EFFECT OF MATERIAL PROPERTIES
ON DESIGN
Dr. Zuhailawati Hussain
EBB 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 propertiesmanufacturing 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.
FUNCTION & CONSUMER
REQUIREMENTS
COMPONENT
DESIGN
MANUFACTURING
PROCESSES
MATERIAL
PROPERTIES
Fig 6.1 Factors should be considered in
component design
PROPERTIES OF
STOCK
MATERIAL
BEHAVIOR OF
MATERIAL IN THE
COMPONENT
GEOMETRY &
EXTERNAL
FORCES
EFFECT OF
FABRICATION
METHOD
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.
2.1 Statistical Parameter
Material properties can be statistically described by
mean value (x), standard deviation () and coefficient of
variation ().
1 n
Mean value, x  x1  x2  x3  ..... X n   xi
n i 1
eq.1
x1, x2, x3 = values of different measurements of the
same material property, n= number of measurement
12
 1
2
Std deviation,   
( xi  x) 

 n  1 i 1

n
eq.2
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
2.2 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 teeth
Changes in cross section causes localized stress
concentration
3.1 Stress Concentration Factor
Severity 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 loading
Kf, fatigue stress concentration factor,
Kf = endurance limit of notch free
endurance limit of notched part
Notch 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 notches
q also depends on component size, increases as size
increase
Kf = Kt, the value of q=1 & materials becomes sensitive
to notches
Kf = 1, q = 0, material is not sensitive to notches
In making a design,
Kt is usually determined from the geometry of
part
q can be specified when the material is selected
Eq 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 avoided
5. Oil holes & similar features should be chamfered
and the bore should be smooth
6. 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 & thinwalled cylinders subjected to compressive axial
loading where failure can take place by buckling.
4.2 Designing for simple axial loading
Component and structure made from ductile material
are usually designed, so that no yield take place under
static loading condition
But, 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
YS
Kt = stress concentration factor, L = applied load
ns = factor of safety,
YS = yield strength
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.
3
50
x
10
N
4
2
2
Area based on yield strength =

2
.
44
x
10
m

244
mm
205 x106 Pa
3
50
x
10
N x 10 m
Area based on deflection =
2

131
mm
212 x109 x18 x10 3 m
Mass = V 
7.8 g
2
x
244
mm
x 10000 mm  19 kg
3
(1x10mm)
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.
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. UTSbrittle, YS-ductile
The factor of safety, ns.
ns = S / S a
S = Nominal strength, Sa = Allowable strength / Design
strength
It is important to define which type of service
condition will the material work on before
calculating the ns.
i. Normal working condition
ii. 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
4.2.2 Determine Dimensions & Shape of Component
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.
Contoh
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
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.
4.3 Designing for Torsional Loading
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).
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 – di4) / 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.
Load
Overhanging – bend downward, negative bending
M
Cantilever – 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=
n sM
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
Example
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