Al4.8
0.375 inches to left of point b as shown in Fig Al4.17, For bending about the centroidal 2 axis without twist the resultant of the internal shear flow system would obviously, due to sym- metry of section about X centroidal axis, lie on the X axis, hence shear center for the given channel section is at point O' in Fig Al4.17 The external load of 100 1b would have to be
located 0.375 inches to the left of the center- line of the channel web if bending of the channel without twist is to occur
A14.6 Shear Stresses for Unsymmetrical Beam Sections
In chapter Al3, which dealt with bending stresses in beams, three methods were presented for determining the bending stresses in beams with unsymmetrical beam sections The bending stress equations for these three methods will be repeated here: - Method 1 The Principal Axis Method My Zp Mey Xp sx-— toot (9) xp 2p Method 2 The Neutral Axis Method os-™m ~ -. - (19) in
Method 3 The Method using section properties about centroidal Z and X axes For brevity this method will be called the k method
O =~ (ky My = Ki My) > (Ka My = K, MQ) 2 (11)
BENDING SHEAR STRESSES SOUND AND OPEN SECTIONS SHEAR CENTER Method 3 The k Method
dy = - (Ks Vy - Ki Vg) BX A ~ (Ke Vz - Ki Vy)
Zaa
EXAMPLE PROBLEM USING THE THREE DIFFERENT METHODS
Fig Al4.26 shows a Zee Section subjected to a 10,000 lb shear load acting through the shear center of the section and in the direction as shown The problem will be to calculate the shear flow qy at two points on the beam section, namely points b and c as indicated on the figure The shear flow at these two places will be cal~ culated by all 3 methods
Since all 3 methods require the use of beam section properties and since the direction
of either the principal axes or the neutral
axis are unknown the Zz
first step in the „ solution regardless of at ¬ which method is used 1s to calculate the a section properties about centroidal X and
Z axes, Table Al4.3 4"
gives the calculations The section has been divided into 4 portions labeled 1, 2, 3 and 4 V = 10000 lb + —! gợi âm Fig Ald 26 TABLE Al4.3 T
Areal Arm | Arm ; Ị
where, Portion) “Tay | cz my oane | az? | ax? fig | ix 1 1 |0.10| 1,$ö| -0 45] -.06525|.,31025|.02025 |.00838 | 000017 Kis XZ = Ix 210,34) 0.701 0 Ö | 068601 6 |.000117.02287 “Ty tf, - ifz’ ° “ly lz ~ lậu a; 0.14] -0.70| 0 0T 06880] 0 |.000117.0Z281 ¡ 4 [0.10] -1.45| 0.43| -,06525| 210251 02025].00638 [000017 | 1 Z | -.4908 | -58770) 04030|.01600 L04377
% SE ene Ty =2ZA27+2 1, = 6035 int, "
In referring back to the derivations of Iz =2ZAx*+ 2 i, = 0574
equations (5), (6) and (7) the above equations lop DA XZ = = 1205 (9), (10) and (11) can be written in terms of *z
beam external shears instead of external bending (Note: in Table 414.3 ty and tg are the mom- moments as follows: - Method 1 The Principal Axis Method Vz, Vx _ p p 4y ST 3Zp4-T—3#%p ê “.——~=~ (12) xp 2p Method 2 The Neutral Axis Method Ÿn TH = ——— (15)
in ents of inertia of each portion about
its own centroidal axis)
SOLUTICN BY PRINCIPAL AXES METHOD (Method 1) Let $ be angle between Principal axes and the X and Z axes From chapter A13,
tan 2 $ =f be Ig - ty
2 (- 0.1305) = 0.4778
Trang 2ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
A4.9
hence 2 $ Z 259 - 22.21 on $ = 12? - 46.1' Now substituting in equation (15)
sin $ = 0.2210 and cos 9 = 0.97527 7348
d› “ - =—=z=rz (1 X 0.1 x 1.5156) The moments of inertia about the principal axes b 0.65316 7 can now be calculated
tx =I, cos* $+ I, sin® $ ~ 2 Ix, sin } cos > - Sa (1 x 0.1} (+ 0.1184) = - 1756 + 2890 —_—* ——A = 0.6035 x 97527 + 0874 X 2210 ~ 2 (~0.1505) = 1134 ib/in x 97527 x 2210 = 63316 in* te =I, sin* + I, cos* ÿ + 2 Ix, sin $ cosq 2 a = 0.6035 x 0.2210 + 0574 x (07507 + 2 (-0,1305) x 97527 x 0.2210 = 0.02782 in* The equation for shear flow q is, Vx -—2: x Tx, lap Vg qz=-—22h2
In Fig Al4.26 the external shear load is 10000 lbs acting in a direction as shown Re- solving this shear load into Z and x components, we obtain,
Vy = 10000 x cos 30° = 6667 1b Vy = 10000 x sin 30° = 5000 ib
Resolving these 2 and x components further into components along the principal axes we obtain, Vg, = 8667 x 97527 - 50000 x 2210 = 7348 1b, 2p Tẹ Calculation of shear flow at point (b) Fig Al4.26)
Fig Al4.27 shows the position of the principal axes as calculated The shear flow at the free edge of the upper portion (1) is zero For the shear flow at point (b), the area to be used in the summations 2 zpA and 2 xXpA is the area of element (1) The arms Zp and Xp can be calculated by simple trigonometry
Fig Al4.27 shows the value of these distances, namely Xp = - 9.1184 and Zp = 1.5136 8667 x 2210 + 5000 x 97527 = 6792 lb (See Fig Al4, 27
Calculation of shear flow at point (c)
For portion (2) area A =1.4x 0.1 = 0.14 Zp = 70 x 97527 = 0.6825 in
Xp = +70 X 2210 2 0.1547
The shear flow at point (c) equals the shear flow at point (b) plus the effect of the portion (2) between points (b) and (c)., nence
dg = 1154 - Ae 0.63316 (0.14 x 0.6825) ~
6792 = =
Sang (0,14 x 0.1547) = 1134 - 1109 - 5285 = - $260 lb./in
The shear stresses at these two points (b) and (c) would equal 3/y = 1154/0.1 and ~ 5260/0.1 or 11340 psi and - 52600 psi respectively,
SOLUTION BY NEUTRAL AXIS METHOD In this solution
it is necessary to find the neutral axis for the given external load-
ing In Fig Al4.28,
the angle 9 is the angle between the plane of loading and the Zn prin~ cipal axis, and this
angle @ equals 30° + 12° - 46" = 42° — 46",
Let a equal angle between Xp *p principal axis and neutral axis n-n
From chapter Al3, we find, (Method 2) _ Ixy tan 9 tan a 5 - I 2p - „ 0.68518 x 0.8245 „ _ m1 952 „0S,
whence, a = ~ 87° - 17" (See Fig A14.28 ?9r location of neutral axis
sin a = 0.9989 , cos a = 0.04742 sin* a, substituting,
Trang 3Al4.10 BENDING SHEAR STRESSES,
— 2
Ty 2 0.63316 x 04742 + 02782 x S585 = 02919
The component of the given external shear load normal to the neutral axis n-n equals
Vy, = 10000 x sin 45° ~ 29" = 7130 1b From equation (13)
Qe ~ Fem A
Shear flow at point (b}: ~
The distance from the neutral axis to the centroid of portion (1) equal Zn "~ 0.0466 in hence _ _ _7150 = đọ “— “GGig (1x O.1) (— 0.0466) = 1185 1b/1m, Shear flow at point (c): - zn # 0.1868 ín 7130 1135 ~ “02518 (1.4 x 0.1) 1868 Qe 1135 - 6390 = ~ 5255 lb./in SOLUTION BY METHOD 3 - (The k Method)
In this method, only the section properties about the centroidal X and Z axes are needed
These properties as previously calculated in
fable Al4.3 are, Ty = 0.6035, I, = 0.0874, Ix, = - 0.1305 The shear flow equation from eq (14) ts, Q=- (Ks Vy - ki Vg) 2 A - (Ky Vg - Ky Vy) ZLaA Ixz ~ 0.1805 X, “————r= TT * Tý lạ = lựý 0.6085 x 0874 ~ (- 0.1308) = 7041305 _ „01768 = ~ 7.406
Resolving the external shear load of 10,000 into x and 2 components, we obtain, Vy = 10000 sin 30° = 5000 1b Vz = 10000 cos 30° = 8667 1b, Substituting values of Vy, Vg and k values in equation (16) we obtain ~ SOUND AND OPEN SECTIONS SHEAR CENTER q = - [34.25 x 5000 ~ (~ 7.406 x 8667) ] 2 xA - [3.257 x 8667 ~ (~ 7.406 x 5000) ] 5 3 A, whence q=- 255488 7 xA - 65258 3 z Á
Shear flow at point (b}
For portion (1), x = ~ 0.45 in., 251.45 in A®ixod1ls.2l Substituting, Gp = - 235438 (- 0.45) 0.1 ~ 65258 x 1.45 x 0.1 = 10597 - 9462 = 1135 lb./in Point (c} Por portion (2), x = 0,220.7, A=1.4x 1 Qe = 1135 - 235438 (0) 14 ~ 65258 x 0.7 x 0.1 = 1125 - 0 - 6395 = - S260 lb./in General Comments
The author prefers solution method number 3 since it avoids the calculation of additional atigles and section properties as required in methods 1 and 2 Furthermore, tn calculating the
shears and moments on the airplane wing, fuse~ lage and other major sturctural units it is con~- venient to refer these shears and moments to the conventional X Y Z axes, and thus these values can be used in method 3 without further resolu- tion From an investigation of many airplane stress analysis reports, it appears that the en~ gineers of most airplane companies prefer to use method 3
Al4.9 Beams with Constant Shear Flow Webs Fig 414.29 shows a z
beam composed of heavy R
flange members and a curved thin web For bending about the X-X axis, the
web on the compressive,
Side of the beam absorbs very little compressive
stress, since buckling of ——— the web will take place “Flange
under low stresses, par- Zz Pig A14.29 ticularly when the curvature of the web is
