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11. Hardware 1
Agenda
1. Areas2. Stress and strain3. Axial loading4. Torsion loading5. Beam loading6. Engineering materials7. Vibration8. Fatigue9. Thermal loading
11. Hardware 2
1. Areas
AreaFirst moment of an areaCentroid of an areaMoment of inertia of an areaParallel axis theoremPolar moment of inertia
1. Areas
11. Hardware 3
Area (1 of 2)
dAA = area =
1
3
2 6
y
x
y = -0.5x + 4
dA = (y-1) dx
(-0.5x + 3)dx = 4A = area =2
6
Area by integrating in x directionArea by integrating in x direction1. Areas
11. Hardware 4
Area (2 of 2)
1
3
2 6
y
x
x = -2y + 8
dA = (x-2) dx
(-2y + 6)dy = 4A = area =1
3
Area by integrating in y directionArea by integrating in y direction1. Areas
11. Hardware 5
First moment of an area (1 of 3)
Qy = first moment of area with respect to the y-axis
= x dA
Qx = first moment of area with respect to the x-axis
= y dA
Definitions of first moment of areaDefinitions of first moment of area1. Areas
11. Hardware 6
First moment of an area (2 of 3)
1
3
2 6
y
x
y = -0.5x + 4
dA = (y-1) dx
(-0.5x + 3) x dx = 13.33Qy =
2
6
QyQy
x
1. Areas
11. Hardware 7
First moment of an area (3 of 3)
1
3
2 6
y
x
y = -0.5x + 4
dA = (y-1) dx
(-2y + 6) y dy = 6.67Qx =
1
3
QxQx
y
1. Areas
11. Hardware 8
Centroid of an area (1 of 2)
xc = Qy/A
yc = Qx/A
Centroids of area in terms of momentsCentroids of area in terms of moments1. Areas
11. Hardware 9
Centroid of an area (2 of 2)
xc = Qy/A = 13.33/4 = 3.33
yc = Qx/A = 6.67/4 = 1.67
Centroid of an area for previous examplesCentroid of an area for previous examples1. Areas
11. Hardware 10
Moment of inertia of an area (1 of 3)
Iy = first moment of area with respect to the y-axis
= x2 dA
Ix = first moment of area with respect to the x-axis
= y2 dA
Definitions of moment of inertiaDefinitions of moment of inertia1. Areas
11. Hardware 11
Moment of inertia of an area (2 of 3)
(-0.5x + 3) x2 dx = 48Iy =
2
6
Iy for previous exampleIy for previous example1. Areas
11. Hardware 12
Moment of inertia of an area (3 of 3)
1
3y
x
y = -0.5x + 7/3
(-0.5x + 4/3) x2 dx = 3.55Iy =
-4/3
8/3
Iy for previous example with axis shifted 10/3 to centroidIy for previous example with axis shifted 10/3 to centroid
-4/3 8/310/3
1. Areas
11. Hardware 13
Parallel axis theorem
Parallel axis theorem -- Iparallel axis = Ic + A d2
From previous example -- 3.55 + 4 x (10/3)2 = 48
Using parallel axis theorem to compute IyUsing parallel axis theorem to compute Iy
1. Areas
11. Hardware 14
Polar moment of inertia (1 of 2)
J = polar moment of inertia = (x2 + y2 ) dA
Definition of polar moment of inertiaDefinition of polar moment of inertia1. Areas
11. Hardware 15
Polar moment of inertia (2 of 2)
r
R
dA=2 rdr
dr
J = r2 2 rdr
0
R
= R4/2
Example of computing J for a circleExample of computing J for a circle1. Areas
11. Hardware 16
2. Stress and strain
Physical requirementsFree-body diagramsStress and strainHooke’s lawPoisson’s ratioStress concentrationsCombined stresses
2. Stress and strain
11. Hardware 18
Free-body diagrams
A diagram that illustrates all the forces acting on a body
If the forces are balanced, the body does not accelerate; otherwise it does
Free-body diagram should include the cross section of interest
2. Stress and strain
11. Hardware 19
Stress and strain
Stress is force per unit areaNormal stress
• Area is normal to force = F/A
