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 2305 Aircraft Structures Laboratory II MANUAL NOTE BOOK Name :……………………………………………………………. Reg. No. :……………………………………………………………. Semester :……………………………………………………………. Academic Year :……………………………………………………………. Dhanalakshmi Srinivasan Engineering College, Perambalur – 621 212. AUC R2008 AIRCRAFT STRUCTURES LAB – II AE2305 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 – elastic materials and study on vibration of beams. LIST OF EXPERIMENTS 1. Unsymmetrical bending of Z-section beams 2. Shear centre location for open channel sections 3. Shear centre location for closed D-sections 4. Constant strength beam 5. Flexibility matrix for cantilever beam 6. Beam with combined loading 7. Calibration of Photo- elastic materials 8. Stresses in circular discs and beams using photo elastic techniques 9. Determination of natural frequencies of cantilever beams 10. Wagner beam – Tension field beam TOTAL: 45 PERIODS LIST OF EQUIPMENT (for a batch of 30 students) Sl. No. Name of the Equipment Qty Experiments Number 1 Beam Test set –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 – elastic apparatus 1 7,8 8 Amplifier 2 9 9 Exciter 2 9 10 Pick – up 2 9 11 Oscilloscope 2 9 12 Wagner beam 1 10 13 Hydraulic Jack 1 10 Gurunath K |V Semester [2014 – 2015] No.CONTENTS Sl. Date Name of the Experiment Page Marks Sign . Completed / Not Completed Average Marks Signature of Staff in Charge . 8. 3. 7. Unsymmetrical Bending of Beams. 5. Perambalur – 621 212. Constant Strength Beam. 2. 4. Shear Centre Location for Closed Sections. Determination of Natural Frequencies of Cantilever Beams.Dhanalakshmi Srinivasan Engineering College. Stresses in Circular Discs using Photo elastic Techniques Gurunath K – AE 2305 Aircraft Structures Laboratory II | 1 . Structural Behaviour of a Semi-Tension Field Beam (Wagner Beam). 6. Beam with Combined Loading. AUC R2008 Aircraft Structures Laboratory II List of Experiments 1. Shear Centre Location for Open Sections. Perambalur – 621 212. AUC R2008 Shear center Extension Piece WH Wa 2Wa = WV Wa Determination of Principal axes Gurunath K – AE 2305 Aircraft Structures Laboratory II | 2 .Dhanalakshmi Srinivasan Engineering College. Actually. FORMULA USED: 1. the plane of bending and the plane of loading need not necessarily are the same. 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. THEORY: 𝑀 The well-known flexure formula 𝜎 = 𝐼 𝑦 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. 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.  Two hooks are attached to the extension pieces to apply the vertical load WV. Theoretical calculation 𝑡𝑎𝑛 2𝜃 = 2𝐼𝑥𝑦 (𝐼𝑦 − 𝐼𝑥 ) Where.  A steel support structure to mount the channel section as cantilever.  Two dial gauges (to be mounted vertically and horizontally as in figure). At the free end extension pieces are attached on either side of the web to facilitate vertical loading.  A string and pulley arrangement to apply the horizontal load WH. Perambalur – 621 212. Therefore. 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 𝜃 of one of the principal axes to OX is given by 2𝐼𝑥𝑦 𝑡𝑎𝑛 2𝜃 = (𝐼 −𝐼 ) …………………… Eqn (1) 𝑦 𝑥 The experimental determination of the principal axes 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. 𝜃 𝐼𝑥 𝐼𝑦 𝐼𝑥𝑦 → → → → Inclination of one of the principal axes Moment about X axis Moment about Y axis Product of inertia Gurunath K – AE 2305 Aircraft Structures Laboratory II | 3 . The determination of the principal axes experimentally is described here.: 01 Unsymmetrical Bending of Beams Date: AIM: To determine the principal axes of an unsymmetrical section.Dhanalakshmi Srinivasan Engineering College. If Ix. even if the applied load passes through the centroid and/or the shear centre of the section. AUC R2008 Exercise No. This enables the determination of displacements u and v. AUC R2008 TABULATION: WV Sl. = Horizontal Load WH Height. h = Breadth. b Dial gauge readings u v = Thickness.Dhanalakshmi Srinivasan Engineering College. Perambalur – 621 212. t 𝑾𝑯 𝑾𝑽 𝒖 𝒗 = Remarks 01 02 03 04 05 06 07 08 09 10 11 Gurunath K – AE 2305 Aircraft Structures Laboratory II | 4 . No. 3. 