2010 Construct 1 Exercise Answers 100413

8/6/2019 2010 Construct 1 Exercise Answers 100413

1/29

1

Answers

Exam Exercise

Aircraft structures

Constructieleer II

2011 Marcel van Varik


2/29

2

Failure modes


3/29

3

Qu

stions

Follo

ing qu

stions

n p

f

bly b

ns

d

o ding to

thods in Niu (op

n boo

x

)

2A, ns

Niu, CH 9.0, (B)

s

(Axi l lo d)

2B, ns

Niu, CH 9.0, (C) s

(T ns

s

lo d)

2C, ns

Niu, CH 9.0, (D) s

(Obliqu

lo d)

Oth

is

h

ft

lt

n ti

s

p

s

nt

d

2A Di

nsioning lug

ith xi l lo d.

Lug: t1 = 0.25 in, t2 = 0.5 in, a = 0.75 in, W = 1.5 in

Mat 2024-T351(Niu, fig 4.3.3) 3.000-4.000 in, A-value,L-dir

Ftu = 57000 psi

Fty = 43000 psi

Fsu = 34000 psi

Fbru = 106000 psi (e/D = 2.0)

Fbry = 84000 psi (e/D = 2.0)

Pax = 15000 lbs (ult load!!)

Bush: D= 0.5 in, Db = 0.65 in

Mat Clad 2024-T3 (Niu, fig 4.3.3)Fsu = 41000 psi (B-value)

Pin(Niu, fig 4.3.6)

Ftu = 160000 psi

i. lo tion of f ilu od s S1 = lug t nsion f ilu

ulti

t (n tt st

ngth)

S2 = lug t nsion f ilu

yi ld (n tt st

ngth)

S3 = lug sh

f ilu

ulti

t

S 4 = b

ing p

ssu

ulti

t

S 5 = b ing p ssu yi ld


4/29

4

ii Calculation of M onl 4 required!

M 1 = lu

tension failure ultimate nett stren

th

M ! 2 = lu"

tension failure#

ield $ nett stren"

th%

M & 3 = lu'

shear failure ultimate

She( )

line: a = too long and conse ) vativeestimate is r = a D/2

A better way is to draw to e0

act dimensions and meas 1 re the length.

M 2 4 = lu3

bearin3

pressure ultimate

M 4 5 = lu5

bearin5

pressure6

ield

Failure mode location Units Applied(Limit)

Allow SF MS

MS1 lug tension failure ultimate psi 23530 57000 1.5, 1.15 0.41

MS2 lug tension failure yield psi 23530 43000 1.0, 1.15 0.59

MS3 lug shear failure ultimate psi 23530 34000 1.5, 1.15 -0.17

MS4 lug bearing pressure ultimate psi 30769 106000 1.5, 1.15 1.00

MS5 lug bearing pressure yield psi 30769 84000 1.0, 1.15 1.37


5/29

5

2B 7 imensionin8

lu8

with transverse load 9

See book.

2C 7 imensionin8

bolt in lu8

9

Lu8

: t1=0.25 in, t2=0.5 in

Mat 2024-@

3(Niu, fig 4.3.3A 3.000-4.000 in, A-value, L-dir

a =0.75 in, W = 1.5 in

Bush: D =0.5 in (inside B , Db=0.65 in (outside B ,

Mat 7075 T62024-T3C

Niu D fiE

4F 3 F 5G

Fsu =44000 psi (B-value B

H in AII I-4340C

Niu D fiE

4F 3F 6G

P

-valueC

nothinE

else availableG

Ftu = 260000 psi

Fty = 215000 psi

Fcy = 240000 psi

Fsu = 156000 psi

D0=0.5 in (outside B , Di =0.28 in (inside B

Q ax= 15000 lbs (ultimate load!!)

