Aircraft Structures II Laboratory Manual with model calculations and graphs

AIRCRAFT STRUCTURES II LABORATORY MANUAL

N.TAMILSELVAM ASSISTANT PROFESSOR AERONAUTICAL ENGINEERING ACE-HOSUR

ACE/AERO/STRUCTURES II LAB MANUAL BY N.TAMILSELVAM

SHEAR CENTRE LOCATION OF OPEN S ECTION (C, Z & I SECTION) Abstract:

Due to external shear stress/forces on the beam, shear stresses are induced on the cross-section of the beam. If the resultant of the internal force system and applied force at any section do not coincide, a torque is developed and therefore the section undergoes twist. In order that the section may not twist, the applied shear force must pass through the shear centre of the open section. This experimental setup helped students to find out the location of the shear centre of the open channel.

Aim: To determine the shear centre for the given open channel section

Apparatus Required: (i) Beam of channel section.. (ii) Known Weights with loading hook and. (iii) Dial gauge with magnetic base.

Theory: Shear centre is defined as that unique point on the beam cross section where the application of the shear force does not produce the beam to twist. The channel section has a horizontal axis of symmetry and the experiment is to determine its location on axis since angular twisting moment causes a twist, it is simply a matter of placing the shear force at a point on the cross section where the shear force and the shear centre combine to produce a zero tensional moment about all points on a plane of beam cross section

2


Procedure: 1) Mount two dial gauges on the flange at a known distance, preload the dial gauge and set it to zero 2) Place a total 500 gram load at A and place 100 grams load at B note down the dial gauge readings 3) Now remove one load piece from hook A and place it on B. this became the total vertical load on this section remains 600 grams 4) Record the dial gauge reading 5) Transfer all the load pieces one by one to hook B, and note down the dial gauges readings 6) For every load case calculate the algebraic different between the dial gauge readings as measure of angle of twist () suffered by the section 7) Plot a graph with ‘e’ value on x-axis and d1 – d2 value on y-axis, the meeting point of the straight line with x-axis determine the shear centre

Formulae Used: For theoretical

 = ℎ(13^26) ℎ For experimental

e

exp =

AB (WA – WB) / (2WT)

Where:

h

=

height of the specimen

b

=

width of the specimen

AB

=

distance between load point

WA

=

load at point A

WB

=

load at point B

WT

=

total load

3


Table:

a. b. c. d.

Length of the section (L) = Breath of section (b) = Height of the section (h) = Thickness of the section (t) =

….. mm …..mm …..mm …..mm

Tabulation:

Sl.no

WA

WB

D1

D2

1 2 3 4 5

Result: The shear centre for the open section Theoretically

=

……………………

Experimentally

=

……………………

4

D1-D2

e exp


MODEL TABULATION AND CALCULATION

Let consider,

a. Breath of section (b) b. Height of the section (h) c. AB

= 50mm = 50mm = 280mm

Tabulation:

Sl.no

WA

WB

D1

D2

D1-D2

e exp

1

500

100

395

55

340

93.33

2

400

200

350

132

218

46.66

3

300

300

300

282

18

0

4

200

400

281

324

-43

-46.66

5

100

500

230

330

-100

-93.33

For Experimental calculation

e

exp =

AB (WA – WB) / (2WT)

= 280(500 – 100)/2 x 600 = 93.33mm

5


For Theoretical calculation

)  == (−^ 50^ 

50(−  )

= - 30mm

Ans

From the Graph, when ( D1-D2 ) = 0 the experimental shear centre for the section is

-29mm

Ans

6


Model Graph for Open section

7


DEFLECTION OF BEAMS AIM

To determine the deflection of beams with various end conditions, experimentally compare with the theoretical values and draw the deflection curve. APPARATUS REQUIRED 

Dial Gauges



Weights



Weights Hanger



Meter Scale



Beam Specimen and



Beam setup

THEORY

The deflection of an Aircraft wing can considerably change the lift distribution on a wing. Hence estimation of deflections and restricting the deflection with in the allowable limits are important for Aerospace structural engineer. From the elementary theory of bending, the differential equation of the deflection curve is

 =   (1)   From the above equation, the deflection curve for a simply-supported beam with a concentrated load at the center is

 =     =    

(2)

The deflection curve for cantilever beam with a tip load is (3)

The above equations are used to calculate the deflection theoretically at various points.

