Bruhn - Analysis And Design Of Flight Vehicles Structures Episode 1 Part 7 potx

The method in general consists of finding the magnitude and location of the elastic weight for each member of a truss due to a strain from a given truss leading or condition and apply

ÄA7.16 Truss Deflection by Method of Elastic

Weights

If the deflection of several or all the

joints of a trussed structure are required, the

metnod of elastic welghts may save considerable

lime over the method of virtual work used in

previous articles of this chapter The method

in general consists of finding the magnitude and

location of the elastic weight for each member

of a truss due to a strain from a given truss

leading or condition and applying these elastic

weights as concentrated loads on an tmaginary

beam The dending moment on this imaginary beam

due to this elastic loading equals numerically

the deflection of the given truss structure

Consider the truss of diagram (1) of Fig

A7,49, Diagram (2} shows the deflection curve

for the truss for a AL shortening of member bc,

all other members considered rigid This de-

flection diagram can be determined by the vir-

tual work ex pression 6 = uAL Thus for deflec-

tion of joint 0, apply a unit vertical load act-

ing down at joint 0 The stress m in bar be due

to this unit load = 2 2P $ ep = 4P Therefore Yr or

Sq = lpg 4P The deflection at other

đt lower chord joints could

be found in a similar

manner by placing a unit load at these joints

Diagram (2) shows the resuiting deflection curve,

This diagram is plainly the influence line for

stress in bar be multiplied by ALpc-

Diagram (3) shows an imaginary beam loaded

with an elastic load 4Loc acting along a verti-

r cal line thru joint 0, the moment center for ob-

taining the stress in bar be The beam reac~

tions for this elastic loading are also given

Diagram (4) shows the beam bending moment dta-

gram due to the elastic load at point 0 It is

noticed that this moment diagram is identical to

the deflection diagram for the truss as shown in

diagram (2)

The elastic weight of a member is therefore

equal to the member deformation divided by the

arm r to its moment center, If this elastic

load is applied to an imaginary beam correspond-

ing to the truss lower chord, the bending moment

on this imaginary beam will equal to the true

truss deflection,

Diagram 5, 4 and 7 of Fig a7.49 gives a

similar study and the results for 4 aL lengthen-

ing of member CK The stress moment center for

this diagonal member lies at point 0, wnich lies

outside the truss The elastic weight AL at

TL point O' can be replaced by an equivalent system

at points O and « on the imaginary beam as shown

in Diagram (6) These elastic loads produce a

bending moment diagram (Diagram 7) tdentical to

the deflection diagram of diagram (5),

Table A7.7 gives a summary of the equations

for the elastic weights of truss chord and web

members together with their location and sign

wend Moment Curve 4)

for Elastic Loads

Ty,

(7) Moment Curve for Elastic Loads Fig A7.48

Table A7.7 Equations for Elastic Weights Elastic- Weight for Chord Members

(See Member ab}

er by the method of moments The sign.of the elastic weight w for a chord member is plus tf 1t tends to produce downward de~

flection of its point of application Thus fora simple truss compression in top chord or tension

in botton chord produces downward or positive on

' 9

Psq=SE

For a truss diagonal member the elastic weights P & Q have opposite signs and are as-

sumed to be directed toward each other or

away according as the member is in compres-

sion or tension In flg a, P is greater

than Q and P is located at the end of the di-

agonal nearest the moment center 0 Downward

elastic weights are plus

Q acts opposite to P at far end of chord memb-

er cut by index section 1-1 used in finding

stress in ab by method of sections

Xeaner|Length | Arex | Load | PL « Ab làm Ì Elastie Point ef |

L A Pp | AE Es29x106 | + | Yt., fw = AL) tion lapplica-

&7,17 Solution of Example Problems

The method of elastic weights as applied to truss deflection can be best explained by the

sOlution of several simple typical trusses

Example Problem 38

Fig A?.50 shows a simply supported truss symmetrically loaded Since the axial deforma-

tions 1n all the members must be found, the

first step is to find the loads in all the memb-

Fig A7.51 shows the elastic weights obtained from Tables 8 and 9 applied to an imaginary beam whose span equals that of the given truss

These elastic weights are the algebraic sum of the elastic weights acting at each truss joint

Defl at A = (.008412+ 00185)15 = 007262 x15

„1097 Defl at b = 109 + (,O007262 - 000507)15 = 109 +

„006755 x 15 = 209"

