STATICALLY AS, 18 SECOND, in the form of eq (14): "“ ọ 806 1.182] [S£1 Qa 1 -1.864 ~1.729 asf = Ja ?* 2% * 1 1.06 Q 416 de 9 9 1.00
Note that in this case [#10] consisted of only one column, {nasmucn as there was only a single external load
In art A7.9 the strain energy was written
= Ls | fad {39}
where [145] is the matrix of member flexibility coefficients (Art A7.10) If now eq (14) and its transpose are used to substitute into (15) the expressicn becomes (note the use of (1,3), (r,s), and (m,n) interchangeably) 2 *(| đ J[SĂ]*Lô_J[Sx]}* *E4z]*(Ezl{s)! Eal{)} Multiplying out 7° | Pa_| [es] [ái] sa] {Pa} * * Ls _| ec Ges] Boal? } L*_] fx] Eg] se] (5)
The reader may satisfy himself that th "cross product" term in the middle of the above result is correct by observing that, because of the symmetry of tua}
LJ [2] Ess) Esa] {Fa} L?a_| [Bes] :3] Ess) ¢%s}
The various matrix triple products occurring above are assigned the following symbols, each having the interpretation given (compare with
eq (24) of Art A7.9)
(ea) * [xs] 2:2) Era] ~ the mitrix
INDETERMINATE STRUCTURES
of externali-point influence coefficients in the "cut" structure: deflection at point m per uit load at point n
17)
rl l1 iy El tủ lo
of influence coefficients relsting relative Placements at the "cuts" to external loads: displacement at cut r per unit icad at point n
dis-
(18)
of influence coefficients relating relative dis— placements at the "cuts" to redundant loads at the "cuts": displacement at cut r per unit re- dundant force at cut s
With the above notation one may Write
=| Pa | ml {Pap * ® Le | [en] {Pap
Now according to the Theorem of Least Work
SU/a_ or = 0 for continuity Then, differentiating eq (19):
ap 70 (Sn) {Pah rs]{as} ~~ 20
This last result may be verifted by writing eq (19) out in expanded form, differentiating and then recombining in matrix form Rearrangec, eq (20) gives
ee cee (21)
Geel {*} >> Bal {’}
Eq (21) 1s a set of simultaneous equations for the redundant internal forces a.» Ig It may be compared with eq (6) of Art A8.6, to which it corresponds £q (21) may be solved directly from the form there displayed or its solution may be obtained by computing [s1 the inverse of the matrix of coeffictants, giving
(:}>- Est] Gal tm}
The matrix product matri - 0z] [ra gives the =* tt values of the redundant forces for unit values of the external loads
symbol [sn] so that
(+) Ball}
Trang 2ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
If now eq
(14) one gets (w
m for n)
(ecb » [Ea] {?a}> [Ecz) [Ear] Be] 9}
= ( {cal - Ez) sé] Beal ) {=} -( Hal* Gel Bal {3}
The matrix set off in parentheses above, gives the internal force distribution per unit
value of the external loads It is given the symbol - (ca) = Bin] ~ (Ber Ber*] [Son] - 08 = [ss] * Ex] Ps] is substituted into eq (22) {th exchange of r for s and tự so that
Eqs (23), (24) constitute the major re- sult, inasmuch as they present the means for computing the internal force distribution in a redundant structure
Example Problem 13
The doubly redundant beam of Fig A8.27 (a) is to be analyzed for the bending moment dis—
tribution The beam is loaded by couples over the supports as shown El constant —P2 E— L ——>>— L —— al 4 (b) %4( (o> ag a Py ì {a) Fig A8 27 Solution:
The choice of internal gensralized forces ts shown in Fig A€.27 (b>) The appropriate member flexibility coefficients were arranged in matrix formas (ref Art A7.10 for coefficient expressions) A8.19 a L/2 9 9 9 9 zt tay cà aI 0 Lz b/g 90 tựy 2
The moments qạ and q, were taken as the redund- ants With these set equal to zero, the internal force distributions due to application of unit values of P, and P, were determined, giving
W 9
[1m] = 0 a
9 9
With the applied loads set equal to zero, unit values of the redundants were applied yielding (r =) (2) (4) -1lA 9 1 9 [2:z] TÍ tựu ~ 1L 9 1
Note that redundant load qạ was applied as a self~equilibrating internal couple, acting on
both beam halfs
The following matrix products were formed:
Trang 3
A8 20, STATICALLY INDETERMINATE STRUCTURES
The inverse of fa ] was found (ref though the matrix of member flexibility coe+
appendix) rs fficients was axnanded to 4 6 x 6, the
coefficients for qd, and q, were zero Thus,
B~1-%|2 4 s List 4
Next, the unit redundant load distribution was found (eq 22) Ea Balta 3 J -.286 ~.428 " [143 ~.286 Finally, the true unit stress distribution was computed (ea) > sa] * Ge] Ps] Lo} lly 9 -.286 -.428 2/9 9] | 1 0 145 -.286 hịo 3 wy 9 0] | 9 o_| ì 1 2 1| 1.286 | 428 = 2 |¬ 286L | ~.428L 5 |~ 429 | 858 4| 148L |~.286L
(The tabular form of presentation of the matrix Gi above, {s used here only to indicate Clearly the functioning of the subscript nota- tional scheme In general, it should be unnec- essary to call out the subscripts in this fashion excepting for the larger matrices, for the handling of which, the tabular form may prove helpful.)
