ANALYSIS AND DESIGN OF FUIGHT VEHICLE STRUCTURES wo r
Pig C9.4 Shear Buckling Coefficient for Long
Simply Supported Curved Plates _Kg 7B ta Fser * 12 (1 - Ve*) Ve*) ® r a r Zp = Bel = Ver) */# Ne (From Ref 2) — 4 E- 2 Ẹ ỗ 40E- - 2 Ks 32 ao i a 20— Os mp E—— << s =| [ E | | a rc i Ị | : i \ —_ I 1 ¿l2 L1 hú Voth if L ch i chị Bề 1 My Zp Ly
Trang 4BUCKLIN ULTIMATE STREN G STRENGTH 0 C9.8 GTH OF STIFF where Ro = fo/Foon
applied internal pressure external inward pressure that would buckle the cylinder of
which the curved panel is a
section Found by use of Flg C8.16 in Chapter ca,
Rp = ratio of
In using Equation C9.5, the value of is negative as inward pressure is opposite the inward acting outward pressure, Rp to C9.7 Shear Buckling Stress of Curved Sheet Panels with Internal Pressure,
As in the case of monocoque cylinders, internal pressure increases the shear buckling Stress of the curved sheet panel Brown and Hopkins (Ref 5) solved the classical
equilibrium equations to determine the effect of radially outward pressure upon the shear buckling stress of curved panels and obtained fair agreement with test data by Rafel and Sandlin (Ref 3) The test data also correlates with the interaction curve used for the effect or internal pressure upon cylinders in torsion (see Chapter C8) The interaction equation 1s;
Rg* + Rp = bo + eee ee LL C9.6
where Rg = Ÿs/P So
~ applied internal pressure
Rp = ratio of external inward pressure that
would buckle the cylinder of
which the curved sheet panel
1s a section Found by use of Fig C8.16 of Chapter C3, In using Equation c9.6, Rp is given a minus sign C9.8 Example Problems PROBLEM 1
Fig C9.6 illustrates a circular fuselage Section with longitudinal stringers represented by the small circles The area of each
stringer is 15 sq in The skin thickness is +04 inches All material is aliminun alloy with Ey = 10,700,000 The fuselage frame Spacing (a) 1s 15.75 inches The fuselage
Section is subjected to the fo
system:- llowing load (causing compression on My = 600,000 in.1b top hai?) Vg 7 5175 lbs {acting up) T = Torsional moment = 210,000 1n.1bs, counterclockwise) {acting
F CURVED SHEET PANELS
ENED CURVED SHEET STRUCTURES,
Fig C9.6
The problem is to determine whether skin
Panels (A) and (B) will duckle under the given combined loading on the fuselage section,
Solution
To find the bending stresses, the moment of inertia of the cross Section about axis y-y is necessary, which axis is also the neutral axis since all material is effective The moment of inertia will equal 4 times that due to material in one quadrant,
Trang 5ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES _ Tử E ty: Foor = 12 (1 -Ue) ‘0! 14_n* x 10,700,000 , 04,8 =——————_-z+— (——z T2 LÊ ‘S.ag) = 2880 pst =
To find the shear buckling stress, we use Fig C9.2 Z is the same as calculated above or 32.9 Thus from Fig C9.2 we read for a/p = 3, that Kg = 20
204
_ 20 2* x 10,700,000 ae
Por =~ Te TT — 3%) Gags) ~ 12800 pst
The bending stress at midpoint of Panel {A) will be calculated:
fp 2 M/ly = (600,000 x19.7)/1725 = 6850 pst
Thus stress ratio Re = fo/Fo,, = 6850/7850
= 874
Shear stress on Panel (A) due to torsion is, fg = T/2at, where A is inclosed area of
fuselage cell
fs = 210,000/2 x m x 20° x 04 = 2090 psi The panel is also subjected to shear
stress due to transverse shear of Vz, = 5i75 lbs The shear flow equation is,
A4 $175 -
a= 7) BZA = > Trap ZA = - 5 IZA
The shear flow will be zero on Z axis
The shear flow at top edge of Panel (A) will
be due to effect of one-half the area of stringer (1)
Gina 7 3 X 0075 x 20 = 4.5 1b./1n
Area of skin between stringers (1) and (2)
18 5.25 X 04 = 21 =A,
Distance from centroid of Fanel (A) from
neutral axis 1s Zr sin a/a a= 15° which gives Z = 19.8 in Thus q,_, = 4-5+3x 21419.8 = 17 1b./1n, Then average shear flow on panel is (17 + 4.5)/2 = 10.75 Then shear stress = q/t = 10.75/.04 = 269 psi
This shear stress has the same sense or direction as the torsional shear stress so we add the two to obtain the true shear or:
9.7
ts (total) = 269 + 2090 = 2359 psi
Then shear stress ratio Rg = ts/Faop =
2359/11200 = 21
The interaction equation for combined compression and shear 18;
Ry + Rg* = 1
„B74 + ,217 = 918 This is less than 1.0 so
panel will not buckle 2
1.8 5 Taya + fara? + ax ele 7 1 = «08
Consider Skin Panel (8)
Arm 2 to midpoint of panel = 15.88 in
fo
Re
Mz/I = 600,000 x 15.88/1725 = S520 psi
8520/7850 = 704
The torsional shear stress is the same on all panels or f, = 2090 pei as previously cal-
culated
Shear flow q due to transverse shear load:
a> 732A = 3 ZA
Calculation of 352A at upper edge of panel:
.075 x 20 + 15 (19.3 + 17.34) 7.0
For stringers =
For skin: Area = 2x 5.25 x 045 42
Vertical distance 2 to centroid of skin portion
=rsina/a a= 30% The result is Z'= 19.1
in The ZA = 19.1 x 42 = 8.03, Total IZA =
8.03 + 7.0 2 18.03, Thon q = 3 EZA 73 x 15.06
= 45.09 1b./1n
A similar calculation for shear flow at lower edge of Panel (B) would give q equal 55.0 Thus average shear flow on panel ts (55 +
45.09)/2 = 50.04 Then f, = a/t = 50.14/.04 =
1251 psi The total shear stress fg on panel
then equals 1251 + 2090 = 3341 psi Rg = fs/Faor = 3341/11200 = 299
Substituting in interaction equation Ry +
Rg* = 704 + 299* = 793, The result is less than 1,0, thus panel will not buckle
Trang 6BUCKLING STRENGTH OF CURVED SHEET PANELS C9 8
General Commen:
In general the compressive stress is the
dominant factor in causing the panel buckling Thus to increase the buckling stress of the
panels and also to give a more effective
Stringer arrangement to carry the bending moment, the stringers should be spaced closer
in the top and bottom regions of the cross-
section and with increased spacing as the neutral axis 1s approached,
PROBLEM 2
The fuselage section in Problem 1 is sub- jected to an internal outward pressure of 5 psi What would be the compressive buckling stress of a panel and also the shear buckling stress with this internal pressure existing
Solution
From Art C9.6, the interaction equation is,
From Problem 1, the compressive buckling
stress Poop = 7850 psi
To evaluate Rp; the external inward radial acting pressure that would cause buckling of a eircular cylinder having 2 radius equal to that of the curved sheet panel must be determined Use is made of Fig C8.16 of Chapter C8 to determine the backing stress under such a loading The lower scale parameter for Fig 08.16 is, = „k5 20x 04 — VỊ - ,35 = 296 From Fig C8.16, we read Ky = 19 ¬ Ấy nE ty Foor = 12 very SL 19 n? x 10,700,000 , 04 = 2 - an (178 = 1215 pst -
The external radial pressure to produce this buckling stress is, DO Fe tớ = 1215 x 04/20 = 2.43 psi 5.0 | Rp # ety = 2.06 Subt in interaction equation with 4 minus for Rp, ULTIMATE STRENGTH OF STIFFENED CURVED SHEET STRUCTURES 2 e 78507 ” 2.06 = 1, or f, = 13750 pst
Thus the internal outward pressure of 5
psi increases the axial compressive buckling stress from 7850 to 13750 pst Shear Buckling Stress Under 5 psi Internal Radial Pressure From Art C9.7, the interaction equation is, Re? + Ry = 1
From Problem 1, Pop or our panel was 11200 psi The value of Rp is determined as above or Rp = 2.06 Subt in interaction equation: fg* Foe - 2.06 = 1 “Ser fg* = 11200" (3.06), or fg = 19500 psi
The internal pressure of 5 psi thus in- creases the shear buckling stress from 11200 to 19500 psi PART 2 ULTIMATE STRENGTH OF STIFFENED CYLINDRICAL STRUCTURES C9.9 Introduction
A cylindrical structure composed of a thin skin covering and stiffened by longitudinal stringers and transverse frames or rings is a common type of structure for airplane fuselages, missiles and various types of space vehicles,
and such structures are often referred to as
the semi-monocoque type of structure The design of 4 semi-monccoque structure involves the solution of two major problems, namely, the
stress distribution in the structure under
Trang 7
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES In general, thin curved sheet panels buckle
under relatively low compressive stress and thus {f design requirements specified no buckling of the sheet under limit or design loads, the sheet would have to be relatively thick or the stringers placed very close to- gether and the fuselage or body structure would be unsatisfactory from a strength weight standpoint In missile structures, internal pressurization increases the buckling stress greatly, thus the buckling weakness of thin Sheet is improved, but keeping a structure pressurized under all operating conditions nas
