

F

A

L

Simplifications:

Support is absolutely rigid

Joints are absolutely rigid

Com-ponents Weight


Development of Structural ModelDevelopment of Structural Model

Too complex structural model voluminous calculation work

Too simple structural model large uncertainty of analysis results

Development of a structural model is based on experience

Real StructureSimlified Mechanical

System

Bar is deformable

Supports are absolutely rigid

Joints are absolutely rigid

Ideal Mechanical System

Parameters with negligible

influence are eliminated

(Saint Venant’ Principle)

STRUCTURAL MODEL = graphical representation of an ideal mechanical system together with relevant dimensions and other data



2.2. Influences of Axial Loading2.2. Influences of Axial Loading


TENSION

COMPRESSION

Axial DeformationAxial Deformation

F

The bar lengthens due to axial load

L

L

Axially loaded bar

Axial Deformation = change of the bar’ length(bar lengthening and/or shortening)

Sign convention:

Strength of Materials deals with elastic deformations only

Lengthening is positive

Shortening is negative

bar (or part of it) lengthens by L (+)

bar (or part of it) shortens by L (-)


F F

Bar’ Load State — Tension or CompressionBar’ Load State — Tension or Compression

Axially loaded uniform and straigth bar

Bar’ tension or compresion = (simple) load state of a bar, where:

Cross-sections remain parallel to each other and perpendicular to the axis

Axis remains straight Axis remains straigth

L (+)

Length changes

L (-)

Length changes

Length of the bar changes (total length may not change in some cases)

Axis of the bar remains straigth Cross-sections of the bar remain parallel to each other and

perpendicular to the axis


Transverse DeformationTransverse Deformation

Tensioned bar lengthens this is accompanied by the cross-section

area reduction

strain axial

strain transversePoisson’ ratio:

Compressed bar shortens this is accompanied by the cross-

section area increase

L L

D

D1

F F

Tension

FF

L

DD

1

L

Compression



2.3. Internal Forces Analysis2.3. Internal Forces Analysis


F

A

Bracket

Axis Bushing

Plate

Bar

From the Theory of Internal Forces:

Internal Force in a Solid BodyInternal Force in a Solid Body

• Resultant action of material particles’ interactions prevents the bar fracture, if the value of load:

• Structure of materials is regarded homogenous

Internal forces

F

Zoom

FInternal (Resisting) Force - resists the

deformations

= forces, acting between the parts of a loaded solid, that:

hold it together keep it from deforming freely

F/A U

U is material’ ultimate strength

strength of that

component must be analysed

Active Load

Reaction

External Force (F)

External Force

- deforms the component


F

Internal Force and the Method of SectionInternal Force and the Method of Section

How the combination of external loads influences the

material’ internal state?

How „violantly“ the particles are influenced inside the material?

METHOD OF SECTION = method in strength analysis for determination of internal forces’ combination and

calculation of respective values

SectionB

What internal forces are

present HERE?

External Force

Lower segment of section B

F

B

Requirement of

Equilibrium:

FNF B0

NBInternal Force


Right On

RInt, 0FF

Bar in Equilibrium

F1

F2

F3

F4

F5

Abstractly Sectioned Bar

Method of Section in GeneralMethod of Section in GeneralBasic Idea of the METHOD of SECTION:

Left On

LInt, 0FF

RInt,LInt, FF

Right On

RInt,

Left On

LInt,

FF

FF

The choice of a segment is free(use the one, which gives a simpler analysis)

a segment, that is sectioned from the bar in equlibrium, must remain in equilibrium

Assumption:

Conclusion: the influences and values of internal forces can be calculated using the equilibrium equations of that segment

0F

Section F3

F4

F5Right Segment (R)

FInt,LFInt,R

Secton’ SurfaceF1

F2

Left Segment (L)


External ForceF2

Superposition of ForcesSuperposition of Forces

The influence of the loads’ system

Changes of the internal forces of a component (and deformations) due to ADDED loads

F1

F2

External Force

DO NOT DEPEND on the loads that were applied before

N2

Internal Force

External ForceF1

Loaded Component

Internal Force

N1

= The sum of the influences of separate loads

N = N1 + N2

Internal Force


ASSUMPTION:The function of an internal force in between two concentrated loads is continuous

Axial Force Diagram — Concentrated Loads (1)Axial Force Diagram — Concentrated Loads (1)

A component is divided into parts with no change of loads

One section must be made in each such part

100 40 190 40

Free Body Diagram

F1 = 30 N F2 = 70 N F3 = 130 N F4 = 160 N F5 = 130 N

Which internal forces and their values act in this component???