small On the tension side, the wed will be
more effective, but if the flange areas are relatively large, the proportion of the total bending tensile stress carried by the web is small as compared to that carried by the tension flange Thus for beams composed of individual flange members connected by thin webs it is often assumed that the flanges develop the en~
therefore means that the shear flow 1s constant
Trang 4
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES over a particular web In other words in the
shear flow equation q = es zaaA, if the area x
ot the web is neglected then q is constant be~ tween flange members
RESULTANT OF CONSTANT SHEAR FLOW FORCE SYSTEMS Fig Al4.29 shows a beam assumed to be carrying a downward shear load (not shown) and to cause, bending about axis x-x without twist Assuming the two flanges develop the entire bending resistance, the shear flow q {s con- stant on the wed and acts upward along the web to balance the assumed external downward load The resultant of this resisting shear flow
force system will give the lateral position of the shear center for this beam section The problem then is to find the resultant of the shear flow system
Let q = load per inch along web (constant) Let R = resultant of the q force system From elementary mechancis,
R=vVðqg+ 3 q2 , where ay and qy are the x and y components of the q forces along the web Since q is constant, 2 qy is zero, hence,
Equation (17) states that the magnitude of the resultant of a constant flow force system is equal to the shear flow q times the straight line distance between the two ends of the shear flow system
Since & qy is zero, the direction of the resultant 1s parallel to the straight line joining the ends of the web
The location of the resultant force is found by using the principle of moments, namely, that the moment of the resultant about any point must equal the moment of the original force system about the same point In Fig 414.29 assume point (0) as a moment center
Then Re =qLr but R=gqgh
In equation (18) the term Lr is equal to twice the area (A), where area (A) is the en-
closed area formed by drawing straight Lines
from moment center (0) to the ends of the shear flow force system Thus
The shear center thus lies at a distance e
to the left of point (0), and the external shear load would have to act through this point if
twisting were to be eliminated
Al4 11 EXAMPLE PROBLEM - RESULTANT OF A CONSTANT PLOW
FORCE SYSTEM
Fig Al4.30 shows a constant flow force system thru points ABC DE with q # 10 lb per inch The resultant of this force system is required Rez 200¢ e £18.93 —‡p—r sn sm +a! m—L 5" =2 ~ - iy Fig A14.30 SOLUTION: -
Draw closing line between the beginning
and end points of force system (line AE) The
length h of this closing line is 20 inches Prom eq (17) R = qh = 10 x 20 * 200 1b
The direction of the resultant is parallel to line AE or horizontal in this problem To find the location of R take moments about any point such as (0) Draw lines from point (0) to
points A and E The enclosed area (A) equals Sx10+5x10+ 5nx5*+10x 50 = 189.3 sq in
From eq (19)
2A „2x 189.5 „
on 20 18.93 In
Pig Al4.30 shows the resultant of 200 lb acting at a distance e from (0) and parallel to lime AE
Al4,10 Example Problems for Beams with Constant Shear Flows Between Flange Members
EXAMPLE PROBLEM 1 Beam Section Symmetrical About One Axis
Trang 5Al4 12
Sheet to form the weds and walls The flange members are numbered a to h and the areas of each are given on the figure It will be as- sumed that the webs and walls develop no bending resistance and thus the shear flow between ad- jacent flange members will be constant The problem is to determine the shear center for the beam section
SOLUTION: -
Since the beam section is symmetrical about the X axis, the centroidal X and Z axes are also principal axes, since the product of
inertia Ix, 1s zero
The vertical position of the beam section centroid due to symmetry is midway between the upper and lower flanges
To find the horizontal position of the centroid, take moments of the flange areas about the left end or line be:
0.4x158+0.2x10+0.2x5
aA 36
= 5.625 tn
The moments of inertia for the section about the centroidal x and y axes are: ~ Ty = 2A 2* = (0.8 x 5*) 2 = 40 Int 2 I, = 0.8 x 5.625 +0.2%x 0.625" +0.2Xx 2 4.375 + 0.4 x 9.375® = 64.4 int, HORIZONTAL POSITION OF SHEAR CENTER: -
The horizontal position of the shear center will coincide with the centroid of the shear flow system due to bending about axis xx with- out twist For simplicity, to eliminate large decimal values for shear flow values an ex-
ternal shear load Vz = 100 1b will be assumed
and the internal resisting shear flow system will be calculated for this external loading From equation (8) Vv; ay * 1 3 2A, substituting values of V„ and x Ix 100 Wy =~ GPszaAT-2522A4
We could start the solution at either of two points (a) or h since these points are free edges and thus dy 1s zero In this solution, we will start at the free edge at point (a} and go counterclockwise around the beam section The area of each flange member has been concen- trated at a point coinciding with the centroid of each flange area In solving for the q val- ues the subscript y will be omitted, and sub- scripts using the flange letters will be used
BENDING SHEAR STRESSES SOUND AND OPEN SECTIONS, SHEAR CENTER
in order to indicate at wnat point the shear flow { is being calculated
đạp 7 - 2.52, 2A 2 -2.5%5 xX 0.1 * = 1.25 1b./in
The first letter of the subscript refers to the flange member where the shear flow q is being calculated and the second ietter indicates on which adjacent side of the particular flange member Hence q,, means the shear flow at
flange (a) but on the side toward (b)
đạp = Na = - 1.25 (Since no additional flange
area is added, and thus
shear flow ts constant on sheet ab Abe = Wa ~ 2-5 Ty ZA =- 1.25 -2.5x%5x 0.4 = - 6.25 lb./in đẹp * + 6-25 Ved = Ich - 2-5 UG ZA - 6.25 - 2.5 x (- 5 x 0.4) = - 1.25 Gag = Ugg — 25 3g ZA = - 1,25 - 2.5 (- 5 x 0.1) =0 ded = dag = 0 dep =O - 2.52, 2A50-2.5 (-5 x 0.1) = 1.25 dre F der * 1.25 Qgg = 1.2 ~ 2.5 Bp 2A (+5 x 0.2) = 3.75 Ger = Gpg = 3-75 Qgh = 3.78 - 2.5 2, ZA 1.25 - 2.5 „ 3.75 - 2.5 x5 x 0.2 1.25 Qng = dgn = 1.25 1.25 - Z.5 šn 2A = 1/25 - 2.5 X5xXO.1 9 (ehecks free edge at h) Sha,
The sign or sense of each shear flow is for the shear flow tn the y direction as explained in the derivations of the shear flow equations The procedure now is to determine the sense of the shear flow in the plane of the cross-section or in the xz plane It is only necessary to determine this sense at the beginning point, that is in sheet panel ab The surest way to deter- mine this sense is to draw a simple free body
sketch of flange member (a) as illustrated in Pig Al4.31 The shear flow on the cut face is Qy(ab) = - 1.25 and
this value is shown
on the free body By + simple rule given at a the end of Art Al4.6,
the shear flow in the plane of the cross~
Trang 6
ANALYSIS AND DESIGN OF section is also directed toward the common boundary line and thus 4x(ap) has a sense as shown in Fig Ald.31 The sense of the shear flow on the cross-section will now continue in this direction until the sign changes in the origional calculation, which means therefore the shear flow sense will reverse Fig Al4.22 shows a plot of the shear flow pattern with the sense indicated by the arrow heads Vg? 100% ~1.25 1.25 8 3.75 1.25 1.25 Fig A14.32
The results will be checked to see if static equilibrium exists relative to 2 Fy and EP, = 0 3 F„ = 100 (ext lead) - 79 x 6.25 - 10 x 3.75 + .28 x 0.5 x 4 - 1,25 x 0.5 x 4= O0 (check), -5x%1.25+5x1.25-5x1.25+5% 1.25 = 0 (check) tt 3x
The shear flow force system in Fig Al4.32
causes the section to bend about axis xx with-
out twist The resultant of this system is 100
lb acting down in the 2 direction The posi-
tion of this resultant will thus locate the
lateral position of the shear center
Equating the moments of the shear flow system about some point such as (c) to the mo~
ment of the resultant about the same point we
obtain:
100 a = 10 x 3.75 x 15 - 1.25 x0.5x2x5- 1.985 xO.5x 2x 10 + 1.25xX0.5X2x 15
hence e # 562.5/100 = 5.625 lnches Thus the shear center lies on a vertical line 5.625 inches to right of line oc
