Chapter 6 Effect of material properties on...

Materials and Process Selection for Engineering Design: Mahmoud Farag

Chapter 6

Effect of material properties on design


Chapter 6: Goal and objectives

The goal of this chapter is to illustrate how the material-

related factors affect the design. The main objectives are

to get better understanding of the material properties that

need to be considered when designing for:

1. static strength, stiffness, and toughness

2. fatigue resistance

3. high-temperature conditions

4. hostile environments

5. wear resistance

Design for static strength I

Designing for simple axial loading


When the component is subjected to uniaxial stress, yielding will take place

when the local stress reaches the yield strength of the material.

The critical cross sectional area, A, of such component can be estimated as:

YSnLKA t / (6.1)

where Kt = stress concentration factor

L = applied load

n = factor of safety

YS = yield strength of the material

Design for static strength II

Designing for torsional loading


The critical cross sectional area of a circular shaft subjected to torsional

loading can be determined from the relationship:

max//2 nTKdI tp (6.2)

where d = shaft diameter at the critical cross section

τmax = shear strength of the material

T = transmitted torque

Ip = polar moment of inertia of the cross section

= π d4/32 for a solid circular shaft

= π (do4 - di

4)/32 for a hollow circular shaft of

inner diameter di and outer diameter do

Design for static strength III

Designing for bending


When a relatively long beam is subjected to bending, the bending

moment, the maximum allowable stress, and dimensions of the cross

section are related by the equation:

YSnMZ /)( (6.3)

where M = bending moment.

Z = section modulus = I/c.

I = moment of inertia of the cross section with respect to

the neutral axis normal to the direction of the load.

c = distance from center of gravity of the cross section to

the outermost fiber.

Design example 6.1 Designing a cantilever beam I

Determine the dimensions of a cantilever beam of length 1 m and

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

The cantilever is not to deflect more than 50 mm for every 1000 N

increment of load at its tip.

The beam material is steel AISI 4340 with

YS 1420 MPa and UTS 1800 MPa.

What is the maximum permissible load?

Assume a suitable factor of safety.


Design example 6.1 Designing a cantilever beam II

Answer:

Deflection (y) of a cantilever beam under a load (L) acting on its tip is

given by the relationship: y = (L l3)/(3 E I),

where l = the length of the cantilever, E = elastic modulus= 210 GPa

I = [b(2b)3]/12 = (L l3)/2 y E. This gives b -14.77 mm

Taking a factor of safety n = 1.5

Z = (nM)/YS = 1.5 x L x 1000/1420 = 14.77 (2 x 14.77)3 / 12 x 14.77

= 2148 mm3

The safe value of L = 2033 N.


Design example 6.2 – Selection of a material

for a stiff and light tie rod I

Select one of the following materials for the manufacture of the tie

rods of a suspension bridge.

The rod is 10 m long and carries a tensile load of 50 kN. The rod

must not yield and the maximum extension is 18 mm.


Material YS

(MPa)

E

(GPa)

Specific

gravity

Area based

on yield

strength

(mm2)

Area

based on

deflection

(mm2)

Mass

(kg)

ASTM A675 Grade 60 205 212 7.8 244 131 19

ASTM A572 Grade 50 345 211 7.8 145 131 11.3

ASTM A717 Grade 70 485 211 7.8 103 131 10.2

Maraging steel Grade 200 1400 211 7.8 36 131 10.2

Al 5052-H38 259 70.8 2.7 193 392 10.6

Cartridge brass 70% hard temper 441 100.6 8.0 113 276 22.1

Design example 6.2 – Selection of a material

for a stiff and light tie rod II

Solution

For the present case, calculations of the area will be carried out twice:

• Area based on yield strength = load/YS

• Area based on deflection = (load x length)/(E x deflection)

• The larger of the two areas will be taken as the design area and will

be used to calculate the mass.

The results of the calculations are shown in the table and show that

steel A717 grade 70 and maraging steel grade 200 give the least

mass. As the former steel is more ductile and less expensive it will

be selected.



Designing for stiffnessExamples of how low stiffness of plastics is overcome by increasing

the second moment of area of the critical cross-section

Design example 6.3 – Substituting HDPE for stainless steel

in a fork

What design changes are required when substituting HDPE for stainless steel in

making a fork for a picnic set while maintaining similar stiffness.

