ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
*loy and Igy = moment of inertia of each portion
about their own X and Y centrold- al axes Location of centroidal axes:- y = Zay = 9 = 767" ZA x = Zax = = 392" XÃ
Transfer moment of inertia and product of iner- tia from reference X and Y axes centroidal axes:- Ty = 1x ~ ay? HỆ aly ~ Ax? to parallel -955 - 875 x 7677 = 440 = „291 - 875 x 3927= 157 1132 - 875 x 767 x 392 = Ixy = ZAxy — ARF = - 150
Calculate angle between centroidal X and Y¥ axes and prinetpal axes through centroidal as fol- lows :~ tan 2 Ø=2 Ïyy =2(-.159_ =-.30 =1.06 Ĩy-ly ,167-.44Ó -,265 3 Ø=469 ~ 4Q" f= 23° - 20' Calculate moments of inertia about centroidal follows :- Ø+Ïy sin? Ø- 2Ïyy sin Ø cos ố =.44x ,518?+ 157x 39657-2(-.150) x 3965 x 916 = 504 int Ty sin? $+Ty cos* principal axes as Typ =I cos* lyp Ø+2Ïxy sing cos fd 44x ,35965 + 187 x 9167+ 2(-.150) x 3965 x 918 = 092 1n,* Má ; T R — + le y Fig A3.12 Example Problem
Fig AS.13 shows a typical distributed flange - 2 cell - wing beam section The upper
and lower surface its stiffened by 2 and bulb angle sections Determine the moment of inertla af the section about the principal axes
Solution:
The properties of the cross-section depend
upon the effective material which can develop resisting axial stresses The question of ef- fective material is taken up in later chapter
Table 9 shows the calculations for the moment of
inertia about the assumed rectangular reference
axes XX and YY (see Fig A3.15) The cross-
A3,11
section has been broken down into 16 stringers
as listed in column 1 For the top surface, a
width of 30 thicknesses of the 032 skin is as-
sumed to act with the stringers and a width of
25 thicknesses of the 04 skin (see Col 3) On
the lower surface, the skin nalf way to adjacent
stringers is assumed acting with each stringer, or the entire skin is effective Colwnn 4 gives
the combined area of each stringer unit and is
considered as concentrated at the centroid of
Trang 2SECT, NÓ A3.12 Typ =Ĩy sin* Ø+ïy cos* Ø+2 Ĩ„ cos = 186.46 x.2 ~36.41 x ,9595 x NOMINAL DIMENSIONS B T CENTROIDS, Table À3, 10 Properties of Zee Sections SECTION ELEMENTS ix ty | Px | Ow Inches Inches CENTER OF GRAVITY, MOMENTS OF INERTIA A3.14 Section Properties of Typical Aircraft Structural 3ections Table A3.:0
give the section properties of a few structural shapes common to aircraft Use of these tables
Trang 3ANALYSIS AND DESIGN OF FLIGHT VEHICLE th —Y tr Table A3 12
i\ vị: Ver Properties of Extruded Aluminum
Trang 4A3.14 CENTROIDS, CENTER OF GRAVITY, MOMENTS OF INERTIA Table A3 16
SECTION PROPERTIES OF TYPICAL AIRCRAFT EXTRUDED SECTIONS
‘sect.| Dimensions Are: Properties about x-x! Properties about _Y-¥ ị j No, | « | B , ¢ + tị tạ 1 OR in, L_Axis i Axis 4 l
Trang 6A3 LỆ CENTROIDS A3.15 Problems
Fig A3 15
(1) Por the section of Fig A3.14 determine
the moment of inertia about each of the princi- pal axes Xp and Zp-
(2) Calculate the moment of inertia of the
section in Fig A3.15 about the principal axes
Fig Ag 17
(3) Fig AS.16 illustrates a box type beam section with six longitudinal stringers De-
termine the moment of inertia of the beam sec-
tion about the principal axes for the follow~
ing assumptions:—
(a) Assume the beam ts bending upward
putting the top portion in compression and the
lower portion in tension, Therefore, neglect
sheet on the top side since it has very little resistance to compressive stresses The sheet
on the bottom side is effective since it is in tension For simplicity neglect the vertical webs in the calculations
(>) Reverse the conditions in (a) thus placing top side in tension and lower side in
compression
(4) For the three stringer single cell box beam section in Fig A3.17, calculate the mo- ments of inertia about the principal axes As-
sume 411 eb or wall material ineffective o _— L9 CENTER OF GRAVITY, MOMENTS OF INERTIA (a) F calculate
cipal axes assuming only effsctive materia
Fig A3.19
A3.19 snows a wing Seam section
with a cut-out on the lower surface Determine
the moments of inertia about the orin xeS assuming the eight stringers ars the tive material (7} Fig A3.20 shows
flanze beam The 7 flange upper
face of beams nave an area of 3 sq in each
and those on the dottom skin 0.2 sq in each
The bottom skin is 03 inches in thickness Compute the moments of inertia about the
principle axes assuming that the flange members and the bottom skin comprise the effective material
\+————Cutout for door Fig A3.21
(8) Fig A3.21 shows the cross-section of
a Small fuselage The dasned line represents a cut-out in the structure due to a dcor Aã-
sume
Q.1 sq in Consider fuselage skin in,
Calculate the moment of inertia of
section about the principal axes
a
Trang 7
CHAPTER A⁄
GENERAL LOADS ON AIRCRAFT A4.1 introduction
Before the structural design of an airplane
can be made, the external loads acting on the airplane in flight, landing and take-off con- ditions mist be known The complete determin- ation of the air loads on an airplane requires a thorough theoretical knowledge of aerodynamics, since modern aircraft fly in sub-sonic, trans-
sonic and super-sonic speed ranges Further- more, there is a wide range of wing configur-
ations, such as the straight tapered wing, the swept wing and the delta wing, and many of
these wings often include leading and trailing edge devices for promoting better lift or con-
trol characteristics The presence of power plant nacelle units, external fuel tanks, etc
are units that effect the airflow around the wing and thus effect the magnitude and distri- bution of the air forces on the wing Likewise,
the fuselage or airplane body itself influences the airflow over-the wing The theoretical cal-
culation of the airloads on the airplane is too
large a subject to be covered in a structures book and it is customary in college aeronautical curricula to provide 4 separate course for this subject
In most airplane companies the loads on the airplane are determined by a group of en-
gineers assigned to the Structures Analysis
Section and this group is often referred to as the Aircraft Load Calculation group While the work of this group is primarily based on the
use of aerodynamics, it is that phase of aero~ dynamics which is conserved with determining
the magnitude and distribution of the air loads om the airplane so that the airplane structure can be properly designed to support these air forces safely and efficiently The engineering
department of an airplane company has a distinct
or separate aerodynamics section, but in general their responsibility is the use of the subject
of aerodynamics to insure or guarantee the per-
formance, stability and control of the airplane A basic general over-all kowledge of the loads on aircraft is desirable in the study of aircraft structural theory, and hence this chapter attempts to give this information In a later chapter dealing with wing design, this subject will be further expanded
A4.# Limit ar Applied Loads Design Loads,
Because an airplane is designed to carry
out a definite job, there result many types of
aircraft relative to size, configuration and
performance For example, a commercial trans-
port like the Douglas DC-8 is designed to doa job of transporting a cartain number of pass-
engers safely, effictently and comfortably over
various distances between airports, On the other hand the Air Force Fighter type of air- craft has a job of shooting down enemy aircraft
or protecting slower friendly aircraft To do
this job efficiently requires a far different configuration as compared to the DC-8 transport Furthermore the Fighter type airplane must be
maneuvered far more sharply to do its required Job as compared to the DC-8 in doing its re-
quired job
In general the magnitude of the air forces on an airplane depend on the velocity of the
airplane and the rate at which this velocity is
changed in magnitude and direction (acceleration)
The magnitude of the flight acceleration factor
may be governed by the capacity of the human body to withstand these acceleration inertia forces without injury which is the situation in
a fighter type of airplane On the other hand the maneuvering accelerations for the DC-3 are not dictated by what the human body can with- stand, but are determined by what is necessary to safely transport passengers from one airport
to another
Designing the airplane structure for loads greater than the airplanes suffers in the per- formance of its required job, obviously will add considerable weight to the airplane and decrease its performance or over-all efficiency relative to the job it is designed to do