7. Mount two dial gauges on the tip section to measure the horizontal and vertical deflections of a point on it. at the chosen point. Calculate the inclination 𝜃 using Eqn (1). Repeat the procedure and check for consistency in measurements. 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). Experimental calculation 𝑊𝑉 𝜃𝑒𝑥𝑝 = tan−1 ( ) 𝑊𝐻 Where. 𝑢 𝑊𝐻 𝑣 𝑊𝑉 5. 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. Calculate the inclination of one of the principal axes to the web as 𝜃 = tan−1 (𝑊 𝑉 ) 𝐻 where WV and WH correspond to the point of intersection.Dhanalakshmi Srinivasan Engineering College. AUC R2008 2. 2. including two hooks of each 200 gm). Gurunath K – AE 2305 Aircraft Structures Laboratory II | 5 . RESULT: Thus the principal axis of an unsymmetrical section has been determined. 4. Perambalur – 621 212. 𝑊 6. Read u and v the horizontal and vertical deflections respectively. 𝑊𝑉 → Vertical load of weight 𝑊𝐻 → Horizontal load of weight PROCEDURE: 1. Apply the vertical load WV (about 2.4 kg. Dhanalakshmi Srinivasan Engineering College. Dial gauge AUC R2008 Dial gauge Wa + Wb = WV Wa Wb Determination of Shear Center Gurunath K – AE 2305 Aircraft Structures Laboratory II | 6 . Perambalur – 621 212. This point is known as shear centre. Note the dial gauge readings (nominally.  Two loading hooks each weighing about 200 gm. The aim of the experiment is to determine its location on this axis if the applied shear to the tip section is vertical (i. Theoretical calculation 𝑒= 3𝑏 ℎ [6 + ( )] 𝑏 Where. over the top flange.: 02 Shear Centre of Open Sections Date: AIM: To determine the shear centre of an open section. The location of this shear centre is important in the design of beams of open sections when they should bend without twisting. This enables the determination of the twist. hooks also weigh a 200 gm each). a known distance apart. if any.  A steel support structure to mount the channel section as cantilever. say two kilograms load at A (loading hook and nine load pieces will make up this value). all other sections of the beam do not twist. Experimental calculation From the graph ‘e’ versus (d1-d2) PROCEDURE: 1. FORMULA USED: 1. APPARATUS REQUIRED:  A thin uniform cantilever beam of channel section as shown in the figure. h → height of the flange b → width of the flange 2. THEORY: For any unsymmetrical section there exists a point at which any vertical force does not produce a twist of that section. experienced by the section.. along the direction of one of the principal axes of the section) and passes through the shear centre tip. Mount two dial gauges on the flange at a known distance apart at the free and of the beam.  Two dial gauges are mounted firmly on this section.e. Place a total of. 2.Dhanalakshmi Srinivasan Engineering College. Gurunath K – AE 2305 Aircraft Structures Laboratory II | 7 . At the free end extension pieces are attached on either side of the web to facilitate vertical loading. A thin walled channel section with its web vertical has a horizontal axis of symmetry and the shear centre lies on it. Set the dial gauge readings to zero. as they are weak in resisting torsion. Perambalur – 621 212. AUC R2008 Exercise No. L = Height. Perambalur – 621 212. b = Thickness. AUC R2008 TABULATION: Length. h WV Distance between the two hook sections (AB) = (Wa+ Wb) = Breadth. t = Dial gauge readings Sl. No. Wa Wb (d1-d2) d1 d2 = 𝒆= 𝑨𝑩(𝑾𝒂 − 𝑾𝒃 ) 𝟐𝑾𝑽 01 02 03 04 05 06 07 08 09 10 11 Gurunath K – AE 2305 Aircraft Structures Laboratory II | 8 .Dhanalakshmi Srinivasan Engineering College. 