A constant unknown shear flow ay Vy and qs for cells 1 and 2 and 3 respectively will be added to the static flexural shear flow so as to make the angular twist 9 of each FLIGHT VEHICLE
Trang 1ANALYSIS AND DESIGN OF Cell (1) Final stresses ~ Fig, AlS.35
Al5.10 Three Cell - Multiple Flange Beam Symmetrical
About One Axis
Pig AlS.36 shows a 3-cell box beam sub-
jected to an external shear load of 1000 Lbs
as shown The section is symmetrical about
axis XX The area of each stringer is shown
in parenthesis at each stringer point The
internal shear flow system which resists the
external load of 1000 Ibs will be calculated
assuming that the webs and walls take no
bending loads, or, the stringers are the only
effective material in bending The moment of
inertia about the XX axis of effec.lve material
equals 250 ins (Note: this beam section is
tdentical to the two cell beam of Fig A15.30
plus the leading edge cell (3)
The system is statically indeterminate, to
the third degree, since the value of the shear
flow q at any point in each cell is unknown
The value of the shear flow will be as-
sumed at a point in each cell and the flexural
shear flow for bending about the XX axis will
be determined consistent with this assumption
A constant unknown shear flow ay Vy and qs
for cells (1) and (2) and (3) respectively
will be added to the static flexural shear
flow so as to make the angular twist 9 of each
FLIGHT VEHICLE STRUCTURES
Al5.15 cell the same, since if any twisting takes place, all cells must suffer the same amount
Furthermore, for equilibrium, the moment of the internal shear flow system plus the moment
of the external shear load must equal zero
For bending about axis XX, the flexural Shear flow will be assumed as zero at a point just to the left of stringer a in cell (3) and
just to the left and right of stringer c in
cells (1) and (2) respectively One might con- Sider the cells as cut at these three points
Fig Al5.37 shows the flexural shear flow under these assumptions Since the leading edge cell (3) has no stringers and the covering is con- Sidered ineffective in bending, the shear flow will be zero on the leading edge portion since the shear flow was assumed zero just to the left
of stringer a, The resulting flexural shear flow for the 3 call section will therefore be identical to Fig A15.31 and the calculations for the flexural shear flow will be identical to those in Art A15.7
AED
Trang 2A15 16 SHEAR FLOW IN
Cell (3)
šqL 2@A „ở -.—
Taking moments of the internal shear flow
systems of Fig A15.37 and A15.36 and the ex-
ternal load of 1000 lbs about stringer a and
Ma, 10 x20x10+10x30x20-5x 1000 78.6q,+ 200q, +200q, = 0
3000 + 78.6q, + 200q, +200q,=0 - - (4)
Solving equations (1) (2) (3) and (4) for the
unknown 4,, dg, 4d, and 9G we obtain:
q, = - 2.12 1b./in
Qa = - 7.09 1D./in
4, = 714.5 lb./in
Gs- 19.9 Adding these constant shear flows to the flex-
ural shear flow of Pig Al5.37, we obtain the
true internal resisting shear flow as shown in
(1) using multiple interior webs, the detri-~
mental effect of shear deformation on bending strass distribution is decreased; (2) the fail safe characteristic of the wing is increased because the wing is-made statically indeter- minate to a high degree and thus failure of individual units due to fatigue or shell fire can take place without greatly decreasing the over-all ultimate strength of the wing;
{3) the ultimate compressive strength of wing flange units is usually increased because column action is prevented by the multiple weds which attach to flange members
In Chapter A6, Art A6.13, the method of successive approximation was presented by de- termining the resisting shear flow system when
a multiple cell beam was subjected to a pure torsional moment This method of approach has now been extended to determine the resisting shear flow when the beam is subject to flexural bending without twist*, Using these two methods the shear flow in a beam with a relatively large number of cells can be determined rather rapidly
as compared to the usual method of solving a mumber of equations
PHYSICAL EXPLANATION OF THE METHOD Fig A15.40 shows a 3-cell beam carrying and external shear load V acting through the shear center of the beam section but as yet un- known in location In other words, the beam bends about the symmetrical axis X-X without twist The problem is to determine the internal resisting shear flow system for bending without
