All, 26 r by : Culumn distribution factor 2B-a Fig All 56 we lesan 1/j52.5 Tis constant, L/J=2.0 we Tir #tg A11,5T 31.3” 00374} 0
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 points THE MOMENT DISTRIBUTION METHOD là 2`L@) * m „ 80 tan ti 100" x—>—% ae vấn exttT Tit SÓT „ 7052, 10061 05,5187 itr t : œ LÊN À!*xcabane ,À 58 N ‘ struts Ao Fig All TY L Spruce ` Pusekj ‹* a ` „ en 6" Axial Beam Loads tL i 211758 | -9571 |-8565|-7559 | ont —— | | A A B 1= 17.21 13 22 Fig Ali 59 } ATSYLOGE-TOOE Tae “3B 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-
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
in Fig All.60 If the axial 1oads were neglec- = 2.12 He
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
cent From Fig Al1.46 C.0 Factor = 66
rig also 2T wa wen se s*i8 | For Member AC P = 10400#, I/L = 0029, I = 137 AY 29,000, 000 x 137 _ L_ 47.20 _ J T6460 = 19.58, 7* i575 7 2.42 Example Proolem
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 Fig A11 62 ~72000 Solution:-
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- necessary „0029 x 1.18 = 00542 ,668 ——> 31600 Fig A11.63 ? 4go00 - =72000
Zfect on End Moments Due to Translation of Joint ¢ Due to Axial Loads
The movement of joint C normal to each member will be calculated by virtual work
(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 Table Syl Sugg Xem | L/AE 3 dạ TẾ t z= ac | 0000048 10400 | i.se | ogez | 1.6 | 0818 ñC | 00000244 | -27150 | -1.60 {1060 _| -1.89 | ,1250 [ i 202, „207,
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 3THE Ali 28 “1.68 Fig ALL 64 Ug tlt Cc €
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 For Member BC:- R= 202/40= 00505 K=I/L= 0083, 2ê-aœ=1.08 for L/J = 2.12 Hence, fixed end moments due to translation of joint Œ equals- ay 6X 0083 x 00505 x 29,000, 000 _ Hạc = Mop = 1.08 ` For Member AC R= 207/47.2= 00438 K= 0029, 28-2 = 92 L⁄4=2.42 and member is in tension 6x 00228 x 0029 x 29,000,000 _ „92 -6750"£ when hence Mac = Mog = 2400" #
The signs are minus because inspection shows 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 (c) -6750 (f) {b)+ 66 -» 4020 -2730 ~8180 (f) +6095 ~ 655 Fig Ali 65
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 4HÀ ` 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 Aas Bat M Tội eS beam Lo Et constant n° Mg Fig All 66 Lẻ ————n - Fig ALL 67 Ma 2 = 2EL LB M diagram Fig A11 68 M/EI Diagram 2 -4{.- L L
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
re
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
= 72{.667L) + 1(.233L) - L_ a——=L
x ~aertl -1
Thus the centroid of the total moment weight ¢ which equals one lies at point A or a distance
L to left of B
In using the elastic center method to de- termine the stiffness and carry-over factor for 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
Trang 5
All 30 THE MOMENT DISTRIBUTION METHOD
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 A Lr B '§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 Art All.4, The bending moment at 8 in Fig All.7 equals, = _ BL, SEI ,L _ ZEỊ TU tr () = SẼ
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 Moment Diagram
Fig All 71 Fig All 72
forces at (Q) to cancel the unit rotation at a
which was assumed in Fig All.71 Te #* 0.6368r 1 Fig, Al1.73
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 Whence, = - 1 Iy = 2978r"t, dut t Sr 2978r* Hence I, = “r— 1„ =7 nm v 2 251
Solving the equations ?or t red
at (0), remembering Ốg = 1 and located at point A = 3 ốs _ = (1} _ -Er Mo 5 - mấy “ "+ 3Øay „ (1) (-.6366r} =I 3Øsy „ Tx Tasers =- Plas 21 El Xo
Trang 6
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES end A The ratio of th point B to that at point A will the carry- “an¬* ˆ * factor 6366 r Aa —— ——— ‘Tre Bending moment at A, Fig Al1.74 = aL + B1 251 Ma = SE - 2.18 (O.6366r) - Grr -0.318E1 1.5651 Bending moment at 3 > my =TÓ:31ÊEl _ 1.26E1, 9.656E1 „ - puoyp EL # ĩ T T
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 i strated in Fig Al1.75 and only 3 equations or static equill orivm Even when
the 2nd moments Fig, Al1.75 2 a 2 s one unknown, Ha \ H ly the hori~ zy B namely the nor Ma UVa Mp zontal reaction VB at one of the beam ends ALL GL The method of how to handle this remaining redundant force can test be explained by pre~=
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
relative to shape, 122
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" 136 Fig All 76 SOLUTION
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
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 on Fig All.76, , 2.314x2, ‘ep 30 For all straight members, the = KL DEF „1540 stiffness factor equals 4EI1/L Hence, = = = 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: - JOINT A EK = 0,153 + ,0770 + ,O500 = 0.2605 Let D equal symbcl for distribution factor Dago 7 +0770/0.2603 = 0.296 Dig * 1883/0.2603 = 0.512 Dap = -05C0/0.2603 = 0.192 JOINT FP EK = 0500+ 0.400+ 0.1540 = 0.6040 Da, = 0500/.6040 = 082 Di 3 = 4000/.6040 = 663 D vag ‘ED = 1540/,6040 = 255 Carry-Over Factors: -
For the straight members the carry-over factor is 0.5 and the moment sign is the same 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 THE MOMENT DISTRIBUTION METHOD distributed moment tence the Zixed end rs is zero
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
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 19,1" | Fig Atl 78 ——ã mm e „ 18.1 xn (0/1) mx 20, 60 1 2 ¥ = 14.49"
(NOTE:~ 19.1 equals distance from line AC to centroid of arch member ABC.)
