7. hardware1 agenda r1. areas r2. stress and strain r3. axial loading r4. torsion loading r5. beam...

7. Hardware 1

Agenda

1. Areas2. Stress and strain3. Axial loading4. Torsion loading5. Beam loading6. Engineering materials7. Vibration8. Fatigue9. Thermal loading

7. Hardware 2

1. Areas

AreaFirst moment of an areaCentroid of an areaMoment of inertia of an areaParallel axis theoremPolar moment of inertia

1. Areas

7. 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

7. 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

7. 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

7. Hardware 6


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

7. Hardware 7


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

7. 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

7. 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

7. 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

7. Hardware 11


(-0.5x + 3) x2 dx = 48Iy =

2

6

Iy for previous exampleIy for previous example1. Areas

7. Hardware 12


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

7. 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

7. 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

7. 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

7. Hardware 16

2. Stress and strain

Physical requirementsFree-body diagramsStress and strainHooke’s lawPoisson’s ratioStress concentrationsCombined stresses


7. Hardware 17

Physical requirements

WorksDoesn’t break


7. 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


7. 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


7. 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

7. 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

7. 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

7. 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


7. 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

7. 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

7. Hardware 26

Combined stresses (1 of 7)

x

xy

xy xy

xy

x

y

y


7. Hardware 27


xxy

xy

y

nt

n

ds

dx

dy


7. Hardware 28


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


7. Hardware 29


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


7. Hardware 30


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}


7. Hardware 31


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}


7. Hardware 32


min max

min

max

(x, -xy )

(y, xy )2

0

Mohr’s circleMohr’s circle2. Stress and strain

7. 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


7. 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

7. Hardware 35

3. Axial loading

Axial forceTransverse forceDeformationStrain energySpring constant

3. Axial loading

7. Hardware 36

Axial force (1 of 2)

An axial force is force in the direction of the axis of the body

3. Axial loading

7. 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

7. Hardware 38

Transverse force (1 of 2)

A transverse force is force perpendicular to the axis of the body

3. Axial loading

7. 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

7. Hardware 40

Deformation

Elastic deformation = = Lo = Lo / E = Lo F / E A, where

• F = axial force• A = area of cross section

3. Axial loading

7. Hardware 41

Strain energy

Total strain energy = UU = 1/2 F = F2 Lo / 2E A

3. Axial loading

7. 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

7. Hardware 43

4. Torsion loading

TorsionShear

4. Torsion loading

7. 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

7. 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

7. 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

7. Hardware 47

5. Beam loading

Shear and momentStressDeflectionFailures

5. Beam loading

7. 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

7. Hardware 49


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

7. Hardware 50


2F/L

L

FV

M FL/2

DiagramsDiagrams

7. 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

7. 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

7. 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

7. 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

7. 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

7. 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

7. 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

7. 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

7. 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

7. 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

7. 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

7. 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

7. Hardware 63

Deflection (7 of 7)

Stiffness = force/deflection

Stiffness = spring constant = force/deflectionStiffness = spring constant = force/deflection

5. Beam loading

7. 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

7. 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

7. Hardware 66

6. Engineering materials

PropertiesConcreteAluminumIron and steelCopperWoodPlastics


7. 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

7. 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

7. 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


7. 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

7. 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

7. 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

7. 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

7. 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


7. 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

7. 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

7. 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

7. 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

7. 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

7. Hardware 80

Aluminum (5 of 5)

Treatments• F -- as fabricated• H -- heat treated• O -- soft, after annealing• T --heat treated

TreatmentsTreatments6. Engineering materials

7. 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


7. Hardware 82


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

7. Hardware 83


E = 30,000,000 psiG = 12,000,000 psi

Mechanical propertiesMechanical properties6. Engineering materials

7. 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

7. 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

7. 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

7. 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

7. 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

7. 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

7. 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

7. 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


7. 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

7. 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

7. 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

7. Hardware 95

7. Vibration

DefinitionEffects of vibrationEquivalent springsSingle-degree-of freedom (SDOF)Converting stressesIsolation

7. Vibration

7. 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

7. Hardware 97

Effect of vibrationVibration can cause the following

problems• Structural failures• Reduced life• Pointing and stabilization errors• Stress

7. Vibration

7. 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

7. 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

7. 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

7. 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)

7. Hardware 102

Damping coefficient, Damping coefficient,

SDOF

= damping coefficient = c/cc

c = cc = 2 sqrt(km) = 2 n

7. Vibration

7. 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

7. 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

7. 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

7. 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

7. 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

7. 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

7. Hardware 109

8. Fatigue

Fatigue failuresS-N curveAccelerated life

8. Fatigue

7. 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

7. 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

7. 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

7. 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

7. 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

7. Hardware 115

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

7. Hardware 116


= 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

7. Hardware 117


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

7. Hardware 118

9. Thermal loading

Coefficient of expansion (COE)Thermal stressesFailures

9. Thermal loading

7. Hardware 119

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

7. 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

7. Hardware 121

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

7. 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

7. Hardware 123

Failures

Circuit cards can have thermal failures if the COE of the card material doesn’t match what’s on the card

Objects can distort wildly if they are made of materials with different COEs

9. Thermal loading

7. hardware1 agenda r1. areas r2. stress and strain r3. axial loading r4. torsion loading r5. beam...

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