To particularly insure safety in the air-
transportation, along with uniformity and sf- ficiency of design, the government aeronautical
agencies (civil and military) have definite re-
quirements for the various types of aircraft relative to the maguitude of loads to be used in the structural design of aircraft In referring in general to these specified aircraft loads two
terms are used as follows:-
Limit or Applied Loads
The terms limit and applisd refer to the
same loads with the civil agencies (C.A.A.}
using the term limit and the military agencies using the term applied
Limit loads are the maximum loads antici- pated on the airplane during its lifetime of
service
The airplane structure shall be capable cf supporting the limit loads without suffering detrimental permanent deformations At all loads up to the limit loads the deformation of the
Structure shall be such as not to interfere with the safe operation of the airplane
Ultimate or Design Loads
These two terms are used in general to mean
Trang 8
A4.2
the same thing Ultimate or Design Loads are
equal to the limit loads multiplied by a factor of safety (F.S.) or
Design Loads = Limtt or Applied Loads times F.S In general the over-all factor of safety is 1.5 The government requirements also specify that these design loads be carried by the
structure without failure
Although aireraft are
undergo greater loads than the specified limit
loads, @ certain amount of reserve strength against complete structural fatlure of a unit ts
necessary in the design of practically any ma- chine or structure This {s due to many factors
such as:~ (1) The approximations involved in aerodynamic theory and also structural stress analysis theory; (2) Variation in physical
properties of materials; (3) Variation in fab-
rication and inspection standards Possibly the most important reason for the factors of safety for airplanes is due to the fact that practically every airplane is limited to the maximum velocity it can be flown and the maxi- mum acceleration it can be subjected to in flight or landing Since these are under the control of the pilot it is possible in emerg- ency conditions that the limit loads may be
slightly exceeded but with a reserve factor of
safety against failure this exceeding of the limit load should not prove sericus from an airplane safety standpoint, although it might cause permanent structural deformations that might require repair or replacements of small units or portions of the structure
Loads due to airplane gusts, are arbitrary in that the gust velocity is assumed Al- though this gust velocity ts based on years of experience in measuring and recording gust forces in flight all over the world, it is quite possible that during the lifetime of an air-
plane, turbulent conditions near storm areas or
over mountains or water areas might produce air gust velocities slightly greater than that Specified in the load requirements, thus the factor of safety insures safety against failure
if this situation would arise
not supposed to
The broad general category of external
loads on conventional aircraft can be broken down into such classifications as follows:-
Due to Airplane Maneuvers (under the control of the pilot) (1) Air Loads Due to Air Gusts (not under control of pilot) Landing on Land (wheel or ski type)
(2) Landing Loads 4 Landing on water
Arresting (Landing on Air- craft Carriers) GENERAL LOADS ON AIRCRAFT Thrust Poy (3) Power Plant Loads Torque c | A ce Off (4) Take off Loads auxiliary short thrust units Hoisting Airplane Towing Airplane Beaching of Hull type Airplane Fuselage Pressurizing Special Loads
(5) Wetght and Inertia Leads
In resolving external loads for str analysis purposes, it is convenient to
set of reference axes, The reference axes
X¥Z passing through the center of avity of the airplane as illustrated in Fig A4&.0 are those normally used in stress analysis work as well 2s
for aerodynamic calculations For convenience the reference axes are often referred to ancther
origin other than the airpiane 9.7 Fig A4.0
A4.4 Weight and Inertia Forces
The term weight is that constant force, pro- portional to its mass, which tends to draw every Dhysical body toward the center of the earth An airplane in steady flight (uniform velocity)
is acted upon by a system of forces in squilib- rium, namely, the weight of the airplane, ‘he air forces on the complete airplane, and the power plant forces The pilot can change this bal- anced steady flight condition by changing the
engine power or by operating ‘he surface controls
to change the direction of the airplane velocity
These unbalanced forces thus cause the airplane to accelerate or de-accelerate
Inertia Forces For Motion of Pure Translation
of Risid Bodies
# the unbalanced forces acting on a rigid
body cause only a change in the magnitude of the
velocity of the body, but not its direction, the motion is called translation, and from basic Physics, the accelerating Zorce F = Ma, where M is the mass of the body or W/g In Fig A4.1 the unbalanced force system © causes the rigid body to accelerate to the r + ig A¢.2 shows
Trang 9
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES F = unbalanced external force Effective Force =5ma 2 Ma ¿a Maa Motion — —!*|~j Motion ~|~ —* 1 † w Assume T £ Friction tw Zero, mya
Fig A4.1 Fig A4.2
a force on each mass particle of mia, maa, etc.,
thus the total effective force is ima = Ma If these effactive forces are reversed they are re-
ferred to as inertia forces The external
forces and the inertia forces therefore forma force system in equilibrium
Prom basic Physics, we have the following
relationships for a motion of pure translation if the acceleration is constant:- V=Veo =at=-=~=~-= —¬-—_-—-—~~ (1) 8 =Vot +#at7 -+ + ++ (2) ye -vo7 = fas e ee - - ee (3) where, s = distance moved in time t vo = initial velocity
v= final velocity after time t
Inertia Forces on Rotating Rigid Bodies
A common airplane maneuver 1s a motion along a curved path in a plane parallel to the XZ plane of the airplane, and generally referred to as the pitching plane A pull up from steady flight or a pull out from a dive causes an air- plane to follow a curved path Fig A4.3 shows an airplane following a curved path If at point A the velocity is increasing along its
path, the airplane is being subjected to two accelerations, nasely, a,, tangential to the
curve at point A and equal in magnitude to Center of Curvature 3° Flight Path ME w*= Mv#/t Fig A4.3 A4.3
a, = Fa, and a, = Tw" an acceleration normal to r
the flight path at A and directed toward the
center of rotation (0) From Newton’s Law the
affective forces due to these accelerations are:-
FL, = MPo* = tvs xxx nh nem (4)
Tỳ eMPa - + + e757 ee ee (5)
where w = angular velocity at the point A a= angular acceleration at point A r = radius of curvature of flight path
at point A
The inertia forces are equal and opposite
to these effective forces as indicated in Fig
A4.3 These inertia forces can then be con-
Sidered as part of the total force system on the
airplane which ts in equilibrium
If the velocity of the airplane along the path is constant, then a, = Zero and thus the inertia force tỳ = 0, leaving only the normal
inertia force FL
If the angular acceleration is constant, the following relationships hold
O~+ We =Qat - + 5 - ee eee 8) § = 0t + at? - 2 x— (7) @? = @.7 = 200 + +e 2 (3)
where @ = angle of rotation in time t
w = initial angular velocity in rad/sec w = angular velocity after time t
In Fig A4.3 the moment T, of the inertia
forces about the center of rotation (0) equals
Mra(r)= Mr’a The term Mr*is the mass moment
of inertia of the airplane about point (o) Since an airplane has considerable pitching moment of inertia about its own center of gravity axis, it should be included ‘Thus by the
parallel axis
T, = Ioa + I, gt
where I, = Mr? and I, g* moment of inertia of
airplane about Y axts through c.g of airplane Inertia Forces For Pitching Rotation of Airplane
about Y Axis Through c.g airplane,
In flight, an air gust may strike the hori- zontal tail producing a tail force which has 4 moment about the airplane c.g In some landing
conditions the ground ðr water forces do not
pass through the airplane c.g., thus producing a moment about the airplane c.g These moments cause the airplane to rotate about the Y axis through the c.g
Therefore for this effect alone the center
Trang 10À4.4