Defl at B = 209 + (,006755 - 002347)15 = 209

„004408 x 15 = 275" +

ANALYSIS AND DESIGN OF FUIGHT VEHICLE STRUCTURES Defl atc 2275 + (004408 ~ 0027)15 = 301"

The slope of the elastic curve at the truss joint

points equals the vertical shear at these points

for the beam of Fig A7.51

Example Problem 39

Find the vertical deflection of the Joints

of the Pratt truss as shown in Flg 47.52 The

member deformationsaL for each member due to the

given loading are written adjacent to each memb-

er, Table A7.10 gives the calculation of member

elastic weights Fig A7.5a shows the imaginary

beam loaded with the elastic weights from Table

Á7.1O The deflections are equal numerically to

the bending moments on,this beam

Elastic Weight Chord Members

Find the vertical joint deflections for

unsymmetrically loaded truss of Fig A7.54

AL deformations for all members are given on

Figure Table A7.11 gives the calculation of

the

The

the

AT, 35 beam loaded with the elastic weights from Table 47,11 Table A7.12 gives the calculation for the joint deflections

Elastic Weight of Web Members

Wember ALi ry |P =ôL t |dotnd rạ | Q = AL [Joint Bo

cr 10,60 c 11.26 00755 3 J0 10.33 J 16.93 |-7 00126 D

Panel | Panel Shear / J Shear Potnt

error Sxample Problem 41 1g A7.56 shows 4 simply supported truss with cantilever overhang on each end.’ This sim- plified truss is representative of a cantilever

the elastic weights, their signs and points of

application Fig A7.55 shows the imaginary wing beam the fuselage attachment points deing

DEFLECTIONS O

AT 36

st sa and e", The AL deformation in each truss

member due to the given external loading is giv-

en on the f f The complete truss elastic

loading will be determined igure, With the elastic

loading known the truss deflections from various

reference lines are readily determined

awe eee Cd oe ef E e Dac c BDA * Sig AT.87

Elastic Weight Loading 1s °

Table AT,13

Rlastic Weights of Chord wembers

k 00838| ¢ .00670| €

00781 | đ 00728| D -.00792| E ~.0020 | £

Table A7.13 gives the calculations for the mag-

nitude of the member elastic weights The signs

and also the Joint lecations for locating the

elastic loads are also given Combining alge-

braically the elastic weights for each joint from

Table A7.13 the beam elastic loading as shown in

Fig A7.57 1s obtained

Let it first be required to determine the

vertical deflection of Joint f relative to the

truss support points at e and e',

To determine the deflections of the truss

betwevn the supports 6 ade! it is only neces-—

sary to consider the slastic weight loading be-

tween these points Fig a7.58 shows the portion

of the imaginary beam of Fig A7.57 between these

points The deflection at f relative to line ee

1s equal to the bending moment at ¢ for the por-

tion of the imaginary beam between joints 6 and

e’ and simply supported at these points

Since the cantilever portion is not fixed at

@ since the restraint is cetermined by the truss between e' and e, this fact must be taken into account in loading the cantilever portion, The reactions on the beam of Fig A7.58 represent the slope at e due to the elastic loading between

e and e' This elastic reaction in acting tn the reverse direction is therefore applied as a load

to the imaginary beam between s and 2 as shown in

a and free at e The bending moment at any point

on this beam equals the magnitude of the vertical deflection at that point

Thus to find the deflection of the truss end (Joint a) we find the bending moment at point a

of the imaginary beam of Fig A7.59

Hence deflection at a = IMg (calling counterclock- wise positive) = (.01922 - 00312) 80 + O00248 x

(From Span ee') Fig AT 59

Mg = 2.077" (student should make calcula- tons) Previously the deflection of f with re- spect to e was found to be -.0586" Thus de- flection of a with respect to point e = 2.077 +

0586 = 2,135" which checks value found above

Let it be required to find the deflection of

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES For this problem we need only to consider

the elastic loads between points > and d as

loads on a simple beam supported at b and d

(See Fig A7.61) The deflection at c with re-

Spect to a line bd of the deflected truss

equals the bending moment at point c for the

loaded beam of Fig A7.61

{1) Find vertical and horizontal deflec-

tion of joint B for the structure in Fig A7.62

Area of AB = 0.2 sq in and BC = 0.3 E =

10,000,000 psi

(2) For the truss in Fig A?7.63, calculate

the vertical deflection of joint Cc Use AE

for each member equal to 2 x 10’

