Chapter1 Stress

Normal Stress 1

Normal Stress (1.1-1.2A)MAE 314 – Solid MechanicsYun Jing

2

Review of statics

This is a structure which was designed to support a 30kN load, it consists of a boom AB and of a rod BC. The boom and rod are connected by a pin at B and are supported by pins and brackets at A and C. (1) Is there a reaction moment at A? why? (2) What is the reaction force in the vertical direction at A? (3) What is the internal force in AB? (4) What is the internal force in BC?

Two-force member

Normal Stress 3

Normal Stress 4

Statics Review

Pins norxnmoment

Normal Stress 5

Statics Review

• Ay andCy cannotbedeterminedfromtheseequations.

kN30

0kN300

kN40

0

kN40

m8.0kN30m6.00

yy

yyy

xx

xxx

x

xC

CA

CAF

AC

CAF

A

AM

• SolveforreactionsatA&C:

Normal Stress 6

Statics Review

• Results: kN30kN40kN40 yx CCA

0

m8.00

y

yB

A

AM• Considerafree‐bodydiagramfortheboom:

kN30yC

substituteintothestructureequilibriumequation

Seesection1.1intextforcompletestaticanalysisandreviewofmethodofjoints.

Bx

Normal Stress 7

Normal Stress 8

Introduction to Normal Stress

Normal Stress 9

Introduction to Normal StressStress Forceperunitarea

AF

Normal Stress 10

Ifstressvariesoveracross‐section,theresultantoftheinternalforcesforanaxiallyloadedmemberisnormal toasectioncutperpendiculartotheaxis.Thus,wecanwritethestressatapointas

WeassumetheforceFisevenlydistributedoverthecross‐sectionofthebar.InrealityF resultantforceovertheendofthebar.


AP

AF

aveA

0

lim

A

ave dAdFAP

SignconventionTensile memberisintension

Compressive memberisincompression

Units force/areaEnglish: lb/in2 psi

kip/in2 ksi

SI: N/m2 Pa PascalkN/m2 kPaMPa,GPa,etc.

Normal Stress 11


00

Tensile

Compressive

Normal Stress 12

Homogenous:materialisthesamethroughoutthebar Cross‐section:sectionperpendiculartolongitudinalaxisofbar

Prismatic:cross‐sectiondoesnotchangealongaxisofbar

Definitions and Assumptions

F’ F

A

Prismatic Non-Prismatic

Normal Stress 13

Uniaxialbar: abarwithonlyoneaxis

NormalStress σ : stressactingperpendiculartothecross‐section.

Deformationofthebarisuniformthroughout. UniformStressState

Stressismeasuredfarfromthepointofapplication.

Loadsmustactthroughthecentroid ofthecross‐section.


Normal Stress 14

Theuniformstressstatedoesnotapplyneartheendsofthebar.

Assumethedistributionofnormalstressesinanaxiallyloadedmemberisuniform,exceptintheimmediatevicinityofthepointsofapplicationoftheloadsSaint‐Venant’sPrinciple .


“Uniform” Stress


Normal Stress 15

Introduction to Normal StressAuniformdistributionofstressisonlypossibleifthelineofactionoftheconcentratedloadsPandP’passesthroughthecentroidofthesectionconsidered.Thisisreferredtoascentric loading.

If a two-force member is eccentrically loaded, then the resultant of the stress distribution in a section must yield an axial force and a moment.

Normal Stress 16

Thestressdistributionsineccentricallyloadedmemberscannotbeuniformorsymmetric.

Normal Stress 17

Howdoweknowallloadsmustactthroughthecentroidofthecross‐section?

LetusrepresentP,theresultantforce,byauniformstressoverthecross‐section sothattheyarestaticallyequivalent .


Normal Stress 18

Momentsduetoσ:

SetMx Mx andMy My


Ay

Ax

dAxM

dAyM

AA

AA

xdAA

dAxP

x

ydAA

dAyP

y

11

11

Equationsforthecentroid

MM xx MM yy

Normal Stress 19

Example Problem Canthestructureweusedforourstaticsreviewsafelysupporta

30kNload? Assumetheentirestructureismadeofsteelwithamaximumallowablestressσall 165MPa.

