Mechanical Engineering Design Project

MECH 390

Lecture 5

Multiple Random Variable• Formulation 3 - Considering more variations of the rod

• Design variable A is deterministic

• P and Y are the random variables P = 20000, SP = 2000

• In formulation 4, all A, Y and P are considered random variables

• Simplified formulations of mean and standard deviation of multiple random variables for

common relationships from appendix 5A

Multiple Random Variable• Concentrated load p is applied to cantilever at angle

• Both are normally distributed with Pmean and SP ; mean

and S

• express the SD of bending moment in terms of Pmean

SP mean S and L

• M = (P cos ) L

• From table

• SM = L ((Pmean2(Scos)

2 + Cos mean2 SP

2)1/2

• Using equations in appendix we can get Scos =

SinmeanS; when = mean

• SM = L ((Pmean2(Sin2mean * S

2 + Cos2mean

SP2)1/2

𝑆𝑐𝑜𝑠 =𝑑 𝑐𝑜𝑠𝑑 𝑚𝑒𝑎𝑛

𝑋 𝑆in appendix 5A 𝑆𝑌 =𝑑 𝑦

𝑑𝑥 𝑥 𝑥𝑚𝑒𝑎𝑛𝑋 𝑆𝑥 so

Determining probability of failure• Formulation 3 - Considering more variations of the rod

• Design variable A is deterministic at 0.5in2

• P and Y (mean and SD are 50000 and 6000) are the random variables P = 20000, SP = 2000

• Two random variables with distributions as shown in figure. Simplifying this by

introducing new variable which represent both distributions

• For A = 0.5

Determining probability of failure• We introduce a random variable Q as Q = y-

• Q is the margin of safety implying failure when Q = 0 or Q > 0 is safe

• Since both are normal distribution

• Substituting the values

Determining probability of failure

Designing to a Specified probability of failure• Formulation 3 - Considering more variations of the rod

• P and Y (mean and SD are 50000 and 6000) are the random variables P = 20000, SP = 2000

• What deterministic area A will give a probability of failure less than .005

• Let us use the Q

• Pr (z < z*) = .005

• From the table, we can see that z* = -2.575

• Substituting the values of mean and SD of

y and p values we can get A as

• Substituting the values of both A in to find

which one is appropriate, we get A as

0.617

Designing to a Specified probability of failure• Formulation 4 - Considering more variations of the rod

• P and Y (mean and SD are 50000 and 6000) are the random variables P = 20000, SP = 2000

• A will also be a random variable hence mean and S can be

• mentioned in terms of mean and Sd of P and A

• Substituting this in the equation

Designing to a Specified probability of failure• If standard deviation SA is assumed as KAmean

where K is a numerical constant (defined in

tolerance)

• For ymean of 50000, Sy of 6000, Pmean of 20000

and SP of 2000; with K as 0.1 we get

• Amean as 0.237 and 0.647

• Substituting both values in the equation and

finding which one gives the positive results,

• We get A as 0.647 in2

Designing to a Specified probability of failure• DC circuit shown, resistance 10±1ohms

• Mean value of volt Emean=120V

• What should be the tolerance on voltage such that there is

less than 1.2% chance the current will exceed 13Amps (all

variables are normally distributed

• We have to Select SE such that Pr (I>13) = .012

• Z* = (I*-Imean)/SI

• Imean = Emean/Rmean = 120/10 = 12 Amps

• I* = 13 Amps

• Pr(Z<-Z*)=.012; -Z* from the table is -2.25

Designing to a Specified probability of failure• Pr(Z<-Z*)=.012; -Z* from the table is -2.25

• Z* = (I*-Imean)/SI ; SI = (I*-Imean)/Z*

• SI = 13-12/2.25 = 0.444Amps

• From SI we can find SE

• SE = (Imean2SR

2 + Rmean2SI

2)1/2 (E=IR)

• Tolerance on Resistance is 1 ohm and SR can be

1/3rd of the that as .33ohm

• SE2 = 122*0.332 + 102*0.442

• SE = 5.91 Volts; so tolerance on Volts is 5.91*3 =

17.7

• Hence the required voltage E = 120±17.7V

Reliability• Probability of failure under tension independent of time was considered

• More like static analysis

• We will look at failure when load is time dependent as in

• Piston on automobile engine

• Bridge due to vehicular traffic

• Electrical connection due thermal expansion etc

• In these, time becomes a factor and an independent variable for analysis

• Number of miles on a tire

• Number of Take off and landing cycles on a flight

• System failing in time as integral in probability density function

Reliability• It can be simplified by using the complement of failure

probability as the dependent variable

• Reliability as a component of failure probability R(t)

• The probability that the system not failing by time t

• Both failing by time and not failing by time covers all

probabilities we can write

• R(t) + F(t) = 1

• We saw adding random variables complicate probabilistic

determination, if we add time, it will complicate further.