its difficulties
In a semi-monocoque body, the longitudinal stringers provide efficient resistance to compressive stresses and buckled sheet panels can transfer shear loads by diagonal tension field action, thus the buckling of the sheet panels is not an important factor in limiting the ultimate strength of the over-all
structural unit When buckling of the skin panels takes place, a stress redistribution over the entire structure takes place, thus it is important to know when skin buckling begins Furthermore, design requirements may often specify that no skin buckling should take place under -a certain percent of the limit or design loads The equations and design curves in Part 1 of this chapter can be used to determine the buckling stress of curved sheet
panels under various stress systems
(2) Panel Instability
The internal rings or frames in a semi-
monocoque structure such as a fuselage divide the longitudinal stringers and their attached skin into lengths called panels If these frames are sufficiently rigid, a semi-monocoque structure if subjected to bending will fail on the compression side as {llustrated in Fig c9.7a The stringers act as columns with an effective length equal to the panel length which is the ring spacing Initial failure thus occurs in a single panel and thus is referred to as a panel instability failure In general, this type of failure occurs in most practical aircraft and aerospace semi-monocoque
structures because the rings are sufficiently stiff to promote this type of failure Since
the inside of a fuselage carries various loads, such as passengers, cargo, etc., the rings must act as structural units to transfer such loads
to the shell skin, thus requiring rings of
considerable strength and stiffness Even
lightly loaded frames must be several inches
deep to provide conduits required in various
installations to pass through the web of the ring cross-section, thus providing a relatively stiff ring for supporting the stiffeners in
their column action When the skin buckles under shear and compressive stresses, the skin
C9.9 panels transfer further shear forces by semi-
diagonal tension field action which produces
additional axial loads in stringers and also bending which must be considered in arriving at the panel failing strength This subject
is treated in detail in Part 2 of Chapter Cll
(3) General Instability
In general instability, failure is not confined to the region between two adjacent frames or rings but may extend over a distance
of several frame spacings as illustrated in
Fig C9.7b for a stiffened cylindrical shell in bending In panel instability, the trans- verse stiffeners provided by the frames on rings is sufficient to enforce nodes in the stringers at the frame support points as illustrated in Fig C9.7a Any additional
stiffeners in excess of this amount does not
contribute to additional buckling strength General instability may thus occur when the stiffeners of the supporting frames is less
than this minimum value
C9.11 The Determination of the Stresses ina
Stiffened Cylindrical Structure Under
External Loads
The stresses in a stiffened cylindrical structure such as used in typical fuselage or missile design can be fairly accurately
determined by the modified beam theory as pre- sented 1n Chapter A20 A more rigorous approach is given in Chapter AS involving matrix formu~ lation but this approach requires the use of a large electronic computor to handle the required calculations For details of applying the modified beam theory, the reader should refer to Articles A20.3 and A20.4 of Chapter A20 In the example problem solution as given in Article A20.4, the effective area to use for the curved sheet was based on the ratio of the buckling stress of the curved panel to the bending compressive stress on the panel due to bending of the entire effective cross-section of the fuselage under the design loads In the example problem as given in Table A20.2, a conservative buckling compressive stress equal to 3 Et/r was used for the curved panel and no consideration of the effect of shear stress on the compressive buckling stress was considered
A more accurate procedure would be to cal- culate the effective area of the curved panels taking into account the influence of combined compression and shear on the buckling strength of the panel Thus tn Table A20.2 on page AZ0.5 of Chapter AZO, the shear stress on each curved panel should also be calculated and then the buckling streas of the panel under the combined compression and shear calculated
Trang 8
ANALYSIS AND DESIGN OF F and shear should be calculated using Equations C9.1 and C9.2, The buckling stress under combined compression and shear is given by the interaction equation:- Re + Rg* = 1, where Re = Ÿfe/Feer› Rg = fs/Fsqp- The expression for margin of safety ts, = 2 -~1 Re + V Re? + 4Rg”
Let f, be the compressive stress that will
buckle the curved sheet panel when subjected to combined compression and shear when the ratio of the applied compressive and shear stresses
in a constant Then,
M.S
s 2
fos fo (Qo ƯNG + đàng
These Ÿ¿ values should then replace the
values in column 5 of Table A20.2,
The author has noted that one aerospace company in their missile design uses only 90 percent of the theoretical buckling stress in computing the effective area of the buckled curved panels This correction assumes that the curved sheet fails to hold the buckling stress as the fuselage section as a whole is
further loaded and the curved sheet suffers more buckling distortion
C9.12 Panel Instability Strength
‘ Panel instability means failure of the
stringer and its effective skin between two adjacent frames The bending of the stiffened
shell as a whole produces a compressive load
or stress on the striuger The semi-tension field action of the skin after buckling
produces an additional compressive load on the stringer and also a bending moment
The compressive stress due to bending of
the stiffened shell as a whole is found by the
methods discussed in Article C9.11 The additional stringer loads due to semi-tension
field action are determined by the theory and procedure given in Part 2 of Chapter C11
These calculated stringer loads are then compared to the stringer strength to determine whether a positive margin of safety exists The local crippling and column strength of a
stringer plus its effective skin can be found by the theory and analysis methods given in Chapter C7 The bending strength of the stringer cross-section can be found by the theory and analysis method given in Chapter cz The strength of the stringer in combined
compression and bending 1s found by use of the proper interaction equation
LIGHT VEHICLE STRUCTURES C8.11
C9.13 Calculation of General Instability
“ great deal of theoretical and experi-
mental work has been done on the subject of
general instability of stiffened shells The general goal in the design of such structures
is to insure the frames have sufficient
stiffness so as to prevent the type of failure illustrated in Fig C9.7b or, in other words ,
to insure the type of failure illustrated in
Pig C9.7a which is panel instability
7 Shauley (Ref 6) has derived an expression for the required frame stiffness to prevent general instability failure of a stiffened shell in pure bending (EI)‡ # Cy MD*%⁄L In a study of available test data, Ce was found to be 1/16000 Thus, (E1)£ = MD*/16000L
where, E = modulus of elasticity
moment of inertia of frame section
diameter of stiffened shell frame spacing
bending moment on shell
Kor
oH "
Recker (Ref 7) in a comprehensive study of most published theoretical and experimental
material relative to the general instability of stiffened shells, summarizes the results of his studies as given in Table C9.1
Bending
For the case of bending, the constant of
4.80 in the equation given in Table C9.1 is for the condition where the frames are attached to
the skin between the stringers For frames not
attached to the skin between stringers, the
constant should be 3.25
The effective sheet width for use with the stringers may be found from the equation,
Wi B= 0.5 (Fog /Fo)3/*
where, We = effective width of skin per side of stringer (in.)
b = stringer spacing (in.)
