strength of material (shrinked)

8/11/2019 Strength of Material (Shrinked)

1/102

Strength of Materials


2/102


3/102


4/102


5/102


6/102


7/102


8/102


9/102


10/102


11/102

Modulus TypesModulus: Slope of the stress-strain curve

Initial Modulus: slope of the curve drawn at the origin. Tangent Modulus: slope of the curve drawn at the tangent of the curve at some point.

Secant Modulus: Ratio of stress to strain at any point on curve in a stress-straindiagram. It is the slope of a line from the origin to any point on a stress-straincurve.

Stress

Strain

Initial Modulus

Tangent Modulus

Secant Modulus


12/102


13/102

Various regions and points on the stress-strain curve.

Strain ( ) (e/Lo)

41

2

3

5

ElasticRegion

PlasticRegion

StrainHardening Fracture

ultimatetensilestrength

Elastic regionslope=Youngs(elastic) modulus yield strength

Plastic regionultimate tensile strengthstrain hardeningfracture

necking

yieldstrength


14/102

1.True elastic limit based on micro strain measurements at strains onorder of 2 x 10 -6 in | in. This elastic limit is a very low value and isrelated to the motion of a few hundred dislocations.

2.Proportional limit is the highest stress at which stress is directlyproportional to strain.

3.Elastic limit is the greatest stress the material can withstand withoutany measurable permanent strain remaining on the complete release ofload. With increasing sensitivity of strain measurement, the value of theelastic limit is decreased until at the limit it equals the true elastic limitdetermined from micro strain measurements. With the sensitivity ofstrain usually employed in engineering studies (10 -4 in | in), the elasticlimit is greater than the proportional limit.

1.The yield strength is the stress required to produce a small-specifiedamount of plastic deformation.The usual definition of this property is the offset yield strength determined by the stress corresponding to the intersection of the stress-strain curve and a line parallel to the elastic part of the curve offset by a

specified strain (Fig. 1).


15/102


16/102


17/102

Stainless Steel E= 28.5 million psi (196.5 GPa) Aluminum E= 10 million psi Brass E= 16 million psi Copper E= 16 million psi Molybdenum E= 50 million psi

Nickel E= 30 million psi Titanium E= 15.5 million psi Tungsten E= 59 million psi Carbon fiber E= 40 million psi Glass E= 10.4 million psi Composites E= 1 to 3 million psi Plastics E= 0.2 to 0.7 million psi

Stress- strain diagrams for various materials


18/102

Universal Testing Machine


19/102

Area in red indicates intensity of stress


20/102

Equipment to measure Stress-Strain Strainometers: measures dimensional changes that

occur during testing extensometers, deflectometers, and compressometers measurechanges in linear dimensions.

load cells measure load data is recorded at several readings and the results averaged,

e.g., 10 samples per second during the test.


21/102

M ater ial Proper ties

There a 5 properties typically used to describe a materials behavior and capabilities:

1. Strength2. Hardness3. Ductility4. Brittleness5. Toughness


22/102

The ability to resist deformation and maintainits shape

1. Strength

-Given in terms of the yield strength, s y, or theultimate tensile strength, s ult


23/102

The ability to resist indentation, abrasion, and wear

2. Hardness

STRENGTH and HARDNESS are related! A high-strength material is typicallyresistant to wear and abrasion...

- For metals, this is determined with the RockwellHardness or Brinell tests that measure indentation/

penetration under a load


24/102

Material Brinell Hardness

Pure Aluminum 15

Pure Copper 35

Mild Steel 120

304 Stainless Steel 250

Hardened Tool Steel 650/700

Hard Chromium Plate 1000

Chromium Carbide 1200

Tungsten Carbide 1400

Titanium Carbide 2400

Diamond 8000

Sand 1000

A comparison of hardness of some typical materials:


25/102

The ability to deform before ultimate failure

3. Ductility

Ductile materials can be pulled or drawn into pipes, wire,and other structural shapes

Ductile materials include copper, aluminum,and brass


26/102

The inability to deform before ultimate failure

4. Brittleness

Brittleness is the LACK of ductility ...

