Transcript Document

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.
Designing For Static Strength
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 and / or fracture.
Component must be strong enough to support the load & may
require stiffness to ensure deflections do not exceed certain limits.
Stiffness , important 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.
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
Designing for simple axial loading
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
The factor of safety, ns.
n s = S / Sa
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
Designing for Static Strength
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
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.
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.
Designing for Torsional Loading
Critical cross sectional area can be calculated,
for circular shaft, 2Ip = KtnsT
d
ζmax
where Kt = stress concentration factor
ns = factor of safety
Moment, 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.
For ductile material, design shear stress = yield / 2N (steady torsion,
N = 2, so ζd = yield / 4)
Designing for Torsional Loading
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.
G = E / (2(1 + ν)) where ν = Poisson’s ratio.
Usual practice is limit the angular deflection in shafts to
about 1 degree, i.e π/180 rad, in length of 20 times the
diameter.
Stiffness of part differ depending on shape of the crosssection – circular section has higher rigidity compared to
other structural shapes, i.e I-beams,wide-flange beams
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.
Designing for Bending
Load
Overhanging – bend downward, negative bending
M
Cantilever – one fixed end, provide support
& moment produced by load
Designing for Bending
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.
Designing for Bending
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.
Designing for Bending
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 ) 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 = 2148 mm3 = ( nM ) / YS = ( 1.5 x L x l ) / 1420
The safe value of L = 2033 N