* "The Analysis of Shear Distribution for Multi-Cell Beams
in Flexure by Means of Successive Numerical Approxi-~
mations." By D, R SAMSON, Journal of the Royal Aeronautical Society, Feb, 1954
Trang 3ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES twist In this example, it is assumed that the
bending moment is resisted entirely by the
flange members as represented by the small
circles on the figure, which means that the
Shear flow will be constant between the flange
members
The first step in the solution is to make
the structure statically determinate relative
to shear flow stresses for bending without
twist In Fig AlS5.41 imagine each cell cut
at points a, b and c as shown For the given
shear load V, the static shear flow qg can be
calculated, assuming the modified section bends
about axis X with no twist Fig AlS.41 shows
the general shape of this static shear flow
pattern,
tv
The static shear flow Qg acting on each cell will cause each cell to twist Since zero twist 1s necessary a constant shear flow q! to cell (1), q, to cell (2), and q} to céil (3) must be added as shown in Fig Al5.42, and the magnitudes of such value as to make the twist
of each cell zero However, the cells are actually not separate but have a common web be- tween adjacent cells, thus the shear flow q acts on web 2-1 which 1s part of cell (1), and thus causes cell (1) to twist Likewise cell (3) 1s twisted by qi and cell (2) oy both q%
and q¿ Therefore to cancel this additional
cell twist, we must add additional constant shear flows q}, q2 and q° as showm in Fig
Al5.43, and considering each cell separate again However, since the celis are not separ- ate these additional shear flows effect the
twist of adjacent cells through the common web
As before this disturbance in cell twist is
again cancelled or made zero by adding further
closing shear flows q?, qq q as shown in Fig Als.44, This procedure is repeated until
the closing shear flows become negligible In
general the converging of this system is quite rapid and only a few cycles are necessary to give the desired accuracy of results
The total closing shear flows q,, 4, and
q, are then equal to -
DERIVATION OF EQUATIONS FOR USE IN SUCCESSIVE APPROXIMATION METHOD
Pig Al5.46 shows cell (2) of the 3-cell
Qs is the static beam shown in Fig A15.45
shear flow and q,, 4, and q, are the re- dundant or unknown Shear flows, Since cell (2) does not twist under these shear zlows
we can write in general,
Fig Al5-46
Trang 4whereas the subscripts i-s and a-s implies
summation only along webs 1-2 or 3-2 respect~
ively L is the length of a sheet panel and t
which must act as a constant shear flow around
cell (2) to cancel the twist due to qg The
resulting value of this first term will be
given the term qa)
The second and third terms in (3) repre- sent the constant closing shear flows required
in cell (2) to cancel the twist of cell (2)
due to the influence of q, and q, in the ad-
jacent cells acting on the common webs between
the cells The ratio in equation (3) before
4, will be referred to as the carry over
influence factor from cell (1) on cell (2) and
will be given the symbol C,.,, and the ratio
before q, in equation (3), the carry over
influence factor from cell (3) to cell (2) and
it will be given the symbol C,_, Thus
equation (3) can now be written as,
de = Mat Credit Coeds
As explained above, q', is the value of the necessary closing shear flow for zero twist
when the adjacent cell shear flows are zero
Hence first approximations to the final shear
flows in each cell can be taken as neglecting
the effect of adjacent cells, or in other words
each cell is considered separate Hence the
first approximations are,
approximation In a similar manner corrections
qi and qf are made to the approximations for
q, and q, Therefore as a third estimate for
qạ, these further corrections should be added
Ge = ger Cie (alti) +Cs2(ah+95) - - - (8)
Thus by repeating the above procedure, 2 power series of the carry over influence factor
is obtained In general the convergency is rapid and only a relatively few cycles or oper- ations are needed for sufficient accuracy for final shear flows A solution of a problem will now be given to show how the necessary operations form a very Simple routine