- The opposite of ductility, brittle materials deform little before ultimately fracturing

- Brittle materials include glass and cast iron


27/102

The ability to absorb energy

5. Toughness

Toughness and Ductility/brittleness are related!

Brittle things. ...are not tough!

- Material Toughness (slow absorption)- not a readily observable property- Defined by the area under the stress-strain curve

- Impact Toughness (rapid absorption)- Ability to absorb energy of an impactwithout fracturing


28/102

StiffnessStiffness is a measure of the materials ability to resist deformation under load asmeasured in stress.

Stiffness is measures as the slope of the stress-strain curve

Hookean solid: (like a spring) linear slope

steel

aluminumiron

copper

Stiffness is usually measured by the Modulus of Elasticity(Stress/strain)

Steel is stiff (tough to bend).


29/102

Analysis and Designof Beams for Bending


30/102

Introduction

Beams - structural members supporting loads atvarious points along the member

Objective - Analysis and design of beams

Transverse loadings of beams are classified asconcentrated loads or distributed loads

Applied loads result in internal forces consistingof a shear force (from the shear stressdistribution) and a bending couple (from thenormal stress distribution)

Normal stress is often the critical design criteria

Requires determination of the location andmagnitude of largest bending moment


31/102

Classification of Beam Supports


32/102


33/102


34/102

Shear and Bending Moment Diagrams

Determination of maximum normal and

shearing stresses requires identification ofmaximum internal shear force and bendingcouple.

Shear force and bending couple at a point aredetermined by passing a section through the

beam and applying an equilibrium analysis onthe beam portions on either side of thesection.

Sign conventions for shear forces V and V

and bending couples M and M


35/102

Quantifying Bending Stress

Compression

Tension

Sagging condition

Neutral Axis

y

A

B

A

B

Bending Stress :M : Bending MomentI : 2 nd Moment of area of the cross sectiony : Vertical distance from the neutral axis

: tensile ( +) or compressive( -) stress

y

L i di l B di S


36/102

Quantifying Bending Stress

Hogging condition y

Compression

Tension

Neutral Axi s

AB

A

B

Neutral Axis : geometric centroid of the cross section ortransition between compression and tension

Logitudinal Bending Stress


37/102

Torsion


38/102

ContentsTorsional Loads on Circular

Shafts

Net Torque Due to InternalStresses

Axial Shear Components Shaft Deformations

Shearing Strain

Stresses in Elastic Range Normal Stresses

Torsional Failure Modes

Angle of Twist in Elastic Range

Design of Transmission Shafts

Stress Concentrations

Torsion of Noncircular Members


39/102

Torsional Loads on Circular Shafts Stresses and strains of circular

shafts subjected to twisting couplesor torques

Generator creates an equal andopposite torque T

Shaft transmits the torque to thegenerator

Turbine exerts torque T on the shaft


40/102

Net Torque Due to Internal Stresses Net of the internal shearing stresses is an

internal torque, equal and opposite to theapplied torque,

Although the net torque due to the shearingstresses is known, the distribution of the stressesis not

Unlike the normal stress due to axial loads, thedistribution of shearing stresses due to torsionalloads can not be assumed uniform.

Distribution of shearing stresses is staticallyindeterminate must consider shaft

deformations


41/102

Axial Shear Components Torque applied to shaft produces shearing

stresses on the faces perpendicular to theaxis.

The existence of the axial shear components isdemonstrated by considering a shaft made upof axial slats.

The slats slide with respect to each other whenequal and opposite torques are applied to theends of the shaft.

Conditions of equilibrium require theexistence of equal stresses on the faces of the

two planes containing the axis of the shaft


42/102

From observation, the angle of twist of theshaft is proportional to the applied torque andto the shaft length.

Shaft Deformations

When subjected to torsion, every cross-sectionof a circular shaft remains plane andundistorted.

Cross-sections of noncircular (non-axisymmetric) shafts are distorted whensubjected to torsion.

Cross-sections for hollow and solid circularshafts remain plain and undistorted because acircular shaft is axisymmetric.