(note: Niu uses also ult loads)

=0.01 in

i 9 Location and failure modes

MS1: pin tensileyield strength (due to bending)

MS2: pin shear strength ult

MS3: pin tensile ult strength (due to bending)


6/29

6

ii R VLS

iii T M U for relevant locations and failure modes V Calculation of bendin

W

moment V

(ultimate load) (Niu, page327 and 328)

Calculation of stresses due to bendinX

Y ult load

Where:

(Niu, Appendix C 8)

M a 1: pin tensileb

ield strenc

th d due to bendinc

e

M f 2: pin shear ult streng

th

(Niu, Appendix C 8)


7/29

7

M h 3: pin tensile ult streni

th p due to bendini

q


Allow SF MS

MS1 pin tensile yield stress bending psi 169675 215000 1.0, 1.15 0.10

MS2 pin shear stress ult psi 37092 156000 1.5, 1.15 1.43

MS3 pin tensile ultimate stress bending psi 169675 260000 1.5, 1.15 -0.41


8/29

8

3A D r tr s t

inr t

init u

l r xpr v

t r d lifr ofw

hr r

l u xis

Miners rule

otherwise structure fails due to fatigue

N1*, N2* = number of allowable cyclic loads.

Safety Factor for life = 8.0

Load

cycle

n fmin(ksi) fmax(ksi) Allowable

cycles

1 6000 * 8 -85*0.6 185*0.6 ~1000

2 10000 * 8 0 185*0.6 ~107

Read allowable cycles from table below.

Miner

Hence x the structure fails.


9/29

9

3B Euler buckliny

Beam: R0=0.65 in, Ri=0.5 in, L =20 in

Mat Al7075-T6(Niu, fig 4.3.5) 0.750-1.499 in

Fty =72000 psi, A-value, L-dir

E = 10.4E+6 psi (conservative) or Ec= 10.7E+6 psi

Pax=35000 lbs (Limit load!!)Pinned and rotating ends

Take care:

I Both edges aresimplesupported. (fig. 10.2.1) c=1

II Left edge is clamped. The other one is free. C=2.05

III Both sides are clamped. C=4

Solution I

(Niu, Appendix C 8)

(Niu, Appendix C 8)

Euler equation (Niu. Eq. 10.2.1 & 10.2.2)

The beam fails on Euler buckling

MS nr Failure mode Units Limit load Allowable SF MS

1 Yield compression psi 64576 72000 1.0 0.11

2 Euler buckling lbs 35000 23377 1.5 -0.55


10/29

10

3C restressed bolt

Cyclic sinusoidal Loadsvary between 0 and 6500 lbs. kp =0.2

Fb,p=0,75*215= 161 ksi

Estimation of prestress Pb,p=75%*Fty*Abolt=0.75*215000**(0.25/2)^2=7925 lbs

Thesame applies to the bolt load, when there is no prestress.

Substitute Load with prestress Pb= Pb,p+Kp*P =7915 + 0.2*6500= 9215 lbs.

Min

bolt

load

(lbs)

Min

bolt

load

(ksi)

Max

bolt

load

(lbs)

Max

bolt

load

(ksi)

Without

prestress

0 0 6500 132

Withprestress

7915 161 9215 188

Valuesubstituted in modified Goodman diagram (assumesurface finish = handpolish unnotched).

Min 0, max 132 105

cycles, Amplitude 61->65

Min 161, max 188 > 107

cycles, Amplitude 10->15

C Consequences of prestress.

yFatigue life has increased with a factor 10

2

.y Structure has morestiffness due to prestress.

.


11/29

11

4A Di

nsioning i

t d joint.

i. Dist ibution of lo ds in l ps nd i tsThe simplest way is to distribute the load e ually over the rivets, due to the fact that the rivets are

not infinitely stiff.

In the method hereafter the assumption is made that the rivets are infinitely stiff.

Load Plimit = 420 lbs is distributed over 2 rivets. Mat1 = Mat2 E1=E2

Load in lap 1 = t1/(t1+t2)*420 = 0.04/0.09*420 = 187 lbs => Load in rivet A = 420 - 187 = 233 lbs

Load in lap 2 = t2/(t1+t2)*420 = 0.05/0.09*420 = 233 lbs => Load in rivet B = 420 - 233 = 187 lbs

Rivet A = LHS rivet; Rivet B = RHS rivet (see picture)

MS ri

tAin0.4in hAl7075-T6sh t mat rial(lap1)

Both rivets are the same. Rivet A is loaded heavier and is critical in shear. (No tension considered).