Shear stress• Area is parallel to force = F/A
Strain is elongation expressed as a fraction or percentage of basis
2. Stress and strain
11. Hardware 20
Hooke’s law (1 of 2)
Relationship between stress and strain = E
• E = modulus of elasticity = normal strain
= G • G = shear modulus = shear strain
Definition of Hooke’s lawDefinition of Hooke’s law2. Stress and strain
11. Hardware 21
Hooke’s law (2 of 2)
Lo = 1
E
G
normal shear
Hooke’s law for normal and shear stressHooke’s law for normal and shear stress2. Stress and strain
11. Hardware 22
Poisson’s ratio (1 of 2) = Poisson’s ratio = ratio of lateral strain
to axial strainx = 1/E (x - y)
y = 1/E (y - x)
0 < < 0.5• liquids = 0.50• aluminum = 0.32 - 0.34• steel = 0.26 - 0.29• brass = 0.33 - 0.36• rubber = 0.49
Definition of Poisson’s ratioDefinition of Poisson’s ratio2. Stress and strain
11. Hardware 23
Poisson’s ratio (2 of 2)Problem
• x = 22,000 psi
• y = -14,000 psi
• E = 30,000,000 psi = 0.3
Solutionx = 1/ 30,000,000 (22,000 - 0.3 (-14,000 ))
= 0.00087y = 1/ 30,000,000 (-14,000 - 0.3 22,000 )
= 0.00069Example using Poisson’s ratioExample using Poisson’s ratio
2. Stress and strain
11. Hardware 24
Stress concentrations (1 of 2)Concentrations occur wherever there is a
discontinuity or non-uniformity in an object• Stepped shafts• Plates with holes and notches• Shafts with key ways
Stress occurs at discontinuitiesStress occurs at discontinuities2. Stress and strain
11. Hardware 25
Stress concentrations (2 of 2)
Concentrations can be thought of as streamlines• Where there are concentrations, the streamlines are closer together
Near concentrations, stress increases over what would normally be calculatedc = k , where 1 k 3
Stress can be visualized as streamlinesStress can be visualized as streamlines2. Stress and strain
11. Hardware 28
Combined stresses (3 of 7)
n ds (1) - x dy (1) sin - y dx (1) cos + xy dy (1) cos + xy dx (1) sin = 0
n = x dy/ds sin + y dx/ds cos - xy dy/ds cos - xy dx/ds sin
dx/ds = cos dy/ds = sin
n = x sin2 + y cos2 - 2 xy sin cos
n = (x + y )/2 + (y - x )/2 cos 2 - xy sin 2n = (x + y )/2 + (y - x )/2 cos 2 - xy sin 2
2. Stress and strain
11. Hardware 29
Combined stresses (4 of 7)
nt ds (1) + x dy (1) cos - y dx (1) sin + xy dy (1) sin - xy dx (1) cos = 0
nt = - x dy/ds cos + y dx/ds sin - xy dy/ds sin + xy dx/ds cos
nt = (y - x )/2 sin 2 + xy cos 2 nt = (y - x )/2 sin 2 + xy cos 2
2. Stress and strain
11. Hardware 30
Combined stresses (5 of 7)
Set d n /d = 0
tan 2 = - xy /[(y - x )/2]
Minimum and maximum axial stress
n = (x + y )/2 sqrt { [(y - x )/2]2 + xy2}
2. Stress and strain
11. Hardware 31
Combined stresses (6 of 7)
Set d nt /d = 0
tan 2 = [(y - x )/2]/ xy
Minimum and maximum shear stress on a plane at 45 degrees to normal stress
nt = sqrt { [(y - x )/2]2 + xy2}
2. Stress and strain
11. Hardware 32
Combined stresses (7 of 7)
min max
min
max
(x, -xy )
(y, xy )2
0
Mohr’s circleMohr’s circle2. Stress and strain
11. Hardware 33
Failures (1 of 2)
P
P
P
Y
U
UU
ductile
ductile without yield
brittle
Stress vs strain for ductile, ductile-without-yield, and brittle materials
Stress vs strain for ductile, ductile-without-yield, and brittle materials
2. Stress and strain
11. Hardware 34
Failures (2 of 2)Definitions
• Elastic limit -- Maximum stress at which all strain disappears when stress goes away
• Proportional limit (P) -- Maximum stress for which stress is proportional to strain