the c.g of airplane, or F =o Thus F, and Fy
equal zero and thus the only inertia force for the pure rotation is I, 8 a, (a couple} and thus the moment of this inertia couple about the
C8 = TF log
AS explained before if the inertia forces
are included with all other applied forces on the airplane, then the airplane is in static equilibrium and the problem is handled by the static equations for equilibrium
A4.5 Air Forces on Wing
The wing of an airplane carries the major portion of the air forces In level steady flight the vertical upward force of tne air on the wing, practically equals the weight of the
airplane, The term airfoil is used when re-
ferring to the shape of the cross-section of a wing Figs A4.4 and A4.5 tllustrate the air pressure intensity diagram due to an air- Angle of Attack = 120 Fig A4.4 Angle of Attack = - 60 Fig A4.5
stream flowing around an airfoil Shape for both a@ positive and negative angle of attack The shape and intensity of this diagram is in- fluenced by many factors, such as the shape of the airfoil itself, as the thickness to chord Tatio, the camber of the top and bottom sur- faces etc A normal wing is attached to a fuselage and it may support external power plants, wing tip tanks etc, Furthermore the normal wing is usually tapered in planform and thickness and may possess leading and trailing slots and flaps to produce high lift or control effects The airflow around the wing is affected by such factors as listed
above and thus wind tunnel tests are usually necessary to obtain a true picture of the air forces on a wing relative to their chordwise spanwise distribution
Resultant Air Force Center of Pressure
—=—_Sss eee Veber Of rressure
It 1s conventent when dealing with the balancing or equilibrium of the airplane as a whole, to deal with the resultant of the total
GENERAL LOADS ON AIRCRAFT
air forecs on the wing For example, consider the two air pressure intensity diagrams in Figs
À4.6 and A4.7, These distributed force systems
can be replaced by their resultant (R), which of course must be known in magnitude, direction and location The location is specified by a term called the center of pressure which is ‘th
point where the resultant R intersects the air-
foil chord line As the angla of attack is changed the resultant air force changes in mag- nitude, direction and center of pressure location S“ TY, Fig A4.7 Fig A4.6 Lift and Drag Components of Resultant Air Force
Instead of dealing with the resultant force R, it is convenient for both aerodynamic and stress analysis considerations to replace the resultant by its two components perpendicular and parallel to the airstream Fig A4.8 il- lustrates this resolution into lift and drag components AERODYNAMIC | center FLIGHT DINEGTION FLIGHT DIRECTION
Fig A4.8 Fig A4.9
Aerodynamic Center (a.c.) Since an air~
plane flies at many different angles of attack,
it means that the center of pressure changes for
the many flight design conditions It so hap~ Pens, that there is one point on the airfoil that the moment due to the Lift and Drag forces is constant for any angle of attack This point is called the aerodynamic center (a.c.) and its approximate location is at the 25 percent of chord measured from the leading edge Thus the resultant R can be replaced by a lift and drag force at the aerodynamic center plus a wing moment Ma,c, 4S illustrated tn Fig A4.3
A4.6 Forces on Airplane in Flight
Fig A4.10 illustrates in general the main forces on the airplane in an accelerated flight
Trang 11ANALYSIS AND DESIGN OF
Fig A4.10
{T = engine thrust
L total wing lift plus fuselage lift D total airplane drag
Ma = moment of L and D with reference to wing a.c (aerodynamic center)
W = weight of airplane
Tụ = inertia force normal to flight path Tp = imertia force parallel to flight path
ly = rotation inertia moment
EB = tail load normal to flight path
For a horizontal constant velocity flight condition, the inertia forces ts Tp» and ¬ would be zero For an accelerated flight con- dition involving translation but not angular acceleration about its own c.g axis, the inertia moment i would be zero, but TL and Tp would have values
Equations of Equilibrium For Steady Flight From Fig A4.10 we can write:
IF, = 0, D+ W sin OT cos B = 0
mF, = 0, L-W cos oT sin B-E=0
My = 0, -Mg-la- db + Tc cos Bt Be = 0 Equations of Equilibrium in Accelerated Flight
IP, =o, D+ Wsin@-Tcos B-1I) =o
ar, = 0, L-wWeos@+Tsinp-1I,-E=0
MM, +0, - Ms - La - Db + Te cos Pt Ee +1 =o [ Forses - Plus ts up and toward tail
Stens Moment - Clockwise is positive Distances from c.g ta force ~
Plus is up and toward tail 4
A4.7 Load Factors,
The term load factor normally given the
FLIGHT VEHICLE STRUCTURES A4.5
tiplying factor by which the forces on the air- plane in steady flight are multiplied to obtain a static system of forces equivalent to the dy- namic force system acting during the accelera~ tion of the airplane Fig A4.11 illustrates 2 L ˆ total lift (Wing & Tail) TzD Fig A4 forces in sents the Therefore celerated shows the
steady horizontal flight L repre- total airplane lift (wing plus tail) L=W Now assume the airplane is ac- upward along the Z axis Fig A4.12 additional inertia force Na 6 acting
downward, or opposite to the direction of
acceleration The total airplane lift L for the z \ T=D Fig A4.12 { ww got
unaccelerated condition in Fig 44.11 must be
multiplied by a load factor n to produce static
equilibrium in the 2 direction w = Thus, nb -W- 5a, ° Since L = W Hence 1 =1+ 4g % 8
An airplane can of course be accelerated along
the X axis as well as the 4 axis, Thus in
Fig A&@.13 the magnitude of the engine thrust T ts greater than the airplane Drag D, which nạk L_ =s+— D —~ÿarnW T Ww Fig A4.13 T is greater than D Wa, g
causes the airplane to accelerate forward It is convenient to express the inertia force in the X
symbol (n) can be defined as the numerical mul-~
* The bar through letter Z has no significance
ing without bar Same mean-
direction in terms of the load factor ay and the
Trang 12A4.6 welght W of the alrplane, nence wỈ wef 4
aw z ay (See Fig Al4.13)
uF, = O0, whence T~p~n zo
4 _ T-D
Hence n= n
Therefore the loads on the airplane can be dis-
cussed in terns of load factors
limit load facters are the maximum load
that might occur curing the service of the par- ticular airplane These loads as iiscussed in
rt A@.2 must de taken by the airplane struc-
ture without appreciable permanent deformation The design load factors are equal to the limit load factors multiplied by the factor of
safety, and these design loads must be carried
by the structure without rupture or collapse, or in other words, complete failure
A4.8 Design Flight Requirements for Airplane
The Civil and Military Aeronautics Author- ities issue requirements which specify whe
design conditions for the various classification of airplanes Generally speaking, any airplane
flight altitude can be defined by stating the
existing values of load factors (acceleration)
and the airspeed (or more properly the dynamic
pressure)
The accelerations on airplane are pro- duced from two causes, namely, maneuvers and air
gusts The accelerations due to maneuvers are subject to the control of the pilot who can manipulate the controls so as not to exceed 4 certain acceleration In highly maneuverable military airplanes, an accelerometer is in- cluded in the cockpit instruments as a guide to
limit the acceleration factor For commercial
airplanes the maneuver factors are made high enough to safely take care of any maneuvers that would be required in the necessary flight opera- tions of the particular type of airplane These limiting maneuver factors are based on years of operating experience and have given satisfactory results from a safety standpoint without pen alizing the airplane from a weight design con-
sideration
The acceleraticns due to the airplane
striking an air gust are not under the control of the pilot since it depends on the direction and velocity of the air gust From much ac- cumulated data obtained by instaliing accelero—
meters in commercial and military aircraft and
flying them in all types of weather and leca-
tions, it has been found that a gust velocity of
30 ft per second appears sufficient
The speed or velocity of the airplane like-
wise effects the loads on the airplene The
higher the velocity the nigher the aerodynamic wing moment Furthermore the gust acceler- ations increase with airplane velocity, thus it