Solution:

E for stainless steel = 210 GPa , E for HDPE = 1.1 GPa

The narrowest cross section of the stainless steel fork is rectangular of 0.6 x 5 mm

I for the stainless steel section = 5 x (0.6)3/12 = 0.09 mm4

• EI for stainless steel = 210 x 0.09 = 18.9, EI for HDPE design = 1.1 x I

• I for HDPE design = 17.2 mm4.

• A channel section of thickness 0.5 mm, web height 4 mm, and width 8 mm

• I = [8 x (4)3 - 7 x (3.5)3]/12 = 17.7 mm4, which meets the required value.

• Areas of the stainless steel and HDPE sections = 3 mm2 and 7.5 mm2

• ρ of stainless steel is 7.8 and that of HDPE is 0.96.

• Relative weight of HDPE/Stainless steel = (7.5 x 0.96)/(3 x 7.8) = 0.3


Designing with high-strength low toughness materials I

High-strength materials are less tolerant of defects than the

traditional lower-strength, tougher materials.

A crack-like defect can safely exist in a low-strength ductile material,

it can cause catastrophic failure in a high-strength low-toughness

material.

Such defects can be the result of:

• Initial flaws in the material, e.g. inclusions and cavities.

• Production deficiencies, e.g. welding defects.

• Service conditions, e.g. fatigue cracks or stress corrosion

cracks.

Fail-safe design : The structure is sufficiently damage tolerant to

allow defects to be detected before they develop to a dangerous

size.


Designing with high-strength low toughness materials II

Guidelines for design

In designing with high-strength low-toughness materials, the

interaction between fracture toughness of the material, the

allowable crack size, and the design stress should be considered.

The condition for failure under plane strain conditions, where a

crack of length 2a exists in a thick infinitely large plate, is given

by:

KI = KIC = Y σf (π a)1/2 (6.8)

Where σf is the fracture stress

Y is a dimensionless factor which is a function of crack geometry.


Design example 6.4 – Designing with KIC

If the minimum detectable crack in the beam described in example

6.1 is 3 mm, what is the maximum permissible load? KIC of the

beam material is 87.4 MPa m1/2.

Answer

• KIC = 87.4 = Y σf (π a)1/2

• Taking Y = 1 and a = 1.5 mm, then σf = 1273.2 MPa.

For a beam in bending, Z = (nM)/YS

• L = (σf x Z)/(n x l) = (1273.2 x 2148)/(1.5 x 1000) = 1824 N

Conclusion

• This load is less than the “safe load” calculated in example 6.1,

which means that had the presence of cracks been ignored, failure

would have taken place.


Leak-before-burst

The approach of leak-before-burst utilizes the fracture mechanics

approach in designing pressure vessels and similar structures.

If a vessel containing pressurized gas or liquid has a growing crack,

the toughness should be sufficiently high to tolerate a defect size

which will allow the contents to leak out before it grows

catastrophically.

For leakage to occur, the crack must grow through the vessel wall

thickness, t.

This means that the crack length 2a is about 2t.

This condition is satisfied when (KIC/Y σ) 2 is larger than π t.

Example 6.5 illustrates the use of the leak-before-burst approach in

design.


Design example 6.5 – Designing of a pressure vessel for

leak-before-burst conditions I

Design a pressure vessel of the following specifications:

• Internal pressure, p = 16 MN/m2 , Internal diameter, D = 400 mm

• Welded joints are inspected using NDT which detect surface

cracks > 3 mm.

Solution

Consider fist the use of AISI 4340 . When tempered at 260oC it has

YS1640 MPa and KIC = 50.0 MPa m1/2 .

Treating the pressure vessel as a thin wall cylinder, the wall

thickness, t, is given as t = p D/2 Sw

where Sw is the working stress.


Design example 6.5 – Designing of a pressure vessel for

leak-before-burst conditions II

Taking a factor of safety of 2, the working stress is 820 MPa and the

wall thickness is 3.9 mm.

The critical surface crack size can then be calculated as:

KIC = 50 = Y σf (π a)1/2 = Y x 820 (π a)1/2

Taking Y as 1, for σf /YS = 0.5, the critical crack length is found to

be 2.37 mm.

This length is too small to be detected by the available NDT

technique and does not satisfy the condition for leak-before-burst,

which makes the selected steel unsuitable.


Design example 6.5 – Designing of a pressure

vessel for leak-before-burst conditions III

Taking the same steel, AISI 4340, but tempering at 425oC gives a

yield Strength of 1420 MPa and KIC of 87.4 MPa m1/2.

Following the same procedure as before, t = 4.5 mm and the critical

crack length is 9.66 mm which can be detected by the available

NDT technique and also satisfies the condition for leak-before-

burst.