Calculation of moment of inertia qt - Member ABC — ore _ mrd* Lora tT s = _ + _ x 4.612 = 10100 v 60 - Member AC = 23 x 14.49% = 6310 I, total = 16410 Total elastic weight of structure equals Zds/l = (mn x 30/1) + 60/2 = 124,24,
Tne next step in the solution is to draw
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
Trang 9
À11.34 THE MOMENT DISTRIBUTION METHOD
Os, = 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
———— The shear reaction 38,1
3Ø; = 83590 at A and C can be € 1259
found by taking x
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 now be computed 2-72 Bs ng: Xp = 2 bt [105000 (4.61) + 70590 x (+14.49) - 90000 (-14.49) ] / 16410 = 46 1b ~32590 TiT.2a = - 673 in.lb >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 accuracy
% point B on member ABC: -
Mp = 1094~-670-46x15.51 = -289 in lb
The horizontal reaction at A will not produce any bending on member AC, thus the
values in Fig All.79 are the true moments
The axial load in member
Fig ALL.80 equals Xo or AC Dy Statics
from a tension load of
Bending Moment in Lower Arch Member :
The horizontal react be fcund berore tr
arch can be found
Trang 10ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES SZ-y_=837550X15.7§+15000x15 76-328000(-3.34) 056 —~ 15 76 a Fig Ail 83 Bending moment at F on memoer FED, Mp = -6962 + 4650 -29415.76 = -6962 in.1b Mg = +6962 + 4650+ 294 x15.76 = 2340 in.lb
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 B 259 Member ABC ` 1084 A` 4 c 1094 Member AC c E 1258 N Member CD Member FD Member FED Figs All.84 Final Bending Moment Curves on Each Member of Frame À11.35 #xample Problem 2 External Loads on Curved Members
Fig AlL.85 shows a frame which has ex- ternal loads applied to the curved member as well as to the straight members 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
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
moment equation is
MsP(r-r cos a)
For the beam portion between the two 1606 lb loads, the bending moment is constant and equals, M=P(r~=r cos 209) qa) Fig All 87 Static Moment Curve Calculation of ds Values
The Zg values equal the area of the M/I curves, The moment curve in Fig Al1.87 has been broken down into three parts labeled (1)
Œ') and (2)
Bs, + Bs 4 2L E (4 - sin a}
a{ 10002 x 302 1,5 {524 — = 28800
The vertical distance ¥ from line AC to centroid of Zs, and @5, values ts, F = Ph -cos — „ 50(1-.867-0.5/3) _ a-sina 524 ~ 50 10 in 5 3, * 9 (L- cos a) 8 = 120°, a @ 209, 2 os, £21 (1 _ 867) = 167300 Vertical distance ¥ from line AC to centroid of Sg, equals, - a
‡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 THE MOMENT DISTRIBUTION METHOD € total elastic weigh moments of inertia elastic center TOTAL ELASTIC WEIGHT = 2 ds/1 =7—> = ul œ np œ ol
Distance ¥ from line AC to elastic
arch ABC equals 0.6366r = 6366 x 3 1x = 0.2978r3/1 š 2978 x 505/1 Fig, A11.88 Calculation of redundant forces at elastic cen- ter = 2 fg, -(28800+ 167300) _ aan Họ *-S q1 "— 2,85 Tu in.1b t, 2B Gay 167800 x5.7 + 98800(-9.1) 9 2st = : Ix 5555 325.3 Ib Yo = ~ 5 ost 5 O (because x = 0} y
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
at A and c
Mạ = Mẹ = Mẹ + Xoy = - 3122 + 129.3 (- 19,2)
= - 652 1n.1b,
Moment Distribution Process
Having determined the fixed end moments, distribution and carry-over facters, the moment
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
Trang 12ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES All, 37 4000''+ B “52 (-652) = carry over from © ⁄ Static Mom Curve i Low = „366 a ot gis ge sin gig se aie Số SN Ấn TU HC TẾ cr 3 a) renal F—2513 gegggasace ag 2513 Bass S958 1g Static Mom Curve Fig All.90 ~1/8 wL* = 4500 | Bg = 2513 x 60/2 = 75390 i 4 gin? aig gle slo zig hy BET STATS a3 : 1Ø, = 218290 - T500 ft A11,.88 5340 ~1500 I Calculation of elastic center location and -2870° xX? 2060 CN istance ÿ from line AC to elastie cenber = -401 * 313 19.1xnx50/1.5 _ 1202 _ 12.97" KH - nxd0 , 60 92.85 -80 5 2 36 at ‘ - TOTAL ELASTIC WEIGHT = 92.83 wa a 1„ = +£Ÿ75 X39”, RX22 (s,13)2 + SỐ x 12,07" 1.5 1.5 2 Tne carry-over moments are: - = 12715 #rom š to D0 # ,5x (-5540) = -2670
Fig, 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
Trang 13Ail, 38 THE MOMENT DISTRIBUTION