8


PROCEDURE

1. The beam is simply supported using the knife edges; weight hanger and dial gauge are to be kept at the middle of the beam. 2. Apply the concentrated load of 0.1kg in the beam. 3. Note down the reading from the dial gauge and increase the load by 0.1kg. 4. The deflections at this point are also calculated theoretically using equation (2) and (3). 5. The deflection curve is drawn on a graph sheet with proper scale. 6. The experiment can be repeated by fixing the specimen as cantilever beam. 7. For cantilever beam, note down the length of the beam using meter scale. TABULATION

For Cantilever and Simply-supported Beam

S.No.

Dial Gauge Reading

Load



P (kg)

1

Theoretical deflection

2 3 4 5

GRAPH

The draw the graph between loads vs. deflection. APPLICATION

I. II.

Aircraft wing is idealized as cantilever beam in the design of Aircraft (Box beam). Aircraft fuselage is idealized as a free-free beam in the design of Aircraft.

9


RESULT

Deflection curves are drawn for the given beam. Experimental values are compared with the theoretical values

MODEL TABULATION AND CALCULATION For Cantilever beam

Let consider, Length of the beam (L) = 60 cm Breadth of the beam (b) = 12.5cm Width of the beam (d) = 6mm Moment of inertia (I =

  

) = 225mm4

Theoretical deflection =

S.No.

Dial Gauge Reading

Load P (kg)




1

0.1

150

149

2

0.2

299

298

3

0.3

448

449

4

0.4

597

597

5

0.5

746

747

Calculation


 = (0.1 x 9.81 x 600 ) / (3 x 210 x 10 x 225)  3

= 1.49mm = 149

10

3


For Simply Supported beam (load and deflection at mid point)

Let consider, Length of the beam (L) = 60 cm Breadth of the beam (b) = 12.5cm Width of the beam (d) = 6mm Moment of inertia (I =

   48

) = 225mm4


S.No.

Dial Gauge Reading

Load P (kg)




1

0.1

10

09

2

0.2

18

18

3

0.3

28

28

4

0.4

37

37

5

0.5

46

46

Calculation


 = (0.1 x 9.81 x 600 ) / (48 x 210 x 10 x 225) 48 3

= 0.09mm = 09

11

3


Model Graph 800 700 600

n 500 io tc 400 le f e D300 200

100 0 0.1

0.2

0.3

Load (kg)

12

0.4

0.5


Maxwell Reciprocal Theorem Aim: To verify clerk Maxwell’s reciprocal theorem

Apparatus Required:





Maxwell’s Reciprocal Theorem apparatus,



Weights, Hanger, Dial Gauge,



Scale,



Knife edges

Theory: Maxwell theorem in its simplest form states that deflection of any point A of any elastic Structure due to load P at any point B is same as the deflection of beam due to same load Applied at A

It is, therefore easily derived that the deflection curve for a point in a structure is the same as the deflected curve of the structure when unit load is applied at the point for which the Influence curve was obtained.

13


Procedure: 1. Apply a load either at the centre of the simply supported span or at the free end of the beam, the deflected form can be obtained. 2. Measure the height of the beam at certain distance by means of a dial gauge before and after loading and determine the deflection before and after at each point separately. 3. Now move a load along the beam at certain distance and for each positions of the Load, the deflection of the point was noted where the load was applied in step1.This 4. Deflection should be measured at each such point before and after the loading, separately. 5. Plot the graph between deflection as ordinate and position of point graph drawn in step 2 and 3.These are the influence line ordinates for deflection of the Beam.

Table:

1

Load at A (g) 100

Deflection at B (mm) .42

Load at B (g) 100

Deflection at A (mm) .42

2 3 4

200 300 400

.97 1.33 2.43

200 300 400

.96 1.33 2.43

SlNo

Precaution: i) Apply the loads without any jerk. ii) Perform the experiment at a location, which is away from any disturbance iii) Ensure that the supports are rigid.

14


Result: From the results in the Table, it is found that the deflection at A is equal to the deflection at B and Hence the Maxwell reciprocal theorem is verified.

CONSTANT STRENGTH BEAM

Aim: To determine the stress at various locations along a length of a constant strength beam to show that they are equal and compare with theoretical values.

Theory: A beam in which section modulus varies along the length of the beam in the same proportion as the bending moment is known as constant strength beam. In this case the maximum stress remains constant along the length of the beam. Consider, b = width of the beam h = depth of the beam l = length of the beam Section Modulus Z = bh

3

σ = M/Z = constant Let the width of the beam be constant and the depth varies Then σ= M/Z = 6Px/bh2 = 6PL/bh02 =constant h2 = h02x/L

(1)

15


Apparatus required:



A constant strength beam



Strain gauges



strain indicator



weights with hook.

