structures lab finl copy

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AIRCRAFT STRUCTURES LAB (06AEL57) =============================================================== =============================================================== DEPT. OF AERONAUTICAL ENGINEERING, DSCE 1 Experiment – 1 Deflection of a Simply Supported Beam Aim: To determine the deflection of a simply supported beam. Apparatus: Beam Test Set-Up with Load cells, steel scale, caliper, flat beam, load indicator and dial gauge. Scale Gear box Pointer Hand wheel Load Load indicator Simply supported beam Dial gauge Fig. 1: Schematic layout of a beam setup Theory: A beam shown in fig-2 shows the section which is simply supported at the ends and is subjected to bending about its major axis with a concentrated load anywhere in the beam. The beam is provided with strain gauge, the deflection of the beam can be determined whenever the load is applied on the beam. Strain gauge values may be noted for several further works

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Page 1: Structures Lab Finl Copy

AIRCRAFT STRUCTURES LAB (06AEL57)

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=============================================================== DEPT. OF AERONAUTICAL ENGINEERING, DSCE 1

Experiment – 1

Deflection of a Simply Supported Beam

Aim: To determine the deflection of a simply supported beam.

Apparatus: Beam Test Set-Up with Load cells, steel scale, caliper, flat

beam, load indicator and dial gauge.

Scale

Gear box

Pointer

Hand wheel

Load

Loadindicator

Simply supportedbeam

Dialgauge

Fig. 1: Schematic layout of a beam setup Theory:

A beam shown in fig-2 shows the section which is simply supported at

the ends and is subjected to bending about its major axis with a concentrated

load anywhere in the beam. The beam is provided with strain gauge, the

deflection of the beam can be determined whenever the load is applied on

the beam. Strain gauge values may be noted for several further works

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.

Deflections are given by following expressions; students may derive these

expressions using Unit Load Method or Castigliano`s Theorem.

Yx = W b [ (L2 –b

2) x –x

3] /(6 E I L) for 0<x<a

Yx = W b [ X3 – L/b (x-a)

3 - (L

2-b

2) x ] /(6 E I L) for a < x <L

Where,

W is the load placed at a distance `a` from the left support in Newton

L = span of the beam in mm

Yx = deflection at any point distance x from left end

I = moment of inertia of the beam in mm4 (Ixx)

E= Young’s modulus in N/mm2

Procedure:

Find the moment of inertia of beam from the following expression:

(1/12) b1 d1 3, where b1 is width of beam and d1 is depth.

Place the beam supporting from two wedge supports. The load position can

be varied. Set the load cell to read zero in the absence of load. Set the

deflection gauge to read zero in the absence of load.

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Load the beam with 0.25 Kg. Note deflections before and after the load point

through deflection gauge. Increase the load to 1.5 Kg and repeat the

experiment.

Find the deflections from the formula and verify.

Tabular column

Load

(N)

Ixx

mm4

L

mm

a

mm

b

mm

x

mm

Theoretical

value mm

Experimental

value mm

CAUTION : NEVER EXCEED 3 Kg LOAD ON THE BEAM

Conclusion: Deflection for three different loads for simply supported beam

is calculated and also verified by experimental values. The mean deflection

is ------units.

Example:

Determine the deflection of a simply supported beam loaded with W

=50,000 Newton, Young’s Modulus E =2 x105 N/mm

2; Second moment of

Inertia Ixx =7332.9x 104

mm4; & the load is placed at a distance a = 4800mm;

and the span of the beam L = 60000mm. Find the deflection at x =3000mm

from the left end.

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=============================================================== DEPT. OF AERONAUTICAL ENGINEERING, DSCE 4

Yx = W b [ (L2 –b

2) X –X

3] /(6 E I L)

= 50000 x 1200x [ (60002 - 1200

2) 3000 -3000

3] /(6x2.0x10

5x 7332.9

x 104

x 6000)

Yx = 8.714 mm

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Experiment – 2

Verification of Maxwell’s Reciprocal Theorem

Aim: To verify the Maxwell’s theorem for the structures system

Apparatus: Beam Test Set-Up with Load cells, steel scale, caliper, flat

beam, load indicator and dial gauge.

Scale

Gear box

Pointer

Hand wheel

Load

Loadindicator

Simply supportedbeam

Dialgauge

Fig. 3: Schematic layout of a beam setup

Theory:

The displacement at point i, in a linear elastic structure, due to

concentrated load at point j is equal to the displacement at point j due to a

concentrated load of same magnitude at point i.

The displacement at each point will be measured in the direction of the

concentrated load at that point. The only other restrictions on this statement,

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in addition to the structure being linear elastic and stable, is that the

displacement at either point must be consistent with the type of load at that

point. If the load at a point is a concentrated force, then the displacement at

that point will be a translation, while if the load is moment, then the

displacement will be rotation. The displacement at any point will be in the

same direction as the load at that point and its positive direction will be in

the same direction as the load.

This theorem often referred to as Maxwell’s reciprocal displacement

theorem.

This can be proved through Unit Load Method i.e.; the deflection at A due

to unit load at B is equal to deflection at B due to unit load at A.

δ = ∫ M m dx / EI

where,

M = Bending Moment at any point x due to external load

m= Bending Moment at any point x due to unit load applied at the point

where deflection is required

let, mxA = Bending Moment at any point x due to unit load at A

mxB = Bending Moment at any point x due to unit load at B

When unit load (external load) is applied at A, M=mxA

To find deflection at B due to unit load at A, apply unit load at B

Then m=mxB

Hence, δBA = ∫ M m dx / EI δ = ∫ mxA mxB dx / EI ------(1)

Similarly, when unit load (external load) is applied at B, M=mxB

To find deflection at A, then m=mxA

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Hence

δAB = ∫ M m dx / EI = ∫ mxA mxB dx / EI ------------------(2)

Comparing eqn. (1) and eqn. (2)

δAB = δBA-------------------------------------------------------------------------------------------- (3)

The external load (W) can be taken as a multiple with unit load, therefore

this load W will appear as multiple with mxA in eqn. (1) & as multiple with

mxB in eqn. (2). Thereby resulting in

W δAB = W δBA------------------------------------------------------ (4)

A beam shown in figure below which is simply supported at the ends

and is subjected to bending about its major axis with a concentrated load

anywhere in the beam.