Cross-section 30 mm x 50 mm

Normal Stress 20

Example Problem Twocylindricalrodsareweldedtogetherandloadedasshown.Find

thenormalstressatthemidsectionofeachrod.

mmd

mmd

30

50

2

1

Shearing and Bearing Stress 21

Shearing and Bearing Stress (1.2C-1.2E, 1.4)MAE 314 – Solid MechanicsYun Jing


What is Shearing Stress?

Welearnedaboutnormalstress σ ,whichactsperpendicular tothecross‐section.

Shearstress τ actstangential tothesurfaceofamaterialelement.

Normal stress results in a volume change.

Shear stress results in a shape change.


Where Do Shearing Stresses Occur? Shearingstressesarecommonlyfoundinbolts,pins,and

rivets.

Free Body Diagram of Bolt

Boltisin“single”shear

ForcePresultsinshearingstress

ForceFresultsinbearingstresswilldiscusslater


Wedonot assumeτ isuniformoverthecross‐section,becausethisisnotthecase.

τ istheaverageshearstress.

Themaximumvalueofτ maybeconsiderablygreaterthanτave,whichisimportantfordesignpurposes.

Shear Stress Defined

AF

AP

ave


Double Shear

Free Body Diagram of Bolt Free Body Diagram of Center of Bolt

AF

A

F

AP

ave 22

Boltisin“double”shear


Bearing Stress Bearing stress is a normal stress, not a shearing stress.

Thus,

where Ab = projected area where bearing pressure is appliedP = bearing force

tdP

AP

bb

Single shear case

Would like to determine the stresses in the members and connections of the structure shown.

Must consider maximum normal stresses in AB and BC, and the shearing stress and bearing stress at each pinned connection

From a statics analysis:FAB = 40 kN (compression) FBC = 50 kN (tension)

Example

Rod & Boom Normal Stresses

Pin Shearing Stresses


Pin Bearing Stresses

Rod & Boom Normal StressesThe rod is in tension with an axial force of 50 kN.

The boom is in compression with an axial force of 40 kN and average normal stress of –26.7 MPa.

The minimum area sections at the boom ends are unstressed since the boom is in compression.

MPa167m10300

1050

m10300mm25mm40mm20

26

3,

26

NAP

A

endBC

At the flattened rod ends, the smallest cross-sectional area occurs at the pin centerline,

At the rod center, the average normal stress in the circular cross-section (A = 314x10-6m2) is BC = +159 MPa.

Pin Shearing StressesThe cross-sectional area for pins at A, B,

and C,

262

2 m104912mm25

rA

MPa102m10491N1050

26

3,

A

PaveC

The force on the pin at C is equal to the force exerted by the rod BC,

The pin at A is in double shear with a total force equal to the force exerted by the boom AB,

MPa7.40m10491

kN2026,

A

PaveA

Divide the pin at B into sections to determine the section with the largest shear force,

(largest) kN25

kN15

G

E

P

P

MPa9.50m10491

kN2526,

A

PGaveB

Evaluate the corresponding average shearing stress,


Pin Bearing Stresses

To determine the bearing stress at A in the boom AB, we have t = 30 mm and d = 25 mm,

MPa3.53mm25mm30

kN40

tdP

b

To determine the bearing stress at A in the bracket, we have t = 2(25 mm) = 50 mm and d = 25 mm,

MPa0.32mm25mm50

kN40

tdP

b


Equilibrium of Shear Stresses Consideraninfinitesimalelementofmaterial.Applyasingleshear

stress,τ1.

Totalshearforceonsurfaceis τ1 bc.

Forequilibriuminthey‐direction,applyτ1 on ‐ surface.

Formomentequilibriumaboutthez‐axis,applyτ2 ontopandbottomsurfaces.

Momentequilibriumequationaboutz‐axis:

Thus,ashearstressmustbebalancedbythreeotherstressesfortheelementtobeinequilibrium.

