Aircraft Structures Lab II Manual for AUC R2008_S

DHANALAKSHMI SRINIVASAN ENGINEERING COLLEGE (Approved by AICTE, New Delhi & Affiliated to Anna University, Chennai – 600 025) (NBA Accredited and ISO 9001:2008 Certified Institution) PERAMBALUR - 621 212.

DEPARTMENT OF AERONAUTICAL ENGINEERING AE 2 3 0 5

Airc Ai rcrr aft af t Stru St rucc ture tu ress L ab abor orat ator ory y II MANUAL NOTE BOOK

Name

:

Reg. No.

:

Semester

:

Academic Year

:

…………………………………………………………….

…………………………………………………………….

…………………………………………………………….

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Dhanalakshmi Srinivasan Engineering College, Perambalur – 621 212.

AE2305

AUC R2008

AIRCRAFT STRUCTURES LAB – II

LTPC 0032

OBJECTIVE To experimentally study the unsymmetrical bending of beams, find the location of shear centre, obtain the stresses in circular discs and beams using photoelastic techniques, calibration of photo – photo – elastic elastic materials and study on vibration of beams. LIST OF EXPERIMENTS 1. Unsymmetrical Unsymmetr ical bending of Z-section Z-sectio n beams 2. Shear centre centre location location for open channel sections 3. Shear centre location for closed closed D-sections D-sections 4. Constant strength beam 5. Flexibility matrix for cantilever cantilever beam 6. Beam with combined loading 7. Calibration of PhotoPhoto- elastic elastic materials 8. Stresses in circular circular discs and beams using using photo elastic techniques techniques 9. Determination Determination of natural frequencies of cantilever cantilever beams 10. Wagner beam – beam – Tension Tension field beam

TOTAL: 45 PERIODS LIST OF EQUIPMENT (for a batch of 30 st udents) Sl. No.

Name of the Equipment

Qty

Experiments Number

1

Beam Test set – set –up up

2

1, 2, 3,4, 5

2

Unsymmetrical ‘Z’ section beam

1

1

3

Channel section beam

1

2

4

Closed ‘D’ section beam

1

3

5

Dial gauges

12

1,2,3

6

Strain indicator and strain gauges

One set

4,5,6

7

Photo – Photo – elastic apparatus

1

7,8

8

Amplifier

2

9

9

Exciter

2

9

10

Pick – Pick – up

2

9

11

Oscilloscope

2

9

12

Wagner beam

1

10

13

Hydraulic Jack

1

10

|V Semester [2014 – 2015] Gurunath K |V

CONTENTS Sl. No.

Date

Name of the Experiment

Page

Marks

Sign

Completed / Not Completed

Average Marks Signature of Staff in Charge


AUC R2008

Aircraft Structures Laboratory II

List of Experiments

1. Unsymmetrical Bending of Beams.

2. Shear Centre Location for Open Sections.

3. Shear Centre Location for Closed Sections.

4. Constant Strength Beam.

5. Beam with Combined Loading.

6. Structural Behaviour of a Semi-Tension Field Beam (Wagner Beam).

7. Determination of Natural Frequencies of Cantilever Beams.

8. Stresses in Circular Discs using Photo elastic Techniques

Gurunath K – AE

2305 Aircraft Structures Laboratory II | 1


AUC R2008

Shear center

Extension Piece

WH Wa

2Wa = WV

Wa

Determination of Principal axes

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AUC R2008

Exercise No.: 01

Unsymmetrical Bending of Beams Date:

AIM:

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

The well-known flexure formula

=  

based on the elementary theory of bending

of beams assumes that the load is always applied through one of the principal axes of the section. Actually, even if the applied load passes through the centroid and/or the shear centre of the section, the plane of bending and the plane of loading need not necessarily are the same. Therefore, a knowledge of the location of the principal axes is required for the determination of the stress distribution in beams (of any arbitrary cross section) using flexure formula. The determination of the principal axes experimentally is described here.