is customary to limit the particular airplane to the GENERAL LOADS ON AIRCRAFT a definite maximum mercial airplanes ‘t reasonabis zlide spee
take care of reasonasle fligh A4.9 Gust Load Factors
when a sharp ede
in a direction normal to the thrust Line (X axis), a sudden change taxes place in the wing
angle of attack with no sudden change in 2iro speed The normal force coefficient (C, ) can
ơ @ assumed to vary linearly with the 2: attack Thus in F fa), let point ( sant the normal airplane force 2€ €e
necessary to maintain level flight w velocity Vand point (Z) the value c a sharp edged sugt KU has caused 4a 8 den change
4a in the angle of attack without change in ¥ The total incrsase in the airplane load in the Z
direction can therefore se expressed 2y the ratio C, at B
a4 A
From Fig (b} for small angles, Aa = KU/V and from Fig (a), acy =m da, where m = the A Slope of the airolane normal force curve (c, A per radian) v Fig b Fig a
The load factor increment due to the gust KU can then be expressed (728) KUVS 2N where
U = gust velocity in ft./sec
gust correction factor depending on wing loading (Curves for K are provided
by Civil Aeronautics Authorities)
ÿ = ‘ndicated air speed in miles per hour, 8 = wing area in sq ?t
W = gross weight of airplane
Trang 13
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
If U is taken as 30 ft./sec and m as the
changs in tạ with respect to angle of attack A
in absolute units per degree, equation (4) re- duces to the following
Therefore the gust load factor n when air- plane is flying in horizontal altitude equals 3KmV 1+ ee ee ee ee ee lL ee nade M8 2 (c) and when airplane {s in a vertical altitude 3KmV ne=+ == -++-+ -+ (D) ~ W/S
A4.10 Iustration of Main Flight Conditions Velocity-Load Factor Diagram
As indicated before the main design flight
conditions for an airplane can be given dy
stating the limiting values of the acceleration
and speed and in addition the maximum value of
the applied gust velocity As an 111ustration, the design loading requirements for a certain
airplane could be stated as follows: "The
proposed airplane shall be designed for applied positive and negative accelerations of + §.0g and -3.5g respectively at all speeds from that corresponding to hoax up to 1.4 times the maximum level flight speed Furthermore, the airplane shall withstand any applied loads due
to 2 30 2t./sec gust acting in any direction
up to the restricted speed of 1.4 times the maximum level flight speed A design factor of safety of 1.5 shall be used on these applied
Loads”
In graphical form these design require~
ments can be represented by plotting load fac-
tor and velocity to obtain a diagram which is
generally referred to as the Velocity-accelera- tion diagram The results of the above speci- fication would be simtlar to that of Pig A4.i4 Thus, the lines AB and CD represent the re-
stricted positive and negative maneuver load
factors which are Limited to speeds inside line
BD which is taken as 1.4 times the maximum
level flight speed in this illustration These
restricted maneuver lines are terminated at
points A and ¢ by their intersection with the maximum c values of the airplane At speeds between A and B, the pilot must be careful not
to exceed the maneuver accelerations, since in
general, it would be possible for him to man-
ipulate the controls to exceed these values
At speeds below A and C, there need be no care
of the pilot as far as loads on the airplane are concerned since a maneuver producing C,
max
A4.7 would give an acceleration less than the lim~
ited values given by lines AB and CD
The positive and negative gust accelera- tions due to a 30 ft./sec gust normal to flizht path are shown on Fig a4.14 In this example
diagram, a positive gust is not critical within
the restricted velocity of the airplane since the gust lines intersect the iine BD below the line AB For a negative gust, the sust load
factor becomes critical at velocities between F and 2 with a maximum acceleration as given by
point E
For airplanes which have a relatively low required maneuver factor the zust accelerations may be critical for both positive and negative accelerations xamination of the cust equation indicates that the most lightly loaded condition (smallest gross weight) produces the highest gust lead factor, thus tnvolving only partial pay load, fuel, etc
On the diagram, the points A and 8 corre- spond in general to what is referred to as high angle of attack (H,A,A,) and low angle of attack
(L.A.A.) respectively, and points C and D the
inverted (H.A.A.) and (L.A.A.+ conditions re-
spectively
Generally speaking, if the airplane is de- Signed for the air loads produced by the veloc-
ity and acceleration conditions at points A, 8,
£, F, and C, it should be safe from a structural strength standpoint if flown within the specified limits regarding velocity and acceleration
Basically, the flight condition require-
ments of the Civil Aeronautics Authority, army,
and Navy are based on consideration of specified
velocities and accelerations and a consideration
of gusts, thus a student understanding the basic
discussion above should have no difficulty un- derstanding the design requirements of these
three government agencies
For stress analysis purposes, all speeds
are expressed as indicated air speeds The
“indicated” air speed is defined as the speed
which would be indicated by a perfect air-speed
indicator, that is, one that would indicate
true air speed at sea level under standard at- mospheric conditions The relation between the actual air speed V, and the indicated air speed
Y, ts given by the equation = [P vy Va Va Po where V, = indicated airspeed V, = actual airspeed
standard air density at sea level
đensity of air in which V, is attained
Trang 14A4.8 n = Acceleration in Terme of : ig Ad 14
44.11 Special Flight Design Conditions
There are many other flight conditions ‘which may be critical for certain portions of
the wing or fuselage structure Most airplanes are equipped with flaps, to decrease the land-
ing speed and such flaps are lowered at speeds
at least twice that of the minimm landing
speed Since the flapped airfoil has different values for the magnitude and location of the airfoil characteristics, the wing structure must be checked for all possible flap conditions within the specified requirement relative to maximum speed at which the flaps may be oper- ated Generally speaking, the flap conditions Willi effect only the wing portion inboard of the
flap and it is usually only critical for the rear beam web or shear wall and for the top and
bottom walls of the torsion box This is due to
the fact that the deflection flap moves the
center of pressure considerably aft thus pro-
ducing more shear load on the rear shear wall as
well as torsional moment on the conventional cantilever box metal beam
The airplane must likewise be investigated
for aileron conditions Operation of the ailer- oms produce a different air lcad on each side
of the airplane wing which produces an angular rolling acceleration of the airplane Further- more, the deflected ailerons change the mag~
nitude and location of the airfoil character~ istics, thus calculations must be carried out to determine whether the loads in the aileron con- ditions are more critical than those for the
normal flight conditions
For angular acceleration resulting from pitching moments due to air gusts on the tail,
GENERAL LOADS ON AIRCRAFT
the loads on the wing should de checked for cases where the engines are attached to tne
and are located Zorward of the leading edge
In cases where the landing gear is attached to wing or when the ?uel and engines are carried
ih and on the wing, the loads produced on the
wing structure in a landing condition may de eritical for some portions of the wing structure inboard of landing gear and engine attachment
points
wing
A4.12 Example Problems Involving Accelerated Motion of Rigid Airplane
AS previously explained, it is general practice to place the airplane under accelerated conditions of motion into a condition of static
equili>rium by adding the inertia forces to the applied force system acting on the airplane It 1s usually assumed that the airplane is a rigid body Several example problems will be pre- sented to illustrate this general procedure Example Problem 1
Fig A4.15 illustrates an airplane landing on 4 Navy aircraft carrier and being arrested by a cable pull T on the airplane arresting nook, If the airplane weight is 12,000 lbs and the airplane is given a constant acceleration of 3.5¢
(112.7 ft/sec?), find the hook pull T, the wheel