• So empirical methods are often used. (experience, test data

or other analytical approach)

Per Unit failure rate• How numerical R(t) can be determined

• P1 & P2 are two populations of transistors

• P1 has 10000 units and fails at 50/hour

• P2 has 1000 units of the same kind and experience same service

• Since number of P2 is 1/10th of P1, we expect 5/hour to fail

• If we mention the same thing without using size of the population as a descriptor, it

would be great

• Both P1 and P2 can be expressed in a per unit failure rate of .005

• The test can be run simultaneously on P1 and P2 or P1 first while removing the

failures, the sample size reduces to P2

Per Unit failure rate• This can used to determine the per unit failure rate h(t) as the ratio of the rate at which

failures occurring at time t to the size of the population at time t

• Expressed as

• Where Nf (t) and Ns (t) are numbers failed and survived at time t respectively

• From the appendix 5B, we can rewrite the equation for reliability as

• F(t) is Nf(t)/N0 and R(t) is Ns(t)/N0

• Where N0 is the initial sample volume

• This allows us to define reliability as a function of per unit failure rate

Per Unit failure rate

• For most products the per unit failure rate h(t) consists of 3 phases known

as the bathtub curve

• Early period

• h(t) reduces due to

improvements in

design/mfg

• Mature period

• h(t) is only due

random failures

• Wear out period

• h(t) increases due

to wear

Per Unit failure rate

• Our interest is in mature life where h(t) is relatively constant

• Let h(t) = where is a constant found from test data or empirical evidence

• Equation reduces to

• For a system with constant h(t) reliability reduces exponentially over time

Mean Time to failure

• MTTF is the average time object will be in service before failure occurs or

• Since t cannot be negative, it can be reflected in terms of reliability as

• For a constant failure rate of , MTTF = 1/

• For systems with constant pre failure rate, MTTF is inverse of the failure

rate

• For the problem with = 2X10-3 the mttf is

System Reliability

• Reliability is analyzed for single component

• But most system consists of multiple components, and each having their

own reliability

• What is the relationship between individual component reliability to the

overall system reliability?

• Depends how the components are connected to form the system

• Systems in series

• Systems in parallel

System Reliability

System Reliability – Series systems

• Electrically speaking, failure of single components fails the entire system

• For series system of n components the reliability is

• Each component reliability is less than one, and system reliability is always

lesser than the least reliable component

• In a special situation where If each component i has h(t)i = i

• which simplifies to

• where system per unit failure rate is defined as

• Constant which is sum of per unit failure rates of all the components in the system

System Reliability – Parallel systems• Electrically speaking, failure of system happens if every component fails

• For parallel system of n components the reliability is

• Two component system with constant per unit failure rates

• expands to

• Parallel systems have variable failure rates even if components have constant failure rate

• Reliability of parallel system is greater than the reliability of the least reliable component

in the system

• Though, parallel systems are not preferred due to high cost

• It is justified to add redundancies (parallel systems) is only when cost of failure is high

• Sometimes, safety regulations also require certain redundancies

23

Reliability Function

•Probability density function of failures f(t) = e-t for t > 0

•Probability of failure from (0, T)

F(t) = 1 – e-T

•Reliability function

R(T) = 1 – F(T) = e-T

24

Series Systems

RS = R1 R2 ... Rn

1 2 n

25

Parallel Systems

RS = 1 - (1 - R1) (1 - R2)... (1 - Rn)

1

2

n

26

Series-Parallel Systems

• Convert to equivalent series system

A B

C

C

D

RA RB RC

RD

RC

A B C’ D

RA RB RD

RC’ = 1 – (1-RC)(1-RC)

Closure • Probabilistic design allows us to design systems to any desired probability of failure

• Design can account for randomness in loading, material properties, dimensions, etc

• More sophisticated than FOS

• But complicated even for simple rod that we studies. It will become more difficult as problem complexity increases

• Several assumptions regarding standard deviation, normal distribution etc…

Example• Aluminum tube of d=1.5” and L=20” used as simply supported column subject to axial

load. Pmean=1000lb and SP=50lb. Find the thickness of tube so that probability of buckling is less than.0001? Tolerance on the wall thickness is ±.001”; both d and L are deterministic. Assume column as thin walled tube

• For thin walled tubes (t<<d), the moment of inertia

• Buckling occurs when the critical load reaches

• Substituting the value of I we get

• Pcr = C. t where

• Given the values of E, d and L we can find

• C=3.27X105

Example• There are two random variables in this equation P and t with each probability distribution.

