as ts customary in design of such wing beams the axial load in span AB will be taken as equal to the average load or -9571-8565-7553 /3=-8565#, The beam bending moments at the support po
Trang 1
#tg A11,5T
Comparing these results with those of Problem #3
we find moment at Cc {s increased 15.7 percent
and that at D is decreased 8.5 percent For
larger values of L/j the difference would be
greater
Example Problem #13
Fig All1.58 shows the upper wing of a bi-~
plane The wing beams are continuous over 3
spans The distributed air loads on the front
beam are shown in the figure, also the axial
loads on front beam induced by the lift and drag
truss The bay sections of the spruce are shown
in Fig Al1.58 The moment of inertia in each
span will be assumed constant, neglecting the
influence of tapered filler blocks at strut
Span-AB,A'B' Span-AA'
Fig Al1.59 shows the total beam axial loads in the various portions of the beam by adding the values shown in Fig Al1.S38 The outer span AB due to the axial loads induced by drag wires of drag truss is subjected to a varying load as ts customary in design of such wing beams the axial load in span AB will be taken as equal to the average load or (-9571-8565-7553) /3=-8565#, The beam bending moments at the support points will be determined by the moment distribu~
tion method
Calculation of Factors:~
Span AB I=17.21, L=100, I/L = i72
£ = 1300,000, P = 8565
- ^/1366,000 x 17.27 000 x 17.21 _ -
J -vỆ- —— 885 — = 51.2 1⁄4 = 100/81.1 = 1.96
From Fig All.47 when L/j = 1.96, correction factor = 86 for fixed far,end and 52 for pinn~
ed far end Therefore
Kap = 52 x 172 = 0895
Kpa = 86 x 172 = 148
From Fig Al1.46
factors are:-
C.O.p, = 62, and zero from A
is considered in its true state or
» When L/J = 1.96, the carry over
to B since 3
freely sup-
ported From Fig All.48 when L/j = 1.96 aqua-
tion for fixed end moment for uniform load =
ÿ.O, Factor +585
Trang 2ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
À11,27
Fixed end moment = wL*/11.35 = 31 x 847/11.25 = For member 3C
18500"# P = - 27150, I/L = 2083 I= 332
me moment distribution process is given je (22000, 000 38 218.82, L/Jj “tr
ted, the bending moment, at support A and A’ ˆ
would be 19480, thus the axial influence in- From Fig All.47, stiffness factor = 82 x 0083
creases the moment at A approximately 7.5 per- = ,00681
rig also 2T wa wen se s*i8 | For Member AC
Pig AlL.S61 shows a triangular truss com-
posed of two members fixed at A and B and rigid-
ly joined at C to the axle bar Let it be re-
quired to determine the end moments on the two
members considering the effect of axial loads on
joint rotation and translation
The magnitude of the axial loads in the
members 1s influenced by the unknown restraining
moments at A and B To obtain a close approxi-
mation of the axial loads, the end moments in
the two members will be determined without con-
sideration of axial loads Thus the external
joint moment of 4 x 18000 = 72000 in 1b atc
is distributed between the two members as shown
in Fig All.62 With the member end moments
known the axial loads and shear reactions at A
and B can be found by statics The resulting
axial loads are
Pac = 10400# and Pua = - 27150# and tne
shear reactions, Sg = SeO# and Sg 20008
With the axial loads known tne modified
peam factors can be determined
Stiffness factor c.c, rfactor +39
Fig Al1.63 shows the moment distribution solu-
tion which includes the effect of axial loads on
joint rotation, Comparing the results in Figs
All.62 and 63, the moment Mog of 24050 is 29 per-
cent larger than that in Fig 62, and the moment
at B is 18 percent larger The effect on the ax-
ial loads of these new final moments will be
quite small, and thus further revision is un-
(Reference Chapter A7) Fig Al1.64 shows the
virtual loading of 1# normal to each member at Cc
The Table shows the calculation of the normal de- flections av Cc
Thus the deflection of joint Ở normal to
BC equals 202" in the dir on assumed for