• Yield point (Y) -- Stress at which strain increases without increase of stress. Most materials don’t have a yield point
• Ultimate strength (U) -- Maximum stress that can be applied
Definitions for previous stress-vs-strain curvesDefinitions for previous stress-vs-strain curves2. Stress and strain
11. Hardware 35
3. Axial loading
Axial forceTransverse forceDeformationStrain energySpring constant
3. Axial loading
11. Hardware 36
Axial force (1 of 2)
An axial force is force in the direction of the axis of the body
3. Axial loading
11. Hardware 37
Axial force (2 of 2)
F = 1040 sin(30o)-300F = 1040 sin(30o)-300
1040 lb
30o8’
6’
A A
32o10’414 lb
300 lb
550 lb
5’ 10’
1040 lb
30oA300 lb
A
H
F
M
3. Axial loading
11. Hardware 38
Transverse force (1 of 2)
A transverse force is force perpendicular to the axis of the body
3. Axial loading
11. Hardware 39
H = 1040 cos(30o)H = 1040 cos(30o)
Transverse force (2 of 2)
1040 lb
30o8’
6’
A A
32o10’414 lb
300 lb
550 lb
5’ 10’
1040 lb
30oA300 lb
A
H
F
M
3. Axial loading
11. Hardware 40
Deformation
Elastic deformation = = Lo = Lo / E = Lo F / E A, where
• F = axial force• A = area of cross section
3. Axial loading
11. Hardware 42
Spring constantStiffness, or spring constant, is the ratio
of force to the displacement caused by the force
Stiffness = k = F/ Stiffness is an important consideration in
accommodating vibration
3. Axial loading
11. Hardware 44
Torsion
Torsion is twistUsually, the cross section of bar warps in
torsionThe shear stress at a point on a boundary
is parallel to the boundaryThe shear stress at a corner is zero
4. Torsion loading
11. Hardware 45
Shear (1 of 2)
L
r
= r /L
= G = G r /L
T = dA r = G /L r2 dA = G J /L
= TL/(JG)
= Tr/J
4. Torsion loading
11. Hardware 46
Shear (2 of 2)
Problem Solid bar Diameter = 2 inches G = 12,000,000 psi T = 50,000 in-lbf L = 4 inches
Find Maximum shear Maximum twist
J = r4 /2 = 1.57 in4
= Tr/J = 50,000 x 1/1.57 = 31,800 psi
= TL/(JG) = 50,000 x 4/(1.57 x 12,000,000) = 0.6o 4. Torsion loading
11. Hardware 48
Shear and moment (1 of 3)
Shear at any point on a beam is the sum of all forces from the point to the left end
V=dM/dxUpward loads are positive
ShearShear5. Beam loading
11. Hardware 49
Shear and moment (2 of 3)
Moment at any point on a beam is the sum of all moments and couples from the point to the left end
Clockwise moments are positive
M= V dx
Maximum moments occur where V=0
MomentMoment5. Beam loading
11. Hardware 51
Normal stress = = My/INormal stress = = My/I
Stress (1 of 6)
M M
y
dA
= C1y
M = dA y = C1y2dA = C1I
c
5. Beam loading
11. Hardware 52
Stress (2 of 6)M M + dM
dxy
c
dx t + M/I y1dA - (M+dM) y1dAc
y
c
y
= [dM/dx y1dA / (It)c
y
Shear stress = =QV / (It)Shear stress = =QV / (It)
5. Beam loading
11. Hardware 53
Stress (3 of 6)
Normal bending stress is greatest at top and bottom of beam, and zero on the centroidal axis
Shear stress is zero at top and bottom, and maximum on centroidal axis
Location of maximum and minimum stressesLocation of maximum and minimum stresses
5. Beam loading
11. Hardware 54
Stress (4 of 6)
1”3”
3”
1” 1”
A A’
2” 2”Find maximum normal stress and shear stress at AA’
Problem statementProblem statement
2”
M= 500 in-lbfV= 700 lbf
5. Beam loading
11. Hardware 55
Stress (5 of 6)y
x
I = y2 b dy = y3 / 3 b = bh3/12
h/2
h/2
b/2 b/2
-h/2
h/2