Foor = compressive buckling stress for curved skin panel
Fo = compressive stress at bending general instability (psi)
Trang 9BUCKLING STRENGTH OF CURVED SHEET PANELS
ULTIMATE STRENGTH OF STIFFENED CURVED SHEET STRUCTURES
Table C9.1
(Ref 7)
THEORETICAL GENERAL INSTABILITY STRESSES OF ORTHOTROPIC CIRCULAR CYLINDERS
(Results are based on the assumption that spacings of longitudinal stiffeners and circumferential frames are uniform and small enough to permit
assumption that cylinder acts as orthotropic shell) Loading Moderate-length cylinders Long cylinders Fo = gE (Ipt)*/*/Rtg Bending - ø =4 80[(b⁄4)(Os/0004/t)2(os/b)^] *⁄+ External radial t, fo or hydrostatic Fy = 5.515 G44 (2/4 Fy = 3E (Pz/R)^ pressure f t 1 Torsion Fer = 3.46 (8)/* ch lấn Pep = 1.784 C54
Fy = Compressive stress at bending general instability (psi)
Fy = Circumferential normal stress under external Pressure at general instability (psi}
FsT = Shearing stress at torsional general instability {psi) b= Stringer spacing (in }
d= Frame spacing (in } R= Cylinder radius (in }
¢ = Skin thickness (in )
Ag = Stringer area (in 7) Ag = Frame section area (in 7) ts = Distributed stringer area = Ag/b tg = Distributed frame area = Ag/d
if = Bending moment of inertia of frame section (in *) I = Distributed bending moment of inertia of frame = ff/d
Trang 10
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES For the frames the effective skin width
should be taken as the total frame spacing (d)
For inelastic stresses, the use of the secant modules appears to be applicabie on the basis
of limited test data
External Radial or Hydrostatic Pressure
The effective skin width to be used in
computing the stringer and frame section
properties may be determined from the following equation
We We
~s ft Be 2 05 Pog /Fe) / ell le c9.9
The subscript s refers to stringer and f
refers to frame The term d is the frame
spacing
Torsion
The effective skin to be used can be
determined from the following equation:—
Weg M
ey 70.5 (Fegp/Pat)*/* - 09.10
where F Ser is the shear buckling stress
for the curved skin panel Fst, the torsional
shear stress at torsional general instability
(psi)
The equations for torsion as given in Table C9.1 would not apply to shells in which
there is a strong tension field that could
introduce appreciable secondary stresses in
the frame The reader should refer to Part 2
of Chapter Cll for a treatment of this subject involving the effect of semi-tension field action in the skin panels
Transverse Shear General Instability From Ref 7, it 1s stated that a con- servative shear general instability shearing stress may be made by utilizing the relation
Fg = 0.85 Fg¢ 09.11
where Fg is the transverse Shear stress under
transverse shear general instability
General Instability in Combined Torsion and
Bending
From (Ref 7} the following interaction
relation may be used to compute the permissible combinations of applied torsion and bending Moments to a stiffened cylinder
C9, 13 where M = applied moment
Mg = moment causing bending general in- :
stability acting alone I T = applied torsional moment
To = torsional moment causing torsional general instability acting alone
General Instability in Combined Transverse
Shear and Bending
(Ref 7) concludes there is no interaction
for this combination of transverse shear and bending loads General instability occurs only
for either type of loading acting alone and thus doth loadings may be examined separately
C9.14 Buckling of Spherical Plates Under Uniform External Pressure
Classical Theory using 0.3 for Poisson’s
ratio gives the following buckling stress for
a perfect spherical shell subjected ta 4 uniform external pressure:-
Avatlable test data on practical shells show this theoretical buckling stress to be much too high Thus to satisfy experimental
results, reduced values must be used The buckling equation which is similar to that for
curved plates, under external pressure (from
Ref 2) is,
Kp tt E tye
For = 1B (1 - BA a
Fig C9.8 shows curves for determining the
buckling coefficient Kp and shows how test data falls considerably below the theoretical buckling
curve Equation C9.14 is for buckling stresses
below proportional limit stress of material
Report AS-D-S68 of the Astronautics Division
of General Dynamics Corp from a statistical
study of test data gives the following equations for the buckling stress of spherical shells under
Trang 11BUCKLING STRENGTH OF CURVED SHEET PANELS cg 14 79/027 200 400 10%
Fig C9 (Ref 2} Test Data for Spherical Plates under External
Pressure Compared with Empirical Theory kg 2B ca cr”TE-Pa5 44) Fer = 0.0908 Ft cr“ r(sin q) 1⁄9 For explanation of angle a see Fig A When œ = 90 or 2 hemisphere (sin a}#/* is one The test range covered was 792 a = 900 175 < r⁄% < 2000
The equations are for buckling stresses below the proportional limit stress of the material (1) (2) (3) (4) ULTIMATE STRENGTH OF STIFFENED CURVED SHEET STRUCTURES, PROBLEMS
The fuselage cross-section as given in Fig C9.6 of example problem 1 is changed by increasing the skin thickness to 05 inches, The design loads are increased to the
following values:
My = 700,000 in 1b., Vz = 6040 lbs
T = 245,000 in Ib
Will any of the skin panels buckle under this combined loading
The fuselage section as given in problem 1
above is subjected to an internal outward
pressure of 6 psi What would be the com- pressive and shear buckling stress for the skin panels under this internal pressure
REFERENCES
Schildcrout &@ Stein Critical Combinations of Shear and Direct Axial Stress for Curved
Rectangular Panels NACA T.N 1928
Gerard 2 Becker Handbook of Structural
Stability Part III Buckling of Curved Plates and Shells NACA T.N 2885,
Rafel & Sandlin fect of Normal Pressure
on the Critical Compression and Shear Stress
or Curved Sheet NACA WRL-57
Rafel Effect of Normal Pressure on the Critical Compressive Stress of Curved Sheet
NACA WRL-258
Brown & Hopkins The Initial Buckling of a Long and Slightly Curved Panel under Combined Shear and Normal Pressure R&M
No 2766, BRITISH ARC (1949)
Shanley F.R Simplified analysis of
General Instability of Stiffened Shells in
Pure Bending Jour of Aero Sciences, Vol 16, Oct 1949
Becker H Handbook of Structural Stability Part VI Strength of Stirren ed Curved Plates and Shells NACA T.N 3786
Trang 12CHAPTER C10 DESIGN OF METAL BEAMS WEB SHEAR RESISTANT (NON-BUCKLING) TYPE PART 1 FLAT SHEET WEB WITH VERTICAL STIFFENERS C10.1 Introduction
The analysis and design of a metal beam
composad of flange members riveted or spot-
Nelded to wed members is a frequent problem in airclane structural design In this chapter, the general theory zor beams with non-buckling weds is considered In Chapter Cll, the more
common case where the beam web wrinkles and
zoes over into a semi~tenston field condition is considered The advantages and disadvantages
of the nom-buckling and the buckling or semi-
tension field web are discussed in Chapter Cll The general beam theory as given in this chapter
is basic to that given in Chapter Cll, thus the student should study this chapter before Cll
gyration of the beam section as large 45 possible, and at the same time maintain a flange section which will have a high local crippling or crush-
ing stress Furthermore, the flange sections for large cantilever beams which are frequently used in wing design should be of such shape as to permit efficient tapering or reducing of the
section as the beam extends outboard This
tapering of section should also be considered
from a fabrication or machining standpoint “he most efficient flange from a strength/weight
Trang 13C10 2 DESIGN OF METAL BEAMS WEB SHEAR RESISTANT (NON- BUCKLING) TYPE,
metal covered wing construction Sections {a},
(5), (e) and (d) are typical beam flange Sections for wide box beams where additional
stringers or sxin stiffeners are also used to
Provide bending resistance These flange
Sections are generally of the extruded type a ugh such sections as (b) and (c) are frequently made ?rom sheet stock, These flange
Sections are almost always used with a beam web
composed of flat sheet, which is stiffened by vertical stiffeners riveted or Spot-welded to the web or st1f?ened by beads or flanged lightening holes
figs (e) and (f) illustrate two types of flange sections used in truss beams which land themselves readily to connection with truss web members Beam flange sections (z) and (nh) are typical sections for wing construction tn which
no additional spanwise stringers are used In
section (g) tapering of sectional area is pro- vided by first omitting the reinforcing plates,
and then gradually decreasing the extruded