Allowable ultimate load 3/32 inch rivet = 217 lbs (Niu, fig 9.2.10). MS = 217/(233*1.5)-1< 0

Allowable ultimate load 1/8 inch rivet = 368 lbs (Niu, fig 9.2.10).

MS4 = 368/(1.5*233)1 = 0.05 choose D = 1/8 inch

Mat1 = Mat2 = Al-7075-T6

(Niu, fig 9.2.10)

w = 0.425 in. (plate width)

P = Plimit = 420 lbs A = 1 in. (fastener distance)

t1 = 0.04 in. t2 = 0.05 in.

MS 3

MS 4MAT 1

7075-T6

PLimitPLimit

tLS13971AD

MAT 2

7075-T6

MS 2

MS 1

w

1


12/29

12

ii Calculate mar ins of safet Ftu=80000 psi, Fty=72000 psi Niu, fig 4.3.4. 7075-T6B-value (0.04-0.125)

M 1: Nett stren

th lap 1 ultimate

Lap 1

-1

M 2: Nett stren

th lap 1

ield

Lap 1 -1

M 3: Ult shear of rivet 1 onl

LS13971AD: MS =388/(1.5*233) 1 =0.11 Niu, fig.9.2.10

M 4: Joint stren

th rivet 1 lap1

See MS4 in question A. MS=0.05

M 5: Ultimate stren

th of undisturbed sheet at attachment point of applied load

-1=

Niu, fig.9.2.10

Mj 6: Yield strenk

th of undisturbed sheet at attachment point of applied load

-1=

Niu, fig.9.2.10

Usually only MS3 and MS4 are calculated. With logical reasoning MS5 and MS6 can be left behind.


Allow SF MS

MS1 Nett strength lap 1 ultimate lbs 420 960 1.5 0.52

MS2 Nett strength lap 1 yield lbs 420 864 1.0 1.06

MS3 Ult shear of rivet1 only lbs 233 388 1.5 0.11

MS4 Joint strength rivet 1 lap1 lbs 233 368 1.5 0.05

MS5 Ultimate strength of undisturbed

sheet at attachment point ofapplied load

lbs 420 1360 1.5 1.16

MS6 Yield strength of undisturbed sheetat attachment point of applied load

lbs 420 1224 1.0 1.91


13/29

13

4B Analysisofril

m ts with availabl m t m stdata

Lap1:t1 = 0.080 in

Lap2 n t2 = 0.071 in w = 1.55 in

Mat Clad 2024-T42 (Niu,fig4.3.2)

P= Plimit= 1400 lbs

Ftu

= 62000 psi

Fty = 38000 psi

Fbru = 118000 psi (e/D = 2.0)

Fbry = 61000 psi (e/D = 2.0)

Rivet:MS20426 DD, D = 0.25 in

Fsu = 41000 psi (B-value)(Niu,fig9.2.8)

Mat 2024-T3

The simplest way is to distribute the load eo ually over the rivets, due to the fact that the rivets are

not infinitely stiff.

Assumption n Rivets are infinitely stiff

iii. Distributionofloadsinlaps and rivetsLoad is distributed over 2 rivets. Plimit = 1400 lbs. Mat1 = Mat2,

Load between rivet A and rivet B in

PLap1 = t1/(t1+t2)*1400=0.080/0.151*1400 = 742 lbs PrivetB = 1400 - 742 = 658 lbs

PLap2 = t2/(t1+t2)*1400=0.071/0.151*1400 = 658 lbs PrivetA = 1400 - 658 = 742 lbs

B Locations of ms and failuremodes (see MS summary sheet)

MS1 Ultimate bolt shear

PP

t1

t2

1 1

2


14/29

14

MS 2 Ultimate tensilestrength lap 2

MS 3 Yield tensilestrength lap 2

MS 4 Bearing ultimate lap 2 (e/D =2.0)

MS 5 Bearing yield lap 2 (e/D =2.0)