CALCULATION OF VERTICAL POSITION OF SHEAR CENTER
For convenience as before, we will assume a shear load Vy, = 100 lb and compute the re- sisting shear flow system to resist this load in vending about axis zz without twist The resultant of this shear flow system will give
the vertical location of the shear center The shear flow equation is,
q = -Kyyqae 100 ExAz- LSSEKA
Ỷ ĩ 2 64.4
FLIGHT VEHICLE STRUCTURES Ald 13
Trang 7Al4.14 BENDING SHEAR STRESSES
Checking to see if 2 Fz =O and 2 Fy = 0: ~ ZF, = 3.592 x 10 - 3.589 x 10+ 5 x O971 - 5 x 0O971 + 5% 674 ~ 5 x 674 + 5 x O971 - 15 x 674 - 5 x 6.504 + 5 x 7.087 = 0 (check) 3 Fy = 100 + Š x 0971 +5 x 674 - 5 (7.087 + 7.184 + 6.504) = 0 (check)
The resultant R of the internal shear flow system is a horizontal force of 100 1b acting toward the left To find the location of the resultant take moments about a point 0.5 inch below point (c) Re = 2M 100 2 = 5 (.0971 + 674) 11 + 10 x 3.589 x 15 + 5 x 0971 x5 - (.5 xX 674 X 10) + ,5 x 674 x 15 + ỗ x 6.504 X 15 + 6 X 674 X 10 ~ (.5 X .0971 x 6) 100 e = 643 @ = 643/100 = 6.43 inches
Pig Al4.35 shows the resulting shear center location for the given beam section
EXAMPLE PROBLEM 2 Unsymmetrical Beam Section Fig Al4.36 shows a four flange beam sec- tion The areas of each flange are shown ad-
jacent to flange The external shear load V equals 141.14 lb and acts in a direction as shown The problem is to find the line of action of V so that section will bend without twisting x Fig Al4.36 SOLUTION: ~ To solve this problem, method (3) will be used
To locate centroidal x and z axes: -
zg.Ă5AX _1xX12+0.5x8_ Bese RE SS in,
z= BAS = {0:5 + 0:5) 16 5 5.333 in
SOUND _AND OPEN SECTIONS, SHEAR CENTER Calculation of Ix, I„ and lyz — l„ “3 Á 27 = 5 (5.3334 + 2.6679) + 1 , (5.535° + 6.667*%) = 90,667 lạ =ZA X® = 1 (10.667%) + 2 (5.333%) = 170.667 Tyg = 2A x2 = 1 xX 6.667 (- 5,333) + 1 (- 5.333) (+ 5.333) + 0.5 x 10.667 x 2.667 + 0.5 x 10.667 (- 5.333) = - 21.333 The constants k,, kg and K, are now determined — _ Txz -_ - 21.545 - Ke 1x ly ~ lx 90.667 x 170.667 - 21.21 2 - 21.386 _ TTEme =- ‹00142 ce = =, = = Iz = 120-867 Dạy = 0.011: *Ð Tự lạ - Ix 15019 k 2 gs Ix 90.867 ô0 sacoar 9 đ Ty 1z - Ge 15019
Trang 8ANALYSIS AND DESIGN OF deg = 6-249 - 0.7487 x 0.5 x 10.667 - 1.276 x 0.5 (= 5.333) = 5.680 dao = Igg = 5-880 ga = 5-660 ~ 0.7457 x 5 x 10,667 - 1.278 x 0.5 X 2.667 = 5.680 - 3.977 + 1.704 = 0 (checks free edge at d where qy must be zero.)
Fig A14.37 shows the resulting shear flow re- sisting pattern The sense of the shear flow in kề ake TZ Fig.Al4.37a q =a oy q= 5.680 VE141.14 ⁄ ee 141.14 q2 8.249 c Fig Al4.37
the plane of the cross-section is determined in web at flange member (a) by the simple free body diagram of stringer (a) in Fig Al4.37a
Check 3 Fy and 3 Fz to see if each equals 100
E Py = - 6.249 x 16 = - 99.99 (checks Vy = 100) RF, =- 12x 4.544 ~ 8 x 5.68 = - 99.94 (checks
Vz = 100)
The resultant of the internal resisting shear flow system equals V¥ 1002 + 1007 = 141.14
1b To locate this resultant we use the prin- ciple of moments Taking point (b) as a moment
center,
141.14 e = 8 x 5.68 x 16 - 4.544 x đề n 212
hence e “TT 12 = 1.50 inch
Therefore external load must act at a distance @ = 1.50" from (pb) as shown in Fig Al4.37 The load so located will pass thru shear center of section To obtain the shear center location, another loading on the beam can be assumed, and where the line of action of the resultant of the resisting shear flow system intersects the re- sultant as found above would locate the shear center as a Single point If the shear center location is desired it is convenient to assume a unit V, and Vy acting separately and find the horizontal and vertical locations of the shear center from the 2 separate shear flow force systems
LIGHT VEHICLE STRUCTURES Ald, 15
Al4.i11 Shear Center Location By Using Neutral Axis
Method
In a beam subjected to bending there is a definite neutral axis position for each differ- ent external plane of loading on the beam The shear flow equation with respect to the neutral axis is,
QW Fo In Š Znà TT xxx TT (20)
where, Vy = Shear resolved normal to neutrai axis
In = Moment of inertia about neutral axis
Zy = Distance to neutral axis
In finding the shear center location of an unsymmetrical section, it 1s convenient to as- sume that the Z and X axes are neutral axis and find the shear flow system for bending about each axis by equation (20) The resultant of each of these shear flow force systems will pass through the shear center, thus the intersection
of these two resultant forces will locate the shear center
Example Problem
The same beam section as used in the previous article (see Fig Al4.36) wiil be used to illustrate the neutral axis method
Fig Al4.38 shows the section with the centroidal axis drawn in The X axis will now
Fig Al4 38 Fig Al4.39
be assumed as the neutral axis for an external plane of loading as yet unknown We will further assume that when this unknown external loading is resolved normal to the X neutral axis, that it will give a value of 100 lb., or Vz = 100 From the previous article Iy = 90.667
Trang 9Al4 16 BENDING SHEAR STRESSES, SOUND AND OPEN SECTIONS SHEAR CENTER OFy = 1.47 x 16 5 23,52 lb 2Fz 2 7.35x12+8%1.47 = 99,96 lb (check) R =V T007122.5827 = 103 lb, tan @ = 25.52/100 = ,2352 ° 100 R= 103 hence 9 = 13° - 16' Let e = distance from 33.52 resultant R to point b Equating moments of resultant about (b) to that of shear flow system about (b), 10Se = -7.35 x 6n*+1.47x8x16 = o4 ese 1057 -6.25 in
Fig Al4.40 shows the location of the resultant We know the shear center lies on the line of action of this resultant Thus we
Fig Al4.41
Fig A14.40
must obtain another resultant force which passes through the shear center before we can definitely locate the shear center Therefore we will now assume that the 2 centroidal axis is a neutral axis and that a resolution of the external load system gives a shear Vx 2 100 1b, - - 9y“ ~Tế 3 XA, I, đạp * — Tro sayX (~6.388)1 = 3.126 1b,/tn 170,667 - 100 - đục = 5.125 -xop xe (-6.385)1 = 6.25 1b./irÌ 100 170.66 ded * 6.25 ~ (0.5) (10.667) = 3,13 Fig Al4.41 shows the-shear flow results 3#„ = -6.251x l6 = -100 lb Šf, = -8x3.13+12x3.125 = 12.5 lb = = R = V0" F10.aF = 100.8 2100.8 = 12.5 _ 125[ 2 tan 9 “8 - 2125 T00 Take moments about (b) and let (e} equal distance to resultant R 100 8¢ e 2.125 x721x6*+ 3,15 x8 x lầ 753/100.8 = 7.47 in
Fig Al4.40 shows the position of this re- sultant force Where it intersects the previous resultant force gives the shear center location Al4.12 Problems
(1) Pig Al4.42 fan artird
shows the cross- bob section of a wood | T beam glued to- L gether on lines a a-a and a-b, The beam is subjected 4” to a vertical shear V >= 2400 lb Determine shearing stress 3 © 1” on sections a~a Zo + and a-b Find maximum shearing stress on beam section Fig Al4.42 Fig Al4.43 (2) Fig Ala.43 shows a Zee sec- tion loaded by a 1000 1b load acting through the shear center as shown, Find the shear stress at sections a-a and b-b by three different methods Fr¬ TP = ajar | <= + 1 1/16 i b-@-b 1000 ns we " Zi0m 210w Pp 7 Neglect Web in Resisting Bending 220" z.10" | Fig Al4.46 “on OF Fig A14.44 ‡ 10" Fig Al4.45
(3) Determine the shear flow diagram and the
shear center location for bending about hori- zontal centroidal axis for the beam sections as given in Figs Al4.44 to Al4.46 Fig Ai4.47 s0 4$ =6 =6 †}*z" 159" | BP Bámaeekans| am iL Ụ b1,0 -4¢
.10n," 00! op" fi 10" | Fig A144
Trang 10ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES sume flange members develop entire bending stress resistance HH ng 4 Fig Al4, 49 b Psp 4
{5) Determine the shear center for the beam
section of Fig Al4.49 Assume only the 8 stringers as being effective in bending Area of stringers (a) and (b) = 2 sq in seach All other stringers 1 sq in each
Fig Al4 50
(6) Determine the shear center for the un- symmetrical beam section of Fig Al4.50 As- sume sheet connecting the four stringers as
ineffective Areas of stringers shown on Fig À14.17 c d Fao 50" Fig Al4 52 Fig, Al4, 51
(7) In Pig Al4.51, the shell structure {s sub-
jected to a torsional moment M = 50,000 in lb The shell skin shown dashed is cut out, thus the torsional moment is resisted by the constant
shear flow on the two curved sheet elements ac,