43/102

Shearing Strain Consider an interior section of the shaft. As a

torsional load is applied, an element on theinterior cylinder deforms into a rhombus.

Shear strain is proportional to twist and radius

It follows that

Since the ends of the element remain planar,the shear strain is equal to angle of twist.


44/102

Stresses in Elastic Range

Recall that the sum of the moments fromthe internal stress distribution is equal tothe torque on the shaft at the section,

The results are known as the elastic torsionformulas,

Multiplying the previous equation by theshear modulus,

From Hookes Law, , so

The shearing stress varies linearly with theradial position in the section.


45/102

Torsional Failure Modes

Ductile materials generally fail inshear. Brittle materials are weaker intension than shear.

When subjected to torsion, a ductilespecimen breaks along a plane ofmaximum shear, i.e., a plane

perpendicular to the shaft axis.

When subjected to torsion, a brittle

specimen breaks along planes perpendicular to the direction inwhich tension is a maximum, i.e.,along surfaces at 45 o to the shaftaxis.


46/102


47/102

Design of Transmission Shafts Principal transmission shaft

performance specifications are:- power- speed

Determine torque applied to shaft at

specified power and speed,

Find shaft cross-section which will notexceed the maximum allowableshearing stress,

Designer must select shaft

material and cross-section tomeet performance specificationswithout exceeding allowableshearing stress.


48/102

Stress Concentrations The derivation of the torsion formula,

assumed a circular shaft with uniformcross-section loaded through rigid end

plates.

Experimental or numerically determined

concentration factors are applied as

The use of flange couplings, gears and pulleys attached to shafts by keys inkeyways, and cross-section discontinuitiescan cause stress concentrations


49/102

Torsion of Noncircular Members

At large values of a/b , the maximumshear stress and angle of twist for otheropen sections are the same as arectangular bar.

For uniform rectangular cross-sections,

Previous torsion formulas are valid foraxisymmetric or circular shafts

Planar cross-sections of noncircularshafts do not remain planar and stressand strain distribution do not varylinearly


50/102

Plastics


51/102


52/102

Why Plastics?


53/102

Why Plastics?


54/102


55/102

Source: Automotive Plastics Report2000

Plastics Applications

2000 2010

Segment M Lb. M Lb. M Lb.

Interior 1,688 2,021 +333

Body 1,181 1,601 +420

Underhood 388 627 +239Chassis 961 1,195 +234

Total 4,217 5,444 1,226


56/102

Specific Gravity

Steel 7.8

Aluminum 2.6

Magnesium 1.75

Plastics 0.9 - 1.6


57/102

Automotive Plastics Basicsfor

Exteriors


58/102


59/102

Selection Considerations

Physical Properties

Chemistry

Process Characteristics

Relative Part or System Cost


60/102

Most Common of the Approx. 60Commercial Families of Plastic Matls

AcrylonitrileButadiene Styrene(ABS)

Acetal (POM) Acrylics (PMMA) Fluoropolymer

(PTFE)

Ionomer Nylon (PA)

Phenolic Polycarbonate (PC)


61/102

Most Common of the Approx. 60Commercial Families of Plastic Matls

Polyester (PBT, PET) Polyester Thermoset

(SMC, BMC) Polyethylene (PE) Polyphenyleneoxide

(PPO) Polypropylene (PP)

Polystyrene (PS) Polyurethane (PUR)

Polyvinylchloride(PVC)

Styrene Acrylonitrile(SAN)

Vinyl Ester


62/102


63/102

Source: Automotive Plastics Report2000

Plastics Applications

2000 2010

Segment M Lb. M Lb. M Lb.

Interior 1,688 2,021 +333

Body 1,181 1,601 +420

Underhood 388 627 +239Chassis 961 1,195 +234

Total 4,217 5,444 1,226


64/102

Primary Processing Methods

Blow Molding Calendaring

Casting Compression Molding Extrusion

Reaction InjectionMolding Injection Molding

Powder or SlushMolding

Thermoforming Filament Winding Pultrusion Resin Transfer

Molding Rotational Molding


65/102

Fatigue (Failure under fluctuating / cyclic stresses)


66/102

Fatigue occurs when a material is subjected to alternatingstresses, over a long period of time.