(3) For the truss of Fig A7.64 determine

the horizontal deflection of joint £ Area of

each truss member = 1 sq in., E = 10,000,000

psi

(4) Determine the vertical deflection of

joint & of the truss in Fig A7.64

(5) Determine the deflection of joint D

normal to a line joining joint Cre of the truss

joint c for the truss in Fig A7.65 due to the

load at joint B Members a, b, c and h have

areas of 20 sq in each Members d, e, f, &

and 1 have areas of 2 sq in each EF

AT 37 (7) For the truss in Fig A7.66 calculate the deflection of joint C along the direction CE

as 2 sq in each and those carrying compression

as 5 sq in each § = 30,000,000 psi

{$) For the truss in Fig 47.68, determine the horizontal displacement of points C and B

Fig AT 68 Fig AT.T0

(10) For the truss in Fig A7.69, find the vertical deflection of joint D Depth of truss = 180" Width of each panel ig 180", The area of each truss member 1s indicated by the number on each bar in the figure £ = 30,000,000 psi Al-

so calculate the angular rotation of bar DE

(11) For the truss in Fig 47.70, calculate the vertical and horizontal displacement of joints A and B Assume the cross-sectional area for members in tension as 1 sq in each and those in compression as 2 sq in B = 10,300,000 pai

(12) For the truss in Fig A7.70 calculate the angular rotation of member AB under the given truss loading

(14) For the beam in Fig A7.72 find the de-

Also the slope of the Assume EI equals to

flection at points A and E

elastic curve at point c

5,000,000 lb in sq

AT.38 DEFLECTIONS

29#/in,

Fig, A7, 73 Reaction

(15) Fig A7.73 illustrates the airloads on

a flap beams ABCDE The flap beams is supported

at B and D and a horn load of 500# Is applied at

Cc The beam is made from 4 1"~.049 aluminum

alloy round tube I = 01659 in*; = = 10,300,000

psi Compute the deflection at points C and E

and the slope of tha elastic curve at point B

the deflections at points C and D in terms of EI

which is constant Also determine slopes of tha

elastic curve at these same points

(17) For the cantilever beam of Fig A7.75

detarmine the deflections and slopes of the

elastic curve at points A and B Take EI as

Fig AT 78 Fig A7.77

(18) For the loaded beam in Fig A7.76 de- termine the value of the fixed end moments Ma

and Mg EI is constant Also find the deflec-

tion at points ¢ and D in terms of EI

{19) In Fig A7.77 determine the magnitude

of the fixed end moment Ma and the simple sup-

Calculate the vertical deflection and the angu~

lar rotation of point aA

(21) For the curved beam in Fig A7.79 find

the vertical deflection and the angular rota-

tion of point A Take EI as constant

„ {22) For the loaded curved beam of Pig

A7.80, determine the vertical deflection and the

angular rotation of the point A Take EI as

EI = 12,000,000

(24) Determine the vertical deflection of point A for the structure in Pig A7.82 EI =

‘A 100#

(25) The cantilever beam of Fig A7.83 is loaded normal to the plane of the paper by the two loads of 100¢ each as shown Find the de- flection of point A normal to the plane of the paper by the method of virtual work The rec- tangular moment of inertia for the tube is 0.0277 in* §E = 29,000,000

(26) The cantilever landing gear strut in Fig A7.84 18 subjected to the load of SOO# in the drag direction at point A and also a torsion-

al moment of 2000 in lb at A as shown De- termine the displacement of point A in the drag direction The tube size for portion CB is 2"- -083 and for portion BA, 2"-.065 round tube

Material is steel with E = 29,000,000 psi

LJ

[atz] {a3} compute the strain energy in the truss of Fig A7.63 (Problem 2) The member flexibility coefficient for a member under uniform axial load 1s L/AE (see Pig A7.3S5a)

Ans U = 22.4 lb.in, (28) Using matrix equation (23) compute the strain energy in the beam of Pig A7.71l

Note: the choice of generalized forces should

be made so as to permit computation of the member flexibility coefficients by the equations

of p A7.19 Ans U = 3533 Ib in

(29) Re-solve the vroblem of example problem 25 for a2 stepped cantilever beam whose (27) Using the matrix equation 2u =

I doubles at point "2" and doubles again at "3", (Heaviest section at dutit-in end.)