shape by machining until only 2 small angle remains In section (n) a gradual decrease in Section area is produced oy milling out the center portion to form an 4 section and then cutting this section down fi 11ÿ to a simple
angle,
C10.3 Allowable Flange Design Stresses
The calculating of the stresses tn the beam flanges is in general not a difficult procedure
if the usual assumptions are made in the flexure
theory The question as to what flange stresses
will cause failure is the dif#icult one from 2
theoretical standpoint The only sure way to determine the design allowable {s to make
suffictent static tests of specially designed test beams
For Deam sections as {llustrated in Fig
€1Ô,1, the following tests are usually necessary tor forming the basis of design allowable
stresses
(1) A test subjecting the beam to pure
bending
{2) A test of a short length of beam in
vending so that failure will secur in xeb
instead of 2lange,
ở) A test of a short length of beam tn
cupression to obtain local crippling
Several tests of beams in combined
Sending and compression, using different ratios of bending te compressive loads
Enough data m the above tests can usually be obtained to give rather complete
allowable stress curves or the design of the
beam For the approximate design of beam flanges, the method given in this chapter can
be used
The beam flange sections (a), {o), (¢) and (d) of Fig Cl0.2 are stabilized by the sheet
covering and also by the beam web; thus compres- sive tests of short lengths to obtain crippling stress, and a test of a length equal to the wing rib spacing should give suffictent information on which to base design allowables, Several iengths of the flange section for the truss of beam should be tested in compression to
obtain the column curve since the distance
between panel points of the truss will vary
type
Sefore designing any test beams, the structural designer would like to know approxi- mately what his proposed beam flange sections will carry from a stress Standpoint, since tt
is desirable to make test specimens closely approximate to the sections to be finally used in the completed structure For most of the Sections of Fig, Cl0.2, the ultimate stresses can be calculated approximately by the methods of Chapter C7 For heavy sections similar to
(g) and (h) of Fig C10.2, where the ultimate
stresses fall far above the yield strength of the material, and where some Darts of the section buckle before other parts, and also
where two different kinds of material are used
in the same flange section, a logical procedure in trying to calculate ultimate Strength of the
Section would be to make use of the stress- strain diagrams of the materials
C10.4 Use of Stress-Strain Diagrams in Computing Beam
Flange Bending Allowable Design Stresses In the beam type of wing construction, where the flange material is concentrated over
the web members instead of distributed over the
surface in the form of stringers, the allowable ultimate compressive stress which can be
developed 1s considerably above the yleld strength of the material since the flange is
composed of a section with thick elements whicn
promotes a high crippling stress and since the beam flange is stabilized in both vertical and horizontal directions 5y the web and skin cover= ing respectively, the infiuence of column action is negligible Fig (a) {llustrates this type of beam The general flexure formula assumes that stress is Proportional to strain whieh 1s
correct for stresses below the proportional limit of the beam material, but the ultimate
resisting stresses for the flange of? a beam such as in Fig (a) {s far above the proportional limit, thus the actual stress đ stribution is more like the dashed line in Fiz (b) instead
of the triangular distribution as assumed In the
commen beam theory Thus to obtain beam fiber
Stresses above the proportional iimit, it is
necessary to consider strain and e stress which accompanies such a strain, which relation-
Trang 14
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Skin Assumed stress 7“ distribution in 7 N beam formula 4 Actual stress “ distribution ự fig b above yield point
ship can be obtained from the stress-strain diagram of the material A straight line
distribution for strain, that is, plane sections remains plane after bending is 4 reasonable one, and verified by tests
In a beam in bending, one side is in
tension and the other in compression The
tensile and compressive stress-strain diagrams
for materials like aluminum alloy are different
above the proportional limit, and the same unit
strain will cause different stresses on the two
sides of the beam In frequent cases of large beam design, the beam flange may be composed of two kinds of material and certain portions for attaching to skin or web may buckle befors the ultimate strength of the section as a wnole is
obtained The solution for the ultimate
internal resisting moment for such a beam re- quires that consideration be given to the stress-strain diagram of the various materials and units making up the beam section This general method of approach in studying the ultimate interrial resisting moment of a beam section can best be explained by an example problem
Example Problem
Fig Cl0.3 shows the cross-section of a beam in a metal covered wing The main flange members are composed of heavy 248T extruded
shapes The extrusions are reinforced by 24SRT sheet strips The beam web ts made from 248T
alclad material The problem is to calculate
the ultimate internal resisting moment for this
beam section
Fig Cl0.4 shows the stress~strain diagram
for these various materials, The 1/8 inch thick
outstanding legs of the extrusions act as 4 plate stiffened on three sides and free on the fourth These legs will buckle at a stress of 35,000 psi in compression as determined by the methods given in Chapter C7 The stress-strain diagram of the two outstanding legs will be horizontal at 35,000 psi as shown in Fig C10.4 Although the legs buckle, they will tend to hold the buckling stress under farther flange strain In Fig C10.5, each beam section has
been broken down into narrow horizontal strips
designated trom (a) to (w) Only that portion
of the web in way of flange has been considered in this example Fig Cl0.Sa shows the strain @istribution assumed for the trial solution A heavy aluminum ailoy section in compression will
C10 3 usually fail at a strain between 008 to O1 inch per inch if column failure is prevented In Fig C10.5a, the compressive unit strain at the upper beam fiber has been taken as 008"/"
The neutral axis of the section has been assumed at 1.25" above center line of beam
Taking zero strain at this point the beam section strain 27^2* as as shown in Fig 10.54
For equilibrium the total compressive bend- ing stresses above the neutral axis must equal the total tensile bending stresses below the
neutral axis
Tables C10.1 and Cl0.2 give the detail calculations for calculating the resisting
moment The bending unit stress in column (5)
is obtained from Fig C10.4 using the unit
strain in column (4) If the summation of column (6) in each Table is the same, the
assumed location of the neutral axis is correct The total ultimate resisting moment for this section equals 1032000 + 1360110 = 2395000 in 1b Using the ordinary beam formula with properties about the geometric neutral axis as given in Fig C10.3 and taking extreme fiber stresses of 46000 and 52000 psi which corres- pond to stresses as per strain diagram of Fig C10.5a, the internal resisting moment would equal, M = f I/e = 46000 x 436/969 = 2070000 in lb as compared to 2392000 in 1b by the more logical solution, which is a difference of
16 percent, The discrepancy would still be larger 1f the outstanding leg (a) did not buckle, since the more exact solution only allowed 35000 psi on this element whereas the general beam formula stresses 1t to 46000 psi
Trial and Error Approach
The location of the neutral axis is un-
known, thus the calculations in Tables C10.1 and
2 are for the final assumption which is the true location of the neutral
axis The general pro-
cedure would be as
follows: Assume neutral and Trial N A axis as center line axis » Above €,
of beam, and find total
axial load on each side / of axis The resuits will /
usually show excess load be x of,
, *F distance
+
on one flange For the
next trial move neutral
Excess
Load
from to axis so as to give excess true N.A
load on other flange ai l
Plot the results as indi- fe 18 Trial Ne
cated in Fig Cl10.5b to
obtain true location of
