ANALYSIS AND DESIGN OF F influence coefficients Ans, 21.8 27.2 27.5 18.3 1 27.2 68.5 65.4 26.4 ['m] = 3 27.5 65.4 68.4 26.2 18.3 26.4 26.2 20.6 (12) Re-solve the doubly redundant beam of Example Problem B, page A8.3 by matrix methods The redundant reactions should be given "q" symbols (See Example Problem 12a; page A8.20)
(13) Re-solve Example Problem 5, page AS.12 by matrix methods For simplicity, make your choice of generalized forces ineluding those designated as X and Y in the example so that Figs A8.21 and A8.22 can be used to give the Zir loadings
(14) By matrix methods re-solve Example Problem 4, p A8.12 using 3 equal bay divisions along the panel (3 times redundant) Use the same structural dimensions as in Example Prob- lem 20, p A8.36 Compare the results with those obtained from the formulas developed in Example Problem 4
(15) Show that the matrix equation
eq (21) is modified to cover the initial stress
LIGHT VEHICLE STRUCTURES A8, 43
problems of Art A8.8 by wr1ting
(rd (x = Ln (4 - [el 4}
where A; is the initial imperfection associated with force qị Refer to the argument leading
to eq (11) of Art as.8
(16) Using the equation of problem (15), above, re-solve Example Problem 6, p Â8.14
(17) Using the equation of problem (15)
above, re-solve Example Problem 7, p A8.14
(18) Using the matrix methods of Art A8.13, re~solve Example Problem 9, p A8.15
(19) For the doubly symmetric four flange box beam of Example Problem 15, p A8.24, de-
termine the redundant stresses da, q7 and qia lf
one flange is heated to a temperature T, uniform spanwise, above the remainder of the structure Ans da 2081 ~Eaf Gr >= To 1599 đa 741 REFERENCES
See references at the end of Chapter A-7
Douglas DC-8 airplane Photograph showing simulated aerodynamic load being applied to
Trang 2
AB 44 STATICALLY INDETERMINATE STRUCTURES
Trang 3
CHAPTER A9 BENDING MOMENTS IN FRAMES AND RINGS BY ELASTIC CENTER METHOD A9,1 Introduction
In observing the inside of an airplane fuselage or seaplane null one sees 4 large num- ber of structural rings or closed frames Some appear quite light and are essentially used to maintain the shape of the body metal shell and to provide stabilizing supports for the longi- tudinal shell stringers At points where large load concentrations are transferred between body and tail, wing power plant, landing gear, etc., relatively heavy frames will be observed In null construction, the bottom structural fram- ing transfers the water pressure in landing to the bottom portion of the null frames woich in turn transfers the load to tne hull shell
In general the frames are of such shape ani the load distribution of such character that these frames or rings undergo bending
forces in transferring the applied loads to the other
=
a iB
- b
resisting portions of the airplane body e cending forces produce frame stresses in
which are of major importance in the sth proportioning of the frame, and thus a reasonable close approximation of such benc-
ing forces is necessary
Such frames are statically indeterminate relative to internal resisting stress and thus consideration must be given to section and physical properties to obtain a solution of the distribution of the internal resisting forces General Methods of Analysis:
There are many methods of applying the principles of continuity to obtain the solution gor the redundant forces in closed rings or frames and bents The author prefers the one
is generally erred to as the "alastic usea it for many years {nl nm The method was orizi-
The main difference to most other methods al coint called
2 yuller-Breslau, H., Die Neveren Hethden der Festigkeitslehre und der Statik der Baukon-
struktionen Leipzig, 1886 A9
Assumptions
In the derivations which follow the distor- tions due to axial and shear forces are neglect- ed In general these distortions are small compared to frame bending distortions anc thus the error is small
In computing distortions plane sections are assumed to remain plane after bending This is not strictly true because the curvature of tne frame chantes this linear distribution of 5end~ ing stresses on 2 frame cross~section (Correc- tions for curvature influence are given in
Chapter AlS
Furthermore it is assumed that stress is proportional to strain Since the airplane stress analyst must calculate the ultimate strength of a frame, this assumption obviously does not hold with heavy frames where the rup- turing stresses for the frame are above the pro~ portional limit of the frame material
This chapter will deal only with the the~ oretical analysis for bending moments in frames and rings by the elastic center methed Prac- tical questions of body frame design are covered
in a later chapter
Trang 4A9.2 BENDING MOMENTS
A9.2 Derivation of Equations, Unsymmetrical Frame
Fig AQ9.1 shows an unsymmetrical curved beam fixed at ends (4) and (B) and carrying some external loading Pi, Pa, etc This structure is statically indeterminate to the third degree because the reactions at (A) and
(B) have three unknown elements, namely, magni- tude, direction and line of action, making a total of six unknowns with only three equations of static equilibrium available
Fig AQ 2
In Pig A9.2 the reaction at (A) has been replaced by its 3 components, namely, the forces Ky and Ys and the moment M, and the structure
is mow treated as a cantilever beam fixed at end (B) and carrying the redundant loads at (A) and the known external loading P,, Pz, etc Because joint (A) is actually fixed it does not suffer translation or rotation when structure ts loaded, thus the movement of end (A) under the loading system of Fig A9.2 must be zero Therefore, three equations of fact can be written stating that the horizontal, vertical and angular deflection of soint (A) must equal
zero,
IN FRAMES AND _ RINGS
Consider a small element ds of the curved beam as shown in Fig A9.2 Let M, equal the bending moment on this small element due to the given external load system The total vending moment on the element ds thus equals,
M=Msg + My - Xay + Ygx
(Moments which cause tension on the inside fibers of the frame are regarded as posi- tive moments )
The following deflection equations for point (A) must equal zero:-
@ +0, (angular rotation of (A) = zero} dy = 0, (movement of (A) in x direction = 0)
dy 20 {movement of (A) in y direction = 0} from Chapter Ă7, which dealt with deflection theory, we have the following equations for the movement of point (A):-
In equation (2) the term m is the Dending moment on a element ds due to a unit moment applied at point (A) (See Fig A9.3) The bend-
ing moment is thus
equal one or unity A
for all ds elements (a SS
of frame thit moment Fig A9.3 Then substi- tuting in equation (2) and using value of M from equation (1) we obtain - i xds 1.ds l.yds = 9e z8 + Mạ ST” - XAĐ ĐT + MZ = 0 whenee;
a = PSE + mynd NUIẾT + YUẾT =o (5)
In equation (3) the term m represents the bending moment on a element ds due to a unit load applied at point (A) and acting in the x direction, as illustrated in Fig Ag.4
The applied unit y
load has a positive 1 ds sign as it has been ‘A +
assumed acting toward wien
the right The Ị B