The bending moment at AC or ABC equals the values in plus those due to My and Xg in Thus at point A on member ABC, Mạng = 584-2356 + 181 x 12.97 = 564 in 1b, which is the same as found by the moment distri- bution process At point Boon arch: - Mg from Fig 411.90 = 584+ 4000 = 4584 Mg from Fig Al1.91 = -2356 +181 x (30 - 12.97) = -S476 Hence Mp = 4584-5476 = ~892 in.1lb STRUCTURES WITH JOINT DISPLACEMENTS
For unsymmetrical frames or for frames
loaded unsymmetrically, the assumption that only joint rotation occurs may give resulting
moments considerably in error Joint dis—
placement can be handled in a manner as
previously explained and illustrated for frames composed of straight members
The distribution of the skin supporting forces on the frame boundary are usually taken as following the shear flow distribution for the shell in bending as explained in a later chapter All 20 Problems
(1) Determine the bending moment diagram for the loaded structures shown in Figs 1 to 6 100 „mẹ w= 101b/in, mer t_i6"— Ler a4 Fig.1 Fig 2 200} gr g"=|100 + w=201h/in 200 Hộ 1 ` thot et ý a 13 VÀ = = ĐÁ bạt háo 21-4 T=F lễ —— Ễ _— Fig.4 Fig 3 lở 100 200 w=l2lb/in + ‡ ¿1L ij 1= 1.5 T=2 T=2 Len 12" 1g" —L 167 13" Fig 6
(2) Pind the bending moments at all 4#oints
and support points of the loaded structures in Figs, 7 to 10 METHOD TA oS 7 Nsã.1852T1-7 g5 = To" 20" 200 [ 400e— 1-8 i400 So 1=4 ~ 1=4| 10 _ Fig 7 Fig 8 100 400 600 goo _ «100 ¡182 10 i we ey ey Pin 4 Lev 1=z4 | I=6 ụ=3 am lel a , : -/-——- 16" —-—+ T=2 k | ¡Pin Tm p= 24 4 20g Le Fig 10 Fig 9 (3) ala ee 12 Ib/in eh LP a eT a tt A 4B “ Dư RE Ƒ 20} — 49" ——}-— 32" —t 40" +206 Fig 11
In the loaded beam of Fig 11, the supports
Band © or the slevator beam deflect 0.1 inch
more than supports C and D Compute the re- sultant bendins moments 2t the supports and finde reactions E&I = 320,000 lb.in.sq
(4) Fig 12 illustrates a continuous 3 span wing beam, carrying a uniform air load of 20 ib./in Determine the beam bending moments at 4 strut points A and B Take
lạg=17in., Iga! = 20 in., and E = 1.3x10° in Neglect effect of support deflection due to
strut axial deformation, «| wd 1 F#— 110 I 100" ——>— 110——-s0) { san Fig, 12
Trang 14ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES CHAPTER AI2
SPECIAL METHODS - SLOPE DEFLECTION METHOD
Al2.1 General The slope-deflsction method its another widely used special method for analyz- ing all types of statically indeterminate
beams and rigid frames In this method as in
the moment distribution method all joints are
considered rigid from the standpoint that all
angles between members meeting at a joint are
assumed not to change in magnitude as loads are applied to the structure In the slope
deflection method the rotations of the joints
are treated as the unknowns For a member bounded by two end joints, the end moments can
be expressed in terms of the end rotations Furthermore for static equilibrium the sum of
the end moments on the members meeting at a
joint must be equal to zero These equations of static equilibrium provide the necessary conditions to handle the unknown foint rota-
tions and when these unknown joint rotations
are found, the end moments can be computed from the slope-ceflection equations The ad- vantages of the slope deflections method will
be discussed at the end of the chapter after the method has been explained and applied to problem solution
Al2,2 Derivation of Slope Deflection Equation
The problem is to determine the relation-
ships between the displacements of the end supports of a beam and the resulting end
moments om the beam
Fig Al2.la shows a beam restrained at
ends land 2 It is assumed unloaded and of constant cross-section or moment of inertia This beam is now displaced as shown in Fig b,
namely, thet the ends are rotated through the angles @ and 6 pius a vertical displacement
ad, and d, of its ends from its original posi-
tion, which produces a relative deflection A
of its two ends The angle % representing the
swing of the member equals A/L
The problem is to derive equations for M;¿ and Ma in temas of the end slepes 6: and 82, the length of the 5eam and BI Figs
d and e illustrate how the total beam deflection in Fig b is broken dewn into three separate deflections In Flg ¢, the end (2) 1s con~ Sidered fixed and a moment Mi is applied to
rotate end (1) through an angle @,, In Fig
end (1) is considered fixed and a moment Mf
apolied at (2) to rotate end (2) through an