Procedure: The constant strength beam is fixed as a cantilever strain gauges are fixed near the root, at 3L/4 and L/2.The strain gauges are fixed both on the top and bottom surfaces at each location to increase the circuit sensitivity of the gauge circuit. Hence half bridge used in the strain indicator to measure the strain indicator to measure the strain at each location. (Strain=strain meter reading x2). The strain gauge readings are noted for every 2kg at locations A, B, C and tabulated as given below. Strain gauge resistance

= 350 ohm

Gauge factor

=2

Young s modulus

= 70 GPa

S.NO

Weight (Kg)

ɛ

A

ɛ

B

ɛ

Theoretical calculations: At point A:

Distance of point A from loading point

= 16

C

ɛ

ΣA =E Ax2 (MPa)

ɛ

ΣB =E Bx2 (MPa)

ɛ

ΣC =E Cx2 (MPa)


Depth of beam

=

Width of beam

=

Moment of inertia

=

Moment

=

Σ=(Mh/2)/I

=

17


At point B:

Distance of point B from loading point

=

Depth of beam

=

Width of beam

=

Moment of inertia

=

Moment

=

Σ = (Mh/2)/I

=

At point c:

Distance of point C from loading point

=

Depth of beam

=

Width of beam

=

Moment of inertia

=

Σ = (Mh/2)/I

=

Result: The experimental values of the stress are compared with the theoretical values.

18


S.NO

1 2 3 4 5

Weight kg

ɛ

2.30 4.20 6.31 8.28 10.31

A

20 39 58 75 89

ɛ

B

21 41 60 76 92

ɛ

A

21 41 60 76 92

ɛ

ΣA =E x2 MPa 2.8 5.46 8.12 1.05 1.24

A

ɛ

Σc =E CX2 MPa

2.94 5.74 8.4 1.06 1.28

2.94 5.74 8.4 1.06 1.28

Theoretical calculations: At point A:

Distance of point A from loading point

= 47cm

Depth of beam

= 78mm

Width of beam

= 5mm

Moment of inertia

= 14.0625cm4

Moment for load of 11.25

= 11.25x9.81x47=5187.0375

Σ= (Mh/2)/I

= 14.38

At point B:

Distance of point B from loading point

= 35.5cm

Depth of beam

= 65mm

Width of beam

= 5mm

Moment of inertia

= 9.1542

Moment for a load of 11.25

= 11.25x9.81x35.5=3917.868

Σ = (Mh/2)/I

= 13.9

19

ɛ

ΣB=E Bx2 MPa


At point c:

Distance of point C from loading point

=

24cm

Depth of beam

=

54mm

Width of beam

= 4mm

Moment of inertia

= 4.9626

Moment for a load of 11.25

= 11.25x9.81x24=2648.7

Σ =(Mh/2)/I

= 14.41

Result: The experimental values of the stress are compared with the theoretical values.

20


UNSYMMETRICAL BENDING OF BEAMS

Aim: To determine the principal axes of an unsymmetrical section.

Theory: The well known flexure formula σ= (My/I) based on the elementary theory of bending of beams assumes that the load is always applied through one of the principal axes of a section. Actually, even if the applied load passes through the centroid and or the shear center of the section the plane of bending and the plane of loading need not necessarily are the same. Therefore, knowledge of the location of principal axes is required for the determination of the stress distribution in beams using flexure formula. The determination of the principal axes is experimentally is described here. If IX, IY and IXY are the moment’s product of inertia of any section about an arbitrary orthogonal centroidal axes OX and OY then the inclination ϴ of one of the principal. Tan2ϴ=2IXY/ (IY-IX) The experimental determination of the principal axes of the given section is based on the fact that when the load passes through the shear center and is the direction of one of the principal axes of section.

Apparatus required: 1. A thin uniform cantilever Z section as shown in fig. At the free end extension pieces are attached on either side of the web. 2. Two dial gauges (to be mounted vertical and horizontal as in fig). This enables the determination of displacement u and v. 3. Two hooks are attached to the extension pieces to apply the vertical load WV. 21


4. A string and pulley arrangement to apply the horizontal load W H . 5. A steel support structure to mount the channel section as cantilever

Procedure: 1. Mount the two dial gauges on the tip section to measure the horizontal and the vertical deflections on the point. 2. Apply the vertical load(about 2.4kg, including two hooks of each 200gms) 3. Read u and v horizontal and vertical deflections respectively. 4. Increase the load W H in steps of about 300gm (for first case 100gm+200gm hook) repeat the procedure and check for consistency in measurements 5. Plot the graphs (u/v) vs.(W H/WV) and find the curve intersection with o

straight line of srcin 45 6. Calculate the inclination one of the principal axes of a web as Ɵ=tan1(WV/WH) 7. Calculate the inclination Ɵ using Equ (1)

