EFFECT OF MATERIAL PROPERTIES ON DESIGN

Dr. Zuhailawati HussainEBB 224 Rekabentuk bahan

Kejuruteraan

1. Factors affecting the behaviour of materials in components

Design should take into account function, material properties & the manufacturing processes. , Fig 6.1

Secondary relationships: material properties-manufacturing processes, function-manufacturing processes, and function-material properties

Behavior of component material depends on influence of properties of stock material, geometry & external forces, & effect of fabrication method, Fig. 6.2.

Fig 6.2 also shows secondary relationship.

COMPONENT DESIGN

FUNCTION & CONSUMER REQUIREMENTS

MATERIAL PROPERTIES

MANUFACTURINGPROCESSES

Fig 6.1 Factors should be considered in component design

BEHAVIOR OF MATERIAL IN THE

COMPONENT

PROPERTIES OFSTOCK

MATERIAL

EFFECT OFFABRICATION

METHOD

GEOMETRY &EXTERNAL

FORCES

Fig 6.2 Factors should be considered in anticipating the behavior of material in the component

2. Statistical variation of material properties

In practice, material properties are seldom homogenous or isotropic, as they are sensitive to variations in parameter such as :

Composition, heat treatment & processing condition.

Others that influence material behavior

Surface roughness, internal stress, sharp corners and stress raiser.

Variation in material properties can be solved by choosing factor of safety. Factor of safety accommodates unknown influences that affect the life of component under service condition.

Material properties can be statistically described by mean value (x), standard deviation () and coefficient of variation ().

x1, x2, x3 = values of different measurements of the same material property, n= number of measurement

n

iin eqx

nXxxxxvalueMean

1321 1.

1.....,

2.1 Statistical Parameter

v = /x (eq. 3)

Coefficient variation (dimensionless quantity) is useful in assessing the relative variability from different source.

Smaller values of standard deviation & coefficient of variation indicate more homogenous material

2.)(1

1,

21

1

2 eqxxn

deviationStdn

ii

2.2 Normal distribution of properties

Frequently, all material properties are normally distributed.

From the normal distribution, confidence limits can be determined

The probabilities of obtaining values at plus or minus 1,2 or 3 standard deviation from the mean value are 68.26%, 95.46% & 99.73% respectively, Fig. C.6

Probability that a value would occur outside the range represented by plus or minus two std deviation from the mean is 4.54%, or the confidence level that the value will be found within the limits is 95.46%.

Assuming an experimental data ( no. of sample > 100) is following a normal distribution, the standard deviation is approximated as

= (max value of property – min value)/6 (eq.4)

Example 1

If the range of strength of an alloy is given as 800 – 1200 MPa, and the mean value can be taken as 1000 MPa

Solution

The mean strength can be taken as 1000 MPa

The standard deviation can be estimate as

= ( 1200 – 800 )/6 = 66.67 MPa

The coefficient of variation v is then

v = 66.67/1000 = 0.0667

•If results are obtained from a sample of

•25 test, eq. 4 is divided by 4

•5 test, eq. 4 is divided by 2

3. Effect of component geometry

In almost engineering components and machine have to incorporate design features which introduce changes in their cross-section.E.g. oil holes, bolt heads, gear teethChanges in cross section causes localized stress concentration

3.1 Stress Concentration FactorSeverity of stress concentration depends on the geometry of the discontinuity and nature of the material.Stress concentration factor, Kt = Smax/Sav

Smax= maximum stress at discontinuity and Sav,= nominal stress.

Kt, value depends only on geometry of the part.

Under static load, Kt gives an upper limit to the stress concentration value & applies only to brittle & notch sensitive materials. Should considered when using high-strength, low ductility, case-hardened & / or heavily cold work materials.

3.2 Stress concentration in fatigue. Stress concentration also should be considered in components that are subject to fatigue loadingKf, fatigue stress concentration factor,

Kf = endurance limit of notch free

endurance limit of notched partNotch sensitivity factor, q = Kf – 1 (eq 8.8)

Kt – 1

q varies from 0 to 1, as appproach to 1 the material becomes sensitive to the presence of notchesq also depends on component size, increases as size increase

Kf = Kt, the value of q=1 & materials becomes sensitive to notchesKf = 1, q = 0, material is not sensitive to notches

In making a design, Kt is usually determined from the geometry of partq can be specified when the material is selectedEq 8.8 is solved for Kf

also depend on component size thus stress-raiser are dangerous in larger masses.

