STRENGTH OF MATERIALSdata.vk.edu.ee/RDER/RDER32/Tugevusopetus I/MHE0011... · 2015-09-01 ·...

Priit Põdra 4. Strength of Components under Torsional Load 1

STRENGTH OF MATERIALSSTRENGTH OF MATERIALS

4. Strength of Components under Torsional Load

4. Strength of Components under Torsional Load

4.1 Structural Model

for Torsion

4.2 Action of

Twisting Loading

4.3Internal Forces

Analysis

4.4Stresses due to Twisting Action

4.5 Shear Stress

Analysis

Mehhanosüsteemide komponentide õppetool

4.6Strength

Calculations



4.1. Structural Model for Torsion

4.1. Structural Model for Torsion

Priit Põdra

M

Structural Model of the Shaft under Torsion

Lx

P

M

4. Strength of Components under Torsional Load 3

Structure and its Structural ModelStructure and its Structural Model

F1

f1

F2

f2

Structure to be Analysed

Friction in the bearings

Rotation imbalance

Body vibrations

Shaft bending and twisting

Loads variate

There are many different influencers and phenomenaSome of them are important — some of them are not

Most Important Load in this case = Twisting Moment



4.2. Action of Torsional Loading

4.2. Action of Torsional Loading


Torsional DeformationTorsional Deformation

Torsional Deformation = rotation of bar' (shaft) cross-sections about the bar axis

Angle of Twist , [rad] =angle of radius relative to its

initial position

Shear Strain (relative angle of twist) , [rad] = angle of generatrix relative to its initial

position

M

Circular Shaft under Torsional Load

Rectangular Shaft under Torsional Load

M


Bar’ Load State --- Pure TorsionBar’ Load State --- Pure Torsion

Pure Torsion is applicable only for circular shafts/bars

Bar’ Torsion = (simple) load state of a bar, where: Cross-sections rotate about the bar axis relative to each other;

Axis remains straigth and the bar’ lebgth does not change;

Cross-sections remain paralleel to each ohter and perpendicular to

the axis;

Cross-sections remain planar and theyr shape does not change.



4.3. Internal Forces Analysis4.3. Internal Forces Analysis


Internal Force of a Twisted BarInternal Force of a Twisted Bar

Internal force resists material deformation and rupture

F1

f1

F2

f2


Zoom

M

M

External Force = Twisting Moment(deforms the bar)

Internal Force = Resists the deformation – this is the counter-action to the twisting moment


Internal TorqueInternal Torque

External Force (Load) = Twisting Moment Cross-Section

Particles’ Interaction(Internal) Torque = Resultant

of Particles’ Interaction

Torque T, [Nm] = Resultant of particles interaction on a given (internal) section due to external twisting action

M

T


Sign Convention for TorqueSign Convention for Torque

M

M

Section

MM

Section

Shafts Under Twisting Action

MT (+)

Section

M

T (-)

Section

Clockwise = Positive

Looking from the side, that was removed

Counter-Clockwise = Negative

Looking from the side, that was removed


Diagram of Torque – Concentrated Loads (1)Diagram of Torque – Concentrated Loads (1)

How are the internal forces distributed in this shaft???Construct the diagramm of Torque distribution

x

xM1 = 2 kNm

M2 = 5 kNm

M3 = 1 kNm

M4 = 4 kNm

Section I

Section II Section III

NB! M = 0

Method of Sections is used in order to analyse internal

forces

The structure must be in equilibrium

Internal force distribution due to concentrated loads is

consistent between these loadsThree sections

must be analysed



M1 = 2 kNm

M2 = 5 kNmM3 = 1 kNm

M4 = 4 kNm

Free Body Diagram

Section I Section IILeft

M1 = 2 kNm

TI

Section I

Condition of Equilibrium (I)

kNm20 1I MM T

The value of torque TI

does not depend on the location of the section in between

M1 and M2

M1 = 2 kNm

M2 = 5 kNm

TII

Left Section II

Condition of Equilibrium (II)

kNm3250 12II MMM T

The value of torque TII does not depend on the location of the section in between M2 and M3

Priit Põdra

M1 = 2 kNm

M2 = 5 kNm

M3 = 1 kNm

M4 = 4 kNm

2

3

4



Diagram of Torque T, kNm

M1 = 2 kNm

M2 = 5 kNmM3 = 1 kNm

M4 = 4 kNm

Free Body DiagramSection III

Right Section III

x

M4 = 4 kNm

TIII

Condition of Equilibrium (III)

kNm40 4III MM T

The value of torque TIII

does not depend on

the location of the section

in between M3 and M4

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


Diagram of Torque – Line Distributed Load (1)Diagram of Torque – Line Distributed Load (1)

b =

3 m

m

Structural Model

How are the internal forces distributed in this component???Construct the diagramm of Torque distribution