TABLE-1

Specimen dimensions Vertical load W V s.no

WH

Dial gauges readings U V

WH/WV

u/v

1 2 3

Ɵexp=

Ɵtheo =

22

remarks


WV = (weight of hook +1kg) x2 =2.4kg U = horizontal displacement. displacement

s.no 1 2 3 4 5 6 7 8 9 10 11

WH 0 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.5 2.8 3.0

V=

U 22 42 49 56 66 75 86 96 105 115 124

b= 52mm

V 39 44 46 50 54 59 64 68 73 78 81

h=78mm

From the graph Ɵexp

Ɵexp =

=

tan-1(WV/WH)

=

tan-1(1/1.7) 30.5

0

23

WH/WV 0 0.125 0.25 0.375 0.5 0.625 0.75 0.875 1.0 1.125 1.25

t=2.5mm

vertical

U/V 0.564 0.954 1.065 1.12 1.27 1.343 1.411 1.433 1.474 1.53 1.578


IXX

IYY =

IXY

=

0.25X783/12+5.2X0.25X3.92

=

49.432cm4

Ɵ =

/12x2+5.2x0.25x2.63x2

=

23.434cm4

=

5.2X0.25X (-2.6) X3.9+5.2X2.6X9-3.9)

= tan2Ɵ

3

0.25x5.2

-26.364CM4

= 31.87

2IXY (IYY-IXX) O

Model graph:

UNSYMMETRICAL BENDING 4.5 4 3.5 3

V2.5 W / H W 2 1.5 1 0.5

0 10

20

30

U/V

24

40

50


COMBINED LOADING Aim: To determine the principal stresses and the principal planes of a hollow circular shaft due to combined loading.

Theory: The most common combined load system encountered in structural design is probably that are due to the bending and torsion. In an aircraft wing the lift acting at the center pressure produces a torque about the elastic axis varying bending moment along the wing span. To understand their combined effect a simpler specimen, namely a hollow cylinder is subjected to a bending and torsion. For an elastic structure Σ =M Y/ I Σ =

Bending stress (N/m2)

M=

bending moment (N-m)

Y=

distance from the layer to the neutral axis (m)

I=

moment of inertia (m4)

I=

π (do4-di4)/64

The shear stress due to torsion Τ= shear stress (N/m2) T= torque applied (N-m) r= radius of shaft (m) j= polar moment of inertia (m4) j=π (do4-di4)/32

25


±(−)  +()2 ±√ 2+ ⋯ [ ⋮  ±√ 2+ ⋯ 2 ⋱ ⋮ ]

σ1,2= σx+σy/2

σ1,2=σ/2

(1)

(τ)2

(2)

σ1,2= 16do/π(do4-di4)

tan2Φ=2τxy/σx-σy

=2τ/σ

(3)

(4)

Φ =principal angle

Apparatus required: Hollow circular shaft fixed as a cantilever, weight hanger with slotted weights, strain gauges, connecting wires, strain indicator and micrometer.

Procedure: Two strain gauges are fixed near the root of the tube fixed as a cantilever, one on the top fiber and the at the bottom to measure the bending strain. Another strains gauge is fixed at the same location of the neutral axes at 45 0 to measure the shear strain. Similarly three more strain gauges are fixed at the middle of the length to verify the various locations of the tube. The strain gauges at the top and the bottom of the tube are connected to half bridge the strain indicator. The outside diameter of the tube is measured using venire calipers. Weights are added to the hook attached to the lever in steps of work kg and strain gauge readings are noted from the train indicator for each load. From the strains the bending stress, shear stress is calculated and hence principal stresses and principal angle are calculated. These values are compared with theoretical values.

Table: Young’s modulus of the tube

=

Outside diameter of the tube

=

Thickness of the tube

=

Length of the tube

=

26


Strain gauge resistance

=

Gauge factor

=

Distance of strain gauges near the root from tip

=

Distance of the strain gauges at the middle from tip

=

Distance from the center of the tube to the center of hook

=

Weight of hook

=

s.no

W kg

ɛ

ɛ

A

Result: Bending stress of root (A)

=EX

Shear stress at (B)

=EX

Principal stress at the root

=

Principal angle at the root

=

Principal stress at the root (theatrical)

=

Principal angle at the root (theoretical)

=

Repeat the calculations for the mid-section.