reaction 2, and the distance (4) between the line of action of the hook pull and the airplane c.g If the landing velocity {s 60 M.P.H what is the stopping distance
‘W = 12000 Ib
Solution: -
On contact of the airplane with the arrest- ing cable, the airplane is decelerated to the right relative to Fig A4.15 The motion is pure
translation horizontally The inertia force is
wa, (12000 & g
The inertia force acts opposite to the direction of acceleration, hence to the left as shown in Fig A&.15,
The unknown forces T and R can now be solved
for by using the static equations of equilibrium
Ma = } 3.5g = 42000 1b,
IF, = ~42000 + T cos 10° = 0 hence, T = 42700 1b
Trang 15ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES hence, R = 19420 1b To find the distance (d) take moments about ¢.g of airplane, Me = 19420 x 24 - 42700 d = 0 hence, d 210.9 in Landing velocity Vo = 60 M.P.H = 88 ft/sec V* = Vo* = gas Subt: - 9° = 887 = 2(+112.7) 3 hence stopping distance s = 34.4 ft Example Problem 2
An airplane equipped with float is cata- pulted into the air from a Navy Cruiser as 11~ lustrated in Fig, A4.16 The catapulting force P gives the airplane a constant horizontal ac-
celeration of 3¢(96.6 ft/sec*), The gross
weight of airplane 9000 lb and the catapult track is 35 ft long Find the catapulting
force P and the reactions Ri and Ra from the
catapult car The engine thrust {s 900 lb What is airplane velocity at end of track run? 900 Bf “ls ‘Thrust _Line a Ma 581-177 z Fig Ad 16 Solution: -
The forces will be determined just after the beginning of the catapult run, where the car velocity {is small, and thus the lift on the
airplane wing and the airplane drag can be
neglected
Horizontal inertia force acting toward the air~ Plane tail equals, 9000 ( z) Ma = 3.0g = 27000 1b, From statics: - ba = -900 - P + 27000 = 0, henca P = 26100 lb To find R, take moments about point A; U1, = 9000 x 55 + 27000 x 78 - 900 x 83 ~ 65Re =0 A4, 8 hence, Ra = 29800 1b (up) aF, = 29800 - 9000 + Ri = 0 hence, Ri = ~ 20800 1b (acting down)
The velocity at end of
be found from the following catapult track can equation v* -V,* = gas Y* - o=2x 96.6 x25 or Vv = 82 ft/sec = 56 M.P.H Example Problem 3
Assume that the transport airplane as il- lustrated in Fig A4.17 has just touched down in landing and that a braking force of 35000 lb on the rear wheels is being applied to bring the airplane to rest The landing horizontal veloc-
ity is 85 M.P.H (125 ft/sec} Neglecting air
forces on the airplane and assuming the propeller forces are zero, what are the ground reactions R, and Rg What is the landing run distance with the constant braking forcs? W = 100, 000 lb, Ll ( "OL ^ 35000 h—— 38 —4 Ri Fig A4.17 Solution: -
The airplane is being decelerated horizon-— tally hence the inertia force through the air-
plane c.g acts toward the front of the airplane
Trang 16A4 10 To find Re take moments about point (A) mM, = 100,000 x 21 ~- 35000 x 9 + GB Re = 0 Ra = 47000 1b (2 wheels) IF, = 47000 - 100,000 + Ri = 0 Ri = 53000 lb Example Problem 4
The airplane in Fig A4.18 weighs 14,000 lbJ It is flying horizontally at a velocity of 500
M.P.H (752 ft/sec) when the pilot pulls it up-
ward into a curved path with a radius of curva~
ture of 2500 ft Assume the engine thrust and
airplanes drag equal, opposite and colinear with
each other (not shown on Fig A4.18) Find: - (a) (d) (e) Acceleration of airplane ín Z dí~ rection
Wing Lift (L) and Tail (T) forces
Airplane Load factor May Engine Thrust tex Fig A4 18 Solution: ~ a _Ý*°_ 7% Acceleration 38: “=” 00 or 214.5/32.2 = 6.67g (upward)
The inertia force normal to the flight path and acting down equals
= 214.5 ft/sec?
Ma, = (299) 6.67g = 93700 1b,
Placing this force on the airplane through the
c.g promotes static equilibrimm, hence to find
tail load T takes moments about wing aerody- namic center (c.p.) mM, p hence - (14000 + 93700) 8 + 210 T = 0 T = 4100 1b (down) To find Wing Lift (L) use IF, = = 4100 ~ 14000 - 93700 + L = 0 L = 111800 lb GENERAL LOADS ON AIRCRAFT i Airplane Load Factor = Airplane Lit 111800 - 4100 =——"la000 7? Example Problem 5
Assume the airplane as used in example problem 4 is in the same attitude as used in
that axample problem Now the airplane is further maneuvered by the pilot suddenly push- ing the control stick forward so as to give the airplane a pitching acceleration of 4 rad/secZ,
(a) Pind the
assuming change
(b) Find the forces on the jet engine which weighs 1500 1b, and whose ¢.g location is
shown in Fig A4.19
Assume moment of inertia ly (pitching) of the
airplane squals 300,000 1b sec?, in
inertia forces and the tail load T, the lift force on the wing does not 98700 1p 111, 800 Ib m—" 2101 T a Fig A4.19 Solution: -
Fig A4.19 shows a free body of the air- plane with the lift and inertia forces as found in Problem 4
The additional inertia force due to the angular acceleration a = 4 rad/sec* equals,
Ta = 300000 x 4 = 1,200,000 in ib which acts clockwise or counter to the direction
of angular accaleration
Trang 17ANALYSIS AND DESIGN OF henca; ase -ò {99200 ng g8 = 7.1 g8 ft/sec* - a The c.g of the engine 1s 50 inches aft of the airplane c.g as shown in Fig A4.19, The force on the engine will be its own weight of 1500 1b., and the inertia forces due to a, and a Inertia force due to a, equals, ta, = (3°) 7.1g = 10650 1b Inertia force due to angular acceleration a equals, = 1500 + -
Mra = asx * 50 x 4 = 778 1b (down)
Then the resultant force on the engine equals 1500 + 10630 + 778 = 12908 1b (down) Note if the engine had been forward of the air-
plane ¢.g., the inertia force of 778 1b would act upward instead of downward
In calculating the inertia force on 4 certain airplane item due to angular acceler- ation, the equation F = Mra assumes that the particular {tem nad negligible mass moment of
inertia about {ts own centroidal Y axis In
the case of a large item this centroidal mass moment of inertia may be appreciable and should be included in the ly of airplane
Then to find the inertia force for such
an item the equation F = Mra should be modified
to be
F= (I C.ke a)/r where r = distance or arm
of item from airplane ¢.g to c.g
1 cee = mass moment of inertia of item about
airplane c.g equals I, + Mr*
where I, 1S mass moment of inertia of item about its own centroidal Y axis
F = inertia force in lbs normal to radius r Example Problem 6
Fig A4.20 Shows a large transport air- plane whose gross weight is 100,000 ib The
airplane pitching mass moment of inertia ly = 40,000,000 lb sec? in
The airplane is making a level landing
with nose wheel sligntly off ground The re-
action on the rear wheels is 319,000 1b in- clined at such an angle to give a drag com- ponent of 100,000 1b and a vertical component of 300,000 1b FLIGHT VEHICLE STRUCTURES A411 Find: (a) The inertia forces on the air- plane
(>) The resultant load on the pilot whose weight {s 180 lb and whose
location 1s shown in Fig A4.20 mm 319000 Ib Fig A4.20 Solution: - The wing lift will be example problem
The inertia forces on forces Ma, and ‘a, and the
Trang 18A4.12 GENERAL LOADS ON AIRCRAFT Calculations of resultant load on pilot: - š = 572" | Ma c.g of pilot — we SP x a, = 2.08 a= le x Ts a = c.g air- = 0.85 plane % = 180° Fig A421
Fig 44.21 shows the airplane c.g accelerations The forces on the pilot consist of the Pilots weight of 1980 lb and the various inertia forces as indicated in the figure
= (280 = tact
ta, = (4°) 1s = 120
= {180 - *
Ma, = CC) 2.0g = 360
The inertia force due to the angular ac~ celeration a acts normal to the radius arm between the airplane c.g and the pilot For convenience this normal force will be replaced by its s and x components > 180 FP, = Maa “ cv 12% 40 X (0.95 = 17 Ib yw eM = 180 - Pa cia = ge ie ® 372 x 0.95 = lel 1b, Total forces in x direction on pilot equals 180 ~ 17 = 163 1b Total force in g direction = 360 + 180 - 161 = 379 lb Hence Resultant force R, equals v 379" + 16a" = 410*
44.13 Effect of Airplane Not Being a Rigid Body The example problems of Art A4.12 as—
sume that the airplane is a rigid body (suffers no structural deformation} On the basis of
this assumption the applied loads on the air-~ plane in either flight or landing conditions are placed in equilibrium with the tnertia forces which occur due to the acceleration of the airplane It is obvious that an airplane structure like any other structure is not a Tigid body, particularly a cantilever wing which undergoes rather large bending deflections in both flight and landing conditions Figure A4&.21 shows a composite photograph taken of a test wing for the Boeing B-47 airplane The maximum upward and downward deflections shown