Trang 15C10 4 DESIGN OF METAL 4-1/8x1-3/4 Plates BEAMS WEB SHEAR RESISTANT (NON-BUCKLING) TYPE, 1 Si \zaset 1⁄8 -80 1⁄42; 2 24SRT +h = ~80 wat § COMPRESSION À248-T | -40 Extrusion | —*—.051 - 24S-T -30 Alclad Sheet N.A at time UPPER FLANGE 20 of Failure —¬ + Fig C10.5 s s-i0 "¬ “ 88'S) = Ịs 5 2 Ht 2 9 ' N.A at Stresses B = g z—— [4 10 below Prop a Limit Ễ h (I = 436 in.*) 18" \== 20 LOWER FLANGE: š .008/% 5 30 24ST Extrusion paarsy 1.28 40 t LN axis 20.75 —F7¢€ 50 60 1⁄8 Br °=1n.giVÐr 7= 000818y 2-1/8x 1-3/4 Plates 24 S-RT Fig C10.3 - Fig, C10 5a 70 - Table C10.1 Table C10 2
Eeeiating Mement of Comp Bending Stresses about Neutral Aniw Resisting Moment of Tensile Bend Stresses about Neutral Agie Strip|é 2 area} y « arwa/Uait strain|tosunit| Load on|Mom about StriplA = realy = arm [Unit strain |f;sunit|Load ou [Mame rt Port, Jof strip fo strap | f= 00876y| stress Strip sỉ ia ‘rom M.A, cA la te Ay Port of strip|to strip| {= 00087éy|/strese strip =| ae tt hla te ay al „3125 ‹00798 |.35000*| 10940 99200 11.5623] ,o1014 32050 | '16280]- 4 i Ho ' ng 458090 373 ng b 11.375 -00997 32000 4870 $5450 ẩ | d| 31250 200734 |-45150 | 5640 47200 zl§ 32,225 r0 | $1030 | 4865| 54100 sie} l1280 00712 |-45000 | 5620 45700 3l 1ô tH 33900 Tân 32900 | alt] 6250 z +ỂD0691 90868 |-44700 | 27900 | 220000 |~44300 | “2770 21150 al 10°373 ` *90810 * 51750 | 39%3 318000 31600 i Sin 00646 -44000 2750 20300 » |e 10,125 „00887 51700 2425 24550 Nya +00825 [43300 | 2710 19300 gy 9.875 „00868 31600 2420 23900 | 3 š 200541 00603 | -43100 | 2680 |-42700 | 287g +4 17700 Aja J 9.275 9.825 „00844 .00822 51500 51400 2415 2420 22600 23250 | 1 00558 -42000 2625 16720 * 9,125 „00800 51300 2405 21950 ; a)” Sia ° .090603 00846 _ |~40700 00825 | 40100 |~39700 s19 512 508 3640 3480 3840 ais sả 9.875 4.879 00778 „00866 31200 49600 2400 433 21300 8250 <a 9-625 00844 49400 929 5050 ị 8 p 00541 -38000 497 3290 & |e 9.375 00822 49300 828 5890 Ñ 8.375 00559 |~38800 492 3130 3 9,128 00800 49200 627 3720 z 3.4125 09728 |-55500 | 13230 | 101800 Sq 4,875 09778 49100 525 5400 dj#| 21875 | 3.1873 00718 |~55200 | 12080 38700 gir 19,8128} 90948 67500 | 14770] 1sa600 ; Eat! 2t875 | 4.0625 00706 |~55100 | 12050 97100 LẦN 10,6875| -0083g 67300 | 14?20| 157200 | lui "21878 | 1.6873 99674 l-54400 | 1lieoo 91450 + 10.5625 | _ 00926 87100 | 14700} 186300 | _; TOTALS i tT 128490 TOTALS T2E277/ T, “Buckling stress C10.5 Flange Strength (Crippling)
In many cases of beam design, the flange is braced laterally because it is part of a cell construction and the sheet covering which Is fastened to the beam flange vrovides a contin- uous lateral bracing to the flange or prevents lateral bending-column action for the beam
flange The beam wed prevents column bending
of the flange in the web direction, thus the flange fails by local crippling action and the crippling stress is determined by the methods of Chapter C7 In many cases the beam loads are relatively small and thus the area required for the flange may be relatively small, which means a flange shape with elements of small thickness, thus the failing local strength may be in the
Trang 16
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES In many cases, such as a frame ina
fuselage, the inner flange of the frame cross- section is free from a lateral brace, thus provisions must be taken to brace the flange laterally or the flange must be designed for lateral column action This subject is dis- cussed further in Chapter D3
C10.6 Web Strength Stable Webs
A stable web beam is one that carries its
design load without buckling of the web, or in other words it remains in its initially flat condition The design shear stress is not greater than the buckling shear stress for the
individual web panels and the web stiffeners
have sufficient stiffness to keep the web from
buckling as a whole
In generai, a thin web beam with web stiffeners designed for non-buckling is not used widely in flight vehicle structures as
its strength/weight ratio is relatively poor In built-in or integral fuel tanks, it is often
desirable to have the beem webs and the skin
undergo no buckling or wrinkling under the
design loads in order to give better insurance
against leaking along riveted wed panel
boundaries
The student should realize that the buckling web stress is not a fatling stress as the web will take more before collapse of the
web takes place, thus in general the web is not loaded to tts full capacity for taking load
and the web stiffeners are only designed for
sufficient stiffness to prevent web buckling and not for the full failing strength of the
web
Equation for Wed Buckling Shear Stress Equation CS.5 of Chapter CS gives the buckling stress of a flat sheet panel under shear loads The equation fs
n* kg EB
Fsor Ets ? - (c10.1)
The curves of Fig C5.13 and C5.11 of Chapter C5 are used to solve this equation and the reader should refer to Art C5.8 of Chapter
cS for a review of sheet buckling under shear
loads
Equation for Wed Buckling Bending Stress
If the wed does not buckle it will be
subjected to longitudinal bending stresses of compression and tension with zero stress at the beam neutral axis Thus in general, the beam 18 Subjected to a combined shear and bending
stress system
C10 5 Equation C5.7 of Chapter CS gives the
bending buckling stress It 15,
_ ip 17% EB tia
Poor “Beis Wey) 777 T (20-2)
In solving this equation, Figs C4.15 and
cS.8 of Chapter CS are used
Buckling of Web Panels Under Combined Shear and Bending
From Art CS.11 of Chapter CS, the inter-
action equation for a flat sheet panel under
combined bending and shear is, {c10.3) Rp? + Rg? = 2 The expression for margin of safety is, 1 —
C10.7 Web Bending and Shear Stresses
Since the web is designed not to buckle under the design loads, the web will be effective
in taking bending stresses and the following well kmown equation applies
- + + ee ee (c10.5)
where Iy is the moment of inertia of the beam section and the web is included
In Art A14.3 of Chapter Al4, the well known flexural shear stress equation was derived, namely,
Vv
fs “ Tp / yaa
Since the term / ydA is maximum for 4a `
section at the neutral axis, the shearing stress
in a beam will be maximum at the neutral axis
In general, the weds of aircraft beams are
relatively thin; thus the term / ydA for the
Web 1s quite small so that the shearing stress intensity over the web 1s approximately uniform
Thus a simple formula fg = V/nb has been widely
used for calculating the maximum web shearing stress, In this equation h is a distance which
will make the shear stress fg check the maximum
value of the shear stress as given by equation c10.5
A simple consideration of the internal stresses on a small element cut from a beam in
pending and shear will indicate what value of
h to use in the simplified shear equation f, =
V/nb
Trang 17C10 8 Cx=Caạ-C› € Al Cc Ca | c 0 — x số Dp By TT (a) T=T,-T, fe dx fsbdx rye Fig €10.6 ~L lo fgbdx
bending moment which produces compression on the upper portion The bending moment on section (AB) is M and on section (CD) M+ aM,
thus the vectors representing the stress on
face (CD) are drawn longer than on face (AB)
The beam element is also subjected to a total shear force ¥ on each face C, and C,
represent the resultant of the total compressive forces on each face and T, and T, the resultant oy the tensile stresses Fig (b) shows the
same free body but with the tensile and
compressive stresses on each face replaced by a simple force C and T which tends to move the block with the same results as the system of Fig (a) In Fig (c) the beam element of Figs
(bd) has been cut along the neutral axis, and a
force applied to the cut faces equal to fg bdx, where fg equais the horizontal shear stress
intensity
Writing the equilibrium equation, that the sum of the horizontal forces on the upper portion must equal zero, we obtain
C - fg bdx = 0, hence C = fg bdx and likewise for the lower -T + fg bdx = 0,
hence T = fg bax
Fig (d) shows the free body of Fig (b); put with C and T replaced by their above equivalent values Taking moments about point (0) „ nh - Vdx = 0, hence the an between
` Thus to obtain a value f, equal to maximum value given by equation (a) use
effective arm (n) equal to the distance
DESIGN OF METAL BEAMS, WEB SHEAR RESISTANT (NON- BUCKLING) TYPE
the bending stress centroids For a rectangular section the effective arm is obviously sequal to 2/3 the beam height, but for the common beam sections as illustrated in Figs C10.11 or C10.12,
the distance between bending stress centroids
4g not so obvious particularly if the web is considered effective 1n bending A close
approximation of the effective arm (h) and a