27

ɛ X2 A

ɛ X4 B

B

ɛ

A


Table: Young’s modulus of the tube

=70Gpa

Outside diameter of the tube

=50mm

Thickness of the tube

=3mm

Length of the tube

=55cm

Strain gauge resistance

= 350ohm

Gauge factor

=2

Distance of strain gauges near the root from tip

=50cm

Distance of the strain gauges at the middle from tip

=25cm

Distance from the center of the tube to the center of hook

=26cm

Weight of hook s.no

=0.25kg W (kg)

ɛ

1 2 3 4 5

2.28 4.31 6.29 8.27 10.26

A

ɛ

L=52cm 23 39 54 68 82

B

L=28cm 13 21 29 38 46

Result: Load:

=10.26kg =70x109x82x10-6x2=11.41Mpa

Bending stress of root (A)

=70x109x16x10-6x4=44.80

Shear stress at (B) Principal stress at the root

=11.87Mpa, -0.95Mpa

Principal angle at the root

=16.89o

Theoritical calculations: I =12.28cm4

J= 24.56cm4

M= 10.26X9.81X50 = 5032.53N-cm 28

ɛ

C

L=38.5 6 8 11 13 16


T=10.26X9.81X26

Σ =Mdo/2I

=12.35Mpa

Τ = Td o/2J

=3.71Mpa

=2616.915N-cm

Principal stress at the root

=10.89Mpa, -0.65Mpa

Principal angle at root

=14.860

Principal stress and directions are calculated using equ (2) & (4).

29


STRUCTURAL BEHAVIOUR OF A SEMI TENS ION FIELD BEAM

Aim: To investigate and study the behavior of a semi-tension beam.

Theory: The development of a structure in which buckling of the web is permitted with the shear loads being carried by diagonal tension stresses in the web is a striking the design of aerospace structures from the standard structural designs. Such as beam design for bridges and buildings. As thin sheets are week in compression, the webs of the Wagner beam will buckle at a low value of the applied vertical load. The phenomena of buckling may be observed by noting the wrinkles that appears on the thin sheet. As applied load further increased, the stress in the compression direction does not increase, however the stress increases in the tension direction. This method of carrying the shear load permits the design of relatively thin webs because high allowable stress in tension. Theory developed by the Wagner .the diagonal tensile stress σt in the thin web given by the expression.

Σ t =2W/dt sin2α Where

(1)

W = shear load D = distance between the c.g of the flanges T =thickness of the web Α =angle at which wrinkling occurs Tan4α =1+td/2AF

/1+tb/AS

(2)

b= distance between stiffeners AF = area of flange As = area of stiffener Z = distance of strain gauge on the flange from the tip. 30


Stress in the flange σF = W/2AF tan α

+WZ/AFd

(3)

Stress in the stiffener σS =Wb/ASdtan α

(4)

Apparatus required: A stiffened thin-webbed cantilever beam held on a suitable frame, strain gauges, strain indicators, hydraulic jack, and load cell and load indicator.

Procedure: The wrinkling angle is calculated using the equation (1) and the strain gauge is fixed at the angle in the web. Strain gauges also fixed on the flanges and a stiffener to measure their respective stresses. The load is applied gradually in steps of 100kg using the hydraulic jack. Fr each loads the load indicator reading, strain indicator reading corresponding to each strain gauge noted. Precaution is taken so that the beam does not undergo any permanent deformation. Hence the beam is loaded about the wrinkling load.

The strain gauges are connected in Quarter Bridge and the strain indicator readings are to be multiplied by 4 to obtain the actual strain.

t=

b=

AF =

AS =

S.NO 1 2 3 4 5

W (kg)

ɛ

t

α=

d=

ɛ

E=

F

ɛ

Σt=4E

S

ɛ

t

ΣF =4E

ɛ

F

Σs =4E

ɛ

Result: Σt, Σf and σs values are calculated theoretically using equations (1),(3) and (4) and compared with the experimental values given in the table. 31

AS


t = 1mm α =43.90

b = 230mm

= 366mm2 AF

S.NO 1 2 3 4 5

A

Wkg 100 200 300 400 500

ɛ

t

19 35 53 75 96

S

ɛ

F

25 45 68 89 119

d = 290mm

=366mm2

ɛ

S

Σt=4E

11 17 23 28 34

E =70Gpa

ɛ

5.32 9.8 14.8 21.00 26.88

t

ΣF =4E

ɛ

7.00 14.00 19.04 24.940 33.32

F

Σs =4E

ɛ

3.08 4.780 6.44 7.84 9.52

Result: Σt, Σf and σs values are calculated theoretically using equations (1), (3) and (4) and compared with the experimental values given in the table.

32

S


HINGED BAR SUSPEBDED BY TWO WIRES OF DIRRERENT MATERIALS

Aim: To determine the forces in the two wires of different materials attached to the hinged bar experimentally.