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES (30) For the truss of Fig A7 95 determine

the influence coefficient matrix relating

vertical deflections due to loads Py, Pa, P Ps

and P, applied as showm Member areas are

shown on the figure

{31} For the truss of problem (30) deter-

mine which of the following two loading con-

ditions produces the greatest deflection of

(32) Determine the matrix of influence

coefficients relating drag load (positive aft),

braking torque (positive nose up) and moment

in the V-S plane (positive right wing down) as

applied to the free end of the gear strut

(33) Find the deflection of the load

applied to the cantilever panel of Fig AT 86,

(Assume the web does not buckle) Use matrix

notation, ans 6 = 7.94 x 107% inches

AT 39

References for Chapters A7, À8,

TEXT BOOKS ON STRUCTURAL THEORY

"advanced Mechanics of Materials", F Seely and

J O0 Smith, 220 ca., John Wiley, New York

"Advanced Strength of Materials", J P Den Hartog, McGraw-Hill, N Y

"Theory of Slasticity", S Timoshenko, McGraw—

Hill, N Y

TEXT BOOKS WITH MATRIX APPLICATIONS Wenle, L and Lansing, W., A Method for Reducing

the Analysis of Complex Redundant Structures te

"Slementary Matrices", R A Grazer, W J a Routine Procedure, Jou of Aero, Sot

Duncan and A R Collar, Cambridge University Vol 1982

"Introduction to the Study of Aircraft Vibration| Matrix Transformation with &r gard to ~ and Flutter", R Scanlan and 2 Rosenbaum, Semi Monocoque Structures, Journ of Aero Sci

Langefors, 3., Matrix Methods for 5 Structures, Journ of Aero Sci ¥

in Structural Analysis”, Journ of Appl Mechs ros , Argyris, J and Kelsey, S , + +

Structural Analysis, Aircra?

Cet 1554, et seg,

Many problems involving calculation of deflections are encountered in the structural design of a large modern airplane such as the Douglas DC-8,

CHAPTER A8

STATICALLY INDETERMINATE STRUCTURES

ALFRED F SCHMITT

A8.00 Introduction,

A statically indeterminate (redundant)

problem 1s one in which the equations of static

equilibrium are not sufficient to determine the

internal stress distribution Additional re-

lationships between displacements must be

written to permit a solution

The "Theory of Slasticity"® shows that all

structures are statically indeterminate when _

analyzed in minute detail The engineer how-

ever, 1s often able to make a number of as-—

sumptions and coarse approximations which render

the problem determinate In addition, auxillary

aids are available such as the Engineering

Theory of sending (2) ana the constant-shear~

flow rules of thumb (q = T/2A) (see Chaps A-5S,

A-6 and A-13 through A-15) While these latter

are certainly not laws of "statics", the en-

gineer employs them often enough so that prob-

lems in which they are used to obtain stress

distributions are often thought of as being

"determinate"

It is frequently the case in aircraft

structural analysis that, in view of the re~

quirements for efficient design, one cannot ob-

tain a determinate problem without sacrificing

necessary accuracy The Theory of Elasticity

assures the existence of a sufficient number of

auxiliary conditions to permit a solution in

such cases

This chapter employs extensions of the

methods of Chapter A-7 to effect the solution

of typical redundant problems Special methods

of handling particular structural configurations

are shown in later chapters

48.0 The Principle of Superposition,

The general principle of superposition

states that the resultant effect of a group of

loadings or causes acting simultaneously is

equal to the algebraic sum of the effects acting

separately, The principle is restricted to the

condition that the resultant effect of the

several loadings or causes varies as a linear

function Thus, the principle does not apply

when the member material is stressed above the

Proportional limit or when the member stresses

are dependent upon member deflections or de-

formations, as, for example, the beam-colum, a

member carrying bending and axial loads at the

same time

A8.1 The Statically Indeterminate Problem

Several characteristics (and interpreta~

tions thereof) of the statically indeterminate

problem may be pointed out These characteris~

tics are individually useful in forming the bases for methods of solution

There are more members in the structure

than are required to support the applied loads

If n members may be removed (cut) while leaving

sa stable structure the original structure is said

to be "n-times redundant”