distance y to the ds ỹ
element is a plus Fig A9.4
Trang 5ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES measured upward from axis x-x through (A)
However the bending moment on the ds element shown is negative (tension in top fibers), thus the value of m= - (1) y = -y The minus sign
is necessary to give the correct bending moment sign
Substituting in Equation (3) and using M from Equation (1):-
2
Âc = 815 - tuyỆt ‹ xyr OS _ va35 5» os (6)
In equation (4) the term m represents the
bending moment on a element ds due to a unit
load at point (A) acting in Y direction as
illustrated in Fig A9.5 Hence, m = l(x) =x Substituting in equation (4) and y ds using M from Le equation (1), we _ x a obtain, 1# Fig A9.5 - sxds xds ‘ds ay = SES + yee - xr + x7ds _ „ YAZ TT Z0 TT nen —— (7)
Equations 5, 6, 7 can now be used to solve for the redundant forces My, X4 and Yy With these values known the true bending moment at any point on structure follows from equation
(1)
REFERRING REDUNDANTS TO ELASTIC CENTER
For the purpose of simplifying equations Š, 6, 7, let it be assumed that end A is attached to a inelastic arm terminating at a paint (0) as tllustrated in Fig A9.6 The point (0) coincides with the centroid of the ds/EI values for the structure Reference axes x and y will now be taken with point (0) as the origin The redundant reactions will now be placed at point (o) the end of the
elastic Fig A9.6
inelastic bracket, as shown in Fig A9.6 Since point A suffers no movement in the actual structure, then we can say that point (o) must undergo no movement since (0) is connected to point (A) by a rigid am
Thus equations 5, 6, and 7 can be re- written using the redundants X,, Yj, and M, in
Ag, 3
place of Xa, Yy and Mg respectively
The axes x and y through the point (o) are centroidal axes for the values ds/EI of the structure This fact means that the summations-~
yas Xds _
5 ar 7° and 5 Er 7°
The expressions 2 x"ds/EI, & y*ds/EI and 2 xyds/El also appear in equations 6 and 7 These terms will be referred to as elastic
moments of inertia and product of inertia of the frame about y and x axes through the elastic
center of the frame, and for simplicity will be
given the following symbols
xfds _ y 24s „ xyds _
Bar “ly: 3 gr Flee BAGS hy
Equations 5, 6 and 7 will now be rewritten using the redundant forces at point (0) = > Mgds ds 6 = 2% + Mod EFF 0 hence, ~ Meas My = eqs ee -e eee (8) „ = -E ST + Xoly - Yolyy = 0
The term MgZ yds/EI is zero since % yds/EI is zero, thus My drops out when substituting in Equation (6)
- ~~ (9)
= M„dsx <
dy = By - Xolyy + Yyly = 0 - = - -(10)
The term Mcds/EI represents the angle change between the end faces of the ds element when acted upon by a constant static moment Mg This angle change which actually is equal in value to the area of the M,/BI diagram on the element ds will be given the symbol @,, that is, as = Msds/BI with this symbol substitution, equations 8, 9, 10 can now be rewritten as follows: -
Mo * ag7r - Tt (4)
~Z Osy + Xp], - Yolyy 20 > - (12) ZB BgX - Xolyy + Ygly FO - - = (23)
Trang 6
A8.4 BENDING MOMENTS IN FRAMES AND RINGS
Xốxy ~ 3Øsx (4) Table A9.1
Xo 2 NET Trẻ am (4) 7 i iy
~ 2% Por-|,,.ds jdist dist -
Tx (a xX L) tien | XÍT from from we) wy Fwy trig twx2viy lex} yay 2z 2 wy 3250 lwx=1440 ~ [205% - 35sy (]] AB |Šs10| 15 | -12 [-120/+180 230% 2750 |iys 20 9= 0 Yg = oF 1 (2 - Tx) 1ccee (15) 2 12x3 2x3 y Tư y ) BC P=12| 30 | 0 | 0 | 360 wy2z10800 J|wx2z0 24 5g fie 28 2576
A9.3 Equations for Structure with Symmetry About One Axis 1x? 79533" at: through Elastic Center
30 wy222250 |wx2=1440
I7 the strueture 1s such that either the x | | CD |fÿ71012 j 12| 120| 1850| „ rap tự rÍn a0
or y axis through the elastic center is a axis
of symmetry than the product of inertia sum! 32 I 0 | 660 16800 3456 ixyds/EI = lyy = zero Thus making the term
lxy = 0 in equations ll, 14 and 15 we obtain, S8 5,
Họ “sgặiy TT TT TT Tan (16)
— (a7)
Yo 228s 2 y TT ST (18) A3.4 Example Problem Solutions
One Axis of Symmetry Structures with at least Example Problem 1 Fig A9.7 shows a rectang- ular frame with 1 fixed supports at points A and | D, and carrying a single load as | shown The l problem is to de- |
termine the bend-
ing moment dia~ |
gram under this ! loading | 30" The first Fig A9.7 step in the Solution is to
find the location of the elastic center of the frame and the elastic moments of inertia 1x and ly
Due to symmetry of the structure about the Y axis the centroidal ¥ axis 1s located midway between the sides of the frame, and thus the elastic center (0) lies on this axis
Table A9.1 shows some of the necessary calculations to determine the location of the elastic center and the elastic moments of inertia The reference axes used are x*=x?
and y-y,
are the elastic moment of inertia of each sort {on of the frame about its centroidal x and y exes Since I is con- Stant over each pertion the centroidal moment of inertia of each portion is identical to that of a rectangle about its centroidal axis
The terms 1, and
To explain for member AB:- Referr.ng to Fig a, tx = gon = ae $x 30s = 750 “7 Tf | LÍ Referring to Fig b, au + Fig a
t Ÿ = one =i x 20x iis 09 (negligibie) “12 12 a:
The distance from the a two reference axes to the — elastic center can now be i calculated: - Lf ds230sb it - Ị ÿ = BY = $89 = 20,625 in | tes ¬- OL "Hy mjrạ?h x= Sy “ap 70 Fig b
Having the moment of inertia 2bout axis x*x* we can now find its value about the centroidal axis xx of che frame, by use of the parallel axis theorem
Ix “ ly - 3⁄4(ÿ*) = 16800 ~ 32 x 20.6262= 2188
ly = Tự ~ Ơ3w(X?®) = 3456 ¬ 32(o) = 3456
Trang 7ANALYSIS AND DESIGN OF PLIGHT VEHICLE STRUCTURES
uM #2 25 _ Area of static M/I diagram
0 548/1 Total elastic weight of structure Moment of static M/I diagram about x2 2Bsy - x axis 9 Tx Elastic moment or inertia about x axis Moment of static M/I diagram about 1, = ak „ y axis 9 ly Elastic moment of inertia about y axis
Thus to solve these three equations we must assume a static frame condition consistent with the given frame and loading In general there are a number of static conditions that can be chosen For example in this problem we might select one of the statically determinate
conditions illustrated in Fig A9.8 cases 1 to oO 10# Ỳ 10 i | 7 " eon TTT | a Case 1 Case 2 Base "Case I “case 5 Fig A9.8
To illustrate the use of different static conditions, three solutions will be presented with each using a different static condition Solution No 1
In this solution we will use Case 3 as the static frame concition The bending moment on the frame for this static frame condition is given in Fig A9.9 The equations
45% ig diagram 32.5 Ms/I diagram Bi Cc B, Cc toa | ¬- I i Mg is positive, tension Fie Assl | | on inside of frame, apap ` * bk Tt "2 3% Fig A9, 10
for the redundants require J, the area of the
Ms/I diagrem, Fig 49.10 shows the Mj/I curve xăxch is obtained by dividing the values in
Fig, A9.9 by the term 2 which is the moment of inertia of member BC as given in the problem
Since the equations for Xg and Y, require the moment of the Ms/I diagram as a load about axes through the elastic center of frame, the