Problem -- compute I for rectangleProblem -- compute I for rectangle
5. Beam loading
11. Hardware 56
Stress (6 of 6)
1”3”
3”
1” 1”
A A’
2” 2”
I = 93x4/12 - 43x2/12 = 232.3
2”
Q = 2x4x2 - 2x1x1.5 = 13
M= 500 in-lbfV= 700 lbf
= My/I = 500x3/232.3 = 6.5 psi
=QV / (It) = 13x700/(232.3x2) = 19.6 psi
Problem -- solutionProblem -- solution
5. Beam loading
11. Hardware 57
Deflection (1 of 7)
=My / I
= /E = My / (EI) = y d /dxy
d d
dx
Rx
d /dx = d/dx (dy/dx) = d2y/dx2
y’’ = d2y/dx2 = M / (EI)
y’’’ = V / (EI)
y= deflectiony’ = slope
y’’ = M / (EI)y’’’ = V/ (EI)
y= deflectiony’ = slope
y’’ = M / (EI)y’’’ = V/ (EI)
5. Beam loading
11. Hardware 58
Deflection (2 of 7)
simple supportbuilt-in supportfree endhinge
y y’ y’’ V M
0 0 0 0 0 0 0
Boundary conditionsBoundary conditions
5. Beam loading
11. Hardware 59
Deflection (3 of 7)
unit load =w / unit-length
Ly
xF = wL
M = wL2/2
Problem: Find the equation for y for the cantilever beam
Problem -- statementProblem -- statement
5. Beam loading
11. Hardware 60
Deflection (4 of 7)
y= deflectiony’ = slopey’’ = M / (EI)y’’’ = V/ (EI)
y’’ = - wx2/ (2EI) = - w/(EI) x2/2
y’ = w/(EI) [-x3 /6 + C1] C1 = L3/6
y = w/(EI) [-x4 /24 + L3/6 x + C2] C2 = - L4/8
y = w/(EI) [-x4 /24 + L3/6 x + - L4/8]
Problem -- solutionProblem -- solution
5. Beam loading
11. Hardware 61
Deflection (5 of 7)
Other methods• Moment area method• Strain energy method• Conjugate beam method• Table-look up method
Other methods for computing deflectionOther methods for computing deflection
5. Beam loading
11. Hardware 62
Deflection (6 of 7)
Superposition -- When the deflection of the beam is small, the deflection due to several forces acting together is the sum of the deflections of the individual forces acting alone
Superposition -- Deflection as the sum of deflectionsSuperposition -- Deflection as the sum of deflections
5. Beam loading
11. Hardware 63
Deflection (7 of 7)
Stiffness = force/deflection
Stiffness = spring constant = force/deflectionStiffness = spring constant = force/deflection
5. Beam loading
11. Hardware 64
Failures (1 of 2)
Elastic failure -- Deflection is elastic, but there’s no yielding. Cracks and misalignment may occur
Local buckling• Vertical buckling -- Whole beam buckles• Web crippling -- Local buckling• Both can be avoided by using stiff3ners
5. Beam loading
11. Hardware 65
Failures (2 of 2)
Lateral buckling• Unsupported member rolls out of
normal plane• Requires support along compression
flangeRotation
• Plastic failure in which yield occurs and gives appearance that beam is hinged at ends and in center
5. Beam loading
11. Hardware 66
6. Engineering materials
PropertiesConcreteAluminumIron and steelCopperWoodPlastics
6. Engineering materials
11. Hardware 67
Properties (1 of 5)
Tensile strength -- ability to withstand pulling
Compressive strength -- ability to withstand squeezing
Torsional strength -- ability to withstand twisting
Stiffness -- ability to withstand bending
Strength propertiesStrength properties6. Engineering materials
11. Hardware 68
Properties (2 of 5)
Ductility -- ability to be stretchedBrittleness -- tendency to break with no
deformationHardness -- ability to withstand scratches,
dents, and cutsToughness -- energy required to breakDensity -- mass / volume
Strength properties (continued)Strength properties (continued)6. Engineering materials
11. Hardware 69
Properties (3 of 5)
