It was the world’s worst airline disaster involving people on the groundon the ground. Eight minutes after the Boeing 747 cargo plane, EL AL Flight 1862, lost g ,two engines, it crashed into a block of flats inblock of flats in Amsterdam. Was anything l t f thlearnt from the tragedy?

Fact File:

th

Fact File:

WHEN: October 4th 1992

WHAT: A Boeing 747-200, EL AL flight 1862, crashes into a 12-storey flat in a residential area in a suburb of Amsterdam minutes afterarea in a suburb of Amsterdam, minutes after taking off from Schiphol airport. At least 47 people are killed including all three crewpeople are killed, including all three crew members of the cargo plane, and 26 injured.

El Al flight 1862, a Boeing 747-200 freighter, with 3 crewmembers and one passenger on board took off fromcrewmembers and one passenger on board, took off from runway 01L at Schiphol Airport, Amsterdam.

f f7 minutes after , engine 3 and its pylon separated from aircraft, damaged parts of leading edge of right wing. Engine 3 then struck engine 4, causing this engine and its g g , g gpylon to depart from the wing. Both engines fell into a lake.

In general, modern aircraft have adequate control capability with only 2 engine functioning. El Al could have been too heavy to maintain straight and level flight with both engines gone.gone.

Due to damage on hydraulic system, flaps on right wing did t i Thi lt d i th l t i i ht Ai ftnot raise. This resulted in the plane turning right. Aircraft was

in a gradual descent.

Aircraft crashed into an 11-floor apartment building in a suburb of Amsterdam, 13 km east of Schiphol. Impact was centered at apex of two connected and angled blocks of apartments. Fragments of aircraft and building were p g gscattered over an area of 400x600 m.

…the aftermath…..

Fatigueg

Static or quasi-static loading in i i ti iengineering practice is rare.

More common and realistic to have repeated, fluctuating and rapidly applied l d lti i f il b f tiloads, resulting in failure by fatigue.

Fatigue failure occurs as a brittle fracture produced by cyclic stresses usually b l l ti li it f t i lbelow elastic limit of materials.

Unlike biological fatigue, damage here is cumulative and unrecoverable even when load is removed.

Fatigue consideration is important because:

(1) 80% - 90% of service failures in engineering components arise from fatigue.g g p g

(2) Strength and shape of machine elements ( ) g punder fatigue loading must be determined on basis of fatigue performance of the material, not on static propertiesnot on static properties.

N t f f tiNature of fatigue

Characterized as a progressive failure phenomenon; proceeds by initiation, propagation of cracks to unstable size and finally failure.

Process may be as follows:

Extrusion

Intrusion

Fracture surface of a typical ductileFracture surface of a typical ductile metal subjected to cyclic loads

Turbine shaft: fracture startedstarted from welding-arc gstrike A, propagated along Balong B, final failure at C

Austenitic steel Nimonic steel

Ductile striations

Al-alloy (7178) Inconel 718

1 striation marking corresponds to 1 cycle of fatigue loadingfatigue loading

Fatigue Loading Zero mean stress

(completely reversed)

C t t lit d Constant amplitude Fixed frequency

σmax

σa

σm = 0Time

σ i

σmin

σmax = maximum stress in the cycle

σ i = minimum stress in the cycle σmin = minimum stress in the cycle

σm = mean stress = ½(σmax+ σmin)

σa = alternating stress amplitude = ½(σ σ )= ½(σmax - σmin)

= range of stress = σmax - σming max min

R = stress ratio = σmin / σmax

Stress

Time

More realistic stress time spectrum Such More realistic stress time spectrum. Such quasi-random pattern may be encountered in refueling, taxi, take off, gusts, maneuversin refueling, taxi, take off, gusts, maneuvers and landing of aircrafts.