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. II. 6.Dhanalakshmi Srinivasan Engineering College. GRAPH: Plot ‘e’ versus (d1-d2) curve and determine where this meets the ‘e’ axis and locate the shear centre. Record the dial gauge readings. Perambalur – 621 212. Gurunath K – AE 2305 Aircraft Structures Laboratory II | 9 . Calculate the distance ‘e’ of the line of action from the web thus: 𝑒=( 𝐴𝐵(𝑊𝑎 −𝑊𝑏 ) 2𝑊𝑉 ) 5. In that event the number of readings taken will reduce proportionately. the twist of the section is zero for this location of the resultant vertical load).e. 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. For every load case calculate the algebraic difference between the dial gauge readings as the measure of the angle of twist 𝜃 suffered by the section. PRECAUTIONS: I. 4.. Repeat the experiments with identical settings several times to ensure consistency in the readings. 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. RESULT: The shear centre obtained experimentally is compared with the theoretical value. This means that the total vertical load on this section remains 2 kg. Do not therefore exceed the suggested values for the loads. Theoretical location of the shear centre 𝑒= 3𝑏 ℎ 𝑏 [6+( )] * Though a nominal value of 2 kg for the total load is suggested it can be less. Plot 𝜃 against ‘e’ and obtain the meeting point of curve (a straight line in this case) with the ‘e’-axis (i. Now remove one load piece from the hook at A and place another hook at B. For the section supplied there are limits on the maximum value of loads to obtain acceptable experimental results. 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 determines the shear centre. The dial gauges must be mounted firmly. AUC R2008 3. Beyond these the section could undergo excessive permanent deformation and damage the beam forever. Dhanalakshmi Srinivasan Engineering College. Perambalur – 621 212.Closed section Gurunath K – AE 2305 Aircraft Structures Laboratory II | 10 . Dial gauge Wa AUC R2008 Dial gauge Wb Determination of Shear Center. The location of this shear centre is important in the design of beams of closed sections when they should bend without twisting.. Gurunath K – AE 2305 Aircraft Structures Laboratory II | 11 . Similarly the wing strut of a semi cantilever wing is a closed tube of aerofoil section. Mount two dial gauges on the flange at a known distance apart at the free and of the beam. At the free end extension pieces are attached on either side of the web to facilitate vertical loading. where the lift produces a torque about the shear centre. Set the dial gauge readings to zero. Now remove one load piece from the hook at A and place another hook at B. A thin walled ‘D’ section with its web vertical has a horizontal axis of symmetry and the shear centre lies on it. all other sections of the beam do not twist. AUC R2008 Exercise No. 3.e.Dhanalakshmi Srinivasan Engineering College. APPARATUS REQUIRED:  A thin uniform cantilever beam of ‘D’ section as shown in the figure.  Two loading hooks each weighing about 200 gm. 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.: 03 Shear Centre of Closed Sections Date: AIM: To determine the shear centre of a closed section. The shear centre is important in the case of a closed section like an aircraft wing. This point is known as shear centre. say two kilograms load at A (loading hook and nine load pieces will make up this value). This enables the determination of the twist. a known distance apart. 4. For every load case calculate the algebraic difference between the dial gauge readings as the measure of the angle of twist 𝜃 suffered by the section. Record the dial gauge readings. This means that the total vertical load on this section remains 2 kg. Place a total of. if any. over the top flange. Note the dial gauge readings (nominally. Perambalur – 621 212.  A steel support structure to mount the channel section as cantilever. hooks also weigh a 200 gm each). experienced by the section.  