Thermal conductivity -- ability to carry heat
Thermal expansion -- ability to expand when heated and contract when cooled
Thermal propertiesThermal properties
6. Engineering materials
11. Hardware 70
Properties (4 of 5)Electrical resistance -- ability to carry
electricityMagnetic properties -- ability to be
magnetized
Electrical and mechanical propertiesElectrical and mechanical properties6. Engineering materials
11. Hardware 71
Properties (5 of 5)
Transparency -- ability to pass lightReflection -- Ability to bounce lightRadiation and absorption -- Ability to give
off and take in heat or light
Optical propertiesOptical properties6. Engineering materials
11. Hardware 72
Concrete (1 of 4)
Concrete is a mixture of sand, aggregate, and cement
Hardening is a chemical reaction as opposed to loss of water
DescriptionDescription6. Engineering materials
11. Hardware 73
Concrete (2 of 4)
Portland cement is the most common• Type 1 normal -- most common• Type 2 modified -- moderate sulfate
resistance; used in hot weather • Type 3 high-early-strength -- develops
strength quickly; high shrinkage • Type 4 low-heat -- used in massive dams• Type 5 sulfate-resistant -- used in alkaline
soils
Portland cementsPortland cements6. Engineering materials
11. Hardware 74
Result = 4.1 ft3/sackResult = 4.1 ft3/sack
Concrete (3 of 4)
Amount of concrete is the sum of the volumes of cement, sand, aggregate, and water
Example• Weight mixture of cement/sand/aggregate
= 1.0/1.9/2.8 • 7 gallons of water/94-pound bag of cement• Cement = 195 pcf• Sand and aggregate = 165 pcf• 7.48 gallons of water/ft3
6. Engineering materials
11. Hardware 75
Concrete (4 of 4)
Density -- 140 - 160 lbm. ft3; 150 lbm/ ft3 nominal
E -- 1,000,000 - 5,000,000 psi using secant method
Strength of new concrete can be improved by keeping cool
Rebar and post-tension cables improve strength by minimizing tensile loading
PropertiesProperties6. Engineering materials
11. Hardware 76
Aluminum (1 of 5)
Most abundant metal in earth’s crust; most widely used
Corrosion resistant because surface oxide forms naturally
Almost as good as copper as conductor of electricity and heat
Machines easilyCannot be magnetized
General propertiesGeneral properties6. Engineering materials
11. Hardware 77
Aluminum (2 of 5)E = 10,000,000 psiG = 4,000,000 psiDensity = 165 lbm/ft3 Melting point = 660 oC
Mechanical propertiesMechanical properties6. Engineering materials
11. Hardware 78
Aluminum (3 of 5)
Most aluminum is alloyed with other elements• 1xxx -- commercially pure• 2xxx -- copper• 3xxx -- manganese• 4xxx -- silicon• 5xxx -- magnesium• 6xxx -- magnesium and silicon• 7xxx -- zinc• 8xxx -- other
DesignationsDesignations6. Engineering materials
11. Hardware 79
Aluminum (4 of 5)
Silicon• >3% -- improves fluidity; good for
casting• >12% -- improves hardness and wear
resistanceCopper
• Makes harder• Increase conductivity but decreases
corrosion resistance• Makes harder and improves corrosion
resistance AlloysAlloys6. Engineering materials
11. Hardware 80
Aluminum (5 of 5)
Treatments• F -- as fabricated• H -- heat treated• O -- soft, after annealing• T --heat treated
TreatmentsTreatments6. Engineering materials
11. Hardware 81
Iron and steel (1 of 3)
Pure iron -- Too soft and ductile for useMild-carbon steel (0.1% - 0.3% carbon)
• Less ductile, higher tensile strengthMedium-carbon steel (0.3% - 0.7% carbon)