If Ix, Iy and Ixy are the moments and product of inertia of any section about an arbitrary orthogonal centroidal axes OX and OY then the inclination of one of the principal axes to OX is given by

 2= (−)

…………………… Eqn (1)

The experimental determination of the principal a xes of a given section is based on the fact that when the load passes through the shear centre and is in the direction of one of the principal axes of the section, the entire section under the load deflects in the direction of the load only. APPARATUS REQUIRED:     

A thin uniform cantilever ‘Z’ section as shown in figure. At the free end extension pieces are attached on either side of the web to facilitate vertical loading. Two dial gauges (to be mounted vertically and horizontally as in figure). This enables the determination of displacements u and v. Two hooks are attached to the extension pieces to apply the verti cal load W V. A string and pulley arrangement to apply the horizontal load W H. A steel support structure to mount the channel section as cantilever.

FORMULA USED:

1. Theoretical calculation

Where,

 

→→ →→

 2= (2)

Inclination of one of the principal axes Moment about X axis Moment about Y axis Product of inertia

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TABULATION:

WV

Sl. No.

=

Height, h

=

Breadth, b

Dial gauge readings

Horizontal Load WH

u

v

=



Thickness, t



=

Remarks

01 02 03 04 05 06 07 08 09 10 11

Gurunath K – AE



 =tan− 

2. Experimental calculation

Where,

AUC R2008

 →→

Vertical load of weight Horizontal load of weight

PROCEDURE:

1. Mount two dial gauges on the tip section to measure the horizontal and vertical deflections of a point on it. 2. Apply the vertical load W V (about 2.4 kg, including two hooks of each 200 gm). 3. Read u and v the horizontal and vertical deflections respectively, at the chosen point. 4. Increase the load WH in steps of about 300 gm (for the first case 100 gm + 200 gm hook) from zero to a maximum of about 3 kg noting down in each case the values of u and v. Repeat the procedure and check for consistency in measurements. 5. Plot the graphs

  vs

and find the intersection of this curve with a straight line

through the origin at 45°. (Note: The X and Y scales must be chosen to be same for the graph). 6. Calculate the inclination of one of the principal axes to the web as



where WV and WH correspond to the point of intersection. 7. Calculate the inclination using Eqn (1).

=tan− 


 =tan− 

2. Experimental calculation

Where,

AUC R2008

 →→

Vertical load of weight Horizontal load of weight

PROCEDURE:

1. Mount two dial gauges on the tip section to measure the horizontal and vertical deflections of a point on it. 2. Apply the vertical load W V (about 2.4 kg, including two hooks of each 200 gm). 3. Read u and v the horizontal and vertical deflections respectively, at the chosen point. 4. Increase the load WH in steps of about 300 gm (for the first case 100 gm + 200 gm hook) from zero to a maximum of about 3 kg noting down in each case the values of u and v. Repeat the procedure and check for consistency in measurements. 5. Plot the graphs

  vs

and find the intersection of this curve with a straight line

through the origin at 45°. (Note: The X and Y scales must be chosen to be same for the graph). 6. Calculate the inclination of one of the principal axes to the web as



where WV and WH correspond to the point of intersection. 7. Calculate the inclination using Eqn (1).

=tan− 

RESULT: Thus the principal axis of an unsymmetrical section has been determined.