Laboratory fatigue testingy g g

StressFerrous alloys and titanium

Stress

Non-ferrous alloys

Endurance limit σEndurance limit, σe

Fatigue strength at N cycles

N >107 cycles

Number of cycles to failure

When no better information is available: σ = 0 5 σ (f t l ith 1400MP d BHN 400) σe = 0.5 σult (for steels with σult <1400MPa and BHN <400)

σe = 700MPa (for steels with σult >1400MPa)

σ = 0 4 σ lt (for cast steel and cast iron) σe 0.4 σult (for cast steel and cast iron)

σe = 0.38σult (for magnesium alloys)e ult ( g y )

σe = 0.45σult (for nickel or copper based alloys)

σ = 0 38σ (f l i i ll t σe = 0.38σult (for aluminium alloys up to a tensile strength = 280MPa)

σe = 0.16σult (for cast aluminium alloys up to tensile strenght = 350MPa)g )

Endurance limit for reversed axial loading du a ce o e e sed a a oad g(for polished, un-notched specimen) = 15% lower that for reversed bending. % gHence, for steel:

σe (axial) = 0.85 x 0.5σult = 0.43σult

For reversed torsional loading of polished un notched steel specimens:un-notched steel specimens:

σ = 0 58σ (bending) = 0 29σσe = 0.58σe (bending) = 0.29σult

Many factors influence fatigue strength of materials:

Material Effects: Chemical composition; material variation*; size and shape*; under-stressing/over-stressing etcstressing, etc.

Manufacturing Effects: Fretting fatigue; Fretting g g g gcorrosion; heat treatment; method of manufacturing*; stress concentration*; surface treatments*, etc.

Environmental Effects: Corrosion; superimposed static stress*; temperature*; varying amplitudes*; exposure to nuclear radiation etcnuclear radiation, etc.

Miscellaneous Effects: Surface fatigue; combined stresses*; etc.

Experimental data

Experimental data

Gerber parabolae

Goodman line

Soderberg line

00 yp ultmean

For general use under high-cycleFor general use under high cycle fatigue, Goodman relationship (or modified Goodman range-of-stress diagram) is recommended.

For ease of use and safer design, Soderberg’s linear relationship may be employedemployed.

Note that both relationships are linearNote that both relationships are linear and hence easier for calculations.

Experimental datae

e

Soderberg lineFS

Safe

0

region

FSyp

Working stress lineyp ult

mean

Soderberg working stress line:Soderberg working stress line:g gg g

K a

e

ypfm

yp KFS

e

Uncertainty is accounted for by using factor of Uncertainty is accounted for by using factor of safety (FS).

Depending upon the factor of safety used, Soderberg’s approach may be unnecessarily

ticonservative.

Stress concentration factor:

6070

50

40

30

20

105 106 107

80

6060

40

20

0105 107106104 10 101010

St t ti f tStress concentration factor:

Holes, grooves, notches, shape changes, key ways, interference fits, etc give stress y gconcentration.

Define a theoretical or geometrical stress concentration factor, Kt as:

Kt = Actual stress / Nominal stress

s

Str

ess

Stre

ss

omin

al

Act

ual S

N A

= Point of maximum stress

Theoretical stress-concentration Theoretical stress-concentrationTheoretical stress concentration factor for a flat bar with shoulder fillet in tension

Theoretical stress-concentration factor for a round bar with shoulder fillet in tension

Theoretical stress-concentration Theoretical stress-concentrationTheoretical stress concentration factor for a shaft with shoulder fillet in bending.

Theoretical stress-concentration factor for a grooved shaft in bending.

For ductile materials, static loading:no need to consider stressno need to consider stress concentration factor.

For brittle materials, fatigue loading:stress concentration factor must be considered.

Under fatigue and dynamic loading, define fatigue stress concentration factor:

Kf = (Endurance limit without stress t ti ) / (E d li itconcentration) / (Endurance limit

with stress concentration)

To relate Kf to Kt , use notch sensitivity factor, q:

q = (Kf - 1) / (Kt – 1)

When q = 0, Kf = 1, material is not notch sensitive.

When q = 1, Kf = Kt , material is fully notch sensitive.

In case of doubt in design, always make Kf = Kt(i 1) t i ti d i(ie., q = 1) to give conservative design.

For reversed bending or reversed axial loads

N h i i i h f l d 2024 T hNotch-sensitivity charts for steels and 2024-T wrought aluminium alloys. For large notch radii, use q for r = 4mm.