This means that the steel in this tempered condition is acceptable for

making the pressure vessel from the fracture mechanics point of

view.

Other factors like availability of material, weldability, weight of

vessel, and cost have to be taken into consideration before this

steel is finally selected.


Designing against fatigue

The fatigue behavior of materials is usually described by means of the

S-N diagram which gives the number of cycles to failure, N, as a

function of the maximum applied alternating stress, Sa .

Such values cannot be directly used for design purposes because the

behavior of a component under fatigue loading does not only depend

on the endurance limit of the material, but also on several other factors

including:

• size and shape of the component or structure;

• type of loading and state of stress;

• stress concentration and surface finish;

• operating temperature and service environment;

The influence of the above factors can be accounted for by modifying the

endurance limit of the material using a number of derating factors.



Table 6.2 Effect of surface condition and environment on the fatigue strength of steels

UTS

MPa (ksi)

Fatigue strength as % of maximum endurance limit

Mirror

polish

Polished Machined 0.1 mm

notch

Hot

worked

surface

Under

fresh

water

Under

salt

water

280 (40) 100 95 93 87 82 72 52

560 (80) 100 92 88 77 63 53 36

840 (122) 100 90 84 66 47 37 25

1120 (162) 100 88 78 55 37 25 17

1400 (203) 100 88 72 44 30 19 14

1540 (223) 100 88 69 39 30 19 12

Fatigue endurance-limit modifying factors

Derating factors are used to account for the parameters that affect the

behavior of components in service under fatigue loading as

follows:

Se = ka kb kc kd ke kf kg kh Se' (6.10)

Se = endurance limit of the material in the component

Se„ = endurance limit of the material from laboratory fatigue test

ka = surface finish factor, kb = size factor

kc = reliability factor, kd = operating temperature factor

ke = loading factor, kf = stress concentration factor

kg = service environment factor,

kh = manufacturing processes factor



Table 6.3 Effect of surface finish and UTS on surface finish factor (ka) for steels

UTS

MPa (ksi)

Forged

Ra=500-125

Hot rolled

Ra=250-63

Machined or

cold drawn

Ra=125-32

Ground

Ra=63-4

Polished

Ra<16

420 (60) 0.54 0.70 0.84 0.90 1.00

700 (100) 0.40 0.55 0.74 0.90 1.00

1000 (143) 0.32 0.45 0.68 0.90 1.00

1400 (200) 0.25 0.36 0.64 0.90 1.00

1700 (243) 0.20 0.30 0.60 0.90 1.00


Table 6.4 Effect of environment and UTS on the service environment factor (kg) for steels

UTS

MPa (ksi)

kg under fresh water kg under salt water

280 (40) 0.72 0.52

560 (80) 0.52 0.36

840 (122) 0.37 0.25

1120 (162) 0.25 0.17

1400 (203) 0.19 0.14

1540 (223) 0.19 0.12

Design example 6.6-Design an axle for fatigue resistance I

Calculate the diameter of the two rear axle shafts for a truck of mass

6000 kg.

2/3 of the mass of the truck is supported by the rear axles, which can

be treated as cantilever beams each of one meter length with the

load acting on its end. The material is 4340 steel of UTS 952 MPa

and the endurance ratio is 0.56.

Solution

The endurance limit is 532 MPa.

Take the surface finish factor ka as 0.68,

Assuming 3 inch diameter shaft, the size factor kb is 0.8.

The reliability is taken as kc = 0.752.

The loading factor is ke = 1, as the shaft is loaded in bending.

Stress concentration factor kf = 0.7 Materials and Process Selection for Engineering Design: Mahmoud Farag

Design example 6.6-Design an axle for fatigue resistance II

Se = 532 x 0.68 x 0.8 x 0.752 x 1.0 x 0.7 = 152.3 MPa.

• Load acting on each shaft = 6000 x 2/3 x 1/2 x 9.8 = 19600 N.

• Bending moment M = 19600 x 1 = 19600 N.m.

• Stress acting on the surface of the shaft = S = M/Z = 32 M/π D3

D3 = (32 x 19600 x 1000)/(π x 152.3) = 1,310,860 mm3

• D = 109.5 mm.


Effect of mean stress on fatigue behavior


Incorporating mean and alternating

stresses in design

In designing for fatigue under mean and alternating stresses,

separate factors of safety for the static and alternating strengths

of the material, nm and na can be used.

