## Introduction To Fuselage Stress Analysis

In general the purpose of an airplane Is to transport a commercial payload or a military useful load. The commercial payload of a modern airliner may be 100 or more passengers and their baggage. These passengers must be transported safely and comfortably. For example,, an airliner flies at high altitudes where temperatures may be far below zero and where the air density is such as not to sustain human life. These facts mean that the body which carries the passengers must be...

## 4

C7.27 Experimentally determined coefficients for failure in wrinkling mode. (Rei. 2) Fig. C7. 28 (Rei. 2) Experimentally determined values of effective rivet offset. The rivets are 3 32 diameter Brazier head type AN456, 2117-T3 naterlal spaced at 0.75 inches. Area of Z stlffener 0.252 in. Area of skin for bs 2 inches 2 x .064 Total area of sheet and stlffener 0.380 in.a Crippling Stress of Stlffener Acting Along. Fcs(ST)' Since we have a Z type of stlffener, equation C7.6 and Fig. C7.9...

## The Work Of The Aerospace Structures Engineer

The first controllable human flight in a heavier than air machine was made by Orvllle Wright on December 17, 1903, at Kitty Hawk, North Carolina. It covered a distance of 120 feet and the duration of flight was twenty seconds. Today, this initial flight appears very unimpressive, but it comes into its true perspective of importance when we realize that mankind for centuries has dreamed about doing or tried to do what the Wright Brothers accomplished in 1903. The tremendous progress accomplished...

## Ae

A8.4 Example Problems - Trusses With Single Redundancy. Fig. A3.5 shows a single bay pin connected truss. The truss is statically determinate with respect to external reactions, but statically indeterminate with respect to Internal member loads, since at any Joint there are 3 unknowns with only t vo equations of statics available for a concurrent force system. The truss Is therefore redundant to the first degree. The general procedure for solution Is to make the truss statically determinate by...

## 03

Vom -ZqL (10.85 x5)2 (0.85x5)2 t ( .03 ) T 3 ) A15.10 Three Cell - Multiple Flange Beam. Symmetrical About One Axis. Fig. A15.36 shows a 3-cell box beam subjected to an external shear load of 1000 lbs. as shown. The section is symmetrical about axis XX. The area of each stringer is shown in parenthesis at each stringer point. The Internal shear flow system which resists the external load of 1000 lbs, will be calculated assuming that the webs and walls take no bending loads, or, the stringers...

## Torsion

J - 4 a 0- 13 + Q-iaA8 a0 jAa3 1 _____ (31) I aOJ_aia + ai3aao + aaoaoij A6.12 Example Problems of Torsional Stresses in Multiple-Cell-Thin-Walled Tubes. Example 1 - Torsional Stresses in Un-symmetrical Two-C ell-Tube. Flg. A6.19 shows a typical 2-cell tubular section as formed by a conventional airfoil shape, and having one Interior web. An external applied torque T of 83450 in. lb. is assumed acting as shown. The internal shear resisting pattern is required. Aa 387.4 sq. in. A 493.2 sq. in....

## Hi

Forces and moment at point 1 of a beam cross-section. They can be resolved into a normal force N and a shear for S plus a moment Mi as shown in Fig. AS.35 where, N H cos a + V sin a S - V cos a - H sin a Later on when the beam section is being designed it may be found that the neutral axis lies at point 0 instead of point 1. Fig. A5.36 shows the forces and moments referred to point 0, with M0 being equal to Mi - Ne. The loads which cause only bending of a beam are located so that their line of...

## Mechanical And Physical Properties Of Metallic Materials For Flight Vehicle Structures

G n rai Explanation, it would require several hundred pages to list the properties of the many materials used in flight vehicle structural design. The metallic materials presented in this chapter are those most widely used and should be sufficient for the use of the student in his structural analysis and design problems. All Tables and Charts in this chapter are taken from the government publication Military Handbook, MIL-HDBK-5, August, 1962. Metallic Materials and Elements for Flight Vehicle...

## 002782

(0.14 x 0.1547) 1134 - 1109 - 5285 The shear stresses at these two points (b) and (c) would equal Q t 1134 0.1 and - 5260 0.1 or 11340 psi and - 52600 psl respectively, SOLUTION BY NEUTRAL AXIS METHOD. (Method 2) It Is necessary to find the neutral axis for the given external loading. In Fig. A14.28, the angle 9 Is the angle between the plane of loading and the z principal axis, and this angle 0 equals 30 + 12 J7 ,0 46 - 46 42 - 46. isV ' Let a equal angle between xp principal axis and neutral...

## O

NO. 2 View Looking Inside of Rear Portion of Fuselage of Beechcraft Twin-Bonanza Airplane. PHOTO. NO. 3 Fuselage Construction of Boeing 707 Jet Airliner. (FOR GENERAL DETAILS OF DOUGLAS DC-8 FUSELAGE CONSTRUCTION SEE PAGE A15. 32) A20. 3 Stress Analysis Methods. Effective Cross-Section. It Is common practice to use the simplified beam theory in calculating the stresses in the skin and stringers of a fuselage structure. If the fuselage is pressurized, the stresses in the skin due to this...

## Bjv

H effective web depth distance between centroids of flange-web connection rivets. The total beam shear V which equals the resistance of both web and flange, equals Vw I The difference between (17) and (18) gives the shear carried by flanges. Up to the buckling stress of the web plate, the shear flow is assumed to be of constant intensity over the effective web depth. When the web buckles, it is assumed that the web maintains the diagonal critical compressive stress that produced the buckling of...

## 1250

To find take moments about point (a') ZMa' + 2679 x .5 - 2534 x 11.5 - 316 X 0.375 - 45 x 22 + 700 x 12 - 11500 x 39 + 460,500 - q c (665) 0 whence qbc s - 13.66 lb. in. and In the above moment equation the moment of the shear flow qjjC about point (a1 ) equals qtc times twice the area of the cell or 665. ZFX - 2679 + 2584 + 316 - 22 q - 22 x 13.66 - 700 0 whence q 35.45 lb. in. with sense as assumed. 2FZ - 1296 - 1250 + 45 + 35.45 X 0.5 -11.5 x 13.66 + 11500 - 11 qab 0 The loads on the...