Two dial gauges are mounted firmly on this section. along the direction of one of the principal axes of the section) and passes through the shear centre tip. 2. THEORY: For any unsymmetrical section there exists a point at which any vertical force does not produce a twist of that section. Calculate the distance ‘e’ of the line of action from the web thus: 𝑒=( 𝐴𝐵(𝑊𝑎 −𝑊𝑏 ) 2𝑊𝑉 ) 5. The aim of the experiment is to determine its location on this axis if the applied shear to the tip section is vertical (i. PROCEDURE: 1. Transfer carefully all the load pieces and finally the hook one by one to the other hook noting each time the dial gauge readings. t = Distance between the two hook sections (AB) Dial gauge readings Sl. Wa Wb (d1 . Perambalur – 621 212.d2) d1 d2 𝒆= = 𝑨𝑩(𝑾𝒂 − 𝑾𝒃 ) 𝟐𝑾𝑽 01 02 03 04 05 06 07 08 09 10 11 Gurunath K – AE 2305 Aircraft Structures Laboratory II | 12 . L = WV = (Wa+ Wb) Height. No. h = Thickness.Dhanalakshmi Srinivasan Engineering College. AUC R2008 TABULATION: Length. GRAPH: Plot ‘e’ vs (d1-d2) curve and determine where this meets the ‘e’ axis and locate the shear centre. In that event the number of readings taken will reduce proportionately. the twist of the section is zero for this location of the resultant vertical load). Plot 𝜃 against ‘e’ and obtain the meeting point of curve (a straight line in this case) with the ‘e’-axis (i.. The dial gauges must be mounted firmly. Perambalur – 621 212. 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. Beyond these the section could undergo excessive permanent deformation and damage the beam forever. Gurunath K – AE 2305 Aircraft Structures Laboratory II | 13 . II. RESULT: The shear centre obtained experimentally is compared with the theoretical value. 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. PRECAUTIONS: I. For the section supplied there are limits on the maximum value of loads to obtain acceptable experimental results.Dhanalakshmi Srinivasan Engineering College. Repeat the experiments with identical settings several times to ensure consistency in the readings. This determines the shear centre. Do not therefore exceed the suggested values for the loads. * Though a nominal value of 2 kg for the total load is suggested it can be less. AUC R2008 6.e. AUC R2008 Constant Strength Beam Gurunath K – AE 2305 Aircraft Structures Laboratory II | 14 . Perambalur – 621 212.Dhanalakshmi Srinivasan Engineering College. Eqn (1) APPARATUS REQUIRED: A constant strength beam in which the depth varies as in Eqn (1) and made of aluminium. Strain gauges. 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. 𝑍= 𝑏ℎ3 12 ℎ 2 = 𝑏ℎ 2 6 𝑀 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑍 Let the width of the beam be constant and the depth varies. C and tabulated as given below. 3𝐿 Strain gauge resistance Gauge factor Young’s modulus.25 kg is to be added). 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. strain indicator and weights with hook. = = E = 350 ohm 2 70 GPa Gurunath K – AE 2305 Aircraft Structures Laboratory II | 15 . B. The strain gauge readings are noted for every 2 kg at locations A. In this case the maximum stress remains constant along the length of the beam.. Perambalur – 621 212. Then 𝜎= 𝜎= ℎ2 = 𝑀 6𝑃𝑥 6𝑃𝐿 = 2= = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑍 𝑏ℎ 𝑏ℎ02 ℎ02 𝑥 𝐿 ……………. b→ width of the beam L→ length of the beam h → depth of the beam Section modulus. 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. One such structure is constant strength beam. PROCEDURE: The constant strength beam is fixed as a cantilever strain gauges are fixed near the root. AUC R2008 Exercise No. 𝐿 at 4 and 2.Dhanalakshmi Srinivasan Engineering College.: 04 Constant Strength Beam Date: AIM: To determine the stress at various locations along the length of a constant strength beam to show that they are equal and compare with theoretical values. 