When the component contains a stress concentration, separate stress

concentration factors Kt, and Kf are also used .

The factors can be incorporated with the Goodman line as follows:

• The Soderberg line can be used by substituting YS for UTS.

• Design example 6.7 illustrates the use of this approach in design.


1// eafamtm SSKnUTSSKn

Design example 6.7 – Design of a cantilever beam for

fatigue behavior I

The cantilever beam in example 6.1: length 1 m and rectangular

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

The material is steel AISI 4340 with

YS 1420 MPa, UTS 1800 MPa., and endurance ratio 0.56.

The load on the beam in is fluctuating around a mean value of 1000

N, what is the maximum alternating load that can be applied

without causing fatigue failure?

Use a suitable factors of safety.


Design example 6.7 – Design of a cantilever beam for

fatigue behavior II

Answer:

• From example 6.1, Z = 2148 mm3,

• Mean stress = Sm = M/Z = 1000 x1000/2148 = 465.3 MPa

• Endurance limit = Se = 1800 x 0.56 = 1008 MPa.

• Taking a factor of safety of 1.5 for the static load and 3 for the

dynamic load,

(nmKtSm/UTS)+(naKfSa/Se) = 1

= (1.5 x 465.3/1800)+(3xSa/1008)

Sa = 205.7 MPa,

Alternating load = (205.7 x 2148)/1000 = 440 N


Cumulative fatigue damage

According to the Palmgren-Miner rule, if cyclic stressing occurs at a

series of stress levels S1, S2, S3,....Si, each of which would

corresponds to a failure life of N1, N2, N3,....Ni if applied singly.

Then, the fraction of total life used at each stress level is the

actual number of cycles applied at this level n1, n2, n3,....ni divided

by the corresponding life.

The part is expected to fail when the cumulative damage satisfies the

relationship:

n1/N1 + n2/N2 + n3/N3+.....+ ni/Ni = C (6.13)

C is usually in the range 0.7 to 2.2 and can be taken as unity.


Other fatigue design criteria

Safe-life, or finite-life, design assumes that the component is free

from flaws but the stress level is higher than the endurance limit.

This means that fatigue crack initiation is inevitable and the

component life is estimated as the number of stress cycles to initiate

such a crack.

Fail-safe design assumes that cracks that form in service will be

detected and repaired before they can lead to failure. This requires

materials with high fracture toughness and a reliable NDT program.

Damage-tolerant design is an extension of the fail-safe criterion; it

assumes that flaws exist in components before they are put in

service. Fracture mechanics techniques are used to determine

whether such cracks will grow large enough to cause failure before

they are detected during a periodic inspection.


Designing under high temperature conditions

Under creep conditions, the service life of a component at a given

service temperature is usually determined by the creep rate and

applied stress.

For design purposes, creep properties are usually presented on plots

which yield reasonable straight lines.

Common methods of presentation include :

• log-log plots of stress vs. steady state creep rate

• stress vs. time to produce different amounts of total strain

In addition to creep, the other factors which must be taken into

consideration include:

• Metallurgical and microstructural changes at high temperature

• Oxidation and hot corrosion



Variation of stress with steady-state creep rate at various

temperatures


Variation of stress with time to produce different amounts of total

strain at a given temperature. The uppermost curve is the stress-

rupture which occurs at different total strains to failure

Design example 6.8 - Designing of a cantilever

beam to resist creep

If the cantilever describes in example 6.1 is made of 5% Cr, o.5% Mo

steel and is to serve at 300 oC, what is the maximum allowable load

that can be endured at least 100,000 hrs?

Analysis

The allowable design stress for 5% Cr - 0.5% Mo steel according to the

ASME Boiler Code is shown in Fig. 6.6.

• From the figure, the allowable stress at 300 oC is about 90 MPa.

• Maximum allowable load = (90 x 2148)/1000 = 193.3 N

Conclusion

Comparing the above result with example 6.1 shows that the maximum

allowable load at 300 oC is less than 10% of the safe load at room

temperature.


Larson-Miller

parameter (LMP)

LMP correlates creep data

at different times and

temperatures.

LMP = T (C + log tr)

T = test temp. in Kelven

tr = time to rupture in hrs.

C = Larson-Miller constant

Usually taken to be 20


Design example 6.9 – Designing a turbine blade to

resist creep using LMP

If a turbine blade made of Nimonic 105 alloy had a life of 10,000 hrs

at 150 MPa at 810 oC, what is the expected life at the same stress

but at service temperature 750 oC ?