the unit load, and she deflection of C normal to
AC 2207 ont
set:
Trang 3
THE Ali 28
The fixed end moments due to support de-
flection equals M = 6RId/L* as derived in Art
All.7? However, the secondary moments due to
axial’ loads times lateral deflection modify this
equation ‘“James" has shown that the modified
equation is,
6E1d ¬ _4
Meixed end“ fetegeay if Kp and R= 7, than
yn = SR ee-a ’
Fig All.56 shows a plot of 2B-a against L/j
that the moments tend to rotate ends of members
counter clockwise (Ref art Al1.2)
Fig Al1.65 shows the moment distribution for these moments The magnitude of the moment
at B is 6.7 percent of that in Fig Al1.63 and
10.4 percent at Joint C, however, it is reliev-
ing in this example
MOMENT DISTRIBUTION METHOD
All1.15 Secondary Bending Moments in Trusses with
Rigid Joints
Often in airplane structural design trusses with rigid joints are used Rigid joints are
produced because of welding the members together
at the truss joints or by the use of gusset plates bolted or riveted to each member at 4 truss joint When a truss bends under loading the truss joints undergo different amounts of
movement or deflection Since the truss members
are held rigidly at the joints, any joint dis-
placement will tend to bend ‘he truss members
The dending moments produced in the truss mem-
bers due to the truss joint ceflections are generally referred to as secondary moments an
the stresses produced by these moments as
secondary stresses
Since fatigue strength 1s becoming more important in aircraft structural design, the question of secondary stresses becomes of more {Importance than in the part The moment distri- bution method provides a simple and rapid method for determining these secondary moments in truss members due to truss deflection The general
procedure would te as follows: -
(1) Pind the horizontal and vertical displace-
ments of each truss joint due te the criti- cal design condition (See Chapter A7 for methods of finding truss deflections)
From these displacements, the transverse
deflection of one end of a member with
respect to the other can de found for each
russ member,
(3 Compute the fixed end moments on each member due to these transverse displacements
(4 Calculate stiffness and carry-over factors
for each truss member
a Calculate distribution factors for each truss joint
œ Carry out the moment distribution process
to find the secondary moments at the ends
of each member
(7) Calculate the stresses due to these second-
ary moments and combine with the primary
axial stresses in the truss members due to
truss action with pinned truss foints
A11,16 Structures with Curved Members
The moment distribution method plied to continuous structures which t curved as well as straight member:
equations for finding the sti7ne
carry over f Straight Đers,
Trang 4
HÀ ` ĐC ncen—
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
in Chapter AQ provides a rapid and simple
method for determining these values for curved
members The use of the elastic center method
in determining the value of stiffness and
carry over factors will now be explained
Ali 17 Structures with Curved Members
Before considering a curved member 4
straight member of constant EI will be con-
sidered Fig All.66 shows a beam freely
supported at end A and fixed at end B A
moment My is applied at end A of such magni-
tude as to turn end A through a unit angle of
one radian as illustrated tn Fig All.66 3y
definition, the necessary moment Mg to cause
this unit rotation at A 1s referred to as the
stiffness of the beam AB In Art All.4 It
was shown that this required moment was equal
in magnitude to -4EI1/L It was also proved
that this moment at A produced a moment at
the fixed end B of 2EI/L or a moment of one
hal’ the magnitude and of opposite sign ta
at at A Fig 411.67 shows the bending
ment diagram which causes cne radian rota-
sion of end A Fig 411.68 shows the M/E1
diagram, which equals the moment diagram
divided by EI which nas deen assumed constant