MS 6 Nett strength Ultimate lap 2

MS 7 Nett strength yield lap 2

Margins from test Data MMPDS01 (former MIL- DBK-5)

MS 8 Joint strength ult (Niu, fig 9.2.8)

MS9 Joint strength yield (Niu, fig 9.2.8)

MS10 Rivet shear strength (Niu, fig 9.2.8)


15/29

15


Allow SF MS

MS1U

lt bolt shear lbs 742 960 1.5 0.81MS2 Ult tensile strength lap 2 psi 12721 62000 1.5 2.25

MS3 Yield tensile strength lap 2 psi 12721 38000 1.0 1.99

MS4 Bearing Ult lap 2 psi 41802 118000 1.5 0.88

MS5 Bearing Yield lap 2 psi 41802 61000 1.0 0.46

MS6 Nett strength ultimate Lap 2 psi 15168 62000 1.5 1.73

MS7 Nett strength Yield Lap 2 psi 15168 38000 1.0 1.51

MS8 Joint strength ult lbs 742 1424 1.5 0.28

MS9 Joint strength yield lbs 742 902 1.0 0.22

MS10 Rivet shear strength lbs 742 2120 1.5 0.90

Conclusion: The margins ofsafety (MS8, MS9 and MS10) calculated with testdata (Niu fi 9 2 8

are

much lower than those for the Margins of Safety based on material properties(Niu, 4.3.2). For MS8-

MS10 is bending assumed. For MS1-MS7 no bending is assumed. Both analysis cases are not

comparable. The assignment bendsslightly. MS8-MS10 are conservative and safe.

4C Testsample single lapjoint

T1.

Plaat Nagelt1 = 0.080 in.

t2 = 0.071 in. MS202426 DD in. (diameter)

w = 1.55 in.

P = Plimit = 1400 lbs Testdata fig. 9.2.8

Mat = Clad 2024-T42 (Alloy sheet and plate)(fig. 4.3.2)


16/29

16

We take lap 2 (0,071 in.) because this is the thinnest plate. Therefore we take the values for a thickness

between 0,063 and 0,249 in. You can find this information in fig. 4.3.2 on page 69.

i) Here we use A-values because it is a critical lap.


17/29

17

ii) Here we use A-Values LT from fig. 4.3.2 on page 69

According to eq. 9.1.1:

Local stress is 4 times higher than average stress, because it is a unsupported joint.

iii) Diameter rivet: in. t 2 = 0,071 in.

In fig. 9.2.8 on page 284 and 285 we find the following allowable: Ultimate Strength: 1424lbs.

Yield Strength: 902lbs.

Here we dont use a fitting factor, because it is often tested.


18/29

18

Nr Failure mode Units Limit load Allowables Safety

factors

MS

i i

MS1 Shear Ult Bolt lbs 1400 2120 1.5;1.15 -0,12

MS2 Nett strength Ult Lap 2 psi 15168 60000 1.5;1.15 1,29

MS3 Nett strength Yield Lap 2 psi 15168 36000 1.0;1.15 1,06

MS4 Bearing Ult Lap psi 78873 114000 1.5;1.15 -0,16

ii i

MS5 Tens stress Ult psi 50886 60000 1.5;1.15 -0,32

MS6 Tens stress Yield psi 50886 38000 1.0;1.15 -0,38

iii

MS7 Joint strength ult lbs 1400 1424 1.5 -0,32

MS8 Joint strength yield lbs 1400 902 1.0 -0,64


19/29

19

5A imensionin

bolted joint

Lap 1: t1=0.25 in, w =2.0 in

Lap 2: t2=0.40 in, w =2.0 in

S =2.5 in

Mat Al2024-T351, t =0.250-0.499 inB-value, L-direction (Niu, fig 4.3.3)

Ftu = 66000 psi

Fty = 50000 psi

Fbru= 100000 psi (e/D = 1.5)

Fbry= 76000 psi (e/D = 1.5)

Fbru= 122000 psi (e/D =2.0)

Fbry= 90000 psi (e/D =2.0)