3.3 Guidelines for design.Stress concentration can be a source of failure in many cases, esp. when designing with high strength materials & under fatigue loading.The following design guidelines should be observed if the effect of stress concentration are to be kept to a minimum:

1. Abrupt changes in cross-section should be avoided.Fillet radii or stress-relieving groove should be

provided. Fig. 11.3(a)

2. Slot and grooves should be provided with generous run-out radii and with fillet radii in all corners.

Fig. 11.3(b)

3. Stress relieving grooves or undercut should be

provided at the end of threads and splines. Fig. 11.3(c)4. Sharp internal corners and external edges

should be avoided5. Oil holes & similar features should be

chamfered and the bore should be smooth6. Weakening features like bolt and oil holes,

identification marks, and part number should not be located in highly stressed areas.

7. Weakening features should be staggered to avoid the addition of their stress concentration effects.

Fig. 11.3(d)

4. Design aspect-loading condition

Static loading – load is applied gradually and remains applied throughout part’s life.Repeated loading – applied and removed several times (repetitive) during life. Fail by fatigue at stress lower than yield strength. Higher design factor is needed.Impact – require large design factor. (i) sudden load causes stresses much higher than computed. (ii) require part to absorb energy of the impact.Static loading but at high T – consider creep, microstructural changes, oxidation & corrosion & influence of method of fabrication on creep.

4.1 Designing For Static Strength

Design based on the Static strength Ability to resist short-term steady load at moderate T. Measured in terms as yield strength, UTS, compressive strength & hardness.

Aimed at avoiding yielding of the component in the case of soft, ductile materials and at avoiding fracture in the case of strong, low toughness materials. Component must be strong enough to support the service load & may require enough stiffness to ensure deflections do not exceed certain limits.

Stiffness, important in machine elements to avoid misalignment and maintain dimensional accuracy.

Elasticity (Young’s M) important when designing struts, columns & thin-walled cylinders subjected to compressive axial loading where failure can take place by buckling.

Component and structure made from ductile material are usually designed, so that no yield take place under static loading conditionBut, when the component is subjected to uniaxial stress, yielding take place

When local stress reaches the yield strength of the material

Critical cross-sectional area, A ;A = KtnsL

YSKt = stress concentration factor, L = applied loadns = factor of safety, YS = yield strength

4.2 Designing for simple axial loading

In some cases the stiffness of the component, rather than its strength, is the limiting factor.In such cases, limits are set on the extension in the component. See example.

Example: It is required to select a structural material for the manufacture of the tie rods of suspension bridge. A representative rod is 10 m long & should carry a tensile load of 50 kN without yielding. The max. extension should not exceed 18 mm. Which one of the materials listed in Table 2.7 will give the lightest load ?

Calculations of the area will be carried out twice:

1. Area based on yield strength = load

YS

2. Area based on deflection = load x length

E x deflection

The larger of the two areas will be taken as the design area & will be used to calculate the mass.

Area based on yield strength =

Area based on deflection =

Mass =

The results show that steel A717 grade 70 & maraging steel grade 200 give the least mass.

The former steel is more ductile & less expensive it will be selected.

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4.2.1 Factor of safety, ns.• Is applied in designing component to ensure it will

satisfactory perform its intended function• To get the strength of material at allowable stress.

• The definition, strength of material depends on the type of material and loading condition. E.g. UTS-brittle, YS-ductile

• The factor of safety, ns. ns = S / Sa

S = Nominal strength, Sa = Allowable strength / Design strengthIt is important to define which type of service condition will the material work on before calculating the ns.i. Normal working conditionii. Limit working condition, such as overloading

Brittle material such as gray cast iron do not exhibit yielding, so design based on ultimate strength.General - % of elongation in a 2-in gage length is less than 5%

Brittle. e.g Al – except for castings, all is ductile.Ductile material – large plastic deformation, so unfit for intended use. Design based on yield strength.Design stress guidelines :

Ductile material, δd = Sy / 2

Brittle material, δd = UTS / 6

Given – (1) Magnitude & type of loading & (2) material condition.Determine yield & ultimate strength & % elongation of material. Decide ductile or brittle.Specify design factor (factor of safety).Compute design stress.Write equation for expected max stress. For direct normal stress, δmax = F/A

Set δmax = δd & solve for required cross-sectional area. Determine minimum required dimension.

4.2.2 Determine Dimensions & Shape of Component

Soalan :Satu struktur penyokong mesin akan dibebankan

dengan beban tegangan statik 16kN. Dicadangkan, dihasilkan

dalam bentuk rod empat segi daripada keluli gelek panas, AISI 1020.

Berikan dimensi yang sesuai untuk keratan rentas rod

tersebut.