For reversed torsion loads.

N t h iti it f t i l i dNotch-sensitivity curves for materials in reversed torsion. For large notch radii, use q for r = 4mm.

Example:

Fi d th i d f th f ti Find the area required for the safe continuous operation of a uniform bar in tension if Pmax = 220 kN and Pmin = 90 kN. The material used has σult =kN and Pmin 90 kN. The material used has σult 620 MPa and σyp = 410 MPa. Take the factor of safety to be 1.5 and let σe = ½ σult .

What would the area be if the bar is not uniform in cross section but has two widths as shown?in cross-section but has two widths as shown?

r

PP D dD/d = 2

/d 0 23r/d = 0.23

Solution:220

σe = ½ σult = 310 MPa220

Pm = 155 kN, Pa = 65 kN

σ = 155/A σ = 65/A90

σm = 155/A σa = 65/A

ypfyp K

310654101155410 N

ae

mFS

22

3

101065

3104101155

5.1410

cmN

AA

A = Area of section = 8.82 cm2

Solution:

For shouldered plate:

D/d = 2 and r/d = 0.23

Kf = Kt = 1.75

3106541075.1155410 N

22103105.1 cmAA

A = Area of smaller section = 11.17 cm2

Soderberg equation may be written in different fforms.

PKP afyp

m

)( For axial loading:

FS

Ayp

e

FSwhere P = load

A = cross-sectional area

For simple bending:MKM

I afe

ypm

)(

For simple bending:

FSC yp

where I = moment of inertiaM = bending momentM = bending moment C = distance from neutral axis to extreme fibre

TKTJ af

ypm

)(

For simple torsion:

FSCJ

yp

fe

Which using Tresca’s criterion:

FS

yp21

TKT yp )(

yp2

TKT

CJ

yp

afe

ypm

)2/(

)(

FS

Design of shafts

BearingShaft

Bearing

Gear

Sectional view of a gear box.

Note how shafts are shafts are supported.

Note where and how bearings bearings and gears are used

Application of Soderberg approach to design of shafts:

For combined bending and torsion of circular shaft, Soderberg equation may be written as:shaft, Soderberg equation may be written as:

J

22

yp C

J

FS

22

af

e

ypmaf

e

ypm TKTMKM

For the general case of hollow shaft:

)1(16

434

324

32R

oio DDD

DDCJ

162

oDC

where ratio of diameters, DR = inner diameter, Di / outer diameter, Do

Hence:

R , i , o

2234 )1( afmafm

oR TKTMKMFSDD ypyp

16

)( afmafmyp

R ee

Most shafts carry steady torque and load is fixed in magnitude and direction thenfixed in magnitude and direction, then

Mm = 0, Ma = M, Tm = T, and Ta = 0Mm 0, Ma M, Tm T, and Ta 0

H Hence:

23 FSD 22

4 )1(16TMK

DFSD

fRyp

oe

yp

22

4

3

)1(32TMK

DFSD

fo yp

4 )1(32 D fRyp

e

F

FR Torque does not vary, FR

F

q y,ie., non-fatigue, hence Ta = 0, Tm = T

FT

Completely reversed case: σm = 0, hence Mm = 0, Ma = applied bending moment, M

Most shafts carry steady torque and load is fixed in magnitude and direction thenfixed in magnitude and direction, then

Mm = 0, Ma = M, Tm = T, and Ta = 0Mm 0, Ma M, Tm T, and Ta 0

H Hence:

23 FSD 22

4 )1(16TMK

DFSD

fRyp

oe

yp

22

4

3

)1(32TMK

DFSD

fo yp

4 )1(32 D fRyp

e

Most shafts are solid. For hollow shaft, suggest:

D = D /D = 0 6DR = Di /Do = 0.6

Hollow shaft has advantage of :

Lighter than solid shaft of similar diameterdiameter

H t t t t i ifHeat treatment is more uniform

Shape of end keyway is important to strength f h ftof shaft.

For press fitted members, in absence of better information, Kf = 2.50.