Bolt: AN Steel bolt (Ftu= 125 ksi) 7/16 in

Singleshear = 11250 lbs(Niu fiz 9{ 2{ 7|

Assumption:Bolts are infinitelystiff

i Distribution of loads on bolts

Load is distributed over 3 bolts. Plimit= 10000 lbs. Mat1 = Mat2,

Assumption: theelongation between 2 bolts is thesame for lap1 and lap2 (S1=S2)

Hence: P1 + P2= P P2= P P1,

s1=s2, E1= E2, A1= w*t1, A2= w*t2

P1=0.385 * P =3850 lbs and P2= 10000-3850=6150 lbs

Evaluation: Loads follow thestiffest way.

Draw N-lines for both laps.

Plaat

Plaat

PP

W

t1

t2


20/29

20

Forces in bolts are calculated with difference of loads in lap at left side and right hand side of bolts

F1= 6150 lbs, F2 = 0 and F3 = 3850 lbs

ii. MarginsofsafetyMS1:Bolt1ultshear

Bolt 1 -1 (Niu,fig9.2.7)

MS2:Bearingultimate lap1(e/D=2.0)

MS3:lap1ultstrength

MS4:lap1yieldstrength

Lap 1

MS5Lap1:Nettstrength Ultimate

MS6Lap1:Nettstrength Yield


21/29

21

Failure mode location Units Limitload

Allow SF MS

MS1 Bolt1 shear lbs 11250 6150 1.5 0.22

MS2 Lap1 bearing ult (e/D=2.0) bolt1 psi 56228 122000 1.5 0.45

MS3 Lap1 ult strength psi 20000 66000 1.5 1.20

MS4 Lap1 yield strength psi 20000 50000 1.0 1.50

MS5 Plate1 nett strength ult lbs 10000 25781 1.5 0.72MS6 Plate1 nett strength yield lbs 10000 19531 1.0 0.95

Conclusion:Bolts2 and 3 are less loaded than bolt 1. Since lap2 has more thickness than lap1, also

the MS for this lap are higher than for lap 1.


22/29

22

Vraa}

stuk 5B Krachtverdelin}

bij bouten in een rij ~

Lap 1:

Mat: Clad 2024-T42

Alloysheet and plate

(Niu fi

4 3 2

S-values (fail safe), L-dir

Ftu =59 ksi

Fty =36 ksi

Fsu=35 ksi

Fbru= 112 ksi (e/D =2.0)

Fbry=58 ksi (e/D =2.0)

E = 10700 ksi

Lap 2:

Mat: Clad 7075-T6

Alloysheet and plate

(Niu fi

4 3 4

B-values (fail safe), L-dir

Ftu =79 ksi

Fty = 72 ksi

Fsu = 45 ksi

Fbru= 148 ksi (e/D =2.0)

Fbry= 120 ksi (e/D =2.0)

E = 10700 ksi

AN steel bolts:

Bolt dia = in

(Niu fi

9 2 7

Ftu = 125 ksi

Singleshear = 11250lbs

Tension ult. =13600 lbs

Alser geen B-value gegeven is, dan wordt er een S-value genomen. Lap 1 en Lap2 hebben identieke

afmetingen.

P= Plimit= 14000 lbs MS11 and MS21 MS22, MS23

MS24, MS25

MS12, MS13


23/29

23

6,0

8,0

10,0

3,0

0,300,50,7

A CB

LAP 2

Topview

1,2 1,6

LAP 1P limit P limit

1,2

MS24, MS25 MS16 andMS26 MS14, MS15

MS17 and MS27

Assume

Fi 1 B

v

i

i

(A

i i i

h

)

i. ( 2 pts ) Neem aan dat de bouten afschuif flexibel zijn

Geef de boutkrachten FA, FB en FC?

There are 3 fasteners and through every fastener goes 1/3 of Plimit.

ii. ( 6 pts ) Neem aan dat de bouten afschuif stijf zijn

Bereken de bout afschuifkrachten.

Increasing length of both laps must be the same. For example between fastener A and B.

