2010 Construct 1 Exercise Answers 100413
Transcript of 2010 Construct 1 Exercise Answers 100413
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Answers
Exam Exercise
Aircraft structures
Constructieleer II
2011 Marcel van Varik
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Failure modes
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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
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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
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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)
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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)
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M h 3: pin tensile ult streni
th p due to bendini
q
Failure mode location Units Applied(Limit)
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
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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.
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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
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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.
.
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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
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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.
Failure mode location Units Applied(Limit)
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
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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
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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)
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Failure mode location Units Applied(Limit)
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)
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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.
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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.
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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
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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
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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
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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.
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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
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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:
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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:
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9:
With 7 and 9:
Normal load trough lap2.
Load trough the fasteners will be:
Fi 1L
1
Fi 2 L 2
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+
+
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
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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
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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
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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
Fig. 4.3.4 on page71
Fig. 9.2.7 on page284
y Paragraph 9.7