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. Hence half bridge used in the strain indicator to measure the strain at each location (strain = strain meter reading x 2). Dhanalakshmi Srinivasan Engineering College. AUC R2008 TABULATION: Sl. Perambalur – 621 212. Weight (Kg) 𝜺𝑨 𝜺𝑩 𝜺𝑪 𝝈𝑨 = 𝑬𝜺𝑨 × 𝟐 (MPa) 𝝈𝑩 = 𝑬𝜺𝑩 × 𝟐 (MPa) 𝝈𝑪 = 𝑬𝜺𝑪 × 𝟐 (MPa) 01 02 03 04 05 Gurunath K – AE 2305 Aircraft Structures Laboratory II | 16 . No. Gurunath K – AE 2305 Aircraft Structures Laboratory II | 17 . AUC R2008 THEORETICAL CALCULATION At point A: Distance of point A from loading point = Depth of the beam = Width of the beam = Moment of inertia = Moment = 𝜎= 𝑀ℎ = 2𝐼 At point B: Distance of point B from loading point = Depth of the beam = Width of the beam = Moment of inertia = Moment = 𝜎= 𝑀ℎ = 2𝐼 At point C: Distance of point C from loading point = Depth of the beam = Width of the beam = Moment of inertia = Moment = 𝜎= 𝑀ℎ 2𝐼 = RESULT: The experimental values of the stress are compared with the theoretical values. Perambalur – 621 212.Dhanalakshmi Srinivasan Engineering College. AUC R2008 Gurunath K – AE 2305 Aircraft Structures Laboratory II | 18 . Perambalur – 621 212.Dhanalakshmi Srinivasan Engineering College. σ → M → y → I → 𝐼 𝑦 Bending stress (N/m2) Bending moment (N-m) Distance of the layer from the neutral axis (m) Moment if Inertia (m4) 𝐼= 𝜋(𝑑𝑜4 − 𝑑𝑖4 ) 64 The shear stress due to torsion 𝜏= 𝑇𝑟 𝐽 Where.2 = ± √( ) + (𝜏)2 2 2 𝜎1. Where. 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. namely a hollow cylinder is subjected to a bending and torsion. AUC R2008 Exercise No. τ T R J Shear stress (N/m2) Torque applied (N-m) radius of the shaft (m) Polar moment of Inertia (m4) → → → → 𝐽= 𝜋(𝑑𝑜4 − 𝑑𝑖4 ) 32 𝜎1.Dhanalakshmi Srinivasan Engineering College.: 05 Combined Loading Date: AIM: To determine the principal stresses and principal planes of a hollow circular shaft due to combined loading.2 = 𝜎𝑥 +𝜎𝑦 2 𝜎𝑥 −𝜎𝑦 2 ± √( 2 ) + (𝜏𝑥𝑦 ) 𝜎 2 𝜎 𝜎1. Perambalur – 621 212.2 = 16𝑑𝑜 𝜋(𝑑𝑜4 − 𝑑𝑖4 ) [𝑀 ± √𝑀2 + 𝑇 2 ] 2 …………… (1) …………… (2) …………… (3) Gurunath K – AE 2305 Aircraft Structures Laboratory II | 19 . To understand their combined effect a similar specimen. THEORY: The most common combined load system encountered in structural design is probably that are due to bending and torsion. 𝑀 𝜎= For an elastic structure. Weight (Kg) 𝜺𝑨 𝜺𝑩 𝜺𝑪 01 02 03 04 05 Gurunath K – AE 2305 Aircraft Structures Laboratory II | 20 . Perambalur – 621 212. 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.Dhanalakshmi Srinivasan Engineering College. since the tension and compression get added up. one on the top fibre and the other at the bottom to measure the bending strain. 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. Perambalur – 621 212.Dhanalakshmi Srinivasan Engineering College. Similarly three more strain gauges are fixed at the middle of the length to verify the result at various locations of the tube. The outside diameter of the tube is measured using vernier callipers. tan 2∅ = 2𝜏𝑥𝑦 𝜎𝑥−𝜎𝑦 = 2𝜏 …………… 𝜎 AUC R2008 (4) Where. connecting wires. weight hanger with slotted weights. Another strain gauge is fixed at the same location on the neutral axis at 45° to measure the shear strain. The strain gauge at 45° is connected to the quarter bridge of the strain indicator to measure the shear strain. These values are compared with theoretical values. strain indicator and micrometre. strain gauges. From the strains the bending stress. shear stress are calculated and hence principal stresses and principal angle are calculated. ∅ → Principal angle APPARATUS REQUIRED: Hollow circular shaft fixed as a cantilever. 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 = Gurunath K – AE 2305 Aircraft Structures Laboratory II | 21 . NOTE: For half bridge the strain readings are multiplied by 2 and for Quarter Bridge by 4 to get the actual strains. 