procedure which ts common practice 1s to assume
(n) equal to the distance between the centroids
of the wed-flange rivets The student should
take several example beam sections and check
this assumption for (n) using Eq (C10.6) against the exact values by Eq (c10.5)
Some structural designers make assumptions as to the proportion of the total vertical beam shear which is carried by the beam web For example, it 1s sometimes assumed that the web
takes the entire beam shear, or it may take only
90 percent The percentage of the shear load carried by the web depends of course upon the
size of the flange sections and the form of the
web section For example, in Fig C10.7, the game flange 1s connectad by a web which is attached to the flanges in two ways as illus-
trated in Figs (a) and (Db) In Fig (a), the shear flow on portions (AB) and (CD) help resist
the external shear, whereas the shear flow on these portions in Fig (b) act in the same direction as the external shear load; thus causing the shear load on the web bo be greater
than the external shear load (See Chapter Alé4
and A15 for general discussion of shear flow in
open and closed sections)
C10.8 Shear Resistance Provided by Sloping Flanges ‘ A large majority of the beams in airplane wing and tail surfaces have sloping flanges be- cause of the taper of the structure in both planform and thickness This sloping of the beam flanges relieves the beam web of con-
sidering shear load and should not be neglected Pig ClO.@ shows a beam (abcd) carrying a Load system P,, Pz, etc The top flange is sloping as shown If both flanges were extended, they would intersect at point (0)
Trang 18ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES T P,P, Py Fig C10 8 Then C = Ty = M/h (h = distance between flange stress centroids)
The vertical component Ty, of the load T in the upper flange equals Ty n/Lo, but Ty = M/n,
hence
Let Vp = shear load carried by beam flanges
Then Vp = W/o - 277 rr (¢10.7)
Thus the shear component carried by the
axial loads in the sloping flange members equals
the bending moment at the section being con- sidered divided by the distance from the section to the point of intersection (0) of the flanges
The above derivation was based on the assumption that the entire resisting moment M was developed by the flanges with the web effective in bending, the moment developed by
the web should be subtracted from the total bending moment M Let I = Ip, + Iweo — xy lWab The moment developed by web = M TT? where
M = total bending moment on section
In airplane construction the centilever
beam with sloping flange members is the common
case, and the shear resistance by the flange
axial loads 18 an important factor which should not be neglected if an efficient structure is
desired from a strength-weight standpoint
10.9 Effect of Variable Moment of Inertia on Flexural Shear Stress Distribution
The fundamental shear stress equation C10.5 as derived in Chapter Al4 applies strictly to beams of constant moment of inertia For air-
plane beams the common case 1s one with variable
moment of inertia; thus the stresses obtained by
equation C10.5 are incorrect, although the
discrepancy in most cases is not large The student should realize this fact in studying
the shear flow picture in tapered wing
structures See Art Al5.15 of Chapter Als
C10.7
C10.9a Flange Discontinuities
From a weight saving standpoint, it is necessary to taper flange sections in order to approach a beam of constant strength relative to the applied loads
Fig a illustrates how such tapering of
the flange section may produce local eccentric flange loads Between sections (1) and (2) the upper flange tapers in side view as shown which Pi @ Fig a @ F
displaces the flange neutral axis as shown Agsiming there is no change in bending moment over the beam portion as shown, the force F
must be greater than F since resisting am a is less than d For equilibrium this moment
due to F, and F not being colinear must be balanced in some manner If the flanges are
rigidly connected to web and stiffeners, this moment can be balanced by an additional shear
stress on the web panei between points (1, 2,
3 and 4) as illustrated in Fig b Thus in
cases of rather abrupt changes in flange
sections which produce the eccentricity as
111ustrated the web and stiffeners should be
checked for the additional shear flow load on the web If such displacements in the flange neutral axts occur in the plan view, the skin covering should be investigated for the additional shearing stresses
€10.10 Stiffener Size to Use with Non- Buckling
Web
A web stiffener is used to decrease the
size of web panel; thus when buckling of the web starts, the stiffener tends to keep buckles from extending across the stiffener or causes the sheet to buckle in two panels instead of one Mr H Wagner in a paper presented before a meeting of the A.S.M.E£ in 1930 offered the
following expression as the required moment of
inertia of a stiffener to be used with a shear resistant web
_ 2.294 V bwt/*
Wee ee TT TT TT (19.8)
where
ly = moment of inertia of stiffener
a = center line distance between stiffeners
Trang 19DESIGN OF METAL BEAMS WEH SHEAR RESISTANT (NON-BUCKLING) TYPE a 8 ˆ & š $3 3 8, 2 .# aon ci vs er fa S aa 5 as Sh a 33-8 Š & Sot 3 Bộ 3 ie — s9 S18 Cae š S5 a 38-88 0-8 bes Ÿ we ý ga g @: IS _ Zan gah 2M ce - Ree § Saks Ss rs š Bea g 254283 4 ¬- sử SŠ 3 og 5 quát số ở CÁ Cm ET Roh ZZzn a Ba SEP S8 Tết ức co Ga, y — 2 = BS › Sắc § Šäsas3§ £ 2 ” Š 5: § 4s RE có : TIS SH mà Š%wd Sổ 3 oe EAS tts Zea aot tae” BS a ° eter gh 2 a PG X
zee 5Š BF eee Besets Sẽ
6a? uae o xã 8 2 — sh§o Sẽ Nees mead islet lai ST
2-2
S52 2 oe es 253 SE
522-6 9E op eRe tsetse ee ee ee Pes Es 8 4.3 Se
6485 Sea gecagetrie
Saag 2 gi<esy gdreenie
AE Grom me ayy SSS cee SS
Am SESS ZTE eos 92a sẽ 0b sưa ST = oa Sa oC age ES: - SS32 25 acs: eRe Sasa: a BOB Ste S° 1000 100 Vo= vertical Shear at section t = wed thickness
& = modulus of elasticity
A more recent criteria for stiffener
stiffness (iy) for both flat and curved webs i$ given by the curve in Fig, ©10.9 When the stiffener {s used puraly as such and not as a
means to transfer a concentrated external lead
to the beam eb, the question arises as to what
4s the minimum number of fasteners required in
attaching the stiffener to the web For non- ocuckling webs, two criteria are suggested:~
The stiffener should be attached to the
flange at each end
The rivet pitch (spacing) should according
to (Ref L) be at the most equal to 1/4 times the stiffener spacing, or 1/4 the
web height if this is smaller, in order ta justify the assumption of simple support at the edges of the web panel Normal practice uses more rivets
C10, 11 Notes on Beam Rivet Design
Exceot for
may de@ extruded very small beam sections which aS one Diece, the usual beam
consists of separate wed and flange members, fastened together by rivets, belts, spot welding or continuous welding in the design of such
beams {t is thus necessary to know what loads
the rivets, bolts, etc., are subjectad to in order to provide the proper connective strength It is quite easy to substituta in simple formulas to find the loads on beam flange rivets, how- ever, the Student should be sure that he under- Stands the fimdamental ceam action behind these formulas
Fle C10.10 illustrates a beam portion
equal in length to the flange rivet pitch p
The beam section at (a&')} is subjected to 4 bending moment M and M + 4M at section (cc')
The vending stress distribution on the beam faces is indicated by the stress triangles In tnis example section it will be assumed that the wab takes no bending stresses
Let Pr equal the total pull on the flange angles due to bending stresses at section AB due to bending moment M Then total pull on flange angles at section oD due to a moment M + AM on beam section CC’ equals Pp + áPp,
Trang 20ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES ra p ._— 1 ——» P pre ADP x = Bw F iF ` ! " \ i it iy M:âM/|V — V AM Nụ, \ "%7 z h " / ul / \ ụ ut po Bt p wePet, te op sph =——., Fig C10 10
the ?lange angles bo the web Thus load on rivet equals APy The same reasoning applles to the rivet holding the bottom flange angles to the web
The beam portion of Fig Cl10.10 as a whole is in equilibrium under the bending stresses
and the shear load V on each face
Taking moments about the intersection of the lower flange bending load and the beam face
cc',
[Pr - (Pp + aPy)| hn’ +V¥p sa - = (C10.9)
hence,
APp =P we +e -e eee (c10.10)
Equation C10.10 says that the change in flange axial load in a distance p equals the vertical shear on beam times rivet pitch p divided by the distance between flange bending