Gurunath K – AE



Dial gauge

AUC R2008

Dial gauge

Wa + Wb = WV

Wa

Wb

Determination of Shear Center

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AUC R2008

Exercise No.: 02

Shear Centre of Open Sections Date: AIM:

To determine the shear centre of an open section. THEORY:

For any unsymmetrical section there exists a point at which any vertical force does not produce a twist of that section. This point is known as shear centre. The location of this shear centre is important in the design of beams of open sections when they should bend without twisting, as they are weak in resisting torsion. A thin walled channel section with its web vertical has a horizontal axis of symmetry and the shear centre lies on it. The aim of the experiment is to determine its location on this axis if the applied shear to the tip section is vertical (i.e., along the direction of one of the principal axes of the section) and passes through the shear centre tip, all other sections of the beam do not twist. APPARATUS REQUIRED:  

 

A thin uniform cantilever beam of channel section as shown in the figure. At the free end extension pieces are attached on either side of the web to facilitate vertical loading. Two dial gauges are mounted firmly on this section, a known distance apart, over the top flange. This enables the determination of the twist, if any, experienced by the section. A steel support structure to mount the channel section as cantilever. Two loading hooks each weighing about 200 gm.

FORMULA USED:

1. Theoretical calculation

Where, h b

→→

= 63ℎ

height of the flange width of the flange

2. Experimental calculation From the graph ‘e’ versus (d 1-d2)

PROCEDURE:

1. Mount two dial gauges on the flange at a known distance apart at the free and of the beam. Set the dial gauge readings to zero. 2. Place a total of, say two kilograms load at A (loading hook and nine load pieces will make up this value). Note the dial gauge readings (nominally, hooks also weigh a 200 gm each).

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AUC R2008

TABULATION:

Length, L =

Height, h

WV

Distance between the two hook sections (AB)

= (Wa+ W b)

=

Breadth, b

=

Thickness, t

=

Dial gauge readings Sl. No.

Wa

Wb

(d1-d2) d1

=

d2

 =   

01 02 03 04 05 06 07 08 09 10 11

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AUC R2008

3. Now remove one load piece from the hook at A and place another hook at B. This means that the total vertical load on this section remains 2 kg. Record the dial gauge readings. 4. Transfer carefully all the load pieces and finally the hook one by one to the other hook noting each time the dial gauge readings. This procedure ensures that while the magnitude of the resultant vertical force remains the same its line of action shifts by a known amount along AB every time a load piece is shifted. Calculate the distance ‘e’ of the line of action from the web thus:

=−  

5. For every load case calculate the algebraic difference between the dial gauge readings as the measure of the angle of twist suffered by the section. 6. Plot against ‘e’ and obtain the meeting point of curve (a straight line in this case) with the ‘e’-axis (i.e., the twist of the section is zero for this location of the resultant vertical load). This determines the shear centre.



Theoretical location of the shear centre

*

= +

Though a nominal value of 2 kg for the total load is suggested it can be less. In that event the number of readings taken will reduce proportionately. GRAPH:


AUC R2008

3. Now remove one load piece from the hook at A and place another hook at B. This means that the total vertical load on this section remains 2 kg. Record the dial gauge readings. 4. Transfer carefully all the load pieces and finally the hook one by one to the other hook noting each time the dial gauge readings. This procedure ensures that while the magnitude of the resultant vertical force remains the same its line of action shifts by a known amount along AB every time a load piece is shifted. Calculate the distance ‘e’ of the line of action from the web thus:

=−  

5. For every load case calculate the algebraic difference between the dial gauge readings as the measure of the angle of twist suffered by the section. 6. Plot against ‘e’ and obtain the meeting point of curve (a straight line in this case) with the ‘e’-axis (i.e., the twist of the section is zero for this location of the resultant vertical load). This determines the shear centre.



Theoretical location of the shear centre

= +

*


Plot ‘e’ versus (d1-d2) curve and determine where this meets the ‘e’ axis and locate the shear centre. PRECAUTIONS:

I.

II.

For the section supplied there are limits on the maximum value of loads to obtain acceptable experimental results. Beyond these the section could undergo excessive permanent deformation and damage the beam forever. Do not therefore exceed the suggested values for the loads. The dial gauges must be mounted firmly. Every time before taking the readings tap the set up (not the gauges) gently several times until the reading pointers on the gauges settle down and do not shift any further. This shift happens due to both backlash and slippages at the points of contact between the dial gauges and the sheet surfaces and can induce errors if not taken care of. Repeat the experiments with identical settings several times to ensure consistency in the readings.