angie 6, In Fig e, doth ends are considered fixed and end (2) ts displaced downward 4 e cy cá, is Fig Al2 la —————L ——————¬ 1 (3) a EI Constant | Fig c + XS 9 Fig doy — X) ø M" —— —— Tig e mm ' 1 a bực Figs Al2.1
distance A which causes end moments MY and M%
In deriving the slope deflection equations each
of the three beam deflections will be considered
separately and the results are then added to give the final equations Fig Al2.2 shows Fig c repeated 2 ‡U Mu: wh Pìg.A12.4 MP, T w a Hà
The applied moment Mi, ts positive (tension in bottom fibers) Since end (2) 1s assumed fixed
an unknown moment Mi, is produced at (2) Fig
Al2.3 shows the bending moment diagram made up
of two triangles M', being unknown will be
Trang 15
Al2.2 SPECIAL METHODS -
assumed also positive, as the algebraic solu~
tion will determine the true sign of MẠ,
Two moment area theorems will be used in deriving the slope-~deflection equations,
namely (1) that the change in slope of the beam elastic curve between two points on the beam 1S equal in magnitude to the area of the
M/EI diagram between the two points, and (II)
the deflection of a point (A) on the elastic
curve away from a tangent to the elastic curve
at (B) is equal in magnitude to the first moment about A of the M/EI diagram between (A)
and (B) acting as a load
In Pig Al2.2 since 8, = 0, then 6, will equal the area of the M curves divided ty ZI
s,= GŒG1⁄2)„ GL/2) „ OM eML _ _ (y
+ EI EI 21
The deflection of end (1) away froma tangent at (2) equals zero, thus we take
Moments of the moment diagram about (1) and equate to zero MAEVE, fh Pin ZElj3 \2E1/ 53 whence -M = Mi/2 Fig
diagram Al2.4 shows the true shape of moment
Substituting results in equation (2) in equation (1) we obtain, MiL 4516 = fb tom AR Lk 5 aro Mà L (3) Then from equation (2) “am 22 (4) #ig A12 5 Fig A126 a a My 3 Fig Al2.7 SLOPE DEFLECTION METHOD
g Al2.6 shows the moment diagram for bending mement -M§ which rotates
an angle 6, when other end (1) 1S g AlZ.5}, Ina similar manner as
ed befors,
Taking moments of M/EI diagram about (2),
MUL, (MENS g ZE1/j5 281/3
or, -M}S MI - (3
whence,
MUL and Mis 20194
Hi REL
Then from equation (5)
Fig Al2.8 shows the third sart of
peam deflection, namely, support (2) is c>ed
downward a distance 4 assuming both ends fixed, MẸ(| #ig A12.8 Fig.A12.8 Mặt uy Fig Al2, 10
The change in slope of the elastic curve
between the ends equal zero, thus »y the moment
Trang 16ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES A12.3 `“ `" Gẽ) Mụn; = 2K (29, +62) tp, TT T— (18a) Mau = 2k (20, +6,) +Mp) - (18b) The comdined ect of deflection Q, O2 À can now be obtained by adding the separ- results Minas Mỹ + CHỦ = SE (26, +0, = 34/L) and Ø = A/L Then, Mụ_„= 2K (29,+0,-5Ø) ~ - - -~— (13) and Mg, ~2K (20,+9,-38) - (14)
These end moments due to the distortion of the supports must be added to the moments due to any applied loads on the beam when
considered as fixed ended Let these fixed end moments be labeled Mp, and Mp, Then
equations (13) and (14) can be written includ-
ing these moments
Mae = 2K (26 + 62-38) - Mp, - (15)
Mạn, = -2K (26,+9,-30} - Mp, - - -
MCDIFIzD SIGN CONVENTION
Equations (15) and (16) have been developed using the conventional signs for
bending moments In general there are ad~ vantages of using a statical sign convention
as was used in the moment distribution method in Chapter All Therefore, the following sign convention will be used in this chapter: ~
(1) The rotation of a joint of member is posi-
tive if it turns in a clockwise direction
mend moment is considered positive if it nés to rotate the end of the member ockwise or the joint counterclockwise ¬ œ This adopted opted for Art 412.2)
sign convention is the same g moment distribution method
xnen equations (15} and (15) are revised
for this new sign convention they become: = My = 2K (20, +6,-30) +Mp - - (17) = 2K (29;+9.-3Ø) tp, T~ (18) te as the EI/L and
For no yielding or transverse movement of
surports, A = 0 and equations (17) and (18)
A12.3 Hinged End Slope Deflection Equation
Consider that end (2) of beam 1-2 is free- ly supported or in other words hinged This
means that M,., 1s zero Thus equation (18) can be equated to zero Mai = 2K (26, +6, -30) +Mp, = 0 whence 2KQ, = -KO, + 5KZ - 5MP, Substituting this value in equation (17) Mie = 5K(9 -Ø) +Mp,.„ =0.8Mp,_,— - -(19) A12.4 Exampie Problems Problem 1