and bd Determine the value of the shear flow (8) Determine the moment of the constant flow force system in Fig, Al4,52 about point (0)
Also find the resultant of this force system priate 8 pio 1,50" 1.0 0.8 1 T 10 20" 1.2 0.7 9.9 1.0) F——z—] T 18 —+ 8 Fig.Ai4.53 ° Khu 54 1
(9) In Fig Al4.53, the four stringers a, b, ¢ and d have the same area Assume the webs in- effective in resisting bending stresses De- termine the distance (e) to product bending about the horizontal axis without twist
(10) For the wing cell beam section in Fig Al4.54, determine the location of the shear
center Assume webs and walls ineffective in bending
Boeing Tôi: Jet jet Stratoliner
Fuselage being Assembled |
Trang 12CHAPTER A 15
SHEAR FLOW IN CLOSED THIN - WALLED SECTIONS
A151 Introduction The wing, fuselage and em= pennage structure of modern aircraft is essen- tially a single or multiple cellular beam with thin webs and walls The design of such structures involves the consideration of the distribution of the internal resisting shear stresses This chapter introduces the student to the general problems of shear flow distri- bution Chapter Al4 should be covered before taking up this chapter
Al5.2 Single Cell Beam Symmetrical About One Axts All Material Effective in Resisting Bending
Stresses,
Fig Al5.1 shows a single cell rectangular beam carrying the load of 100 1b, as shown The problem is to find the internal resisting shear flow pattern at section abcd 100# ! I i, ' a r ———————— —~ / 08 a s } b nỗ Ị “ iw \ 05x 10" - - Y ° — „08 {fo * DỤ d1, s 50m tr x TH oe Ụ Fig Al5-1 Solution 1
Due to symmetry of material the X cen- troidal axis lies at the mid-height of the beam The shear flow equation requires the value of ly, the moment of inertia of the section about the X axis
ly = aex.15x10°+ 2 [20 x.06 x 6 * ] = 62,5 in.t
From Chapter Al4, the equation for shear
flow is,
ay = _ ZÁ j j {Tên nh (1)
This equation gives the change in shear flow between the limits of the summation In
open sections we could start the summation ata free surface where q would be zero, thus the summation to any other point would give the true shear flow qy Ina closed cell there is no free end, therefore the value of Qy is unknown for any point
Equation (1) gives the shear distribution for bending about the X axis githout twist
The general procedure is to ải i a value of
the shear flow dy at some point and then find the shear flow pattern for bending without twist under the given external load The cen~ troid of this internal shear flow system will be the location where the external shear load _ should act for bending without twist Since the given external shear would have a moment about this centroid, this unbalanced moment must be made zero by adding a constant shear
flow system to the cell
To tllustrate we will assume dy to be zero at point 0 on the web ad = Vz = = Io = 0 The term 73 = 100/62.5 = 1.6 Gao = -1.6 252A 2 91.6 x2.5x5x0.1 = -2 Ib/in dục = -Ø-1.6 32⁄4 = -2=1,6X5x20x.05 = ~10 ' Qo! =10~1.6 Sọ ZÀZ~10~1.6X2,5X6x 06 = ~11 đẹo: E ~11~1,6 Bộ, 2Á # -11-1.6 x (-2.5)5 x 05 = -10 dea = -10-1.6 29 2 = -10~1.6(-5)20 x 05 = -2 God = -2-1.8 25 ZA #-2~1.6(-2.5)5x0.150
Fig Al5.2 shows a plot of the shear flow results, On the vertical web the increase in shear is parabolic since the area varies directly with distance z
The intensity of qy and qg in the plane of the cross-section is equal to the values of q found above which are in the y direction The sense of dy and qd, is determined as explained in detail in Art Al4.6 of Chapter A14,
Trang 13ALS 2 SHEAR FLOW
Fig AlS~2
Fig Al5.2 also shows the resultant shear flow force on each of the four walls of the cell The resultant shear force on each equals the area of the shear flow diagram on each position For example, 2+ 10 Qap = (= 3+" )e20 = 120 1h Qda, $x 2x 102 6.67 1h, 10 x 10 + 0.667 x 1 x 10 = 106.7 Qe
The internal shear flow force system as given in Fig A15.2 will now be checked forw equilibrium with given external shear loading of 100 lb as shown in Fig Al5.1
SF, = 100 (external) + 6.67 ~ 106.67 =0 (check)
SFy = 120 - 120 = 0 (check)
Equilibrium of moments must also be Satisfied Take moments of all forces about point d The external load has no moment about d
mMg = 120 x 10 + 106.67 x 20 = 3344 in Ib clockwise
Thus we have an unbalanced moment which
must de made zero if we are to have equil-
ibrium,
The unbalanced shear flow of 3334 in.1b, can be balanced by adding a constant shear in a gounter-clockwise direction around the €gll The value of this balancing shear flow would equal,
M _—
Br Bx B00 = “gốc 1D/1H, qs-
(A equals area of cell = 10 x 20)
IN CLOSED THIN-WALLED SECTIONS
SHEAR CENTER
Adding this constant shear flow to that of Fig 415.2 we obtain the final shear flow pattern of A15.3 ¬6 34 #/in Fˆ 1 66 #/in ~8 34 #/in, ` 2.66 #/in, 1.66 #/in —6 34 #/in Pig A15-3
If this constant shear flow of -8.34 was not added then the external load of 100 1b would have to be re-located so that it passed through the centroid of the shear flow pattern in AlS.2 To find this centroid location, we can equation the moment of the internal shear flow pattern about some point to the moment of the resultant of the system about the same point 7 Resulitant R z V Ÿfy 7+ hy T= (1007+ 0 = 100 acting down Take moments about point d Let X equal distance to resultant \ RX = IM, (internal system) 100 X = 120 x 10 + 106.867 x 20
hence X = 33.33 in Thus the external load
would have to be moved 33.33 inches to right if the pattern of A15.2 would hold it in equilibrium
Since we assume q = zero at point 0, this means
an open cell with the free end at 0 would bend without twist if the external load was moved the
distance X
The student should work this same problem by assuming q, at some other point is zero in- stead or point O as assumed in the above solu- tion
Solution No 2 Shear Center Method
In this solution, we determine the centroid of the internal shear flow system for bending of the closed section about axis X without twist This point is called the shear center The external shear load can then be resolved into a shear force acting through the shear center plus a torsional moment about the shear center,
We start the solution, exactly as in solu- tion 1, dy assuming the shear flow q = 0 at point 0 Ina sense we are cutting the cell at
The resulting This open O and making it an open section
Trang 14
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Al5,3
section will bend without twist if the ex- 10, 9" 9.1 ternal shear load acts through the shear oe xt 2 100% 7]
center of the open section 4.36 #/ in
3.64 #/in The closed section will be assumed to
bend without twist, and the resulting snear flow pattern will be determined
The equation for angular twist ¢@ (See Chapter A6) per inch length of beam is,
9 =~: , were L equals the length
of a web or wall
1, qL or 2AQ = a ›_— to
this equation represents the total shearing strain around the cell which must be zero for no twist of cell Since G is constant, we can assume it as unity as only relative values of strain are needed in the solution Thus the total shearing strain 6 around cell is proportional to,
The right hand side of
Using the values of q from Fig A15.2 and substituting in (2), pape a (225, (10+ 2) 20 , 10x5 t Sxl 2 „05 „95 (11 ~ 10)(.667 x 5) = + OS 2 = 7000
Since the section and shear flow pattern is symmetrical about the X axis, the substi- tution above is written for one-half of the cell and the results multiplied by 2 When the shear flow q varies over a portion the average shear flow 1s used in the above substitution
If the cell is not to twist the relative twist of 7000 must be cancelled by adding a constant shear flow q around the cell to give a total shear strain of -7000
The shearing strain for a constant q equals, xe 5, 20, 5p 2 — sale 06 7 Se) = 7000 hence q = 2000 5 6.36 1b/in
Fig A15.4 shows the resultant shear flow pattern if the constant shear flow of -6.236 ts added to the shear flow pattern of Al5.2 64 #/in Lr ells a 4, 36 #/in hex 7.9%
The location of the resultant of the shear flow force system of Pig A15.4 will locate the horizontal position of the shear center Due to symmetry of the section about the X axis, the vertical position of the shear center will be on the X axis, because for bending about the Z axis, the shear flow would be symmetrical and thus, the resultant would coincide with the X axis