RESULT:

The shear centre obtained experimentally is compared with the theoretical value.

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Dial gauge

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Dial gauge

Wa

Wb

Determination of Shear Center- Closed section

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AUC R2008

Exercise No.: 03

Shear Centre of Closed Sections

Date: AIM:

To determine the shear centre of a closed section. THEORY:

For any unsymmetrical section there exists a point at which any vertical force does not produce a twist of that section. This point is known as shear centre. The location of this shear centre is important in the design of beams of closed sections when they should bend without twisting. The shear centre is important in the case of a closed section like an aircraft wing, where the lift produces a torque about the shear centre. Similarly the wing strut of a semi cantilever wing is a closed tube of aerofoil s ection. A thin walled ‘D’ section with its web vertical has a horizontal axis of symmetry and the shear centre lies on it. The aim of the experiment is to determine its location on this axis if the applied shear to the tip section is vertical (i.e., along the direction of one of the principal axes of the section) and passes through the shear centre tip, all other sections of the beam do not twist. APPARATUS REQUIRED:  

 

A thin uniform cantilever beam of ‘D’ section as shown in the figure. At the free end extension pieces are attached on either side of the web to facilitate vertical loading. Two dial gauges are mounted firmly on this section, a known distance apart, over the top flange. This enables the determination of the twist, if any, experienced by the section. A steel support structure to mount the channel section as cantilever. Two loading hooks each weighing about 200 gm.

PROCEDURE:

1. Mount two dial gauges on the flange at a known distance apart at the free and of the beam. Set the dial gauge readings to zero. 2. Place a total of, say two kilograms load at A (loading hook and nine load pieces will make up this value). Note the dial gauge readings (nominally, hooks also weigh a 200 gm each). 3. Now remove one load piece from the hook at A and place another hook at B. This means that the total vertical load on this section remains 2 kg. Record the dial gauge readings. 4. Transfer carefully all the load pieces and finally the hook one by one to the other hook noting each time the dial gauge readings. This procedure ensures that while the magnitude of the resultant vertical force remains the same its line of action shifts by a known amount along AB every time a load piece is shifted. Calculate the distance ‘e’ of the line of action from the web thus:

=−  

5. For every load case calculate the algebraic difference between the dial gauge readings as the measure of the angle of twist suffered by the section.

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AUC R2008

TABULATION:

Length, L

=

WV

= (Wa+ W b)

Height, h

=

Thickness, t Distance between the two hook sections (AB)

Dial gauge readings Sl. No.

Wa

=

Wb

(d1 - d2) d1

d2

=

 =   

01 02 03 04 05 06 07 08 09 10 11

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AUC R2008



6. Plot against ‘e’ and obtain the meeting point of curve (a straight line in this case) with the ‘e’-axis (i.e., the twist of the section is zero for this location of the resultant vertical load). This determines the shear centre. *


Plot ‘e’ vs (d1-d2) curve and determine where this meets the ‘e’ axis and locate the shear centre. PRECAUTIONS:

I.

II.



AUC R2008



6. Plot against ‘e’ and obtain the meeting point of curve (a straight line in this case) with the ‘e’-axis (i.e., the twist of the section is zero for this location of the resultant vertical load). This determines the shear centre. *


Plot ‘e’ vs (d1-d2) curve and determine where this meets the ‘e’ axis and locate the shear centre. PRECAUTIONS:

I.

II.


RESULT:

The shear centre obtained experimentally is compared with the theoretical value.

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Constant Strength Beam

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Exercise No.: 04

Constant Strength Beam Date: AIM:

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

The aerospace structures engineer is constantly searching for types of structures which will save structural weight and still provide a structure which is satisfactory from a fabrication and economic standpoint. One such structure is constant strength beam. A beam in which section modules 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.