1:


24/29

24

With Hooke

s law:

(valueF in Lbs)

2: =distance between A and B (fasteners arestiff)

With 1 and 2: between A and B

3:

4: ( )

5: With 3,4 and 5:

6:

But also counts: 7: 8: Take for A the averagesurface

= W-0,6=0,4 *

With 6 and 8:


25/29

25

9:

With 7 and 9:

Normal load trough lap2.

Load trough the fasteners will be:

Fi 1L

1

Fi 2 L 2


26/29

26

+

+

Lap 1

Lap 2

Loc Area E w-D (w-D)*h D*h DA 1.2 0.3 0,36 10700 0,7625 0,22875 0,13125 7/16

B 1.2 0.5 0,6 10700 0,7625 0,38125 0,21875 7/16

C 1,2 0.7 0,84 10700 0,7625 0,53375 0,30625 7/16

( As earlier mentioned assume: )

iii. ( 2 pts) Kies bout diameter, zodat RF>1 voor de meest kritische bout

(See book fig. 9.2.7 page284)

Most loaded bolt =5600lbs. limit load

This is 1,15*1,5*5600=9660 lbs ultimate load

A diameter which will beenough for the fastener is The maximum shear strength of this bolt is 11250 lbs.

Loc Area E w-D (w-D)*h D*h DA 1.2 0.7 0,84 10700 0,7625 0,53375 0,30625 7/16

B 1.2 0.5 0,6 10700 0,7625 0,38125 0,21875 7/16

C 1.2 0.3 0,36 10700 0,7625 0,22875 0,13125 7/16


27/29

27

iv. (10 pts ) Bereken Margins ofsafetyvoor verschillende typen bezwijkvormen. Geef de

locatie aan in de tekening en vul MS samenvatting in.

For calculations ,see next page!

*Geef ook het type gekozen allowable aan

Lap1 :

At location C: ; here MS14 is calculated


28/29

28

At location A lap 2 is thinner. Therefore calculate MS values lap 2.

Lap2:

At location A:=0.3 ; here MS24 is calculated

nr Failure mode/lapnr/mat nr Lapnr Units Limit load Allowable

/A/B*

L/L

/ST*

Safet

factors

MS

MS11 Bolt Ult shear 1 Lbs 5600 S, L 1,5; 1,15 0,16

MS12 Lap Ult tensilestrength 1 Lbs 14000 S, L 1,5; 1,15 1,05

MS13 Lap Yield tensilestrength 1 Lbs 14000 S, L 1,0; 1,15 1,48

MS14 Lap Bearing Ult 1 Lbs 5600 S, L 1,5; 1,15 0,52

MS15 Lap Bearing Yield 1 Lbs 5600 S, L 1,0; 1,15 1,76

MS16 Lap nett strength ultimate 1 Lbs 5600 S, L 1,5; 1,15 2,26

MS17 Lap nett strength Yield 1 Lbs 5600 S, L 1,0; 1,15 1,98


29/29

nr Failure mode/lapnr/mat nr Lapnr Units Limit load Allowable

S/A/B*

L/LT/ST*

Safet

factors

MS

MS21 Bolt Ult shear 2 Lbs 5600 B, L 1,5; 1,15 0,16

MS22 Lap Ult tensilestrength 2 Lbs 14000 B, L 1,5; 1,15 1,75

MS23 Lap Yield tensilestrength 2 Lbs 14000 B, L 1,0; 1,15 0,61

MS24 Lap Bearing Ult 2 Lbs 5600 B, L 1,5; 1,15 1,01

MS25 Lap Bearing Yield 2 Lbs 5600 B, L 1,0; 1,15 1,45

MS26 Lap nett strength ultimate 2 Lbs 5600 B, L 1,5; 1,15 3,37

MS27 Lap nett strength Yield 2 Lbs 5600 B, L 1,0; 1,15 4,97

Information from the book to answer thequestions:

y The figures can be found in the book:

Fig. 4.3.2 on page69



y Paragraph 9.7

2010 Construct 1 Exercise Answers 100413

Documents

Transcript of 2010 Construct 1 Exercise Answers 100413