Example Problem 13a
The redundant beam problem of Fig 48.27 is to be re-solved using the redundant reactions as the unknowns
Solution:
The support reactions under the loads P, and P, were given ‘the symbols qs and qa, respec~ tively, positive up These forces did not enter into the strain energy expression so that, al~ L?/z L/¿ 0 8 00 9 9 L/⁄2 9 0 9 0 1⁄4 1 0 6 9 0 0 # 9 9 9 ° 90 00
With the redundants q, and qd, set equal to zero, successive applications of unit external couples P, and P, gave the stress distribution 0 Ộ 1 9 = 9 9 [ta “Ta 1 9 9 9 Cc
and, with P, and P, zero, successive applica- ions of unit redundant forces qd and qd gave “1 õ L ọ -l +l (rr | = fan L 1 9 Ba L9 Then, multiplying out per eqs (17) and (18) [1 5 Gala f > fis 8] [ors] = a [*
The inverse was found:
[ers-*] = SEI [Sil Finally,
Trang 4ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES AB 21 0 -.50 8 oo SN,“ / Fe) ` ý \0 y V 6 of \o of 0 9 h3 \V/.25 \/.25 \/ 0 a Vo 0 Vo 0 " 1/714 -.428 (Compare with solution to Example Problem 13) mple Problem 14
The continucus truss of Fig Ag.28 1s teice redundant it is desired to analyze it for stress distributions under 4 vartety of loading conditions consisting of concentrated vertical Loads applied at the four external points indicated i 2 3 4 § § T v2 © = = s^*ev*v vợ y1 v VN ưNG By 8 —" Boe + "n i › » + + Try @ 'Rạ tr ® try Ñ 8 panets at 20" = 160" ———— Fig A8 28 Sclution:
The internal generalized forces (q;;4;)
employed were the axial loads in the various members These were numbered from one to thirty-one as shown on the figure The member flexibility coeffictents in this case were of
the forma, = Mết: (Ref Fig A7.35a) The coeffictents ar¢ written as a colum matrix be- low (They were employed as the diagonal ele- ments of a Square matrix in the matrix multi- plications, but are written here as a column to conserve space,)
Member loads q, and q, were selected as redundants With q, and q, Set equal to zero
("out"), unit loads were applied successively at external loading points one through four, the four stress distributions thus found being arranged in four columns giving the matrix
Bia] (below) By way of illustration, the
loading figure used to obtain the second colum
of [Ea] is shown in Fig À8,29,
Next, unit forces were applied successively at the redundant cuts "three“ and “five” as shown in Figs AS.30a and A8.30b These loads were
Trang 5
À8 22 STATICALLY INDETERMINATE STRUCTURES
NOTE: VOIDS INDICATE ZEROES Multiplying out gave 136 -8,Ô0 Cairnicnir bi Ba: GOED Ba “8.0 -15.0 9 9 9 9 ~l5.0 ~8,0 The inverse of [rs] was found (ref appendix) ga g [14s s.õ E— [se sa]
Next; the values of the redundant forces, for unit values of the applied loads, were found per eq (22)
[Pea] - Ea!) Gea]
O59 -111 0065 0035 ˆ 20035 O065 111 «059
The calculation was completed as per eq
(23) to give [Sa] ; the values of the member
forces for unit applied external loads
31
Example Problem 14a
Fig A@.31 snows the two bays of 4 steel tubular tail fuselage truss which is loaded by tail air loads to be resolved into three con- centrated loads applied as shown The fuselage Dulkhead at the attach-points station (A ?-K) is heavy enough so that it may be assumed to be rigid in its own plane, Hence, the truss may be analyzed as if cantilevered from A-E-F-K as shown All members are steel tubes, their lengths and areas being tabulated below
Solution:
Trang 6ANALYSIS AND DESIGN OF Fig A8.31 MEMBER | NUMBER 38 1 LENGTH 35.4 43 347 64 go S8 + 9 14 | poly rac fia]e, for] os] oo] of oy bai Dh afas)F| Ol Gy co} ml|E-ltolr-lei 6| ụ1 c, ile oO œ 0
The structure was three times redundant, In @ space framework or p joints, 3p-6 inde- perncent equations of statics may ce written (o 2 0), Here, however, stress ¢stails in the plane ARKF are to be sacrificed; six equa- tions are lost thereby since only net forces in
ons in this Slane can >e summed -6 521 equations; 24 member unknown Members £2, 23 and 24 ere cut
The next step was to compute ths unit
stress distributions [Ft | and fee] Rather FLIGHT VEHICLE STRUCTURES A8 23 than apply successive unit loads and forces at points m= 1, 2, 3 and r = 22, 23, 24 and then carry each loading through the structure, an- otner procedure, often better adapted to large complex structures, was employed
In the method used, the equations of static equilibrium were written for sach of the seven Joints." Summation of forces in three directions on each joint gave 21 equations in 24 q’s (the unknowns} and the three applied loads P,, P, and P; Some typical equations obtained were: From 2 F on Joint Cc -21368 Q,, - 21368 q., + 18334 q, - -19334 q, = -P, at + Ga = 0 3ua + Qe = -1,0236 Py From £ F on Joint B „081759 qi - 1854 aa > Os —.07617 dis 7 5227 day a - Qe - 91593 a, = 0 1410 qi + Gis + 4146 Gay = -.8523 dua
And so forth, for the other joints Note that in each case the equaticns were arranged with the applied loads (P,) and the redundant q’s (qa2, d2s,; daa) grouped on the right hand side of the equal sign This ar- rangement was observed for all 21 equations, after which the equations were placed in matrix
“
1,3 = 1,2, 24 n 21,2,3
= 22,23,24
(Note that there were 24 equations here, the additional three equations being the ident- ities
Gea = daa das = Gas
đau “ đa )
On the right hand side of the above matrix equation the matrices are shown "partitioned™ Tne first three columns of (D] are the coeffic-
ients or Pn and the last three are the
Trang 7
AB, 24
coefficients of the a5:
The matrix equation wes solved for the q, a by finding the inverse of fy ] (see appendix), Thus , where đ«is:l:z i2] Thus, the unit stress distributions in the determinate structure were found by a procedure having as its main advantage the reduction of the work to a routine mathematical operation, In the conduct of this work appropriate stand- ardized techniques may be employed *
The result in this case was
+ 2 3
-ð4\6 | 2:727 | 5.571
The member flexibility coe?ficients Ly, were arranged as the diagonal elements of the
matrix [5] - Then, multiplying out according te eqs (17) and (18), 177.9 -1137.0 -6430.3 = 195.0 - 151.7 -2263.0 -154.4 161.6 2273.3 2970 1150 ~1148 (rs } 2 1150 1221 -1025 -1148 -1035 1224 * See references in appendix STATICALLY INDETERMINATE STRUCTURES [Pen] =-{4.3* | [z]z:2*|¬ =~ 31.4 176 Finally, the complete unit stress distribution was obtained as [Esa] = [sa] * [ce] Ea] Example Problem 15
The doub symmetric four flanges idealized box beam cf Fig 48.32 is to de analyzed for Stresses due to load application 2t the six points indicated Flange arsas* t per linearly from root to tip while sheet thicknesses are constant in sach panel The beam ts ounted rigidly at the root, providing full re nt
——-.——Ầ_—
* These are the “effective areas", being the flange area plus adjacent effective cover sheet area plus one-sixth of the