Theory: A statically determinate force system is one in which the values of all the external forces acting on the body can be determined by the equation of statics along more unknown forces than the equations of the equilibrium such a case of the force system is said to be statically in determinate. A hinged bar suspended by two wires of different materials in an indeterminate system. it will have one unknown force in each wire and a vertical unknown reaction at the hinged support. To determine the known forces ‘DETERMINATION Method’s used, in which considers the deformation of the system. The procedure to be followed in analyzing. An indeterminate system. Enough equations involving deformations must be written so that the total number of equations from both statics and deformations is equal to the number of unknown forces involved. The present experiment is aimed at finding these unknown forces using simple dial gauges.

Apparatus required: Loading frame, hinged bar suspended steel and copper wires of equal length and the diameter, dial gauges (2 Nos.), spirit level, weights and hook for placing the weights.

Procedure: The hinged bar is suspended horizontally using spirit level. Two dial gauges are placed exactly below the point of the attachment of each wire. Weights are placed in the weight hanger in steps of 2kgs up to 10kgs and the corresponding dial gauges reading are noted and tabulated as given below. The load carried by each wire is calculated using the theory from ‘strength of materials’ as shown below and compared with the values obtained from the dial gauge readings.

33


Experimental calculations: =

L1 =L2 = L = 577mm (π/4) d2

d1 = d2= 1mm

E1 = 2000Gpa S.no 1 2 3 4 5

Δ1 24 37 46 57 68

A2

Δ2 11 17 23 30 35

Δ1 = 44

Δ2 =

P1 =AE1/L1Δ1

=

(Πx100x10 3x44x10-2)/4x577

P2 = A2E1/L2Δ2

=

-2 (Πx200x10 3 x23x10)/4x577

= =

59.9N 62.64N

3.99

FROM THEORY P1 = P/2

=

E2 =100Gpa

P in kg 2 4 6 8 10

For a load 10-2 =8kg

P1/P2 =

A1

AND P2 =P/2

Hence the ratio of forces from the experiment and theory coincides.

34

23

=6.1kgf =6.4kgf


VERIFICATION OF RECIPROCAL THEOREM AND PRINCIPLE OF SUPERPOSITION

Aim: (i) (ii)

To verify the reciprocal theorem To verify the principle of superposition

Theory: According to reciprocal theorem, the deflection of the beam at point 1, due to the load p at point 2 is equal deflection of the beam at point 2, when same load is placed at point 1.

Consider a simply supported beam shown in figure. Y = (pbx/6EIL) (L2 –b2-x2) Yx =d =

(Pbx/6EIL) (L 2-b2-d2)

This expression does not change if we substitute’s’ for ‘b’ and ‘b’ for’d’ This means the if the load p is moved from the point c to the D, the deflection measured at ‘c’ is equal to the deflection point D.

The principle of superposition states that the deflection of the point in a structure due loads P1 AND P2 is equal to the sum of deflections of the same point due to the load p1. Load p2 applied separately 35


Apparatus required: (i)

The beam is simply –supported using the knife edges and the weight hanger is kept at L/4. Apply a load of 0.5kg and measure the deflection at 3L/4 using the dial gauges. Load is applied gradually in steps of 0.5kg up to 2.5kg and the corresponding dial gauge reading is noted at 3L/4. Now the dial gauge and the weight hanger are interchanged and the deflections are noted for loads 0.5kg to 2.5kg. The readings are tabulated.

(ii)

The beam is simply- supported using the knife edges weight hanger is kept at L/4. Apply a load of 0.5kg and measure the deflection at L/2 using the dial gauge. Load is applied. Load is applied gradually in steps of 0.5kg up to 2.5kg and the corresponding dial gauge reading is noted at 3L/4. Now the dial gauge and the weight hanger is interchanged and the deflections are noted for loads 0.5kg to 2.5kg. The readings are tabulated.

The experiment can be reaped by fixing the specimen as cantilever beam. For cantilever apply load in steps of 2.0kg and L =40-50cm. similarly for aluminum specimen apply load in steps of 2.0kg.

36


Table:-1

S.NO

Load at A

Deflection at C

Load at C

Deflection at A

PA kg

Δc

Pckg

ΔA

1

0.5

0.5

2

1.0

1.0

3

1.5

1.5

4

2.0

2.0

5

2.5

2.5

Table-2:

S.NO

Load at A

Defln at B

Load at C

Defln. At

Load at A

Defln. At

PA kg

ΔB

PC kg

B

and C

B

ΔB

(PA+PC)kg

ΔB

1

0.5

0.5

1.0

2

1.0

1.0

2.0

3 4

1.5 2.0

1.5 2.0

3.0 4.0

5

2.5

2.5

5.0

Result: (i)

From the results in the table 1, it is found that ΔCA = Δ AC and hence the reciprocal theorem is verified.