Fig Al2.11 shows a two span continuous beam fixed at ends (1) and (3) and carrying lateral loads as shown The bending moments at points 1, 2 and 3 will be determined Relative
values of I are shown for each member 100 1b, —12— 1 fig A12.11 3 SOLUTION: Calculation of fixed end moments due to applied loads - Mp,_, = -PL/8 = 100x24/8 = =300 in 1b, Mp,_, = PL/8 = 300
The signs as shown are determined from the fact that the end moment at (1) tends to rotate the end of the member counterclockwise which is a positive moment according to our adopted sign conventicn By similar reasoning the fixed in
moment at (2) is positive because the moment
tends to rotate end of member clockwise For span 2-3: -
Mp,_, = wl?/12 = 24x24%/12 = ~1152 in.ib Mp _„ = 1152
K Values K = EI/L, Since = is constant it
will se considered unity FOR BEAM 1-2, K
Trang 17
Al2.4 SPECIAL METHODS - SLOPE DEFLECTION METHOD
Substituting in slope-deflection equations 18a
and 18b (% = 0 since supports do not trans~ late.) Beam 1-2 6, = 0 because of fixity at (1) Mie = 2K (26 +0.) +Mr, Mi-a = 2x1 (0+ 62) -300 Mi-g = 26,.-300 - (a) Ma- = 2K (26.+6i) +Mp, Maa = 2xI (20,+0) +300 My = 40,+300 - (>) Beam 2-3 6, = 0, because of fixity at (3) Macs = 2K (202403) + Mp, Mes = 2x2 (26, +0) -~1152 Meus = 86,-1152 - (e) Mẹ „ “ 2K (29, +Ôa;) +Mp, MạT-a Z 2x2 (O0+9Q;)+ 1152 My, 7 40,+1152 - (4) At joint (2) IM = 0 for static equilibrium Hence My + Mg, 70 Substituting from (b) and (c) 40, + 300+ 80, -1152 = 0 whence 6, = 71
With &, known the final end moments can be found by substituting in equations (a), (b), (c) and (4) as follows: - M +s 2x71-300 =-158 in.lbd My 4x71+300 = 584 My, = 8x71-1152 =~-584 My = 4x71+1152 = 1456
Changing the resulting moment signs to the
conventional sign practice, gives M,2 5-158, Hạ, = Mạ „ = -884, a~s ~1436 Mee Example Prcblem 2 Lan ¿ au | 9> 12 lb/in "` #toên (ĐT Lszc TỔ) oe Z4) T 224 | 1 Kel K Fig Al2.12
Fig Al2.12 shows a loaded 3 span beam fixed at points (1) and (2) This fixity at
Trang 18ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
Solving equations (g} and (h) for 9, and 4,
gives,
@a = 26.15, 95 > - 19.9
Substituting in equations (a) to (f)
to find final end moments inclusive i 28„ -20Q = 2x 26,15 - 300 1ì 247.7 in.lb M thung Mụ_, = 46, +00 = 4x 26.15 + 500 = 404.6 M, „ = 89, +48, -554 2 8X26.15+4 (=19.9) - 534 = - 404.6 M,_a = 86, +49, +534 = 8 (-19.9) +4x26.15 * 384 = 479.4 Ms = 46 -400 = 4 (-19.9) -400 = 479.4 My_s = 20, +400 = 2 (-19.9) +400 = 360.2
(Mote: - Student should convert to conventional
moment signs and draw complete bending moment diagram)
Zxample Problem 3 Beam with Simple Supports Fig 412.13 shows a loaded 3 span contin- uous beam with ninged su2ports at points (1)
and (4), which means that My and My, = 0
The moments at supports (2) and (3) will be determined Fig Al2.13 SOLUTION: end moments: - Fixed Mp,_, =-PL/8 = 100x20/8 =-250, Mp,_, = 250 Mp Pang = MPs = Mp. 3.0 (No load on span 2-3) Mip,_, = PL/@ 200 x20/8 = -500, My = 500 Slope-deflection equations: - Since supports at (1) JOINT (2) Al2.5
or free to rotate we use equation (19) in writ-
ing equations for Ma and Ms_
Mani = KOa+Mp, | -0.5 Mp,_, (f is zero) Maa = 3x1x@2+250-0.5 (-250) = 3@.+375 -~~ - (a) KH My_ = 2x x9; ~500~0.6 (500) = 305 -70 ~ - (d) Using equations 18a and 18b and substituting, Mas = 2K(26,+6,) + se s 2%0.5(20,+0,) +0 = 266, - - (c) Mans Mya = 2K(26, + 8.) + Mp, —a Myra = 2%0.5(205 +82) +0 = 205+ O2- ¬ (4) For statical equilibrium of joints: - Ma-i +Ma-s = 0 Subt 36, +325+292+90; = 0 or 662+6,+375 = 0 - rr rn (e) JOINT (5) My „+M¿ „ = 0 20,79, +38, -750 = 0 or 56,+9,-750250 - (£) Solving (e) and (f) for 6, and 9, gives, @,=- 109.4, 6, 5172
The end moments at (2) and (3) can now be round
from equation (a) to (d) inclusive Ma = 30, +375 = 3 (-109.4) +375 = 4 Mas = 20279, 5 2 (-109.4) +172 = 46.8 My_, = 26,+9, = 2x172+ (-109.4) = 234 Mg, = 36, -750 = 3x172~750 = - 234 A12.3 Loaded Continuous Beam with Yielding Supports The movable control
namely the elevator, rud
attached at several fin and wins
surfaces 2 car
thus th poorting pot
Trang 19A12 6 SPECIAL METHODS -~-
SLOPE DEFLECTION METHOD
supports, The slope deflection equations can
take care of this support displacement as
illustrated in the following example problem Example Problem 3 tb/in 4 tb/in Ann oo kL Lip litre 2 3 + ĩ mì _m am =m.— “a 10†— | Ị | Pl£ A1214 tube ve + T 12 3/16" | deflected supports Fig A12 15