Deflections δx at any distance `x` from left support are given by following

expressions; reference may be made to experiment no. 1.

δx = W b [ (L2 –b

2) x –x

3] /(6 E I L) for 0<x<a

δx = W b [ X3 – L/b (x-a)

3 - (L

2-b

2) x ] /(6 E I L) for a < x <L

Where,

W is the load placed at a distance `a` from the left support in Newton

b=distance of load from right side support

L = span of the beam in mm

Yx = deflection at any point distance x from left end

I = moment of inertia of the beam in mm4 (Ixx)

E= Young`s modulus in N/mm2

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The Maxwell’s Reciprocal Displacement theorem is very useful in the

analysis of statistically indeterminate structures for evaluating the flexibility

coefficients.

The displacement relationship can be expressed at point i and j

δ i,B =f i,j W--------------------------------------------------- (5)

δ j,A =f j,i W--------------------------------------------------- (6)

Where, f i,j is the displacement at point i due to a unit load at point j and f

j,i is the displacement at point j due to a unit load at point i. If we now

substitute these expressions in Betti`s (Maxwell-Betti) law and cancel out

the term W on each side, we obtain

f i,j = f j,i

δ21 δ11

P

Fig-4: Simply supported Beam loaded at position 1 & 2

δ12 δ22

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The theorem can be restated as the displacement at point i, in an elastic

structure, due to a unit load at point j is equal to the displacement at point j

due to unit load at point i.

Procedure:

Find the moment of inertia of beam from the following expression:

(1/12) b1 d1 3

where b1 is width of beam and d1 is depth. Place the beam supporting from

two wedge supports. The load position can be varied. Set the load cell to

read zero in the absence of load. Set the deflection gauge to read zero in the

absence of load.

Load the beam with 0.25 Kg. Note deflections at any point through

deflection gauge. Interchange the load location with the point of deflection

measurement and repeat the readings. Increase the load to 1.5 Kg and repeat

the experiment. Find the deflections from the formula and verify.

CAUTION: NEVER EXCEED 3 Kg LOAD ON THE BEAM .

Tabular column

Load

(N)

Ixx

mm4

L

mm

a

mm

b

mm

x

mm

Theoretical

value mm

δBA or δAB

Experimental

value mm

δBA or δAB

Conclusions: Maxwell’s reciprocal theorem is verified from the results i.e,

δ21=δ12. Deflection at 1, 2 are equal with interchange of loads (same

magnitude and direction).

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Experiment – 3

Determination of Young’s Modulus using strain gauges

Aim: To determine the Young’s modulus of a simply supported beam or a

cantilever beam.

Apparatus: Beam Test Set-Up with load cells, Cantilever beam with

calibrated rosette strain gauge, strain measuring equipment and load

indicator.

Gear box

Strainindicator Load

indicator

Pointer

Scale

Load

Hand wheel

Cantileverbeam

straingauge

c1 c2 c3

c2

c1

c3

Strain guagearrangement

Fig. 5: Schematic layout of beam setup for Young’s modulus determination

Theory:

A beam shown in fig-4 shows the section which is simply supported at

the ends and is subjected to bending about its major axis with a concentrated

load anywhere in the beam. The beam is provided with strain gauge, the

deflection of the beam can be determined wherever the load is applied on to

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the beam. An available cantilever beam can also be utilized for this

experiment.

The strain gauge is at a fixed position in the beam and load position

can be varied. A strain gauge is mounted on a free surface, which in

general, is in a state of plane stress where the state of stress is with regards to

a specific xy rectangular rosette. Consider the three element rectangular

rosette shown in fig-6, which provides normal strain components in three

directions spaced at angles of 450. If a xy coordinate system is assumed to

coincide with the gauge A and C then εx= εA and εy= εc. Gauge B provides

information necessary to determine γxy. Once εx, εy and γxy are known, then

Hooke’s law can be used to determine σx, σy, and τxy. However in this case

the requirement is to determine Young`s Modulus (E), which can be

determined from equation (1) below.

)()1( 2 yxx v

v

Eεεσ +

−= ; )(

)1( 2 xyy vv

Eεεσ +

−= ;

xyxyv

Eγτ

)1(2 +=

εx = εA; εy = εc; γxy = 2 εB - εA -εc ;

Sl.No Load (N)

εA εB εC εx εy γxy ν=εy/εx

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M/I = σ/y --------------------------------------------------------------------------1

also

Shear Modulus xy

yxG

γ

τ=

------------------------------------------------2

Young’s Modulus E = 2G (1+ν) ------------------------------------------------3

Procedure: Mount the cantilever beam at the left support of beam test set-

up. Connect the strain gauges wires with the strain measuring equipment.

Use the following color codes

εA Red wires ch1

εB Yellow wires ch3

εc Black wires ch2

Set the load cell to read `zero` value in the absence of load. Set the three

strains to read `zero` in the absence of load. Now Load the beam with 0.25

Kg at some point and record the strains in three directions. Record the load

value at the load cell.

Repeat the experiment with load value of 1Kg. Compute the values of

Poisson’s ratio from:

ν=εy/εx

CAUTION: NEVER EXCEED 3 Kg LOAD ON THE BEAM

Tabular Column:

Sl.No Load

(N) εA

εB εC εx εy γxy ν=εy/εx

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Find Young `s modulus through formulas above.

