Aircraft Structure-I Manual

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    Ex. No. 1 Buckling of columns

    Aim:

    To determine the critical load of a column using south-well plot.Theory:

    The need to make use of materials with high strength to weight ratio in aircraft design has

    resulted in using of slender structure components that fail more often by instability the simplestexample is a slender column. Ideal column under small compressive load is slightly disturbed and

    return to original position after removal for particular loading, it takes neighboring column

    equilibrium position this is neutral equilibrium. The instability occurs at Euler load or critical load.

    The ideal column deflection occurs suddenly, but in actual column it appears as

    soon as load applied.

    South well should have a relation between applied load and correspondingdeflection, which can be used to determine critical load, eccentrically by a graphical

    procedure without destroying the specimen.

    The well known formula for critical load is,

    Where c is a const. depending on the end condition of the column

    E- Youngs modulus of the material,

    I- Moment of Inertia, L- Length of section.

    For a column which is always imperfect, the deflections are determinate at all loads. For

    example the deflection of a simply supported beam ( SSB ) column at its middle due to load

    P can be written as,

    It can be written as .

    Apparatus Required:

    Column testing apparatus

    Specimen

    Screw Gauge

    Vernier Caliper

    WeightsProcedure:

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    The given column is aligned on the column testing apparatus with its longitudinal axis

    vertical.

    The dial gauge is placed at the mid point of the column.

    New loads are applied gradually in steps.

    The corresponding deflections of a dial gauge are noted and tabulated.

    New . Ration is calculated and the graph of Vs... Plotted. The inverse slope of the load curve gives the critical load of the material for given

    dimensions.

    Result:

    Thus the crippling load was determined experimentally and the theoretical value wasverified for mild steel aluminum and Brass.

    Ex. No:2 Determination of fracture strength and fracture pattern of ductile material

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    Aim:

    To determine the fracture strength and study of fracture pattern of the given mild steel

    specimen.Apparatus Required:

    UTM (Universal Testing Machine )

    Mild steel Specimen

    Formula Used:

    Ultimate Stress = Ultimate load / Cross section of the specimen

    Procedure:

    Fix the mild steel specimen in the UTM.

    Apply the tensile load gradually.

    Carefully observe the various stages of specimen after it is getting stretched.

    Note down the load at which fracture occurs. This is called ultimate load and

    corresponding stress is called ultimate stress. Draw the fracture pattern of the mild steel specimen.

    Types of fracture in tension:

    There are two kinds of fracture to the distinguished in tension of a single critical

    specimen with a material such as rock salt, we have brittle fracture without substantial plasticdeformation and fracture occurs when the magnitude of normal n anyone of the principle planes

    reach critical values. This is called cohesive fracture. Single critical specimens of metal usually

    show large plastic deformations along certain crystal planes. This is known as shear fracture.

    The relation between resistance to separation and resistance to sliding do not remainconstant for the same material. It depends on temperature of specimen at which the test made.

    In case of polycrystalline specimen there are two kinds of fracture as given as,

    Brittle fracture Shear fracture

    In the first case fracture occur practically without plastic deformation over a cross sectionperpendicular to the cross sectional axis of the specimen. In the second case the fractured

    occurs with plastic deformation.

    These are two kinds of fracture can again be forwarded based on the two characteristics the

    resistance to sliding and the resistance to separation. The up and low fracture occurs only after

    a considerable uniform stretching and subsequent local reduction of the cross sectional of the

    specimen.

    The stress distribution in the neck has been increased and it will be found that near thetensile force in longitudinal fiber has directional indication by arrows. The horizontal

    component produces radial and tangent stresses so as to the decimal elements having maximumshear stresses the constant tension test of plastic flow axially. The specimen has maximum

    stress distribution as shown in figure by shaded area. The magnitude of Imax and Imin depends on

    radius of minimum cross section and radius of curvature R. of the neck and are given byformula.

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    Result:

    The fracture pattern of the given mild steel was observed

    The ultimate strength of mild steel specimen = ..