## Solution 3y Method

Bolt Shear Strength ar.d Bolt Bending Strength are calculated in the same manner as in Method 1 and thus the calculations will not be repeated. is the tension efficiency factor to take care of stress concentration due to the hole and is determined from Fig. D1.12. Table D1.3 says to use curve number 1 for all steels. To use Fig. D1.12 requires the ratio W D - (1.1875 .525) 3 1.9. Then from Fig. D1.12 we read Kt .98, whence, Pu .93 x 125000 x (1.1875 - .625).375 25500 lb. Fig. D1.12 says that a...

## M

Solving equations (c) and (d), we obtain, qt -62.8 lb. in., qa * -317 lb. In. These values are listed In columns (12) and (13) of Table A19.6. The final or resultant shear flow qr on any sheet panel equals the sum of (14) of Table A19.6. Fig. A19.32 shows the potted shear flow pattern. Comparing this figure with Fig. A19.30 shows the effect of adding the leading edge cell to 'the single cell of the previous problem. A19.15 Bending Strength of Thick Skin - Wing Section Figs. 1 and k of Fig....

## Continuous Structures Moment Distribution Method

The moment distribution method was originated by Professor Hardy Cross.* The method Is simple, rapid and particularly adapted to the solution of continuous structures of a high degree of redundancy, where It avoids the usual tedious algebraic manipulations of numerous equations. Furthermore, it possesses the merit of giving one a better conception of the true physical action of the structure in carrying its loads, a fact which is usually quite obscure in some methods of...

## 2

Qod -z-1.6 2 ZA -2 -1.6(-2.5)5x0.1 0 Fig. A15.2 shows a plot of the shear flow results. On the vertical web the increase in shear is parabolic since the area varies directly with distance z. The intensity of qx and qz in the plane of the cross-section is equal to the values of q found above which are in the y direction. The sense of qx and q2 is determined as explained in detail in Art. A14.6 of Chapter A14. Fig. A15.2 also shows the resultant shear flow force on each of the four walls of the...

## Aluminum Alloys

Heat Treated, t , 020 to . 062 In. 2024-T6 Clad Sheet & Plate, Heat Treated, t s 0.063 in. 2024-T6 Clad Sheet & Plate, Heat Treated, t -e 0.063 in. 2024-T81 Clad Sheet, Heat Treated, t -r 0.064 to. 6061-T6 Sheet, Heat Treated & Aged, t 0.25 in. 7075-T6 Bare Sheet b Plate, t s 0. 50 in. 7075-T6 Hand Forging , Area - 16 aq. in. 7075-T6 Clad Sheet & Plate, t S 0. 50 in. 7079-T6 Hand Forcings, t < 6.0 in. 2 2 2 2 1 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2...

## Equilibrium Of Colinear Force System

A collnear force system Is one where all forces act along the same line or in other words, the direction and location of the forces Is known but their magnitudes are unknown, thus only magnitude needs to be found to define the resultant of a collnear force system. Thus only one equation of equilibrium is available, namely where moment center 1 is not on the line of action of the force system A2. 3 Structural Fitting Units for Establishing the Force Characteristics of Direction and Point of...

## Example Problem Illustrating Transfer Cf Concentrated Load To Sheet Panel

A21.8 shows a cantilever beam composed of 2 flanges and a web. A concentrated load of 1000 lb. is applied at point (A) in the direction shown. Another concentrated load of 1000 lb. is applied at point (E) as shown. To distribute the load of 1000 lb. at (A), a horizontal stiffener (AB) and a vertical stiffener (CAD) are added as shown. A fitting would be required at (A) which would be attached to both stiffeners. The horizontal component of the 1000 lb. load which equals 500 lb. is taken by...

## 05

A3.16 illustrates a box type beam section with six longitudinal stringers. Determine the moment of Inertia of the beam section about the principal axes for the following assumptions - (a) Assume the beam is bending upward putting the top portion in compression ana 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...

## 35000

To find R take moments about point (A) 2M, 100,000 x 21 - 35000 x 9 + 38 Ra 0 A The airplane In Fig. A4.IB weighs 14,000 lb It is flying horizontally at a velocity of 500 tt.P.H. (732 ft sec) when the pilot pulls It upward into a curved path with a radius of curvature of 2500 ft. Assume the engine thrust and airplane drag equal, opposite and colinear with each other (not shown on Fig. A4.18). (a) Acceleration of airplane in Z direction (b) Wing Lift (L) and Tail (T) forces , -, , r, 4. Airplane...

## 3

Tension which checks value previously found. The y and z components of the strut reaction at C will then be, These values are indicated on Fig. A19.14. 2Fy - 7226 + Ay 0, hence Ay 7226 lb. In finding the reaction Ay previously, the value was 5085 lb. The difference Is due to the drag bending moment which tends to put a tension load on front spar and compression on the rear spar. Fig. A19.15 shows the air drag load of 4 lb. in. The bending moment on panel at a distance y from...

## Typical Welded Steel Tube Terminals

Irminatin < *ttding at offffasito wnds of Satisfactory a And eendinon mt am section of njC*. jmrmj 5< 7T foe Tory coveting ryin* mount* If any A < xm not fr ft SO* 'h< S jomt Tt mord* rt* twwk strwqth f ryfi*. Satisfactory ufi to XTantp,t of b**d A**id 5czmfr> ci*Y v*ntff r+a ttef rhw ae f j cost > y prwrf twd for wo* r& nqws of A irminatin < *ttding at offffasito wnds of Satisfactory a And eendinon mt am section of njC*. jmrmj Safsfaewry a w.dw hmgw a rew na. tnd > t gmtrmiy...