Trang 8ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES against warding of the root cross section due
%o any torsion loadings F— 30" —— ưa in? NỔ (Typ.) 10" À— or 20" (Typ.) bers Pig, A8.32 Solution:
The generalized forces employed are shown on the exploded view, Fig 43.33 No forces were shown Zor the lower surface, its nembers and forces being equal to those of the upper surface because of symmetry In the [14] matrix this fact was accounted fer by doubling the member flexibility coefficients for the members of the upper surface NI ue yas — \ Qs die Aas \ a } | Gas ` Xg T% “ds k b 4a đa r - ý qa q; Fig A8.33 A8 25 {ne member Z1exibility coefficients were collected matrix form es shown below Note that entries Tor 444, Gee, Seo 2nd Gioyio were collected from <wo stringers each (as well as being doubled as discussed above) Coefficients
for these tapered stringers were computed from the formulas of art A7.10
NOTE: VOIDS INDICATE ZEROES & = 1 Cover sheet shear flows qa, q; and dia were selected as redundants With these set equal to zero, and with unit loads placed successively at
loading points “one” through "six", the [im |
matrix was obtained when the cover sheets are "cut" the two webs act independently as plane- web beams The details of the stress calculation
for such beams are similar to those of Zxample Problem 21, Art A7.7 and are not shown ners
NOTE: VOIDS INDICATE ZEROES
Calculations for [zr] were made by suc-
cessively assigning unit values to the redund~ ants 42, d+ and dia The calculations are illustrated by the exploded view of the end bay in Fig A8.34 showing the calculation in that part of the structure for qa = 1 Note that da, = 1 was applied as a self-equilibrating pair of shear flows acting one on each side of the "cụt", The ribs were considered rigid in their own planes
Put qdag=l
Trang 9A8 26 Fig A8.34 @, = -1.0 qd, = 1.0 From equilibrium of 220 rip: Qạ 2 +(.5625 ~ 25) = -,2125 Qe = 23125 qa, = O, by hypothesis
So on, into the next bay In the idealization
used here, the ribs have zero stiffness normal to their own plane so that the axial flange loads are transmitted directly to the flange ends of the adjacent bay (see da, 47, de, dio of Fig A8.33) 1.00 : “| 9 244 1.72 = 1955 NOTE: VOIDS INDICATE ZEROES
STATICALLY INDETERMINATE STRUCTURES
The follicwing matrix products Per eq (18): were 2ormed: 13774 ,1789 0520 a = 10° 97222 [ers | = Be 1789 ,2678 ,07522 20520 07322 1254 Per eq (17): 7 C3356 ,05556 -,009218 009318 ¬,001605 001605 -Ÿ.02324 ,02524 -,01156 01156 -,CO2259 202258 ~,008216 ,008216 ~.QO5150 ,005150 -.002278 ,002278 The inverse of Ips | was found (ref appendix) - 3.862 -2.562 - 1137 -_¬ x8 | a 2.522 [Šss `] * tạ 2.562 6.157 ~2.52 - 1137 -2.521 9.499 Then ~ -T [sn] =~ [Pre] [Fra] 10699 =.0699 ,005S8 -.00598 = | 03593 -.03593 «01568 -.01568 206016 -.00016 203409 -,03409 00403 -.00403 „01872 -,01872 01578 ~,01 73 | Finally, the K1
true sfresses were (per eq 23)
[a] - Fez] Pra] 4 5 6 H 0060 |~.Q092|_.0002 -,01 - - : ~.Š + : - : +610
The reader will observe that the result displays the "bending stresses due to torsion"; that
near root tion
is, the buildup of axial flange stresses the root of a beam under torsion when the
Trang 10ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES A8, 27
found by superposing stresses for Pi, = 1 and 144 -15,0 ° 9
Pa = -l Thus, one finds that under this con- 1Í -¬s.o 48.0 9
dition there 1s a root flange load of ESE SE1E17Ẻ mà =- mn] =! Sai} fors} fem] ==] 77° : 9
Gig = 3.47 ~ 2.76 = 0.71 lbs, 9 9 -15.0 144
for a torque of 15 inch lbs
.472 888 0525 ,028
A8.11 Redundant Problem Deflection Calculations by Deflections (and in particular, the matrix Matrix Methods ~ [se] [srs"] [sm] + “E | 10825 0975 1.665 .888 7 1 888 1.665 oes .0525 „0975 .888 0525 472
of influence coefficients) are readily computed from the results of Art AG.10
Assume that the redundant forces q >have
ceen determined from eq (22) The total de-
flection of external loading points is then
easily computed as the sum of, ONE, the de- flection due to external loads acting on the
"cut" structure, [mn] {2} (ref eq 16) and,
TwO, the deflections due to the redundant forces acting on this same cut structure,
[ens | {4s} (ref eq 17) Thus,
{a} Ges} {"}° Esltss}-
Substituting from eq (22}
(5}* Geol (%}- Gal Es] Balto} ‘(Fal CEI El te}
The matrix expression set off in paren- theses above, giving as it does the deflections for unit values of the applied loads, is the matrix of influence coefficients Let = - _ -_— [un] * [San] ~ [ss] [$:5"] [Pra] (35) so that {2} Bon) {Pa} Example Probiem 16
Determine the matrix of influence coeffic- ients for the redundant truss of example problem 14, Solution: From the products were previous work the following matrix formed Finally, the sum of the above two matrices gave 144 “15.9 - 052 - 0z8 - 15.9 36.3 - 098 = 052 =i - 052 - 098 36.3 -15.9 - 028 - ,052 «15.9 144 Example Problem 17
Determine the matrix of influence coefficients for the box beam of example problem 15
Solution:
An alternate procedure to that shown by eq (25) was followed The influence coeffic- lent matrix was formed es in Chapter A~7, Art A7.9, by the product
Fea] = Pa] Fc] Bs]
This product was formed readily, {nasmuch as [Ea] was available from example problem 15, The result was
A8.12 Precision and Accuracy in Redundant Stress Caiculations
Matters of precision are dependent upon the number of significant figures obtained and re-
tained in dealing with the geometry of the structure and in the care with which arithmetic operations are performed In the discussion to follow tt 1s assumed that all due caution ts exercised with regard to the precision of the work
Matters of accuracy have to do with the num- ber of significant figures finally obtained in the answer as influenced by the manner of formu-
Trang 11
A8, 28 STATICALLY INDETERMINATE STRUCTURES
lation of the problem ‘The accuracy of the re- Sult may be affected by a muuber o? factors, two of the most tmportant of which are discussed here
Two factors influencing accuracy often are considered together under the neading "choice of Tsdundants”, They are:
$ — the number of significant figures which may be retained in the solution for the inverse of the redundant force coefficient ma- trix, [°rs (eq 21)
@@ ~ the magnitude of the rsduncant forces relative to the size of the determinate forces These two factors are concerned respective ly with the left hand side and the right hand
side of eq 31, viz
(Be]{% p= Era] {Fa}
they are discussed in detail below
® ~ ACCURACY OF INVERSION oF [rs | ;
THE CONDITION OF THE MATRIX
The characteristic of the matrix [rs |
Wnich determines the accuracy with which its in- verse can be computed is its condition The condition of the matrix is an indication of the magnitude of elements off the main diagonal