Conclusion: Young’s modulus for cantilever beam with point load is

evaluated as ------Units.

Example:

Sl No. Load

(N) εA εB εC εx εy γxy ν=εy/εx

1 10N 92µ 46µ -18µ 92µ -18µ 18µ 0.1957

2 20N 64µ 304µ -18µ 64µ -18µ 14µ 0.2188

3 30N 37µ 19µ -10µ 37µ -10µ 11µ 0.2973

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Experiment – 4

Poisson`s ratio (ν) Determination

Aim: To determine the Poisson’s ratio of cantilever beam.

Apparatus: Beam Test Set-Up with load cells, Cantilever beam with

calibrated rosette strain gauge, strain measuring equipment and load

indicator.

Theory: A cantilever beam is subjected to bending about its major axis with

a concentrated load anywhere in the beam. The beam is provided with

rosette strain gauge.

A calibrated strain gauge rosette is fixed at a location with-in the span

of the beam, and load position can be varied. Calibration has been done to

read strains in microns (µ). Consider the three element rectangular rosette

shown in fig-7, which provides normal strain components in three directions

spaced at angles of 450. If an xy co-ordinate system is assumed to coincide

with the gauge A and C then εx= εA and εy= εc.

Gauge B provides information necessary to determine shear strain

(γxy). Once εx, εy and γxy are known, then Hooke’s law can be used to

determine σx, σy, and τxy. Subsequently principal stresses can be determined.

Poisson’s ratio ( ν ) can be determined from:

ν=εy/εx

Fig. 7

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)()1( 2 yxx v

v

Eεεσ +

−= ; )(

)1( 2 xyy vv

Eεεσ +

−= ;

xyxyv

Eγτ

)1(2 +=

εx = εA; εy = εc ; γxy = 2 εB - εA -εc

Principal stress axes:

Principal stress axes is located with the angle θ according to :

tan 2θ = (2 εB - εA -εc) / (εA -εc)

Principal Stresses are given by following expressions:

σ1 = A + √ B2 + C

2 , σ2 = A - √ B2 + C

2

τmax = ( σ1- σ2 )/ 2 =√ B2 + C

2

Where, A= (σx+ σy)/2 , B = (σx - σy)/2 and C = τxy

E= Young’s modulus in N/mm2

Strains are in microns

Procedure: Mount the cantilever beam at the left support of beam test set-

up. Connect the strain gauges wires with the strain measuring equipment.

Use the following color codes

εA Red wires channel 1

εB Yellow wires channel 3

εc Black wires channel 2

Set the load cell to read `zero` value in the absence of load. Set the three

strains to read `zero` in the absence of load. Now Load the beam with 0.25

Kg at some point and record the strains in three directions. Record the load

value at the load cell.

Repeat the experiment with load value of 1.5 Kg. Compute the values of

Poisson’s ratio from: νννν=εεεεy/εεεεx

CAUTION: NEVER EXCEED 3 Kg LOAD ON THE BEAM.

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Students may determine stresses using formulas above. MATLAB program

is provided at the end of this manual to compute stresses from given strain

data.

Tabular Column:

Sl.No Load

(N) εA

εB εC εx εy γxy

ν=εy/εx

Conclusion: Poisson’s ratios for three different loads are evaluated and

mean Poisson’s ratio for a cantilever beam is=-------

Example: A typical data is shown for the purpose of computing shear strain

and Poisson’s ratio is given below.

Sl.No load εA εB εC εx εy γxy ν=εy/εx

1 10N 92µ 46µ -18µ 92µ -18µ 18µ 0.1957

2 20N 64µ 304µ -18µ 64µ -18µ 14µ 0.2188

3 30N 37µ 19µ -10µ 37µ -10µ 11µ 0.2973

Example: A typical data is shown for the purpose of computing stresses,

principal stresses, and planes of principal stresses.

Sl

No εA εB εC

σx

Mpa σy Mpa

τxy

Mpa

Principle

stress

σ1 Mpa

Principle

Stress

σ2 Mpa

1 200µ 900µ 1000µ 105.58 230.09 46.69 342.04 -6.371

Normal, shear stress,& principal stress

Angle of principal

stress

Sl

No σx

Mpa

σy

Mpa

τxy

Mpa

σ1

Mpa

σ2

Mpa θ1 θ2

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1 5.6787 0.8159 0.3879 6.8156 -0.242 64.65 -86.00

2 12.064 1.1000 0.9842 13.299 -0.651 51.45 -85.62

3 17.944 1.2762 1.5695 19.347 -1.273 41.80 -85.04

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Experiment - 5

Buckling Load of Slender Eccentric Columns and Construction of

South Well plot

Aim: Practical columns have some imperfections in the form of initial

curvature and the buckling of loads of such struts is of real practical value.

The experiment aims at measuring the buckling loads of columns and

construction of South Well Plot. The imperfection amounts to initial

curvature, which shows up in this plot.

Apparatus: WAGNER beam set-up, hinged supports, load cells, Long

column with initial curvature, mounted dial for deflection measurements.

Support

Gear box Hand wheel

Load

Loadindicator

Dialindicator

Eccentricloading

Column

Fig. 8: WAGNER beam set-up for buckling load

Theory: Consider a pin ended strut AB of length L, whose centroidal longitudinal

axis is initially curved as shown in fig (9). Under the application of the end

load P, the strut will have some additional lateral displacement y at any

section.

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In this case, bending moment at any point is proportional to the change in

curvature of the column from its initial bent position y- δ0. The equation for

curvature of column is as follows.