→→ →

b width of the beam L length of the beam h depth of the beam Section modulus,

 

= = 

=  =  =  = 6ℎ = ℎ6  = 

Let the width of the beam be constant and the depth varies. Then

ℎ =  

…………….. Eqn (1)

APPARATUS REQUIRED:

A constant strength beam in which the depth varies as in Eqn (1) and made of aluminium. Strain gauges, strain indicator and weights with hook. PROCEDURE:

at



The constant strength beam is fixed as a cantilever strain gauges are fixed near the root, and



. The strain gauges are fixed both on the top and bottom surfaces at each location

to increase the circuit sensitivity of the strain gauge circuit. Hence half bridge used in the strain indicator to measure the strain at each location (strain = strain meter reading x 2). The beam loaded gradually in steps of 2 kg up to 10 kg by placing the weights slowly in the hook near the tip of the cantilever (loading hook weight 0.25 kg is to be added). The strain gauge readings are noted for every 2 kg at locations A, B, C and tabulated as given below. Strain gauge resistance Gauge factor Young’s modulus,

= = E =

350 ohm 2 70 GPa

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TABULATION:

Sl. No.

Weight (Kg)

 





  =  ×  = ×  = × (MPa)

(MPa)

(MPa)

01

02

03

04

05

Gurunath K – AE


Dhanalakshmi Srinivasan Engineering College, Perambalur – 621 212. THEORETICAL CALCULATION At point A:

Distance of point A from loading point

=

Depth of the beam

=

Width of the beam

=

Moment of inertia

=

Moment

=

= 

=

At point B:

Distance of point B from loading point

=

Depth of the beam

=

Width of the beam

=

Moment of inertia

=

Moment

=

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THEORETICAL CALCULATION At point A:

Distance of point A from loading point

=

Depth of the beam

=

Width of the beam

=

Moment of inertia

=

Moment

=

= 

=

At point B:

Distance of point B from loading point

=

Depth of the beam

=

Width of the beam

=

Moment of inertia

=

Moment

=

= 

=

At point C:

Distance of point C from loading point

=

Depth of the beam

=

Width of the beam

=

Moment of inertia

=

Moment

=

= 

=

RESULT:

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

Gurunath K – AE



Gurunath K – AE

AUC R2008



AUC R2008

Exercise No.: 05

Combined Loading Date: AIM:

To determine the principal stresses and 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 bending and torsion. In an aircraft wing the lift acting at the centre of pressure produces a torque about the elastic axis and varying bending moment along the wing span. To understand their combined effect a similar specimen, namely a hollow cylinder is subjected to a bending and torsion.

=  

For an elastic structure, Where, σ M y I

→→ →→

Bending stress (N/m2) Bending moment (N-m) Distance of the layer from the neutral axis (m) Moment if Inertia (m4)

  ( −    =  ) The shear stress due to torsion

= 

Where, τ T R J

→→ →→           = 32 

Shear stress (N/m2) Torque applied (N-m) radius of the shaft (m) Polar moment of Inertia (m4)

  +  −      , =  ±    ()     , =  ±   , = ( −) [±√  ] Gurunath K – AE

……………

(1)

……………

(2)

……………

(3)



AUC R2008

TABULATIONS:

Young’s modulus of the tube

=

Outside diameter of the tube

=

Thickness of the tube

=

Length of the tube

=

Strain gauge resistance

=

Gauge factor

=

Distance of the strain gauges near root from tip

=

Distance of the strain gauges at the middle from tip

=

Distance from the centre of the tube to the centre of the hook

=

Weight of the hook

=

Sl. No.