(upper left to lower rfgnt) relative to those on The smalier are the relative sizes of elements off the main diagonal, the better is the condition of the matrix A weJl conditioned matrix is more accurately inverted than 4 poorly conditioned one.” Two extreme cases are Tow given for {liustration:
a) the diagonal matrix Its off-diagonal elements are Zéro, so that it is ideally con-+ Gitioned Thus, the inverse of 2 0 Ộ 97 9 9 0 «3 1⁄2 6 9 9 ly 9 9 5 -1⁄2
b) a matrix all of whose elements are equal in each row, All the elements off the main diagonal are equal to those on.” The determin- ant of such a matrix ts Zero and hence its
inverse cannot be found (rat condition is terridls,
é 2 2
3 -3 +3
In the matrix ¬ the off-diagonal ele- ments are measures of the
coupling of one redundant
The strength of
determinate
FOR EXAMPLE, the doubly redundant beam of Fig, a8.35(a) may be made statically determinate by "cutting" any two constraints Pole LL =— 1, — Q—= (4) Fig Aa 35
Fig A8.35(D) shows the choice of ganerai- ized forces, Only two (a, and qa) are required to describe the strain energy, but the central support reactions were also given symbols as it
was destred to conSider them in the discussion Then 2⁄34 1⁄4 0 0 Ve 2/3 0 0 E:;]*# 0 0 0 9 9 0 °
Trang 12ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Multiplying out, Lee] = Get] Gud] Bie] - r,$=3,4 iss Ld
The condition of this matrix is poor Poysically, a unit load at point "3" causes almost as much deflection at point "4" as at "3" itself The cros§~coupling is large
SECOND, suppose the moments q, and q, had Deen chosen as redundants The "cut structure” in this case being visualized as in Fig
A8.35(d) Application of unit redundant forces a = Land q, = 1 gave 1 +2 1 9 2 9 1 gir] = 2/1 I: GPa 12a 4 Et Multiplying out, Ga] -EIEDE-« T,ạ S= 1,2 lr ey a m
The condition of this redundant matrix ts obviously better than that obtained with the first choice of redundants There is less cross-coupling between the redundant forces
Thus the analyst, by choice of redundants, determines the condition of the matrix The choice may be critical in the
case of a highly redundant structure, for it may prove im- possible to invert a large, ill-conditioned matrix with the
limited number of significant figures available from the initial data, The following statements and rules-of-thumb may be useful in the treatment of highly redundant problems
redundants for whi t
ninimumi-zero, in tact Theoretically ¢ hen, “Sy proper choice of redundants, the matrix fa vs |
may be reduced to a diagonal matrix (ideally conditioned)
(1a) The choice of reduncants which gives zero cross-coupling (orthogonal functions”) is not readily round in gene In Some special
structures, such as rings and frames, orthogonal AB 29 redundants may be chosen simply In most struc- tures, nowever, the additional labor involved in seeking an orthogonal set of redundants is not warranted
(2) In choosing a set of redundants which
Will yield a well~conditioned redundant matrix, it_is best to make such “cuts” as will leave a statically determinate structure retaining as many of the characteristics of the original structure as Dossible Thus, one may consider that the structure of Fig AS.35(d) retains more of the features of the original continuous—deam
structure than does that of Fig a8.35(c)
(3) The degree to which one redundant in~ fluences another (extent of cross-coupling) can de visualized by observing how much their indi- vidual unit-load diagrams "overlap™
ọ elaborate on this point, refer once again to the above {illustrative example The cross-coupling of qs, with q, may be expected to be large if their unit-load diagrams are drawn as below in Pig A8.36(a) This deduction follows easily if it is recalled that the dummy-unit load equation for such 4 cross-coupling term is of
Mam, ax* ET
Strictly, the comparison is with the terms the form | » obviously large for ms,
Tạo
[she =
gr? at
elements of the matrix [rs]
which form the on-diazonal 2 2L —1 ms my 2 mạ yi mạ fa} ®) Fig A8, 36 Study of the unit-load diagrams
redundant choice qi, da (Fig 35d) r that the cross-coupling should be small hers
inca BX sạn
since an intecral of the form
have a contribution from the center stan only Hence, the J By St is obviously considerably
m2 m2 ay
smaller than J BS or fs se Thus, a
visual insrection of Fig A.36 reveals thet qi, Qạ 15 4 detter choice of re cundavee than is Iss Dae
* see eq, (8) Art 48.7
Trang 13A8.30
(4) Combinations of redundants may be sm- ployed to yield new unit- uncant stress dis- tributions which do not “overlap” as axtensively as do those of the individual redundants origin-~ ally chosen
For example, suppose that in the previous illustrative problem, the choice qs, dq for re- dundants had deen made originally, leading to the unit-load diasrams of Fig A@.37(a) In-
Spection of the diagrams leads one to anticipate a strong degree of cross-coupling and hence a new set of redundants is scught Rather than return to the structure to choose new "cuts", combinations of the ms and m, diagrams are looked for which will have less "overlap" and hence less cross-coupling
It is ebserved by inspection that two new stress distributions wnich have the desired property may be formed from the ms, m, diagrams by proper combination Thus, if one half the m, diagram is subtracted from the m, diagram to
Torm oné few stréss distribution and one nalr
the results are as shown in Fig A.37b Thera
1s obviously less “overlap” of the diagrams for these new combinations 2 cP el m3 im n 2 + ™| $L—> myn Mạ L orth “TỐ {a) {b) Fig A&.37
New unit-redundant-force stress distributions (b) obtained by
combining previous distributions (2)
In this way two new unknowns are introduced by linear combination In matrix notation, the old
stress distributions ra are transformed to a new set [815] by forming C2] - Ez] Ere] i 3 2 -¬ dự, 3 9 1 matrix [ro] is written in The transformation
STATICALLY INDETERMINATE STRUCTURES
such 3 way that the vroduct [Sir] [›] does exactly what {s desired, viz., "take one times the qd, column minus one half the q, column to give the first column cf the new distribution
Do
ct
8 ", Then "taxe uír ne half of the [=2] Then "taxe ininus one £ qs
the
column plus one times the q, colum to give second column of the new distribution [2:01 " tt us Multiplying out in the above example ' ton Tc ° ' i [Ea] - : 8 ¬š 1 B ' Now form the matrix of redundant coefficients [#10] * for the new unknowns (subscripts 9, Ơ; [210] ) [Pee] * (Fs:] Fea] Ese) „1a jt + “247 {2 4
The condition of this matrix is greatly improved over that obtained for as, qd, alone (previously computed), viz.,
[ers] = #2081 E { (r,s =
(That the [50 | matrix obtained here happens 3, #)
to be similar to that obtained for qa., q, ina previous example, 15 coincidental.)