Fig Al2.14 is representative of an ele-
vator beam attached to deflecting stabilizer structure at the five reaction points as indicated in the figure The elevator beam
1s a round aluminum tube 1-1/4 - 049 In size The moment of inertia (I) of this tube equals 0.03339 and the modulus of elasticity (5) of
the material equals 10 million psi The air load on the elevator beam is variable as indicated in Fig Al12.14
Fig Al2.15 shows the shape of the de- flecting supporting structure, which means that supports (2) and (3) move downward through the distances indicated on the figure The problem will be to determine the bending moment at the supports under the combined action of transverse loading and settling of
supports SOLUTION: ~
Calculation of fixed end moments: -
For a trapeqoidal loading as shown tn Fig a, the fixed end moments are, Le Mire = gy (Su+ av) dị, ut Lm : | Me = = (5u + 3v) * TH L— Substituting in these equations for the loading
values as shown in Fig Al2.4, Fig a 2 Mpg = HS xS+1) 2-426 mv, = 40" =\ fr „ “ sọc (6x5+1.5) = 440 a Mp s+ 60 = 42° (5%3.5+1) = 494 40" | s = = M = +1 = 5d Mpg = Gov (9 x3.5 41.5) = 507
The beam has a constant section, hence,
K for all spans equals,
K = EI/L = (10,000,000 x 0.02339) 40 = 8347
Calculation of % values
Span 2-3 The settlement of support (2} with respect toa support (3) = (0.5- 11375) =
+3125 inches = 4
@ =-A/L =- 3125/40 =-0.007912 rad
(Since (2) moves counterclockwise with respect
to (3) the sign of @ is negative Sran d-4 4 = 1875 inches
8 =-A/L =- 1875/40 =~ ,004688 rad
Since the beam, external loading and support Settlement is symmetrical about support (4), the
Slope of the beam elastic curve at (4) is hori-
zontal or zero and therefore 6 = 0 Thus only
one-half of the structure need be considered in the solution Due to the fact that (2) is 4
Trang 20ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES whence, 667760, + 166940, +572 20 - (8)
Solving equations (1) and (5) for 6, and
9, we obtain 6, = 0.00439 and 6, =~ 0,00967 Substituting in equations (2), (3) and (4)
gives the moments at these points Maa = 33388 (-.00967) + 16694 x 00439 + 831 = 582 in.lb ty Ms, = 33883 {-.00967) - 259 = -5a2 My 16694 (~.00967) + 742 = 580
Converting these signs to the conventional signs would give Ma =-50, Ms, =-S82 and
M, = -580
Al2.6 Statically indeterminate Frames Joint Rotation Only
The slope deflection equations can be used in solving all types of framed structures
In frames as compared to straight continuous beams, the axial leads in the members are usually more important, however the influence
on joint displacement due to axial deformation of members is relatively small and 1s there- fore usually neglected in most simple framed
structures To illustrate the slope-deflection
method as applied to frames, the structure shown in Fig, Al2.i6 will be solved Example Problem 1 100 i” | fe a ge a 1 L= 20 3 1 = 20 Kei L = 24 L= 24 100 ——*|[ = 40 I = 40f*— 100 K 31,67 K= 1.67 Ls 20 1 = 20 Kzl 3TTTTTTTTTTTTTTTT € 18 lh/in Fig A12 16 Fig Al2.16 shows a closed rectangular
frame sudjected to loadings on its four sides
whicn nold the frame in equilibrium The bending moments at the frame joints will be determined SOLUTION: ~ FIXED END MOMENTS: - Al2.7 Mp,_, = (100x8x12")/20"+ (100x12x8")/ 20" =~ 480 Mp,_, = 480 Mp _, = 100x24/8 = 300 Mp i = - 300 Mp,_, = 10x20°/l2 = 334 Mp,_, = - 334
Due to symmetry of loading we know that
Trang 21Al2.8 SPECIAL METHODS -
Subt in equation (a) Mi-e = 2x 26.3 -480 =~+428 in.15 Subt in equation (đ) Ms-—« = 2(-13,94) +334 = 306 in.lb Fig Al2.17 shows the resulting bending moment diagram A12.T Frames with Joint Dispiacements
In the previous example problems, the conditions were such that only joint rotations
took place, or if any transverse joint dis- placement took place as in the example problem of Art, Al2.5, these displacements were known, or in other words the value of @ in the slope-
deflection was known
In many practical frames however, the joints suffer unknown displacements as for
example in the frame of Pig Al2.18 The term Ø # A/L 1n the slope~deflection equation was derived on the basis of transverse displace- ment of the member ends when both ends were fixed Thus in Fig Al2.18 under the action of the external loads, will sway to the right as magnified by the dashed lines Neglecting any joint displacement due to axial deforma-