area of the M,/I diacram will be concentrated
at the centroid of the diagram and along the centerline of the frame, or more accurately
Ag.5 along the neutral axis of the frame members
In Fig 49,10 the area of the M,/I diagram equals @g = 22.5 x 24/2 = 270 The centroid by simple calculations of this triangle would fall 10 inches from B Fig A9.11 now shows the frame with its Mg/I or its @; load @, is Øs= 270 Z“H Lan positive since Mg 1s
Ba c positive The next
step is to solve the equations for the x -¬¬gE—~-+ x redundant at the
20,625" elastic center The Signs of the distances A ' D + x and y from the axes
' x and y are conven-
Fig AQ 11 tonal Thus, = 728s - (270) Zas/t * Gi (roa Table Aộu | ~ “9-487 in-1d Xo = aise C= 0.7989 1D, Ắ—5Ø<X „ -(270)(-2) _ Yo a se = 0.1562 1b
Fig A9.12 shows these values of the re- dundants acting at the elastic center,
Fig A9.12 Fig A9 13
The bending moments due to these redundant forces will now de calculated My = - 8.437 - 1562 x 12 + 7939 x 20.625 2 6.06 tn.1b Mg = - 8.487 - 7939 x 9.375 - 1562 x 12 = ~17.76 1n.15, Mc = - 3.487 - 7939 x 9.375 + 1562 x l2 = -14.00 in.1b Mp = - 8.437 + 7939 x 20,625 + 1862 x 125 9.81 tn.1d
Trang 8AS 6 BENDING MOMENTS Adding this bending moment diagram to the static bending diagram of Pig A9.9 we obtain the final bending moment diagram of Fig A9.13
The final bending moments could also be obtained by substituting directly in equation (1) using subscript (o) instead of (A) Thus,
MaMg + My = Xo¥ + ¥ox
For example, determine bending moment at point 8 For point B, x substituting in (19) = -12 and y = 9.375, Mg 20 =O + (-8.437) = 7939 x 9.375 + 1562 (-12) = -17,75 as previously found Mp AT POINT D x = le, y = 20.625, Mg = 0 Mp = 0 + (-8.437) - 7939 (-20.625) + +1562 x 12 = 9.81 in.ib Solution No 2
In this solution we will use Case 4 (See
Fig A9.8) as the assumed static condition,
that is two cantilever beams with half the external load or 5 lb acting om each canti- lever Fig A9.14 shows the static bending moment diagram and Pig .A9.15 the M,/I diagram g0s=~45 32405 Se S# sĩ -sebăi Lee<ẤLso any Pfs Z2 TNN =| sim| |F109 6¬ 2 “4 lef eas is ads | B |e Pig A9 14 & co 2 32-30) /Os=-doo E2 Ca tot ! I ^^ ly -30 -80 ‘ 10 sig ages
Fig 49.15 also shows the results of calculating the Bs Value for each pertion of the Mg/I diagram and its centroid location Substituting in the equations for the re~ dundants we obtain, ~ 28g _ = (+45-405 Mo * Sas7r = B00 900) = 51.56 In,ib, xạ = 208¥ = (-45-405)9.375+ (-900-300) (-5.625) ox 2188 = 0.7939 1b, IN FRAMES AND RINGS ~3Øsx „ -L—+5(~19)-405x6-300(~12)~900x12] Yo 5 Ty 3e = 2.656 1b The final moments at any point can now be found by equation (19) Consider point B:~ Mg = -3O trom Fig a9.14 x= -12, y = 9.375 Subt in (19) Mp = -30+51.56-.7939x9.375+2.656(-12) -17.75 in.lb which checks first solution uk Consider point D:- Mg = +90, x = 12, y = -20.625 Subt in (19) Mp = -90+51.56 - 7939(-20.625) + 2.656 x 12 9,80 ín.lb, Solution No 3
In this solution we will use Case 5 (Fig A9.8) as the assumed static condition, namely a frame with 3 hinges at points A, D and 5, as tllustrated in Fig A9.16
Before the bending moment dia- gram can be caleu~ lated the reacticns at A and D are necessary 30" To find Vp take moments about point A Hp?l SM, = 1x6 - 24Vp baer 5 Vpz2 5 4 =o hence, Vp = 2.5 To find Vg take EFy = 0 = ~10 + 2.5 + V, TA ¥y A =o hence Vy = 7.5
To find Hp take moments about hinge at E of all forces on frame to right side of = and
equate to zero,
Trang 9ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES The frame static bending moment ciazram can now be calculated and drawn as shown in Fig AQ.17 ~30 -80 -30 |(2) (5) -30 CMC) Pin e ì 9.375 m——————> x22 = (8) Ds? +150 | ep Os?~150 QU - — — —*= ———x \ ys es › elastic center i J \ i | ' 20 625 20 Ị j y i Fig A9,17 D cPi OPin
The moment diagram is labeled in 6 parts 1 to 6 as indicated by the values in the small circles on each portion Most of the calcu- lations from this point onward can be done conveniently in table form as illustrated in Table A9.2 Table AQ 2 i 2 3 4 § 6 7 8
Mom z 1 „|x dist.| y = dist
Dia- | Area] ro ®s “| trom Y | from X
gram| of Beam a | axis to | axis to Osx | Oey Por- | Mom Section} 1 C.g of | C.g of tion | Dia % Øs 1 ~450 3 ~1580| -12.00 | -0.625 | 1800| 93 2 - 60 2 = 30) -10.67 9.375 320|~281 3 15 2 7.8) - 6.67 9.375 |- 50} 70 4 45 2 22.51 - 4.00 9.375 j~ 90) 211 5 -180 2 5 | sum |
In order to take moments of the Js values in column (4) of the table, the centrotd of each portion of the diagram must be determined For example, the centroid of the two triangular bending moment portions marked 1 and 6 is
«867 x 30 from the lower end or 20 inches as shown in Fig A9.17 Thus the distances x
na y from this Zs location to the y and x - 90 8,00 9.375 |- T20|-844 ~480 3 ~189 12.00 | -0.625 |-1800| 93 ~390 ~ 540|-888 A9.7T axes through the frame elastic center are then calculated 45 -12 and -0.625 inches respect- ively = 7 2ds - (-390) _ My = Fae = YS = 12.10 tn, vv, X= St = = -0.203 1b - -ỔsX „ - (-540) _ s2 1 Yo z = ase = 9.1562 1b,
The final moments at any point can now be found by use of equation (19), namely MSM +My - Xyy + 5x Consider point B:- x =-12, y= 9.375, Ms = -30 Substituting - Mg = -30+12.19-(-0.203)9.375+0.1562(-12) = =17.76 in.lb.(checks previous solu- tions) Consider point D:- ` x=12, yˆ-20.625, Mẹ ^o Subt in (19) Mp = 0+12.19-(-0.203) (-20.625)+0.1562x12 = 9,82 in.lb (checks previous solu- tions)
The final complete bending moment diagram would of course de the same as draym in Fig AQ.13 for the results of Solution No 1 Example Problem 2
Trang 10A9.8 BENDING MOMENTS IN FRAMES AND RINGS ƒB iB ẽ | : i Ữ ! E.c.* —=cut ‘A Di on the s 8), these resultis ( iF portions of
Fig A9.19 Fig Ag 20
Mg diagram for w icading Mg diagram for Pa load
Tne Tirst step is to find the location of the slastic center of the frame Due to
syunet TY of ?rame about the y axis, the elastic enter will be on a y axis through the middle or the frame The vertical distance ¥
measured from 2 axis through AD equais, = (3) Ÿ 24+ vi2x2
? rẽ ——=
1=
The next step is to determine the elastic moment of inertia of the frame about x and y axes through the elastic center of the frame
Moment of inertia about x axis = Iy t- Members AB and DC, 1 =($ xậ x 14.88) ø = 653.9 x\5*5 - ~ ` ( xặx 9.879] 2 = 200.9 Member BC = (30/4) (14,337) = 1540.0 (30/2) (3.67) = 1402.7 Ty = 3797.5 Moment of inertia about y axis = ly is Members AD and 3c, = 3 x + x 30% = 563 1 1 = ia *ox 308 = 126 Members AB, CD, ly ( 3 Je = 3600 Table A9.3
Portion | Area Og > = dist.ly = dist.|
Dlagremi Boruen Ha Axis Ans | 2s% | Gạy () TA 1 i - 1440/3|- 480] ~15 10.33 7200- 4958 2 ~ 720014|~ 1800 9 14.33 0ị~ 25794 3 ~ 14403J~ 4801 15 10.33 /- 7200|- 4958 4 144013) 480} 15 - 5.67 7206|- 2726 5 3600|2| 1800! 5 ~ 9.67 9000{/- 17406 6 ~ 45000|4|~11250 T1, a m đ3 |= 84375|~161212 7 ~108000/3/-36000 15 2.33 |~5400001- 83880 8 ~ 67500|21-33750) 5 ~ 9.67 |-188750| 326363 9 3240/3} 1080 15 - 3.67 16200|- 3964 10 5400/2) 2700 5 - 9.67 13500l~ 26109 sum |- 77700} ~T47225|- 4845 Solving for the redundants at canter, = = 285 ~(-77700) _ 38.5 2£0ie in.lp -(-747225) _ 5289 ~ h hă th bea ơ