## A7

A7.16 Truss Deflection by Method of Elastic Weights If the deflection of several or 11 the Joints of a trussed structure are required, the method of elastic '.'.'eights nay save considerable 3line 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 loading or condition and applying these elastic weights as concentrated...

## Solution

The beam formula for bending stress at any point is a* Mxz Ix. To solve this equation we must have the effective moment of ir.ertia of the beam cross-section. The bottom surface being in tension 'under the given design bending moment is entirely effective, however the top surface has a variable effectiveness since the skin, stringers and corner flange members have different ultimate failing stresses. From equation (1) the effective width of sheet to use with each rivet line depends on the...

## L

M is positive when it tends to cause compression on the upper fibers of the beam at the section being considered. Reference AC1C 493 Niles, Airplane Design Newell and Niles Airplane Structures For Table of many other loadings, see NACA T.M. 985. NATURAL SINES. COSINES, AND TANGENTS F ANGLES IN RADIANS

## 0 j y

In equation (A), for translation deflections, T twisting moment at any section due to applied twisting forces, t torsional moment at any section due to a virtual unit 1 lb. force applied at the point where deflection is wanted and applied in the direction of the desired displacement, (in lbs lb) 3 shearing modulus of elasticity for the material, (also 0) J polar moment of inertia of the circular cross-section. In equation (B), for rotational deflections, 9 angle in twist at any section due to...

## 2741 3088 3474 3860 4825 6176 7720

A Bauinc rmiu* ar baaad od uus computed uaing the nomin* hol* diamat*i* ipceiflad in tabic 8.1.1.1.1(d). Table Dl. 10 Unit Bearing Strengths for Pin Size Indicated lb.a Bearing strength of sheet for rivet size indicated, lb Re& rin values are based on areas computed a sin the nominal pin diameters indicated. Slice rivets are same size, all rivets are assumed to share equally in resisting H and V loads. Load on each rivet due to Hc,g. 3000 5 1333 acting in H direction and to the right. Load...

## The Work Qf The Special Projects Group

In general, all the various technical groups have a special sub-group which are working on design problems that will be- encountered In the near or distant future as aviation progresses. For example, in the structures Group, this sub-group might be studying such problems as (1) how to calculate the thermal stresses in the wing structure at super-sonic speeds (2) how to stress analyze a new type of wing structure (3) what type of body structure is best for future space travel ana what kind of...

## Els

E Is the eccentricity between stiffener junctions due to the flanges telng carved rather than straight. The same applies to the inner flange, except that .it is in compression in this discussion. When the outer flange is in compression and the Inner one in tension, all of the loadings are reversed, as in Fig. 03.28(c) and (d). The web is then subjected to tension loads and there 13 no collapsing problem. No stiffeners are required, ether than for the normal reasons. However, in practice, most...

## General Loads On Aircraft

Before the structural design of an airplane can be made, the external loads acting on the airplane in flight, landing and take-off conditions must be known. The complete determination 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. Furthermore, there is a wide range of wing configurations, such as the straight tapered wing, the swept wing and the delta wing, and many of...

## 90

A Welded after heat treatment or normalized after v- ld. b Gusset9 or plate inserts considered 0 taper with center line. The heat-treatable alloys, commonly referred to as the strength alloys, such as 2014, 2024, 7075, cannot be welded with the oxyacstylene torch without destroying their mechanical propertlss, which are not restored even if heat-treated after welling. These alloys are generally classed as unweldable. Constant research is rolng on to develop aluminum, alloys that have relatively...

## 1

This tarn usually expressed in percent represents the additional ultimate strength of the structure over that strength required to carry the ultimate loads. Cl. 14 Required Strength of Flight Structures. The flight vehicle structure shall be 'designed to have sufficient strength to carry simultaneously the limit loads and other accompanying environmental phenomena for each design condition without undergoing excessive elastic or plastic deformation. Since most...

## References

Critical Combinations of Shear and Direct Axial Stress for Curved Rectangular Panels. NACA T.N. 1928. (2) Gerard & Becker, Handbook of Structural Stability. Part III. Buckling of Curved Plates and Shells. NACA T.N. 3883. (3) Rafel & Sandlin. Effect of Normal Pressure on the Critical Compression and Shear Stress of Curved Sheet. NACA WRL-57. (4) Rafel. Effect of Normal Pressure on the Critical Compressive Stress of Curved Sheet. NACA WRL-258. (5) Brown &...

## Weight Distribution

Table A5.3 gives the weight and balance calculations for all items attached to fuselage or carried in ivs fuselage, except the wing and items attached to the wing as the front landing gear and the fuel. In order to obtain a close approximation to the true shears and moments on the fuselage due to the dead weight inertia loads, it is necessary to distribute the weights of the various items as given in Table A5.3. Fig. A5.42 shows a side view of the airplane with the center of gravity locations...

## Materials

Curve 1 - 2014-T6 and 7075-T6 Die Forging (L) 4130 and 6630 Steel 2014-T6 and 7075-T6 Plate 2 0. 5 (L, LT) 7075-T8 Bar and Extrusion (L) 2014-T6 Hand Forged Billet 144 in.a (L) Curve 2 - 2014-T6 and 7075-T6 Plate 0. 5 in. i in. (L, LT) 7075-T6 Extrusion (LT, ST) 2Q14-TB Hand Forged Billet a- 144 in.a (L) 2Q14-T6 Hand Forged Billet s 36 in.a (LT) 2014-T6 and 7075-T6 Die Forgings (LT) Curve 3 - 2024-T4, 2024-T2 Extrusion (L, LT, ST) Curve 4 - 2014-T5 and 7075-T6 Plate - I in. (L, LT) 2024-T4 Bar...

## Bending Moments Beam Column Action

A beam-column is a member subjected to transverse loads or end moments plus axial loads. The transverse loading, or 9nd moments, produces bending moments which, in turn, produce lateral bending deflection of the member. The axial loads produce secondary bending moments due to the axial load times this lateral deflection. Compressive axial loads tend to increase the primary transverse bending moments, where as tensile axial loads tend to decrease them. Beam-column members are quite common in...