Fig AlS-~4
Fig Al5.4 also shows the resultant shear load Q on each portion of the cell wall which equals the area of the shear flow diagram for various portions as shown
SF, = - 56.9 - 43.1 = - 100, which balences the external load of 100 lb
Take moments about axis XX and side ad 3y = 0 by observation
point 0, the intersection of
= Hie = 43.1 x20 + (~25.8 + 16.58)10 „
100 7.90 1n
Hence the shear center lies on the X axis, 7.9" from side ad
The moment of the external load of 100 lb about shear center equals 100 x 7.9 = 790 in lb clockwise The moment of the internal shear flow of Fig AlS.4 1s zero thus we have an un- balanced moment of 790 Therefore for equil- {orium of moments we must add a constant shear flow q around cell to develop ~790 in.lb or
- M ~ 790 _
q = ar * Boo Bx 7 L‹98 1b/1n,
Adding this constant shear flow to that of Fig AlS.4, we obtain the final shear flow pattern which will be identical to that in Fig Al5.3 or the results of solution 1
A15,3 Single Cell - 2 Flange Beam Constant Shear Flow Webs As discussed in Art Al4.9, the common
Trang 15ALS, 4 SHEAR FLOW
walls and webs which are stiffened by members usually referred to as flange members A simplifying common assumption is to assume that the flange members alone develop the resistance to the bending moment This
assumption therefore means that the shear flow is constant between flange members
Fig Al5.5 shows a single cell beam with two flanges at A and B Find the internal shear flow force system when the beam carries the external load of 100 lb as shown plot! V=100# i 6.21 { ! Tf " hị [1⁄2 yon ý | z {4 ron? ay Lod 1210411 lơ, i, a web | " 6.21 Skin’\ Al 7 Fig A1S-8 Fig Al5S-7 028-7 Lựa Fig AlS-5
It is assumed that the two flanges develop the entire bending stress resistance This means that shear flow is constant on each
web Let q, and q, be the constant shear flow
as shown in Fig AlS.6 The sense or direction
of these shear flows will be assumed as indi-
cated by the arrow heads in the figure The force system in the plane of the eross-section has 2 unknowns, namely q, and
q, and thus we can solve for q, and q, by
simple statics
To find q, take moments about point A
mM, = 100 x 10 = q, (2 x 80.52) = 0 hence q, = 1000/lb 1.04 = 6.21 lb/in The sign of q, comes out plus, thus the
assumed sense of q, a8 shown in Pig Al5.6 is correct
In the above equation of moments about
point A, the moment of the constant flow system q, about A, equals q, times double the enclosed area formed by lines running from A to the-ends of the web which carries the shear flow q, In this case the area is the area of the cell, or 80.52 sq in
To find the remaining unknown q, We use the aquilibrium equation, Wz = O05 - 100-10 x 6.21 + dq, = 0 IN CLOSED THIN-WALLED SECTIONS SHEAR CENTER,
hence q, = 16.21 lb/in (sign comes out positive,
hence assimed sense of q, 18 correct)
Fig AlS.7 shows a plot of the internal shear flow resisting system
A single cell beam having only two flanges
can resist only external loads which are parallel to the line AB, and thus a two flange box beam is not used very often in atreraft structure When the bending moment in a plane at right angles to line AB is small, the resistance of the curved panel to compressive bending stresses may be sufficient to resist such external bend-
ing moments and thus be satisfactory
Al5,4 Shear Center of Single Cell - Two Flange Beam Let it be required to find the shear center of the beam as given in Pig Al5.5 In other words where would the external load have to be
placed so that the beam would bend without twist 16,21 T.95 q 4 constant q
Fig A15-8 Fig Al5~9 Fig A15-16
Fig 415.8 shows the resulting shear flow system in resisting the 100 1b external load acting as shown in Fig Al5.5 This shear flow system will cause the cell to twist Therefore we add a constant shear flow q to the cell to produce zero twist (Fig AlS.9) The centroid of the combined shear flow system will then locate the lateral location of the shear center
To find q we must write an expression
which measures the twist when subjected to the Shear flows of Figs A15.8 and 9 and equate the Tesult to zero, and then solve for the one un-
kmown q
Trang 16ANALYSIS AND DESIGN _ 16.21 x 10 _ 8,21 x 24:28 ,„ 109 5= 04 „086 04 ,„ 22.28g _ es 9 hence q # 8.26 1b.,in
Adding this constant shear flow to that of Fig Al5.9, we obtain the shear flow of Fig Al5.10,
The lateral position of the shear center
is given by the location of the resultant of
the shear flow system of Fig Al5.10 The resultant R = y EF," + IF"
IFz = 0
a # 10 x 7.95 + 2.05 + 10 = 100 1b, 1
Therefore R = 100 1b
Equate moments of R about point A to moment of shear flow system about A Re = My u 100e = 2.05 (2 x 80.52) H 9 3.30 in
Thus shear center lies 3.30 inches to
left of line AB (See Fig A15.10)
Thus if the given external load of 100 lb acts through the shear center, it will
produce the shear flow system of Fig A15.10
However it acts 13.3 to right of shear center, hence it produces a clockwise moment of 100 x 13.3 = 1330 in.1b on the cell For equilibrium, this moment must be balanced by a constant resisting shear flow around cell which will produce 4 moment of -1330
c.- 1520 _„
The required q “+ V 80.82 =8 26 1b.in which if added to the shear flow system of Al5.10 will give the true shear flow system
of A1S.8
Thus having the shear center location, the external load system can be broken down
into a load through the shear center plus a
moment about the shear center The shear flow due to each is then added to give the true resisting shear flow
It should be noticed that the web or skin thickness does not influence the magnitude of the shear flow system in a single cell beam A change in thickness, however, effects the unit shearing stress and therefore the shear-
ing strain and thus in computing angular twist
OF FLIGHT VEHICLE STRUCTURES
A15.5 of the cell, the web and wall thickness does influence the amount of twist for a given torsional load In the shear center solution, it {1s known what portion of the shear flow is due to torque or pure twist, and also that due to bending without twist, which fact is some— times of importance
TORSIONAL DEFLECTION OF CELL
The angular twist as given by the final shear flow pattern of Fig A15.8 equals 2QAG = ZqL/t, whence
After finding the shear center location, we found that the external load had a moment of 1330 in 1b about shear center, which was resisted by a constant shear flow of ~ 8.26 1b/in The angular twist under this pure torque shear flow should therefore give the same result as equation (3) above
- 8.26% 24.28 _ 8.26X10„_ 1oogg 20aG = 7085 0
which checks the result of equation (3) A15.5 Single Cell-Three Flange Beam Constant Shear Flow Webs 100 ' 100 Ib A235 z st L—z ib đạp x Woe ip OY Fig A15-12 9.13 / gt + 7 an | f_ 1 _ c Bl f — —! *2, 639 Area Cell ” F— = 128, 54 sq in Fig Ai5 13 -Fig A15.11
Fig Al5.11 shows a single cell beam with three flange members, A, B and C, carrying the external load as shown A three flange box if
the flanges are not located in a straight line
can take bending in any direction and therefore ts often used in design because of its simpli- city
For such a structure, there are six un- knowns, namely, the axial load in each stringer and the shear flow q in each of the three sheet panels that make up the cell For a space structure, we have six static equations of equilibrium, thus a three flange single cell
Trang 17A15.6 SHEAR FLOW
beam can be solved by statics if we assume that the three flange members develop all the bending stress resistance, thus producing constant shear flow webs,
Fig AlS.12 shows the cross section ABC The three unknown resisting shear flows have been assumed with a positive sign (Clockwise flow is positive shear flow) These three unknown shear flows can be determined by statics
To find ea take moments about point B and equate to zero,
Mp = loc x 5 - 25 x 10 + q.,(128.54 x 2) = 0 _ _ 250
257.08
To find qa, take 2Fz = 0
hence Gcoq = =- 0.972 1b/in IF, 100 = 10 x 0.972 - 10agy = 0 đạp Z 9.13 1b/1n To find qụẹ take š3P„ # O šfx # - lỗ X 0.972 ~ 25 ~ 15qpẹ = 0 hence qục # ~ 2.659 1b/1n
The signs of deg and Apc Came out negative, hence the sense of the shear flow on these cell wall portions is opposite to that assumed in Fig AlS.12 The resulting shear flow pattern is plotted in Fig A15.13