Under fluctuating / cyclic stresses, failure can occur at loadsconsiderably lower than tensile or yield strengths of material under astatic load.

Estimated to causes 90% of all failures of metallic structures(bridges, aircraft, machine components, etc.)

Fatigue failure is brittle-like (relatively little plastic deformation) -even in normally ductile materials. Thus sudden and catastrophic!

Examples: springs, turbine blades, airplane wings, bridges and bones


67/102

Fatigue does not always lead to failure

Failure can occur if the stress surpasses theendurance limit of the material[Endurance Limit (Sn): Is the stress value below which an infinite number

of cycles will not cause failure]

Steel will not fail if the endurance limit is not passed

Aluminum will eventually fail regardless of theendurance limit


68/102


69/102

Cycl ic Stresses There are three common ways in which stresses may be applied:

axial (tension or compression), torsional (twisting),or flextural (bending)Examples of these are seen in Fig. 1.

Figure 1 Visual examples of axial stress, torsional stress, and flexural stress.
http://www.sv.vt.edu/classes/MSE2094_NoteBook/97ClassProj/anal/kelly/fatigue.htmlhttp://www.sv.vt.edu/classes/MSE2094_NoteBook/97ClassProj/anal/kelly/fatigue.html


70/102


71/102


72/102


73/102


74/102

TheS-NCurve :


75/102

The S N Curve :A very useful way to visualize time to failure for a specific material is with the S-Ncurve.The " S-N " means str ess v/s cycles to f ailur e , which when plotted uses the stress

amplitude,s

a plotted on the vertical axis and the logarithm of the number of cycles tofailure.An important characteristic to this plot as seen in Fig. 2 is the fatigue limit

Figure 2 A S-N Plot for an aluminum alloy
http://www.sv.vt.edu/classes/MSE2094_NoteBook/97ClassProj/glossary.htmlhttp://www.sv.vt.edu/classes/MSE2094_NoteBook/97ClassProj/anal/kelly/fatigue.htmlhttp://www.sv.vt.edu/classes/MSE2094_NoteBook/97ClassProj/anal/kelly/fatigue.htmlhttp://www.sv.vt.edu/classes/MSE2094_NoteBook/97ClassProj/glossary.html


76/102

Other important terms are - fatigue strength and fatigue life .

The stress at which failure occurs for a given number of cyclesis the fatigue str ength .

The number of cycles required for a material to fail at a certainstress in fatigue li fe .

Signif icance of the fatigue l imit : If the material is loaded below this stress, then it will not fail,

regardless of the number of times it is loaded.

Material such as aluminum, copper and magnesium do notshow a fatigue limit, therefore they will fail at any stress and numberof cycles.
http://www.sv.vt.edu/classes/MSE2094_NoteBook/97ClassProj/glossary.htmlhttp://www.sv.vt.edu/classes/MSE2094_NoteBook/97ClassProj/glossary.htmlhttp://www.sv.vt.edu/classes/MSE2094_NoteBook/97ClassProj/glossary.htmlhttp://www.sv.vt.edu/classes/MSE2094_NoteBook/97ClassProj/glossary.htmlhttp://www.avweb.com/articles/metfatig/figure7.gif


77/102

S-N Cur ve for F er rous v/s non-fer rous metals
http://www.avweb.com/articles/metfatig/figure7.gif


78/102


79/102


80/102


81/102


82/102


83/102


84/102


85/102

f


86/102

Stages of creep


87/102

1. Instantaneous deformation :mainly elastic.

2. Primary/transient creep :Slope of strain vs. time decreases with time: work-hardening

3. Secondary/steady-state creep :Rate of straining is constant: balance of work-hardening and recovery .

4. Tertiary:

Rapidly accelerating strain rate up to failure:formation of internal cracks, voids, grain boundary separation, necking, etc.