## Ll

C5. 7 Chart of Nondimensional Compressive Buckling Stress for Long Hinged Flanges. 7f (Eg EJfl - Ue3) (1 - V Fig. C5. 8 Chart of Nondimensional Compressive Buckling Stress for Long Clamped Flanges and for Supported Plates with Edge Rotational Restraint. 7 -(Eg 2E) l + 0.5 l + (3Et Es) 1 a (l - Ve') 1 - V Ve - 0.3. Find, the buckling stress absolution We use Fig. C5.8 since it covers the boundary conditions- of our problem. The parameter for bottom scale is, For a b 9 3 3, we find kc from...

## Info

0 .002 .004 .006 .008 .010 .012 STRAIN - IN. IN. 0 .002 .004 .006 .008 .010 .012 STRAIN - IN. IN. For a rather complete study and comparison of the various effective widths theories as compared, see article by Gerard (Ref. 7). Equation C7.15 is in general conservative for higher b t ratios. The effective sheet area is considered to act monolithically with the stlffener. However, if the rivets or spot welds that fasten the sheet to the stlffener are spaced too far apart, the sheet will buckle...

## Pcf

On the top and bottom edges respectively have been assumed with the sense as shown. Taking moments about point E, ZMr 300 x3 - 12 x lOq 0, whence q 20 lb. in. ZFx 300 - 20 x10 - lOqa 0, whence qa 50 lb. in. PHOTO. A21.1 Type of Wing Ribs Used in Cessna 180 Model Airplane, a 4 Piace Commercial Airplane. PHOTO. A21.1 Type of Wing Ribs Used in Cessna 180 Model Airplane, a 4 Piace Commercial Airplane. PHOTO. A21. 3 Rib Construction and Arrangement in High Speed, Swept Wing, Fighter Type of...

## A2

Double Pin - Universal Joint lutings Since single pin fitting units can resist applied moments about axes normal to the pin axis, a double, pin joint as illustrated above is often used. This fitting unit cannot resist moments about y or z axes and thus applied forces such as P and Q must have a line of action and direction such as to pass through the center of the fitting unit as illustrated in the figure. The fitting unit can,, however, resist a moment about the x axis or In other words, a...

## A51

Unbalanced force, which means that the resisting moment Mj in the form of a couple, as shown In Fig. A5.5, or 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 portion of the beam is called the External Vertical Shear, or commonly referred to as the beam Vertical Shear and is represented by the term V. From equation (1), the Vertical Shear at any section of a beam, can...

## Statically Indeterminate Structures Special Method The Column Analogy Method

The Column Analogy- method, is a method that is widely used by engineers in determining the bending moments in a bent or ring type structure. The method considers only distortions due to bending of the structure. The numerical work in using the column analogy method is practically identical to that carried out in applying the elastic center method of Chapter A9. AlO. 2 General Explanation of Column Analogy Method. Fig. A10.1 shows a short column loaded in compression by a load P...

## Oop 000

C2.17 we find Fc V0-T .74 A very common aluminum alloy in aircraft construction is 2014-T6 extrusions. Let it be required to determine the allowable stress Fc for our member when made of this material. Since we have not presented column curves for this material, we will use Fig. C2.17. From Table Bl.l, for our material, ws find n 18.5, Ec 10,700,000 and F0 T 53,000 From Fig. CI.17 for 3 1.16 and n 18.5, we read FC FQ.T .71, hence Fc .71 x 53,000 37,600. The result shows that the...

## Problem 5

This is a typical problem involving the rivet loads in a sheet-stringer type of construction as Illustrated in Fig. D1.4G. Before the rivet size and spacing at the points (1) to (10) can be determined, the rivet loads at these points must be known. The shear flow in direction and magnitude on the webs and skin are shown on the figure and are in lbs. per inch. These values represent the results in one of the flight conditions. The structural designer must look at all the shear flows In the...

## 640

A21.48 obtained in solution method (1). Fig. A21.60 shows the corrective shear flows of Fig. A21.53 applied tc bay (2). On the bottom skin the corrective shear flew is shown on the boundary of the cut-out. These shear flows cause differential bending of the front and rear beams in bay (2). If we make the assumption that the beam end suffer no rotation, the bending moment is zero at midpoint of the bay and thus the flange loads at points a, b, c and d of bay (1)...

## 160

Let q be the constant flow reaction of the cell skin on the rib perimeter which is necessary to hold the rib in equilibrium under the applied air loads. Take moments about some point such as the lower flange (1). 2M -213.2x10 + 3x15x7.5 + 2x139.3 q s 0 whence, q 1232 273.6 4.42 lb. In. With the applied forces on the rib known, the shears and bending moments at various sections as desired can be calculated. For example, consider a section B-B, 2.5 from the leading edge. Fig. A21.24. 3x2.5x 1.25...

## A2016

S 10,500,000 psl. The ultimate compressive strength of stringer plus its effective skin is 35000 psl.x For effective sheet width use w l.9t (E asiOT-For buckling strength of curved panels use cr 53 Et r. Determine the ultimate bending moment that the fuselage section will develop for bending about horizontal neutral axis. Use linear stress distribution. Follow procedure as given in example problem in Art. A20.4. Fig. A20.13 shows the cross-section of a...

## The Column Analogy Method

The term 20s represents the area of the static Mg I curve. (E has teen assumed constant ana therefore emitted). Let the term 20s be called the elastic lead and give It a new symbol P. The term 2ds I equals the elastic weight of the frame and equals the sum of the length of each member times Its width which equals l l. Let this total frame elastic weight be given a new symbol A, In the expressions for Yo anc X0 the terms 20sx and 20sy represent the moment of the static K I curve acting as a load...