• So we define a random variable whose function is a combination of P and t

• Q = Pcr- P where Q<0 represent failure

• So we need to choose t such that Pr(Q<0)=.0001

• Writing the Q in standard normal form of Z

• Express Qmean and SQ in terms of P and t

• Qmean = Pcr mean – Pmean

• Qmean = C tmean – Pmean

Example• Substituting the values of Qmean and SQ in the equation of Z

• For Pr (Q<0)= .0001; Z from the table is -3.72

• Only unknown here it tmean

• Substituting the numerical values of C, St, SP, and Pmean we get

• tmean = 4.42*10-3” C=3.27X105

Example• A two bar symmetric pin connected truss shown is subject to

nominally vertical load of 1000lb acting at point A. misalignment of load occurs randomly about the vertical axis of symmetry as a normal distribution with a mean of mean=0° and S =10°. Select a CSA such that the probability of truss failure is .001 (Steel with Y=30,000psi)

• Forces on the truss in terms of misalignment

Example• Adding these

• We get

• Analyzing only one side due to symmetry,

• Introducing the new variable Q such that

• So the standard variable Z is

• Probability of failure from table Z = -3.09 (from equation of probability density or table)

• This also means Q=0 (for failure)

Example• This also means Q=0 (for failure); so

• But we know

• And mean of Y is 30000psi

• We know from

• So we can rewrite the Z equation as

• Since SY is deterministic, we can find SQ (or SQ2 due to symmetry)

• From the appendix we know that S(sin+cos )=S at =0 which is the mean

• So

Example• Substituting this value of SQ

• in this equation

• we get

• Solving for Cross section Area A gives

• A= 0.0363in2

Example• Steel (E= 30X106psi) cantilevel of L=10ft is required to support a concentrated load on free

end with Pmean= 500lb and SP=100lb. If the CSA is rectangular with width (b) of 2” design depth (h), such that the probability of deflection less than 3” is 99.9%

• Given

• Required

• Deflection of the cantilever at free end is

• So deflection is

• where

• C can be calculated from the given values of L E and b as 0.1152

Example• For deflection, there is only one random variable P and we need to select h

such that

• Converting to the standard normal variable

• We can express mean and S in terms of Pmean and SP

• Substituting this in the Z equation we get

• From the table we can get the value of Z as 3.09 when is max

Example• Using the value of Z as 3.09 and

• Solving for h we get

• Knowing Pmean as 500lb, SP as 100lb, C calculated as 0.1152 and max as 3 in

• We can find h as 3.14 in

Example• A straight circular rod (E = 30X106psi) subject to tensile load P. Both P and L are normally

distributed with Pmean SP; Lmean and SL. Find an expression for CS area A of the rod such that the probability of rod stretching more than 0.5” under load is 1/100. Assume SA=.05Amean

• Given where P, L and A are random variables. We need to select Amean

• Such that

• Introducing the Z variable then

• So the probability of Pr(Z<Zallow) = .01

• From table, Zallow is -2.33 so

Example• Right hand side of the equation can be rewritten using appendix as

• Using the table for random variables we can rewrite this as

• Since there are three variables, we can deal with P/A as

• Then from table for random variables we get

• Substituting and into the equation below

• We get

Example• We know from table 5.2 and

• We also know that SA is K*Amean so

• Substituting these values in the equation gives

• This can be solved to give and Amean as

Example• A simplified model of thermal power plant is shown in the figure with per unit failure rate

of individual components. What is the reliability of the plant at 1000 hours

• Given

• We can calculate

Feed water pumps

10-4hr-1

Boilers

10-3hr-1

Steam turbines

5*10-3hr-1

Electric Generators

10-5hr-1

Example• A gearbox is suppose to have a reliability of 99.98%. 3 vendors offer gear sets that meets

the functions of the gear box. The reliabilities and costs are tabulated.

• What combination of gear sets will achieve the reliability at the least cost. You can use as many gear sets from each vendor or combination of vendors as needed. Gears are arranged in Parallel

• The system reliability of gearbox with n sets is

• R=1-(1-R1) (1-R2)…. (1-Rn)

• Explore 2 sets, 3 sets, and then more until least cost option is achieved

Vendor Reliability Cost

A 0.95 50

B 0.93 40

C 0.90 35

System RA RB RB RSystem CostA CostB CostC CostSystem

A+A 0.9975 100

A+A+A 0.9999 150

A+A+B 0.9998 140

A+A+C 0.9998 135

A+B+B 0.9998 130

A+B+C 0.9997 125

Example

• For options involving two gear sets, the most reliable combination does not meet the 99.98% reliability, so no need to try other combinations

• For 3 gear sets, option 2 has highest reliability, but we progressively reduce cost, until, the moste reliable and cheapest option is achieved in option 5

• Going for 4 gear sets, requires at least a cost of 140 (35X4), which is expensive than the reliable 3 gear set option that we secured.

System RA RB RB RSystem CostA CostB CostC CostSystem

A+A 0.9975 100

A+A+A 0.9999 150

A+A+B 0.9998 140

A+A+C 0.9998 135

A+B+B 0.9998 130

A+B+C 0.9997 125