The component is loaded by the uniform line distributed twisting moment mx = hp, [Nm/m]

Equivalent Structural Model

Section

Just one sections must be analysed



Condition of Equilibrium

Nm10001001,001,0101000 3 xxxLphxLT xmM

0 then m,1,0

)( Nm100 then ,0 If

xTL

phLxTx

Section

T

Right

M = mx(L - x)

T

Line distributed load can here be replaced

by an equivalent concentarted load

According to the Principle of Saint-Venant

The value of torque T depends linearly on the location of the section



100

Diagram of Torque T, Nm

Uniform Line Integrated Load

NB! Each uniform line distributed load load is represented by a ramp

on the internal force diagram mxL



4.4. Stresses due to Twisting Action

4.4. Stresses due to Twisting Action


Shear Stresses in GeneralShear Stresses in General

Shear Stress (aka Tangential Stress):

• acts along the section surface (perpendicular to the norma);

• shows the intensity of internal forces due to the in-plane relative displacements of material layers.

Torsion Cross-Sections are rotated about bar axis

Pure Shear Cross-Sections are linearly

shifted

Torsional Stress

(Pure) Shear Stress


Shear Stress Sign ConventionShear Stress Sign Convention

Shear Stress is considered positive, when it is directed positively on the positive section (or negatively on the negatiive section);

Shear Stress is considered negative, when it is directed negatively on the positive section (or positively on the negatiive section).

xy

Shear Stress

Section Normal

Stress Direction y

z

x

xy (-)

xy (+)

xz (+)xy (-)

Positive Section

y

z

x

xy (+)

xz (+)

xy (-)

Negative Section

xz (+)

Normal „flows“ into section

Normal „flows“ out of section


Shearing Sections of Pure TorsionShearing Sections of Pure Torsion

Shear Stresses on Cross-Sections

EFBC

C

BF

E

Shear Stresses on Cross-Sections

BC EF

C

BF

E

Element in Equilibrium

BC EF

BC

EC

Requirement of Equilibrium

BCEF is NOT in Equilibrium

BCEF is now in Equilibrium

Pure Torsion

C

B F

E

Element of Twisted Shaft

Other sections, EC and BF, must also

have shear stresses

EFBCECBF


Principle of Complementary Shear StressesPrinciple of Complementary Shear Stresses

A shear stress is always accompanied by a balancing shear stress across the section at right angles – they both have the same sign and the same magnitude

(they both are directed towards to or outwards from the common edge of these planes)

(This is valid for every loading case)

Shear stresses always act IN PAIRS:

For each shear stress there is a COMPLEMENTARY

shear stress

xy is accompanied by equivalentyx

yz is accompanied by equivalent zy

xzzx

zyyz

yxxy

y

z

xzy (+)

yz (+)yx (-)

xy (-)

xz (+)zx (+)

Principle of Complementary Shear Stresses

zx is accompanied by equivalent xz


Complementary Stresses under TorsionComplementary Stresses under Torsion

M

T

Cross-Section

Longitudinal Section

Torsional Shear Stress

Complementary Shear Stress

Complementary (to the torsional shear stress, acting on the cross-section) shear stress is acting on the

longitudinal sections of a twisted shaft (of circular cross-section)!!!


Stress State of a Twisted Circular BarStress State of a Twisted Circular Bar

Requirement of Equilibrium Sectioned stress element must also be in equilibrium!!!

C

BF

E

Pure Shear Element in Equilibrium

BF

Sectioned Element in Equilibrium

Known are: the sections of pure shear of a bar (shape BCEF, where normal stresses are absent)

Must be determined: the section (with an angle relative to BCEF), where maximum normal stresses are acting

(Find , where = max)

Pure Torsion


Stresses at the Inclined SectionStresses at the Inclined Section

0sincos 0

0cossin 0

2

1

NQQF

NQQF

y

x

Conditions of EquilibriumStresses on

Inclined Section

2sin

2cos

Stress state of a section of a twisted bar depends on, how that section is located

BF

Surface Areas of Sections

A0

A0cot

A0

sin

BF

Internal Forces at Sections

Q

Q1

Q2

N

x

y

sin/

sin/

cot

0

0

02

01

AQ

AN

AQ

AQ

All stress distributions are considered uniform


Normal and Tangential Stresses at Pure TorsionNormal and Tangential Stresses at Pure Torsion