    Ex. No: 3 Determination of South-well Plot

    Aim:

    To determine the critical load of column using south well plot

    Apparatus required: Column testing apparatus

    Specimen

    Dial gauge

    Vernier caliper

    Weights

    Theory:

    The need to make use of materials with high strength to weight ratio in aircraft design has

    resulted in using of slender structure components that fail more often by instability ( buckling )then by excessive stress..

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    Ex. No: 4 Determination of Youngs Modules for various engineering Material

    Aim:To find the defection of given simply supported beam with central point load and to

    calculate Youngs modules for given material.

    Apparatus Required:

    Simply supported beam Dial gauge

    Magnetic stand

    Weight pan with weights

    Measuring scale

    Formula Used:Deflection of simply supported beam with central point load

    Y = wl3/ 48 EI

    E = wl3/ 48 yI

    Where,

    E- Youngs Modules in N/mm2I- Moment of Inertia in mm4

    l- Length of the beam in mm

    w- Load in NProcedure:

    First considering mild steel as test specimen, the weight pan is placed at its center.

    The first load is placed and the deflection for beam is noted.

    Various steps of loading are done and different values of deflection are noted.

    Now, loads are removed in steps and values are noted.

    Length, breadth and depth of mild steel is measured.

    Youngs Modules is calculated using formula. Same procedure is repeated wooden beam

    Result:The values of Youngs modules for given,

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    Mils steel beam =..

    Wooden Beam = ..

    Ex. No : 5 Verification of Maxwell Reciprocal theorem

    Aim:

    To verify the Maxwells reciprocal theorem using given beamApparatus Required:

    Dial; gauge

    Vernier caliper

    Measuring scale

    Weight pan with weights Simply supported Beam

    Maxwell Reciprocal theorem:

    In an beam, the deflection at any point D due to a point load w at a point C other than

    D in the beam will be same as the deflection at c due to the load at D.Procedure:

    First the points D and C are marked on the beam at equal distances from the ends.

    The loads are first applied at a appoint D and the deflections corresponding to the load

    are measured at point C

    Similarity readings are noted for unloading the loads

    The average deflections are calculated

    Note the loads and dial gauge is interchanged between the points and deflections at D arenoted.

    A graph between applied load and deflection is drawn.

    Result:

    The Maxwell Reciprocal theorem was verified for given beam

    Slopes of two graphs ate found to be same.

    Ex. No: 6 Deflection of beam subjected to eccentric point load

    Aim:To find the deflection of simply supported beam when a point load is applied eccentrically.

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    Apparatus Required:

    Simply supported Beam

    Dial gauge with magnetic stand

    Weight span with weights

    Measuring scale

    Formula Used:Deflection of simply supported beam with eccentric point load is given by,

    Yc = Wa2b2 / 3EIl

    Where,E= 1.8 *105N/mm2

    I = bd3 / 12 mm4

    b= breadth of the beam in mm

    d= depth of the beam in mm

    I= Moment of inertia in mm4E= Youngs Modules N/ mm2

    Procedure:

    Measure the length, breadth, depth of the given beam.

    The youngs Modules value can assumed to be 1.8 *105N/mm2, through our earlyequipment on beams.

    Mark a point on the beam at a distance a from one end and b from other end.

    Add weights and note the deflection at marked while loading and unloading.

    Also find the theoretical equation value for the given load by using formula and compare

    the experimental value with the theoretical value.

    Result:

    The deflection of the simply supported beam with eccentric point load has been found out

    thus and has been compared with theoretical value

    Ex. No: 7 Riveted Joints

    Aim:

    To study about the riveted joints.

    Lap Joints and Butt Joints:It of often necessary to joint together two members of the structures. This can be done by

    means of rivets. This joints may be either a lap joint or butt joint.In a lap joint a part of one member overlaps apart of the other and the two part of

    one are then jointed by rivets. In a butt joint, the edges of the two members butt against each other

    ( ie the ends touch each other ). Either placed one cover plate or two cover plate one on each side isplaced on the joints and the cover plate is used then the joints is called as single cover butt joints. If

    two cover plates are used then the joints is called double cover butt joint.