Analisis :Objektif – tentu dimensi keratan rentas rod.Diberi – F = 16 kN = 16 000 N beban statik.Bahan – AISI 1020 HR, Sy = 331 MPa, 36%

elongation

Contoh

Jawapan :Analisis – Biar δ = δd = Sy / 2 bahan mulur.

Analisis tegasan – δ = F/A, maka A diperlukan A = F / δd

Tetapi A = a2 (a = dimensi sisi square)Dimensi minimum yang dibenarkan a = A

Maka Tegasan, δd = Sy/2 = 331 MPa/2

= 165.5 MPa (N/mm2)Keratan rentas, A = F/ δd = (16 000 N) =

96.7mm2

(165.5 N/mm2)Dimensi minimum a = A = 96.7 mm2 = 9.83 mm.Dimensi minimum a = 10 mm.

Torsional – loading of a component / part that tends to cause it to rotate or twist.When torque is applied, shearing stress is developed & torsional deformation occurs, resulting in an angle of twist of one end of part relative to the other.Material must have sufficient rigidity for the part to perform properly in service.

Torque = T = F x d where F = applied force & d = distance from

action of force to axis of the part.

Power = torque x rotational speed (n in rad/s).

4.3 Designing for Torsional Loading

Torsional shear stress, ζmax = Td / 2Ip

where T = applied torque, d = diameter & Ip = polar moment of inertia of the cross section.

Critical cross sectional area of a circular shaft can be calculated, 2Ip = KtnsT

d ζmax

where Kt = stress concentration factor, ζmax=Torsional

shear stress, ns = factor of safety,

Ip = polar moment of inertia of the cross section,

T = Torque

Ip = πd4 / 32 for solid circular shaft

Ip = π(d04 – di

4) / 32 for hollow circular shaft of inner di & outer d0.

ASTME code of practice ; allowable value of shear stress of 0.3 yield or 0.18 UTS, which is smaller.For ductile material, design shear stress = yield / 2N (steady torsion, N = 2, so ζd = yield / 4)

Torsional rigidity of component is usually measured by the angle of twist, θ, per unit length. For circular shaft, θ is given by,

θ = T / GIp

where G = modulus of elasticity in shear = E / (2(1 + ν)) where ν = Poisson’s ratio.

Usual practice is to limit the angular deflection in shafts to about 1 degree, i.e π/180 rad, in length of 20 times the diameter.

4.4. Designing for Bending

Beam – component that carries load transversely, that is, perpendicular to its long axis.Loading – normal concentrated load, inclined concentrated load, uniformly distributed load, varying distributed load & concentrated moments.Moment – an action that tends to cause rotation of an object. Can be produced by a pair of parallel forces acting in opposite directions, called couple.Beam types ; simple, overhanging, cantilever, compound & continuous.Bending moments – internal moments cause bending.

Overhanging – bend downward, negative bending

Load

MCantilever – one fixed end, provide support & moment produced by load

F

F

Compound – two or more parts extending in different directions.

Continuous – extra support or both ends fixed, require different approaches to analyze forces & moments.

Relation between bending moment, max allowable stress & dimensions given by ;

Z = nsM

YS where Z = section modulus = I/c

c = distance from center of gravity of cross section to the outermost fiber/beam.

I = moment of inertia of cross section with respect to neutral axis normal to direction of load.

M = bending moment & YS = max allowable stress.

ns = factor of safety.

When load is placed on a beam, the beam is bent and every portion of it is moved in a direction parallel to the direction of the load.

The distance that a point on the beam moves/ deflection depends

i. Its position in the beam

ii. Type of beam

iii. Type of support

Example

Question :

Determine the diameter of a cantilever beam of length 1 m

and rectangular cross section of depth-to-width ratio 2:1.

The cantilever is expected not to deflect more than 50 mm

for every 1000 N increment of load at its tip. The material

used in making the beam is steel AISI 4340 with a yield

strength of 1420 MPa and UTS 1800 MPa. What is the max

permissible load ? Assume a suitable factor of safety.

Example

Solution :The deflection ( y ) of a cantilever beam under a load L acting on its

tip is given by relationship :

y = ( Ll3 ) / ( 3 EI )

where l = the length of cantilever

E = elastic modulus of the cantilever material = 210 Gpa

I = the second moment of area of the cross-section

From fig 4.3

I = b x ( 2b )3 = 1000 x 1 x 1000 12 50 x 3 x 210 x 109

Where b is the width of the beam; b = 14.77 mm

Taking a factor of safety n = 1.5 and using eq.

Z= I/c, c=H/2=b

Z = 2148 mm3 = ( nM ) / YS = ( 1.5 x L x l ) / 1420

The safe value of L = 2033 N

Example