## By

McCOMBS (DESIGN SPECIALIST - CHANCE VOUGHT CORP.) In the design and fabrication of an airplane the major components receive a thorough review and evaluation. Many of the smaller parts, however, are designed at the last minute and, not receiving so much attention, sometimes have faulty details. It is these which frequently lead to trouble In service and in tests. This chapter represents an attempt to point out some of the more common details that seem, somehow, to be overlooked from...

## 10

E 101 psi 30.0 G, 10 psi 11.0 C, Btu (lb)(F) 0.11 (32 to 212 F). K, Btu UhrHft'HFVft . 10.3 (at 300 F) 11.2 (at 500 F) 13.1 (at 900 F). a, 10- in. in. F 6.0 (70 to 200 F) 6.1 (70 to 400 F) 6.5 (70 to Vendors guaranteed minimums (or and e. ' Test direction longitudinal these properties not applicable to the short transverse (thickness) direction. Fig. B2. 23 Effect of temperature on the compressive yield strength (Fcy) of 17-4 PH (H9Q0) stainless steel. Fig. B2. 24 Effect of temperature on the...

## Ib

A15.93 shows a circular single cell beam with 8 flange members. The area of each flange member is 0.1 sq. In. throughout the beam length. For the given 400 lb. external loading determine the resisting shear flow pattern at section A-A using the AP method over a distance of 25 inches between sections A-A and B-B. Assume cell wall Ineffective in resisting bending stresses. (21) Fig. A15.92 shows a single cell-o flange tapered beam carrying a 1000 lb. load as shown Calculate resisting...

## 111800 4100 14000

Acceleration a 5 214.5 ft seca s r iiouu The Inertia fores normal to the flight path Placing this force on the airplane through the e.g. promotes static equilibrium, hence to find tail load T takes moments about wing aerodynamic center (c.p.) SM _ - (14000 + 93700) 8 + 210 T 0 2Fz - 4100 - 14000 - 93700 + L 0 Assume the airplane as used in example problem 4 is In the same attitude as used in that example problem. Now the airplane is further maneuvered by the pilot suddenly pushing the control...

## Wri

For ultimate strength, fc could be carried up to FCy, conservatively, in the typical case. In order to realize the maximum strength and stiffness, the load in the net section must be applied in the corner1'. This is to prevent stresses due to bending out of the plane of the remaining leg. This requires that a minimum of 2 fasteners be provided to receive the load at the Joint. The reasoning here is the same as discussed in Art. 33.2 concerning minimum type shear clips, and the fastener loads...

## Doors And Windows

The various cutouts in the shell of a pressurized cabin require special consideration if an excessive weight penalty is to be avoided. Consider the panel removed from the pressurized cylinder of Fig. A16-14a. Following a common practice in dealing with cutouts, we determine what forces the panel-to-be-removed applies to the main structure around the border of the cutout, and then superpose a set of equal but opposite, self-equilibrating stresses to cancel these. The cutout border is then...

## Structural Layout Of Wing

A19.1Q and A19.ll shows the wing dimensions and general structural layout of a . monoplane wing with cr.e external brace strut. The wing panel is attached to fuselage by single pin fittings at points A and B with pin axes parallel to x axis. The mating lugs of the fittings at point A are made snug fit but those at B with some gap, thus drag reaction of wing loads on fuselage is resisted entirely at fitting A. Since the fittings at A and 3 cannot resist moments about x axis, it is necessary...

## Analysis Of Wing Structures

Fig. (a) shows the bending moment diagram due to the landing gear reaction alone. The internal resistance to this bending moment cannot be uniform on a beam section adjacent to section a-a because of the shear strain in the sheet panels or what is called shear lag effect. To approximate this stringer effectiveness, a shear lag triangle of length 3b is assumed, and the same procedure as discussed in the previous article on cut-outs is used in finding the longitudinal...

## Example Problem

A rectangular beam section is 0.25 Inches wide and 1.5 Inches deep. What yield ana ultimate bending moment will the section develop when made from 7075-T6 extruded aluminum alloy. Solution The modulus of bending stress is given by equation (3), 2Q 2 X 0.75 x .25 X .375 x .1408 ( ) .25xl.59(l .75) ' 938 The value of k could also be found in Fig. C3.7. Material is 7075-T6 aluminum alloy. From Fig. C3.17, Ftu 75000, Fty 65000. To find the yield bending strength, the value of fm in equation (3),...

## Qb

Stress-strain curves for three different materials. Material (A) is strong but brittle, whereas material (c) is weak and ductile, and material (B) represents average strength and ductility. However, all three materials have the same modulus of toughness since the areas under all three curves is the same. When a material is stressed, It will deform in the direction of the stress and also at right angles to it. For axial loading and for stress below the proportional limit stress, the ratio of the...

## 5

Some aircraft companies have specific strength data and practices for the design of Joggled members. This should, of course, be consulted by the designer, if available. Some companies use a 6 1 Joggle length to depth ratio, others use a 3 1 ratio, or both may be used. Strength or stiffness data for one ratio should not be used blindly for another. Some of this data Indicates that when, In the case of angle members, the depth of the joggle Is to the order of the thickness of the joggled leg or...

## Local Buckling Stress For Composite Shapes

Thin flat sheet is inefficient for carrying compressive loads because the buckling stresses are relatively low. However, this weakness or fault can be greatly improved by forming the flat sheet into composite shapes such as angles, channels, zees, etc. Most of the many composite shapes can also be made by the extruding process. Formed or extruded members are widely used In Flight Vehicle Structures, thus methods of calculating the compressive strength of such members is necessary. C8. 2...

## Thin-wall Shear Stress Flow At Cross Section By Excel

A14.5 Maximum Shear Stresses for Simple Cross-Sections. Table A14.2 gives the value of the maximum shear stress on a few simple sections and where It occurs on the cross section, (e is distance from neutral axis to point of maximum shear stress). V equals the shear load normal to the neutral axis and It acts along the centerline axis, thus no twisting on the section. A Is the total cross-sectional area. The maximum shear stress Is given In terms of the average shear stress which equals v a*...