Al5.12 Example Problem Solution Problem No 1
Fig AlS.47 shows 2 cellular beam with five cells The flange areas and the.web and Wall thicknesses are labeled on the figure
{The problem will be to determine the internal shear flow pattern when resisting an external shear load of Vz = 1000 lbs without twist of the beam Having determined this shear flow system the shear center location follows as 4 simple matter
Fig A15.48 shows the assumed static condition for determining the shear flow system
in carrying a Vz load of 1000 ib without twist The static condition is that all webs except the right end web have been imagined cut as indicated thus making the shear flow q, at these points zero
In this example problem it will be assumed that the flange members develop all the bending stress resistance, which assumption makes the shear flow constant between adjacent flange members
The total top flange area equals 5.5 in.*, and also the total bottom flange area, Due to symmetry the centroidal X axis lies at the mid-depth point
Hence, Iy 3 (5.5x5#)2 5 275 in
dg = ¬ IZA = = mm SZA = -đ.656 72A
Starting at the lower left hand corner, the static shear flow dg will be computed going counter-clockwise around beam
đạp # ~ 3.836(=5)2 z 36.36 1b./in
dbe = 36.36 +3.636(-5)1 = 54.55 1b./in
Qcd # 54.55 -3.636(-5)0.5 = 63.64 Continuing in like manner around the beam, the values of qg as shown in Fig, A15.48 would 5e obtained
The solution from this point onward 1s made in table form as shown in Table Al15.2 which should be located Below a drawing of the
Trang 5ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
cellular beam as illustrated, and the numbers
in the Table should be lined up with respect
to the cells as indicated
The solution as presented in Table A15.2
is carried out in 17 simple steps The first
step as given in row 1 of the Table is to
compute for each cell the value for ZqgsL,
t where qg is the static shear flow on each
sheet panel of a cell; L the length of the
panel and t its thickness Values for dg are
taken from Fig A15.48,
For example, for Cell 1
3 ag-—E 5 2(86.36 x 10}-Š~_ = 18180
The sign 1s positive because qg is posi-
tive (Clockwise shear flow on a cell is
positive.) Row 1 in the Table shows the values
as calculated for the 5 cells
The second step as indicated in row 2 of
the Table is to calculate the value of the
expression = L/t for each cell
For example, for Cell l,
The third step as indicated in row 3 is to
calculate the value for the L/t of the common
wed between two adjacent cells
For example, for web db’ between cells
(1) and (2),
€) a 7 TOR * 200
The fourth step is to determine the
carry over factor from one cell to the adjacent
cell The results are recorded in row 4 of
the Table
Referring to equation (3) for general
xplanation, the carry over factor from cell
is considered separate or independent of the
other This constant closing shear flow q'
equals,
L 5qsƑ q' *s~ T” The minus sign is necessary
=
$ because the twist under the static shear flow must be canceled The values for q' are re- corded in row 5 of the Table
For example for cell (1),
ca _ 16180 _ _ —- - 21.258 For cell (2),
' 27275 _
4q ~ 950 = ~ 28.71 Steps 6 to 13 as recorded in rows 6 to 13
of the Table are identical in operation, namely, the carry over influence from one cell to the adjacent cell because of the common web between the cells As a closing shear flow is added to each cell to make the cell twist zero when they are considered separate, this result is contin- ually disturbed because of the common wed
Gradually these corrective shear flows become smaller and smaller until the cells reach their true state and possess Zero twist In the Table, arrows have been used for two cycles
to help clarify the operations
For example in row 6, the carry over shear flow from cell (1) to cell (2) 1s,
- 6.700 x 2105 = + 1.414 From cell (2) to cell (1), the carry over value
is (-4.480 - 8.330) 0.2336 = - 3.000 From cell (2) to cell (3},
(~4,480 - 8.330) 0.250 = - 3.216
Trang 6Web Data o” X 1.064 (1) 05 (2) 04 (3) 04 @) 03 — (5) 03| X
Á Carry Over Factor (C) .3105 | 2336 .250 | 2633 200 | 250 250] , 266
5 Ist Approx q’ 2 ~2qg L/t/S L/t} -21 238 -28, 71 ~31,82 ~38, 79 -63 910