Stresses at Inclined Sections

There is a section, inclined by 45 about the purely torsioned bar’ cross-section, where the shear is absent and the tension is maximum

2sin

2cosPure Torsion

= 0 = /6 = 30 = /4 = 45 = /3 = 60 = /2 = 90

= = 0,5 = 0 = 0,5 = = 0 = 0,87 = = 0,87 = 0


Bar’ failure Modes under TorsionBar’ failure Modes under Torsion

Pure Torsion

Failure at the Cross-Section (Steel)

Failure at the Longitudinal Sections (Wood)

Failure at the Inclined Section (Cast iron)

45

Shear strength is lower than

tensile strength

Shear strength along the fiber is lower than that across the fiber

Shear strength is higher than

tensile strength



4.5. Shear Stress Analysis4.5. Shear Stress Analysis


Possibility of Shear Stresses on the Edge of SectionPossibility of Shear Stresses on the Edge of Section

Tangential Component tComplementary

Stress’n NB! There could be no stresses at the

external surfaces????

NB! Remember the principle of Complementary Shear Stresses

Assumed Shear Stress at the external corner of Cross-Section

A

C

t1

t2

’t1 ’t2

A

C

t n

’n

Assumed Shear Stress at the edge of Cross-Section

can be divided

Normal Component n Tangential Component t1

Tangential Component t2Complementary

Stress’t1Complementary

Stress’t2

Both appears at the external

surface of a barAppears at the

external surface of a bar

Priit Põdra

A

C

t

Section under torsional/shearing action

External CornerTangent to the Edge


Special Features of Shear Stress DistributionSpecial Features of Shear Stress Distribution

There could be no stresses at the bar’ external surfaces

Shear stress at the edge of a section can only be directed along

the edge tangent at that point

There could not be any shear stress acting at the

external corner point of the edge (there always = 0)

Assuming, that:The shape of the bar stays intact

The shape of the cross-section stays intact


Torsional Stresses of a Circular BarTorsional Stresses of a Circular Bar

Bar deforms/twists due to the loading

Equal elements of an uniform bar deform equally

(The bar deforms uniformly)

The shape and volume of elements do not change

Shear strain values (and deformations) are

proportional to the values of respective radii

Hooke’s Law for Shear:

G ~

T

Element of Twisted Bar

~

Priit Põdra

Circle

max

T

max

T

T

Ring

Torsional Stress Distribution for Circular Sections

Torque-Resultant Equation

AA

xzxy dAdAyzT

Txy

Section of Twisted Bar

z

y

M

xz

A


Torsional Stress Distribution of a Circular SectionTorsional Stress Distribution of a Circular Section

~Torsional stress value at each point is proportional to its radius

KConstant

A

dAKT 2

0

42

32I

DdA

A

0I

TK

Torsional Stress Distribution of a Circular Section

0I

T

Section Area Polar Moment of Inertia


Maximum Torsional Stress of a Circular SectionMaximum Torsional Stress of a Circular Section

This function is lineaar with repect

Maxumum torsional stressmax is acting at the edge of the circle, where = D/2

Maximum Torsional Stress of a Circular Section:

0max

0max W

T

I

T

D

IIW 0

max

00

2

Circle’ Polar Moment of Resistance

3max

16

D

T

Torsional Stress Distribution of a Circular Section

0I

T

Diagram of is a straigth line, that crosses the bar’ axis

C

D

16

323

0

4

0

DW

DI

max

T

max

Diagram of


Torsional Stresses of Non-Circular BarsTorsional Stresses of Non-Circular Bars

3Bmax

20

b

T

Equilateral TriangleEllipseRectangle

BbD

2h

Bmax

KhbK

T

NB! Bernoulli’s hypothesis is not valid – plane sections will not remain plain

B

D

b

h

T

B

D

B

D

b

a

T

B

D

2D

2Bmax

16

16

ba

Tab

T

b

TB

B

BB

B

h/b 1 1,2 1,5 2 3 5 10 Kh 0,208 0,219 0,231 0,246 0,267 0,291 0,312 0,333

Kb 1 0,93 0,86 0,79 0,75 0,74 0,74 0,74



4.6. Strength Calculations4.6. Strength Calculations

Priit Põdra

For Design Materials in General

UU

YY

1...5,0

6,0...56,0


Design Stress for TorsionDesign Stress for TorsionDesign Stress for Torsion = shear stress limit value, that was

considered safe for a particular task of torsion

Slim

materials brittlefor

strength ultimate

materials ductilefor strength yield

U

Y

S

S

Shear Stress-Strain Diagram for Steel

Yield Strength

Ultimate Strength

U

Y

Fracture

For Steels in General

UU

YY

7,0...6,0

6,0...56,0

Design Factor for that task


Strength Analysis for Circular BarStrength Analysis for Circular Bar

Strength condition should be applied at the section of highest torsion, i.e.

where T = max

0

max W

T

Strength Condition for Torsion

M


MT

Section

Strength condition must be fulfilled in ALL points of the loaded component!!!