The student should realize the thickness of the wall elements does not influence the shear flow distribution if we assume the three flanges develop the entire resistance to the bending moment
A15.6 Shear Center of Single Cell-Three Flange Beam,
Constant Shear Flow Webs
Let it be required to determine the shear center location for the beam in Pig AlS.ll The shear center is a point on the beam cross-
section through which the resultant external Shear must act if the cell is to bend without twist
The shear center location will be deter- mined in two steps, first its horizontal
location and then its vertical location, Calculation of horizontal location:- We will assume any vertical shear load, as the example, the same vertical shear as used in the problem of art Al5.4, namely, a 100 1b load acting five inches from A, as illustrated in the following Fig A1l5,14,
IN_ CLOSED THIN-WALLED SECTIONS SHEAR CENTER, 100 Z1 5 Z4 Hoe “ ct J
Fig Als-14 Fig A15-15 Fig A15~16
The three unknown resisting shear flows will be assumed with the sense as indicated by the arrow heads
To find dg, take moments about 8 IMg = 100 x 5 - dg, (128.54 x 2) 5 0 Gag = 1.945 ld/in IP, = - 15 x 1,945 + 15qy, = 0 Qe = 1.945 BFz = 100 ~ 10 x 1.945 ~ l0qay 20 đạp = 8.085 1b/1n,
The algebraic signs of the unknown q value all come out positive, thus the assumed direction of shear flows in Fig A15.14 is correct
To make the cell twist zero, we must add a constant shear flcw q to the cell (see Fig 415.15) The relative twist under the shear
flow of Figs 14 and 15 will be equated to zero „.SL „ _ 1:945X20.71 _ 1.945415 , 8.055 x10 t 708 005 — ĐC 20.71g „ 10q „154 _ TT * tor * Toes = 9
Whence, q = 0,322 lb/in with sense as assumed in Fig Al5.15, Adding this constant Shear flow to that of Fig Al5,.14 we obtain the shear flow system of Fig 415.16 The resultant R of this shear flow system {s obviously - 100
1b.; Since the external load was 100 lb The location of this resultant R will therefore locate the horizontal position of the shear center Equate moment of resultant R about point B to the moment of the shear flow system about B, whence,
100e = 1.623 (128.54 x 2) or e 2 417/100 = 4.17 in,
(Fig A15.16) rom line AB
Trang 18ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES assumed acting on the cell Since we used a
25 lb load in the example problem of Art 15.5, we will assume the same load in this solution Pig Al5.17 shows the loading and the assumed directions of the three unknown shear flows A 25 od Tea q R25 + JŠ fab! 4 a} = ST 18 qch_ 4 L Cc B 1.018 B
Flg A15-17 Fig A15-18 Fig Al5-19 Solving for the three unknown shear flows in Pig AlS.17 Bg = - 25 x 10 + dog (128.54 x 2) = 0 cq = 0.972 1b/in IFy = - 25 + 972 x 15 + l5dgy = 0 = 0.695 e ơ u IF, = 10 x 0.972 ~ l0day = 0 = 0,972 , ơ 4
A constant shear flow q is now added to cell to make twist zero (Fig Als.18)
Writing ZqL/t for both loadings and equating to Zero:~ + 1 972 Zqb/t = 0.972 x20.71 - 9 see 5 + 9 we 20.71q 10q 15 _ *“og* Toa * Tége= 9 whence, q = - 0.324 1b/1n
Adding this constant shear flow to that of Fig A15.17 we obtain the values in Fig A15.18
R (the resultant) = 25 lb,
Equating moment of resultant about 8 to moment of shear flow system about B,
Re = IMg
256 = 0.649 (128.54 x 2) Therefore @ = 6.65 inches
Thus shear center lies 6.65 inches above 2, and 4,17 inches to left of B as previously
found
Al5.7 A15,7 Single Cell-Multiple Flange-One Axis of
Symmetry
Fig A15.20 shows a single cell beam with 8 flange members, carrying a 100 lb shear load, The resisting shear flow system will be calcu- lated 1004 TƯ, na 20 z Fig A15-20
The beam section which is symmetrical about the X axis is identical to the beam section relative to flange material which was used in example problem 1 of Art 414.10
SOLUTION: =
Assuming the 8 flanges develop all the bending stress resistance, the shear flow will ‘ therefore be constant between flanges Since
the beam section is a closed one the value of the shear flow q at any point is unknown Thus we will imagine the top cover cut between flange members a and h, thus making q Zero in this
panel due to the free end at the cut We now find the internal resisting shear flow system for bending of this open section about axis x-xK under a external shear load Vz = 100 lb
Trang 19A15.8 SHEAR FLOW
If 2Fy and 2Fz are considered for equil- ibrlưm of external and internal loads, they will be found to equal zero
To check IMy = zero take moments about some point such as C
MM, 7+ 100 x 7.5 + 10x 3.75 x 15 - 1.25 x5 ~ 1.25 x 10 + 1.25 x 15 = - 187.5 in.lb Therefore to make IM, = 0, a constant shear flow q equal to M/2A = (187.5/2 x 11 x 15} = 0.57 lb./in is required Adding this con- stant shear flow to that in Fig A1l5.21, we obtain the final shear flow pattern of Fig 415.22 This final pattern is not much different from that of Fig A15.21, the
reason being that the location of the imaginary cut to make q equal zero, was not far from the true fact, since the final q in this panel was only 0.57, If we had started the solution by assuming the wed be cut or qn, = 0, then the correction constant flow that would be needed to satisfy IMy = 0 would have come out q = -5.88, since tats is the final q in web bc Since the shear flow which is a load on the cell wall influences the required thickness of sheet required, it is good practice to try to place the imaginary cut at a point where the shear is near zero, so that preliminary estimates in routine design relative to shell thickness required will be based on shear flow values that are near the final values 87 1,82 5.88 4.92 1⁄28 1.25 -88 1.82 1.82 Fig A15-22 A15,8 Single Cell - Unsymmetrical - Multiple Flange Example Problem 1
Fig, Al5.23 shows a 4 flange unsym-
metrical single cell beam carrying two external loads as shown This beam is identical to the one uséd in example problem 1 of Art Al3.8 which dealt with bending stresses
IN CLOSED THIN-WALLED SECTIONS, SHEAR CENTER By 260004 #30" eo" E— mà jt Fig Al5-23
Solution 1 Using Section Properties and Ex- ternal Shears with Reference to Centroidal Axes (The K Method) In Art Al3.8 the calculations for this beam section gave - ly = 81.18, I, = 183.58, Ix, = - 21.33 Fig Al5.23a shows the location of x and z cen- troldal axes Fig, Al5-23a
In Art, Al4.9 of Chapter Al4, the method of solution was referred to as the K method, The shear flow equation (see Eq 14 of Art Al4.8) 18, dy = -(K, Van, Vz) ŸXÁ ~(KQÝz-k Vy) 2ZÁ - ~21.33 = K: — 1xz/lylzˆ lụa” " ĐT T8 X 185.50 - 21.58" 7 -21.25 _ Bois 77 001775 Ka # 1z/1xlz - Ixz* = 153.58/12016 = ,01279 % “ = 1x/1Ixlz - Tyg? = 81.18/12016 = 00674 1
For the given beam loading the external shear loads at section abcd are,
Trang 20ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Substituting in the equation for ay 48 given above, dy = - [00674 (-1600) - (-.00177 x 6009) |ZxA ~ [.01279 x 6000 - (~.00177)(~1500)]52A whence Qy = 0.16 2xA - 73,91 BZA - - - - = - In using equation (4) to compute the shear flow pattern we will imagine top panel ab cut, thus making the shear flow dy = zero
in this panel Subt in (4) ~ ~ =~ đạc =0 + 16 x 1 (-6.333) - 73.91 x1 x 6.074 = ~ 449.78 1b/1n, ded = - 449.78 + 16 x 8 (-5.333) - 73.91 x B8 % (-5.926) = ~ 100.05 đạp = ~ 100.05 + 16 x 4 x 10.667 ~ 73.91 X 4 (-5.926} = 75.80 tua = 75.80 + 16 x 5 x 10.667 - 75.91 x 5 x 2,074 = 0
Fig Al5.24 shows the plotted shear flow results, This pattern satisfies IFz = 0 and iF, = 0, To check equilibrium of moments 6000#= Vz, Fig Al5-24 Fig Al5-25 about a y axis point d Assume a y axis going through Mg = 6000 x 8 ~ 1600 x 6 - 449.78 x 12 x 16 = - 51160 in.1b
Thus for equilibrium a moment of plus 51160 in.1b is required This is produced by adding a constant shear flow q around the cell walls, where „1 „ _õ1169 4“ 2A ”Z x 180 (A = area of cell = 160) = 159,88 lb.in Al5.9 Adding this volume of q to those in Fig Al5.24 we obtain the final shear flow resisting
pattern in Fig Al5.25
Solution 2 Principal Axes Method
The shear flow system can of course be found by referring section properties and ex- ternal shear loads to the principal axes of the beam section The equation for shear flow is
(see Eq 15 of Chapter Al4),
(The subscript p refers to principal axes.)