stress centroids 4Pp is also the horizontal
shear flow produced by flange angles over 4
length of p The horizontal shear (q) per inch
due to flange would be
q=V/⁄n' we ee eee ee eee (010.11)
The general expression for shear flow (see chapter Al4) is,
ast/ yas (cl0.12)
Eq (C10.11) is easier to use since the termy I and / ydA of Zq (G10.12) are not re-
quired, and the distance h can be estimated closely without calculation, and will be greater
than the distance between the centroids of the
flange areas
Equation (C10.11) was derived on the
assumption that the beam web took no bending stresses In general this is not true or only partially true With the web taking bending,
C10, 9
equation (C10.9) would be wrong because the
moment of the bending stresses on the web about our moment center is not included Thus to make the simplified equation (C10.11), check the exact result as given by equation (C10.12), the distance (h} would have to be greater than the distance between the flange stress centroids
In fact, taking (h) equal to beam depth would
not be far off from the results of, equation (C10.12)
C10.12 Loads on Rivets Attaching Reinforcing Plates to
Flange Member
The beam flange is commonly composed of a
main unit plus several reinforcing plates or parts which are held to the main unit by rivets or spot welds Fig C10,12 {illustrates typical beam flanges The basic section of the upper flange 1s reinforced by the plates (a) and {b), and the lower flange by the plates (c) and’ (d) The purpose of the rivets is to keep the rein- forcing plates from sliding along the flange tee section due to the bending of the beam; thus making the plates effective in bending This
horizontal force tending to slide reinforcing
plates and which is prevented by the rivets in shear is given by the fundamental shear flow equation (C10.12), namely
4 ~T⁄/ ydA, which
equals the horizontal shear per inch along the
beam To find the load
or rivet section 1-1 of
the upper flange of Fig
C10.11, the term / ydA
equals the area of the
plate (a) times the
distance from its centroid to the neutral
axis For shear load on rivet at section (2-2),
the term / ydA equals the
area of plate (b) times distance to neutral axis On the lower flange rivet
section (3-3) is critical
since both reinforcing plates are on same side, and the entire shear flow produced by plates
(c) and (4) must be resisted by rivet section (3-3) The term / ydA would thus equal area of
both plates times the distance to the neutral
axis of the beam
Fig 10.11
A simplified method which yields good re-
sults is based on the relative areas of the
units of the total flange To design connection of flange to beam wed, the total horizontal shear q produced by entire flange is always
necessary and involves the use of the entire
flange area in the shear flow calculation
Trang 21
C10, 10
The shear flow produced by a reinforcing plate is then taxen as proportional to the area of the plate over the total flange area times the total flange shear flow Using simplified equation (C10.11), we can write Y ap q" Nà where eee ee et ee eee (c10.13) ap area of plates under consideration Ap q 3 V and nt
au total area of flange
load per inch on rivet same as explained before For rivet loads in beams with sloping flanges the shear ¥ is the net shear as explained before in discussing web shear
stresses
C10.13 Web Splices
Usually in designing a sheet girder beam,
1E 18 necessary from a weight saving standpoint
‘to use several web sheet thicknesses, which means web splices Fig C10.12 illustrates typical web splices Fig (a) is typical fora feta tr a he Let ba Htl++ + thet | tt ltt ++'#+ let! +e +! + tư tt sales Fig C10 12
comparatively heavy web which prevents joggling
of web as in the case of Fig (b) In the case
of Fig (b), the lap is usually made under a web stiffener which provides a support for the web in driving the rivets through the thin web sheets
Loads on Web Splice Rivets
The web 1s subjected to shear loads and for stable webs, the web undergoes bending stresses
For rivet design it is usually assumed that
the web Shear stress is constant over the depth
Thus the vertical component of load on each web Splice rivet is the same or
DESIGN OF METAL BEAMS WES SHEAR RESISTANT (NON-BUCKLING) TYPE
re (cl0.14)
where
V = net web shear
nm = number of rivets in splice If butt *
splice, n, equals the number of rivets on one side of splice
In bending the splice rivets must transfer
the bending moment due to the bending moment M developed by the web The largest rivet load
on a rivet due to bending will be on the most
remote rivet, e.g., rivet (a) at distance Ta from center of rotation of the rivet group Then load on this rivet due to web moment equals
ir* = moment of rivet group which equals
the sum of the squares of the distances of the rivets from the center of rotation of bolt
group
The resultant load on the critical rivet will equal the vector sum of the values of
equations (C10.14) and (15)
Since in most cases only two rows of rivets are used in a web splice, a close approximation
for the moment load on critical rivet can be
written by using the vertical distances (y) to the rivets instead of the radial distances (r),
the resulting force acting in the horizontal direction Hence Ma mm (010.18) The resultant combined load on critical rivet is RV Ry," + Ry 7-H (c10.17)
The student should review Chapter Dl for more detailed information on rivet loads due to moment loads on riveted connections
When the web of one beam is joined to that of another beam using shear "clips," a special problem may sometimes arise regarding the adequacy of the clips This design problem is discussed in Chapter Dg
C10 14 Example Rivet Problem
Trang 22keg lie Detail of + Flange Angles O51 NLA 1 gt ST m † Web Depth = 6-7/8" T Flg C10.18 ension Flange Solution:
The loads on the rivets will be calculated vy the "exact" and also by the simplified approximate equations
The exact shear flow equation is Vv
qe yaa
The first step will be the calculation of
the moment of inertia of the beam section about the neutral axis The bending loads are such
as to put the upper flange in compression The moment of inertia will be calculated about the centerline axis of the beam section and then transferred to the neutral axis Table 10.3 gives the detailed calculations Table C10.3 Part area Y ‘ay | Ay? lạ" 1x Ar?+ lại Upper angles | 1780| 3.301 ,588| 1.84 | 008 1.8948 PEE Frintore-| 1093) 3,532) 386] 1.36 | negl 1,360 Ị 1/13 x ,051 [te „3510| 0 | 0 o |z 6.8753 - 1,380 \ 1.38 [Rewer avgiew | 1780|-3,301|~.5aa/ l.e4 | 008 1,848 G T man ~,044 |-3,063| 087|~ 208i - | -0.z08 ‘sums 71943 1453 [#43 y = fax + 483 0 57 i * iG Iya, = 6.43 - 7943 x 577 „ 6.17 in.4 "Io = moment of inertia about its own centroidal azis, |
Rivet load on upper flange rivets which
attach angles to web:
Rivet pitch = 1-1/8 inch
Horizontal shear load per 1-1/8 inch ‘distance equals 4= 1.125 7 / VÓA C10.11 / ydA = first moment of flange area about = angles = 2x 089 x 2.751 = 486 Reinforcing plate = 1.75 x 0625 x 2.961 = 324 Total = 810 Substituting in equation (A) 3000 = q2 1.125 x 6.17 x 810 = 443 1b
Shear flow by simplified equation q =a
h' = beam depth = 7.062" (see Art 10.10) hence
3000
q21.125x 7.062 7 478 ib which is con-
servative compared to 443 by exact expression
The web 1s attached to angles by 1/8 diameter 2117-T3 alumimm alloy rivets and are
acting in double shear
From Chapter Dl:
Double shear strength of 1/8 - 2117-TS rivet 22x 388 = 776 1b
Bearing strength on 2024-T3 - 051 web plate = 630 1b., thus dearing 18 critical and
630
oee1ls
Margin of safety > 142 „42 and with
the approximate method the margin of safaty
would equal (630/478) - 1 = 3l Check of Cover Plate Rivets:~
Rivet spacing = 1.5" with two rows of rivets By exact equation, load on two rivets q “1.5 / y$k 3000 1.5 x S17 * Load per unit = 236/2 = 118 lb -1093 x 2.961 = 236 lb By simplified formula: a v “plate 2000 „ 1093 a?& W arrange 1.5 5 (SS 7.062 ~ 12873 X +Sh5q) 1-5 = 243 lb = 122 lb./rivet
The rivets are in single shear which is
critical and equals 388 lb as given above
388 _ 1 = 2.28
Hence margin of safety = 118
ys ¬+ t
Trang 23C10 12 Flange Rivets 1/8 Dia At, 1" Spacing 2700 lb 2400 ib 2400 Case Ib, DESIGN OF METAL BEAMS WEB SHEAR RESISTANT (NON-BUCKLING) TYPE
Web Stiffeners 1/2x1/2x.04 Angles, 25" | 28" ——— DỤ Ld || ch Bay 1 - Web =.057 | Bay 2 ~ Web = 072 Bay 3 - Web = 072 -+-ltE===sp ~E- lRErzxrzrù Bay 4 - Web = 057 Fig C10 14 Sketch of Beam (See Fig C10 15 for Cross-section of Beam)