## So

Shear, or x 6 180, we obtain 2205 in.lb. as the maximum moment. This can be checked by taking the area of the shear diagram between the point of zero shear and point 3 210 Fig. A5.12 illustrates a landing gear oleo strut ADEO braced by struts BD and CE. A landing ground load of 15000 lb. is applied through the wheel axle as shown. Let it be required to find the axial load in all members and the shear and bending moment diagram for the oleo strut. Pins at Points B, C, E. No Vertical Resistance...

## Example Problems

A6.3 shows a conventional control stick-torque tube operating unit. For a side load of 150 lbs. on stick grip, determine the shearing stress on aileron torque tube and the angle of twist between points A and B. Torsional moment on tube AB due to side stick force of 150 150 x 26 3S00 in. lb. The resistance to this torque is provided by the 150 Fig. A6.4 illustrates an aileron control surface, consisting of a circular torque tube (1-1 4 - .043 in size) supported on three hlr.ge brackets ana...

## Introduction To Wing Stress Analysis

A19.1 Typical Wing Structural Arrangement For aerodynamic reasons, the wing cross-section must have a streamlined shape commonly referred to as an airfoil section. The aerodynamic forces in flight change in magnitude, direction and location. Likewise in the various landing operations the loacs change in magnitude, direction and location, thus the required structure must he one that can efficiently resist loads causing combined tension, compression, bending and torsion. To provide torsional...

## Buckling Of Flat Sheets Under Combined Loads

The practical design case involving the use of thin sheets usually involves a combined load system, thus the calculation of the buckling strength of flat sheets under combined stress systems is necessary. The approach used involves the use of inter-action equations or curves (see Chapter CI, Art. CI.15 for explanation of inter-action equations). C5. 9 Combined Bending and Longitudinal The interaction equation that has been widely used for combined bending and longi- Fig. CS. 13 (Ref. 1) Chart...

## Buckling Strength Of Monocoque Cylinders

This chapter presents Information on the buckling strength of circular cylinders under compressive, bending and torsional loads acting separately and in combination, without and with internal pressure. Some information on the buckling strength of conical cylinders is presented. Before the advent of the high speed aircraft and particularly the missile and space vehicle, the use of the unstiffened cylinder or monocoque type of structure was quite limited. However, the arrival of the space age has...

## 072

(1) Determine the resisting shear flow pattern for the loaded single cell beam, as shown in Fig. A15.77, Assume load P zero in this problem. Assume all material effective in bending. Make two solutions, one of them involving the use of the shear center. (2) Same as problem (1) but add load P 1000 lb. (3) Fig. A15.78 shows an unsymmetrlcal single cell beam loaded as shown. Assume all material effective in bending. Determine resisting shear flow diagram. (4) Figs. A15.79 and A15.80 show two...

## Welded Connections

Since the overall structure of an airplane, missile or space vehicle cannot be fabricated as a single continuous unit, such structures Involve many structural parts which must be fastened together. For certain materials and types of structural units, welding plays an important role in Joining or connecting structural units. Research is constantly going on to develop better welding machines and weldin techniques and also to develop new materials that can be welded without producing a detrime...

## Fittings And Connections Bolted And Riveted

The Ideal flight vehicle structure would be the single complete unit of the sane material involving one manufacturing operation. Unfortunately the present day types of materials and their method of working dictates a composite structure. Furthermore general requirements of repair, maintenance and stowage dictate a structure of several main units held to other units by main or primary fittings or connections, with each unit incorporating many primary and secondary connections Involving fittings,...

## Shear Flow Problem In Three Cell Thin Section

And since T 2 qA, then also, ds_ t where is angle of twist in radians per unit length of one inch cf tube. For a tube length ------------ A6.9 Expression for Torsional Moment in Terms of Internal Shear Flow Systems for Multiple Ceil Closed Sections. Fig. A6-.16 shows the internal shear flow pattern for a 2-cell thin-walled tube, when the tube Is subjected to an external torque. qA, q3 and q3 represent the shear load per inch on the three different portions of the cell walls. For equilibrium of...

## Effect Of Temperature On 7075 Aluminum Alloys

B2.75 Effect of exposure at elevated temperatures on the room-temperature tensile yield strength (Fty) Of 7075-T6 aluminum alloy (all products). Fig. B2. 72 Effect of temperature on the ultimate tensile strength (Ftu) of 7075-T6 aluminum alloy (all products) Fig. B2. 73 Effect of temperature on the tensile yield strength (Fty) of 7075-T6 aluminum alloy (all products). Fig. B2. 74 Effect of exposure at elevated temperatures on the room-temperature ultimate tensile strength (Ftu) of 7075-T6...

## S5

Determine the crippling stress for the formed section shown in Fig. e if material Is aluminum alloy 2024-T3. FCy 40,000, Ec 10,700,000. The section is divided into 6 angle units by the dashed lines In Fig. e. They are numbered (1) to (3) since we have symmetry. The procedure will be to find the failing load for each angle and add up the total for the 6 angle units. The crippling stress will then equal this total load divided by the section area. b' t (a + b) 2t (.375- .02) + (.5- .02) .OS...

## Problem

Fig. (i) Illustrates a welded plate fitting unit fastened to a round steel tube. 3oth fitting plate and tube are steel Ftu 95000. VTnat is the maximum design load which the fitting car. be subjected to if a fit Sing factor of 1.2 is used. Fitting is not subjected to vibration or rotation o.> hinge -in. (2) Same as Problem 1 but tube and fitting is heat-treated after welding to 7 -150,000. A -.051, 2024-T3 aluminum sheet carries an ultimate tension load of 00 lbs. per inch It is spliced by a...