Build the axial force diagram!

Section I Section II Section III Section IV


x

Left

F1 = 30 N

Section II

F2 = 70 N

F1 = 30 N

Section I

x

Left

Condition of Equilibrium

xI



xI = (0 … 100) mm

NI

N300 1I FNF

Does not depend on the value of xII

Does not depend on the value of xI(+)Tension

xII

NII

xII = (100 … 140) mm

N10070300 21II FFNF (+)

Tension



Right

F4 = 160 N

Section III

xF5 = 130 N

NIII

xIII

xIII = (140 … 330) mm


N301301600 54III FFNF

Does not depend on the value of xIII

(–)

Compression

x

F5 = 130 N

RightSection IVxIV

NIV

xIV = (330 … 370) mmCondition of Equilibrium

N1300 5IV FNF

Does not depend on the value of xIV

Tension(+)


100 40 190 40

Free Body Diagram

F1 = 30 N F2 = 70 N F3 = 130 N F4 = 160 N F5 = 130 N

Axial Force Diagram — Concentrated Loads (4)Axial Force Diagram — Concentrated Loads (4)Internal Force Diagram = Graph of an internal force along the bar’ axis

NB! Each concentrated load is represented by a step on the internal force diagram

30

100

30

130

N diagram, N

F1

F2

F3F4

F5

Rules for Internal Force Diagrams:

• the axes may not be shown

• the values are shown on specific locations along the diagram

• positive values are marked on the value axis positive side

• signs of the values may be shown on the diagram

• diagram may be painted or cross-hatched

• name and unit of a parameter are given in the title of diagram


Axial Force Diagram — Line Distributed Load (1)Axial Force Diagram — Line Distributed Load (1)

L=

100

0 m

m

Structural Model

Build the axial force diagram!

Weight of a vertical uniform bar is continuous axial line distributed load

gAp

= 7800 kg/m3 material’ density

g = 9.81 m/s2 gravity acceleration

A = 1 cm2 bar’ cross-section area

Section

x

Internal force changes due to gravity along the bar’ axis

Total weigth of the bar is: gALF G

Higher the position of a section:

• more material is located below the section• higher is the internal force due to the gravity of that material

Assuming the continuous px, there is a need for just one section

Which internal forces and their values act in this component due to its mass???


Structural Model

Axial Force Diagram — Line Distributed Load (2)Axial Force Diagram — Line Distributed Load (2)

Condition of equilibrium

xFNF xx 7.650 G

x

Section

Lower

FGx = Agx

Nx

If x =0, then Nx = 0

, then Nx = 7,65 N1 m

p = Ag = = const

AgL

7.65N diagram, N

xxgxAF x 7.659.81780010 4G

Tension

(+)Nx = Agx

Depends linearly on x

NB! Each constant line distributed load is

represented by an inclined straight line on the internal

force diagram


Segment of the Bar

xd

x

Internal Force Formula for a Bar with Line Distributed LoadInternal Force Formula for a Bar with Line Distributed Load

Bar with Axial Line Distributed Load

px = const

Internal force is an integral of line distributed load

x

y

xd

x

Sections

p = f(x) const

N(x + dx) = Nx + dN

N(x) = Nx


F = 0 (Nx + dN) – pxdx – Nx = 0

xx Ndx

dNp pdxN



2.4. Normal Stress Analysis2.4. Normal Stress Analysis


STRESS

Stress as a Distributed Internal ForceStress as a Distributed Internal Force

NORMAL STRESS

FExternal Force

F

SHEAR STRESS

External Force

Intensity of an internal force (acting on an internal surface)

orDimension

total of internal force for the unit of internal surface area

or density of internal force on the component’ internal surface

2m

NPa

Internal force and stress in direction of

a normal

Normal of internal surface

(axis of the bar)

Normal of internal surface

(axis of the bar)

Internal force and stress in transverse

direction to the normal


F

Cross-section

Normal Stress due to Axial LoadingNormal Stress due to Axial Loading

Hooke’s Law: E

The bar deforms due to laoding (lengthens)

Lx

L

x

Bernoulli’s Hypothesis:

All cross-sections of the bar remain flat

Cross-sections remain parallel and perpendicular to the axis

The axis

remains straigth

i points

const

x

xi L

L consti

The normal strains of all points on the cross-section area are equal

Axially Loaded Bar


Diagram of Cross-Section Normal Stress

Axial force is a resultant of stresses over an internal surface

Stress

Normal stress distribution

x

Distribution of Stress here is UniformDistribution of Stress here is Uniform

A

dAN

constA

N

Internal Force

= resultant of a cross-section stress

E

A

Axial Force-Resultant equation

N/ANormal stress distribution for an

axial load case is uniform

= const

= const

const

N

diagram


F

Section

F

Compression

A

F

Section

F

Tension

A

Sign Convention for a Normal StressSign Convention for a Normal Stress

Tensile stress is positive Compressive stress is negative

N (+)

)(A

N

Tension

N (-)

)(A

NCompression



2.5. Shear Stress Analysis2.5. Shear Stress Analysis


F

Internal Forces on the INCLINED SECTION of a tensioned bar

Stresses on Inclined SestionsStresses on Inclined Sestions

Section I

F

Section I

N

Internal Force on the CROSS-SECTION of a tensioned bar

Section II

N

A Cross-section area

A Inclined section area:

CROSS-SECTION normal stress

A

F

A

N

INCLINED SECTION normal stress

2cosA

F

A

N

INCLINED SECTION shear stress

sin22A

F

A

Q

cos

AA

Inclined section’axial force:

From equilibrium conditions:

N = Fcos Inclined section’shear force:

Q = Fsin

F

Section IIQ


Maximum Shear Stress in Axially Loaded BarMaximum Shear Stress in Axially Loaded Bar

Normal stress in axially loaded bar

2cosA

F has max value in the section, where = 0

This is a cross-section

Shear stress in axially loaded bar

sin22A

F has max value in the section, where = 45º

Maximum shear stress values of the axially loaded bar act on the

internal surfaces, that are inclined for 45º with respect of the cross-

section

A

FMaxThat’s why the strength anaysis must

be made in component’ cross-section

A

F

2Max

Maximum shear stress value of an axially loaded bar is 2x smaller than that of the maximum normal stress



2.6. Strength Calculations2.6. Strength Calculations

Priit Põdra

A

B C

D

E’

E

F/A’

Steels Stress-Strain Diagram

Yield Strength

Ultimate Strength

2. Strength of Components under Axial Loading 30

Malm

Surve

Tõmme

Value of Design (Permissible) StressValue of Design (Permissible) Stress

Design Stress

Ductile material

S

σσ Lim

In general:

TensionCompr σσ

= stress limit value, that was considered safe for a particular task

Material’ limit state stress

Design factor for the task

Steel

Brittle materialCast Iron

Ultimate Strengthfor compression

Ultimate Strengthfor tension

S

σσ U Strength Ultimate

SLim

Limit state stress

S

σσ Y strength Yield

SU Strength Ultimate

SY strength Yield

σ


Strength conditions based on SAFETY FACTORS

STRENGTH CONDITION in general:

For Tension

Strength Conditions for Axially Loaded BarStrength Conditions for Axially Loaded Bar

TensionTension Actual tensile

stress in CROSS-SECTION

the highest stress values in the material of a loaded bar should not exceed the

respective values of design stress

Design Stress for Tension

For Compression

ComprCompr

Actual compressive stress in CROSS-

SECTION

Design Stress for

Compression

or

the value of safety factor for strength must not be less than that of te respective

design factor

For Shear

Actual shear stress in 45º

INCLINED SECTION Design Stress

for Shear

Actual Factor of Safety Stress Actual

Stress StateLimit MaterialS

Design Factor

STension/Compr/Shear [S] Tension/Compr/Shear


Priorities of Strength Conditions for Axially Loaded BarPriorities of Strength Conditions for Axially Loaded Bar

General rule: ALL strength conditions must be fulfilled in ALL points of the loaded component

For structural materials

USUALLY :

Y = (0.56 … 0.6)Y

U = (0.5 … 1)U

Yield Strength for Shear

Tensile Yield Strength

Ultimate Stress for Shear

Tensile Ultimate Strength

Strength Condition for Shear

A

N

2Max

For different design factors USUALLY:

[S]Tension/Compr/Shear = const

0.5...12

0.56...0.62Max

Max

MaxMax

Strength Condition for Tension/Compr

1...2

1.12...1.2

Max

Max

A

NMax

Design stress for shear USUALLY:

[] = (0.56 … 0.6)[]

[] = (0.5 … 1)[]

for a ductile material

for a brittle material

For an axially loaded bar USUALLY:

The strength condition for tension/compression has HIGHER PRIORITY, tha that for shear


A

N

F

Tension

Strength CalculationsStrength Calculations

The strength condition must be fulfilled in all points of a component

N (+)