The section properties about the principal
axes were computed for this same beam section on page Al3.5 of Chapter Al3 The values are:-
Ixp = 75.38, Igp = 159.34
Fig 415.26 which was also taken from page A13.5 shows the location of the principal axes and the distances from the four flange members to the principal axes, fren Tr _| 9.75, —T var _r* # eB? 5 926 $90 Id Xp N Fig Al5-26
Trang 21Al15 10 SHEAR FLOW
Assume dy = 0 in top panel ab đạc = -71.21 x 4.45 x 1 + 19.59 x (-6.74) x 15 448,92 Geq = ~448.92 - 71.21 (-7.12).8 + 19.59 {-3.58).8 = ~ 99.41 Qdb = - 99,41 ~ 71.21 (~2,90).4 + 19.59 x 11.80 x 4 = 75.45 Ib/in, qua * 75.65 - 71,21 (4.82).5 + 19.59 x 9.75 x 5 = 0 (check)
These shear flows are practically the
Same aS obtained in solution no 1 as recorded in Fig Al5.24 Discrepanctes are due to slide rule accuracy
For equilibrium of moments ; take moments
about (b),
Mẹ # 6000 x 8 ¬ 1800 x 8 - 448,92 x 12 x 16 = -50993 1n.Ib,
A constant shear flow q around cell must be added to produce 50993 in.lb for equil- ibrium This balancing shear flow is,
-.MH „ 50993 _
"Ba 160 7 159.35 1b/in *3x which is the same as in solution no l Example Problem 2
Fig 415.27 illustrates a typical single cell wing beam with miltiple flange members The external shear load on this beam section is Vz = 1000 and Vy = 400 located as shown The internal shear flow resisting pattern will be calculated
This beam section is the same as that
used in example problem 5 of Chapter A3, where
the calculations of the section properties were made, The results were: Ty = 186.5, I, = 431.7, Iyg = 36.41 Vz= 1000 1b ek L—>—Z—*#*— & | £ 7 5s 9 Vx=4004 4 5 = ix x 10 11 3 13 14 1816 z Fig AiS-27 IN CLOSED THIN-WALLED SECTIONS SHEAR CENTER Solution The K method of solution will be used _I _ 36.41 Kì FT TT = 186:2X451.7-58.412 = 36.41 = Seek = 0004598 I 431.7 _ ke" Titg-lgg? *7oies 7 005452 - 1 _ 186.5 _ - Ke = = gorge? 7 7eTam = -008855 (KVx-k Vz )EXA = (K Vz-k Vy) B28 ay Substituting in above equation, Gy = =(.002555 x 400 ~ 0004598 x 1000)5XA —(,005452 x 1000 = 0004598 x 400) 2zA Gy = -0.4825 3xA ~5.268 3ZA ¬ - - ~ ~ (7)
Since the value of the shear flow is un= known at any point on the cell walls, it will be assumed that the cell wall is cut between flange members 1 and 10, thus making q zero on the sheet panel numbered (1-10) Then using equation (7) the shear flow is calculated in
going clockwise around the panel Columns 1 to
10 of Table Al5.1 show the calculations in solving equation (7) For explanation on how to determine sense of shear flows qa, in Columns 9, 10, and 11, review Art Al4.6 of Chapter A14
The shear flow values in column 11 would be the results if the external loads as given were so located as to act through the centroid of this shear flow force system Since they do not we will solve for the unbalanced moment on the beam section about point (0) the cen- troid of the beam cross-section The moment of the shear flow force on a sheet panel between any two adjacent flange members is equal in magnitude to q times double the enclosed area
formed by drawing lines from the moment center (0) and the ends of the particular sheet panel Fig 416.28 illustrates this explanation The value m in column 12 of the table lists the double areas of these various triangular areas,
Taking moments of all forces both external and internal about point (0),
IM, = 1000x2+400x3+17123 = 20323 in.1b (17123 equals summation of column 43) Thus for equilibrium a negative moment of -20323
Trang 22ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES AVS, 11 Table A15 1 8 3XA 0,14 | 4.396[ 0.615 |~17.411~2.437 3.81 constant negative shear flow around ¢ell where Magnitude equals q 2S = = 20.6 lb /in {493 = area of cell)
Adding this constant shear flow to that in columm 11, we obtain the final shear flow in column 14 Fig Al5.29 shows true shear flow pattern Values from col 14 Table Al5.1 46.98 Am KỶ - 1.88 Fig A15-29 9 10 11 12 14 Gy 3-0 4822] gy = =5, 268 | Sxz * 8 =-(col 9+ m ‘ Final =
(šXA) (32A) cà toy 78% Em ane 1 30.6
A15.9 Two Cell-Multiple Flange Beam Symmetrical About One Axis
Fig Al5.30 shows a two cell cantilever beam with 10 flange stringers The cross-sec- tion is constant Let it be required to deter- mine the internal shear flow in resisting the
1000 1b load acting as shown For simplifi- cation, the top and bottom sheet covering and the three vertical webs will be considered in- effective in taking bending flexural loads Since the beam section is symmetrical about the X axis, the beam will bend about this axis in
resisting the given external load The moment of inertia of the section about the X axis equals 250 in * 1000# Fig A15-30 Solution 1 (Without use of shear center}
Trang 23A15 12 SHEAR FLOW
shear flow at any point in each cell is unknown Therefore, to make the flexural shear flow Statically determinate, a value for the shear flow q in each cell will be assumed at some point, and the flexural shear flow for each cell will then be calculated, consistent with the assumed conditions These resulting static shear flow systems will, in general, produce 4 different total shearing strain around the perimeter of each cell, or in other words, produce a different cell twist ince full continuity exists between cells, this condition cannot exist, and therefore an unknown constant shear flow of q, in cell (1) and q, in cell (2) must be added tO make the twist of both cells identical This fact gives us the basis for one equation and the other equation necessary for the solution of the two unimowns q, and a, is given by the requirement of equilibrium, namely, that the moment of the external and the internal shear forces about any point in the plane of the cross section must equal zero In Fig 415.31 the flexural shear flow has been assumed as Zero just to the left of stringer c in cell (1) and just right of stringer c in cell (2) The balance of the flexural shear system consistent with this assumption is calculated as follows: ram 1 1m „ ri #/in, 04/1 2n 2WEZ 504/1a, Z wy ấn (2) ác tu» A A 4 Ta c -⁄ E inwen L 10z/nn, Flg A15-3L The general shear flow equation is, ay = ~ 2% gq = — 1000 Bad = ~4E2A Ty 250
Cell (1) Starting in panel cb where the shear has been assumed zero and proceding counter-clockwise around cell
đẹp = zero (assumed) &
dba = -402A = 0-4%5%X0.5 = =10 1b /in Qaat 2 -10-4x5x2 = -50 lb /in
datb! = -50-4x (-5) x2 = +10 1b./in Quet @ -10-4x (-5) x0.5 = 0 lb /in We cannot proceed deyond stringer c' be-
cause there are two connecting weds with un-
own shear flows we can zet around this difficulty by going back to stringer e, where the shear flow on each side of c was assumed
IN CLOSED THIN-WALLED SECTIONS
SHEAR CENTER,
zero, Thus the shear flow in the vertical web ec’ is determined by the stringer ¢ alone, namely
'
đọc! = ~SÏ 2Á # ~4X1x6 = -80 1b./1A,