## Buckling Under Shear Loads

C5. 7 Buckling of Flat Rectangular Plates Under Shear Loads. The critical elastic shear buckling stress for flat plates with various boundary conditions is given by the following equation Where (b) is always the shorter dimension of the plate as all edges carry shear. ks is the shear buckling coefficient and is plotted as a function of the plate aspect ratio a b in Fig. C5.ll for simply supported edges and clamped edges. If buckling occurs at a stress above the proportional limit stress, a...

## 3316

Symmetrical About One Axis. Fig. A15.30 shows a two cell cantilever beam with 10 flange stringers. The cross-section is constant. Let it be required to determine the internal shear flow In resisting the 1000 lb. load acting as shown. For simplification, the top and bottom sheet covering and the three vertical webs will be considered ineffective In taking bending flexural loads. Since the beam section is symmetrical about the X axis, the beam will bend about...

## Ifg

(6) Find shear center location for beam in Fig. A1S.81 if the 3 flanges provide the entire bending resistance. (7) Find the Internal resisting shear flow pattern for the 3 flange-single cell beam of Fig. A15.82. Assume webs or walls ineffective in bending. (8) Determine shear center location for beam of Fig. A15.82. Webs and walls are Ineffective In bending.

## Loads And Stresses On Ribs And Frames

For aerodynamic reasons the wing contour in the chord direction must be maintained without appreciable distortion. Unless the wing skin Is quite thick, spanwise stringers must be attached to the skin in order to increase the bending efficiency of the wing. Therefore to hold the skin-stringer wing surface to contour shape and also to limit the length of stringers to an efficient column compressive strength, internal support or brace units are required. These structural units...

## 575 W

K* gust correction factor depending on wing loading (Curves for K are provided by Civil Aeronautics Authorities). V indicated air speed In miles per hour. * NACA Technical Note 2964 (June 1953), proposes that the alleviation factor K should be replaced by a gust factor, Kg 0. 88 Mg (5. 3 + Mg). In this expression Mg is the airplane mass ratio or mass parameter, 2 W apcgs, in which c is the mean geometric chord in feet and g the acceleration due to gravity. If U Is taken as 30 ft. sec. and m as...

## Ultimate Strength Of Stiffened Cylindrical Structures

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 a semi-monccoque structure Involves the so lux ion of two major problems, namely, the stress distribution in the structure under various external loadings and the check of...

## 04

The rivet or spot weld spacing is made such as to prevent lnter-rivet buckling. Thus the column area will be as shown in Fig. c, namely, the stiffener area plus the area of the sheet for the effective width w. Since the effective sheet width w is a function of the stiffener stress and since the stiffener stress is a function of the radius of gyration, the design procedure is of the trial and error category. Assume the effective sheet width is based on column strength of 2 stiffener acting...

## 190t iEFst

Some early experiments by Newell indicated the constant 1.90 was too high and for light stringers a value of 1.7 was more realistic, thus 1.7 has been widely used In Industry. If we assume the stiffness of the stiffener and Its attachment to the sheet as developing a fixed or clamped edge condition for the sheet, then Fcr 6.35E(t br or w s 2.52t VE Fg For general design purposes, it is felt that 1.9 or equation C7.15 is appropriate for determining the effective width w. If stiffener is...

## A A

Fc-19 'j 'g 191 j > g 19 Average g 18. 83 a) Y-stiffened panel. Fig. C7.22 Method of cutting stiffened panels to determine g. article. This sheet buckling does not deform the flange of the stlffener to < vhlch the sheet is attached. However, if the rivet or spot weld spacing is such as to prevent inter-rlvet buckling of the sheet, then failure often occurs by a larger wrinkling of the sheet as Illustrated in Fig. C7.26. The larger wrinkle shape subjects to flange of the stlffener to which...

## P

The A and 9 curves are recommended Tor practical design. The A level curve is recommended ior use for those structures where a single failure would result In catastrophic loss or injury to personnel. The stress level at the small end of the cone should be checked to preclude the possibility of an early failure precipitated by inelastic stresses. C8.19 Additional Design Buckling Curves for Thin-Walled Conical Shells. Figs. C3.25, 26, 27 gives curves for determining the allowable buckling stress...

## Ship Structur Plats Name

METHOD 1 QF BOLT & LUG STRENGTH ANALYSIS Failure in Tension. Fig. D1.8 indicates how a fitting plate can pull apart due to tension stresses on a section through the centerline of the bolt hole. Both the male and female parts of the fitting must transfer the load past the centerllne of the hole, thus both parts must be considered in the design of a fitting. Equating the allowable load Pu to the ultimate resisting tensile stresses at points (a) and (b) Fig. Dl.S, we obtain, where Ftu ultimate...

## Ultimate Torsional Strength Of Round Tubes

In Article A6.2 of Chapter A6, the torsion formula for circular sections, fs Tr J, was derived. This equation assumes the maximum shear stress on the cross-section of a round bar or tube does not exceed the proportional limit of the material, or the stress variation is linear as shown in Fig. C4.15 and this situation exists under the flight vehicle limit loads. Before a round bar made of ductile material fails In torsion, the shear stresses fall in the inelastic or plastic range and thus the...

## 260 270 110

* Properties for annealed condition also applicable to annealed AISI 303, 303, 304, 321. and 34 . 1 Only annealed condition applicable to plate. See table 2.8.1.1 (b). NOTE. Yield strength, particularly in compression, and modulus ol elasticity in the longitudinal direction may be raised appreciably by thermal 3 tress-relieving treatment in the range MO* to 800 T. A IS I 301 STAINLESS STEEL (Cont.) 5 Fig. B2. 39 Effect of temperature on the ultimate tensile strength (Ftu) of AISI 301...

## Other Types Of Nonbuckling Beam Webs

At this point It should be noted that the web design discussed in Part 1 required a large number of parts (the stlffeners) to achieve lightness. To keep manufacturing costs down, the number of parts must be kept to a minimum. A balance, or economic compromise must be found between manufacturing expense and weight penalty. This can be arrived at since in every airplane design some dollar penalty can be assigned to every extra pound of weight. The exact figure will depend upon the type of...