F

Section

Tension

F

Compression

(–)N

F

Section

Compression

TensionTension

ComprCompr

Allowable load

calculation

NA

AN

Check of adequate strength

Dimensioning


IPE-section

A

Rectangular section

b

A

h

Hollow circular section

D

A

d

D

A

Circular section

Dimensioning — ExamplesDimensioning — Examples

diagram diagram diagram diagram

4

2DA

N

D4

4

22 dDA

21

4

c

ND

h

bc

bhA

c

Nh

N

A

Suitable section shall be chosen from the product catalogue

according to A

D

dc


Strength conditions:

Strength Calculation — Concentrated Loads (1)Strength Calculation — Concentrated Loads (1)

100 40 190 40

Free Body Diagram

F1 = 30 N F2 = 70 N F3 = 130 N F4 = 160 N F5 = 130 N

30

100

30

130

N diagram, N

Calculate the diameter of this uniform bar!

Material:

D

A

Circular section

A

N

diagram

S235 EN 10 025

Design factor for compression: [S]Compr = 10

Yield Strength Y = 235 MPa

Highest tension

Highest compression

Design factor for tension: [S]Tension = 4

Critical part for tensin: NTõmme = 130 N

Critical part for compr.: NCompr = 30 N Tension/Compr []Tension/Compr

or STension/Compr [S]Tension/Compr

From previous



Dimensioning:

MPa2323.510

235

Compr

YCompr

S

• design stress for

COMPRESSION in this task:

• for adequate strength in COMPRESSION, Surve []Surve:

mm1.5m0.001281023

30446

Compr

Compr

N

D

MPa5858.74

235

Tension

YTension

S

• design stress for TENSION in this task:

• for adequate strength in TENSION, Tõmme []Tõmme :

mm2.0m0.001681058

130446

Tension

Tension

N

D

Str

eng

th C

alcu

lati

on

fo

r C

om

pre

ssio

nS

tren

gth

Cal

cula

tio

n

for

Ten

sio

n

• for adequate strength in all cases of an uniform bar: mm2.01.5;2.0max D



Check of adequate strength:

• max value of tensile stress in the component, if D = 2.0 mm, is:

MPa58σMPa42Pa1041.30.002

13044Tension

622

TensionTension

D

N

• i.e., the min safety factor value for tension is:

4S5.55.5942

235Tension

Tension

YTension

S

• max value of compressive stress in the component, if D = 2.0 mm, is:

MPa23σMPa9Pa109.50.002

3044Compr

622

ComprCompr

D

N

• i.e., the min safety factor value for compression is:

10S2626.19

235Compr

Compr

YCompr

S

Component strength for TENSION is adequate

Component strength for COMPRESSION is adequate

Solution: The diameter of that straigth uniform bar must be 2 mm


Strength Calculation — Line Distributed Load (1)Strength Calculation — Line Distributed Load (1)

Material: S355 EN 10 025Yield StrengthY = 355 MPa

Structural Model

7.65LN diagram

L

Design Factor: [S] = 3

Max axial force due to bar weigth — critical cross-section: NMax = 7.65L

Strength condition:

= 7800 kg/m3 material density

g = 9.81 m/s2 gravity acceleration

A = 1 cm2 bar’ cross-section area

p = Ag = 7.65 N/m = = const

Square Section

10

A

10

epüür

A

N Tension []Tension

or S [S]

From previous

How long could be the uniform bar of square cross-section?


Strength Calculation — Line Distributed Load (2)Strength Calculation — Line Distributed Load (2)

Bar Length Calculation:

MPa118118.33

355

Tension

YTension

S

• design stress for TENSION in this task :

• max TENSILE stress in the points of the bar’ holding cross-section:

LL

A

Ν76500

10

7.654-

MaxTension

• for the adequate strength, Tension []Tension :

61011876500 L m15001542.476500

10118 6

L

Check of Adequate Strength:

• max value of tensile stress in the component, when L = 1500 m, is:

MPa118MPa115Pa10114.715007650076500 Tension6

Tension L

• i.e., the min safety factor value for tension is:

3S3.03.08115

355

Tension

Y

S

The strength of a component is adequate

Solution: The length of that uniform bar could be up to 1500 m



THANK YOU!THANK YOU!

Questions, please?

Mehhanosüsteemide komponentide õppetool

Documents

STRENGTH OF MATERIALS - data.vk.edu.eedata.vk.edu.ee/RDER/RDER32/Tugevusopetus I/MHE0011 Strength of...STRENGTH OF MATERIALS 2.1. Structural Model. Priit Põdra 2. Strength of Components