M =P(y+y0) = - EI d2y/dx

2 … (1)

P/EI =k2

0

22

2

2

ykykdx

yd−=+ … (2)

Assuming initial curvature (yo) to be sinusoidal, satisfying the equation

)/sin(00 Lxy πδ= … (3)

Where δ0 equals to the initial displacement at the centre of the strut

The general solution of this differential equation is

Y

P

y

P

δ0

Fig-9

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)/sin(/

sincos222

0

2

LxkL

kkxBkxAy π

πδ

−++= … (4)

If the ends are hinged, for the end conditions, then:

y=0 at x =0 and x =L

This results in values of constants: A=B=0.

The resulting equation is as below:

=L

x

kL

ky

ππ

δsin

2

2

2

0

2

… (5)

Substituting Euler’s buckling load (Pe)

Pe =

and k 2 = P/EI

−=

L

x

P

Pey

πδsin

1

0 … (6)

for x=L/2, at center of column, the deflection at center yc

1

0

−=

P

Pey c

δ …(7)

The value Pe represents the buckling load for perfectly straight strut. In the

relation for deflection (y), the additional lateral displacement of the strut,

that the effect of end load P is to increase (y) by a factor 1/ (pe /p)-1; shown

by equation (6). When P approaches Pe, the additional displacement at mid

length of the strut is expressed by eqn. (7).

The load deflection relationship of eqn. (5) is the basis of South Well plot

technique for extrapolating for the elastic critical load from experimental

measurement.

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Pe

PPe

P

−=

10δ

δ … (8)

Rearranging the above equation we get

δ = (Pe / P) δ - δ0 … (9)

or

δ = (δ/ P) Pe - δ0

The linear relationship between δ/P and δ shown in figure below can be

experimentally determined. Thus if a straight line is drawn which best fits

the points determined from the experimental measurements of P and δ , the

reciprocal of the slope of this line gives an estimate of the magnitude of δ0 of

the initial curvature that can be determined from the intercept on the

horizontal axis.

δ

Fig.10: South Well

Plot

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Procedure: Set up the two hinge supports on the WAGNER beam at the top

and bottom supports. Fix the column in the supports.

Set the load reading to zero in load cell. Determine the center of column. Set

–up the deflection dial gage for reading the column deflections at the center

of column. Set the deflection dial gage reading to zero.

Apply the vertical load in steps of 10 Kgs. each, in four steps (20Kg, 30Kg,

40Kg...) and record the deflections at each step of load.

Tabular Column:

Load (P)

N

Deflection (δ) mm δ/P

mm/N

Draw South Well Plot (δ/P vs δ ).

Determine the slope and estimate Pe. Find out the initial deflection of

column.

Conclusion: The buckling loads are computed for the column and the

SOUTHWELL plot is constructed. Slope is= --------- and Pe= ----------units.

CAUTION: NEVER EXCEED 100 Kg OF LOAD ON THE MILD

STEEL COLUMN AND 75Kg ON THE ALUMINIUM COLUMN.

Example: 1

A slender strut, 1800mm long, and of rectangular section 30mm x 12mm

transmits a longitudinal load P acting at the centre of each end. The strut was

slightly bent about its minor principal axis before loading. If the P is

increased form 500N to 1500N, the deflection at the middle of the length

increases by 4mm. Determine the amount of deflection before loading.

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Find also the total deflection and the maximum stress when P is 2000 N.

Take E = 2.15 x 105 N/mm

2.

I = 1/12 x 30 x 123

= 4320 mm4

Let Pe = Π2 EI /L

2 = Π2

2.15 x 105 x 4320 / (1800 X1800) =2830 N

Let δ1 be the central deflection when P =500 N and let δ2 be the central

deflection when P =1500 N

Substituting in 0δδ xPPe

Pec −

=

δ1 = 2830/ (2830-500) x δ0 … (1)

δ2 = 2830/ (2830-1500) x δ0 … (2)

from eqns.1 & 2

δδδδ0 = 4.381 mm

P =2000 N

δc = 2830/ (2830-2000) x 4.381 = 14.94 mm

A = 30 x 12 = 360mm2

Z = 1/6 x 30 x 122

Mc = P x δc = 2000 x 14.94 = 298880 N-mm

σ 0= P/A = 2000/360mm2

σb = Mc/Z = 29880/ 720 =41.5 N/mm2

σ max =σ0+ σb = 5.55 + 41.5 =47.05 N/mm2

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Experiment - 6

Shear failure of Bolted and Riveted Joints

Aim: To determine the ultimate shear stress in a bolt.

Apparatus: WAGNER beam set-up, bolt for failure analysis, vernier caliper

and load indicator.

Support

Gear box Hand wheel

Load

Loadindicator

Single shear Double shear

Bolt

Load

P(vertical )N

P(vertical )N

Fig. 11: WAGNER beam set-up for shear failure of bolted joint

Theory: Riveted and bolted connections are common in structural

assemblies. Following are modes of failure in riveted joints:

� Tension failure in a plate

� Shearing failure across one or more planes of rivet

� Bearing failure between plate and the rivet

� Plate shear or shear out failure in the plate

In a riveted joint, the rivets may themselves fail in shear. The tendency is to

cut through the rivet across the section lying in the plane between the plates

it connects.

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If the load is transmitted through bearing between the plate and the shank of

the rivet producing shear in the rivet, the rivet is said to be in shear.

When the load is transmitted by shear in only one section of the rivet, the

rivet is said to be in single shear. When the loading of the rivet is such as to

have the load transmitted in two shear planes, the rivet is said to be in double

shear. When load is transmitted in more than two planes, the rivet is said to

be in multiple shear.

Rivets and bolts subjected to both shear and axial tension shall be so

proportioned that the calculated shear and axial tension do not exceed the

allowable stresses τ υf and σ tf and the flowing expression does not

exceed a specified value.

(τ υf ,cal / τ υf + σ tf,cal / σ tf )------------------------------------- (1)

Shearing failure of the rivet: In a riveted joint, the rivets may themselves fail

in shear. The tendency is to cut through the rivet across the section lying in

the plane between the plates it connects.