• Very tough, high tensile strengthHigh-carbon steel (0.7% - 1.3% carbon)
• Very hard and brittle, used for cutting tools
General propertiesGeneral properties6. Engineering materials
11. Hardware 82
Iron and steel (2 of 3)
Stainless steel• Iron with chromium• Chromium prevents oxidation
Cast iron (3% carbon, 2% silicon)• Very brittle, poor tensile strength• Can be cast and machined
General properties (continued)General properties (continued)6. Engineering materials
11. Hardware 83
Iron and steel (3 of 3)
E = 30,000,000 psiG = 12,000,000 psi
Mechanical propertiesMechanical properties6. Engineering materials
11. Hardware 84
Copper (1 of 5)
Copper• Pure metal; third most important in volume• Ductile• Corrosion resistance with green oxide• Cuts, saws, and machines well
Copper propertiesCopper properties6. Engineering materials
11. Hardware 85
Copper (2 of 5)
Brass• Copper and zinc alloy• Greater hardness and tensile strength• Good conductivity and anti-corrosion
properties
Brass propertiesBrass properties6. Engineering materials
11. Hardware 86
Copper (3 of 5)Bronze
• Copper and tin alloy• Increases fluidity; improves casting• Improved salt-water corrosion
resistance• Tin is more expensive than zinc
Bronze propertiesBronze properties6. Engineering materials
11. Hardware 87
Copper (4 of 5)Copper-lead alloy
• Lead insoluble in copper• Forms tiny, soft particles• Good wear ; used in bearings
Copper-lead propertiesCopper-lead properties6. Engineering materials
11. Hardware 88
Copper (5 of 5)Copper-silicon alloy
• Improves mechanical properties of copper
• Very hard• High strength and corrosion resistance
-- boilers
Copper-silicon propertiesCopper-silicon properties6. Engineering materials
11. Hardware 89
Wood (1 of 2)
Wood can be thought of as a bundle of tubes
Designation of hardwood and softwood is a botanical designation • Hardwood -- from deciduous trees• Softwood --from conifers
DescriptionDescription6. Engineering materials
11. Hardware 90
Wood (2 of 2)
Tensile strength along grain is highCompressive strength along grain is about
half of tensileTensile and compressive strength across
grain is poorMost softwoods cut easily; majority of
hardwoods machine betterLow water content is important to strength
and durability
PropertiesProperties6. Engineering materials
11. Hardware 91
Plastics
Plastics -- Materials that behave like putty somewhere in the manufacturing process
Two types of plastics• Thermoplastic -- can be softened by heating
again• Thermosetting -- cannot be softened by
heating
General propertiesGeneral properties6. Engineering materials
11. Hardware 92
PlasticsThermoplastic plastics
• High-density polyethylene -- soft, chemical resistant
• Low-density polyethylene -- opaque or transparent; good insulator; most commonly used plastic
• Polypropylene -- tough, high-impact resistance, rigid, low-density
Thermoplastic plasticsThermoplastic plastics6. Engineering materials
11. Hardware 93
Plastics
Thermoplastics (continued)• Polyvinyl chloride (PVC) -- tough, stiff and
flexible forms• Acylics -- sheet material• Nylon -- fiber, gears, bearings• Polystyrene -- insulator, shock absorber
Thermoplastic plastics (continued)Thermoplastic plastics (continued)6. Engineering materials
11. Hardware 94
PlasticsThermosetting plastics
• Phenol formaldehyde (Bakelite) -- thermal and electrical insulator