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. PROCEDURE: Two strain gauges are fixed near the root of the tube fixed as a cantilever. AUC R2008 Wagner Beam Gurunath K – AE 2305 Aircraft Structures Laboratory II | 22 . Perambalur – 621 212.Dhanalakshmi Srinivasan Engineering College. .......Dhanalakshmi Srinivasan Engineering College... As the applied load is further increased.. b AF AS → → → Distance between stiffeners Area of flange Area of stiffener Gurunath K – AE 2305 Aircraft Structures Laboratory II | 23 . the stress in the compression direction does not increase. 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 example of the departure of the design of aerospace structures from the standard structural design methods in other fields of structures... the webs of the Wagner beam will buckle at a low value of the applied vertical load... however the stress increase in the tension direction. AUC R2008 Exercise No. Eqn (1) Where...... According to the theory developed by Wagner.. The phenomena of buckling may be observed by nothing the wrinkles that appear on the thin sheet.. This method of carrying the shear load permits the design of relatively thin webs because of high allowable stresses in tension.. Eqn (2) Where... Perambalur – 621 212. the diagonal tensile stress σt in the thin web is given by the expression 𝜎𝑡 = 2𝑊 𝑑𝑡 sin 2𝛼 . As thin sheets are weak in compressions.: 06 Structural Behaviour of a Semi-Tension Field Beam Date: AIM: To investigate and study the behaviour of a semi-tension field beam. W d t α → → → → Shear load Distance between the CG of the flanges Thickness of the web Angle at which wrinkling occurs 4 tan 𝛼 = 𝑡𝑑 ) 2𝐴𝐹 𝑡𝑏 1+( ) 𝐴𝑆 1+( . 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... such as beam design for bridges and buildings... Weight (Kg) 𝜺t 𝜺F 𝜺S 𝝈t = 4E 𝜺t 𝛼= 𝝈F = 4E 𝜺F 𝝈S = 4E 𝜺S 01 02 03 04 05 Gurunath K – AE 2305 Aircraft Structures Laboratory II | 24 . AUC R2008 TABULATION: t = b = d = AF = AS = E = Sl.Dhanalakshmi Srinivasan Engineering College. No. Perambalur – 621 212. .. Eqn (3) . hydraulic jack.. Eqn (4) APPARATUS REQUIRED: A stiffened thin-webbed cantilever beam held in a suitable frame. Stress in the Flange 𝜎𝐹 = Stress in the stiffener 𝜎𝑆 = 𝑊 2𝐴𝐹 tan∝ 𝑊𝑏 𝐴𝑆 𝑑 + tan ∝ 𝑊𝑧 𝐴𝐹 𝑑 AUC R2008 . The load is applied gradually in steps of 100 kg using the hydraulic jack... load cell and load indicator... Gurunath K – AE 2305 Aircraft Structures Laboratory II | 25 . RESULT: σt. For each load the load indicator reading. Precaution is taken so that the beam does not undergo any permanent deformation.. σF and σS values are calculated theoretically using Eqns (1).(3) and (4) and compared with the experimental values given in the table. strain indicator reading corresponding to each strain gauge is noted. 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...... Perambalur – 621 212.... 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..Dhanalakshmi Srinivasan Engineering College.. strain indicator. strain gauges. Hence the beam is not loaded up to wrinkling load..... 7.2.1. 7. A cantilever beam Fig.Dhanalakshmi Srinivasan Engineering College.2 depicts of cantilever beam under the free vibration. The cantilever beam under free vibration Fig. Perambalur – 621 212. Gurunath K – AE 2305 Aircraft Structures Laboratory II | 26 . and Fig.1 shows of a cantilever beam with rectangular cross section. AUC R2008 Fig. 7. 7. which can be subjected to bending vibration by giving a small initial displacement at the free end. established by its mass and stiffness distribution. E is the modulus of rigidity of beam material.. ρ is the material density.: 07 Determination of Natural Frequencies of Cantilever Beams Date: AIM: To obtain natural frequencies of a cantilever beam up to the second mode. the fundamental natural frequency) is called as the fundamental (or the first) mode. The displacements at some points may be zero. and hence. In actual practice there is always some damping (e. THEORY: Free vibration takes place when a system oscillates under the action of forces inherent in the system itself due to initial disturbance. m = ρA(x). The system has infinite number of degrees of freedom and infinite number of natural frequencies. and when the externally applied forces are absent. the equation of motion can be written as (Meirovitch. In case of continuous system the properties of the system are the function of spatial coordinates.) present in the system which causes the gradual dissipation of vibration energy. This virtual experiment is based on a theme that the actual experimental measured vibration data are used. 