We can now continue around cell (2) starting with stringer c©" where we were previously stopped _ Feta! = Apter + de! - “HỖ mà =O0~-20-4x (-5) 120 Gate’ 7 0 - 4x (-5) 0.5 = 10 lb /in đạ'a = 10 - 4x (-5) 1 = 30 lb /in dạ'd S 30 ~ 4X 5 x1 = 10 1h.⁄1n, đạc 2 10 - 4 x 5 x 0.5 = 0, which checks the assumed value of q = 0 in panel cd
The shear flows in cell (2) could of course been found by starting in panel ed where the shear has been assumed zero and proceeding clockwise around cell as for example
We 20+ 4x5 x 5 = -10 lo./in
dee! = +10 ~4x5x1 2 -30 lb /in
Geta! = +30 - 4x (+5) 1 # ~10 1b,/1n, datg? * -10 - 4x (-5) 0.5 = 0 lb./in, The magnitude of the results are the same as previously calculated but the signs are opposite As emphasized previously the shear flow calculated together with its Sign is in the y direction or qy‹ The direction of the shear flow along the cell walls in the xz plane can be determined by drawing simple free body diagrams as illustrated in Chapter Al4 but it 1s simpler to use the automatic ruls of Art Al4,11 To illustrate, the solution started in panel cb and proceeded counter- clockwise around cell (1) 8y the rule this means that the shear flows in ‘he xz plane will have the same sign as dy Since the signs of Qy were negative, the direction of she shear Plows in the xz plane will be counter-clockwise
around cell (1)
To obtain the shear flow in the vertical
web cc’ 1t was necessary to start atc and go
toward c' This direction 1s counter-clockwise with respect to cell (2) or clockwise with res-
pect to cell (1), Gy for this panel from the equation was found to be -20, If we consider web cc" as part of cell (1) then the direction in solving the summation EZA was closkwise with respect to cell (1) and oy our rule the shear flow in the Z direction will have the opposite
Trang 24ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
Fig A15.32 illustrates the unknown con- stant shear flow systems q, and q, which must act on cells (1) and (2) respectively to pro- duce the same cell twist when added to the snear flow system of Pig A1S.31 The sense
of q, and q, has been assumed clockwise or
positive in each cell ⁄⁄⁄⁄2` & 214 f uUUau 1 4 At w | @ 19 te ght + zz flg A15-32
The equation for the angular twist © per unit length of beam is,
2AG9 = zak
using the shear flow values in Figs Al5.31 and 32, the angular twist of each cell will be calculated by substituting in the above equation For cell (1) ~10x5_ 50x10 _ 10x5, 20x10, 109, , „gỗ 0S ” 70S Sx Pan ~ 590, Hence, 24,60, =1200q, - 333q, 24,69, = For cell (2) ~ 10x5x2,30x10 20x10 3x10, 10, BAGO, og * ag ag * age * To Al5 13 Since there is continuity between cells, 0, = 9, Also since area of each cell is the same,
A, =A, Equating (1) and (2),
1633q, - 18@3g, - 10840 = 0-~ - (3)
One other equation is necessary to solve for
unknowns q, and q, , and it is given by the
moments of the external and internal shear
forces about any point in the plane of the cross section, which must equal zero for equilibrium
Take moments about point (b) of tne shear flow system of Figs Al5.31 and A1S.32 and also the external shear load of 1000 1b., which in this case has no moment about our assumed moment center WM 2 =60X10x§ + Z0 x10 K§ + 1Ơ x 30 x 15 + 200q, + 200g, = 0 hence, 200g + 200g, +3000 =O - (4) Solving equations (3) and (4) for a, and q,, we obtain a, = -4.07 1b./in qq = -10.80 1b./in The final or true internal shear flow sys- tem then equals that of Fig A15.31 plus that of
Pig Al5.32 when q, = -4.07 and q, = -10.80 1b./
in., which gives the shear flow diagram of Fig 415.33 4.07 10,8 38 ~ MOT? Ae -10.8 (Values in 1b,/1n.) Fig Al5-33
Solution 2 (By use of shear center)
In this solution, we find the flexural
shear flow for bending about axis X-X without
twist The centroid of this internal shear system locates the shear center The moment of the external shear load about the shear center produces pure torsion on the 2 cell beam Thus, adding the shear due to this pure torsion to that of pure bending, we obtain the final re~ sisting internal shear flow
In bending about axis X-X without twist, the shearing strain for each cell as given by equations (1) and (2) must equal zero Hence:
1200 4, -333q, - 6670 = 0 - (5)
-B83q, + 1250q, +4170 =0 - (6)
Trang 25À15.14 SHEAR FLOW
Solving equations (5) and (6) for q,and q,, we obtain gq, = - 2.0 1b./in q, = 5.00 1b./in Therefore, taking these values of q, and q, in Fig Al5.32 and adding the results to that of Fig AlS.31, we obtain the shear flow pattern of Fig Al5.34 which is the shear flow system for bending without twist about X axis The centroid of this shear system locates the shear ceilter
In Fig A15.34,
ZV = 0 = -10x45-10x28-10x27 = - 1000 1b., which checks the external shear of 1000 1b, XH = 0 by observation of Fig Al5.34 V= 10004 +2 -5 + s.c a7 ~4 (Values in lb./in.) Fig A15-34
To find the horizontal position of the centrotd of the shear flow in Pig Al5.34 take moments about point ‘a:
Mg = 10x27 x10 +10x29x20
+5x8x10-5x2x10 = 8600 in.1b,
hence, x = sees = 8.6" to the righv of web aa'
The external shear load of 1000 1b acts §" to the right of aa', and therefore causes a moment about the shear center equal to (8.6 -5.0) 1000 = 3600 in.lb To resist this torsional moment, a constant torsional shear flow 9 (1) and qy(a) must act on cells (1) and (2) respectively
can be found The values of at 1) and ae £ foun
of Art AS by using equations (16) and (17
of Chapter AS Thus 1 Suoậy + 3; Ã ®(1) *“^¿|———————— |! bi To N a, A* + a,A,? (28 +a+ 3B} 100+ 48x 200 (10,10, 10} 10 (Tas ‘Toa *Tog) 100" +x 200 8+ (10°, 30 10) 70a |n =] 158200 |, (aoe as tag eae = 2) ase200 In « „002847 9 (1) =-5
Since the external torque equals 3600 in.Ib.,
IN CLOSED THIN-WALLED SECTIONS
SHEAR CENTER
the resisting internal š
equal -3600 Therefore, rque must therefore Qt (1) = -00254(-3600) = -9.17 1b./in Solving ?or qy (2) 1 2, LÂy +2 ,à “=> IT (2) 31190000 (18 10, 40) 100 + TC x 200 1] G03" 03" 05) P=, 002487 2 51120000 hence: dt(2) = 00245 x-3600 = -8.85 1b/1n Therefore, if we add to the shear flow sys- tem of Fig Al5.24, a constant shear flow of -9.17 1b./in to cell (1) and -8.85 lb./in to cell (2), we will obtain the true internal re- sisting shear flow of Fig A15.35, which checks solution 1, any discrepancy being due to slide rule accuracy, -14.17 WH} : ‡ Y 2 Ze ° ‡ „ 3 3 Si ¡2 2 a =1 = % Fa aT oes Fig A15-35 Torsional Deflections
The angular twist of each cell is the same The value of the angular twist 9 per unit length
of the beam can be found using the shear flow pattern of Fig AlS.35 which is the true result- ant shear flow, or the pure torsional shear flows
9Ÿ đr(1) = -9.17 1b./in and 4 (2) = 8.85 1b./in
may be used {f destred,
The results will, of course, be the same For example:
For cell (1) due to dg (1) = 79.17 1d /in,