Solution

LMP = T (C + log tr)

where T = is temperature in degrees Kelvin or degrees Rankin

tr = time to rupture in hours, the log is to the base 10

C= the Larson-Miller constant, often taken to be 20.

LMP = (810 + 273) (20 + 4) = (750 + 273) (20 + log tr)

Life at the new service temperature is about 255,637 hours.


Life under combined fatigue and creep loading

Similar to the cumulative fatigue reasoning, life can be predicted

under combined fatigue and creep loading:

t1/tr1 + t2/tr2 + t3/tr3 +..... + n1/N1 + n2/N2 + n3/N3 +..... = 1

where

t1, t2, t3 ... are the times spent by the part under stress levels 1, 2, 3,

tr1, tr2, tr3, ... are the rupture lives of the part under stress levels 1, 2, 3,

n1,n2, n3,.. are the number of cycles at stress levels 1, 2, 3,

N1,N2, N3,.. are fatigue lives at stress levels 1, 2, 3,


Designing for hostile environment

Avoid galvanic corrosion by ensuring that dissimilar metals are not

in electrical contact. Use insulating washers, gaskets, or paint.

Eliminate crevices formed by the geometry of the structure, e.g.

rivet joints and areas of metal under deposits or dirt.

Ensure full drainage of rain water and liquids in reservoirs .

Provide ventilation in closed tanks and sections to avoid

condensation, or sweating.

Avoid sharp corners and convex surfaces which tend to have

thinner coatings or are subject to coating cracks.

Eliminate features that cause turbulence and rotary motion in

moving liquids or gases.

Prevent fretting corrosion by avoiding slippage or relative motion

between surfaces that are not meant to move.



Features of design against corrosion I


Features of design against corrosion II

Designing with specific materials

A successful design with a specific material must use its points of

strength and avoid its points of weakness.

• In metallic materials, ductility and toughness decrease as the

strength increases.

• Electrical and thermal conductivity of metallic materials may be

important factors is some designs.

• Polymers are light but their stiffness and impact strength are

much lower than metals. They are also insulators .

• Ceramics are corrosion resistant and strong in compression but

are very brittle and sensitive to mechanical and thermal shocks.

They are also insulators.

• Composite materials generally have high strength/weight ratio

and their properties can be tailored to suite the type of loading.



Chapter 6: Summary I

A successful design should take into account

• function,

• material properties, and

• manufacturing processes.

The relationship between material properties and design is complex because the behavior of the material in the finished product can be quite different from that of the stock material.

The stiffness of components under bending is proportional to the elastic constant of the material (E) and the moment of inertia of the cross section (I).


Chapter 6: Summary II

Fail-safe design requires a structure to be sufficiently damage tolerant to allow defects to be detected before they develop to a dangerous size.

In the case of high-strength materials, the interaction between fracture toughness, crack size, and the design stress should be considered.

Leak-before-burst is a useful design criterion in the case of pressure vessels and similar products. The toughness of the material used in such applications should be sufficiently high to tolerate a defect size which will allow the contents to leak out before it grows catastrophically.

Chapter 6: Summary III

In designing for fatigue, consider the fatigue strength of the material as well as size and shape of the component or structure, type of loading and state of stress, stress concentration, surface finish, operating temperature, service environment, method of fabrication.

Safe-life, or finite-life, design assumes that the component is free from flaws but the stress level in certain areas is higher than the endurance limit of the material. In this case, fatigue-crack initiation is inevitable and the life of the component is estimated on the basis of the number of cycles to initiate such a crack.



Chapter 6: Summary IV

The Larson-Miller parameter can be used when creep data are

incomplete and have to be supplemented or extended by

interpolation or extrapolation.

Many of the causes of failure in hostile environment can be avoided

by observing appropriate design rules such as

• avoiding electrical contact between dissimilar metals,

• sealing crevices,

• provision of appropriate ventilation and drainage of liquids

• avoiding sharp corners in coated components,

• preventing turbulence in moving liquids,

• preventing slippage between surfaces that are not meant to

move.


Chapter 6: Summary V

A successful design with a specific material must use its points of

strength and avoid its points of weakness.

• In metallic materials, ductility and toughness decrease as the

strength increases.

• Polymers are light but their stiffness and impact strength are

much lower than metals.

• Ceramics are corrosion resistant and strong in compression but

are very brittle and sensitive to mechanical and thermal shocks.

• Composite materials generally have high strength/weight ratio

and their properties can be tailored to suite the type of loading.

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