A centrifugal blower rotating at 600 rpm is to

Example: A centrifugal blower rotating at 600 rpm is to

have a capacity of 12m3/sec. A belt drive is to be used to connect the blower to a 15 kW 1750be used to connect the blower to a 15 kW, 1750 rpm electric motor. The blower shaft is to be machined from hot-rolled 1020 steel in which σyp = 300 MPa, σe = 180 MPa and σult = 500 MPa. The only part of the shaft under significant t i th h t hi h th llstress is the overhung part on which the pulley

is mounted. The belt forces resolve into torque of 240 Nm and a force of 2 5 kN on the shaftof 240 Nm and a force of 2.5 kN on the shaft. The figure shows the location of the bearings, the steps in the shaft and the plane in which the p presultant belt force and torque act.

The ratio of the journal diameter to the overhung h ft di t i t b i t l 1 2 ishaft diameter is to be approximately 1.2, ie.,

D/d = 1.2 and the fillet is to be a full fillet, ie., r = (D-d)/2 = 0 1d Specify the dimensions of d D(D-d)/2 = 0.1d. Specify the dimensions of d, D and r for the shaft having a factor of safety of 3.

100

2510

25

Di t f h ft i i d l it l thDiameter of shaft is varied along its length so that different standard components, like bearings could be accommodatedbearings, could be accommodated.

Gi F S 3 300 MP

Solution: Given: F.S. = 3, σyp = 300 MPa

σe = 180 MPa DR = 0T = 240 Nm Belt force = 2 5 kNT = 240 Nm Belt force = 2.5 kN

There are two points at which failure may occur: There are two points at which failure may occur:

(a) at the end of the keyway, M = 2.5 kN x 65x10-3 m to the keyway

100

10

25

(b) at the shoulder fillet,M = 2.5 kN x 75x10-3 m to the fillet.

100

25

10

25

From Soderberg equation:

23 FSD 22

4

3

)1(32TMK

DFSD

fRyp

oe

yp

Si ll t b id K d M t t Since all terms beside Kf and M are constant, Do

3 is proportional to KfM.

The critical section will be the one with larger KfM The critical section will be the one with larger KfM.

100

25Which section has larger KfM ?10

25g f

( )(b)

Si t h iti it f t i d d t

(a)(b)

Since notch-sensitivity factor q is dependent upon absolute dimensions of notch which are unknown only basis for comparison will be K Munknown, only basis for comparison will be KtM.

Once dimensions are determined, design can , gthen be checked using Kf and q factor.

100

2510

25

( )(b)

At d f kAt h ld fill t

(a)(b)

At end of keyway:

K M = 1 6 x 2500 x

At shoulder fillet:

K M = 1 62 x 2500 x KtM = 1.6 x 2500 x 65x10-3

KtM = 1.62 x 2500 x 75x10-3

= 260 Nm= 303.75 Nm

For D/d = 1.2 and r/d = 0.1and r/d 0.1

Kt = 1.62

Theoretical stress-concentrationTheoretical stress concentration factor for a shaft with shoulder fillet in bending.

Since KtM at shoulder fillet is larger (303.75 Nm) design will be done by consideringNm), design will be done by considering shoulder fillet as the critical section.

Using Soderberg equation:

3

223

32TMKFSd

fyp

e

yp

yp

223003

240753033

d 180300

6 24075.3031030032

d = 3.85 x 10-2 m = 38.5 mm

From the calculated d, r and D can now be found:

r = 0.1d = 3.85 mmD = 1.2d = 46.2 mm

With r known possible now to determine With r known, possible now to determine notch sensitivity factor q.

q = 0.80

For σult = 500 MPaand r = 3 85 mm

r = 3.85 mm

and r = 3.85 mm

Since q = (K - 1) / (K – 1) Since q = (Kf - 1) / (Kt – 1)

Kf = 1 + 0.80 (1.62 –1) = 1.50f ( )

Summary: Kf = 1.50 and Kt = 1.62Summary: Kf 1.50 and Kt 1.62

If desired a new value of d may be calculated If desired, a new value of d may be calculated using Kf = 1.50; but in view of the fact that any error introduced by using Kt instead of Kfy y g t ferrs on the safe side, redesign may not be necessary.