Once a transformation has been performed, leading to a new unit redundant matrix [210] > the problem may be completed in the "2, ơ
system" The appropriate equations obtain- ed from eqs (14), (21), (23) and (25) simply
Trang 14ANALYSIS AND
52I(x)*~ISe(} TT? so Gals GO GD Ba] 7-7-7 sai
z!8.,||®(¡||5ạn| TT TT TTT— (aob) (Goal = [Bor] ss] [Eo]
The final unit load distribution is
(Gsm) = [21a] ~ al [oa*] [Boal SY
and the matrix of influence coefficients is given by
GaGa - BGA ©
@ © - THE MAGNITUDE OF THE REDUNDANT FORCES; SIZE OF ELEMENTS IN [mn]
eqs (14) or (29) reveals forces act in the nature of determinate stress distri-
assumed (the [Em] } It is
Inspection of that the redundant corrections to the butions originally
apparent that if these corrections are large, chen any inaccuracies in the redundants (arising from the difficulties inherent in accurately
vert —
inverting [rs | or [00] will have an im
portant effect on the accuracy or the final stress distribution Thus, as a matter of general practice, it is desirable to keep the magnitudes of the redundant forces as small as
possible
It follows immediately that one should use for the determinate stress distribution one re~ quiring a minimum of correction, i.@., Select as the determinate stress distribution one which approximates the true stress distribution as closely as possible
The rule Siven in connection with the "choice of redundants" (rule 2, above) is an aid in making a good selection for the determinate stress distribution If, as suggested, a deter- minate structure is obtained by making "cuts" which leave a system naving properties similar
to the original, the stress distribution obtain- ed therein by statics should be a fair approxi- mation to the final true stress distribution
However, it is even more important to realize that the determinate stress distribu-
ion only need be in static equilibrium with the external applied loads and that it may be determined with the aid of any appropriate DESIGN OF FLIGHT VEHICLE STRUCTURES A8, 31
auxiltfary rules, test information, or even ;n~ tuitive guess-work which leads to a distribution close to the final true distribution There is no need to set the redundant (cut) member forces equal to zero in establishing the determinate distribution, Instead, reasonable approximate values may be employed for them (The correc- tions to these values become the unknown redundants /)
Mathematically, the magnitudes of the re~ dundant forces are directly dependent upon the
magnitudes of the elements in the matrices [Sra] or fan] on the right-hand side of eqs (21) or
(30) (The right-hand side of such a set of simultaneous equations {s called the non-homo~
geneous part.) Thus, the relative merits or
several possible determinate stress distributions may be judged by forming the matrix product [Sen] with each and comparing results
FOR EXAMPLE, if the doubly redundant struc- ture of Example Problem 3, Art A8.5 were formu-
lated in matrix form the [Sir] and [2]
matrices would be (see Fig 48.38 for numbering scheme )
Fig A838
42
q 3 Generalized force numbering scheme in illustrative prob- lem qg and dq selected as 4P s{* redundants 432 9 9 © 1 0 720 9 eu “3 9 9 402 9 9 9 9 312 „806 1,154 -1.564 -1.729 ir] = 1.0 9 9 1.0
Several possible determinate stress distri- butions (2s | will now be tried FIRST, the
stress distribution obtained by statics alone in
the "cụt" structure (aq, = 4, = 0)
Q
Trang 15AB 32
(This is certainly a poor approximation to the final stress distribution) Multiplying out
Gal ESET}
SECOND, a stress distribution in which loads of 0.40 lbs were guessed at for qi and qs (Ina two times redundant structure any two forces May be assigned values arbitrarily while satis- fying static equilibrium.) da, a, were found 71126] 1 ~1245 | ET cuT by statics Thus, 0.40 _ ) 0.258 Si = 9 9,40 GUESS 0.0672 Multiplying out one finds 9.49 a = [Sm] GUESS -101 *
Note that a reasonable guess at a stress distribution resulted in non-nomogenous terms only one-tenth as great as those obtained by use of the "cut" distribution The magnitudes of redundants are correspondingly reduced
THIRD, as a matter of interest, the true stress distribution, obtained in Example Prob- lem 3, was used The result:
ƒ „450
.2582 |, 16 [1
ae = 51260 (3 [rn] 7) _ ig f=
maug |-280 TRUE
The mon-homogeneous terms are practically zero, as they should be* The redundant forces would be zero also
*The demonstration here suggests a useful check upon the
final result of a redundant stress calculation After ob-
taining the final true stresses [Gim] (= [Gjn]), one forms
[220] aoe * (6 feu] [4]
and compares the result element-by-element with the matrix previously computed,
[ere] + fee} [20] fe
The "true-matrix" elements ought to be zero, or
nearly 0, if [Gim] is error-free
STATICALLY INDETERMINATE STRUCTURES
Example Problem 13 It is desired &
the calculation in the ? the box beam, Example Problem 15 It 2 be assumed for this purpose that the initial data of that problem were sufficiently preci 9 warrant an increase of accuracy the accuracy of Solution:
The first step taken was an examinati che unit-redundant stress distributions sb with the choice praviously made of
as redundants, The three unit-red: ct Qa, distributions were represented zraphically as in Fig A839 -18.73 18,79 26.44 26.44 “S471 31.71 ita} Fig A8 39
Unit-redundant-force stress distributions -
axial flange forces
Trang 16ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES 0 [Exo] = 5 59.58 1,9285 | - ~_ ,8230 1 „Ô10 =_ ,026 ‘ 2026 ~31.71 21.71
NOTE: VOIDS DENOTE ZEROES
Fig A8.39a shows plots of the new distri- butions, these having greatly reduced "overlap"
{compare with Fig, 48.39)
<3L.71 31.71
Fig A8.39a
New unit redundant stress distributions, (sio]
The new redundant coefficient matrix [sa] o
was obtained by multiplying out per eq (30a)
2584 -.01132 ,0000178 a] = 10" J|¬.01132 2566 -, 03124
[a] = 28 ° .00001758 ¬.03124 1254
This matrix is very well conditioned, being a considerable improvement over that of Gas
found originally in Example Problem 15, viz., A8 33 73774 ,1788 ,0520 = 10 789.267 [3,5] it 1789 2678 07322 * 0820 07522 ,1254
It remained to select a determinate stress distribution which would reduce the magnitude of the redundants For this purpose, the engin- eering theory of bending was employed to compute Stress distributions satisfying equilibrium for each application of a unit external load The result was (refer to Example Problem 30, Art A7,11),
NOTE: VOIDS DENCTE ZEROES
N Next the matrix [eon] was obtained by oa b
multiplying out per eq (30d)
95 -3495 -1841 1841 0 9
[on | =i |1167 -1167 1554 -1554 -1450 1450
E
1082 -1082 1445 -1445 1802 ~1802 This result compared very favorably with [crm | previously obtained, the elements being from one- half to one-tenth as large
The solution may be carried to completion in
the "9,0 system" using the matrices [te]: [00]
[Zim] *8¢ [Son] in eqs (31) ane (32)