Fig A9.21 shows th acting at the elastic dal shows the 5ending morent a
redundant forces The calculations ;
a0 ®
Trang 11
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES -84 4153 —15— saat 14.3 I E LÔ ĐẤU 22 E] Fig a9 22 1 | T EB : sas L41.28 om -HẾ 4124 ụ pt Fig AS 21 My = 2018 - 141.28 x 15 - 1,22 x 9,67 = -118 Mg = 2018 - 141.28 x 15 + 1.22 x 14.33 = ~84 Mo = 2018 + 141.28 x 15 + 1.22 x 14.33 = 4153 Mp = 2018 + 141.28 x 15 = 1.22 x 9.67 = 4124
Combining the bending moment diagrams of
Flgs A9, 18, 19, 20 with Fig A9.22 would
give the true or final oending moment diagram Example Problem 3 Circular Ring
Fig A9.23 shows a circular ring of constant cross-section subjected to a sym- metrical loading as shown The problem is to determine the bending moment diagram 50 50# +8 a Center 1 Bracke! TECK Ng car! \ \ e rT Mg@el7Ts A ~/ H at 2 > Fig A9.25 Moments due to redundant forces Fig A9.26
Solution Due to symmetry of the ring structure the elastic center falls at the center of the ring Since the ring has been assumed with constant cross-section, a relative value of one will be used for 1
Total elastic weight of ring =
ds _ x (36)
BERT = 113.1
The elastic moment of inertia about x and y axes through center point of ring are the same for each axis and equal
Ix = ly = or® = 1 x 1e* = 18300
The next step in the solution is to assume a static ring condition and determine the static
(Mg) diagram In general it is good practice to try and assume a static condition such that the M, diagram is symmetrical about one or if possible about both x and y axes through the elastic center, thus making one or both of the redundants Xo and Yo zero and thus reducing considerably the amount of numerical calcula- tion for the solution of the problem
In order to obtain symmetry of the Mg diagram dnd also the Mg/I diagram since I is constant, the static condition as shown in Fig A9.24 is assumed, namely, a pin at (e) and rollers at (f) The static bending moment at points (a), (b), (c) and (d) are the same
magnitude and equal,
Mg = 50(18 - 18 cos 459) = 265 in.1b The sign is positive because the pending moment
produces tension on the inside of the ring The next step is to determine the Js, 05x and Zgy values
Øs is the area of the M,/I diagram, how- ever since I is unity it is the area of the My diagram The static M, diagram of Fig A9.24
is divided into similar portions labeled (1) and (2) Hence
zi.) = area of portion (1) = Pr’(q -sin a), where P = 50 lb and a = 45°
Substituting and multiplying by 4 since there are four portions labeled (1).,
Os(1) = 4[ 50 + 18*(0.786 ~ 0.707) | = S082
The area of portion labeled (2) equals,
Pr*e(1 - cos a)
Since there are two areas (2) we obtain,
Bs (a)=2[50 x 18* x2 (1~0.707) ] = 16000
Hence, total Bs = 15000 + 5052 = 20052
Trang 12Ag 10 BENDING MOMENTS with the center or elastic center of the frame,
the terms 29,x and Z%sy will be zero
Substituting ta determine the value of tne redundants at the elastic center we obtain, = 7s - (20052) Mo *Sas7i *~Tigp 7 7 177 inet Xo = ỔaY - 20052(0) „ Tx T8200 ~3Ø2x „ ¬(20052) (0) = =^“ 2” * oe es Yq Ty 18300 9
Fig A9.25 shows the values acting at the elastic center and the bending moment diagram produced by these forces Adding the bending moment diagram of Fig A$.25 which is a constant value over entire frame of -177 to the static moment diagram of Pig A9.24 gives the final bending moment diagram as shown in Fig A9.26, Example Problem 4 Huil Frame
Fig A9.27 shows a closed frame subjected to the loads as shown The problem is to determine the bending moment diagram d,:/r8es2 Tig A8, 30 Fig, A9.29 lcan2 IN FRAMES AND RINGS Solution:
The first step is to determine the elastic center of the frame and the elastic moments of inertia Table A¢.4 shows the calculations A reference axis x'-x' has Deen selected at the midpoint of the side AB Since a static frame condition has been Selected to make the Mg diagram symmetrical about y axes through the elastic center {see Fig A9.28), it ts not necessary to determine ly since the redundant Yq Will De zero due to this symmet Table A9.4 Length ds! Member} ds |t Ws T] 9 |wy | ig = ixe wy? BDB' [94.25 1.5| 82.82r „ 493080 |5400+151200 = 156600 " 248.1 AB 60.0 [1.0] 60.0 9 QO |18000 + 0 = 18000 A'B' |60.0 |1.0 60.0] 0 |0 |18000+0 = 18000 AC 38.4 12.0 18.2| ~42 806922 + 33850 = 34772 ca’ |38.4 [2.0 19.2] -42 |-89064922 + 33850 = 34772 Totals 221.2 1468 262140
In the last column of Table 49.4 the tern iy 1s the moment of inertia of a particular member about its own centroidal x axis, Thus for member BDB; D „ ore ty * T = ,3X ð0°/1,5 2 54C0 For members AC and CA , ly ® yg dL at
= Aedes axeae=e22 i2*2
Let ¥ = distance from X'X' ref axis to
centroidal elastic axis X-x H3 Swy 21468 3w ` Z21.2 2 6.54 in By parallel axis theorem, Ix = Ig = 6.847 (Ew) = 262140 - 6.84" (222.2) = 252400
Trang 13ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Ag it general shape of static moment curve as shown
in the figure
Consider member BDB',
The bending moment curve will be con= sidered in two parts, namely (1) and (2) The tern #5 represĩnts the area of the Ms/I diagram Thus for portion (1) and (1') Ổg(v) + Ốg(v') # aE ta «sin a) a =3 S000 80" (9,524 - 0.8) 3 172800 The vertical distance from the line 5B’ to centroid of Mg curve for portion (1) and (1') is, ˆ 2 r(1 = cos a - =e) 30(1 ~ 0.847 - 234) a-sina 0.524 - 0.5 yr = 10 tn
For portion (2) of the Mg diagram the area of the M,/I diagram which equals @, is
pea a
5g (4) : i (1 - cos a} £000 xs x2.1 q-0,367)
= 1,007,000 Consider member AC,
From free body diagram of bottom portion of frame (Fig 49.31) the equation for bending i 5000 momen! — /4880 6000 ` Sy Fig A9 344, on member AC equals: My = 4690 x -100 x* Area of M/I curve between A and C when I = 2 38.4 1⁄2 (4690 x - 100 x#}cx - 3690 x3 100 x2] 78-4 = 1/2 |e _ 00 = 784000 9 Distance to centroid of M/I curve along line ac from A 38.4 1/2 (4690 x 100 x*)xdx 9 x= 78400 38,4 —.„„[ 4690 x° lọ6 x - # =1⁄2 =e = *] = 21.77 _ 9 78400
Vertical distance from line AA' to centroid =
21.77 x 24/36.4 # 13.4", The static moment
weight for afc! {ts same as for AC, thus
ac + By'gt 2 2 x 784000 = 1568000
Pig, A9.29 shows the frame with the moment weights 23 located at the centroids, together
with the redundant forces Mg and X, at the elas-
tic center It makes no difference where the frame is cut to form our residual cantilever, if one of the cut faces is attached to elastic center and the other ts considered fixed With the elastic properties and moment weights known the redundants can be solved for: Ma 225 = (1007000 + 172800 + 156800) _ 9 Tw 21 12440" 22sy - Xa ° =ẽ<S^S* = “TY =1007000 + 48,16 + 172800 x 35.36 + 156800 x ~5.24 252400 = ~98,6z£
Yo is zero because of the symmetrical frame and loading The final or true bending moment at any point equals Ms Mg + My + Xp¥ Thus for point B Mg = 0 + (-12420) - (-98.5 x 23.36) = -10130"° For point C Mg at point C = 4690 x 38.4 ~ 200x (38.4) */2 = 22700 Hence Mg = 32700 ~ 12430 - (~98.5 x ~60,64) = 14300"
Fig, AQ.30 shows the general shape of the true frame bending moment diagram