1

2

2

1

bacabc )()( 21

21


Equilibrium of Shear Stresses

1

2

2

1

Face Direction Shear Stress

+ + +

+ - -

- - +

- + -

Whatdoesthistellus? Shearstressesonopposite parallel facesofanelementareequal in

magnitudeandopposite indirection. Shearstressonadjacent perpendicular facesofanelementareequal in

magnitudeandbothpointtowardsorawayfromeachother. Signconventionforshearstresses

Positiveface– normalisin x,y,orzdirection Negativeface‐ normalisin ‐ x,y,orzdirection

+

+-

-


σx stressinx‐direction appliedintheplanenormaltox‐axis σ y stressiny‐direction appliedintheplanenormaltoy‐axis σ z stressinz‐direction appliedintheplanenormaltoz‐axis τxy stressiny‐direction appliedintheplanenormaltox‐axis τ xz stressinz‐direction appliedintheplanenormaltox‐axis τ zy stressiny‐direction appliedintheplanenormaltoz‐axis Andsoon…

Define General State of Stressy

z x


Thereare9componentsofstress:σx,σ y,σ z,τxy,τ xz,τ yx,τ yz,τ zx,τ zy

Asshownpreviously,inordertomaintainequilibrium:τ xy τ yx,τ xz τ zx,τ yz τ zy

Thereareonly6independentstresscomponents.

Define General State of Stressy

z x


Example Problem AloadP 10kipsisappliedtoarodsupportedasshown

byaplatewitha0.6in.diameterhole.Determinetheshearstressintherodandtheplate.


Example Problem LinkABisusedtosupporttheendofahorizontalbeam.IflinkABis

subjecttoa10kipscompressiveforcedeterminethenormalandbearingstressinthelinkandtheshearstressineachofthepins.

ind

int

inb

1

41

2

Oblique Planes and Design Considerations 42

Oblique Planes and Design Considerations (1.3, 1.5)

MAE 314 – Solid MechanicsYun Jing


Stress on an Oblique Plane Whathavewelearnedsofar?

Axialforcesinatwo‐forcemembercausenormalstresses.

Transverseforcesexertedonboltsandpinscauseshearingstresses.


Stress on an Oblique Plane Axialforcescausebothnormalandshearingstressesonplaneswhich

arenotperpendiculartotheaxis.


Stress on an Oblique Plane Axialforcescausebothnormalandshearingstressesonplaneswhich

arenotperpendiculartotheaxis.

Consideraninclinedsectionofauniaxialbar.

TheresultantforceintheaxialdirectionmustequalPtosatisfyequilibrium.

Theforcecanberesolvedintocomponentsperpendiculartothesection,F,andparalleltothesection,V.

TheareaofthesectioniscosPF sinPV

cos/cos 00 AAAA


Stress on an Oblique Plane Wecanformulatetheaveragenormalstressonthesectionas

Theaverageshearstressonthesectionis

Thus,anormal force appliedtoabaronaninclinedsectionproducesacombinationofshear andnormalstresses.

2

00

coscos/

cosAP

AP

AF

cossincos/

sin

00 AP

AP

AV


Stress on an Oblique Plane Sinceσ andτ arefunctionsofsineandcosine,weknowthe

maximumandminimumvalueswilloccuratθ 00,450,and900.

At θ=±900 σ=0

At θ=±450 σ=P/2A0

At θ=00 σ=P/A0 (max)

At θ=±900 τ=0

At θ=±450 τ=P/2A0 (max)

At θ=00 τ=0

2

0

cosAP

cossin0A

P

Oblique Planes and Design Considerations

Stress on an Oblique Plane Whatdoesthismeaninreality?


Design Considerations Fromadesignperspective,itisimportanttoknowthe

largestloadwhichamaterialcanholdbeforefailing.

Thisloadiscalledtheultimateload,Pu.

Ultimatenormalstressisdenotedasσu andultimateshearstressisdenotedasτu.

APu

u

APu

u


Design Considerations Often the allowable load is considerably smaller than the

ultimate load.

It is a common design practice to use factor of safety.

all

u

PP

loadallowableloadultimate

SF ..

all

u

stressallowablestressultimate

SF

..all

u

stressallowablestressultimate

SF

..


Example Problem Twowoodenmembersaresplicedasshown.Ifthemaximum

allowabletensilestressinthespliceis75psi,determinethelargestloadthatcanbesafelysupportedandtheshearingstressinthesplice.


Example Problem Aloadissupportedbyasteelpininsertedintoahanging

woodenpiece.Giventheinformationbelow,determinetheloadPifanoverallfactorofsafetyof3.2isdesired.

mmd

mmc

mmb

MPa

MPa

tensioninMPa

steelu

woodu

woodu

12

55

40

145

5.7

)(60

_

_

_

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Chapter1 Stress