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    The common line along with the edges of the main plates meet is called ht e length of the

    joint as shown by dotted lines in the figure. A line of rivets parallel to the length of joints is called

    row of the rivets. The row nearest to the joints shall be tuned the inner most row while the one forthus from the joint shall be termed the outer most row.

    If there is only one row of rivets in the lap joints or one row on each side in butt joints.Then the joint is called as single riveted joints. If there are two rows of rivets is a lap joints or two

    rows on either in a butt joint may be triple riveted double riveted and so on.

    The distance center to center between two consecutive rivets in the same row is calledPitch (P) of the rivets. Some times the pitch is different for different rows in such cases the least at

    the inner most row and is called the short pitch. While it is greater at the outer most row and it is

    called Long pitch. The pitch of the intermediate row is called middle pitch. The distance center to

    center of two consecutive rows is called the back pitch. When a group of rivets is such that thesame pattern is repeated along the length of the joints. It is called the repeating section. The length

    of the repeating section is equal to the long pitch as far as possible lap joints and single cover butt

    joins should be avoided since the forces acting on those points, which possess distortion of the

    joint as shown in Figure (i).Failure of riveted Joints:

    A riveted joints may fails is one of the following ways:

    Rivet may be shared off.

    The rivets are the part of the plate is contact with it, may be crushed due to the bearing

    pressure between the two.

    The main plate or cover may tear across a row of rivets.

    The main plate may tear diagonally between ht rivets of two successive rows. This kind of

    failure can be avoided by keeping the back pitch at least equal to 13/4 times the diameter ofthe rivets.

    The plate may tear or shear between the edge and the nearest rivet. This kind of failure can

    be avoided by keeping the distance of center of the nearest rivet form the edge of the plateat least 13/4 times of the diameter of the rivet.

    Since failure by last two methods can be easily avoided. It is necessary to examine failure

    by methods of first three methods in the analysis and design of s riveted joints.

    Rivets:

    Ex. No: 8 Determination of Youngs modules value of given Aluminum usingmechanical extensor meters.

    Aim:

    To find the youngs modules vales of Given Aluminum beam by using mechanical strain

    Gauges.Apparatus required:

    1. Cantilever Aluminum beam

    2. strain gauges3. strain Indicator

    4. Weights

    5. measuring scale6. vernier caliper

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    Formula Used:

    Strain = pl6/bt2E

    M / I = F/te = pl / bt2E*6

    Piezoelectric = 6pl / bt2.E

    E = 6pl / bt2

    .eStrain gauges:

    If a metal conductor is stretched or compressed, its resistance changes ie there is a change

    in the value of resistively of the conductor when it is strained and this property is called Piezoelectric effect. Therefore resistance strain gauges are used for measuring strain and associated

    stress in experiment stress analysis.

    Digital strain Indicator:

    It is a compact and portable read out instrument designed for strain measurement by meansof strain gauges ranged form 100 to 1000ohms. This instrument provides highly accurate and

    reliable values directly on the meter..

    Procedure:

    Connect the cantilever beam to the strain indicator using four power cable Plug the main chord to the 230V/50Hz AC main.

    Switch on the instrument.

    Turn the functional switch to read position. Adjust zero potentiometer until the display read

    0 0 0.

    Turn the functional switch to GF position. Adjust GF potentiometer to read 500 on thedisplay.

    Turn the function to CALL Position. Adjust CALL potentiometer until display reads 1000.

    Once again turn the functional switch to read position to ensure whether it is showing 0 0 0.

    Apply the load the load in terms of 100g.

    Display directly shows the micro strain for the corresponding load.

    The reading what we have obtained form the strain indicator is the value of 1Amps. So toget value for 4Amps we have to multiply the value by 4.

    The youngs modules value for the given Aluminum found out.

    Result:

    The youngs modules vales for the given Aluminum beam is found out as ..