Trang 14Ag 12 BENDING MOMENTS half of a symmetrical null frame
in an actual seaplane, Tne main external load on such frames is the water pressure on the hull bottom plating The hull bottom stringers transfer the bottom pressure as concentrations on the frame bottom as shown The resistance to this bottom upward load on the frame is provided by the hull metal covering which exerts tangen~ tial loads on the frame contour The question as to the distribution of these resisting forces is discussed in later chapters In this problem the resisting shear flow in the null sheet has been assumed constant between the chine point and the upper heavy longeron Fer analysis purposes the frame has deen divided into 20 strips The centroid of these strips “ located on the neutral axis of the frame sec~- tions are numbered 1 to 20 in Fig A9.21 The tangential skin resisting forces are shown as concentrations on frame strips #6 to #16 Ôn the figure these tangential loads have 5een replaced by their horizontal and vertical com~ ponerits The sum of the vertical components should equal the vertical component of the bottom water pressure
that was used
Table A9.S5 shows the complete calculations for determining the bending moment on the frame Calumns 1 to 7 give the calculations for the ĩlastic properties of the frame, namely the elastic weight of the frame; the elastic center location, and the elastic centroidal moment of inertia about the horizontal ventroidal elastic axis A reference horizontal axis X'X' has been selected as shown All distance recorded in the table have been obtained by sealing fri a large drawing of the frame
The static condition assumed for comput ing the Ms moments is a double symmetrical canti- lever beam as illustrated in Fig A9.32 The frame is cut at the top to form the free end of the cantilever beams, and the fixed end has been taken at the centerline bottom frame section The static bending moment diagram will be symmetrical about the y axis through the elastic center of the frame and thus the redundant Yg at the elastic center will be zero since the term 2@sx will be zero The calculations for determining the static moments Mg in Column 8 of Table A9.5 are not shown The student should refer to Art A5.9 of Chapter AS to refresh his thinking relative to bending moment calculations on curved beams
Columis 9 and 10 give the calculations of the @s values (area of Mg/I diagram) and the
first moment (%.x values) The summations of
columns 3, 9, 10 permits the solution for the redundants Mg and X_ as shown below the table, The final bending moment M at any point on the
frame is by simple statics equals, M= Mg + My - Xọy,
IN| FRAMES AND_ RINGS
The moment of inertia of the frame
cross sections are given in column
Trang 15ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Ag 13 Table A9.5 Calculation of Frame Elastic Properties Calculation of Moment Weights and Solution of Redundants Col 1 2 3 4 5 8 T 8 9 10 11 12 13
Elastic | Arm {| wy' w yt Arm Static Moment Os y Mo ~Xoy *Total
: a= Weight to to Moment ; Weight Moment = zB Ref Axis Dg = M 3 as weds | Axis xx Mg Ms w = a t | xx zy TOOT ds sự 1 7.6 „065 | 117 00 26.1 | 3050 79700 17.2 g 8 9 -9020 16650 7630 2 7,5 ,063 | 117.00 24.4 | 2850 69500 15.5, 0 0 9 - 15020 6000 3 T6 065 | 117.00 21.4 | 2497 53300 12.3 0 0 9 * 12100 3080 4 7,6 065 | 117.00 17.3 | 2025 35020 3.4 0 8 Q “ 8130 - 890 5 7.6 065 | 117.00 12,0 | 1403 16820 3.1 Q Q 9 Bị 3000 -6020 6 6.4 25 25 60 5.9 181 892 | - 3.0 1220 31.2 | - 93.7 ‘ =_2910 | -10710 T7 §.4 125 25.60 9 of 9 = 8.9 2880 71.6 | - 637 " ~_8610 | -14840 8 6.4 25 25.60 | - 6,3 | -161 1017 | -13.2 6230 159.3 | -_ 2420 - -~14720 | -17510 9 6.4 225 25.60 | -12.6 | -323 4060 | -21.5 11260 286.5 | - 6180 - ~20830 j ~18590 10 6.4 125 25.60 | -19.1 | -488 9320 | -28.0 18470 473.0 - 13220 a 27100 | -t7650 11 §.4 „30 21.30 | -23.5 | -543 13850 | -34.4 29020 618.0 | - 21230 a 33300 | -13300 12 6.4 34 18.82 | -32.0 | -602 19250 | -40.9 43180 814.0 | - 33100 " ~39700 | - 5540 13 6.4 34 18.82 | -38.3 | -720 27600 | -47.2 58980 1110.0 | ~ 52400 a 45600 4360 14 6.4 34 18.82 | -44.6 | -840 57500 | -33.5 77330 1455.0 | - 77800 „ -$1700 16610 15 6.4 34 18.82 | ~50.8 | -959 48750 | -59.9 99030 1865.0 | -111300 Ầ ~ 57900 32110 Chine ~63.0 | 111700 ~61100 41600 16 7.0 | 12.0 0 58 „4 | - 32 1780 | -64 3} 148700 86.7 | - 5570 " 62200 78500 | 17 7.0 | 12,0 0 58 „4 | = 34 1980 | -67.3 | 221400 128.3 | - 8650 * 65100 | 147300 18 7,0 | 18.0 0,39 4 | - 24 1470 | -70,3 | 271200 105.8 | ~ 7440 " ~6ö8100 | 194100 18 7,0 | 34.0 0.21 „4 | l3 840 | -72.1 1 308600 64.9 | - 4680 _ -70000 | 229600 20 7.0 |20.0 9.14 8 - 9 388 | -73.7 | 327900 45.9 | = 3385 Bị -71400 | 247500 Keel -75.0 | 332900 79020 | -72600 | 2512380 Totais 811,48 7228 | 423237 7315.0 | -348106 y = 7228/8211, 48 = 8,9" Mẹ > ~ŠØs > 81x 1900) z -00204 aw 5T Iyx = 423237 - 811.48 x 8,92 = 358940 = 20sy- E 348106 x 1000") « _968¢ Xo rSpads [- SR or | 7 79084 “Total Moment M at any station = Ms + Mo - Xoy
Columns 11 and 12 record the values of
Mg and -X oy for each station point For
example, the value of ~Xoy for station (1) equals - (-968 x 17.2) = 16650 and for station
(20) = -(-968(-73.7)] = -721400
Fig, AG.d4 snows the shape of the final moment curve as the result of the values in column 13
A9.5 Unsymmetrical Structures Example Problem Solutions,
ixamole Problem 1
49.24 shows an unsymmetrical frame a | 77 A Fig A9 34 ne loads as snowm Determine ti
Trang 16
A9 14 BENDING MOMENTS
The distance X from the line AB to the
elastic center is, (15)9 + 6x 6 + 10 x 12 _ x= =5 =2 lại x= 3T 5,082 1n, The distance ¥ from line BC to elastic center equais, pei X7.5+ 6x 0+10K5 ve 2 = 5.242 1n,
The elastic moments of inertia ly and Tự and the preduct of inertia Ixy are required Ix # G3) 15)(2.258)* + G2l(s 249) + Gor + (10)(0.242)” = 606,51 (2° 2x12 (10)(6.968)* = 942.96 = (15)(5.0g2)" + +B) (0 968)® + Ixy = (15)(-5.032)(-2.258) + chú, 968) (5.242) + (10){6.968) (0.242) = 217,7
The next step is to assume some static frame condition and draw the static bending moment diagram Fig A9.35 shows that the
frame has been assumed cut near point C which gives two cantilever beams The dending moment diagram in three parts for this static condition is also shown on Fig A9.35 12-2160 71440 (1) cụt 1 + 3 in + BZ 3 DỊ 9.788 ki 3258 C, sa=~21600 ạ_ xao ~1080 Mg diagram | (Ósz-4860 A 5.032 Fig AS 36 I ⁄ aoe Fig A9.35
The Ss values which equal the area of the Mg diagram divided by the I values of the particular portion will be calculated,
Ốs, = 3(-1440)/2 = -2160
Ase = -1440 x 15 = -21600 Og, # 4.5(-1080) = -4860
ŠØs = ~2160 21600 - 4860 = -28620
IN FRAMES AND RINGS
Fig, 49.36 shows the centroid locations of
the Js; values along the center line of the
frame The moment of these Zs values about the
x and ÿ axes will now be calculated,
ZØsx = (-2160) (3.032) + (-21800) (+5.032) + (74860) (291088) = 139700
3Øzy = (-2160)(5,242) + (~21200)(~2.258) +
(+4860) (-6.758) = 70280
The values of che redundants at the elastic