-COROLLARY-

In an n-times redundant structure the mag- nitude of the forces in n members may be assigned arbitrarily while establishing stresses in equi- librium with the applied loads Thus, in Fig

A8.1 (a singly redundant structure), the internal

force distribution of (a) is in equilibrium with the external loads for any and all values of X, the force in member BD

in static equilibrium with the applied loads, (>), with one

zero-resuitant stress distribution, (c), superposed

Only the system (b) is actually required to equilibrate the external loads (corresponding to X= 0) Note that the system (c) has zero ex- ternal resultant

fd 0í a11 the possible stress (force) dis-

tributions satisfying static equilibrium the one correct solution is that one which results

in Kinematically possible strains (displace- ments), i.e retains continuity of the struc- ture

Thus, for example, there are an infinite

mumber of bending moment distributions satis- tying static equilibrium in Fig, A8.1 (4) since

M, can assume any value Of these, only one will result in the zero deflection of the right

hand beam tip necessary to maintain structural

continuity witn the support at that point

A8.1

Singly redundant beam with root bending

moment M, undetermined by statics

-COROLLARY-

if n member loads have been assigned ar~

bitrarily while establishing equilibrium with

the external loads, relative movements of the

elements will result, violating continuity at n

points n zero-resultant stress (force) dis-—

tributions may then be superposed to reduce the

relative motions to zero The resulting stress

distribution is the correct one

A8.2 The Theorem of Least Work

A theorem extremely useful in the solution

of redundant problems may be obtained from

Castigliano’s Theorem Consider first the

problem of redundant reactions such as in a

beam over three supports (Fig 48.2) One of

the reactions cannot be obtained by statics,

If the unknown reaction (say that on the far

right) 1s given a symbol, Ry, then the remaining

reactions and the bending moments may be de-

termined from statics The strain energy U may

then be written as a function of Ry, 1 ;

Usf (Ry) Next form

ệU _

aR * OR,

This 1s the deflection at Ry due to Ry But

this must also be zero, Since the support is

rigid Hence

£q (1) is true for all redundant reactions

occuring at fixed supports Because it corre-~

Spends to the natnematical condition for the

minimum of a function, eq (1) {s said to state

the Theorem of Least Work In words; "the rate

of change of strain energy with respect to a fixed redundant reaction is zero”

A8.2.1 Determination of Redundant Reactions by Least Work

Example Problem A

By way or illustration, the problem posed

by Fig A8.2 was carried to completion The bending moment was given 5y ÍX, y, ® measured

from the left ends of the three beam divisions)

Us 3 a | (500+Rx) ®c®4x

°

+ ser SORE + (Ry = s00ly] dy

Differentiating under the integral s

Determine the redundant fixed end moments

for the beam of Fig A8.2{(a)

+ EI * constant

a pee Mp

Fig AS, Za

A doubly redundant beam with two reactions given

two arbitrary values

Solution:

The redundant end moments were designated

as My and Mg for the left and right beam ends

respectively and were taken positive as shown,

The moment equations for the two beam portions

(x from left end, y from right) were

MH + 2PL ~ 3L

A8.2.2 Redundant Stresses by Least Work

The Theorem of Least Work may be applied to the problem of determining redundant member forees within a statically indeterminate structure Thus, in an n-times redundant structure if the redundant member forces are assigned symbols X,

Y, 8, + - - etc., the values which these forces

must assume for continuity of the structure are such that the displacements associated with these forces (the discontinuities) must be zero

Hence, by an argument parallel to that used for redundant reactions, one writes,

ẩ ON

A8.4

In words, "the rate of change of strain enerzy

with respect to the redundant forces {s zero”

Eqs (2), like eq (1}, are statements of the

Tnsorem of Least Work They provide n equations

for thé n-times rédundant structures The simul-

taneous solution of these equations yields the

desired solution of the problen

Example Problem C

The cantilever beam and cable system of

Fig a8.3(a) is singly redundant Find the

member loadings by use of the Least Work

The tensile load in the cable was treated

as the redundant load and was given the symbol

X (1g A8.2(b)) The strain energies con-

sidered were those of flexure in portions AC,

CD and BC and that of tension in the cable AB

mergies due to axial forces in the beam port-

ions were considered negligible

The bending moment in BC (origin at 3) was

XS) veS(= 2\AE/,

X and hence could not enter the croblem Dif- ferentiating under the integral sign

526