37


(ii)

From the results in the table 2, it is found that the sum of the deflections in the columns three and five is equal to the deflections in the column seven and hence the principle of superposition is verified.

BUCKLING OF COULMNS

Aim: To determine a critical load of column using south well plot.

Theory The need to make use of materials with high strength-to-weight ratio in aircrafts design has resulted in the use of slender structural components that fail more than often by instability than by excessive stress. The simplest example of such structural component is a slender column. Ideal column under the small compressive equilibrium position returns to it srcinal equilibrium position. Further increase the load does not alter the situation. Until a stage is reached. This is neutral equilibrium position for the column position also. The column is said to have failed due its instability. The load beyond which the column is unstable is called the Euler load or the critical load. In an ideal column deflection appear suddenly at the critical load whereas in actual column due to imperfections present the deflection starts appearing as soon as the loads are applied. South well shown that there exists a relation between a compressive load. Later deflations which can be utilized profitably to determine the critical load and the eccentricity of the column by the graphical procedure without actually destroying the test specimen. The well known formula for critical load of an uniform slender column is Pcr =C (π2EI)/L2 Where C is constant depending on the end conditions of the column, E is the youngs modulus of the material of the column, I is the least moment of inertia of the section and L is the length.

38


As mentioned earlier the lateral deflection of the ideal column is indeterminate at the critical load. But for the deflations ‘Δ’ of a simple supported column at its midpoint due to load P can be written as

Δ = e/

[/)1 ⋮ ⋱⋯ ⋮ ] ⋯

Rewriting the equation in the following form, Δ

=Pcr(Δ/P)-e

It is seen that the plot of the ratio (Δ/P) against Δ is a straight line and that and its inverse slope gives the critical load and the intercept on (i)(i)(i) abovecthe Δ – axis gives the magnitude of the eccentricity.

Apparatus required: 1. 2. 3. 4. 5.

Column testing apparatus as shown in the figure. Specimens Dial gauge Venire calipers Weights

Procedure: The ends of the given column specimen are carefully prepared to provide knifeedge supports. The specimen is then mounted on the column testing apparatus. The columns should be placed with the longitudinal axis aligned vertically at the mid-point of the given column, a dial gauge is placed. Loads are then gradually applied. A plot of Δ/p versus Δ is obtained. This is called the south well plot, which should be a straight line. The inverse of the slope of this line will be indicatedthe value of the critical buckling load, pcr of the given hinged- hinged column as shown in the theory.

39


Table: Specimen dimensions’ =

s.no 1 2 3 4

d=

L=

Δ

P

9 10

Pcr =π2EI/L2 Ptheo =

Pexp =

40

Δ/P


SOUTH WELL PLOT LOAD FRAME WEIGHT = 4kg

s.no 1 2 3

P 4+1=5 4+2=6 4+3=7

Δ 12 16 20

Δ/P 2.4 2.6 2.8

4 5 6 7 8 9 10

4+4=8 4+5=9 4+6=10 4+7=11 4+8=12 4+9=13 4+10=14

27 34 43 55 67 81 100

3.3 3.7 4.3 5 5.5 6.2 7.1

Material: aluminum L =1 b= 14.5mml d = 8mm I = bd3/12 I = 618.6mm4 P = (π2EI)/L2 E =70X10 9Pa P = 432.6N Pcr the= 432.6/9.81

= 44.09kg

Pcr exp= tan- 1 θ = (25.5/1.5)

=17.0kg

41


(i) (ii) (iii)

Aircraft landing gears fixed as well as retractable are columns. Buckling of stringers in the aircraft is very important in the design of aircraft The wing struts of executive aircrafts are beam- columns.

Model graph:

SOUTH WELL PLOT 4.5 4 3.5 3

P 2.5 / Δ

2 1.5 1 0.5 0 10

20

30

40

tanƟ

42

50


CALIBRATION OF PHOTO ELASTIC MODEL MATERIAL BY USING A BEAM SUBJECTED TO PURE BENDING Aim: To find out material fringe value of photo elastic model material .

Apparatus required: 12” diffused light transmission polar scope universal loading frame.

Formula: fσ = 12PLY/h3 N Where. P = Load applied L = Distance of point of load application Y = Distance of fringe from neutral axis h = Depth of the beam model

Procedure: (i)

Load the beam at four points by applying known force

(ii)

Use white lights with circular polar scope arrangement and identify fringe order

43


(iii)

The trace of the material surface is identified by zero order fringe.