T

W

WT

0

0

Check of adequate strength

Allowed load calculation

Dimensioning


Dimensioning of Circular BarDimensioning of Circular Bar

max

T max

Diagram of D

16

3

0

DW

316

T

D

Circular Cross-Section

max

T max

Diagram of D

Ring Cross-Section

d

43

0 116

cD

W

341

16

c

TD

D

dc

All cross-section points have

torsional stress

The most critical points of a circular section are located on its edge

All points of the edge have equal criticality


Strength Analysis for Rectangular BarStrength Analysis for Rectangular Bar

Strength condition should be applied at the section of highest torsion, i.e. where T = max

Strength Condition for Torsion

Strength condition must be fulfilled in ALL points of the loaded component!!!

Check of adequate strength

Allowed load calculation

Dimensioning

2

hmax hbK

T

2hhbKT

Rectangular Shaft under Torsional Load

M

SectionT

M

b

h


Dimensioning of Rectangular BarDimensioning of Rectangular Bar

Rectangular Cross-Section

b

h

T

max

Diagrams of

b

hc

3

h

2

K

Tch

Square Cross-Sectiona

T

max

Diagrams of max

3

208.0 T

a The most critical points of a rectangular section are located on its edge, in the

middle of the longest sides

Symmetrical points of the edge have equal criticality


Dimensioning – Concentrated Loads (1)Dimensioning – Concentrated Loads (1)

M1 = 2 kNm

M2 = 5 kNm

M3 = 1 kNm

M4 = 4 kNm

2

3

4

Diagram of T, [kNm]

Shaft Cross-Section and Torsional Stress Distribution

From previous

Material:

Steel, [] = 60 MPa

Strength Condition

Calculate the required diameter of the twisted uniform circular shaft!

B C E G


Dimensioning – Concentrated Loads (2)Dimensioning – Concentrated Loads (2)

Dimensioning:

• highest value of torque is acting in between sections E and G of the shaft: TEG = 4 kNm;

• higest value of torsional stress max is acting at all points of the edges of every sections in that intervall;

• in order to fulfill the condition of torsional strength (max ≤ []), the value of shaft diameeter must obey:

mm70m0698,01060

10416163

6

3

3

TD

Check of Adequate Strength:

• maximum torsional stress value in the shaft, when D = 70 mm:

MPa60MPa4,5907,0

10416163

3

3max

D

T Strength condition is fulfilled

The diameeter of that shaft must be at least: D = 70 mmAnswer:


Check of Adequate Strength – Line Distributed Load (1)Check of Adequate Strength – Line Distributed Load (1)

100

Uniform Line Integrated Twisting Load

Check the strength of the twisted uniform rectangular bar!

Material:

Steel, [] = 90 MPa

Strength Condition

From previous

Bar Cross-Section and Torsional Stress Distribution

B

C



Check of adequate strength:

• highest value of torque is acting at the section B (where the bar is attached to the support) and G of the shaft: Tmax = TB = 100 Nm;

• higest value of torsional stress max is acting at the both middle points of the longest edges of that section;

• condition of strength in this case is:

• value of the coefficient Kh must be determined based on the ratio of h and b:

2

hmax hbK

T

3,333,33

10

b

hc



• for the required value of Kh, linear interpolation is used:

if h/b = 3,3, then:

• fulfilment of the condition of torsional strength (max ≤ []) is then checked:

312,0 then ,5/ if

267,0 then ,3/ if

h

h

Kbh

Kbh

274,0267,0312,035

33,3267,0

hK

MPa90MPa2,12003,001,0274,0

1002

hmax

hbK

T

Strength condition is fulfilled

The strength of that bar is adequateAnswer:



THANK YOU!THANK YOU!

Questions, please?

Mehhanosüsteemide komponentide õppetool

STRENGTH OF MATERIALSdata.vk.edu.ee/RDER/RDER32/Tugevusopetus I/MHE0011... · 2015-09-01 ·...

Documents

Transcript of STRENGTH OF MATERIALSdata.vk.edu.ee/RDER/RDER32/Tugevusopetus I/MHE0011... · 2015-09-01 ·...