are for design loads, which in general are 1.5 times ths applied loads It would not be correct
to say-that the wing deflections under the ap-
plied loads for these two High Angle of attack
conditions would be 2/3 the deflections shown in the photograph since under the design loads a
considerable portion of the wing would oe stressed beyond the elastic limit of the material or into the plastic range where the stiffness modulus is
OEFLECTED
POSTION
Pig A4, 21
considerably less than the modulus of elasti- city, hence the deflections under the applied
loads would be somewhat less than 2/3 those
shown in the photograph This photograph thus
indicates very strikingly that a wing structure
is far from being a rigid body -
Static loads are loads which are gradually applied and cause no appreciable shock or vi~ bration of structure On high speed aircraft, air gusts, flight maneuvers and landing re~ actions are applied quite rapidly and thus can be classed as dynamic loads Therefore when these dynamic loads strike a flexible (non- rigid) airplane cantilever wing, a rather large wing deflection 1s produced and the wing tends
to vibrate This vibration therefore causes additional accelerations of the mass units of the wing which means additional inertia forces
on the wing Furthermore if the time rate of application of the external applied forces approaches the natural bending frequencies of the wing, the vibration excited can produce
Trang 19ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Up until World War II practically all air-
planes were assumed as rigid bodies for struc- tural design purposes During the war failure
of aircraft occured under load conditions which the conventional design procedure based on rigid
body analysis, indicated satisfactory or safe
stresses The failures were no doubt due to
dynamic overstress because the airplane is not a rigid body
Furthermore, airplane design progress has resulted in thin wings and relatively large wing spans, and in many cases these wings carry
concentrated masses, such as, power plants,
bombs, Wing tip fuel tanks etc, Thus the flexibility of wings have increased which means the natural bending frequencies have decreased This fact together with the fact that airplane
speeds have greatly increased and thus cause
air gust loads to be applied more rapidly, or the loading is becoming more dynamic in char-
acter and thus the overall load effect on the
wing structure is appreciable and cannot be
neglected in the strength design of the wing General Dynamic Effect of Air Forces on
Wing Loads
- The critical airloads on an airplane are caused by maneuvering the airplane by the pilot or in striking a transverse air gust A trans-
port airplane does not have to be designed for
sharp maneuvers producing high airplane accel~ erations in its job of transporting passengers, thus the time of applying the maneuver loads is considerably more than a fighter type airplane pulling up sharply from high speeds
Fig A&.22 shows the result of a pull-up maneuver on the Douglas D.C,3 airplane at 180
M.P.H relative to ioad factor versus time of
application of load As indicated the peak
load of load factor 3.25 was obtained at the end of one second of time Fig A4 22 © hà Q Load Factor -5 1 Pull-up of DC-3 Airplane at 180 mph 15 2
The author estimates the natural trequency of the D.c.3 wing to be around 10 to 15 cycles per second, thus a loading time of 1 second against
a time of 1/10 or 1/15 for half a wing deflec- tion cycle indicates that dynamic overstress
Should not be appreciable In general, it can
be sald that dynamic over-stress under maneu- vering loads on transport airplanes is not as great as from other conditions such as air
gusts or landing
Dynamic Effect ‘of Air Gusts
The higher the air gust velocity and the
higher the airplane velocity, the less the time
A4 13 for applying the load on the wing when striking the air gust
NACA Technical Note 2424 reports the flight test results on a twin-engine Martin transport airplane Strain gages were placed at various points on the wing structure, and strains were read, for various gust conditions for which the
normal airplane accelerations were also recorded
Then slow pull-up maneuvers were run to give similar airplane normal accelerations The wing had a natural frequency of 3.9 cps and the air-
plane speed was 250 M.P.H Two of the con-
clusions given in this report are: - (1) The bending strains per unit normal acceleration under air gusts were approximately 20 percent higher than those of slow pull-ups for all mea- suring positions and flight conditions of the
tests, and (2) The dynamic component of the wing
pending strains appeared to be due primarily to excitation of the fundamental wing bending mode
These results thus indicate that air gusts apply a air load more rapidly to a wing than a maneuver load giving the same airplane normai acceleration for a commercial transport type of airplane, and thus the dynamic strain effect on
the wing is more pronounced for gust conditions
Figs A4,23, 24 and 25 show results of dy- namic effect of air gusts on a large wing as de-
termined by Bisplinghoff* The results in these
figures show that dynamic effects tend to con- siderably increase wing forces on some portions of the wing and decrease it on other portions Fig A4 23 Comparative Shear Distribution ——— Dynamic Analysis —~+-— Rigid Airplane Analysis 1.0 Fig A4.24 Comparative Bending Moment Distribution B Moment Fig Ad 25 Comparative Distri- bution of Torque Abou Elastic Axis ‘ing Tip = General Data:- Wing Span = 189 (t, Gross Wt = 184000 Ib Airlane Vel =260 mph 8 2 4 6 8 10 Fraction of Semi-Span
Trang 20A4.14 GENERAL LOADS It has also been found that landing loads
applied through the conventional landing gear or
by water pressure on a flying boat are applied rapid enough to be classed 4s dynamic loads and such loads applied to wings of large span pro~
duce dynamic stresses which cannot ce neglected
in the safe design of such structures A4,14 General Conclusions on Influence of Dynamic
Loading on Structural Design of ‘Airplane
The advent of the turbo-Jet and the rocket
type engines has opened up a range of possible airplane airspeeds hardly dreamed of only a few
years ago, and already trans-sonic and super-
sonic speed airplanes are a common development
From an aerodynamic standpoint such speeds have dictated a thin airfoil section which has thus
promoted a high density wing Thus for air- planes with appreciable wing spans like M1li- tary bombers and near future jet commercial transports, which usually carry large concen- trated masses on the wing such as engines, fuel tanks etc., the assumption that the airplane is a rigid body is not sufficiently accurate enough because the dynamic stresses are appreciable
The calculation of the dynamic loading om the wing requires that the mass and stiffness distribution of the wing structure be known
Since these factors are not known when the
structural design of a wing is started, the general procedure in design would be to first base the design on the assumption that the wing is a rigid body plus correction factors based on
past design experience or available research in- formation to approximately take care of the in-
fluence of the elastic wing on the airplane aerodynamic characteristics and the build up dynamic inertia forces With the wing thus in- itidally designed by this procedure, it then can be checked by a complete dynamic analysis and modified as the results dictate and then re- calculated for the modified elastic wing This procedure is now practical because of the avail-
ability of high speed computors
A4.15 PROBLEMS
(1) The airplane in Fig A4.26 is being
launched from the deck of an aircraft carrier by
the cable pull T which gives the airplane a for- ward acceleration of 2.25g The gross weight of the airplane is 15,000 lb
(a) Find the tension load T in the launching!
cable, and the wheel reactions Ri and Ra«
(b) If the flying speed is 75 M.P.H., what
launching distance is required and the
launching time t?
(2) Assume the airplane of Fig A4.26 1s
landing at 75 M.P.H on a rumway and brakes are applied to the rear wheels squal to ,4 of the
vertical rear wheel reaction What is the nort- ON AIRCRAFT W = 15000 Ib — 90" —4 Fig Ad, 26
zontal deceleration and the stopping distance for the airplane?