Weight (Kg)

 





01

02

03

04

05

Gurunath K – AE



Where,

tan2∅=  =  ∅ →

……………

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(4)

Principal angle

APPARATUS REQUIRED:

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

Two strain gauges are fixed near the root of the tube fixed as a cantilever, one on the top fibre and the other at the bottom to measure the bending strain. Another strain gauge is fixed at the same location on the neutral axis at 45° to measure the shear stra in. Similarly three more strain gauges are fixed at the middle of the length to verify the result at various locations of the tube. The strain gauges on the top and bottom of the tube are connected to half bridge circuit in the strain indicator to increase the circuit sensitivity, since the tension and compression get added up. The strain gauge at 45° is connected to the quarter bridge of the strain indicator to measure the shear strain. The outside diameter of the tube is measured using vernier callipers. Weights are added to the hook attached to the lever in steps of 2 kg and the strain gauge readings are noted from the strain indicator for each load. From the strains the bending stress, shear stress are calculated and hence principal stresses and principal angle are calculated. These values are compared with theoretical values. NOTE:

For half bridge the strain readings are multiplied by 2 and for Quarter Bridge by 4 to get the actual strains.

RESULT:

Bending stress at the root (A)

=

Shear stress at (B)

=

Experimental:

Principal stresses at the root

=

Principal angle at the root

=

Theoretical:

Principal stresses at the root

=

Principal angle at the root

=

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Wagner Beam

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Exercise No.: 06

Structural Behaviour of a Semi-Tension Field Beam Date:

AIM:

To investigate and study the behaviour of a semi-tension field beam. THEORY:

The development of a structure in which buckling of the web is permitted wit h the shear loads being carried by diagonal tension stresses in the web is a s triking example of the departure of the design of aerospace structures from the standard structural design methods in other fields of structures, such as beam design for bridges and buildings. The first study and research on this new type of structural design involving diagonal semi-tension field action in beam webs done by Wagner and hence Wagner beam. As thin sheets are weak in compressions, 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 nothing the wrinkles that appear on the thin sheet. As the applied load i s further increased, the stress in the compression direction does not increase, however the stress increase in the t ension direction. This method of carrying the shear load permits the design of relatively thin webs because of high allowable stresses in tension. According to the theory developed by Wagner, the diagonal tensile stress σt in the thin web is given by the expression

  = 

Where, W d t α

Where, b AF AS

→→ →→

Eqn (1)

Shear load Distance between the CG of the flanges Thickness of the web Angle at which wrinkling occurs

 +  tan = +

→→ →

...............

...............

Eqn (2)

Distance between stiffeners Area of flange Area of stiffener

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TABULATION:

t

=

b

=

d

=

AF

=

AS

=

E

=

Sl. No.

Weight (Kg)



t



F



S

=

      t = 4E

t

F=

4E

F

S=

4E

S

01

02

03

04

05

Gurunath K – AE



Stress in the Flange

Stress in the stiffener

 = ∝    =  tan∝

AUC R2008

............ Eqn (3) ............ Eqn (4)

APPARATUS REQUIRED:

A stiffened thin-webbed cantilever beam held in a suitable frame, strain gauges, strain indicator, hydraulic jack, load cell and load indicator. PROCEDURE:

The wrinkling angle is calculated using the Eqn (1) and a strain gauge is fixed at this angle in the web. Strain gauges are also fixed on the flanges and a stiffener to measure their respective stresses. The load is applied gradually in steps of 100 kg using the hydraulic jack. For each load the load indicator reading, strain indicator reading corresponding to each strain gauge is noted. Precaution is taken so that the beam does not undergo any permanent deformation. Hence the beam is not loaded up to w rinkling load. The readings are tabulated as given below The strain gauges are connected in Quarter Bridge and hence the strain indicator readings are to be multiplied by 4 to obtain the actual strain.