A8.13 Thermal Stress Calculations by Matrix Methods
The thermal stress problem is conventently formulated in matrix notation by an extension of the techniques presented above
Trang 17AS 34
Gal} Baleyee
In light of the physical interpretations given
{ 9| b
to ("rs | and [ern] ) the equation is seen to be a matrix-form statement of the condition for continuity at the redundant cuts, viz., the dis- Placement at each cut caused by the redundant forces plus the displacement at each cut caused by the external loads, must be equal to zero, To modify this equation for thermal stresses, the appropriate expressions for thermal dis~ Placements at the cuts must be added Following the argument used in Art A8.9 one writes
{20} eal{%9}* Beal {fa} °
where ad is the displacement at the rth cut due to thermal straining in the determinate structure The explicit form for this term will be derived below Rewritten,
[ers] {4p =~ Era]{Pa$- {Saf (88) Eq (33) 1s a modified form of eq (21), giving as its solution the redundant forces in an indeterminate structure under the application of both external loads and a temperature distri~ bution
To derive an explicit expression for Sap the virtual work concept may be employed to ad-
vantage Thus, following the argument of the "virtual work" derivation for deflections (Arts A7.7, A7.8), the thermal deflection at the rth
redundant cut must be equal to the total in- ternal virtual work done by the rtburedundant
force virtual stresses (due to a unit rthure
dundant force) moving through distortions caused by thermal strains
Since the internal stress distribution {s expressed in terms of the internal generalized
forces q› Sự it 18 convenient to employ these q’s in writing the virtual work of straining If one lets ‘iq be the displacement of internal generalized force 4 due to thermal straining, then the virtual work done by a single general~ ized force is 4 Âm -Ổ The quantity don will be called the member thermal distortion The total virtual work throughout the structure is ob- tained by summing, giving the deflection at the
rth cut as the matrix product
Sa Ley {is}
where By is the value of the 3 due to a unit (virtual) load at cut r., Note that the term 4 iT
STATICALLY INDETERMINATE STRUCTURES
may consist of the sum of several contributions should ay act on more than one member,
a é Because rT
dundant cuts, eq (34) is expanded by writing
{°ro}* Ere] {ts}
[2] is, of course, the transpose of [21r > is desired at each of the re-
the unit redundant force stress distribution in the determinate structure Substitution of eq
(35) into eq (33) gives
[rs] {9s b= (rn {Papo [Ea] {ir} ~ (56) Solution of eq (36) gives the values of the redundants Ggs after which the problem may be completed in the usual fashion, viz,
{u}- Bel{sy:
It 18 obvious that the use of combinations
of redundants (the "p,o" system of Art A8.12) is possible One makes a direct substitution or
[15] for tr} [on | for Em} etc into eq
(36)
MEMBER THERMAL DISTORTIONS
It remains to establish the forms for Âm, Thermal strains on an infinitestinal ele- ment of homogeneous material can cause uniform normal extensions only, so that no shear strains develop Hence oniy normal (as opposed to Shear) virtual stresses need be considered in computing the internal virtual work Note that normal stresses associated with flexure must be included It follows that only virtual work in axially loaded bars and in beams {n flexure need de considered Hence Sen is zero for all ay
which are shear flows on panels or torques on shafts
BARS
The general expression for the virtual work done by virtual axial loads u in a bar under varying temperature T is W= [ u:atTdx where a is the material thermal coefficient of expansion
* tt will be convenient later to designate by Gsq the solution
Trang 18ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES A8 45
Several specific cases are now treated BEAMS
A bar under linearly varying load with ` ¬ For a beam the general form of expression hưu varying temperature is shown in Fig for the virtual work done during thermal strain~
tn this case ing by a top-to-bottom-surface temperature dif-
mm Neen ference OT, varying linearly over the beam depth,
a - 44 is (see Art A7.8, Ex Prob 24) us qy +——— a % 7 UU, son |a #8 h Ty = Ty ‘ Te Ty +X Fig A8, 40 L where
Then, assuming a constant m = virtual moment (positive for compress- lon on top fiber) L a,-4 T, -T oT = T, -7, dea ae EEE) BE a sortom ~ Trop h h = beam depth * Applied to the case of Fig 48.42 one gets wap ett Tị + Ø1; 8 s = a at ol ——— q 6 6 J L a 4 - 4 OT, = oT, This expression may be put in the form Wee a ° ay + ———— x] L oTy + L Yj1‹ We Way Bip ty Oop = ob (58T; + ÔT „ 8L (6T, + 267; where br 6 aR ee aT, +T 6T, Mg Fob Ed B1; ‘ qy | | 4 T, + eT yx dy tot Hd Cre 4 J § h 2 constant Tig A8.42
Note that variation in the cross sectional
area of the bar does not affect the distortions | or
3m Đụ Weq,4,,+q4,4
The alternate choice of generalized forces - 1 “it 4 UST for the bar under varying axial load is shown where
1n 71g A8.41 By a
derivation similar to q Aen 2 ab (=: + )
that above one finds im q "nh 6
Wea Ayn 7 4, Âm ý CHỦ A _ ab OT, + 2673 Te 6
where Fig a8 41
- Spectal forms of the thermal distortion
T + expressions for beams of varying depth may be
Am ma iT sử 3 derived readily as required
Sxample Problem 19
dem =aL® TT VỊ The upper surface of the beam of Fig A8.43
4T 6 is subjected to a temperature 6T above that of the lower surface, varying linearly 2s shown (l.e.,
The simpler cases of uniform load (ay =u ðTo
= constant) and uniform T Œ, = Ty) follow im-
mediately dy specialization of the above forms x
For example, for a bar under constant lead Oy = const
ay = 3a and constant temperatures, Tụ = Ty =T, on L —— L —¬- L ¬ one nas A,,, = a LT
Trang 19A8 36
are equal at the left end and the right end) Determine the assuming @ is constant the temperatures differ by OT, at center reactions Solution:
In the illustrative example of Art A8.12 this structure was analyzed by employing as generalized internal forces the bending moments Gd, and qe over the central supports (see Fig A8.35) The two central reactions were denoted by q, and q, The matrix of redundant coeffic- tents, considering qi and qe as redundants (the better choice, it will be recalled) was (rer Art A8.12) Ga-& F ] The unit redundant load distribution for q, = 1 and qd, = 1 was r i 1 2 Ll] oi 9 2 0 1 & = 2 1 ES Hi: 1 2 ey ñ|TL
Member thermal distortions were computed (Note that oT was negative according to the convention adopted earlier) a, 24g 6% aL (Esmraxen) fp = tý CS oe SN SS L 36 3 wu (M222 357 +h yf ea" a 6 2 i 2x3 OT, + 4 -# 3 _ =-§% Sor, Then from eq (36) 2 1 aLét
L a.tla © lo- ii -o!:
BEI 68 EL lig da ate Ø1 i - : - 2aL „ OTe °o | STATICALLY INDETERMINATE STRUCTURES 2zI a OT tap 1 2 q.,Ì E1 vốf, Qe h
The final stress distribution was