(3) The flying patrol boat in Fig 44.27
makes a water landing with the resultant bottom water pressure of 250,000 lb as shown in the figure Assume lift and tail loads as shown The pitching moment of inertia of the airplane
is 10 million lb sec.* in Determine the air-
plane pitching acceleration what is the total load on the crew member who weighs 200 1b and
is located tn a seat at the rear end of the null? 1000 1b Pig Ad 27 26000 Ib
(+), The jet-plane in Fig A4.28 1s diving at
a speed of 600 M.P.H when pilot starts a 8g pull-out Weight of airplane is 16,000 lb Assume that engine thrust and total airplane
drag are equal, opposite and colinear
(a) Find radius of flight path at start of
pull-out
(b) Find inertia force in Z direction
{c) Find lift L and tail load T
Fig A4 28
Trang 21CHAPTER A5
BEAMS - SHEAR AND MOMENTS
A5.1 Introduction
In general, a structural member that sup- ports loads perpendicular to its longitudional axis is referred to as a beam The structure of
aircraft provides excellent examples of beam units, such as the wing and fuselage Very
seldom do bending forces act alone on a major
aircraft structural unit, but are accompanied by axial and torsional forces However, the bend-
ing forces and the resulting beam stresses due to bending of the beam are usually of primary importance in the design of the beam structure A5.2 Staticaliy Determinate and Statically Indeterminate
Beams
A beam can be considered as subjected to known applied loads and unknown supporting re- actions If the distribution of the applied
known loads to the supporting reactions can de
determined from the conditions of static equil-
ibrium alone, namely, the summation of forces and moments equal zero, then the beam is con- sidered as a statically determinate beam, How~
ever, if the distribution of the known applied
leads to the supporting beam reactions is in- fluenced by the behavior of the beam material during the loading, then the supporting reactions cannot be found by the statical equilibrium
equations alone, and the beam ts classified as a statically indeterminate beam To solve such 4
beam, other conditions of fact based on the
team deformations must be used in combination with the static equilibrium equations
AS.3 Shear and Bending Moment
A given beam is subjected to a certain ap~
plied known loading The beam reactions to hold
the beam in static equilibrium are then calcu- lated by the necessary equations of static equi- librium, namely: -
mV = 0, or the algebraic summation of all verti-
cal forces equal zero
SH = 0, or the algebraic summation of ail hort-
zontal forces equal zero
0, or the algebraic summation of all the
moments equal zero
With the entire beam in static equilibrium, it follows that every portion of She beam must
likewise be in static equilibrium Now consider
the beam in Fig AS.1 The known applied load of P = 100 lb ts held in equilibrium by the two
reactions of 25 and 75 lbs as shown and are calculated from simple statics (Beam weight ts
neglected in this problem) Now consider the
beam as cut at section a-a and consider the A5.1 P = 100 Ib 30" L— tê vy PTT 7 4 ‘ 25 a E—18"—*hg ip, Fig A5.1 , 100 = P ks + 10 —3 a 8 a Pị | ig A8 2 a Fig A5.3 a 15 T8 100 =1 100 = P a Pf al 5 2 c¬ s @ lp cf c| _ 7 iJ 2 L ts R 15
Fig A5.4 Fig A5.5
right side portion as a free body in equilibrium as shown in Fig A5.2 For static equilibrium,
SV, SH and IM must equal zero for all forces and
moments acting on this beam portion Consider- ing 2V = 0 in Fig A5.3: -
5V = 78 - 100 5= - 25 1b, - (2) thus, under the forces shown, the force system is unbalanced in the V direction, and therefore an internal resisting force Vy equal to 25 lb must have existed on section a-a to produce
equilibrium of forces in the V direction AS.3 shows the resisting shear force, Vy =
25 lb, which must exist for equilibrium, Considering IM = 0 in Fig 45.3, take
moments about some point O on section a-a, Fig
Mo = - 75 x 15 + 100 x 5 = - $25 in.1b{2) or an unbalanced moment of ~ 625 tends to ro- tate the portion of the beam about section e-a,
A counteracting resisting moment M = 625 must exist on section a-a to provide equiltorium Fig AS.4 shows the free Dody with the Vy and
My acting
Now ZH must equal zero The external
forces as well as the internal resisting shear vy have no norizontal components Therefore, the internal forces producing the resisting
Trang 22A5.2
unbalanced force, which means that the resisting
moment My in the form of a couple, as shown in Fig A5.5, or My = Cd or Td and T must equal C to make 2H = 0
The tendency of the loads and reactions acting on a beam to shear or move one portion of a beam up or down relative to the adjacent por-
tion of the beam is called the External Vertical
Shear, or commonly referred to as the beam Vert- ical Shear and is represented by the term V
From equation (1), the Vertical Shear at
any section of a beam can be defined as the al- gebraic sum of all the forces and reactions acting to one side of the section at which the Shear is desired If the portion of the beam to the left of the saction tends to move up rela- tive to the right portion, the sign of the Vertical Shear is taken as positive shear and negative if the tendency is opposite Or in other words, if the algebraic sum of the forces is up on the left or down on the right side, then the Vertical Shear is positive, and nega= tive for down on the left and up on the right, From equation (2), the Bending Moment at
any section of a beam can he defined as the al-
gebraic sum of the moments of ail the forces acting to either side of the section about the Section, 1f this bending moment tends to pro- duce compression (shortening} of the upper fib-
ers and tension (stretching) of the lower’ fibers’ of the beam, the bending moment is classed as a positive bending moment, and negative for the
reverse condition
A5.4 Shear and Moment Diagrams
In aircraft design, a large proportion of
the beams are tapered in depth and section, and also carry 4 variable distributed load Thus,
to design or check the various sections of such
beams, it is necessary to have a complete pic-
ture as to the value of the vertical shear and
bending moment at all sections along the beam If these values are plotted as ordinates from a base line, the resulting curves are referred to
as Shear and Moment diagrams A few example
Shear and Moment diagrams will be plotted, to refresn the students knowledge regarding these
diagrams
Example Problem 1
Draw a shear and bending moment diagram for the beam shown in Fig A5.6 Neglect the weight of the beam In general, the first step is to determine the reactions To find Rp, take moments about point A my = - 4 x 500 + 1000 x 5 + 300 x 13 - 10Rg = 0 hence Rg = 690 lb EV = = 500 + Ra - 1000 - 300 + 6905 0 hence Ra = 1110 lb BEAMS SHEAR AND MOMENTS 500 Ib 1000 lb 309 10, a 2,3 als 6,7 at T 212 ai SỊT 8| “ mg 3 it — RẠ = 1110 Ib, Rgl= 680 Ib Flg AB.6
Calculations for Shear Diagram: -
We start at the left end of the beam Considering a section just to the right of the
500 lb load, or section 1-1, and considering the portion to the left of the section, the
Vertical Shear at 1-1 = IV =~500 (negative, down on left.} l +610 Ib +610 lb +300 Ib, = 300 Ib 2 l
- 500 Ib = 500 Ib ~ $90 Tb ~ $90 Ib
Fig AS 7 (Shear Diagram) 1050 in Ib -900 in Ib ~2000 in lb, Fig A5.8 (Bending Moment Diagram) Next, consider section 2-2, just to left of reaction Ra ZV = ~ S00, or same as at section 1-1 Next, consider section 3-3, just to right of Ra 3Ÿ = - 500 + 1110 = 610 (positive, up on
left side of saction)
Next, consider section 4-4, just to left of 1000 load ZV = - SOO + 1110 = 610 (same as at section 33) , Section 5-5, to right of 1000 load: EV = + 500 + 1110 - 1000 = ~ 390 (down on left) -
Check this shear at section 5-5 by using the portion of the beam to the right of 5-5 as a free body
ZV = ~ 300 + 690 = 390, which checks (sign of shear is minus, because ZV is up on
Trang 23ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
Section 6-8:
IV = - 300 (positive shear)
Fig AS.7 shows the plotted values on the shear
diagram
Calculation of the Moment Diagram
Start at section 1-1, and consider the forces to the left only:
m=-s500x0=0
Since sections 2-2 and 3-3 are only a differen- tial distance apart, assume a section just above
Ra and consider the forces on the left side only:
IM = - 500 x 4 = - 2000 In lb (Negative moment, because of tension in the top fibers)