In analysing this possible manner of failure, one must always note whether a

rivet acts in single shear or double shear. In the latter case, the two cross-

sectional areas of the same rivet resist the applied force. The shearing stress

is assumed to be uniformly distributed over the cross-section of the rivet.

Let Pus = pull required, per pitch length, for shear failure

fs = ultimate shear strength of the rivet material

d = gross diameter of the bolt

Resisting area of the rivet section = (π/4) d2 in single shear

and = 2 x (π/4) d2 in double shear

Pus = (π/4) d2 fs for single shear

Pus = 2x (π/4) d2 fs for double shear

Procedure:

� Set-up the WAGNER beam test set-up.

� Note the diameter of the bolt.

� Place a bolt in the slot.

� Set the load cell to read zero.

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� Apply the load gradually.

� Note the load reading at the point of shear failure of mild steel bolt.

� Change the loading for double shear and calculate the ultimate shear

stress from the formula below:

fs = Pus / [2x (π/4) d2

] ---------------------------------(2)

Conclusion: The ultimate shear stress for single shear= ----------units

The ultimate shear stress for double shear= ---------units.

CAUTION: NEVER EXCEED 3 TONS OF LOAD ON THE BEAM

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Experiment - 7

Bending Modulus of a Sandwich Beam

Aim: To determine the Modulus in Bending of a cantilever sandwich

beam.

Apparatus: Beam Test Set-up, load cells, Sandwich Beam of symmetrical

section with strain gage installed, Strain measuring device.

Fig. 12 Sandwich beam

Theory: Beams that are made of more than one material are called

`composite beams`. Sandwich beam is one such example. It consists of

following:

� Two thin layers of strong material, called faces, placed at top and

bottom.

� Thick core, consisting of light weight, low strength material. The core

simply serves as a filler or spacer. Sandwich construction is used where

light weight combined with high strength and high stiffness is needed.

If both the materials are rigidly joined together, they will behave like a unit

piece and the bending will take place about the combined axis. Such

composite beams can be analysed by the same bending theory that is

applicable for beam of single material.

On the other hand, if both the materials have been simply placed one above

the other, they will bend about their respective axes. However, in both the

cases, the total amount of moment of resistance will be equal to the sum of

moments of resistance of individual sections.

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The position of neutral axis may not be the centroid of the section. The

criterion of strain compatibility has to be used, i.e. strain in two materials, at

a given vertical distance from the neutral axis, has to be same.

A cantilever sandwich beam provided with strain gauge can be used to

do experimental assessment of strains. The strain gauge is at a fixed position

in the beam and load position can be varied. A strain gauge is mounted on a

free surface, which in general, is in a state of plane stress where the state of

stress is with regards to a specific xy rectangular rosette. Consider the three

element rectangular rosette shown in fig-13, which provides normal strain

components in three directions spaced at angles of 450. If a xy coordinate

system is assumed to coincide with the gauge A and C then εx= εA and εy=

εc. Gauge B provides information necessary to determine γxy. Once εx, εy

and γxy are known, then Hooke’s law can be used to determine σx, σy, and

τxy. However in this case the requirement is to determine Modulus in

bending (E). An alternative approach is to measure deflection under load

condition.

Fig. 13

)()1( 2 yxx v

v

Eεεσ +

−= ; )(

)1( 2 xyy vv

Eεεσ +

−= ;

xyxyv

Eγτ

)1(2 +=

εx = εA; εy = εc; γxy = 2 εB - εA -εc;

Sl.No Load

(N) εA εB εC εx εy γxy

ν=εy/εx

Also, σx = M/Z , Z = I/ Y (1)

σ2 = (M - σ1 Z1 ) / Z2 (2)

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σ2 = m σ1 , m = E2 / E1 (3)

Suffix `1` & `2` refer two different sections of beam

Procedure: Mount the cantilever beam at the left support of beam test set-

up. Connect the strain gauges wires with the strain measuring equipment.

Use the following color codes

εA Red wires channel1

εB Yellow wires channel3

εC Black wires channel2

Set the load cell to read `zero` value in the absence of load. Set the three

strains to read `zero` in the absence of load. Now Load the beam with 2.5 Kg

at some point and record the strains in three directions. Record the load

value at the load cell.

Repeat the experiment with load value of 5 Kg. Compute the values of

Poisson’s ratio from:

ν=εy/εx

Tabular Column:

Sl.No Load

(N)

εA

εB εC εx εy γxy

ν=εy/εx

Find Modulus in bending through formulas above.

Conclusion: Bending modulus of sandwich beam is= ---------units.

CAUTION: NEVER EXCEED 15 Kg LOAD ON THE BEAM.

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Experiment - 8

Verification of Superposition Theorem

Aim: To verify the theorem of superposition.

Apparatus: Beam Test Set-Up with multiple loading capability (atleast two

load points required), atleast two load cells, cantilever strain gauged beam,

strain measuring equipment.

Gear box

Strainindicator Load

indicator

Pointer

Scale

Load

Hand wheel

Cantileverbeam

straingauge

c1 c2 c3

c2

c1

c3

Strain guagearrangement

Fig. 14: Schematic layout of beam setup

Theory:

Many times, a structural member is subjected to a number of forces

acting not only at the ends, but also at the intermediate points along its

length. Such a member can be analyzed by the application of the principal of

superposition; the resulting strain will be equal to the algebraic sum of the

strains caused by individual forces acting along the length of member.

The strain gauge is at a fixed position in the beam and load position

can be varied. A strain gauge is mounted on a free surface, which in

general, is in a state of plane stress where the state of stress is with regards to

a specific xy rectangular rosette. Consider the three element rectangular

rosette shown in fig-2, which provides normal strain components in three

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directions spaced at angles of 450. If a xy coordinate system is assumed to

coincide with the gauge A and C then εx= εA and εy= εc. Gauge B provides

information necessary to determine γxy.