tion, the upper end of each member will move through the same displacement A, Then we can
write,
Bea > Minas Bis = W/Laaes Bee F 4/Le—s
Fig, A12 18
SLOPE DEFLECTION METHOD
Due to the fixity at ja
(S), @iea 3 Os-a 3 Osa = 0
unknowns sre Sa-1, S4—3, Sas
There are three statical joint equations of equilibrium available, namely,
Mau +Ma, = 0, Mati = 0, and
Mew + Mos 2 9
Another equation is necessary because
four unknewns This additionai equation is obtained by applying the equations of statics
to the free bodies of the vertical members Fig Al2.19 shows the free bodies M ¬ i \2 T 4 nu TT 6 An | Pi | Lod ! “HT Lena | Lave tice | | dy + là mg | | Lis i Hs ies wi Ma~s dự Ma~« Mi wa Fig Al2.19
Treating each member separately, we Take moments about the upper end and equate to zero
and then solve for the horizontal reaction at
the bottom end The results are,
Đi Q Hạo, Ta aS Ling toe
With these horizontal reactions known the
static equation ZH = 0 can now de applied to the frame as a wnole, which gives,
Py +H, -H, -H, = 0
Equation (20) is generally referr
the bent or shear equation and is an equation
that supplies the necessary extra condition to
take care of the additional unknown 4
Al2.3 Example Probiems of Frames with Unknown
Joint Displacement
Example Problem 1
Fig Al2.20 shows an unsymmetrical frame
with unsymmetrical loading The Sending moment curve will be determined
SOLUTICN: -
Trang 22ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES member is given on ‘she figure The relative K = I/L values are aiso {ndicated adjacent to each member Relative values of % due to sidesway of frame: + The angles @ are proportional to the A/L for each member where A ts Same for members
(1-2) and (3-4) and tak ae
zero for member (2-4) T¡ II Hướni
(See Fig Al2.10) Ị ! í x Hence, Ƒ no > ca Fig A12.20 1 1 Ốa.v= is = ,0667Ø Fig A12 10 1 Ø ° To = O,10Ø Z “ 0 FIXED END MOMENTS: ~ Member 1-2 Mp,_, 7 (48x6"x9)/15* =-69,.12 in.1d, Mp,_, = (48x97x6)/15* = 103.68 Member 2-4, MP, „ =~ (96x12)/8 #- 144 Mp, = 144
Member 3-4 has no lateral loads and thus fixed end moments are zero
The slope-deflection equations are: -
Mina = 2K(26, +8, ~30) +Mp ~ - (17)
Mụ_, = 2X(29,+9, -3Ø) +Mp - - =¬ (18)
Writing above equations for each member
and noting that 6, and 9, are zero Decause frame is fixed at support points, gives, Mì na = 2xO.667(0 + 9„ -ö x 0667Ø) ~ 689.12 Mia "ụ w Dp " i a hộ œ a ` a i a oO be ' ‘ t ' m Me, = 2X 0.667(28, +0-3X „0667Ø) + 105.68 Mj = 2.6679, =O.2667Ø + 105.68 - ¬ - (>) A12.9 Mang = 2X1.667(20 + 0, -0) - 144 Ma_, = 5.6670.+5,.2330,-144 - () Mya = 2x1,667(26, +6, -0) +144 Mya = 6.6676, +3.3336, + 144 - ~ - - - (4) Mạ Z 2x1(0*+9, -3x0.18) +0 My, = 20,-0.68 -+ (a) Mas = 2x1(26,+0-3x0.1f) +0 Mi, = 40,-0.68 - (t) Equilibrium equations: - JOINT (2) Mat tMa = 0 Substituting: - 2.6679, - 0.26679 + 103.68 + 6.6676, + 3.3236, -144=0 whence, 9.5359, - 0.26678 + 3.3330, = 40.32 BQe-s rr ert ~ (g) JOINT (4), MyatMes = 0 6.6676, + 3.3350, + 144+ 49, -0.68 = 0 whence, 10.6676, + 3.3330, -0.60+144 0 - (nh) Writing the bent equation (See Eq 20) H, +H, -48 5 0-+-+ - (4)
In order to get H, and H, in terms of end moments see Fig Al2.21 Man Maes TF Ti 4 10 | ] + 3H How Hy Ms-4
Mi~a Fig Al2.21
Trang 23Al2, 10
Me tM + 10H, = 0
whence, Hy =~ Us=a*Meus) sao =)
Substituting values of Mạ_„ and M,~.s in above equations, gives, Hs, =~0,.60, + 0.28 -~+ -~ (k) Substituting values of H, and H, in equation (1), gives, ~0,26676, - 68, + 15558- 31.12 =0- ~ (1)
Solving equations (g), (h}, (1) for the
unknowns 6,, 8, and @ gives,
$= 196.9, 6, = 12.17, a, = - 6.22 The final end moments can now be ound by substituting these values in equations (a) tơ (f) inclusive M,_„ = 1.355 x12,17 - 2667 x 196.9 ¬ 69.12 =~105.4 In,lb Wa-¿ 2 2.667 x12.17 ¬ 2667 x 196.9 + 103.68 =~83.6 6.667 x 12.17+3.233 (-6.22) — 144 = 83,6 Maw Mya = 6.667 (~6.22) + 3,335 x 12.17 + 144 “ 142,9 4 (-8.22) ~0.6x196.9 =~ 142.9 Mas Ma = 2 (6.22) - 0.6 x 196.9 =- 130.45
Fig Al2.22 shows the bending moment diagram The student should also draw the shear diagram, find axial load in members
Since all these loads are needed when the
Fig Al2,22 strength of the frame members are computed
and compared against the stresses caused by
the frame loading Example Problem 2,
Fig Al2.23 shows a gable frame, carry-
SPECIAL METHODS - SLOPE DEFLECTION METHOD