{% p+ Esa] {*a}* Bez) {} Solving „267 +933 20+ Ela Of, 1 0 “ 9 11|.267 2 1 ~ £ „955 i 12 L a +267 _ B1 qốP, 933 SP ih ,599/L ~1.60/L Thus the reactions were as = 856 “L EHo - EI a 6T (negative indi- de = “1.60 cates DOWN) Example Problem 20
The symmetric sheet-stringer panel of Fig 48,44(a) 1s to be analyzed for thermal stresses developed by heating the two outside stringers to a uniform temperature T above the center stringer Assume G = 0,.38S5E
Solution:
The panel was divided for convenience into three bays The numbering and placing of gener- alized forces {s shown in Pig, A6.44(b)
Transverse members (ribs, not shown on Figure) were considered rigid in their own planes - a Satisfactory assumption for symmetric panels Because of symmetry only one half the panel was handled All member flexidility coefficients and thermal distortions were doubled where appropri- ate
Trang 20ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES A8.3T Then —.Ð 9 © —.ỗ Q 9 ~ 5 1.0 9 9 The matrix of member thermal distortions kh = 9 1.0 9 was ° ọ 1.0 40 ~ 025 9 9 40 20 2025 - 025 9 9 9 025 - ,025 {eo} =aT 5 6 Multiplying out per eq 16; 9 9 178.2 13.8 9
Here, for example Ayp = 2.x 2x 10 = 40 (doubled Grs}=2L | 15.8 172.2 15.8
once because qi acts on two stringer ends and “ 9 12.8 86.1 doubled again to account for the otner half of and
the panel)
peeps Is, poe 75 0 0 20 0 0 ~c2 -.06 07] {40
[Ect]{sw}s+t |9 HE 6 9 20 0 9 ~o25 ces! | 40 a | 9 0 ~ã 0 0 2.0 0 0 ~.085 20| as to ia 9 502 ° 3 tryp sọ" | de | Q T T af cm, 1% Q | * | : ạ PS Hưng Tư ợm) œ6
(a) Then the redundant equations, per eq (36),
Fig A8 44 were
Trang 21A8 38
The result compares favorably with “exact”
Solutions made under the same assumptions (ref
NACA TN 2240) as far as the stringer loads are concerned The shear flow result is not very
Satisfactory, primarily due to the use of too
few "bays" in this analysis Example Problem 21
The uniform Tour-flange box beam of Fig A8.46a is to be analyzed for the thermal
Stresses developed upon heating one flange to a temperature T, uniform Spanwise, above the other three flanges, 4 N ` NA N TA ` ỒN ` > er Ñ ¿7 \ N AAS 5 t Ara AQ Se WE Ash o Fig A8 46 ` — Solution: Qo)
The beam was divided into four equal bays giving a four times redundant problem Four Self-equilibrating (zero-resultant ) independent stress distributions were taken as the unknowns , these being shown in Pig A8.46(b}) Such zero- resultant stress distributions are the only ones possible in a structure having no applted loads*,
The matrix of member flexibility coerfic- lents was formed by collecting coefficients from the several members Unit redundant stress dis- tributions were prepared, taking qi, đa, da and 3; 3S the redundants and Setting these equal to unity successively, „S00 E0]: ke) 11667" +500 667K" „1867k where k* = Gt / AB (b +c) ——
* Since external loads are not to be applied it follows from
statics that the generalized forces for adjacent webs and cover sheets are equal, as are the loads in front and rear Spar caps at any given station, Hence an economy of numbering in the generaiized force scheme is possible,
Much labor is saved in the handling of data when the same
symbol can be employed on several members whose loads are known to be equal STATICALLY INDETERMINATE STRUCTURES w L/2 be Lực b/2 fal ty Lye Ly Lig + L/ạ Lực 1/2 Lựa ‘ Multiplying out per eq (18),
Pe + ess] ees 375 us|
„82 se * 83 «375 12s
[ea] - He
ae +375 co ae =-
1128 2228 1128 (ERED + 083s For a specific case let «*L*7 = 1 The inverse was computed to >e 1.070 |~ 5291|~ 2337|~ Q6587 E *] : Gt = 5291 1,566 |- 3613|¬ 1018 Si" ie +e) {2337 | seis 1.497 [2 cisae ~ 06587|= 1018]~ ,1934| 1.752
Member thermal distortions (Am) were computed for loads dae, Ga, Ga And Ga
collected from the one heated flange and were only (Note that if two ad-
jacent flanges are heated equally, one aust Set the corresponding dan
Trang 22ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
{San} ~ [Xe] ni] te}
whieh, when multiplied out gave
4,079 GtaLT ) 1.938
{sq} “"i6 (e+e) 1 8662
12413 |
Finally the complete set of thermal stresses
were (still for the case L*k* = 1) 4,079 2.039 L 1.938 3.008 L »8562 3.437 Lo „2413 3.558 L GtaLT {asrb=[21s] {eer} = - ES sy
The result compares favorably with an exact solution (NACA TN 2240) insofar as flange stresses are concerned, an error of less than 1 percent being present at the root Shear flow values in the sheet are less satisfactory due to the (relatively) crude assumption of constant shear in*each beam quarter
A8.14 Thermal Deflections by Matrix Methods A STATICALLY DETERMINATE STRUCTURES
The problem of the thermal deflections of @ statically determinate structure was con- sidered earlier in Art, A7.8 in non-matrix form It should be apparent from the deriva- tion, that the matrix method presented for the calculation of redundant~cut thermal deflec-— tions may be applied equally well to the
problem of computing the thermal deflections of external points of a determinate structure Thus
th}? [Ôn
where Sup is the thermal deflection of external point m, and [Sas | is the transpose of [Sim]
the unit-applied load stress distribution (compare with eq 35)
B REDUNDANT STRUCTURES
In the case of the redundant structure, additional strains are present due to the thermal stresses set up; the effect of these strains upon the deflection of external points must be included in the calculation
The appropriate equation is most easily derived by visualizing the action in two
stages FIRST, the redundant structure is
Ag 39 "cut" making it determinate, after which the temperature distribution is applied producing thermal deflections
{en} [Eo] {os}
at the external points (compare with eqs (35)
and (27) ), Simultaneously the redundant cuts
will experience relative displacements
SECOND, the redundant cuts are restored to zero displacement by the application of redund- ant forces (this problem was solved in Art A813) The a, are given dy eq (36); they pro~ duce additional deflections at points m
{om} Eal{s}
The total deflection of point m ts then
{er} [Ear] {*:0}* Ena] {see} Bs]* [ao] Ea] ,
{Sa}? [Su]! [Sai] sa) Ess] {4s}
{Sun }= [Eat] (ss}" Eu(sz}) - (38)
The matrix quantity in parentheses is the total strain (thermal plus "mechanical")
For derinite reasons the equation for thermal deflections has been left in the form of eq (38) rather than the more polished forms which might be obtained by substitution from eq
(36), FIRST, the Ijey? the thermal stresses, will probably have been solved for previously and will be readily available in explicit form -SECOND,
and far more important from a labor saving stand- point, the unit load distribution [Em] (wnose transpose 1s used in eq, 38) may be any conven- jent stress distribution satisfying statics in the simplest of "cut" structures One need not
even use the same Sim distribution (and same
choice of "cuts™) as employed in the redundant thermal stress calculation; a more conventent choice of cuts may be employed’ In principle, any stress distribution statically equivalent to