• Urea formaldehyde -- replacement for bakelite• Melamine formaldehyde -- heat resistant, used
for tableware• Polyester resin -- polymerizes at room
temperature, reinforced with fiber glass, used for molding
Thermosetting plasticsThermosetting plastics6. Engineering materials
11. Hardware 95
7. Vibration
DefinitionEffects of vibrationEquivalent springsSingle-degree-of freedom (SDOF)Converting stressesIsolation
7. Vibration
11. Hardware 96
Definition
Vibration -- a to-and-fro motion when displaced from equilibrium
Harmonic vibration -- vibration that repeats itself in time
Random vibration -- Vibration that never seems to repeat itself
Shock -- A sudden non-periodic disturbance
Reference -- MIL-STD-810
7. Vibration
11. Hardware 97
Effect of vibrationVibration can cause the following
problems• Structural failures• Reduced life• Pointing and stabilization errors• Stress
7. Vibration
11. Hardware 98
Equivalent springs
Spring constant = force/deflection
N springs in parallel ke = ki
N springs in series 1/ ke = 1/ki
i=1
N
i=1
N
7. Vibration
11. Hardware 99
SDOF
body mass
spring, k shock absorber, c
body displacement
x1(t)
vibratorvibrator
displacementx2(t)
m d2x1/dt2 = -c(dx1/dt - dx2/dt) - k (x1 - x2)m d2x1/dt2 = -c(dx1/dt - dx2/dt) - k (x1 - x2)
7. Vibration
11. Hardware 100
Undamped natural frequency, n Undamped natural frequency, n
SDOF
Assume no vibrator or damping
m d2x1/dt2 = - k x1
Assume x1 = A sin t dx1/dt = A cos t d2x1/dt2 = -A 2 sin t
-m A n2 sin n t + k A sin n t = 0
n = undamped natural frequency sqrt(k/m)
7. Vibration
11. Hardware 101
Critical damping, cc Critical damping, cc
SDOFAssume no vibrator
m d2x1/dt2 = -c dx1/dt - k x1
Assume x1 = A est
dx1/dt = A s est d2x1/dt2 = A s2 est
s2 + c/m s + k/m = 0
s = -c/(2m) sqrt { [c/(2m)]2 - k/m}
cc = critical damping when radical = 0 , cc = 2 sqrt(km)
11. Hardware 102
Damping coefficient, Damping coefficient,
SDOF
= damping coefficient = c/cc
c = cc = 2 sqrt(km) = 2 n
7. Vibration
11. Hardware 103
SDOF equation in terms of and n SDOF equation in terms of and n
SDOF
m d2x1/dt2 = -c(dx1/dt - dx2/dt) - k (x1 - x2)
d2x1/dt2 +c/m x1/dt + k/m x1 = c/m dx2/dt) + k/m x2
d2x1/dt2 + 2 n x1/dt + n 2 x1 = 2 n dx2/dt + n 2 x2
7. Vibration
11. Hardware 104
SDOF
d2x1/dt2 + 2 n x1/dt + n 2 x1 = 2 n dx2/dt + n 2 x2
x2 = X2 ej t
x1 = X1 ej ( t - )
Q = X1/ X2
= sqrt{ [ 1 + 2 2(/n )2] / [ (1 - (/n )2 )2 + (2 (/n ))2 ]
Transmissibility, Q Transmissibility, Q
7. Vibration
11. Hardware 105
SDOF
0.1
1.0
10
100
0.1 1.0 10
/n
Transmissibility, Q
= 0.01, Q = 50
= 0.5
Transmissibility vs /n Transmissibility vs /n
7. Vibration
11. Hardware 106
Converting stresses (1 of 2)
Converting random to harmonic Converting random to harmonic
a harmonic = ced sqrt ( Prandom fn/(2Q) )), where
a harmonic = harmonic acceleration
Q = peak transmissibility fn = natural frequency
Prandom = Power spectral density of random input at fn; assumes is small
ced = equivalent damage coefficient
7. Vibration
11. Hardware 107
Converting stresses (2 of 2)
Converting harmonic to static Converting harmonic to static
astatic = a harmonic Q, where a = acceleration
Example
a harmonic = 3g
Q = 10
astatic = 3 x 10 = 30g
7. Vibration
11. Hardware 108
Isolation
In mechanical systems• Keep deflections low• Avoid natural frequencies that align
with the driving force by changing spring constant and mass
7. Vibration
11. Hardware 110