6, q”*= CQq! (Carry over) - 6 100.” -4 480 | ~8.3802<7.170 |-9 7006 364 ~17, 560% —>- 9.700
T, q'" = Cq" (Carry over) - 3, 000” ~^~1, 414 | <4 430-3.218 |<5.075Z%<3.372 |-ôuB60*7 ^- 3.07Ệ
8 ETC (Carry over) = 1, 362 ~0.631|-2.518 -1.460 [-1.489_-1.837 | -1, 592 - 1.488
CALCULATION OF SHEAR CENTER LOCATION
SSE UPR EAR CENTER LOCATION
in Fig Al5.47 let X = distance from lett end of beam to shear center
upper left corner of the shear flow forces in Figs,
1000X = 10(86 56+54.56+6 „64+72 ,75+81 „88 )10 + 100 x 10 x50 - 2 x 100(55,51 + + 84.39), hence 1000x = 19472 or x = 19.47 inches
Taking moments about
A15.48 and A15.49 and equating to 1000 x
$2.45 + 63.43 + 73.91
Trang 7ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
In row 13 of the Table, the carry over
values are so small that the process is termin-
ated The final constant shear flow that must
be added to each cell to cancel the twist due
to the static shear flow equals the algebraic
sum of the values from the beginning row 5 to
row 14 The results are shown in row 14 of the
Table
The results in row 14 are obtained after a
‘considerable number of multiplications and
additions of numbers, thus it is easy to make a
numerical mistake To check whether any
appreciable mistakes have been made, we take
the values in row 14 and consider these values
of constant shear flow in each cell as that
causing zero twist if cells cre separate Then
bringing the cells together, through the common
webs causes a disturbance in twist and this is
made zero by the carry over values, This step
in the Table is referred to as a reiteration
and is indicated in row 15 Then adding the
values in row 15 to the initial approximation
q' in row 5, which value is repeated in row 16,
we obtain the final value of q in row 17 The
values in row 17 are practically the same
magnitude as in row 14, thus no appreciable
mistakes have been made If the difference
was appreciable, then a second, and if needed,
even a greater number of reiterations should
be carried out In the Table a second reiter-
ation is shown in rows 18, 19, 20 and the
results in row 20 are practically the same as
in row 17
Tt will be assumed that the solution was
stopped after first iteration, and thus the
values in row 17 are the constant shear flows
that must be added to the static shear flows
to produce bending without twist Fig A15.49
shows these final closing constant shear flows
Adding these values to those in Fig A15.48 we
obtain the final shear flows in Fig Al5.49,
The lateral location of the shear center
for this given 5 cell beam coincides with the
centroid of the shear flow force system in
Pig 415.50 The calculations for locating
the shear center are given below Fig A15.50
Solution 2 of Problem 1
In solution 1, the assumed static condi-
tion involved cutting all vertical webs except
the right end wed Thus the static beam
section became an open channel section and the
resulting static shear flows must obviously be
far different than the ?inal true shear flow
values, since the webs always carry the greater
Shear flows in bending without twist This
fact is indicated by the relatively large num-
ber of steps required in Table Al1S.2 to reach
a state where successive corrections were small
enough to give a desired accuracy of final re~
Al5, 21 condition where the static shear flows in the webs should be much closer to the final values and thus hasten the convergency in the succes~
sive approximation procedure
Thus in Fig Al5.51, we have assumed the top panel in each cell as cut to give the static condition The static shear flow is now con- fined to the vertical webs and zero values for top and bottom sheet, Table AlS.3 shows the calculations for carrying out the successive approximations and needs no further explanation
It should be noticed that after the first ap- proximation was made in row 5, only three carry over cycles were needed in rows 6, 7 and 8 to obtain the same degree of accuracy as required
in 8 cycles in Table A15.2 for solution 1 Fig
A15.52 shows the final shear flows which equal the constant shear flows in each cell from row
9 of Table added to the static shear flows in Fig Al5.S1 These values check the results of solution 1 as given in Fig A15.50, within slide rule accuracy In Table A1l5.3 no reiteration steps were given The student should make it a practice to use such checks