C10,15 Example Problem Strength Check of Beam
Fig Cl0.14 shows a built up (I) section
beam, simply supported and carrying its three concentrated loads as shown as the design load
for the beam
Check on Bending Strength of Beam
Since the beam cross-se.*1on is constant, the critical section is at the midpoint of the beam where the bending moment is:
Myax 7 3750 x 50 ~ 2400 x 25 = 127500 in.1by
As indicated in Fig C10.14, the beam ts riveted to the sheet covering On the upper
flange wnich is in compression under the given
loading, a certain effective sheet width will act with the beam flange This effective Sheet width depends upon the beam flange stress
which is unknown as yet As a preliminary
value, the assumption that a width of skin sheet equal to 30 t as acting with each rivet
line is a reasonable one On the tension or
bottom side, the entire skin sheet is effective
or 2 inches to each side of the beam which is
the distance half way to the first skin
stiffener on each side
Pig C10.15 shows the details of the effective cross-section at midspan of the beam, Three rivet holes are assumed in the tension
flange Table Cl0.4 shows the calculations for the section properties, first about centerline reference axis, then transferred to the neutral
axis
Bending stress at midpoint of horizontal
leg of upper angles:- In a, = 10.44
fp = My/I = 127500 x 3.614/10.44 = 44200 psi The assumption of 30t = 0.75 inch as the
skin effective width W will be checked From Chapter c7, Lot ¥ B/Feq w —3/4—> &~3F4—s| —.025 7 + T† Sheet TÌ L— 1⁄4 214 7 NLA TT 4 Upper Flange Angle | —f-4 — N.A, : +———- Material 2014-T6 Extrustion ——_ Lower Flange Angle 3 “Pa Té Area =.089 i = 004 087 3⁄4 NAL = 7-7/8" 7 1 Rivet Holes tL + ¬ - NLA 1/16 j 020 "199 ———— EE Fig C10, 15 Table C10 4 Area a a Part ‘A y Ay? | Ay | To flo+ ay Upper angles - 264 | 3.786} 1.000 | 3.786] 012 3 798 Upper skin - 0375! 4.012) 0, 1504| 0.604) - 9.604 Web 4490! 0 8 1/12 x} 2.310 -072x 7 875 = 2.31 Lower angles | 1780|-3.801/-0.676 | 2.570! 008 3.578 Lower skin „080 |-4.01 |-0.321 | 1.2901 - 1.290 Skin rivet holes |-, 0206|-3 9589| 0.082 |-0.323| - -0.323 Web flange M rivetholes |- 0227/-3 562] 0.081 |-0.288| - -0, 288 SUM 9249 0 316 9.980 ÿ =.318/, 9249 ~.34 là Ạ, 9.98 <.945 x 34” = 8 87 ín.*
Section with 072 web:-
Ty, a, ™ 10,44 (N.A Location Same )
Trang 24
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Fgt = flange compressive stress which
44200 psi Then,
w21.90 x 025 Vv 10,700,000/44200 =
«74 in Since 30t or 0.75 tnches was assumed for
w, the error is quite small and no revision ts necessary
Calculation of Allowable Failing Compressive Stress for Flange
Since the flange is braced tn two
directions at right angles to each other by the skin and the web, column action is prevented and the allowable stress will be the crippling stress of the flange unit
Chapter C7 gives methods of calculating
the crippling stress In this example problem the Needham method will be used and the
crippling stress for a single angle will be calculated, 2s (.75 - 3/64)/3/82 = 7.55 From Fig C7.5 of Chapter C7, we obtain Feg/V Foye = 068 For 2014-Té Ext., Poy = 57000, E, = 10,700,000 Whence Feg = 068 x ¥ 57000x 10,700,000 = 83,000 psi
Since the angle legs which attach to web are riveted together, the crippling stress for these legs would be higher than for the angie legs connecting to the skin, thus the value of 53,000 1s somewhat conservative The margin of safety for the upper flange is (53,000/46,700) - 1 = 14 Check of Bottom Flange Tensile Strengthi- fy = My/I = 127,500 x 4.35/10.44 = 53,000 psi The material for lower skin = 2024-T3
Sluminum alloy, which has an ultimate tensile strength of Fy, = 65,000 psi; and the extruded
angles have a Fey = 60,000 Thus M.S
{60000/83000) - 1 „1ä
AS a practice problem for the student, the ultimate bending resisting moment should be calculated by the method of Article Cl0.4 and compared with the above margins of safety The
result would show a greater margin of safety C10 13
Check of Web Buckling Stress
The maximum shearing stress occurs at the support point and equals 3750 lbs The web thickness at this point 1s 057 and the web stiffener spacing is 3.57 inches The maximum Shearing stress on the web by the simplified equation ts, fg = V/nt = 3750/7.125 x 057 = 9230 psi By the more exact equation, ăn _ 9.87 x 057 +0375 x 3.673 + 0S7 x 3.6 x 1.8) = 9450 psi fy =e / yaa = (264 x 3.446 +
Since the bending moment in the first web panel adjacent to beam end support ts practically
zero, the web can be considered as subjected to
shear stress only
The buckling stress is given by equation C10.1,
gen ®
> = 3.57 inches
a > distance between rivet lines = 7.125 a/> = 7.125/3.57 = 2.0 Assuming simply
Supported edge conditions, we obtain Kg = 6.4 from Fig CS.11 of Chapter ¢5
p= 22%6.4x 10,700,000 (982,
Ser ET - 3) (tụ
This stress 1s below the proportional limit stress of the material so no plasticity
correction is necessary or 7g = 1.0
* = 15,800 pst
The margin of safety against buckling is therefore (15,800/9450) - 1 = 67 This value is somewhat conservative as boundary condition for web panel 1s no doubt larger than for simply Supported, which condition was assumed
Check of 057 Web at End of Bay 1 ee ee a ee OF Bay ie
The web will be more critical at this
point because the beam is still subjected to an external shear load of 3750 lbs., plus a bending
moment of 3750 x 23.22 = 87000 in.lb at the
midpoint of the end panel
Since the web 1s clamped between the flange
angles, buckling of the web will occur adjacent to the lower end of the upper flange angles or
at a distance of 2.91 inches from neutral axis
fy = My/I = 87000 x 2.91/9.87 = 25600
Trang 25
C10, 14 DESIGN OF METAL BEAMS, WEB SHEAR RESISTANT (NON-BUCKLING) TYPE ¬ _ f2 Kp 5 Zya
"oor “ 12 (1 - Ve ) &
bd = 6.5 in taken as distance between
adges of flange angles since web is clamped in between flange angles a = stiffener spacing =
3.57 The a/b = 55 From Fig C5.15 of Chapter G5:~
Kp for simply supported edges = 36 Ky for clamped edges = 50
We will assume an average value of ky = 43
s71) x 4õ x 10,700,000 (.087)s
"dor Te (1 - 3* 3.5) 7 32200 pst
The shear stress at this section will be
same as at support since the external shear
load is the same Thus fg = 9450 and Faor =
15800
The interaction equation for combined pending and snear is, Rp? + Rg* = 1 Ry = fo/Poep Rg = fs/Fsor " 25600/32200 = 795 9450/15800 = 398 L i WS * Tire ng 7 + Vote" + oa" 7 M.S = ,01
Check of 072 Web at Centerline of Beam
The bending moment at the midpoint of the web panel adjacent to the beam centerline is
123,200 in 1b The shear load on the panel is 1350 lb fp = My/I = 123200x2.91/10.44 = 24200 psi 2072 27X43 x 10,700,000 ae Poor Te (r= 3h (6,5) © 54800 pst
This value is in the inelastic stress
range so correction must be made for plasticity ™ne curves of Fig C5.8 of Chapter C5 will be used For 2024-TS sheet, Fạ,; Z 39000 and m=11.5
The value of the lower scale parameter in
Fig 05.8 will be Fy _/Fo,7 = 51,800/59000 =
1.32 Prom Pig 05.8 for n = 11.5 we read Foo /Fo.r = 93 Therefore, Poop m 39000 X 93 = 36300 psi The shear stress is, V 1 #z—/y =———"—./: = 27 fs Tr / ¥oA = Igvary 77g (b-5L) = 2710 psi n*x 6.4.x 10,700,000 072 Poop 7ST SY (B57) | * 25500 pst Rs au 2710/25300 = 107 34200/36300 = 94 1
mS = Type «ae 1 = «Ot
Check of Stiffness of the Web Stiffeners
Rp
The web is stiffened by 1/2 x 1/2 x 04 angles on one side of the web From Equation
(C10.8), the moment of inertia required of web stiffener to prevent web buckling is, = 2.298 ch Bw 4/3 lạp =T£ Ga „ 8:29 X3.B7 , 5750 x 7.875 /s 057 3S x 10,700,000 = ,00050 The moment of inertia of stiffener cross-
section is 00094, thus stiffness is satis~ factory
The required moment of inertia will also be checked using curves in Fig C10.9 The lower scale parameter in Fig C10.9 is dfn ¥ Kg
where kg as used in buckling equation must be
multiplied byn7/l = 3* = 905 Thus, đ/h V kg = 5.57/7.125 v 6.4x 905 = 308 From Fig C10.9 we read 1.1 for Iy/dt* Whence Ty 2 1.1x3.57x 057° = ,000735 (Stiffener O.K.) Check of Rivets Attaching Web to Flange Angles End Bay V = 3750 lb., t = -057 The horizontal shear load q per inch by approximate formula C10.11 is q = V/h' = 3750/8 = 468 lb per inch By the most exact C10.12 equation, ates ydA =-Š E2 (.264x3.446 + 0375 I 9.87 x 3.673) q = 398
The single shear strength of 1/8 diameter rivets made from 2117-T3 aluminum alloy is 388 lbs Since rivets are in double shear and 1 inch spacing, rivet shear strength is 2x 388 = 776 lbs versus the design load of
398 lbs
The bearing strength of a 1/8 diameter rivet on 0S7 web is 790 lbs as against load