center can now be calculated using equations Œ1), (14), (18), ly -5ŠỦs - (-28620) =~ỈÖs_„ = (-28620) „ s2z In, 1p, lọ “TS S7T 31 " Bey - Bdsx “XU Xp Soe Tr— 1x x + 3 Bi Ỹ _ 70290 ~ 189700 ( ng) = a = 68.46 lo (606.51) (942 96)’ - ~| 28x ~ Isr dey | 606.51 (1 ~ Yo = ( T1 ) = ự xty ~ | 139700 - 70290 ae 7) aera = -122,36 1b, 942.96 (1- ($06.51) (942.96)
The bending moment at any
(19) equais, point from eq
M=Mg + Mo + Xoy + Yox, for example, Consider point A X 3 -5.032, y = -9.758, Mg = -2520 Ma = +2520 + 923 - 68.46 (-9,758) + (-132.36)(+5.032) = -263 tn 1b, Point B x = -5.032, y = 5.242, = -1440 Mg = +1440 + 923 ~ 58,46 x 5.242 + (~182.56) (-5.032) = - 209 in ib Mo = 0 + 923 ~ 68.46 x 5.242 + (-132.36) (6.968) = -357 in lb,
Trang 17ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Ă9, l5
8xample Problem 2 Fig A9.37 shows an un- Total Iy = 6451.07 symmetrical closed frame The bending moment
diagram will be determined under the given Calculation of ly -
frame loading = 30 os, 1 =
Member AB, ly = Te* 15% + ia * 30(0.667)* = 4500.7 67 - Member BC, ly = 3 xB x 30° = 791.7 Member CD, Iy = 2° x 15" +~4-x 20(1)* = 4801.7 ply Px TẾ = Member AD, ly #4 x-+x 30° = 1125.0 poy 1222 ° i x : = ; HD Parson Total ly = 10919 m—101¬ 10>—¬ x me fT te Calculation of Ixy - D
30'—————~ Member AB, Ixy = 22 (-15) (3,343) = ~1002.9
iy! Fig AQ 37 a 67 (13.343) (0) - vă xế? Member BC, Ixy = TS 3 Solution: - (30) (10) = -283.92 The elastic weight of frame equals = 20 5 ay) = gs 30 | 31.67 , 20, 30 gn Member CD, Ixy =~T (15) (-1.657} = -197,1 PT Sy te tT te F 65-58 Member AD, Iyy = 0
The location of the elastic axes will be mn - am determined with reference to assumed axes Total Ixy = -1753.9
x'x' and y'y! as showm on Fig A9.37, The next step in the solution ts to assume a static frame condition and draw the Mg dla-
5)+ FP) (30)+ >) (15) gram fig, AS.38 shows the assumed static
x = 15" | condition, namely pimned at point A and supported
as rollers at point D The bending moment
30 31.67 diagram is drawn in parts as shown
— 15) 4) (259+ B2) (10)+ EY (0) 150"
ys DEREE] = (3) Due to Pg
11,687" 600! (3) Due to Py
- - 600
These distances X and ÿ loeate the x and ‘pl |
y elastic axes 2s shown in Fig A9.37 yor | \
The elastic moments of inertia and the Due to Pi" | c
elastic product of inertia will now de cal- \
culated 4 Fig a9 38 Calculation of Iy - 1 1=} : x5 ` Member GD, Ix=‡ x + (8.48°+ 11.6575) = 721,58 1 NT nue to Pg and Py ~500 =50U r = /61.67)(13.3442%) 1 (41.67) Member BC, Ix = ee +i nê (10) = 1967.48 Member AB The next step is to compute the value of
+ = (20,74, - 2) a @s for each portion of the moment diagram 43 Member AD, Ix F ) 1,887 + 12 * 30 x is the area of the Ms/I diagram For reference
(0.5) ° = 2038.50 TT of the M, diegram have been labeled
Trang 18Ad 16 BENDING MOMENTS IN FRAMES AND RINGS Portion the 2 Portion B Portion wa 1 _— i288 1 a3 Portion —Ì-——2 ti, %—1 số, ~~ 4.814 11.657 _— Ai TT—P—
The individual Ø; values are now con- pote 15
centrated at the c.g of each of the Mg dia- Fig a9 40 Fig A9.41
grams Fig A9.39 shows the location of the Consider point A: - Mg = 0
5 values with respect to the x and y axes
through the elastic center My = 0 - 124.8 - 4.874 x 15 + 31.07 x 11.657 = 167 in ib, 523167 \ Point B Mg = 600 Đse6000+—r Ì i 15 | 13.343! Mg = 600 - 124.8 - 4,674 x 15 - 31.07 x 18.343 ¡8343| [1 = -165 in.1b t #—‡†——~—L——.-—x : | : Point c Mg = 0 | 11/887 | i i | Me Q - 124.8 - 31.07 x 8.345 + 4.674 x 15 a
pp toy O8*9000 = Sid ind
Fig AQ 39 Point D M; = 0
3Øsx = 6000(-15) + 3167(-5) = -105835 Ó - 124.3 + 4.674 x 15 + ổ1.07 X 11.557
= tn.1b
Jốsy = 6OCO(8,543) + 3167(15) + 307 in.1b
3958(13.343) -5000(-11.657) = 208659
The values of the redundants at the elastic center can now be calculated = = Bes Mo * 3a57t = 2s ¥ = 2x CEL) =——————m' -tx 1x (1 ah = ~124.8 in.lb 208659 - (~105685) (~2283:3) - 10819) _ = : = 51,07 1h gas [1 - 1755-32) — Tươ18 x S11 -[ 05x - sy bx | ee ty (1-8) -[ -10se8 - 208659 (pra [2 _ _(-1762.9 =) 10918 x 6451 §74 1b 10919 Fig A9.41 shows the true bending moment diagram
49.6 Analysis of Frame with Pinned Supports
Fig A9.42 shows a rectangular frame and loading This frame 1s identical to example problem 1 of Art A9.3, except it is pinned at points A and D instead of fixed
The first step will be to determine the elastic weight of the frame, the elastic center location and the elastic moments of inertia about axes through the elastic center
The term ds/EI of a Deam element of length ds represents the angle change between its two end laces when the element is acted upon by a unit moment In this chapter this term has
been called the elastic weight of the element
Trang 19If a support is pinned or hinged it nas no resistance to rotation and thus a unit moment acting on a hinge would have infinite angle change or rotation and therefore a hinge or pin possesses infinite elastic weight 10 Ib, | Ps 4 Ị Ị 6 04 133 L230 | i i Mẹ Xd i ị i? 4 ay ho A B Yd ee y Fig A942 Fig AQ 44 Fig AQ 43
Due to symmetry of structure about the centerline y axis the elastic center will lie on this axis, Since the two hinges at A and B have infinite elastic weight, the centroid or elastic center of the frame will obviously lie midway between A and B Pig A9.43 shows the elastic center 0 connected to the point A by a rigid bracket,
ads/EI for frame 1s infinite because of the hinges at A and B
The elastic moment of inertia about a y axis through elastic center is infinite since the hinge supports have infinite elastic weight Ix is calculated 25 follows: - For Portion AB =} x+x 30° = 3000 For Portion CD =+x——x 30° = 3000 cate cafe ale oj 1 For Portion BC = 24 xXx 30*= 10800 1y = 16800
EFig A9,44 showsa the static frame con- dition assumed to obtain the Mg values
The value of Zs for member BC equals the
area of the Mg curve divided by I for BC, hence @, = 45 x 24 x 1 = 270 The centroid
2 2
of this @, vaiue is 10 inches from point B The redundant forces at the elastic center can
now be solved for =~ -3Zzx „ ~(270)10 _„ Yo = Ty infinity ~ 9 -3Øsy „ 270 x 50 _ x 9 % 6 0.462 1b AQ.17 Fig, AQ 45 Fig A9.45 shows the bending moment dia- Fig A9 46
gram due to the redundant X, Adding this diagram to the original static diagram gives the final bending moment curve in Fig A9.46
AQ.7 Analysis of Frame with One Pinned and One Fixed Support
Fig A9.47 shows the same frame and loading as in the previous example but point D is fixed instead of hinged
1p The support D
has zero elastic weight and the pin
| at A has infinite
elastic weight, therefore the elastic center of the frame lies at point A The total elastic Fig Ag 47 y weight of frame is infinite because of pin at A
The elastic moments of inertia will be calculated about x and y axes through A
Iy = 16800 (Same as previous example)
ly = Qo xas* expe 2) = 8064 Ixy = GE x 15 x 24) + (BE x 30 x 12) = 7920
The static frame condition will be assumed the same as in the previous example problen, hence gg = 270 and acts 10" from B (Fig A9.48) 52270 =— x5 “¬18.1 ~11.82x4 -11.92 | ji0| : 2 if i Ế 28.45 arf | toy | | ĩ _ q 7 Mo i ậ ; a Lio AQT xO ˆ 80384 -18 6.18 Yo [.2582# Fig A9.50
Fig AQ 48 Fig A9 49
Trang 20CHAPTER AiO
STATICALLY INDETERMINATE STRUCTURES
SPECIAL METHOD - THE COLUMN ANALOGY METHOD