Al5.13 Example Problem 2
All Material Effective in Bending Resistance
The general trend in supersonic wing struc- tural design is toward a large number of cells and relatively thick skins, thus in general, ail cross-sectional material of the wing is effective
in resisting bending stresses and thus the shear Tlow varies in intensity along the walls and webs of the beam cells Fig A15.53 shows a ten cell beam with web and wall thicknesses as shown It will be assumed that all beam mater- fal is effective in bending The shear flow resisting system for bending about the horizontal axis without twist will be determined The cen- troid of this’ system will then locate the shear center,
Fig AlS.54 shows the static condition that has been assumed, namely, that the upper sheet panel in each cell has been imagined cut at its midpoint, thus making the static shear flow zero
at these points The static shear flow values
dg are shown on Fig AlS.54 To explain how they were calculated, a sample calculation will
For convenience an external shear load
Vz = 8750 lb will be assumed acting on this beam section
=a =o =
Hence, q Te 224 mẽ 3ZA Z~100 32A
Trang 8A15 22 SHEAR FLOW
Now consider Fig A15.55 which shows a
sketch of cell (1) plus half of cell (2), As
previously explained the upper c9ll panels were
assumed cut at their midpoints (a) and (m),
(a) in cell (1) where the
shear flow 18 zero and going counter-clockwise
around the cell,
as follows: - the static shear flows are
12 L/t for each cell
Final Shear Flows
Solution I (See Fig A15-50)
Example Problem 2 Ten Cell Beam - A11 Matertal Sf?ective in Bending
Top Skin = 125 Inches Thick
{
Fig, A15-53 3 oF mF @ HF w F 6œ Ss | FF @m a @ §| (@ Š| ans ĩ
Bottom Skin = 128 Inches Thick m—5'—¬
-13 gay Bey +1867 — S156) —— 168) — sey cise tse ot
8 9
-3 1
~11.28
=10.42 -i1.31 ~1,1 9g
-9.07 Ũ
~4 25 -~4.27
(Note: Shear Flow at Ends of Webs Equal Sum of Shear Flows in Adjacent Skin Panels )
TẾ T3 ‹34.04 T2.98+12.43 83,55 -55,39 100.61
Trang 9ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
At point (f) there are two other connecting
sheet panels so we cannot proceed past this
joint in calculating the shear flow in the two
With two other webs intersecting at joint
(f) the shear flow summation cannot continue
past (f), hence we go to point (m) in cell (2)
where shear flow is zero due to the assumed
Now at Joint (n) we have the shear flow of 78
magnitude on each top panel, thus the shear
flow in the vertical web at (h) equals the sum
of these two shear flows or 156
A15, 23 Proceeding to (g)
Qg = dng + 1002%2A = 156+ 100 (1.25 x2.5x h
.094) = 156 + 29.4 = 185,4 lb./in
tg = 185.4+1002°ZA = 185.4+100 (1.25) g (2.5 x 094) = 186 Ib /in
atk = 4¢g-Gfe *-156+78 =-78
Iker =-78- 1007574 2-78-100 (-2.5)(2.5x 0,125) =~78+78 =0
Fig AlS.56 shows a plot of these calcu-
lated values The arrows give the sense of the
Clockwise shear flow in a cell is positive shear flow Since an interior web 1s part of two ad- jJacent cells, the sign of the shear flow on vertical webs is referred to the left hand cell
in order toa determine whether sense is positive
or negative
Having determined the static shear flows which will be referred to as dg, we can now start the operations Table Al5.4 The first horizontal row gives the calculations of the twist of each cell under the static shear flows, which {s relatively measured by the term
Sag-E tor each cell
With all material effective in bending the shear flow varies along each sheet Fig A15.56 shows this variation on the sheet panels of cell (1) The term Zgg L/t is nothing more than the
area of the shear flow diagram on each sheet
divided by the sheet thickness To illustrate,
consider cell (1) in Fig A15.56
Upper sheet panel: -
š UL - -(0+78) 2.5 +(0+78) 2.5 9
ize
Trang 10
ALS 24 SHEAR FLOW IN
For lower sheet panel: -
fase O (Same figure as for upper sheet)
For left hand vertical web: -
Treating the shear flow diagram as 4
rectangle with height 78 and a parabola with
height 98-78 = 20,
Bag hs 27142
For right hand web of cell (1),
The shear flow diagram 15 likewise made
up of a rectangle and a parabola
L„ 186x5 + 70.084 0,667
248 + (186.5 ~ 156) 5x 9.094 = 9342
Therefore for entire ceil (1)