Consider the section under the 1000 in 1b, load: aM to left = ~ 600 x 9 + 1110 x 5 = 1050 in lb (positive moment, compressing the top fibers) Check by considering the forces to the right: IM right = 300 x 8 - 690 x 5 = = 1050 in.1b Next, consider a section over Ry: aM right = 300 x 3 = 900 in 1b, moment, tension in top fibers} At Section 8-8: mM right = 300x050
Fig AS.& shows the plotted values
From the above results it may be noticed that nhen the bending moment is obtained from the forces that lie to the left of any section, the bending moment is positive when it is clock- wise If obtained from the forces to the right,
it is positive, when counter-clockwise The student should sketch in the approximate shape
of the deflected structure and determine the signs from whether tension or compression exists in the upper and lower fibers
(Negative
Example Problem 2
Calculate and draw the shear and moment
diagrams for the beam and loading as shown in Fig AS5.9 First, determine the reactions, Ra and Rg: - IM, = 36 x 10 x 18 + 120 x 9 ~ đ6Rp = O, hence Rg = 210 1b 5V = - 120 - 36 x 10 + 210 + Ry = 0, hence Rg = 270 lb Shear Diagram: -
The vertical shear just to the right of the reaction at A is equal to 270 up, or positive, This is plotted as line AE in Fig AS.10 The vertical shear at section C just to the left of the load and considering the forces to the left cof the section = 270 - 9 x 10 = 180 1b up, or positive The vertical shear for any section between A and C at a distance x from A is:
A5.3 Vy = 270 - 10x, and hence, the shear de-
creases at aconstant rate of 10 1lb./in from 270
at A to 180 atc
The vertical shear at section D, just to the right of load is, Vp = BVyery = 270 - 10 x 9 - 120 = 60 up, or positive 120#=P or a a w= 10 lb /in c ape, lam 3 SA AS,g Rg=210 270 Ib BAe Moment Diagram
The vertical shear between points D and B, when x ig the distance of any section between D and
B from A:
Vpp = 270 - 120 = 10x
At point B, x = 36:
hence Vg = 270 - 120 - 10 x 56 = - 210 1b., which checks the reaction Rp
Since the Vertical Shear decreases at a
rate of 10 lb/in from D to B, it will be 6"
from D to a point where the shear is zero, since the shear at D fs 60 lb
This point could also be located by equa- ting equation (1) to zero and solving for x as follows:
150 no
O = 270 - 120 - 10x, ox aay 21s trom
A
If the shear diagram has passed through zero under the concentrated load, then the method of equating the shear equation to zero and solving for x could not be used, thus in general, it is best to draw a shear diagram to find when shear is Zero Fig A&S.10 shows the plotted shear diagram
Moment Diagram: -
At section A just to the right of reaction Ra the bending moment, considering the forces to the left, 1s zero, since the arm of Rag 1S Zero
The bending moment at any section between A and C, at a distance x from the left reaction
Ry, is,
Trang 24Àã,.4
The equation for the bending moment between
D and B (x greater than 9) is 2 My = Rx - P (x-9) ~ 5 TT (3) = 270 x ~ 120 (x9) + WES = 1080 + 150 x - Sx* + - - - (4) At section C, x = 9", substitute in equation (4) My, = 1080 + 150 x 9 ~S x 9* = 2025 in.lb (positive, compression in top fibers) is" M = 1960 + 150 x 15 - 5 x 157 = 2205
Thus, by substituting tn equation (2) and
(4) the moment diagram as plotted in Fig AS.11 is obtained
At the point of zero shear, x =
A5.5 Section of Maximum Bending Moment
The general expression for the bending mo- ment om the beam of example problem 2 is from equation (3):
wx?
My = Rax - P (x-9) - TC
Now, the value of x that will make Ma
maximum or minimm is the value that will make the first derivative of M, with respect to x
equal to Zero, or
By Get Ra- Pou
Therefore, the value of x that will make My 2 maximm or minimm may de found from the equation
Ra ~ P-we = 0
But, observation of this equation indicates
that the term Ra - P - wx is the shear for the section at a distance x from the left reaction
Therefore, where the shear 1s zero, the bending moment is maximm Thus, the Shear diagram
which shows where the shear is zero is a con- venient medium for locating the points of maxi-
mum bending moment
A5.6 Relation Between Shear and Bending Moment
Equation (5) can also be written = av,
since the right hand portion of equation (5) is
equal to the shear
Hence, dM = Vax -+ - - Which means that the difference dM between the bending moments at two sections that are a distance dx apart, is equal to the area Vdx under the shear curve between the two sections Thus, for two sections x, and Xa; BEAMS SHEAR AND MOMENTS Ma Xa aM = M Xã Vdx
Thus, the area of the shear diagram between any two points equals the change in Dending moment Detween these two points,
To tllustrate this relationship, consider the shear diagram in example problem 2 (Fig AS.10), The change in bending moment detween the left reaction Ra and the load is equal to the area of the shear diagram between these two points, or 270 + 180 2 moment at
x 9 5 2025 in.lb Since the bending the left support is zero, this change therefore equals the true moment at a section
under the load P,
Adding to this the area of the small tri-
angle between point D and the point of zero 60
=—x6e 5 180, we obtain 2205 in,1b,
as the maximm moment This can be checked by
taking the area of the shear diagram between the point of zero shear and point B=
210 z= * 21 = 2208 in lb,
shear, or
Example Problem 3
Fig 45.12 111ustratas a landing gear oleo strut ADEO braced dy struts BD and CEH A land~ ing ground load of 15000 1b is applied through the wheel axle as shown Let it be required to
find the axial load in all members and the shear
Trang 25ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES 11550/cos Z0° or 122320 1b Hạ = 11550 x15/26 = 6660 To find Hy take moments about point 2, Mp = 11550 x 15 - 6660 x 10 - 15000 x 0.5 x 32 + 10 Hạ =0 hence, Ha = 13340 lb To find Vp take iFy = 0, aFy = 15000 cos 30° + 11550 - Vg = 0 hence, Vg = 24550 1b
The axial load in member 3D therefore equals
24550/cos 30° = 28360 lb (compression) The
reaction Hp therefore equals 28360 x sin 30 = 14180 1b, “
Fig AS.13 shows the oleo strut as a free body with the reactions at A, D and E as calcu- lated Fig AS.13 also shows the axial load, vertical shear and bending moment diagrams
The bending moments due to applied loads without regard to bending’deformation of the beam are usually referred to as the primary bending moments If a member carries axial loads additional bending moments will be pro-
duced due to the axial loads times the lateral
deflection of the beam, and these bei ting mo- ments are usually referred to as sec ondary bend~
ing moments (Arts A23-30 covers the calcula-
tion of secondary moments) 13340 8680 | 13990 rØ 11550 \4 16 r 18 10 a 7500 aasso M4180 7 Axial Load 13000 Ib Diagram ~=Compression 13340 Shear Diagram Ữ ZA -840 -7500 Ib Bending Moment Diagram Fig AS 13 ~ 120000 ~133440 in Ib Example Problem 4
Fig AS.14 shows a beam loaded with both transverse and longitudional loads This beam
loading is typical of interior beams in the air-
plane fuselage which support all kinds of fixed equipment The reactions for the beam are at points A and B Required: - Shear and bending moment diagrams > | |.—Bracket— 7 rt” ' rf PP ơn | hon Bracket { | , je — Íb — te = aes Bp ‡ tl gn 7a av er este 500 Ib Wal VB Fig AS 14 SOLUTION: -
Calculations of reactions at A and B: -
To find Vp take moments about point A, 2M, = - 500 x 7 - S00 x 6 + 1000 x 20 + 1000 x sin 45° x 10 + 1000 cos 45° x 2 - 22 Vg = 0 hence, Vg = 999.3 1b (up) To find Vy take ZV = 0, ZV = 999.3 - 1000 - 1000 sin 45° - 500 + Va = 0 hence, V, = 1207.8 lb (up) To find Hy take ZH = 0, tH = - 500 + 1000 cos 45° - Hp = 0, hence Hg = 207.1
with the exception of the 1000 lb load at 45°, @11 loads are applied to brackets which in
turn are fastened to the beam Therefore the
next step is to find the reaction of the loaded brackets at the beam centerline support points
The load at EB and the reaction Hg at B will be also referred to beam centerline
Pig AS.15 (a,b,c,d) show the cantilever
brackets as free bodies The reactions at the base of these cantilevers will be determined
These reactions reversed will then be the applied
loads to the beam at points C, D, F and E, 500 T la rl 500 Ss Pìế Ec<s0i LY ” Hp=500 1008 5" z sen #
MGA We=800 Mp=4000 Hinde vn a
Fig.a Fig.b Fig.c Fig d Fig A5.15
For bracket at C, to find Ho take ZH = 0, or obviously Hy = 500 1b In like manner use SV = 0 to find Va = 500 lb To find Mg take
moments about point C IMc = - 500 x 2+ 500 x 8 ~ Mo = 0, hence Mo = 3000 in lb The student
should check the reactions at the base of the
(See Fig b,c)