The total moment weight @ as explained
in Chapter A@ equals the area of the M/EL
diagram Thus @ for the M/SI diagram in
Fig All1.68 equats,
2 +2 xe tí mai
In other words the total elastic moment
weight @ equals one or unity
The location of this total moment weignt
@ will coincide with the centroid of the M/51
All, 29
diagram Thus to find the distance x from B
to this centroid we take moments of the M/EI diagram about B and divide oy g the total area
of the M/aI diagram Thus
a straight beam, we assume that the bending
moment curve due to a moment applied at A is of
such magnitude as to turn the end A through an angle of l radian As shown above, the moment
weight g for this loading ts unity or 1 and its
centroid location is at A Then by the elastic center method we find the moment required to
turn end A back to zero rotation The value of
this moment at A will then equal the stiffness factor of the beam AB In order to simplify
the equations for the redundant forces the elastic center method refers them to the elastic
center From Chapter A9 the equations for the
redundant forces at the elastic center for a
structure symmetrical about one axis are: -
Fig All.69 shows beam of Fig All.66 ra-
placed py a beam with the reaction at end A
replaced oy 4 rigid bracket terminating at point (0) the elastic center of the beam, which due to symmetry of the beam lies at the mid-
point of the beam The elastic moment loading
is fg = 1 and its location is at A as shown in
Fig A1l.69,
2x —ÿ—
Fig All.69 Solving for redundants at (0) by equations (1), (2) and (3)
te gl ele:
Mo = y-as7ir * T7et Eữ/L
Trang 5
ễƑ_ 5 x SEI is removed and end A is connected by a rigid
Yo = ly 12 E1 Le L* bracket terminating at the elastic center of
the structure xe will now find the required
Xạ ^ Bas ye QO, because y the vertical
x
istance of gg load to x axis through elastic
center (0) is zero
Fig All.70 shows these forces acting at
the elastic center
EYL
'§E/L2 Fig Al1,70 The bending moment at end A equals
El GEL jy _ 4E1
`
By definition the stiffness factor is the
moment at A which is required to turn end A
through an angle of 1 radian Thus 481/L
is the stiffness factor and this result
checks the value as previously derived in
Hence the carry-over factor from A to B
is 5 and the carTy-over moment is of opposite
sign to that of the moment at A
All, 18 Stiffness and Carry-Over Factors For Curved Members,
Pig All.71 shows a curved member, namely,
@ half circular are of constant BI cross~
Section The end B {s fixed and the end A is
reely supported A moment Mg is applied at
A of such magnitude as to Cause a rotation at
A of 1 radian as illustrated in the Fig
Fig All.72 shows the general shape of the
bending moment curve which ts statically in-
determinate In Fig All.73 the support at A
Diagram
forces at (Q) to cancel the unit rotation at a
which was assumed in Fig All.71
The total area of the M/SI curve for the
curve in Fig All.72 if calculated would equal one or unity as explained in detail for a straight member The centroid of this M/ZI diagram would if calculated fall at point A
Thus in Fig All.73 we apply a unit Bg load at
A and find the redundant force at 0 Due to
symmetry of structure about a vertical or +
axis the elastic center lies on tht vmnetr1ea1l
axis The vertical distance from @ line AB
to elastic center equals y = 6366r, (See page A3.4 of Chapter A5),
The elastic moments of inertia Ty and ly can be calculated or taken from
sources such as the table on 9a
Hence I, = “r—
1„ =7 nm
Solving the equations ?or t red
at (0), remembering Ốg = 1 and located at point
?ixed at far end 3 and freely supported at near
Trang 6
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
end A The ratio of th
ore tne stiffness factor for a nalf-
ular arch of constant £I is 2.314 EI/r
faa " a sy
The carry-over factor equals the ratio of
Mg to My or (+1.042 SI/r)/(+2.314 ZI/r) =
It srould be noticed that the carry-
over moment has the same sign as the applied