2qạ + = ~7142 + 9342 = 2200, which is the
value in row (1) of Table Al5.4 under cell (1)
The results for the other nine cells as calcu-
lated in a similar manner are recorded in row
1 of the Table The procedure as followed in
the remaining rows of Table 4 is the same as
explained in detail for example problem 1 In
JTable A1S.4 only one reiteration 18 carried
through as the values in the bottom or last
row are practically the same as arrived at after
the fifth carry over cycle Adding the con-
stant shear flow values in the last row in the
Table to the static shear flow values in Fig
Al5.54 we obtain the final shear flow values
of Fig AlS.57 The resultant of this shear
flow pattern is a force of 8750 acting down
in the Z direction Its location would be
through the centroid of the shear flow force
system Let x equal distance from upper left
hand corner of beam to line of action of shear
flow resultant force
Taking moments of shear flow force system
of Fig AlS.54 plus the constant shear closing
values in each cell as given in the last row
of Table Al5.4 and equating to 8750 X, we
obtain;
Due to uniform static shear flow on each webs —
M = (156 x6) (5 + 10 + 16 + 20 + 26 + 20 + 35 +40 +45) +78x5x50 = 175000
Due to parabolic static shear flow in each web:
M = (29.4x5x0.667)(S+10+15+20+ 25)
*+ 25.5x 5 x0,667 (50 + 55) + 22.5 x 5 x 0,667 (40 + 45) +20x5x 0.667 x50 = 22520 Due to constant closing shear flows as
CLOSED THIN-WALLED SECTIONS
Referring to the final shear flow values
in Table A15.57, it will be noticed that the final results are not much different from the assumed static shear flows with the possible exception of the two end webs If we had assumed all the webs cut except one to form the static condition, then Table Al5.4 would have required several times as many carryover cycles
to obtain the same accuracy of final results
A15,14 Use of Successive Approximation Method for Multiple
Celi Beams when Subjected to Combined Bending and
Torsional Loads
The internal shear flow resisting force system for a beam subjected to bending and twist- ing loads at the same time is carried out in two distinct steps and the results are added to given the true final shear flow system First, the shear flow resisting system is found for being without twist as was explained in this chapter, The results of this first step locates the shear center The external load system is then transferred to the sheat center, which normally would produce a torsional moment about the shear center The internal resisting shear flow system to balance this torsional moment is then handled by the successive approximation method as explained and illustrated in detail
in Art A6.13 of Chapter A6
A15.15 Shear Flow in Cellular Beams with Variable Moment
of Inertia
The previous part of this chapter dealt entirely with beams of constant moment of iner- tia along the flange direction In airplane wing and fuselage structures, the common case
is a beam of non-uniform section in the flange direction In cases where the change of the cross-sectional areas is fairly well distributed between the various flange members which make up the beam cross section, the shear flow results
as given by the solution for beams of constant moment of inertia are not much in error For beams where this is not the case, the shear flow results may be considerably different from the actual shear flows This fact will be illus- trated later by the solution of a few example problems
A15,16 The Determination of the Flexural Shear Flow
Distribution by Considering the Changes in Flange
Loads, (The AP Method.) Fig AlS.S8 shows a single cell distributed
Trang 11ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
flange beam Consider the beam acts as a canti-
lever beam with the bending moment existing at
section (A) being greater than that existing at
section (B) and that the bending moment produces
compression on the upper surface By the use of
the flexural stress equations, the vending
stress on each stringer can be found, which if
multiplied by the stringer area gives the
stringer axial load Thus at beam section (B),
let P,, P,, P,, etc represent the axial loads
The external bending
Py Be p> Pa Ps
due to a bending moment M
Fig A15-59 Fig Al5-60 Fig Al5-61
moment at section (A) ts M + aM, hence the