ing the symmetrical distributed load as shown W = 12 lb/in 10" ị 20" fy - 1 #10] Fig À12.23 L= 20") —_ 1 5 TTm
The dending moment diagram under the given load-
ing will be determined
SOLUTION: -
Relative Stiffness (K values)
Members 1-2 and 4-5, K = 10/20 = 0.5
Members 2-3 and 4-3, K = 20/22.4 = 0.852 Fixed end moments:
Mey = wif/l2 = (12x 207)/12 =-400 in.1b,
Mp, = 400, Mp | = 400, Mp = - 400
Relative values of g: -
Due to the sloping members, the relative
transverse deflections of each member are not as obvious as when the vertical members are con-
nected to horizontal members as in the previous example, In this example, joints (1} and (5)
are fixed Because of symmetry of frame and
loading, joints (2) and (4) will move outward the same distance 4 as indicated in Fig, A12.24,
Furthermore, due to symmetry, joint (3) will
undergo vertical movement only,
In Fig, Al2.24, draw a line from point (2')
parallel to 2-3 and equal in length to 2-3 to
locate the point (3") Erect a perpendicular to 2'-3" at 3" and where it intersects a vertical through (3) locates the point (3') the final location of joint (3) The length or 2'-3'
equals 2-3, In the triangle 3-3'-3", the side
3'~3" = VBE, The relative values of % which
are measured by A/L for sach member can now 5e calculated
1 5-2
Trang 24ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES A1211
M,-a “= 2xO.ð(-29„ +O~3 x ,05Ø) +O Maus F720, -0,159 -+ (g) Mau, = 2K(29, +9, - 2Ø) TP Mong = 2X0.5(0-02 -3 x 059) +0 Meu =-0,-0.158 - (a) Fig A12.24 Ber =- 0.19 Bae = 05f
Substitution in Slope Deflection Equations: -
We Kriow that 9, = 6s = 0, due to fixity
at joints (1) and (5) Also 6, = 0 due to
symmetry or only vertical movement of joint (3) Furthsrmere due to symmetry 9, = - 6, My = 2K(26, +9, - 3%) +Mp Mig = 2x0.5(0+6, + 05 x30) +0 Mi.76,+0.158 + + -+- (a) Mai = 2K(20a +9, ~3Ø) + Mp Ma = 2xX0.5(20, + 05x3} +0 Ma = 20, +0,15f- - (b) x a | ° a t K(20, +6; -5Ø) tMp — Wạ_¿ = 2x0,862(26; + O~= ð 0.1) ~400 3.6686, -0,5352$-400 - ~ - = (c) Mo = 2K(26, +, - 30) + Mp, Maa = 2x 892(0+ 0, -3x0.18+ 400 Myung = 1.7348, ~ 0.53526 + 400 - - - - (4) Moog = 2K(09, +98, - 308) + MeL 2(0-6, +3x 0.19) -400 Mgug =-1.7846, +0,.53529-400 - - = (a) z 2K(26, + -3Ø) + Hy _ M,_, = 2x Ê92(-26, + Ó +3 x 9.1) + 400 ~n Œ) x =-3.5680, + 9.53520 1 + 400 My, = 2k(29, +9, ~3Ø) + ase JOINT (2) Equilibrium equation Ma.tMas = 0, substituting - 262 + 0,150 + 3.5686, - 0.53528 -400 = 0 whence, 5,5686, -0.38529~+400 > 0 - - - - (1)
The joint equilibrium equations at (3) and (4) will not provide independent equations be-
cause in the previcus substitutions in the
slope-deflection equations 6, was made equal to zero and & was equal to-~@,
Shear Zquaticns: -
Due to symmetry the horizontal reactions
at points (1} and (5) are equal and opposite
and therefore in static balance Since we have
two unknowns 6, and @, we need another equation to use with equation (1) This equation can be
obtained by stating that the horizontal reaction
Ha, on member 2-1 at end (2) must be equal and
Trang 25Al2, 12 Shear equation, Her = Has whence, li o 20 Substituting values for the end moments: - 8, + 0.158 + 20, + O,15đ = +5688, - O45 20 10 525 = 400 + 1,7849;, =0 10 B + 400 + £300 whence, 0.38520, ~0.1228+24020 - (4) Solving equations 1 and j for 6, and @, sives @ = 2805, 6, = 265.8
Substituting these values in equations (a} to (n), the end moments are obtained as follows: - Mie = 6, + 0.159 = 265.8+ 0,15 x 2805 = 666.5 in.lb May = 20 + 0.158 = 531.6 + 420.7 = 952 Macs = 3.5680 - 0.53529 - 400 3.568 x 265.3 - 0.5352 x 2805 = 400 = - 952 Mao 5 1.7846, - 0.53529 + 400 = 1.784 x 265.8 ~ 0,552 x 2805 + 400 = 628 Ms, =- 1.7848, + 0.53529 - 400 =~ 629 M,_„ = 952 My, 7-952 952 686.5 đu Fig Al2.27 SPECIAL METHODS - SLOPE DEFLECTION METHOD mattsr known tr calculated, A12.9 Comments on Slope-Deflection Method The exampl that the metho: rapid
Thus for structures reduncancy, the slope- De considered as 0osei51,
sclution The solving of the equat
method are readily programmed for solution
high speed computing machinery
A12.10 Problems,
{1} Determine the bending moments at surport points 4, B, C, D, for the continuous team shown in Fig 412.28 100 100 |mrLsn~Lm re Flø A12, 28
(2) Same as problem (1) but consider support a as freely supported instead of fixed
am “or
(3) Determine the bending moment diag