clement at A as compared to the opposite sign
for straight members In other words, there
are two points of inflection in the elastic
curve for the curved arch as compared to one
for the straight member
The fixed end moments on a curved member
any external loading can be determined
te rapidly oy the elastic center method as
t d in Chapter AQ and thus the ex-
planation will not be repeated here
Th
that wh 2 student should realize or understand e n n moments on a straight member
ructure are found from the
m process, the remaining
tically determinate, whereas
moer’in 2 continuous structure, crowing end moments does not make the
curved member statically determinate, since
we dave six unknowns at the two supports as
orivm Even when
the 2nd moments Fig, Al1.75
senting some example problem soluticns
Al1.19 Exampie Problems Continuous Structures
Involving Curved Members
Example Problem 1
B
Fig Al1.76 shows T21
2 frame consisting of both straizht and ws 10#/in
curved members Al~ Levey eriib il
though simplified ˆ Leo" e
this frame is somewhat
representative of a
fuselage frame with two 1a80" L
Cr9SS members,-one be- Tel 1
tween A and C to support
installations above
cabin ceiling and the
other between F and D
to support the cabin
floor loads The frame F
supporting forces are assumed provided by the fuselage skin as shown by the arrows
on the side members E Eccentricity of these
skin supporting forces
relative to neutral
axis of frame member is neglected in this simpli- fied example problem, since the main purpose of this example problem is to tllustrate the appli-
cation of the moment distribution method to
solving continuous structures involving curved members
w= 50#/in,
1111L111Li11 160"
Due to symmetry of structure and loading,
no translation of the frame joints takes place due to frame sidesway
The ’rame cross members AC and FD prevent horizontal movement of joints A, C, F, and D
due to bending of the two arches Any horizontal
movement of these joints due to axial deforma~
tion is usually of minor importance relative to
causing bending of frame members, Therefore it can be assumed that the frame joints suffer rotation only and therefore the moment distri-
bution method is directly applicable
Calculation of stiffness (K} values for each member of frame: -
Trang 7A11.32
Substituting: -
< - 2.31e x1 _
Rago = Koga = “3g
stant and therefore omitted since
ues are needed for the K va
dnertia of the memper is given
The relative moment of
cross-section of each frame
= = = z
Ko = KV, 5 4x 2/60 = 0.1353 Kop = Kop =4x6/60 = 0.4000
Ky = Ky = 4x1/80 = 0.0500 Kop = Kya = 4x1/80 = 0.0500
DISTRIBUTION ACTORS AT ZACH JOINT: -
as the distributed moment when the sig con~
vention adopted in this chapter is used
For a nalf circular arch of constant I,
the carry-over factor was derived in the pre-
vious article and was found to be 0.452 The
sign of this carry-over moment was the same
Sign as the distributed moment at the ot
end of the beam However, using the sisn
convention 4s adopted Zor the moment distr
Dution in this book, the carry-over factor
‘would be minus or of orposite sis So the
applied external loadings
moments on the curved me:
member AC, fixed end mement equals
10 x 607/12 = 3000 in lb and Zor
member FD = 50 x 60712 = 15000 in.lb,
ces on the 2g along
fixed eng zero
Sines the supporting skin
sice members have Seen assumed ac
centerline of frame members, the moments on members AF and CD are
Moment Distribution Process: ~
Fig All.77 shows the calculations in carrying cut the successive cycles of the
moment distribution process Due to symmetry
carry over from C = -.452 X C(-880) |
Trang 8
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES the (CI at each foint The process is started
by placing the fixed end moments with due re-
gard to sign at the ends of members AC and FD,
namely, -3000 at AC, 3000 at CA, -15000 at FD
and 15000 at DF We now unlock joint A and
find an unbalanceé moment of -3000 which means