Stress in the Flange

Stress in the stiffener

 = ∝    =  tan∝

AUC R2008

............ Eqn (3) ............ Eqn (4)

APPARATUS REQUIRED:

A stiffened thin-webbed cantilever beam held in a suitable frame, strain gauges, strain indicator, hydraulic jack, load cell and load indicator. PROCEDURE:

The wrinkling angle is calculated using the Eqn (1) and a strain gauge is fixed at this angle in the web. Strain gauges are also fixed on the flanges and a stiffener to measure their respective stresses. The load is applied gradually in steps of 100 kg using the hydraulic jack. For each load the load indicator reading, strain indicator reading corresponding to each strain gauge is noted. Precaution is taken so that the beam does not undergo any permanent deformation. Hence the beam is not loaded up to w rinkling load. The readings are tabulated as given below The strain gauges are connected in Quarter Bridge and hence the strain indicator readings are to be multiplied by 4 to obtain the actual strain.

RESULT:

and σS values are calculated theoretically using Eqns (1),(3) and (4) and compared with the experimental values given in the table. σt, σF

Gurunath K – AE



AUC R2008

Fig. 7.1. A cantilever beam

Fig. 7.2. The cantilever beam under free vibration

Fig. 7.1 shows of a cantilever beam with rectangular cross section, which can be subjected to bending vibration by giving a small initial displacement at the free end; and Fig. 7.2 depicts of cantilever beam under the free vibration.

Gurunath K – AE



AUC R2008

Exercise No.: 07 Determination of Natural Frequencies of Cantilever Beams Date:

AIM:

To obtain natural frequencies of a cantilever beam up to the second mode, experimentally; and to observe the response of the system subjected to a small initial disturbance and virtualization of the experiment. This virtual experiment is based on a theme that the actual experimental measured vibration data are used. THEORY:

Free vibration takes place when a system oscillates under the action of forces inherent in the system itself due to initial disturbance, and when the externally applied forces are absent. The system under free vibration will vibrate at one or more of its natural frequencies, which are properties of the dynamical system, established by its mass and stiffness distribution. In case of continuous system the properties of the system are the function of spatial coordinates. The system has infinite number of degrees of freedom and infinite number of natural frequencies. In actual practice there is always some damping (e.g., the internal molecular friction, viscous damping, aerodynamical damping, etc.) present in the system which causes the gradual dissipation of vibration energy, and it results gradual decay of amplitude of the free vibration. Damping has very little effect on natural frequency of the system, and hence, the calculations for natural frequencies are generally made on the basis of no damping. Damping is of great importance in limiting the amplitude of oscillation at resonance. The relative displacement configuration of the vibrating system for a particular natural frequency is known as the Eigen function in continuous system. The mode shape corresponding to lowest natural frequency (i.e. the fundamental natural frequency) is called as the fundamental (or the first) mode. The displacements at some points may be zero. These points are known as nodes. Generally nth mode has (n-1) nodes (excluding end points). The mode shape changes for different boundary conditions of a beam. MATHEMATICAL ANALYSIS:

For a cantilever beam subjected to free vibration, and the system is considered as continuous system in which the beam mass is considered as distributed along with the stiffness of the shaft, the equation of motion can be written as (Meirovitch, 1967),

   =

1

Where, E is the modulus of rigidity of beam material, I is the moment of inertia of the beam cross-section, Y(x) is displacement in y direction at distance x from fixed end, ω is the circular natural frequency, m is the mass per unit length, m = ρA(x), ρ is the material density, x is the distance measured from the fixed end.

Gurunath K – AE



AUC R2008

Fig. 7.3. The first three undamped natural frequencies and mode shape of cantilever beam

Gurunath K – AE



AUC R2008

We have following boundary conditions for a cantilever beam.

 =0, =0,   =0          =,  =0,  =0  =0  = 

For a uniform beam under free vibration from equation (1), we get

where,

The mode shapes for a continuous cantilever beam i s given as

   ={sin cos  sin sinh  sinh    coshcoscosh} where,

=1,2,3…∞  =   =   =1.875,4.694,7.885

A closed form of the circular natural frequency and boundary conditions can be written as,

where,

, from above equation of motion

RESULT:

First Natural Frequency

 = 

Second Natural Frequency

 = 

Gurunath K – AE

Third Natural Frequency

 = 



AUC R2008

Fig. 8.1. Disk in compression

Fig. 8.2. Stress distribution along horizontal diameter

Gurunath K – AE



AUC R2008

Exercise No.: 08 Stresses in Circular Discs using Photo elastic Techniques Date:

AIM:

To obtain the stresses in circular discs and beams using photo elastic techniques.