𝑑2 𝑑2 𝑌(𝑥) {𝐸𝐼(𝑥) } = 𝜔2 𝑚(𝑥)𝑌(𝑥) 2 2 𝑑𝑥 𝑑𝑥 (1) Where. and the system is considered as continuous system in which the beam mass is considered as distributed along with the stiffness of the shaft. m is the mass per unit length.e. the calculations for natural frequencies are generally made on the basis of no damping. which are properties of the dynamical system. viscous damping. and it results gradual decay of amplitude of the free vibration. MATHEMATICAL ANALYSIS: For a cantilever beam subjected to free vibration. etc.g. and to observe the response of the system subjected to a small initial disturbance and virtualization of the experiment. The system under free vibration will vibrate at one or more of its natural frequencies. Gurunath K – AE 2305 Aircraft Structures Laboratory II | 27 . Y(x) is displacement in y direction at distance x from fixed end. The mode shape corresponding to lowest natural frequency (i.Dhanalakshmi Srinivasan Engineering College. These points are known as nodes. Generally nth mode has (n-1) nodes (excluding end points). experimentally. 1967). Damping has very little effect on natural frequency of the system. I is the moment of inertia of the beam cross-section. ω is the circular natural frequency. The relative displacement configuration of the vibrating system for a particular natural frequency is known as the Eigen function in continuous system. Perambalur – 621 212. The mode shape changes for different boundary conditions of a beam. aerodynamical damping. the internal molecular friction. AUC R2008 Exercise No. Damping is of great importance in limiting the amplitude of oscillation at resonance. x is the distance measured from the fixed end. Dhanalakshmi Srinivasan Engineering College.3. AUC R2008 Fig. Perambalur – 621 212. 7. The first three undamped natural frequencies and mode shape of cantilever beam Gurunath K – AE 2305 Aircraft Structures Laboratory II | 28 . from above equation of motion and boundary conditions can be written as. 4.694.2.3 … ∞ 𝑎𝑛𝑑 𝛽𝑥 𝐿 = 𝑛𝜋 A closed form of the circular natural frequency 𝜔𝑛𝑓 . Perambalur – 621 212.885 RESULT: First Natural Frequency 𝜔𝑛𝑓 = 𝛼𝑛2 √ 𝐸𝐼 𝑚𝐿4 Second Natural Frequency 𝜔𝑛𝑓 = 𝛼𝑛2 √ 𝐸𝐼 𝑚𝐿4 Third Natural Frequency 𝜔𝑛𝑓 = 𝛼𝑛2 √ 𝐸𝐼 𝑚𝐿4 Gurunath K – AE 2305 Aircraft Structures Laboratory II | 29 . 𝑎𝑡 𝑎𝑡 𝑥 = 0. 𝐸𝐼 𝜔𝑛𝑓 = 𝛼𝑛2 √ 4 𝑚𝐿 where. 𝑑2 𝑌(𝑥) = 0. 𝑌(𝑥) = 0.875. 7. 𝛽4 = 𝜔2 𝑚 𝐸𝐼 The mode shapes for a continuous cantilever beam is given as 𝑓𝑥 (𝑥) = 𝐴𝑥 {(sin 𝛽𝑥 𝐿 − sinh 𝛽𝑥 𝐿)(sin 𝛽𝑥 𝑥 − sinh 𝛽𝑥 𝑥) + (cos 𝛽𝑥 𝐿 − cosh 𝛽𝑥 𝐿)(cos 𝛽𝑥 𝑥 − cosh 𝛽𝑥 𝑥)} where. we get 𝑑 4 𝑌(𝑥) − 𝛽 4 𝑌(𝑥) = 0 𝑑𝑥 4 where.Dhanalakshmi Srinivasan Engineering College. 𝑛 = 1. 𝑑𝑌(𝑥) =0 𝑑𝑥 𝑥 = 𝑙. 𝑑𝑥 2 𝑑 3 𝑌(𝑥) =0 𝑑𝑥 3 For a uniform beam under free vibration from equation (1). 𝛼𝑛 = 1. AUC R2008 We have following boundary conditions for a cantilever beam. AUC R2008 Fig. 8. Disk in compression Fig. 8. Perambalur – 621 212.1.Dhanalakshmi Srinivasan Engineering College. Stress distribution along horizontal diameter Gurunath K – AE 2305 Aircraft Structures Laboratory II | 30 .2. the maximum difference 𝜎1 − 𝜎2 occurs at the centre. The horizontal and vertical normal stresses along the x axis are principal stresses because the shear stress 𝜏𝑥𝑦 vanishes due to symmetry about the x axis. the solutions for the normal stresses along the horizontal diameter are (after Dally and Riley 1991). 𝜎𝑥 is positive. At this point. Perambalur – 621 212.: 08 Stresses in Circular Discs using Photo elastic Techniques Date: AIM: To obtain the stresses in circular discs and beams using photo elastic techniques.2. Also. We therefore take 𝜎1 = 𝜎𝑥 and 𝜎2 = 𝜎𝑦 so as to render 𝜎1 − 𝜎2 ≥ 0. 