a plus 3000 is needed for static balance
Joint A is therefore balanced by distributing
»512 x 3000 = 1536 to AC, 296 x 3000 = 880 to
ABC, and 192 x 3000 = 576 to AF Short hori-
zontal lines are then drawn under each of these
distributed values to indicate that these are
balancing moments Carry-over moments are
immediately taken care of by carrying over to
joint F, 5 x 576 = 288 From A to C the
earry-over moment would be 5 x 1536 = 768 and
therefore the carry-over from C to A would be
-5(-1536) = -768 which is recorded at A as
shown, For the arch member ABC, the carry-over
moment from A to C would be -0.452 x 880 =
-401 (not shown) and therefore from C to A =
-0.452 x (-880) = 401 as shown at joint A in
the figure for arch member CBA Joint C in
the figure has been balanced once for the
purpose of helping the student understand the
sign of the carry-over moments which flow to
the left side from the right side of the frame
After balancing joint A and taking care
of the carry-over moments, we imagine A as
fixed again and preceed to joint F where we
find an unbalanced moment of -15000 + 288 =
-14712, thus plus 14712 ts necessary for
balancing The balancing distribution 1s 255
x 14712 = 3750 to FED, 663 x 14712 = 9750 to
FD and 062 x 14712 = 1212 to FA The carry-
over moments are 5 x 1212 = 606 to A, 452 x
3750 = 1695 from D to F by way of the arch
member and -.5 x 9750 = -4875 from D on member
FD We mow go dack to joint A which has been
unbalanced by the carry-over moments and repeat
the balancing and cerrying-over cycle In the
complete solution as given in Figure All.77
each joint A and F was balanced five times
The final bending moments at the ends of the
members at each joint are shown below the
double short lines
The arch member ABC has 3 unknown forces
gach end A and C or a total of 6 unknowns,
3 equations of static equilibrium avail~
able plus the mown values of the end moments
at A and C as found from the moment distribu-
tion process, the arch member is still static-
ally indeterminate to one degree Thus the
norizontal reaction at A or C as provided dy
the axial load in member AC must be found 5e~
fore the bending moments on arch ABC can be
calculated
The first step in this problem is to find
the elastic center of the frame portion com-
posed of members ABC and AC, as shown in Fig
ALL.78, and then find tne elastic moments of
Al1,33 inertia about x and y axes through the elastic canter
the bending moment curves on this frame portion
due to the given load on member AC and the end
moments as found by the moment distribution
process in Fig All.77 It is composed of three parts labeled (1) to (3) in Fig All.79, Portions {1) and (2) are due to the end moments
and portion (3) due to the distributed lateral load on member AC
Fig Al1.79
The next step is to find the a, (area of
M/I curve} for each portion and its centroid
location
Trang 9Os, = 1094 x mn x 30/1 = 103000 46 1b, The member AC also suffers an axi
due to the shear reactions at
Da, = 60 x 2353/2 = 70590 FA and DC Fig All.él shows
side member FA and DC with the
Bs, = (-.667 x $0 x 4500)/2 = -90000 found in Sig.All.77
3Ø; = 83590 at A and C can be € 1259
Fig All.80 shows the %5 values concen- moments about lower C:C
trated at their centroid locations and referred | ends Thus for FD, :
to the x and y axes through the elastic center | Ry = (1259+ 1791)/30 : ‘ |
= 38.1 ib Likewise a uF D
Peugtions react on cross member ^C in the vor Fig All.81
= 103000 opposite directions
T x thus giving 2 compression load of 38.1 lb in
@,,= 70590 | 14.49 member AC, which must be added to the tension
As — —— load of 46 1d from the arch
“a, = -90000 Fig A11.80 the final load in the cross-member
The frame has been imagined cut at A and the
arch end at A has been connected by a rigid
bracket to the elastic center The redundants
at the elastic center which will cancel any
relative movement of the cut faces at A can
>3 Bek = 0 because x = zero
¥
Yo =