Fatigue failuresIn many industries, more structures break from fatigue
than from static loadFailure start as small cracksCrack propagates until remaining area can survive the
loadImmediate failure followsNote that fatigue failures occur below the proportional
limit
8. Fatigue
11. Hardware 111
S-N curve (1 of 4)
Plot of stress (S) at rupture against the number of cycles at failure (N)
103 104 105 106 107 108
N
120,000
80,000
40,000
0
, psi
Example S/N curve
8. Fatigue
11. Hardware 112
S-N curve (2 of 4)
Some materials reach a stress called the endurance limit below which they don’t fail
103 104 105 106 107 108
N
120,000
80,000
40,000
0
proportional limit
no failure
Monel metal
Endurance limit
S/N curve for material with endurance limit
8. Fatigue
11. Hardware 113
S/N curve (3 of 4)
Other materials don’t have an endurance limit
103 104 105 106 107 108
N
120,000
80,000
40,000
0
proportional limit
aluminum alloy
S/N curve for material without endurance limit
8. Fatigue
11. Hardware 114
S/N curve (4 of 4)
It is common to designate the strength at 500,000,000 cycles as being the endurance level for materials such as many light metals that don’t have an endurance limit
Endurance limit when there is no endurance limit
8. Fatigue
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Accelerated life (1 of 3)
Most equipment must survive a long time, such as 20 years, in use
Testing survival over such a long period isn’t practical
A common practice is to accelerate the time to failure by raising the stress level
Reason for accelerated life
8. Fatigue
11. Hardware 116
Accelerated life (2 of 3)
= c N-1/b = equation of S/N curve, where = stress at failure• c = constant• -1/b = slope of S/N curve
c = test / (Ntest)-1/b = required / (Nrequired)-1/b
test = required (Nrequired / Ntest) 1/b
Assume PSD 2
PSD test = PSDrequired (Nrequired / Ntest) 2/b
Lifetime can be verified in shorter time
Calculation
8. Fatigue
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Accelerated life (3 of 3)
b = 1/8Nrequired = 40,000 hours
PSD test = PSDrequired(40,000/1) 2/8
PSD test = PSDrequired (14)
Raising stress by 3.8 or PSD by 14 allows 40,000 hour life to be tested in one hour
Example
8. Fatigue
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9. Thermal loading
Coefficient of expansion (COE)Thermal stressesFailures
9. Thermal loading
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COE (1 of 2)
Coefficient of linear expansion = L = Lo(T2 - T1)
Coefficient of area expansion = A = Lo(T2 - T1)
2 Coefficient of volumetric expansion =
V = Lo(T2 - T1)
3
9. Thermal loading
11. Hardware 120
COE (2 of 2)
Aluminum 12.8 10-6 /oFCast iron 5.6 10-6 /oFConcrete 6.7 10-6 /oFPyrex 1.8 10-6 /oFSteel 6.5 10-6 /oFTitanium alloy 4.9 10-6 /oFTungsten 2.4 10-6 /oF
9. Thermal loading
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Thermal stresses (1 of 2)
Thermal stress is handled like strain in an applied load
If a material is heated and not allowed to expand, stress can be calculated using Hooke’s law = E
= (T2 - T1)
9. Thermal loading
11. Hardware 122
Thermal stresses (2 of 2)
Problem• L = 10 feet• A= 2 in2• Material = steel• Temperature at installation = 40o F• Current temperature
Solution = (T2 - T1)= 12.8 10-6 (70 - 40) = 384 10-6
= E = (10 10+6) x (384 10-6) = 3840 psi
9. Thermal loading
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