THEORY:

A common calibration specimen is the circular disk of diameter D and thickness loaded in diametral compression (Fig. 8.1).



ℎ

    ≥0

The horizontal and vertical normal stresses along the x axis are principal stresses because the shear stress vanishes due to symmetry about the x axis. Also, is positive, while is negative. We therefore take and so as to render . From theory of elasticity, the solutions for the normal stresses along the horizontal diameter are (after Dally and Riley 1991).



where

 =  =   2 1  = ℎ 1  1 13   1 6  = ℎ 1 =  = 2   =0   = ℎ8 ℎ =  = ℎ8  = 8  ℎ ℎ ℎ 

These stresses are plotted in Fig. 8.2. Along the horizontal diameter, the maximum difference occurs at the centre, that is, at . At this point,

Combining this result with the basic photo elastic relation gives

or

Notice that the specimen thickness does not appear in this equation. The reason is that the relative retardation is proportional to , but for a given force P, the stresses are inversely proportional to . The net effect is a result for that is independent of .

Gurunath K – AE

ℎ



AUC R2008

Fig. 8.3. Light-field isochromatics in a diametrally loaded circular disk.

Fig. 8.4. Enhanced image using only the green component of the light used in Fig. 8.3 .

Gurunath K – AE



AUC R2008

A photograph of a light-field isochromatic pattern for a diametrally loaded disk made of PSM-1 is shown in Fig. 8.3. For this specimen, the diameter D was 63.50 mm (2.500 in.), and the load was 1.33 kN (298 lb). The value of N at the centre of the disk, as seen in Fig. 8.3, is approximately 7.0. Therefore the fringe constant for this material is approximately

  = 8  = 0.8635 13307 =761.94  

A more accurate way of determining using this specimen is to record several readings of increasing load P as the fringe value N at the centre takes on integer or half-integer values. The saddle shape of the central fringe allows rather precise determi nation of N for this purpose. Then P is plotted as a function of N, and the best straight-line fit of the data is used to determine the ratio P / N to be used in fringe constant equation. The disk in diametral compression is a favourite specimen for calibration because it is simple to fabricate (at least with a template); it is easy to load; it is not likely to fracture; it produces a fringe pattern that, in the region of interest, is insensitive to edge imperfections; and it is simple to analyse. It also tests the limit of fringe density that can be recorded photographically, by producing very large fringe orders in the vicinity of the contact regions. The image in Fig. 8.3 was obtained with a mercury-vapour light source, which is rich in green light, but which also contains other colours as well. A Tiffen #58 Green filter was placed on the 35 mm camera to reduce the transmission of the other colours. Kodak Gold ASA 200 colour film was used to record the image at f/8 with an exposure of 0.7 s. This arrangement results in distinct fringes up to about N = 15, which is adequate for most work in photo elasticity. A somewhat enhanced image is shown in Fig. 8.4. This image was produced from the same negative as that used to produce Fig. 8.3. However, only the green component of the red-green-blue (RGB) digital scan was retained, and this component was then enhanced by increasing the contrast. Fringe orders up to about N = 20 can be discerned in this figure. The low-order values of N are marked along the horizontal diameter. Note that the centre of the specimen is the location of a true saddle point in the function : to the left and right of this point, the function decreases, and above and below this point, the function increases. Such saddle points are common in photo elastic patterns.

 

RESULT:

The value of the stresses in Circular disc are computed using Photo elastic technique. The stress distribution in a disk in diametral compression is unique in that the fringe number N is equal to zero everywhere along the unloaded boundary.

Gurunath K – AE


Aircraft Structures Lab II Manual for AUC R2008_S

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