Fig-15

εx = εA; εy = εc; γxy = 2 εB - εA -εc

Sl.No Load

(N) εA εB εC εx εy γxy

Procedure:

Mount the cantilever beam at the left support of beam test set-up. Connect

the strain gauges wires with the strain measuring equipment. Use the

following color codes:

εA Red wires channel1

εB Yellow wires channel3

εC Black wires channel2

Set the load cell to read `zero` value in the absence of load. Set the three

strains to read `zero` in the absence of load. Now Load the beam with 0.5 Kg

at some point from vertical and record the strains in three directions. Record

the load value at the load cell. Record the point of loading.

Remove this load.

Set the load cell to read `zero` value in the absence of load. Set the three

strains to read `zero` in the absence of load. Now Load the beam with load

load value of 1 Kg. from vertical at same point and record the strains in three

directions. Record the load value.

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Remove this load.

Set the load cell to read `zero` value in the absence of load. Set the three

strains to read `zero` in the absence of load. Now Load the beam with 1.5 Kg

at same point from vertical as done earlier.

Record the strains in three directions. Record the load values in the two load

cells. Record the vertical load position.

Compute the values of γxy from the formula:

γxy = 2 εB - εA -εc

CAUTION: NEVER EXCEED 3 Kg LOAD ON THE BEAM.

Tabular Column:

Table 1: Strains due to individual vertical loading

Sl.No Vertical

Load

(N)

Load

position

mm

εA

εB εC εx εy γxy

Table 2: Strains due to combined loading

Sl.No Vertical

Load

(N)

Vertical

Load

position

mm

εA

εB εC εx εy γxy

Add the values of strains in table 1 and compare with value in Table 2.

Conclusion: The strains obtained are compared between individual and

combined loading and hence the principle of superposition is verified.

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Experiment – 9

Vibration of Cantilever Beam

Objective: To determine the lateral or transverse vibration of a cantilever

beam when the beam is fixed at one end and free at the other end.

Introduction: A beam which is cantilevered of span L with uniform mass w/g per

unit run is shown in figure-1and fixed at one end. Assume a deflection

function and obtain the first approximation for the fundamental frequency

with the origin at the free end.

L is the span of the beam in mm

E is the Young’s Modulus of the beam in N/mm2

I =bd3/12 is the Moment of Inertia of the beam mm

4

w = Weight per unit length

g = Acceleration due to gravity

The fundamental frequency of the cantilever beam

ω2= 12.39 E I / wl

4

The natural frequency for the transverse vibration of a uniform beam fixed at

one end and free at the other end.

The four roots of the equation are:

β1 L= 1.8751

β2 L= 4.6941

β3 L= 7.8548

β4 L= 10.9955

Fig. 16: Cantilever beam

L

P

H

x

A B

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4

22

AL

EInn

ρπω =

…. (1)

Example: Determine the three lowest natural frequency for the system

shown in fig-1

Given m=10kg,E=200x109 N/m2; ρ =7800 kg/m3 ; A =2.6 x 10-3 m2 ;

L=1m ; I= 4.7 x10-6

m4.

43

622

1106.27800

107.4102001

9

xxx

xxxn

= πω =215.3 φ2

φ1=1.423; φ2=4.113; φ3=7.192

ω1 = 486.1 rad/ s; ω2 = 3642 rad/s; ω3

= 1.114 x 10

4 rad/s.

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Experiment-10

Wagner Beam (Tension Field)

Objective: The objective of this experiment is to determine the constant k of the

Wagner beam.

Support

Gear box Hand wheel

Load

Strainindicator

Loadindicator

c1 c2 c3

c2 c3

c1

Strain guagearrangement

Wagner beam

Fig. 17: WAGNER beam set-up for tension field

Introduction: In the analysis of wing beams of airplanes, the designer is faced with several

problems which, in general are not present in civil engineering structural

design. The civil Engineer endeavors to make the web sheet of all beams

thick enough so that the web will not buckle before the design load is

reached on the structure.

Buckling is a case of failure and the shearing stress causing buckling

determines the allowable maximum shear that can be applied. The critical

buckling shear stress is given by

2

2

2

)1(12

−=

b

tEcr µ

πτ

…1

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If the sheet is very thin, buckling stress given by equation (1) is extremely

low and, in the interest of making efficient use of all available material. The

aircraft engineer raises the question as to how much additional shear can be

carried by such a buckled plate before:

� Some portion of the sheet has a total stress equal to the yield point of

the material, thus giving rise to permanent deformations; or

� The ultimate strength is reached.

If now, we assume that the web plate in the beam is very thin, the above

discussion of the principal stress patterns takes on new significance. The

normal stresses, we see that one of them is a compressive stress against

which thin plates have a very low resistance. The tendency will then be for

the plate to buckles in a direction perpendicular to the compressive stress at

a value of the applied shear which becomes less and less as the web becomes

thinner and thinner, the limiting case being for a sheet of zero thickness.

In this case, the sheet buckles upon the application of a shear load and can

only resist shear by means of the tensile stresses at 45degresss. With such a

pattern, it is obvious that the tensile stresses will tend to pull the two beam

flanges together, thus necessitating vertical members to counteract this

tendency (as shown dotted in fig. 18).

Wagner assumes that the web buckles immediately upon the application of

shear load and that the only stresses resisting the shear forces are the tensile

stresses which act approximately 45 degrees.

Considering infinitely rigid parallel span flanges and vertical stiffners the

following equations for this limiting case is shown as below:

=

)2sin(

12

ασ

ht

Pt …2

Axial force in the tension flange:

)cot(2

αP

h

PxFt −= …3

Axial force in the compression flange

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)cot(2

αP

h

PxFc −−= …4

Axial force in the vertical stiffeners

)tan(αh

dPFc −= …5

The limiting case of a web having no compressive strength has been treated

by Wagner and beams approximating this are known as Wagner beam.