Parameters of creep behavior


88/102

The stage secondary/steady-state creep is of longest duration and the steady-statecreep rate is the most important parameter of the creep behavior in long-life applications.Another parameter, especially important in short-life creep situations, is time to rupture, or the

rupture lifetime, tr. t / s . e . = e &

d ff


89/102

Creep: stress and temperature effects With increasing stress or temperature:

The instantaneous strain increasesThe steady-state creep rate increasesThe time to rupture decreases


90/102

All f hi h


91/102

Alloys for high-temperature use (turbines in jet engines, hypersonic airplanes, nuclear reactors, etc.)

Creep is generally minimized in materials with: High melting temperature High elastic modulus Large grain sizes (inhibits grain boundary sliding)

Following materials are especially resilient to creep: Stainless steels

Refractory metals (containing elements of high melting point,like Nb, Mo, W, Ta) Superalloys (Co, Ni based: solid solution hardening and

secondary phases)

5.6 Design Philosophy


92/102

Loading known and geometry specified Specifyfactor of safety, N, and determine material.

Loading known and material specified Specifyfactor of safety, N, and determine requiredgeometry.

Loading known and material and geometryspecified Determine factor of safety Is itsafe??

Design

AnalysisAlso check de f lec t ion ! !


93/102

5.8 Failure Theories


94/102

Uniaxial: Bi-axial:or

1. Maximum Normal Stress

2. Modified Mohr

3. Yield strength

4. Maximum shear stress

5. Distortion energy

6. Goodman

7. Gerber

8. Soderberg

Ductile or Brittle

Dynamic or Static

FatigueLoading

Theory to use depends on:

StaticLoading

FailureTheory:

When to Use? Failure When: Design Stress:


95/102

1. MaximumNormal Stress

Brittle Material/ UniaxialStatic Stress

2. Yield Strength

(Basis for MCH T213)

Ductile Material/

Uniaxial Static NormalStress

3. Maximum ShearStress (Basis forMCH T 213)

Ductile Material/ Bi-axial Static Stress

4. Distortion Energy(von Mises)


5. GoodmanMethod

Ductile Material/Fluctuating NormalStress (Fatigue Loading)

Ductile Material/Fluctuating Shear Stress(Fatigue Loading)

Ductile Material/Fluctuating CombinedStress (Fatigue Loading)

Failure Theories for STATIC Loading

Uniaxial: Bi-axial:or


96/102

FailureTheory:

When Use? Failure When: Design Stress:


97/102

1. MaximumNormal Stress

Brittle Material/ UniaxialStatic Stress

2. Yield Strength

(Basis for MCH T213)

Ductile Material/

Uniaxial Static NormalStress

3. Maximum ShearStress (Basis forMCH T 213)


4. Distortion Energy(von Mises)


5. GoodmanMethod

a. Ductile Material/Fluctuating NormalStress (Fatigue Loading)

b. Ductile Material/Fluctuating Shear Stress(Fatigue Loading)

c. Ductile Material/Fluctuating CombinedStress (Fatigue Loading)

Comparison of Static Failure Theories:

Shows no failure zones


98/102

Maximum Shear most conservative

Th G d Di diff b f il


99/102

The Goodman Diagram - note difference between no failurezone and safe zone

s m

s a

General Comments:


100/102

1. Failure theory to use depends on material (ductile vs. brittle) and type of loading (static ordynamic). Note, ductile if elongation > 5%.

2. Ductile material static loads ok to neglect Kt (stress concentrations)

3. Brittle material static loads must use Kt

4. Terminology:

Su (or Sut) = ultimate strength in tension

Suc = ultimate strength in compression

Sy = yield strength in tension

Sys = 0.5*Sy = yield strength in shear

Sus = 0.75*Su = ultimate strength in shear

Sn = endurance strength = 0.5*Su or get from Fig 5-8 or S-N curve

S n = estimated actual endurance strength = Sn(C m) (C st ) (C R) (C s)

Ssn = 0.577* Sn = estimated actual endurance strength in shear

5.9 What Failure Theory to Use:


101/102


102/102

THANK YOU

strength of material (shrinked)

Documents