Al0.1 General The Column Analogy method is a method that is widely used by engineers in ce~
termining the bending moments in 4 bent or ving type structure The method considers only distortions due to bending of the structure
The numerical work in using the column analogy method 1s practically identical to that carried out in applying the elastic center method of Chapter AQ
A10.2 General Explanation of Column Analogy Method Fig Al0.1 shows a short column loaded in compression by 2 load P located at distances
(a) and (b) from the principal axes x and y of the column cross-section B Nv Fig Al0.1
To find the bearing stress between the supporting base and the lower end of the column, it is convenient to transfer the load P to the column centroid plus moments about the principal axes Then if we let o equal the
vearing stress intensity at some point a distance x and y from the yy and xx axes, we can write
Where A is the area of the column cross~ section and Pa and Pb the moment of the load P about the xx and yy axes respectively If we let Pa = My and Pb = My; the above equation can be written,
P+ Myy + Myx
aoe eee ccc ì
TA tx y
*Method of Analysis due to Prof Hardy Cross, See "The Column Analogy” Univ Il Eng Expt Sta Bull 215,
Now assume we have a frame whose centerline length and shape is identical to that of the column section in Fig Al0.1 The width of each portion of this frame will be proportional to 1/EI of the member Fiz A10.2 shows this assumed frame Furthermore, assume that end B of the frame ts fixed and that a rigid bracket is attached to the end A and terminating at point (0) the elastic center of the frame frame is subjected to an external loading, Wiz Wa, SEC,
The
Fig A10.2
This cantilever structure will suffer bend- ing distortion under the external load system Wi, Wa, etc., and point (Q) will be displaced Point (0) can be brought back to its original undeflected position by applying a couple and two forces at (0), namely, Mo, Xq and Y, as
shown in Fig Al0.2 Since point (0) is attached
to frame end A by the rigid bracket these three forces at the elastic center (0) will cause point A to remain stationary or in other words to be fixed Therefore, for the frame in Fig Al0.2 fixed at A and B, the moment and two forces acting at the elastic center cause the static- ally indeterminate moments M; when resisting 4 given external loading causing static moments Mg The final true bending moment M at a point on the frame than equals
H21 + My
Trang 21A10 2
Mo = 228s = ~ 28sx „ ZØay
= fase Yo 7 Ty Ee FoF tx
The tern Ig, represents the area of the static Mg/I curve (BE has been assumed con- stant and therefore omitted) Let the term SOs be called the elastic lead and give ita new symbol P The term ids/i equals the elastic weilsht of the frame anc equals the sim
2 the length of each member times its width which equals 1/I, Let this total frame elastic weight be given a new symbol A,
In the expressions for Yo anc Xo the terms 205x and Sfsy represent the mement of the static M/I curve acting as 2 load abcut
the y and x axes respectively passing through the frame elastic center Therefore let If5x
be given a new symbol My and 3Øsy 2 new symbol My With these new symbols, equation
(2) can now be rewritten as follows: -
Comparing equations (1) anc (3) we see they are simtlar In other words, the inde- terminate bending moments M, in a frame are analogous to the column bearing pressures o, hence the name "Colum Analogy” for the method using equation (1) With this seneral ex- planation, the method can now be clearly ex- plained by giving several example problem solutions
Ă10.3 Frames with One Axis of Symmetry
From the previous discussion we can write, (1) The cross-section of the analogous colum
consists of an area, the shape of which is the same as that of the given frame and the thickness of any part equals 1/EL of that part
(2) The loading applied to the top of the analogous column is equal to the Mg/f1 diagram, where Mg is the statical moment
in any basic determinate structure de-
rived from the given frame If
causes bending compression on the inside face of the frame it is a positive bending moment and the analogous load P on the column acts downward
(3) The indeterminate bending moment M, at a
given frame point equals the bese © pressure at this same point on the
analogous column Thus the indeterminate moment at any point on the frame equals {from eq 3), =P, Mey , Myx nh (4) THE COLUMN ANALOGY METHOD fixed supports at ¬oint moments 4y points B termined by the ame and i problem 1 of on was 11 10 1b agra Hs : | I | L230" 1230) i ị =3 133 † | | i ' al am - D h 4 1.5 : 2.5 Fig A10.3 Fig Al0.4 SOLUTICN NO 1 s e 9
shape as shown in Fig Al0.3 as the cr section of a snort column (see Fig Ald equal to 1/EI of the member cross-section Since © is constant, it will be made unity and the widths will then squal 1/1
Fig Ald 5 Fig, Ald 6 The first step in the calculations is to compute the area (A) of the column cross~section in Fig, Al0O.5 and the moments of inertia of the column cross-section about x and y centroidal axes
area Aa 3S wo of 8 + t+ 30 „ 24 37 3
Trang 22and I, would be identical to the complete calculations given in Art AS.4, and Table A9.1 where this same problem is solved by the elastic center methed These calculations will not be repeated here The results were, ly = 3188 and ly = 3456
Since the frame is statically indetermin- ate the next steo {s to assume a static frame condition consistent with the given frame and loading Fig Al0.4 shows the condition assumed for this solution, namely, pinned at point A and a pin with rollers at point D The static M, diagram is therefore as shown in Fig Al0.4 We now load the column cross- section with the M,/I diagram as a load as shown in Fig ALO.S The static moment sign is positive because the static condition causes tension on the inside face of the frame In the column analogy method a positive Mg/T loading 1s a cownward or compressive load on top of column, and therefore a negative M,/I value would be an upward or tension load on
the column ,
Equation (3) requires the values of the My and My the moment of the M,/I diagram as 4
load about the x and y axes Equation (3) also requires the total column load P which equals the area of the Mg/I diagram
For this problem the value of P Fig A10.5 equais,
from
PF 22.5 x 24/2 > 270
The centroid of this triangular loading is 2 inches to left of y axis Fig Al0.6 shows the column section with this resultant load P
‘de now use equation (3) to find the in- determinate moments My which are equal to the pase pressures on the column Equation (3) involves bending moments My and My and distances x and y, all of which must have
signs The signs will be determined as follows: When moment My produces compression on base on that portion above x axis, then My is positive
Waen moment My produces compression on base on that portion to right of y axis, then My is positive
A distance y measured upward from x axis is positive; measured downward is negative
A distance x measured to right from y axis is positive, to left negative