## X

Fig. 2.10 shows a 2 bay trass supported at points A and 3 ana carrying a known load system P, Q,, All members of the truss are connected at their ends by a common pin at each Joint. The reactions at A and 3 are applied through fittings as indicated. The question is whether the structure is statically determinate. Relative to external reactions at A ana 3 the structure is statically determinate because the type of support produces only one unknown at A and two unknowns at 3, namely, V3 ana Ha as...

## 129

(I0 of (2) about its x centroldal axis is negligible) px f* , Area A 2.5 x 2.75 - 2 x .75 x 1.25 - px v 3.03 4.375 .83 in. Calculation of Iy Portion (1) -(1.25 X .75 x .875a)2 -(1.25 X .753 12)2 - 1.52 Portion (2) -( .25 X .625 X ,833)4-4( .25 x 1.25V36) - .438 Column strength is considerably influenced by the end restraint conditions. For failure by bending about the x-x axis, the end restraint against rotation is zero as the single fitting bolt has an axis parallel to the x-x axis and thus c...

## S

Kur the slress relieved tciri Kr-T35l (plate only), till mines lifr I In- -T4 tenner a ijily with Ilk' eiceinlon ol K. which inay he some hu( In v. i-r I lent Iri'uliil by user refers m nil muli-riiil solution I lent 11 ruled hy I lie user nyuriUi-ss o( I ho prior temivr uf (lie material. * ft ietlftcuilon mlnlmiims lor chid nuiirrlul n. MM loch ihlel. unil heavier are lor (lie core inuteiiul. liiuMiiui'li us ii round lest si .crimen is relinked for testlnir The values clveii hero lor...

## Physical Explanation Of The Method

A15.40 shows a 3-cell beam carrying and external shear load V acting through the shear center of the beam section but as yet unknown in location. In other words, the beam bends about the symmetrical axis X-X without twist. The problem is to determine the internal resisting shear flow system for bending without The Analysis of Shear Distribution for Multi-Cell Beams in Flexure by Means of Successive Numerical Approximations. By D. R. SAMSON. Journal of the Royal Aeronautical Society, Feb....

## Joint F

IK .0500 + 0.400 + 0.1540 0.6040 For the straight members the carry-over factor is 0.5 and the moment sign is the same as the distributed moment when the sign convention adopted in this chapter is used. For a half circular arch of constant I, the carry-over factor was derived in the previous article and was found to be 0.452. The sign of this carry-over moment was the same sign as the distributed moment at the other end of the beam. However, using the sign convention as adopted for the moment...

## Rsf Erenc Es

Strength of Metal Aircraft Elements March, 1355. (2) Military Handbook MIL-HDBK-o. August, 1S62 Metallic Materials and Elements for Flight Vehicle Structures. (3) Cozzone, MeIcon, and Hobllt Analysis of Lugs and Shear Pins Made of Aluminum and Steel Alloys. Product Engineering Vol. 21, May, 1350. (4) Me icon and Hobllt Developments in the Analyses of Lugs and Pins. Product Engineering. June, 1353. ( 5) Various Structural Manuals of Various Arcraft Companies.

## 1200

28 28 4> - 28 - 28 (7) Same as problem (6), but Instead of a cantilever truss use a simply supported truss with supports at points (A) and (B). (8) Fig. 5 shows a front beam and front lift strut in an externally braced monoplane. The wing beam and lift strut are in the same vertical plane. The ultimate design loads on the beam for the critical conditions are w - 50 lb. in. and w - -30 lbs. per inch. Minus means load is acting down. (a) Design a streamline tube to act as the lift strut....

## End Fixity Coefficient Of Fuselage Truss

(1) Fig. 1 shows a portion of a steel tubular fuselage of a small airplane. The critical tension and compression load is shown adjacent to each truss member. Assuming and end fixity coefficient c 2 for all members, select tube sizes for all members of the truss. The minimum size to be used Is 3 4 - .035. The top and bottom longerons should be spliced at least once using telescoping sizes. The material is alloy steel Ftu s 95000 and truss is welded. (2) For the cantilever welded steel tubular...

## Fuselage Stress Analysis

Substituting K values in equation for av, a - .00262 x-146310 - (-.0001245 x 611800) y - Q00156 x611300 -(-.0001248 X-146310Q Z a j 307.0 y -936.1 z (plus ab Is tension) -19.4 (516.6 x 286 - 19.4 a) -19.4 147620 -.0001315 CTb - 70035 c-116800-(-.0001315 x 49215oT y - Qc01936 x 432150 -(-.0001315x 116320 Z Column (12) in Tables A20.7 and. A20.S gives the results of solving the equations for a . Since an external load of 1500 lb. is acting normal to the sections and through the section centroids,...

## Problems

(1) A round tube 1-1 2 inches in diameter has a wall thickness of .095 inches. It is made of aluminum alloy whose stress-strain curve is shown in Fig. C3.2. If the maximum unit strain in compression is limited to .008, what bending resisting moment win the section develop. Note, since stress-strain curve has different shape in tension and compression, neutral axis does not coincide with center line axis, thus use trial and error method. (2) Same as Problem (1) but use a square tube with 1-1 4...

## A44

Thus Fn and Ft equal zero and thus the only inertia fores for the pure rotation Is I a, (a couple) and thus the moment of this inertia couple about the 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 The wing of an airplane carries the major portion of the air forces. In level steady flight the vertical upward force of the air on the...

## Referring Redundants To Elastic Center

For the purpose of simplifying equations 5, 6, 7, let it be assumed that end A is attached to a inelastic arm terminating at a point (o) as illustrated in Fig. A9.6. The point (o) coincides with the centrold of the ds EI values for the structure. Reference axes x and y will now be taken with point (o) as the origin. The redundant reactions will now be placed at point (o) the end of the inelastic bracket, as shown in Fig. A9.6. Since point A suffers no movement In the actual structure, then we...