(iv)

Shift the monochromatic light and carefully measure the distance y from the nutral axis.

(v)

The middle section of the beam subjected to a pure bending

(vi)

moment Assuming that only σx is acting

(vii) Calculate value of fσ and find out the average value .

Observation: (I)

Load applied =16.35kg

(II)

Distance of point of load application =113cm

(III)

Depth of beam model 3cm

Tabular Column:

S.NO

Load on

Load on

Fringe

Distance

pan W kg

model

order

from

P kg

Distance

NA = Y

fσ

kg/cm

Average Kg/cm

from NA = Y 1

10

16.35

1

0.25cm

14.30

2

10

16.35

2

0.4cm

11.44

44

12.71


Model Calculation: fσ = 12pLY/h3 .N fσ = 12x16.35x3x0.25/(2.9)3x1 fσ = 14.30kg/cm

Result:

Model of the beam subjected to pure bending can be used to calibrate the photo elastic material and average fσ is found out. CALIBRATION OF PHOTO ELASTIC MODEL MATERIAL BY USING CIRCULAR DISC UNDER DIAMETER COMPRESSION

Aim: To calibrate photo elastic model by using circular disc under diametrical compression.

Apparatus required: Circular disc prepared out of photo elastic model material.

1.

2. Universal loading frame 3. 12” diffused light transmission polar scope

Procedure: 1. Load the disc in universal loading frame under diametrical compression 2.

The distance x and y must be initially measured.

3. Apply light load and plane polar scope arrangement 4. Observe the fringe part ten and note the iso clinic reading for the point of interest p

on the disc model 45


5. In this case as the point of interest p which is the center of the disc, isoclinic reading

automatically becomes zero. 6. Now apply known value of the load at the end of the lever and set the circular polar

scope arrangement. 7. Use white light and identify the fringe order at the point p 8. Increase the load in steps and note down the fractional fringe order at the center point 9. After measuring the diameters of the disc proceed to calculate material fringe value

Formula:

fσ = 8p/πDN Where, P = load on the model D = diameter of disc in cm N = fractional fringe order at the center

46


Calculation:

(I)

Distance

x=

(II)

Distance

y=

(III)

Diameter of disc =

Tabular column:

s.no

Load

Load on the

Fractional

Fringe value

(w)

model

fringe order

fσ

P =wy/x kg

N At the center

47


Model calculation:

(I)

Distance

x = 31.7cm

(II)

Distance

y = 113cm

(III)

Diameter of disc = 6cm

Tabular column:

s.no

Load

Load on the

Fractional

Fringe value

(w)

model

fringe order

fσ

P =wy/x kg

N At the center

1

4

12.5

0.491

10.8

2

6

18.75

0.720

10.9

3

8

25.00

0.9

11.78

48


(1)

P = WY /X

P = 100X4/32 P = 12.5kg

(2)

N= fringe order

N = 0+0.494 = 0.494

(3)

fσ =

8p /π 6 x 0.494

fσ = 10.8 kg /cm 2

Result: Circular disc model is easy to prepare to load hence is suitable for calibration of photo elastic model The average value of fσ calculates is in good arrangement with the value specification

49


VIBRATION MEASUREMENTS Aim:

To determine the acceleration, velocity and displacement of the given specimen.

Apparatus required: 

Vibration shaker



Signal generator



Frequency indicator

Procedure: (i)

Increase the voltage of the signal generator approaches and setup the frequency.

(ii)

The frequency may be set at 150 and 200

(iii)

The shaker oscillates with the above frequency and the given test piece u placed on the shaker.

(iv)

On the display signal and generate instrument measure velocity, acceleration and displacement.

(v)

With the given frequency, verify the displacement, velocity and acceleration.

Tabular column:

S.NO

Frequency Hz

Displacement

Acceleration

mm

m/s2

50

Velocity m/s


MODEL CALCULATION:

Tabular column:

S.NO

Frequency Hz

Displacement

Acceleration

mm

m/s2

Velocity m/s

1

20

0285

2.5

1.23

2

30

0.342

4.8

2.53

3

50

0.356

11

3.82

Velocity (v) = acceleration x100/2π x fs x = 2.5x100/2π x 20 x = 14cm/s

√2

√2

Displacement = A X1000 X2/ (2π X20)2 = (2.5) X1000 X2/ (2π X20)2 =0 .316mm

Result:

Thus the velocity, acceleration and displacement of the given test piece was found out. 51


52

Aircraft Structures II Laboratory Manual with model calculations and graphs

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