From Fig, Al0.6: = P = 270 Al0.3 2530 (positive because =Z 870x9.375 = compressive on column portion base pressure is above x axis} My == (270x%2) =-540 (negative because base pressure is tension on cclumn portion to right of y axis) Substitution in equation (3) Frame Point A x 2-12", y = = 20.625" =P, Myy , Myx Mp eat te nom 270 | 2580 (-20.625) , (-540)(-12) sê 3188 3456 8,44 =16,38 + 1,88 =—6,06 1n.Ib Le ~ " + " The true bending moment from equation (5), MS Mg - My
Mg = 0, see Fig A10.5
whence, MA = O-(-6.06) = 6.06 in.1b Frame Point 8B x =- 12", y = 9,325" My = 220 4 2830%9.375 , (2540) (-12) 1— s2 2188 5%456- = 8.44+7,44+1.88 = 17.77 Mg = Mg-My = O- (17.77) 5-17.77 in.db Frame Point ¢ x=12, y= 9,325 M, = 270, 2580x9.325 , (-540)12 1” 52 3188 3456 = 8.44+7.44~1.,88 = 14.0 Mo = Mg ~M, = O- (14.0) =~14.0 in.lb Frame Point D x=12, y = -20.625 ~ 270 2580(-20.625) , (-540)12 My ~ Se —— iss 3188 “3456 456 = 8,44 -— 16,8 -1.88 = - 9.82 Mp = Mg -My = O- (-9.82) = 9.92 1in.1b
Fig 410.7 shows tne final bending moment dia- gram, which of course checks the solution by the elastic center method in
Trang 23Al0.4
Solution No, 2
In this solution a different static ‘frame condition will be assumed as Shown in Fig AlO.8, namely the frame is cut under the load and one-half the 10 1b load will be assumed as going to each cantilever part Fig Al0.9
shows the static moment diagram and Fig
AlO.10 the static Mg/I diagram with centroid locations of each portion of dlagram which 55 Ụ -i0n Tito BrTg——€ 4 cố Fig A10.8 3) Fig, Alo 3 45 LOSES a 146) 6.25} 5.625
Fig A10, 10 Fig Alo 11 are numbered (1), (2), (3) and (4) The area of each of these portions will represent 4 load P,, Pg, etc on the column tn Fig, Al0.11 Since the static moment is negative on each portion the load on the column section will be upward Py =-10(30) =-300, P, =-45x9 =-405 P, =-15(6)/2 3-45, P, =-30x30 =-900 = 3P =~ 300-45 ~ 405 - 900 =-1650 From Fig AlO.11 My =- (45 +405) 9.575 + (400 + 900) 6.635 = 2551 My = 300 x 12 + 45 x 10 ~ 405 x 6 ¬ 900 ~ 12 = -9180 =P Muy , Myx my Spee ae POINT A x =- 12, y = - 20,625 = 21650 , 2531(-20.625) „ (-9120)(~12) tị“ a2 3188 3456 z ~36.06 THE COLUMN ANALOGY METHOD Mạ = Ms -Hị =—50¬ (~36.06) = 6,06 1n.Ib which checks solution 1 Frame Point B x2-12, y = 9.375 = 71650 | 2531 x 9.325 {-9180)(-12) My 5 + TH rẻ ng =- 12.23 Mp = Mg -My =-30 - (-12,.33) 2-17.77 in.lb Frame Point Cc x=12, y # 9.375 =~ 1950 M = +76.0 » 2831 %9.375 „ Í- 9189) 12 3188 Mg = Mg -My =-90- Frame Point D 71650 | 32 = -99.82 MHỊi “—z— Mp = My ~My =-90- (-99.82) = 9.82 in.lb Thus solution 2 checks solution 1 The student should solve this problem using other static conditions
Al0.4 Unsymmetrical Frames or Rings
In applying the column analogy method to unsymmetrical frames and rings, the moment of the M/EI diagram must be taken about principal axes and the moments of inertia with respect to principal axes
Trang 24ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES As previously done in Art Al0.2 let, My 5 25x and My = 2@ey Furthermore, let, My = My (POL) and My = My == Loy? and k = (1 - xiy ) Then substituting in equations (a) and (b) we obtain, My -M! My - Mt ko Ses to From equation (2) we have, My = Mẹ + Xoy + Tox
8ubstituting values of Xe and Ÿo into this equation, we obtain as the equation for My the indeterminate moment in the column analogy method, the following ~
-P, (My-My)y „ (My -My)x
My Ft Sa ay - (8)
The true moment is the same as for the symmetrical section, namely,
MF Mg -My
Thus the solution of an unsymmetrical frame by the column analogy method follows the same procedure as for a symmetrical section except that equation (6) is used instead of equation (3),
A10.5 Example Problem - Unsymmetrical Section
Fig Al0.12 shows a loaded unsymmetrical frame fixed at points A and 0 Required, the true bending moments at points A, B, C and D This problem is identical to example problem 1 of Art, AQ.S where a solution was given by the elastic center method 240 lb ape" : Ter Fig A10 12 Fig, A10.13 A10 5 SOLUTION:
Fig 410.13 shows the cross-section of the analogous column The length of each member of the column section being the same as in Fig Al0.12, and the width of each portion being 1⁄1 The first step in the solution is to find the
column section properties
The total area (A) of column section equais
(15/1) + (10/1) + (12/2) = 31
The other properties required are: - ly, ly and Ixy
The calculations to determine the location of the centroidal x and y axes and the above properties about these axes would be identical to that given in example problem 1 of Art 49.5 of Chapter AQ, whera this same problem was solved by elastic center method The results were Iy = 606.51, Ly = 942.96 and Iyy = 217.74 The location of the axes were as shown in Fig A10,15,
The next step in the solution is to choose a static frame condition and determine the static (Mg) diagram Fig Al0.14 shows the assumed static condition, namely, two cantilevers because of the frame cut as shown The figure also shows the static bending moment diagram made up of three portions labeled (1), (2) and
Trang 25A10,6
These loads act on the centerline of the frame members and through the centroid of the geometrical moment diagram shapes These centroid locations are indicated by the neavy dots in Fig Al0.14 and their locations ars given with respect to the centroidal x and ÿ axes The loads P,, Pa and Ps are now placed on the column in Al0.13, acting upward because they are negative
We now find the moments My and My which equal the moments of the loads P about the centroidal axes My 3 ~ 2160 x 5,242 + 21600 x 2.258 + 4860 X% 6.758 = 70290 My = 2160 x 3.032 + 21600 x 5.032 + 4860 x 5,032 = 139700 To solve equation (6) we must have the terns i: i, and k I = 70290 x 217.74/606.51 My a } /606.8 25234 My = My CC?) = 139700 x 217.74/942.96 = 32258 k=(q~11g ° yee 217.74 ) Tly SÙS.5T x 042.06 9171 Substituting values in equation (6) we obtain, M, = 728620 , (70290 -32258)y | 1 SL O.9171 x 606.51 (139700 - 25254)x 0.9171 x 942.596 wherice, My == 923 + 68.37y +122.36x - = - (7) For Frame Point A, x #~§.032, y =-9.758 My = - 923 + 68.37(-9.758) + 132.36 (-3.032) 2 ~ 923 - 667.15 ~666.0 = - 2256 Ma = Mg -My =~ 1440-1080 - (2256) = ~ 264 in.lb For Frame Point 8 xX 3-5.082, y = 5.242 My = - 923 + 68.37 x 5.242 + 132,36 (-5.032) — 923 + 358.39 + 666.0 =~ 1230.6 Mg = Mg -Mg >- 1440 - (-1230.6) = -209,4 1n,1b, THE COLUMN ANALOGY METHOD Frame Point 5 x = 6,968, y= 5,242 = 923 + 68.37 x 5.242 + 122.36 x 6,968 = 357.7 & 1t Mg -My = O~- (357.7) =-357.7 in.ld Frame Point D x = 6.968, y =~4.758 My 3-923 + 68.37 (-4.752) + 132.36 x 6.968 = ~326 Mp 5 Mg -My = O-(-326) 5 326 tn.1b,
The above results check the solution o? this same problem by the elastic center method
in Art A9.5 of Chapter Ag, ‘ie Student should solve this problem by choosing other static frame conditions,
Al0.6 Problems
(1) Determine the dending moment diagram “or the loaded structures of Figs AlO.15 to Al0.20 , 400 400 [Fi5Trl§ +154 _ w=ld Ib/in L=30 t I =10 | L = 24 | 24 h =6 ÌI =6 mm mm Fig A10 16 200 300 Lb, đm Pee My 1 20 iL 20 L= 20 ! h =8 Lxz5 —2'——¬ Fig A10, 18 400 1 + % “ <7 10 Đệ Ey 200 200 Fig Aid 20 Fig A10 19