stringer axial loads at section (A) will equal
P, + AP,, P, + AP,, P, + AP, etc These
stringer axial loads are shown on Fig 415.58
Imagine the upper sheet panel 2, 2', 3,
3' 1s cut along line (a-a) Furthermore con-
sider stringer number (3) cut out and shown as
a free body in Fig A15.59 Lat dy be the
average shear flow per inch over the distance
don the sheet edge ob It has been assumed
positive relative to sense along y axis
For equilibrium of this free body,
or dy a
Therefore starting at any place where the value of ay is known, the change in the average shear flow to some other section equals
oP
3y #~Ã*m- TT =~x ren T— (1)
If the summation is started where dy 1s zero then equation (1) will give the true average shear flow dys
Fig 415.61 shows sheet panel (3,3' ,4,4') isolated as afree body Taking moments about corner 4’ and equating to zero for equilibrim,
IM, =-489.15 yo qxbd # 0 whence, qy = AP,/d
Thus for rectangular sheet panels between flange members the shear flow gy or q, equals the average shear ay
The same rules as previously presented to determine the sense of gy or dg after having ay can be used and will not be repeated here
To show that equation (1) reduces to the
Shear flow equation previously derived and used, consider a beam with constant cross-section and take a beam length d = 1 inch Then,
AM = Vạd = Vz(1) # Vy
aP = ca =z 28 (where A = area of
From equation (1) dy = - 24P Substituting value of AP found above,
ay = ~ qe BZA ee eee (3) which is the Shear flow equation previously derived for beams with constant moment of inertia
A15.17 Example Problem te Compare Results in Using
Equations (1) and (3)
Fig Al5.62 shows a square single cell bean with six flange stringers Between points B and
C, the beam has 4 constant flange section which
is shown in Section B-B The numerals beside each stringer represent the area of the stringer
Between points B and D, the flange material tapers uniformly with the flange material at point A as indicated in Section A-dA It should
be noticed that the increase in flange area is
Trang 12
A18, 26 SHEAR FLOW IN CLOSED THIN-WALLED SECTIONS, SHEAR CENTER
1000# 1000# | with the shear flow system of Fig A15.6%3, which
x-p |Â iB ' t therefore is the final shear flow system for
st a wy = 3 + a Solution No 2 Considering AP Loads in Plenge
SECTION A-A SECTION B-B TO C-C Bending moment at section AA = 1000x560 = 60000
Fig A15-62 Bending moment at section BB = 1000x 30 = 30000
and c The shear flow on section A-A will be
computed using equation (3) which applies only
to beams with constant section and also by
equation (1) which applies to beams with vary-
ing moment of inertia
Solution 1 Using Shear Flow Equation for
Beams of Constant Cross-Section
(Equation 3)
Since qy at any point on the cell is un- known, it will be assumed that the upper sur-
face on Section A~A is cut through the midpoint
of flange stringer (a), thus making the shear
flow qy equal to zero on this free surface
One-half of stringer (a) thus acts with each
side of the top surface In this solution the
webs and walls will be assumed ineffective in
resisting bending stresses, thus the shear flow
is constant between adjacent stringers
Starting at midpoint of stringer (a) and
going counter-clockwise around cell,
Shear flow could be calculaved, but due to sym-
metry enough values have been found for the
shear flow to draw the complete shear flow pic-
ture for bending about the X axis when it is
assumed that one-half of the area of stringer
(a) acts with each adjacent wed Fig A15.63
shows the resulting shear flow diagram The
resultant of this shear flow pattern is a
1000 1b force in the Z direction and its
location through the midpoint of the box since
the flow ts symmetrical The external load of
1000 1d, also acts through the midpoint of the
cell hence the external load ts in equilibrium
Considering Section B-B:
Bending stress intensity at midpoint of
stringers by the flexural formula:
oly = 300005
Op Fh Ts 1000 psi
Axial load in each of the
e = 1000 x 1 = 1000 lb stringers a, b, and Considering Section A-A:
puted by equation (1)
It will be assumed that one-half of the AP load in stringer (a) will flow to each adjacent web, However, there is no 4P load in stringer (a) hence dap = qq 7 0 Then from equation (1),
Qpp = 0-2 s 0 - 2980 = ~ 80 1b./1n.