2 2𝑃 1 − 𝜁 2 𝜎1 = ( ) 𝜋ℎ𝐷 1 + 𝜁 2 1 2 2 6𝑃 (1 − 𝜁 ) (1 + 3 𝜁 ) 𝜎2 = − (1 + 𝜁 2 )2 𝜋ℎ𝐷 where 𝜁= 𝑥 2𝑥 = 𝑅 𝐷 These stresses are plotted in Fig. 8. Along the horizontal diameter. but for a given force P. The reason is that the relative retardation is proportional to ℎ. From theory of elasticity. Gurunath K – AE 2305 Aircraft Structures Laboratory II | 31 .Dhanalakshmi Srinivasan Engineering College. AUC R2008 Exercise No. THEORY: A common calibration specimen is the circular disk of diameter D and thickness ℎ loaded in diametral compression (Fig. 8. 𝜎1 − 𝜎2 = 8𝑃 𝜋ℎ𝐷 Combining this result with the basic photo elastic relation gives 𝑁𝑓𝜎 8𝑃 = 𝜎1 − 𝜎2 = ℎ 𝜋ℎ𝐷 or 𝑓𝜎 = 8 𝑃 𝜋𝐷 𝑁 Notice that the specimen thickness ℎ does not appear in this equation.1). that is. at 𝜁 = 0. The net effect is a result for 𝑓𝜎 that is independent of ℎ. the stresses are inversely proportional to ℎ. while 𝜎𝑦 is negative. 8. 8. Gurunath K – AE 2305 Aircraft Structures Laboratory II | 32 .3. Light-field isochromatics in a diametrally loaded circular disk.3. Perambalur – 621 212. AUC R2008 Fig.Dhanalakshmi Srinivasan Engineering College. Enhanced image using only the green component of the light used in Fig. Fig.4. 8. Then P is plotted as a function of N. is insensitive to edge imperfections. The value of N at the centre of the disk. Therefore the fringe constant for this material is approximately 𝑓𝜎 = 8 𝑃 8 1330 𝑁 = = 761. 8. Such saddle points are common in photo elastic patterns. but which also contains other colours as well. 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 2305 Aircraft Structures Laboratory II | 33 . the function decreases. RESULT: The value of the stresses in Circular disc are computed using Photo elastic technique. 8.50 mm (2. it produces a fringe pattern that. The saddle shape of the central fringe allows rather precise determination of N for this purpose.3. and above and below this point.3 was obtained with a mercury-vapour light source. as seen in Fig. is approximately 7. It also tests the limit of fringe density that can be recorded photographically. A Tiffen #58 Green filter was placed on the 35 mm camera to reduce the transmission of the other colours.500 in.94 𝜋𝐷 𝑁 𝜋(0. AUC R2008 A photograph of a light-field isochromatic pattern for a diametrally loaded disk made of PSM-1 is shown in Fig. it is not likely to fracture.Dhanalakshmi Srinivasan Engineering College. which is adequate for most work in photo elasticity.3.635) 7 𝑚 − 𝐹𝑟𝑖𝑛𝑔𝑒 A more accurate way of determining 𝑓𝜎 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. Perambalur – 621 212.7 s.4.). it is easy to load. the function increases. 8. Fringe orders up to about N = 20 can be discerned in this figure. A somewhat enhanced image is shown in Fig. The disk in diametral compression is a favourite specimen for calibration because it is simple to fabricate (at least with a template). in the region of interest. 8. The low-order values of N are marked along the horizontal diameter. This arrangement results in distinct fringes up to about N = 15. the diameter D was 63. 8. which is rich in green light. and it is simple to analyse. For this specimen. and the best straight-line fit of the data is used to determine the ratio P / N to be used in fringe constant equation. Kodak Gold ASA 200 colour film was used to record the image at f/8 with an exposure of 0. and the load was 1. only the green component of the red-green-blue (RGB) digital scan was retained. Note that the centre of the specimen is the location of a true saddle point in the function 𝜎1 − 𝜎2 : to the left and right of this point. The image in Fig.3.0. However. This image was produced from the same negative as that used to produce Fig. and this component was then enhanced by increasing the contrast.33 kN (298 lb). by producing very large fringe orders in the vicinity of the contact regions.