The bending moment at any point on ABC
or AC equals that due to Mo and X, plus the
moments in Fig A11,79,
For example,
At point A on member ABC,
Ma = 1094 - 670+46x14.49 = 1091 (should
be 1094 since moment as found in Fig Al1.77
is correct one Small error due to slide rule
values in Fig All.79 are the true moments
The axial load in member
Fig ALL.80 equals Xo or AC Dy Statics
Bending Moment in Lower Arch Member :
The horizontal react
be fcund berore tr arch can be found
as yor the upper arc
Trang 10
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES SZ-y_=837550X15.7§+15000x15 76-328000(-3.34)
The axial load in member FD = 294 lb
compression plus 38.1 lbs tension due to shear
reaction from side members at points FP and D
or a resultant load of 255.9 1b compression
‡g, A11.84 shows the final bending moment
diagram on each member of the frame
supporting forces have been assumed as acting
uniformly along the side members AD and CE
The problem will be to determine the bending moments at frame points ABCDE
1000 B 1000
Fig Al1.85
133, Le 60 SOLUTION: ~
Calculation of stiffness factors K
Member ABC Kapo = 2.314 I/r = 2.314x1.5/30 =
1158 Kac 8 41/L = 4 x 2/60 = 1333 Kap = 4I/L = 4 x 1/60 = 0667 Kpg = 4I/L = 4 x 3/60 = 2000
Calculation of Distribution Factors D
JOINT A, 2K = 1156+ 1883+ 0667 = 3158 Dapc = 1158/.3158 = 366
Dac = 1Z53/.3158 = 4Z2
Dap *# O0667/.3188 = 212 JOINT D, šK = 0667+ 2000 = 2667 Dpg = -0667/.2667 = 25
Dpg = 2000/.2667 = 75
The carry-over factor for ABC is -0.452 as pre~
viously derived for a nalf circle arc, and 0.5 for the straight members
Calculation of Fixed ind Moments
Trang 11All 36
Curved member ABC
The fixed end moments on this curvec member due to the external loads will be de-
termined by the elastic center method The
assumed static frame condition will be an arch
pinned at A and supported on rollers at B
(See Fig All.86)
Fig Al1.67
shows the general 1000 =P 8 1000 = P
shape of the static Ị aa ị
moment.curve For
the frame portion / kK 10
between the re- poe 800 _ c
actions and the load
points, the bending Fig Ail 86
The vertical distance ¥ from line AC to
centroid of Zs, and @5, values ts,
‡x er Sin 9⁄2 „ 2x8 kiệt = 24.8 in
Fig 411.88 shows the %, values and their lo-
cation with respect to x and y axes through
the arch elastic center
The elastic center method requires the
Distance ¥ from line AC to elastic
arch ABC equals 0.6366r = 6366 x 3 1x = 0.2978r3/1 š 2978 x 505/1
The fixed end moments at ends A and C wiil
equal the moment due to the redundant foress ¬
and Y, since the static moment assumed was zer
distribution can now be carried out rig
All.89 shows the solution The first cycle will
be explained Starting at jot {A) the un-
balanced moment 1s +3000 - 652
joint is balanced by distributing = x 3652
= 1543 to AC; 364 x 1543 = 1555 to ABC and 212 x 3652 = 776 to AD The carry-over moment
from C to A = S(-1543) 3 -772; from © on member CBA to A = - 1452 (1332) = 601; and from
A to a = öX 776 = 388 Now proce eding to
joint 5 the unbalanced moment is -7500 + 388 =
-7112, Tne joint is balanced by distributing
+75 x 7112 2 5340 to DE and the remainder 25
per cent = 1772 to Da
Trang 12Fig, All.91 shows the elastic center loca~
To A from D = S (1772) = 986 tion and the fg values together with their
centroid locations
The first cycle is now completed Five more
cycles are carried out tn Fig A11.89 in order
to obtain reasonable accuracy of results The
final end moments are listed below the double
short lines
On arch member ABC the end moments of 584
are correct However before the end moments
at any other point on the arch can de found
the horizontal reaction on the arch at A must
be determined This reaction will de de-
termined by the elastic center method
Fig Al2.90 shows the oending moment
curves for members ABC and aC as made up of
5 parts labeled 1 to 5
Calculation of 0g values which squal area
9Ÿ sach moment curve divided by I of member
Solving for redundant forces at elastic center,