For those who are interested to check:

KfM = 1.50 x 2500 x 75x10-3 = 281.25 Nm (N t thi i till l th K M f k f(Note, this is still larger than KfM for keyway of 260 Nm, ie., shoulder is still critical section)

Using Soderberg equation:

22180300

6

3

24025.28110300

332

d 18061030032

d 3 77 10 2 37 7 C t d = 3.77 x 10-2 m = 37.7 mm. Compare to 38.5 mm, the difference may be small (2%).

Example:

The railroad freight-car axle is being designed to carry a rated load of 110kN. g yThe axle is to be turned from a normalized 1040 steel forging in which σyp = 340 MPa g g ypand σe = 260 MPa. If d2 /d1 = 1.25 and d3/d2 = 1.1, determine the various dimensions of the axle.

Bearing, d1 d2 d3 Center line of track

250 50 100 600250 50 100 600

Rail

Fillet radius = 19

Wheel press fitted onto axleRail p

Only half of freight-gcar is shown

Solution: Given: σyp = 340 MPa σe = 260 MPa

d2 /d1 = 1.25 d3 /d2 = 1.1

r 19 mmr = 19 mm

Load acting on one wheel = 110 / 2 = 55 kNLoad acting on one wheel 110 / 2 55 kN

Due to uncertain environmental conditions and Due to uncertain environmental conditions and danger to human life and costly property damage in case of failure, a large factor of g , gsafety is used. Take FS = 4.0.

Even though d2 /d1 and d3 /d2 ratios are known r/d ratio is not and so K cannot beknown, r/d ratio is not and so Kt cannot be determined.

It is necessary to assume a trial shaft diameter; then to check the assumption used.

d1 d2 d355 kN

A BA

175

125

55 kN

225

55 kN

(a) Consider section at A:

T = 0; M = 55x103x125 Nmm

Assume d1 = 75 mm, then r/d = 19/75 = 0.25; and D/d = d2 /d1 = 1.25; this gives Kt = 1.33.

If q = 1, then Kf = 1.33.

125105533.1 331

MKd f

426032

FSe

d1 = 112 16 mm d1 = 112.16 mm

This is different from assumed value of 75 mm.

Second assumption is necessary.

Assume d1 = 115 mm, then r/d = 19/115 = 0 165; and D/d = d /d = 1 25; this gives K =0.165; and D/d = d2 /d1 = 1.25; this gives Kt = 1.45.

260125105545.1

32

331

MKde

f

4FSe

d1 = 116.03 mm (this gives Kt ≈ 1.45)

This is close to assumed value of 115 mm. Not This is close to assumed value of 115 mm. Not necessary to re-calculate. Hence d2 = 1.25d1 = 145.04 mm.

(b) Consider section at B:(b) Consider section at B:

T = 0; M = 55x103x175 Nmm

Assume d2 = 145 mm, then r/d = 19/145 = 20.131; and D/d = d3 /d2 = 1.10; this gives Kf = 1.50.

33

260175105550.1

32

332

MKde

f

4FS

d2 = 131.27 mm (this gives Kf ≈ 1.47)

d2 here is less than d2 in part (a) of 145.04 mm. Hence, d2 here is not critical. , 2

d3 = 1 1d2 = 1 1 x 145 = 159 5 mm d3 1.1d2 1.1 x 145 159.5 mm.

(c) Consider section at wheel seat:

T = 0; M = 55x103x225 Nmm

Because wheel is press fitted onto axle, Kf = 2 5 Therefore:2.5. Therefore:

225105552 33 MKd

4260

22510555.232

333

FS

MKde

f

4FS

d3 = 169 24 mm d3 = 169.24 mm

Compared with part (b), it can be seen that d3 = 169.24 mm is the critical dimension, and NOT 159.5 mm from part (b).

Final design based on d3 = 169.24 mm :

d3 = 169.24 mm,

d2 = d3 /1.1 = 153.85 mm

d1 = d2 /1.25 = 123.08 mm