The basic assumptions of this theory is that total shear force in the web

can be divided into a shear force carried by shear in the sheet and the

shear forced carried by diagonal tension;

ncrk )1(τ

τ−=

Fig.18: Wagner’s Beam Tension Field beam

0.2 0.2 0.2

0.1

5 KN

0.012

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Fig.19

A strain gauge is mounted on a free surface, which in general, is in a state of

plane stress where the state of stress with regards to a specific xy rectangular

rosette since each gauge element provides only one piece of information, the

indicated normal strain at the point in direction of the gauge. Consider the

three element rectangular rosette shown in fig-20, which provides normal

strain components in three directions spaced at angles of 450.

If an xy coordinate system is assumed to coincide with the gauge A and C

then

εx= εA and εy= εc. Gauge B provides information necessary to determine γxy.

Once εx, εy and γxy are known, then Hooke’s law can be used to determine

σx, σy, and τxy.

25000

500 1000 500 1000 1000 1000 1000

325000

qs = 104 N/m

275000 25000

500 2000 500 1000 1000 1000 1000

250e06

N/m 0.5e06

N/m

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Fig. 20

)()1(

2 yxx

Eυεε

µσ +

−= ; )(

)1(2 xyy

Eυεε

µσ +

−= ;

xyxy

µτ

)1(2 +=

εx = εA; εy = εc; γxy = 2 εB - εA -εc;

Sl

No εA εB εC

σx

MPa

σy

MPa

τxy

MPa

σ1

MPa

σ2

MPa

1 200µ 900µ 1000µ 105.58 230.09 46.69

Example: 1

Apply a load of given intensity on the Wagner beam

Step: 1 Consider an axially loaded aluminium panel simply supported edges and

rivet lines .Each stringer has an area of 12mm x 8mm and E=0.588e11 N/m2

and determine the compressive load i) when the sheet first buckles ii)

Stringer stress is 58.84x106 N/mm2.

Critical Buckling Stress:

2

1

=b

tKEcσ

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σc= 3.62 * 0.588 x 10 11

x (0.002/0.2)2 = 21.2856 x 10

6 N/m

2

A= 0.002 x 0.6 = 1.2 x 10-3

m2

P = A x σσσσc1 = 1.2 x 10-3

m2 x21.2856 x 10

6 N/m

2 = 25,542.72 N

Post Buckling Strenth:

uσE

0.85twe1 = we1= 0.85* 0.002√(0.5884e11/58.58e06 = 0.05374

m

A = (3.0*0.053774)* 0.002 = 6.4488 e-04 αm2;

P = 58.84e06 * 6.4488e-04 = 37945.05289 N

The ratio of Post buckling strength to critical buckling stress is 1.48.

Step: 2 Determination of k

E= 58.8e109 N/αm2; K= 5.34 (Simply supported beam) K= 10.38 (Fixed

end)

The ratio of longer dimension/ shorter dimension is called aspect ratio =2.0

ht

V=τ

= 50000/ 0.1* 0.002=250 e06 N/m2

2

=b

tKEcrτ =5.34 * 0.588 x 10

11 x (0.002/0.1)

2 = 125.5968 x 10

6

N/m2

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)1(τ

τ crk −= = (1-125.5986e06/250e06) = 0.5024

Step 3: Schematic distribution of load

ht

Pσ s =

= 50000/(0.1*0.002) =250e06

N/m2

Assume buckling angle =450

(Normal Stress in Y-direction) yσ =sσ )tan(α = 250 e

06 tan (45) =250e

06

N/m2

(Normal Stress in x-direction) σx =sσ )cot(α = 250 e

06 cot (45) =250e

06 N/m2

=

)2sin(

12

ασ

ht

Pt = 2*50000*/(0.1*0.002)*(1/sin90) =500e

06

N/m2 ;( Tension in the beam at 45 degrees angle)

q s = σs * t = 250e06

* 0.002 = 0.5e06 N/m2; (Shear flow in the beam)

Fc_start = Ft_start = )cot(2

αP

h

PxFt −= = 0-50000/2*cot (45)

= 25000 N

Fc_end = Ft_end = )cot(2

αP

h

PxFt −= = (50000*0.6/0.1)-50000/2*cot

(45) = 325000,275000 N respectively. The stress distribution is shown in

figure.

Force in the vertical stiffener: )tan(αh

dPFc −=

= -50000*(0.2/0.1)*tan45 = - 0.100e06 N/m2

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Appendix -1 Unsymmetrical bending of angle section

%function

[xbar,ybar,Ixx,Iyy,Ixy,Iuu,Ivv,Iuv,alpha_uv]=us_ang

le01(thick1,breadth1,thick2,breadth2,x1_bar,x2_bar,

y1_bar,y2_bar)

function

[xbar,ybar,Ixx,Iyy,Ixy,Iuu,Ivv,Iuv,alpha_uv]=us_ang

le01(thick1,breadth1,thick2,breadth2,x1_bar,x2_bar,

y1_bar,y2_bar)

area1=thick1*breadth1;

area2=thick2*breadth2;

total_area=area1+area2;

Nrx=area1*x1_bar+area2*x2_bar;

Nry=area1*y1_bar+area2*y2_bar;

xbar=Nrx/total_area;

ybar=Nry/total_area;

t1=1/12;

tx1=(xbar-x1_bar);

ty1=(ybar-y1_bar);

tx1_2=(xbar-x1_bar)*(xbar-x1_bar);

ty1_2=(ybar-y1_bar)*(ybar-y1_bar);

tx2=(xbar-x2_bar);

ty2=(ybar-y2_bar);

tx2_2=(xbar-x2_bar)*(xbar-x2_bar);

ty2_2=(ybar-y2_bar)*(ybar-y2_bar);