the unit applied lead(s) may 5e used for Sim |
in eq (38) (See p A8.9)
Trang 23AB 40
Example Problem 22
Compute the rotation occurring at the right hand end of the beam of Fig A6.43,
Solution:
To compute the rotation a unit couple was applied (positive counterclockwise) at the right hand end An additional generalized force, called qs, also was added at that point Then, SE (~) Ae Z6T, +207) 4aL6%, ar 2 = § oh and, using some results from Example Prob 19, fue
Note that q, and q,, the intermediate support reactions, were omitted from consideration They do not enter into any expression for the
internal virtual work of the structure; or, equally, they are not used to describe the strain energy of the structure Hence they are mot included in writing the total strain
(Their Ap ere Zero.) qLô"', on 1=1, 2,5 The member flexibility ccefficient matrix was 4 1 £0 1 4 1 [Esa] = ar fs1,2,5
From Example Problem 19, the true thermal stress distribution was
{ar} Ela 67, {|
a ọ i1 =1, 2, §
(qe was zero by inspection)
To determine [in } a unit couple qa, =l
was applied Taking as the "cut" structure one where q, = 4, = 0 one has simply featelfhs LateLe =a Then substituting into eq (338), „267| BỊ a8 =a) Loo Lf 3 + = ~ GST Jey, Liss ony on 4 6El g1 veo STATICALLY INDETERMINATE STRUCTURES qLôTy |9 9 1 — ~.289 qLơƠT; By = - ,289 ? a
It is apparent that the values of the re- dundant moments d., Ga could nave deen chosen
ar5itrarily in {5a} above, without affecting
the result Here, clearly, ftiqb for any “cut™ ts
structure visualized will lead to the same re- sult, q, deing equal to unity in all cases Example Problem 23
“~~ Compute the axial movement of the free end of the central stringer of the panel of Fig AB.44,
Solution:
An additional generalized force, dio, was added axially to the free end of the central stringer (ref Fig A8.44b) Then 177.6} 44, 44.,4| 177.6| 44 44.4 88 OEE 31,202 21,200 Í44.% bạ: Đề SooooocobBÐĐ
Trang 24
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES AS 41 Toen, substituting into eq (38), ret Lf [a 13,34 11,81 13.46 6.656 84.24 6.833 «5928 2.402
It remained to find the determinate stress dis- tribution for qi, = 1
As the determinate distribution {8in} the stresses due to a unit load dio = 1 were com~ puted in the “cut” structure obtained when di = 42243 = 0 (Note that this is a differ- ent choice of redundant cuts from that employed in computing the thermal stresses in Example Problem 20.) Thus fu HHOOOC 2, 2?22OOCOC and finally, Bm =œT| 00011100 28.19 26.55 13.34 11.81 ‘ 13.46 8.656 -84.24 6.855 +5928 2.402 - = 54.2 œ7 (ANSWER )
AS 4 matter of interest, a different unit stress distribution was employed with a differ- ent choice of “cuts" If the forces q,, qd, and Ga are set equal to zero ("cut"), application of
a unit load qi, = 1 gives (writing the transpose
sĩ {81m >)
[ Ea: {7| 59 +50 60900 025 00 1 |
Then multiplying out for the thermal deflection:
Ômm = 54.3 œT (same answer)
It is apparent from the above result that the simplest determinate ("cut") structure should be used to compute [#in]? it is completely
adequate
CLOSURE
The general thermal stress problem is complicated by the fact that the material prop-~ erties E, G-and a vary with temperature The problem created thereby is primarily one of book- keeping - computing the member flexibility caeffi- clents (ay) ane the member thermal distortions
(Am) for a structure whose properties vary from
point to point with the temperature The vari- ations of E, G and a with T will, of course, have to be xnown from test data,
Two additional complications, not considered here, are the lowering of the yield point with heating (and the attendant increased likelinood of developing inelastic strains) and the phenom- enon of "creep" (the time-dependent development of inelastic strains under steady loading)
Should it prove necessary to analyze for thermal stresses under more than one temperature distribution, the member thermal distortion matrix {Sun } wey be generalized easily into a
rectangular form such as
Cia] = Era] Fa]
where KH is the member thermal distortion as- sociated with force ay from thermal loading con-
dition R The matrix E8.) of member thermal coefficients consists of the constant coefficients
Trang 25AB 42
A&.15 Problems
Note: Problems (1) through (S$) below may be worked by either the Least Work or Dummy Unit Load Methods The student will de well advised to try some problems both ways for comparison 5000# A 5) B 9 a3 s Ø8! 100% D (1.5) cb 100" ¬ Fig a Fig b
(1) Determine the load in all the members of the loaded truss shown in Fig (a) Values 1n ( ) on members represent the cross-sectional area in sq in for that member All members of same material
(2) For the structure in Pig (b), deter- Mine the load in each member for a 700# load at
joint B Areas of members are given by the values in ( ) on each member All members made of same material 300% Ƒ—— 60" — 307 4 a Ấ w 1) gOS al aoe ^ @) @» Nip | : Fig c ỊP j 1000# 1000#
(3) For the loaded truss in Fig ©, deter- mine the axial load in all members, Values in parenthesis adjacent to members represent rela- tive areas £ is constant for all members
—-~
26"
(4) For structure in Pig d, calculate the axial loads in all the members Values in paren~ thesis adjacent to each member represent relative
areas 1 18 constant or same for all members 5000# Đ c | 60” Fig e } B — 120" STATICALLY INDETERMINATE STRUCTURES
(S) For the "King post” truss in Fig e, calculate the load in member BD Members AB, BC and BD have area of 2 sq in each The con- tinuous member ADC has an area of 9.25 sq in and moment of inertia of 214 tn* = is same for all members
500#
(6) In Fig f, AB is a steel wire both 0.50 Sq in, area The steel angle frame CBD has 4 4 sq in cross section Determine the load in member AB = 30,000,000 psi
(7) In Fig g find the loads in the two tle
rods BD and CE Ig¢ = 72 in.*; Ang = 0.05 sq
in Age = 0.15 sq in E is same for all members + Làn er A 180" his —— 18 ị 1 50, 000% A de [u>— Fig h Fig i
(8) Por the structure in Fig h, determine the reactions at points A, B Members CE and ED are steel tie rods with areas of 1 sq in each, Member AB is a wood beam with 12" x 12" cross
section Esteei = 30,000,000 pst Bwood =
1,300,000 psi
(9) Por the structure in Fig i, determine the axial loads, bending moments and shears in the various members The structure is continu- ous at joint D Members AB, BC are wires The member areas are AB = 1.2, BC = 0.6; CD = 6.0;
BDE = 10.0 The moment of inertia for members CD = 60.0 in.*; for BDE = 140 in.*
(10) Re-solve Example Problem 10, using as redundants the two restraints (couple and transverse force) Solved
way the problem is doubly redundant as no ad- vantage is made of the symmetry of the structure
(11) Add two additional members, diagonals FB and ZC (each with areas 1.0 in?) to the truss of Fig A7.85, chapter A-7 Find the matrix of
p À8,15 at one end