Ixx_1=t1*breadth1*thick1^3;

Iyy_1=t1*thick1*breadth1^3;

Ixy_1=area1*tx1*ty1;

Ixx_2=t1*breadth2*thick2^3;

Iyy_2=t1*thick2*breadth2^3;

Ixy_2=area2*tx2*ty2;

Ixx1=Ixx_1+area1*ty1_2;

Iyy1=Iyy_1+area1*tx1_2;

Ixx2=Ixx_2+area2*ty2_2;

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Iyy2=Iyy_2+area2*tx2_2;

Ixx=Ixx1+Ixx2;

Iyy=Iyy1+Iyy2;

Ixy=Ixy_1+Ixy_2;

raddeg=180/pi;

tmpnr=Ixy;

tmpdr=(Iyy-Ixx)*0.5;

tmp=tmpnr/tmpdr;

alpha_uv=0.5*atan(tmp)*raddeg;

term1=0.5*(Ixx+Iyy);

term2=0.5*sqrt((Ixx-Iyy)*(Ixx-Iyy)+4.0*Ixy*Ixy);

Iuu=term1+term2;

Ivv=term1-term2;

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Appendix-2

Rectangular strain rosette Compuation of stresses sig_x and sig_y and gamma_xy

ea=input('ea the strain in a-directon: \n')

eb=input('eb the strain in b-directon: \n')

ec=input('ec the strain in c-directon: \n')

%E=input('E the Youngs strain E=58.8e09 N/m2: \n')

%mu=input('mu the poissons ratio:(0.33) \n')

E=58.8e09;

mu=0.33;

ex=ea*10^-06

ey=ec*10^-06

exy=(2*eb-ec-ea)*10^-06

k1=E/(1-mu*mu);

sig_x=k1*(ex+mu*ey)

sig_y=k1*(ey+mu*ex)

tau_xy=0.5*(E/(1+mu))*exy

con1=(sig_x+sig_y);

con2=con1*con1;

con3=4*tau_xy*tau_xy;

sig_p1=0.5*(con1+sqrt(con2+con3))

sig_p2=0.5*(con1-sqrt(con2+con3))

theta_p1=atan((sig_p1-sig_x)/tau_xy)

theta_p2=atan((sig_p2-sig_x)/tau_xy)

thetp1=180/pi*theta_p1

thetp2=180/pi*theta_p2

valuexy=sig_x+sig_y

value12=sig_p1+sig_p2

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Appendix -3 Flexibility matrix for cantilever beam %The program to compute the deflection and rotation of a cantilever beam % Def_x= P*x*x(3*L-x)/6EI Rotx=P*x(2*L-x)/2EI,where

x is from the fixed end;

x=1.5 %m;

L=3 %m;

E=2e08 %kN/m2;

I=1.70797e-4 %m4;

P=10 %kN

sd_x=0.05*x

sd_L=0.05*L

sd_E=0.05*E

sd_I=0.05*I

sd_P=0.15*P

u1=unifrnd(0,1,100000,1);

x1=norminv(u1).*sd_x+x;

const=P/(6*E*I);

x2=(3.*L-x1);

def_x=const*x2.*x1;

def_x1=def_x.*x1;

hist(def_x1)

mean_x1=mean(def_x1)

std_x1=std(def_x1)

median_x1=median(def_x1)

sk_x1=skewness(def_x1)

cov_x1=std_x1/mean_x1

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=============================================================== DEPT. OF AERONAUTICAL ENGINEERING, DSCE 46

VIVA QUESTIONS:

1. Explain Castigliano`s theorem and its verification through any

experimental set-up.

2. Explain Maxwell`s reciprocal theorem and its verification though

experiment.

3. What is Betti`s theorem?

4. Explain South well plot.

5. What are various modes of failures of riveted joints?

6. What is the value of maximum loading allowed in the existing

experimental beam test set-up?

7. What is the value of maximum loading allowed in the existing Wagner

beam set-up?

8. What is bending modulus?

9. What is a sandwich beam?

10. What is strain compatibility?

11. Explain unit load method for determining the deflection of beams.

12. How Young’s modulus and Poisson’s ratio can be determined from

beam test set-up?

13. Explain Superposition theorem.

14. What are mode shapes and types of modes in vibrations?

15. Define DOF.

16. What is a Rosette strain gauge?

17. Explain unsymmetrical bending of beams.

18. What is the purpose of Stiffeners, What is the purpose of Longirons?

19. What stresses are taken up by the web and the flange of the spars?

20. Explain how bending and torsion is taken up by the wing structure.

21. What is a torsion box?

22. Explain how bending and torsion is taken up by the fuselage structure.

23. What is shear center and how it can be determined through

experimental set-up?

24. What is Flexibility matrix?

25. What is Power Spectral Density?

26. What are Principal stress axes?

27. Define simple stress and simple strain.

28. Define plain stress, plain strain.

29. What is a stress Tensor? Stress at a point in a material will have how

many components? Segregate normal and shear components of stress

tensor.

Page 47: Structures Lab Finl Copy

AIRCRAFT STRUCTURES LAB (06AEL57)

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=============================================================== DEPT. OF AERONAUTICAL ENGINEERING, DSCE 47

30. What is a Wagner Beam test set-up? What are Wagner assumptions?

31. Explain the Energy methods used for structural analysis.

32. Draw SFD and BMD for i) SSB with UDL ii) CB with UDL.

33. Draw the orthographic views of i) civil airframe ii) military airframe

(minimum 3 views i.e. front, top and side view).

34. Differentiate b/w i) stringer ii) spar and iii) longiron based on the type

of loading they undergo in general.

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